TechxGenus/Typst-Train · Datasets at Hugging Face

repo stringlengths 26 115	file stringlengths 54 212	language stringclasses 2 values	license stringclasses 16 values	content stringlengths 19 1.07M
https://github.com/lxl66566/my-college-files	https://raw.githubusercontent.com/lxl66566/my-college-files/main/信息科学与工程学院/机器视觉实践/报告/4/4.typ	typst	The Unlicense	#import "../template.typ": * #show: project.with( title: "4", authors: ("absolutex",), ) = 机器视觉实践四 == 实验目的图像补光算法与界面设计 + 实现图像补光算法（增加局部图像亮度，保持对比度，避免失真） + 设计图像补光界面，从左到右补光图像逐步替换原图像的可视化效果。 == 实验代码这是一份单文件 html 代码，使用时需要在本地启动一个 http server. #include_code("../src/brightness/index.html") 具体的原理是定义了三个补光因子，分别对应暗部，中间色调，亮部。对每个像素的每个 RGB 进行遍历，并将其乘以对应的补光因子。界面设计就直接调用前端的 img-comparison-slider 库，实现了可拖拽的亮暗对比。 == 实验结果与心得体会 #figure( image("result.png", width: 100%), caption: [补光界面效果图], ) 可以看出补光效果较好，特别是对白色背景的物体有着巨大提升。不过对于本身是黑色的物体，即使乘以补光因子，其亮度值还是过小，无法看到明显的效果。
https://github.com/MobtgZhang/sues-thesis-typst	https://raw.githubusercontent.com/MobtgZhang/sues-thesis-typst/main/paper/chapters/ch02.typ	typst	MIT License	= 公式、图表等表示方法 == 公式方便快捷写入公式是Typst相对于Word编辑器最为主要的优势之一，特别是熟练掌握之后，在输入公式的时候具有非常大的提升效果。 Typst 中的公式分为两类，包括有#text("行内公式",fill:rgb(255,0,0))和#text("行间公式",fill:rgb(255,0,0))，例如这是一个行内公式 $f(x)=1/(sqrt(2pi)sigma)dot exp(-((x-mu)^2)/(2sigma^2))$。下面举例几个行间公式 $ f(x)&= 1/(sqrt(2pi)sigma)dot exp(-((x-mu)^2)/(2sigma^2)) $ 例如，定义一个分段函数 $ f(x)&= cases( -x^3 + x + 8 &"," x<=2 \ 1/2 x^2 &"," 2<x<=10\ x+ 10 &"," x>=10 ) $ 也可以定义一个多行的连等的等式，定义如下所示 $ cos(2x) &= cos^2x - sin^2x &= 2cos^2x - 1 &= 1 - 2 sin^2x $ 可以将多个等式对齐写在同一个语句块当中，例如麦克斯韦方程组积分形式： $ cases( integral.cont_l H dot d l &= integral.double_S J dot d S + integral.double_(S)(diff D)/(diff t) dot S \ integral.cont_(l) EE dot d l &= - integral.double_(S)(diff BB)/(diff t) dot d S \ integral.cont_(S) BB dot S &= 0 \ integral.cont_(S) D dot d S &= integral.triple_(V) rho VV ) $ 微分形式： $ cases( nabla times H &= J + (diff D)/(diff t) \ nabla times E &= - (diff B)/(diff t) \ nabla dot B &= 0 \ nabla dot H &= rho ) $ 带有矩阵定义的公式： $ H = -mu dot B = -gamma B_(o) S_(z) = (gamma B_(o))/2 mat( 1, dots.h.c ,1 ; dots.h.c , dots.down , dots.h.c ; 1 & dots.h.c & 1 ) $ 在求解凸优化问题的时候，问题研究最后求解归结为以下的方程形式： $ arg_(x_(j)) min_(j=1,dots.h.c,N) sum_(j=1)^(N)c_(j)x_(j) $ $ s.t. cases( sum_(j=1)^(N)a_(i j)x_(j)=b_(i) & "," i=1 "," dots.h.c "," m \ x_(j)>=0 ) $ <equ:matrix> 在文章当中每一个公式的后面均可以添加一个label的标签，这样就可以应用公式了，例如#ref(<equ:matrix>)就是刚刚我们表达的矩阵表达式。
https://github.com/ryuryu-ymj/mannot	https://raw.githubusercontent.com/ryuryu-ymj/mannot/main/examples/showcase.typ	typst	MIT License	#import "/src/lib.typ": * #set page(width: auto, height: auto, margin: (x: 2cm, y: 1cm), fill: white) #set text(24pt) #show: mannot-init $ mark(1, tag: #<num>) / mark(x + 1, tag: #<den>, color: #blue) + mark(2, tag: #<quo>, color: #red) #annot(<num>, pos: top)[Numerator] #annot(<den>)[Denominator] #annot(<quo>, pos: right, yshift: 1em)[Quotient] $
https://github.com/figarofuga/Typst-template	https://raw.githubusercontent.com/figarofuga/Typst-template/main/CJD/CJD.typ	typst		= Introduction CJD(Creutz-feldt jacob disease)はRapid progression dementiaの中の代表的疾患である。 Rapid progression dementiaの定義は決まっていないが、大体2年以内に進行する認知症と言われている。最も重要な事は、CJDは感染性がある上に治療法が確立していない事である。いけた！ - 例えば、結核の事 - 例えば、NTMの事 = 症例 = いつ疑う？ = 身体所見・時間経過 = 検査 == 頭部MRI == 脳波 == 髄液検査 === RT-Quick == 具体的には？ドライアイスを固める == 病理解剖について当院では不可能ブレインバンク東京精神 1. いろんなことがある。 2. そういうものだ。例えばreferenceはこうやる。
https://github.com/michalrzak/muw-typst-template	https://raw.githubusercontent.com/michalrzak/muw-typst-template/main/README.md	markdown		# About The repository contains a Typst template for a "_Diplomarbeit_" / "_Masterarbeit_" / "_Master thesis_" at the Medical University of Vienna (Meduni Wien). [Typst](https://github.com/typst/typst) is a typesetting system, offering a fun alternative to LaTeX. The template should follow all guidelines outlined in [Leitfaden](https://ub.meduniwien.ac.at/fileadmin/content/OE/ub/dokumente/Leitfaden_Studierende_Hochschulschriften_MedUni_Wien.pdf) provided by the Medical University of Vienna. Even though the template automates most necessary things, allowing you to concentrate on writing the thesis, I still recommend skimming the document before starting to get an overview of all requirements. # Known issues 1. The template is only designed around writing the thesis is English. This can be easily changed by editing the template. If you do the work a PR would be appreciated :). # Acknowledgement Medical University of Vienna logo from: <https://commons.wikimedia.org/wiki/File:Meduni-wien.svg>
https://github.com/SkiFire13/master-thesis	https://raw.githubusercontent.com/SkiFire13/master-thesis/master/chapters/2-background.typ	typst		#import "../config/common.typ": * = Background <section-background> In this chapter we give an overview of the theoretical background used in the rest of this thesis. We will first recap some basic notions on order theory with special focus on (complete) lattices. Then we will define what a system of fixpoint equations over complete lattices is and what is its solution, along with a number of related concepts in order theory. We will then give a brief introduction to parity games and describe how to characterize the solution of a system of fixpoint equations using a parity game, with some care for efficiency issues. Finally we will introduce two algorithms used to solve parity games which we will be exploiting later on. #include "./background/1-lattices.typ" #include "./background/2-tuples.typ" #include "./background/3-fixpoint-system.typ" #include "./background/4-applications.typ" #include "./background/5-parity-games.typ" #include "./background/6-game-characterization.typ" #include "./background/7-strategy-improvement.typ"
https://github.com/jgm/typst-hs	https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/compiler/import-14.typ	typst	Other	// Unresolved import. // Error: 23-35 unresolved import #import "module.typ": non_existing
https://github.com/GartmannPit/Praxisprojekt-II	https://raw.githubusercontent.com/GartmannPit/Praxisprojekt-II/main/Praxisprojekt%20II/PVA-Templates-typst-pva-2.0/appendix.typ	typst		#let showAppendix() = [ = Anhang // change heading for other languages // Write your appendix after the heading within the block Add your appendix in _appendix.typ_ \ // delete me #lorem(200) // delete me ]
https://github.com/elteammate/typst-compiler	https://raw.githubusercontent.com/elteammate/typst-compiler/main/src/typesystem-parser.typ	typst		#import "typesystem-lexer.typ": * #import "typesystem-def.typ": * #let typesystem_parse = { (tokens) => { let token_mapping = if type(ts_token_kind.punc_colon) == "string" {let res = (:); res.insert(ts_token_kind.punc_colon, 0); res.insert(ts_token_kind.ty_array, 1); res.insert(ts_token_kind.ident, 2); res.insert(ts_token_kind.ty_bool, 3); res.insert(ts_token_kind.punc_gt, 4); res.insert(ts_token_kind.ty_int, 5); res.insert(ts_token_kind.ty_arguments, 6); res.insert(ts_token_kind.punc_comma, 7); res.insert(ts_token_kind.ty_none_, 8); res.insert(ts_token_kind.ty_content, 9); res.insert(ts_token_kind.ty_object, 10); res.insert(ts_token_kind.ty_dictionary, 11); res.insert(ts_token_kind.ty_any, 12); res.insert(ts_token_kind.ty_tuple, 13); res.insert(ts_token_kind.alias, 14); res.insert(ts_token_kind.punc_lt, 15); res.insert(ts_token_kind.ty_string, 16); res.insert(ts_token_kind.ty_float, 17); res.insert(ts_token_kind.ty_function, 18); res} else {let res = range(ts_token_kind.len()); res.at(ts_token_kind.punc_colon) = 0; res.at(ts_token_kind.ty_array) = 1; res.at(ts_token_kind.ident) = 2; res.at(ts_token_kind.ty_bool) = 3; res.at(ts_token_kind.punc_gt) = 4; res.at(ts_token_kind.ty_int) = 5; res.at(ts_token_kind.ty_arguments) = 6; res.at(ts_token_kind.punc_comma) = 7; res.at(ts_token_kind.ty_none_) = 8; res.at(ts_token_kind.ty_content) = 9; res.at(ts_token_kind.ty_object) = 10; res.at(ts_token_kind.ty_dictionary) = 11; res.at(ts_token_kind.ty_any) = 12; res.at(ts_token_kind.ty_tuple) = 13; res.at(ts_token_kind.alias) = 14; res.at(ts_token_kind.punc_lt) = 15; res.at(ts_token_kind.ty_string) = 16; res.at(ts_token_kind.ty_float) = 17; res.at(ts_token_kind.ty_function) = 18; res} let callbacks = (((o,_0,ts,_2)=>mk_type(o,..ts)),((t,_0,ts,_2)=>mk_type(t,..ts)),((f,_0,ret,_1,ts,_2)=>mk_type(f,ret,..ts)),(a => a),((i)=>"any"),(a => a),(a => a),((n,_,t)=>((name:n,ty:t),)),(a => a),((ts,_,t)=>ts+(t,)),((i)=>"tuple"),(()=>()),(a => a),((i)=>"object"),((i)=>"array"),((i)=>"arguments"),((i)=>"none"),((i)=>"float"),((ts,_,n,_1,t)=>ts+((name:n,ty:t),)),(a => a),((d,_1,t,_2)=>mk_type(d,t)),((i)=>"bool"),((i)=>"int"),((i)=>"content"),(a => a),((i)=>"function"),((t)=>(t,)),((a)=>panic("Not implemented (type alias resolution)")),((i)=>i),(a => a),((a,_1,t,_2)=>mk_type(a,t)),((i)=>"string"),(()=>()),(a => a),((i)=>"dictionary"),) let table = ((0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,1,0,0,0,),(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,57,0,0,0,0,0,),(0,0,0,0,-23,0,0,-23,0,0,0,0,0,0,0,0,0,0,0,-23,0,0,0,0,0,),(0,0,0,0,-27,0,0,-27,0,0,0,0,0,0,0,0,0,0,0,-27,0,0,0,0,0,),(0,0,0,0,-30,0,0,-30,0,0,0,0,0,0,0,0,0,0,0,-30,0,0,0,0,0,),(0,0,0,0,-29,0,0,-29,0,0,0,0,0,0,0,0,0,0,0,-29,0,0,0,0,0,),(0,0,0,0,-2,0,0,-2,0,0,0,0,0,0,0,0,0,0,0,-2,0,0,0,0,0,),(0,0,0,0,-16,0,0,-16,0,0,0,0,0,0,0,0,0,0,0,-16,0,0,0,0,0,),(0,0,0,0,-32,0,0,-32,0,0,0,0,0,0,0,0,0,0,0,-32,0,0,0,0,0,),(0,0,0,0,-6,0,0,-6,0,0,0,0,0,0,0,0,0,0,0,-6,0,0,0,0,0,),(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,54,0,0,0,0,0,0,0,0,0,),(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,51,0,0,0,0,0,0,0,0,0,),(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,46,0,0,0,0,0,0,0,0,0,),(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,40,0,0,0,0,0,0,0,0,0,),(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,16,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-8,0,0,-8,0,0,0,0,0,0,0,0,0,0,0,-8,0,0,0,0,0,),(0,30,19,26,-3,24,27,-3,20,21,32,28,25,31,0,0,22,23,29,0,18,0,17,0,0,),(0,0,0,0,35,0,0,36,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(33,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-7,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-19,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-12,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-18,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-13,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-31,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-14,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-20,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-10,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-21,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-25,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-22,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,34,0,0,0,),(0,0,0,0,-28,0,0,-28,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-35,0,0,-35,0,0,0,0,0,0,0,0,0,0,0,-35,0,0,0,0,0,),(0,0,37,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(38,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,39,0,0,0,),(0,0,0,0,-17,0,0,-17,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,11,0,8,-24,6,9,-24,2,3,14,10,7,13,15,0,4,5,12,0,0,42,0,41,0,),(0,0,0,0,43,0,0,44,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-9,0,0,-9,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-34,0,0,-34,0,0,0,0,0,0,0,0,0,0,0,-34,0,0,0,0,0,),(0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,45,0,0,0,),(0,0,0,0,-26,0,0,-26,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,47,0,0,0,),(0,0,0,0,0,0,0,48,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,11,0,8,-24,6,9,-24,2,3,14,10,7,13,15,0,4,5,12,0,0,42,0,49,0,),(0,0,0,0,50,0,0,44,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-33,0,0,-33,0,0,0,0,0,0,0,0,0,0,0,-33,0,0,0,0,0,),(0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,52,0,0,0,),(0,0,0,0,53,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-5,0,0,-5,0,0,0,0,0,0,0,0,0,0,0,-5,0,0,0,0,0,),(0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,55,0,0,0,),(0,0,0,0,56,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-15,0,0,-15,0,0,0,0,0,0,0,0,0,0,0,-15,0,0,0,0,0,)) let arg_count = (4,4,6,1,1,1,1,3,1,3,1,0,1,1,1,1,1,1,5,1,4,1,1,1,1,1,1,1,1,1,4,1,0,1,1,) let goto_index = (21,21,21,21,20,21,21,22,21,23,20,23,21,20,20,20,20,20,22,21,21,20,20,20,24,20,23,21,20,21,21,20,22,21,20,) let cast_table = (_ => none,_=>ptype.array,x=>x.text,_=>types.bool,_ => none,_=>types.int,_=>types.arguments,_ => none,_=>types.none_,_=>types.content,_=>ptype.object,_=>ptype.dictionary,_=>types.any,_=>ptype.tuple,x=>x.text,_ => none,_=>types.string,_=>types.float,_=>ptype.function,) let stack = (0, ) let ast_stack = () let cur_token = 0 for i in range(9999) { for j in range(9999) { let state = stack.last() let terminal = if cur_token < tokens.len() { token_mapping.at(tokens.at(cur_token).kind) } else { 19 } let action = table.at(state).at(terminal) if action == 57 { assert(ast_stack.len() == 1) return ast_stack.first() } else if action > 0 { stack.push(action) ast_stack.push(cast_table.at(terminal)(tokens.at(cur_token))) cur_token += 1 } else if action < 0 { let rhs = () for _ in range(arg_count.at(action)) { let _ = stack.pop() rhs.push(ast_stack.pop()) } let rule = callbacks.at(action) ast_stack.push(rule(..rhs.rev())) let goto_state = table.at(stack.last()).at(goto_index.at(action)) if goto_state > 0 { stack.push(goto_state) } else { panic("Expected shift action") } } else { panic("Parsing error at state: " + repr(stack) + " and token: " + repr(if cur_token < tokens.len() { tokens.at(cur_token) } else {"EOF"}) + " at: " + repr(cur_token) ) } } } panic("too complex") } }
https://github.com/donghoony/KUPC-2023	https://raw.githubusercontent.com/donghoony/KUPC-2023/master/description.typ	typst		#import "colors.typ" : KUPC_GREEN, PALE_RED #import "problem_info.typ" : constructTitle #import "problems.typ" : contest_problems #import "emoji/lib.typ" : * #let mono(s, color: black) = {text(font: "Inconsolata", fill: color)[#s]} #let bf(s) = {text(weight: "bold")[#s]} // 줄바꿈은 #linebreak()를 중간에 넣으면 됩니다. // 페이지 넘김은 문제 내부에서 ()를 새로 만들어 주세요. // monospace 문자열은 #mono("abc")와 같이 쓸 수 있습니다. // 미리 정의되지 않은 operation의 경우에는 #math.op("MEX")와 같이 쓰면 됩니다. 대부분은 정의돼 있으니 그냥 $cos$ 와 같이 쓰면 됩니다. // 시간복잡도는 $cal(O)(N log N)$ 와 같이 써주세요. log가 자체적으로 함수가 있습니다. // 이모지는 그냥 넣으면 됩니다. 지원하지 않는 이모지는 깨집니다. 이모지는 https://github.com/polazarus/typst-svg-emoji 를 사용했습니다. // 그래프도 작성 가능합니다. https://www.graphviz.org/docs/graph/ 를 참고해서 ```render <여기에 그래프를 작성하세요>``` #let cell(num, color: black, fill: none, stroke: none) = { rect( height: 50pt, width: 50pt, stroke: stroke, fill: fill, )[#align(center + horizon)[#text(fill: color)[#num]]]; } #let descriptions = ( // 2A ( ( [- 주어진 문자열에서 $1$이 연속으로 등장하는 구간을 알아내는 방법은 무엇일까요?], [- 여러 가지 방법이 있지만, 구현이 간단한 방법을 소개합니다.], [- #mono(1) 이후에 #mono(0) 이 등장한다면, 검은 줄의 구간이 끝나는 것을 의미합니다.], [- 주어진 문자열을 훑으면서, #mono("s[i] = 1")이면서 #mono("s[i+1] = 0")인 #mono("i")의 개수를 셉시다.], [- 편의를 위해, 주어진 문자열에 #mono(0) 을 추가해도 답은 같습니다.], [- 시간복잡도는 $cal(O)(N)$ 입니다.], ), ), // 2B/1A ( ( [- 건덕이와 건구스는 자신의 차례에 #bf("반드시") 움직여야 합니다.], [- 건덕이와 건구스는 다음과 같은 네 가지 종류로 움직입니다: #mono("RR"), #mono("RL"), #mono("LL"), #mono("LR")], [- 어떻게 움직이더라도 두 플레이어 사이 간격의 #bf("홀짝성")#sub(mono("parity"))은 변하지 않습니다.], ), ( [- 전장의 길이가 주어지는 순간 승자는 결정되며, 승자는 상대를 향해 전진합니다.], [- 패자의 경우 앞으로 가면 패배로 더 빠르게 도달하므로, 가능하다면 뒤로 후퇴합니다.], [- 전장을 벗어나도록 이동할 수 없으므로, #bf("두 플레이어간 간격은 단조감소")합니다.], ), ( [- 간격을 짝수로 만드는 사람은 결국 두 플레이어 사이의 거리를 $0$으로 만듭니다.], [- 플레이어 사이의 거리가 $0$이 되는 순간, 다음 차례의 플레이어는 공격할 수 있습니다.], [- 주어진 전장의 길이가 짝수라면, 선공이 먼저 둘 사이의 간격을 홀수로 만듭니다.], [- 홀수라면, 선공이 둘 사이의 간격을 짝수로 만듭니다.], [- 따라서 전장의 길이가 짝수인 경우, 건덕이가 승리합니다.], [- 주어진 전장의 길이의 홀짝성을 판단하므로 시간복잡도는 $cal(O)(1)$입니다.], [],[], [#emoji.arm.muscle $N times N$ 전장에서 각 플레이어가 $(1,1)$, $(N,N)$에서 시작하면 누가 승리할까요?] ), ), // 2C ( ( [- 바닥수 $N$이 주어졌을 때, 바닥수가 $N$인 길이 $L$의 원래 수를 구해야 합니다.], [- 길이 $L$의 모든 수에 대해서 바닥수가 되는지 확인하는 것은 오래 걸립니다.], [- 곱셈의 성질을 활용해서 쉽게 문제를 해결해 봅시다.] ), ( [- 곱셈의 항등원을 이용합시다. $1$은 여러 번 곱해도 $1$입니다.], [- $1$을 $L-1$번 적은 뒤, $N$를 뒤에 덧붙인 수의 바닥수는 $N$입니다.], [- 원래 수는 $0$으로 시작하지 않음에 주의합시다.] ), ), // 2D/1B ( ( [- 학생의 키가 같은 쌍이 존재하지 않는다면, 모든 학생이 참여할 수 있습니다.], [- 모두가 한 쪽을 바라보고, 바라보는 방향으로 키가 작아지도록 줄세우면 됩니다.], [- 학생의 키가 같은 쌍이 존재하는 경우에는 몇 명까지 참여할 수 있을까요?], ), ( [- 두 명의 키가 서로 같다면, 서로 반대 방향을 보고 서면 됩니다.], [- 세 명의 키가 서로 같다면, 한 명은 어느 쪽을 보든 점프할 타이밍을 놓칠 수 있습니다.], [- 따라서 키가 같은 사람은 #bf("최대 두 명")까지만 참여할 수 있습니다.], [- 줄을 세울 수 있는지의 여부를 물어보았으므로, 각 사람들의 키에 대해서 참여할 수 있는 사람의 수를 세어 줍시다.], [- 시간복잡도는 $cal(O)(N)$입니다.], [],[], [#emoji.arm.muscle 올바르게 줄 세우는 방법 중 하나를 구할 수 있을까요?] ), ), // 2E ( ( [- 왼쪽 절반의 문자와 오른쪽 절반의 문자의 교환을 통해 팰린드롬을 만들어 봅시다.], [- 연산을 통해 임의의 위치에 존재하는 두 문자를 교환할 수 있다면, #linebreak() 모든 문자를 원하는 곳에 배치할 수 있습니다. 과연 가능할까요?#v(0.5em)], [- 편의상 배열의 길이가 짝수라고 가정합시다. 교환하는 경우의 수는 두 가지입니다. ], [#h(2em) #emoji.ast 두 문자가 서로 다른 절반에 속하는 경우], [#h(2em) #emoji.ast 두 문자가 같은 절반에 속하는 경우], ), ( [- 두 문자가 서로 다른 절반에 속하는 경우, 두 원소를 직접 교환할 수 있습니다.], [- 두 문자가 서로 같은 절반에 속하는 경우에는 두 원소를 직접 교환할 수 없습니다.], [- 이때, 반대쪽 절반에 속하는 원소 한 개를 임시로 사용해 교환할 수 있습니다.], [ #let cell(num, color: black, fill: none) = { rect( height: 50pt, width: 50pt, stroke: none, fill: fill, )[#align(center + horizon)[#text(fill: color)[#num]]]; } #set table(align: center, stroke: gray + 0.5pt, inset: 0pt, columns: 4) #align(center)[ #table( )[#cell(1, color: blue, fill:rgb("ddd"))][#cell(2)][#cell(3, color: PALE_RED)][#cell(4, color: PALE_RED, fill: rgb("ddd"))] ] #v(-0.8em) #align(center)[ #table( )[#cell(4, color: PALE_RED, fill:rgb("ddd"))][#cell(2)][#cell(3, color: PALE_RED, fill:rgb("ddd"))][#cell(1, color: blue)] ] #v(-0.8em) #align(center)[ #table( )[#cell(3, color: PALE_RED, fill:rgb("ddd"))][#cell(2)][#cell(4, color: PALE_RED)][#cell(1, color: blue, fill:rgb("ddd"))] ] #v(-0.8em) #align(center)[ #table( )[#cell(1, color: blue)][#cell(2)][#cell(4, color: PALE_RED)][#cell(3, color: PALE_RED)] ] ] ), ( [- 문자열의 길이가 홀수인 경우에는 가운데 글자를 교환할 수 없음에 유의합시다.], [- 해당 문자를 세지 않고, 짝수 길이의 문자열이라고 생각하면 됩니다.], [- 팰린드롬이 되기 위해서는 양 쪽에 같은 수의 문자가 존재해야 합니다.], [- 알파벳이 문자열에 짝수번 등장한다면 양 쪽에 골고루 분배할 수 있습니다.], [- 시간복잡도는 $cal(O)(N)$입니다.], ), ), ( ( [- 두 개의 말뚝과 깃대 하나를 골라 만들 수 있는 삼각형 넓이 중 최댓값을 구합시다.], [- 단, 넓이가 $R$ 이하여야 합니다.], [- 간단한 방법은 모든 말뚝 쌍에 대해서, 모든 깃대를 탐색하는 방법입니다.], [- 해당 풀이는 $cal(O)(N^2M)$으로, 제한 시간 안에 해결할 수 없습니다.], ), ( [- 만약 밑변을 고정한다면, 주어진 깃대를 정렬한 뒤 이분탐색을 사용할 수 있습니다.], [- 가능한 모든 밑변을 구하는 데에는 $cal(O)(N^2)$입니다.], [- 이분 탐색에 깃대를 활용하기 위해 깃대를 정렬하는 데 $cal(O)(M log M)$입니다.], [- 하나의 밑변 길이에 대해서, $R$ 이하인 최대 넓이를 구하는 데 $cal(O)(log M)$이 듭니다.], [- $cal(O)(M log M + N^2 log M)$ 은 충분히 통과할 수 있습니다.] ), ), ( ( [- 스위치를 누르면 $3$초동안 두 배의 점수를 얻을 수 있습니다.], [- 음수의 점수도 두 배로 계산됨에 유의해야 합니다.], [- 그리디하게 스위치를 누르는 전략은 오답을 받습니다.], [- 스위치를 누른 뒤, $i$초 시점에 스위치의 효과가 $j$초 남았다는 것을 활용합시다.], ), ( [- $i$초에 $j=0$이라면, $i-1$초에 $j=0$이거나 $j=1$인 경우 중 최댓값을 가져옵시다.], [- $j=1$이라면 아직 스위치의 효과가 남아있으므로 $i-1$초의 $j=2$를 가져옵니다.], [- $j=2$인 경우도 마찬가지로 $i-1$초의 $j=3$을 가져옵니다.], [- $j=3$은 지금 스위치를 누른 상황으로, $i-1$초의 $j=0$이거나 $j=1$인 경우에서 가져올 수 있습니다.], [- 스위치의 효과가 적용되는 경우에는 배열의 값의 두배를 더해줍시다.], [- 시간복잡도는 $cal(O)(N)$입니다.] ) ), ( ( [ #let cell(num, color: black, fill: none, stroke: none) = { rect( height: 50pt, width: 50pt, stroke: stroke, fill: fill, )[#align(center + horizon)[#text(fill: color)[#num]]]; } #set table(align: center, stroke: rgb("add8e6") + 1pt, inset: 0pt, columns: 5) #align(center)[ #table()[#cell(1)][#cell(2)][#cell(3, stroke: red)][#cell(4)][#cell(5)][#cell(5)][#cell(2)][#cell(1, stroke: red)][#cell(4)][#cell(6)][#cell(0)][#cell(2)][#cell(4,fill: rgb("ddd"), stroke: red)][#cell(2)][#cell(1)][#cell(0)][#cell(0)][#cell(2)][#cell(1)][#cell(7)] ] ], [- 일감호입니다. #emoji.fishing], [- 무게추 $3$을 달아 $3$의 힘으로 낚싯대를 휘두르면 회색 칸에 찌가 도달합니다.], [- 빨간 색 테두리에 해당하는 칸에 존재하는 $8$마리의 물고기가 사로잡힙니다.] ), ( [ #set table(align: center, stroke: rgb("add8e6") + 1pt, inset: 0pt, columns: 5) #align(center)[ #table()[#cell(1)][#cell(2, stroke: red)][#cell(3, fill: rgb("add8e6"))][#cell(4)][#cell(5)][#cell(5)][#cell(2, fill: rgb("ddd"), stroke: red)][#cell(1, fill: rgb("add8e6"))][#cell(4)][#cell(6)][#cell(0)][#cell(2)][#cell(4, fill: rgb("add8e6"))][#cell(2)][#cell(1)][#cell(0)][#cell(0)][#cell(2)][#cell(1)][#cell(7)] ] ], [- 낚싯줄을 한 바퀴 감아올리면 찌의 위치가 바뀝니다.], [- 바뀐 칸에서 물고기 $4$마리가 사로잡힙니다.], [- 이를 미끼가 일감호를 벗어날 때까지 반복합니다.], ), ( [ #let cell(num, color: black, fill: none, stroke: none) = { rect( height: 50pt, width: 50pt, stroke: stroke, fill: fill, )[#align(center + horizon)[#text(fill: color)[#num]]]; } #set table(align: center, stroke: rgb("add8e6") + 1pt, inset: 0pt, columns: 5) #align(center)[ #table()[#cell(1, fill: rgb("add8e6"))][#cell(2, fill: rgb("add8e6"))][#cell(3, fill: rgb("add8e6"))][#cell(4)][#cell(5)][#cell(5)][#cell(2, fill: rgb("add8e6"))][#cell(1, fill: rgb("add8e6"))][#cell(4)][#cell(6)][#cell(0)][#cell(2)][#cell(4, fill: rgb("add8e6"))][#cell(2)][#cell(1)][#cell(0)][#cell(0)][#cell(2)][#cell(1)][#cell(7)] ] ], [- 쿼리가 주어질 때마다 다음과 같은 계단 모양에 칠해진 수의 합을 구하는 문제입니다.], [- 간단하게 생각하면, 매 쿼리마다 계산하는 $cal(O)(N M Q)$ 풀이가 있습니다 #emoji.face.explode.], [- 이는 시간 초과입니다. 어떻게 빠르게 해결할 수 있을까요?] ), ( [- 여러 방법 중 하나를 소개합니다.], [- 2차원 배열에서 직사각형 모양의 부분합은 $cal(O)(N M + Q)$에 구할 수 있습니다.], [- 쿼리가 직사각형의 부분합을 구하게끔 일감호를 재배치할 수 있을까요?], ), ( [ #set table(align: center, stroke: rgb("add8e6") + 1pt, inset: 0pt, columns: 3) #align(center)[ #grid(columns: 3)[ #align(center + horizon)[ #table()[#cell(1, fill: rgb("add8e6"))][#cell(2,fill: rgb("add8e6"))][#cell(3,fill: rgb("add8e6"))][#cell(5)][#cell(2,fill: rgb("add8e6"))][#cell(1,fill: rgb("add8e6"))][#cell(0)][#cell(2)][#cell(4,fill: rgb("add8e6"))]] ][#align(center + horizon)[#h(3em)->#h(3em)]][ #align(center + horizon)[ #table()[#cell(0, fill: rgb("add8e6"))][#cell(0, fill: rgb("add8e6"))][#cell(3, fill: rgb("add8e6"), stroke: PALE_RED)][#cell(0,fill: rgb("add8e6"))][#cell(2,fill: rgb("add8e6"), stroke: PALE_RED)][#cell(1,fill: rgb("add8e6"), stroke: PALE_RED)][#cell(1, fill: rgb("add8e6"), stroke: PALE_RED)][#cell(2, fill: rgb("add8e6"), stroke: PALE_RED)][#cell(4,fill: rgb("add8e6"), stroke: PALE_RED)][#cell(5, stroke: PALE_RED)][#cell(2, stroke: PALE_RED)][#cell(0)][#cell(0, stroke: PALE_RED)][#cell(0)][#cell(0)]] ] ] ], [- 가능합니다.], [- 시간복잡도는 2차원 부분합과 같은 $cal(O)(N M + Q)$입니다.], ), ( [- 아래쪽으로 누적합을 구해준 뒤, 오른쪽 아래 대각선 방향으로 누적합을 구해나가도 정답을 구할 수 있습니다.], [],[], [ #let c1 = rgb(240,229,235); #let c2 = rgb(212,236,220) #let c3 = rgb(252,249,218) #let c4 = rgb(252,223,215) #set table(align: center, stroke: rgb("add8e6") + 1pt, inset: 0pt, columns: 3) #align(center)[ #grid(columns: 5)[ #align(center + horizon)[ #table()[#cell(1, fill: c1)][#cell(2,fill: c2)][#cell(3,fill: c3)][#cell(5, fill: c1)][#cell(2,fill: c2)][#cell(1,fill: c3)][#cell(0, fill: c1)][#cell(2, fill: c2)][#cell(4,fill: c3)]] ][#align(center + horizon)[#h(1em)->#h(1em)]][#align(center + horizon)[ #table()[#cell(1, fill: c1)][#cell(2,fill:c2)][#cell(3)][#cell(6,fill:c3)][#cell(4,fill: c1)][#cell(4,fill: c2)][#cell(6)][#cell(6, fill:c3)][#cell(8,fill: c1)]]][#align(center + horizon)[#h(1em)->#h(1em)]][ #table()[#cell(1)][#cell(2)][#cell(3)][#cell(6)][#cell(5)][#cell(6)][#cell(6)][#cell(12)][#cell(13)] ] ] ], ), ), ( ( [- 주어진 등굣길을 그래프로 모델링해 봅시다.], [- 각 칸에 쓰인 수만큼 양쪽 방향으로 떨어진 칸으로 향하는 간선을 만들 수 있습니다.], [- 최대 두 번 방향을 반전할 수 있으므로, 같은 칸이라도 $0$번 반전했을 때와 $2$번 반전했을 때는 다르게 생각해야 합니다.], ), ( [- 정점을 세 개로 나누어 아래와 같은 그래프 모형을 생각할 수 있습니다.], [#align(center)[#image("images/time.png", width: 70%)]] ), ( [- 문제의 정답은 $1$번 정점에서 학교에 해당하는 정점들까지의 최장거리와 같습니다.], [- 구축한 그래프는 사이클 없는 방향 그래프입니다.], [- $1$번 정점에서 위상정렬을 시행한 뒤, 정렬한 순서대로 정답을 갱신해나가면 됩니다.], [- 학교에 처음 도착한 시간을 늦추는 것에 유의합시다. 학교에서 되돌아가는 간선은 존재할 수 없습니다.], [- 시간복잡도는 $cal(O)(N)$입니다.], ), ( [- 더 쉬운 풀이로, 단순하게 오른쪽으로 한 번, 왼쪽으로 한 번, 다시 오른쪽으로 한 번 DP를 훑어도 해결할 수 있습니다.], [- 아직 나에게 오는 경로가 없거나 이미 학교에 도착한 경우, 칸에 숫자가 $0$이 쓰인 경우 등을 잘 처리해야 합니다.], [- 시간복잡도는 위와 같은 $cal(O)(N)$이지만, 상수가 작아 더 빠르게 통과합니다.], ) ), ( ( [- 주어진 문자열을 하나의 집합으로 생각합니다.], [- 예를 들어 문자열이 #mono("0011223344")라면 대응되는 집합은 ${0,1,2,3,4}$입니다.], [- 숫자는 $10$개 이므로 만들어질 수 있는 집합의 개수는 $2^10 = 1024$개입니다.], [- 이제 집합 두 개를 합쳐 크기가 $K$인 집합의 개수를 세면 됩니다.], ), ( [- 그럼 크기가 $K$인 집합의 개수는 어떻게 셀까요?], [- 만들어질 수 있는 $1024$개의 집합 중 $A union B$의 원소의 개수가 $K$개인 두 집합 $A$, $B$를 순서를 고려하지 않고 뽑습니다.], [- $A=B$일 때는 정답에 $(#math.op("cnt")\(A\) times (#math.op("cnt") (A) - 1))/2$를 더합니다.], [- $A!=B$일 때는 정답에 $#math.op("cnt") (A) times #math.op("cnt") (B)$를 더합니다.], [- 시간복잡도는 $cal(O)(N + 1024^2 times 10)$입니다.], [- 비트집합을 사용하면 $cal(O)(N + 1024^2)$으로 줄일 수 있지만, 사용하지 않아도 됩니다.] ), ), ( ( [- 우선 언제 #mono("NO", color:PALE_RED) 를 출력해야할 지 생각해봅니다.], [- #math.op("MEX")의 정의에 의해, 길이가 $N$인 수열의 #math.op("MEX")는 $N+1$보다 클 수 없습니다.], [- $A$는 순열이기 때문에 $B_N = N+1$입니다.], [- $S_i = {A_1, A_2, dots, A_i}$라고 할 때, $S_i subset S_(i+1)$이므로 #linebreak() $B_i > B_(i+1) (1 <= i <= N-1)$일 수 없습니다. 즉 $B$는 #bf("단조증가")합니다.], [- 위 세 가지 규칙을 지키지 않는다면 #mono("NO", color:PALE_RED)를 출력해야 합니다.], [- 그 이외의 경우에는 모두 가능할까요?], ), ( [- 몇 가지 관찰을 해 봅시다.], [- 만약 $A_i = K$라면, $i<=j$일 때 $B_j = K$일 수 없습니다.], [- $K$를 포함하는 집합의 #math.op("MEX")는 정의상 $K$일 수 없기 때문입니다.], [- 결국 $B_j = K$일 때, $A_i = K (j<i)$입니다.] ), ( [- $B_i != B_(i+1) (1 <= i <= N-1)$인 경우에 주목합니다.], [- #math.op("MEX")값이 $B_(i+1)$이기 위해서는 해당 집합이 $B_i$를 가지고 있어야 합니다.], [- 이전 관찰과 종합하면 $B_i != B_(i+1)$일 때, $A_(i+1) = B_i$가 됨을 알 수 있습니다.], ), ( [- 이제 남은 값을 배치하는 방법은 여러 가지가 있습니다.], [- 그 중 하나는 작은 수부터 빈 곳에 채워 넣는 것입니다.], [- 왜 이 방법이 가능할까요?], ), ( [- 채워야하는 곳은 $B_(i-1) = B_i$인 $A_i$입니다.], [- 해당 방법으로 빈 곳을 배치하면 $B_i < A_i$입니다.], [- #math.op("MEX")값이 $B_i$인 배열에 $B_i$보다 큰 값을 배치하더라도 #math.op("MEX")값은 변하지 않습니다.], [- 그리고 $B_(k-1) != B_k$인 $k$에 대해 총 $k$개의 공간에 $B_k$보다 작은 $B_k - 1 (<= k)$개의 수를 배치하므로 작은 수부터 배치하면 원하는 결과를 얻을 수 있습니다.], ), ( [- $B_i != B_(i+1)$일 때, $A_(i+1) = B_i$ #v(1em)], [#align(center)[#table(columns:5)[#cell(1)][#cell(2)][#cell(2)][#cell(4)][#cell(6)]]], [#v(-0.5em) $ B $ ], [#align(center)[#table(columns:5)[#cell("?")][#cell(1, color:PALE_RED)][#cell("?")][#cell(2, color:PALE_RED)][#cell(4, color:PALE_RED)]]], [#v(-0.5em) $ A $ ], ), ( [- 남은 값을 작은 수부터 빈 곳에 채워 넣습니다. #v(1em)], [#align(center)[#table(columns:5)[#cell(1)][#cell(2)][#cell(2)][#cell(4)][#cell(6)]]], [#v(-0.5em) $ B $ ], [#align(center)[#table(columns:5)[#cell(3, color:PALE_RED)][#cell(1)][#cell(5, color:PALE_RED)][#cell(2)][#cell(4)]]], [#v(-0.5em) $ A $ ], [- 시간복잡도는 $cal(O)(N)$입니다.], ) ), ( ( [- 하나의 등불이 다른 등불을 앞지르는 횟수가 곧 소원을 비는 횟수입니다.], [- 상대적으로 오른쪽의 등불이 상승하는 정도가 더 크면, #linebreak()왼쪽의 등불을 앞지르는 순간이 존재합니다.], [- 길이가 $S$인 구간에서 $i<j$ 이면서 $A_i < A_j$인 $(i, j)$ 쌍의 개수를 구해야 합니다.], ), ( [- 구간 내 원소에 대해서 모든 쌍을 찾는 방법을 먼저 생각해 봅시다.], [- 하나의 구간에 대해 $cal(O)(N^2)$이 걸리므로, 모든 구간에서 계산한다면 $cal(O)(N^3)$으로 제한시간 내에 풀 수 없습니다.], [- 앞지르는 쌍의 개수를 더 빠르게 구할 수 있을까요?], ), ( [- Merge Sort를 응용하면 문제를 조금 더 효율적으로 풀 수 있습니다.], [- 두 배열을 합치는 과정에서, 양쪽 배열은 정렬된 상태입니다.], [- 왼쪽 절반 배열의 원소가 더 작다면, 해당 원소보다 뒤에 존재하는 원소 모두가 오른쪽에서 고른 원소보다 작습니다.], [- 하나의 구간에서 모든 쌍의 개수를 $cal(O)(N log N)$에 구할 수 있습니다.], [- 여전히 모든 구간을 확인하는 데 $cal(O)(N^2 log N)$으로, 더 빠른 방법이 필요합니다.], ), ( [- 문제의 특성을 파악해서 빠르게 풀어 봅시다.], [- $k$번 등불부터 $S$개 고른 구간을 $T_k$라고 합시다.], [- $T_k$ 구간과 $T_(k+1)$ 구간의 공통된 구간은 $[k+1, k+S-1]$입니다.], [- 공통된 구간에서 발생하는 앞지르는 쌍의 개수는 $k$번째 등불과 $k+S$번째 등불에 영향받지 않습니다.], [- 해당 구간의 연산을 매 구간마다 반복하지 않으면 효율적으로 풀 수 있습니다.], ), ( [- $T_k $구간에서 다음 구간으로 넘어갈 때 아래의 값들을 빼고 더해줘야 합니다.], [#h(2em) #emoji.ast $k$ 번째 등불을 제거했을 때 감소하는 앞지르는 쌍의 개수], [#h(2em) #emoji.ast $k+S$ 번째 등불을 추가할 때 증가하는 앞지르는 쌍의 개수], [], [- 세그먼트 트리를 활용하면 앞지르는 쌍의 개수를 빠르게 구할 수 있습니다.], [#h(2em)#emoji.ast $k$번째 등불을 제거할 때, $A_k$보다 상승하는 정도가 큰 등불의 개수], [#h(2em)#emoji.ast $k+S$번째 등불을 추가할 때, $A_(k+S)$보다 상승하는 정도가 작은 등불의 개수] ), ( [- 주어진 등불 상승 속도의 범위가 최대 $10^9$이므로 좌표압축을 먼저 진행해야 합니다.], [- 시간복잡도는 $cal(O)(N log N)$입니다.], ), ), ( ( [- BFS의 순회 순서대로 정점에 번호를 붙이겠습니다. 정점을 적절히 배치해서 순회 순서의 차를 최대화해야 합니다.], [- 루트 바로 아래에 $k$개의 정점이 존재한다고 합시다.], [- 트리의 $i(1 <= i <= N)$번 정점을 루트로 하는 서브트리의 크기를 $S_i$라고 합시다.], [- $sum_(i=2)^(k+1) S_i$ 는 루트 노드를 제외한 트리의 크기이므로 $N-1$입니다.] ), ( [- $D_2 = B_2 = 2$이고, $2 < i <= k+1$인 $i$에 대한 $D_i$와 $B_i$값은 아래와 같습니다.], [#h(2em) #emoji.ast $D_i = sum_(j=2)^(i-1)S_j + 2$], [#h(2em) #emoji.ast $B_i = i$#v(1em)], [- $sum_(i=2)^(k+1)\|D_i-B_i\|=&\|2-2\|+\|S_2+2-3\|+\|S_2+S_3+2-4\|+dots#linebreak() & + \|sum^(k)_(j=2)S_j+2-(k+1)\| #linebreak() &=\|S_2-1\| + \|S_2+S_3-2\| + dots + \|sum^(k)_(j=2)S_j-(k-1)\|$], [- 여기에서 서브트리의 크기는 1 이상이므로, 절댓값 내부의 식을 빼내서 식을 정리할 수 있습니다.] ), ( [- $sum_(i=2)^(k+1)\|D_i-B_i\| &= (S_2 - 1) + (S_2+S_3-2) + dots + (sum_(j=2)^(k)S_j-(k-1))#linebreak() &= (k-1)S_2 + (k-2)S_3 + ... + S_k - sum_(i=1)^(k-1)\i$], [- 위 식이 최대가 되도록 $S_i$를 배분하는 것은 $S_2=N-k$, 나머지는 $1$로 두는 것입니다.], [- $2$번 정점을 루트로 하는 서브트리에서 노드를 배치하는 형태는 $sum_(i=1)^N\|D_i-B_i\|$를 최대화하는 데 영향을 미치지 않습니다.], [- 따라서 $k$값 중, 문제의 정답을 최대화하는 값을 찾아야 합니다.], ), ( [#v(-2em)#align(center)[#image("images/bfsdfs.png", width: 50%)]], [- 순회 순서의 차를 보면, $N-k-1$개에 해당하는 #math.op("BFS") 순서는 $k+1$씩 밀려납니다.], [- 반대로, $k-1$개의 #math.op("DFS") 순서는 $N-k+1$씩 밀립니다.], [- 따라서 $(k-1) times (N-k+1) + (k+1) times (N-k-1)$의 최댓값을 구해야 합니다.] ), ( [- 식을 펼쳐 계산하면 $k=N/2$인 경우에 최대가 되며, 언급한 대로 트리를 구축해주면 됩니다.], ) ), ( ( [- 팰린드롬 애너그램에서 보였듯, $N$이 짝수인 경우 항상 가능합니다.], [- 홀수인 경우에는 한가운데의 원소가 자신의 자리에 있을 때에만 가능합니다.], [- 어떻게 최소한의 연산으로 원하는 배열을 만들 수 있을까요?] ), ( [- 다음과 같은 배열을 봅시다.], [#align(center)[#table(columns:6)[#cell(2, color:PALE_RED)][#cell(1,color:PALE_RED)][#cell(5,color:blue)][#cell(3,color:blue)][#cell(6,color:blue)][#cell(4,color:blue)]]], [- 이 배열이 정렬되기 위해서는 어떻게 해야 할 지 생각해 봅시다.], [- 각 원소가 자신의 자리로 돌아가기 위해서, 현재 적혀있는 수에 해당하는 칸으로 간선을 그어 봅시다.], [- 배열에서 사이클을 분리해낼 수 있습니다. 사이클의 종류는 총 세 가지입니다.], [- 왼쪽 절반 또는 오른쪽 절반으로만 이루어져 있는 사이클, 교차하는 사이클], [- 각각을 $L$, $R$, $M$ 사이클이라고 정의합시다.] ), ( [- $L$ 사이클과 $R$ 사이클은 그 자체로 교환이 불가합니다.], [- $M$ 사이클로 만들어서 정렬하는 과정이 필요합니다.], [- $1$회 교환을 통해서 원소가 한 개 늘어난 $M$ 사이클을 만들 수 있습니다.], [- 이때, $L$과 $R$ 사이클이 둘 다 존재한다면, 두 사이클의 원소를 서로 교환해서 하나의 $M$ 사이클을 만들 수 있습니다.], [- 이는 각 사이클을 $M$ 사이클로 만들었을 때보다 교환을 $1$번 적게 할 수 있으므로, 가능하다면 $L$, $R$ 사이클 간 교환을 통해 $M$ 사이클을 만드는 것이 이득입니다.], ), ( [- 이제 $M$ 사이클을 올바르게 배치하는 데 드는 비용을 알아봅시다.], [- 크기가 $2$인 $M$ 사이클을 정렬하는 데에는 $1$번의 교환이면 충분합니다.], [- 크기가 $i$인 $M$ 사이클을 정렬하는 데 걸리는 비용이 $k$라고 합시다.], [- 크기가 $i+1$인 $M$ 사이클을 정렬할 때, 세 개 이상의 원소를 동시에 교환할 수 있는 방법이 없으므로 $1$번의 교환을 통해 크기가 $i$인 $M$ 사이클을 만드는 것이 최선입니다.], [- $i+1$ 개의 원소를 가지는 $M$ 사이클은 $k+1$번 교환을 통해 정렬할 수 있습니다.], [- $i=2$일 때 $k=1$이므로, 크기가 $N$인 $M$ 사이클은 $N-1$번의 교환을 통해 정렬할 수 있습니다.] ), ( [- 따라서 $L$, $R$, $M$ 사이클의 개수를 각각 구해준 뒤, $L$, $R$ 사이클 쌍들을 $M$ 사이클들로 만들어 줍시다.], [- 남는 $L$, $R$ 사이클은 크기가 $1$ 증가한 $M$ 사이클이 됩니다.], [- 기존 $M$ 사이클은 각 사이클의 원소의 개수보다 $1$ 작은 횟수로 정렬이 가능합니다. ], [- $M$ 사이클을 구성하는 왼쪽/오른쪽 원소의 개수가 같은 쪽으로 가도록 교환하면 $M$ 사이클을 유지하면서 크기를 $1$씩 줄여나갈 수 있습니다.], [- 사이클의 개수, 종류, 구성하는 원소 등을 조합해 위에서 설명한 대로 답을 도출할 수 있습니다.], [- 시간복잡도는 $cal(O)(N)$입니다.] ) ), ) #let create_page(index) = { set list(marker: text(fill:KUPC_GREEN)[✓]) for pg in descriptions.at(index) { [ #constructTitle(contest_problems.at(index), size: 2em, bookmark:false) #v(5em) ] for desc in pg { set text(size: 2em) pad(left: -1.5em)[#desc] v(0em); } pagebreak(weak: true) } }
https://github.com/connachermurphy/typst-cv	https://raw.githubusercontent.com/connachermurphy/typst-cv/main/works_in_progress.typ	typst	MIT License	#let items = ( [#quote[A project in progress, sure to revolutionize the field.]], ) #list(..items)
https://github.com/goshakowska/Typstdiff	https://raw.githubusercontent.com/goshakowska/Typstdiff/main/tests/test_working_types/super_script/super_script_inserted.typ	typst		First#super[super text] Normal text#super[inserted super text] Second#super[super text]
https://github.com/Ciolv/typst-template-bachelor-thesis	https://raw.githubusercontent.com/Ciolv/typst-template-bachelor-thesis/main/chapter.typ	typst		// Has to be imported wherever acronyms are used #import "acronyms.typ": ac,acl,acs,acsp,acp,aclp,acronyms = This chapter is included from another file <chap:useful-guides> Let's see in @lst:hellew-wörld how nicely syntax highlighting works in #link("https://typst.app", "Typst").\ I really like it! #figure( caption: "Rust sample", )[ #set par(leading: 0.75em) #set align(left) // We really don't want our code to be centered line by line... ```rust fn main() { println!("Hello, world!"); } ``` ]<lst:hellew-wörld> Awesome, isn't it? \ Well, if you're coming from a nice Markdown editor, there is nothing new with it, but in comparison to L#super(size: 0.8em,baseline: -0.2em)[A]T#sub(size: 0.8em, baseline: 0.2em)[E]X? No extra package required! \ \ By the way, do you like @harry? = Test <chap:test>
https://github.com/kaarmu/splash	https://raw.githubusercontent.com/kaarmu/splash/main/src/palettes/okabe-ito.typ	typst	MIT License	/* Color scheme by <NAME> and <NAME> * * Source: https://jfly.uni-koeln.de/color/ * Accessed: 2023-06-16 */ #let okabe-ito = ( black : rgb("#000000"), orange : rgb("#E69F00"), sky-blue : rgb("#56B4E9"), bluish-green : rgb("#009E73"), yellow : rgb("#F0E442"), blue : rgb("#0072B2"), vermilion : rgb("#D55E00"), reddish-purple : rgb("#CC79A7"), )
https://github.com/jgm/typst-hs	https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/meta/footnote-01.typ	typst	Other	// Test space collapsing before footnote. A#footnote[A] \ A #footnote[A]
https://github.com/typst/webapp-issues	https://raw.githubusercontent.com/typst/webapp-issues/main/README.md	markdown		# Web App Issues Official issue tracker for [Typst's web app.][app] Here, you can report bugs or send feature requests for the official web app. If you want to report a bug with the Typst language or compiler instead, please [open an issue here][compiler]. ## FAQ - Will the web app remain free? \ The web app will always have a free tier. However, there is also a paid plan with additional features like Comments and Git sync. - Is the web app open source? \ The web app is not open source. We think open-sourcing the compiler and keeping the app proprietary is a fair division that allows the project to sustain itself long-term. - Can I self-host the web app? \ Yes! We offer a paid on-premises version for organizations. If this is interesting for you or your team, please reach out to us at <<EMAIL>>. - Will there be a desktop app? \ We plan to provide a desktop version of the web app down the road, but can't give a timeframe for this at the moment. - Where do you store my data / is my data safe? \ Your data is stored in a Microsoft Azure data center in Germany and encrypted at rest. We can access your documents, but do so only to fix problems on your request or to enforce our terms of service. Read our [privacy policy] for more details. [app]: https://typst.app [compiler]: https://github.com/typst/typst [privacy policy]: https://typst.app/privacy [paid plan]: https://typst.app/pricing
https://github.com/luiswirth/numpde-slides	https://raw.githubusercontent.com/luiswirth/numpde-slides/main/src/week03.typ	typst		#import "setup.typ": * #show: this-template #let pathemph(a, b) = [ #text(fill: white.darken(60%))[#a]#b ] #titleslide("03") #pagebreak() #githubref #pagebreak() = Setting Hilbert space $V_0$ \ Symmetric Positive-Definite Bilinear form $a: V times V -> RR$ \ Continuous Linear form $l: V -> RR$ \ Gives rise to linear variational problem $ u in V_0: quad a(u, v) = l(v) quad forall v in V_0 $ Notice that here we are considering only the vector space and not the affine space. Meaning that a conversion is needed, if the problem was originally posed on an affine space. Is will be done using the offset function trick. $hat(u) = u_0 + tilde(u)$ #pagebreak() Finite dimensional subspace #text(size: 60pt)[$ V_(0, h) subset V_0 \ dim V_(0,h) < oo $] Discrete linear variational problem (DVP) $ u_h in V_(0,h): quad a(u_h, v_h) = l(v_h) quad forall v_h in V_(0,h) $ #pagebreak() #cetz.canvas(length: 3cm, { import cetz.draw: * let box_text_size = 16pt let lvp = text(box_text_size)[$ u in V_0: quad a(u, v) = l(v) quad forall v in V_0 $] let dvp = text(box_text_size)[$ u_h in V_(0,h): quad a(u_h, v_h) = l(v_h) quad forall v_h in V_(0,h) $] let cmin = text(box_text_size)[$ u = argmin_(v in V_0) J(v) $] let dmin = text(box_text_size)[$ u_h = argmin_(v_h in V_(0,h)) J(v_h) $] set-style( mark: (fill: white, scale: 2), line: (stroke: white), circle: (stroke: white), stroke: (thickness: 0.4pt, cap: "round"), content: (padding: 5pt) ) rect((0, 0), (3.0, 1), stroke: (paint: white, thickness: 1pt),name: "rect0") rect((5, 0), (9, 1), stroke: (paint: white, thickness: 1pt),name: "rect1") rect((0, -2), (3.0, -1), stroke: (paint: white, thickness: 1pt),name: "rect2") rect((5, -2), (9, -1), stroke: (paint: white, thickness: 1pt),name: "rect3") content("rect0", lvp) content("rect1", dvp) content("rect2", cmin) content("rect3", dmin) line("rect0", "rect1", mark: (end: "stealth"), name: "line0") line("rect2", "rect3", mark: (end: "stealth"), name: "line1") content(("line0.start", 50%, "line0.end"), align(center)[Galerkin\ Discretization], anchor: "south") content(("line1.start", 50%, "line1.end"), align(center)[Ritz\ Discretization], anchor: "south") content(("rect0", 50%, "rect2"), text(40pt, sym.arrow.t.b.double)) content(("rect1", 50%, "rect3"), text(40pt, sym.arrow.t.b.double)) }) #pagebreak() $ frak(B)_h = {b_h^1, dots, b_h^N} quad N = dim V_(0,h) \ V_(0,h) = "span" frak(B)_h \ u in V_(0,h) ==> u = sum_(i=1)^N mu_i b_h^i quad u tilde.eq vvec(mu) in RR^N \ v in V_(0,h) ==> v = sum_(i=1)^N nu_i b_h^i quad v tilde.eq vvec(nu) in RR^N $ #pagebreak() $ u_h in V_(0,h):&& quad bilf(a)(u_h, v_h) &= linf(l)(v_h) quad &&forall v_h in V_(0,h) \ vvec(mu) in RR^N:&& quad bilf(a)(sum_(j=j)^N mu_j b_h^j, sum_(i=1)^N nu_i b_h^i) &= linf(l)(sum_(i=1)^N nu_i b_h^i) quad &&forall avec(nu) in RR^N \ vvec(mu) in RR^N:&& quad sum_(j=1)^N mu_j sum_(i=1)^N nu_i bilf(a)(b_h^j, b_h^i) &= sum_(i=1)^N nu_i linf(l)(b_h^i) quad &&forall avec(nu) in RR^N \ vvec(mu) in RR^N:&& quad sum_(i=1)^N nu_j (sum_(i=1)^N mu_i bilf(a)(b_h^j, b_h^i) - linf(l)(b_h^i)) &= 0 quad &&forall avec(nu) in RR^N \ vvec(mu) in RR^N:&& quad sum_(i=1)^N mu_i bilf(a)(b_h^j, b_h^i) - linf(l)(b_h^i) &= 0 quad &&forall i in {1,dots,N} \ vvec(mu) in RR^N:&& quad sum_(i=1)^N mu_i bilf(a)(b_h^j, b_h^i) &= linf(l)(b_h^i) quad &&forall i in {1,dots,N} \ vvec(mu) in RR^N:&& quad amat(A) vvec(mu) &= vvec(phi) quad&& $ with $ &vvec(mu) &&= [mu_1, dots, mu_N]^transp &&in RR^N \ &amat(A) &&= [bilf(a)(b_h^j, b_h^i)]_(i,j=1)^N &&in RR^(N times N) \ &vvec(phi) &&= [linf(l)(b_h^i)]_(i,j=1)^N &&in RR^N $ LSE!!! $ amat(A) vvec(mu) = vvec(phi) $ The solution can then be recovered by $ u_h = sum_(i=1)^N mu_i b_h^i $ #pagebreak() Concrete 1D Galerkin Discretization $ u in H^1_0 (]a,b[): quad integral_a^b (dif u)/(dif x)(x) (dif v)/(dif x)(x) dif x = integral_a^b f(x) v(x) dif x quad forall v in H^1_0 (]a,b[) $
https://github.com/soul667/typst	https://raw.githubusercontent.com/soul667/typst/main/PPT/MATLAB/touying/docs/docs/utilities/oop.md	markdown		--- sidebar_position: 1 --- # Object-Oriented Programming Touying provides some convenient utility functions for object-oriented programming. --- ```typst #let empty-object = (methods: (:)) ``` An empty class. --- ```typst #let call-or-display(self, it) = { if type(it) == function { return it(self) } else { return it } } ``` Call or display as-is. --- ```typst #let methods(self) = { .. } ``` Used to bind self to methods and return, very commonly used.
https://github.com/jgm/typst-hs	https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/math/frac-05.typ	typst	Other	// Test associativity. $ 1/2/3 = (1/2)/3 = 1/(2/3) $
https://github.com/sitandr/typst-examples-book	https://raw.githubusercontent.com/sitandr/typst-examples-book/main/src/snippets/code.md	markdown	MIT License	# Code formatting ## Inline highlighting ```typ #let r = raw.with(lang: "r") This can then be used like: #r("x <- c(10, 42)") ``` ## Tab size ```````typ #set raw(tab-size: 8) ```tsv Year Month Day 2000 2 3 2001 2 1 2002 3 10 ``` ``````` ## Theme See [reference](https://typst.app/docs/reference/text/raw/#parameters-theme) ## Enable ligatures for code ```typ #show raw: set text(ligatures: true, font: "Cascadia Code") Then the code becomes `x <- a` ``` ## Advanced formatting See [packages](../packages/code.md) section.
https://github.com/FriendlyUser/IntroductionToTypst	https://raw.githubusercontent.com/FriendlyUser/IntroductionToTypst/main/README.md	markdown	Apache License 2.0	# IntroductionToTypst Source Code for introduction to Typst Article
https://github.com/mumblingdrunkard/mscs-thesis	https://raw.githubusercontent.com/mumblingdrunkard/mscs-thesis/master/src/computer-architecture-fundamentals/anatomy-of-an-in-order-pipelined-processor.typ	typst		== Anatomy of an In-order Pipelined Processor <sec:pipelined-processor> This first major optimisation of the microarchitecture is based on the observation that all instructions have required steps in common, and that the components used in each step are usually different. This introduces the concept of the _pipelined processor_. A classic processor pipeline may look like: instruction fetch (IF), instruction decode (ID), operand fetch (OF), execute (EXE), memory access (MEM), and writeback (WB). Each stage focuses on a specific stage of performing an instruction, like an assembly line where each step adds a new part to the product. IF fetches the instruction from memory. ID decodes the instruction and determines which control signals to set later in the pipeline. OF fetches the values to be used later in the pipeline. EXE performs the operation on some of the values. MEM performs memory access with an address computed in EXE. WB writes the values back to the register file so that they can be used by later instructions. Between each stage, there is a big register called a _pipeline register_ that holds values and control signals that tells the various stages what work to perform. The values come from outputs of previous stages, and are used as inputs in the current stage. Each stage takes only one cycle to complete This is a form of _instruction-level parallelism_ (ILP): the observation that a processor can work on many instructions at the same time because each instruction requires different parts of the processor at any given time. @fig:pipelined-cpu shows a high-level overview of a pipelined processor. In this version, the ID and OF stages are merged together, meaning values are read out of the register file at the same time the instruction is being decoded. Each stage is separated by a pipeline register. #figure( ```monosketch ┌───────────────────────────┐ │ ┌────────┬────┐ │ │ │ ┌───┐ │ │ │ ┌──┐ ║ ┌──▼──┐ ║ ┌▼─▼┐ ╟┘┌─▼─┐ ╟┘┌──┐ │ │IF├▶╟▶│ID/OF├▶╟▶│EXE├▶╟▶│MEM├▶╟▶│WB├─┘ └─▲┘ ║ └─────┘ ║ └───┘ ╟┐└───┘ ║ └──┘ └─────────────────────┘ ```, caption: [A high-level overview of a pipelined microarchitecture], kind: image, )<fig:pipelined-cpu> The connection from the EXE/MEM pipeline register to IF is to allow branch and jump instructions to change the PC. The connection from WB to OF is there to allow the WB stage to write values back to the register file which is traditionally stored where operand fetch is performed. === Overlapping Execution A simple pipelined processor like this can perform any single instruction in just five cycles, which is a good deal better than the shared bus architecture. However, the real trick is to overlap execution of multiple consecutive instructions. When an instruction moves from IF into ID, the IF stage is freed up and can start fetching the next instruction. This holds for every stage. Because of this, a pipelined processor can finish executing one instruction every cycle. ==== Hazards With overlapping pipelining comes execution _hazards_. Hazards arise when instructions depend on results from older instructions that have not yet completed. For some of these, the results are ready and available in pipeline registers even if they are not yet written to the register file. In this case, the values can be _forwarded_ to the stages where they are needed. The connections from the EXE/MEM and MEM/WB pipeline registers to EXE and from MEM/WB to MEM are there for forwarding. When a stage detects that an instruction in either of these later stages is going to write to one of its own source registers, it will use the value from the pipeline registers instead. Some hazards cannot be dealt with by only forwarding. For example: when one instruction reads from memory, and the following instruction depends on the result in the EXE stage, the procsessor has to _stall_ for a cycle. ==== Branches All instructions that enter IF after a branch have a dependency on the branch. The simplest thing is to stall IF until the branch instruction has left EXE and potentially modified the PC. A possible step up is to assume that the branch condition will resolve to "False" and to keep fetching. If the assumption turns out to be correct, three cycles have been saved. If the assumption turns out to be wrong, the results of the incorrectly fetched instructions must be _squashed_ (ignored). The next step up is to observe patterns in branch instructions and predict the outcome with more accuracy to prevent squashing too often. This is the founding basis of _branch prediction_, a form of _speculation_.
https://github.com/typst/packages	https://raw.githubusercontent.com/typst/packages/main/packages/preview/unichar/0.1.0/ucd/block-1CD0.typ	typst	Apache License 2.0	#let data = ( ("VEDIC TONE KARSHANA", "Mn", 230), ("VEDIC TONE SHARA", "Mn", 230), ("VEDIC TONE PRENKHA", "Mn", 230), ("VEDIC SIGN NIHSHVASA", "Po", 0), ("VEDIC SIGN YAJURVEDIC MIDLINE SVARITA", "Mn", 1), ("VEDIC TONE YAJURVEDIC AGGRAVATED INDEPENDENT SVARITA", "Mn", 220), ("VEDIC TONE YAJURVEDIC INDEPENDENT SVARITA", "Mn", 220), ("VEDIC TONE YAJURVEDIC KATHAKA INDEPENDENT SVARITA", "Mn", 220), ("VEDIC TONE CANDRA BELOW", "Mn", 220), ("VEDIC TONE YAJURVEDIC KATHAKA INDEPENDENT SVARITA SCHROEDER", "Mn", 220), ("VEDIC TONE DOUBLE SVARITA", "Mn", 230), ("VEDIC TONE TRIPLE SVARITA", "Mn", 230), ("VEDIC TONE KATHAKA ANUDATTA", "Mn", 220), ("VEDIC TONE DOT BELOW", "Mn", 220), ("VEDIC TONE TWO DOTS BELOW", "Mn", 220), ("VEDIC TONE THREE DOTS BELOW", "Mn", 220), ("VEDIC TONE RIGVEDIC KASHMIRI INDEPENDENT SVARITA", "Mn", 230), ("VEDIC TONE ATHARVAVEDIC INDEPENDENT SVARITA", "Mc", 0), ("VEDIC SIGN VISARGA SVARITA", "Mn", 1), ("VEDIC SIGN VISARGA UDATTA", "Mn", 1), ("VEDIC SIGN REVERSED VISARGA UDATTA", "Mn", 1), ("VEDIC SIGN VISARGA ANUDATTA", "Mn", 1), ("VEDIC SIGN REVERSED VIS<NAME>", "Mn", 1), ("VEDIC SIGN VISARGA UDATTA WITH TAIL", "Mn", 1), ("VEDIC SIGN VISARGA ANUDATTA WITH TAIL", "Mn", 1), ("VEDIC SIGN ANUSVARA ANTARGOMUKHA", "Lo", 0), ("VEDIC SIGN ANUSVARA BAHIRGOMUKHA", "Lo", 0), ("VEDIC SIGN ANUSVARA VAMAGOMUKHA", "Lo", 0), ("VEDIC SIGN ANUSVARA VAMAGOMUKHA WITH TAIL", "Lo", 0), ("VEDIC SIGN TIRYAK", "Mn", 220), ("VEDIC SIGN HEXIFORM LONG ANUSVARA", "Lo", 0), ("VEDIC SIGN LONG ANUSVARA", "Lo", 0), ("VEDIC SIGN RTHANG LONG ANUSVARA", "Lo", 0), ("VEDIC SIGN ANUSVARA UBHAYATO MUKHA", "Lo", 0), ("VEDIC SIGN ARDHAVISARGA", "Lo", 0), ("VEDIC SIGN ROTATED ARDHAVISARGA", "Lo", 0), ("VEDIC TONE CANDRA ABOVE", "Mn", 230), ("VEDIC SIGN JIHVAMULIYA", "Lo", 0), ("VEDIC SIGN UPADHMANIYA", "Lo", 0), ("VEDIC SIGN ATIKRAMA", "Mc", 0), ("VEDIC TONE RING ABOVE", "Mn", 230), ("VEDIC TONE DOUBLE RING ABOVE", "Mn", 230), ("VEDIC SIGN DOUBLE ANUSVARA ANTARGOMUKHA", "Lo", 0), )
https://github.com/Enter-tainer/typstyle	https://raw.githubusercontent.com/Enter-tainer/typstyle/master/tests/assets/typstfmt/138-non-converge-block-comment.typ	typst	Apache License 2.0	#let test_func() = { ( /* test */ ) }
https://github.com/lyzynec/orr-go-brr	https://raw.githubusercontent.com/lyzynec/orr-go-brr/main/04/main.typ	typst		#import "../lib.typ": * #knowledge[ #question(name: [Give the first-order necessary conditions of optimality for a general optimal control problem for a nonlinear discrete-time system over a finite horizon. Namely, give the general two-point boundary value problem, highlighting the state equation, the co-state equation and a stationarity equation. Do not forget to include general boundary conditions.])[ The augmented cost function is $ J'_i (bold(x)_k, bold(u)_k, bold(lambda)_(k+1)) = phi.alt(bold(x)_N, N) + sum_(k=1)^(N-1) [L_k (bold(x)_k, bold(u)_k) + bold(lambda)_(k+1)^T [bold(f)_k (bold(x)_k, bold(u)_k) - bold(x)_(k+1)]] $ To null the gradient as $gradient J'_i = 0$ we have to satisfy $ x_(k+1) &= gradient_lambda_(k+1) H_k(bold(x)_k, bold(u)_k, bold(lambda)_(k+1)) = f_k (x_k,u_k) "(state equation)"\ lambda_k &= gradient_bold(x)_k H_k(bold(x)_k, bold(u)_k, bold(lambda)_(k+1)) "(costate equation)"\ 0 &= gradient_bold(u)_k H_k(bold(x)_k, bold(u)_k, bold(lambda)_(k+1)) "(stationarity equation)"\ 0 &= gradient_bold(x)_i H_k(bold(x)_k, bold(u)_k, bold(lambda)_(k+1)) dot upright(d) bold(x)_i, "for fixed " bold(x)_i = bold(r)_i => upright(d) bold(x)_i = bold(0) \ 0 &= (gradient_bold(x)_N phi.alt - bold(lambda)_N)^T dot d x_N = cases("fixed " bold(x)_N &=> bold(x)_N = bold(r)_N => upright(d) bold(x)_N = bold(0), "free " bold(x)_N &=> upright(d) bold(x)_N != 0 +> bold(lambda)_N = gradient_(bold(lambda)_N) phi.alt)\ $ where $ H(bold(x)_k, bold(u)_k, bold(lambda)_(k+1)) = L_k (bold(x)_k, bold(u)_k) + bold(lambda)_(k+1)^T bold(f)_k (bold(x)_k, bold(u)_k) $ As starting point is probably given, the boundary condition could be something like $ bold(x)_0 = bold(r)_0\ $ for the end state, it could be given $ bold(x)_N = bold(r)_N $ or could be subject to optimization as $ bold(lambda)_N = gradient phi.alt(bold(x)_N) $ ] #question(name: [Give the first-order necessary conditions of optimality for a linear and time invariant (LTI) discrete-time system and a quadratic cost function over a finite horizon. Namely, give them in the format displaying the state equation, co-state equation and stationarity equation. Show and discuss also two types of boundary conditions.])[ With cost function $ J = 1/2 bold(x)_N^T bold(S)_N bold(x)_N + 1/2 sum_(k=0)^(N-1) [bold(x)_k^T bold(Q) bold(x)_k + bold(u)_k^T R u_k] $ the Hamiltonian $H$ function would be $ H_k = 1/2 (bold(x)_k^T bold(Q) bold(x)_k + bold(u)_k^T bold(R) bold(u)_k) + bold(lambda)_(k+1)^T (bold(A) bold(x)_k + bold(B) bold(u)_k) $ the equations would become #align(center)[#grid(columns: 2, row-gutter: 10pt, column-gutter: 10pt, align: left, [state:], $bold(x)_(k+1) = bold(A) bold(x)_k + bold(B) bold(u)_k$, [costate:], $bold(lambda)_k = bold(Q) bold(x)_k + bold(A)^T bold(lambda)_(k+1)$, [stationarity:], $bold(0) = bold(R) bold(u)_k + bold(B)^T bold(lambda)_(k+1)$, [boundary:], $bold(x)_0 = bold(r)_0$, [boundary:], $bold(0) &= (bold(S)_N bold(x)_N - bold(lambda)_N)^T upright(d) bold(x)_N$, )] There is other type for the last boundary condition, as it could also be given as $ bold(x)_N = bold(r)_N $ From the stationarity equation we can extract optimal control as $ bold(u)_k = - bold(R)^(-1) bold(B)^T bold(lambda)_(k+1) $ Than we can subtitute the optimal control into the rest to obtain state and costate values. ] #question(name: [Give a qualitative characterization of the solution to the fixed final state LQ-optimal control problem over a finite horizon, that is, you do not have to give formulas but you should be able to state among the highlights that the control is open-loop and that reachability of the system is a necessary condition.])[ - the reslt will be offline precalculated control signal, meaining it is open--loop - the control will be proportional to $bold(r)_N - bold(A)^N bold(x)_0$ meaning the difference between desired end--time state and state in which the system would end up without any control (this one is pretty reasonable) - the control will be inversly proportional to _reachability Gramian_ $ bold(G)_(0,N,bold(R)) = sum_(i=0)^(N-1) bold(A)^(N-i-1) bold(B) bold(R)^(-1) bold(B)^T (A^T)^(N-i-1) $ if the _reachability Gramian_ is singular, it means the state is not reachable, barring us from calculating optimal control for reaching it (again pretty reasonable) ] #question(name: [Give a qualitative characterization of the solution to the free final state LQ-optimal control problem over a finite horizon, that is, you do not have to give formulas but you should be able to state among the highlights that the control is closed-loop, namely, a time-varying linear state feedback and that the feedback gains can be computed by solving a difference Riccati equation.])[ - the problem results in time varying linear state feedback, noteworthy is the fact that the state feedback gain increases drasticcaly when nearing the horizon - the Riccati equation looks like this $ bold(S)_k = bold(Q) + bold(A)^T bold(S)_(k+1) (bold(I) + bold(B) bold(R)^(-1) bold(B)^T bold(S)_(k+1))^(-1) bold(A) $ - this Riccati equation was constructed with the assumption that the final state boundary equation $ bold(S)_N bold(x)_N = bold(lambda)_N $ holds for any $k$ not just $N$ $ bold(S)_k bold(x)_k = bold(lambda)_k $ ] #question(name: [Discuss how solution to the free final state LQ problem changes as the horizon is extended to infinity. Emphasize that the optimal solution is given by a constant linear state feedback whose gains are computed by solving a discrete-time algebraic Riccati equation (DARE). What are the conditions under which a stabilizing solution is guaranteed to exist? What are the conditions under which it is guaranteed that there is a unique stabilizing solution of DARE?])[ - the solution is simmilar to the previous one with the exception that, as we never reach the horizon, the state feedback gain remains constant - DARE (discrete time algebraic Riccati equation) assumes that in the steady state (as $k -> oo$) $ bold(S)_(k+1) approx bold(S)_k $ it, works because this assumption is correct - for stabilizing solution to exist, there are two conditions - system $(bold(A), sqrt(bold(Q)))$ is stabilizable #footnote[Non--controllable parts of system are stable] - system $(bold(A), sqrt(bold(Q)))$ is detectable #footnote[Non--observable patrts of system are stable] - unique stablilizing solution requires that in addition to existance of stabilizing solution, the system $(bold(A), sqrt(bold(Q)))$ is detectable ] ] #skills[ #question(name: [Design an LQ--optimal state feedback controller for a discrete--time linear system both for a finite and an infinite horizon, both for regulation and for tracking.])[] ]
https://github.com/LDemetrios/Typst4k	https://raw.githubusercontent.com/LDemetrios/Typst4k/master/src/test/resources/suite/layout/limits.typ	typst		// Test how the layout engine reacts when reaching limits like // zero, infinity or when dealing with NaN. --- issue-1216-clamp-panic --- #set page(height: 20pt, margin: 0pt) #v(22pt) #block(fill: red, width: 100%, height: 10pt, radius: 4pt) --- issue-1918-layout-infinite-length-grid-columns --- // Test that passing infinite lengths to drawing primitives does not crash Typst. #set page(width: auto, height: auto) // Error: 58-59 cannot expand into infinite width #layout(size => grid(columns: (size.width, size.height))[a][b][c][d]) --- issue-1918-layout-infinite-length-grid-rows --- #set page(width: auto, height: auto) // Error: 17-66 cannot create grid with infinite height #layout(size => grid(rows: (size.width, size.height))[a][b][c][d]) --- issue-1918-layout-infinite-length-line --- #set page(width: auto, height: auto) // Error: 17-41 cannot create line with infinite length #layout(size => line(length: size.width)) --- issue-1918-layout-infinite-length-polygon --- #set page(width: auto, height: auto) // Error: 17-54 cannot create polygon with infinite size #layout(size => polygon((0pt,0pt), (0pt, size.width)))
https://github.com/kalxd/morelull	https://raw.githubusercontent.com/kalxd/morelull/master/doc.typ	typst		#import "@local/morelull:0.5.0": * #show: morelull.with(标题: "这里写上一个标题") = 大大的标题写在这里。 #t 这里又是另一个#underline[故事]了。 #t 一块硬盘,容量 1T,作为应用#underline[数据盘],一般性的#underline[程序]都放在上面,包括所产生的数据;另一块容量 4T,作为媒体数据盘,数字媒体、个人数据都在上面。 #t 这里又加起新的段，你看看效果。 ```rust fn main() { println!("hello world") } ``` #t 这里再启动一个新的东西。
https://github.com/JunzheShen/SJTU-Resume-Template-in-Typst	https://raw.githubusercontent.com/JunzheShen/SJTU-Resume-Template-in-Typst/main/README.md	markdown		# SJTU Resume Template in Typst This repo provides a Typst version of a SJTU resume template called 蓝色梦想. Note that there are minor differences between this version and the official Microsoft Word version, feel free to adjust some parameters to make make the two versions more alike. This repo is is built upon [OrangeX4/Chinese-Resume-in-Typst](https://github.com/OrangeX4/Chinese-Resume-in-Typst) Detailed instructions to use this repo will be updated when I have some time to kill.
https://github.com/TypstApp-team/typst	https://raw.githubusercontent.com/TypstApp-team/typst/master/tests/typ/meta/figure.typ	typst	Apache License 2.0	// Test figures. --- #set page(width: 150pt) #set figure(numbering: "I") We can clearly see that @fig-cylinder and @tab-complex are relevant in this context. #figure( table(columns: 2)[a][b], caption: [The basic table.], ) <tab-basic> #figure( pad(y: -6pt, image("/files/cylinder.svg", height: 2cm)), caption: [The basic shapes.], numbering: "I", ) <fig-cylinder> #figure( table(columns: 3)[a][b][c][d][e][f], caption: [The complex table.], ) <tab-complex> --- // Testing figures with tables. #figure( table( columns: 2, [Second cylinder], image("/files/cylinder.svg"), ), caption: "A table containing images." ) <fig-image-in-table> --- // Testing show rules with figures with a simple theorem display #show figure.where(kind: "theorem"): it => { let name = none if not it.caption == none { name = [ #emph(it.caption.body)] } else { name = [] } let title = none if not it.numbering == none { title = it.supplement if not it.numbering == none { title += " " + it.counter.display(it.numbering) } } title = strong(title) pad( top: 0em, bottom: 0em, block( fill: green.lighten(90%), stroke: 1pt + green, inset: 10pt, width: 100%, radius: 5pt, breakable: false, [#title#name#h(0.1em):#h(0.2em)#it.body#v(0.5em)] ) ) } #set page(width: 150pt) #figure( $a^2 + b^2 = c^2$, supplement: "Theorem", kind: "theorem", caption: "Pythagoras' theorem.", numbering: "1", ) <fig-formula> #figure( $a^2 + b^2 = c^2$, supplement: "Theorem", kind: "theorem", caption: "Another Pythagoras' theorem.", numbering: none, ) <fig-formula> #figure( ```rust fn main() { println!("Hello!"); } ```, caption: [Hello world in _rust_], ) --- // Test breakable figures #set page(height: 6em) #show figure: set block(breakable: true) #figure(table[a][b][c][d][e], caption: [A table]) --- // Test custom separator for figure caption #set figure.caption(separator: [ --- ]) #figure( table(columns: 2)[a][b], caption: [The table with custom separator.], )
https://github.com/noamzaks/Barvazim	https://raw.githubusercontent.com/noamzaks/Barvazim/main/writeups/bsides-2024/forensics/skibidi.typ	typst		// Category: Forensics #import "../../template.typ": writeup #show: writeup.with(ctf: "BSides", exercise: "Skibidi", date: datetime(day: 29, month: 6, year: 2024)) Every docx file is a ZIP archive. After unzipping the `Skibidi.docx` file, we can go over the different files included. Most seem uninteresting, however, the `img.xml` seems suspicious - why is an image an XML file? And the content is especially suspicious: ```xml <root> <person firstname="<KEY>" lastname="<KEY> city="Haifa" country="Israel" firstname2="<KEY>" lastname2="<KEY> email="<EMAIL>" /> <random>6</random> <random_float>89.838</random_float> <bool>true</bool> <date>1986-09-28</date> <regEx>helloooooooooooooooooooooooooooooooooooooooooooo world</regEx> <enum>generator</enum> <elt>Alyssa</elt><elt>Flory</elt><elt>Ulrike</elt><elt>Teriann</elt><elt>Reeba</elt> <Ulrike> <age>56</age> </Ulrike> </root> ``` Well, the firstname, lastname, firstname2 and lastname2 appear to be Base64 data because they are only letters digits, and they end with the equal signs. By now we have arrived at the solution of this challenge: ```python >>> import base64 >>> base64.b64decode("Qn<KEY>") + base64.b64decode("M2YzNXQzZF90aDNfc2sxYjFkaX0=") b'BsidesTLV2024{w3_d3f35t3d_th3_sk1b1di}' ```
https://github.com/danisltpi/seminar	https://raw.githubusercontent.com/danisltpi/seminar/main/template/ausarbeitung.typ	typst		#import "template.typ": project #import "@preview/cetz:0.2.2" #set text(lang: "de", region: "de") #set par(leading: 1em) // #show outline: set par(leading: 2em) #show heading: it => [#pad(bottom: 1em, top: 1em)[#it]] #show: project.with( title: "Fibonacci Heaps", subtitle: "Seminararbeit", study_program: "Informatik (INFB)", institution: "Hochschule Karlsruhe", date: datetime.today().display("[day].[month].[year]"), examiner: "Prof. Dr. rer. <NAME>", logo: "hka.svg", authors: ((name: "<NAME>", matriculation_number: "79663"),), ) // #show outline.entry.where(level: 1): it => { // strong(it) // } // #outline(indent: auto) #pagebreak(weak: true) = Motivation Ein Fibonacci-Heap ist eine Datenstruktur, die Prioritätswarteschlangen implementiert und werden vielfältig eingesetzt. Sie besteht aus mehreren Heaps (Binärbäume) \ \ #align(center)[ #cetz.canvas({ import cetz.draw: * rect((-1, -1), (1, 1)) circle((0, 0), fill: yellow, stroke: blue) line((0, 0), (2, 1)) line((0, 0), (1.5, -1)) })] == Dijkstra-Algorithmus = Laufzeit Daraus ergibt sich eine Laufzeit von $O(n log n)$ für das Hinzufügen eines Knotens bzw. $Theta(n log n)$ = Das ist ein Test #lorem(100) \ #figure( image("../assets/circle.svg", width: 100%), caption: [Beispielhafte Struktur eines Fibonacci Heaps, bestehend aus 5 Teilbäumen #lorem(50)], gap: 4em, ) <image> = Erklärung Wie man in der vorherigen Abbildung erkennen kann: @image Dies ist besonders gut #lorem(1000) @cormen_introduction_2009 #pagebreak(weak: true) #set page(header: []) #bibliography("literatur.bib", style: "ieee", title: "Literatur")
https://github.com/eLearningHub/resume-typst	https://raw.githubusercontent.com/eLearningHub/resume-typst/main/main.typ	typst	Apache License 2.0	#let portfolio = yaml("portfolio.yaml") #let settings = yaml("settings.yaml") #show link: set text(blue) #show heading: h => [ #set text( size: eval(settings.font.size.heading_large), font: settings.font.general ) #h ] #let sidebarSection = {[ #par(justify: true)[ #par[ #set text( size: eval(settings.font.size.contacts), font: settings.font.minor_highlight, ) Email: #link("mailto:" + portfolio.contacts.email) \ Phone: #link("tel:" + portfolio.contacts.phone) \ LinkedIn: #link(portfolio.contacts.linkedin)[mikhail-liamets] \ GitHub: #link(portfolio.contacts.github)[caffeintazedgaze] \ #portfolio.contacts.address ] #line(length: 100%) ] = Summary #par[ #set text( eval(settings.font.size.education_description), font: settings.font.minor_highlight, ) An experienced software engineer with a confident grasp of infrastructure, system design, and DevOps, now seeking opportunities to excel in the realms of solution architecture. Open to roles ranging from software engineering to DevOps/SRE. ] = Education #{ for place in portfolio.education [ #par[ #set text( size: eval(settings.font.size.heading), font: settings.font.general ) #place.from – #place.to \ #link(place.university.link)[#place.university.name] ] #par[ #set text( eval(settings.font.size.education_description), font: settings.font.minor_highlight, ) #place.degree #place.major ] ] } = Skills #{ for skill in portfolio.skills [ #par[ #set text( size: eval(settings.font.size.description), ) #set text( // size: eval(settings.font.size.tags), font: settings.font.minor_highlight, ) #skill.name #linebreak() #skill.items.join(" • ") ] ] } ]} #let mainSection = {[ // #par[ // #set align(center) // #figure( // image("images/Kodak 20 Zanvoort Lumi.jpg", width: 6em), // placement: top, // ) // ] #par[ #set text( size: eval(settings.font.size.heading_huge), font: settings.font.general, ) #portfolio.contacts.name ] #par[ #set text( size: eval(settings.font.size.heading), font: settings.font.minor_highlight, top-edge: 0pt ) #portfolio.contacts.title ] = Experience #{ for job in portfolio.jobs [ #par(justify: false)[ #set text( size: eval(settings.font.size.heading), font: settings.font.general ) #job.from – #job.to \ #job.position #link(job.company.link)[\@ #job.company.name] ] #par( justify: false, leading: eval(settings.paragraph.leading) )[ #set text( size: eval(settings.font.size.description), font: settings.font.general ) #{ for point in job.description [ #h(0.5cm) • #point \ ] } ] #par( justify: true, leading: eval(settings.paragraph.leading), )[ #set text( size: eval(settings.font.size.tags), font: settings.font.minor_highlight ) ] ] } ]} #{ grid( columns: (2fr, 5fr), column-gutter: 3em, sidebarSection, mainSection, ) }
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("SIGNWRITING TOUCH SINGLE", "So", 0), ("SIGNWRITING TOUCH MULTIPLE", "So", 0), ("SIGNWRITING TOUCH BETWEEN", "So", 0), ("SIGNWRITING GRASP SINGLE", "So", 0), ("SIGNWRITING GRASP MULTIPLE", "So", 0), ("SIGNWRITING GRASP BETWEEN", "So", 0), ("SIGNWRITING STRIKE SINGLE", "So", 0), ("SIGNWRITING STRIKE MULTIPLE", "So", 0), ("SIGNWRITING STRIKE BETWEEN", "So", 0), ("SIGNWRITING BRUSH SINGLE", "So", 0), ("SIGNWRITING BRUSH MULTIPLE", "So", 0), ("SIGNWRITING BRUSH BETWEEN", "So", 0), ("SIGNWRITING RUB SINGLE", "So", 0), ("SIGNWRITING RUB MULTIPLE", "So", 0), ("SIGNWRITING RUB BETWEEN", "So", 0), ("SIGNWRITING SURFACE SYMBOLS", "So", 0), ("SIGNWRITING SURFACE BETWEEN", "So", 0), ("SIGNWRITING SQUEEZE LARGE SINGLE", "So", 0), ("SIGNWRITING SQUEEZE SMALL SINGLE", "So", 0), ("SIGNWRITING SQUEEZE LARGE MULTIPLE", "So", 0), ("SIGNWRITING SQUEEZE SMALL MULTIPLE", "So", 0), ("SIGNWRITING SQUEEZE SEQUENTIAL", "So", 0), ("SIGNWRITING FLICK LARGE SINGLE", "So", 0), ("SIGNWRITING FLICK SMALL SINGLE", "So", 0), ("SIGNWRITING FLICK LARGE MULTIPLE", "So", 0), ("SIGNWRITING FLICK SMALL MULTIPLE", "So", 0), ("SIGNWRITING FLICK SEQUENTIAL", "So", 0), ("SIGNWRITING SQUEEZE FLICK ALTERNATING", "So", 0), ("SIGNWRITING MOVEMENT-HINGE UP DOWN LARGE", "So", 0), ("SIGNWRITING MOVEMENT-HINGE UP DOWN SMALL", "So", 0), ("SIGNWRITING MOVEMENT-HINGE UP SEQUENTIAL", "So", 0), ("SIGNWRITING MOVEMENT-HINGE DOWN SEQUENTIAL", "So", 0), ("SIGNWRITING MOVEMENT-HINGE UP DOWN ALTERNATING LARGE", "So", 0), ("SIGNWRITING MOVEMENT-HINGE UP DOWN ALTERNATING SMALL", "So", 0), ("SIGNWRITING MOVEMENT-HINGE SIDE TO SIDE SCISSORS", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE FINGER CONTACT", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE FINGER CONTACT", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE SINGLE STRAIGHT SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE SINGLE STRAIGHT MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE SINGLE STRAIGHT LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE SINGLE STRAIGHT LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE SINGLE WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE DOUBLE STRAIGHT", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE DOUBLE WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE DOUBLE ALTERNATING", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE DOUBLE ALTERNATING WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CROSS", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE TRIPLE STRAIGHT MOVEMENT", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE TRIPLE WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE TRIPLE ALTERNATING", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE TRIPLE ALTERNATING WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE BEND SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE BEND MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE BEND LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CORNER SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CORNER MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CORNER LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CORNER ROTATION", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CHECK SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CHECK MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CHECK LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE BOX SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE BOX MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE BOX LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE ZIGZAG SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE ZIGZAG MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE ZIGZAG LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE PEAKS SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE PEAKS MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE PEAKS LARGE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ROTATION-WALLPLANE SINGLE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ROTATION-WALLPLANE DOUBLE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ROTATION-WALLPLANE ALTERNATING", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ROTATION-FLOORPLANE SINGLE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ROTATION-FLOORPLANE DOUBLE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ROTATION-FLOORPLANE ALTERNATING", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE SHAKING", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ARM SPIRAL SINGLE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ARM SPIRAL DOUBLE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ARM SPIRAL TRIPLE", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL AWAY SMALL", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL AWAY MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL AWAY LARGE", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL AWAY LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL TOWARDS SMALL", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL TOWARDS MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL TOWARDS LARGE", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL TOWARDS LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN AWAY SMALL", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN AWAY MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN AWAY LARGE", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN AWAY LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN TOWARDS SMALL", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN TOWARDS MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN TOWARDS LARGE", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN TOWARDS LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE SINGLE STRAIGHT SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE SINGLE STRAIGHT MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE SINGLE STRAIGHT LARGE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE SINGLE STRAIGHT LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE SINGLE WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE DOUBLE STRAIGHT", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE DOUBLE WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE DOUBLE ALTERNATING", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE DOUBLE ALTERNATING WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CROSS", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE TRIPLE STRAIGHT MOVEMENT", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE TRIPLE WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE TRIPLE ALTERNATING MOVEMENT", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE TRIPLE ALTERNATING WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE BEND", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CORNER SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CORNER MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CORNER LARGE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CHECK", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE BOX SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE BOX MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE BOX LARGE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE ZIGZAG SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE ZIGZAG MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE ZIGZAG LARGE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE PEAKS SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE PEAKS MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE PEAKS LARGE", "So", 0), ("SIGNWRITING TRAVEL-FLOORPLANE ROTATION-FLOORPLANE SINGLE", "So", 0), ("SIGNWRITING TRAVEL-FLOORPLANE ROTATION-FLOORPLANE DOUBLE", "So", 0), ("SIGNWRITING TRAVEL-FLOORPLANE ROTATION-FLOORPLANE ALTERNATING", "So", 0), ("SIGNWRITING TRAVEL-FLOORPLANE ROTATION-WALLPLANE SINGLE", "So", 0), ("SIGNWRITING TRAVEL-FLOORPLANE ROTATION-WALLPLANE DOUBLE", "So", 0), ("SIGNWRITING TRAVEL-FLOORPLANE ROTATION-WALLPLANE ALTERNATING", "So", 0), ("SIGNWRITING TRAVEL-FLOORPLANE SHAKING", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE QUARTER SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE QUARTER MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE QUARTER LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE QUARTER LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE HALF-CIRCLE SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE HALF-CIRCLE MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE HALF-CIRCLE LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE HALF-CIRCLE LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE THREE-QUARTER CIRCLE SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE THREE-QUARTER CIRCLE MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE HUMP SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE HUMP MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE HUMP LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE LOOP SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE LOOP MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE LOOP LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE LOOP SMALL DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WAVE CURVE DOUBLE SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WAVE CURVE DOUBLE MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WAVE CURVE DOUBLE LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WAVE CURVE TRIPLE SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WAVE CURVE TRIPLE MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WAVE CURVE TRIPLE LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE THEN STRAIGHT", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVED CROSS SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVED CROSS MEDIUM", "So", 0), ("SIGNWRITING ROTATION-WALLPLANE SINGLE", "So", 0), ("SIGNWRITING ROTATION-WALLPLANE DOUBLE", "So", 0), ("SIGNWRITING ROTATION-WALLPLANE ALTERNATE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE SHAKING", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE HITTING FRONT WALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE HUMP HITTING FRONT WALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE LOOP HITTING FRONT WALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WAVE HITTING FRONT WALL", "So", 0), ("SIGNWRITING ROTATION-WALLPLANE SINGLE HITTING FRONT WALL", "So", 0), ("SIGNWRITING ROTATION-WALLPLANE DOUBLE HITTING FRONT WALL", "So", 0), ("SIGNWRITING ROTATION-WALLPLANE ALTERNATING HITTING FRONT WALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CURVE HITTING CHEST", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE HUMP HITTING CHEST", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE LOOP HITTING CHEST", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WAVE HITTING CHEST", "So", 0), ("SIGNWRITING ROTATION-WALLPLANE SINGLE HITTING CHEST", "So", 0), ("SIGNWRITING ROTATION-WALLPLANE DOUBLE HITTING CHEST", "So", 0), ("SIGNWRITING ROTATION-WALLPLANE ALTERNATING HITTING CHEST", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WAVE DIAGONAL PATH SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WAVE DIAGONAL PATH MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WAVE DIAGONAL PATH LARGE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CURVE HITTING CEILING SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CURVE HITTING CEILING LARGE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE HUMP HITTING CEILING SMALL DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE HUMP HITTING CEILING LARGE DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE HUMP HITTING CEILING SMALL TRIPLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE HUMP HITTING CEILING LARGE TRIPLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE LOOP HITTING CEILING SMALL SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE LOOP HITTING CEILING LARGE SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE LOOP HITTING CEILING SMALL DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE LOOP HITTING CEILING LARGE DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE WAVE HITTING CEILING SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE WAVE HITTING CEILING LARGE", "So", 0), ("SIGNWRITING ROTATION-FLOORPLANE SINGLE HITTING CEILING", "So", 0), ("SIGNWRITING ROTATION-FLOORPLANE DOUBLE HITTING CEILING", "So", 0), ("SIGNWRITING ROTATION-FLOORPLANE ALTERNATING HITTING CEILING", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CURVE HITTING FLOOR SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CURVE HITTING FLOOR LARGE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE HUMP HITTING FLOOR SMALL DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE HUMP HITTING FLOOR LARGE DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE HUMP HITTING FLOOR TRIPLE SMALL TRIPLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE HUMP HITTING FLOOR TRIPLE LARGE TRIPLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE LOOP HITTING FLOOR SMALL SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE LOOP HITTING FLOOR LARGE SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE LOOP HITTING FLOOR SMALL DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE LOOP HITTING FLOOR LARGE DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE WAVE HITTING FLOOR SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE WAVE HITTING FLOOR LARGE", "So", 0), ("SIGNWRITING ROTATION-FLOORPLANE SINGLE HITTING FLOOR", "So", 0), ("SIGNWRITING ROTATION-FLOORPLANE DOUBLE HITTING FLOOR", "So", 0), ("SIGNWRITING ROTATION-FLOORPLANE ALTERNATING HITTING FLOOR", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CURVE SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CURVE MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CURVE LARGE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CURVE LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE CURVE COMBINED", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE HUMP SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE LOOP SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE WAVE SNAKE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE WAVE SMALL", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE WAVE LARGE", "So", 0), ("SIGNWRITING ROTATION-FLOORPLANE SINGLE", "So", 0), ("SIGNWRITING ROTATION-FLOORPLANE DOUBLE", "So", 0), ("SIGNWRITING ROTATION-FLOORPLANE ALTERNATING", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE SHAKING PARALLEL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE ARM CIRCLE SMALL SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE ARM CIRCLE MEDIUM SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE ARM CIRCLE SMALL DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE ARM CIRCLE MEDIUM DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE ARM CIRCLE HITTING WALL SMALL SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE ARM CIRCLE HITTING WALL MEDIUM SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE ARM CIRCLE HITTING WALL LARGE SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE ARM CIRCLE HITTING WALL SMALL DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE ARM CIRCLE HITTING WALL MEDIUM DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE ARM CIRCLE HITTING WALL LARGE DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WRIST CIRCLE FRONT SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE WRIST CIRCLE FRONT DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE WRIST CIRCLE HITTING WALL SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE WRIST CIRCLE HITTING WALL DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE FINGER CIRCLES SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE FINGER CIRCLES DOUBLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE FINGER CIRCLES HITTING WALL SINGLE", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE FINGER CIRCLES HITTING WALL DOUBLE", "So", 0), ("SIGNWRITING DYNAMIC ARROWHEAD SMALL", "So", 0), ("SIGNWRITING DYNAMIC ARROWHEAD LARGE", "So", 0), ("SIGNWRITING DYNAMIC FAST", "So", 0), ("SIGNWRITING DYNAMIC SLOW", "So", 0), ("SIGNWRITING DYNAMIC TENSE", "So", 0), ("SIGNWRITING DYNAMIC RELAXED", "So", 0), ("SIGNWRITING DYNAMIC SIMULTANEOUS", "So", 0), ("SIGNWRITING DYNAMIC SIMULTANEOUS ALTERNATING", "So", 0), ("SIGNWRITING DYNAMIC EVERY OTHER TIME", "So", 0), ("SIGNWRITING DYNAMIC GRADUAL", "So", 0), ("SIGNWRITING HEAD", "So", 0), ("SIGNWRITING HEAD RIM", "Mn", 0), ("SIGNWRITING HEAD MOVEMENT-WALLPLANE STRAIGHT", "Mn", 0), ("SIGNWRITING HEAD MOVEMENT-WALLPLANE TILT", "Mn", 0), ("SIGNWRITING HEAD MOVEMENT-FLOORPLANE STRAIGHT", "Mn", 0), ("SIGNWRITING HEAD MOVEMENT-WALLPLANE CURVE", "Mn", 0), ("SIGNWRITING HEAD MOVEMENT-FLOORPLANE CURVE", "Mn", 0), ("SIGNWRITING HEAD MOVEMENT CIRCLE", "Mn", 0), ("SIGNWRITING FACE DIRECTION POSITION NOSE FORWARD TILTING", "Mn", 0), ("SIGNWRITING FACE DIRECTION POSITION NOSE UP OR DOWN", "Mn", 0), ("SIGNWRITING FACE DIRECTION POSITION NOSE UP OR DOWN TILTING", "Mn", 0), ("SIGNWRITING EYEBROWS STRAIGHT UP", "Mn", 0), ("SIGNWRITING EYEBROWS STRAIGHT NEUTRAL", "Mn", 0), ("SIGNWRITING EYEBROWS STRAIGHT DOWN", "Mn", 0), ("SIGNWRITING DREAMY EYEBROWS NEUTRAL DOWN", "Mn", 0), ("SIGNWRITING DREAMY EYEBROWS DOWN NEUTRAL", "Mn", 0), ("SIGNWRITING DREAMY EYEBROWS UP NEUTRAL", "Mn", 0), ("SIGNWRITING DREAMY EYEBROWS NEUTRAL UP", "Mn", 0), ("SIGNWRITING FOREHEAD NEUTRAL", "Mn", 0), ("SIGNWRITING FOREHEAD CONTACT", "Mn", 0), ("SIGNWRITING FOREHEAD WRINKLED", "Mn", 0), ("SIGNWRITING EYES OPEN", "Mn", 0), ("SIGNWRITING EYES SQUEEZED", "Mn", 0), ("SIGNWRITING EYES CLOSED", "Mn", 0), ("SIGNWRITING EYE BLINK SINGLE", "Mn", 0), ("SIGNWRITING EYE BLINK MULTIPLE", "Mn", 0), ("SIGNWRITING EYES HALF OPEN", "Mn", 0), ("SIGNWRITING EYES WIDE OPEN", "Mn", 0), ("SIGNWRITING EYES HALF CLOSED", "Mn", 0), ("SIGNWRITING EYES WIDENING MOVEMENT", "Mn", 0), ("SIGNWRITING EYE WINK", "Mn", 0), ("SIGNWRITING EYELASHES UP", "Mn", 0), ("SIGNWRITING EYELASHES DOWN", "Mn", 0), ("SIGNWRITING EYELASHES FLUTTERING", "Mn", 0), ("SIGNWRITING EYEGAZE-WALLPLANE STRAIGHT", "Mn", 0), ("SIGNWRITING EYEGAZE-WALLPLANE STRAIGHT DOUBLE", "Mn", 0), ("SIGNWRITING EYEGAZE-WALLPLANE STRAIGHT ALTERNATING", "Mn", 0), ("SIGNWRITING EYEGAZE-FLOORPLANE STRAIGHT", "Mn", 0), ("SIGNWRITING EYEGAZE-FLOORPLANE STRAIGHT DOUBLE", "Mn", 0), ("SIGNWRITING EYEGAZE-FLOORPLANE STRAIGHT ALTERNATING", "Mn", 0), ("SIGNWRITING EYEGAZE-WALLPLANE CURVED", "Mn", 0), ("SIGNWRITING EYEGAZE-FLOORPLANE CURVED", "Mn", 0), ("SIGNWRITING EYEGAZE-WALLPLANE CIRCLING", "Mn", 0), ("SIGNWRITING CHEEKS PUFFED", "Mn", 0), ("SIGNWRITING CHEEKS NEUTRAL", "Mn", 0), ("SIGNWRITING CHEEKS SUCKED", "Mn", 0), ("SIGNWRITING TENSE CHEEKS HIGH", "Mn", 0), ("SIGNWRITING TENSE CHEEKS MIDDLE", "Mn", 0), ("SIGNWRITING TENSE CHEEKS LOW", "Mn", 0), ("SIGNWRITING EARS", "Mn", 0), ("SIGNWRITING NOSE NEUTRAL", "Mn", 0), ("SIGNWRITING NOSE CONTACT", "Mn", 0), ("SIGNWRITING NOSE WRINKLES", "Mn", 0), ("SIGNWRITING NOSE WIGGLES", "Mn", 0), ("SIGNWRITING AIR BLOWING OUT", "Mn", 0), ("SIGNWRITING AIR SUCKING IN", "Mn", 0), ("SIGNWRITING AIR BLOW SMALL ROTATIONS", "So", 0), ("SIGNWRITING AIR SUCK SMALL ROTATIONS", "So", 0), ("SIGNWRITING BREATH INHALE", "So", 0), ("SIGNWRITING BREATH EXHALE", "So", 0), ("SIGNWRITING MOUTH CLOSED NEUTRAL", "Mn", 0), ("SIGNWRITING MOUTH CLOSED FORWARD", "Mn", 0), ("SIGNWRITING MOUTH CLOSED CONTACT", "Mn", 0), ("SIGNWRITING MOUTH SMILE", "Mn", 0), ("SIGNWRITING MOUTH SMILE WRINKLED", "Mn", 0), ("SIGNWRITING MOUTH SMILE OPEN", "Mn", 0), ("SIGNWRITING MOUTH FROWN", "Mn", 0), ("SIGNWRITING MOUTH FROWN WRINKLED", "Mn", 0), ("SIGNWRITING MOUTH FROWN OPEN", "Mn", 0), ("SIGNWRITING MOUTH OPEN CIRCLE", "Mn", 0), ("SIGNWRITING MOUTH OPEN FORWARD", "Mn", 0), ("SIGNWRITING MOUTH OPEN WRINKLED", "Mn", 0), ("SIGNWRITING MOUTH OPEN OVAL", "Mn", 0), ("SIGNWRITING MOUTH OPEN OVAL WRINKLED", "Mn", 0), ("SIGNWRITING MOUTH OPEN OVAL YAWN", "Mn", 0), ("SIGNWRITING MOUTH OPEN RECTANGLE", "Mn", 0), ("SIGNWRITING MOUTH OPEN RECTANGLE WRINKLED", "Mn", 0), ("SIGNWRITING MOUTH OPEN RECTANGLE YAWN", "Mn", 0), ("SIGNWRITING MOUTH KISS", "Mn", 0), ("SIGNWRITING MOUTH KISS FORWARD", "Mn", 0), ("SIGNWRITING MOUTH KISS WRINKLED", "Mn", 0), ("SIGNWRITING MOUTH TENSE", "Mn", 0), ("SIGNWRITING MOUTH TENSE FORWARD", "Mn", 0), ("SIGNWRITING MOUTH TENSE SUCKED", "Mn", 0), ("SIGNWRITING LIPS PRESSED TOGETHER", "Mn", 0), ("SIGNWRITING LIP LOWER OVER UPPER", "Mn", 0), ("SIGNWRITING LIP UPPER OVER LOWER", "Mn", 0), ("SIGNWRITING MOUTH CORNERS", "Mn", 0), ("SIGNWRITING MOUTH WRINKLES SINGLE", "Mn", 0), ("SIGNWRITING MOUTH WRINKLES DOUBLE", "Mn", 0), ("SIGNWRITING TONGUE STICKING OUT FAR", "Mn", 0), ("SIGNWRITING TONGUE LICKING LIPS", "Mn", 0), ("SIGNWRITING TONGUE TIP BETWEEN LIPS", "Mn", 0), ("SIGNWRITING TONGUE TIP TOUCHING INSIDE MOUTH", "Mn", 0), ("SIGNWRITING TONGUE INSIDE MOUTH RELAXED", "Mn", 0), ("SIGNWRITING TONGUE MOVES AGAINST CHEEK", "Mn", 0), ("SIGNWRITING TONGUE CENTRE STICKING OUT", "Mn", 0), ("SIGNWRITING TONGUE CENTRE INSIDE MOUTH", "Mn", 0), ("SIGNWRITING TEETH", "Mn", 0), ("SIGNWRITING TEETH MOVEMENT", "Mn", 0), ("SIGNWRITING TEETH ON TONGUE", "Mn", 0), ("SIGNWRITING TEETH ON TONGUE MOVEMENT", "Mn", 0), ("SIGNWRITING TEETH ON LIPS", "Mn", 0), ("SIGNWRITING TEETH ON LIPS MOVEMENT", "Mn", 0), ("SIGNWRITING TEETH BITE LIPS", "Mn", 0), ("SIGNWRITING MOVEMENT-WALLPLANE JAW", "Mn", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE JAW", "Mn", 0), ("SIGNWRITING NECK", "Mn", 0), ("SIGNWRITING HAIR", "Mn", 0), ("SIGNWRITING EXCITEMENT", "Mn", 0), ("SIGNWRITING SHOULDER HIP SPINE", "So", 0), ("SIGNWRITING SHOULDER HIP POSITIONS", "So", 0), ("SIGNWRITING WALLPLANE SHOULDER HIP MOVE", "So", 0), ("SIGNWRITING FLOORPLANE SHOULDER HIP MOVE", "So", 0), ("SIGNWRITING SHOULDER TILTING FROM WAIST", "So", 0), ("SIGNWRITING TORSO-WALLPLANE STRAIGHT STRETCH", "So", 0), ("SIGNWRITING TORSO-WALLPLANE CURVED BEND", "So", 0), ("SIGNWRITING TORSO-FLOORPLANE TWISTING", "So", 0), ("SIGNWRITING UPPER BODY TILTING FROM HIP JOINTS", "Mn", 0), ("SIGNWRITING LIMB COMBINATION", "So", 0), ("SIGNWRITING LIMB LENGTH-1", "So", 0), ("SIGNWRITING LIMB LENGTH-2", "So", 0), ("SIGNWRITING LIMB LENGTH-3", "So", 0), ("SIGNWRITING LIMB LENGTH-4", "So", 0), ("SIGNWRITING LIMB LENGTH-5", "So", 0), ("SIGNWRITING LIMB LENGTH-6", "So", 0), ("SIGNWRITING LIMB LENGTH-7", "So", 0), ("SIGNWRITING FINGER", "So", 0), ("SIGNWRITING LOCATION-WALLPLANE SPACE", "So", 0), ("SIGNWRITING LOCATION-FLOORPLANE SPACE", "So", 0), ("SIGNWRITING LOCATION HEIGHT", "So", 0), ("SIGNWRITING LOCATION WIDTH", "So", 0), ("SIGNWRITING LOCATION DEPTH", "So", 0), ("SIGNWRITING LOCATION HEAD NECK", "Mn", 0), ("SIGNWRITING LOCATION TORSO", "So", 0), ("SIGNWRITING LOCATION LIMBS DIGITS", "So", 0), ("SIGNWRITING COMMA", "Po", 0), ("SIGNWRITING FULL STOP", "Po", 0), ("SIGNWRITING SEMICOLON", "Po", 0), ("SIGNWRITING COLON", "Po", 0), ("SIGNWRITING PARENTHESIS", "Po", 0), (), (), (), (), (), (), (), (), (), (), (), (), (), (), (), ("SIGNWRITING FILL MODIFIER-2", "Mn", 0), ("SIGNWRITING FILL MODIFIER-3", "Mn", 0), ("SIGNWRITING FILL MODIFIER-4", "Mn", 0), ("SIGNWRITING FILL MODIFIER-5", "Mn", 0), ("SIGNWRITING FILL MODIFIER-6", "Mn", 0), (), ("SIGNWRITING ROTATION MODIFIER-2", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-3", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-4", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-5", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-6", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-7", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-8", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-9", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-10", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-11", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-12", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-13", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-14", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-15", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-16", "Mn", 0), )
https://github.com/herbhuang/utdallas-thesis-template-typst	https://raw.githubusercontent.com/herbhuang/utdallas-thesis-template-typst/main/content/abstract_de.typ	typst	MIT License	Note: Insert the German translation of the English abstract here.
https://github.com/Zeta611/simplebnf.typ	https://raw.githubusercontent.com/Zeta611/simplebnf.typ/main/examples/system-f.typ	typst	MIT License	#import "../simplebnf.typ": * #set page( width: auto, height: auto, margin: .5cm, fill: white, ) #bnf( Prod( $e$, delim: $→$, { Or[$x$][variable] Or[$λ x: τ.e$][abstraction] Or[$e space e$][application] Or[$λ τ.e space e$][type abstraction] // @typstyle off Or[$e space [τ]$][type application] }, ), Prod( $τ$, delim: $→$, { Or[$X$][type variable] Or[$τ → τ$][type of functions] Or[$∀X.τ$][universal quantification] }, ), )
https://github.com/GYPpro/DS-Course-Report	https://raw.githubusercontent.com/GYPpro/DS-Course-Report/main/Rep/01.typ	typst		#import "@preview/tablex:0.0.6": tablex, hlinex, vlinex, colspanx, rowspanx #import "@preview/codelst:2.0.1": sourcecode // Display inline code in a small box // that retains the correct baseline. #set text(font:("Times New Roman","Source Han Serif SC")) #show raw.where(block: false): box.with( fill: luma(230), inset: (x: 3pt, y: 0pt), outset: (y: 3pt), radius: 2pt, ) // #set raw(align: center) #show raw: set text( font: ("consolas", "Source Han Serif SC") ) #set page( // flipped: true, paper: "a4", // background: [#image("background.png")] ) #set text( font:("Times New Roman","Source Han Serif SC"), style:"normal", weight: "regular", size: 13pt, ) #let nxtIdx(name) = box[ #counter(name).step()#counter(name).display()] #set par( // first-line-indent: 1cm ) #set math.equation(numbering: "(1)") // Display block code in a larger block // with more padding. #show raw.where(block: true): block.with( fill: luma(240), inset: 10pt, radius: 4pt, ) #set math.equation(numbering: "(1)") #set page( paper:"a4", number-align: right, margin: (x:2.54cm,y:4cm), header: [ #set text( size: 25pt, font: "KaiTi", ) #align( bottom + center, [ #strong[暨南大学本科实验报告专用纸(附页)] ] ) #line(start: (0pt,-5pt),end:(453pt,-5pt)) ] ) /----/ = 基于双向链表的`linkedList` \ #text( font:"KaiTi", size: 15pt )[ 课程名称#underline[#text(" 数据结构 ")]成绩评定#underline[#text(" ")]\ 实验项目名称#underline[#text(" ") 基于双向链表的`linkedList` #text(" ")]指导老师#underline[#text(" 干晓聪 ")]\ 实验项目编号#underline[#text(" 01 ")]实验项目类型#underline[#text(" 设计性 ")]实验地点#underline[#text(" 数学系机房 ")]\ 学生姓名#underline[#text(" 郭彦培 ")]学号#underline[#text(" 2022101149 ")]\ 学院#underline[#text(" 信息科学技术学院 ")]系#underline[#text(" 数学系 ")]专业#underline[#text(" 信息管理与信息系统 ")]\ 实验时间#underline[#text(" 2024年6月13日上午 ")]#text("~")#underline[#text(" 2024年7月13日中午 ")]\ ] #set heading( numbering: "1.1." ) = 实验目的实现一个双向列表类，在类中实现增、删、改、查的方法并完成测试 = 实验环境 \ #h(1.8em)计算机：PC X64 操作系统：Windows + Ubuntu20.0LTS 编程语言：C++：GCC std20 IDE：Visual Studio Code #pagebreak() = 程序代码 == `linkedList.h` #sourcecode[```cpp // #define _PRIVATE_DEBUG #ifndef LINKED_LIST_HPP #define LINKED_LIST_HPP #ifdef _PRIVATE_DEBUG #include <iostream> #endif namespace myDS { template<typename VALUE_TYPE> class linkedList{ protected: class linkedNode { public: VALUE_TYPE data = VALUE_TYPE(); linkedNode * next = nullptr; linkedNode * priv = nullptr; linkedNode() { } linkedNode(VALUE_TYPE _data){ next = nullptr; priv = nullptr; data = _data; } linkedNode(VALUE_TYPE _data,linkedNode * priv) { next = nullptr; priv = priv; data = _data; } ~linkedNode() { #ifdef _PRIVATE_DEBUG // if(this->next != nullptr) // std::cout << "Unexpected Delete at :" << this->data // << " with next:" << this->next->data << "\n"; #endif } linkedNode * linkNext(linkedNode * _next) { next = _next; _next->priv = this; return this->next; } linkedNode * linkPriv(linkedNode * _priv) { priv = _priv; _priv->next = this; return this->priv; } void insertNext(linkedNode * _inst){ if(_inst == nullptr) return; if(this->next == nullptr) linkNext(_inst); else { _inst->next = this->next; this->next->priv = _inst; _inst->priv = this; this->next = _inst; } } void deleteNext() { if(this->next == nullptr) return; else { linkedNode * tmp = this->next; this->next = this->next->next; this->next->priv = this; tmp->next = nullptr; delete tmp; } } }; private: class _iterator { private: linkedNode _ptr; public: enum __iter_dest_type { front, back }; __iter_dest_type _iter_dest; _iterator(linkedNode _upper ,__iter_dest_type _d) { _ptr = _upper; _iter_dest = _d; } VALUE_TYPE & operator() { return _ptr->data; } VALUE_TYPE operator->() { return _ptr; } myDS::linkedList<VALUE_TYPE>::_iterator operator++() { if (_iter_dest == front) _ptr = _ptr->next; else _ptr = _ptr->priv; return this; } myDS::linkedList<VALUE_TYPE>::_iterator operator++(int) { myDS::linkedList<VALUE_TYPE>::_iterator old = this; if (_iter_dest == front) _ptr = _ptr->next; else _ptr = _ptr->priv; return old; } // myDS::linkedList<VALUE_TYPE>::_iterator operator+(size_t _n) // { // if (_iter_dest == front) // _upper_idx += _n; // else // _upper_idx -= _n; // _ptr = &((_upper_pointer)[_upper_idx]); // return this; // } bool operator==( myDS::linkedList<VALUE_TYPE>::_iterator _b) { if (&(_b) == _ptr) return 1; else return 0; } bool operator!=( myDS::linkedList<VALUE_TYPE>::_iterator _b) { if (&(_b) == &(_ptr->data)) return 0; else return 1; } }; linkedNode * head = new linkedNode(); linkedNode * tail = new linkedNode(); int cap = 0; public: linkedList(){ head->linkNext(tail); } ~linkedList(){ clear(); delete head; delete tail; } void push_back(VALUE_TYPE t) { tail->data = t; tail->linkNext(new linkedNode()); tail = tail->next; cap ++; } void push_frount(VALUE_TYPE t) { head->data = t; head = (head->linkPriv(new linkedNode())); cap ++; } void clear() { linkedNode * deletingObject; while(tail->priv != head) { deletingObject = tail; tail = tail->priv; delete deletingObject; } cap = 0; delete head; delete tail; tail = new linkedNode(); head = new linkedNode(); head->linkNext(tail); } std::size_t erase(VALUE_TYPE p) { linkedNode * ptr = head; int ttl = 0; while(ptr->next != tail) { if(ptr->next->data == p){ ptr->deleteNext(); ttl ++; } else ptr = ptr->next; } cap -= ttl; return ttl; } std::size_t size() {return cap;} bool erase(linkedList<VALUE_TYPE>::_iterator p) { myDS::linkedList<VALUE_TYPE>::_iterator ptr = this->begin(); linkedNode * cur = head; while(ptr != p) { cur = cur->next; ptr ++; if(cur == tail) return 0; } cur->deleteNext(); cap --; return 1; } myDS::linkedList<VALUE_TYPE>::_iterator begin() { enum myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type _FRONT = myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type::front; return myDS::linkedList<VALUE_TYPE>::_iterator(head->next,_FRONT); } myDS::linkedList<VALUE_TYPE>::_iterator rbegin() { enum myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type _BACK = myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type::back; return myDS::linkedList<VALUE_TYPE>::_iterator(tail->priv,_BACK); } myDS::linkedList<VALUE_TYPE>::_iterator end() { enum myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type _FRONT = myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type::front; return myDS::linkedList<VALUE_TYPE>::_iterator(tail,_FRONT); } myDS::linkedList<VALUE_TYPE>::_iterator rend() { enum myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type _BACK = myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type::back; return myDS::linkedList<VALUE_TYPE>::_iterator(head,_BACK); } myDS::linkedList<VALUE_TYPE>::_iterator get(std::size_t p) { linkedNode * ptr = head->next; while(p --) ptr = ptr->next; enum myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type _FRONT = myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type::front; return myDS::linkedList<VALUE_TYPE>::_iterator(ptr,_FRONT); } VALUE_TYPE & operator[](std::size_t p) { linkedNode * ptr = head; while(p --) ptr = ptr->next; return ptr->next->data; } #ifdef _PRIVATE_DEBUG void innerPrint() { std::cout << "--Header[" << head << "]: " << head->data << "\n"; std::cout << "--Tail[" << tail << "]: " << tail->data << "\n"; std::cout << "-----------\n"; std::cout << "cur:" << cap<< "\n"; auto ptr = head; do { std::cout << "[" << ptr << "] ->next:" << ptr->next << " ->priv:" << ptr->priv << " \|\|data:" << ptr->data << "\n"; ptr = ptr->next; }while(ptr != nullptr); } #endif }; } #endif ```] == `_PRIV_TEST.cpp` #sourcecode[```cpp #define DS_TOBE_TEST linkedList #define _PRIVATE_DEBUG #include "Dev\01\linkedList.h" #include <iostream> #include <math.h> #include <vector> using namespace std; using TBT = myDS::DS_TOBE_TEST<int>; void accuracyTest() {//结构正确性测试 TBT tc = TBT(); for(;;) { string op; cin >> op; if(op == "clr") { //清空 tc.clear(); } else if(op == "q") //退出测试 { return; } else if(op == "pb")//push_back { int c; cin >> c; tc.push_back(c); } else if(op == "pf")//push_frount { int c; cin >> c; tc.push_frount(c); } else if(op == "at")//随机访问 { int p; cin >> p; cout << tc[p] << "\n"; } else if(op == "delEL")//删除所有等于某值元素 { int p; cin >> p; cout << tc.erase(p) << "\n"; } else if(op == "delPS")//删除某位置上的元素 { int p; cin >> p; cout << tc.erase(tc.get(p)) << "\n"; } else if(op == "iterF") //正序遍历 { tc.innerPrint(); cout << "Iter with index:\n"; for(int i = 0;i < tc.size();i ++) cout << tc[i] << " ";cout << "\n"; cout << "Iter with begin end\n"; for(auto x = tc.begin();x != tc.end();x ++) cout << (x) << " ";cout << "\n"; cout << "Iter with AUTO&&\n"; for(auto x:tc) cout << x << " ";cout << "\n"; } else if(op == "iterB") //正序遍历 { tc.innerPrint(); cout << "Iter with index:\n"; for(int i = 0;i < tc.size();i ++) cout << tc[tc.size()-1-i] << " ";cout << "\n"; cout << "Iter with begin end\n"; for(auto x = tc.rbegin();x != tc.rend();x ++) cout << (x) << " ";cout << "\n"; // cout << "Iter with AUTO&&\n";."\n"; } else if(op == "mv")//单点修改 { int p; cin >> p; int tr; cin >> tr; tc[p] = tr; } else if(op == "") { } else { op.clear(); } } } void memLeakTest() {//内存泄漏测试 TBT tc = TBT(); for(;;){ tc.push_back(1); tc.push_back(1); tc.push_back(1); tc.push_back(1); tc.clear(); } } signed main() { // accuracyTest(); memLeakTest(); } ```] = 测试数据与运行结果运行上述`_PRIV_TEST.cpp`测试代码中的正确性测试模块，得到以下内容： ``` pb 1 pb 2 pb 3 pb 4 pf 3 pb 3 iterF iterB delEL 3 iterF delPS 1 clr pb 1 pb 2 iterF delPS 0 delEL 2 iterF pb 1 pb 2 pb 3 pb 4 pf 3 pb 3 iterF --Header[0x662720]: 0 --Tail[0x662770]: 0 ----------- cur:6 [0x662720] ->next:0x662540 ->priv:0 \|\|data:0 [0x662540] ->next:0x662590 ->priv:0x662720 \|\|data:3 [0x662590] ->next:0x6625e0 ->priv:0x662540 \|\|data:1 [0x6625e0] ->next:0x662630 ->priv:0x662590 \|\|data:2 [0x662630] ->next:0x662680 ->priv:0x6625e0 \|\|data:3 [0x662680] ->next:0x6626d0 ->priv:0x662630 \|\|data:4 [0x6626d0] ->next:0x662770 ->priv:0x662680 \|\|data:3 [0x662770] ->next:0 ->priv:0x6626d0 \|\|data:0 Iter with index: 3 1 2 3 4 3 Iter with begin end 3 1 2 3 4 3 Iter with AUTO&& 3 1 2 3 4 3 iterB --Header[0x662720]: 0 --Tail[0x662770]: 0 ----------- cur:6 [0x662720] ->next:0x662540 ->priv:0 \|\|data:0 [0x662540] ->next:0x662590 ->priv:0x662720 \|\|data:3 [0x662590] ->next:0x6625e0 ->priv:0x662540 \|\|data:1 [0x6625e0] ->next:0x662630 ->priv:0x662590 \|\|data:2 [0x662630] ->next:0x662680 ->priv:0x6625e0 \|\|data:3 [0x662680] ->next:0x6626d0 ->priv:0x662630 \|\|data:4 [0x6626d0] ->next:0x662770 ->priv:0x662680 \|\|data:3 [0x662770] ->next:0 ->priv:0x6626d0 \|\|data:0 Iter with index: 3 4 3 2 1 3 Iter with begin end 3 4 3 2 1 3 delEL 3 3 iterF --Header[0x662720]: 0 --Tail[0x662770]: 0 ----------- cur:3 [0x662720] ->next:0x662590 ->priv:0 \|\|data:0 [0x662590] ->next:0x6625e0 ->priv:0x662720 \|\|data:1 [0x6625e0] ->next:0x662680 ->priv:0x662590 \|\|data:2 [0x662680] ->next:0x662770 ->priv:0x6625e0 \|\|data:4 [0x662770] ->next:0 ->priv:0x662680 \|\|data:0 Iter with index: 1 2 4 Iter with begin end 1 2 4 Iter with AUTO&& 1 2 4 delPS 1 1 clr Unexpected Delete at :4 with next:16187728 pb 1 pb 2 iterF --Header[0x6625e0]: 0 --Tail[0x662680]: 0 ----------- cur:2 [0x6625e0] ->next:0x662540 ->priv:0 \|\|data:0 [0x662540] ->next:0x662630 ->priv:0x6625e0 \|\|data:1 [0x662630] ->next:0x662680 ->priv:0x662540 \|\|data:2 [0x662680] ->next:0 ->priv:0x662630 \|\|data:0 Iter with index: 1 2 Iter with begin end 1 2 Iter with AUTO&& 1 2 delPS 0 1 delEL 2 1 iterF --Header[0x6625e0]: 0 --Tail[0x662680]: 0 ----------- cur:0 [0x6625e0] ->next:0x662680 ->priv:0 \|\|data:0 [0x662680] ->next:0 ->priv:0x6625e0 \|\|data:0 Iter with index: Iter with begin end Iter with AUTO&& ``` 可以看出，代码运行结果与预期相符，可以认为代码正确性无误。运行`_PRIV_TEST.cpp`中的内存测试模块，在保持CPU高占用率运行一段时间后内存变化符合预期，可以认为代码内存安全性良好。 #image("1.png")
https://github.com/lucifer1004/leetcode.typ	https://raw.githubusercontent.com/lucifer1004/leetcode.typ/main/problems/p0015.typ	typst		#import "../helpers.typ": * #import "../solutions/s0015.typ": * = 3Sum Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`. Notice that the solution set must not contain duplicate triplets. #let _3sum(nums) = { // Solve the problem here } #testcases( _3sum, _3sum-ref, ( (nums: (-1, 0, 1, 2, -1, -4)), (nums: (0, 1, 1)), (nums: (0, 0, 0)), (nums: range(-10, 20, step: 3)), (nums: range(-10, 10)) ) )
https://github.com/polarkac/MTG-Stories	https://raw.githubusercontent.com/polarkac/MTG-Stories/master/stories/003_Gatecrash.typ	typst		#import "@local/mtgset:0.1.0": conf #show: doc => conf("Gatecrash", doc) #include "./003 - Gatecrash/001_Gruul Ingenuity.typ" #include "./003 - Gatecrash/002_The Fathom Edict.typ" #include "./003 - Gatecrash/003_The Absolution of the Guildpact.typ" #include "./003 - Gatecrash/004_Persistence of Memory.typ" #include "./003 - Gatecrash/005_The Burying, Part 1.typ" #include "./003 - Gatecrash/006_The Greater Good.typ" #include "./003 - Gatecrash/007_The Guild of Deals.typ" #include "./003 - Gatecrash/008_Experiment One.typ" #include "./003 - Gatecrash/009_Fblthp.typ" #include "./003 - Gatecrash/010_The Burying, Part 2.typ" #include "./003 - Gatecrash/011_Bilagru Will Come for You.typ" #include "./003 - Gatecrash/012_The Hard Sell.typ" #include "./003 - Gatecrash/013_Behind the Black Sun.typ"
https://github.com/Starkillere/TIPE-detection-informations-cachees	https://raw.githubusercontent.com/Starkillere/TIPE-detection-informations-cachees/main/journnaldebord.typ	typst		#align(center)[ = Carnet de recherche _par <NAME> \ Initialisation : 12/03/2024 \ MÀJ : 12/03/2024 _ ] #set heading(numbering: "1.1.1 :") = 12/03/2024 : Définition du sujet. == Définition du problème La stéganographie désigne l'art de dissimler de l'information de manière subtile. Tout la sécurité de cette méthode de dissimulation réside dans la non connaissance des observateur non averties, de la présence d'une information cachée. La variante informatique de ce procéder consite dans la dissimulation des donnée dans le coprs d'autres donnée. Si la stéganographie permet de transférer des donnée à l'abrie du regards des observateurs non averties, nous pouvons toujours nous demander si il n'est pas possible d'affaiblir la sécurité de cette méthode de dissimulation. Autrement dit est-il possible de distinguer le bruit d'une information cachée ? == Idée d'orientation Il existe un champs de recherche à part entiere qui s'interresse à la distinction entre donnée pur et donnée issue d'une procéssuce stéganographique qui se nomme #link("https://fr.wikipedia.org/wiki/St%C3%A9ganalyse")[Stéganalyse] === Méthodes de distinction - Analyse statistique : \ Les données qui contiennent simplement du bruit peuvent avoir des caractéristiques statistiques différentes de celles qui cachent des informations. Vous pourriez étudier des mesures telles que l'entropie, la distribution des valeurs de pixels, les corrélations spatiales, etc. - Analyse de la fréquence : \ Les images qui cachent d'autres images peuvent avoir des motifs de fréquence différents de ceux des images contenant seulement du bruit. Les techniques de transformée de Fourier ou d'ondelettes peuvent être utiles pour analyser ces différences. - Analyse visuelle : \ Même si les données semblent similaires visuellement, il peut y avoir des artefacts ou des modèles non perceptibles à l'œil nu. Vous pourriez explorer des techniques de traitement d'image avancées pour mettre en évidence ces différences. - Apprentissage automatique : \ Vous pourriez également explorer des approches basées sur l'apprentissage automatique, où vous entraînez un modèle à différencier les deux types de données à partir d'un ensemble d'exemples étiquetés. = L'analyse statistique = 02/04/2024 : Phénomène aléatoire == Entropie de Shannon == Théorie de l'information = 02/04/2024 : Meqsure #line(length: 500pt) = 12/09/2024 - Implémentation Ocaml - Implémentation ocaml des algorithme pour la stéganographie image et = 19/09/2024 - git init et recher documentaire. == Définition de la problèmatique Problématique : Est-il possible de créer un algorithme de stéganalyse géneraliste, i.e un algorithme qui n'a pas connaissance du mode de dissimulation utilisé ? Les differents modes de dissimulation : - Système de substitution : remplacer une partie de la cover (1) par des donnée de l'information à dissimulée. - Transformation des paramètre de la cover : modification des paramètre physique de la cover en fonction de l'information à dissimulé (ex: fréquence) - Même choses avec le spectre. - Méthode statistique : modifier la distribution statistique de la cover en fonction de la stégo. - Techniques de distortion : stocker des informations par distorsion du signal et mesurer l'écart par rapport à la couverture originale lors de l'étape de décodage - Méthodes de génération de couverture : encoder les informations de manière à cacher un secret la communication se crée. Objectif : Trouver un invariant de dissimulation ! = 26/09/2024 : Prolongement par continuité de la semaine dernière (lecture 10) - problèmatique : Est-il possible d'identifier un paternel, une caractéristique propre aux données issues du processus de stéganographie ? == Protocole : - Étudier les differentes méthodes de stéganograpbhie (substitution, Transformation, spectre, statistique, distortion, géneration de cover) - Étudier la réponse stéganalyse à ses algorithmes - Identification d'invariant de dissimulation == 03/10/2024 : Définition formelle de l'information : [ \ l1 \|(1,0,1) (1,0,1)\| \ l2 \|(0,0,0) (0,0,0)\| \ ] \ Une information est une matrice de tuple de taille n de nombre binaire sur. - Cas de base : - Information vide (null) : #pad(x:20pt)[ On note $epsilon$ l'information vide de taille $\|epsilon\| = 0$ \ $epsilon = mat()$] - Information de base : #pad(x:20pt)[ $forall space (b_n) in BB^NN$ fini $L = mat((b_0b_1...b_n))$ de taille $\|L\| = n+1$ ] - Notation #pad(x:20pt)[ - On note $cal(M)_(n,p,l) (BB^NN)$ l'ensemble des information de matrice dans $cal(M)_(n,p) (BB^NN)$ dont les tuple sont de $l$ élément. ] - Operation sur les informations : - Taille d'une information : $"Soit" L in cal(M)_(n,p)(BB^NN) "une information"$, la taille de $L$ est noté $\|L\| = n×p$ - Caractéristiques d'une information : $"Soit" L in cal(M)_(n,p)(BB^NN) "une information" $ - Union/Intersection : $"Soit" L_1 "et" L_2 "deux information de taille" n$ - $L_1 union L_2 =$ = 10/10/2024 : Définition du repertoir documentation/prototypage == Définition formelle de l'information = Vocabulaire (MAJ 12/03/2024) + donnée pur : donnée de cachant pas d'autres données issue d'un processuce stéganographique. + cover : suport pour la dissimulation d'information cachée. + stego : information à cachée. + = Lecture en attente : + #link("https://utt.hal.science/hal-02470070/document")\ + #link("https://fr.wikipedia.org/wiki/Entropie_de_Shannon")\ + #link("https://fr.wikipedia.org/wiki/Th%C3%A9orie_de_l%27information")\ + #link("https://hal.science/hal-00394108/document")\ + #link("https://greenteapress.com/thinkdsp/thinkdsp.pdf")\ + #link(" http://tinyurl.com/thinkdsp08")\ // REP- pour les algo de traitement de signale + #link("https://fr.wikipedia.org/wiki/Algorithme_de_Knuth-Morris-Pratt")\ + #link("https://theses.hal.science/tel-00706171v2/file/RCogranne_soutenance.pdf") + #link("https://repository.root-me.org/St%C3%A9ganographie/FR%20-%20Analyse%20st%C3%A9ganographique%20d%27images%20num%C3%A9riques.pdf") + #link("https://d1wqtxts1xzle7.cloudfront.net/11025045/22359536_lese_1-libre.pdf?1363619886=&response-content-disposition=inline%3B+filename%3DA_survey_of_steganographic_techniques.pdf&Expires=1726758425&Signature=UWNEvv4JIxHsL-iZcX-PzwvRlbmce0~unnnAUFS2lB~tsuJUbrH1Mzt4ZnO~D1Dhn9DKUo0jtG-BZnkuZYYz5iSvTUuJHJJqcZ65yceho5qgmi7Jpv9OnJsNLxnqAjhHp~frVhRI3yYvhmZRsOL0gdCCCy6O5Bb9XcylGMKZA5k8SZq0Jqme~XdEXRGESCvJy69F2bQ5K~X5IF9j5VaYj7WMOj~n-QC8DG2cJBk-1GRz5NbPu5Udq4R1U-pr2GvYZKJJmqnb7MQoutftG~9-jS~WMxnag3IlAe8g~vlz87mWWLxGle-6fbBg1I-EOa63b3fzUVsFY2bLQo0WgwqNMQ__&Key-Pair-Id=<KEY>")
https://github.com/Myriad-Dreamin/typst.ts	https://raw.githubusercontent.com/Myriad-Dreamin/typst.ts/main/fuzzers/corpora/bugs/hide-meta_01.typ	typst	Apache License 2.0	#import "/contrib/templates/std-tests/preset.typ": * #show: test-page #set text(8pt) #outline() #set text(2pt) #hide(block(grid( [= A], [= B], block(grid( [= C], [= D], )) )))
https://github.com/0x1B05/english	https://raw.githubusercontent.com/0x1B05/english/main/grammar/content/名词.typ	typst		#import "../template.typ": * = 名词名词短语：限定词+形容词+名词，每一个部分都有可能缺失。限定词的部分，如果后面的名词是复数或者不可数，就可以采用零冠词 #tip("Tip")[ 句子中的名词，哪怕只有一个字，也要当作由三个部分组成的名词短语来看。 ] == 名词 === 可数名词度量衡单位： - a pound - 20 feet - an hour - ... === 不可数名词 ==== 物质名词没有固定形状。（前面可以采用零冠词） - water(水) - air(空气) - gold(金) - paper(纸) #tip("Tip")[ 如果看到物质名词前面加了a或者末尾加了s，那已经不是物质名词了，已经当作可数的普通名词来用了。 ] #example("Example")[ - I have a _paper_ to write tonight.(一份报告) - Drinking _a couple of beers_ a day won't do you any harm.(几瓶啤酒) ] ==== 抽象名词 - cowardice - ugliness - wisdom - eternity #tip("Tip")[ 如果看到抽象名词前面加了a或者末尾加了s，已经当作可数的普通名词来用了，意义往往不同。 ] #example("Example")[ - Your sister is _a real beauty._ ] ==== 动名词少数的动名词可以当可数名词使用： - There were _three weddings_ at this restaurant yesterday. ==== 专有名词一个名词只代表单一的一个对象，属于不可数（因为只有一个对象） - London - <NAME> == 复合名词三种写法： - 两个名词写在一次成为一个单字。（the dishwasher) - 两个名词之间有个连字符。（the pole-vaulter) - 两个名词分开。（the flower shop）即使使永远复数的名词（pants,trousers,glasses,scissors...)放在复合名词当形容词使用的时候也要改为单数，例如_his trouser pocket_ 但是_a clothes hanger(衣架)_，如果改成单数，就成了布架，所以这时候就需要复数了。还有一些习惯采用复数的形容词。 - a sports car - the admissions office == 副词加在形容词前面加强语气：_that rather old jacket_ == 名词短语的省略省略之后要让读者非常清楚想要表达的意思。
https://github.com/kaarmu/splash	https://raw.githubusercontent.com/kaarmu/splash/main/doc/util.typ	typst	MIT License	#let code(body) = { set text(weight: "regular") show: box.with( fill: luma(240), inset: 0.4em, radius: 3pt, baseline: 0.4em, ) raw(body) } #let get-color-value(color) = { let s = repr(color) let m = s.match(regex("(.)\\((.)\\)")) let p = (name: m.captures.at(0), value: m.captures.at(1).replace("\"", "", count: 2)) text(fill: luma(200))[ #raw(p.name) #h(1fr) #raw(p.value) ] } #let make-title(title: none, author: none, date: none, description: none) = [ #set align(center) = #title #v(1em) #text(style: "italic", description) #v(1em) / Author: #author / Date: #date #v(3em) ] #let section( title: none, description: none, cols: 2, col-gutter: 2em, row-gutter: 2pt, do-page-break: true, name: none, colors, ) = { heading(level: 3, title + if name != none [ --- #code(name); ]) v(.5em) if description != none { description } let arr = () for (name, color) in colors.pairs() { let blk = rect( stroke: none, )[ #set align(horizon) #code(name) #h(1fr) #box( width: 3em, height: 1em, fill: color, stroke: luma(230), radius: 2pt, baseline: 0.25em, ) #linebreak() #get-color-value(color) ] arr.push(blk) } grid( row-gutter: row-gutter, column-gutter: col-gutter, columns: (1fr,) * cols, ..arr, ) if do-page-break { pagebreak(weak:true) } }
https://github.com/jomaway/typst-teacher-templates	https://raw.githubusercontent.com/jomaway/typst-teacher-templates/main/examples/exam/cover.typ	typst	MIT License	#import "@local/ttt-exam:0.1.0": cover-page #let details = toml("details.toml") #let meta = ( class: details.exam.class, subject: details.exam.subject, kind: "sa", dates: ( gehalten: details.exam.date, zurückgegeben: none, eingetragen: none, ), comment: none, total_points: 70, ) #set text(lang: "de", font: "Rubik", weight: 300) #set strong(delta: 200) #cover-page(..meta)
https://github.com/liuguangxi/erdos	https://raw.githubusercontent.com/liuguangxi/erdos/master/Problems/typstdoc/figures/p151.typ	typst		#import "@preview/cetz:0.2.1" #cetz.canvas({ import cetz.draw: * let fill-color = blue.lighten(80%) circle((0, 0), radius: 0.5, fill: fill-color, name: "c1") circle((-2, 1.5), radius: 0.5, fill: fill-color, name: "c2") circle((2, 1.5), radius: 0.5, fill: fill-color, name: "c3") circle((-3, 3), radius: 0.5, fill: fill-color, name: "c4") circle((-1, 3), radius: 0.5, fill: fill-color, name: "c5") circle((1, 3), radius: 0.5, fill: fill-color, name: "c6") circle((3, 3), radius: 0.5, fill: fill-color, name: "c7") circle((-2, 4.5), radius: 0.5, fill: fill-color, name: "c8") circle((2, 4.5), radius: 0.5, fill: fill-color, name: "c9") circle((0, 6), radius: 0.5, fill: fill-color, name: "c10") line("c2", "c1", mark: (end: "stealth", fill: black)) line("c3", "c1", mark: (end: "stealth", fill: black)) line("c4", "c2", mark: (end: "stealth", fill: black)) line("c5", "c2", mark: (end: "stealth", fill: black)) line("c6", "c3", mark: (end: "stealth", fill: black)) line("c7", "c3", mark: (end: "stealth", fill: black)) line("c8", "c4", mark: (end: "stealth", fill: black)) line("c8", "c5", mark: (end: "stealth", fill: black)) line("c9", "c6", mark: (end: "stealth", fill: black)) line("c9", "c7", mark: (end: "stealth", fill: black)) line("c10", "c8", mark: (end: "stealth", fill: black)) line("c10", "c9", mark: (end: "stealth", fill: black)) })
https://github.com/TechnoElf/mqt-qcec-diff-presentation	https://raw.githubusercontent.com/TechnoElf/mqt-qcec-diff-presentation/main/content/data.typ	typst		#let unclip(res) = { res.filter(r => not r.clipped).enumerate().map(((i, r)) => { r.i = i r }) } #let sort-by-circuit-size(res) = { res.sorted(key: r => r.total-circuit-size).enumerate().map(((i, r)) => { r.i = i r }) } #let filter(res) = { res.filter(r => r.equivalence-rate > 0.35).enumerate().map(((i, r)) => { r.i = i r }) } #let filter-rev(res) = { res.filter(r => r.equivalence-rate-rev > 0.35).enumerate().map(((i, r)) => { r.i = i r }) } #let results-r1-b5q16-cprop = csv("../resources/results-r1-b5q16-cprop-smc.csv", row-type: dictionary) #let results-r1-b5q16-cmyersrev-pmismc = csv("../resources/results-r1-b5q16-cmyersrev-pmismc-smc.csv", row-type: dictionary) #let results-r1-b5q16-cmyersrev-p = csv("../resources/results-r1-b5q16-cmyersrev-p-smc.csv", row-type: dictionary) #let results-r1-b5q16-cmyers-p = csv("../resources/results-r1-b5q16-cmyers-p-smc.csv", row-type: dictionary) #let results-r1-b5q16-cpatience-p = csv("../resources/results-r1-b5q16-cpatience-p-smc.csv", row-type: dictionary) #let results-r1-b5q16-cmyers-pmismc = csv("../resources/results-r1-b5q16-cmyers-pmismc-smc.csv", row-type: dictionary) #let results-r1-b5q16 = results-r1-b5q16-cprop.enumerate().map(((i, r)) => { let cmyersrev-pmismc = results-r1-b5q16-cmyersrev-pmismc.find(r2 => r2.name == r.name) let cmyersrev-p = results-r1-b5q16-cmyersrev-p.find(r2 => r2.name == r.name) let cmyers-p = results-r1-b5q16-cmyers-p.find(r2 => r2.name == r.name) let cmyers-pmismc = results-r1-b5q16-cmyers-pmismc.find(r2 => r2.name == r.name) let cpatience-p = results-r1-b5q16-cpatience-p.find(r2 => r2.name == r.name) let num-gates-1 = float(r.numGates1) let num-gates-2 = float(r.numGates2) let total-circuit-size = num-gates-1 + num-gates-2 let num-qubits-1 = int(r.numQubits1) let num-qubits-2 = int(r.numQubits2) ( name: r.name, i: i, clipped: not ((r.finished == "true") and (cmyersrev-pmismc.finished == "true") and (cmyersrev-p.finished == "true") and (cmyers-pmismc.finished == "true") and (cmyers-p.finished == "true")), total-circuit-size: total-circuit-size, circuit-size-difference: calc.abs(num-gates-1 - num-gates-2), total-qubit-count: num-qubits-1 + num-qubits-2, qubit-count-difference: calc.abs(num-qubits-1 - num-qubits-2), equivalence-rate: float(cmyers-pmismc.diffEquivalenceCount) / total-circuit-size, equivalence-rate-rev: float(cmyersrev-pmismc.diffEquivalenceCount) / total-circuit-size, cprop: ( mu: float(r.runTimeMean) ), cmyersrev-pmismc: ( mu: float(cmyersrev-pmismc.runTimeMean) ), cmyersrev-p: ( mu: float(cmyersrev-p.runTimeMean) ), cmyers-p: ( mu: float(cmyers-p.runTimeMean) ), cmyers-pmismc: ( mu: float(cmyers-pmismc.runTimeMean) ), cpatience-p: ( mu: float(cpatience-p.runTimeMean) ), ) }) #let results-r1-b5q16-hist = { let min = calc.log(0.001) let max = calc.log(20) let bins = 15 let bins-mu = range(bins + 1).map(x => calc.pow(10, min + x * (max - min) / bins)) let cprop-mu = bins-mu.slice(1).map(_ => 0) let cmyersrev-pmismc-mu = bins-mu.slice(1).map(_ => 0) let cmyersrev-p-mu = bins-mu.slice(1).map(_ => 0) let cmyers-p-mu = bins-mu.slice(1).map(_ => 0) let cmyers-pmismc-mu = bins-mu.slice(1).map(_ => 0) let cpatience-p-mu = bins-mu.slice(1).map(_ => 0) for r in unclip(results-r1-b5q16) { for b in range(bins) { if bins-mu.at(b) <= r.cprop.mu and r.cprop.mu < bins-mu.at(b + 1) { cprop-mu.at(b) += 1 } if bins-mu.at(b) <= r.cmyersrev-pmismc.mu and r.cmyersrev-pmismc.mu < bins-mu.at(b + 1) { cmyersrev-pmismc-mu.at(b) += 1 } if bins-mu.at(b) <= r.cmyersrev-p.mu and r.cmyersrev-p.mu < bins-mu.at(b + 1) { cmyersrev-p-mu.at(b) += 1 } if bins-mu.at(b) <= r.cmyers-p.mu and r.cmyers-p.mu < bins-mu.at(b + 1) { cmyers-p-mu.at(b) += 1 } if bins-mu.at(b) <= r.cmyers-pmismc.mu and r.cmyers-pmismc.mu < bins-mu.at(b + 1) { cmyers-pmismc-mu.at(b) += 1 } if bins-mu.at(b) <= r.cpatience-p.mu and r.cpatience-p.mu < bins-mu.at(b + 1) { cpatience-p-mu.at(b) += 1 } } } let scientific(val) = { let exp = calc.floor(calc.log(val)) [$#(calc.round(val / calc.pow(10, exp), digits: 2)) dot 10 ^ #exp$] } ( bins-mu: bins-mu.slice(0, -1).zip(bins-mu.slice(1)).map(((s, e)) => [$<$ #scientific(e)]), cprop: ( mu: cprop-mu ), cmyersrev-pmismc: ( mu: cmyersrev-pmismc-mu ), cmyersrev-p: ( mu: cmyersrev-p-mu ), cmyers-p: ( mu: cmyers-p-mu ), cmyers-pmismc: ( mu: cmyers-pmismc-mu ), cpatience-p: ( mu: cpatience-p-mu ) ) }
https://github.com/zurgl/typst-resume	https://raw.githubusercontent.com/zurgl/typst-resume/main/templates/main.typ	typst		#import "../metadata.typ": * #import "commun.typ": * #import "letter/main.typ": * #import "resume/section.typ": * #import "resume/entry.typ": * #import "resume/skills.typ": * #import "resume/header.typ": * #import "resume/footer.typ": * #import "@preview/fontawesome:0.1.0": *
https://github.com/Floffah/documents	https://raw.githubusercontent.com/Floffah/documents/main/lib/template.typ	typst	MIT License	// Feature inspiration taken from Ilm (MIT) - https://github.com/talal/ilm // The project function defines how your document looks. // It takes your content and some metadata and formats it. // Go ahead and customize it to your liking! #let project( title: "", authors: (), date: none, logo: none, formal: false, // Whether to display a maroon circle next to external links. external-link-circle: true, // Display an index of figures (images). figure-index: ( enabled: false, title: "", ), body, ) = { // Set the document's basic properties. set document(author: authors.map(a => a.name), title: title) set page(numbering: "1", number-align: center) let font = "Source Sans Pro" if formal { font = "Libertinus Serif" } set text(font: font, lang: "en", size: 12pt) // Set paragraph spacing. set par(spacing: 1.2em) set heading(numbering: "1.a.i") // See ILM (MIT) - https://github.com/talal/ilm/blob/main/lib.typ show link: it => { it // Workaround for ctheorems package so that its labels keep the default link styling. if external-link-circle { if type(it.dest) == str { sym.wj h(1.6pt) sym.wj super(box(height: 3.8pt, circle(radius: 1.2pt, stroke: 0.7pt + rgb("#993333")))) } else if type(it.dest) == label { sym.wj h(0.6pt) sym.wj super(box(height: 3.8pt, text("#", stroke: 0.2pt + rgb("#0284c7")))) } } } // Title page. // The page can contain a logo if you pass one with `logo: "logo.png"`. v(0.6fr) if logo != none { align(right, image(logo, width: 26%)) } v(9.6fr) text(1.1em, date) v(1.2em, weak: true) text(2em, weight: 700, title) // Author information. pad( top: 0.7em, right: 20%, grid( columns: (1fr,) * calc.min(3, authors.len()), gutter: 1em, ..authors.map(author => align(start)[ #author.name \ #author.affiliation ]), ), ) v(2.4fr) pagebreak() // Table of contents. outline(depth: 3, indent: true) pagebreak() // Main body. set par(justify: true) set text(hyphenate: false) set list(marker: ([•], [◦], [‣], [⁃])) // Utils let ignore(content) = {} body pagebreak() bibliography(("../references.yml", "../zotero.bib"), style: "institute-of-electrical-and-electronics-engineers") // See ILM (MIT) - https://github.com/talal/ilm/blob/main/lib.typ let fig-t(kind) = figure.where(kind: kind) let has-fig(kind) = counter(fig-t(kind)).get().at(0) > 0 if figure-index.enabled { show outline: set heading(outlined: true) context { let imgs = figure-index.enabled and has-fig(image) if imgs { // Note that we pagebreak only once instead of each each // individual index. This is because for documents that only have a couple of // figures, starting each index on new page would result in superfluous // whitespace. pagebreak() } if imgs { outline(title: figure-index.at("title", default: "Index of Figures"), target: fig-t(image)) } } } }
https://github.com/Yzx7/public_study_files	https://raw.githubusercontent.com/Yzx7/public_study_files/main/Monografía FIEE/template.typ	typst		// The project function defines how your document looks. // It takes your content and some metadata and formats it. // Go ahead and customize it to your liking! #let project(title: "",t-tipo:"", t-para:"", authors: (),cursoName:"",docenteName:"", fecha:"",add:(("")),logo: none, body) = { // Set the document's basic properties. set document(author: authors, title: title) set page( margin: (left: 35mm, right: 35mm, top: 30mm, bottom: 30mm), numbering: "1", number-align: end, ) set text(font: "New Computer Modern", lang: "es") show math.equation: set text(weight: 400) set heading(numbering: "1.1.") // Title page. // The page can contain a logo if you pass one with `logo: "logo.png"`. align(center)[ #image("LogoUNMSM icon.png", height: 60pt) #v(10pt) #block(width: 100%)[ #text(size: 16pt, weight: "extrabold")[ Universidad Nacional Mayor de San Marcos ] #text(size: 11pt, weight: "extrabold")[ Universidad del Perú. Decana de América ] #text(size: 12pt, weight: "extrabold")[ Facultad de Ingeniería Electrónica y Eléctrica ] #text(size: 10pt, weight: "extrabold")[ // EAP: Ing. Telecomunicaciones ] ] ] v(1fr) align(center)[ #text(1.5em, weight: 700, title) #v(20pt) #text(1.5em, weight: 700, t-tipo) #v(20pt) #text(1.4em, t-para) #v(1fr) // Author information. // #pad( // top: 0.7em, //grid( // columns: (1fr,) * calc.min(3, authors.len()), //gutter: 1em, //..authors.map(author => align(start, strong(author))), // ), //) #text( weight: 700)[Autores:] #pad( bottom: 10pt, grid( columns: (1fr), gutter: 1em, ..authors.map(author => align(center, author)), ) ) ] align(center)[ #grid( gutter: 12pt, ..add.map((ait) => grid(columns: 2, gutter: 4pt, text(weight: 700)[#ait.at(0):], ait.at(1) ) ), ) ] if docenteName != "" { align(center)[ #text( weight: 700)[Docente:] #docenteName ] } if cursoName != "" { align(center)[ #text(weight: 700)[Curso:] #cursoName ] } if fecha != "" { align(center)[ #text( weight: 700)[Fecha:] #fecha ] } align(center)[ #v(1.4fr) #text(size: 1.2em, weight: 700)[Lima, Perú] // #text(size: 1.2em, weight: 700)[30 de junio] #text(size: 1.2em, weight: 700)[2024] ] pagebreak() // Table of contents. // outline(depth: 3, indent: true,target: heading.where(outlined: true) ) // pagebreak() // outline(depth: 3, indent: true,target: figure.where(kind: image), title: "Índice de Figuras" ) // pagebreak() // outline(depth: 3, indent: true,target: figure.where(kind: "table"), title: "Índice de tablas" ) // pagebreak() // Main body. set par(justify: true) set text(hyphenate: false) body }
https://github.com/JakMobius/courses	https://raw.githubusercontent.com/JakMobius/courses/main/mipt-os-basic-2024/sem01/utils.typ	typst		#import "@preview/cetz:0.2.2" #let draw-compiler-lifecycle(arr) = { let margin = -1.5 let arrow-top = 5.5 let arrow-bottom = 4.5 let arrow-shortage = 1.4 let anchor-prev = 0 let x = 0 let i = 0 cetz.draw.set-style(mark: (end: ">"), stroke: 3pt + black) for step in arr { let background-color = color.mix((step.color, 20%), (white, 80%)) let stroke-color = color.mix((step.color, 50%), (black, 50%)) let text-color = stroke-color let has-code = step.at("code", default: none) != none let lower-boundary = 0 if has-code { lower-boundary = 1.4 } let y = 0 if calc.rem(i, 2) == 0 { y = 6 } cetz.draw.content( (x, y + 4), (x + step.width, y), padding: 0, )[ #box( fill: background-color, radius: 20pt, width: 100%, height: 100%, stroke: 1pt + stroke-color, ) ] cetz.draw.content( (x, y + 4), (x + step.width, y + lower-boundary), padding: 0, )[ #set text(fill: text-color, size: 18pt) #box( width: 100%, height: 100%, inset: (left: 7pt, top: 7pt, right: 7pt, bottom: 0pt), )[ #align(center + horizon)[ #step.text ] ] ] if has-code { cetz.draw.content( (x, y + lower-boundary), (x + step.width, y), padding: 0, )[ #set text(fill: text-color, font: "Monaco", size: 18pt) #box( width: 100%, height: 100%, inset: (left: 7pt, top: 0pt, right: 7pt, bottom: 7pt), )[ #align(center + horizon)[ #step.code ] ] ] } let anchor = x + step.width / 2 - arrow-shortage if x != 0 { if calc.rem(i, 2) == 0 { cetz.draw.line((anchor-prev, arrow-bottom), (anchor, arrow-top)) } else { cetz.draw.line((anchor-prev, arrow-top), (anchor, arrow-bottom)) } } anchor-prev = x + step.width / 2 + arrow-shortage x = x + margin + step.width i = i + 1 } }
https://github.com/viniciusmuller/ex_typst	https://raw.githubusercontent.com/viniciusmuller/ex_typst/main/README.md	markdown	Apache License 2.0	# ExTypst Elixir bindings and helpers for the [`typst`](https://github.com/typst/typst) typesetting system. Check [Typst's documentation](https://typst.app/docs) for a quick start. # Usage ```elixir # Write typst markup template = """ = Current Employees This is a report showing the company's current employees. #table( columns: (auto, 1fr, auto, auto), [No], [Name], [Salary], [Age], <%= employees %> ) """ # Create some data defmodule Helper do @names ["John", "Nathalie", "Joe", "Jane", "Tyler"] @surnames ["Smith", "Johnson", "Williams", "Brown", "Jones", "Davis"] def build_employees(n) do for n <- 1..n do name = "#{Enum.random(@names)} #{Enum.random(@surnames)}" salary = "US$ #{Enum.random(1000..15_000) / 1}" [n, name, salary, Enum.random(16..60)] end end end # Convert it to a nice-looking PDF {:ok, pdf_binary} = ExTypst.render_to_pdf(template, employees: ExTypst.Format.table_content(Helper.build_employees(1_000)) ) # Write to disk File.write!("employees.pdf", pdf_binary) # Or maybe send via email Bamboo.Email.put_attachment(email, %Bamboo.Attachment{data: pdf_binary, filename: "employees.pdf"}) ``` You can see the generated PDF [here](./examples/employees.pdf). ## Security Please note that currently ExTypst is experimental and content added to templates is not escaped. ## Installation If [available in Hex](https://hex.pm/docs/publish), the package can be installed by adding `ex_typst` to your list of dependencies in `mix.exs`: ```elixir def deps do [ {:ex_typst, "~> 0.1"} ] end ``` Documentation can be generated with [ExDoc](https://github.com/elixir-lang/ex_doc) and published on [HexDocs](https://hexdocs.pm). Once published, the docs can be found at <https://hexdocs.pm/ex_typst>.
https://github.com/gianzamboni/cancionero	https://raw.githubusercontent.com/gianzamboni/cancionero/main/wip/cuentamedusa.typ	typst		#import "../theme/project.typ": *; #cancion("Cuentamedusa","Valbé")[ #seccion[A] Pensemos otra vez Si faltara vivir Quizás el mundo No termina para ti Después de tanto Que se ha hablado de sufrir No tengas tiempo De mirarte sonreír Un día caerás Un día caerás Pero hasta entonces no tendremos la necesidad De respirar mejor Y hablar de la sinceridad Y rescatarnos de la generalidad Y no hablo de traición Y no hablo de traición Pero quisiera que la vida la pase mejor Y menos de crueldad Y menos desamor Y menos desarraigo A nuestro corazón #seccion[Repetir: A] Cada vez, que pienso en el mar Me sale la voz Cada vez que miro el brillo de tu bien Y de tu mal De tu bien y de tu mal #seccion[C] Cuentame dusa como Has hecho para colorear la sal Cuentame dusa lagrimas de mar Cuentame dusa como Has hecho para colorear la sal Cuentame dusa lagrimas de mar Cuentame dusa Como has aprendido a nadar Como has aprendido a nadar #seccion[Repetir: C] Como has aprendido a nadar ]
https://github.com/teamdailypractice/pdf-tools	https://raw.githubusercontent.com/teamdailypractice/pdf-tools/main/typst-pdf/examples/example-01.typ	typst		In this report, we will explore the various factors that influence fluid dynamics in glaciers and how they contribute to the formation and behaviour of these natural structures.
https://github.com/jgm/typst-hs	https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/compiler/ops-invalid-16.typ	typst	Other	// Error: 3-28 cannot divide these two relative lengths #((10% + 1pt) / (20% + 1pt))
https://github.com/LEXUGE/typzk	https://raw.githubusercontent.com/LEXUGE/typzk/main/graph.typ	typst	MIT License	#import "@preview/diagraph:0.2.5" #let digraphState = state("typzk_digraphState", (graph: (:), hierarchy: (), labels: (:), clusters: (:))) #let deep-merge-pair(dict1, dict2) = { let final = dict1 for (k, v) in dict2 { if (k in dict1) { if type(v) == "dictionary" { final.insert(k, deep-merge-pair(dict1.at(k), v)) } else { final.insert(k, dict2.at(k)) } } else { final.insert(k, v) } } return final } #let deep-merge(..args) = { let final = args.pos().first() for dict in args.pos() { final = deep-merge-pair(final, dict) } return final } #let node_descend(hierarchy, identity, payload) = { let graph = (:) if hierarchy.len() != 0 { let h = hierarchy.remove(0) graph.insert(h, node_descend(hierarchy, identity, payload)) } else { graph.insert(identity, payload) } return graph } // TODO: Allow setting extra options #let node(identity, prefix: "node_", desc: none, links: (), back_links: (), body) = { let prefixed_identity = prefix + identity let edges = () for dst in links { edges.push(identity + "->" + dst) } for orig in back_links { edges.push(orig + "->" + identity) } digraphState.update(state => { state.graph = deep-merge(state.graph, node_descend(state.hierarchy, identity, edges)) let linked_desc = if type(desc) == content { [#link(label(prefixed_identity), desc)] } else { [#link(label(prefixed_identity), identity)] } let new_label = (:) new_label.insert(identity, linked_desc) state.labels = deep-merge(state.labels, new_label) return state }) [#body #label(prefixed_identity)] } // TODO: Allow setting extra options #let subgraph(identity, desc: none, prefix: "cluster_", body) = { let prefixed_identity = prefix + identity digraphState.update(state => { state.hierarchy.push(prefixed_identity); if type(desc) == content { let new_desc = (:) new_desc.insert(prefixed_identity, [#link(label(prefixed_identity), desc)]) state.clusters = deep-merge(state.clusters, new_desc) } return state }) // This seems to work: link to the first element of the body [#body #label(prefixed_identity)] digraphState.update(state => { state.hierarchy.pop(); return state }) } #let heading_to_label(prefix: "cluster_") = { let prefixed_identity = prefix let lvls = counter(heading).get() for x in lvls { prefixed_identity += str(x) + "_" } return prefixed_identity } // Create subgraph using heading // NOTE: Seems like state update call must be wrapped in a content block, otherwise it will not take effect. #let heading_subgraph(args, prefix: "cluster_") = { let desc = args.body let prefixed_identity = heading_to_label(prefix: prefix) let lvls = counter(heading).get() digraphState.update(state => { assert(lvls.len() - state.hierarchy.len() <= 1, message: "heading levels must only increment by 1 at maximum"); while lvls.len() <= state.hierarchy.len() { state.hierarchy.pop(); } if lvls.len() > state.hierarchy.len() { state.hierarchy.push(prefixed_identity); } if type(desc) == content { let new_desc = (:) new_desc.insert(prefixed_identity, [#link(label(prefixed_identity), desc)]) state.clusters = deep-merge(state.clusters, new_desc) } return state }) } // Assemble our graph state into DOT language // All statements must end with colon #let marshal(graph) = { let s = "" let e = () for (k, v) in graph { if type(v) == "dictionary" { let (nodes, edges) = marshal(v) // This label statement would be overriden when `clusters` has matching identity. // NOTE: This label statement is necessary as somehow otherwise the label replacement will not take effect. // In addition to that, replaced label will be pushed around, so the only workaround is to set all subgraph with empty label and replace using `clusters` later. let label_stmt = "label=\"\";" s += ("subgraph " + k + " {" + label_stmt + nodes + " };") e += edges } else if type(v) == "array" { s += (k + ";") e += v } } return (s, e) } // Generate the final graphviz code #let gen_graphviz(graph, extra: "", path: ()) = { let scoped_graph = if path.len() == 0 { graph } else { let g = graph for name in path { g = g.at("cluster_" + name) } g } let (nodes, edges) = marshal(scoped_graph) let s = "digraph {" + extra s += nodes s += if edges.len()!=0 { edges.join(";") + ";" } s += "}" return s } // Use path to indicate the subgraph to render #let render_graph(extra: "", path: ()) = context { let state = digraphState.final() let s = gen_graphviz(state.graph, extra: extra, path: path) return diagraph.render(s, labels: state.labels, clusters: state.clusters) }
https://github.com/tiankaima/typst-notes	https://raw.githubusercontent.com/tiankaima/typst-notes/master/feebf7-2023_fall_TA/recitations/recitation_1.typ	typst		#import "../utils.typ": * = 习题课 1 ```plain Time: Week 1, 09.17 Sun. 19:00 ~ 20:30 ``` 摘要: 归纳/初等不等式/可数/习题选讲 == 归纳公理 #statement[ $ cases( 0 in S, n in S => n+1 in S ) quad => quad S = NN $ ] === 例题 #homework[ $forall n in NN, f(n) = n^4 + 2n^3 + 2n^2 + n$ 证明 $6 \| f(n)$ (《数学基础选讲》程艺 P3) ] #homework[ $a_1, a_2, ..., a_n (n>=2)$都是正数且$a_1 + a_2 + ... + a_n < 1$,求证: $ 1 / (1- sum_(k=1)^n a_k) > product_(k=1)^n (1+a_k) > 1 + sum_(k=1)^n a_k $ ] #v(6cm) #pagebreak() == 初等不等式 === Cauchy 不等式 #statement[ $ forall a_1, a_2, ..., a_n, b_1, b_2, ..., b_n in RR quad (sum_(k=1)^n a_k^2)(sum_(k=1)^n b_k^2) >= (sum_(k=1)^n a_k b_k)^2 $ $ <=> x, y in RR^n quad abs(x dot.c y) <= abs(x) abs(y) $ ] #proof[ 考虑二次函数: $ (a_1 x + b_1)^2 + (a_2 x + b_2)^2 + ... + (a_n x + b_n)^2 \ = sum_(k=1)^n a_k^2 x^2 + 2 sum_(k=1)^n a_k b_k x + sum_(k=1)^n b_k^2 $ 由于二次函数恒大于等于0, 所以判别式小于等于0, 即: $ Delta / 4 = (sum_(k=1)^n a_k b_k)^2 - (sum_(k=1)^n a_k^2)(sum_(k=1)^n b_k^2) <= 0 quad qed $ ] === Bernoulli 不等式 #statement[ $ forall x in RR, quad x >= -1 quad n in NN, n >= 2\ (1+x)^n >= 1 + n x $ ] #proof[ $n=2$ 时, $ (1+x)^2 = x^2 + 2x + 1 >= 1 + 2x $ 假设 $n=k$ 时成立, 则 $n=k+1$ 时: $ (1+x)^{k+1} = (1+x)^k (1+x) >= (1+k x)(1+x) = 1 + (k+1)x + k x^2 >= 1 + (k+1)x $ ] === HM-GM-AM-QM 不等式 #statement[ $ forall a_1, a_2, ..., a_n in RR_+\ n / (1 / a_1 + 1 / a_2 + ... + 1 / a_n) <= ( a_1 a_2 ... a_n )^(1 / n) <= (a_1 + a_2 + ... + a_n) / n <= sqrt((a_1^2 + a_2^2 + ... + a_n^2)/n) $ 其中几个平均值的定义为: - 算数平均值(#strong[A]rithmetic #strong[M]ean): $ "AM" = (a_1 + a_2 + ... + a_n)/n $ - 几何平均值(#strong[G]eometric #strong[M]ean): $ "GM" = (a_1 a_2 ... a_n)^(1/n) $ - 调和平均值(#strong[H]armonic #strong[M]ean): $ "HM" = n/(1/a_1 + 1/a_2 + ... + 1/a_n) $ - 平方平均值(#strong[Q]uadratic #strong[M]ean): $ "QM" = sqrt((a_1^2 + a_2^2 + ... + a_n^2)/n) $ ] #caption[ $"AM"$和$"GM"$的不等式有着很显然的几何意义: 在$RR^2$上, $ (a_1+a_2) / 2 >= sqrt(a_1 a_2) quad => quad 2(a_1+a_2) >= 4sqrt(a_1 a_2) $ 意味着, 面积相同的矩形,正方形的周长最小.注意到相同的结论可以直接推广到$RR^n$上. ] #proof[ $"AM"$和$"GM"$的不等式,可通过上面这种性质做一个巧妙的证明: $ alpha = "AM" = 1 / n (a_1 + ... + a_n) $ 若 $a_1, a_2, ..., a_n$ 不全相等, 则存在 $a_i < alpha < a_j$, 对这两项做如下替换: $ a_i^prime &= alpha \ a_j^prime &= a_i + a_j - alpha $ 替换后满足: $ "AM"^prime &= "AM" = alpha \ "GM"^prime &= "GM" / (a_i dot.c a_j) * (a_i^prime dot.c a_j^prime) > "GM" $ 注意到 $"GM"^prime > "GM"$是严格的. 重复上述过程, 直到所有的 $a_i$ 都等于 $alpha$(容易说明只需要至多$n-1$步), 此时: $ "AM" = "GM"^((n)) > "GM"^((n-1)) > ... > "GM"^(prime) > "GM" $ 证毕. #caption[ 上述证明同时说明,等号成立当且仅当 $a_1 = a_2 = ... = a_n$. 此时无需经过变换,$"AM" = "GM"$. 只要经过变换,就有严格的不等式成立. ] #homework[ $"AM"$和$"GM"$的不等式,也可以通过归纳法证明. ] ] #proof[ 首先说明$"AM"$和$"QM"$的不等式: $ forall a_1, a_2, ..., a_n in RR_+\ (a_1 + a_2 + ... + a_n) / n <= sqrt((a_1^2 + a_2^2 + ... + a_n^2)/n) $ 是Cauchy不等式的推论.令 $x = (a_1, a_2, ..., a_n), y = (1, 1, ..., 1)$, 则: $ (a_1 + a_2 + ... + a_n) = x dot.c y <= abs(x) abs(y) = sqrt(a_1^2 + a_2^2 + ... + a_n^2) dot.c sqrt(n) $ 简单变换即可得到上述不等式. #line(length: 100%, stroke: 0.2pt) 接下来说明$"HM"$和$"QM"$的不等式是由$"AM"$和$"GM"$的不等式推出的: $ forall a_1, a_2, ..., a_n in RR_+ quad (a_1 a_2 ... a_n)^(1 / n) <= (a_1 + a_2 + ... + a_n) / n $ 在上述不等式中将 $a_i$ 替换为 $1/a_i$, 则: $ forall a_1, a_2, ..., a_n in RR_+ quad => quad (1 / a_1 1 / a_2 ... 1 / a_n)^(1 / n) &<= (1 / a_1 + 1 / a_2 + ... + 1 / a_n) / n \ &arrow.b.double \ (a_1 a_2 ... a_n)^(1 / n) &>= n / (1 / a_1 + 1 / a_2 + ... + 1 / a_n) $ ] #pagebreak() == 可数 === 集合间映射 === 集合的基数 === 可数 ==== 性质 - 有限个可数集的并是可数的, 可数个可数集的并是可数的 - 如果集合A是可数的, 集合B是有限的, 那么A x B是可数的: $ A times B := {(a, b) \| a in A, b in B} $ - $abs(2^X) = 2^abs(X)$ ==== 不可数集 - $2^NN$, Cantor对角线方法 - $RR$, 也类似于Cantor对角线方法 #pagebreak() == 习题选讲 #pagebreak()
https://github.com/AU-Master-Thesis/thesis	https://raw.githubusercontent.com/AU-Master-Thesis/thesis/main/sections/3-methodology/study-1/algorithm.typ	typst	MIT License	#import "../../../lib/mod.typ": * === Algorithm <s.m.algorithm> // #jonas[I do believe all this is new to you?] // This section should explain the GBP inference algorithm as it has been implemented in the simulation tool. // 1. Fixed update schedule // 2. Chained systems // 3. Manual stepping and pausing // Introduce the Robot Mission concept? #[ #show regex("\b(UpdatePrior\|CurrentlyConnected\|Connect\|Disconnect\|InternalGBP\|ExternalGBP)\b") : set text(theme.mauve, font: "JetBrainsMono NF") As explained in the background section #nameref(<s.b.ecs>, "Entity Component System"), an #acr("ECS") architecture has been used as the foundation for #acr("MAGICS") through the Bevy Engine@bevyengine. The #acr("GBP") inference process has been implemented as a series of 7 systems that are executed in a fixed update schedule. This schedule is one of the default schedules provided by Bevy with a configurable frequency#footnote[This happens in the #acr("MAGICS") source code in the #crates.gbpplanner-rs crate at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/simulation_loader.rs#L564", "src/simulation_loader.rs:564")], which is also exposed to #acr("MAGICS") through the `simulation` table in `config.toml`#footnote[Example at #repo(org: "AU-Master-Thesis", repo: "gbp-rs") at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/config/simulations/Intersection/config.toml#L79", "config/simulations/Intersection/config.toml:79")] file. These 7 systems are listed here and marked #boxed[GBP-X]. #[ #set enum(numbering: box-enum.with(prefix: "GBP-")) #grid( columns: (1fr, 1fr), [ + `update_robot_neighbours` + `delete_interrobot_factors` + `create_interrobot_factors` + `update_failed_comms` ], [ #set enum(start: 5) + `iterate_gbp` + `update_prior_of_horizon_state` + `update_prior_of_current_state` ], ) ] The algorithm is written out in @a.m.algorithm, where `CurrentlyConnected` is a way to retrieve which robots are currently set as being connected to a specific robot, $R_i$. Before the main loop, find which robots are within communication distance at the current timestep, $N(R_i)$. Then in the main loop while the simulation is _running_, create an interrobot factor between all robots within communication distance if it does not already exist with `Connect`, and delete the interrobot factor if the robot is no longer within communication distance with `Disconnect`. After this, the internal and external #acr("GBP") iterations are run with `InternalGBP` and `ExternalGBP`, respectively. #algorithm( caption: [GBP For Robot $R_i$@gbpplanner] )[ #let comment(content) = text(theme.overlay2, content) #show regex("\b(UpdatePrior\|CurrentlyConnected\|Connect\|Disconnect\|InternalGBP\|ExternalGBP)\b") : set text(size: 0.85em) // #show regex("--(.?)\b") : set text(theme.crust, font: "JetBrainsMono NF", size: 0.85em) #show regex("(while\|for\|do\|end)") : set text(weight: "bold") // #set par(first-line-indent: 0em) #let ind() = h(2em) Input:* $R_i$ \ \ #comment[Retrieve the robots that were previously set to be connected to $R_i$] \ $C(R_i) #la "CurrentlyConnected"(R_i)$ \ \ $"UpdatePrior"(#m.x _0, delta_t)$ \ $"UpdatePrior"(#m.x _(K-1), delta_t)$ \ \ $N(R_i) #la {R_j \| norm(R_i - R_j) < r_C}$ \ \ while _running_ do \ #ind()for $R_j in N(R_i) \\ C(R_i)$ do \ #ind()#ind()$"Connect"(R_i, R_j)$ \ #ind()end \ #ind()for $R_j in C(R_i) \\ N(R_i)$ do \ #ind()#ind()$"Disconnect"(R_i, R_j)$ \ #ind()end \ #ind()$"InternalGBP"(M_I)$ \ #ind()$"ExternalGBP"(M_E)$ \ #ind()end \ end ]<a.m.algorithm> Following is an explanation of the systems, and their responsibilities, along with which part of the original #gbpplanner implementation they correspond to. Furthermore, each system is related to @a.m.algorithm when relevant: #let space = v(0.5em) #set par(first-line-indent: 0em) #set enum(numbering: box-enum.with(prefix: "GBP-")) + `update_robot_neighbours`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L1247", "src/planner/robot.rs:1247")] utilizes the #acr("ECS") to mutably query for all entities that have a `RobotConnections` component, and then consequently update them with all robots within communication range. #space Parity with `Simulator::calculateRobotNeighbours` in #gbpplanner. Corresponds to the setting the internal data of `RobotConnection` to that of $N(R_i)$. + `delete_interrobot_factors`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L1271", "src/planner/robot.rs:1271")] removes all interrobot factors from the factor graph that are no longer relevant due to the updated robot connections. #space Parity with half of the responsibility of `Robot::updateInterrobotFactors` in #gbpplanner. This corresponds to the `Disconnect` part of the algorithm. + `create_interrobot_factors`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L1326", "src/planner/robot.rs:1326")] creates new interrobot factors for all robot connections that are not already represented in the factor graph. #space Parity: with the other half of the responsibility of `Robot::updateInterrobot``Factors` in #gbpplanner. This corresponds to the `Connect` part of the algorithm. + `update_failed_comms`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L1478", "src/planner/robot.rs:1478")] updates the communication status of all robots, based on the configurable parameter `communication_failure_rate` under the `robot``.communication` table in `config.toml`. #space Parity with `Simulator::setCommsFailure` in #gbpplanner. This does not have a correlary in @a.m.algorithm. + `iterate_gbp`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L1654", "src/planner/robot.rs:1654")] iterates the #acr("GBP") algorithm for all robots in the simulation. That is, it has the responsibilitity for the 4 inference steps; _variable update, variable to factor message passing, factor update, and factor to variable message passing_. #space Parity with `Simulator::iterateGBP` in #gbpplanner. This corresponds to the `InternalGBP` and `ExternalGBP` part of the algorithm. + `update_prior_of_horizon_state`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L2042", "src/planner/robot.rs:2042")] updates the prior of the horizon state for all robots in the simulation. This is the pose factor anchoring earlier mentioned in @s.m.factors.pose-factor. #space Parity with `Robot::updateHorizon` in #gbpplanner. + `update_prior_of_current_state`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L2174", "src/planner/robot.rs:2174")] updates the prior of the current state for all robots in the simulation. Again, this is where the effect of the pose factor anchoring takes place. #space Parity with `Robot::updateCurrent` in #gbpplanner. ] #let r = ( A: text(theme.lavender, weight: "bold", "A"), B: text(theme.mauve, weight: "bold", "B") ) Through these steps the lifecycle of the interrobot factors has been allured to. This lifecycle is visualized in @f.interrobot-lifecycle, where two robots #r.A and #r.B approach each other. When they are within communication range, interrobot factors are created. The messaging happens through these factors is the communication that would happen wirelessly in a real-world implementation. Furthermore, when one of the robots' radio fails, the interrobot factors that are maintained by that robot are simply deactivated instead of removed. This has been done as an optimization, instead of deallocating, for then possibly reallocating in the next timestep. Finally, when the robots are no longer within communication range, the interrobot factors are deallocated. To summarize for two robots, $A$ and $B$, with variable $v_n^A$ and $v_n^B$, connected by interrobot factors $f_(i_n)^A (v_n^A, v_n^B)$ and $f_(i_n)^B (v_n^A, v_n^B)$. There are four possible states the pairing between the two can be in. + The communication medium of both $A$ and $B$ are inactive, preventing the factors and variable from exchanging messages. + The communication medium of $A$ is active, preventing $B$ from exchanging messages with $A$ during external message passing. + The communication medium of $B$ is active, preventing $A$ from exchanging messages with $B$ during external message passing. + The communication medium of both $A$ and $B$ are active, allowing the factors and variable to exchange messages between each other during external message passing. These four states correspond to the states shown at timestep $t_(n+1)$ to $t_(n+4)$ in @f.interrobot-lifecycle. #figure( block(breakable: false, include "figure-interrobot-lifecycle.typ", ), caption: [ #let comms = { let l1 = place(dy: -0.35em, line(length: 1em, stroke: (thickness: 2pt, paint: theme.teal, dash: "dashed", cap: "round"))) let l2 = place(dy: -0.35em, line(length: 1em, stroke: (thickness: 2pt, paint: theme.surface0, dash: "dashed", cap: "round"))) box(inset: (x: 2pt), outset: (y: 2pt), l1 + l2 + h(1.6em)) } Interrobot factor, $f_i$, lifecycle. On A) the two robots, #r.A and #r.B, are approaching each other, but not within communication range, shown with dashed circles #inline-line(stroke: (paint: theme.teal, thickness: 2pt, dash: "dashed", cap: "round")) #inline-line(stroke: (paint: theme.overlay0, thickness: 2pt, dash: "dashed", cap: "round")). On B) both robots are within communication range, and interrobot factors are created symmetrically between robots #r.A and #r.B. On C) and D) one of the two robots' radio has failed, resulting in the corresponding interrobot factors being inactive. On E) the robots are no longer within communication range, and the interrobot factors are removed. ] )<f.interrobot-lifecycle>
https://github.com/Rhinemann/mage-hack	https://raw.githubusercontent.com/Rhinemann/mage-hack/main/src/chapters/Talents.typ	typst		#import "../templates/interior_template.typ": * #show: chapter.with(chapter_name: "Talents") = Talents #show: columns.with(2, gutter: 1em) In addition to Hinder each character has a handful of SFX, reflecting special capabilities associated with their many different abilities, these are called talents. These A PC also has at least one Limit. A Limit is a special type of SFX that imposes a disadvantage on your character in order to earn them #spec_c.pp or another reward. Whenever you gain a Talent or Limit, you can rename it to better suit your character. #block(breakable: false)[ == Sample Talents When you create a new character, they gain two of the following SFX of your choice (in addition to the _Hinder_ SFX all characters receive): // TODO write descriptions ] / Adaptable: Step down and double one die of your choice in your pool. / All-Out Attack: Spend a #spec_c.pp to target multiple opponents when you roll to inflict #smallcaps[Hurt]. For each additional target, add #spec_c.d6 and keep an extra effect die. / Brilliant Under Pressure: Spend a #spec_c.pp to add your #smallcaps[Rattled] or #smallcaps[Tired] to your roll to create an asset. If the action succeeds, step down the stress you used. / Combat Veteran: When your roll to inflict #smallcaps[Hurt] or #smallcaps[Rattled] during a battle includes #smallcaps[Firearms] or #smallcaps[Weaponry], step down the largest die in your pool to add #spec_c.d8. If your roll succeeds, step up your effect die. / Distracting Presence: When you roll to inflict #smallcaps[Unsound] by distracting someone, add #spec_c.d6 and step up your effect die. / Strong Empathy: Step up #smallcaps[Empathy] on a roll to create an asset related to trust, reading people, or reassurance. / Energetic: When you would take #smallcaps[Tired] stress, spend a #spec_c.pp to step down the stress you take. If this steps the stress down below #spec_c.d6, you take no stress at all. / Flash of Insight: When you fail a test to obtain information, you may spend a #spec_c.pp or take #smallcaps[Unsound] #spec_c.d6 to obtain that information by other means. / Have a Little Faith: When you would take #smallcaps[Rattled] stress, spend a #spec_c.pp to step down the stress you take. If this steps the stress down below #spec_c.d6, you take no stress at all. / Hinder: Roll #spec_c.d4 instead of #spec_c.d8 for a distinction to earn a #spec_c.pp. / In Harm's Way: When another character near you takes stress, you can step down the stress they would take, then take #spec_c.d6 stress of the same type yourself. / Impossible to Ignore: Spend a #spec_c.pp to target multiple opponents when you roll to inflict #smallcaps[Unsound]. For each additional target, add #spec_c.d6 and keep an extra effect die. / Inspiring Leadership: Add a #spec_c.d6 and step up your effect die when you roll #smallcaps[Social] skills to create assets for allies. / Keen Intellect: Add a #spec_c.d6 and step up your effect die when you roll #smallcaps[Mental] skills to create an asset related to recalling or researching information. / Master Plan: Spend a #spec_c.pp to add a die to your pool equal to the largest complication anyone has in the scene. After the roll fails or succeeds, step down that complication. / Misdirection: When you use #smallcaps[Social] skills on a roll related to escape, deception, or stealth, step down the largest die in your pool to add #spec_c.d8. If your roll succeeds, step up your effect die. / Outmaneuver: When your roll to inflict #smallcaps[Hurt] or #smallcaps[Tired] while outdoors includes #smallcaps[Athletics], step down the largest die in your pool to add #spec_c.d8. If your roll succeeds, step up your effect die. / Peacemaker: If you have #smallcaps[Hurt] stress inflicted by another character in the scene when you roll to de-escalate a conflict, double #smallcaps[Empathy] in your dice pool. If the roll still fails, take #smallcaps[Rattled] #spec_c.d6. / Push Through It: Before you roll a dice pool including a #smallcaps[Physical] skill, spend a #spec_c.pp to recover #smallcaps[Hurt] and step up a #smallcaps[Physical] attribute for that roll. Take #smallcaps[Tired] #spec_c.d6 stress if the roll succeeds, or #spec_c.d8 if it fails. / Reassuring Comrade: Step up or double #smallcaps[Empathy] in your dice pool when helping others recover #smallcaps[Rattled]. You can also spend a #spec_c.pp to step down your own or a nearby character's #smallcaps[Rattled]. / Reckless Gambit: When you roll dice, add a die to your pool equal to the largest stress or complication anyone has in the scene. Take a complication at #spec_c.d6 if the roll succeeds, or #spec_c.d8 if it fails. / Reliable Memory: Spend a #spec_c.pp to reroll a dice pool focused on memory or recall that included #smallcaps[Intelligence]. / Skill Focus: When your pool includes a specialty, you can replace two dice of equal size with one die one step larger. / Sudden Yet Inevitable: When someone betrays you or deceives you, or you betray or deceive someone, spend a #spec_c.pp to create a #spec_c.d8 asset related to having planned for it. / Team Player: When you witness an ally rolling a heroic success, you can step down your own or another witness's #smallcaps[Rattled]. / Tough: When you would take #smallcaps[Hurt] stress, spend a #spec_c.pp to step down the stress you take. If this steps the stress down below #spec_c.d6, you take no stress at all. / Trained Physician: Step up or double #smallcaps[Medicine] in your dice pool when helping others recover #smallcaps[Hurt]. You can also spend a #spec_c.pp to step down your own or a nearby character's #smallcaps[Hurt]. / Undaunted Determination: Step up or double #smallcaps[Stamina] for one roll. If the roll fails, take #smallcaps[Tired] stress equal to the largest die in your pool. / Vicious Contempt: When you roll to inflict #smallcaps[Rattled] with mockery or contempt, add #spec_c.d6 and step up your effect die. / Vigilant Eye: Spend a #spec_c.pp to double #smallcaps[Investigation] in a pool related to following a trail, aiming at a distant target, or spotting something far off. / Watch It All Burn: Add a die to your pool equal to the largest stress or complication anyone has in the scene and step up your effect die. Succeed or fail, take #smallcaps[Unsound] #spec_c.d6. You can spend points to take one or more of these SFX at character creation or later #block(breakable: false)[ === Magick Talents (Supernatural Talents) These are the examples of magick-related and supernatural SFX available to mages to inspire the players and Storytellers. ] / Advanced Necromancy: Spend a #spec_c.pp to use both Matter and Life when your action is related to animating the dead. / Area Effect: When your effect targets an area or a number of creatures, spend a #spec_c.pp to add a #spec_c.d6 and keep an additional effect die for each additional target past the first. / Conjunctional Effects Mastery: When performing a conjunctional effect add two or more Spheres to a dice pool and step each Sphere down by one for each additional Sphere beyond the first. / Destructive Proclivities: When you include a Sphere to destroy an object, spend a #spec_c.pp to step up your effect die. / Enchant Patterns: When your effect includes Prime #spec_c.d6 or higher spend Quintessence to inflict #spec_c.d6 Hurt stress. / Fast Casting: When your action includes a Sphere, you can gain a #spec_c.d6 complication to inflict #spec_c.d6 Hurt stress. / Instrument Arsenal: Spend a #spec_c.pp to create a #spec_c.d8 Instrument asset for a particular type of magick. / Paradox Contaminating: When your action includes Prime #spec_c.d12 or higher, you can inflict Paradox on a target besides yourself. / Paradox Transmitting: When your action includes Prime #spec_c.d12 or higher and you have the Paradox Contaminating SFX you can recover one Paradox die level for each Paradox die inflicted. / Primal Channeling: When your action includes Prime #spec_c.d10 or higher you can recover one Quintessence die for each step of Stress you inflict. / Pushing Through: Whenever you take stress caused by a Sphere, spend a #spec_c.pp to step it down. At the end of the session, if you still have stress on that Sphere, step it up. / Quick Curse: When your action includes a Sphere, you can gain a #spec_c.d6 to keep a second effect die as a complication on a nearby character. / Reckless Casting: Step up or double any Sphere for one roll. If the roll fails, add your Sphere die to the Paradox pool. / Rein In: When you include a #spec_c.d10 or #spec_c.d8 Sphere, gain a #spec_c.pp and step it down. You can recover it by activating an opportunity rolled by the GM. / Swift Warding: When your action includes a Sphere, you can gain a #spec_c.d6 Moving Too Fast complication to keep a second effect die as a Magical Aegis asset. / Talent for Growth: When you succeed at a test including a Sphere, spend a #spec_c.pp to create a Watch And Learn #spec_c.d8 asset. Anyone can use this asset alongside a #spec_c.d4 or #spec_c.d6 Sphere. #block(breakable: false)[ ==== Sphere Talents Sphere talents are a special case of supernatural talents that are unlocked automatically as you advance your understanding of magick. ] / Sphere Perception: Step up your lowest die on any roll to perceive any phenomena under the purview of Sphere or create a related asset. You unlock this talent at Sphere rating #spec_c.d4 for that Sphere. / Sphere Manipulation: On rolls to create an asset that can be produced by a #spec_c.d6 or lower Sphere rating, add #spec_c.d6 and step up your effect die. You unlock this talent at Sphere rating #spec_c.d6 for that Sphere. / Sphere Control: Spend a #spec_c.pp to create a #spec_c.d8 asset that can be produced by a #spec_c.d8 or lower Sphere rating. You unlock this talent at Sphere rating #spec_c.d8 for that Sphere. / Sphere Command: Spend a #spec_c.pp to step up or double your Sphere die on a roll for an effect that can be accomplished by a #spec_c.d10 or lower Sphere rating. You unlock this talent at Sphere rating #spec_c.d10 for that Sphere. / Sphere Mastery: Take #spec_c.d6 appropriate stress or complication to double your Sphere die for for a roll. On a failure, step up the same stress or complication you took to activate. You unlock this talent at Sphere rating #spec_c.d12 for that Sphere.
https://github.com/mem-courses/calculus	https://raw.githubusercontent.com/mem-courses/calculus/main/homework-1/week0.typ	typst		#import "../template.typ": * #show: project.with( title: "微积分 Homework #0", authors: ( (name: "<NAME>", email: "<EMAIL>", phone: "3230104585"), ), date: "September 15, 2023", ) = Pre Problem 1 #prob[已知数列 ${x_n}$ 满足 $ cases( x_1 = a, x_(n+1) = 1/2 x_n^2 (3 - x_n), ) quad quad (a "为常数",quad a in (0,1) union (1,2) union (2,3)) $ 问当 $n>=1$（或 $n>=2$）时，${x_n}$ 是否为单调数列（需说明理由或给出论证过程）] 设 $ f(x) = 1/2 x^2 (3-x) $ 则有 $ f'(x) = 1/2( 2x(3-x) + (-x^2)) = 3/2x (2 - x) $ = xmr's Problem 1 #prob[设 $x,y in RR$，求 $(x-y)^2 + (2x-5)^2 + 4y^2$ 的最小值．]
https://github.com/EpicEricEE/typst-based	https://raw.githubusercontent.com/EpicEricEE/typst-based/master/src/base16.typ	typst	MIT License	/// Encodes the given data as a hex string. /// /// Arguments: /// - data: The data to encode. Must be of type array, bytes, or string. /// /// Returns: The encoded string (lowercase). #let encode(data) = { if data.len() == 0 { return "" } for byte in array(bytes(data)) { if byte < 16 { "0" } str(int(byte), base: 16) } } /// Decodes the given hex string. /// /// Arguments: /// - string: The string to decode (case-insensitive). /// /// Returns: The decoded bytes. #let decode(string) = { let dec(hex-digit) = { let code = str.to-unicode(hex-digit) if code >= 48 and code <= 57 { code - 48 } // 0-9 else if code >= 65 and code <= 70 { code - 55 } // A-F else if code >= 97 and code <= 102 { code - 87 } // a-f else { panic("Invalid hex digit: " + hex-digit) } } let array = range(string.len(), step: 2).map(i => { 16 * dec(string.at(i)) + dec(string.at(i + 1)) }) bytes(array) }
https://github.com/typst/packages	https://raw.githubusercontent.com/typst/packages/main/packages/preview/silky-letter-insa/0.1.0/lib.typ	typst	Apache License 2.0	// SHORT DOCUMENT : #let insa-short( author : none, date : datetime.today(), doc ) = { set text(lang: "fr") set page( "a4", margin: (top: 3.2cm), header-ascent: 0.71cm, header: [ #place(left, image("logo.png", height: 1.28cm), dy: 1.25cm) #place(right + bottom)[ #author\ #if type(date) == datetime [ #date.display("[day]/[month]/[year]") ] else [ #date ] ] ], footer: [ #place( right, dy: -0.6cm, dx: 1.9cm, image("footer.png") ) #place( right, dx: 1.55cm, dy: 0.58cm, text(fill: white, weight: "bold", counter(page).display()) ) ] ) doc }
https://github.com/rxt1077/it610	https://raw.githubusercontent.com/rxt1077/it610/master/markup/slides/git.typ	typst		#import "/templates/slides.typ": * #import "@preview/fletcher:0.5.1" as fletcher: diagram, node, edge #import fletcher.shapes: diamond #show: university-theme.with( short-title: [git], ) #title-slide( title: [git], ) #alternate( title: [What is git?], image: licensed-image( file: "/images/git.svg", license: "FAIRUSE", title: [git logo], url: "https://git-scm.com/downloads/logos", width: 100%, ), text: [ - a version control system - command line based - keeps track of files and changes to them - works locally but can "push" to a remote ] ) #slide(title: [What do people think of git?], align(center, licensed-image( file: "/images/git-bingo.png", license: "CC BY-NC-SA 4.0", title: [git discussion bingo], url: "https://wizardzines.com/comics/git-discussion-bingo/", author: [<NAME>], author-url: "https://wizardzines.com/", )) ) #alternate( title: [Why do we need it?], image: [ #set text(size: 16pt) #diagram( node-shape: rect, node((0, 0), [Collaboration], stroke: red, fill: red.lighten(80%)), node((1, 0), [Open Source], stroke: orange, fill: orange.lighten(80%)), node((2, 0), [Scalable], stroke: yellow, fill: yellow.lighten(80%)), node((0, 1), [Distributed], stroke: green, fill: green.lighten(80%)), node((1, 1), [git], stroke: blue, shape: diamond, fill: blue.lighten(80%)), node((2, 1), [Workflow], stroke: purple, fill: purple.lighten(80%)), node((0, 2), [Integrity], stroke: yellow, fill: yellow.lighten(80%)), node((1, 2), [Branching], stroke: red, fill: red.lighten(80%)), node((2, 2), [History], stroke: orange, fill: orange.lighten(80%)), edge((1, 1), (0.25, 0.25), "->"), edge((1, 1), (1, 0.25), "->"), edge((1, 1), (1.75, 0.25), "->"), edge((1, 1), (0.25, 1), "->"), edge((1, 1), (1.75, 1), "->"), edge((1, 1), (0.25, 1.75), "->"), edge((1, 1), (1, 1.75), "->"), edge((1, 1), (1.75, 1.75), "->"), ) ], text: [ - make things less brittle - keep track of things - find out who changed things ] ) #alternate( title: [How is it used?], image: licensed-image( file: "/images/xkcd-git.png", license: "CC BY-NC 2.5", title: [Git], url: "https://xkcd.com/1597/", author:[<NAME>], author-url: "https://xkcd.com/about/", ), text: [ - create a repo: `git init` (local or remote, GitHub can do this for you) - clone the repo: `git clone` - add files you want tracked: `git add` - commit changes: `git commit` - push changes: `git push` ] ) #slide(title: [How do I collaborate with git?])[ #diagram( node-shape: circle, node-fill: njit-red, edge-stroke: 8pt + njit-blue.lighten(50%), spacing: (4.8em, 4em), label-size: 0.8em, node((0, 0), " "), node((1, 1), " ", fill: njit-blue), node((2, 1), " ", fill: njit-blue), node((3, 1), " ", fill: njit-blue), node((4, 1), " "), node((5, 1), " "), node((6, 0), " "), edge((0, 0), (6, 0), label: [main], stroke: 8pt + njit-red.lighten(50%)), edge((0, 0), (1, 1), label: [fork], bend: -30deg, label-angle: auto), edge((1, 1), (3, 1), label: [user makes changes], label-side: right), edge((3, 1), (4, 0), label: [pull request], stroke: (paint: njit-blue.lighten(50%), thickness: 8pt, dash: ("dot", 0.5em)), bend: -30deg, label-angle: auto, label-side: left, label-sep: 0.7em, ), edge((3, 1), (5, 1), label: [maintainer adds commits], label-side: right), edge((5, 1), (6, 0), label: [merge], bend: -30deg, label-angle: auto), ) \ - fork a repo (create a branch) - make your changes - submit a PR - PR gets merged (hopefully) ] #alternate( title: [Why are we talking about git in a sysadmin class?], image: licensed-image( file: "/images/git-push-git-paid.svg", license: "CC BY-NC 4.0", title: [git-push-git-paid.svg], url: "https://github.com/rxt1077/it610/blob/master/markup/images/git-push-git-paid.svg", author: [<NAME>], author-url: "https://using.tech", ), text: [ - configurations a typically lots of little files we need to track - Docker Compose projects can be tracked in git - Kubernetes projects can be tracked in git - git helps with change management ] ) #alternate( title: [git services], image: block(breakable: false)[ #set align(center) #set text(8pt) #grid(columns: (1fr, 1fr), rows: (40%, 40%), gutter: 40pt, image("/images/github-logo.svg", height: 100%), image("/images/gitlab-logo.svg", height: 100%), image("/images/sourcehut-logo.svg", height: 100%), image("/images/radicle-logo.svg", height: 100%), ) GitHub, GitLab, SourceHut, and Radicle logos are used under fair use. ], text: [ - #link("https://github.com")[GitHub] - #link("https://gitlab.com")[GitLab] - #link("https://sr.ht")[SourceHut] - #link("https://radicle.xyz")[Radicle (p2p, v1.0 just came out!)] ], )
https://github.com/catppuccin/typst	https://raw.githubusercontent.com/catppuccin/typst/main/tests/background/test.typ	typst	MIT License	#import "/src/lib.typ": catppuccin, themes, get-palette #set page(width: auto, height: auto) #let perms = () #for theme in themes.values() { for code-block in (true, false) { for syntax in (true, false) { perms.push((theme: theme, code-block: code-block, syntax: syntax)) } } } #for p in perms [ #pagebreak(weak: true) #show: catppuccin.with(p.theme, code-block: p.code-block, code-syntax: p.syntax) = #get-palette(p.theme).name - Code block: #p.code-block - Code syntax: #p.syntax ```typ #import "/src/lib.typ": catppuccin, themes, get-palette #let perms = () #for theme in themes.values() { for code-block in (true, false) { for syntax in (true, false) { perms.push((theme: theme, code-block: code-block, syntax: syntax)) } } } #for p in perms [ #show: catppuccin.with(p.theme, code-block: p.code-block, code-syntax: p.syntax) = #get-palette(p.theme).name == Code block: #p.code-block == Code syntax: #p.syntax ] ``` ]
https://github.com/SillyFreak/typst-packages-old	https://raw.githubusercontent.com/SillyFreak/typst-packages-old/main/scrutinize/README.md	markdown	MIT License	# Scrutinize Scrutinize is a library for building exams, tests, etc. with Typst. It has three general areas of focus: - It helps with grading information: record the points that can be reached for each question and make them available for creating grading keys. - It provides a selection of question writing utilities, such as multiple choice or true/false questions. - It supports the creation of sample solutions by allowing to switch between the normal and "pre-filled" exam. Right now, providing a styled template is not part of this package's scope. Also, visual customization of the provided question templates is currently nonexistent. See the [manual](docs/manual.pdf) for details. ## Example A rendered version of this example can be found in the [gallery](gallery/). ```typ #import "@preview/scrutinize:0.2.0": grading, question, questions #import question: q #import questions: free-text-answer, single-choice, multiple-choice, set-solution, unset-solution // toggle this comment or pass `--input solution=true` to produce a sample solution // #questions.solution.update(true) #set table(stroke: 0.5pt) #context [ #let total = grading.total-points(question.all()) The candidate achieved #h(3em) out of #total points. ] = Instructions #with-solution(true)[ Use a pen. For multiple choice questions, make a cross in the box, such as in this example: #pad(x: 5%)[ Which of these numbers are prime? #multiple-choice( (([1], false), ([2], true), ([3], true), ([4], false), ([5], true)), ) ] ] #show heading: it => [ #it.body #h(1fr) / #question.current().points ] #q(points: 2)[ = Question 1 Write an answer. #free-text-answer(height: 4cm)[ An answer ] ] #q(points: 1)[ = Question 2 Select the largest number: #single-choice( ([5], [20], [25], [10], [15]), 2, // 0-based index ) ] ```
https://github.com/typst/packages	https://raw.githubusercontent.com/typst/packages/main/packages/preview/fh-joanneum-iit-thesis/1.1.0/template/chapters/4-background.typ	typst	Apache License 2.0	#import "global.typ": * = Background #lorem(45) #todo([ In the background section you might give explanations which are necessary to read the remainder of the thesis. For example define and/or explain the terms used. Optionally, you might provide a glossary (index of terms used with/without explanations). #v(1cm) Hints for equations in Typst: The notation used for #textbf([calculating]) of #textit([code performance]) might typically look like the one in @slow and @veryslow, which demonstrates what (very) slow algorithms mean. $ O(n) = n^2 $ <slow> $ O(n) = 2^n $ <veryslow> Hints for footnotes in Typst: As shown in #footnote[Visit https://typst.app/docs for more details on formatting the document using typst. Note, _typst_ is written in the Rust programming language.] we migth discuss the alternatives. Hints for formatting in Typst: + You can use built-in styles: + with underscore (\_) to _emphasise_ text + forward dash (\`) for `monospaced` text + asterisk (\) for strong* (bold) text You can create and use your own (custom) formatting macros: + check out the custom style (see in file `lib.typ`): + `\#textit` for #textit([italic]) text + `\#textbf` for #textbf([bold face]) text ])
https://github.com/tingerrr/chiral-thesis-fhe	https://raw.githubusercontent.com/tingerrr/chiral-thesis-fhe/main/src/core/component/title-page.typ	typst		#import "/src/core/kinds.typ" as _kinds #import "/src/core/authors.typ" as _authors #import "/src/utils.typ" as _utils // TODO: proper handling of more than one author // TODO: stable positioning // TODO: use subtitle #let make-title-page( title: "Mustertitel", subtitle: none, author: "Mustermann, Max", supervisors: ( "Prof. Dr. <NAME>", "Prof. Dr. <NAME>", ), field: "Angewandte Informatik", date: datetime(year: 1970, month: 01, day: 01), id: "AI-1970-BA-999", kind: _kinds.report, _fonts: (:), ) = { set align(center + top) stack( align(right, image("/assets/images/logo-fhe.svg", width: 45%)), 5em, text(16pt, font: _fonts.sans, strong[ #kind.name \ #field ]), ..if _kinds.is-thesis(kind) { (1em, [Nr. #id]) }, 5em, text(32pt, font: _fonts.sans, strong(title)), 3.4em, text(16pt, strong(_authors.format-author(author, email: false))), 2.5em, text(18pt)[Abgabedatum: #_utils.format-date(date)], ) if _kinds.is-thesis(kind) { place(center + bottom, text( 18pt, supervisors.map(_authors.format-author.with(email: false)).join(linebreak()), )) } pagebreak(weak: true) }
https://github.com/frectonz/the-pg-book	https://raw.githubusercontent.com/frectonz/the-pg-book/main/book/097.%20badeconomy.html.typ	typst		badeconomy.html Why to Start a Startup in a Bad Economy Want to start a startup? Get funded by Y Combinator. October 2008The economic situation is apparently so grim that some experts fear we may be in for a stretch as bad as the mid seventies.When Microsoft and Apple were founded.As those examples suggest, a recession may not be such a bad time to start a startup. I'm not claiming it's a particularly good time either. The truth is more boring: the state of the economy doesn't matter much either way.If we've learned one thing from funding so many startups, it's that they succeed or fail based on the qualities of the founders. The economy has some effect, certainly, but as a predictor of success it's rounding error compared to the founders.Which means that what matters is who you are, not when you do it. If you're the right sort of person, you'll win even in a bad economy. And if you're not, a good economy won't save you. Someone who thinks "I better not start a startup now, because the economy is so bad" is making the same mistake as the people who thought during the Bubble "all I have to do is start a startup, and I'll be rich."So if you want to improve your chances, you should think far more about who you can recruit as a cofounder than the state of the economy. And if you're worried about threats to the survival of your company, don't look for them in the news. Look in the mirror.But for any given team of founders, would it not pay to wait till the economy is better before taking the leap? If you're starting a restaurant, maybe, but not if you're working on technology. Technology progresses more or less independently of the stock market. So for any given idea, the payoff for acting fast in a bad economy will be higher than for waiting. Microsoft's first product was a Basic interpreter for the Altair. That was exactly what the world needed in 1975, but if Gates and Allen had decided to wait a few years, it would have been too late.Of course, the idea you have now won't be the last you have. There are always new ideas. But if you have a specific idea you want to act on, act now.That doesn't mean you can ignore the economy. Both customers and investors will be feeling pinched. It's not necessarily a problem if customers feel pinched: you may even be able to benefit from it, by making things that save money. Startups often make things cheaper, so in that respect they're better positioned to prosper in a recession than big companies.Investors are more of a problem. Startups generally need to raise some amount of external funding, and investors tend to be less willing to invest in bad times. They shouldn't be. Everyone knows you're supposed to buy when times are bad and sell when times are good. But of course what makes investing so counterintuitive is that in equity markets, good times are defined as everyone thinking it's time to buy. You have to be a contrarian to be correct, and by definition only a minority of investors can be.So just as investors in 1999 were tripping over one another trying to buy into lousy startups, investors in 2009 will presumably be reluctant to invest even in good ones.You'll have to adapt to this. But that's nothing new: startups always have to adapt to the whims of investors. Ask any founder in any economy if they'd describe investors as fickle, and watch the face they make. Last year you had to be prepared to explain how your startup was viral. Next year you'll have to explain how it's recession-proof.(Those are both good things to be. The mistake investors make is not the criteria they use but that they always tend to focus on one to the exclusion of the rest.)Fortunately the way to make a startup recession-proof is to do exactly what you should do anyway: run it as cheaply as possible. For years I've been telling founders that the surest route to success is to be the cockroaches of the corporate world. The immediate cause of death in a startup is always running out of money. So the cheaper your company is to operate, the harder it is to kill. And fortunately it has gotten very cheap to run a startup. A recession will if anything make it cheaper still.If nuclear winter really is here, it may be safer to be a cockroach even than to keep your job. Customers may drop off individually if they can no longer afford you, but you're not going to lose them all at once; markets don't "reduce headcount."What if you quit your job to start a startup that fails, and you can't find another? That could be a problem if you work in sales or marketing. In those fields it can take months to find a new job in a bad economy. But hackers seem to be more liquid. Good hackers can always get some kind of job. It might not be your dream job, but you're not going to starve.Another advantage of bad times is that there's less competition. Technology trains leave the station at regular intervals. If everyone else is cowering in a corner, you may have a whole car to yourself.You're an investor too. As a founder, you're buying stock with work: the reason Larry and Sergey are so rich is not so much that they've done work worth tens of billions of dollars, but that they were the first investors in Google. And like any investor you should buy when times are bad.Were you nodding in agreement, thinking "stupid investors" a few paragraphs ago when I was talking about how investors are reluctant to put money into startups in bad markets, even though that's the time they should rationally be most willing to buy? Well, founders aren't much better. When times get bad, hackers go to grad school. And no doubt that will happen this time too. In fact, what makes the preceding paragraph true is that most readers won't believe it—at least to the extent of acting on it.So maybe a recession is a good time to start a startup. It's hard to say whether advantages like lack of competition outweigh disadvantages like reluctant investors. But it doesn't matter much either way. It's the people that matter. And for a given set of people working on a given technology, the time to act is always now.Russian TranslationChinese TranslationJapanese Translation
https://github.com/Myriad-Dreamin/typst.ts	https://raw.githubusercontent.com/Myriad-Dreamin/typst.ts/main/fuzzers/corpora/layout/pagebreak-parity_01.typ	typst	Apache License 2.0	#import "/contrib/templates/std-tests/preset.typ": * #show: test-page #set page(width: auto, height: auto) // Test with auto-sized page. First #pagebreak(to: "odd") Third
https://github.com/yhtq/Notes	https://raw.githubusercontent.com/yhtq/Notes/main/常微分方程/main.typ	typst		#import "../template.typ": proof, note, corollary, lemma, theorem, definition, example, remark, proposition, der, partialDer #import "../template.typ": * // Take a look at the file `template.typ` in the file panel // to customize this template and discover how it works. #show: note.with( title: "常微分方程", author: "YHTQ", date: none, logo: none, ) = 前言 - 教师：李伟固 - 作业：王子康 <EMAIL>，单周周三前交含有单个未知函数的方程称为常微分方程，往往形如： $ f(x, y， y', y'', ..., y^((n))) = 0 $ 其中 $n$ 为方程的阶数，它的位置类似于代数方程中的次数。当然阶数更高的方程更加复杂。若 $f(x, y， y', y'', ..., y^((n))) = 0$ 中的 $f$ 是除去 $x$ 的多元线性函数（每个变量的次数都是1），则称为线性微分方程，否则称为非线性微分方程。线性微分方程的形式为： $ sum_(i=0)^n a_i (x) y^((i)) = 0 $ 这类方程有比较完善的理论。 + 通解与特解形式上，定义若 $y=h(x)$ 代入 $f(x, y， y', y'', ..., y^((n))) = 0$ 是恒等式，则称 $y=h(x)$ 是方程的解。许多时候研究解的存在性与唯一性已经足够困难了。称微分方程的通解为含有 $n$ 个独立常数的解，对常数任意取值都可以获得一个解。相对的，称没有未定常数的解为特解，它就是某个固定的函数。一般而言，$n$ 阶的微分方程大约有 $n$ 阶的通解，通解包含了大多数的解，但很多时候并不是所有的解。 #example[][ 微分方程 $y' = y^(1/3)$ 的一个通解为： $ cases((2/3 x + C)^(3/2) quad 2/3x + C >0, 0 quad 2/3x + C <= 0) $ 但 $y = 0$ 显然是一个解，并不能被通解包含。 ] + 初值问题（柯西问题）往往给定一个初值条件便可在通解中减少一个自由变元。当给定足够的初值时以及合理的条件时，微分方程便有唯一解。以后我们会研究在何种条件下它确实有解/有唯一解。 + 定性问题自然的，解微分方程是非常复杂的，简单易用的初等解法非常有限，且往往只能解一些特殊的方程。大部分微分方程相关的研究都是在避免解出具体的解的情况下进行的，例如直接通过方程本身分析它是否有周期解，有界解等等，或者通过方程大致描述解对应曲线的形状。类似 $f(x) + f'(x-1) = 0$ 的微分方程有时称为时滞微分方程，它与常微分方程的研究方法大不相同。 = 常微分方程初等解法尽管可以证明，绝大多数常微分方程没有初等解，更不可能被初等方法解出，但这些解法仍然十分重要。 == 恰当形式与积分因子 #definition[对称形式][ 若 $y$ 是关于 $x$ 的函数，则一阶常微分方程均可化简为： $ der(y, x) = f(x, y) $ 形式上与 $f(x, y) dif x - dif y = 0$相当。严格来说这不是微分方程，但是它使用颇为方便，而且可以写成对称的形式： $ p(x, y) dif x + q(x, y) dif y = 0 $<proper> 从而统一研究了 $x, y$ 作为自变量的形式，因此也将@proper 称为微分方程的对称形式。 ] #definition[恰当形式][ 若存在一个可微函数 $u(x, y)$ 使得： $ dif u = p(x, y) dif x + q(x, y) dif y\ <=> cases( partialDer(u, x) = p(x, y), partialDer(u, y) = q(x, y) ) $ 则称@proper 为恰当形式。 ] #theorem[][ 设： $ d u(x, y) = p(x, y) dif x + q(x, y) dif y $ 则 $u(x, y) = C$ 产生的可微隐函数为@proper 的通解 ] #proof[ 只证明 $x$ 作为自变量的情形。设： $ u(x, y(x)) = c $ 求导立得： $ partialDer(u, x) + partialDer(u, y) der(y, x) = 0\ p(x, y) + q(x, y) der(y, x) = 0 $ 这就表明 $y(x)$ 是原方程的解 ] 这启示我们，形式上我们确实可以将 $der(y, x)$ 视作分式进行处理 #theorem[][ 设 $p, q$ 在区域 $D$ 上 $C^1$ 且 $ p(x, y) dif x + q(x, y) dif y $ 恰当，则： $ partialDer(p, y) = partialDer(q, x) $ 若 $D$ 是单连通区域，则反之也成立 ] #proof[ - 设 $dif u = p(x, y) dif x + q(x, y) dif y$，则： $ partialDer(p, y) = partialDer(partialDer(u, x), y) = partialDer(partialDer(u, y), x) = partialDer(q, x) $ （注意到 $u$ 是 $C^2$ 的，因此可以交换偏导数的次序） - 反之若上式成立且 $D$ 单连通，这就是格林公式的直接推论，取（与道路无关的）曲线积分： $ u(x, y) = integral_(x_0, y_0)^(x, y) p(x, y) dif x + q(x, y) dif y $ 容易验证 $u$ 满足要求 ] #example[][ $ (3x^2 + 6 x y^2) dif x + (6 x^2 y + 4 y^3) dif y = 0 $ 可以验证这是恰当形式，下面我们具体求出 $u$ - 由于 $partialDer(u, x) = 3x^2 + 6 x y^2$，可得： $ u = integral (3x^2 + 6 x y^2) dif x = x^3 + 3 x^2 y^2 + C(y) $ - 进一步 $ partialDer(u, y) = 6 x^2 y + C'(y) = 6 x^2 y + 4 y^3 => C'(y) = 4 y^3 => C(y) = y^4 + C_1 $ - 因此 $u = x^3 + 3 x^2 y^2 + y^4 + C_1$，这就是通解 ] #example[][ $ (x^2 + 2 x y - y^2) dif x + (x^2 - 2 x y - y^2) dif y = 0\ dif (1/3 x^3 - 1/3 y^3) + 2x y dif x + x^2 dif y - (y^2 dif x + 2 x y dif y) = 0\ dif (1/3 x^3 - 1/3 y^3) + dif (x^2 y) - dif (y^2 x) = 0\ dif (1/3 x^3 - 1/3 y^3 + x^2 y - y^2 x) = 0 $ ] 一旦微分方程的形式恰当，解出微分方程是十分容易的。然而事实上，一个微分方程可能对应众多的对称形式。设 $f(x, y) !=0$，则： $ p(x, y) dif x + q(x, y) dif y = 0 <=> p(x, y) f(x, y) dif x + q(x, y) f(x, y) dif y = 0 $ 选取不同的 $f(x, y)$ 可能改变对称形式的恰当性 #definition[][ 设非零可微函数 $f(x, y)$ 满足： $ p(x, y) f(x, y) dif x + q(x, y) f(x, y) dif y = 0 $ 是恰当形式，则称 $f(x, y)$ 为@proper 的积分因子 ] #example[][ $ y dif x - x dif y = 0 $ 并不是恰当形式，但： $ mu = 1/x^2, 1/y^2, ... $ 都是其积分因子\ 事实上： $ dif(y/x) = 1/x dif y - y/x^2 dif x = 1/x^2 (y dif x - x dif y) = 0 $ ] 我们当然希望对一般的对称形式找到合适的积分因子，具体而言，是要求： $ partialDer(p f, y) = partialDer(q f, x) <=>\ f (partialDer(p, y) - partialDer(q, x)) = q partialDer(f, x) - p partialDer(f, y) $<int_factor> 这是一个一阶线性偏微分方程，不幸的是该类偏微分方程的通用解法只能归结于求解反推过程的常微分方程，因此一般的积分因子是没有通用的解法的。所幸，在一些特殊情形下，我们是可以求得积分因子的。我们直接对 $f$ 施加额外条件： - $f$ 与 $y$ 无关，原方程变为： $ f partialDer(p, y) = f partialDer(q, x) + q partialDer(f, x) $<ori_res> 这是常微分方程： $ partialDer(f, x) =f (partialDer(p, y) - partialDer(q, x)) / q $ 观察可得 $(partialDer(p, y) - partialDer(q, x)) / q$ 应与 $y$ 无关。同时此时我们只需解 $f$，$y$ 可以视作定值，因此上述方程可以化简为： $ (dif f) / f = (partialDer(p, y) - partialDer(q, x)) / q dif x\ ln \|f\| = integral (partialDer(p, y) - partialDer(q, x)) / q dif x + C $<res> 由于 $f$ 是非零的，因此 $f$ 的符号是确定的，这就解出了 $f$\ 不难发现只要 $(partialDer(p, y) - partialDer(q, x)) / q$ 与 $y$ 无关，@res 便可给出一个与 $y$ 无关的 $f$ 满足@ori_res，进而代入@int_factor 应当也成立。这表明 $(partialDer(p, y) - partialDer(q, x)) / q$ 与 $y$ 无关是可以找到此形式的积分因子的充要条件。 #theorem[一阶线性微分方程][ 一阶线性微分方程： $ y' = p(x) y + q(x) $ 的对称形式为： $ (p(x) y + q(x)) dif x - dif y = 0 $ 计算: $ (partialDer(p(x) y + q(x), y) - partialDer(1, x)) / (-1) = - p(x) $ 积分因子 $f$ 满足： $ ln \|f\| = integral - p(x) dif x + C\ \|f\| = e^C e^(-integral p(x) dif x) = A e^(-integral p(x) dif x) $ 因此取 $f = e^(-integral p(x) dif x)$ 即可，进而原方程可解 ] 不幸的是，这样能够找到的积分因子仍然非常有限。实践上寻找积分因子更多只能靠直接观察形式 #example[][ 寻找一个曲线使得从定点 $(c, 0)$ 射出的光线经曲线反射后与 $x$ 轴平行： - 通过几何方法可以列出方程 $y/x = (2y)/(1-y'^2)$ - 不妨设 $y' > 0, y> 0$，解得： $ y' = - x /y + sqrt(1 + (x/y)^2)\ y dif y = (sqrt(x^2 + y^2)-x)dif x\ x dif x + y dif y = sqrt(x^2 + y^2) dif x\ (x dif x + y dif y)/sqrt(x^2 + y^2) = dif x\ dif(sqrt(x^2 + y^2)) = dif x\ sqrt(x^2 + y^2) = x + C $ - 平方可得方程为： $ y^2 = 2 x C + C^2 = 2C(x+1/2C) $ ] #theorem[][ 假设 $p, q, f, g in C^1$，$f, g$ 都是对称形式： $ 0 = p(x, y) dif x + q(x, y) dif y := w $ 的积分因子，且 $f/g$ 不是常数，则 $f/g = C$ 是原方程的通解，其中 $C$ 为常数 ] #proof[ 设： $ f w = d u_1\ g w = d u_2 $ 不妨设 $partialDer(u_2, y) !=0$，则由隐函数定义， $u_2 = C$ 可以解出 $y = y(x)$，此时有： $ 0 = der(u_1(x, y(x)) w(x, y(x)), x)\ = w(x, y)der(u_1(x, y(x)), x) + u_1(x, y)der(w(x, y(x)), x)\ = w(x, y) $ ] 理论上讲，所有常微分方程的初等解法都可以化为恰当方程。但是实践上我们更多采用具体的方法求解。 == 分离变量法 #theorem[分离变量法][ 设微分方程形如： $ der(y, x) = f(x)g(y) $ - 若 $g(y) = 0$，则 $f(x) = C$ 就是原方程的解 - 否则，化为： $ 1/g(y) der(y, x) = f(x) $ 设 $G(y) = integral 1/(g(y)) dif y$，则: $ G'(y) = 1/g(y) der(y, x) = f(x) => G(y) = integral f(x) dif x + C $ ] #example[][ 求解 $der(y, x) = y^2 cos x$: - 首先 $y = 0$ 是它的一个解 - 其次，假设 $y$ 不恒为零，则： $ 1/(y^2) dif y = cos x dif x\ -1/y = sin x + C\ y = -1/(sin x + C) $ ] #remark[][ 这种形式下我们已经默认了 $y$ 是因变量而 $x$ 是自变量，因此不能让 $x$ 恒等于 $cos x$ 的零点。如果不考虑函数关系只考虑解曲线，则这也是一个特殊的解 ] #example[][ $der(y, x) = y^(1/3)$: - $y = 0$ 当然是解 - 否则，我们先证明若 $y(x_0) = 0$，则当 $x < x_0$ 时 $y = 0$： - 否则，假设 $x_1 < x_0, f(x_1) !=0$，不妨设其大于零。取： $ x_2 = inf(Inv(y)(0) sect [x_1, x_0]) $ 注意到上式右侧是闭集，因此 $y(x_2) = 0 => der(y, x) \|_(x = x_2)= 0$\ 此外，$der(y, x)$ 在 $[x_1, x_2]$ 之间恒正，这意味着 $y(x)$ 在 $[x_1, x_2]$ 之间单调递增，矛盾！ - 之后，我们分离变量 $ 1/y^(1/3) dif y = dif x\ 3/2 y^(2/3) = x + C\ y = plus.minus (2/3 x + C)^(3/2) $ - 综上，原方程的通解为： $ y = cases((2/3 x + C)^(3/2) quad 2/3x + C >0, 0 quad 2/3x + C <= 0) $ ] #example[][ 下面是几种常见形式 + $der(y, x) = f(x)y$ - $y = 0$ 当然是解 - 否则： $ 1/y dif y = f(x) dif x\ ln \|y\| = integral f(x) dif x + C\ y = A e^(integral f(x) dif x) $ 事实上，可以证明若 $y(x_0) = 0$，则 $y = 0$，因此可以去掉绝对值 + $der(y, x) = g(y/x)$\ 令 $y = x u$，方程化为： $ der(y, x) = u + x der(u, x) = g(u)\ 1/(g(u) - u) dif u = 1/x dif x\ integral 1/(g(u) - u) dif u = ln \|x\| + C\ $ + $der(y, x) = (a_1 x + b_1 y + c_1)/(a_2 x + b_2 y + c_2)$\ - 若 $c_1 = c_2 = 0$，上下同时除以 $x$ 即化为上一种情形 - 否则，试图做线性替换：设 $A = mat(a_1, b_1;a_2, b_2), alpha = vec(c_1,c_2)$ $ vec(x, y) = B vec(x', y') + beta $ 我们希望： $ A vec(x, y) + alpha = A(B vec(x', y') + beta) + alpha = A B vec(x', y') $ 也就是： $ A beta + alpha = 0 $ 该线性方程有解时，一定可以通过线性替换化为上一种情形\ - 否则 $"rank"(A) = 1$，不妨设 $(a_2, b_2) = k (a_1, b_1)$，令： $ u = a_1 x + b_1 y $ 原方程化为： $ der(u, x) = a_1 + b_1 der(y, x) = a_1 + b_1 (u + c_1)/(k u + c_2) $ 这时右侧已无 $x$，直接分离变量积分即可 + $y f(x y) dif x + x g(x y) dif y = 0$\ 令 $x y = u$，方程化为： $ u/x f(u) dif x + g(u)(dif u - u/x dif x) = 0\ $ 这也是分离变量的形式 ] #example[最速降线][ 求 $a -> b$ 的一条曲线，使得小球从 $a$ 无摩擦的滑下的时间最短\ 严格来说这个问题需要用到变分法，不过抛开高级理论，利用光线折射定律的相关想法也可以不严格的求解。\ 可以证明，光线从 $a$ 点到 $b$ 点经过若干种介质时，入射角的正弦与介质中传播的速度是定值。我们依次来对曲线建立微分方程： - 在曲线上任意一点 $(x, y)$ 的速度表示为： $ v = sqrt(2 g y) $ （能量守恒） - “入射角”也即与垂直方向夹角满足 $sin alpha = sqrt(1+y'^2)$ 因此可以建立方程： $ sqrt(2 g y)/ sqrt(1+y'^2) = C\ 2 g y = C^2 (1+y'^2)\ 1 + y'^2 = 2/C^2 g y\ y' = sqrt(2/C^2 g y - 1)\ 1/sqrt(2/C^2 g y - 1) dif y = dif x\ integral 1/sqrt(2/C^2 g y - 1) dif y = x + C\ $ ] == 一阶线性微分方程 #theorem[常数变易法][ 对于一阶线性微分方程： $ der(y, x) + p(x) y = q(x) $ 一般假设 $p, q$ 都连续且 $p(x)$ 不恒为零我们可以先解出齐次方程： $ der(y, x) + p(x) y = 0 $ 它的解为： $ y = C e^(-integral p(x) dif x) := C v(x) $ 做变量替换： $ y = u(x) v(x) $ 代入发现： $ q(x) - p(x) u(x) v(x) = der(u(x) v(x), x) = v(x) der(u(x), x) + u(x) der(v(x), x) $ 但是注意到： $ der(v(x), x) + p(x) v(x) = 0 $ 上式化简为： $ q(x) = v(x) der(u(x), x)\ $ 待解的函数是 $u(x)$，这是可分离变量的 ] #example[贝努利方程][ 形如： $ der(y, x) = p(x) y + q(x) y^n $ 的方程称为贝努利方程。除去 $y = 0$ 的解外，可以做变量替换： $ t = y^(1-n)\ dif t = -(1-n) y^(-n) dif y\ dif y = -1/(1-n) y^(n) dif t\ $ 上式化简为： $ y^(-n)der(y, x) = p(x) y^(-n+1) + q(x)\ -1/(1-n) der(t, x) = p(x) t + q(x)\ $ 这就化为了一阶线性微分方程 ] #example[][ 对于形如： $ der(y, x) = p(x)y^2+q(x)y+r(x) $ 的微分方程，一般没有初等解法。但是假设我们猜出了一个解 $y = f(x)$，可以做变量替换： $ y = u + f(x)\ dif y = dif u + f'(x) dif x\ $ 代入原方程后，仍然是该方程的形式。然而注意到 $u = 0$ 一定是一个解，最后一定会消掉零次项。代入发现： $ dif u + f'(x) dif x = (p(x)(u+f(x))^2+q(x)(u+f(x))+r(x))dif x\ dif u = (p(x)(u^2 + 2u f(x)) + q(x) u ) dif x\ der(u, x) = p(x)u^2 + (2 p(x) f(x) + q(x))u $ 这是贝努利方程，进而可解 ] #theorem[][ 设 $a, b in RR, a != 0$，方程： $ der(y, x) = a y^2 + b x^m $ 当且仅当 $m = 0, - 2, - (4k)/(2k plus.minus 1)$ 时可积分求解 ] #proof[ 无妨假设 $a = 1$\ - $m = 0$ 时显然可解 - $m = -2$ 时，做变量替换： $ z = x y\ dif z = x dif y + y dif x\ dif y = 1/x dif z - y/x dif x\ $ 代入得： $ 1/x dif z - y/x dif x = (y^2 + b x^(-2)) dif x\ x dif z - x y dif x = ((x y)^2 + b) dif x\ x dif z -z dif x = (z^2 + b) dif x $ 这是可分离变量的 - $m = -(4k)/(2k plus.minus 1)$ 必要性由刘维尔给出，颇为复杂 ] 特别的，$der(y, x) = x^2 + y^2$ 是不可积分求解的，这也说明了求解微分方程是非常之困难的。 == 一阶隐方程 #definition[][ 形如： $ F(x, y, y') = 0 $ 的方程称为一阶隐方程 ] 一般而言它的求解当然是非常困难的，但是对于特定的几类，我们可以尝试求解： #theorem[][ 对于以下几类隐方程我们给出一些思路： - 可以化为： $ y = f(x, y') $<ori> 的方程。\ 此时两边对 $x$ 求导得： $ y' = partialDer(f(x, y'), x) = f'_1(x, y') + f'_2(x, y')y'' $ 这是关于 $y', x$ 的显式微分方程。\ 若能解出 $y' = p(x)$，代入@ori 就是原方程的解\ 若只能解出 $x = q(y')$，代入@ori 就是原方程以 $y'$ 为参数的参数表达\ 否则，若只能解出 $g(x, y') = 0$，方程的解只能表达为： $ cases( y = f(x, y'), g(x, y') = 0 ) $ - 可以化为： $ x = f(y, y') $ 两边对 $y$ 求导得： $ 1/y' = partialDer(f(y, y'), y) = f'_1(y, y') + f'_2(y, y')der(y', y) = f'_1(y, y') + f'_2(y, y') y''/y' $ 做类似讨论即可 - 可以化为 $f(x, y') = 0$，只需要考虑它的参数形式： $ x = u(t)\ y' = v(t) $ 则： $ dif y = v(t) dif x = v(t) u'(t) dif t $ 两边积分即得 $y$ - 可以化为 $f(y, y') = 0$ 的形式。类似的，如果能找到参数方程： $ y = u(t)\ y' = v(t) $ 则： $ dif y = u'(t) dif t\ dif y = v(t) dif x\ v(t) dif x = u'(t) dif t\ dif x = (u'(t))/v(t) dif t $ 积分即可\ 注意需要补充 $v(t) = 0$ 的解 ] #example[][ $y'^3+2x y' - y = 0$ 两边直接对 $y$ 求导得： $ 3y'^2 y'' + 2y'+ 2x y'' - y' = 0\ (3y'^2 + 2x)y'' = -y'\ (3P^2 + 2x)P' = -P\ der(P, x) =- P/(3P^2+2x)\ 2P der(P, x) = -(2P^2)/(3P^2+2x)\ der(P^2, x) = -(2P^2)/(3P^2+2x)\ $ 这是方程右侧对于 $P^2$ 是齐一次的，是我们可解的方程\ $ der(x, P^2) = -3/2 - x/P^2 $ 令 $u = x/P^2, dif u = (P^2 dif x - x dif P^2)/P^4$ $ der(u, P^2) = 1/P^2 der(x, P^2) - u/P^2 = -1/P^2(3/2+u) -u/P^2\ (dif u)/(3/2 + 2u) = - 1/P^2 dif P^2\ ln \|3/2 + 2u\| = - ln 1/P^2 + C\ 3/2 + 2u = C P^2\ $ ] #example[Clairaut][ $y = x y' + f(y')$\ 求导得: $ y' = y' + x y'' + f'(y') y''\ x y'' + f'(y') y'' = 0\ $ 讨论： - $y'' = 0$，此时原方程是一族直线 $y = x c + f(c)$ - $x = - f'(y')$，此时代回得 $y = - y' f'(y') + f(y')$，这是以 $y'$ 为参数的特解当 $f'' != 0$ 时，直线族恰好是特解的切线族，这是因为此时 $f'$ 有反函数，在特解中可以解出： $ y' = u(x)\ y = x u(x) + f(u(x)) $ 此时，直线： $ y = x u(x_0) + f(u(x_0)) $ 恰好就是特解于 $(x_0, y_0)$ 点的切线（显然直线过 $(x_0, y_0)$ 且斜率 $u(x)$ 当然就是该点处的导数）事实上，我们并没有说明这些就是原方程的所有解。后续我们会证明在 $f'' != 0$ 的情况下，确实这些囊括了原方程解的所有情况（但可以构造出其他的解，例如将切线-特解-切线拼接起来也是一个解） ] == 应用举例至此，我们已经给出了所有一阶微分方程的初等解法。尽管方法有限，但是它们已经能够解出了许多重要的方程 #example[两个物种的生态方程][ 一战期间，生态学家发现随着人类捕鱼量下降，两类鱼中以其他鱼为食的鱼的比例上升，而以植物为食的鱼的比例下降。他们提出了一个模型：\ 设以鱼为食的鱼的数量为 $x = x(t)$, 以植物为食的鱼的数量为 $y = y(t)$，令 $r_x, r_y$ 是两者的增长率。显然 $r_x$ 应该随 $y$ 递增，而当 $y$ 较小时应该有 $r_x < 0$，进而设： $ r_x = sigma y - lambda $ 类似的原因，设： $ r_y = mu - delta x $ （上面的常数应该都是正数）\ 这给出了微分方程组： $ x'/x = sigma y - lambda\ y'/y = mu - delta x $ 两式相比： $ y'/x' x/y = (mu - delta x)/(sigma y - lambda) $ 注意到 $y'/x' = der(y, x)$，这就消掉了 $t$ $ y' (sigma y - lambda)/y = (mu - delta x)/x\ sigma y - lambda ln y = mu ln x - delta x + C\ sigma y + delta x - lambda ln y - mu ln x = C\ $ 虽然我们难以继续计算，但我们可以做定性研究，可以发现它的解都是周期解。\ 同时，我们还可以求出平均值： $ (dif x)/x = (-lambda + sigma y) dif t\ integral_T (dif x)/x = integral_T (-lambda + sigma y) dif t\ integral_T (dif ln(x)) = integral_T (-lambda + sigma y) dif t\ 0 = integral_T (-lambda + sigma y) dif t\ 0 = -lambda T + sigma integral_T y dif t\ (lambda)/sigma = 1/T integral_T y dif t $ 这就是 $y$ 在一个周期内的平均值\ 当捕捞量减少时，两个物种的增长率都增加，相当于 $mu$ 增大而 $lambda$ 减少，可以看到 $x$ 的平均值会增加，而 $y$ 的平均值会减少 ] #example[$n$ 体问题][ $n$ 体问题是指万有引力下 $n$ 个大质量质点相互作用的问题。设有 $n$ 个质点 $p_i$，分别有： - 坐标 $P_i = (x_i, y_i, z_i)$ - 质量 $m_i$ - 牛顿第二定律和万有引力定律： $ m_i (dif^2 P_i)/(dif t^2) = sum_(j != i) G m_i m_j (P_j - P_i)/norm(P_j - P_i)^3 $ 不妨设 $G = 1$，化简为： $ m_i (dif^2 P_i)/(dif t^2) = sum_(j != i) G m_i m_j (P_j - P_i)/(sqrt((x_i - x_j)^2 + (y_i - y_j)^2 + (z_i - z_j)^2)^3) $ 记 $U = sum_(i != j)( m_i m_j )/norm(P_i -P_j)$，化简为： $ m_i (dif^2 P_i)/(dif t^2) = partialDer(U, P_i) $ 记 $q_i = der(P_i, t)$（速度），则： $ m_i der(q_i, t) = partialDer(U, P_i) $ 令 $p = autoVecN(p, n), q = autoVecN(q, n)$，则： $ dif /(dif t) vec(p, q) = vec(q, m partialDer(U, p)) $ 可以发现： $ sum_i m_i der(q_i, t) = sum_i partialDer(U, q_i) = sum_i (sum_(j != i) m_i m_j (P_i - P_j)/norm(P_i - P_j)^3) = 0 $ 这就是在说质心不动，也就是动量守恒，进而： $ sum_i m_i q_i = C_1\ integral sum_i m_i q_i dif t = C_1 t + C_2\ sum_i m_i p_i = C_1 t + C_2 $ 质心以匀速直线运动类似的，可以验证角动量守恒，能量守恒\ 然而可以证明，除了这些以外不再有与之独立的代数的首次积分，因此代数的首次积分就是这些接下来我们解二体问题，由之前提到的守恒量可设质心恒定在原点，运动都在平面上，可以化简到： $ cases( (dif^2 x)/(dif t^2) = -m x/(x^2 + y^2)^(3/2), (dif^2 y)/(dif t^2) = -m y/(x^2 + y^2)^(3/2) ) $ 其中，1 对应 $ m = (m_2^3)/(m_1 + m_2)^2 $ 2 对应 $ m = (m_1^3)/(m_1 + m_2)^2 $ 角动量守恒给出： $ x der(y, t) - y der(x, t) = c_1\ $ 能量守恒给出： $ (der(x, t))^2 + (der(y, t))^2 - (2m)/(sqrt(x^2 + y^2)) = c_2 $ 做极坐标换元： $ x = r cos theta\ y = r sin theta $ 代入角动量守恒化简得： $ c_1 = r^2 der(theta, t)\ c_1 (t_2 - t_1) = integral_(t_1)^(t_2) r^2 dif theta\ $ 上式右侧是扫过的扇形面积，这实际上就是开普勒第二定律\ 代入能量守恒化简得： $ (der(r, t))^2 = c_1 + 2 m Inv(r) - c_1^2 r^(-2) \ der(r, t) = sqrt(c_2 + m^2/c_1^2 - (c_1/r - m/c_1)^2)\ der(theta, t) = c_1/r^2\ der(r, theta) = r^2/c_1 sqrt(c_2 + m^2/c_1^2 - sqrt(c_1/r - m/c_1)^2) $ 这是可分离变量的，最终可以化简得到： $ r = p/(1 + e cos (theta - theta_0)) $ 其中 $e = c_1/m sqrt(c_2 + m^2/c_1^2), p = c_1^2/m$\ 这一定是一条圆锥曲线，这就是开普勒第一定律\ 再次带回： $ (p/(1 + e cos (theta - theta_0)))^2 dif theta = c_1 dif t\ c_1 T = integral_(0)^(2pi) p^2/(1 + e cos (theta - theta_0))^2 dif theta\ = (2p^2 pi)/sqrt((1 - e^2)^3) $ 计算可得： $ T^2/(p^3 /(1-e^2)^3) = (4 pi^2)/m $ 当 $m_1$ 很大时，$m$ 几乎就是 $m_2$，这也就是太阳系中的开普勒第三定律二体问题还有著名的反问题，假设任何星球围绕太阳的有界轨道都是椭圆，且万有引力形如： $ f = m_1 m_2 f(norm(p_1 - p_2))(p_1 - p_2) $ 那么数学上可以证明万有引力只能是正比于距离或者反比于距离的平方，前者会导致太阳位于轨道中心，后者太阳将位于焦点 ] == 二阶自洽微分方程 #theorem[二阶自洽微分方程的解法][ 称形如 $ (dif^2 x)/(dif t^2) + f(x) = 0 $ 为二阶自洽方程，引进 $F(x)$ 使得 $F'(x) = f(x)$，令 $y = x', H(x, y) = y^2/2 + F(x)$，有： $ der(H, t) = y der(y, t) + F'(x) der(x, t) = y (dif^2 x)/(dif t^2) + f(x) der(x, t) = 0 $ 换言之，参数方程 $(x(t), y(t))$ 表示的函数就是 $H$ 的等高线，可以解得： $ y = plus.minus sqrt(2(C - F(x)))\ (dif x)/(dif t) = plus.minus sqrt(2(C - F(x)))\ plus.minus 1/(sqrt(2(C - F(x)))) dif x = dif t $ 两边直接积分及得原方程的解 ] #example[弹簧][ 弹簧方程 $(dif^2 x)/(dif t^2) + a x = 0, a > 0$ 就是上面的形式，因此可以解得： $ t + C' = plus.minus integral 1/sqrt(2(C - 1/2 a x^2)) dif x $ 计算化简得到 $ x = C_1 cos a t + C_2 sin a t\ y = - a C_1 sin a t + a C_2 cos a t $ 容易发现所有解都以 $(2 pi)/a$ 为周期，且 $(0, 0)$ 为所有轨道的中心。由于所有轨道都是相同周期的，因此也称为等时中心。线性系统中一定是等时的 ] #example[单摆方程][ 单摆方程 $(dif^2 x)/(dif t^2) + a sin x = 0, a > 0$ 也是上面的形式，因此可以解得： $ dif t = 1/sqrt(2(C - a cos x)) dif x $ 这是椭圆积分，我们无法写出它的初等形式\ 转而考虑： $ H(x, y) = y^2/2 - a cos x + a $ 的等高线。注意到 $H$ 关于 $x$ 以 $2 pi$ 为周期，可以把它想象成定义在圆柱面上的函数。容易发现这里 $(0, 0)$ 也是中心，但不是等时中心，事实上，在等高线： $ y^2 - a cos x = C $ 上，有： $ 1/4 T = integral_0^(pi/2) der(t, x) dif x = integral_0^(pi/2) 1/sqrt(2(C - a cos x)) dif x $ 显然不可能与 $C$ 无关 ] #theorem[][ 设二阶自洽微分方程满足 $f(x)$ 连续且 $x f(x) > 0$，引进 $F(x)$ 并不妨设 $F(0) = 0$，则对： $ H(x, y) = y^2/2 + F(x) $ + $H(0, 0)$ 是 $H$ 的最小值点 + 在 $(0, 0)$ 附近，$H(x, y) = C$ 给出一个闭合曲线 + $H = h$ 交 $x$ 轴于 $x_1(h), x_2(h)$ 两点，$x_1 < x_2$ + 原点是等时中心当且仅当 $(x_2(h) - x_1(h))/sqrt(h)$ 是与 $h$ 无关的常数 ] #proof[ + 由题设将有 $F(x) >= 0$ 因此显然 + 直接解出 $y$ 即可 + 由 $F(x)$ 的单调性及 $F(0) = 0$ 显然 + 这个证明并不用到高级的知识，但极具技巧性\ 令 $s = F(x)$，则 $f(x) dif x = dif s$，有： $ 1/2 T_h = integral_(x_1)^(x_2) der(t, x) dif x = integral_(x_1)^(x_2) 1/sqrt(2(h - F(x))) dif x\ = integral_0^(h) 1/sqrt(2(h - s)) 1/(f_+(x)) dif s - integral_0^(h) 1/sqrt(2(h - s)) 1/(f_-(x)) dif s\ 1/2 integral_0^H T_h/sqrt(H-h) dif h = integral_0^H ( integral_0^(h) 1/sqrt(2(H - h)(h - s)) 1/(f_+(x)) dif s)dif h - integral_0^H ( integral_0^(h) 1/sqrt(2(H - h) (h - s)) 1/(f_-(x)) dif s)dif h\ = integral_0^H ( integral_s^H 1/sqrt(2(H - h)(h - s)) 1/(f_+(x)) - 1/sqrt(2(H - h) (h - s)) 1/(f_-(x)) dif h)dif s\ = pi/(sqrt(2)) (integral_0^H 1/(f_+(x)) - 1/(f_-(x)) )dif s\ = pi/(sqrt(2)) (integral_0^H 1/(f_+(x)) - 1/(f_-(x)) )dif F(x)\ = pi/(sqrt(2)) (integral_0^H f(x)/(f_+(x)) - f(x)/(f_-(x)) )dif x\ = pi/(sqrt(2)) (x_2(H) - x_1(H))\ $ - 假设周期与 $H$ 无关，则有： $ T_0 sqrt(H) = pi/sqrt(2) (x_2(H) - x_1(H))\ T_0 = pi/sqrt(2) (x_2(H) - x_1(H))/sqrt(H) $ 表明上式右侧与 $H$ 无关，这也给出了周期的计算公式 ] #corollary[][ 若原点是等时中心且 $f(x)$ 是奇函数，则 $f(x)$ 一定线性 ] = 微分方程的解理论 == Peano 定理 #theorem[Peano][ 设： $ f(x, y) in C(DD), DD = {abs(x - x_0) <= a, abs(y -y_0) <= b} $ 方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 在 $[x_0 - h, x_0 + h]$ 上有解，其中： $ h = min{a, b/(max_DD abs(f))} $ 若 $y, f$ 是向量值函数，结论也是类似的 ]<peano> 本节的目标就是证明它。\ #lemma[][ 有界区间上等度连续的一致有界的函数列有一致收敛的子列 ] #proof[ #lemmaLinear[][ 设 $f_n: I -> RR$ 对于每个 $x in I, f_n (x)$ 都有界 $M_x$，$I$ 是有限区间，则任取 $I$ 的一个可数子集 $E$，存在 $f_n$ 的子列使得这个子列在 $E$ 上收敛 ] #proof[ 设 $E = {x_1, x_2, ..., x_n}$，注意到 $f_n (x_1)$ 是有界序列，有收敛子列 $f_(n_1) (x_1)$\ 再考虑 $f_(n_1) (x_2)$ ，它当然也有收敛子列，记为 $f_(n_2) (x_2)$\ 不断进行下去，我们得到了若干子列 $f_(n_k) (x_k)$。\ 用对角线法，发现： $ f_(n_n) (x_n), n = 1, 2, 3, ... $ 在每个 $x_k$ 上都收敛，证毕 ] #lemmaLinear[][ 设 $I$ 是有限区间，$f_n : I -> RR$ 等度连续，且在 $I$ 的稠密子集上收敛，那么它在 $I$ 上一致收敛 ] #proof[ 利用柯西收敛原理，任取 $epsilon > 0$，存在 $delta >0$，使得： $ forall n in NN, forall x_1, x_2, abs(x_1 - x_2) < delta => abs(f_n (x_1) - f_n (x_2)) < epsilon $ 对于这个 $delta$，我们可以取出 $E$ 中有限个点 $E'$ 使得： $ union_(x in E') B(x, delta) = I $ 由于这里只有有限个 $x$，当然 $f_n\|_E'$ 将一致收敛，因此可设存在 $N$ 使得： $ forall m, n > N, forall x in E', abs(f_m (x) - f_n (x)) < epsilon $ 因此，有： $ forall m, n > N, forall x in I, exists x' in E'\ abs(x - x') < delta &=> abs(f_m (x) - f_n (x)) \ &< abs(f_m (x) - f_m (x')) + abs(f_m (x') - f_n (x')) + abs(f_n (x') - f_n (x)) < 3 epsilon $ ] 在引理中取 $E = QQ sect I$，它是 $I$ 的稠密子集，因此结论成立 ] 回到 @peano 的证明，这个方法是所谓的欧拉折线法 #proof[ 微分方程可以转化为积分方程: $ y(x) = y_0 + integral_(x_0)^(x) f(t, y(t)) dif t $ 只需证明它在 $[x_0, x_0 + h]$ 上有解，另一侧类似。\ 将其 $n$ 等分，设分点为 $x_0, x_1, ..., x_n$，按照以下定义分段线性函数： $ (x_(i-1), y_(i-1)) 到 (x_i, y_i) "的斜率由" f(x_(i-1), y_(i-1)) "给出" $ 需要验证这些点落在 $DD$ 中，显有： $ y_k - y_(k-1) = f(x_(k-1), y_(k-1)) (x_k - x_(k-1)) $ 两边求和： $ y_k - y_0 &= sum_(i = 1)^(k) f(x_(i-1), y_(i-1)) (x_i - x_(i-1))\ abs(y_k -y_0) &<= sum_(i = 1)^(k) abs(f(x_(i-1), y_(i-1)) (x_i - x_(i-1))) \ &<= max_DD abs(f) sum_(i = 1)^(k) (x_i - x_(i-1)) <= max_DD abs(f) abs(x_k - x_0) \ &<= h max_DD abs(f) <= b $ 这也体现了 $h$ 为何如此定义\ 设得到的函数为 $g_n (x)$，验证它们： - 一致有界：显然 - 等度连续：注意到所有 $g_n (x)$ 均满足： $ abs(g_n (x) - g_n (y)) <= max_DD abs(f) abs(x - y) $ 因此也成立由引理，它有一致收敛子列。不妨设其收敛，只需证明极限函数 $g(x)$ 满足积分方程\ 事实上（注意到 $f$ 是有界闭集上的连续函数，进而一致连续，从而 $f(t, g_n (x))$ 也一致收敛）： $ g(x) - integral_(x_0)^x f(t, g(t)) dif t = lim_(n->+infinity) g_n (x) - integral_(x_0)^x f(t, g_n (t)) dif t $ 有： $ g_n (x_0 + k/(n h)) &= sum_(i = 1)^(k) f(x_i, y_i) (x_i - x_(i-1))\ integral_(x_0)^(x_0 + k/(n h)) f(t, g_n (t)) dif t = sum_(i = 1)^(k) integral_(x_(i-1))^(x_i) f(t, g_n (t)) dif t &= sum_(i = 1)^(k) f(xi_i, g_n (xi_i)) (x_i - x_(i-1))\ g_n (x_0 + k/(n h)) - integral_(x_0)^(x_0 + k/(n h)) f(t, g_n (t)) dif t &= sum_(i = 1)^(k) (f(x_i, y_i) - f(xi_i, g_n (xi_i))) (x_i - x_(i-1))\ &=sum_(i = 1)^(k) (f(x_i, g_n (x_i)) - f(xi_i, g_n (xi_i))) (x_i - x_(i-1)) $ 任意 $epsilon > 0$ ，令 $delta, delta'$ 满足： - $abs(x-x') < delta, abs(y-y') < delta => abs(f(x, y) - f(x', y')) < epsilon$ - $abs(x-x') < delta' => abs(g_n (x) - g_n (x')) < min(delta, epsilon), forall x, forall n$ 取 $N$ 充分大使得 $forall n > N$： - $1/(n h) < delta' => abs(g_n (x) - g_n (x_0 + k/(n h))) < epsilon, g_n (x_i) - g_n (xi_i) < delta$ - $1/(n h) < delta => abs(x_i - xi_i) < delta$ - $1/(n h) $ ] #lemma[Gronwall][ 设 $f, g in C[a, b], g(x) >= 0$，则有： $ f(x) <= C + integral_a^x f(t) g(t) dif t => f(x) <= C e^(integral_a^x g(t) dif t) $ ] #proof[ 令 $Phi(x) = integral_a^x f(t) g(t) dif t$，则有： $ Phi' <= (C + Phi)g(x) $ 设 $h = e^(integral_a^x g(s) dif s), Phi = u h$，有： $ u' h + u h' <= (C + u h)g\ u' h <= C g\ u' <= C g/h\ u(x) - u(a) <= C integral_a^x g(s)/h(s) dif s\ (Phi(a) = 0 => u(a) = 0)\ u(x) <= C integral_a^x g(s)/h(s) dif s = C integral_a^x g(s) e^(-integral_a^x g(s) dif s) dif s = C (1 - 1/h(x))\ $ 从而： $ f(x) <= C + Phi(x) = C + u h <= C h $ 证毕 ] #corollary[][ 若在 Gronwall 不等式中，有 $C <= 0, f(x) >= 0$，则 $f(x) = 0$ ] #theorem[][ 在@peano 中，若 $f$ 是 Lipschitz 连续的，则解是唯一的。更进一步，构造出的折线列 $g_n (x)$ 本身即一致收敛 ] #proof[ 假设 $y = phi(x), y= psi(x)$ 是两个解，则： $ abs(phi(x) - psi(x)) = abs(integral_(x_0)^x f(t, phi(t)) dif t - integral_(x_0)^x f(t, psi(t)) dif t) <= L abs(integral_(x_0)^x abs(phi(t) - psi(t)) dif t) $ - 当 $x > x_0$ 时，上式即为： $ abs(phi(x) - psi(x)) <= L integral_(x_0)^x abs(phi(t) - psi(t)) dif t $ 直接利用 Gronwall 不等式可得 $abs(phi(x) - psi(x)) <= 0 => abs(phi(x) - psi(x)) = 0$ - 当 $x < x_0$ 时，类似可以证明结论成立综上，$phi(x) = psi(x)$ ] 事实上，在一维的情形下，我们可以找到更弱的条件 #theorem[][ 设 $f$ 满足 Osgoad 条件，也即： $ exists F: [0, +infinity] -> [0, +infinity)\ integral_(0)^(t) 1/F(r) dif r = +infinity\ abs(f(x, y) - f(x, z)) <= F(abs(y - z)) $ 则上述方程的解存在且唯一 ] #proof[ 设 $y_1, y_2$ 是两个不同的解，令 $r(x) = y_1 (x) - y_2(x)$\ 注意到 $r(x_0) = 0$，若 $r$ 不恒为零，不妨设 $x_1$ 是零点且 $r(x) > 0 ,forall x in (x_1, x_2]$\ 此时： $ der(r, x) = y'_1 - y'_2 = f(x, y_1) - f(x, y_2) <= F(abs(y_1 - y_2)) = F(r(x))\ => r'/F(r(x)) <= 1\ => integral_(s)^(t) r'/F(r(x)) dif x <= t -s\ => integral_(r(s))^(r(t)) 1/F(r) dif r <= t - s $ 令 $s -> x_1$，上式左侧趋于无穷，而右侧有限，矛盾！ ] #theorem[][ 设 $f(x, y)$ 满足 $f(x, y)$ 关于 $y$ 单调下降，则解也存在并唯一 ] #proof[ 设 $y_1, y_2$ 是两个不同的解，令 $r(x) = y_1 (x) - y_2(x)$\ 注意到 $r(x_0) = 0$，若 $r$ 不恒为零，不妨设 $x_1$ 是零点且 $r(x) > 0 ,forall x in (x_1, x_2]$\ 此时： $ der(r, x) = y'_1 - y'_2 = f(x, y_1) - f(x, y_2) <= 0 $ 而 $r(x_1) = 0$，上式表明 $r$ 单调下降，进而 $r(x) <= 0$，因此只能 $= 0$，矛盾！ ] 本节利用的欧拉折线不仅在理论上很重要，在数值计算上也广泛用于数值求解微分方程，理论依据就是本节的定理 == Picard 定理 Picard 定理表面上和 Peano 定理的结论类似但条件更强，但是它的方法非常重要，因此仍有必要专门学习 #theorem[Picard][ 在 @peano 中，再补充 $f$ Lipschitz 连续，则方程的解存在 ] #proof[ 仍然转化为积分方程: $ y(x) = y_0 + integral_(x_0)^(x) f(t, y(t)) dif t $ 总体思想是所谓的不动点法，递归定义： - $phi_0 (x) = y_0$ - $phi_n (x) = y_0 + integral_(x_0)^(x) f(t, phi_(n-1) (t)) dif t$ + 首先验证定义合理，需要保证所有的值不超出区域。 - $x$ 的合理性是显然的 - 假设 $phi_(n - 1) subset overline(U(y_0, b))$，则： $ abs(phi_n (y_1) - y_0) <= integral_(x_0)^(x) abs(f(t, phi_(n-1) (t))) dif t\ <= max_DD abs(f) h <= b $ 这就验证了 $phi_(n) subset overline(U(y_0, b))$ + 接下来证明它一致收敛 - 首先归纳证明 $abs(phi_n (x) - phi_(n-1)(x)) <= (M L^(n-1))/(n!) abs(x -x_0)^n$ $ abs(phi_n (x) - phi_(n-1)(x)) = abs(integral_(x_0)^(x) f(t, phi_(n-1) (t)) - f(t, phi_(n-2) (t)) dif t)\ <= integral_(x_0)^(x) abs(f(t, phi_(n-1) (t)) - f(t, phi_(n-2) (t))) dif t\ <= integral_(x_0)^(x) L abs(phi_(n-1) (t) - phi_(n-2) (t)) dif t $ 利用归纳条件计算即可 - 其次： $ (M L^(n-1))/(n!) abs(x -x_0)^n <= (M L^(n-1))/(n!) h^n $ 上式右侧只需利用比较判别法知求和收敛，从而容易利用维尔斯特拉斯判别法知道原级数 $sum (phi_n (x) - phi_(n-1)(x))$ 一致收敛 + 最后我们得到了极限函数 $phi(x)$，显然应当满足： $ phi(x) = y_0 + integral_(x_0)^(x) f(t, phi(t)) dif t $ 这就是原方程 ] #remark[][ 这个定理的证明有以下优点： - 可以推广到抽象的函数空间，例如 Banach 空间 - 假设 $f$ 是解析函数，则构造的 $phi_n$ 都是解析函数，进而可以证明 $phi$ 也是解析的 ] == 解的延拓之前我们给出了至少在某个局部，标准形式的微分方程是有解的。然而我们往往需要在更大的定义域上找到解 #theorem[][ 设 $G$ 是开区域，$f(x, y)$ 在 $G$ 上连续，$(x_0, y_0) in G$\ 对含 $(x_0, y_0)$ 的任意有界闭区域 $G_1 subset G$，过 $(x_0, y_0)$ 点的微分方程的解 $y=f(x)$ 都可以延拓到 $G_1$ 的边界上 ] #proof[ 由题设，$G_1$ 中的点到 $G$ 的边界的距离有正下界 $d$\ 因此可以找到闭集： $ G_2 = {(x, y) \| abs(x - x') <= delta, abs(y - y') <= delta, exists (x, y) in G_1} subset G $ 再令： $ M = max_(f in G_2) abs(f) +1\ delta'_0 = delta_0 / M $ 利用之前的存在性定理，可以找到 $U(x_0, delta'_0)$ 内的解。再次利用存在性定理可以继续前进，且步长为 $delta'_0$ 不变，因此经过有限步之后，我们就得到了直到 $G_1$ 的边界上的解 ] #example[][ - 设微分方程 $der(y, x) = x^2 + y^2$ 容易验证过任何一点的解存在且唯一。然而，可以证明任何一个解都只在有界区间上成立事实上，任取 $y(x)$ 在区间 $I$ 上是方程的一个解，假设 $I$ 无界，不妨取为 $[0, +infinity]$，则： $ y' = x^2 + y^2 >= 1 + y^2\ y'/(1+y^2) >= 1\ integral_(1)^(t) y'/(1+y^2) dif x >= x - 1\ integral_(1)^(y(t)) 1/(1+y^2) dif y >= x - 1\ $ 然而令 $t -> +infinity$，上式左侧有界而右侧无界，矛盾！这不与解的延拓定理矛盾，容易想到它在 $x$ 方向有垂直渐近线，而在 $y$ 方向可以进行无限延伸 - 给定微分方程： $ der(y, x) = (x^2 + y^2 +1) sin (pi y) $ 可以验证它也满足存在唯一的条件，并且 - $y = k, k in ZZ$ 当然是一个解 - 对于其他的解，显然必须夹在上下两个水平直线之间（否则若相交，则交点处解将不唯一），因此在 $x$ 方向必将可以无限延伸 - 给定微分方程： $ der(y, x) = (y-3)(y+1) e^(x+y)^2 $ 容易验证满足存在唯一的条件 - 显然 $y = 3, -1$ 是两个解 - 若 $y_0 in [-1, 3]$，过该点的解同样与 $y = 3, -1$ 不相交，进而导数总为负数，单调下降，因此在 $-infinity, +infinity$ 方向都有极限。观察等式可知极限只能分别为 $3, -1$ - 若 $y_0 in [3, +infinity]$ ，负向仍然有水平渐近线，而正向类似刚才的将会存在垂直渐近线这些例子都说明，只要延拓定理的条件成立，最终得到的解一定在某个方向上无界 ] == 比较定理给定两个微分方程，我们希望从形式看出解的大小关系，由如下定理： #theorem[第一比较定理][ 设 $f, F$ 都连续，且 $f < F$，两个微分方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 )\ cases( der(Y, X) = F(X, Y), Y(X_0) = Y_0 ) $ 在区间 $I$ 上各有解 $y, Y$，则有： $ y < Y, forall x > x_0\ y > Y, forall x < x_0 $ ] #proof[ 显然 $Y'(x_0) = F(x_0, y_0) > y'(x_0)$，而 $Y(x_0) = y(x_0)$，因此可取 $delta$ 使得： $ forall x in (x_0, x_0 + delta), Y(x) > y(x)\ forall x in (x_0 - delta, x_0), Y(x) < y(x) $ 设 $S = {x in I \| Y(x) = y(x) }$，仿照之前的过程可以看出 $S$ 中每个点都孤立，然而任取 $x_1 > x_2 in S$，设 $g(x) = Y(x) - y(x)$，既然： $ g(x_1) = g(x_2) = 0 $ 且 $g$ 在 $x_2$ 右侧严格单增，在 $x_1$ 左侧严格单减，因此其间一定有 $S$ 中的点。反复进行，这将与 $S$ 中每个点都孤立矛盾\ 这意味着 $S$ 中恰有一点，只能是 $x_0$，故结论当然成立 ] #corollary[][ 设 $f in C(a, b), abs(f) <= A(x) abs(y) + B(x), forall x in (a, b)$，则微分方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 所有的解都可以延拓到 $(a, b)$ ] #proof[ #lemma[][ 两个微分方程： $ der(y, x) = A(x) abs(y) + B(x)\ der(y, x) = - A(x) abs(y) - B(x) $ 过任意点的解存在唯一，且解的存在区间都是 $(a, b)$ ]<linear_all_range> #proof[ 存在唯一性是容易的。对于存在区间，只证明第一个方程\ 显然它的解单调递增，若解的存在区间不是 $(a, b)$ ，则必然在某点 $x_0 in (a, b)$ 处的左/右极限为 $+\/minus infinity$，不妨设为左极限为正无穷\ 这意味着存在 $delta, (x_0 - delta, delta)$ 上时有： $ der(y, x) = A(x) abs(y) + B(x) = A(x) y + B(x) < M_1 y + M_2 $ 但稍加计算可得 $der(y, x) = M_1 y + M_2$ 的解不可能在有界区间内趋于无穷，进而该方程的解由比较定理被控制在有界区间，矛盾！ ] 回到原题，显然有： $ -(A(x) abs(y) + B(x) + 1)<= der(y, x) = f < A(x) abs(y) + B(x) + 1 $ 由比较定理，每个有限区间上的解当然被控制在有界区间，证毕 ] #definition[][ 给定微分方程 $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 称其解 $f, g$ 为最大解，最小解，若任取 $y$ 为一个解，都有： $ g <= y <= f $ ] #theorem[][ 设 $D = {abs(x-x_0) < delta_x, abs(y - y_0) < delta_y}$，则存在 $tau > 0$ 使得方程 $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 在 $x in [x_0 , x_0 + tau]$ 上有最大/最小解 ] #proof[ 设 $y_n$ 是微分方程： $ cases( der(y, x) = f(x, y) + 1/n, y(x_0) = y_0 ) $ 由 @peano，每个方程的解都在： $ [x_0 - h_n, x_0 + h_n] "where" h_n = min{delta_x, delta_y/(max_DD abs(f_n))} $ 存在，显然 $h_n$ 有公共正下界，取 $tau$ 是一个正下界，则这些方程都在 $[x_0 - tau, x_0 + tau]$ 上有界\ // 我们希望找到一些一致收敛性。事实上，这些函数： // - 等度连续，既然： // $ // y'_n (x) <= max_DD abs(f) + 1.n <= max_DD abs(f) + 1 // $ // - 一致有界，既然导数有一致界且都定义在有限区间上事实上，这些解一致收敛。这是因为若设 $forall n, m > N, phi(x) = y_n (x) - y_m (x) $，则： $ abs(der(phi, x)) = abs(1/n - 1/m) < 1/N\ abs(phi(x) - phi(0)) = abs(integral_(x_0)^(x) der(y, x) dif s) <= integral_(x_0)^(x) abs(der(y, x)) dif s <= (x - x_0)/N <= tau/N $ 柯西原理给出了一致收敛性，因此我们可以取极限，而转化为积分方程，它们的极限就是原方程的解，而容易验证这个解一定是原方程的最大解 ] #remark[][ 我们当然可以将两端的最大解拼接，得到双边的最大解 ] #theorem[][ 设 $f, g$ 是微分方程： $ cases( der(y, x) = f(x, y) , y(x_0) = y_0 ) $ 在 $I$ 上的最大/最小解，则任取 $x_1 in I, y_1 in [g(x_1), f(x_1)]$，存在方程的解 $h$ 使得 $h(x_1) = y_1$ ] #proof[ #lemmaLinear[][ 设 $g_1, g_2$ 在区间 $I$ 上均满足 $g'_i (x) = f(x, g_i (x))$，则 $max(g_1, g_2), min(g_1, g_2)$ 也满足 ] #proof[ 只证 $max$，$min$ 类似\ 设 $h = max(g_1, g_2)$，验证上面的等式只需逐点验证即可： - 任取 $x in I$，讨论： - 若 $g_1(x) > g_2(x)$，则有 $g_1, g_2$ 连续性知存在 $x$ 的开邻域 $B$ 使得： $ h\|_B = g_1 $ 此时由于 $g_1$ 在该点是微分方程的解，$h$ 当然也是 - 若 $g_1(x) < g_2(x)$，同理 - 若 $g_1(x) = g_2(x)$，注意到： $ g'_1(x) = f(x, g_1(x)) = f(x, g_2(x)) = g'_2(x) := d $ 因此： $ abs((h(x + Delta x) - h(x))/(Delta x) - d) &<= abs((h(x + Delta x) - g_1 (x + Delta x))/(Delta x)) + abs((g_1 (x + Delta x) - g_1 (x))/(Delta x) - g'_1 (x))\ &<= abs((g_2(x + Delta x) - g_1 (x + Delta x))/(Delta x)) + abs((g_1 (x + Delta x) - g_1 (x))/(Delta x) - g'_1 (x))\ &<= abs(((g_2(x + Delta x) - g_1 (x + Delta x)) - (g_2 (x) - g_1 (x)))/(Delta x)) + abs((g_1 (x + Delta x) - g_1 (x))/(Delta x) - g'_1 (x))\ &<= abs(g'_2 (xi) - g'_1 (xi)) + abs((g_1 (x + Delta x) - g_1 (x))/(Delta x) - g'_1 (x))\ &"（对" g_2(x) - g_1(x) 在 x, x + Delta x "上利用微分中值定理）" $ 其中 $xi$ 在 $x, x + Delta x$ 之间。注意到由微分方程，$g_1, g_2$ 当然连续可导，进而当 $Delta x -> 0$ 时上式 $-> 0$，这就保证了： $ h'(x) = d = f(x, h(x)) $ 也即 $h$ 在该点满足微分方程得证 ] 考虑微分方程： $ cases( y' = f(x, y) , y(x_1) = y_1 ) $ 它在 $x_1$ 的某个邻域内有解 $h$，又可不妨设 $h$ 在 $I$ 内某个极大的区间 $I_1$（也即所有可延拓闭区间的并）上是原方程的解，并设： $ h_1 = max(min(h, Z), W), x in I_1 $ 将满足 $h'_1 (x) = f(x, h_1 (x))$，同时也不难验证 $h_1(x_1) = y_1$，因此是过该点的微分方程的解\ 另一方面，显然有： $ max(min(h, Z), W) >= W\ max(min(h, Z), W) <= max(Z, W) = Z $ 因此 $h_1$ 是区间上有界的微分方程的解从而： - 若 $x_0 in I_1$，由上面的条件和引理知 $h_1$ 就是要找的解 - 否则，设 $I_1$ 闭包中距 $x_0$ 最近的点为 $x_2$，注意到 $h_1$ 在 $I_1$ 上有界，因此由延拓定理要么意味着存在 $x'$ 使得 $abs(h_1(x') - y_0) = b$，要么 $h_1$ 一定可以至少延拓到 $x_2 := x'$，然而已经假设 $h$ 不能再延拓，因此该点附近一定取 $Z$ 或 $W$ ，无论如何一定有： $ exists x', h_1(x') = Z(x') 或 W(x') $ 无论何者，都意味着 $h_1$ 可以与 $Z$ 或 $W$ 的其中之一拼接起来，这就是所要找的解。 ] #corollary[][ 若微分方程的单侧的最大解，最小解均存在且不相等，则该侧微分方程有无穷多解，这无穷多解被控制在最大/最小解之间 ] #example[][ - 微分方程： $ cases( der(y, x) = y^(1/3), y(0) = 0 ) $ 通解为： $ y = cases( 0 quad x <= -C\ plus.minus (2/3 x + C)^(3/2) quad x > C ) $ 在 $0$ 右侧的最大解为 $y = (2/3 x)^(3/2)$，最小解为 $y = -(2/3 x)^(3/2)$ - 微分方程： $ cases( der(y, x) = abs(y)^(1/2), y(0) = 0 ) $ 通解为： $ y = cases( 1/4(C_1 - x)^2 quad x <= C_1, 0 quad C_1 <= x <= C_2, 1/4(x - C_2)^2 quad x >= C_2 ) $ ] #theorem[第二比较定理][ 设 $f, F$ 都连续，且 $f <= F$，两个微分方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 )\ cases( der(Y, X) = F(X, Y), Y(X_0) = Y_0 ) $ 则： - 在初值右侧，前者的最小解小于等于后者的任意解;在初值左侧，前者的最大解大于等于后者的任意解 - 在初值右侧，后者的最大解大于等于前者的任意解;在初值左侧，后者的最小解小于等于前者的任意解 ] #proof[ 仿照上面的逼近过程再使用第一比较定理即可 ] == 奇解 #definition[奇解][ 设 $Gamma = {(x, y) \| y = f(x)}$ 是微分方程 $F(x, y') = 0$ 的一个解，若对任意的 $(x_, y) in Gamma$，在该点的任何一个邻域里存在过 $(x_0, y_0)$ 的不同于 $Gamma$ 的解，且该解与 $Gamma$ 在该点相切，则称 $Gamma$ 是奇解 ] #example[][ - $y = x y' - 1/2 y'^2$，它的解为： $ y = 1/2 x^2\ y = c x - c^2/2 $ 注意到下面的直线族恰好是上面抛物线的所有切线，因此抛物线当然就是一个奇解 - $y^2 + y'^2 = 1$\ 令 $y = cos theta, y' = sin theta, x = u(theta)$，有： $ (-sin theta)/(u') = sin theta\ sin theta = 0 或 u' = -1 $ 因此解为： $ y' = 0 => y = plus.minus 1\ u' = -1 => x = -theta + C => y = cos(x - C) $ 可以看出 $y = plus.minus 1$ 是奇解 ] #theorem[奇解的必要条件][ 设 $G subset RR^3, F(x, y, y') in C^0, partialDer(F, y), partialDer(F, y')$ 存在且连续。假设 $y = phi(x)$ 是奇解，则： $ cases( F(x, phi(x), phi'(x)) = 0, partialDer(F, y') (x, phi(x), phi'(x)) = 0 ) $ ] #proof[ 第一个式子是显然的。对于第二个式子，假设存在 $x_0$ 使得 $partialDer(F, y) (x_0, phi(x_0), phi'(x_0)) != 0$，设 $(x_0, y_0, P_0) = (x_0, phi(x_0), phi'(x_0))$，在该点对 $F$ 利用隐函数定理，存在某个开邻域使得： $ F(x, y, P) = 0 <=> P = f(x, y) $ 且 $f$ 是连续函数且关于 $y$ 连续可导，将在 $(x_0, y_0)$ 的某个邻域内 Lipschitz 连续，可以利用解的存在唯一性定理可知该邻域内微分方程的解存在唯一，与 $phi$ 是奇解显然矛盾 ] 上面的定理给出了奇解的一个必要条件。这里有两个等式，若可以化简直到可解出 $y$，便可能求出原方程的奇解。更进一步，若能消去 $y'$ 得到： $ Delta (x, y) = 0 $ 则奇解就在该曲线之中，这条曲线被称为判别曲线 #example[][ - $ y = P x ln x + (x p)^2 - y\ $ 可以解得判别曲线为 $y = -1/4 (ln x)^2$，容易验证它是一个解，稍后会判断它是否是奇解 - $y'^2 + y - x = 0$\ 可以解得判别曲线为 $y = x$，但不是原方程的解，因此原方程没有奇解 - $y'^2 - y'^2 = 0$，它的判别曲线是 $y = 0$ 是解，然而容易看出它的所有解是: $ y = C e^(plus.minus x) $ 以及它们可能的拼接，然而仅有 $y = 0$ 唯一一个解可能经过 $x$ 轴，因此过 $(x_0, 0)$ 的解都是存在且唯一的 ] #theorem[奇解的充分条件][ 设 $F(x, y, y') in C^infinity$，假设： $ cases( F(x, phi(x), phi'(x)) = 0, partialDer(F, y') (x, phi(x), phi'(x)) = 0 ) <=> y = psi(x) $ （事实上，也就是 $Delta(x, y)$ 是原方程的解）\ 若： $ cases( (diff^(k+l) F)/(diff y^k diff p^l) (x, psi(x), psi'(x)) = 0\, forall 0 <=k <= m-1\, 0 <= l <= n-1, (diff^m F)/(diff y^m) (x, psi(x), psi'(x)) != 0, (diff^n F)/(diff p^n) (x, psi(x), psi'(x)) != 0 ) $ 其中 $n, m in NN, n > m$。如果： - $m, n$ 之一为奇数，或 - $m, n$ 均为偶数，且 $(diff^m F)/(diff y^m) (diff^n F)/(diff p^n) < 0$ 则 $psi$ 是奇解 ] #proof[ #lemma[][ 设 $f(x, y) : C^infinity (RR^n times RR -> RR)$，设： $ f(x, y) = sum_(k=0)^n f_k (x) y^k + R_(n+1) (x, y) y^n $ 是关于 $y$ 的泰勒展开，则 $R_n (x)$ 也是 $C^infinity$ ] #proof[ 利用积分余项： $ f(x, y) - sum_(k=0)^n f_k (x) y^k = y^n/n! integral_0^1 (1-t)^n (diff^n f(x, y t))/(diff t^n) dif t $ 容易看出结论成立 ] #lemma[][ 考虑微分方程 $u' = plus.minus A(x, u) abs(u)^alpha, 0< alpha <1 $ 其中有： - $A$ 连续且在 $u !=0$ 时连续可导 - $A$ 有正下界 $c_0$，正上界 $c_1$ 则 $u = 0$ 是奇解 ]<lemma0_is_odd> #proof[ 不妨设符号取正，此时方程的解当然递增。\ 我们的目标是在 $x_0$ 处构造一个原方程的非平凡解。先考虑右侧，注意到： $ u' = A(x, u)u^alpha\ u^(-alpha) u' = A(x, u)\ $ 设 $v = u^(1 - alpha), v' = (1-alpha) u^(-alpha) u'$，则： $ 1/(1-alpha) v' = A(x, v^(1/(1-alpha))) $ ] 回到原方程，做换元 $y = psi(x) + u$，方程变成： $ H(x, u, u') = F(x, psi(x) + u, psi'(x) + u') = 0 $ 只需证明 $u = 0$ 是奇解即可。事实上，$H$ 满足如下条件： $ cases( (diff^(k+l) F)/(diff y^k diff p^l) (x, 0, 0) = 0\, forall 0 <=k <= m-1\, 0 <= l <= n-1, (diff^m F)/(diff y^m) (x, 0, 0) != 0, (diff^n F)/(diff p^n) (x, 0, 0) != 0 ) $ 可以想象，$H$ 泰勒展开后的形式非常简单。事实上，有： $ H(x, u, u') = u^m H_1 (x, u, u') + u'^n H_2 (x, u, u') $ 其中 $H_1, H_2$ 都无穷阶可导，且： $ H_1 (x, 0, 0) = (diff^m F)/(diff y^m) (x, psi(x), psi'(x)) != 0\ H_2 (x, 0, 0) = (diff^n F)/(diff p^n) (x, psi(x), psi'(x)) != 0 $ 因此可以（在某个小邻域内）不妨设 $H_1, H_2 !=0 $，方程化为： $ u^m H_1 (x, u, u') + u'^n H_2 (x, u, u') = 0 $ 我们当然希望进行开方，可以验证在假设的条件下（即 $m, n$ 之一为奇数，或 $m, n$ 均为偶数，且 $(diff^m F)/(diff y^m) (diff^n F)/(diff p^n) < 0$），我们将确实可以开方，对开方后的函数利用 @lemma0_is_odd 即可 ] #corollary[][ 设 $F(x, y, y') in C^2$，且： $ cases( F(x, phi(x), phi'(x)) = 0, partialDer(F, y') (x, phi(x), phi'(x)) = 0 ) <=> y = psi(x) $ 若： $ (diff^2 F)/(diff p^2) (x, psi(x), psi'(x)) != 0\ partialDer(F, y) (x, phi(x), phi'(x)) != 0 $ 则 $psi$ 是奇解 ] #proof[ 就是在上面的定理中取 $n = 2, m = 1$ 的结果，至于条件可以放松的原因可以在上面的证明过程中仔细验证 ] == 包络 #definition[包络][ 设 $k_c, c in C$ 是光滑曲线族，称光滑曲线 $gamma$ 为 $k_c$ 的包络，如果 $forall (x, y) in gamma, exists k_c$ 使得 $k_c$ 与 $gamma$ 相切，且在该点的任意开邻域内 $k_c != gamma$ ] #example[][ - 给定一族曲线 $y = (x-c)^2 + 1$，当然 $y = 1$ 是其一个包络 - 给定一族直线 $y = c x - c^2 / 4$，它就是 $y = x^2$ 的切线族，自然 $y = x^2$ 是包络 ] #theorem[][ 假设 $F(x, y, y') = 0$ 有通解 $u(x, y, c) = 0$，假设 $partialDer(u, y) != 0$。设 $u(x, y, c) = 0$ 关于 $c$ 构成的曲线族有包络 $gamma := y = phi(x)$，则 $y = phi(x)$ 是奇解 ] #proof[ - 首先证明 $gamma$ 是解。当然 $gamma$ 上每点都可以找到解 $u(x, y, c_0) = 0$ 与之在该点相切。由假设条件和隐函数定理，可以反解出： $ u(x, y, c_0) = 0<=> y = psi(x) $ 当然 $phi(x)$ 与 $psi(x)$ 相切但不相同，这表明该点处 $gamma$ 的函数值和导数值满足微分方程，因此 $gamma$ 是解 - 其次，过 $gamma$ 每点都可以找到与之相切的不同解，这就表明它是奇解 ] #theorem[][ 假设 $Gamma$ 是曲线族 $k_c: v(x, y, c) = 0$ 的包络，并且 $Gamma$ 与 $k_C$ 切于 $(f(c), g(c))$，其中 $f, g in C^1$，则 $Gamma$ 满足方程： $ cases( v(x, y, c) = 0, v'_c (x, y, c) = 0 ) $ 如果这个方程组中能消去 $c$，则称消去后的曲线为 $C$ 判别曲线 ] #proof[ 任取 $(x_0, y_0) in Gamma sect k_c$，则首先第一个方程当然成立\ 在 $v(f(c), g(c), c) = 0$ 中，两边对 $c$ 求导得： $ v'_x (f(c), g(c), c) f'(c) + v'_y (f(c), g(c), c) g'(c) + v'_c (f(c), g(c), c) = 0 $ 只需证明 $vec(v'_x (f(c), g(c), c), v'_y (f(c), g(c), c) ) dot vec(f'(c), g'(c)) = 0$，不妨设他们都不为零\ 然而注意到 $vec(f(c), g(c))$ 事实上是 $Gamma$ 在局部上的一个参数方程，$vec(f'(c), g'(c))$ 将成为 $Gamma$ 的切向量。\ 同时，$vec(v'_x (f(c), g(c), c), v'_y (f(c), g(c), c) )$ 事实上是 $v(x, y, c) = 0$ 在 $(f(c), g(c))$ 处的法向量，由于 $Gamma$ 是包络当然两者正交 ] #theorem[][ 假设曲线族 $k_c: v(x, y, c) = 0$ 的 $C$ 判别曲线确定了一条 $C^1$ 曲线 $x = x(c), y = y(c)$，且满足非退化条件： $ (x'(c), y'(c)) != (0, 0)\ (partialDer(v,y)(x(c), y(c), c), partialDer(v,x)(x(c), y(c), c)) != (0, 0) $ 并且该曲线在曲线上任意一点的局部都不在原曲线族中，则它是包络 ] #proof[ 任取 $c$，在 $P(c) = (x(c), y(c))$ 的局部都可以利用隐函数定理反解出 $v(x, y, c) = 0$ 的解，可以验证这个解与 $Gamma$ 在该点相切，因此是包络 ] #example[][ 求一条曲线，满足其任意一点的切线两个截距的倒数平方和为 $1$ 事实上，容易想到满足这个条件的曲线一定是一族直线的包络，这族直线满足截距的倒数平方和为 $1$，进而形如 $ a x plus.minus sqrt(1 - a^2) y = 1 $ 只需找到： $ v(x, y, c) = c x plus.minus sqrt(1 - c^2) y - 1 = 0 $ 的包络，计算： $ partialDer(v, x) = c\ partialDer(v, y) = plus.minus sqrt(1 - c^2)\ partialDer(v, c) = x minus.plus y c/sqrt(1 - c^2) $ 计算其 $c$ 判别式，发现就是单位圆 $x^2 + y^2 = 1$\ 当然，单位圆和满足要求的直线的任意光滑拼接都满足要求 ] = 解对参数和初值的依赖性 == $n$ 维欧式空间的微分方程首先，叙述高维的常用定理，设 $R: abs(x - x_0) <= a, norm(y - y_0) <= b$ #theorem[Picard][ 设 $f in C(R, RR^n), abs(f(x, y) - f(x, z)) <= L norm(y - z)$，则微分方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 在 $[x_0 - h, x_0 + h]$ 上有唯一解，其中 $h = min{a, b/(max_R norm(f))}$ ] #theorem[Peano][ 设 $f in C(R, RR^n)$，则微分方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 在 $[x_0 - h, x_0 + h]$ 上有解，其中 $h = min{a, b/(max_R norm(f))}$ ] #definition[$n$ 维线性方程组][ 设微分方程组： $ cases( der(y, x) = A(x) y + B(x), y(x_0) = y_0 ) $ 满足 $A, B$ 都在 $(a, b)$ 上是连续函数，则称其为 $n$ 维线性方程组 ] #proposition[][ 对于任意初值，$n$ 维线性方程组的解在 $(a, b)$ 内存在唯一 ] == 解对初值和参数的依赖性微分方程的初值当然依赖于初值。出于多种考虑，我们当然希望这种依赖有某种连续性，这就是本章的核心内容。 #example[][ 考虑微分方程： $ y' = y^(1/3), y(0) = epsilon(epsilon < 0) $ 的最大解为： $ cases( -(2/3 x + epsilon^(2/3))^(3/2) quad x >= 0, 0 quad x < 0 ) $ 然而令 $epsilon -> 0$，它与 $0$ 处的最大解相去甚远 ] 这个例子表明最大/最小解的连续依赖性往往并不成立，我们往往只研究微分方程解唯一时的初值依赖性\ 此外，我们说明初值依赖性与参数依赖性之间是一致的。事实上，给定参数方程： $ cases( y' = f(x, y, lambda), y(x_0) = y_0(lambda) ) $ 令 $Y = vec(y, lambda)$，它等价于： $ cases( Y' = vec(f, 0), Y(x_0) = vec(y_0(lambda), lambda) ) $ 反过来，也可以通过平移将初值吸收进参数 #theorem[连续依赖性][ 给定一族微分方程： $ cases( y' = f(x, y), y(x_0) = y_0 ) $<ori-equation> （其中 $f$ 连续）的解存在唯一，并设其解的存在区间为闭区间 $I$ \ 设 $phi(x, xi, eta)$ 满足： $ cases( partialDer(phi, x) = f(x, phi(x, xi, eta)), y(xi) = eta ) $ 则： $ lim_((s, xi, eta) ->(x, x_0, y_0)) phi(s, xi, eta) = phi(x, x_0, y_0), forall x in I $ ] #proof[ 用反证法。如若不然，则则存在点列使得： $ (s_n, xi_n, eta_n) -> (x, x_0, y_0)\ d(phi(s_n, xi_n, eta_n), phi(x, x_0, y_0)) > epsilon $ 注意到有积分方程： $ phi(x, xi_n, eta_n) = eta_n + integral_(xi_0)^(x) f(s, phi(s, xi_n, eta_n)) dif s $ 因此这族关于 $x$ 的函数等度收敛一致有界，不妨就假设一致收敛到 $psi(x)$，两边取极限将有： $ lim_(n -> infinity) phi(x, xi_n, eta_n) = y_0 + integral_(x_0)^(x) f(s, lim_(n -> infinity) phi(x, xi_n, eta_n)) dif s $ （还要利用 $f$ 一致连续/有界）\ 表明 $lim_(n -> infinity) phi(x, xi_n, eta_n)$ 就是@ori-equation 的唯一一个解 $phi(x, x_0, y_0)$，与假设矛盾！ ] #corollary[][ 给定一族微分方程： $ cases( y' = f(x, y, lambda), y(x_0) = y_0 ) $ 对于任意 $x_0, y_0, lambda$ 的解都存在唯一，则设 $phi(x, x_0, y_0, lambda)$ 是上述方程的解，它将是连续函数 ] #theorem[光滑依赖性][ $ cases( y' = f(x, y, lambda), y(x_0) = y_0 ) $ 其中 $f$ 连续，对 $y, lambda$ 是 $C^1$ 的，且对于任意 $x_0, y_0, lambda$ 的解都存在唯一，则设 $phi(x, x_0, y_0, lambda)$ 是上述方程的解，它将是 $C^1$ 的 ] #proof[ 由定义，$phi$ 关于 $x space C^1$ 是显然的。对于 $y, lambda$，前面叙述了 $lambda, y_0$ 互相转换，只需证明关于 $lambda$ 连续可导就好，进一步，不妨假设方程就是： $ cases( y' = f(x, y, lambda), y(0) = 0 ) $ 为了证明结论，构造 Picard 序列： $ phi_0 = 0\ phi_(n) (x) = integral_(0)^(x) f(s, phi_(n-1), lambda) dif s $ 由于这里关于 $y$ 已经 $C^1$，Lipschitz 条件当然成立，因此它一致收敛到原方程的解。同时，容易归纳得到 $phi_n$ 关于 $lambda$ 都是 $C^1$ 的，计算导数： $ partialDer(phi_n (x), lambda) = integral_(0)^(x) partialDer(f, y) (s, phi_(n-1), lambda) partialDer(phi_(n-1), lambda) + partialDer(f, lambda) (s, phi_(n-1), lambda) dif s $ 这里如果 $lambda, y$ 是高维的，无非采用梯度/向量导数即可。\ 由分析学结论，只需证明上面的序列一致收敛。不妨设： $ norm(partialDer(f, y)), norm(partialDer(f, lambda)) <= M $ 将有： $ norm(partialDer(phi_1 (x), lambda)) <= integral_(0)^(x) norm(partialDer(f, lambda) (s, phi_(n-1), lambda)) dif s <= alpha norm(x)\ norm(partialDer(phi_2 (x), lambda)) <= integral_(0)^(x) alpha^2 norm(x) + alpha dif s <= alpha^2 norm(x)^2/2 + alpha norm(x)\ $ 可以归纳证明它一致有界\ 设： $ V_(k, n) = norm(partialDer(phi_(k+n) (x), lambda) - partialDer(phi_(k) (x), lambda)) $ 将有： $ V_(k+1, n) = norm(integral_(0)^(x) (partialDer(f, lambda) (s, phi_(k+n), lambda) - partialDer(f, lambda) (s, phi_(k+1), lambda)) \ + (partialDer(f, y) (s, phi_(k+n), lambda)partialDer(phi_(k+n), lambda) - partialDer(f, y) (s, phi_(k+1), lambda)partialDer(phi_(k+1), lambda) )dif s)\ <= integral_(0)^(x_0) norm(partialDer(f, y) (s, phi, lambda)) v_(k, k+1) dif s\ + d_(k, n) $ 其中 $d_(k, n)$ 在 $k$ 充分大时一致收敛于零，因此 $exists E_n$ 单调下降使得 $d_(k, n) <= E_n$，原式化为： $ V_(k+1, n) <= alpha integral_(0)^(x_0) V_(k, k+1) dif s + E_n $ 归纳计算将可得到 $V_(k, n)$ 关于 $k$ 一致收敛于零，由柯西法则知结论成立最后，我们还要考察关于 $x_0$ 的偏导数。这里我们无法直接吸收，因为我们没有假设 $f$ 关于 $x$ 可求偏导，这里 $lambda,y_0$ 无关紧要，不妨设方程为： $ cases( y' = f(x, y), y(x_0) = 0 ) $ 仍然构造 Picard 序列： $ phi_0 = 0\ phi_(n) (x) = integral_(x_0)^(x) f(s, phi_(n-1)) dif s $ 类似的，计算： $ partialDer(phi_n, x_0) = -f(x_0, phi_(n-1)) + integral_(x_0)^(x) partialDer(f, y) partialDer(phi_(n-1), x_0) dif s $ 接下来的计算是完全类似的 ] #remark[][ 这里我们不对关于 $x$ 的光滑性有很高要求，当然是因为微分方程的解相当于 $x$ 的积分，当然会提高光滑性。 ] #corollary[][ $ cases( y' = f(x, y, lambda), y(x_0) = y_0 ) $ 其中 $f$ 连续，对 $y, lambda$ 是 $C^k$ 的（$k$ 可能为无穷），且对于任意 $x_0, y_0, lambda$ 的解都存在唯一，则设 $phi(x, x_0, y_0, lambda)$ 是上述方程的解，它将是关于 $y_0, lambda$ 是 $C^k$ 的，关于 $x, x_0$ 是 $C^1$ 的 ] #proof[ $C^1$ 刚刚已经证明，同时也可以将 $y_0$ 吸收，化为： $ cases( y' = f(x, y, lambda), y(0) = 0 ) $ 化为积分方程： $ y = integral_(0)^(x) f(s, y, lambda) dif s $ 可以直接求导： $ partialDer(y, lambda) = integral_(0)^(x) partialDer(f, y) (s, y, lambda) partialDer(y, lambda) + partialDer(f, lambda) (s, y, lambda) dif s $ 设 $p = partialDer(y, lambda)$，将有： $ p' = partialDer(f, y) (x, y, lambda) p + partialDer(f, lambda) (x, y, lambda) $ 这是关于 $p$ 的线性微分方程，而右侧关于 $lambda$ 是 $C^(k-1)$，关于 $y$ 是 $C^infinity$ 的，不断利用定理即可 ] #remark[][ 上面的定理换成关于 $y, lambda$ 解析也对，既然 Picard 序列中每一项的解析，而解析函数的一致收敛极限也是解析的 ] #corollary[][ 在上面的定理中将条件换成 $f$ 对所有参数都 $C^k$，则 $phi$ 就是 $C^k$ 的 ] #proof[ 此时不妨将所有系数吸收进参数，只需研究方程： $ cases( der(y, x) = f(x, y, x_0, y_0, lambda), y(0) = 0 ) $ 前面已经证明关于 $x_0, y_0, lambda$ 都是 $C^k$ 的，只需考虑关于 $x$ 的，然而： $ partialDer(phi, x) = f(x, phi, x_0, y_0, lambda) $ 前面已经证明 $phi$ 是 $C^1$ 的，上式表明 $phi$ 将是 $C^2$ 的，继而归纳可得 $phi$ 是 $C^k$ 的 ] #corollary[解对初值和参数的光滑依赖性最终版本][ 在上面的定理中将条件换成 $f$ 对所有参数都 $C^(k-1)$，且对 $y, lambda$ 是 $C^k$ 的，则 $phi$ 就是 $C^k$ 的 ] #proof[ 已经证明了 $k = 1$ 时情景，同时 $phi in C^(k-1)$ 也已经成立，同样可以直接看到关于 $x$ 是 $C^k$ 的，并且注意到： $ partialDer(phi, x) = f(x, phi, lambda)\ partialDer(partialDer(phi, x), x_0) = partialDer(f, y) partialDer(phi, x_0) $ 令 $Y = partialDer(phi, x_0)$ ，将有微分方程： $ Y' = partialDer(f, y) Y $ 上式右端关于所有参数都是 $C^(k-1)$ 的，因此 $Y$ 关于 $x_0$ 也是 $C^(k-1)$，这就证明了对 $x_0$ 的可微性\ ] #example[][ $ cases( y' = y + mu (x^2 + y^2), y(0) = 1 ) $ 试求 $partialDer(phi, mu)\|_(mu = 0)$ 事实上，容易看出 $f$ 是解析函数，继而它的解都解析，我们直接求偏导并交换顺序： $ partialDer(partialDer(phi, x), mu) = partialDer(phi, mu) + (x^2 + phi^2) + 2 mu phi partialDer(phi, mu) $ 设 $u = partialDer(phi, mu)$，它满足微分方程： $ u' = u + (x^2 + phi^2) + 2 phi u mu = (1 + 2 phi mu) u+ x^2 + phi^2 $ 这是关于 $u$ 的一阶线性微分方程\ 同时，注意到 $mu = 0$ 时方程是好解的，因此： $ phi(x, 0) = e^x $ 代入得： $ u' = u + x^2 + e^(2 x) $ 解出 $u(x, 0) = e^(2 x) - x - 1$ ] == 局部变换本章的内容是从理论上研究微分方程。 #definition[自治系统][ 自治系统是指形如： $ cases( der(x, t) = f(x), x(t_0) = x_0 ) $ ] 局部变换是希望将一个自治系统的解在局部变换为另一个常微分方程的解。 #definition[][ - 称 $x_0$ 是常点，如果 $f(x_0) != 0$ - 称 $x_0$ 是奇点/平衡点，如果 $f(x_0) = 0$ ] #theorem[][ 设 $U$ 是微分同胚，$x = U(y)$，将微分方程： $ der(x, t) = f(x) $ 变成 $ der(y, t) = g(y) $ 则： $ g(y) = Inv((U')) f(x) $<local-transform> 这里 $U$ 指微分同胚的雅可比矩阵 ] #proof[ $ der(U y, t) = f(x)\ U' der(y, t) = f(x)\ U' g(y) = f(x) $ ] #definition[][ 设两个微分方程： $ der(x, t) = f(x)\ der(y, t) = g(y) $ 之间，存在 $C^k$ 的微分同胚使得满足@local-transform ，则称两个方程 $C^k$ 等价。\ 类似的，若在 $x_0$ 处的某个邻域存在 $C^k$ 的微分同胚使得满足@local-transform ，则称两个方程在 $x_0$ 处局部 $C^k$ 等价。 ] #theorem[常点的 $C^k$ 分类/拉直定理][ 设微分方程 $der(x, t) = f(x)$ 满足 $f(0) != 0$，则方程在 $0$ 处局部 $C^k$ 等价于： $ der(y, t) = Y(y), Y(y) = vec(1, 0, dots.v, 0) $ ] #proof[ #let vc = $vec(1, 0, dots.v, 0)$ 设 $f_i (x)$ 是分量，不妨设 $f_1 (0) != 0$\ 显然第二个方程的解就是： $ psi = y_0 + t vc $ 设 $phi(t, x_0)$ 是原方程的解，令: $ U(0) = 0\ U(y) = phi(y_1, 0, y_2, dots, y_(n)) $ 为了验证微分同胚，计算： $ abs(der(U, y)) = abs(partialDer(phi, t) der(y_1, y) + partialDer(phi, x_0) der((0, y_2, ..., y_n)^T, y))\ = abs(f(phi) der(y_1, y) + partialDer(phi, x_0) der((0, y_2, ..., y_n)^T, y)) $ 此外，将有： $ U(psi(t, y)) = U(t+y_1, y_2, ..., y_n) = phi(t+y_1, 0, y_2, ..., y_n) $ 求导再令 $t = 0$ 可得： $ U'(y) Y(y) = f(phi(y_1, 0, y_2, ... y_n)) = f(U(y)) $ 证毕 ] 说明常点的局部等价分类是非常简单的，接下来我们考虑奇点处的局部等价分类 #theorem[][ 线性微分方程 $x' = A x, x' = B x$ 在 $0$ 处局部 $C^k$ 等价当且仅当 $A, B$ 相似 ] #proof[ 任取微分同胚 $U$ 在 $0$ 处泰勒展开计算即可 ] 对于一般的方程，我们当然希望通过泰勒展开将其化为线性方程。然而方程 $x' = A x + o(x)$ 当然不总是与 $x' = A x$ 等价，至少我们需要 $A$ 非退化。在什么条件下可以将其化为线性方程是上个世纪常微分方程研究的重要课题之一 #theorem[][ 设 $f(x) = A x + o(x)$ 是 $C^infinity$ 的，而 $A$ 的特征根满足非共振条件： $ sum_(i = 1)^n m_i lambda_i != 0, (m_i) in ZZ^n - {0} $ 则方程 $x' = f(x)$ 在 $0$ 处局部 $C^infinity$ 等价于 $x' = A x$ ] = 线性微分方程 == 一般线性微分方程方便起见，这里约定对向量/矩阵的求导/积分都是逐元素进行的 #definition[][ 设 $x in RR^n, t in RR$\ 称微分方程： $ der(x, t) = A(t) x + B(t) $<linear-equation> 为一般线性微分方程，如果 $A, B$ 都关于 $t$ 连续 ] #proposition[][ - 一般线性微分方程在任何点处的解都存在唯一 - 每个解总是在大范围存在 ] #proof[ - 就是 @peano - 就是 @linear_all_range ] 一般而言，当 $A(t)$ 不是常数时，方程是无法写出解的。我们的目标是研究解空间的性质。 == 齐次线性微分方程 #definition[][ 在 @linear-equation 中，若 $B(t) = 0$，则称之为齐次微分方程 ] #proposition[][ - 齐次线性方程的解构成线性空间 - 齐次线性方程的解要么恒为零，要么恒不为零 - 齐次线性方程的若干解线性相关当且仅当在存在某点，它们在该点线性相关 ]<homogeneous-linear> #proof[ - 简单验证即可 - 注意到 $x = 0$ 是平凡解，结合解的唯一性立得 - 注意到若干解的线性组合还是解，利用上一条性质立得 ] #theorem[][ $n$ 维齐次线性微分方程的解空间恰为 $n$ 维线性空间。换言之，若可以找到 $n$ 个线性无关的解，则它们生成的线性空间恰为解空间，这称为方程的通解 ] #proof[ 令 $e_i$ 是 $RR^n$ 中一组标准基，令 $x_i (t)$ 是方程： $ cases( der(x, t) = A(t) x, x(t_0) = e_i ) $ 的一个解，断言它们就是原方程解的基 - 首先，它们线性无关，既然它们在 $t_0$ 处线性无关，利用 @homogeneous-linear 即可 - 其次，任取原方程的初值为 $x(t_0) = x_0$，令： $ x(t) = sum_i x_0^i x_i (t) $ 容易看出 $x(t)$ 也是符合该初值的解，由唯一性这就是唯一一个解证毕 ] #definition[][ 若矩阵 $X$ 是齐次线性微分方程的解，也即： $ der(X, t) = A(t) X $ 则称之为矩阵解。显然矩阵是解当且仅当每一列都是原方程的一个解，继而该矩阵的列秩至多为 $n$，恰为 $n$ 时称之为基本解矩阵或者基解阵 ] #proposition[][ 矩阵解 $X$ 是基础解矩阵当且仅当在某个点上有 $det(X) != 0$ ] #proof[ 就是 @homogeneous-linear ] #theorem()[][ 设 $Phi(t)$ 是基解阵，则 $X(t)$ 是基解阵当且仅当存在可逆矩阵 $C$ 使得： $ Phi(t) C = X(t) $ ] #proof[ 注意到求导是线性的，因此： $ der(Phi(t) C, t) = der(Phi(t), t) C = A(t) Phi(t) C $ 因此 $Phi(t) C$ 确实是解，计算行列式可得它是基础解矩阵另一方面，设 $X(t)$ 是基础解矩阵，任取 $t_0$ 并设有： $ Phi(t_0) C = X(t_0) $ 注意到 $ Phi(t_0), X(t_0)$ 都是可逆矩阵，$C$ 当然存在\ 此时 $X(t) - Phi(t) C$ 是原方程有零点的解，就是零 ] #proposition()[Liouville][ 设 $x_i$ 是 $n$ 个解，设 $det(x_1, x_2, ..., x_n) := W(t)$，则： $ W'(t) = tr(A(t)) W(t) $ ] #proof[ 容易发现： $ W'(t) = sum_i det(x_1, x_2, ..., x'_i, ..., x_n) = sum_i det(x_1, x_2, ..., A x_i, ..., x_n) $ #lemmaLinear[][ $ sum_i det(x_1, x_2, ..., A x_i, ..., x_n) = tr(A) det(x_1, x_2, ..., x_n) $ ] ] == 非齐次线性微分方程 #proposition[][ 设线性微分方程： $ x' = A(t) x + B(t) $ 则： - 任意两个解的差是对应齐次线性微分方程 $x' = A(t) x$ 的解 - 设 $gamma(t)$ 是一个特解（任意一个解），$X$ 是对应齐次微分方程的解空间，则原方程的所有解为 $gamma(t) + X$ ] #proposition[常数变易法][ 设 $Phi(t)$ 是 $x' = A(t) x$ 的基本解矩阵，则 $x' = A(t) x + f(t)$ 的解可以通过常数变易： $ x = Phi(t) C(t)\ Phi'(t) C(t) + Phi(t) C'(t) = A(t) Phi(t) C(t) + f(t)\ Phi(t) C'(t) = f(t)\ C'(t) = Inv(Phi(t)) f(t)\ C(t) = integral_(t_0)^(t) Inv(Phi(s)) f(s) dif s $ 这给出了可行的 $C$ ]<constant-variation> 上述命题表明，解一般线性微分方程的困难根本上来源于求齐次线性微分方程基础解矩阵的困难 == 常系数线性微分方程为了叙述方便，我们先给出矩阵幂级数，指数等定义 #definition[][ - 本节中定义矩阵的模为 $sum_(i, j) abs(a_(i j))$ 或 $max_(abs(x) = 1) abs(A x)$，它们都满足 $abs(A B) <= abs(A) abs(B)$ - 形式定义： $ e^A = sum_(n = 0)^infinity A^n/n! $ 注意到矩阵绝对收敛只需要每个分量绝对收敛，同时 $ sum_(n = 0)^infinity abs(A^n/n!) <= sum_(n = 0)^infinity abs(A)^n/n! = e^(abs(A)) < +infinity $ 满足性质： - 若 $A B = B A$ 则 $e^(A + B) = e^A e^B$ - $det(e^A) = e^(tr(A)) > 0$（证明需要若当标准型） ] #theorem[][ 给定常系数微分方程 $y' = A y + f(x)$ ，则 $e^(A x)$ 是方程的基础解矩阵 ] 以上结论非常漂亮，结合之前的理论我们可以求解出方程的通解。唯一的问题是按照定义求出 $e^(A x)$ 并不容易。 #example[][ 由约当标准型，可设： $ A = P D P^(-1)\ D = sum_(d=1)^(d_max) sum_(lambda in Lambda) sum_s lambda I + J_(d s) $ 其中 $d$ 是约当块维度，$s$ 代表不同的块。这些块都在不同的位置上，乘积为零，因此： $ D^n = (sum_(d) sum_(lambda in Lambda) sum_s lambda I + J_(d s))^n\ = sum_(d) sum_(lambda in Lambda) sum_s (lambda I + J_(d s))^n\ = sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^n C_n^i lambda^(n-i) J_(d s)^i\ = sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(min {d, n}) C_n^i lambda^(n-i) J_(d s)^i\ $ 注意到 $n > d$ 时上式的求和项数已与 $n$ 无关，$n < d$ 仅有有限个，因此可以求出 $D^n$ 的通式，利用： $ e^(A x) = sum_(k=0)^infinity A^k/k! x^k = P (sum_(k=0)^infinity D^k/k! x^k) Inv(P)\ = P (sum_(k=0)^infinity 1/k! (sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(min {d, k}) C_k^i lambda^(k-i) J_(d s)^i x^k)) Inv(P)\ = P (sum_(k=0)^infinity 1/k! (sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(min {d_max, k}) C_k^i lambda^(k-i) J_(d s)^i x^k)) Inv(P)\ = P (sum_(k=0)^d_max 1/k! (sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(n) C_k^i lambda^(k-i) J_(d s)^i x^k) + sum_(k=d_max + 1)^infinity 1/k! (sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(d_max) C_k^i lambda^(k-i) J_(d s)^i x^k)) Inv(P)\ = P (sum_(k=0)^d_max 1/k! (sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(k) C_k^i lambda^(k-i) J_(d s)^i x^k) + sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(d_max) J_(d s)^i sum_(k=d_max + 1)^infinity 1/k! C_k^i lambda^(k-i) x^k ) Inv(P)\ $ 上式第二项可以求得，第一项仅有有限项当然，这里的计算是极其麻烦的。有些时候，我们可以用技巧大大简化，例如： - 若 $A$ 可对角化，则 $d_(max) = 0$ 上式简化为： $ P (sum_(lambda in Lambda) sum_s sum_(k=0)^infinity 1/k! lambda^(k) x^k J_(0 s) ) Inv(P) = P e^(D x) Inv(P) $ - 若可求得 $A$ 的零化多项式，可以尝试降次 - 若 $A$ 仅有一个特征值 $lambda$ 则有： $ e^(A x) = e^(lambda x) e^((A - lambda) x) $ 后者第二项是幂零矩阵，可以直接求出 ] #example[][ 大部分时候如果只为了求方程的解，没必要完整算出 $e^(A x)$ ，既然若设： $ e^(A x) = P e^(D x) Inv(P) $ 则 $e^(A x) P = P e^(D x)$ 也是基础解矩阵，它会更加好求，既然 $D$ 是对角/约当矩阵，$e^(D x)$ 是容易求的，而 $P$ 恰由特征向量/循环子空间的基构成。 ] #example[][ 有些形式较好的方程并不用使用上面的一般方法求解，例如： $ cases( x' = y + z, y' = x + y, z' = x + z ) $ 三式相加，立得 $(x + y + z)' = x' + y' + z' = 2 (x + y + z), x+ y + z = C_1 e^(2t)$，代回得 $x' = e^(2 t) - x$ 解出即可 ] == 高阶线性微分方程 #theorem[][ 给定高阶线性微分方程： $ x^((n)) + a_(n-1)(t) x^((n-1)) + dots + a_1(t) x' + a_0(t) x = f(t) $ 可做换元，令： $ x_0 = x\ x_1 = x'\ ...\ x_n = x^((n)) $ 方程变为多元线性微分方程： $ vec(x_0, x_1, dots.v, x_(n-1), x_n)' = mat(0, 1, ...,0, 0; 0, 0, ...,0, 0; dots.v, dots.v, ..., dots.v, dots.v; 0, 0, ..., 0, 1; -a_0 (t), -a_1 (t), ..., -a_(n-1) (t), 1) vec(x_0, x_1, dots.v, x_(n-1), x_n) + vec(0, 0, dots.v, 0, f(t)) $ 因此，可以利用线性方程组的理论知道： - 方程的解是齐次线性方程的解空间 $n$ 维线性空间加上一个特解 - 给定初值 $x(0), x'(0), ..., x^((n))(0)$ 则方程的解唯一存在，且解是大范围的 ] #lemma[Wronskian][ 假设 $x_1 (t), x_2 (t), ..., x_n (t)$ 都 $n-1$ 阶可导，则称： $ Det(x_1, x_2, ..., x_n; x'_1, x'_2, ..., x'_n; dots.v, dots.v, ..., dots.v; x_1^((n-1)), x_2^((n-1)), ..., x_n^((n-1))) = W(t) $ 为朗斯基行列式。若这些函数线性相关，则朗斯基行列式恒为零。反之结论一般是不成立的，但若这些函数是齐次线性微分方程的解，则反命题也成立。 ] #proof[ 设 $sum_i c_i x_i (t) = 0$，对于任意 $t$ 反复求导可得： $ mat(x_1, x_2, ..., x_n; x'_1, x'_2, ..., x'_n; dots.v, dots.v, ..., dots.v; x_1^((n-1)), x_2^((n-1)), ..., x_n^((n-1)))vec(c_1, c_2, dots.v, c_n) = 0 $ 线性相关性相当于上面的齐次线性方程有非零解，因此行列式为零，证毕。至于反命题，一个简单的反例是 $x^2, x abs(x)$，而都是同一个齐次方程的解的情形就是 @homogeneous-linear ] #example[][ 假设我们已求出齐次方程的解，根据 @constant-variation 当然可以通过常数变易求出线性方程的解。实际这么做时，我们要遇到对： $ sum_i c_i (t) x_i (t) $ 求 $n$ 阶导的问题，由于我们只需要一个解即可，不妨添加条件： $ sum_i c'_i (t) x_i (t) = 0\ sum_i c'_i (t) x'_i (t) = 0\ ...\ sum_i c'_i (t) x_i^((n-1)) (t) = 0 $ 这样求导的形式就很简单： $ (sum_i c_i (t) x_i (t)) ' = sum_i c_i (t) x'_i (t) \ (sum_i c_i (t) x_i (t)) '' = sum_i c_i (t) x''_i (t) \ ...\ (sum_i c_i (t) x_i (t)) ^((n-1)) = sum_i c_i (t) x^((n-1))_i (t)\ (sum_i c_i (t) x_i (t)) ^((n)) = sum_i c_i (t) x^((n))_i (t) + sum_i c'_i (t) x^((n-1))_i (t) $ 最终方程变成： $ sum_i a_i (sum_i c_i (t) x_i (t)) ^((i)) = f(t)\ a_n sum_j c'_i (t) x^((n-1))_i (t) + sum_i a_i (sum_j c_j (t) x^((i))_j (t)) = f(t)\ a_n sum_j c'_i (t) x^((n-1))_i (t) = f(t) $ 这是因为中间凑出了齐次方程的解，因此可以直接消去。加上之前的假设方程变成关于 $c'_i$ 的线性方程组，求解即可。 ] #lemma[][ 对于二阶齐次线性微分方程： $ x'' + a(t) x' + b(t) x = 0 $ 设有两无关解 $x_1, x_2$ 不难验证若令： $ W = Det(x_1, x_2; x'_1, x'_2) = x_1 x'_2 - x'_1 x_2 $ 则有： $ W' &= x'_1 x'_2 + x_1 x''_2 - x''_1 x_2 - x'_1 x'_2 = x_1 x''_2 - x''_1 x_2 \ &= - x_1 (a(t) x'_2 + x_2) + x_2 (a(t) x'_1 + x_1) \ &= - a(t) (x_1 x'_2 - x'_1 x_2) \ &= - a(t) W $ 可以直接解出 $W$，此时假设 $x_1$ 已知，方程变成了关于 $x_2$ 的一阶线性微分方程，是可解的。这表明对于二阶齐次线性微分方程，若已知一个解，另一个解可以通过一阶线性微分方程求出。 ] #theorem[][ 对于二阶常系数齐次线性微分方程： $ x^((n)) + a_(n-1)(t) x^((n-1)) + dots + a_1(t) x' + a_0(t) x = 0 $ 设： $ p(lambda) = lambda^n + a_(n-1) lambda^(n-1) + dots + a_1 lambda + a_0 $ 有 $s$ 个不同根 $lambda_i$ ，重数分别为 $m_1, m_2, ..., m_s$，则： $ x^k e^(lambda_i x) forall k <= m_i, forall i = 1, 2, ..., s $ 恰构成一组基础解 ] #proof[ 设 $L(x) = x^((n)) + a_(n-1)(t) x^((n-1)) + dots + a_1(t) x' + a_0(t) x$，则有： $ L(e^(lambda x)) = p(lambda) e^(lambda x) $ 因此当然 $p(lambda) = 0$ 时 $e^(lambda x)$ 是解。进一步，两边对 $lambda$ 求导得： $ L(x e^(lambda x)) = (lambda p(lambda) + p'(lambda)) e^(lambda x) $ 若 $lambda$ 是至少二重根，上式也是零，反复进行即得结论。它们的线性无关性是显然的 ] = 幂级数解法 == 一般幂级数本章中 $y$ 允许多元函数 #lemma[][ 设微分方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 其中 $f$ 在 $x_0$ 附近解析，则它的解存在唯一，且是解析函数。 ] #proof[ 前面 Picard 序列的证明中给出了这个推论 ] 理论上来说，幂级数展开并比对系数可以将一般的微分方程化为代数方程。然而一般的情形仍然难以计算，最常见的情形是对线性方程做展开。 #example[][ - $der(y, x) = y - x$，令 $y = sum_i a_i x^i$，计算得： $ sum_(i >= 1) i a_i x^(i-1) = sum_i a_i x^i - x $ 有： $ a_1 = a_0\ 2 a_2 = a_1 - 1\ (i+1) a_(i+1) = a_i\ $ 可以递推解得 $a_i$ - $y'' - 2 x y' + 4 y = 0$，令 $y = sum_i a_i x^i$，计算得： $ sum_i (i+1)(i+2)a_(i+2)x^i - 2 sum_i i a_(i) x^i - 4 sum_i a_i x^i = 0 $ 得到一般的递推关系： $ (i+1)(i+2)a_(i+2) = 2 i a_i + 4 a_i\ (i+1) a_(i+2) = 2 a_i $ - $y'' = x y$，计算得： $ sum_i (i+1)(i+2)a_(i+2)x^i = sum_i a_(i-1) x^i $ 有： $ a_2 = 0\ (i+1)(i+2)a_(i+2) = a_(i-1) $ 可以解得： $ a_(3 k + 2) = 0\ a_(3 k) = ((3k - 3)!!!)/((3k) !) a_0 $ ] #remark[][ 对于形如： $ u(x) der(y, x) = v(x, y) $ 的微分方程，如果 $u(x) > 0$，将其除掉即可得到解的解析性。但若 $u(x)$ 有零点情形未必。例如： $ cases( x^2 der(y, x) = y - x, y(0) = 0 ) $ 若其有解析解，比对系数发现一定有 $a_n = n!$，但是这个幂级数不收敛，因此是荒谬的。下节的目标便是处理这种方程。 ] == 广义幂级数 #definition[广义幂级数][ 称： $ sum_(n=0)^(+infinity) a_n x^(n + alpha), alpha in RR $ 为广义幂级数。 ] #theorem[][ 设二阶微分方程： $ y'' + p(x) y' + q(x) y = 0 $ 其中 $p, q$ 可能以 $0$ 为奇点，但 $x p, x^2 q$ 都在 $0$ 处解析且不全为零，则它在 $0$ 附近有广义幂级数解 ] #proof[ 方程等价于： $ x^2 y'' + x (sum_i a_i x^i) y' + (sum_i b_i x^i) y = 0 $ 设 $y = sum_(n=0)^(+infinity) c_n x^(n + alpha)$，代入得： $ x^(alpha)(sum_(n=2)^(+infinity) c_n (n+alpha)(n+alpha-1) x^(n) \ + (sum_(n=1)^(+infinity) c_n (n+alpha) x^(n))(sum_(i=0)^infinity a_i x^i) \ + (sum_(n=0)^(+infinity) c_n x^(n))(sum_(i=0)^infinity a_i x^i)) = 0 $ ] #example[贝塞尔方程][ 方程： $ y'' + 1/x y' + (x^2 - n^2) / x^2 y = 0 $ 称为贝塞尔方程，由上面的定理它在 $0$ 附近有广义幂级数解，并且计算可得 $n$ 是正整数时解是整函数。 ] = 微分方程定性理论：边值问题 == Sturm 比较定理本节我们的研究的是形如： $ y'' + p(x) y' + q(x) y = 0 $<obj-def> 其中 $p, q$ 是某个区间 $J$ 上的连续函数 #lemma[][ @obj-def 的非零解都是简单零点（导数非零），进一步都是孤立零点 ] #proof[ 如若不然，设 $x_0$ 处函数值和导数值均为零，由存在唯一性这将导致解恒为零，矛盾！ ] #lemma[][ 设 $f, g$ 是 @obj-def 的两个非零解，且都有零点： - $f, g$ 线性相关当且仅当有相同的零点集 - $f, g$ 线性无关当且仅当零点相间，也即每两个相邻零点构成的开区间内有对方的零点 ] #proof[ - - 若 $f, g$ 相关则 $lambda f + mu g = 0$，不难发现 $lambda, mu$ 非零，因此零点集相同 - 若零点集相同，考虑 Wronskian 行列式： $ W(x) = Det(f, g; f', g') $ 由条件可知存在一个共同零点，则在该点处 $W(x) = 0$，之前的结论表明 $f, g$ 线性相关 - 设 $x_1, x_2$ 是 $f$ 的相邻零点，不妨假设在 $(x_1, x_2)$ 上有 $f > 0$ - 设 $f, g$ 无关，则： $ W(x) = Det(f, g; f', g') $ 定号，进而 $W(x_1) W(x_2) > 0$，而： $ W(x_1) = -g(x_1) f'(x_1)\ W(x_2) = -g(x_2) f'(x_2) $ 之前证明了 $f'(x_1), f'(x_2) != 0$，不难发现一定有 $f'(x_1) > 0, f'(x_1) < 0$，上式表明 $g(x_1), g(x_2)$ 异号，当然就有介质定理。同时其间只能有一个零点，否则若有两个可以反过来找到 $f$ 的零点，与 $x_1, x_2$ 相邻矛盾！ - 反之，若零点相间结论由前一条结论知零点集不同，当然无关 ] #remark[][ 上面的引理中需要留意零点的存在性，例如 $f, g$ 无关而 $f$ 仅有一个零点，此时 $g$ 的零点个数可能是 $0, 1, 2$，与定理都不矛盾 ] #theorem[比较定理][ 设有两个微分方程： $ y'' + p(x) y' + q(x) y = 0 $<eq-1> $ y'' + p(x) y' + r(x) y = 0 $<eq-2> 且满足： $ r(x) >= q(x) $ 设 $f, g$ 分别是@eq-1 和@eq-2 的两个非零解，$x_1, x_2$ 是 $f$ 的两个相邻零点，则 $g$ 在 $[x_1, x_2]$ 上有零点。 ]<compare-two> #proof[ 不妨假设 $f$ 在 $(x_1, x_2)$ 上恒正，有： $ f'(x_1) > 0, f'(x_2) < 0 $ 假设 $g(x)$ 在 $[x_1, x_2]$ 上无零点，不妨设其恒正。令： $ W(x) = Det(f, g;f', g') = f g' - f' g\ W'(x) = Det(f, g;f'', g'') = Det(f, g;-p(x)f' - q(x) f, -p(x) g' - r(x) g)\ =Det(f, g;-p(x)f', -p(x) g' - (r(x) - p(x)) g)\ = - p(x) W(x) - f g (r(x) - p(x)) $ 注意到一定有： $ f g(r(x) - p(x)) >= 0 $ 由一阶方程的比较定理，有： $ B e^(-p(x))<= W(x) <= A e^(-p(x)) $ 其中 $A e^(-p(x))$ 是： $ cases( W'(x) = -p(x) W(x), W(x_1) = - g(x_1) f'(x_1) ) $ 的解，而 $B e^(-p(x))$ 是： $ cases( W'(x) = -p(x) W(x), W(x_1) = - g(x_2) f'(x_2) ) $ 的解，然而不难发现 $- g(x_1) f'(x_1) < 0, - g(x_2) f'(x_2) > 0$ 导致 $A < 0, B > 0$，这是荒谬的！ ] #remark[][ 上面的定理实际上是说 $y$ 前面的系数表示解振荡的频率，系数越大振荡越快。因此有如下对于振动的研究： ] #definition[][ 设@obj-def 的有至少两个零点的非零解为振动解，有无穷多个零点的非零解为无穷解 ] #example[][ - 考虑方程： $ y'' + p(x) y' + r(x) = 0 $<eq-3> 其中 $r(x) <= 0$，注意到它可以对： $ y'' + p(x) y' = 0 $<eq-4> 利用比较定理。假如@eq-3 有振动解，则由 @compare-two 可知@eq-4 的任意一个解都有零点，然而@eq-4 有解 $y = 1$ 无零点，矛盾！因此@eq-3 不可能有振动解 - 考虑方程： $ y'' + q(x) y = 0 $ 其中 $q(x) >= m > 0$，则它任意非零解无限振动，且相邻零点间距离不超过 $pi / sqrt(m)$\ 首先证明对于任何 $a$, $[a, a + pi / sqrt(m)]$ 的区间都有方程的解即可。考虑方程： $ y'' + m y = 0 $ 后面的方程有解： $ y = sin (sqrt(m) (x - a)) $ 以 $a, a + pi / sqrt(m)$ 为零点，利用 @compare-two 立得结论。当然这样的区间有无穷多，因此方程的解有无穷多个零点。\ 注意这个结论不能加强到 $q(x) > 0$，例如： $ y'' + 1/(4 x^2) y = 0, x in [1, +infinity] $ 这是欧拉方程，可以解得： $ y = sqrt(x) (c_1 + c_2 ln x) $ 当然至多只有一个零点 - 考虑微分方程： $ y'' + q(x) y = 0 $ 且存在非负整数 $n$ 使得： $ n^2 < q(x) < (n+1)^2 $ 则方程的任意非零解不是 $2 pi$ 周期的。\ 如若不然，设 $f$ 是非零的 $2 pi$ 周期解，代入方程不难发现 $q(x) f$ 也是以 $2 pi$ 周期的，而 $f$ 的零点孤立，由 $q$ 的连续性知它应该以 $2 pi$ 为周期，因此有最大最小值。设： $ m^2 <= q(x) <= M^2 $ 仿照上面的例子，利用 @compare-two 可以导出 $f$ 的两个相邻零点间的距离在 $pi/M, pi/m$ 之间\ 设 $f$ 在一个周期上有 $2n$ 个零点，零点分别为： $ t_0 < t_1 < ... < t_(2 n) = t_0 + 2 pi $ 对距离求和，可得： $ 2n/M pi <= 2 pi <= 2n/m pi\ m <= n <= M $ 然而由条件 $m, M$ 夹在两个连续整数之间，与 $n$ 是整数矛盾！ - 考虑微分方程： $ y'' + q(y) = f(x) $ 且存在非负整数 $n$ 使得： $ n^2 < q'(y) < (n+1)^2 $ 则方程至多有一个 $2 pi$ 周期解，否则设 $y_1, y_2$ 是两个 $2 pi$ 周期解，容易得到： $ (y_1 - y_2)' + q(y_1) - q(y_2) = 0 $ 令 $p(x) = (q(y_1) - q(y_2))/(y_1 - y_2)$，则 $y_1 - y_2$ 成为微分方程： $ y'' + p(x) y = 0 $ 的 $2 pi$ 周期解，然而上面的结论表明这样的方程没有 $2 pi$ 周期解，矛盾！ ] == 边值问题 #definition[][ 设有微分方程： $ cases( (p(x) y')' + (lambda r(x) + q(x)) y = 0, k y(a) + l y'(a) = M y(b) + N y'(b) = 0 ) $ 其中 $p, q, r in C[a, b], p, r > 0, (K, L) != 0, (M, N) != 0$ 该微分方程非平凡解的存在性称为二阶微分方程的边值问题。若对 $lambda = lambda_0$ 方程有非零解 $phi$，则称 $lambda_0$ 为一个特征值，$phi$ 为对应的特征函数。 ]<boundary-value> #remark[][ 这个叫法是因为若设： $ A y = ((p y')' + q y)/(-r) $ 则它是线性算子，而边值问题实际上是求解 $A y = lambda y$ 的问题 ] #example[][ - 考虑方程： $ cases( y'' + lambda y = 0, y'(0) = y'(l) = 0 ) $ - 当 $lambda = -a^2 < 0$ 时，方程的通解为： $ y = linearCombinationC(e^(a x), e^(-a x)) $ 计算发现若要符合边值，将有： $ cases( C_1 = C_2, C_1 = -C_2 ) $ 导出零解 - 当 $lambda = 0$ 时方程的解是线性函数，因此每个常函数都是特征函数 - 当 $lambda = a^2 > 0$，通解为： $ linearCombinationC(cos a x, sin a x) $ 计算得： $ cases( C_2 = 0, - C_1 sin a l + C_2 cos a l = 0 ) $ 当且仅当 $a$ 为 $(n pi)/l$ 时有非零解综上，特征值为 $(n^2 pi^2)/l^2, forall n in NN$ ]<sin-cos-example> @boundary-value 中的形式当然可以再简化，通过一些简单的线性变换，可设 $a = 0, b = 1$，进一步由 $p > 0$，设： $ t = 1/(c_0) integral_(0)^(x) 1/(p(s)) dif s \ tilde(y) (t) = y(x(t)) $ $x(t)$ 存在是因为 $t$ 关于 $x$ 单调上升，有反函数。不难计算得： $ p(x) der(y, x) = 1/(c_0) der(tilde(y), t)\ der(p(x) der(y, x) , x) = 1/(c_0^2 p(x)) tilde(y)'' $ 总之，方程化为了 $ cases( y'' + (lambda c_0^2 p(x(t))r(x(t)) + c_0^2 p(x(t)) q(x(t))) y = 0, K y(0) + L /(c_0 p(a)) y'(0) = M y(1) + N /(c_0 p(b)) y'(1) = 0) $ #lemma[][ @boundary-value 中的问题可以化归为以下标准形式： $ cases( y'' + (lambda r(x) + q(x)) y = 0, y(0) cos alpha - y'(0) sin alpha = y(1) cos beta - y'(1) sin beta = 0 ) $<standard-form-boundary-value> 其中 $r, q$ 连续，$r > 0, alpha, beta in [0, pi)$ ] #theorem[Sturm-Liouville][ @standard-form-boundary-value 给出的边值问题有可数多个特征值，且所有特征值形成链： $ lambda_0 < lambda_1 < ... < lambda_n < ... $ 其中 $lambda_n -> +infinity$\ 设 $phi_n (x)$ 是 $lambda_n$ 对应的特征函数，则 $phi_n$ 恰有 $n$ 个零点 ]<sturm-liouville-theorem> #proof[ 令 $y = phi(x, lambda)$ 是满足： $ y(0) = sin alpha, y'(0) = cos alpha $ 的解。@standard-form-boundary-value 中第一个边值条件已经满足，只需找到 $lambda$ 满足第二个方程。显然 $y$ 不是零解，$y, y'$ 不同时为零，因此可做极坐标变换： $ cases( phi = rho cos theta, phi' = rho sin theta ) $ 此时，第二个边值条件等价于 $theta(1) = beta + 2 k pi$\ 计算： $ der(theta, x) = der(arctan phi/phi', x) = (phi'^2 - phi phi'' )/(phi^2 + phi'^2) = (phi'^2 + (lambda r + q) phi^2 )/(phi^2 + phi'^2)\ = cos^2 theta + (lambda r + q) sin^2 theta $<eq-order1> 结合 $theta(0, lambda) = 0$，这是关于 $theta$ 的一阶微分方程，接下来的讨论都是关于这个方程的。 - 上式对 $lambda$ 求偏导得到的是关于 $lambda$ 的一阶线性微分方程，可以解得： $ partialDer(theta, lambda) = integral_(0)^(x) e^(integral_(t)^(x) E(s, lambda) dif x) r(t) sin^2 theta dif t $ （其中 $E(s, lambda) = (lambda r(s) + q(x) - 1) sin(2 theta)$）由条件知被积函数恒正，从而 $theta$ 关于 $lambda$ 严格递增 - 观察方程可以发现 $theta = 0$ 时 $theta' = 1$，且当 $lambda$ 充分小时，只要 $theta$ 不太小，就有 $theta' < 0$，我们猜测并证明以下结论： - $theta(x) > 0$\ 如若不然，假设 $x_0$ 是最小的零点，之前的论述表明 $theta'(x_0) = 1 > 0$，而 $theta(0) = alpha > 0$，利用介质定理将可构造出更小的零点，矛盾！ - $ lim_(lambda -> +infinity) theta(x) = +infinity, forall x_0 $ 设 $h = min r > 0$，将有： $ theta' >= (lambda h + q) sin^2 theta + cos^2 theta $ 取 $lambda$ 充分大，可设 $lambda h + q > n + 1$，将有： $ theta' >= 1 + n sin^2 theta $ 由比较定理，只需证明 $theta' = 1 + n sin^2 theta$ 的解充分大即可。事实上： $ x = integral_(alpha)^(theta) 1/(1 + n sin^2 t) dif t $ 观察不难发现 $n$ 充分大时 $theta$ 也应该充分大 - $ lim_(lambda -> -infinity) theta(x) = 0, forall x_0 $ 任取 $epsilon > 0$，当 $lambda$ 充分小时可以证明，$theta$ 落在线段 $(0, alpha) -> (1, epsilon)$ 时，一定有 $theta' < 0$，因此 $theta$ 被限制在该折线和 $y = 0$ 之间，当然意味着 $theta(1) < epsilon$，证毕以上论断表明，$theta(1, lambda) = beta + 2 k pi$ 对于每个 $k in NN$ 都恰有一解，且每个解都逐个增大。此外，特征函数为： $ phi(x) = rho(x) sin (theta_(lambda_k) (x)) $ 由介质定理，至少： $ theta_(lambda_k) (x) = j pi $ 对于 $j = 1, 2, ..., k$ 都有一个解 $x_k$，并且注意到 $forall x with theta(x) = i pi, i in ZZ$，都有： $ phi'(x) = cos^2 theta(x) + (lambda r(x) + q(x)) sin^2 theta(x) = 1 $ 换言之： $ theta_(lambda_k) (x) = j pi $ 只能有一个解（否则两个相邻解的导数值必然反号）\ 同时，对于 $j > k$， $theta_(lambda_k) (x) = j pi$ 将无解，否则由于 $theta(0), theta(1) < j pi$，至少产生两解，与上面的论述矛盾！ ] == 特征函数系的正交性 #lemma[][ 在 @sturm-liouville-theorem 中，每个特征值对应的特征函数彼此相关（特征空间只有一维） ] #proof[ 假设 $phi, psi$ 是 $lambda_n$ 的特征函数，将有： $ Det(phi(0), phi'(0);psi(0), psi'(0)) = 0 $ 由 Wronskian 行列式的结论，这意味着 $phi, psi$ 线性相关 ] #theorem[][ 在 @sturm-liouville-theorem 中，每个特征值对应的特征函数彼此正交，也即： $ integral_(0)^(1) r(x) phi_m phi_n dif x = 0, m != n $ ] #proof[ 有两个方程： $ cases( phi''_n + (lambda_n r + q)phi_n = 0, phi''_m + (lambda_m r + q)phi_m = 0 ) $ 两式分别乘以 $phi_m, phi_n$ 并相减，得到： $ phi''_n phi_m - phi''_m phi_n + (lambda_n - lambda_m) r phi_m phi_n = 0 $ 只需计算： $ &integral_(0)^(1) phi''_n phi_m - phi''_m phi_n dif x \ &= integral_(0)^(1) (phi'_n phi_m - phi_n phi'_m)' dif x\ &= (phi'_n phi_m - phi_n phi'_m)\|_0^1\ &= 0 $ 证毕 ] #remark[][ 有了正交性，对于任意的以上形式的二阶线性微分方程，我们都可以考虑在特征函数系上做傅里叶展开。事实上，@sin-cos-example 说明了通常 $sin, cos$ 产生的傅里叶展开是本章定理的一种特殊情况。 ] = 一阶偏微分方程 == 首次积分 #let yv = $bold(y)$ #let xv = $bold(x)$ #definition[][ 在微分方程： $ der(yv, x) = f(x, yv), f in C^1 $ 中，称 $H(xv, yv)$ 为首次积分，若 $H$ 不是常数且在方程的任意解曲线上取常值 ] #example[][ - 考虑方程： $ cases( der(x,t) = -y, der(y, t) = x ) $ 则 $x^2 + y^2$ 就是一个首次积分，既然设 $x, y$ 是一族解，将有： $ (x^2 + y^2)' = 2 x x' + 2 y y' = 0 $ - 一般的，方程： $ cases( der(x,t) = f(y), der(y, t) = g(x) ) $ 则 $integral f(y) - integral g(x)$ 就是一个首次积分，既然： $ (integral f(y) - integral g(x))' = f(y)y' - g(x)x' = 0 $ - 考虑方程： $ cases( der(x,t) = y - x(x^2 + y^2 - 1), der(y, t) = -x - y(x^2 + y^2 - 1) ) $ 注意到： $ (x^2 + y^2)' = 2 x' x + 2 y' y = -x^2 (x^2 + y^2 - 1) - y^2 (x^2 + y^2 - 1) \ = (x^2 + y^2) - (x^2 + y^2)^2 $ 可以解出 $x^2 + y^2$ 进而产生一个首次积分\ 同时： $ x y' - y x' = -x^2 - y^2\ (y/x)' = -1 - (y/x)^2 $ 也可以解出 $y/x$ 产生一个首次积分。事实上，可以发现方程有唯一的平衡点 $(0, 0)$ 和孤立的有界闭解曲线（称为极限环）。对于形如： $ cases( x' = f(x, y), y' = g(x, y) ) $ 其中 $f, g$ 是不超过 $n$ 次的多项式，这个系统称为 $n$ 次系统。$n$ 次系统极限环个数的上界及极限环的分布情况是希尔伯特第十六问题的重要部分，至今二次系统的情况都仍未解决，目前最多举出了四个极限环的例子，并且证明了每个二次系统的极限环个数都是有限的。 ] #lemma[][ $H(x, yv)$ 是 $yv' = f(x, yv)$ 的首次积分当且仅当： $ partialDer(H, x) + partialDer(H, yv) dot der(yv, x) = partialDer(H, x) + partialDer(H, yv) dot f(x, yv) = 0 $ ] #proof[ 由定义显然可得 ] 验证首次积分是非常简单的，然而找到一个首次积分是极其困难的。 #theorem[][ 首次积分可以将原方程降维。具体而言，假设首次积分形如： $ phi(yv) $ 由于 $phi(yv) !=0$，若还有 $der(phi(yv), yv) != 0$，便可在局部找到隐函数消去若干变量 ] #definition[][ 设 $phi_i$ 是 $n$ 个首次积分，如果： $ abs(partialDer((phi_1, phi_2, ..., phi_n), (yv_1, yv_2, ..., yv_n))) != 0 $ 则称 $phi_1, phi_2, ..., phi_n$ 为相互独立的首次积分 ] #theorem[][ $n$ 维自治系统至多 $n$ 个独立的首次积分，且在局部恰有 $n$ 个独立的首次积分。设 $phi_i$ 是 $n$ 个独立的首次积分，则任何首次积分在局部都形如 $h(phi_1, ..., phi_n), H in C^1$ ]<first-integral> #proof[ 第一部分我们不证明，可以参考教材对于第二个结论，设 $c = vec(c_1, dots.v, c_n)$，在任何一个点 $(x_0, c)$ 由存在唯一性可以找到解 $y = phi(x, c)$，显然有： $ partialDer(phi, c)\|_(x = x_0) = id $ 因此由隐函数定理，可以反解出： $ c = psi(x, y) $ 并有： $ der(y, c) = partialDer(phi, c)\ der(c, y) = Inv((partialDer(phi, c))) $ 这个 $psi$ 当然就是局部的首次积分，恰有 $n$ 个独立的分量（既然在 $x_0$ 处上式表明偏导是 $id$）也即 $n$ 个首次积分。第三部分是类似的，也不证明。 ] == 一阶线性齐次偏微分方程 #definition[一阶线性齐次偏微分方程][ 称： $ A(xv) der(u, xv) = f(xv) $ 是一阶线性齐次偏微分方程 ] #definition[特征方程][ 称： $ der(x_i, A_i (xv)) = der(x_j, A_j (xv)) $ 为齐次一阶线性偏微分方程的特征方程，它可以看成关于某个分量的 $n-1$ 阶常微分方程 ] #theorem[][ 设特征方程有 $n-1$ 个独立的首次积分 $phi_i$ ，则原偏微分方程的通解恰为： $ Phi(phi_1, phi_2, ..., phi_(n-1)) $ 其中 $Phi$ 是任何 $C^1$ 函数 ] #proof[ 设 $A_1 (xv) != 0$，特征方程实际上形如： $ der(x_i, x_1) = (A_i (x))/(A_1 (x)) $ 不难发现 $u$ 是偏微分方程的解当且仅当 $u$ 是常数或者是特征方程的一个首次积分，由 @first-integral 立得结论。 ] #example[][ - 考虑方程： $ (x + y) partialDer(u, x) - (x - y) partialDer(u, y) = 0 $ 其特征方程为： $ der(y, x) = (y - x)/(y + x) $ 可以解出一个首次积分： $ (x^2 + y^2)e^(2 arctan y/x) $ 因此原方程的通解为： $ phi((x^2 + y^2)e^(2 arctan y/x) ) $ - $ sqrt(x) partialDer(f, x) + sqrt(y) partialDer(f, y) + z partialDer(f, z) = 0 $ 且 $f(x, y, 1) = x y$，求 $f$\ 类似的可以解出： $ f = phi(sqrt(x) - sqrt(y), sqrt(y) - ln z)\ $ 代入可得： $ x y = f(x, y, 1) = phi(sqrt(x) - sqrt(y), sqrt(y) ) $ 做变量替换： $ cases( u = sqrt(x) - sqrt(y), v = sqrt(y) ) $ 可反解出 $phi$ ] == 一阶拟线性偏微分方程 #definition[一阶拟线性偏微分方程][ 称： $ A(xv, u) der(u, xv) = B(xv, u) $ 是一阶拟线性偏微分方程 ] #theorem[][ #let sumf = sumf.with(lower: $1$, upper: $n$) 对于一阶拟线性偏微分方程，考虑微分方程： $ sumf() A_i (xv, u) der(Phi, x_i) + B(xv, u) der(Phi, u) = 0 $<eq-t> 这是关于 $xv, u$ 的一阶线性齐次偏微分方程，设其解为： $ Phi(xv, u) $ 则 $Phi(xv, u) = 0$ 对应的 $u$ 就是原方程的解。反之，任意原方程的解 $u = phi(xv)$ 都有： $ Phi = phi(xv) - u $ 是@eq-t 的解 ] #theorem[][ 若称： $ der(x_i, A_i (xv)) = der(x_j, A_j (xv)) = der(u, B (xv, u)) $ 为拟线性方程的特征方程，且有 $n$ 个独立的首次积分，则原方程的通解恰为： $ Phi(phi_1, phi_2, ..., phi_n) = 0 $ ]
https://github.com/jgm/typst-hs	https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/compute/calc-12.typ	typst	Other	// Test the `quo` function. #test(calc.quo(1, 1), 1) #test(calc.quo(5, 3), 1) #test(calc.quo(5, -3), -1) #test(calc.quo(22.5, 10), 2) #test(calc.quo(9, 4.5), 2)
https://github.com/herbhuang/utdallas-thesis-template-typst	https://raw.githubusercontent.com/herbhuang/utdallas-thesis-template-typst/main/layout/proposal_template.typ	typst	MIT License	#import "/layout/titlepage.typ": * #import "/layout/transparency_ai_tools.typ": transparency_ai_tools as transparency_ai_tools_layout #import "/utils/print_page_break.typ": * // The project function defines how your document looks. // It takes your content and some metadata and formats it. // Go ahead and customize it to your liking! #let proposal( title: "", titleGerman: "", degree: "", program: "", supervisor: "", advisors: (), author: "", startDate: datetime, submissionDate: datetime, transparency_ai_tools: "", is_print: false, body, ) = { titlepage( title: title, titleGerman: titleGerman, degree: degree, program: program, supervisor: supervisor, advisors: advisors, author: author, startDate: startDate, submissionDate: submissionDate ) print_page_break(print: is_print) // Set the document's basic properties. set page( margin: (left: 30mm, right: 30mm, top: 40mm, bottom: 40mm), numbering: "1", number-align: center, ) // Save heading and body font families in variables. let body-font = "New Computer Modern" let sans-font = "New Computer Modern Sans" // Set body font family. set text( font: body-font, size: 12pt, lang: "en" ) show math.equation: set text(weight: 400) // --- Headings --- show heading: set block(below: 0.85em, above: 1.75em) show heading: set text(font: body-font) set heading(numbering: "1.1") // --- Paragraphs --- let firstParagraphIndent = 1.45em show heading: it => { it h(firstParagraphIndent) } set par(leading: 1em, justify: true, first-line-indent: 2em) // --- Citation Style --- set cite(style: "alphanumeric") // --- Figures --- show figure: set text(size: 0.85em) body pagebreak() bibliography("/thesis.bib") pagebreak() transparency_ai_tools_layout(transparency_ai_tools) }
https://github.com/fenjalien/metro	https://raw.githubusercontent.com/fenjalien/metro/main/tests/num/rounding/round-pad/test.typ	typst	Apache License 2.0	#import "/src/lib.typ": * #set page(width: auto, height: auto, margin: 1cm) #metro-setup(round-mode: "figures", round-precision: 4) #num(12.3) #num(12.3, round-pad: false)
https://github.com/ymgyt/techbook	https://raw.githubusercontent.com/ymgyt/techbook/master/programmings/js/typescript/specification/literal_type.md	markdown		# literal type * primitive型の特定の値だけを代入にする型を表現できる。 ```typescript // xにはなんでもassigneできる let x: number; x = 1; ``` ```typescript // xには1だけを代入できる let x: 1 x = 1; // compile error // x = 2; ``` ## literal typeが利用できるprimitive型 * bool * number * string ```typescript let status: 1 \| 2 \| 3 = 1; ```
https://github.com/bamboovir/typst-resume-template	https://raw.githubusercontent.com/bamboovir/typst-resume-template/main/README.md	markdown	MIT License	# Typst Resume Template A simple resume template for [typst.app](https://typst.app/). Aesthetic style inspired by the following project: - [Awesome-CV](https://github.com/posquit0/Awesome-CV) - [LaTeX Resume](https://github.com/billryan/resume) This is not a perfect clone, the main purpose of this project is to explore and experiment with Typst's typography features. ## [Sample](./resume.pdf) ![awesome-sample](./assets/image/awesome-sample.png) ![latex-sample](./assets/image/latex-sample.png) ## Declaration If you want to see a more realistic example rendered using this template, check [this](https://github.com/bamboovir/typst-resume-template/blob/main/huiming-sun-sde-resume.pdf). This is the resume I built in 2022, may be somewhat out of date and not actively maintained, and is not intended to be an accurate description of any of my current experiences but is intended solely to demonstrate the aesthetics of this template. You are free to take my .typ template and modify it to create your own resume. Please don't use my resume for anything else without my permission, though! ## Development Environment - Install [Typst](https://github.com/typst/typst) - Install [Just](https://github.com/casey/just) ## Build Resume ```bash just build ``` ## Interactive Development Resume ```bash just dev ``` ## Containerized Build ```bash just containerized-build ``` ## GitHub Action for resume build automation - [Resume Build CI Pipeline](https://github.com/bamboovir/typst-resume-template/actions/workflows/build-resume.yml) ## Credit [Typst](https://github.com/typst/typst) is a new markup-based typesetting system that is designed to be as powerful as LaTeX while being much easier to learn and use. [FontAwesome](https://fontawesome.com/) is the Internet's icon library and toolkit, used by millions of designers, developers, and content creators. [Roboto](https://github.com/google/roboto) is the default font on Android and ChromeOS, and the recommended font for Google’s visual language, Material Design. [Source Sans Pro](https://github.com/adobe-fonts/source-sans-pro) is a set of OpenType fonts that have been designed to work well in user interface (UI) environments.
https://github.com/jgm/typst-hs	https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/compiler/recursion-02.typ	typst	Other	// Test capturing with named function. #let f = 10 #let f() = f #test(type(f()), "function")
https://github.com/ParaN3xus/tex2typ	https://raw.githubusercontent.com/ParaN3xus/tex2typ/main/README.md	markdown	MIT License	# tex2typ A tool to rebuild [Typst](https://typst.app/) mathematical formulas from [KaTeX](https://katex.org/) syntax tree. ## Features - Convert LaTeX mathematical formulas to Typst mathematical formulas. - Extremely useful for building Typst formula datasets. ## Differences from [MiTeX](https://github.com/mitex-rs/mitex) - Focuses on ensuring a generally similar visual effect rather than an identical one, aiming to make the formulas look like they were written by a human. - The generated formulas do not rely on any special Typst environments or packages and can be compiled directly with the standard Typst. - We didn't provide any Typst package. ## TODO - [ ] Improve the handling of spaces in TeX formulas. - [ ] Refactor and optimize the code logic to reduce redundancy. - [x] Fix the issue with incorrect delimiter passing when reconstructing functions like `cases` and `vec`. ## Credits This project makes use of the following open-source projects: - [KaTeX](https://github.com/KaTeX/KaTeX): Fast math typesetting for the web. - [mitex](https://github.com/mitex-rs/mitex): LaTeX support for Typst, powered by Rust and WASM. - [im2markup](https://github.com/harvardnlp/im2markup/): Neural model for converting Image-to-Markup. Thanks to the developers and contributors of these projects for their hard work and dedication. ## LICENSE MIT
https://github.com/typst/packages	https://raw.githubusercontent.com/typst/packages/main/packages/preview/outrageous/0.1.0/examples/basic.typ	typst	Apache License 2.0	#import "../outrageous.typ" #set heading(numbering: "1.") #set outline(indent: auto) #page(height: auto, width: 15cm, margin: 1cm)[ #show outline.entry: outrageous.show-entry #outline() ] #page(height: auto, width: 15cm, margin: 1cm)[ #show outline.entry: outrageous.show-entry.with( // the typst preset retains the normal Typst appearance ..outrageous.presets.typst, // we only override a few things: // level-1 entries are italic, all others keep their font style font-style: ("italic", auto), // no fill for level-1 entries, a thin gray line for all deeper levels fill: (none, line(length: 100%, stroke: gray + .5pt)), ) #outline() ] = Introduction #lorem(400) #lorem(400) == What is this About? #lorem(400) #lorem(400) == Why am I Here? #lorem(400) #lorem(400) = The Backstory #lorem(400) #lorem(400) == How it all Started #lorem(400) #lorem(400) === Early Beginnings #lorem(400) #lorem(400) === First Settlements #lorem(400) #lorem(400) = The Consequences #lorem(400) #lorem(400) = Happy Ending #lorem(400) #lorem(400)
https://github.com/HezelTm/TypstDocuments	https://raw.githubusercontent.com/HezelTm/TypstDocuments/main/README.md	markdown	MIT License	# TypstDocuments Repository to store all Typst Documents
https://github.com/lee-flower/Sn-nCr	https://raw.githubusercontent.com/lee-flower/Sn-nCr/main/main.typ	typst	Apache License 2.0	#import "template.typ": * #show: project.with( title: "等差数列的r阶前n项和与组合数的联系", authors: ( ( name: "江林锐", organization: [梅州市梅县区高级中学], email: "<EMAIL>" ), ), abstract: "本文的目的在于：用数学归纳法证明了等差数列的r阶前n项和与组合数之间的联系。", keywords: ( "等差数列r阶前n项和", "组合数", "数学归纳法" ), ) = 定义 == 等差数列${a_n}$的r阶前n项和设数列${a_n}$为等差数列，其通项公式为$a_n = n$，现定义$S_n^(\(r\))$为数列${a_n}$的r 阶前 n 项和，其中 $r in NN$. 而且对于$forall r >= 1$，都有： $ display(S_n^(\(r\)) = sum_(i=1)^(n) S_i^(\(r-1\))) $ 另外，我们规定：$S_n^(\(0\)) = a_n$. == 排列、排列数及其计算公式关于排列及排列数的定义，详见@pailie . 现给出排列数的计算公式： $ A_n^m = n(n-1)(n-2) dots.h.c (n-m+1) = n!/(n-m)! $ 其中，$n, m in NN$且$m <= n$. == 组合、组合数及其计算公式关于组合及组合数的定义，详见@zuhe . 现给出组合数的计算公式： $ binom(n,m) = (A_n^m)/(A_m^m) = n!/(m!(n-m)!) $ 其中，$n, m in NN$且$m <= n$. === 组合数的一个推论与一个性质现给出用于本文证明的组合数的一个推论和一个性质：推论1：对于任意的$n in NN$，有$binom(n,n)=1$. 性质1：对于任意的$n, m in NN_(+)$且$m <= n$，有$binom(n,m) + binom(n,m-1) = binom(n+1,m)$. 对于上述组合数的推论和性质的证明，详见@xingzhi . = 猜想 == 举例由定义1.1可知，当$r=1$时，$S_(n)^(\(1\))$即$S_(n) = sum_(i=1)^(n) a_i = 1/2 n (n + 1)$；又如，当$r=2$时，$S_(n)^(\(2\)) = sum_(i=1)^(n) S_i = 1/6 n(n+1)(n+2)$. #figure( image("f1.svg", width:76%), caption: [ 绿点、紫点、蓝点分别表示数列${a_n}$、${S_n}$及${S_n^(\(2\))}$ ] ) == 猜想等差数列${a_n}$的r阶前n项和的表达式据2.1所举例子，可得如下猜想： #set math.equation(numbering: "(1)") $ S_n^(\(r\)) = 1/(r+1)! product_(i=0)^(r) (n+i) = (n+r)!/((r+1)!(n-1)!) = binom(n+r,r+1) $<chaixiang> = 证明现对所做猜想给出证明：证. (1) 当$r=0$或$r=1$时，显然有 @chaixiang 成立. (2) 假设当$r=k (k in NN_(+))$时，@chaixiang 成立，即 $ S_n^(\(k\)) = binom(n+k,k+1) $ 则据定义1.1，有， $ S_n^(\(k+1\)) & = sum_(i=1)^(n) S_i^(\(k\)) \ & = S_1^(\(k\)) + S_2^(\(k\)) + S_3^(\(k\)) + dots.h.c + S_n^(\(k\)) \ & = binom(1+k,k+1) + binom(2+k,k+1) + binom(3+k,k+1) + dots.h.c + binom(n+k,k+1) $<tuidao3> 由组合数的推论1可知，@tuidao3 等价于 $ S_n^(\(k+1\)) = binom(2+k,k+2) + binom(2+k,k+1) + binom(3+k,k+1) + dots.h.c + binom(n+k,k+1) $<tuidao4> 又由组合数的性质1可知，@tuidao4 等价于 $ S_n^(\(k+1\)) & = binom(3+k,k+2) + binom(3+k,k+1) + dots.h.c + binom(n+k,k+1) \ & = binom(4+k,k+2) + binom(4+k,k+1) + dots.h.c + binom(n+k,k+1) \ & dots.h \ & = binom(n+k,k+2) + binom(n+k,k+1) \ & = binom(n+k+1,k+2) = binom(n+(k+1),(k+1)+1) $ 即当$n=k+1$时， @chaixiang 也成立.\ 由(1)(2)可知， @chaixiang 对任何$r in NN$都成立. 证毕故猜想成立. = 附录 == 排列及排列数<pailie> 排列，一般地，从$n$个不同的元素中取出$m(m<=n)$个元素，按照一定的顺序排成一列，叫做从$n$个元素中取出$m$个元素的一个排列. 特别地，当$m=n$时，这个排列被称作_全排列_. 排列数，排列数指的是从$n$个不同元素中任取$m(m <= n)$个元素排成一列（考虑元素先后出现次序）称此为一个排列，此种排列的总数即为排列数，即叫做从$n$个不同元素中取出$m$个元素的排列数，记作$A_n^m$. == 组合及组合数<zuhe> 组合，一般地，从n个不同的元素中，任取$m(m<=n)$个元素为一组，叫作$n$个不同元素中取出$m$个元素的一个组合. 组合数，从$n$个不同元素中取出$m(m<=n)$个元素的所有组合的个数，叫做从$n$个不同元素中取出$m$个元素的组合数，记作$binom(n,m)$. === 组合数的一个推论与一个性质的证明<xingzhi> 现给出对推论1和性质1的证明. ==== 推论1的证明 #set math.equation(numbering: none) 设$n,m in NN$且$m <= n$，则由组合数的计算公式$ binom(n,m) = A_m^n / A_m^m $知，\ 当$m=n$时，上式为$ binom(n,n) = A_n^n / A_n^n = 1 $. 故推论1得证. ==== 性质1的证明设$n,m in NN_(+)$且$m <= n$，则 $ binom(n,m) + binom(n,m-1) & = n!/(m!(n-m)!) + n!/((m-1)!(n-m+1)!) \ & = n!(1/(m(m-1)!(n-m)!) + 1/((n-m+1)(m-1)!(n-m)!)) \ & = n!/((m-1)!(n-m)!) (1/m + 1/(n-m+1)) \ & = n!/((m-1)!(n-m)!) dot.op (n-m+1+m)/(m(n-m+1)) \ & = (n!(n+1))/(m(m-1)!(n-m+1)(n-m)!) \ & = ((n+1)!)/(m!(n-m+1)!) \ & = binom(n+1,m) $ 故性质1得证. == 一个有趣的数当$n=4,r=20$时，$S_n^(\(r\)) = S_4^(\(20\)) = binom(24,5) = 2024$，恰好是本文的写作年份. 在此，本文作者也祝大家2024新年快乐（虽然已经过去快4个月了）！本文写于2024年3月31日. == 致谢在此感谢钟运辉老师、李嘉才同学、王富平同学在本文写作过程中给予的帮助.
https://github.com/7sDream/fonts-and-layout-zhCN	https://raw.githubusercontent.com/7sDream/fonts-and-layout-zhCN/master/chapters/02-concepts/dimension/dim-3.typ	typst	Other	#import "/lib/draw.typ": * #import "/lib/glossary.typ": tr #let start = (0, 0) #let end = (250, 220) #let base = (100, 70) #let up = 150 #let down = 35 #let width = 74 #let lt = (base.at(0), base.at(1) + up) #let rb = (base.at(0) + width, base.at(1) - down) #let bbox-lt = (93, 178) #let bbox-width = 74 #let bbox-height = 111 #let bbox-ct = (bbox-lt.at(0) + bbox-width / 2, bbox-lt.at(1)) #let line-color = gray.darken(30%) #let graph = with-unit((ux, uy) => { // mesh(start, end, (50, 50), stroke: 1 * ux + gray) rect( lt, end: rb, stroke: 1.4 * ux + line-color, ) let line-stroke = 1 * ux + line-color segment( (0, base.at(1)), (end.at(0), base.at(1)), stroke: line-stroke, ) let arrow-y = base.at(1) - down - 20 let arrow-length = 12 arrow( (base.at(0) + width - arrow-length, arrow-y), (base.at(0) + width, arrow-y), stroke: line-stroke, head-scale: 3, ) arrow( (base.at(0) + arrow-length, arrow-y), (base.at(0), arrow-y), stroke: line-stroke, head-scale: 3, ) txt(tr[horizontal advance], (base.at(0) + width / 2, arrow-y), size: 12 * ux) txt([#tr[baseline]], (0, base.at(1)), size: 12 * ux, anchor: "lb", dy: 2) rect( bbox-lt, width: bbox-width, height: bbox-height, stroke: 1.2 * ux + line-color, ) txt(text(fill: line-color)[#tr[outline]#tr[bounding box]], bbox-ct, size: 12 * ux, anchor: "cb", dy: 2) txt(text(font: ("Fresca",))[b], base, size: 150 * ux, anchor: "lb", dx: -5, dy: -2) }) #canvas(end, start: start, width: 50%, graph)
https://github.com/VisualFP/docs	https://raw.githubusercontent.com/VisualFP/docs/main/SA/design_concept/content/poc/ui.typ	typst		#import "../../../acronyms.typ": * = User Interface <ui> This section describes the features of the #ac("PoC") application #ac("UI"), the high-level implementation, and how functional reactive programming could be applied to VisualFP. == Features The #ac("UI") for the #ac("PoC") application includes two main components, as shown in @ui-empty-editor: A sidebar with pre-defined value blocks and the function editor. #figure( image("../../static/ui_empty_editor.png"), caption: "Undefined function value in the VisualFP UI" ) <ui-empty-editor> The #ac("PoC") allows the construction of a value, the "userDefinedFunction", which starts with a generic type hole. Starting with a generic function type allows more flexible testing. In a completed application version, the user can define the function name and type when creating it. #grid( columns: (60%, 40%), gutter: 5pt, [#figure( image("../../static/ui_editor_drag_lambda.png"), caption: "Dragging lambda block into value definition" ) <ui-dragging-lambda>], [#figure( image("../../static/ui_editor_dropped_lambda.png"), caption: "Updated function definition including a lambda block" ) <ui-dropped-lambda>] ) @ui-dragging-lambda and @ui-dropped-lambda show how a lambda block is inserted into the value definition. To build the value definition, the user drags the lambda block from the sidebar into the type hole. The drop event then triggers the application to insert the lambda block into the function definition and infer the types of the new function definition. This process can be repeated with suiting value blocks until no type hole is left. As the #ac("PoC") is intended to test the concept, only a reset button exists to return to the initial empty definition. In a full version, this would be replaced with the possibility to remove specific blocks from the definition. Finally, the user-built function definitions can be viewed as Haskell code by clicking the "View Haskell" button. @ui-view-haskell shows the Haskell code for the `mapAdd5` function. #figure( image("../../static/ui_view_haskell.png"), caption: "Haskell defintion of mapAdd5 function in VisualFP" ) <ui-view-haskell> == Implementation The #ac("UI") implementation consists of an Electron.js app hosting a Threepenny #ac("UI"). The Electron app is packaged with an executable of the Threepenny #ac("UI") and all #ac("UI") related static files, i.e. #ac("CSS") & JavaScript files. When starting the Threepenny #ac("UI"), the Electron app passes a usable port for the local web server and the file path of the static #ac("UI") files to the Threepenny #ac("UI"). The function editor is the most significant part of the Threepenny #ac("UI") and has two primary responsibilities: - Rendering of function value blocks - Reacting to value block drop events The rendering part creates an #ac("HTML") representation of each block in the value definition and annotates it with #ac("CSS") classes according to its block type. Reacting to the drop events is a bit more complicated. The block values in the application's sidebar carry their names as data transfer data. When the user drops a block value into a type hole, the data transfer data is included in the event data. Unfortunately, the drop events cannot be registered when creating the type hole elements in the rendering part. So, to register the drop event listeners, the IDs of type holes need to be collected upfront. With these IDs, the #ac("HTML") elements added to the #ac("DOM") can be loaded, and the event handlers registered. The drop event handlers always do the same, regardless of the block value that was dropped: 1. Replace the type hole with the dropped value 2. Infer the updated function definition 3. Clear all elements from the function editor 4. Render the inferred function definition == Functional Reactive Programming Threepenny includes an #ac("FRP") library, which follows the concepts described by <NAME> and <NAME>. #ac("FRP") has two main concepts: Events and Behaviors. An Event is defined as a list of occurrences in time. A Behavior represents a value that changes over time. @frp_elliott_hudak While the first intention was to build the #ac("PoC") with an #ac("FRP") architecture, it became clear over time that Threepenny's #ac("FRP") library is not yet ready for more complex use-cases like VisualFP's function editor. The main problem is that no function allows it to merge multiple events. Implementing the #ac("FRP") architecture through Threepenny could be considered again once the #ac("FRP") library is replaced by reactive-banana #footnote("https://github.com/HeinrichApfelmus/reactive-banana"). The author of Threepenny, <NAME>, plans to do that in a future release @threepenny-frp-replacement. Generally, there is no reason why VisualFP couldn't be implemented using #ac("FRP"). In such an implementation, there would be three kinds of events: - "Reset Editor" button is clicked - "View Haskell" button is clicked - A block value is dropped into a type hole. This event combines all events from every type hole in the function definition. The value definition of the user-defined function is a behavior that changes every time a block value is dropped into the value definition. When the value definition changes, the elements displayed in the function editor must also be updated.
https://github.com/justmejulian/typst-documentation-template	https://raw.githubusercontent.com/justmejulian/typst-documentation-template/main/README.md	markdown		# ZHAW typst-documentation-template Typst documentation template for a ZHAW thesis, based on [ZHAW guidelines](https://gpmpublic.zhaw.ch/GPMDocProdDPublic/Vorgabedokumente_ZHAW/Z_RL_Richtlinie_Corporate_Design_Markengrundelemente.pdf). Learn more about Typst [here](https://github.com/typst/typst). This template is based on the [School of TUM thesis-template-typst](https://github.com/ls1intum/thesis-template-typst). ## Installation Install typst. ```bash brew install typst ``` Once you have installed Typst, you can use it like this: ```bash # Creates `main.pdf` in working directory. typst compile main.typ # Watches source files and recompiles on changes. typst watch main.typ ``` I recommend using the skim pdf viewer, which can be installed via brew. Skim automatically reloads the pdf when it changes. ```bash brew install --cask skim ``` ## neovim todo: add how to setup typst-lsp / Treesitter. ## Todo Github action to build https://github.com/marketplace/actions/github-action-for-typst
https://github.com/jgm/typst-hs	https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/visualize/image-08.typ	typst	Other	// Error: 8-18 failed to parse svg: found closing tag 'g' instead of 'style' in line 4 #image("test/assets/files/bad.svg")
https://github.com/yaoyuanArtemis/resume	https://raw.githubusercontent.com/yaoyuanArtemis/resume/main/README-zh.md	markdown	Do What The F*ck You Want To Public License	# typst-cv-miku 这是一个简单、优雅、学术风格的 [typst](https://typst.app/) 个人简历（CV）模板。支持中英文（以及更多）。你可以在 [这里](https://typst.app/project/rbxGsQC-tEkDq0mnNIuxkv) 查看在线演示。 ## 示例 ![cv_1](./assets/cv_1.webp) ![cv_2](./assets/cv_2.webp) ## 使用说明 1. 阅读 [typst](https://typst.app/docs/) 文档。 2. 安装此模板需要的字体： - [kpfonts](https://ctan.org/pkg/kpfonts) - [Source Han Sans](https://github.com/adobe-fonts/source-han-sans) - [Source Han Serif](https://source.typekit.com/source-han-serif/cn/) 3. 根据需要修改 `.typ` 文件. 你可能需要了解 typst 的一些基本语法。 ## 此外 Typst 目前在 Emoji 输出上有一些 [bugs](https://github.com/typst/typst/issues/144)，所以暂时用 SVG 替代，你可以在 [twemoji utils](https://twemoji.godi.se/) 找到更多。小图标来自 Material Icons (Community). ## License Licensed under [WTFPL](http://www.wtfpl.net/).
https://github.com/protohaven/printed_materials	https://raw.githubusercontent.com/protohaven/printed_materials/main/meta-environments/env-branding.typ	typst		// Branding #let color = ( tablegrey: rgb(95%,95%,95%), lightgrey: rgb(65%,65%,65%), midgrey: rgb(50%,50%,50%), darkgrey: rgb(38%,38%,38%), warning: rgb("#900000"), accent: rgb("#6EC7E2"), link: blue, ) #let font = ( title: "Noto Sans", sans: "Noto Sans", link: "Fira Mono", )
https://github.com/Toniolo-Marco/git-for-dummies	https://raw.githubusercontent.com/Toniolo-Marco/git-for-dummies/main/book/roles-duties.typ	typst		= Ruoli e mansioni #let wgc = "Working group coordinator" #let gl = "Group leader" #let gm = "GitHub maintainer" Uno degli aspetti principali del corso è che il progetto viene strutturato per assomigliare il più possibile alla realtà lavorativa, questo si riflette in una struttura gerarchica ben definita dove ognuno ha un ruolo e dei compiti associati, nel 2023 la struttura era: - #wgc - #gl - #gm - Tester - Reporter - Member == #wgc il #wgc è il responsabile di tutta la parte comune del progetto, è lui che ha l'ultima parola su come interpretare le specifiche fornite dal prof, gestire le riunioni, accettare o rifiutare le proposte (anche se è comune indirre delle votazioni), fissare le scadenze e scrivere i report sullo stato del progetto dopo ogni riunione. Viene eletto a maggioranza dai membri presenti durante una delle lezioni, il professore vi dirà in anticipo quando sarà il giorno, solitamente poco dopo la presentazione del progetto. Il primo compito del #wgc è quello di scegliere i suoi collaboratori, ovvero i #gl (solitamente 2), i Tester (solitamente 2/3) e i Reporter (solitamente 2), fatto questo il assieme ai #gm, deve creare l'organizzazione e dare ad essi tutti i permessi così che possano gestirla. == #gl Ogni gruppo ha un responsabile, questo va scelto tra i membri e comunicato al docente entro la data stabilita assieme al nome del gruppo e all'elenco dei partecipanti. Il suo compito è quello di partecipare alle riunione generali con #wgc e di rappresentare gli interessi del gruppo, votando le varie sulle questioni. == #gm Si consiglia di scegliere qualcuno che ha dimestichezza con git e GitHub, meglio se ha la possibilità di hostare delle applicazioni e dimestichezza con Docker. Si occuperà di gestire il repository, l'organizzazione su GitHub, effettuare il setup dell'inviter (o in alternativa aggiungere a mano tutte le persone), creare le action e cosa più importante gestire le issue, le milestones e le pull requests che verranno create, lavorerà in stretto contatto con i principali sviluppatori del Common Crate e coi testers. È compito suo accertarsi che una pr non rompa tutto il codice già presente o in caso contrario, che ci sia un valido motivo, deve controllare che il codice rispetti gli standard decisi e che sia ben documentato. == Tester Solitamente 2/3 persone, si occupano di scrivere i test che il Common Crate deve superare quando si implementa una nuova feature, oltre a scrivere i test che le applicazioni dei gruppi devono superare per verificare che stiano usando il Common Crate nel modo corretto, forzando gli standard decisi dalle specifiche. L'idea è che ad ogni pull request vengano eseguiti i test e in base all'esito, si procede con la revisione manuale. == Reporter Sono coloro che partecipano ad ogni riunione col compito di stilare un resoconto degli eventi, utile per tracciare i progressi e la direzione del progetto, basandosi su esso il #wgc, stilerà il suo report da inviare al docente. Hanno inoltre il compito di scrivere le specifiche man mano che vengono corrette e definite. == Member Sono tutti i membri dei vari gruppi, il loro compito è partecipare all'implementazione del Common Crate, proporre feature e aprire pull requests con i cambiamenti proposti. Il #wgc potrebbe apportare delle modifiche, come decidere di avere più o meno persone per ruolo, oppure crearne di nuovi, quello che possiamo dirvi è che noi ci siamo trovati bene con questa struttura.
https://github.com/Mufanc/hnuslides-typst	https://raw.githubusercontent.com/Mufanc/hnuslides-typst/master/configs.typ	typst		#let slide = ( width: 640pt, height: 360pt, margin: (x: 3em, top: 2em, bottom: 0em), )
https://github.com/DrGo/typst-tips	https://raw.githubusercontent.com/DrGo/typst-tips/main/refs/samples/typst-uwthesis-master/README.md	markdown		# typst-uwthesis This is a [typst](https://typst.app/) template that should (almost) satisfy the [University of Waterloo's thesis formatting requirements](https://uwaterloo.ca/graduate-studies-postdoctoral-affairs/current-students/thesis/thesis-formatting). The resulted document is similar to, but not exactly the same as the [LaTeX template](https://uwaterloo.atlassian.net/wiki/spaces/ISTKB/pages/2666037269/LaTeX+Software+for+Thesis+and+Document+Preparation+and+the+Overleaf+Cloud+Service). I wrote this template for my thesis proposal. At this moment, `project`, `appendix` and `gls` are only documented using the code itself. ## Limitations 1. Equations are not labeled in `section.equ` format. We would either need to "hack" using `set equ` or wait for typst to implement this. 2. Some duplicated headings appear in PDF's outline (not in TOC). This [issue](https://github.com/typst/typst/pull/1566) has been addressed by typst's developer. If you compile from GitHub's `main` branch, instead of using the pre-compiled package, the problem will disappear. We expect the problem will also disappear in the next release of typst. 3. Limited reverse link from bibliography. This needs to be addressed by typst. 3. No reverse link from list of abbreviations. I might find a way to hack this later. ## License Placeholders in the example file might contain copyrighted example texts obtained from UW's IST. By the time you complete you thesis, all copyrighted texts will have been removed. Currently, there should be no legal problem for a UW student to use this template. We have plan to release the template under an Apache license similar to [simple-typst-thesis's](https://github.com/zagoli/simple-typst-thesis) after getting rid of the potentially copyrighted texts. If you need this template soon, feel free to replace the text by non-copyrighted material and submit a pull request.
https://github.com/davystrong/umbra	https://raw.githubusercontent.com/davystrong/umbra/main/src/lib.typ	typst	MIT License	#let version = version((0, 1, 0)) #import "shadow-path.typ": shadow-path
https://github.com/Maso03/Bachelor	https://raw.githubusercontent.com/Maso03/Bachelor/main/Bachelorarbeit/chapters/VR.typ	typst	MIT License	== Virtual Reality (VR) Virtual Reality (VR) ist eine Technologie, die es Benutzern ermöglicht, in eine computergenerierte, dreidimensionale Umgebung einzutauchen. Diese Umgebung kann mit Hilfe von VR-Headsets und anderen Peripheriegeräten erlebt werden. VR findet Anwendung in Bereichen wie Gaming, Ausbildung, Medizin und Architektur. Durch die Schaffung immersiver Erlebnisse kann VR das Lernen und die Interaktion mit digitalen Inhalten erheblich verbessern.
https://github.com/PA055/5839B-Notebook	https://raw.githubusercontent.com/PA055/5839B-Notebook/main/appendix.typ	typst		#import "./packages.typ": notebookinator #import notebookinator: * #import themes.radial.components #import "./utils.typ": get-page-number #import "./glossary.typ" #create-appendix-entry(title: "Glossary")[ #components.glossary() ]
https://github.com/Enter-tainer/typstyle	https://raw.githubusercontent.com/Enter-tainer/typstyle/master/tests/assets/unit/code/param-len.typ	typst	Apache License 2.0	#let get-page-dim-writer() = locate(w_loc => {}) #let get-page-dim-writer(a) = locate(w_loc => {}) #let get-page-dim-writer(a, b) = locate(w_loc => {}) #let get-page-dim-writer(a, b, c) = locate(w_loc => {}) #let get-page-dim-writer( a, b, c) = locate(w_loc => {})
https://github.com/VaranTavers/vspct	https://raw.githubusercontent.com/VaranTavers/vspct/main/vspct.typ	typst	MIT License	// Varan's simple Pseudocode for Typst #let lang = "en" #let pkeywords = ( en: ( if_pre: "if", if_post: "", while_pre: "while", while_post: "", do_start: "repeat", do_end_pre: "until", do_end_post: "", end_pre: "end", end_post: "", else_pre: "else", return_pre: "return", for_pre: "for", for_post: "", in_pre: "In", out_pre: "Out", algorithm_pre: "Algorithm", algorithm_post: "", ), hu: ( if_pre: "Ha", if_post: "akkor", while_pre: "Amíg", while_post: "végezd el", do_start: "Ismételd", do_end_pre: "Ameddig", do_end_post: "", end_pre: "", end_post: "vége", else_pre: "különben", return_pre: "térít", for_pre: "Minden", for_post: "végezd el", in_pre: "Be", out_pre: "Ki", algorithm_pre: "Algoritmus", algorithm_post: "", ) ) #let gets = sym.arrow.l.long #let algCounter = counter("alg") #let pseudo(body, caption: "") = { block( breakable: false, { set align(left) strong[ #pkeywords.at(lang).at("algorithm_pre") #algCounter.display()#pkeywords.at(lang).at("algorithm_post") ] ": " + caption algCounter.step() block( breakable: false, stroke: (top: black, bottom: black), above: 0.4em, inset: (top: 0.5em, left: 1em, right: 1em, bottom:0.5em), par( leading: 0.5em, body ) ) } ) } #let pblock(fname_pre, fname_post, f, body) = { text(weight: "bold", fname_pre)+" "+f+" "+text(weight: "bold",fname_post) block( above: 0.7em, below: 0.5em, inset: (top: 0em, left: 1em), par( leading: 0.5em, body ) ) text(weight: "bold", pkeywords.at(lang).at("end_pre") + " " + fname_pre + " " + pkeywords.at(lang).at("end_post")) } #let pfor(iterator, start, end, cumul: "", body) = { text(weight: "bold", pkeywords.at(lang).at("for_pre"))+" "+iterator+" "+gets+" "+start+", "+end if cumul != "" { ", " + cumul } " " text(weight: "bold", pkeywords.at(lang).at("for_post")) block( above: 0.7em, below: 0.5em, inset: (top: 0em, left: 1em), par( leading: 0.5em, body ) ) text(weight: "bold", pkeywords.at(lang).at("end_pre") + " " + pkeywords.at(lang).at("for_pre") + " " + pkeywords.at(lang).at("end_post")) } #let prepeat(f, body) = { text(weight: "bold", pkeywords.at(lang).at("do_start")) block( above: 0.7em, below: 0.5em, inset: (top: 0em, left: 1em), par( leading: 0.5em, body ) ) text(weight: "bold", pkeywords.at(lang).at("do_end_pre")+ " " + f + " " + pkeywords.at(lang).at("do_end_post")) } #let pwhile(f, body) = pblock(pkeywords.at(lang).at("while_pre"), pkeywords.at(lang).at("while_post"), f, body) #let pif(f, body) = pblock(pkeywords.at(lang).at("if_pre"), pkeywords.at(lang).at("if_post"), f, body) #let pifelse(f, body1, body2) = { text(weight: "bold", pkeywords.at(lang).at("if_pre"))+" "+f+text(weight: "bold",pkeywords.at(lang).at("if_post")) block( above: 0.7em, below: 0.5em, inset: (top: 0em, left: 1em), par( leading: 0.5em, body1 ) ) text(weight: "bold", pkeywords.at(lang).at("else_pre")) block( above: 0.7em, below: 0.5em, inset: (top: 0em, left: 1em), par( leading: 0.5em, body2 ) ) text(weight: "bold", pkeywords.at(lang).at("end_pre")+" "+pkeywords.at(lang).at("if_pre")) + " " + pkeywords.at(lang).at("end_post") } #let preturn(body) = { text(weight: "bold", pkeywords.at(lang).at("return_pre")) + " " + body } #let pin(body) = { text(weight: "bold", pkeywords.at(lang).at("in_pre")) + " " + body } #let pout(body) = { text(weight: "bold", pkeywords.at(lang).at("out_pre")) + " " + body } #let pref(label) = pkeywords.at(lang).at("algorithm_pre") + " " + locate(loc => { algCounter.at( query(label, loc).first().location() ).at(0) }) + " " + pkeywords.at(lang).at("algorithm_post") /* #show figure: it => if rev [ #set align(left) #strong[ #it.supplement #it.counter.display(it.numbering) ]: #it.caption #it.body ] else [ #set align(center) #it.body #emph[ #it.supplement #it.counter.display(it.numbering) ]: #it.caption ] */
https://github.com/LDemetrios/Typst4k	https://raw.githubusercontent.com/LDemetrios/Typst4k/master/src/test/resources/suite/foundations/panic.typ	typst		--- panic --- // Test panic. // Error: 2-9 panicked #panic() --- panic-with-int --- // Test panic. // Error: 2-12 panicked with: 123 #panic(123) --- panic-with-str --- // Test panic. // Error: 2-24 panicked with: "this is wrong" #panic("this is wrong")
https://github.com/VZkxr/Typst	https://raw.githubusercontent.com/VZkxr/Typst/master/Tests/document.typ	typst		Sea $\{X_n\}_{n in NN}$ una cadena de Markov con espacio de estados $EE=\{0,1,2,3,4\}$, distribución inicial $pi^0 = (1,0,0,0,0)$ y matriz de probabilidades de transición $ PP= mat( 0 , 1 , 0 , 0 , 0; frac(1, 4), 0, frac(3, 4), 0, 0; 0, frac(1, 2), 0, frac(1, 2), 0; 0, 0, frac(3, 4), 0, frac(1, 4); 0, 0, 0, 1, 0; ) $ Calcula la distribución estacionaria de $X_n_{n in NN}$ para $n$ par, es decir, utilizando $PP$. #let ofi = [Office] #let rem = [_Remote_] #let lea = [On leave] #table( columns: 6 * (1fr,), table.header( [Team member], [Monday], [Tuesday], [Wednesday], [Thursday], [Friday] ), [<NAME>], table.cell(colspan: 2, ofi), table.cell(colspan: 2, rem), ofi, [<NAME>], table.cell(colspan: 5, lea), [<NAME>], rem, table.cell(colspan: 2, ofi), rem, ofi, ) #table( columns: 5 * (1fr,), table.header( [], table.cell(colspan:2, [Blue chip]), [Fresh IPO], [Penny st'k], ), table.cell( rowspan: 4, align: horizon, [ USD/day ], ), [0.20], [104], [5], [3.17], [108], [4], [1.59], [84], [1], [0.26], [98], [15], [0.01], [195], [4], [7], [ USD/hr ], [57], [2], [3], [6.7] ) #table( columns: (auto, 1fr, auto, auto), table.header( table.cell(colspan:4, align: center, [Índice temático]), ), table.cell(rowspan: 2,[]), table.cell(rowspan: 2, align: center + horizon, [Tema]), table.cell(colspan: 2, align: center, [Horas de curso]), [Teorías], [Prácticas], [1], [Conjuntos de números], [1], [], [2], [La recta real], [1], [1], [3], [Potencias], [1], [1], [4], [Raíces], [1], [1], [5], [Porcentajes], [1], [1], table.cell(colspan: 2, align: right, [Subtotal]), [5], [4], table.cell(colspan: 2, align: right, [Total]), table.cell(colspan: 2, align: center, [9 hrs]) ) #pagebreak() #table( columns: (auto, 1fr), table.header( table.cell(colspan:2, align: center, [Contenido temático]), ), table.cell(rowspan: 2,[]), table.cell(rowspan: 2, align: center + horizon, [Temas y subtemas]), [1], [Conjuntos de números \ #v(.07cm) 1.1 #h(.4cm)Números naturales \ #v(.07cm) 1.2 #h(.4cm)Números enteros \ #v(.07cm) 1.3 #h(.4cm)Números racionales \ #v(.07cm) 1.4 #h(.4cm)Números irracionales \ #v(.15cm)], [2], [La recta real \ #v(.07cm) 2.1 #h(.4cm)Propiedades de los signos \ #v(.07cm) 2.2 #h(.4cm)Suma y resta \ #v(.07cm) 2.3 #h(.4cm)Multiplicación y división \ #v(.07cm) 2.4 #h(.4cm)Jerarquía de operaciones \ #v(.07cm) 2.5 #h(.4cm)Operaciones con fracciones \ #v(.15cm)], [3], [Potencias \ #v(.07cm) 3.1 #h(.4cm)Noción intuitiva de la potencia \ #v(.07cm) 3.2 #h(.4cm)Operaciones con potencias \ #v(.15cm)], [4], [Raíces \ #v(.07cm) 4.1 #h(.4cm)Raíces exactas \ #v(.07cm) 4.2 #h(.4cm)Raíces no exactas \ #v(.15cm)], [5], [Porcentajes \ #v(.07cm) 5.1 #h(.4cm)Representaciones \ #v(.07cm) 5.2 #h(.4cm)Conversiones \ #v(.07cm) 5.3 #h(.4cm)Problemas de aplicación \ #v(.15cm)] ) #pagebreak() #table( columns: (1fr, 1fr), table.header(align(center)[Estrategias didácticas], align(center)[Evaluación de aprendizaje]), [Exposición #h(1fr) (X)], [Exámenes parciales #h(1fr) (X)], [Trabajo en equipo #h(1fr) (#h(.2cm))], [Examen final #h(1fr) (X)], [Lecturas #h(1fr) (#h(.2cm))], [Trabajos y tareas #h(1fr) (#h(.2cm))], [Trabajo de investigación #h(1fr) (#h(.2cm))], [Presentación del tema #h(1fr) (#h(.2cm))], [Prácticas (taller o laboratorio) #h(1fr) (#h(.2cm))], [Participación en clase #h(1fr) (#h(.2cm))], [Prácticas de campo #h(1fr) (#h(.2cm))],[Asistencia #h(1fr) (#h(.2cm))], [Aprendizaje por proyectos #h(1fr) (#h(.2cm))],[Rúbricas #h(1fr) (#h(.2cm))], [Aprendizaje basado en problemas #h(1fr) (#h(.2cm))],[Portafolios #h(1fr) (#h(.2cm))], [Casos de enseñanza #h(1fr) (#h(.2cm))],[Listas de cotejo #h(1fr) (#h(.2cm))], [Otras (especificar) #h(1fr) (#h(.2cm))],[Otras (especificar) #h(1fr) (#h(.2cm))] )
https://github.com/typst/packages	https://raw.githubusercontent.com/typst/packages/main/packages/preview/cetz/0.1.2/src/aabb.typ	typst	Apache License 2.0	#import "vector.typ" /// Compute an axis aligned bounding box (aabb) /// for a list of vectors. /// /// An aabb dictionary has the following keys: /// - low (vector) Min. bounds vector /// - high (vector) Max. bounds vector /// /// - pts (array): List of vectors/points /// - init (dictionary): Initial aabb /// -> dictionary AABB object #let aabb(pts, init: none) = { let bounds = init if type(pts) == array { for (i, pt) in pts.enumerate() { if bounds == none and i == 0 { bounds = (low: pt, high: pt) } else { let (x, y, z) = pt let (lo-x, lo-y, lo-z) = bounds.low bounds.low = (calc.min(lo-x, x), calc.min(lo-y, y), calc.min(lo-z, z)) let (hi-x, hi-y, hi-z) = bounds.high bounds.high = (calc.max(hi-x, x), calc.max(hi-y, y), calc.max(hi-z, z)) } } return bounds } else if type(pts) == dictionary { if init == none { return pts } else { return aabb((pts.low, pts.high,), init: bounds) } } panic("Expected array of vectors or bbox dictionary, got: " + repr(pts)) } /// Get the mid-point of an aabb as vector. /// /// - bounds (AABB): AABB /// -> vector #let mid(bounds) = { return vector.scale(vector.add(bounds.low, bounds.high), .5) } /// Get the size of an aabb as vector /// this is a vector from the aabb's low to high. /// /// - bounds (AABB): AABB /// -> vector #let size(bounds) = { return vector.sub(bounds.high, bounds.low) }
https://github.com/thomasschuiki/thomasschuiki	https://raw.githubusercontent.com/thomasschuiki/thomasschuiki/main/cv/fontawesome.typ	typst		//! typst-fontawesome //! //! https://github.com/duskmoon314/typst-fontawesome // Implementation of `fa-icon` #import "lib-impl.typ": fa-icon // Generated icons #import "lib-gen.typ": *
https://github.com/Vaskozlov/Lectures	https://raw.githubusercontent.com/Vaskozlov/Lectures/main/Алгебра (многочлены).typ	typst		= Многочлены == Производная многочлена $ f(x) = product_(i = 1)^(n)(x - x_i) $ $ f'(x) = sum_(i = 1)^(n) frac(f(x), x - x_i) = f(x) sum_(i = 1)^(n) frac(1, x - x_i) $ $ f'(x) = f(x) sum_(i = 1)^(n) frac(1, x - x_i) => frac(f'(x), f(x)) = sum_(i = 1)^(n)frac(1, x - x_i) $ Если многочлен f(x) имеет корень кратности n, то это значит, что + $f(alpha) = 0$ + $f^((i))(alpha) = 0 "для" forall i in [1,n - 1]$ === Пример (связь производной с функцией и ее корнями) $ f(x) = x^2 - 3x + 2 = (x - 2)(x - 1) $ $ f'(x) = f(x) dot frac(1, x - 1) + frac(1, x - 2) = f(x) dot frac(2x - 3, x ^ 2 - 3x + 2) = f(x) dot frac(2x - 3, f(x)) = 2x - 3 $ $ f'(x) = 2x - 3 $ === Пример (задачи на нахождение суммы) == Простейшие дроби Дробь $frac(f, g) in K[t]$ - простейшая, если $g = p^k$, где $p in K[t]$ – неприводимый многочлен и $"deg"(f) < "deg"(g)$. == Связь разложения на простейшие дроби с интерполяцией === Алгоритм, когда все корни первой степени $ g(x) = (x - a_1) dots (x - a_n) $ $ frac(f(x), g(x)) = sum_(i = 1)^(n) frac(f(a_i), g'(a_i) dot (x - a_i)) $ #pagebreak() == Алгоритм интерполяции Лагранжа $ f(x) = product_(i = 1)^(n)(x - x_i) $ $ L(X) = sum_(i = 0)^(n)y_i dot l_(i)(x) = sum_(i = 0)^(n) y_i dot frac(f(x), f'(x_i) dot (x - x_i)) $ $ l_i(x) = product_(j = 0, j != i)frac(x - x_j, x_i - x_j) $ === Пример #columns(2)[ #align(right)[ $ l_0(x) = frac((x - x_1)(x - x_2)(x - x_3), (x_0 - x_1)(x_0 - x_2)(x_0 - x_3)) = \ = frac((x - 2)(x - 3)(x - 5), -30) $ $ l_1(x) = frac((x - x_0)(x - x_2)(x - x_3), (x_1 - x_0)(x_1 - x_2)(x_1 - x_3)) = \ = frac((x - 0)(x - 3)(x - 5), 6) $ ] #colbreak() #table( columns: 3, [i], [x], [y], [0], [0], [0], [1], [2], [1], [2], [3], [3], [3], [5], [2]) ] $ L(x) = y_0 dot l_0(x) + y_1 dot l_1(x) + y_2 dot l_2(x) + y_3 dot l_3(x) = 0 dot l_0(x) + 1 dot l_1(x) + 3 dot l_2(x) + 2 dot l_3(x) $ == Алгоритм интерполяции по Ньютону $ N = a_0 + a_1 dot (x - x_0) + a_2 dot (x - x_0)(x - x_1) + dots $ Чтобы найти многочлен по точкам, нужно постепенно подставлять значения x, тогда если подставляем $x_i$, то начиная с i будут нули. === Пример #columns(2)[ #align(right)[ $ 2 = a_0 + a_1(1 - 1) + a_2(1 - 1)(2 - 1) + dots $ $ 3 = a_0 + a_1(2 - 1) + a_2 (2 - 1)(2 - 2) + dots $ ] #colbreak() #table( columns: 3, [i], [x], [y], [0], [1], [3], [1], [2], [-10], [2], [3], [5]) ] === Упрощение алгоритма интерполяции Ньюетона при $Delta x = "const"$ Пусть $h = Delta x$, тогда построим табличку, где $Delta^k y_i = Delta^(k - 1)y_(i + 1) - Delta^(k - 1)y_i$, где i - строчка в таблице. #table( columns: 5, align: (center, center, center , center, center), [i], [x], [y = $Delta^0$y], [$Delta^1$y], [$Delta^2$y], [0], [1], [3], [-13], [28], [1], [2], [-10], [15], [], [2], [3], [5], [], [] ) Тогда коэффициент $a_k = frac(Delta^k y_0, k! dot h^k)$
https://github.com/jiamingluuu/mata35-notes	https://raw.githubusercontent.com/jiamingluuu/mata35-notes/main/diff-eqn.typ	typst		#set text(size: 13pt) #set math.equation(numbering: "(1)") #set rect(width: 100%, radius: 8pt, fill: rgb("#f2f2f2"), stroke: 1pt, inset: 12pt) = Differential Equations == Introduction Definition. _Differential equation (DE)_ is an equation involving functions and their derivatives. The study of differntial equations plays a significant role in the modern study of physics, engineering, economics, and etc. To dive deep in the mathematically rigorous dicussion of differnetial equation, it is inevitable to introduce a diverse branches of mathematics, for instance, analysis and numerical methods. In light of the limitation to the person who made the note, we are only going to talk about the most fundamental part of differneitlal equation. == First Order Differential Equations Definition. The _order_ of a differential equation is defined by the highest degree of derivative of the differential equation has. Definition. let $y: RR -> RR$ be a function with variable $t$, and $f: RR^2 -> RR$. A _first order differential equatoin_ has the the form $ (dif y)/(dif t) = f(y, t) $ Example. _Acceleration_ describes the rate of change of velocity of an object. Suppose a ball is dropped from air and undergoes a free fall, that is, only gravity is acting on the ball, its acceleration $a(t) = (dif v)/(dif t)$ is characterized by a first order DE $ (dif v)/(dif t) = g. $ Where $g$ is the _gravitational constant_ There is a special class of DE we are interested to study: separable DE. Definition. A _separable differential equation_ can be written as $ (dif y)/(dif t) = M(y) N(t). $ To solve a separable, we follows the following strategy in general: 1. Re-write the given DE into the form given in (3). 2. Separable varable, so each side of equation only contains one type of variable $ 1/(M(y))(dif y)/(dif t) = N(t). $ 3. Then integrate both sides of the equation at the same time. $ integral 1/(M(y)) (dif y)/(dif t) dif t = integral N(t) dif t. $ Example (Newton's Law of Cooling). Let $T(t)$ be the temperature of an object at time $t$, and $T_s$ be the temperature of surrounding environment. The rate of change of temperature of $T(t)$ is described as $ (dif T)/ (dif t) = k(T(t) - T_s), $ where both $k, T_s$ are a constant in real number. To solve (6), we are going to follow the strategy stated before: $ 1/(T(t) - T_s) (dif T)/(dif t) &= k\ integral 1/(T(t) - T_s) (dif T)/cancel(dif t) cancel(dif t) &= integral k dif t\ integral 1/(T(t) - T_s) dif T &= integral k dif t\ ln abs(T(t) - T_s) &= k t + c\ T(t) - T_s &= e^(k t + c)\ T(t) - T_s = e^(k t) times e^c\ T(t) = A e^(k t) + T_s\ $ #rect[ _Remark._ Notice that it is irrigorous to treat the derivative $(dif T)/(dif t)$ as a fraction and cancel it with the term $dif t$. The definition of derivative involving limit, and is to find the flactuation of the original function in the infidecimal change in $t$, that is $ (dif T)/(dif t) equiv lim_(h -> 0) (T(t + h) - T(t))/(h), $ which is a unity becuase the limit is not gaurantee to be congervent. Whereas when we write the whole term $ 1/(T(t) - T_s) (dif T)/(dif t) dif t, $ it is a 1-form of differential form, which is a smooth section of the co-tangent bundle on a manifold. Under no circumstance should those two notion to be inter-changibly used. ] However, what if the DE is not separable? For instace, how can we solve for a differenial equation has the form $ A(x)(dif y)/(dif x) + B(x) y = C(x), $ where function $A, B, C$ functions over real. In this senario, we need introduce some prior knowledge. Definition. Let $f: RR^n -> RR$ be a differentiable function over $RR$ with respect to variable $x_1, x_2, ..., x_n$. The _total derivative_ $dif f$ can be written as $ dif f = sum^n_(i = 1) (diff f)/(diff x_i) dif x_i $ Example. The total derivative of $f(x, y) = x^2 + y^2$ is given by $ dif f &= (diff f)/(diff x)dif x + (diff f)/(diff y)dif y\ &= diff/(diff x)(x^2 + y^2)dif x + diff/(diff y)(x^2 + y^2)dif y\ &= 2x dif x + 2y dif y\ $ Definition. A differential form $alpha$ is _exact_ if there exists some differential form $beta$ such that $ dif beta = alpha $ Proposition. Let $P, Q: RR^2 -> RR$ be multi-variable functions with respect to variable $x$ and $y$. Then the differential form $ P dif x + Q dif y $ is exact if and only if $ (diff P)/ (diff y) = (diff y)/ (diff x) $ #rect[ _Remark._ The notion of exactness is telling us, if a differential form $alpha$ is exact, then it can be obtained by computing the total derivative of another differential form $beta$. Furthermore, the proposition above provides us an easy, swift way of verifying if a given differential form is exact. The intuition behinds the proposition is the following: if equation (12) were exact, then $P$ and $Q$ are the derivative of same function but with respect to different variable. That is $ P = (diff f)/(diff x), quad Q = (diff f)/(diff y). $ Therefore, $ (diff P)/(diff y) = (diff^2 f)/(diff y diff x) "and" (diff Q)/(diff x) = (diff^2 f)/(diff x diff y), $ which are essentially equal. ] In light the the introductory of differential form and exactness, we can take their advantages in the discussion of solving DEs. Given a first order differential equation with the form: $ A (dif y)/(dif x) + B y = C, $ where $A, B, C: RR -> RR$ are functions over real with respect to variable $x$, we can re-write it by using differential forms $ underbrace(A dif y + B y dif x, alpha) = C dif x. $ Hence, if $alpha$ were exact, then by equation (11) there exists some other form $beta$ such that $ dif beta = alpha. $ It implies that (17) is equivalent to $ dif beta = C dif x. $ And by integrating both sides of (19), we can obtain the solution of our DE (16) in implicit form: $ integral dif beta &= integral C dif x\ beta &= integral C dif x $ Example. Solve the following differential equation $ (4 + t^2) (dif y)/(dif t) + 2 t y = 4t. $ Let's firstly write it using differential form and check if it is exact: $ underbrace((4 + t^2) dif y, P) + underbrace(2t y dif t, Q) = 4t dif t. $ $ (diff P)/(diff t) &= diff/(diff t) (4 + t^2)\ &= 2t\ (diff Q)/ (diff y) &= diff/(diff y)(2t y)\ &= 2t. $ So we can see the left part of (22) is an exact differential form, which is implying that $ P = (diff f)/(diff y), quad Q = (diff f)/(diff y). $ Hence $ f &= integral P dif y\ &= integral 4 + t^2 dif y\ &= 4y + t^2 y + h(t) $ And we can solve the unknown function $h(t)$ by computing the derivative of $f$ with respect to $t$: $ (diff f)/(diff t) &= diff/(diff t)(4y + t^2 y + h(t))\ &= 2t y + h'(t)\ $ Therefore, by using the fact that $ (diff f)/(diff t) &= Q\ 2t y + h'(t) &= 2t y\ h'(t) &= 0 $ And it follows that $h(t) = c$, where $c$ is an aribitrary constant in $RR$. Hence, we have the final answer: $ f = 2y + t^2 y + c $ However, what if the given DE is not exact? Then we need to develop some trick to modify the DE and change to exact. Definition. Given a DE of the from $ A dif y + B y dif x = C $ A _integrating factor_ is an auxiliary function $I(x)$ such that when we multiply it to the both sides of the eqaution, making $ I A dif y + I B dif x $ to be an exact differential form. Proposition. If a DE has the form $ A dif y + B y dif x = C, $ where $A, B, C: RR -> RR$ are function with respect to variable $x$, and $A$ is non-zero over its domain. Then the DE has an integrating factor $ I(x) = e^(integral P(x) dif x), $ where $P(x) = B(x)/A(x)$. Example. Solve the following DE $ x dif y + 2y dif x = 4x^3 dif x $ As you can verify, this DE is not exact, so we are going to find the integrating factor $ I(x) &= e^(integral P(x) dif x)\ &= e^(integral 2/x dif x)\ &= e^(2ln(x))\ &= x^2 $ By multiplying both sides by the integrating factor, we have $ x dif y + 2y dif x &= 4x^3 dif x\ dif y + (2y)/x dif x &= 4x^4 dif x\ x^2 (dif y + (2y)/x dif x) &= x^2 dot 4x^4 dif x\ x^2 dif y + 2y x dif x &= 4x^6 dif x\ d(x^2 y) &= 4x^6 dif x\ integral d(x^2 y) &= integral 4x^6 dif x\ x^2 y &= 4/7 x^7 + c $
https://github.com/flechonn/interface-typst	https://raw.githubusercontent.com/flechonn/interface-typst/main/BD/TYPST/exo4.typ	typst		#show terms: it => { let title = label("Mon Exercice de Traitement du Signal") let duration = label("1h30") let difficulty = label("Facile") let solution = label("0") let figures = label("") let points = label("5") let bonus = label("0") let author = label("Moi") let references = label("") let language = label("Français") let material = label("") let name = label("exo4") } = Exercice 1 Considérez un signal sinusoïdal x(t)=$A⋅sin(2πft+ϕ)$, où AA est l'amplitude, ff est la fréquence en Hz et ϕϕ est la phase en radians. Écrivez une fonction en Python pour calculer la valeur efficace (RMS) de ce signal sur une période donnée. Testez votre fonction avec $A=3$, $f=50$Hz et $ϕ=π4$ sur une période de T=$1/f$. = Solution Solution de l'exercice: Pour calculer la valeur efficace (RMS) d'un signal sinusoïdal $x(t)=A⋅sin(2πft+ϕ)$ nous pouvons utiliser la formule suivante : $"RMS"=sqrt(1/Tintegral_(0)^T x^2 dif x)$ où $A$ est l'amplitude, $f$ est la fréquence en Hz, $ϕ$ est la phase en radians, et $T$ est la période du signal ($T=1/f$). Dans notre cas, avec $A=3$, $f=50$ Hz et $ϕ=4π$ la formule devient : $"RMS"=sqrt(1/Tintegral_(0)^T (3⋅sin(2π⋅50⋅t+4π))² dif x)$ Après résolution numérique de cette intégrale sur une période T, nous obtenons la valeur efficace du signal.
https://github.com/jujimeizuo/ZJSU-typst-template	https://raw.githubusercontent.com/jujimeizuo/ZJSU-typst-template/master/template/toc.typ	typst	Apache License 2.0	#import "font.typ": * #import "utils.typ": * #show heading : it => { set align(center) set text(font:heiti, size: font_size.sanhao) it par(leading: 1.5em)[#text(size:0.0em)[#h(0.0em)]] } #set page(footer: [ #set align(center) #set text(size: 10pt, baseline: -3pt) #counter(page).display( "I", ) ] ) // 图目录 #heading(level: 1, outlined: false)[图目录] #v(2em) #show outline: it => { set heading(numbering: "1.1") set text(font: songti, size: font_size.xiaosi) set par(leading: 1em ) it } #outline( title: none, target: figure.where(kind: image), indent : true, ) #pagebreak() #heading(level: 1, outlined: false)[表目录] #v(2em) #show outline: it => { set text(font: songti, size: font_size.xiaosi) set par(leading: 1em ) it } #outline( title: none, target: figure.where(kind: table), indent : true, ) #pagebreak() // 目录 #heading(level: 1, outlined: false)[目录] #v(2em) #{ set align(right) set text(font: songti, size: font_size.xiaosi) set par(first-line-indent: 0pt) [摘要 ] + [.] * 144 + [ I] set par(leading: 1em) [Abstract ] + [.] * 137 + [ II] set par(leading: 1em) [图目录 ] + [.] * 137 + [ III] set par(leading: 1em) [表目录 ] + [.] * 137 + [ IV] set par(leading: 1em) [目录 ] + [.] * 143 + [ V] } #show outline: it => { set text(font: songti, size: font_size.xiaosi) set par(leading: 1em ) it } #outline( title: none, indent : true, )

https://github.com/lxl66566/my-college-files

https://raw.githubusercontent.com/lxl66566/my-college-files/main/信息科学与工程学院/机器视觉实践/报告/4/4.typ

typst

The Unlicense

#import "../template.typ": * #show: project.with( title: "4", authors: ("absolutex",), ) = 机器视觉实践四 == 实验目的图像补光算法与界面设计 + 实现图像补光算法（增加局部图像亮度，保持对比度，避免失真） + 设计图像补光界面，从左到右补光图像逐步替换原图像的可视化效果。 == 实验代码这是一份单文件 html 代码，使用时需要在本地启动一个 http server. #include_code("../src/brightness/index.html") 具体的原理是定义了三个补光因子，分别对应暗部，中间色调，亮部。对每个像素的每个 RGB 进行遍历，并将其乘以对应的补光因子。界面设计就直接调用前端的 img-comparison-slider 库，实现了可拖拽的亮暗对比。 == 实验结果与心得体会 #figure( image("result.png", width: 100%), caption: [补光界面效果图], ) 可以看出补光效果较好，特别是对白色背景的物体有着巨大提升。不过对于本身是黑色的物体，即使乘以补光因子，其亮度值还是过小，无法看到明显的效果。

https://github.com/MobtgZhang/sues-thesis-typst

https://raw.githubusercontent.com/MobtgZhang/sues-thesis-typst/main/paper/chapters/ch02.typ

typst

MIT License

= 公式、图表等表示方法 == 公式方便快捷写入公式是Typst相对于Word编辑器最为主要的优势之一，特别是熟练掌握之后，在输入公式的时候具有非常大的提升效果。 Typst 中的公式分为两类，包括有#text("行内公式",fill:rgb(255,0,0))和#text("行间公式",fill:rgb(255,0,0))，例如这是一个行内公式 $f(x)=1/(sqrt(2pi)sigma)dot exp(-((x-mu)^2)/(2sigma^2))$。下面举例几个行间公式 $ f(x)&= 1/(sqrt(2pi)sigma)dot exp(-((x-mu)^2)/(2sigma^2)) $ 例如，定义一个分段函数 $ f(x)&= cases( -x^3 + x + 8 &"," x<=2 \ 1/2 x^2 &"," 2<x<=10\ x+ 10 &"," x>=10 ) $ 也可以定义一个多行的连等的等式，定义如下所示 $ cos(2x) &= cos^2x - sin^2x &= 2cos^2x - 1 &= 1 - 2 sin^2x $ 可以将多个等式对齐写在同一个语句块当中，例如麦克斯韦方程组积分形式： $ cases( integral.cont_l H dot d l &= integral.double_S J dot d S + integral.double_(S)(diff D)/(diff t) dot S \ integral.cont_(l) EE dot d l &= - integral.double_(S)(diff BB)/(diff t) dot d S \ integral.cont_(S) BB dot S &= 0 \ integral.cont_(S) D dot d S &= integral.triple_(V) rho VV ) $ 微分形式： $ cases( nabla times H &= J + (diff D)/(diff t) \ nabla times E &= - (diff B)/(diff t) \ nabla dot B &= 0 \ nabla dot H &= rho ) $ 带有矩阵定义的公式： $ H = -mu dot B = -gamma B_(o) S_(z) = (gamma B_(o))/2 mat( 1, dots.h.c ,1 ; dots.h.c , dots.down , dots.h.c ; 1 & dots.h.c & 1 ) $ 在求解凸优化问题的时候，问题研究最后求解归结为以下的方程形式： $ arg_(x_(j)) min_(j=1,dots.h.c,N) sum_(j=1)^(N)c_(j)x_(j) $ $ s.t. cases( sum_(j=1)^(N)a_(i j)x_(j)=b_(i) & "," i=1 "," dots.h.c "," m \ x_(j)>=0 ) $ <equ:matrix> 在文章当中每一个公式的后面均可以添加一个label的标签，这样就可以应用公式了，例如#ref(<equ:matrix>)就是刚刚我们表达的矩阵表达式。

https://github.com/ryuryu-ymj/mannot

https://raw.githubusercontent.com/ryuryu-ymj/mannot/main/examples/showcase.typ

typst

MIT License

#import "/src/lib.typ": * #set page(width: auto, height: auto, margin: (x: 2cm, y: 1cm), fill: white) #set text(24pt) #show: mannot-init $ mark(1, tag: #<num>) / mark(x + 1, tag: #<den>, color: #blue) + mark(2, tag: #<quo>, color: #red) #annot(<num>, pos: top)[Numerator] #annot(<den>)[Denominator] #annot(<quo>, pos: right, yshift: 1em)[Quotient] $

https://github.com/figarofuga/Typst-template

https://raw.githubusercontent.com/figarofuga/Typst-template/main/CJD/CJD.typ

typst

= Introduction CJD(Creutz-feldt jacob disease)はRapid progression dementiaの中の代表的疾患である。 Rapid progression dementiaの定義は決まっていないが、大体2年以内に進行する認知症と言われている。最も重要な事は、CJDは感染性がある上に治療法が確立していない事である。 *いけた！* - 例えば、結核の事 - 例えば、NTMの事 = 症例 = いつ疑う？ = 身体所見・時間経過 = 検査 == 頭部MRI == 脳波 == 髄液検査 === RT-Quick == 具体的には？ドライアイスを固める == 病理解剖について当院では不可能ブレインバンク東京精神 1. いろんなことがある。 2. そういうものだ。例えばreferenceはこうやる。

https://github.com/michalrzak/muw-typst-template

https://raw.githubusercontent.com/michalrzak/muw-typst-template/main/README.md

markdown

# About The repository contains a **Typst template** for a "_Diplomarbeit_" / "_Masterarbeit_" / "_Master thesis_" at the **Medical University of Vienna** (Meduni Wien). [Typst](https://github.com/typst/typst) is a typesetting system, offering a fun alternative to LaTeX. The template should follow all guidelines outlined in [Leitfaden](https://ub.meduniwien.ac.at/fileadmin/content/OE/ub/dokumente/Leitfaden_Studierende_Hochschulschriften_MedUni_Wien.pdf) provided by the Medical University of Vienna. Even though the template automates most necessary things, allowing you to concentrate on writing the thesis, I still recommend skimming the document before starting to get an overview of all requirements. # Known issues 1. The template is only designed around writing the thesis is English. This can be easily changed by editing the template. If you do the work a PR would be appreciated :). # Acknowledgement Medical University of Vienna logo from: <https://commons.wikimedia.org/wiki/File:Meduni-wien.svg>

https://github.com/SkiFire13/master-thesis

https://raw.githubusercontent.com/SkiFire13/master-thesis/master/chapters/2-background.typ

typst

#import "../config/common.typ": * = Background <section-background> In this chapter we give an overview of the theoretical background used in the rest of this thesis. We will first recap some basic notions on order theory with special focus on (complete) lattices. Then we will define what a system of fixpoint equations over complete lattices is and what is its solution, along with a number of related concepts in order theory. We will then give a brief introduction to parity games and describe how to characterize the solution of a system of fixpoint equations using a parity game, with some care for efficiency issues. Finally we will introduce two algorithms used to solve parity games which we will be exploiting later on. #include "./background/1-lattices.typ" #include "./background/2-tuples.typ" #include "./background/3-fixpoint-system.typ" #include "./background/4-applications.typ" #include "./background/5-parity-games.typ" #include "./background/6-game-characterization.typ" #include "./background/7-strategy-improvement.typ"

https://github.com/jgm/typst-hs

https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/compiler/import-14.typ

typst

Other

// Unresolved import. // Error: 23-35 unresolved import #import "module.typ": non_existing

https://github.com/GartmannPit/Praxisprojekt-II

https://raw.githubusercontent.com/GartmannPit/Praxisprojekt-II/main/Praxisprojekt%20II/PVA-Templates-typst-pva-2.0/appendix.typ

typst

#let showAppendix() = [ = Anhang // change heading for other languages // Write your appendix after the heading within the block Add your appendix in _appendix.typ_ \ // delete me #lorem(200) // delete me ]

https://github.com/elteammate/typst-compiler

https://raw.githubusercontent.com/elteammate/typst-compiler/main/src/typesystem-parser.typ

typst

#import "typesystem-lexer.typ": * #import "typesystem-def.typ": * #let typesystem_parse = { (tokens) => { let token_mapping = if type(ts_token_kind.punc_colon) == "string" {let res = (:); res.insert(ts_token_kind.punc_colon, 0); res.insert(ts_token_kind.ty_array, 1); res.insert(ts_token_kind.ident, 2); res.insert(ts_token_kind.ty_bool, 3); res.insert(ts_token_kind.punc_gt, 4); res.insert(ts_token_kind.ty_int, 5); res.insert(ts_token_kind.ty_arguments, 6); res.insert(ts_token_kind.punc_comma, 7); res.insert(ts_token_kind.ty_none_, 8); res.insert(ts_token_kind.ty_content, 9); res.insert(ts_token_kind.ty_object, 10); res.insert(ts_token_kind.ty_dictionary, 11); res.insert(ts_token_kind.ty_any, 12); res.insert(ts_token_kind.ty_tuple, 13); res.insert(ts_token_kind.alias, 14); res.insert(ts_token_kind.punc_lt, 15); res.insert(ts_token_kind.ty_string, 16); res.insert(ts_token_kind.ty_float, 17); res.insert(ts_token_kind.ty_function, 18); res} else {let res = range(ts_token_kind.len()); res.at(ts_token_kind.punc_colon) = 0; res.at(ts_token_kind.ty_array) = 1; res.at(ts_token_kind.ident) = 2; res.at(ts_token_kind.ty_bool) = 3; res.at(ts_token_kind.punc_gt) = 4; res.at(ts_token_kind.ty_int) = 5; res.at(ts_token_kind.ty_arguments) = 6; res.at(ts_token_kind.punc_comma) = 7; res.at(ts_token_kind.ty_none_) = 8; res.at(ts_token_kind.ty_content) = 9; res.at(ts_token_kind.ty_object) = 10; res.at(ts_token_kind.ty_dictionary) = 11; res.at(ts_token_kind.ty_any) = 12; res.at(ts_token_kind.ty_tuple) = 13; res.at(ts_token_kind.alias) = 14; res.at(ts_token_kind.punc_lt) = 15; res.at(ts_token_kind.ty_string) = 16; res.at(ts_token_kind.ty_float) = 17; res.at(ts_token_kind.ty_function) = 18; res} let callbacks = (((o,_0,ts,_2)=>mk_type(o,..ts)),((t,_0,ts,_2)=>mk_type(t,..ts)),((f,_0,ret,_1,ts,_2)=>mk_type(f,ret,..ts)),(a => a),((i)=>"any"),(a => a),(a => a),((n,_,t)=>((name:n,ty:t),)),(a => a),((ts,_,t)=>ts+(t,)),((i)=>"tuple"),(()=>()),(a => a),((i)=>"object"),((i)=>"array"),((i)=>"arguments"),((i)=>"none"),((i)=>"float"),((ts,_,n,_1,t)=>ts+((name:n,ty:t),)),(a => a),((d,_1,t,_2)=>mk_type(d,t)),((i)=>"bool"),((i)=>"int"),((i)=>"content"),(a => a),((i)=>"function"),((t)=>(t,)),((a)=>panic("Not implemented (type alias resolution)")),((i)=>i),(a => a),((a,_1,t,_2)=>mk_type(a,t)),((i)=>"string"),(()=>()),(a => a),((i)=>"dictionary"),) let table = ((0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,1,0,0,0,),(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,57,0,0,0,0,0,),(0,0,0,0,-23,0,0,-23,0,0,0,0,0,0,0,0,0,0,0,-23,0,0,0,0,0,),(0,0,0,0,-27,0,0,-27,0,0,0,0,0,0,0,0,0,0,0,-27,0,0,0,0,0,),(0,0,0,0,-30,0,0,-30,0,0,0,0,0,0,0,0,0,0,0,-30,0,0,0,0,0,),(0,0,0,0,-29,0,0,-29,0,0,0,0,0,0,0,0,0,0,0,-29,0,0,0,0,0,),(0,0,0,0,-2,0,0,-2,0,0,0,0,0,0,0,0,0,0,0,-2,0,0,0,0,0,),(0,0,0,0,-16,0,0,-16,0,0,0,0,0,0,0,0,0,0,0,-16,0,0,0,0,0,),(0,0,0,0,-32,0,0,-32,0,0,0,0,0,0,0,0,0,0,0,-32,0,0,0,0,0,),(0,0,0,0,-6,0,0,-6,0,0,0,0,0,0,0,0,0,0,0,-6,0,0,0,0,0,),(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,54,0,0,0,0,0,0,0,0,0,),(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,51,0,0,0,0,0,0,0,0,0,),(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,46,0,0,0,0,0,0,0,0,0,),(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,40,0,0,0,0,0,0,0,0,0,),(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,16,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-8,0,0,-8,0,0,0,0,0,0,0,0,0,0,0,-8,0,0,0,0,0,),(0,30,19,26,-3,24,27,-3,20,21,32,28,25,31,0,0,22,23,29,0,18,0,17,0,0,),(0,0,0,0,35,0,0,36,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(33,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-7,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-19,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-12,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-18,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-13,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-31,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-14,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-20,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-10,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-21,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-25,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(-22,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,34,0,0,0,),(0,0,0,0,-28,0,0,-28,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-35,0,0,-35,0,0,0,0,0,0,0,0,0,0,0,-35,0,0,0,0,0,),(0,0,37,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(38,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,39,0,0,0,),(0,0,0,0,-17,0,0,-17,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,11,0,8,-24,6,9,-24,2,3,14,10,7,13,15,0,4,5,12,0,0,42,0,41,0,),(0,0,0,0,43,0,0,44,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-9,0,0,-9,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-34,0,0,-34,0,0,0,0,0,0,0,0,0,0,0,-34,0,0,0,0,0,),(0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,45,0,0,0,),(0,0,0,0,-26,0,0,-26,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,47,0,0,0,),(0,0,0,0,0,0,0,48,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,11,0,8,-24,6,9,-24,2,3,14,10,7,13,15,0,4,5,12,0,0,42,0,49,0,),(0,0,0,0,50,0,0,44,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-33,0,0,-33,0,0,0,0,0,0,0,0,0,0,0,-33,0,0,0,0,0,),(0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,52,0,0,0,),(0,0,0,0,53,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-5,0,0,-5,0,0,0,0,0,0,0,0,0,0,0,-5,0,0,0,0,0,),(0,11,0,8,0,6,9,0,2,3,14,10,7,13,15,0,4,5,12,0,0,55,0,0,0,),(0,0,0,0,56,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,),(0,0,0,0,-15,0,0,-15,0,0,0,0,0,0,0,0,0,0,0,-15,0,0,0,0,0,)) let arg_count = (4,4,6,1,1,1,1,3,1,3,1,0,1,1,1,1,1,1,5,1,4,1,1,1,1,1,1,1,1,1,4,1,0,1,1,) let goto_index = (21,21,21,21,20,21,21,22,21,23,20,23,21,20,20,20,20,20,22,21,21,20,20,20,24,20,23,21,20,21,21,20,22,21,20,) let cast_table = (_ => none,_=>ptype.array,x=>x.text,_=>types.bool,_ => none,_=>types.int,_=>types.arguments,_ => none,_=>types.none_,_=>types.content,_=>ptype.object,_=>ptype.dictionary,_=>types.any,_=>ptype.tuple,x=>x.text,_ => none,_=>types.string,_=>types.float,_=>ptype.function,) let stack = (0, ) let ast_stack = () let cur_token = 0 for i in range(9999) { for j in range(9999) { let state = stack.last() let terminal = if cur_token < tokens.len() { token_mapping.at(tokens.at(cur_token).kind) } else { 19 } let action = table.at(state).at(terminal) if action == 57 { assert(ast_stack.len() == 1) return ast_stack.first() } else if action > 0 { stack.push(action) ast_stack.push(cast_table.at(terminal)(tokens.at(cur_token))) cur_token += 1 } else if action < 0 { let rhs = () for _ in range(arg_count.at(action)) { let _ = stack.pop() rhs.push(ast_stack.pop()) } let rule = callbacks.at(action) ast_stack.push(rule(..rhs.rev())) let goto_state = table.at(stack.last()).at(goto_index.at(action)) if goto_state > 0 { stack.push(goto_state) } else { panic("Expected shift action") } } else { panic("Parsing error at state: " + repr(stack) + " and token: " + repr(if cur_token < tokens.len() { tokens.at(cur_token) } else {"EOF"}) + " at: " + repr(cur_token) ) } } } panic("too complex") } }

https://github.com/donghoony/KUPC-2023

https://raw.githubusercontent.com/donghoony/KUPC-2023/master/description.typ

typst

#import "colors.typ" : KUPC_GREEN, PALE_RED #import "problem_info.typ" : constructTitle #import "problems.typ" : contest_problems #import "emoji/lib.typ" : * #let mono(s, color: black) = {text(font: "Inconsolata", fill: color)[#s]} #let bf(s) = {text(weight: "bold")[#s]} // 줄바꿈은 #linebreak()를 중간에 넣으면 됩니다. // 페이지 넘김은 문제 내부에서 ()를 새로 만들어 주세요. // monospace 문자열은 #mono("abc")와 같이 쓸 수 있습니다. // 미리 정의되지 않은 operation의 경우에는 #math.op("MEX")와 같이 쓰면 됩니다. 대부분은 정의돼 있으니 그냥 $cos$ 와 같이 쓰면 됩니다. // 시간복잡도는 $cal(O)(N log N)$ 와 같이 써주세요. log가 자체적으로 함수가 있습니다. // 이모지는 그냥 넣으면 됩니다. 지원하지 않는 이모지는 깨집니다. 이모지는 https://github.com/polazarus/typst-svg-emoji 를 사용했습니다. // 그래프도 작성 가능합니다. https://www.graphviz.org/docs/graph/ 를 참고해서 ```render <여기에 그래프를 작성하세요>``` #let cell(num, color: black, fill: none, stroke: none) = { rect( height: 50pt, width: 50pt, stroke: stroke, fill: fill, )[#align(center + horizon)[#text(fill: color)[#num]]]; } #let descriptions = ( // 2A ( ( [- 주어진 문자열에서 $1$이 연속으로 등장하는 구간을 알아내는 방법은 무엇일까요?], [- 여러 가지 방법이 있지만, 구현이 간단한 방법을 소개합니다.], [- #mono(1) 이후에 #mono(0) 이 등장한다면, 검은 줄의 구간이 끝나는 것을 의미합니다.], [- 주어진 문자열을 훑으면서, #mono("s[i] = 1")이면서 #mono("s[i+1] = 0")인 #mono("i")의 개수를 셉시다.], [- 편의를 위해, 주어진 문자열에 #mono(0) 을 추가해도 답은 같습니다.], [- 시간복잡도는 $cal(O)(N)$ 입니다.], ), ), // 2B/1A ( ( [- 건덕이와 건구스는 자신의 차례에 #bf("반드시") 움직여야 합니다.], [- 건덕이와 건구스는 다음과 같은 네 가지 종류로 움직입니다: #mono("RR"), #mono("RL"), #mono("LL"), #mono("LR")], [- 어떻게 움직이더라도 두 플레이어 사이 간격의 #bf("홀짝성")#sub(mono("parity"))은 변하지 않습니다.], ), ( [- 전장의 길이가 주어지는 순간 승자는 결정되며, 승자는 상대를 향해 전진합니다.], [- 패자의 경우 앞으로 가면 패배로 더 빠르게 도달하므로, 가능하다면 뒤로 후퇴합니다.], [- 전장을 벗어나도록 이동할 수 없으므로, #bf("두 플레이어간 간격은 단조감소")합니다.], ), ( [- 간격을 짝수로 만드는 사람은 결국 두 플레이어 사이의 거리를 $0$으로 만듭니다.], [- 플레이어 사이의 거리가 $0$이 되는 순간, 다음 차례의 플레이어는 공격할 수 있습니다.], [- 주어진 전장의 길이가 짝수라면, 선공이 먼저 둘 사이의 간격을 홀수로 만듭니다.], [- 홀수라면, 선공이 둘 사이의 간격을 짝수로 만듭니다.], [- 따라서 전장의 길이가 짝수인 경우, 건덕이가 승리합니다.], [- 주어진 전장의 길이의 홀짝성을 판단하므로 시간복잡도는 $cal(O)(1)$입니다.], [],[], [#emoji.arm.muscle $N times N$ 전장에서 각 플레이어가 $(1,1)$, $(N,N)$에서 시작하면 누가 승리할까요?] ), ), // 2C ( ( [- 바닥수 $N$이 주어졌을 때, 바닥수가 $N$인 길이 $L$의 원래 수를 구해야 합니다.], [- 길이 $L$의 모든 수에 대해서 바닥수가 되는지 확인하는 것은 오래 걸립니다.], [- 곱셈의 성질을 활용해서 쉽게 문제를 해결해 봅시다.] ), ( [- 곱셈의 항등원을 이용합시다. $1$은 여러 번 곱해도 $1$입니다.], [- $1$을 $L-1$번 적은 뒤, $N$를 뒤에 덧붙인 수의 바닥수는 $N$입니다.], [- 원래 수는 $0$으로 시작하지 않음에 주의합시다.] ), ), // 2D/1B ( ( [- 학생의 키가 같은 쌍이 존재하지 않는다면, 모든 학생이 참여할 수 있습니다.], [- 모두가 한 쪽을 바라보고, 바라보는 방향으로 키가 작아지도록 줄세우면 됩니다.], [- 학생의 키가 같은 쌍이 존재하는 경우에는 몇 명까지 참여할 수 있을까요?], ), ( [- 두 명의 키가 서로 같다면, 서로 반대 방향을 보고 서면 됩니다.], [- 세 명의 키가 서로 같다면, 한 명은 어느 쪽을 보든 점프할 타이밍을 놓칠 수 있습니다.], [- 따라서 키가 같은 사람은 #bf("최대 두 명")까지만 참여할 수 있습니다.], [- 줄을 세울 수 있는지의 여부를 물어보았으므로, 각 사람들의 키에 대해서 참여할 수 있는 사람의 수를 세어 줍시다.], [- 시간복잡도는 $cal(O)(N)$입니다.], [],[], [#emoji.arm.muscle 올바르게 줄 세우는 방법 중 하나를 구할 수 있을까요?] ), ), // 2E ( ( [- 왼쪽 절반의 문자와 오른쪽 절반의 문자의 교환을 통해 팰린드롬을 만들어 봅시다.], [- 연산을 통해 임의의 위치에 존재하는 두 문자를 교환할 수 있다면, #linebreak() 모든 문자를 원하는 곳에 배치할 수 있습니다. 과연 가능할까요?#v(0.5em)], [- 편의상 배열의 길이가 짝수라고 가정합시다. 교환하는 경우의 수는 두 가지입니다. ], [#h(2em) #emoji.ast 두 문자가 서로 다른 절반에 속하는 경우], [#h(2em) #emoji.ast 두 문자가 같은 절반에 속하는 경우], ), ( [- 두 문자가 서로 다른 절반에 속하는 경우, 두 원소를 직접 교환할 수 있습니다.], [- 두 문자가 서로 같은 절반에 속하는 경우에는 두 원소를 직접 교환할 수 없습니다.], [- 이때, 반대쪽 절반에 속하는 원소 한 개를 임시로 사용해 교환할 수 있습니다.], [ #let cell(num, color: black, fill: none) = { rect( height: 50pt, width: 50pt, stroke: none, fill: fill, )[#align(center + horizon)[#text(fill: color)[#num]]]; } #set table(align: center, stroke: gray + 0.5pt, inset: 0pt, columns: 4) #align(center)[ #table( )[#cell(1, color: blue, fill:rgb("ddd"))][#cell(2)][#cell(3, color: PALE_RED)][#cell(4, color: PALE_RED, fill: rgb("ddd"))] ] #v(-0.8em) #align(center)[ #table( )[#cell(4, color: PALE_RED, fill:rgb("ddd"))][#cell(2)][#cell(3, color: PALE_RED, fill:rgb("ddd"))][#cell(1, color: blue)] ] #v(-0.8em) #align(center)[ #table( )[#cell(3, color: PALE_RED, fill:rgb("ddd"))][#cell(2)][#cell(4, color: PALE_RED)][#cell(1, color: blue, fill:rgb("ddd"))] ] #v(-0.8em) #align(center)[ #table( )[#cell(1, color: blue)][#cell(2)][#cell(4, color: PALE_RED)][#cell(3, color: PALE_RED)] ] ] ), ( [- 문자열의 길이가 홀수인 경우에는 가운데 글자를 교환할 수 없음에 유의합시다.], [- 해당 문자를 세지 않고, 짝수 길이의 문자열이라고 생각하면 됩니다.], [- 팰린드롬이 되기 위해서는 양 쪽에 같은 수의 문자가 존재해야 합니다.], [- 알파벳이 문자열에 짝수번 등장한다면 양 쪽에 골고루 분배할 수 있습니다.], [- 시간복잡도는 $cal(O)(N)$입니다.], ), ), ( ( [- 두 개의 말뚝과 깃대 하나를 골라 만들 수 있는 삼각형 넓이 중 최댓값을 구합시다.], [- 단, 넓이가 $R$ 이하여야 합니다.], [- 간단한 방법은 모든 말뚝 쌍에 대해서, 모든 깃대를 탐색하는 방법입니다.], [- 해당 풀이는 $cal(O)(N^2M)$으로, 제한 시간 안에 해결할 수 없습니다.], ), ( [- 만약 밑변을 고정한다면, 주어진 깃대를 정렬한 뒤 이분탐색을 사용할 수 있습니다.], [- 가능한 모든 밑변을 구하는 데에는 $cal(O)(N^2)$입니다.], [- 이분 탐색에 깃대를 활용하기 위해 깃대를 정렬하는 데 $cal(O)(M log M)$입니다.], [- 하나의 밑변 길이에 대해서, $R$ 이하인 최대 넓이를 구하는 데 $cal(O)(log M)$이 듭니다.], [- $cal(O)(M log M + N^2 log M)$ 은 충분히 통과할 수 있습니다.] ), ), ( ( [- 스위치를 누르면 $3$초동안 두 배의 점수를 얻을 수 있습니다.], [- 음수의 점수도 두 배로 계산됨에 유의해야 합니다.], [- 그리디하게 스위치를 누르는 전략은 오답을 받습니다.], [- 스위치를 누른 뒤, $i$초 시점에 스위치의 효과가 $j$초 남았다는 것을 활용합시다.], ), ( [- $i$초에 $j=0$이라면, $i-1$초에 $j=0$이거나 $j=1$인 경우 중 최댓값을 가져옵시다.], [- $j=1$이라면 아직 스위치의 효과가 남아있으므로 $i-1$초의 $j=2$를 가져옵니다.], [- $j=2$인 경우도 마찬가지로 $i-1$초의 $j=3$을 가져옵니다.], [- $j=3$은 지금 스위치를 누른 상황으로, $i-1$초의 $j=0$이거나 $j=1$인 경우에서 가져올 수 있습니다.], [- 스위치의 효과가 적용되는 경우에는 배열의 값의 두배를 더해줍시다.], [- 시간복잡도는 $cal(O)(N)$입니다.] ) ), ( ( [ #let cell(num, color: black, fill: none, stroke: none) = { rect( height: 50pt, width: 50pt, stroke: stroke, fill: fill, )[#align(center + horizon)[#text(fill: color)[#num]]]; } #set table(align: center, stroke: rgb("add8e6") + 1pt, inset: 0pt, columns: 5) #align(center)[ #table()[#cell(1)][#cell(2)][#cell(3, stroke: red)][#cell(4)][#cell(5)][#cell(5)][#cell(2)][#cell(1, stroke: red)][#cell(4)][#cell(6)][#cell(0)][#cell(2)][#cell(4,fill: rgb("ddd"), stroke: red)][#cell(2)][#cell(1)][#cell(0)][#cell(0)][#cell(2)][#cell(1)][#cell(7)] ] ], [- 일감호입니다. #emoji.fishing], [- 무게추 $3$을 달아 $3$의 힘으로 낚싯대를 휘두르면 회색 칸에 찌가 도달합니다.], [- 빨간 색 테두리에 해당하는 칸에 존재하는 $8$마리의 물고기가 사로잡힙니다.] ), ( [ #set table(align: center, stroke: rgb("add8e6") + 1pt, inset: 0pt, columns: 5) #align(center)[ #table()[#cell(1)][#cell(2, stroke: red)][#cell(3, fill: rgb("add8e6"))][#cell(4)][#cell(5)][#cell(5)][#cell(2, fill: rgb("ddd"), stroke: red)][#cell(1, fill: rgb("add8e6"))][#cell(4)][#cell(6)][#cell(0)][#cell(2)][#cell(4, fill: rgb("add8e6"))][#cell(2)][#cell(1)][#cell(0)][#cell(0)][#cell(2)][#cell(1)][#cell(7)] ] ], [- 낚싯줄을 한 바퀴 감아올리면 찌의 위치가 바뀝니다.], [- 바뀐 칸에서 물고기 $4$마리가 사로잡힙니다.], [- 이를 미끼가 일감호를 벗어날 때까지 반복합니다.], ), ( [ #let cell(num, color: black, fill: none, stroke: none) = { rect( height: 50pt, width: 50pt, stroke: stroke, fill: fill, )[#align(center + horizon)[#text(fill: color)[#num]]]; } #set table(align: center, stroke: rgb("add8e6") + 1pt, inset: 0pt, columns: 5) #align(center)[ #table()[#cell(1, fill: rgb("add8e6"))][#cell(2, fill: rgb("add8e6"))][#cell(3, fill: rgb("add8e6"))][#cell(4)][#cell(5)][#cell(5)][#cell(2, fill: rgb("add8e6"))][#cell(1, fill: rgb("add8e6"))][#cell(4)][#cell(6)][#cell(0)][#cell(2)][#cell(4, fill: rgb("add8e6"))][#cell(2)][#cell(1)][#cell(0)][#cell(0)][#cell(2)][#cell(1)][#cell(7)] ] ], [- 쿼리가 주어질 때마다 다음과 같은 계단 모양에 칠해진 수의 합을 구하는 문제입니다.], [- 간단하게 생각하면, 매 쿼리마다 계산하는 $cal(O)(N M Q)$ 풀이가 있습니다 #emoji.face.explode.], [- 이는 시간 초과입니다. 어떻게 빠르게 해결할 수 있을까요?] ), ( [- 여러 방법 중 하나를 소개합니다.], [- 2차원 배열에서 직사각형 모양의 부분합은 $cal(O)(N M + Q)$에 구할 수 있습니다.], [- 쿼리가 직사각형의 부분합을 구하게끔 일감호를 재배치할 수 있을까요?], ), ( [ #set table(align: center, stroke: rgb("add8e6") + 1pt, inset: 0pt, columns: 3) #align(center)[ #grid(columns: 3)[ #align(center + horizon)[ #table()[#cell(1, fill: rgb("add8e6"))][#cell(2,fill: rgb("add8e6"))][#cell(3,fill: rgb("add8e6"))][#cell(5)][#cell(2,fill: rgb("add8e6"))][#cell(1,fill: rgb("add8e6"))][#cell(0)][#cell(2)][#cell(4,fill: rgb("add8e6"))]] ][#align(center + horizon)[#h(3em)->#h(3em)]][ #align(center + horizon)[ #table()[#cell(0, fill: rgb("add8e6"))][#cell(0, fill: rgb("add8e6"))][#cell(3, fill: rgb("add8e6"), stroke: PALE_RED)][#cell(0,fill: rgb("add8e6"))][#cell(2,fill: rgb("add8e6"), stroke: PALE_RED)][#cell(1,fill: rgb("add8e6"), stroke: PALE_RED)][#cell(1, fill: rgb("add8e6"), stroke: PALE_RED)][#cell(2, fill: rgb("add8e6"), stroke: PALE_RED)][#cell(4,fill: rgb("add8e6"), stroke: PALE_RED)][#cell(5, stroke: PALE_RED)][#cell(2, stroke: PALE_RED)][#cell(0)][#cell(0, stroke: PALE_RED)][#cell(0)][#cell(0)]] ] ] ], [- 가능합니다.], [- 시간복잡도는 2차원 부분합과 같은 $cal(O)(N M + Q)$입니다.], ), ( [- 아래쪽으로 누적합을 구해준 뒤, 오른쪽 아래 대각선 방향으로 누적합을 구해나가도 정답을 구할 수 있습니다.], [],[], [ #let c1 = rgb(240,229,235); #let c2 = rgb(212,236,220) #let c3 = rgb(252,249,218) #let c4 = rgb(252,223,215) #set table(align: center, stroke: rgb("add8e6") + 1pt, inset: 0pt, columns: 3) #align(center)[ #grid(columns: 5)[ #align(center + horizon)[ #table()[#cell(1, fill: c1)][#cell(2,fill: c2)][#cell(3,fill: c3)][#cell(5, fill: c1)][#cell(2,fill: c2)][#cell(1,fill: c3)][#cell(0, fill: c1)][#cell(2, fill: c2)][#cell(4,fill: c3)]] ][#align(center + horizon)[#h(1em)->#h(1em)]][#align(center + horizon)[ #table()[#cell(1, fill: c1)][#cell(2,fill:c2)][#cell(3)][#cell(6,fill:c3)][#cell(4,fill: c1)][#cell(4,fill: c2)][#cell(6)][#cell(6, fill:c3)][#cell(8,fill: c1)]]][#align(center + horizon)[#h(1em)->#h(1em)]][ #table()[#cell(1)][#cell(2)][#cell(3)][#cell(6)][#cell(5)][#cell(6)][#cell(6)][#cell(12)][#cell(13)] ] ] ], ), ), ( ( [- 주어진 등굣길을 그래프로 모델링해 봅시다.], [- 각 칸에 쓰인 수만큼 양쪽 방향으로 떨어진 칸으로 향하는 간선을 만들 수 있습니다.], [- 최대 두 번 방향을 반전할 수 있으므로, 같은 칸이라도 $0$번 반전했을 때와 $2$번 반전했을 때는 다르게 생각해야 합니다.], ), ( [- 정점을 세 개로 나누어 아래와 같은 그래프 모형을 생각할 수 있습니다.], [#align(center)[#image("images/time.png", width: 70%)]] ), ( [- 문제의 정답은 $1$번 정점에서 학교에 해당하는 정점들까지의 최장거리와 같습니다.], [- 구축한 그래프는 사이클 없는 방향 그래프입니다.], [- $1$번 정점에서 위상정렬을 시행한 뒤, 정렬한 순서대로 정답을 갱신해나가면 됩니다.], [- 학교에 처음 도착한 시간을 늦추는 것에 유의합시다. 학교에서 되돌아가는 간선은 존재할 수 없습니다.], [- 시간복잡도는 $cal(O)(N)$입니다.], ), ( [- 더 쉬운 풀이로, 단순하게 오른쪽으로 한 번, 왼쪽으로 한 번, 다시 오른쪽으로 한 번 DP를 훑어도 해결할 수 있습니다.], [- 아직 나에게 오는 경로가 없거나 이미 학교에 도착한 경우, 칸에 숫자가 $0$이 쓰인 경우 등을 잘 처리해야 합니다.], [- 시간복잡도는 위와 같은 $cal(O)(N)$이지만, 상수가 작아 더 빠르게 통과합니다.], ) ), ( ( [- 주어진 문자열을 하나의 집합으로 생각합니다.], [- 예를 들어 문자열이 #mono("0011223344")라면 대응되는 집합은 ${0,1,2,3,4}$입니다.], [- 숫자는 $10$개 이므로 만들어질 수 있는 집합의 개수는 $2^10 = 1024$개입니다.], [- 이제 집합 두 개를 합쳐 크기가 $K$인 집합의 개수를 세면 됩니다.], ), ( [- 그럼 크기가 $K$인 집합의 개수는 어떻게 셀까요?], [- 만들어질 수 있는 $1024$개의 집합 중 $A union B$의 원소의 개수가 $K$개인 두 집합 $A$, $B$를 순서를 고려하지 않고 뽑습니다.], [- $A=B$일 때는 정답에 $(#math.op("cnt")$A$ times (#math.op("cnt") (A) - 1))/2$를 더합니다.], [- $A!=B$일 때는 정답에 $#math.op("cnt") (A) times #math.op("cnt") (B)$를 더합니다.], [- 시간복잡도는 $cal(O)(N + 1024^2 times 10)$입니다.], [- 비트집합을 사용하면 $cal(O)(N + 1024^2)$으로 줄일 수 있지만, 사용하지 않아도 됩니다.] ), ), ( ( [- 우선 언제 #mono("NO", color:PALE_RED) 를 출력해야할 지 생각해봅니다.], [- #math.op("MEX")의 정의에 의해, 길이가 $N$인 수열의 #math.op("MEX")는 $N+1$보다 클 수 없습니다.], [- $A$는 순열이기 때문에 $B_N = N+1$입니다.], [- $S_i = {A_1, A_2, dots, A_i}$라고 할 때, $S_i subset S_(i+1)$이므로 #linebreak() $B_i > B_(i+1) (1 <= i <= N-1)$일 수 없습니다. 즉 $B$는 #bf("단조증가")합니다.], [- 위 세 가지 규칙을 지키지 않는다면 #mono("NO", color:PALE_RED)를 출력해야 합니다.], [- 그 이외의 경우에는 모두 가능할까요?], ), ( [- 몇 가지 관찰을 해 봅시다.], [- 만약 $A_i = K$라면, $i<=j$일 때 $B_j = K$일 수 없습니다.], [- $K$를 포함하는 집합의 #math.op("MEX")는 정의상 $K$일 수 없기 때문입니다.], [- 결국 $B_j = K$일 때, $A_i = K (j<i)$입니다.] ), ( [- $B_i != B_(i+1) (1 <= i <= N-1)$인 경우에 주목합니다.], [- #math.op("MEX")값이 $B_(i+1)$이기 위해서는 해당 집합이 $B_i$를 가지고 있어야 합니다.], [- 이전 관찰과 종합하면 $B_i != B_(i+1)$일 때, $A_(i+1) = B_i$가 됨을 알 수 있습니다.], ), ( [- 이제 남은 값을 배치하는 방법은 여러 가지가 있습니다.], [- 그 중 하나는 작은 수부터 빈 곳에 채워 넣는 것입니다.], [- 왜 이 방법이 가능할까요?], ), ( [- 채워야하는 곳은 $B_(i-1) = B_i$인 $A_i$입니다.], [- 해당 방법으로 빈 곳을 배치하면 $B_i < A_i$입니다.], [- #math.op("MEX")값이 $B_i$인 배열에 $B_i$보다 큰 값을 배치하더라도 #math.op("MEX")값은 변하지 않습니다.], [- 그리고 $B_(k-1) != B_k$인 $k$에 대해 총 $k$개의 공간에 $B_k$보다 작은 $B_k - 1 (<= k)$개의 수를 배치하므로 작은 수부터 배치하면 원하는 결과를 얻을 수 있습니다.], ), ( [- $B_i != B_(i+1)$일 때, $A_(i+1) = B_i$ #v(1em)], [#align(center)[#table(columns:5)[#cell(1)][#cell(2)][#cell(2)][#cell(4)][#cell(6)]]], [#v(-0.5em) $ B $ ], [#align(center)[#table(columns:5)[#cell("?")][#cell(1, color:PALE_RED)][#cell("?")][#cell(2, color:PALE_RED)][#cell(4, color:PALE_RED)]]], [#v(-0.5em) $ A $ ], ), ( [- 남은 값을 작은 수부터 빈 곳에 채워 넣습니다. #v(1em)], [#align(center)[#table(columns:5)[#cell(1)][#cell(2)][#cell(2)][#cell(4)][#cell(6)]]], [#v(-0.5em) $ B $ ], [#align(center)[#table(columns:5)[#cell(3, color:PALE_RED)][#cell(1)][#cell(5, color:PALE_RED)][#cell(2)][#cell(4)]]], [#v(-0.5em) $ A $ ], [- 시간복잡도는 $cal(O)(N)$입니다.], ) ), ( ( [- 하나의 등불이 다른 등불을 앞지르는 횟수가 곧 소원을 비는 횟수입니다.], [- 상대적으로 오른쪽의 등불이 상승하는 정도가 더 크면, #linebreak()왼쪽의 등불을 앞지르는 순간이 존재합니다.], [- 길이가 $S$인 구간에서 $i<j$ 이면서 $A_i < A_j$인 $(i, j)$ 쌍의 개수를 구해야 합니다.], ), ( [- 구간 내 원소에 대해서 모든 쌍을 찾는 방법을 먼저 생각해 봅시다.], [- 하나의 구간에 대해 $cal(O)(N^2)$이 걸리므로, 모든 구간에서 계산한다면 $cal(O)(N^3)$으로 제한시간 내에 풀 수 없습니다.], [- 앞지르는 쌍의 개수를 더 빠르게 구할 수 있을까요?], ), ( [- Merge Sort를 응용하면 문제를 조금 더 효율적으로 풀 수 있습니다.], [- 두 배열을 합치는 과정에서, 양쪽 배열은 정렬된 상태입니다.], [- 왼쪽 절반 배열의 원소가 더 작다면, 해당 원소보다 뒤에 존재하는 원소 모두가 오른쪽에서 고른 원소보다 작습니다.], [- 하나의 구간에서 모든 쌍의 개수를 $cal(O)(N log N)$에 구할 수 있습니다.], [- 여전히 모든 구간을 확인하는 데 $cal(O)(N^2 log N)$으로, 더 빠른 방법이 필요합니다.], ), ( [- 문제의 특성을 파악해서 빠르게 풀어 봅시다.], [- $k$번 등불부터 $S$개 고른 구간을 $T_k$라고 합시다.], [- $T_k$ 구간과 $T_(k+1)$ 구간의 공통된 구간은 $[k+1, k+S-1]$입니다.], [- 공통된 구간에서 발생하는 앞지르는 쌍의 개수는 $k$번째 등불과 $k+S$번째 등불에 영향받지 않습니다.], [- 해당 구간의 연산을 매 구간마다 반복하지 않으면 효율적으로 풀 수 있습니다.], ), ( [- $T_k $구간에서 다음 구간으로 넘어갈 때 아래의 값들을 빼고 더해줘야 합니다.], [#h(2em) #emoji.ast $k$ 번째 등불을 제거했을 때 감소하는 앞지르는 쌍의 개수], [#h(2em) #emoji.ast $k+S$ 번째 등불을 추가할 때 증가하는 앞지르는 쌍의 개수], [], [- 세그먼트 트리를 활용하면 앞지르는 쌍의 개수를 빠르게 구할 수 있습니다.], [#h(2em)#emoji.ast $k$번째 등불을 제거할 때, $A_k$보다 상승하는 정도가 큰 등불의 개수], [#h(2em)#emoji.ast $k+S$번째 등불을 추가할 때, $A_(k+S)$보다 상승하는 정도가 작은 등불의 개수] ), ( [- 주어진 등불 상승 속도의 범위가 최대 $10^9$이므로 좌표압축을 먼저 진행해야 합니다.], [- 시간복잡도는 $cal(O)(N log N)$입니다.], ), ), ( ( [- BFS의 순회 순서대로 정점에 번호를 붙이겠습니다. 정점을 적절히 배치해서 순회 순서의 차를 최대화해야 합니다.], [- 루트 바로 아래에 $k$개의 정점이 존재한다고 합시다.], [- 트리의 $i(1 <= i <= N)$번 정점을 루트로 하는 서브트리의 크기를 $S_i$라고 합시다.], [- $sum_(i=2)^(k+1) S_i$ 는 루트 노드를 제외한 트리의 크기이므로 $N-1$입니다.] ), ( [- $D_2 = B_2 = 2$이고, $2 < i <= k+1$인 $i$에 대한 $D_i$와 $B_i$값은 아래와 같습니다.], [#h(2em) #emoji.ast $D_i = sum_(j=2)^(i-1)S_j + 2$], [#h(2em) #emoji.ast $B_i = i$#v(1em)], [- $sum_(i=2)^(k+1)|D_i-B_i|=&|2-2|+|S_2+2-3|+|S_2+S_3+2-4|+dots#linebreak() & + |sum^(k)_(j=2)S_j+2-(k+1)| #linebreak() &=|S_2-1| + |S_2+S_3-2| + dots + |sum^(k)_(j=2)S_j-(k-1)|$], [- 여기에서 서브트리의 크기는 1 이상이므로, 절댓값 내부의 식을 빼내서 식을 정리할 수 있습니다.] ), ( [- $sum_(i=2)^(k+1)|D_i-B_i| &= (S_2 - 1) + (S_2+S_3-2) + dots + (sum_(j=2)^(k)S_j-(k-1))#linebreak() &= (k-1)S_2 + (k-2)S_3 + ... + S_k - sum_(i=1)^(k-1)\i$], [- 위 식이 최대가 되도록 $S_i$를 배분하는 것은 $S_2=N-k$, 나머지는 $1$로 두는 것입니다.], [- $2$번 정점을 루트로 하는 서브트리에서 노드를 배치하는 형태는 $sum_(i=1)^N|D_i-B_i|$를 최대화하는 데 영향을 미치지 않습니다.], [- 따라서 $k$값 중, 문제의 정답을 최대화하는 값을 찾아야 합니다.], ), ( [#v(-2em)#align(center)[#image("images/bfsdfs.png", width: 50%)]], [- 순회 순서의 차를 보면, $N-k-1$개에 해당하는 #math.op("BFS") 순서는 $k+1$씩 밀려납니다.], [- 반대로, $k-1$개의 #math.op("DFS") 순서는 $N-k+1$씩 밀립니다.], [- 따라서 $(k-1) times (N-k+1) + (k+1) times (N-k-1)$의 최댓값을 구해야 합니다.] ), ( [- 식을 펼쳐 계산하면 $k=N/2$인 경우에 최대가 되며, 언급한 대로 트리를 구축해주면 됩니다.], ) ), ( ( [- 팰린드롬 애너그램에서 보였듯, $N$이 짝수인 경우 항상 가능합니다.], [- 홀수인 경우에는 한가운데의 원소가 자신의 자리에 있을 때에만 가능합니다.], [- 어떻게 최소한의 연산으로 원하는 배열을 만들 수 있을까요?] ), ( [- 다음과 같은 배열을 봅시다.], [#align(center)[#table(columns:6)[#cell(2, color:PALE_RED)][#cell(1,color:PALE_RED)][#cell(5,color:blue)][#cell(3,color:blue)][#cell(6,color:blue)][#cell(4,color:blue)]]], [- 이 배열이 정렬되기 위해서는 어떻게 해야 할 지 생각해 봅시다.], [- 각 원소가 자신의 자리로 돌아가기 위해서, 현재 적혀있는 수에 해당하는 칸으로 간선을 그어 봅시다.], [- 배열에서 사이클을 분리해낼 수 있습니다. 사이클의 종류는 총 세 가지입니다.], [- 왼쪽 절반 또는 오른쪽 절반으로만 이루어져 있는 사이클, 교차하는 사이클], [- 각각을 $L$, $R$, $M$ 사이클이라고 정의합시다.] ), ( [- $L$ 사이클과 $R$ 사이클은 그 자체로 교환이 불가합니다.], [- $M$ 사이클로 만들어서 정렬하는 과정이 필요합니다.], [- $1$회 교환을 통해서 원소가 한 개 늘어난 $M$ 사이클을 만들 수 있습니다.], [- 이때, $L$과 $R$ 사이클이 둘 다 존재한다면, 두 사이클의 원소를 서로 교환해서 하나의 $M$ 사이클을 만들 수 있습니다.], [- 이는 각 사이클을 $M$ 사이클로 만들었을 때보다 교환을 $1$번 적게 할 수 있으므로, 가능하다면 $L$, $R$ 사이클 간 교환을 통해 $M$ 사이클을 만드는 것이 이득입니다.], ), ( [- 이제 $M$ 사이클을 올바르게 배치하는 데 드는 비용을 알아봅시다.], [- 크기가 $2$인 $M$ 사이클을 정렬하는 데에는 $1$번의 교환이면 충분합니다.], [- 크기가 $i$인 $M$ 사이클을 정렬하는 데 걸리는 비용이 $k$라고 합시다.], [- 크기가 $i+1$인 $M$ 사이클을 정렬할 때, 세 개 이상의 원소를 동시에 교환할 수 있는 방법이 없으므로 $1$번의 교환을 통해 크기가 $i$인 $M$ 사이클을 만드는 것이 최선입니다.], [- $i+1$ 개의 원소를 가지는 $M$ 사이클은 $k+1$번 교환을 통해 정렬할 수 있습니다.], [- $i=2$일 때 $k=1$이므로, 크기가 $N$인 $M$ 사이클은 $N-1$번의 교환을 통해 정렬할 수 있습니다.] ), ( [- 따라서 $L$, $R$, $M$ 사이클의 개수를 각각 구해준 뒤, $L$, $R$ 사이클 쌍들을 $M$ 사이클들로 만들어 줍시다.], [- 남는 $L$, $R$ 사이클은 크기가 $1$ 증가한 $M$ 사이클이 됩니다.], [- 기존 $M$ 사이클은 각 사이클의 원소의 개수보다 $1$ 작은 횟수로 정렬이 가능합니다. ], [- $M$ 사이클을 구성하는 왼쪽/오른쪽 원소의 개수가 같은 쪽으로 가도록 교환하면 $M$ 사이클을 유지하면서 크기를 $1$씩 줄여나갈 수 있습니다.], [- 사이클의 개수, 종류, 구성하는 원소 등을 조합해 위에서 설명한 대로 답을 도출할 수 있습니다.], [- 시간복잡도는 $cal(O)(N)$입니다.] ) ), ) #let create_page(index) = { set list(marker: text(fill:KUPC_GREEN)[✓]) for pg in descriptions.at(index) { [ #constructTitle(contest_problems.at(index), size: 2em, bookmark:false) #v(5em) ] for desc in pg { set text(size: 2em) pad(left: -1.5em)[#desc] v(0em); } pagebreak(weak: true) } }

https://github.com/connachermurphy/typst-cv

https://raw.githubusercontent.com/connachermurphy/typst-cv/main/works_in_progress.typ

typst

MIT License

#let items = ( [#quote[A project in progress, sure to revolutionize the field.]], ) #list(..items)

https://github.com/goshakowska/Typstdiff

https://raw.githubusercontent.com/goshakowska/Typstdiff/main/tests/test_working_types/super_script/super_script_inserted.typ

typst

First#super[super text] Normal text#super[inserted super text] Second#super[super text]

https://github.com/Ciolv/typst-template-bachelor-thesis

https://raw.githubusercontent.com/Ciolv/typst-template-bachelor-thesis/main/chapter.typ

typst

// Has to be imported wherever acronyms are used #import "acronyms.typ": ac,acl,acs,acsp,acp,aclp,acronyms = This chapter is included from another file <chap:useful-guides> Let's see in @lst:hellew-wörld how nicely syntax highlighting works in #link("https://typst.app", "Typst").\ I really like it! #figure( caption: "Rust sample", )[ #set par(leading: 0.75em) #set align(left) // We really don't want our code to be centered line by line... ```rust fn main() { println!("Hello, world!"); } ``` ]<lst:hellew-wörld> Awesome, isn't it? \ Well, if you're coming from a nice Markdown editor, there is nothing new with it, but in comparison to L#super(size: 0.8em,baseline: -0.2em)[A]T#sub(size: 0.8em, baseline: 0.2em)[E]X? No extra package required! \ \ By the way, do you like @harry? = Test <chap:test>

https://github.com/kaarmu/splash

https://raw.githubusercontent.com/kaarmu/splash/main/src/palettes/okabe-ito.typ

typst

MIT License

/* Color scheme by <NAME> and <NAME> * * Source: https://jfly.uni-koeln.de/color/ * Accessed: 2023-06-16 */ #let okabe-ito = ( black : rgb("#000000"), orange : rgb("#E69F00"), sky-blue : rgb("#56B4E9"), bluish-green : rgb("#009E73"), yellow : rgb("#F0E442"), blue : rgb("#0072B2"), vermilion : rgb("#D55E00"), reddish-purple : rgb("#CC79A7"), )

https://github.com/jgm/typst-hs

https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/meta/footnote-01.typ

typst

Other

// Test space collapsing before footnote. A#footnote[A] \ A #footnote[A]

https://github.com/typst/webapp-issues

https://raw.githubusercontent.com/typst/webapp-issues/main/README.md

markdown

# Web App Issues Official issue tracker for [Typst's web app.][app] Here, you can report bugs or send feature requests for the official web app. If you want to report a bug with the Typst language or compiler instead, please [open an issue here][compiler]. ## FAQ - **Will the web app remain free?** \ The web app will always have a free tier. However, there is also a paid plan with additional features like Comments and Git sync. - **Is the web app open source?** \ The web app is not open source. We think open-sourcing the compiler and keeping the app proprietary is a fair division that allows the project to sustain itself long-term. - **Can I self-host the web app?** \ Yes! We offer a paid on-premises version for organizations. If this is interesting for you or your team, please reach out to us at <<EMAIL>>. - **Will there be a desktop app?** \ We plan to provide a desktop version of the web app down the road, but can't give a timeframe for this at the moment. - **Where do you store my data / is my data safe?** \ Your data is stored in a Microsoft Azure data center in Germany and encrypted at rest. We can access your documents, but do so only to fix problems on your request or to enforce our terms of service. Read our [privacy policy] for more details. [app]: https://typst.app [compiler]: https://github.com/typst/typst [privacy policy]: https://typst.app/privacy [paid plan]: https://typst.app/pricing

https://github.com/luiswirth/numpde-slides

https://raw.githubusercontent.com/luiswirth/numpde-slides/main/src/week03.typ

typst

#import "setup.typ": * #show: this-template #let pathemph(a, b) = [ #text(fill: white.darken(60%))[#a]#b ] #titleslide("03") #pagebreak() #githubref #pagebreak() = Setting Hilbert space $V_0$ \ Symmetric Positive-Definite Bilinear form $a: V times V -> RR$ \ Continuous Linear form $l: V -> RR$ \ Gives rise to linear variational problem $ u in V_0: quad a(u, v) = l(v) quad forall v in V_0 $ Notice that here we are considering only the vector space and not the affine space. Meaning that a conversion is needed, if the problem was originally posed on an affine space. Is will be done using the *offset function trick*. $hat(u) = u_0 + tilde(u)$ #pagebreak() Finite dimensional subspace #text(size: 60pt)[$ V_(0, h) subset V_0 \ dim V_(0,h) < oo $] Discrete linear variational problem (DVP) $ u_h in V_(0,h): quad a(u_h, v_h) = l(v_h) quad forall v_h in V_(0,h) $ #pagebreak() #cetz.canvas(length: 3cm, { import cetz.draw: * let box_text_size = 16pt let lvp = text(box_text_size)[$ u in V_0: quad a(u, v) = l(v) quad forall v in V_0 $] let dvp = text(box_text_size)[$ u_h in V_(0,h): quad a(u_h, v_h) = l(v_h) quad forall v_h in V_(0,h) $] let cmin = text(box_text_size)[$ u = argmin_(v in V_0) J(v) $] let dmin = text(box_text_size)[$ u_h = argmin_(v_h in V_(0,h)) J(v_h) $] set-style( mark: (fill: white, scale: 2), line: (stroke: white), circle: (stroke: white), stroke: (thickness: 0.4pt, cap: "round"), content: (padding: 5pt) ) rect((0, 0), (3.0, 1), stroke: (paint: white, thickness: 1pt),name: "rect0") rect((5, 0), (9, 1), stroke: (paint: white, thickness: 1pt),name: "rect1") rect((0, -2), (3.0, -1), stroke: (paint: white, thickness: 1pt),name: "rect2") rect((5, -2), (9, -1), stroke: (paint: white, thickness: 1pt),name: "rect3") content("rect0", lvp) content("rect1", dvp) content("rect2", cmin) content("rect3", dmin) line("rect0", "rect1", mark: (end: "stealth"), name: "line0") line("rect2", "rect3", mark: (end: "stealth"), name: "line1") content(("line0.start", 50%, "line0.end"), align(center)[Galerkin\ Discretization], anchor: "south") content(("line1.start", 50%, "line1.end"), align(center)[Ritz\ Discretization], anchor: "south") content(("rect0", 50%, "rect2"), text(40pt, sym.arrow.t.b.double)) content(("rect1", 50%, "rect3"), text(40pt, sym.arrow.t.b.double)) }) #pagebreak() $ frak(B)_h = {b_h^1, dots, b_h^N} quad N = dim V_(0,h) \ V_(0,h) = "span" frak(B)_h \ u in V_(0,h) ==> u = sum_(i=1)^N mu_i b_h^i quad u tilde.eq vvec(mu) in RR^N \ v in V_(0,h) ==> v = sum_(i=1)^N nu_i b_h^i quad v tilde.eq vvec(nu) in RR^N $ #pagebreak() $ u_h in V_(0,h):&& quad bilf(a)(u_h, v_h) &= linf(l)(v_h) quad &&forall v_h in V_(0,h) \ vvec(mu) in RR^N:&& quad bilf(a)(sum_(j=j)^N mu_j b_h^j, sum_(i=1)^N nu_i b_h^i) &= linf(l)(sum_(i=1)^N nu_i b_h^i) quad &&forall avec(nu) in RR^N \ vvec(mu) in RR^N:&& quad sum_(j=1)^N mu_j sum_(i=1)^N nu_i bilf(a)(b_h^j, b_h^i) &= sum_(i=1)^N nu_i linf(l)(b_h^i) quad &&forall avec(nu) in RR^N \ vvec(mu) in RR^N:&& quad sum_(i=1)^N nu_j (sum_(i=1)^N mu_i bilf(a)(b_h^j, b_h^i) - linf(l)(b_h^i)) &= 0 quad &&forall avec(nu) in RR^N \ vvec(mu) in RR^N:&& quad sum_(i=1)^N mu_i bilf(a)(b_h^j, b_h^i) - linf(l)(b_h^i) &= 0 quad &&forall i in {1,dots,N} \ vvec(mu) in RR^N:&& quad sum_(i=1)^N mu_i bilf(a)(b_h^j, b_h^i) &= linf(l)(b_h^i) quad &&forall i in {1,dots,N} \ vvec(mu) in RR^N:&& quad amat(A) vvec(mu) &= vvec(phi) quad&& $ with $ &vvec(mu) &&= [mu_1, dots, mu_N]^transp &&in RR^N \ &amat(A) &&= [bilf(a)(b_h^j, b_h^i)]_(i,j=1)^N &&in RR^(N times N) \ &vvec(phi) &&= [linf(l)(b_h^i)]_(i,j=1)^N &&in RR^N $ LSE!!! $ amat(A) vvec(mu) = vvec(phi) $ The solution can then be recovered by $ u_h = sum_(i=1)^N mu_i b_h^i $ #pagebreak() Concrete 1D Galerkin Discretization $ u in H^1_0 (]a,b[): quad integral_a^b (dif u)/(dif x)(x) (dif v)/(dif x)(x) dif x = integral_a^b f(x) v(x) dif x quad forall v in H^1_0 (]a,b[) $

https://github.com/soul667/typst

https://raw.githubusercontent.com/soul667/typst/main/PPT/MATLAB/touying/docs/docs/utilities/oop.md

markdown

--- sidebar_position: 1 --- # Object-Oriented Programming Touying provides some convenient utility functions for object-oriented programming. --- ```typst #let empty-object = (methods: (:)) ``` An empty class. --- ```typst #let call-or-display(self, it) = { if type(it) == function { return it(self) } else { return it } } ``` Call or display as-is. --- ```typst #let methods(self) = { .. } ``` Used to bind self to methods and return, very commonly used.

https://github.com/jgm/typst-hs

https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/math/frac-05.typ

typst

Other

// Test associativity. $ 1/2/3 = (1/2)/3 = 1/(2/3) $

https://github.com/sitandr/typst-examples-book

https://raw.githubusercontent.com/sitandr/typst-examples-book/main/src/snippets/code.md

markdown

MIT License

# Code formatting ## Inline highlighting ```typ #let r = raw.with(lang: "r") This can then be used like: #r("x <- c(10, 42)") ``` ## Tab size ```````typ #set raw(tab-size: 8) ```tsv Year Month Day 2000 2 3 2001 2 1 2002 3 10 ``` ``````` ## Theme See [reference](https://typst.app/docs/reference/text/raw/#parameters-theme) ## Enable ligatures for code ```typ #show raw: set text(ligatures: true, font: "Cascadia Code") Then the code becomes `x <- a` ``` ## Advanced formatting See [packages](../packages/code.md) section.

https://github.com/FriendlyUser/IntroductionToTypst

https://raw.githubusercontent.com/FriendlyUser/IntroductionToTypst/main/README.md

markdown

Apache License 2.0

# IntroductionToTypst Source Code for introduction to Typst Article

https://github.com/mumblingdrunkard/mscs-thesis

https://raw.githubusercontent.com/mumblingdrunkard/mscs-thesis/master/src/computer-architecture-fundamentals/anatomy-of-an-in-order-pipelined-processor.typ

typst

== Anatomy of an In-order Pipelined Processor <sec:pipelined-processor> This first major optimisation of the microarchitecture is based on the observation that all instructions have required steps in common, and that the components used in each step are usually different. This introduces the concept of the _pipelined processor_. A classic processor pipeline may look like: instruction fetch (IF), instruction decode (ID), operand fetch (OF), execute (EXE), memory access (MEM), and writeback (WB). Each stage focuses on a specific stage of performing an instruction, like an assembly line where each step adds a new part to the product. IF fetches the instruction from memory. ID decodes the instruction and determines which control signals to set later in the pipeline. OF fetches the values to be used later in the pipeline. EXE performs the operation on some of the values. MEM performs memory access with an address computed in EXE. WB writes the values back to the register file so that they can be used by later instructions. Between each stage, there is a big register called a _pipeline register_ that holds values and control signals that tells the various stages what work to perform. The values come from outputs of previous stages, and are used as inputs in the current stage. Each stage takes only one cycle to complete This is a form of _instruction-level parallelism_ (ILP): the observation that a processor can work on many instructions at the same time because each instruction requires different parts of the processor at any given time. @fig:pipelined-cpu shows a high-level overview of a pipelined processor. In this version, the ID and OF stages are merged together, meaning values are read out of the register file at the same time the instruction is being decoded. Each stage is separated by a pipeline register. #figure( ```monosketch ┌───────────────────────────┐ │ ┌────────┬────┐ │ │ │ ┌───┐ │ │ │ ┌──┐ ║ ┌──▼──┐ ║ ┌▼─▼┐ ╟┘┌─▼─┐ ╟┘┌──┐ │ │IF├▶╟▶│ID/OF├▶╟▶│EXE├▶╟▶│MEM├▶╟▶│WB├─┘ └─▲┘ ║ └─────┘ ║ └───┘ ╟┐└───┘ ║ └──┘ └─────────────────────┘ ```, caption: [A high-level overview of a pipelined microarchitecture], kind: image, )<fig:pipelined-cpu> The connection from the EXE/MEM pipeline register to IF is to allow branch and jump instructions to change the PC. The connection from WB to OF is there to allow the WB stage to write values back to the register file which is traditionally stored where operand fetch is performed. === Overlapping Execution A simple pipelined processor like this can perform any single instruction in just five cycles, which is a good deal better than the shared bus architecture. However, the real trick is to overlap execution of multiple consecutive instructions. When an instruction moves from IF into ID, the IF stage is freed up and can start fetching the next instruction. This holds for every stage. Because of this, a pipelined processor can finish executing one instruction every cycle. ==== Hazards With overlapping pipelining comes execution _hazards_. Hazards arise when instructions depend on results from older instructions that have not yet completed. For some of these, the results are ready and available in pipeline registers even if they are not yet written to the register file. In this case, the values can be _forwarded_ to the stages where they are needed. The connections from the EXE/MEM and MEM/WB pipeline registers to EXE and from MEM/WB to MEM are there for forwarding. When a stage detects that an instruction in either of these later stages is going to write to one of its own source registers, it will use the value from the pipeline registers instead. Some hazards cannot be dealt with by only forwarding. For example: when one instruction reads from memory, and the following instruction depends on the result in the EXE stage, the procsessor has to _stall_ for a cycle. ==== Branches All instructions that enter IF after a branch have a dependency on the branch. The simplest thing is to stall IF until the branch instruction has left EXE and potentially modified the PC. A possible step up is to assume that the branch condition will resolve to "False" and to keep fetching. If the assumption turns out to be correct, three cycles have been saved. If the assumption turns out to be wrong, the results of the incorrectly fetched instructions must be _squashed_ (ignored). The next step up is to observe patterns in branch instructions and predict the outcome with more accuracy to prevent squashing too often. This is the founding basis of _branch prediction_, a form of _speculation_.

https://github.com/typst/packages

https://raw.githubusercontent.com/typst/packages/main/packages/preview/unichar/0.1.0/ucd/block-1CD0.typ

typst

Apache License 2.0

#let data = ( ("VEDIC TONE KARSHANA", "Mn", 230), ("VEDIC TONE SHARA", "Mn", 230), ("VEDIC TONE PRENKHA", "Mn", 230), ("VEDIC SIGN NIHSHVASA", "Po", 0), ("VEDIC SIGN YAJURVEDIC MIDLINE SVARITA", "Mn", 1), ("VEDIC TONE YAJURVEDIC AGGRAVATED INDEPENDENT SVARITA", "Mn", 220), ("VEDIC TONE YAJURVEDIC INDEPENDENT SVARITA", "Mn", 220), ("VEDIC TONE YAJURVEDIC KATHAKA INDEPENDENT SVARITA", "Mn", 220), ("VEDIC TONE CANDRA BELOW", "Mn", 220), ("VEDIC TONE YAJURVEDIC KATHAKA INDEPENDENT SVARITA SCHROEDER", "Mn", 220), ("VEDIC TONE DOUBLE SVARITA", "Mn", 230), ("VEDIC TONE TRIPLE SVARITA", "Mn", 230), ("VEDIC TONE KATHAKA ANUDATTA", "Mn", 220), ("VEDIC TONE DOT BELOW", "Mn", 220), ("VEDIC TONE TWO DOTS BELOW", "Mn", 220), ("VEDIC TONE THREE DOTS BELOW", "Mn", 220), ("VEDIC TONE RIGVEDIC KASHMIRI INDEPENDENT SVARITA", "Mn", 230), ("VEDIC TONE ATHARVAVEDIC INDEPENDENT SVARITA", "Mc", 0), ("VEDIC SIGN VISARGA SVARITA", "Mn", 1), ("VEDIC SIGN VISARGA UDATTA", "Mn", 1), ("VEDIC SIGN REVERSED VISARGA UDATTA", "Mn", 1), ("VEDIC SIGN VISARGA ANUDATTA", "Mn", 1), ("VEDIC SIGN REVERSED VIS<NAME>", "Mn", 1), ("VEDIC SIGN VISARGA UDATTA WITH TAIL", "Mn", 1), ("VEDIC SIGN VISARGA ANUDATTA WITH TAIL", "Mn", 1), ("VEDIC SIGN ANUSVARA ANTARGOMUKHA", "Lo", 0), ("VEDIC SIGN ANUSVARA BAHIRGOMUKHA", "Lo", 0), ("VEDIC SIGN ANUSVARA VAMAGOMUKHA", "Lo", 0), ("VEDIC SIGN ANUSVARA VAMAGOMUKHA WITH TAIL", "Lo", 0), ("VEDIC SIGN TIRYAK", "Mn", 220), ("VEDIC SIGN HEXIFORM LONG ANUSVARA", "Lo", 0), ("VEDIC SIGN LONG ANUSVARA", "Lo", 0), ("VEDIC SIGN RTHANG LONG ANUSVARA", "Lo", 0), ("VEDIC SIGN ANUSVARA UBHAYATO MUKHA", "Lo", 0), ("VEDIC SIGN ARDHAVISARGA", "Lo", 0), ("VEDIC SIGN ROTATED ARDHAVISARGA", "Lo", 0), ("VEDIC TONE CANDRA ABOVE", "Mn", 230), ("VEDIC SIGN JIHVAMULIYA", "Lo", 0), ("VEDIC SIGN UPADHMANIYA", "Lo", 0), ("VEDIC SIGN ATIKRAMA", "Mc", 0), ("VEDIC TONE RING ABOVE", "Mn", 230), ("VEDIC TONE DOUBLE RING ABOVE", "Mn", 230), ("VEDIC SIGN DOUBLE ANUSVARA ANTARGOMUKHA", "Lo", 0), )

https://github.com/Enter-tainer/typstyle

https://raw.githubusercontent.com/Enter-tainer/typstyle/master/tests/assets/typstfmt/138-non-converge-block-comment.typ

typst

Apache License 2.0

#let test_func() = { ( /* test */ ) }

https://github.com/lyzynec/orr-go-brr

https://raw.githubusercontent.com/lyzynec/orr-go-brr/main/04/main.typ

typst

#import "../lib.typ": * #knowledge[ #question(name: [Give the first-order necessary conditions of optimality for a general optimal control problem for a nonlinear discrete-time system over a finite horizon. Namely, give the general two-point boundary value problem, highlighting the state equation, the co-state equation and a stationarity equation. Do not forget to include general boundary conditions.])[ The augmented cost function is $ J'_i (bold(x)_k, bold(u)_k, bold(lambda)_(k+1)) = phi.alt(bold(x)_N, N) + sum_(k=1)^(N-1) [L_k (bold(x)_k, bold(u)_k) + bold(lambda)_(k+1)^T [bold(f)_k (bold(x)_k, bold(u)_k) - bold(x)_(k+1)]] $ To null the gradient as $gradient J'_i = 0$ we have to satisfy $ x_(k+1) &= gradient_lambda_(k+1) H_k(bold(x)_k, bold(u)_k, bold(lambda)_(k+1)) = f_k (x_k,u_k) "(state equation)"\ lambda_k &= gradient_bold(x)_k H_k(bold(x)_k, bold(u)_k, bold(lambda)_(k+1)) "(costate equation)"\ 0 &= gradient_bold(u)_k H_k(bold(x)_k, bold(u)_k, bold(lambda)_(k+1)) "(stationarity equation)"\ 0 &= gradient_bold(x)_i H_k(bold(x)_k, bold(u)_k, bold(lambda)_(k+1)) dot upright(d) bold(x)_i, "for fixed " bold(x)_i = bold(r)_i => upright(d) bold(x)_i = bold(0) \ 0 &= (gradient_bold(x)_N phi.alt - bold(lambda)_N)^T dot d x_N = cases("fixed " bold(x)_N &=> bold(x)_N = bold(r)_N => upright(d) bold(x)_N = bold(0), "free " bold(x)_N &=> upright(d) bold(x)_N != 0 +> bold(lambda)_N = gradient_(bold(lambda)_N) phi.alt)\ $ where $ H(bold(x)_k, bold(u)_k, bold(lambda)_(k+1)) = L_k (bold(x)_k, bold(u)_k) + bold(lambda)_(k+1)^T bold(f)_k (bold(x)_k, bold(u)_k) $ As starting point is probably given, the boundary condition could be something like $ bold(x)_0 = bold(r)_0\ $ for the end state, it could be given $ bold(x)_N = bold(r)_N $ or could be subject to optimization as $ bold(lambda)_N = gradient phi.alt(bold(x)_N) $ ] #question(name: [Give the first-order necessary conditions of optimality for a linear and time invariant (LTI) discrete-time system and a quadratic cost function over a finite horizon. Namely, give them in the format displaying the state equation, co-state equation and stationarity equation. Show and discuss also two types of boundary conditions.])[ With cost function $ J = 1/2 bold(x)_N^T bold(S)_N bold(x)_N + 1/2 sum_(k=0)^(N-1) [bold(x)_k^T bold(Q) bold(x)_k + bold(u)_k^T R u_k] $ the Hamiltonian $H$ function would be $ H_k = 1/2 (bold(x)_k^T bold(Q) bold(x)_k + bold(u)_k^T bold(R) bold(u)_k) + bold(lambda)_(k+1)^T (bold(A) bold(x)_k + bold(B) bold(u)_k) $ the equations would become #align(center)[#grid(columns: 2, row-gutter: 10pt, column-gutter: 10pt, align: left, [state:], $bold(x)_(k+1) = bold(A) bold(x)_k + bold(B) bold(u)_k$, [costate:], $bold(lambda)_k = bold(Q) bold(x)_k + bold(A)^T bold(lambda)_(k+1)$, [stationarity:], $bold(0) = bold(R) bold(u)_k + bold(B)^T bold(lambda)_(k+1)$, [boundary:], $bold(x)_0 = bold(r)_0$, [boundary:], $bold(0) &= (bold(S)_N bold(x)_N - bold(lambda)_N)^T upright(d) bold(x)_N$, )] There is other type for the last boundary condition, as it could also be given as $ bold(x)_N = bold(r)_N $ From the stationarity equation we can extract optimal control as $ bold(u)_k = - bold(R)^(-1) bold(B)^T bold(lambda)_(k+1) $ Than we can subtitute the optimal control into the rest to obtain state and costate values. ] #question(name: [Give a qualitative characterization of the solution to the fixed final state LQ-optimal control problem over a finite horizon, that is, you do not have to give formulas but you should be able to state among the highlights that the control is open-loop and that reachability of the system is a necessary condition.])[ - the reslt will be offline precalculated control signal, meaining it is open--loop - the control will be proportional to $bold(r)_N - bold(A)^N bold(x)_0$ meaning the difference between desired end--time state and state in which the system would end up without any control (this one is pretty reasonable) - the control will be inversly proportional to _reachability Gramian_ $ bold(G)_(0,N,bold(R)) = sum_(i=0)^(N-1) bold(A)^(N-i-1) bold(B) bold(R)^(-1) bold(B)^T (A^T)^(N-i-1) $ if the _reachability Gramian_ is singular, it means the state is not reachable, barring us from calculating optimal control for reaching it (again pretty reasonable) ] #question(name: [Give a qualitative characterization of the solution to the free final state LQ-optimal control problem over a finite horizon, that is, you do not have to give formulas but you should be able to state among the highlights that the control is closed-loop, namely, a time-varying linear state feedback and that the feedback gains can be computed by solving a difference Riccati equation.])[ - the problem results in time varying linear state feedback, noteworthy is the fact that the state feedback gain increases drasticcaly when nearing the horizon - the Riccati equation looks like this $ bold(S)_k = bold(Q) + bold(A)^T bold(S)_(k+1) (bold(I) + bold(B) bold(R)^(-1) bold(B)^T bold(S)_(k+1))^(-1) bold(A) $ - this Riccati equation was constructed with the assumption that the final state boundary equation $ bold(S)_N bold(x)_N = bold(lambda)_N $ holds for any $k$ not just $N$ $ bold(S)_k bold(x)_k = bold(lambda)_k $ ] #question(name: [Discuss how solution to the free final state LQ problem changes as the horizon is extended to infinity. Emphasize that the optimal solution is given by a constant linear state feedback whose gains are computed by solving a discrete-time algebraic Riccati equation (DARE). What are the conditions under which a stabilizing solution is guaranteed to exist? What are the conditions under which it is guaranteed that there is a unique stabilizing solution of DARE?])[ - the solution is simmilar to the previous one with the exception that, as we never reach the horizon, the state feedback gain remains constant - DARE (discrete time algebraic Riccati equation) assumes that in the steady state (as $k -> oo$) $ bold(S)_(k+1) approx bold(S)_k $ it, works because this assumption is correct - for stabilizing solution to exist, there are two conditions - system $(bold(A), sqrt(bold(Q)))$ is stabilizable #footnote[Non--controllable parts of system are stable] - system $(bold(A), sqrt(bold(Q)))$ is detectable #footnote[Non--observable patrts of system are stable] - unique stablilizing solution requires that in addition to existance of stabilizing solution, the system $(bold(A), sqrt(bold(Q)))$ is detectable ] ] #skills[ #question(name: [Design an LQ--optimal state feedback controller for a discrete--time linear system both for a finite and an infinite horizon, both for regulation and for tracking.])[] ]

https://github.com/LDemetrios/Typst4k

https://raw.githubusercontent.com/LDemetrios/Typst4k/master/src/test/resources/suite/layout/limits.typ

typst

// Test how the layout engine reacts when reaching limits like // zero, infinity or when dealing with NaN. --- issue-1216-clamp-panic --- #set page(height: 20pt, margin: 0pt) #v(22pt) #block(fill: red, width: 100%, height: 10pt, radius: 4pt) --- issue-1918-layout-infinite-length-grid-columns --- // Test that passing infinite lengths to drawing primitives does not crash Typst. #set page(width: auto, height: auto) // Error: 58-59 cannot expand into infinite width #layout(size => grid(columns: (size.width, size.height))[a][b][c][d]) --- issue-1918-layout-infinite-length-grid-rows --- #set page(width: auto, height: auto) // Error: 17-66 cannot create grid with infinite height #layout(size => grid(rows: (size.width, size.height))[a][b][c][d]) --- issue-1918-layout-infinite-length-line --- #set page(width: auto, height: auto) // Error: 17-41 cannot create line with infinite length #layout(size => line(length: size.width)) --- issue-1918-layout-infinite-length-polygon --- #set page(width: auto, height: auto) // Error: 17-54 cannot create polygon with infinite size #layout(size => polygon((0pt,0pt), (0pt, size.width)))

https://github.com/kalxd/morelull

https://raw.githubusercontent.com/kalxd/morelull/master/doc.typ

typst

#import "@local/morelull:0.5.0": * #show: morelull.with(标题: "这里写上一个标题") = 大大的标题写在这里。 #t 这里又是另一个#underline[故事]了。 #t 一块硬盘,容量 1T,作为应用#underline[数据盘],一般性的#underline[程序]都放在上面,包括所产生的数据;另一块容量 4T,作为媒体数据盘,数字媒体、个人数据都在上面。 #t 这里又加起新的段，你看看效果。 ```rust fn main() { println!("hello world") } ``` #t 这里再启动一个新的东西。

https://github.com/JunzheShen/SJTU-Resume-Template-in-Typst

https://raw.githubusercontent.com/JunzheShen/SJTU-Resume-Template-in-Typst/main/README.md

markdown

# SJTU Resume Template in Typst This repo provides a Typst version of a SJTU resume template called 蓝色梦想. Note that there are minor differences between this version and the official Microsoft Word version, feel free to adjust some parameters to make make the two versions more alike. This repo is is built upon [OrangeX4/Chinese-Resume-in-Typst](https://github.com/OrangeX4/Chinese-Resume-in-Typst) Detailed instructions to use this repo will be updated when I have some time to kill.

https://github.com/TypstApp-team/typst

https://raw.githubusercontent.com/TypstApp-team/typst/master/tests/typ/meta/figure.typ

typst

Apache License 2.0

// Test figures. --- #set page(width: 150pt) #set figure(numbering: "I") We can clearly see that @fig-cylinder and @tab-complex are relevant in this context. #figure( table(columns: 2)[a][b], caption: [The basic table.], ) <tab-basic> #figure( pad(y: -6pt, image("/files/cylinder.svg", height: 2cm)), caption: [The basic shapes.], numbering: "I", ) <fig-cylinder> #figure( table(columns: 3)[a][b][c][d][e][f], caption: [The complex table.], ) <tab-complex> --- // Testing figures with tables. #figure( table( columns: 2, [Second cylinder], image("/files/cylinder.svg"), ), caption: "A table containing images." ) <fig-image-in-table> --- // Testing show rules with figures with a simple theorem display #show figure.where(kind: "theorem"): it => { let name = none if not it.caption == none { name = [ #emph(it.caption.body)] } else { name = [] } let title = none if not it.numbering == none { title = it.supplement if not it.numbering == none { title += " " + it.counter.display(it.numbering) } } title = strong(title) pad( top: 0em, bottom: 0em, block( fill: green.lighten(90%), stroke: 1pt + green, inset: 10pt, width: 100%, radius: 5pt, breakable: false, [#title#name#h(0.1em):#h(0.2em)#it.body#v(0.5em)] ) ) } #set page(width: 150pt) #figure( $a^2 + b^2 = c^2$, supplement: "Theorem", kind: "theorem", caption: "Pythagoras' theorem.", numbering: "1", ) <fig-formula> #figure( $a^2 + b^2 = c^2$, supplement: "Theorem", kind: "theorem", caption: "Another Pythagoras' theorem.", numbering: none, ) <fig-formula> #figure( ```rust fn main() { println!("Hello!"); } ```, caption: [Hello world in _rust_], ) --- // Test breakable figures #set page(height: 6em) #show figure: set block(breakable: true) #figure(table[a][b][c][d][e], caption: [A table]) --- // Test custom separator for figure caption #set figure.caption(separator: [ --- ]) #figure( table(columns: 2)[a][b], caption: [The table with custom separator.], )

https://github.com/noamzaks/Barvazim

https://raw.githubusercontent.com/noamzaks/Barvazim/main/writeups/bsides-2024/forensics/skibidi.typ

typst

// Category: Forensics #import "../../template.typ": writeup #show: writeup.with(ctf: "BSides", exercise: "Skibidi", date: datetime(day: 29, month: 6, year: 2024)) Every docx file is a ZIP archive. After unzipping the `Skibidi.docx` file, we can go over the different files included. Most seem uninteresting, however, the `img.xml` seems suspicious - why is an image an XML file? And the content is especially suspicious: ```xml <root> <person firstname="<KEY>" lastname="<KEY> city="Haifa" country="Israel" firstname2="<KEY>" lastname2="<KEY> email="<EMAIL>" /> <random>6</random> <random_float>89.838</random_float> <bool>true</bool> <date>1986-09-28</date> <regEx>helloooooooooooooooooooooooooooooooooooooooooooo world</regEx> <enum>generator</enum> <elt>Alyssa</elt><elt>Flory</elt><elt>Ulrike</elt><elt>Teriann</elt><elt>Reeba</elt> <Ulrike> <age>56</age> </Ulrike> </root> ``` Well, the firstname, lastname, firstname2 and lastname2 appear to be Base64 data because they are only letters digits, and they end with the equal signs. By now we have arrived at the solution of this challenge: ```python >>> import base64 >>> base64.b64decode("Qn<KEY>") + base64.b64decode("M2YzNXQzZF90aDNfc2sxYjFkaX0=") b'BsidesTLV2024{w3_d3f35t3d_th3_sk1b1di}' ```

https://github.com/danisltpi/seminar

https://raw.githubusercontent.com/danisltpi/seminar/main/template/ausarbeitung.typ

typst

#import "template.typ": project #import "@preview/cetz:0.2.2" #set text(lang: "de", region: "de") #set par(leading: 1em) // #show outline: set par(leading: 2em) #show heading: it => [#pad(bottom: 1em, top: 1em)[#it]] #show: project.with( title: "Fibonacci Heaps", subtitle: "Seminararbeit", study_program: "Informatik (INFB)", institution: "Hochschule Karlsruhe", date: datetime.today().display("[day].[month].[year]"), examiner: "Prof. Dr. rer. <NAME>", logo: "hka.svg", authors: ((name: "<NAME>", matriculation_number: "79663"),), ) // #show outline.entry.where(level: 1): it => { // strong(it) // } // #outline(indent: auto) #pagebreak(weak: true) = Motivation Ein Fibonacci-Heap ist eine Datenstruktur, die Prioritätswarteschlangen implementiert und werden vielfältig eingesetzt. Sie besteht aus mehreren Heaps (Binärbäume) \ \ #align(center)[ #cetz.canvas({ import cetz.draw: * rect((-1, -1), (1, 1)) circle((0, 0), fill: yellow, stroke: blue) line((0, 0), (2, 1)) line((0, 0), (1.5, -1)) })] == Dijkstra-Algorithmus = Laufzeit Daraus ergibt sich eine Laufzeit von $O(n log n)$ für das Hinzufügen eines Knotens bzw. $Theta(n log n)$ = Das ist ein Test #lorem(100) \ #figure( image("../assets/circle.svg", width: 100%), caption: [Beispielhafte Struktur eines Fibonacci Heaps, bestehend aus 5 Teilbäumen #lorem(50)], gap: 4em, ) <image> = Erklärung Wie man in der vorherigen Abbildung erkennen kann: @image Dies ist besonders gut #lorem(1000) @cormen_introduction_2009 #pagebreak(weak: true) #set page(header: []) #bibliography("literatur.bib", style: "ieee", title: "Literatur")

https://github.com/eLearningHub/resume-typst

https://raw.githubusercontent.com/eLearningHub/resume-typst/main/main.typ

typst

Apache License 2.0

#let portfolio = yaml("portfolio.yaml") #let settings = yaml("settings.yaml") #show link: set text(blue) #show heading: h => [ #set text( size: eval(settings.font.size.heading_large), font: settings.font.general ) #h ] #let sidebarSection = {[ #par(justify: true)[ #par[ #set text( size: eval(settings.font.size.contacts), font: settings.font.minor_highlight, ) Email: #link("mailto:" + portfolio.contacts.email) \ Phone: #link("tel:" + portfolio.contacts.phone) \ LinkedIn: #link(portfolio.contacts.linkedin)[mikhail-liamets] \ GitHub: #link(portfolio.contacts.github)[caffeintazedgaze] \ #portfolio.contacts.address ] #line(length: 100%) ] = Summary #par[ #set text( eval(settings.font.size.education_description), font: settings.font.minor_highlight, ) An experienced *software engineer* with a confident grasp of *infrastructure*, *system design*, and *DevOps*, now seeking opportunities to excel in the realms of solution architecture. Open to roles ranging from *software engineering* to *DevOps/SRE*. ] = Education #{ for place in portfolio.education [ #par[ #set text( size: eval(settings.font.size.heading), font: settings.font.general ) #place.from – #place.to \ #link(place.university.link)[#place.university.name] ] #par[ #set text( eval(settings.font.size.education_description), font: settings.font.minor_highlight, ) #place.degree #place.major ] ] } = Skills #{ for skill in portfolio.skills [ #par[ #set text( size: eval(settings.font.size.description), ) #set text( // size: eval(settings.font.size.tags), font: settings.font.minor_highlight, ) *#skill.name* #linebreak() #skill.items.join(" • ") ] ] } ]} #let mainSection = {[ // #par[ // #set align(center) // #figure( // image("images/Kodak 20 Zanvoort Lumi.jpg", width: 6em), // placement: top, // ) // ] #par[ #set text( size: eval(settings.font.size.heading_huge), font: settings.font.general, ) *#portfolio.contacts.name* ] #par[ #set text( size: eval(settings.font.size.heading), font: settings.font.minor_highlight, top-edge: 0pt ) #portfolio.contacts.title ] = Experience #{ for job in portfolio.jobs [ #par(justify: false)[ #set text( size: eval(settings.font.size.heading), font: settings.font.general ) #job.from – #job.to \ *#job.position* #link(job.company.link)[\@ #job.company.name] ] #par( justify: false, leading: eval(settings.paragraph.leading) )[ #set text( size: eval(settings.font.size.description), font: settings.font.general ) #{ for point in job.description [ #h(0.5cm) • #point \ ] } ] #par( justify: true, leading: eval(settings.paragraph.leading), )[ #set text( size: eval(settings.font.size.tags), font: settings.font.minor_highlight ) ] ] } ]} #{ grid( columns: (2fr, 5fr), column-gutter: 3em, sidebarSection, mainSection, ) }

https://github.com/typst/packages

https://raw.githubusercontent.com/typst/packages/main/packages/preview/unichar/0.1.0/ucd/block-1D800.typ

typst

Apache License 2.0

#let data = ( ("SIGNWRITING HAND-FIST INDEX", "So", 0), ("SIGNWRITING HAND-CIRCLE INDEX", "So", 0), ("SIGNWRITING HAND-CUP INDEX", "So", 0), ("SIGNWRITING HAND-OVAL INDEX", "So", 0), ("SIGNWRITING HAND-HINGE INDEX", "So", 0), ("SIGNWRITING HAND-ANGLE INDEX", "So", 0), ("SIGNWRITING HAND-FIST INDEX BENT", "So", 0), ("SIGNWRITING HAND-CIRCLE INDEX BENT", "So", 0), ("SIGNWRITING HAND-FIST THUMB UNDER INDEX BENT", "So", 0), ("SIGNWRITING HAND-FIST INDEX RAISED KNUCKLE", "So", 0), ("SIGNWRITING HAND-FIST INDEX CUPPED", "So", 0), ("SIGNWRITING HAND-FIST INDEX HINGED", "So", 0), ("SIGNWRITING HAND-FIST INDEX HINGED LOW", "So", 0), ("SIGNWRITING HAND-CIRCLE INDEX HINGE", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE", "So", 0), ("SIGNWRITING HAND-CIRCLE INDEX MIDDLE", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE BENT", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE RAISED KNUCKLES", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE HINGED", "So", 0), ("SIGNWRITING HAND-FIST INDEX UP MIDDLE HINGED", "So", 0), ("SIGNWRITING HAND-FIST INDEX HINGED MIDDLE UP", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CONJOINED", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CONJOINED INDEX BENT", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CONJOINED MIDDLE BENT", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CONJOINED CUPPED", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CONJOINED HINGED", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CROSSED", "So", 0), ("SIGNWRITING HAND-CIRCLE INDEX MIDDLE CROSSED", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE BENT OVER INDEX", "So", 0), ("SIGNWRITING HAND-FIST INDEX BENT OVER MIDDLE", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE THUMB", "So", 0), ("SIGNWRITING HAND-CIRCLE INDEX MIDDLE THUMB", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE STRAIGHT THUMB BENT", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE BENT THUMB STRAIGHT", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE THUMB BENT", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE HINGED SPREAD THUMB SIDE", "So", 0), ("SIGNWRITING HAND-FIST INDEX UP MIDDLE HINGED THUMB SIDE", "So", 0), ("SIGNWRITING HAND-FIST INDEX UP MIDDLE HINGED THUMB CONJOINED", "So", 0), ("SIGNWRITING HAND-FIST INDEX HINGED MIDDLE UP THUMB SIDE", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE UP SPREAD THUMB FORWARD", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE THUMB CUPPED", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE THUMB CIRCLED", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE THUMB HOOKED", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE THUMB HINGED", "So", 0), ("SIGNWRITING HAND-FIST THUMB BETWEEN INDEX MIDDLE STRAIGHT", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CONJOINED THUMB SIDE", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CONJOINED THUMB SIDE CONJOINED", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CONJOINED THUMB SIDE BENT", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE THUMB HOOKED INDEX UP", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB HOOKED MIDDLE UP", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CONJOINED HINGED THUMB SIDE", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CROSSED THUMB SIDE", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CONJOINED THUMB FORWARD", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CONJOINED CUPPED THUMB FORWARD", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE THUMB CUPPED INDEX UP", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB CUPPED MIDDLE UP", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE THUMB CIRCLED INDEX UP", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE THUMB CIRCLED INDEX HINGED", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB ANGLED OUT MIDDLE UP", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB ANGLED IN MIDDLE UP", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB CIRCLED MIDDLE UP", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE THUMB CONJOINED HINGED", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE THUMB ANGLED OUT", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE THUMB ANGLED", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE THUMB ANGLED OUT INDEX UP", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE THUMB ANGLED OUT INDEX CROSSED", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE THUMB ANGLED INDEX UP", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB HOOKED MIDDLE HINGED", "So", 0), ("SIGNWRITING HAND-FLAT FOUR FINGERS", "So", 0), ("SIGNWRITING HAND-FLAT FOUR FINGERS BENT", "So", 0), ("SIGNWRITING HAND-FLAT FOUR FINGERS HINGED", "So", 0), ("SIGNWRITING HAND-FLAT FOUR FINGERS CONJOINED", "So", 0), ("SIGNWRITING HAND-FLAT FOUR FINGERS CONJOINED SPLIT", "So", 0), ("SIGNWRITING HAND-CLAW FOUR FINGERS CONJOINED", "So", 0), ("SIGNWRITING HAND-FIST FOUR FINGERS CONJOINED BENT", "So", 0), ("SIGNWRITING HAND-HINGE FOUR FINGERS CONJOINED", "So", 0), ("SIGNWRITING HAND-FLAT FIVE FINGERS SPREAD", "So", 0), ("SIGNWRITING HAND-FLAT HEEL FIVE FINGERS SPREAD", "So", 0), ("SIGNWRITING HAND-FLAT FIVE FINGERS SPREAD FOUR BENT", "So", 0), ("SIGNWRITING HAND-FLAT HEEL FIVE FINGERS SPREAD FOUR BENT", "So", 0), ("SIGNWRITING HAND-FLAT FIVE FINGERS SPREAD BENT", "So", 0), ("SIGNWRITING HAND-FLAT HEEL FIVE FINGERS SPREAD BENT", "So", 0), ("SIGNWRITING HAND-FLAT FIVE FINGERS SPREAD THUMB FORWARD", "So", 0), ("SIGNWRITING HAND-CUP FIVE FINGERS SPREAD", "So", 0), ("SIGNWRITING HAND-CUP FIVE FINGERS SPREAD OPEN", "So", 0), ("SIGNWRITING HAND-HINGE FIVE FINGERS SPREAD OPEN", "So", 0), ("SIGNWRITING HAND-OVAL FIVE FINGERS SPREAD", "So", 0), ("SIGNWRITING HAND-FLAT FIVE FINGERS SPREAD HINGED", "So", 0), ("SIGNWRITING HAND-FLAT FIVE FINGERS SPREAD HINGED THUMB SIDE", "So", 0), ("SIGNWRITING HAND-FLAT FIVE FINGERS SPREAD HINGED NO THUMB", "So", 0), ("SIGNWRITING HAND-FLAT", "So", 0), ("SIGNWRITING HAND-FLAT BETWEEN PALM FACINGS", "So", 0), ("SIGNWRITING HAND-FLAT HEEL", "So", 0), ("SIGNWRITING HAND-FLAT THUMB SIDE", "So", 0), ("SIGNWRITING HAND-FLAT HEEL THUMB SIDE", "So", 0), ("SIGNWRITING HAND-FLAT THUMB BENT", "So", 0), ("SIGNWRITING HAND-FLAT THUMB FORWARD", "So", 0), ("SIGNWRITING HAND-FLAT SPLIT INDEX THUMB SIDE", "So", 0), ("SIGNWRITING HAND-FLAT SPLIT CENTRE", "So", 0), ("SIGNWRITING HAND-FLAT SPLIT CENTRE THUMB SIDE", "So", 0), ("SIGNWRITING HAND-FLAT SPLIT CENTRE THUMB SIDE BENT", "So", 0), ("SIGNWRITING HAND-FLAT SPLIT LITTLE", "So", 0), ("SIGNWRITING HAND-CLAW", "So", 0), ("SIGNWRITING HAND-CLAW THUMB SIDE", "So", 0), ("SIGNWRITING HAND-CLAW NO THUMB", "So", 0), ("SIGNWRITING HAND-CLAW THUMB FORWARD", "So", 0), ("SIGNWRITING HAND-HOOK CURLICUE", "So", 0), ("SIGNWRITING HAND-HOOK", "So", 0), ("SIGNWRITING HAND-CUP OPEN", "So", 0), ("SIGNWRITING HAND-CUP", "So", 0), ("SIGNWRITING HAND-CUP OPEN THUMB SIDE", "So", 0), ("SIGNWRITING HAND-CUP THUMB SIDE", "So", 0), ("SIGNWRITING HAND-CUP OPEN NO THUMB", "So", 0), ("SIGNWRITING HAND-CUP NO THUMB", "So", 0), ("SIGNWRITING HAND-CUP OPEN THUMB FORWARD", "So", 0), ("SIGNWRITING HAND-CUP THUMB FORWARD", "So", 0), ("SIGNWRITING HAND-CURLICUE OPEN", "So", 0), ("SIGNWRITING HAND-CURLICUE", "So", 0), ("SIGNWRITING HAND-CIRCLE", "So", 0), ("SIGNWRITING HAND-OVAL", "So", 0), ("SIGNWRITING HAND-OVAL THUMB SIDE", "So", 0), ("SIGNWRITING HAND-OVAL NO THUMB", "So", 0), ("SIGNWRITING HAND-OVAL THUMB FORWARD", "So", 0), ("SIGNWRITING HAND-HINGE OPEN", "So", 0), ("SIGNWRITING HAND-HINGE OPEN THUMB FORWARD", "So", 0), ("SIGNWRITING HAND-HINGE", "So", 0), ("SIGNWRITING HAND-HINGE SMALL", "So", 0), ("SIGNWRITING HAND-HINGE OPEN THUMB SIDE", "So", 0), ("SIGNWRITING HAND-HINGE THUMB SIDE", "So", 0), ("SIGNWRITING HAND-HINGE OPEN NO THUMB", "So", 0), ("SIGNWRITING HAND-HINGE NO THUMB", "So", 0), ("SIGNWRITING HAND-HINGE THUMB SIDE TOUCHING INDEX", "So", 0), ("SIGNWRITING HAND-HINGE THUMB BETWEEN MIDDLE RING", "So", 0), ("SIGNWRITING HAND-ANGLE", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE RING", "So", 0), ("SIGNWRITING HAND-CIRCLE INDEX MIDDLE RING", "So", 0), ("SIGNWRITING HAND-HINGE INDEX MIDDLE RING", "So", 0), ("SIGNWRITING HAND-ANGLE INDEX MIDDLE RING", "So", 0), ("SIGNWRITING HAND-HINGE LITTLE", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE RING BENT", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE RING CONJOINED", "So", 0), ("SIGNWRITING HAND-HINGE INDEX MIDDLE RING CONJOINED", "So", 0), ("SIGNWRITING HAND-FIST LITTLE DOWN", "So", 0), ("SIGNWRITING HAND-FIST LITTLE DOWN RIPPLE STRAIGHT", "So", 0), ("SIGNWRITING HAND-FIST LITTLE DOWN RIPPLE CURVED", "So", 0), ("SIGNWRITING HAND-FIST LITTLE DOWN OTHERS CIRCLED", "So", 0), ("SIGNWRITING HAND-FIST LITTLE UP", "So", 0), ("SIGNWRITING HAND-FIST THUMB UNDER LITTLE UP", "So", 0), ("SIGNWRITING HAND-CIRCLE LITTLE UP", "So", 0), ("SIGNWRITING HAND-OVAL LITTLE UP", "So", 0), ("SIGNWRITING HAND-ANGLE LITTLE UP", "So", 0), ("SIGNWRITING HAND-FIST LITTLE RAISED KNUCKLE", "So", 0), ("SIGNWRITING HAND-FIST LITTLE BENT", "So", 0), ("SIGNWRITING HAND-FIST LITTLE TOUCHES THUMB", "So", 0), ("SIGNWRITING HAND-FIST LITTLE THUMB", "So", 0), ("SIGNWRITING HAND-HINGE LITTLE THUMB", "So", 0), ("SIGNWRITING HAND-FIST LITTLE INDEX THUMB", "So", 0), ("SIGNWRITING HAND-HINGE LITTLE INDEX THUMB", "So", 0), ("SIGNWRITING HAND-ANGLE LITTLE INDEX THUMB INDEX THUMB OUT", "So", 0), ("SIGNWRITING HAND-ANGLE LITTLE INDEX THUMB INDEX THUMB", "So", 0), ("SIGNWRITING HAND-FIST LITTLE INDEX", "So", 0), ("SIGNWRITING HAND-CIRCLE LITTLE INDEX", "So", 0), ("SIGNWRITING HAND-HINGE LITTLE INDEX", "So", 0), ("SIGNWRITING HAND-ANGLE LITTLE INDEX", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE LITTLE", "So", 0), ("SIGNWRITING HAND-CIRCLE INDEX MIDDLE LITTLE", "So", 0), ("SIGNWRITING HAND-HINGE INDEX MIDDLE LITTLE", "So", 0), ("SIGNWRITING HAND-HINGE RING", "So", 0), ("SIGNWRITING HAND-ANGLE INDEX MIDDLE LITTLE", "So", 0), ("SIGNWRITING HAND-FIST INDEX MIDDLE CROSS LITTLE", "So", 0), ("SIGNWRITING HAND-CIRCLE INDEX MIDDLE CROSS LITTLE", "So", 0), ("SIGNWRITING HAND-FIST RING DOWN", "So", 0), ("SIGNWRITING HAND-HINGE RING DOWN INDEX THUMB HOOK MIDDLE", "So", 0), ("SIGNWRITING HAND-ANGLE RING DOWN MIDDLE THUMB INDEX CROSS", "So", 0), ("SIGNWRITING HAND-FIST RING UP", "So", 0), ("SIGNWRITING HAND-FIST RING RAISED KNUCKLE", "So", 0), ("SIGNWRITING HAND-FIST RING LITTLE", "So", 0), ("SIGNWRITING HAND-CIRCLE RING LITTLE", "So", 0), ("SIGNWRITING HAND-OVAL RING LITTLE", "So", 0), ("SIGNWRITING HAND-ANGLE RING LITTLE", "So", 0), ("SIGNWRITING HAND-FIST RING MIDDLE", "So", 0), ("SIGNWRITING HAND-FIST RING MIDDLE CONJOINED", "So", 0), ("SIGNWRITING HAND-FIST RING MIDDLE RAISED KNUCKLES", "So", 0), ("SIGNWRITING HAND-FIST RING INDEX", "So", 0), ("SIGNWRITING HAND-FIST RING THUMB", "So", 0), ("SIGNWRITING HAND-HOOK RING THUMB", "So", 0), ("SIGNWRITING HAND-FIST INDEX RING LITTLE", "So", 0), ("SIGNWRITING HAND-CIRCLE INDEX RING LITTLE", "So", 0), ("SIGNWRITING HAND-CURLICUE INDEX RING LITTLE ON", "So", 0), ("SIGNWRITING HAND-HOOK INDEX RING LITTLE OUT", "So", 0), ("SIGNWRITING HAND-HOOK INDEX RING LITTLE IN", "So", 0), ("SIGNWRITING HAND-HOOK INDEX RING LITTLE UNDER", "So", 0), ("SIGNWRITING HAND-CUP INDEX RING LITTLE", "So", 0), ("SIGNWRITING HAND-HINGE INDEX RING LITTLE", "So", 0), ("SIGNWRITING HAND-ANGLE INDEX RING LITTLE OUT", "So", 0), ("SIGNWRITING HAND-ANGLE INDEX RING LITTLE", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE DOWN", "So", 0), ("SIGNWRITING HAND-HINGE MIDDLE", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE UP", "So", 0), ("SIGNWRITING HAND-CIRCLE MIDDLE UP", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE RAISED KNUCKLE", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE UP THUMB SIDE", "So", 0), ("SIGNWRITING HAND-HOOK MIDDLE THUMB", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE THUMB LITTLE", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE LITTLE", "So", 0), ("SIGNWRITING HAND-FIST MIDDLE RING LITTLE", "So", 0), ("SIGNWRITING HAND-CIRCLE MIDDLE RING LITTLE", "So", 0), ("SIGNWRITING HAND-CURLICUE MIDDLE RING LITTLE ON", "So", 0), ("SIGNWRITING HAND-CUP MIDDLE RING LITTLE", "So", 0), ("SIGNWRITING HAND-HINGE MIDDLE RING LITTLE", "So", 0), ("SIGNWRITING HAND-ANGLE MIDDLE RING LITTLE OUT", "So", 0), ("SIGNWRITING HAND-ANGLE MIDDLE RING LITTLE IN", "So", 0), ("SIGNWRITING HAND-ANGLE MIDDLE RING LITTLE", "So", 0), ("SIGNWRITING HAND-CIRCLE MIDDLE RING LITTLE BENT", "So", 0), ("SIGNWRITING HAND-CLAW MIDDLE RING LITTLE CONJOINED", "So", 0), ("SIGNWRITING HAND-CLAW MIDDLE RING LITTLE CONJOINED SIDE", "So", 0), ("SIGNWRITING HAND-HOOK MIDDLE RING LITTLE CONJOINED OUT", "So", 0), ("SIGNWRITING HAND-HOOK MIDDLE RING LITTLE CONJOINED IN", "So", 0), ("SIGNWRITING HAND-HOOK MIDDLE RING LITTLE CONJOINED", "So", 0), ("SIGNWRITING HAND-HINGE INDEX HINGED", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB SIDE", "So", 0), ("SIGNWRITING HAND-HINGE INDEX THUMB SIDE", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB SIDE THUMB DIAGONAL", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB SIDE THUMB CONJOINED", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB SIDE THUMB BENT", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB SIDE INDEX BENT", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB SIDE BOTH BENT", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB SIDE INDEX HINGE", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB FORWARD INDEX STRAIGHT", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB FORWARD INDEX BENT", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB HOOK", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB CURLICUE", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB CURVE THUMB INSIDE", "So", 0), ("SIGNWRITING HAND-CLAW INDEX THUMB CURVE THUMB INSIDE", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB CURVE THUMB UNDER", "So", 0), ("SIGNWRITING HAND-FIST INDEX THUMB CIRCLE", "So", 0), ("SIGNWRITING HAND-CUP INDEX THUMB", "So", 0), ("SIGNWRITING HAND-CUP INDEX THUMB OPEN", "So", 0), ("SIGNWRITING HAND-HINGE INDEX THUMB OPEN", "So", 0), ("SIGNWRITING HAND-HINGE INDEX THUMB LARGE", "So", 0), ("SIGNWRITING HAND-HINGE INDEX THUMB", "So", 0), ("SIGNWRITING HAND-HINGE INDEX THUMB SMALL", "So", 0), ("SIGNWRITING HAND-ANGLE INDEX THUMB OUT", "So", 0), ("SIGNWRITING HAND-ANGLE INDEX THUMB IN", "So", 0), ("SIGNWRITING HAND-ANGLE INDEX THUMB", "So", 0), ("SIGNWRITING HAND-FIST THUMB", "So", 0), ("SIGNWRITING HAND-FIST THUMB HEEL", "So", 0), ("SIGNWRITING HAND-FIST THUMB SIDE DIAGONAL", "So", 0), ("SIGNWRITING HAND-FIST THUMB SIDE CONJOINED", "So", 0), ("SIGNWRITING HAND-FIST THUMB SIDE BENT", "So", 0), ("SIGNWRITING HAND-FIST THUMB FORWARD", "So", 0), ("SIGNWRITING HAND-FIST THUMB BETWEEN INDEX MIDDLE", "So", 0), ("SIGNWRITING HAND-FIST THUMB BETWEEN MIDDLE RING", "So", 0), ("SIGNWRITING HAND-FIST THUMB BETWEEN RING LITTLE", "So", 0), ("SIGNWRITING HAND-FIST THUMB UNDER TWO FINGERS", "So", 0), ("SIGNWRITING HAND-FIST THUMB OVER TWO FINGERS", "So", 0), ("SIGNWRITING HAND-FIST THUMB UNDER THREE FINGERS", "So", 0), ("SIGNWRITING HAND-FIST THUMB UNDER FOUR FINGERS", "So", 0), ("SIGNWRITING HAND-FIST THUMB OVER FOUR RAISED KNUCKLES", "So", 0), ("SIGNWRITING HAND-FIST", "So", 0), ("SIGNWRITING HAND-FIST HEEL", "So", 0), ("SIGNWRITING TOUCH SINGLE", "So", 0), ("SIGNWRITING TOUCH MULTIPLE", "So", 0), ("SIGNWRITING TOUCH BETWEEN", "So", 0), ("SIGNWRITING GRASP SINGLE", "So", 0), ("SIGNWRITING GRASP MULTIPLE", "So", 0), ("SIGNWRITING GRASP BETWEEN", "So", 0), ("SIGNWRITING STRIKE SINGLE", "So", 0), ("SIGNWRITING STRIKE MULTIPLE", "So", 0), ("SIGNWRITING STRIKE BETWEEN", "So", 0), ("SIGNWRITING BRUSH SINGLE", "So", 0), ("SIGNWRITING BRUSH MULTIPLE", "So", 0), ("SIGNWRITING BRUSH BETWEEN", "So", 0), ("SIGNWRITING RUB SINGLE", "So", 0), ("SIGNWRITING RUB MULTIPLE", "So", 0), ("SIGNWRITING RUB BETWEEN", "So", 0), ("SIGNWRITING SURFACE SYMBOLS", "So", 0), ("SIGNWRITING SURFACE BETWEEN", "So", 0), ("SIGNWRITING SQUEEZE LARGE SINGLE", "So", 0), ("SIGNWRITING SQUEEZE SMALL SINGLE", "So", 0), ("SIGNWRITING SQUEEZE LARGE MULTIPLE", "So", 0), ("SIGNWRITING SQUEEZE SMALL MULTIPLE", "So", 0), ("SIGNWRITING SQUEEZE SEQUENTIAL", "So", 0), ("SIGNWRITING FLICK LARGE SINGLE", "So", 0), ("SIGNWRITING FLICK SMALL SINGLE", "So", 0), ("SIGNWRITING FLICK LARGE MULTIPLE", "So", 0), ("SIGNWRITING FLICK SMALL MULTIPLE", "So", 0), ("SIGNWRITING FLICK SEQUENTIAL", "So", 0), ("SIGNWRITING SQUEEZE FLICK ALTERNATING", "So", 0), ("SIGNWRITING MOVEMENT-HINGE UP DOWN LARGE", "So", 0), ("SIGNWRITING MOVEMENT-HINGE UP DOWN SMALL", "So", 0), ("SIGNWRITING MOVEMENT-HINGE UP SEQUENTIAL", "So", 0), ("SIGNWRITING MOVEMENT-HINGE DOWN SEQUENTIAL", "So", 0), ("SIGNWRITING MOVEMENT-HINGE UP DOWN ALTERNATING LARGE", "So", 0), ("SIGNWRITING MOVEMENT-HINGE UP DOWN ALTERNATING SMALL", "So", 0), ("SIGNWRITING MOVEMENT-HINGE SIDE TO SIDE SCISSORS", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE FINGER CONTACT", "So", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE FINGER CONTACT", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE SINGLE STRAIGHT SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE SINGLE STRAIGHT MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE SINGLE STRAIGHT LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE SINGLE STRAIGHT LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE SINGLE WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE DOUBLE STRAIGHT", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE DOUBLE WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE DOUBLE ALTERNATING", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE DOUBLE ALTERNATING WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CROSS", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE TRIPLE STRAIGHT MOVEMENT", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE TRIPLE WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE TRIPLE ALTERNATING", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE TRIPLE ALTERNATING WRIST FLEX", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE BEND SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE BEND MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE BEND LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CORNER SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CORNER MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CORNER LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CORNER ROTATION", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CHECK SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CHECK MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE CHECK LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE BOX SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE BOX MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE BOX LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE ZIGZAG SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE ZIGZAG MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE ZIGZAG LARGE", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE PEAKS SMALL", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE PEAKS MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-WALLPLANE PEAKS LARGE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ROTATION-WALLPLANE SINGLE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ROTATION-WALLPLANE DOUBLE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ROTATION-WALLPLANE ALTERNATING", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ROTATION-FLOORPLANE SINGLE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ROTATION-FLOORPLANE DOUBLE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ROTATION-FLOORPLANE ALTERNATING", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE SHAKING", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ARM SPIRAL SINGLE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ARM SPIRAL DOUBLE", "So", 0), ("SIGNWRITING TRAVEL-WALLPLANE ARM SPIRAL TRIPLE", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL AWAY SMALL", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL AWAY MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL AWAY LARGE", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL AWAY LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL TOWARDS SMALL", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL TOWARDS MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL TOWARDS LARGE", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL TOWARDS LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN AWAY SMALL", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN AWAY MEDIUM", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN AWAY LARGE", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN AWAY LARGEST", "So", 0), ("SIGNWRITING MOVEMENT-DIAGONAL BETWEEN TOWARDS 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("SIGNWRITING TONGUE MOVES AGAINST CHEEK", "Mn", 0), ("SIGNWRITING TONGUE CENTRE STICKING OUT", "Mn", 0), ("SIGNWRITING TONGUE CENTRE INSIDE MOUTH", "Mn", 0), ("SIGNWRITING TEETH", "Mn", 0), ("SIGNWRITING TEETH MOVEMENT", "Mn", 0), ("SIGNWRITING TEETH ON TONGUE", "Mn", 0), ("SIGNWRITING TEETH ON TONGUE MOVEMENT", "Mn", 0), ("SIGNWRITING TEETH ON LIPS", "Mn", 0), ("SIGNWRITING TEETH ON LIPS MOVEMENT", "Mn", 0), ("SIGNWRITING TEETH BITE LIPS", "Mn", 0), ("SIGNWRITING MOVEMENT-WALLPLANE JAW", "Mn", 0), ("SIGNWRITING MOVEMENT-FLOORPLANE JAW", "Mn", 0), ("SIGNWRITING NECK", "Mn", 0), ("SIGNWRITING HAIR", "Mn", 0), ("SIGNWRITING EXCITEMENT", "Mn", 0), ("SIGNWRITING SHOULDER HIP SPINE", "So", 0), ("SIGNWRITING SHOULDER HIP POSITIONS", "So", 0), ("SIGNWRITING WALLPLANE SHOULDER HIP MOVE", "So", 0), ("SIGNWRITING FLOORPLANE SHOULDER HIP MOVE", "So", 0), ("SIGNWRITING SHOULDER TILTING FROM WAIST", "So", 0), ("SIGNWRITING TORSO-WALLPLANE STRAIGHT STRETCH", "So", 0), ("SIGNWRITING TORSO-WALLPLANE CURVED BEND", "So", 0), ("SIGNWRITING TORSO-FLOORPLANE TWISTING", "So", 0), ("SIGNWRITING UPPER BODY TILTING FROM HIP JOINTS", "Mn", 0), ("SIGNWRITING LIMB COMBINATION", "So", 0), ("SIGNWRITING LIMB LENGTH-1", "So", 0), ("SIGNWRITING LIMB LENGTH-2", "So", 0), ("SIGNWRITING LIMB LENGTH-3", "So", 0), ("SIGNWRITING LIMB LENGTH-4", "So", 0), ("SIGNWRITING LIMB LENGTH-5", "So", 0), ("SIGNWRITING LIMB LENGTH-6", "So", 0), ("SIGNWRITING LIMB LENGTH-7", "So", 0), ("SIGNWRITING FINGER", "So", 0), ("SIGNWRITING LOCATION-WALLPLANE SPACE", "So", 0), ("SIGNWRITING LOCATION-FLOORPLANE SPACE", "So", 0), ("SIGNWRITING LOCATION HEIGHT", "So", 0), ("SIGNWRITING LOCATION WIDTH", "So", 0), ("SIGNWRITING LOCATION DEPTH", "So", 0), ("SIGNWRITING LOCATION HEAD NECK", "Mn", 0), ("SIGNWRITING LOCATION TORSO", "So", 0), ("SIGNWRITING LOCATION LIMBS DIGITS", "So", 0), ("SIGNWRITING COMMA", "Po", 0), ("SIGNWRITING FULL STOP", "Po", 0), ("SIGNWRITING SEMICOLON", "Po", 0), ("SIGNWRITING COLON", "Po", 0), ("SIGNWRITING PARENTHESIS", "Po", 0), (), (), (), (), (), (), (), (), (), (), (), (), (), (), (), ("SIGNWRITING FILL MODIFIER-2", "Mn", 0), ("SIGNWRITING FILL MODIFIER-3", "Mn", 0), ("SIGNWRITING FILL MODIFIER-4", "Mn", 0), ("SIGNWRITING FILL MODIFIER-5", "Mn", 0), ("SIGNWRITING FILL MODIFIER-6", "Mn", 0), (), ("SIGNWRITING ROTATION MODIFIER-2", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-3", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-4", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-5", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-6", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-7", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-8", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-9", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-10", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-11", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-12", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-13", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-14", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-15", "Mn", 0), ("SIGNWRITING ROTATION MODIFIER-16", "Mn", 0), )

https://github.com/herbhuang/utdallas-thesis-template-typst

https://raw.githubusercontent.com/herbhuang/utdallas-thesis-template-typst/main/content/abstract_de.typ

typst

MIT License

Note: Insert the German translation of the English abstract here.

https://github.com/Zeta611/simplebnf.typ

https://raw.githubusercontent.com/Zeta611/simplebnf.typ/main/examples/system-f.typ

typst

MIT License

#import "../simplebnf.typ": * #set page( width: auto, height: auto, margin: .5cm, fill: white, ) #bnf( Prod( $e$, delim: $→$, { Or[$x$][variable] Or[$λ x: τ.e$][abstraction] Or[$e space e$][application] Or[$λ τ.e space e$][type abstraction] // @typstyle off Or[$e space [τ]$][type application] }, ), Prod( $τ$, delim: $→$, { Or[$X$][type variable] Or[$τ → τ$][type of functions] Or[$∀X.τ$][universal quantification] }, ), )

https://github.com/GYPpro/DS-Course-Report

https://raw.githubusercontent.com/GYPpro/DS-Course-Report/main/Rep/01.typ

typst

#import "@preview/tablex:0.0.6": tablex, hlinex, vlinex, colspanx, rowspanx #import "@preview/codelst:2.0.1": sourcecode // Display inline code in a small box // that retains the correct baseline. #set text(font:("Times New Roman","Source Han Serif SC")) #show raw.where(block: false): box.with( fill: luma(230), inset: (x: 3pt, y: 0pt), outset: (y: 3pt), radius: 2pt, ) // #set raw(align: center) #show raw: set text( font: ("consolas", "Source Han Serif SC") ) #set page( // flipped: true, paper: "a4", // background: [#image("background.png")] ) #set text( font:("Times New Roman","Source Han Serif SC"), style:"normal", weight: "regular", size: 13pt, ) #let nxtIdx(name) = box[ #counter(name).step()#counter(name).display()] #set par( // first-line-indent: 1cm ) #set math.equation(numbering: "(1)") // Display block code in a larger block // with more padding. #show raw.where(block: true): block.with( fill: luma(240), inset: 10pt, radius: 4pt, ) #set math.equation(numbering: "(1)") #set page( paper:"a4", number-align: right, margin: (x:2.54cm,y:4cm), header: [ #set text( size: 25pt, font: "KaiTi", ) #align( bottom + center, [ #strong[暨南大学本科实验报告专用纸(附页)] ] ) #line(start: (0pt,-5pt),end:(453pt,-5pt)) ] ) /*----*/ = 基于双向链表的`linkedList` \ #text( font:"KaiTi", size: 15pt )[ 课程名称#underline[#text(" 数据结构 ")]成绩评定#underline[#text(" ")]\ 实验项目名称#underline[#text(" ") 基于双向链表的`linkedList` #text(" ")]指导老师#underline[#text(" 干晓聪 ")]\ 实验项目编号#underline[#text(" 01 ")]实验项目类型#underline[#text(" 设计性 ")]实验地点#underline[#text(" 数学系机房 ")]\ 学生姓名#underline[#text(" 郭彦培 ")]学号#underline[#text(" 2022101149 ")]\ 学院#underline[#text(" 信息科学技术学院 ")]系#underline[#text(" 数学系 ")]专业#underline[#text(" 信息管理与信息系统 ")]\ 实验时间#underline[#text(" 2024年6月13日上午 ")]#text("~")#underline[#text(" 2024年7月13日中午 ")]\ ] #set heading( numbering: "1.1." ) = 实验目的实现一个双向列表类，在类中实现增、删、改、查的方法并完成测试 = 实验环境 \ #h(1.8em)计算机：PC X64 操作系统：Windows + Ubuntu20.0LTS 编程语言：C++：GCC std20 IDE：Visual Studio Code #pagebreak() = 程序代码 == `linkedList.h` #sourcecode[```cpp // #define _PRIVATE_DEBUG #ifndef LINKED_LIST_HPP #define LINKED_LIST_HPP #ifdef _PRIVATE_DEBUG #include <iostream> #endif namespace myDS { template<typename VALUE_TYPE> class linkedList{ protected: class linkedNode { public: VALUE_TYPE data = VALUE_TYPE(); linkedNode * next = nullptr; linkedNode * priv = nullptr; linkedNode() { } linkedNode(VALUE_TYPE _data){ next = nullptr; priv = nullptr; data = _data; } linkedNode(VALUE_TYPE _data,linkedNode * priv) { next = nullptr; priv = priv; data = _data; } ~linkedNode() { #ifdef _PRIVATE_DEBUG // if(this->next != nullptr) // std::cout << "Unexpected Delete at :" << this->data // << " with next:" << this->next->data << "\n"; #endif } linkedNode * linkNext(linkedNode * _next) { next = _next; _next->priv = this; return this->next; } linkedNode * linkPriv(linkedNode * _priv) { priv = _priv; _priv->next = this; return this->priv; } void insertNext(linkedNode * _inst){ if(_inst == nullptr) return; if(this->next == nullptr) linkNext(_inst); else { _inst->next = this->next; this->next->priv = _inst; _inst->priv = this; this->next = _inst; } } void deleteNext() { if(this->next == nullptr) return; else { linkedNode * tmp = this->next; this->next = this->next->next; this->next->priv = this; tmp->next = nullptr; delete tmp; } } }; private: class _iterator { private: linkedNode *_ptr; public: enum __iter_dest_type { front, back }; __iter_dest_type _iter_dest; _iterator(linkedNode * _upper ,__iter_dest_type _d) { _ptr = _upper; _iter_dest = _d; } VALUE_TYPE & operator*() { return _ptr->data; } VALUE_TYPE *operator->() { return _ptr; } myDS::linkedList<VALUE_TYPE>::_iterator operator++() { if (_iter_dest == front) _ptr = _ptr->next; else _ptr = _ptr->priv; return *this; } myDS::linkedList<VALUE_TYPE>::_iterator operator++(int) { myDS::linkedList<VALUE_TYPE>::_iterator old = *this; if (_iter_dest == front) _ptr = _ptr->next; else _ptr = _ptr->priv; return old; } // myDS::linkedList<VALUE_TYPE>::_iterator operator+(size_t _n) // { // if (_iter_dest == front) // _upper_idx += _n; // else // _upper_idx -= _n; // _ptr = &((*_upper_pointer)[_upper_idx]); // return *this; // } bool operator==( myDS::linkedList<VALUE_TYPE>::_iterator _b) { if (&(*_b) == _ptr) return 1; else return 0; } bool operator!=( myDS::linkedList<VALUE_TYPE>::_iterator _b) { if (&(*_b) == &(_ptr->data)) return 0; else return 1; } }; linkedNode * head = new linkedNode(); linkedNode * tail = new linkedNode(); int cap = 0; public: linkedList(){ head->linkNext(tail); } ~linkedList(){ clear(); delete head; delete tail; } void push_back(VALUE_TYPE t) { tail->data = t; tail->linkNext(new linkedNode()); tail = tail->next; cap ++; } void push_frount(VALUE_TYPE t) { head->data = t; head = (head->linkPriv(new linkedNode())); cap ++; } void clear() { linkedNode * deletingObject; while(tail->priv != head) { deletingObject = tail; tail = tail->priv; delete deletingObject; } cap = 0; delete head; delete tail; tail = new linkedNode(); head = new linkedNode(); head->linkNext(tail); } std::size_t erase(VALUE_TYPE p) { linkedNode * ptr = head; int ttl = 0; while(ptr->next != tail) { if(ptr->next->data == p){ ptr->deleteNext(); ttl ++; } else ptr = ptr->next; } cap -= ttl; return ttl; } std::size_t size() {return cap;} bool erase(linkedList<VALUE_TYPE>::_iterator p) { myDS::linkedList<VALUE_TYPE>::_iterator ptr = this->begin(); linkedNode * cur = head; while(ptr != p) { cur = cur->next; ptr ++; if(cur == tail) return 0; } cur->deleteNext(); cap --; return 1; } myDS::linkedList<VALUE_TYPE>::_iterator begin() { enum myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type _FRONT = myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type::front; return myDS::linkedList<VALUE_TYPE>::_iterator(head->next,_FRONT); } myDS::linkedList<VALUE_TYPE>::_iterator rbegin() { enum myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type _BACK = myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type::back; return myDS::linkedList<VALUE_TYPE>::_iterator(tail->priv,_BACK); } myDS::linkedList<VALUE_TYPE>::_iterator end() { enum myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type _FRONT = myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type::front; return myDS::linkedList<VALUE_TYPE>::_iterator(tail,_FRONT); } myDS::linkedList<VALUE_TYPE>::_iterator rend() { enum myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type _BACK = myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type::back; return myDS::linkedList<VALUE_TYPE>::_iterator(head,_BACK); } myDS::linkedList<VALUE_TYPE>::_iterator get(std::size_t p) { linkedNode * ptr = head->next; while(p --) ptr = ptr->next; enum myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type _FRONT = myDS::linkedList<VALUE_TYPE>::_iterator::__iter_dest_type::front; return myDS::linkedList<VALUE_TYPE>::_iterator(ptr,_FRONT); } VALUE_TYPE & operator[](std::size_t p) { linkedNode * ptr = head; while(p --) ptr = ptr->next; return ptr->next->data; } #ifdef _PRIVATE_DEBUG void innerPrint() { std::cout << "--Header[" << head << "]: " << head->data << "\n"; std::cout << "--Tail[" << tail << "]: " << tail->data << "\n"; std::cout << "-----------\n"; std::cout << "cur:" << cap<< "\n"; auto ptr = head; do { std::cout << "[" << ptr << "] ->next:" << ptr->next << " ->priv:" << ptr->priv << " ||data:" << ptr->data << "\n"; ptr = ptr->next; }while(ptr != nullptr); } #endif }; } #endif ```] == `_PRIV_TEST.cpp` #sourcecode[```cpp #define DS_TOBE_TEST linkedList #define _PRIVATE_DEBUG #include "Dev\01\linkedList.h" #include <iostream> #include <math.h> #include <vector> using namespace std; using TBT = myDS::DS_TOBE_TEST<int>; void accuracyTest() {//结构正确性测试 TBT tc = TBT(); for(;;) { string op; cin >> op; if(op == "clr") { //清空 tc.clear(); } else if(op == "q") //退出测试 { return; } else if(op == "pb")//push_back { int c; cin >> c; tc.push_back(c); } else if(op == "pf")//push_frount { int c; cin >> c; tc.push_frount(c); } else if(op == "at")//随机访问 { int p; cin >> p; cout << tc[p] << "\n"; } else if(op == "delEL")//删除所有等于某值元素 { int p; cin >> p; cout << tc.erase(p) << "\n"; } else if(op == "delPS")//删除某位置上的元素 { int p; cin >> p; cout << tc.erase(tc.get(p)) << "\n"; } else if(op == "iterF") //正序遍历 { tc.innerPrint(); cout << "Iter with index:\n"; for(int i = 0;i < tc.size();i ++) cout << tc[i] << " ";cout << "\n"; cout << "Iter with begin end\n"; for(auto x = tc.begin();x != tc.end();x ++) cout << (*x) << " ";cout << "\n"; cout << "Iter with AUTO&&\n"; for(auto x:tc) cout << x << " ";cout << "\n"; } else if(op == "iterB") //正序遍历 { tc.innerPrint(); cout << "Iter with index:\n"; for(int i = 0;i < tc.size();i ++) cout << tc[tc.size()-1-i] << " ";cout << "\n"; cout << "Iter with begin end\n"; for(auto x = tc.rbegin();x != tc.rend();x ++) cout << (*x) << " ";cout << "\n"; // cout << "Iter with AUTO&&\n";."\n"; } else if(op == "mv")//单点修改 { int p; cin >> p; int tr; cin >> tr; tc[p] = tr; } else if(op == "") { } else { op.clear(); } } } void memLeakTest() {//内存泄漏测试 TBT tc = TBT(); for(;;){ tc.push_back(1); tc.push_back(1); tc.push_back(1); tc.push_back(1); tc.clear(); } } signed main() { // accuracyTest(); memLeakTest(); } ```] = 测试数据与运行结果运行上述`_PRIV_TEST.cpp`测试代码中的正确性测试模块，得到以下内容： ``` pb 1 pb 2 pb 3 pb 4 pf 3 pb 3 iterF iterB delEL 3 iterF delPS 1 clr pb 1 pb 2 iterF delPS 0 delEL 2 iterF pb 1 pb 2 pb 3 pb 4 pf 3 pb 3 iterF --Header[0x662720]: 0 --Tail[0x662770]: 0 ----------- cur:6 [0x662720] ->next:0x662540 ->priv:0 ||data:0 [0x662540] ->next:0x662590 ->priv:0x662720 ||data:3 [0x662590] ->next:0x6625e0 ->priv:0x662540 ||data:1 [0x6625e0] ->next:0x662630 ->priv:0x662590 ||data:2 [0x662630] ->next:0x662680 ->priv:0x6625e0 ||data:3 [0x662680] ->next:0x6626d0 ->priv:0x662630 ||data:4 [0x6626d0] ->next:0x662770 ->priv:0x662680 ||data:3 [0x662770] ->next:0 ->priv:0x6626d0 ||data:0 Iter with index: 3 1 2 3 4 3 Iter with begin end 3 1 2 3 4 3 Iter with AUTO&& 3 1 2 3 4 3 iterB --Header[0x662720]: 0 --Tail[0x662770]: 0 ----------- cur:6 [0x662720] ->next:0x662540 ->priv:0 ||data:0 [0x662540] ->next:0x662590 ->priv:0x662720 ||data:3 [0x662590] ->next:0x6625e0 ->priv:0x662540 ||data:1 [0x6625e0] ->next:0x662630 ->priv:0x662590 ||data:2 [0x662630] ->next:0x662680 ->priv:0x6625e0 ||data:3 [0x662680] ->next:0x6626d0 ->priv:0x662630 ||data:4 [0x6626d0] ->next:0x662770 ->priv:0x662680 ||data:3 [0x662770] ->next:0 ->priv:0x6626d0 ||data:0 Iter with index: 3 4 3 2 1 3 Iter with begin end 3 4 3 2 1 3 delEL 3 3 iterF --Header[0x662720]: 0 --Tail[0x662770]: 0 ----------- cur:3 [0x662720] ->next:0x662590 ->priv:0 ||data:0 [0x662590] ->next:0x6625e0 ->priv:0x662720 ||data:1 [0x6625e0] ->next:0x662680 ->priv:0x662590 ||data:2 [0x662680] ->next:0x662770 ->priv:0x6625e0 ||data:4 [0x662770] ->next:0 ->priv:0x662680 ||data:0 Iter with index: 1 2 4 Iter with begin end 1 2 4 Iter with AUTO&& 1 2 4 delPS 1 1 clr Unexpected Delete at :4 with next:16187728 pb 1 pb 2 iterF --Header[0x6625e0]: 0 --Tail[0x662680]: 0 ----------- cur:2 [0x6625e0] ->next:0x662540 ->priv:0 ||data:0 [0x662540] ->next:0x662630 ->priv:0x6625e0 ||data:1 [0x662630] ->next:0x662680 ->priv:0x662540 ||data:2 [0x662680] ->next:0 ->priv:0x662630 ||data:0 Iter with index: 1 2 Iter with begin end 1 2 Iter with AUTO&& 1 2 delPS 0 1 delEL 2 1 iterF --Header[0x6625e0]: 0 --Tail[0x662680]: 0 ----------- cur:0 [0x6625e0] ->next:0x662680 ->priv:0 ||data:0 [0x662680] ->next:0 ->priv:0x6625e0 ||data:0 Iter with index: Iter with begin end Iter with AUTO&& ``` 可以看出，代码运行结果与预期相符，可以认为代码正确性无误。运行`_PRIV_TEST.cpp`中的内存测试模块，在保持CPU高占用率运行一段时间后内存变化符合预期，可以认为代码内存安全性良好。 #image("1.png")

https://github.com/lucifer1004/leetcode.typ

https://raw.githubusercontent.com/lucifer1004/leetcode.typ/main/problems/p0015.typ

typst

#import "../helpers.typ": * #import "../solutions/s0015.typ": * = 3Sum Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`. Notice that the solution set must not contain duplicate triplets. #let _3sum(nums) = { // Solve the problem here } #testcases( _3sum, _3sum-ref, ( (nums: (-1, 0, 1, 2, -1, -4)), (nums: (0, 1, 1)), (nums: (0, 0, 0)), (nums: range(-10, 20, step: 3)), (nums: range(-10, 10)) ) )

https://github.com/polarkac/MTG-Stories

https://raw.githubusercontent.com/polarkac/MTG-Stories/master/stories/003_Gatecrash.typ

typst

#import "@local/mtgset:0.1.0": conf #show: doc => conf("Gatecrash", doc) #include "./003 - Gatecrash/001_Gruul Ingenuity.typ" #include "./003 - Gatecrash/002_The Fathom Edict.typ" #include "./003 - Gatecrash/003_The Absolution of the Guildpact.typ" #include "./003 - Gatecrash/004_Persistence of Memory.typ" #include "./003 - Gatecrash/005_The Burying, Part 1.typ" #include "./003 - Gatecrash/006_The Greater Good.typ" #include "./003 - Gatecrash/007_The Guild of Deals.typ" #include "./003 - Gatecrash/008_Experiment One.typ" #include "./003 - Gatecrash/009_Fblthp.typ" #include "./003 - Gatecrash/010_The Burying, Part 2.typ" #include "./003 - Gatecrash/011_Bilagru Will Come for You.typ" #include "./003 - Gatecrash/012_The Hard Sell.typ" #include "./003 - Gatecrash/013_Behind the Black Sun.typ"

https://github.com/Starkillere/TIPE-detection-informations-cachees

https://raw.githubusercontent.com/Starkillere/TIPE-detection-informations-cachees/main/journnaldebord.typ

typst

#align(center)[ = Carnet de recherche _par <NAME> \ Initialisation : 12/03/2024 \ MÀJ : 12/03/2024 _ ] #set heading(numbering: "1.1.1 :") = 12/03/2024 : Définition du sujet. == Définition du problème La stéganographie désigne l'art de dissimler de l'information de manière subtile. Tout la sécurité de cette méthode de dissimulation réside dans la non connaissance des observateur non averties, de la présence d'une information cachée. La variante informatique de ce procéder consite dans la dissimulation des donnée dans le coprs d'autres donnée. Si la stéganographie permet de transférer des donnée à l'abrie du regards des observateurs non averties, nous pouvons toujours nous demander si il n'est pas possible d'affaiblir la sécurité de cette méthode de dissimulation. *Autrement dit est-il possible de distinguer le bruit d'une information cachée ?* == Idée d'orientation Il existe un champs de recherche à part entiere qui s'interresse à la distinction entre donnée pur et donnée issue d'une procéssuce stéganographique qui se nomme #link("https://fr.wikipedia.org/wiki/St%C3%A9ganalyse")[Stéganalyse] === Méthodes de distinction - *Analyse statistique :* \ Les données qui contiennent simplement du bruit peuvent avoir des caractéristiques statistiques différentes de celles qui cachent des informations. Vous pourriez étudier des mesures telles que l'entropie, la distribution des valeurs de pixels, les corrélations spatiales, etc. - *Analyse de la fréquence :* \ Les images qui cachent d'autres images peuvent avoir des motifs de fréquence différents de ceux des images contenant seulement du bruit. Les techniques de transformée de Fourier ou d'ondelettes peuvent être utiles pour analyser ces différences. - *Analyse visuelle :* \ Même si les données semblent similaires visuellement, il peut y avoir des artefacts ou des modèles non perceptibles à l'œil nu. Vous pourriez explorer des techniques de traitement d'image avancées pour mettre en évidence ces différences. - *Apprentissage automatique :* \ Vous pourriez également explorer des approches basées sur l'apprentissage automatique, où vous entraînez un modèle à différencier les deux types de données à partir d'un ensemble d'exemples étiquetés. = L'analyse statistique = 02/04/2024 : Phénomène aléatoire == Entropie de Shannon == Théorie de l'information = 02/04/2024 : Meqsure #line(length: 500pt) = 12/09/2024 - Implémentation Ocaml - Implémentation ocaml des algorithme pour la stéganographie image et = 19/09/2024 - git init et recher documentaire. == Définition de la problèmatique *Problématique* : Est-il possible de créer un algorithme de stéganalyse géneraliste, i.e un algorithme qui n'a pas connaissance du mode de dissimulation utilisé ? *Les differents modes de dissimulation :* - Système de substitution : remplacer une partie de la cover (1) par des donnée de l'information à dissimulée. - Transformation des paramètre de la cover : modification des paramètre physique de la cover en fonction de l'information à dissimulé (ex: fréquence) - Même choses avec le spectre. - Méthode statistique : modifier la distribution statistique de la cover en fonction de la stégo. - Techniques de distortion : stocker des informations par distorsion du signal et mesurer l'écart par rapport à la couverture originale lors de l'étape de décodage - Méthodes de génération de couverture : encoder les informations de manière à cacher un secret la communication se crée. *Objectif :* Trouver un invariant de dissimulation ! = 26/09/2024 : Prolongement par continuité de la semaine dernière (lecture 10) - *problèmatique : * Est-il possible d'identifier un paternel, une caractéristique propre aux données issues du processus de stéganographie ? == Protocole : - Étudier les differentes méthodes de stéganograpbhie (substitution, Transformation, spectre, statistique, distortion, géneration de cover) - Étudier la réponse stéganalyse à ses algorithmes - Identification d'invariant de dissimulation == 03/10/2024 : Définition formelle de l'information : [ \ l1 |(1,0,1) (1,0,1)| \ l2 |(0,0,0) (0,0,0)| \ ] \ Une information est une matrice de tuple de taille n de nombre binaire sur. - *Cas de base :* - *Information vide (null) : * #pad(x:20pt)[ On note $epsilon$ l'information vide de taille $|epsilon| = 0$ \ $epsilon = mat()$] - *Information de base :* #pad(x:20pt)[ $forall space (b_n) in BB^NN$ fini $L = mat((b_0b_1...b_n))$ de taille $|L| = n+1$ ] - *Notation* #pad(x:20pt)[ - On note $cal(M)_(n,p,l) (BB^NN)$ l'ensemble des information de matrice dans $cal(M)_(n,p) (BB^NN)$ dont les tuple sont de $l$ élément. ] - *Operation sur les informations :* - *Taille d'une information :* $"Soit" L in cal(M)_(n,p)(BB^NN) "une information"$, la taille de $L$ est noté $|L| = n×p$ - *Caractéristiques d'une information :* $"Soit" L in cal(M)_(n,p)(BB^NN) "une information" $ - *Union/Intersection :* $"Soit" L_1 "et" L_2 "deux information de taille" n$ - $L_1 union L_2 =$ = 10/10/2024 : Définition du repertoir documentation/prototypage == Définition formelle de l'information = Vocabulaire (MAJ 12/03/2024) + donnée pur : donnée de cachant pas d'autres données issue d'un processuce stéganographique. + cover : suport pour la dissimulation d'information cachée. + stego : information à cachée. + = Lecture en attente : + #link("https://utt.hal.science/hal-02470070/document")\ + #link("https://fr.wikipedia.org/wiki/Entropie_de_Shannon")\ + #link("https://fr.wikipedia.org/wiki/Th%C3%A9orie_de_l%27information")\ + #link("https://hal.science/hal-00394108/document")\ + #link("https://greenteapress.com/thinkdsp/thinkdsp.pdf")\ + #link(" http://tinyurl.com/thinkdsp08")\ // REP- pour les algo de traitement de signale + #link("https://fr.wikipedia.org/wiki/Algorithme_de_Knuth-Morris-Pratt")\ + #link("https://theses.hal.science/tel-00706171v2/file/RCogranne_soutenance.pdf") + #link("https://repository.root-me.org/St%C3%A9ganographie/FR%20-%20Analyse%20st%C3%A9ganographique%20d%27images%20num%C3%A9riques.pdf") + #link("https://d1wqtxts1xzle7.cloudfront.net/11025045/22359536_lese_1-libre.pdf?1363619886=&response-content-disposition=inline%3B+filename%3DA_survey_of_steganographic_techniques.pdf&Expires=1726758425&Signature=UWNEvv4JIxHsL-iZcX-PzwvRlbmce0~unnnAUFS2lB~tsuJUbrH1Mzt4ZnO~D1Dhn9DKUo0jtG-BZnkuZYYz5iSvTUuJHJJqcZ65yceho5qgmi7Jpv9OnJsNLxnqAjhHp~frVhRI3yYvhmZRsOL0gdCCCy6O5Bb9XcylGMKZA5k8SZq0Jqme~XdEXRGESCvJy69F2bQ5K~X5IF9j5VaYj7WMOj~n-QC8DG2cJBk-1GRz5NbPu5Udq4R1U-pr2GvYZKJJmqnb7MQoutftG~9-jS~WMxnag3IlAe8g~vlz87mWWLxGle-6fbBg1I-EOa63b3fzUVsFY2bLQo0WgwqNMQ__&Key-Pair-Id=<KEY>")

https://github.com/Myriad-Dreamin/typst.ts

https://raw.githubusercontent.com/Myriad-Dreamin/typst.ts/main/fuzzers/corpora/bugs/hide-meta_01.typ

typst

Apache License 2.0

#import "/contrib/templates/std-tests/preset.typ": * #show: test-page #set text(8pt) #outline() #set text(2pt) #hide(block(grid( [= A], [= B], block(grid( [= C], [= D], )) )))

https://github.com/0x1B05/english

https://raw.githubusercontent.com/0x1B05/english/main/grammar/content/名词.typ

typst

#import "../template.typ": * = 名词名词短语：限定词+形容词+名词，每一个部分都有可能缺失。限定词的部分，如果后面的名词是复数或者不可数，就可以采用*零冠词* #tip("Tip")[ 句子中的名词，哪怕只有一个字，也要当作由三个部分组成的名词短语来看。 ] == 名词 === 可数名词度量衡单位： - a pound - 20 feet - an hour - ... === 不可数名词 ==== 物质名词没有固定形状。（前面可以采用零冠词） - water(水) - air(空气) - gold(金) - paper(纸) #tip("Tip")[ 如果看到物质名词前面加了a或者末尾加了s，那已经不是物质名词了，已经当作可数的普通名词来用了。 ] #example("Example")[ - I have a _paper_ to write tonight.(一份报告) - Drinking _a couple of beers_ a day won't do you any harm.(几瓶啤酒) ] ==== 抽象名词 - cowardice - ugliness - wisdom - eternity #tip("Tip")[ 如果看到抽象名词前面加了a或者末尾加了s，已经当作可数的普通名词来用了，意义往往不同。 ] #example("Example")[ - Your sister is _a real beauty._ ] ==== 动名词少数的动名词可以当可数名词使用： - There were _three weddings_ at this restaurant yesterday. ==== 专有名词一个名词只代表单一的一个对象，属于不可数（因为只有一个对象） - London - <NAME> == 复合名词三种写法： - 两个名词写在一次成为一个单字。（the dishwasher) - 两个名词之间有个连字符。（the pole-vaulter) - 两个名词分开。（the flower shop）即使使永远复数的名词（pants,trousers,glasses,scissors...)放在复合名词当形容词使用的时候也要改为单数，例如_his trouser pocket_ 但是_a clothes hanger(衣架)_，如果改成单数，就成了布架，所以这时候就需要复数了。还有一些习惯采用复数的形容词。 - a sports car - the admissions office == 副词加在形容词前面加强语气：_that rather old jacket_ == 名词短语的省略 *省略之后要让读者非常清楚想要表达的意思。*

https://github.com/kaarmu/splash

https://raw.githubusercontent.com/kaarmu/splash/main/doc/util.typ

typst

MIT License

#let code(body) = { set text(weight: "regular") show: box.with( fill: luma(240), inset: 0.4em, radius: 3pt, baseline: 0.4em, ) raw(body) } #let get-color-value(color) = { let s = repr(color) let m = s.match(regex("(.*)\$(.*)\$")) let p = (name: m.captures.at(0), value: m.captures.at(1).replace("\"", "", count: 2)) text(fill: luma(200))[ #raw(p.name) #h(1fr) #raw(p.value) ] } #let make-title(title: none, author: none, date: none, description: none) = [ #set align(center) = #title #v(1em) #text(style: "italic", description) #v(1em) / Author: #author / Date: #date #v(3em) ] #let section( title: none, description: none, cols: 2, col-gutter: 2em, row-gutter: 2pt, do-page-break: true, name: none, colors, ) = { heading(level: 3, title + if name != none [ --- #code(name); ]) v(.5em) if description != none { description } let arr = () for (name, color) in colors.pairs() { let blk = rect( stroke: none, )[ #set align(horizon) #code(name) #h(1fr) #box( width: 3em, height: 1em, fill: color, stroke: luma(230), radius: 2pt, baseline: 0.25em, ) #linebreak() #get-color-value(color) ] arr.push(blk) } grid( row-gutter: row-gutter, column-gutter: col-gutter, columns: (1fr,) * cols, ..arr, ) if do-page-break { pagebreak(weak:true) } }

https://github.com/jomaway/typst-teacher-templates

https://raw.githubusercontent.com/jomaway/typst-teacher-templates/main/examples/exam/cover.typ

typst

MIT License

#import "@local/ttt-exam:0.1.0": cover-page #let details = toml("details.toml") #let meta = ( class: details.exam.class, subject: details.exam.subject, kind: "sa", dates: ( gehalten: details.exam.date, zurückgegeben: none, eingetragen: none, ), comment: none, total_points: 70, ) #set text(lang: "de", font: "Rubik", weight: 300) #set strong(delta: 200) #cover-page(..meta)

https://github.com/liuguangxi/erdos

https://raw.githubusercontent.com/liuguangxi/erdos/master/Problems/typstdoc/figures/p151.typ

typst

#import "@preview/cetz:0.2.1" #cetz.canvas({ import cetz.draw: * let fill-color = blue.lighten(80%) circle((0, 0), radius: 0.5, fill: fill-color, name: "c1") circle((-2, 1.5), radius: 0.5, fill: fill-color, name: "c2") circle((2, 1.5), radius: 0.5, fill: fill-color, name: "c3") circle((-3, 3), radius: 0.5, fill: fill-color, name: "c4") circle((-1, 3), radius: 0.5, fill: fill-color, name: "c5") circle((1, 3), radius: 0.5, fill: fill-color, name: "c6") circle((3, 3), radius: 0.5, fill: fill-color, name: "c7") circle((-2, 4.5), radius: 0.5, fill: fill-color, name: "c8") circle((2, 4.5), radius: 0.5, fill: fill-color, name: "c9") circle((0, 6), radius: 0.5, fill: fill-color, name: "c10") line("c2", "c1", mark: (end: "stealth", fill: black)) line("c3", "c1", mark: (end: "stealth", fill: black)) line("c4", "c2", mark: (end: "stealth", fill: black)) line("c5", "c2", mark: (end: "stealth", fill: black)) line("c6", "c3", mark: (end: "stealth", fill: black)) line("c7", "c3", mark: (end: "stealth", fill: black)) line("c8", "c4", mark: (end: "stealth", fill: black)) line("c8", "c5", mark: (end: "stealth", fill: black)) line("c9", "c6", mark: (end: "stealth", fill: black)) line("c9", "c7", mark: (end: "stealth", fill: black)) line("c10", "c8", mark: (end: "stealth", fill: black)) line("c10", "c9", mark: (end: "stealth", fill: black)) })

https://github.com/TechnoElf/mqt-qcec-diff-presentation

https://raw.githubusercontent.com/TechnoElf/mqt-qcec-diff-presentation/main/content/data.typ

typst

#let unclip(res) = { res.filter(r => not r.clipped).enumerate().map(((i, r)) => { r.i = i r }) } #let sort-by-circuit-size(res) = { res.sorted(key: r => r.total-circuit-size).enumerate().map(((i, r)) => { r.i = i r }) } #let filter(res) = { res.filter(r => r.equivalence-rate > 0.35).enumerate().map(((i, r)) => { r.i = i r }) } #let filter-rev(res) = { res.filter(r => r.equivalence-rate-rev > 0.35).enumerate().map(((i, r)) => { r.i = i r }) } #let results-r1-b5q16-cprop = csv("../resources/results-r1-b5q16-cprop-smc.csv", row-type: dictionary) #let results-r1-b5q16-cmyersrev-pmismc = csv("../resources/results-r1-b5q16-cmyersrev-pmismc-smc.csv", row-type: dictionary) #let results-r1-b5q16-cmyersrev-p = csv("../resources/results-r1-b5q16-cmyersrev-p-smc.csv", row-type: dictionary) #let results-r1-b5q16-cmyers-p = csv("../resources/results-r1-b5q16-cmyers-p-smc.csv", row-type: dictionary) #let results-r1-b5q16-cpatience-p = csv("../resources/results-r1-b5q16-cpatience-p-smc.csv", row-type: dictionary) #let results-r1-b5q16-cmyers-pmismc = csv("../resources/results-r1-b5q16-cmyers-pmismc-smc.csv", row-type: dictionary) #let results-r1-b5q16 = results-r1-b5q16-cprop.enumerate().map(((i, r)) => { let cmyersrev-pmismc = results-r1-b5q16-cmyersrev-pmismc.find(r2 => r2.name == r.name) let cmyersrev-p = results-r1-b5q16-cmyersrev-p.find(r2 => r2.name == r.name) let cmyers-p = results-r1-b5q16-cmyers-p.find(r2 => r2.name == r.name) let cmyers-pmismc = results-r1-b5q16-cmyers-pmismc.find(r2 => r2.name == r.name) let cpatience-p = results-r1-b5q16-cpatience-p.find(r2 => r2.name == r.name) let num-gates-1 = float(r.numGates1) let num-gates-2 = float(r.numGates2) let total-circuit-size = num-gates-1 + num-gates-2 let num-qubits-1 = int(r.numQubits1) let num-qubits-2 = int(r.numQubits2) ( name: r.name, i: i, clipped: not ((r.finished == "true") and (cmyersrev-pmismc.finished == "true") and (cmyersrev-p.finished == "true") and (cmyers-pmismc.finished == "true") and (cmyers-p.finished == "true")), total-circuit-size: total-circuit-size, circuit-size-difference: calc.abs(num-gates-1 - num-gates-2), total-qubit-count: num-qubits-1 + num-qubits-2, qubit-count-difference: calc.abs(num-qubits-1 - num-qubits-2), equivalence-rate: float(cmyers-pmismc.diffEquivalenceCount) / total-circuit-size, equivalence-rate-rev: float(cmyersrev-pmismc.diffEquivalenceCount) / total-circuit-size, cprop: ( mu: float(r.runTimeMean) ), cmyersrev-pmismc: ( mu: float(cmyersrev-pmismc.runTimeMean) ), cmyersrev-p: ( mu: float(cmyersrev-p.runTimeMean) ), cmyers-p: ( mu: float(cmyers-p.runTimeMean) ), cmyers-pmismc: ( mu: float(cmyers-pmismc.runTimeMean) ), cpatience-p: ( mu: float(cpatience-p.runTimeMean) ), ) }) #let results-r1-b5q16-hist = { let min = calc.log(0.001) let max = calc.log(20) let bins = 15 let bins-mu = range(bins + 1).map(x => calc.pow(10, min + x * (max - min) / bins)) let cprop-mu = bins-mu.slice(1).map(_ => 0) let cmyersrev-pmismc-mu = bins-mu.slice(1).map(_ => 0) let cmyersrev-p-mu = bins-mu.slice(1).map(_ => 0) let cmyers-p-mu = bins-mu.slice(1).map(_ => 0) let cmyers-pmismc-mu = bins-mu.slice(1).map(_ => 0) let cpatience-p-mu = bins-mu.slice(1).map(_ => 0) for r in unclip(results-r1-b5q16) { for b in range(bins) { if bins-mu.at(b) <= r.cprop.mu and r.cprop.mu < bins-mu.at(b + 1) { cprop-mu.at(b) += 1 } if bins-mu.at(b) <= r.cmyersrev-pmismc.mu and r.cmyersrev-pmismc.mu < bins-mu.at(b + 1) { cmyersrev-pmismc-mu.at(b) += 1 } if bins-mu.at(b) <= r.cmyersrev-p.mu and r.cmyersrev-p.mu < bins-mu.at(b + 1) { cmyersrev-p-mu.at(b) += 1 } if bins-mu.at(b) <= r.cmyers-p.mu and r.cmyers-p.mu < bins-mu.at(b + 1) { cmyers-p-mu.at(b) += 1 } if bins-mu.at(b) <= r.cmyers-pmismc.mu and r.cmyers-pmismc.mu < bins-mu.at(b + 1) { cmyers-pmismc-mu.at(b) += 1 } if bins-mu.at(b) <= r.cpatience-p.mu and r.cpatience-p.mu < bins-mu.at(b + 1) { cpatience-p-mu.at(b) += 1 } } } let scientific(val) = { let exp = calc.floor(calc.log(val)) [$#(calc.round(val / calc.pow(10, exp), digits: 2)) dot 10 ^ #exp$] } ( bins-mu: bins-mu.slice(0, -1).zip(bins-mu.slice(1)).map(((s, e)) => [$<$ #scientific(e)]), cprop: ( mu: cprop-mu ), cmyersrev-pmismc: ( mu: cmyersrev-pmismc-mu ), cmyersrev-p: ( mu: cmyersrev-p-mu ), cmyers-p: ( mu: cmyers-p-mu ), cmyers-pmismc: ( mu: cmyers-pmismc-mu ), cpatience-p: ( mu: cpatience-p-mu ) ) }

https://github.com/zurgl/typst-resume

https://raw.githubusercontent.com/zurgl/typst-resume/main/templates/main.typ

typst

#import "../metadata.typ": * #import "commun.typ": * #import "letter/main.typ": * #import "resume/section.typ": * #import "resume/entry.typ": * #import "resume/skills.typ": * #import "resume/header.typ": * #import "resume/footer.typ": * #import "@preview/fontawesome:0.1.0": *

https://github.com/Floffah/documents

https://raw.githubusercontent.com/Floffah/documents/main/lib/template.typ

typst

MIT License

// Feature inspiration taken from Ilm (MIT) - https://github.com/talal/ilm // The project function defines how your document looks. // It takes your content and some metadata and formats it. // Go ahead and customize it to your liking! #let project( title: "", authors: (), date: none, logo: none, formal: false, // Whether to display a maroon circle next to external links. external-link-circle: true, // Display an index of figures (images). figure-index: ( enabled: false, title: "", ), body, ) = { // Set the document's basic properties. set document(author: authors.map(a => a.name), title: title) set page(numbering: "1", number-align: center) let font = "Source Sans Pro" if formal { font = "Libertinus Serif" } set text(font: font, lang: "en", size: 12pt) // Set paragraph spacing. set par(spacing: 1.2em) set heading(numbering: "1.a.i") // See ILM (MIT) - https://github.com/talal/ilm/blob/main/lib.typ show link: it => { it // Workaround for ctheorems package so that its labels keep the default link styling. if external-link-circle { if type(it.dest) == str { sym.wj h(1.6pt) sym.wj super(box(height: 3.8pt, circle(radius: 1.2pt, stroke: 0.7pt + rgb("#993333")))) } else if type(it.dest) == label { sym.wj h(0.6pt) sym.wj super(box(height: 3.8pt, text("#", stroke: 0.2pt + rgb("#0284c7")))) } } } // Title page. // The page can contain a logo if you pass one with `logo: "logo.png"`. v(0.6fr) if logo != none { align(right, image(logo, width: 26%)) } v(9.6fr) text(1.1em, date) v(1.2em, weak: true) text(2em, weight: 700, title) // Author information. pad( top: 0.7em, right: 20%, grid( columns: (1fr,) * calc.min(3, authors.len()), gutter: 1em, ..authors.map(author => align(start)[ *#author.name* \ #author.affiliation ]), ), ) v(2.4fr) pagebreak() // Table of contents. outline(depth: 3, indent: true) pagebreak() // Main body. set par(justify: true) set text(hyphenate: false) set list(marker: ([•], [◦], [‣], [⁃])) // Utils let ignore(content) = {} body pagebreak() bibliography(("../references.yml", "../zotero.bib"), style: "institute-of-electrical-and-electronics-engineers") // See ILM (MIT) - https://github.com/talal/ilm/blob/main/lib.typ let fig-t(kind) = figure.where(kind: kind) let has-fig(kind) = counter(fig-t(kind)).get().at(0) > 0 if figure-index.enabled { show outline: set heading(outlined: true) context { let imgs = figure-index.enabled and has-fig(image) if imgs { // Note that we pagebreak only once instead of each each // individual index. This is because for documents that only have a couple of // figures, starting each index on new page would result in superfluous // whitespace. pagebreak() } if imgs { outline(title: figure-index.at("title", default: "Index of Figures"), target: fig-t(image)) } } } }

https://github.com/Yzx7/public_study_files

https://raw.githubusercontent.com/Yzx7/public_study_files/main/Monografía FIEE/template.typ

typst

// The project function defines how your document looks. // It takes your content and some metadata and formats it. // Go ahead and customize it to your liking! #let project(title: "",t-tipo:"", t-para:"", authors: (),cursoName:"",docenteName:"", fecha:"",add:(("")),logo: none, body) = { // Set the document's basic properties. set document(author: authors, title: title) set page( margin: (left: 35mm, right: 35mm, top: 30mm, bottom: 30mm), numbering: "1", number-align: end, ) set text(font: "New Computer Modern", lang: "es") show math.equation: set text(weight: 400) set heading(numbering: "1.1.") // Title page. // The page can contain a logo if you pass one with `logo: "logo.png"`. align(center)[ #image("LogoUNMSM icon.png", height: 60pt) #v(10pt) #block(width: 100%)[ #text(size: 16pt, weight: "extrabold")[ Universidad Nacional Mayor de San Marcos ] #text(size: 11pt, weight: "extrabold")[ Universidad del Perú. Decana de América ] #text(size: 12pt, weight: "extrabold")[ Facultad de Ingeniería Electrónica y Eléctrica ] #text(size: 10pt, weight: "extrabold")[ // EAP: Ing. Telecomunicaciones ] ] ] v(1fr) align(center)[ #text(1.5em, weight: 700, title) #v(20pt) #text(1.5em, weight: 700, t-tipo) #v(20pt) #text(1.4em, t-para) #v(1fr) // Author information. // #pad( // top: 0.7em, //grid( // columns: (1fr,) * calc.min(3, authors.len()), //gutter: 1em, //..authors.map(author => align(start, strong(author))), // ), //) #text( weight: 700)[Autores:] #pad( bottom: 10pt, grid( columns: (1fr), gutter: 1em, ..authors.map(author => align(center, author)), ) ) ] align(center)[ #grid( gutter: 12pt, ..add.map((ait) => grid(columns: 2, gutter: 4pt, text(weight: 700)[#ait.at(0):], ait.at(1) ) ), ) ] if docenteName != "" { align(center)[ #text( weight: 700)[Docente:] #docenteName ] } if cursoName != "" { align(center)[ #text(weight: 700)[Curso:] #cursoName ] } if fecha != "" { align(center)[ #text( weight: 700)[Fecha:] #fecha ] } align(center)[ #v(1.4fr) #text(size: 1.2em, weight: 700)[Lima, Perú] // #text(size: 1.2em, weight: 700)[30 de junio] #text(size: 1.2em, weight: 700)[2024] ] pagebreak() // Table of contents. // outline(depth: 3, indent: true,target: heading.where(outlined: true) ) // pagebreak() // outline(depth: 3, indent: true,target: figure.where(kind: image), title: "Índice de Figuras" ) // pagebreak() // outline(depth: 3, indent: true,target: figure.where(kind: "table"), title: "Índice de tablas" ) // pagebreak() // Main body. set par(justify: true) set text(hyphenate: false) body }

https://github.com/JakMobius/courses

https://raw.githubusercontent.com/JakMobius/courses/main/mipt-os-basic-2024/sem01/utils.typ

typst

#import "@preview/cetz:0.2.2" #let draw-compiler-lifecycle(arr) = { let margin = -1.5 let arrow-top = 5.5 let arrow-bottom = 4.5 let arrow-shortage = 1.4 let anchor-prev = 0 let x = 0 let i = 0 cetz.draw.set-style(mark: (end: ">"), stroke: 3pt + black) for step in arr { let background-color = color.mix((step.color, 20%), (white, 80%)) let stroke-color = color.mix((step.color, 50%), (black, 50%)) let text-color = stroke-color let has-code = step.at("code", default: none) != none let lower-boundary = 0 if has-code { lower-boundary = 1.4 } let y = 0 if calc.rem(i, 2) == 0 { y = 6 } cetz.draw.content( (x, y + 4), (x + step.width, y), padding: 0, )[ #box( fill: background-color, radius: 20pt, width: 100%, height: 100%, stroke: 1pt + stroke-color, ) ] cetz.draw.content( (x, y + 4), (x + step.width, y + lower-boundary), padding: 0, )[ #set text(fill: text-color, size: 18pt) #box( width: 100%, height: 100%, inset: (left: 7pt, top: 7pt, right: 7pt, bottom: 0pt), )[ #align(center + horizon)[ #step.text ] ] ] if has-code { cetz.draw.content( (x, y + lower-boundary), (x + step.width, y), padding: 0, )[ #set text(fill: text-color, font: "Monaco", size: 18pt) #box( width: 100%, height: 100%, inset: (left: 7pt, top: 0pt, right: 7pt, bottom: 7pt), )[ #align(center + horizon)[ #step.code ] ] ] } let anchor = x + step.width / 2 - arrow-shortage if x != 0 { if calc.rem(i, 2) == 0 { cetz.draw.line((anchor-prev, arrow-bottom), (anchor, arrow-top)) } else { cetz.draw.line((anchor-prev, arrow-top), (anchor, arrow-bottom)) } } anchor-prev = x + step.width / 2 + arrow-shortage x = x + margin + step.width i = i + 1 } }

https://github.com/viniciusmuller/ex_typst

https://raw.githubusercontent.com/viniciusmuller/ex_typst/main/README.md

markdown

Apache License 2.0

# ExTypst Elixir bindings and helpers for the [`typst`](https://github.com/typst/typst) typesetting system. Check [Typst's documentation](https://typst.app/docs) for a quick start. # Usage ```elixir # Write typst markup template = """ = Current Employees This is a report showing the company's current employees. #table( columns: (auto, 1fr, auto, auto), [*No*], [*Name*], [*Salary*], [*Age*], <%= employees %> ) """ # Create some data defmodule Helper do @names ["John", "Nathalie", "Joe", "Jane", "Tyler"] @surnames ["Smith", "Johnson", "Williams", "Brown", "Jones", "Davis"] def build_employees(n) do for n <- 1..n do name = "#{Enum.random(@names)} #{Enum.random(@surnames)}" salary = "US$ #{Enum.random(1000..15_000) / 1}" [n, name, salary, Enum.random(16..60)] end end end # Convert it to a nice-looking PDF {:ok, pdf_binary} = ExTypst.render_to_pdf(template, employees: ExTypst.Format.table_content(Helper.build_employees(1_000)) ) # Write to disk File.write!("employees.pdf", pdf_binary) # Or maybe send via email Bamboo.Email.put_attachment(email, %Bamboo.Attachment{data: pdf_binary, filename: "employees.pdf"}) ``` You can see the generated PDF [here](./examples/employees.pdf). ## Security Please note that currently ExTypst is experimental and content added to templates is not escaped. ## Installation If [available in Hex](https://hex.pm/docs/publish), the package can be installed by adding `ex_typst` to your list of dependencies in `mix.exs`: ```elixir def deps do [ {:ex_typst, "~> 0.1"} ] end ``` Documentation can be generated with [ExDoc](https://github.com/elixir-lang/ex_doc) and published on [HexDocs](https://hexdocs.pm). Once published, the docs can be found at <https://hexdocs.pm/ex_typst>.

https://github.com/gianzamboni/cancionero

https://raw.githubusercontent.com/gianzamboni/cancionero/main/wip/cuentamedusa.typ

typst

#import "../theme/project.typ": *; #cancion("Cuentamedusa","Valbé")[ #seccion[A] Pensemos otra vez Si faltara vivir Quizás el mundo No termina para ti Después de tanto Que se ha hablado de sufrir No tengas tiempo De mirarte sonreír Un día caerás Un día caerás Pero hasta entonces no tendremos la necesidad De respirar mejor Y hablar de la sinceridad Y rescatarnos de la generalidad Y no hablo de traición Y no hablo de traición Pero quisiera que la vida la pase mejor Y menos de crueldad Y menos desamor Y menos desarraigo A nuestro corazón #seccion[Repetir: A] Cada vez, que pienso en el mar Me sale la voz Cada vez que miro el brillo de tu bien Y de tu mal De tu bien y de tu mal #seccion[C] Cuentame dusa como Has hecho para colorear la sal Cuentame dusa lagrimas de mar Cuentame dusa como Has hecho para colorear la sal Cuentame dusa lagrimas de mar Cuentame dusa Como has aprendido a nadar Como has aprendido a nadar #seccion[Repetir: C] Como has aprendido a nadar ]

https://github.com/teamdailypractice/pdf-tools

https://raw.githubusercontent.com/teamdailypractice/pdf-tools/main/typst-pdf/examples/example-01.typ

typst

In this report, we will explore the various factors that influence fluid dynamics in glaciers and how they contribute to the formation and behaviour of these natural structures.

https://github.com/jgm/typst-hs

https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/compiler/ops-invalid-16.typ

typst

Other

// Error: 3-28 cannot divide these two relative lengths #((10% + 1pt) / (20% + 1pt))

https://github.com/LEXUGE/typzk

https://raw.githubusercontent.com/LEXUGE/typzk/main/graph.typ

typst

MIT License

#import "@preview/diagraph:0.2.5" #let digraphState = state("typzk_digraphState", (graph: (:), hierarchy: (), labels: (:), clusters: (:))) #let deep-merge-pair(dict1, dict2) = { let final = dict1 for (k, v) in dict2 { if (k in dict1) { if type(v) == "dictionary" { final.insert(k, deep-merge-pair(dict1.at(k), v)) } else { final.insert(k, dict2.at(k)) } } else { final.insert(k, v) } } return final } #let deep-merge(..args) = { let final = args.pos().first() for dict in args.pos() { final = deep-merge-pair(final, dict) } return final } #let node_descend(hierarchy, identity, payload) = { let graph = (:) if hierarchy.len() != 0 { let h = hierarchy.remove(0) graph.insert(h, node_descend(hierarchy, identity, payload)) } else { graph.insert(identity, payload) } return graph } // TODO: Allow setting extra options #let node(identity, prefix: "node_", desc: none, links: (), back_links: (), body) = { let prefixed_identity = prefix + identity let edges = () for dst in links { edges.push(identity + "->" + dst) } for orig in back_links { edges.push(orig + "->" + identity) } digraphState.update(state => { state.graph = deep-merge(state.graph, node_descend(state.hierarchy, identity, edges)) let linked_desc = if type(desc) == content { [#link(label(prefixed_identity), desc)] } else { [#link(label(prefixed_identity), identity)] } let new_label = (:) new_label.insert(identity, linked_desc) state.labels = deep-merge(state.labels, new_label) return state }) [#body #label(prefixed_identity)] } // TODO: Allow setting extra options #let subgraph(identity, desc: none, prefix: "cluster_", body) = { let prefixed_identity = prefix + identity digraphState.update(state => { state.hierarchy.push(prefixed_identity); if type(desc) == content { let new_desc = (:) new_desc.insert(prefixed_identity, [#link(label(prefixed_identity), desc)]) state.clusters = deep-merge(state.clusters, new_desc) } return state }) // This seems to work: link to the first element of the body [#body #label(prefixed_identity)] digraphState.update(state => { state.hierarchy.pop(); return state }) } #let heading_to_label(prefix: "cluster_") = { let prefixed_identity = prefix let lvls = counter(heading).get() for x in lvls { prefixed_identity += str(x) + "_" } return prefixed_identity } // Create subgraph using heading // NOTE: Seems like state update call must be wrapped in a content block, otherwise it will not take effect. #let heading_subgraph(args, prefix: "cluster_") = { let desc = args.body let prefixed_identity = heading_to_label(prefix: prefix) let lvls = counter(heading).get() digraphState.update(state => { assert(lvls.len() - state.hierarchy.len() <= 1, message: "heading levels must only increment by 1 at maximum"); while lvls.len() <= state.hierarchy.len() { state.hierarchy.pop(); } if lvls.len() > state.hierarchy.len() { state.hierarchy.push(prefixed_identity); } if type(desc) == content { let new_desc = (:) new_desc.insert(prefixed_identity, [#link(label(prefixed_identity), desc)]) state.clusters = deep-merge(state.clusters, new_desc) } return state }) } // Assemble our graph state into DOT language // All statements must end with colon #let marshal(graph) = { let s = "" let e = () for (k, v) in graph { if type(v) == "dictionary" { let (nodes, edges) = marshal(v) // This label statement would be overriden when `clusters` has matching identity. // NOTE: This label statement is necessary as somehow otherwise the label replacement will not take effect. // In addition to that, replaced label will be pushed around, so the only workaround is to set all subgraph with empty label and replace using `clusters` later. let label_stmt = "label=\"\";" s += ("subgraph " + k + " {" + label_stmt + nodes + " };") e += edges } else if type(v) == "array" { s += (k + ";") e += v } } return (s, e) } // Generate the final graphviz code #let gen_graphviz(graph, extra: "", path: ()) = { let scoped_graph = if path.len() == 0 { graph } else { let g = graph for name in path { g = g.at("cluster_" + name) } g } let (nodes, edges) = marshal(scoped_graph) let s = "digraph {" + extra s += nodes s += if edges.len()!=0 { edges.join(";") + ";" } s += "}" return s } // Use path to indicate the subgraph to render #let render_graph(extra: "", path: ()) = context { let state = digraphState.final() let s = gen_graphviz(state.graph, extra: extra, path: path) return diagraph.render(s, labels: state.labels, clusters: state.clusters) }

https://github.com/tiankaima/typst-notes

https://raw.githubusercontent.com/tiankaima/typst-notes/master/feebf7-2023_fall_TA/recitations/recitation_1.typ

typst

#import "../utils.typ": * = 习题课 1 ```plain Time: Week 1, 09.17 Sun. 19:00 ~ 20:30 ``` 摘要: 归纳/初等不等式/可数/习题选讲 == 归纳公理 #statement[ $ cases( 0 in S, n in S => n+1 in S ) quad => quad S = NN $ ] === 例题 #homework[ $forall n in NN, f(n) = n^4 + 2n^3 + 2n^2 + n$ 证明 $6 | f(n)$ (《数学基础选讲》程艺 P3) ] #homework[ $a_1, a_2, ..., a_n (n>=2)$都是正数且$a_1 + a_2 + ... + a_n < 1$,求证: $ 1 / (1- sum_(k=1)^n a_k) > product_(k=1)^n (1+a_k) > 1 + sum_(k=1)^n a_k $ ] #v(6cm) #pagebreak() == 初等不等式 === Cauchy 不等式 #statement[ $ forall a_1, a_2, ..., a_n, b_1, b_2, ..., b_n in RR quad (sum_(k=1)^n a_k^2)(sum_(k=1)^n b_k^2) >= (sum_(k=1)^n a_k b_k)^2 $ $ <=> x, y in RR^n quad abs(x dot.c y) <= abs(x) abs(y) $ ] #proof[ 考虑二次函数: $ (a_1 x + b_1)^2 + (a_2 x + b_2)^2 + ... + (a_n x + b_n)^2 \ = sum_(k=1)^n a_k^2 x^2 + 2 sum_(k=1)^n a_k b_k x + sum_(k=1)^n b_k^2 $ 由于二次函数恒大于等于0, 所以判别式小于等于0, 即: $ Delta / 4 = (sum_(k=1)^n a_k b_k)^2 - (sum_(k=1)^n a_k^2)(sum_(k=1)^n b_k^2) <= 0 quad qed $ ] === Bernoulli 不等式 #statement[ $ forall x in RR, quad x >= -1 quad n in NN, n >= 2\ (1+x)^n >= 1 + n x $ ] #proof[ $n=2$ 时, $ (1+x)^2 = x^2 + 2x + 1 >= 1 + 2x $ 假设 $n=k$ 时成立, 则 $n=k+1$ 时: $ (1+x)^{k+1} = (1+x)^k (1+x) >= (1+k x)(1+x) = 1 + (k+1)x + k x^2 >= 1 + (k+1)x $ ] === HM-GM-AM-QM 不等式 #statement[ $ forall a_1, a_2, ..., a_n in RR_+\ n / (1 / a_1 + 1 / a_2 + ... + 1 / a_n) <= ( a_1 a_2 ... a_n )^(1 / n) <= (a_1 + a_2 + ... + a_n) / n <= sqrt((a_1^2 + a_2^2 + ... + a_n^2)/n) $ 其中几个平均值的定义为: - 算数平均值(#strong[A]rithmetic #strong[M]ean): $ "AM" = (a_1 + a_2 + ... + a_n)/n $ - 几何平均值(#strong[G]eometric #strong[M]ean): $ "GM" = (a_1 a_2 ... a_n)^(1/n) $ - 调和平均值(#strong[H]armonic #strong[M]ean): $ "HM" = n/(1/a_1 + 1/a_2 + ... + 1/a_n) $ - 平方平均值(#strong[Q]uadratic #strong[M]ean): $ "QM" = sqrt((a_1^2 + a_2^2 + ... + a_n^2)/n) $ ] #caption[ $"AM"$和$"GM"$的不等式有着很显然的几何意义: 在$RR^2$上, $ (a_1+a_2) / 2 >= sqrt(a_1 a_2) quad => quad 2(a_1+a_2) >= 4sqrt(a_1 a_2) $ 意味着, 面积相同的矩形,正方形的周长最小.注意到相同的结论可以直接推广到$RR^n$上. ] #proof[ $"AM"$和$"GM"$的不等式,可通过上面这种性质做一个巧妙的证明: $ alpha = "AM" = 1 / n (a_1 + ... + a_n) $ 若 $a_1, a_2, ..., a_n$ 不全相等, 则存在 $a_i < alpha < a_j$, 对这两项做如下替换: $ a_i^prime &= alpha \ a_j^prime &= a_i + a_j - alpha $ 替换后满足: $ "AM"^prime &= "AM" = alpha \ "GM"^prime &= "GM" / (a_i dot.c a_j) * (a_i^prime dot.c a_j^prime) > "GM" $ 注意到 $"GM"^prime > "GM"$是严格的. 重复上述过程, 直到所有的 $a_i$ 都等于 $alpha$(容易说明只需要至多$n-1$步), 此时: $ "AM" = "GM"^((n)) > "GM"^((n-1)) > ... > "GM"^(prime) > "GM" $ 证毕. #caption[ 上述证明同时说明,等号成立当且仅当 $a_1 = a_2 = ... = a_n$. 此时无需经过变换,$"AM" = "GM"$. 只要经过变换,就有严格的不等式成立. ] #homework[ $"AM"$和$"GM"$的不等式,也可以通过归纳法证明. ] ] #proof[ 首先说明$"AM"$和$"QM"$的不等式: $ forall a_1, a_2, ..., a_n in RR_+\ (a_1 + a_2 + ... + a_n) / n <= sqrt((a_1^2 + a_2^2 + ... + a_n^2)/n) $ 是Cauchy不等式的推论.令 $x = (a_1, a_2, ..., a_n), y = (1, 1, ..., 1)$, 则: $ (a_1 + a_2 + ... + a_n) = x dot.c y <= abs(x) abs(y) = sqrt(a_1^2 + a_2^2 + ... + a_n^2) dot.c sqrt(n) $ 简单变换即可得到上述不等式. #line(length: 100%, stroke: 0.2pt) 接下来说明$"HM"$和$"QM"$的不等式是由$"AM"$和$"GM"$的不等式推出的: $ forall a_1, a_2, ..., a_n in RR_+ quad (a_1 a_2 ... a_n)^(1 / n) <= (a_1 + a_2 + ... + a_n) / n $ 在上述不等式中将 $a_i$ 替换为 $1/a_i$, 则: $ forall a_1, a_2, ..., a_n in RR_+ quad => quad (1 / a_1 1 / a_2 ... 1 / a_n)^(1 / n) &<= (1 / a_1 + 1 / a_2 + ... + 1 / a_n) / n \ &arrow.b.double \ (a_1 a_2 ... a_n)^(1 / n) &>= n / (1 / a_1 + 1 / a_2 + ... + 1 / a_n) $ ] #pagebreak() == 可数 === 集合间映射 === 集合的基数 === 可数 ==== 性质 - 有限个可数集的并是可数的, 可数个可数集的并是可数的 - 如果集合A是可数的, 集合B是有限的, 那么A x B是可数的: $ A times B := {(a, b) | a in A, b in B} $ - $abs(2^X) = 2^abs(X)$ ==== 不可数集 - $2^NN$, Cantor对角线方法 - $RR$, 也类似于Cantor对角线方法 #pagebreak() == 习题选讲 #pagebreak()

https://github.com/AU-Master-Thesis/thesis

https://raw.githubusercontent.com/AU-Master-Thesis/thesis/main/sections/3-methodology/study-1/algorithm.typ

typst

MIT License

#import "../../../lib/mod.typ": * === Algorithm <s.m.algorithm> // #jonas[I do believe all this is new to you?] // This section should explain the GBP inference algorithm as it has been implemented in the simulation tool. // 1. Fixed update schedule // 2. Chained systems // 3. Manual stepping and pausing // Introduce the Robot Mission concept? #[ #show regex("\b(UpdatePrior|CurrentlyConnected|Connect|Disconnect|InternalGBP|ExternalGBP)\b") : set text(theme.mauve, font: "JetBrainsMono NF") As explained in the background section #nameref(<s.b.ecs>, "Entity Component System"), an #acr("ECS") architecture has been used as the foundation for #acr("MAGICS") through the Bevy Engine@bevyengine. The #acr("GBP") inference process has been implemented as a series of 7 systems that are executed in a fixed update schedule. This schedule is one of the default schedules provided by Bevy with a configurable frequency#footnote[This happens in the #acr("MAGICS") source code in the #crates.gbpplanner-rs crate at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/simulation_loader.rs#L564", "src/simulation_loader.rs:564")], which is also exposed to #acr("MAGICS") through the `simulation` table in `config.toml`#footnote[Example at #repo(org: "AU-Master-Thesis", repo: "gbp-rs") at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/config/simulations/Intersection/config.toml#L79", "config/simulations/Intersection/config.toml:79")] file. These 7 systems are listed here and marked #boxed[*GBP-X*]. #[ #set enum(numbering: box-enum.with(prefix: "GBP-")) #grid( columns: (1fr, 1fr), [ + `update_robot_neighbours` + `delete_interrobot_factors` + `create_interrobot_factors` + `update_failed_comms` ], [ #set enum(start: 5) + `iterate_gbp` + `update_prior_of_horizon_state` + `update_prior_of_current_state` ], ) ] The algorithm is written out in @a.m.algorithm, where `CurrentlyConnected` is a way to retrieve which robots are currently set as being connected to a specific robot, $R_i$. Before the main loop, find which robots are within communication distance at the current timestep, $N(R_i)$. Then in the main loop while the simulation is _running_, create an interrobot factor between all robots within communication distance if it does not already exist with `Connect`, and delete the interrobot factor if the robot is no longer within communication distance with `Disconnect`. After this, the internal and external #acr("GBP") iterations are run with `InternalGBP` and `ExternalGBP`, respectively. #algorithm( caption: [GBP For Robot $R_i$@gbpplanner] )[ #let comment(content) = text(theme.overlay2, content) #show regex("\b(UpdatePrior|CurrentlyConnected|Connect|Disconnect|InternalGBP|ExternalGBP)\b") : set text(size: 0.85em) // #show regex("--(.*?)\b") : set text(theme.crust, font: "JetBrainsMono NF", size: 0.85em) #show regex("(while|for|do|end)") : set text(weight: "bold") // #set par(first-line-indent: 0em) #let ind() = h(2em) *Input:* $R_i$ \ \ #comment[Retrieve the robots that were previously set to be connected to $R_i$] \ $C(R_i) #la "CurrentlyConnected"(R_i)$ \ \ $"UpdatePrior"(#m.x _0, delta_t)$ \ $"UpdatePrior"(#m.x _(K-1), delta_t)$ \ \ $N(R_i) #la {R_j | norm(R_i - R_j) < r_C}$ \ \ while _running_ do \ #ind()for $R_j in N(R_i) \\ C(R_i)$ do \ #ind()#ind()$"Connect"(R_i, R_j)$ \ #ind()end \ #ind()for $R_j in C(R_i) \\ N(R_i)$ do \ #ind()#ind()$"Disconnect"(R_i, R_j)$ \ #ind()end \ #ind()$"InternalGBP"(M_I)$ \ #ind()$"ExternalGBP"(M_E)$ \ #ind()end \ end ]<a.m.algorithm> Following is an explanation of the systems, and their responsibilities, along with which part of the original #gbpplanner implementation they correspond to. Furthermore, each system is related to @a.m.algorithm when relevant: #let space = v(0.5em) #set par(first-line-indent: 0em) #set enum(numbering: box-enum.with(prefix: "GBP-")) + `update_robot_neighbours`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L1247", "src/planner/robot.rs:1247")] utilizes the #acr("ECS") to mutably query for all entities that have a `RobotConnections` component, and then consequently update them with all robots within communication range. #space *Parity* with `Simulator::calculateRobotNeighbours` in #gbpplanner. Corresponds to the setting the internal data of `RobotConnection` to that of $N(R_i)$. + `delete_interrobot_factors`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L1271", "src/planner/robot.rs:1271")] removes all interrobot factors from the factor graph that are no longer relevant due to the updated robot connections. #space *Parity* with half of the responsibility of `Robot::updateInterrobotFactors` in #gbpplanner. This corresponds to the `Disconnect` part of the algorithm. + `create_interrobot_factors`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L1326", "src/planner/robot.rs:1326")] creates new interrobot factors for all robot connections that are not already represented in the factor graph. #space *Parity:* with the other half of the responsibility of `Robot::updateInterrobot``Factors` in #gbpplanner. This corresponds to the `Connect` part of the algorithm. + `update_failed_comms`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L1478", "src/planner/robot.rs:1478")] updates the communication status of all robots, based on the configurable parameter `communication_failure_rate` under the `robot``.communication` table in `config.toml`. #space *Parity* with `Simulator::setCommsFailure` in #gbpplanner. This does not have a correlary in @a.m.algorithm. + `iterate_gbp`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L1654", "src/planner/robot.rs:1654")] iterates the #acr("GBP") algorithm for all robots in the simulation. That is, it has the responsibilitity for the 4 inference steps; _variable update, variable to factor message passing, factor update, and factor to variable message passing_. #space *Parity* with `Simulator::iterateGBP` in #gbpplanner. This corresponds to the `InternalGBP` and `ExternalGBP` part of the algorithm. + `update_prior_of_horizon_state`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L2042", "src/planner/robot.rs:2042")] updates the prior of the horizon state for all robots in the simulation. This is the pose factor anchoring earlier mentioned in @s.m.factors.pose-factor. #space *Parity* with `Robot::updateHorizon` in #gbpplanner. + `update_prior_of_current_state`#footnote[Found in crate #crates.gbpplanner-rs at #source-link("https://github.com/AU-Master-Thesis/gbp-rs/blob/8960686facb7d38c0259141e5b22346c7d745271/crates/gbpplanner-rs/src/planner/robot.rs#L2174", "src/planner/robot.rs:2174")] updates the prior of the current state for all robots in the simulation. Again, this is where the effect of the pose factor anchoring takes place. #space *Parity* with `Robot::updateCurrent` in #gbpplanner. ] #let r = ( A: text(theme.lavender, weight: "bold", "A"), B: text(theme.mauve, weight: "bold", "B") ) Through these steps the lifecycle of the interrobot factors has been allured to. This lifecycle is visualized in @f.interrobot-lifecycle, where two robots #r.A and #r.B approach each other. When they are within communication range, interrobot factors are created. The messaging happens through these factors is the communication that would happen wirelessly in a real-world implementation. Furthermore, when one of the robots' radio fails, the interrobot factors that are maintained by that robot are simply deactivated instead of removed. This has been done as an optimization, instead of deallocating, for then possibly reallocating in the next timestep. Finally, when the robots are no longer within communication range, the interrobot factors are deallocated. To summarize for two robots, $A$ and $B$, with variable $v_n^A$ and $v_n^B$, connected by interrobot factors $f_(i_n)^A (v_n^A, v_n^B)$ and $f_(i_n)^B (v_n^A, v_n^B)$. There are four possible states the pairing between the two can be in. + The communication medium of both $A$ and $B$ are inactive, preventing the factors and variable from exchanging messages. + The communication medium of $A$ is active, preventing $B$ from exchanging messages with $A$ during external message passing. + The communication medium of $B$ is active, preventing $A$ from exchanging messages with $B$ during external message passing. + The communication medium of both $A$ and $B$ are active, allowing the factors and variable to exchange messages between each other during external message passing. These four states correspond to the states shown at timestep $t_(n+1)$ to $t_(n+4)$ in @f.interrobot-lifecycle. #figure( block(breakable: false, include "figure-interrobot-lifecycle.typ", ), caption: [ #let comms = { let l1 = place(dy: -0.35em, line(length: 1em, stroke: (thickness: 2pt, paint: theme.teal, dash: "dashed", cap: "round"))) let l2 = place(dy: -0.35em, line(length: 1em, stroke: (thickness: 2pt, paint: theme.surface0, dash: "dashed", cap: "round"))) box(inset: (x: 2pt), outset: (y: 2pt), l1 + l2 + h(1.6em)) } Interrobot factor, $f_i$, lifecycle. On A) the two robots, #r.A and #r.B, are approaching each other, but not within communication range, shown with dashed circles #inline-line(stroke: (paint: theme.teal, thickness: 2pt, dash: "dashed", cap: "round")) #inline-line(stroke: (paint: theme.overlay0, thickness: 2pt, dash: "dashed", cap: "round")). On B) both robots are within communication range, and interrobot factors are created symmetrically between robots #r.A and #r.B. On C) and D) one of the two robots' radio has failed, resulting in the corresponding interrobot factors being inactive. On E) the robots are no longer within communication range, and the interrobot factors are removed. ] )<f.interrobot-lifecycle>

https://github.com/Rhinemann/mage-hack

https://raw.githubusercontent.com/Rhinemann/mage-hack/main/src/chapters/Talents.typ

typst

#import "../templates/interior_template.typ": * #show: chapter.with(chapter_name: "Talents") = Talents #show: columns.with(2, gutter: 1em) In addition to Hinder each character has a handful of SFX, reflecting special capabilities associated with their many different abilities, these are called talents. These A PC also has at least one Limit. A Limit is a special type of SFX that imposes a disadvantage on your character in order to earn them #spec_c.pp or another reward. Whenever you gain a Talent or Limit, you can rename it to better suit your character. #block(breakable: false)[ == Sample Talents When you create a new character, they gain two of the following SFX of your choice (in addition to the _Hinder_ SFX all characters receive): // TODO write descriptions ] / Adaptable: Step down and double one die of your choice in your pool. / All-Out Attack: Spend a #spec_c.pp to target multiple opponents when you roll to inflict #smallcaps[Hurt]. For each additional target, add #spec_c.d6 and keep an extra effect die. / Brilliant Under Pressure: Spend a #spec_c.pp to add your #smallcaps[Rattled] or #smallcaps[Tired] to your roll to create an asset. If the action succeeds, step down the stress you used. / Combat Veteran: When your roll to inflict #smallcaps[Hurt] or #smallcaps[Rattled] during a battle includes #smallcaps[Firearms] or #smallcaps[Weaponry], step down the largest die in your pool to add #spec_c.d8. If your roll succeeds, step up your effect die. / Distracting Presence: When you roll to inflict #smallcaps[Unsound] by distracting someone, add #spec_c.d6 and step up your effect die. / Strong Empathy: Step up #smallcaps[Empathy] on a roll to create an asset related to trust, reading people, or reassurance. / Energetic: When you would take #smallcaps[Tired] stress, spend a #spec_c.pp to step down the stress you take. If this steps the stress down below #spec_c.d6, you take no stress at all. / Flash of Insight: When you fail a test to obtain information, you may spend a #spec_c.pp or take #smallcaps[Unsound] #spec_c.d6 to obtain that information by other means. / Have a Little Faith: When you would take #smallcaps[Rattled] stress, spend a #spec_c.pp to step down the stress you take. If this steps the stress down below #spec_c.d6, you take no stress at all. / Hinder: Roll #spec_c.d4 instead of #spec_c.d8 for a distinction to earn a #spec_c.pp. / In Harm's Way: When another character near you takes stress, you can step down the stress they would take, then take #spec_c.d6 stress of the same type yourself. / Impossible to Ignore: Spend a #spec_c.pp to target multiple opponents when you roll to inflict #smallcaps[Unsound]. For each additional target, add #spec_c.d6 and keep an extra effect die. / Inspiring Leadership: Add a #spec_c.d6 and step up your effect die when you roll #smallcaps[Social] skills to create assets for allies. / Keen Intellect: Add a #spec_c.d6 and step up your effect die when you roll #smallcaps[Mental] skills to create an asset related to recalling or researching information. / Master Plan: Spend a #spec_c.pp to add a die to your pool equal to the largest complication anyone has in the scene. After the roll fails or succeeds, step down that complication. / Misdirection: When you use #smallcaps[Social] skills on a roll related to escape, deception, or stealth, step down the largest die in your pool to add #spec_c.d8. If your roll succeeds, step up your effect die. / Outmaneuver: When your roll to inflict #smallcaps[Hurt] or #smallcaps[Tired] while outdoors includes #smallcaps[Athletics], step down the largest die in your pool to add #spec_c.d8. If your roll succeeds, step up your effect die. / Peacemaker: If you have #smallcaps[Hurt] stress inflicted by another character in the scene when you roll to de-escalate a conflict, double #smallcaps[Empathy] in your dice pool. If the roll still fails, take #smallcaps[Rattled] #spec_c.d6. / Push Through It: Before you roll a dice pool including a #smallcaps[Physical] skill, spend a #spec_c.pp to recover #smallcaps[Hurt] and step up a #smallcaps[Physical] attribute for that roll. Take #smallcaps[Tired] #spec_c.d6 stress if the roll succeeds, or #spec_c.d8 if it fails. / Reassuring Comrade: Step up or double #smallcaps[Empathy] in your dice pool when helping others recover #smallcaps[Rattled]. You can also spend a #spec_c.pp to step down your own or a nearby character's #smallcaps[Rattled]. / Reckless Gambit: When you roll dice, add a die to your pool equal to the largest stress or complication anyone has in the scene. Take a complication at #spec_c.d6 if the roll succeeds, or #spec_c.d8 if it fails. / Reliable Memory: Spend a #spec_c.pp to reroll a dice pool focused on memory or recall that included #smallcaps[Intelligence]. / Skill Focus: When your pool includes a specialty, you can replace two dice of equal size with one die one step larger. / Sudden Yet Inevitable: When someone betrays you or deceives you, or you betray or deceive someone, spend a #spec_c.pp to create a #spec_c.d8 asset related to having planned for it. / Team Player: When you witness an ally rolling a heroic success, you can step down your own or another witness's #smallcaps[Rattled]. / Tough: When you would take #smallcaps[Hurt] stress, spend a #spec_c.pp to step down the stress you take. If this steps the stress down below #spec_c.d6, you take no stress at all. / Trained Physician: Step up or double #smallcaps[Medicine] in your dice pool when helping others recover #smallcaps[Hurt]. You can also spend a #spec_c.pp to step down your own or a nearby character's #smallcaps[Hurt]. / Undaunted Determination: Step up or double #smallcaps[Stamina] for one roll. If the roll fails, take #smallcaps[Tired] stress equal to the largest die in your pool. / Vicious Contempt: When you roll to inflict #smallcaps[Rattled] with mockery or contempt, add #spec_c.d6 and step up your effect die. / Vigilant Eye: Spend a #spec_c.pp to double #smallcaps[Investigation] in a pool related to following a trail, aiming at a distant target, or spotting something far off. / Watch It All Burn: Add a die to your pool equal to the largest stress or complication anyone has in the scene and step up your effect die. Succeed or fail, take #smallcaps[Unsound] #spec_c.d6. You can spend points to take one or more of these SFX at character creation or later #block(breakable: false)[ === Magick Talents (Supernatural Talents) These are the examples of magick-related and supernatural SFX available to mages to inspire the players and Storytellers. ] / Advanced Necromancy: Spend a #spec_c.pp to use both Matter and Life when your action is related to animating the dead. / Area Effect: When your effect targets an area or a number of creatures, spend a #spec_c.pp to add a #spec_c.d6 and keep an additional effect die for each additional target past the first. / Conjunctional Effects Mastery: When performing a conjunctional effect add two or more Spheres to a dice pool and step each Sphere down by one for each additional Sphere beyond the first. / Destructive Proclivities: When you include a Sphere to destroy an object, spend a #spec_c.pp to step up your effect die. / Enchant Patterns: When your effect includes Prime #spec_c.d6 or higher spend Quintessence to inflict #spec_c.d6 Hurt stress. / Fast Casting: When your action includes a Sphere, you can gain a #spec_c.d6 complication to inflict #spec_c.d6 Hurt stress. / Instrument Arsenal: Spend a #spec_c.pp to create a #spec_c.d8 Instrument asset for a particular type of magick. / Paradox Contaminating: When your action includes Prime #spec_c.d12 or higher, you can inflict Paradox on a target besides yourself. / Paradox Transmitting: When your action includes Prime #spec_c.d12 or higher and you have the Paradox Contaminating SFX you can recover one Paradox die level for each Paradox die inflicted. / Primal Channeling: When your action includes Prime #spec_c.d10 or higher you can recover one Quintessence die for each step of Stress you inflict. / Pushing Through: Whenever you take stress caused by a Sphere, spend a #spec_c.pp to step it down. At the end of the session, if you still have stress on that Sphere, step it up. / Quick Curse: When your action includes a Sphere, you can gain a #spec_c.d6 to keep a second effect die as a complication on a nearby character. / Reckless Casting: Step up or double any Sphere for one roll. If the roll fails, add your Sphere die to the Paradox pool. / Rein In: When you include a #spec_c.d10 or #spec_c.d8 Sphere, gain a #spec_c.pp and step it down. You can recover it by activating an opportunity rolled by the GM. / Swift Warding: When your action includes a Sphere, you can gain a #spec_c.d6 Moving Too Fast complication to keep a second effect die as a Magical Aegis asset. / Talent for Growth: When you succeed at a test including a Sphere, spend a #spec_c.pp to create a Watch And Learn #spec_c.d8 asset. Anyone can use this asset alongside a #spec_c.d4 or #spec_c.d6 Sphere. #block(breakable: false)[ ==== Sphere Talents Sphere talents are a special case of supernatural talents that are unlocked automatically as you advance your understanding of magick. ] / Sphere Perception: Step up your lowest die on any roll to perceive any phenomena under the purview of Sphere or create a related asset. You unlock this talent at Sphere rating #spec_c.d4 for that Sphere. / Sphere Manipulation: On rolls to create an asset that can be produced by a #spec_c.d6 or lower Sphere rating, add #spec_c.d6 and step up your effect die. You unlock this talent at Sphere rating #spec_c.d6 for that Sphere. / Sphere Control: Spend a #spec_c.pp to create a #spec_c.d8 asset that can be produced by a #spec_c.d8 or lower Sphere rating. You unlock this talent at Sphere rating #spec_c.d8 for that Sphere. / Sphere Command: Spend a #spec_c.pp to step up or double your Sphere die on a roll for an effect that can be accomplished by a #spec_c.d10 or lower Sphere rating. You unlock this talent at Sphere rating #spec_c.d10 for that Sphere. / Sphere Mastery: Take #spec_c.d6 appropriate stress or complication to double your Sphere die for for a roll. On a failure, step up the same stress or complication you took to activate. You unlock this talent at Sphere rating #spec_c.d12 for that Sphere.

https://github.com/mem-courses/calculus

https://raw.githubusercontent.com/mem-courses/calculus/main/homework-1/week0.typ

typst

#import "../template.typ": * #show: project.with( title: "微积分 Homework #0", authors: ( (name: "<NAME>", email: "<EMAIL>", phone: "3230104585"), ), date: "September 15, 2023", ) = Pre Problem 1 #prob[已知数列 ${x_n}$ 满足 $ cases( x_1 = a, x_(n+1) = 1/2 x_n^2 (3 - x_n), ) quad quad (a "为常数",quad a in (0,1) union (1,2) union (2,3)) $ 问当 $n>=1$（或 $n>=2$）时，${x_n}$ 是否为单调数列（需说明理由或给出论证过程）] 设 $ f(x) = 1/2 x^2 (3-x) $ 则有 $ f'(x) = 1/2( 2x(3-x) + (-x^2)) = 3/2x (2 - x) $ = xmr's Problem 1 #prob[设 $x,y in RR$，求 $(x-y)^2 + (2x-5)^2 + 4y^2$ 的最小值．]

https://github.com/EpicEricEE/typst-based

https://raw.githubusercontent.com/EpicEricEE/typst-based/master/src/base16.typ

typst

MIT License

/// Encodes the given data as a hex string. /// /// Arguments: /// - data: The data to encode. Must be of type array, bytes, or string. /// /// Returns: The encoded string (lowercase). #let encode(data) = { if data.len() == 0 { return "" } for byte in array(bytes(data)) { if byte < 16 { "0" } str(int(byte), base: 16) } } /// Decodes the given hex string. /// /// Arguments: /// - string: The string to decode (case-insensitive). /// /// Returns: The decoded bytes. #let decode(string) = { let dec(hex-digit) = { let code = str.to-unicode(hex-digit) if code >= 48 and code <= 57 { code - 48 } // 0-9 else if code >= 65 and code <= 70 { code - 55 } // A-F else if code >= 97 and code <= 102 { code - 87 } // a-f else { panic("Invalid hex digit: " + hex-digit) } } let array = range(string.len(), step: 2).map(i => { 16 * dec(string.at(i)) + dec(string.at(i + 1)) }) bytes(array) }

https://github.com/typst/packages

https://raw.githubusercontent.com/typst/packages/main/packages/preview/silky-letter-insa/0.1.0/lib.typ

typst

Apache License 2.0

// SHORT DOCUMENT : #let insa-short( author : none, date : datetime.today(), doc ) = { set text(lang: "fr") set page( "a4", margin: (top: 3.2cm), header-ascent: 0.71cm, header: [ #place(left, image("logo.png", height: 1.28cm), dy: 1.25cm) #place(right + bottom)[ #author\ #if type(date) == datetime [ #date.display("[day]/[month]/[year]") ] else [ #date ] ] ], footer: [ #place( right, dy: -0.6cm, dx: 1.9cm, image("footer.png") ) #place( right, dx: 1.55cm, dy: 0.58cm, text(fill: white, weight: "bold", counter(page).display()) ) ] ) doc }

https://github.com/rxt1077/it610

https://raw.githubusercontent.com/rxt1077/it610/master/markup/slides/git.typ

typst

#import "/templates/slides.typ": * #import "@preview/fletcher:0.5.1" as fletcher: diagram, node, edge #import fletcher.shapes: diamond #show: university-theme.with( short-title: [git], ) #title-slide( title: [git], ) #alternate( title: [What is git?], image: licensed-image( file: "/images/git.svg", license: "FAIRUSE", title: [git logo], url: "https://git-scm.com/downloads/logos", width: 100%, ), text: [ - a version control system - command line based - keeps track of files and changes to them - works locally but can "push" to a remote ] ) #slide(title: [What do people think of git?], align(center, licensed-image( file: "/images/git-bingo.png", license: "CC BY-NC-SA 4.0", title: [git discussion bingo], url: "https://wizardzines.com/comics/git-discussion-bingo/", author: [<NAME>], author-url: "https://wizardzines.com/", )) ) #alternate( title: [Why do we need it?], image: [ #set text(size: 16pt) #diagram( node-shape: rect, node((0, 0), [Collaboration], stroke: red, fill: red.lighten(80%)), node((1, 0), [Open Source], stroke: orange, fill: orange.lighten(80%)), node((2, 0), [Scalable], stroke: yellow, fill: yellow.lighten(80%)), node((0, 1), [Distributed], stroke: green, fill: green.lighten(80%)), node((1, 1), [git], stroke: blue, shape: diamond, fill: blue.lighten(80%)), node((2, 1), [Workflow], stroke: purple, fill: purple.lighten(80%)), node((0, 2), [Integrity], stroke: yellow, fill: yellow.lighten(80%)), node((1, 2), [Branching], stroke: red, fill: red.lighten(80%)), node((2, 2), [History], stroke: orange, fill: orange.lighten(80%)), edge((1, 1), (0.25, 0.25), "->"), edge((1, 1), (1, 0.25), "->"), edge((1, 1), (1.75, 0.25), "->"), edge((1, 1), (0.25, 1), "->"), edge((1, 1), (1.75, 1), "->"), edge((1, 1), (0.25, 1.75), "->"), edge((1, 1), (1, 1.75), "->"), edge((1, 1), (1.75, 1.75), "->"), ) ], text: [ - make things less brittle - keep track of things - find out who changed things ] ) #alternate( title: [How is it used?], image: licensed-image( file: "/images/xkcd-git.png", license: "CC BY-NC 2.5", title: [Git], url: "https://xkcd.com/1597/", author:[<NAME>], author-url: "https://xkcd.com/about/", ), text: [ - create a repo: `git init` (local or remote, GitHub can do this for you) - clone the repo: `git clone` - add files you want tracked: `git add` - commit changes: `git commit` - push changes: `git push` ] ) #slide(title: [How do I collaborate with git?])[ #diagram( node-shape: circle, node-fill: njit-red, edge-stroke: 8pt + njit-blue.lighten(50%), spacing: (4.8em, 4em), label-size: 0.8em, node((0, 0), " "), node((1, 1), " ", fill: njit-blue), node((2, 1), " ", fill: njit-blue), node((3, 1), " ", fill: njit-blue), node((4, 1), " "), node((5, 1), " "), node((6, 0), " "), edge((0, 0), (6, 0), label: [main], stroke: 8pt + njit-red.lighten(50%)), edge((0, 0), (1, 1), label: [fork], bend: -30deg, label-angle: auto), edge((1, 1), (3, 1), label: [user makes changes], label-side: right), edge((3, 1), (4, 0), label: [pull request], stroke: (paint: njit-blue.lighten(50%), thickness: 8pt, dash: ("dot", 0.5em)), bend: -30deg, label-angle: auto, label-side: left, label-sep: 0.7em, ), edge((3, 1), (5, 1), label: [maintainer adds commits], label-side: right), edge((5, 1), (6, 0), label: [merge], bend: -30deg, label-angle: auto), ) \ - fork a repo (create a branch) - make your changes - submit a PR - PR gets merged (hopefully) ] #alternate( title: [Why are we talking about git in a sysadmin class?], image: licensed-image( file: "/images/git-push-git-paid.svg", license: "CC BY-NC 4.0", title: [git-push-git-paid.svg], url: "https://github.com/rxt1077/it610/blob/master/markup/images/git-push-git-paid.svg", author: [<NAME>], author-url: "https://using.tech", ), text: [ - configurations a typically lots of little files we need to track - Docker Compose projects can be tracked in git - Kubernetes projects can be tracked in git - git helps with change management ] ) #alternate( title: [git services], image: block(breakable: false)[ #set align(center) #set text(8pt) #grid(columns: (1fr, 1fr), rows: (40%, 40%), gutter: 40pt, image("/images/github-logo.svg", height: 100%), image("/images/gitlab-logo.svg", height: 100%), image("/images/sourcehut-logo.svg", height: 100%), image("/images/radicle-logo.svg", height: 100%), ) GitHub, GitLab, SourceHut, and Radicle logos are used under fair use. ], text: [ - #link("https://github.com")[GitHub] - #link("https://gitlab.com")[GitLab] - #link("https://sr.ht")[SourceHut] - #link("https://radicle.xyz")[Radicle (p2p, v1.0 just came out!)] ], )

https://github.com/catppuccin/typst

https://raw.githubusercontent.com/catppuccin/typst/main/tests/background/test.typ

typst

MIT License

#import "/src/lib.typ": catppuccin, themes, get-palette #set page(width: auto, height: auto) #let perms = () #for theme in themes.values() { for code-block in (true, false) { for syntax in (true, false) { perms.push((theme: theme, code-block: code-block, syntax: syntax)) } } } #for p in perms [ #pagebreak(weak: true) #show: catppuccin.with(p.theme, code-block: p.code-block, code-syntax: p.syntax) = #get-palette(p.theme).name - Code block: #p.code-block - Code syntax: #p.syntax ```typ #import "/src/lib.typ": catppuccin, themes, get-palette #let perms = () #for theme in themes.values() { for code-block in (true, false) { for syntax in (true, false) { perms.push((theme: theme, code-block: code-block, syntax: syntax)) } } } #for p in perms [ #show: catppuccin.with(p.theme, code-block: p.code-block, code-syntax: p.syntax) = #get-palette(p.theme).name == Code block: #p.code-block == Code syntax: #p.syntax ] ``` ]

https://github.com/SillyFreak/typst-packages-old

https://raw.githubusercontent.com/SillyFreak/typst-packages-old/main/scrutinize/README.md

markdown

MIT License

# Scrutinize Scrutinize is a library for building exams, tests, etc. with Typst. It has three general areas of focus: - It helps with grading information: record the points that can be reached for each question and make them available for creating grading keys. - It provides a selection of question writing utilities, such as multiple choice or true/false questions. - It supports the creation of sample solutions by allowing to switch between the normal and "pre-filled" exam. Right now, providing a styled template is not part of this package's scope. Also, visual customization of the provided question templates is currently nonexistent. See the [manual](docs/manual.pdf) for details. ## Example A rendered version of this example can be found in the [gallery](gallery/). ```typ #import "@preview/scrutinize:0.2.0": grading, question, questions #import question: q #import questions: free-text-answer, single-choice, multiple-choice, set-solution, unset-solution // toggle this comment or pass `--input solution=true` to produce a sample solution // #questions.solution.update(true) #set table(stroke: 0.5pt) #context [ #let total = grading.total-points(question.all()) The candidate achieved #h(3em) out of #total points. ] = Instructions #with-solution(true)[ Use a pen. For multiple choice questions, make a cross in the box, such as in this example: #pad(x: 5%)[ Which of these numbers are prime? #multiple-choice( (([1], false), ([2], true), ([3], true), ([4], false), ([5], true)), ) ] ] #show heading: it => [ #it.body #h(1fr) / #question.current().points ] #q(points: 2)[ = Question 1 Write an answer. #free-text-answer(height: 4cm)[ An answer ] ] #q(points: 1)[ = Question 2 Select the largest number: #single-choice( ([5], [20], [25], [10], [15]), 2, // 0-based index ) ] ```

https://github.com/typst/packages

https://raw.githubusercontent.com/typst/packages/main/packages/preview/fh-joanneum-iit-thesis/1.1.0/template/chapters/4-background.typ

typst

Apache License 2.0

#import "global.typ": * = Background #lorem(45) #todo([ In the background section you might give explanations which are necessary to read the remainder of the thesis. For example define and/or explain the terms used. Optionally, you might provide a glossary (index of terms used with/without explanations). #v(1cm) *Hints for equations in Typst*: The notation used for #textbf([calculating]) of #textit([code performance]) might typically look like the one in @slow and @veryslow, which demonstrates what *(very) slow* algorithms mean. $ O(n) = n^2 $ <slow> $ O(n) = 2^n $ <veryslow> *Hints for footnotes in Typst*: As shown in #footnote[Visit https://typst.app/docs for more details on formatting the document using typst. Note, _typst_ is written in the *Rust* programming language.] we migth discuss the alternatives. *Hints for formatting in Typst*: + You can use built-in styles: + with underscore (\_) to _emphasise_ text + forward dash (\`) for `monospaced` text + asterisk (\*) for *strong* (bold) text You can create and use your own (custom) formatting macros: + check out the custom style (see in file `lib.typ`): + `\#textit` for #textit([italic]) text + `\#textbf` for #textbf([bold face]) text ])

https://github.com/tingerrr/chiral-thesis-fhe

https://raw.githubusercontent.com/tingerrr/chiral-thesis-fhe/main/src/core/component/title-page.typ

typst

#import "/src/core/kinds.typ" as _kinds #import "/src/core/authors.typ" as _authors #import "/src/utils.typ" as _utils // TODO: proper handling of more than one author // TODO: stable positioning // TODO: use subtitle #let make-title-page( title: "Mustertitel", subtitle: none, author: "Mustermann, Max", supervisors: ( "Prof. Dr. <NAME>", "Prof. Dr. <NAME>", ), field: "Angewandte Informatik", date: datetime(year: 1970, month: 01, day: 01), id: "AI-1970-BA-999", kind: _kinds.report, _fonts: (:), ) = { set align(center + top) stack( align(right, image("/assets/images/logo-fhe.svg", width: 45%)), 5em, text(16pt, font: _fonts.sans, strong[ #kind.name \ #field ]), ..if _kinds.is-thesis(kind) { (1em, [Nr. #id]) }, 5em, text(32pt, font: _fonts.sans, strong(title)), 3.4em, text(16pt, strong(_authors.format-author(author, email: false))), 2.5em, text(18pt)[Abgabedatum: #_utils.format-date(date)], ) if _kinds.is-thesis(kind) { place(center + bottom, text( 18pt, supervisors.map(_authors.format-author.with(email: false)).join(linebreak()), )) } pagebreak(weak: true) }

https://github.com/frectonz/the-pg-book

https://raw.githubusercontent.com/frectonz/the-pg-book/main/book/097.%20badeconomy.html.typ

typst

badeconomy.html Why to Start a Startup in a Bad Economy Want to start a startup? Get funded by Y Combinator. October 2008The economic situation is apparently so grim that some experts fear we may be in for a stretch as bad as the mid seventies.When Microsoft and Apple were founded.As those examples suggest, a recession may not be such a bad time to start a startup. I'm not claiming it's a particularly good time either. The truth is more boring: the state of the economy doesn't matter much either way.If we've learned one thing from funding so many startups, it's that they succeed or fail based on the qualities of the founders. The economy has some effect, certainly, but as a predictor of success it's rounding error compared to the founders.Which means that what matters is who you are, not when you do it. If you're the right sort of person, you'll win even in a bad economy. And if you're not, a good economy won't save you. Someone who thinks "I better not start a startup now, because the economy is so bad" is making the same mistake as the people who thought during the Bubble "all I have to do is start a startup, and I'll be rich."So if you want to improve your chances, you should think far more about who you can recruit as a cofounder than the state of the economy. And if you're worried about threats to the survival of your company, don't look for them in the news. Look in the mirror.But for any given team of founders, would it not pay to wait till the economy is better before taking the leap? If you're starting a restaurant, maybe, but not if you're working on technology. Technology progresses more or less independently of the stock market. So for any given idea, the payoff for acting fast in a bad economy will be higher than for waiting. Microsoft's first product was a Basic interpreter for the Altair. That was exactly what the world needed in 1975, but if Gates and Allen had decided to wait a few years, it would have been too late.Of course, the idea you have now won't be the last you have. There are always new ideas. But if you have a specific idea you want to act on, act now.That doesn't mean you can ignore the economy. Both customers and investors will be feeling pinched. It's not necessarily a problem if customers feel pinched: you may even be able to benefit from it, by making things that save money. Startups often make things cheaper, so in that respect they're better positioned to prosper in a recession than big companies.Investors are more of a problem. Startups generally need to raise some amount of external funding, and investors tend to be less willing to invest in bad times. They shouldn't be. Everyone knows you're supposed to buy when times are bad and sell when times are good. But of course what makes investing so counterintuitive is that in equity markets, good times are defined as everyone thinking it's time to buy. You have to be a contrarian to be correct, and by definition only a minority of investors can be.So just as investors in 1999 were tripping over one another trying to buy into lousy startups, investors in 2009 will presumably be reluctant to invest even in good ones.You'll have to adapt to this. But that's nothing new: startups always have to adapt to the whims of investors. Ask any founder in any economy if they'd describe investors as fickle, and watch the face they make. Last year you had to be prepared to explain how your startup was viral. Next year you'll have to explain how it's recession-proof.(Those are both good things to be. The mistake investors make is not the criteria they use but that they always tend to focus on one to the exclusion of the rest.)Fortunately the way to make a startup recession-proof is to do exactly what you should do anyway: run it as cheaply as possible. For years I've been telling founders that the surest route to success is to be the cockroaches of the corporate world. The immediate cause of death in a startup is always running out of money. So the cheaper your company is to operate, the harder it is to kill. And fortunately it has gotten very cheap to run a startup. A recession will if anything make it cheaper still.If nuclear winter really is here, it may be safer to be a cockroach even than to keep your job. Customers may drop off individually if they can no longer afford you, but you're not going to lose them all at once; markets don't "reduce headcount."What if you quit your job to start a startup that fails, and you can't find another? That could be a problem if you work in sales or marketing. In those fields it can take months to find a new job in a bad economy. But hackers seem to be more liquid. Good hackers can always get some kind of job. It might not be your dream job, but you're not going to starve.Another advantage of bad times is that there's less competition. Technology trains leave the station at regular intervals. If everyone else is cowering in a corner, you may have a whole car to yourself.You're an investor too. As a founder, you're buying stock with work: the reason Larry and Sergey are so rich is not so much that they've done work worth tens of billions of dollars, but that they were the first investors in Google. And like any investor you should buy when times are bad.Were you nodding in agreement, thinking "stupid investors" a few paragraphs ago when I was talking about how investors are reluctant to put money into startups in bad markets, even though that's the time they should rationally be most willing to buy? Well, founders aren't much better. When times get bad, hackers go to grad school. And no doubt that will happen this time too. In fact, what makes the preceding paragraph true is that most readers won't believe it—at least to the extent of acting on it.So maybe a recession is a good time to start a startup. It's hard to say whether advantages like lack of competition outweigh disadvantages like reluctant investors. But it doesn't matter much either way. It's the people that matter. And for a given set of people working on a given technology, the time to act is always now.Russian TranslationChinese TranslationJapanese Translation

https://github.com/Myriad-Dreamin/typst.ts

https://raw.githubusercontent.com/Myriad-Dreamin/typst.ts/main/fuzzers/corpora/layout/pagebreak-parity_01.typ

typst

Apache License 2.0

#import "/contrib/templates/std-tests/preset.typ": * #show: test-page #set page(width: auto, height: auto) // Test with auto-sized page. First #pagebreak(to: "odd") Third

https://github.com/yhtq/Notes

https://raw.githubusercontent.com/yhtq/Notes/main/常微分方程/main.typ

typst

#import "../template.typ": proof, note, corollary, lemma, theorem, definition, example, remark, proposition, der, partialDer #import "../template.typ": * // Take a look at the file `template.typ` in the file panel // to customize this template and discover how it works. #show: note.with( title: "常微分方程", author: "YHTQ", date: none, logo: none, ) = 前言 - 教师：李伟固 - 作业：王子康 <EMAIL>，单周周三前交含有单个未知函数的方程称为常微分方程，往往形如： $ f(x, y， y', y'', ..., y^((n))) = 0 $ 其中 $n$ 为方程的阶数，它的位置类似于代数方程中的次数。当然阶数更高的方程更加复杂。若 $f(x, y， y', y'', ..., y^((n))) = 0$ 中的 $f$ 是除去 $x$ 的多元线性函数（每个变量的次数都是1），则称为线性微分方程，否则称为非线性微分方程。线性微分方程的形式为： $ sum_(i=0)^n a_i (x) y^((i)) = 0 $ 这类方程有比较完善的理论。 + 通解与特解形式上，定义若 $y=h(x)$ 代入 $f(x, y， y', y'', ..., y^((n))) = 0$ 是恒等式，则称 $y=h(x)$ 是方程的解。许多时候研究解的存在性与唯一性已经足够困难了。称微分方程的通解为含有 $n$ 个独立常数的解，对常数任意取值都可以获得一个解。相对的，称没有未定常数的解为特解，它就是某个固定的函数。一般而言，$n$ 阶的微分方程大约有 $n$ 阶的通解，通解包含了大多数的解，但很多时候并不是所有的解。 #example[][ 微分方程 $y' = y^(1/3)$ 的一个通解为： $ cases((2/3 x + C)^(3/2) quad 2/3x + C >0, 0 quad 2/3x + C <= 0) $ 但 $y = 0$ 显然是一个解，并不能被通解包含。 ] + 初值问题（柯西问题）往往给定一个初值条件便可在通解中减少一个自由变元。当给定足够的初值时以及合理的条件时，微分方程便有唯一解。以后我们会研究在何种条件下它确实有解/有唯一解。 + 定性问题自然的，解微分方程是非常复杂的，简单易用的初等解法非常有限，且往往只能解一些特殊的方程。大部分微分方程相关的研究都是在避免解出具体的解的情况下进行的，例如直接通过方程本身分析它是否有周期解，有界解等等，或者通过方程大致描述解对应曲线的形状。类似 $f(x) + f'(x-1) = 0$ 的微分方程有时称为时滞微分方程，它与常微分方程的研究方法大不相同。 = 常微分方程初等解法尽管可以证明，绝大多数常微分方程没有初等解，更不可能被初等方法解出，但这些解法仍然十分重要。 == 恰当形式与积分因子 #definition[对称形式][ 若 $y$ 是关于 $x$ 的函数，则一阶常微分方程均可化简为： $ der(y, x) = f(x, y) $ 形式上与 $f(x, y) dif x - dif y = 0$相当。严格来说这不是微分方程，但是它使用颇为方便，而且可以写成对称的形式： $ p(x, y) dif x + q(x, y) dif y = 0 $<proper> 从而统一研究了 $x, y$ 作为自变量的形式，因此也将@proper 称为微分方程的对称形式。 ] #definition[恰当形式][ 若存在一个可微函数 $u(x, y)$ 使得： $ dif u = p(x, y) dif x + q(x, y) dif y\ <=> cases( partialDer(u, x) = p(x, y), partialDer(u, y) = q(x, y) ) $ 则称@proper 为恰当形式。 ] #theorem[][ 设： $ d u(x, y) = p(x, y) dif x + q(x, y) dif y $ 则 $u(x, y) = C$ 产生的可微隐函数为@proper 的通解 ] #proof[ 只证明 $x$ 作为自变量的情形。设： $ u(x, y(x)) = c $ 求导立得： $ partialDer(u, x) + partialDer(u, y) der(y, x) = 0\ p(x, y) + q(x, y) der(y, x) = 0 $ 这就表明 $y(x)$ 是原方程的解 ] 这启示我们，形式上我们确实可以将 $der(y, x)$ 视作分式进行处理 #theorem[][ 设 $p, q$ 在区域 $D$ 上 $C^1$ 且 $ p(x, y) dif x + q(x, y) dif y $ 恰当，则： $ partialDer(p, y) = partialDer(q, x) $ 若 $D$ 是单连通区域，则反之也成立 ] #proof[ - 设 $dif u = p(x, y) dif x + q(x, y) dif y$，则： $ partialDer(p, y) = partialDer(partialDer(u, x), y) = partialDer(partialDer(u, y), x) = partialDer(q, x) $ （注意到 $u$ 是 $C^2$ 的，因此可以交换偏导数的次序） - 反之若上式成立且 $D$ 单连通，这就是格林公式的直接推论，取（与道路无关的）曲线积分： $ u(x, y) = integral_(x_0, y_0)^(x, y) p(x, y) dif x + q(x, y) dif y $ 容易验证 $u$ 满足要求 ] #example[][ $ (3x^2 + 6 x y^2) dif x + (6 x^2 y + 4 y^3) dif y = 0 $ 可以验证这是恰当形式，下面我们具体求出 $u$ - 由于 $partialDer(u, x) = 3x^2 + 6 x y^2$，可得： $ u = integral (3x^2 + 6 x y^2) dif x = x^3 + 3 x^2 y^2 + C(y) $ - 进一步 $ partialDer(u, y) = 6 x^2 y + C'(y) = 6 x^2 y + 4 y^3 => C'(y) = 4 y^3 => C(y) = y^4 + C_1 $ - 因此 $u = x^3 + 3 x^2 y^2 + y^4 + C_1$，这就是通解 ] #example[][ $ (x^2 + 2 x y - y^2) dif x + (x^2 - 2 x y - y^2) dif y = 0\ dif (1/3 x^3 - 1/3 y^3) + 2x y dif x + x^2 dif y - (y^2 dif x + 2 x y dif y) = 0\ dif (1/3 x^3 - 1/3 y^3) + dif (x^2 y) - dif (y^2 x) = 0\ dif (1/3 x^3 - 1/3 y^3 + x^2 y - y^2 x) = 0 $ ] 一旦微分方程的形式恰当，解出微分方程是十分容易的。然而事实上，一个微分方程可能对应众多的对称形式。设 $f(x, y) !=0$，则： $ p(x, y) dif x + q(x, y) dif y = 0 <=> p(x, y) f(x, y) dif x + q(x, y) f(x, y) dif y = 0 $ 选取不同的 $f(x, y)$ 可能改变对称形式的恰当性 #definition[][ 设非零可微函数 $f(x, y)$ 满足： $ p(x, y) f(x, y) dif x + q(x, y) f(x, y) dif y = 0 $ 是恰当形式，则称 $f(x, y)$ 为@proper 的积分因子 ] #example[][ $ y dif x - x dif y = 0 $ 并不是恰当形式，但： $ mu = 1/x^2, 1/y^2, ... $ 都是其积分因子\ 事实上： $ dif(y/x) = 1/x dif y - y/x^2 dif x = 1/x^2 (y dif x - x dif y) = 0 $ ] 我们当然希望对一般的对称形式找到合适的积分因子，具体而言，是要求： $ partialDer(p f, y) = partialDer(q f, x) <=>\ f (partialDer(p, y) - partialDer(q, x)) = q partialDer(f, x) - p partialDer(f, y) $<int_factor> 这是一个一阶线性偏微分方程，不幸的是该类偏微分方程的通用解法只能归结于求解反推过程的常微分方程，因此一般的积分因子是没有通用的解法的。所幸，在一些特殊情形下，我们是可以求得积分因子的。我们直接对 $f$ 施加额外条件： - $f$ 与 $y$ 无关，原方程变为： $ f partialDer(p, y) = f partialDer(q, x) + q partialDer(f, x) $<ori_res> 这是常微分方程： $ partialDer(f, x) =f (partialDer(p, y) - partialDer(q, x)) / q $ 观察可得 $(partialDer(p, y) - partialDer(q, x)) / q$ 应与 $y$ 无关。同时此时我们只需解 $f$，$y$ 可以视作定值，因此上述方程可以化简为： $ (dif f) / f = (partialDer(p, y) - partialDer(q, x)) / q dif x\ ln |f| = integral (partialDer(p, y) - partialDer(q, x)) / q dif x + C $<res> 由于 $f$ 是非零的，因此 $f$ 的符号是确定的，这就解出了 $f$\ 不难发现只要 $(partialDer(p, y) - partialDer(q, x)) / q$ 与 $y$ 无关，@res 便可给出一个与 $y$ 无关的 $f$ 满足@ori_res，进而代入@int_factor 应当也成立。这表明 $(partialDer(p, y) - partialDer(q, x)) / q$ 与 $y$ 无关是可以找到此形式的积分因子的充要条件。 #theorem[一阶线性微分方程][ 一阶线性微分方程： $ y' = p(x) y + q(x) $ 的对称形式为： $ (p(x) y + q(x)) dif x - dif y = 0 $ 计算: $ (partialDer(p(x) y + q(x), y) - partialDer(1, x)) / (-1) = - p(x) $ 积分因子 $f$ 满足： $ ln |f| = integral - p(x) dif x + C\ |f| = e^C e^(-integral p(x) dif x) = A e^(-integral p(x) dif x) $ 因此取 $f = e^(-integral p(x) dif x)$ 即可，进而原方程可解 ] 不幸的是，这样能够找到的积分因子仍然非常有限。实践上寻找积分因子更多只能靠直接观察形式 #example[][ 寻找一个曲线使得从定点 $(c, 0)$ 射出的光线经曲线反射后与 $x$ 轴平行： - 通过几何方法可以列出方程 $y/x = (2y)/(1-y'^2)$ - 不妨设 $y' > 0, y> 0$，解得： $ y' = - x /y + sqrt(1 + (x/y)^2)\ y dif y = (sqrt(x^2 + y^2)-x)dif x\ x dif x + y dif y = sqrt(x^2 + y^2) dif x\ (x dif x + y dif y)/sqrt(x^2 + y^2) = dif x\ dif(sqrt(x^2 + y^2)) = dif x\ sqrt(x^2 + y^2) = x + C $ - 平方可得方程为： $ y^2 = 2 x C + C^2 = 2C(x+1/2C) $ ] #theorem[][ 假设 $p, q, f, g in C^1$，$f, g$ 都是对称形式： $ 0 = p(x, y) dif x + q(x, y) dif y := w $ 的积分因子，且 $f/g$ 不是常数，则 $f/g = C$ 是原方程的通解，其中 $C$ 为常数 ] #proof[ 设： $ f w = d u_1\ g w = d u_2 $ 不妨设 $partialDer(u_2, y) !=0$，则由隐函数定义， $u_2 = C$ 可以解出 $y = y(x)$，此时有： $ 0 = der(u_1(x, y(x)) w(x, y(x)), x)\ = w(x, y)der(u_1(x, y(x)), x) + u_1(x, y)der(w(x, y(x)), x)\ = w(x, y) $ ] 理论上讲，所有常微分方程的初等解法都可以化为恰当方程。但是实践上我们更多采用具体的方法求解。 == 分离变量法 #theorem[分离变量法][ 设微分方程形如： $ der(y, x) = f(x)g(y) $ - 若 $g(y) = 0$，则 $f(x) = C$ 就是原方程的解 - 否则，化为： $ 1/g(y) der(y, x) = f(x) $ 设 $G(y) = integral 1/(g(y)) dif y$，则: $ G'(y) = 1/g(y) der(y, x) = f(x) => G(y) = integral f(x) dif x + C $ ] #example[][ 求解 $der(y, x) = y^2 cos x$: - 首先 $y = 0$ 是它的一个解 - 其次，假设 $y$ 不恒为零，则： $ 1/(y^2) dif y = cos x dif x\ -1/y = sin x + C\ y = -1/(sin x + C) $ ] #remark[][ 这种形式下我们已经默认了 $y$ 是因变量而 $x$ 是自变量，因此不能让 $x$ 恒等于 $cos x$ 的零点。如果不考虑函数关系只考虑解曲线，则这也是一个特殊的解 ] #example[][ $der(y, x) = y^(1/3)$: - $y = 0$ 当然是解 - 否则，我们先证明若 $y(x_0) = 0$，则当 $x < x_0$ 时 $y = 0$： - 否则，假设 $x_1 < x_0, f(x_1) !=0$，不妨设其大于零。取： $ x_2 = inf(Inv(y)(0) sect [x_1, x_0]) $ 注意到上式右侧是闭集，因此 $y(x_2) = 0 => der(y, x) |_(x = x_2)= 0$\ 此外，$der(y, x)$ 在 $[x_1, x_2]$ 之间恒正，这意味着 $y(x)$ 在 $[x_1, x_2]$ 之间单调递增，矛盾！ - 之后，我们分离变量 $ 1/y^(1/3) dif y = dif x\ 3/2 y^(2/3) = x + C\ y = plus.minus (2/3 x + C)^(3/2) $ - 综上，原方程的通解为： $ y = cases((2/3 x + C)^(3/2) quad 2/3x + C >0, 0 quad 2/3x + C <= 0) $ ] #example[][ 下面是几种常见形式 + $der(y, x) = f(x)y$ - $y = 0$ 当然是解 - 否则： $ 1/y dif y = f(x) dif x\ ln |y| = integral f(x) dif x + C\ y = A e^(integral f(x) dif x) $ 事实上，可以证明若 $y(x_0) = 0$，则 $y = 0$，因此可以去掉绝对值 + $der(y, x) = g(y/x)$\ 令 $y = x u$，方程化为： $ der(y, x) = u + x der(u, x) = g(u)\ 1/(g(u) - u) dif u = 1/x dif x\ integral 1/(g(u) - u) dif u = ln |x| + C\ $ + $der(y, x) = (a_1 x + b_1 y + c_1)/(a_2 x + b_2 y + c_2)$\ - 若 $c_1 = c_2 = 0$，上下同时除以 $x$ 即化为上一种情形 - 否则，试图做线性替换：设 $A = mat(a_1, b_1;a_2, b_2), alpha = vec(c_1,c_2)$ $ vec(x, y) = B vec(x', y') + beta $ 我们希望： $ A vec(x, y) + alpha = A(B vec(x', y') + beta) + alpha = A B vec(x', y') $ 也就是： $ A beta + alpha = 0 $ 该线性方程有解时，一定可以通过线性替换化为上一种情形\ - 否则 $"rank"(A) = 1$，不妨设 $(a_2, b_2) = k (a_1, b_1)$，令： $ u = a_1 x + b_1 y $ 原方程化为： $ der(u, x) = a_1 + b_1 der(y, x) = a_1 + b_1 (u + c_1)/(k u + c_2) $ 这时右侧已无 $x$，直接分离变量积分即可 + $y f(x y) dif x + x g(x y) dif y = 0$\ 令 $x y = u$，方程化为： $ u/x f(u) dif x + g(u)(dif u - u/x dif x) = 0\ $ 这也是分离变量的形式 ] #example[最速降线][ 求 $a -> b$ 的一条曲线，使得小球从 $a$ 无摩擦的滑下的时间最短\ 严格来说这个问题需要用到变分法，不过抛开高级理论，利用光线折射定律的相关想法也可以不严格的求解。\ 可以证明，光线从 $a$ 点到 $b$ 点经过若干种介质时，入射角的正弦与介质中传播的速度是定值。我们依次来对曲线建立微分方程： - 在曲线上任意一点 $(x, y)$ 的速度表示为： $ v = sqrt(2 g y) $ （能量守恒） - “入射角”也即与垂直方向夹角满足 $sin alpha = sqrt(1+y'^2)$ 因此可以建立方程： $ sqrt(2 g y)/ sqrt(1+y'^2) = C\ 2 g y = C^2 (1+y'^2)\ 1 + y'^2 = 2/C^2 g y\ y' = sqrt(2/C^2 g y - 1)\ 1/sqrt(2/C^2 g y - 1) dif y = dif x\ integral 1/sqrt(2/C^2 g y - 1) dif y = x + C\ $ ] == 一阶线性微分方程 #theorem[常数变易法][ 对于一阶线性微分方程： $ der(y, x) + p(x) y = q(x) $ 一般假设 $p, q$ 都连续且 $p(x)$ 不恒为零我们可以先解出齐次方程： $ der(y, x) + p(x) y = 0 $ 它的解为： $ y = C e^(-integral p(x) dif x) := C v(x) $ 做变量替换： $ y = u(x) v(x) $ 代入发现： $ q(x) - p(x) u(x) v(x) = der(u(x) v(x), x) = v(x) der(u(x), x) + u(x) der(v(x), x) $ 但是注意到： $ der(v(x), x) + p(x) v(x) = 0 $ 上式化简为： $ q(x) = v(x) der(u(x), x)\ $ 待解的函数是 $u(x)$，这是可分离变量的 ] #example[贝努利方程][ 形如： $ der(y, x) = p(x) y + q(x) y^n $ 的方程称为贝努利方程。除去 $y = 0$ 的解外，可以做变量替换： $ t = y^(1-n)\ dif t = -(1-n) y^(-n) dif y\ dif y = -1/(1-n) y^(n) dif t\ $ 上式化简为： $ y^(-n)der(y, x) = p(x) y^(-n+1) + q(x)\ -1/(1-n) der(t, x) = p(x) t + q(x)\ $ 这就化为了一阶线性微分方程 ] #example[][ 对于形如： $ der(y, x) = p(x)y^2+q(x)y+r(x) $ 的微分方程，一般没有初等解法。但是假设我们猜出了一个解 $y = f(x)$，可以做变量替换： $ y = u + f(x)\ dif y = dif u + f'(x) dif x\ $ 代入原方程后，仍然是该方程的形式。然而注意到 $u = 0$ 一定是一个解，最后一定会消掉零次项。代入发现： $ dif u + f'(x) dif x = (p(x)(u+f(x))^2+q(x)(u+f(x))+r(x))dif x\ dif u = (p(x)(u^2 + 2u f(x)) + q(x) u ) dif x\ der(u, x) = p(x)u^2 + (2 p(x) f(x) + q(x))u $ 这是贝努利方程，进而可解 ] #theorem[][ 设 $a, b in RR, a != 0$，方程： $ der(y, x) = a y^2 + b x^m $ 当且仅当 $m = 0, - 2, - (4k)/(2k plus.minus 1)$ 时可积分求解 ] #proof[ 无妨假设 $a = 1$\ - $m = 0$ 时显然可解 - $m = -2$ 时，做变量替换： $ z = x y\ dif z = x dif y + y dif x\ dif y = 1/x dif z - y/x dif x\ $ 代入得： $ 1/x dif z - y/x dif x = (y^2 + b x^(-2)) dif x\ x dif z - x y dif x = ((x y)^2 + b) dif x\ x dif z -z dif x = (z^2 + b) dif x $ 这是可分离变量的 - $m = -(4k)/(2k plus.minus 1)$ 必要性由刘维尔给出，颇为复杂 ] 特别的，$der(y, x) = x^2 + y^2$ 是不可积分求解的，这也说明了求解微分方程是非常之困难的。 == 一阶隐方程 #definition[][ 形如： $ F(x, y, y') = 0 $ 的方程称为一阶隐方程 ] 一般而言它的求解当然是非常困难的，但是对于特定的几类，我们可以尝试求解： #theorem[][ 对于以下几类隐方程我们给出一些思路： - 可以化为： $ y = f(x, y') $<ori> 的方程。\ 此时两边对 $x$ 求导得： $ y' = partialDer(f(x, y'), x) = f'_1(x, y') + f'_2(x, y')y'' $ 这是关于 $y', x$ 的显式微分方程。\ 若能解出 $y' = p(x)$，代入@ori 就是原方程的解\ 若只能解出 $x = q(y')$，代入@ori 就是原方程以 $y'$ 为参数的参数表达\ 否则，若只能解出 $g(x, y') = 0$，方程的解只能表达为： $ cases( y = f(x, y'), g(x, y') = 0 ) $ - 可以化为： $ x = f(y, y') $ 两边对 $y$ 求导得： $ 1/y' = partialDer(f(y, y'), y) = f'_1(y, y') + f'_2(y, y')der(y', y) = f'_1(y, y') + f'_2(y, y') y''/y' $ 做类似讨论即可 - 可以化为 $f(x, y') = 0$，只需要考虑它的参数形式： $ x = u(t)\ y' = v(t) $ 则： $ dif y = v(t) dif x = v(t) u'(t) dif t $ 两边积分即得 $y$ - 可以化为 $f(y, y') = 0$ 的形式。类似的，如果能找到参数方程： $ y = u(t)\ y' = v(t) $ 则： $ dif y = u'(t) dif t\ dif y = v(t) dif x\ v(t) dif x = u'(t) dif t\ dif x = (u'(t))/v(t) dif t $ 积分即可\ 注意需要补充 $v(t) = 0$ 的解 ] #example[][ $y'^3+2x y' - y = 0$ 两边直接对 $y$ 求导得： $ 3y'^2 y'' + 2y'+ 2x y'' - y' = 0\ (3y'^2 + 2x)y'' = -y'\ (3P^2 + 2x)P' = -P\ der(P, x) =- P/(3P^2+2x)\ 2P der(P, x) = -(2P^2)/(3P^2+2x)\ der(P^2, x) = -(2P^2)/(3P^2+2x)\ $ 这是方程右侧对于 $P^2$ 是齐一次的，是我们可解的方程\ $ der(x, P^2) = -3/2 - x/P^2 $ 令 $u = x/P^2, dif u = (P^2 dif x - x dif P^2)/P^4$ $ der(u, P^2) = 1/P^2 der(x, P^2) - u/P^2 = -1/P^2(3/2+u) -u/P^2\ (dif u)/(3/2 + 2u) = - 1/P^2 dif P^2\ ln |3/2 + 2u| = - ln 1/P^2 + C\ 3/2 + 2u = C P^2\ $ ] #example[Clairaut][ $y = x y' + f(y')$\ 求导得: $ y' = y' + x y'' + f'(y') y''\ x y'' + f'(y') y'' = 0\ $ 讨论： - $y'' = 0$，此时原方程是一族直线 $y = x c + f(c)$ - $x = - f'(y')$，此时代回得 $y = - y' f'(y') + f(y')$，这是以 $y'$ 为参数的特解当 $f'' != 0$ 时，直线族恰好是特解的切线族，这是因为此时 $f'$ 有反函数，在特解中可以解出： $ y' = u(x)\ y = x u(x) + f(u(x)) $ 此时，直线： $ y = x u(x_0) + f(u(x_0)) $ 恰好就是特解于 $(x_0, y_0)$ 点的切线（显然直线过 $(x_0, y_0)$ 且斜率 $u(x)$ 当然就是该点处的导数）事实上，我们并没有说明这些就是原方程的所有解。后续我们会证明在 $f'' != 0$ 的情况下，确实这些囊括了原方程解的所有情况（但可以构造出其他的解，例如将切线-特解-切线拼接起来也是一个解） ] == 应用举例至此，我们已经给出了所有一阶微分方程的初等解法。尽管方法有限，但是它们已经能够解出了许多重要的方程 #example[两个物种的生态方程][ 一战期间，生态学家发现随着人类捕鱼量下降，两类鱼中以其他鱼为食的鱼的比例上升，而以植物为食的鱼的比例下降。他们提出了一个模型：\ 设以鱼为食的鱼的数量为 $x = x(t)$, 以植物为食的鱼的数量为 $y = y(t)$，令 $r_x, r_y$ 是两者的增长率。显然 $r_x$ 应该随 $y$ 递增，而当 $y$ 较小时应该有 $r_x < 0$，进而设： $ r_x = sigma y - lambda $ 类似的原因，设： $ r_y = mu - delta x $ （上面的常数应该都是正数）\ 这给出了微分方程组： $ x'/x = sigma y - lambda\ y'/y = mu - delta x $ 两式相比： $ y'/x' x/y = (mu - delta x)/(sigma y - lambda) $ 注意到 $y'/x' = der(y, x)$，这就消掉了 $t$ $ y' (sigma y - lambda)/y = (mu - delta x)/x\ sigma y - lambda ln y = mu ln x - delta x + C\ sigma y + delta x - lambda ln y - mu ln x = C\ $ 虽然我们难以继续计算，但我们可以做定性研究，可以发现它的解都是周期解。\ 同时，我们还可以求出平均值： $ (dif x)/x = (-lambda + sigma y) dif t\ integral_T (dif x)/x = integral_T (-lambda + sigma y) dif t\ integral_T (dif ln(x)) = integral_T (-lambda + sigma y) dif t\ 0 = integral_T (-lambda + sigma y) dif t\ 0 = -lambda T + sigma integral_T y dif t\ (lambda)/sigma = 1/T integral_T y dif t $ 这就是 $y$ 在一个周期内的平均值\ 当捕捞量减少时，两个物种的增长率都增加，相当于 $mu$ 增大而 $lambda$ 减少，可以看到 $x$ 的平均值会增加，而 $y$ 的平均值会减少 ] #example[$n$ 体问题][ $n$ 体问题是指万有引力下 $n$ 个大质量质点相互作用的问题。设有 $n$ 个质点 $p_i$，分别有： - 坐标 $P_i = (x_i, y_i, z_i)$ - 质量 $m_i$ - 牛顿第二定律和万有引力定律： $ m_i (dif^2 P_i)/(dif t^2) = sum_(j != i) G m_i m_j (P_j - P_i)/norm(P_j - P_i)^3 $ 不妨设 $G = 1$，化简为： $ m_i (dif^2 P_i)/(dif t^2) = sum_(j != i) G m_i m_j (P_j - P_i)/(sqrt((x_i - x_j)^2 + (y_i - y_j)^2 + (z_i - z_j)^2)^3) $ 记 $U = sum_(i != j)( m_i m_j )/norm(P_i -P_j)$，化简为： $ m_i (dif^2 P_i)/(dif t^2) = partialDer(U, P_i) $ 记 $q_i = der(P_i, t)$（速度），则： $ m_i der(q_i, t) = partialDer(U, P_i) $ 令 $p = autoVecN(p, n), q = autoVecN(q, n)$，则： $ dif /(dif t) vec(p, q) = vec(q, m partialDer(U, p)) $ 可以发现： $ sum_i m_i der(q_i, t) = sum_i partialDer(U, q_i) = sum_i (sum_(j != i) m_i m_j (P_i - P_j)/norm(P_i - P_j)^3) = 0 $ 这就是在说质心不动，也就是动量守恒，进而： $ sum_i m_i q_i = C_1\ integral sum_i m_i q_i dif t = C_1 t + C_2\ sum_i m_i p_i = C_1 t + C_2 $ 质心以匀速直线运动类似的，可以验证角动量守恒，能量守恒\ 然而可以证明，除了这些以外不再有与之独立的代数的首次积分，因此代数的首次积分就是这些接下来我们解二体问题，由之前提到的守恒量可设质心恒定在原点，运动都在平面上，可以化简到： $ cases( (dif^2 x)/(dif t^2) = -m x/(x^2 + y^2)^(3/2), (dif^2 y)/(dif t^2) = -m y/(x^2 + y^2)^(3/2) ) $ 其中，1 对应 $ m = (m_2^3)/(m_1 + m_2)^2 $ 2 对应 $ m = (m_1^3)/(m_1 + m_2)^2 $ 角动量守恒给出： $ x der(y, t) - y der(x, t) = c_1\ $ 能量守恒给出： $ (der(x, t))^2 + (der(y, t))^2 - (2m)/(sqrt(x^2 + y^2)) = c_2 $ 做极坐标换元： $ x = r cos theta\ y = r sin theta $ 代入角动量守恒化简得： $ c_1 = r^2 der(theta, t)\ c_1 (t_2 - t_1) = integral_(t_1)^(t_2) r^2 dif theta\ $ 上式右侧是扫过的扇形面积，这实际上就是开普勒第二定律\ 代入能量守恒化简得： $ (der(r, t))^2 = c_1 + 2 m Inv(r) - c_1^2 r^(-2) \ der(r, t) = sqrt(c_2 + m^2/c_1^2 - (c_1/r - m/c_1)^2)\ der(theta, t) = c_1/r^2\ der(r, theta) = r^2/c_1 sqrt(c_2 + m^2/c_1^2 - sqrt(c_1/r - m/c_1)^2) $ 这是可分离变量的，最终可以化简得到： $ r = p/(1 + e cos (theta - theta_0)) $ 其中 $e = c_1/m sqrt(c_2 + m^2/c_1^2), p = c_1^2/m$\ 这一定是一条圆锥曲线，这就是开普勒第一定律\ 再次带回： $ (p/(1 + e cos (theta - theta_0)))^2 dif theta = c_1 dif t\ c_1 T = integral_(0)^(2pi) p^2/(1 + e cos (theta - theta_0))^2 dif theta\ = (2p^2 pi)/sqrt((1 - e^2)^3) $ 计算可得： $ T^2/(p^3 /(1-e^2)^3) = (4 pi^2)/m $ 当 $m_1$ 很大时，$m$ 几乎就是 $m_2$，这也就是太阳系中的开普勒第三定律二体问题还有著名的反问题，假设任何星球围绕太阳的有界轨道都是椭圆，且万有引力形如： $ f = m_1 m_2 f(norm(p_1 - p_2))(p_1 - p_2) $ 那么数学上可以证明万有引力只能是正比于距离或者反比于距离的平方，前者会导致太阳位于轨道中心，后者太阳将位于焦点 ] == 二阶自洽微分方程 #theorem[二阶自洽微分方程的解法][ 称形如 $ (dif^2 x)/(dif t^2) + f(x) = 0 $ 为二阶自洽方程，引进 $F(x)$ 使得 $F'(x) = f(x)$，令 $y = x', H(x, y) = y^2/2 + F(x)$，有： $ der(H, t) = y der(y, t) + F'(x) der(x, t) = y (dif^2 x)/(dif t^2) + f(x) der(x, t) = 0 $ 换言之，参数方程 $(x(t), y(t))$ 表示的函数就是 $H$ 的等高线，可以解得： $ y = plus.minus sqrt(2(C - F(x)))\ (dif x)/(dif t) = plus.minus sqrt(2(C - F(x)))\ plus.minus 1/(sqrt(2(C - F(x)))) dif x = dif t $ 两边直接积分及得原方程的解 ] #example[弹簧][ 弹簧方程 $(dif^2 x)/(dif t^2) + a x = 0, a > 0$ 就是上面的形式，因此可以解得： $ t + C' = plus.minus integral 1/sqrt(2(C - 1/2 a x^2)) dif x $ 计算化简得到 $ x = C_1 cos a t + C_2 sin a t\ y = - a C_1 sin a t + a C_2 cos a t $ 容易发现所有解都以 $(2 pi)/a$ 为周期，且 $(0, 0)$ 为所有轨道的中心。由于所有轨道都是相同周期的，因此也称为等时中心。线性系统中一定是等时的 ] #example[单摆方程][ 单摆方程 $(dif^2 x)/(dif t^2) + a sin x = 0, a > 0$ 也是上面的形式，因此可以解得： $ dif t = 1/sqrt(2(C - a cos x)) dif x $ 这是椭圆积分，我们无法写出它的初等形式\ 转而考虑： $ H(x, y) = y^2/2 - a cos x + a $ 的等高线。注意到 $H$ 关于 $x$ 以 $2 pi$ 为周期，可以把它想象成定义在圆柱面上的函数。容易发现这里 $(0, 0)$ 也是中心，但不是等时中心，事实上，在等高线： $ y^2 - a cos x = C $ 上，有： $ 1/4 T = integral_0^(pi/2) der(t, x) dif x = integral_0^(pi/2) 1/sqrt(2(C - a cos x)) dif x $ 显然不可能与 $C$ 无关 ] #theorem[][ 设二阶自洽微分方程满足 $f(x)$ 连续且 $x f(x) > 0$，引进 $F(x)$ 并不妨设 $F(0) = 0$，则对： $ H(x, y) = y^2/2 + F(x) $ + $H(0, 0)$ 是 $H$ 的最小值点 + 在 $(0, 0)$ 附近，$H(x, y) = C$ 给出一个闭合曲线 + $H = h$ 交 $x$ 轴于 $x_1(h), x_2(h)$ 两点，$x_1 < x_2$ + 原点是等时中心当且仅当 $(x_2(h) - x_1(h))/sqrt(h)$ 是与 $h$ 无关的常数 ] #proof[ + 由题设将有 $F(x) >= 0$ 因此显然 + 直接解出 $y$ 即可 + 由 $F(x)$ 的单调性及 $F(0) = 0$ 显然 + 这个证明并不用到高级的知识，但极具技巧性\ 令 $s = F(x)$，则 $f(x) dif x = dif s$，有： $ 1/2 T_h = integral_(x_1)^(x_2) der(t, x) dif x = integral_(x_1)^(x_2) 1/sqrt(2(h - F(x))) dif x\ = integral_0^(h) 1/sqrt(2(h - s)) 1/(f_+(x)) dif s - integral_0^(h) 1/sqrt(2(h - s)) 1/(f_-(x)) dif s\ 1/2 integral_0^H T_h/sqrt(H-h) dif h = integral_0^H ( integral_0^(h) 1/sqrt(2(H - h)(h - s)) 1/(f_+(x)) dif s)dif h - integral_0^H ( integral_0^(h) 1/sqrt(2(H - h) (h - s)) 1/(f_-(x)) dif s)dif h\ = integral_0^H ( integral_s^H 1/sqrt(2(H - h)(h - s)) 1/(f_+(x)) - 1/sqrt(2(H - h) (h - s)) 1/(f_-(x)) dif h)dif s\ = pi/(sqrt(2)) (integral_0^H 1/(f_+(x)) - 1/(f_-(x)) )dif s\ = pi/(sqrt(2)) (integral_0^H 1/(f_+(x)) - 1/(f_-(x)) )dif F(x)\ = pi/(sqrt(2)) (integral_0^H f(x)/(f_+(x)) - f(x)/(f_-(x)) )dif x\ = pi/(sqrt(2)) (x_2(H) - x_1(H))\ $ - 假设周期与 $H$ 无关，则有： $ T_0 sqrt(H) = pi/sqrt(2) (x_2(H) - x_1(H))\ T_0 = pi/sqrt(2) (x_2(H) - x_1(H))/sqrt(H) $ 表明上式右侧与 $H$ 无关，这也给出了周期的计算公式 ] #corollary[][ 若原点是等时中心且 $f(x)$ 是奇函数，则 $f(x)$ 一定线性 ] = 微分方程的解理论 == Peano 定理 #theorem[Peano][ 设： $ f(x, y) in C(DD), DD = {abs(x - x_0) <= a, abs(y -y_0) <= b} $ 方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 在 $[x_0 - h, x_0 + h]$ 上有解，其中： $ h = min{a, b/(max_DD abs(f))} $ 若 $y, f$ 是向量值函数，结论也是类似的 ]<peano> 本节的目标就是证明它。\ #lemma[][ 有界区间上等度连续的一致有界的函数列有一致收敛的子列 ] #proof[ #lemmaLinear[][ 设 $f_n: I -> RR$ 对于每个 $x in I, f_n (x)$ 都有界 $M_x$，$I$ 是有限区间，则任取 $I$ 的一个可数子集 $E$，存在 $f_n$ 的子列使得这个子列在 $E$ 上收敛 ] #proof[ 设 $E = {x_1, x_2, ..., x_n}$，注意到 $f_n (x_1)$ 是有界序列，有收敛子列 $f_(n_1) (x_1)$\ 再考虑 $f_(n_1) (x_2)$ ，它当然也有收敛子列，记为 $f_(n_2) (x_2)$\ 不断进行下去，我们得到了若干子列 $f_(n_k) (x_k)$。\ 用对角线法，发现： $ f_(n_n) (x_n), n = 1, 2, 3, ... $ 在每个 $x_k$ 上都收敛，证毕 ] #lemmaLinear[][ 设 $I$ 是有限区间，$f_n : I -> RR$ 等度连续，且在 $I$ 的稠密子集上收敛，那么它在 $I$ 上一致收敛 ] #proof[ 利用柯西收敛原理，任取 $epsilon > 0$，存在 $delta >0$，使得： $ forall n in NN, forall x_1, x_2, abs(x_1 - x_2) < delta => abs(f_n (x_1) - f_n (x_2)) < epsilon $ 对于这个 $delta$，我们可以取出 $E$ 中有限个点 $E'$ 使得： $ union_(x in E') B(x, delta) = I $ 由于这里只有有限个 $x$，当然 $f_n|_E'$ 将一致收敛，因此可设存在 $N$ 使得： $ forall m, n > N, forall x in E', abs(f_m (x) - f_n (x)) < epsilon $ 因此，有： $ forall m, n > N, forall x in I, exists x' in E'\ abs(x - x') < delta &=> abs(f_m (x) - f_n (x)) \ &< abs(f_m (x) - f_m (x')) + abs(f_m (x') - f_n (x')) + abs(f_n (x') - f_n (x)) < 3 epsilon $ ] 在引理中取 $E = QQ sect I$，它是 $I$ 的稠密子集，因此结论成立 ] 回到 @peano 的证明，这个方法是所谓的欧拉折线法 #proof[ 微分方程可以转化为积分方程: $ y(x) = y_0 + integral_(x_0)^(x) f(t, y(t)) dif t $ 只需证明它在 $[x_0, x_0 + h]$ 上有解，另一侧类似。\ 将其 $n$ 等分，设分点为 $x_0, x_1, ..., x_n$，按照以下定义分段线性函数： $ (x_(i-1), y_(i-1)) 到 (x_i, y_i) "的斜率由" f(x_(i-1), y_(i-1)) "给出" $ 需要验证这些点落在 $DD$ 中，显有： $ y_k - y_(k-1) = f(x_(k-1), y_(k-1)) (x_k - x_(k-1)) $ 两边求和： $ y_k - y_0 &= sum_(i = 1)^(k) f(x_(i-1), y_(i-1)) (x_i - x_(i-1))\ abs(y_k -y_0) &<= sum_(i = 1)^(k) abs(f(x_(i-1), y_(i-1)) (x_i - x_(i-1))) \ &<= max_DD abs(f) sum_(i = 1)^(k) (x_i - x_(i-1)) <= max_DD abs(f) abs(x_k - x_0) \ &<= h max_DD abs(f) <= b $ 这也体现了 $h$ 为何如此定义\ 设得到的函数为 $g_n (x)$，验证它们： - 一致有界：显然 - 等度连续：注意到所有 $g_n (x)$ 均满足： $ abs(g_n (x) - g_n (y)) <= max_DD abs(f) abs(x - y) $ 因此也成立由引理，它有一致收敛子列。不妨设其收敛，只需证明极限函数 $g(x)$ 满足积分方程\ 事实上（注意到 $f$ 是有界闭集上的连续函数，进而一致连续，从而 $f(t, g_n (x))$ 也一致收敛）： $ g(x) - integral_(x_0)^x f(t, g(t)) dif t = lim_(n->+infinity) g_n (x) - integral_(x_0)^x f(t, g_n (t)) dif t $ 有： $ g_n (x_0 + k/(n h)) &= sum_(i = 1)^(k) f(x_i, y_i) (x_i - x_(i-1))\ integral_(x_0)^(x_0 + k/(n h)) f(t, g_n (t)) dif t = sum_(i = 1)^(k) integral_(x_(i-1))^(x_i) f(t, g_n (t)) dif t &= sum_(i = 1)^(k) f(xi_i, g_n (xi_i)) (x_i - x_(i-1))\ g_n (x_0 + k/(n h)) - integral_(x_0)^(x_0 + k/(n h)) f(t, g_n (t)) dif t &= sum_(i = 1)^(k) (f(x_i, y_i) - f(xi_i, g_n (xi_i))) (x_i - x_(i-1))\ &=sum_(i = 1)^(k) (f(x_i, g_n (x_i)) - f(xi_i, g_n (xi_i))) (x_i - x_(i-1)) $ 任意 $epsilon > 0$ ，令 $delta, delta'$ 满足： - $abs(x-x') < delta, abs(y-y') < delta => abs(f(x, y) - f(x', y')) < epsilon$ - $abs(x-x') < delta' => abs(g_n (x) - g_n (x')) < min(delta, epsilon), forall x, forall n$ 取 $N$ 充分大使得 $forall n > N$： - $1/(n h) < delta' => abs(g_n (x) - g_n (x_0 + k/(n h))) < epsilon, g_n (x_i) - g_n (xi_i) < delta$ - $1/(n h) < delta => abs(x_i - xi_i) < delta$ - $1/(n h) $ ] #lemma[Gronwall][ 设 $f, g in C[a, b], g(x) >= 0$，则有： $ f(x) <= C + integral_a^x f(t) g(t) dif t => f(x) <= C e^(integral_a^x g(t) dif t) $ ] #proof[ 令 $Phi(x) = integral_a^x f(t) g(t) dif t$，则有： $ Phi' <= (C + Phi)g(x) $ 设 $h = e^(integral_a^x g(s) dif s), Phi = u h$，有： $ u' h + u h' <= (C + u h)g\ u' h <= C g\ u' <= C g/h\ u(x) - u(a) <= C integral_a^x g(s)/h(s) dif s\ (Phi(a) = 0 => u(a) = 0)\ u(x) <= C integral_a^x g(s)/h(s) dif s = C integral_a^x g(s) e^(-integral_a^x g(s) dif s) dif s = C (1 - 1/h(x))\ $ 从而： $ f(x) <= C + Phi(x) = C + u h <= C h $ 证毕 ] #corollary[][ 若在 Gronwall 不等式中，有 $C <= 0, f(x) >= 0$，则 $f(x) = 0$ ] #theorem[][ 在@peano 中，若 $f$ 是 Lipschitz 连续的，则解是唯一的。更进一步，构造出的折线列 $g_n (x)$ 本身即一致收敛 ] #proof[ 假设 $y = phi(x), y= psi(x)$ 是两个解，则： $ abs(phi(x) - psi(x)) = abs(integral_(x_0)^x f(t, phi(t)) dif t - integral_(x_0)^x f(t, psi(t)) dif t) <= L abs(integral_(x_0)^x abs(phi(t) - psi(t)) dif t) $ - 当 $x > x_0$ 时，上式即为： $ abs(phi(x) - psi(x)) <= L integral_(x_0)^x abs(phi(t) - psi(t)) dif t $ 直接利用 Gronwall 不等式可得 $abs(phi(x) - psi(x)) <= 0 => abs(phi(x) - psi(x)) = 0$ - 当 $x < x_0$ 时，类似可以证明结论成立综上，$phi(x) = psi(x)$ ] 事实上，在一维的情形下，我们可以找到更弱的条件 #theorem[][ 设 $f$ 满足 Osgoad 条件，也即： $ exists F: [0, +infinity] -> [0, +infinity)\ integral_(0)^(t) 1/F(r) dif r = +infinity\ abs(f(x, y) - f(x, z)) <= F(abs(y - z)) $ 则上述方程的解存在且唯一 ] #proof[ 设 $y_1, y_2$ 是两个不同的解，令 $r(x) = y_1 (x) - y_2(x)$\ 注意到 $r(x_0) = 0$，若 $r$ 不恒为零，不妨设 $x_1$ 是零点且 $r(x) > 0 ,forall x in (x_1, x_2]$\ 此时： $ der(r, x) = y'_1 - y'_2 = f(x, y_1) - f(x, y_2) <= F(abs(y_1 - y_2)) = F(r(x))\ => r'/F(r(x)) <= 1\ => integral_(s)^(t) r'/F(r(x)) dif x <= t -s\ => integral_(r(s))^(r(t)) 1/F(r) dif r <= t - s $ 令 $s -> x_1$，上式左侧趋于无穷，而右侧有限，矛盾！ ] #theorem[][ 设 $f(x, y)$ 满足 $f(x, y)$ 关于 $y$ 单调下降，则解也存在并唯一 ] #proof[ 设 $y_1, y_2$ 是两个不同的解，令 $r(x) = y_1 (x) - y_2(x)$\ 注意到 $r(x_0) = 0$，若 $r$ 不恒为零，不妨设 $x_1$ 是零点且 $r(x) > 0 ,forall x in (x_1, x_2]$\ 此时： $ der(r, x) = y'_1 - y'_2 = f(x, y_1) - f(x, y_2) <= 0 $ 而 $r(x_1) = 0$，上式表明 $r$ 单调下降，进而 $r(x) <= 0$，因此只能 $= 0$，矛盾！ ] 本节利用的欧拉折线不仅在理论上很重要，在数值计算上也广泛用于数值求解微分方程，理论依据就是本节的定理 == Picard 定理 Picard 定理表面上和 Peano 定理的结论类似但条件更强，但是它的方法非常重要，因此仍有必要专门学习 #theorem[Picard][ 在 @peano 中，再补充 $f$ Lipschitz 连续，则方程的解存在 ] #proof[ 仍然转化为积分方程: $ y(x) = y_0 + integral_(x_0)^(x) f(t, y(t)) dif t $ 总体思想是所谓的不动点法，递归定义： - $phi_0 (x) = y_0$ - $phi_n (x) = y_0 + integral_(x_0)^(x) f(t, phi_(n-1) (t)) dif t$ + 首先验证定义合理，需要保证所有的值不超出区域。 - $x$ 的合理性是显然的 - 假设 $phi_(n - 1) subset overline(U(y_0, b))$，则： $ abs(phi_n (y_1) - y_0) <= integral_(x_0)^(x) abs(f(t, phi_(n-1) (t))) dif t\ <= max_DD abs(f) h <= b $ 这就验证了 $phi_(n) subset overline(U(y_0, b))$ + 接下来证明它一致收敛 - 首先归纳证明 $abs(phi_n (x) - phi_(n-1)(x)) <= (M L^(n-1))/(n!) abs(x -x_0)^n$ $ abs(phi_n (x) - phi_(n-1)(x)) = abs(integral_(x_0)^(x) f(t, phi_(n-1) (t)) - f(t, phi_(n-2) (t)) dif t)\ <= integral_(x_0)^(x) abs(f(t, phi_(n-1) (t)) - f(t, phi_(n-2) (t))) dif t\ <= integral_(x_0)^(x) L abs(phi_(n-1) (t) - phi_(n-2) (t)) dif t $ 利用归纳条件计算即可 - 其次： $ (M L^(n-1))/(n!) abs(x -x_0)^n <= (M L^(n-1))/(n!) h^n $ 上式右侧只需利用比较判别法知求和收敛，从而容易利用维尔斯特拉斯判别法知道原级数 $sum (phi_n (x) - phi_(n-1)(x))$ 一致收敛 + 最后我们得到了极限函数 $phi(x)$，显然应当满足： $ phi(x) = y_0 + integral_(x_0)^(x) f(t, phi(t)) dif t $ 这就是原方程 ] #remark[][ 这个定理的证明有以下优点： - 可以推广到抽象的函数空间，例如 Banach 空间 - 假设 $f$ 是解析函数，则构造的 $phi_n$ 都是解析函数，进而可以证明 $phi$ 也是解析的 ] == 解的延拓之前我们给出了至少在某个局部，标准形式的微分方程是有解的。然而我们往往需要在更大的定义域上找到解 #theorem[][ 设 $G$ 是开区域，$f(x, y)$ 在 $G$ 上连续，$(x_0, y_0) in G$\ 对含 $(x_0, y_0)$ 的任意有界闭区域 $G_1 subset G$，过 $(x_0, y_0)$ 点的微分方程的解 $y=f(x)$ 都可以延拓到 $G_1$ 的边界上 ] #proof[ 由题设，$G_1$ 中的点到 $G$ 的边界的距离有正下界 $d$\ 因此可以找到闭集： $ G_2 = {(x, y) | abs(x - x') <= delta, abs(y - y') <= delta, exists (x, y) in G_1} subset G $ 再令： $ M = max_(f in G_2) abs(f) +1\ delta'_0 = delta_0 / M $ 利用之前的存在性定理，可以找到 $U(x_0, delta'_0)$ 内的解。再次利用存在性定理可以继续前进，且步长为 $delta'_0$ 不变，因此经过有限步之后，我们就得到了直到 $G_1$ 的边界上的解 ] #example[][ - 设微分方程 $der(y, x) = x^2 + y^2$ 容易验证过任何一点的解存在且唯一。然而，可以证明任何一个解都只在有界区间上成立事实上，任取 $y(x)$ 在区间 $I$ 上是方程的一个解，假设 $I$ 无界，不妨取为 $[0, +infinity]$，则： $ y' = x^2 + y^2 >= 1 + y^2\ y'/(1+y^2) >= 1\ integral_(1)^(t) y'/(1+y^2) dif x >= x - 1\ integral_(1)^(y(t)) 1/(1+y^2) dif y >= x - 1\ $ 然而令 $t -> +infinity$，上式左侧有界而右侧无界，矛盾！这不与解的延拓定理矛盾，容易想到它在 $x$ 方向有垂直渐近线，而在 $y$ 方向可以进行无限延伸 - 给定微分方程： $ der(y, x) = (x^2 + y^2 +1) sin (pi y) $ 可以验证它也满足存在唯一的条件，并且 - $y = k, k in ZZ$ 当然是一个解 - 对于其他的解，显然必须夹在上下两个水平直线之间（否则若相交，则交点处解将不唯一），因此在 $x$ 方向必将可以无限延伸 - 给定微分方程： $ der(y, x) = (y-3)(y+1) e^(x+y)^2 $ 容易验证满足存在唯一的条件 - 显然 $y = 3, -1$ 是两个解 - 若 $y_0 in [-1, 3]$，过该点的解同样与 $y = 3, -1$ 不相交，进而导数总为负数，单调下降，因此在 $-infinity, +infinity$ 方向都有极限。观察等式可知极限只能分别为 $3, -1$ - 若 $y_0 in [3, +infinity]$ ，负向仍然有水平渐近线，而正向类似刚才的将会存在垂直渐近线这些例子都说明，只要延拓定理的条件成立，最终得到的解一定在某个方向上无界 ] == 比较定理给定两个微分方程，我们希望从形式看出解的大小关系，由如下定理： #theorem[第一比较定理][ 设 $f, F$ 都连续，且 $f < F$，两个微分方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 )\ cases( der(Y, X) = F(X, Y), Y(X_0) = Y_0 ) $ 在区间 $I$ 上各有解 $y, Y$，则有： $ y < Y, forall x > x_0\ y > Y, forall x < x_0 $ ] #proof[ 显然 $Y'(x_0) = F(x_0, y_0) > y'(x_0)$，而 $Y(x_0) = y(x_0)$，因此可取 $delta$ 使得： $ forall x in (x_0, x_0 + delta), Y(x) > y(x)\ forall x in (x_0 - delta, x_0), Y(x) < y(x) $ 设 $S = {x in I | Y(x) = y(x) }$，仿照之前的过程可以看出 $S$ 中每个点都孤立，然而任取 $x_1 > x_2 in S$，设 $g(x) = Y(x) - y(x)$，既然： $ g(x_1) = g(x_2) = 0 $ 且 $g$ 在 $x_2$ 右侧严格单增，在 $x_1$ 左侧严格单减，因此其间一定有 $S$ 中的点。反复进行，这将与 $S$ 中每个点都孤立矛盾\ 这意味着 $S$ 中恰有一点，只能是 $x_0$，故结论当然成立 ] #corollary[][ 设 $f in C(a, b), abs(f) <= A(x) abs(y) + B(x), forall x in (a, b)$，则微分方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 所有的解都可以延拓到 $(a, b)$ ] #proof[ #lemma[][ 两个微分方程： $ der(y, x) = A(x) abs(y) + B(x)\ der(y, x) = - A(x) abs(y) - B(x) $ 过任意点的解存在唯一，且解的存在区间都是 $(a, b)$ ]<linear_all_range> #proof[ 存在唯一性是容易的。对于存在区间，只证明第一个方程\ 显然它的解单调递增，若解的存在区间不是 $(a, b)$ ，则必然在某点 $x_0 in (a, b)$ 处的左/右极限为 $+\/minus infinity$，不妨设为左极限为正无穷\ 这意味着存在 $delta, (x_0 - delta, delta)$ 上时有： $ der(y, x) = A(x) abs(y) + B(x) = A(x) y + B(x) < M_1 y + M_2 $ 但稍加计算可得 $der(y, x) = M_1 y + M_2$ 的解不可能在有界区间内趋于无穷，进而该方程的解由比较定理被控制在有界区间，矛盾！ ] 回到原题，显然有： $ -(A(x) abs(y) + B(x) + 1)<= der(y, x) = f < A(x) abs(y) + B(x) + 1 $ 由比较定理，每个有限区间上的解当然被控制在有界区间，证毕 ] #definition[][ 给定微分方程 $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 称其解 $f, g$ 为最大解，最小解，若任取 $y$ 为一个解，都有： $ g <= y <= f $ ] #theorem[][ 设 $D = {abs(x-x_0) < delta_x, abs(y - y_0) < delta_y}$，则存在 $tau > 0$ 使得方程 $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 在 $x in [x_0 , x_0 + tau]$ 上有最大/最小解 ] #proof[ 设 $y_n$ 是微分方程： $ cases( der(y, x) = f(x, y) + 1/n, y(x_0) = y_0 ) $ 由 @peano，每个方程的解都在： $ [x_0 - h_n, x_0 + h_n] "where" h_n = min{delta_x, delta_y/(max_DD abs(f_n))} $ 存在，显然 $h_n$ 有公共正下界，取 $tau$ 是一个正下界，则这些方程都在 $[x_0 - tau, x_0 + tau]$ 上有界\ // 我们希望找到一些一致收敛性。事实上，这些函数： // - 等度连续，既然： // $ // y'_n (x) <= max_DD abs(f) + 1.n <= max_DD abs(f) + 1 // $ // - 一致有界，既然导数有一致界且都定义在有限区间上事实上，这些解一致收敛。这是因为若设 $forall n, m > N, phi(x) = y_n (x) - y_m (x) $，则： $ abs(der(phi, x)) = abs(1/n - 1/m) < 1/N\ abs(phi(x) - phi(0)) = abs(integral_(x_0)^(x) der(y, x) dif s) <= integral_(x_0)^(x) abs(der(y, x)) dif s <= (x - x_0)/N <= tau/N $ 柯西原理给出了一致收敛性，因此我们可以取极限，而转化为积分方程，它们的极限就是原方程的解，而容易验证这个解一定是原方程的最大解 ] #remark[][ 我们当然可以将两端的最大解拼接，得到双边的最大解 ] #theorem[][ 设 $f, g$ 是微分方程： $ cases( der(y, x) = f(x, y) , y(x_0) = y_0 ) $ 在 $I$ 上的最大/最小解，则任取 $x_1 in I, y_1 in [g(x_1), f(x_1)]$，存在方程的解 $h$ 使得 $h(x_1) = y_1$ ] #proof[ #lemmaLinear[][ 设 $g_1, g_2$ 在区间 $I$ 上均满足 $g'_i (x) = f(x, g_i (x))$，则 $max(g_1, g_2), min(g_1, g_2)$ 也满足 ] #proof[ 只证 $max$，$min$ 类似\ 设 $h = max(g_1, g_2)$，验证上面的等式只需逐点验证即可： - 任取 $x in I$，讨论： - 若 $g_1(x) > g_2(x)$，则有 $g_1, g_2$ 连续性知存在 $x$ 的开邻域 $B$ 使得： $ h|_B = g_1 $ 此时由于 $g_1$ 在该点是微分方程的解，$h$ 当然也是 - 若 $g_1(x) < g_2(x)$，同理 - 若 $g_1(x) = g_2(x)$，注意到： $ g'_1(x) = f(x, g_1(x)) = f(x, g_2(x)) = g'_2(x) := d $ 因此： $ abs((h(x + Delta x) - h(x))/(Delta x) - d) &<= abs((h(x + Delta x) - g_1 (x + Delta x))/(Delta x)) + abs((g_1 (x + Delta x) - g_1 (x))/(Delta x) - g'_1 (x))\ &<= abs((g_2(x + Delta x) - g_1 (x + Delta x))/(Delta x)) + abs((g_1 (x + Delta x) - g_1 (x))/(Delta x) - g'_1 (x))\ &<= abs(((g_2(x + Delta x) - g_1 (x + Delta x)) - (g_2 (x) - g_1 (x)))/(Delta x)) + abs((g_1 (x + Delta x) - g_1 (x))/(Delta x) - g'_1 (x))\ &<= abs(g'_2 (xi) - g'_1 (xi)) + abs((g_1 (x + Delta x) - g_1 (x))/(Delta x) - g'_1 (x))\ &"（对" g_2(x) - g_1(x) 在 x, x + Delta x "上利用微分中值定理）" $ 其中 $xi$ 在 $x, x + Delta x$ 之间。注意到由微分方程，$g_1, g_2$ 当然连续可导，进而当 $Delta x -> 0$ 时上式 $-> 0$，这就保证了： $ h'(x) = d = f(x, h(x)) $ 也即 $h$ 在该点满足微分方程得证 ] 考虑微分方程： $ cases( y' = f(x, y) , y(x_1) = y_1 ) $ 它在 $x_1$ 的某个邻域内有解 $h$，又可不妨设 $h$ 在 $I$ 内某个极大的区间 $I_1$（也即所有可延拓闭区间的并）上是原方程的解，并设： $ h_1 = max(min(h, Z), W), x in I_1 $ 将满足 $h'_1 (x) = f(x, h_1 (x))$，同时也不难验证 $h_1(x_1) = y_1$，因此是过该点的微分方程的解\ 另一方面，显然有： $ max(min(h, Z), W) >= W\ max(min(h, Z), W) <= max(Z, W) = Z $ 因此 $h_1$ 是区间上有界的微分方程的解从而： - 若 $x_0 in I_1$，由上面的条件和引理知 $h_1$ 就是要找的解 - 否则，设 $I_1$ 闭包中距 $x_0$ 最近的点为 $x_2$，注意到 $h_1$ 在 $I_1$ 上有界，因此由延拓定理要么意味着存在 $x'$ 使得 $abs(h_1(x') - y_0) = b$，要么 $h_1$ 一定可以至少延拓到 $x_2 := x'$，然而已经假设 $h$ 不能再延拓，因此该点附近一定取 $Z$ 或 $W$ ，无论如何一定有： $ exists x', h_1(x') = Z(x') 或 W(x') $ 无论何者，都意味着 $h_1$ 可以与 $Z$ 或 $W$ 的其中之一拼接起来，这就是所要找的解。 ] #corollary[][ 若微分方程的单侧的最大解，最小解均存在且不相等，则该侧微分方程有无穷多解，这无穷多解被控制在最大/最小解之间 ] #example[][ - 微分方程： $ cases( der(y, x) = y^(1/3), y(0) = 0 ) $ 通解为： $ y = cases( 0 quad x <= -C\ plus.minus (2/3 x + C)^(3/2) quad x > C ) $ 在 $0$ 右侧的最大解为 $y = (2/3 x)^(3/2)$，最小解为 $y = -(2/3 x)^(3/2)$ - 微分方程： $ cases( der(y, x) = abs(y)^(1/2), y(0) = 0 ) $ 通解为： $ y = cases( 1/4(C_1 - x)^2 quad x <= C_1, 0 quad C_1 <= x <= C_2, 1/4(x - C_2)^2 quad x >= C_2 ) $ ] #theorem[第二比较定理][ 设 $f, F$ 都连续，且 $f <= F$，两个微分方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 )\ cases( der(Y, X) = F(X, Y), Y(X_0) = Y_0 ) $ 则： - 在初值右侧，前者的最小解小于等于后者的任意解;在初值左侧，前者的最大解大于等于后者的任意解 - 在初值右侧，后者的最大解大于等于前者的任意解;在初值左侧，后者的最小解小于等于前者的任意解 ] #proof[ 仿照上面的逼近过程再使用第一比较定理即可 ] == 奇解 #definition[奇解][ 设 $Gamma = {(x, y) | y = f(x)}$ 是微分方程 $F(x, y') = 0$ 的一个解，若对任意的 $(x_, y) in Gamma$，在该点的任何一个邻域里存在过 $(x_0, y_0)$ 的不同于 $Gamma$ 的解，且该解与 $Gamma$ 在该点相切，则称 $Gamma$ 是奇解 ] #example[][ - $y = x y' - 1/2 y'^2$，它的解为： $ y = 1/2 x^2\ y = c x - c^2/2 $ 注意到下面的直线族恰好是上面抛物线的所有切线，因此抛物线当然就是一个奇解 - $y^2 + y'^2 = 1$\ 令 $y = cos theta, y' = sin theta, x = u(theta)$，有： $ (-sin theta)/(u') = sin theta\ sin theta = 0 或 u' = -1 $ 因此解为： $ y' = 0 => y = plus.minus 1\ u' = -1 => x = -theta + C => y = cos(x - C) $ 可以看出 $y = plus.minus 1$ 是奇解 ] #theorem[奇解的必要条件][ 设 $G subset RR^3, F(x, y, y') in C^0, partialDer(F, y), partialDer(F, y')$ 存在且连续。假设 $y = phi(x)$ 是奇解，则： $ cases( F(x, phi(x), phi'(x)) = 0, partialDer(F, y') (x, phi(x), phi'(x)) = 0 ) $ ] #proof[ 第一个式子是显然的。对于第二个式子，假设存在 $x_0$ 使得 $partialDer(F, y) (x_0, phi(x_0), phi'(x_0)) != 0$，设 $(x_0, y_0, P_0) = (x_0, phi(x_0), phi'(x_0))$，在该点对 $F$ 利用隐函数定理，存在某个开邻域使得： $ F(x, y, P) = 0 <=> P = f(x, y) $ 且 $f$ 是连续函数且关于 $y$ 连续可导，将在 $(x_0, y_0)$ 的某个邻域内 Lipschitz 连续，可以利用解的存在唯一性定理可知该邻域内微分方程的解存在唯一，与 $phi$ 是奇解显然矛盾 ] 上面的定理给出了奇解的一个必要条件。这里有两个等式，若可以化简直到可解出 $y$，便可能求出原方程的奇解。更进一步，若能消去 $y'$ 得到： $ Delta (x, y) = 0 $ 则奇解就在该曲线之中，这条曲线被称为判别曲线 #example[][ - $ y = P x ln x + (x p)^2 - y\ $ 可以解得判别曲线为 $y = -1/4 (ln x)^2$，容易验证它是一个解，稍后会判断它是否是奇解 - $y'^2 + y - x = 0$\ 可以解得判别曲线为 $y = x$，但不是原方程的解，因此原方程没有奇解 - $y'^2 - y'^2 = 0$，它的判别曲线是 $y = 0$ 是解，然而容易看出它的所有解是: $ y = C e^(plus.minus x) $ 以及它们可能的拼接，然而仅有 $y = 0$ 唯一一个解可能经过 $x$ 轴，因此过 $(x_0, 0)$ 的解都是存在且唯一的 ] #theorem[奇解的充分条件][ 设 $F(x, y, y') in C^infinity$，假设： $ cases( F(x, phi(x), phi'(x)) = 0, partialDer(F, y') (x, phi(x), phi'(x)) = 0 ) <=> y = psi(x) $ （事实上，也就是 $Delta(x, y)$ 是原方程的解）\ 若： $ cases( (diff^(k+l) F)/(diff y^k diff p^l) (x, psi(x), psi'(x)) = 0\, forall 0 <=k <= m-1\, 0 <= l <= n-1, (diff^m F)/(diff y^m) (x, psi(x), psi'(x)) != 0, (diff^n F)/(diff p^n) (x, psi(x), psi'(x)) != 0 ) $ 其中 $n, m in NN, n > m$。如果： - $m, n$ 之一为奇数，或 - $m, n$ 均为偶数，且 $(diff^m F)/(diff y^m) (diff^n F)/(diff p^n) < 0$ 则 $psi$ 是奇解 ] #proof[ #lemma[][ 设 $f(x, y) : C^infinity (RR^n times RR -> RR)$，设： $ f(x, y) = sum_(k=0)^n f_k (x) y^k + R_(n+1) (x, y) y^n $ 是关于 $y$ 的泰勒展开，则 $R_n (x)$ 也是 $C^infinity$ ] #proof[ 利用积分余项： $ f(x, y) - sum_(k=0)^n f_k (x) y^k = y^n/n! integral_0^1 (1-t)^n (diff^n f(x, y t))/(diff t^n) dif t $ 容易看出结论成立 ] #lemma[][ 考虑微分方程 $u' = plus.minus A(x, u) abs(u)^alpha, 0< alpha <1 $ 其中有： - $A$ 连续且在 $u !=0$ 时连续可导 - $A$ 有正下界 $c_0$，正上界 $c_1$ 则 $u = 0$ 是奇解 ]<lemma0_is_odd> #proof[ 不妨设符号取正，此时方程的解当然递增。\ 我们的目标是在 $x_0$ 处构造一个原方程的非平凡解。先考虑右侧，注意到： $ u' = A(x, u)u^alpha\ u^(-alpha) u' = A(x, u)\ $ 设 $v = u^(1 - alpha), v' = (1-alpha) u^(-alpha) u'$，则： $ 1/(1-alpha) v' = A(x, v^(1/(1-alpha))) $ ] 回到原方程，做换元 $y = psi(x) + u$，方程变成： $ H(x, u, u') = F(x, psi(x) + u, psi'(x) + u') = 0 $ 只需证明 $u = 0$ 是奇解即可。事实上，$H$ 满足如下条件： $ cases( (diff^(k+l) F)/(diff y^k diff p^l) (x, 0, 0) = 0\, forall 0 <=k <= m-1\, 0 <= l <= n-1, (diff^m F)/(diff y^m) (x, 0, 0) != 0, (diff^n F)/(diff p^n) (x, 0, 0) != 0 ) $ 可以想象，$H$ 泰勒展开后的形式非常简单。事实上，有： $ H(x, u, u') = u^m H_1 (x, u, u') + u'^n H_2 (x, u, u') $ 其中 $H_1, H_2$ 都无穷阶可导，且： $ H_1 (x, 0, 0) = (diff^m F)/(diff y^m) (x, psi(x), psi'(x)) != 0\ H_2 (x, 0, 0) = (diff^n F)/(diff p^n) (x, psi(x), psi'(x)) != 0 $ 因此可以（在某个小邻域内）不妨设 $H_1, H_2 !=0 $，方程化为： $ u^m H_1 (x, u, u') + u'^n H_2 (x, u, u') = 0 $ 我们当然希望进行开方，可以验证在假设的条件下（即 $m, n$ 之一为奇数，或 $m, n$ 均为偶数，且 $(diff^m F)/(diff y^m) (diff^n F)/(diff p^n) < 0$），我们将确实可以开方，对开方后的函数利用 @lemma0_is_odd 即可 ] #corollary[][ 设 $F(x, y, y') in C^2$，且： $ cases( F(x, phi(x), phi'(x)) = 0, partialDer(F, y') (x, phi(x), phi'(x)) = 0 ) <=> y = psi(x) $ 若： $ (diff^2 F)/(diff p^2) (x, psi(x), psi'(x)) != 0\ partialDer(F, y) (x, phi(x), phi'(x)) != 0 $ 则 $psi$ 是奇解 ] #proof[ 就是在上面的定理中取 $n = 2, m = 1$ 的结果，至于条件可以放松的原因可以在上面的证明过程中仔细验证 ] == 包络 #definition[包络][ 设 $k_c, c in C$ 是光滑曲线族，称光滑曲线 $gamma$ 为 $k_c$ 的包络，如果 $forall (x, y) in gamma, exists k_c$ 使得 $k_c$ 与 $gamma$ 相切，且在该点的任意开邻域内 $k_c != gamma$ ] #example[][ - 给定一族曲线 $y = (x-c)^2 + 1$，当然 $y = 1$ 是其一个包络 - 给定一族直线 $y = c x - c^2 / 4$，它就是 $y = x^2$ 的切线族，自然 $y = x^2$ 是包络 ] #theorem[][ 假设 $F(x, y, y') = 0$ 有通解 $u(x, y, c) = 0$，假设 $partialDer(u, y) != 0$。设 $u(x, y, c) = 0$ 关于 $c$ 构成的曲线族有包络 $gamma := y = phi(x)$，则 $y = phi(x)$ 是奇解 ] #proof[ - 首先证明 $gamma$ 是解。当然 $gamma$ 上每点都可以找到解 $u(x, y, c_0) = 0$ 与之在该点相切。由假设条件和隐函数定理，可以反解出： $ u(x, y, c_0) = 0<=> y = psi(x) $ 当然 $phi(x)$ 与 $psi(x)$ 相切但不相同，这表明该点处 $gamma$ 的函数值和导数值满足微分方程，因此 $gamma$ 是解 - 其次，过 $gamma$ 每点都可以找到与之相切的不同解，这就表明它是奇解 ] #theorem[][ 假设 $Gamma$ 是曲线族 $k_c: v(x, y, c) = 0$ 的包络，并且 $Gamma$ 与 $k_C$ 切于 $(f(c), g(c))$，其中 $f, g in C^1$，则 $Gamma$ 满足方程： $ cases( v(x, y, c) = 0, v'_c (x, y, c) = 0 ) $ 如果这个方程组中能消去 $c$，则称消去后的曲线为 $C$ 判别曲线 ] #proof[ 任取 $(x_0, y_0) in Gamma sect k_c$，则首先第一个方程当然成立\ 在 $v(f(c), g(c), c) = 0$ 中，两边对 $c$ 求导得： $ v'_x (f(c), g(c), c) f'(c) + v'_y (f(c), g(c), c) g'(c) + v'_c (f(c), g(c), c) = 0 $ 只需证明 $vec(v'_x (f(c), g(c), c), v'_y (f(c), g(c), c) ) dot vec(f'(c), g'(c)) = 0$，不妨设他们都不为零\ 然而注意到 $vec(f(c), g(c))$ 事实上是 $Gamma$ 在局部上的一个参数方程，$vec(f'(c), g'(c))$ 将成为 $Gamma$ 的切向量。\ 同时，$vec(v'_x (f(c), g(c), c), v'_y (f(c), g(c), c) )$ 事实上是 $v(x, y, c) = 0$ 在 $(f(c), g(c))$ 处的法向量，由于 $Gamma$ 是包络当然两者正交 ] #theorem[][ 假设曲线族 $k_c: v(x, y, c) = 0$ 的 $C$ 判别曲线确定了一条 $C^1$ 曲线 $x = x(c), y = y(c)$，且满足非退化条件： $ (x'(c), y'(c)) != (0, 0)\ (partialDer(v,y)(x(c), y(c), c), partialDer(v,x)(x(c), y(c), c)) != (0, 0) $ 并且该曲线在曲线上任意一点的局部都不在原曲线族中，则它是包络 ] #proof[ 任取 $c$，在 $P(c) = (x(c), y(c))$ 的局部都可以利用隐函数定理反解出 $v(x, y, c) = 0$ 的解，可以验证这个解与 $Gamma$ 在该点相切，因此是包络 ] #example[][ 求一条曲线，满足其任意一点的切线两个截距的倒数平方和为 $1$ 事实上，容易想到满足这个条件的曲线一定是一族直线的包络，这族直线满足截距的倒数平方和为 $1$，进而形如 $ a x plus.minus sqrt(1 - a^2) y = 1 $ 只需找到： $ v(x, y, c) = c x plus.minus sqrt(1 - c^2) y - 1 = 0 $ 的包络，计算： $ partialDer(v, x) = c\ partialDer(v, y) = plus.minus sqrt(1 - c^2)\ partialDer(v, c) = x minus.plus y c/sqrt(1 - c^2) $ 计算其 $c$ 判别式，发现就是单位圆 $x^2 + y^2 = 1$\ 当然，单位圆和满足要求的直线的任意光滑拼接都满足要求 ] = 解对参数和初值的依赖性 == $n$ 维欧式空间的微分方程首先，叙述高维的常用定理，设 $R: abs(x - x_0) <= a, norm(y - y_0) <= b$ #theorem[Picard][ 设 $f in C(R, RR^n), abs(f(x, y) - f(x, z)) <= L norm(y - z)$，则微分方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 在 $[x_0 - h, x_0 + h]$ 上有唯一解，其中 $h = min{a, b/(max_R norm(f))}$ ] #theorem[Peano][ 设 $f in C(R, RR^n)$，则微分方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 在 $[x_0 - h, x_0 + h]$ 上有解，其中 $h = min{a, b/(max_R norm(f))}$ ] #definition[$n$ 维线性方程组][ 设微分方程组： $ cases( der(y, x) = A(x) y + B(x), y(x_0) = y_0 ) $ 满足 $A, B$ 都在 $(a, b)$ 上是连续函数，则称其为 $n$ 维线性方程组 ] #proposition[][ 对于任意初值，$n$ 维线性方程组的解在 $(a, b)$ 内存在唯一 ] == 解对初值和参数的依赖性微分方程的初值当然依赖于初值。出于多种考虑，我们当然希望这种依赖有某种连续性，这就是本章的核心内容。 #example[][ 考虑微分方程： $ y' = y^(1/3), y(0) = epsilon(epsilon < 0) $ 的最大解为： $ cases( -(2/3 x + epsilon^(2/3))^(3/2) quad x >= 0, 0 quad x < 0 ) $ 然而令 $epsilon -> 0$，它与 $0$ 处的最大解相去甚远 ] 这个例子表明最大/最小解的连续依赖性往往并不成立，我们往往只研究微分方程解唯一时的初值依赖性\ 此外，我们说明初值依赖性与参数依赖性之间是一致的。事实上，给定参数方程： $ cases( y' = f(x, y, lambda), y(x_0) = y_0(lambda) ) $ 令 $Y = vec(y, lambda)$，它等价于： $ cases( Y' = vec(f, 0), Y(x_0) = vec(y_0(lambda), lambda) ) $ 反过来，也可以通过平移将初值吸收进参数 #theorem[连续依赖性][ 给定一族微分方程： $ cases( y' = f(x, y), y(x_0) = y_0 ) $<ori-equation> （其中 $f$ 连续）的解存在唯一，并设其解的存在区间为闭区间 $I$ \ 设 $phi(x, xi, eta)$ 满足： $ cases( partialDer(phi, x) = f(x, phi(x, xi, eta)), y(xi) = eta ) $ 则： $ lim_((s, xi, eta) ->(x, x_0, y_0)) phi(s, xi, eta) = phi(x, x_0, y_0), forall x in I $ ] #proof[ 用反证法。如若不然，则则存在点列使得： $ (s_n, xi_n, eta_n) -> (x, x_0, y_0)\ d(phi(s_n, xi_n, eta_n), phi(x, x_0, y_0)) > epsilon $ 注意到有积分方程： $ phi(x, xi_n, eta_n) = eta_n + integral_(xi_0)^(x) f(s, phi(s, xi_n, eta_n)) dif s $ 因此这族关于 $x$ 的函数等度收敛一致有界，不妨就假设一致收敛到 $psi(x)$，两边取极限将有： $ lim_(n -> infinity) phi(x, xi_n, eta_n) = y_0 + integral_(x_0)^(x) f(s, lim_(n -> infinity) phi(x, xi_n, eta_n)) dif s $ （还要利用 $f$ 一致连续/有界）\ 表明 $lim_(n -> infinity) phi(x, xi_n, eta_n)$ 就是@ori-equation 的唯一一个解 $phi(x, x_0, y_0)$，与假设矛盾！ ] #corollary[][ 给定一族微分方程： $ cases( y' = f(x, y, lambda), y(x_0) = y_0 ) $ 对于任意 $x_0, y_0, lambda$ 的解都存在唯一，则设 $phi(x, x_0, y_0, lambda)$ 是上述方程的解，它将是连续函数 ] #theorem[光滑依赖性][ $ cases( y' = f(x, y, lambda), y(x_0) = y_0 ) $ 其中 $f$ 连续，对 $y, lambda$ 是 $C^1$ 的，且对于任意 $x_0, y_0, lambda$ 的解都存在唯一，则设 $phi(x, x_0, y_0, lambda)$ 是上述方程的解，它将是 $C^1$ 的 ] #proof[ 由定义，$phi$ 关于 $x space C^1$ 是显然的。对于 $y, lambda$，前面叙述了 $lambda, y_0$ 互相转换，只需证明关于 $lambda$ 连续可导就好，进一步，不妨假设方程就是： $ cases( y' = f(x, y, lambda), y(0) = 0 ) $ 为了证明结论，构造 Picard 序列： $ phi_0 = 0\ phi_(n) (x) = integral_(0)^(x) f(s, phi_(n-1), lambda) dif s $ 由于这里关于 $y$ 已经 $C^1$，Lipschitz 条件当然成立，因此它一致收敛到原方程的解。同时，容易归纳得到 $phi_n$ 关于 $lambda$ 都是 $C^1$ 的，计算导数： $ partialDer(phi_n (x), lambda) = integral_(0)^(x) partialDer(f, y) (s, phi_(n-1), lambda) partialDer(phi_(n-1), lambda) + partialDer(f, lambda) (s, phi_(n-1), lambda) dif s $ 这里如果 $lambda, y$ 是高维的，无非采用梯度/向量导数即可。\ 由分析学结论，只需证明上面的序列一致收敛。不妨设： $ norm(partialDer(f, y)), norm(partialDer(f, lambda)) <= M $ 将有： $ norm(partialDer(phi_1 (x), lambda)) <= integral_(0)^(x) norm(partialDer(f, lambda) (s, phi_(n-1), lambda)) dif s <= alpha norm(x)\ norm(partialDer(phi_2 (x), lambda)) <= integral_(0)^(x) alpha^2 norm(x) + alpha dif s <= alpha^2 norm(x)^2/2 + alpha norm(x)\ $ 可以归纳证明它一致有界\ 设： $ V_(k, n) = norm(partialDer(phi_(k+n) (x), lambda) - partialDer(phi_(k) (x), lambda)) $ 将有： $ V_(k+1, n) = norm(integral_(0)^(x) (partialDer(f, lambda) (s, phi_(k+n), lambda) - partialDer(f, lambda) (s, phi_(k+1), lambda)) \ + (partialDer(f, y) (s, phi_(k+n), lambda)partialDer(phi_(k+n), lambda) - partialDer(f, y) (s, phi_(k+1), lambda)partialDer(phi_(k+1), lambda) )dif s)\ <= integral_(0)^(x_0) norm(partialDer(f, y) (s, phi, lambda)) v_(k, k+1) dif s\ + d_(k, n) $ 其中 $d_(k, n)$ 在 $k$ 充分大时一致收敛于零，因此 $exists E_n$ 单调下降使得 $d_(k, n) <= E_n$，原式化为： $ V_(k+1, n) <= alpha integral_(0)^(x_0) V_(k, k+1) dif s + E_n $ 归纳计算将可得到 $V_(k, n)$ 关于 $k$ 一致收敛于零，由柯西法则知结论成立最后，我们还要考察关于 $x_0$ 的偏导数。这里我们无法直接吸收，因为我们没有假设 $f$ 关于 $x$ 可求偏导，这里 $lambda,y_0$ 无关紧要，不妨设方程为： $ cases( y' = f(x, y), y(x_0) = 0 ) $ 仍然构造 Picard 序列： $ phi_0 = 0\ phi_(n) (x) = integral_(x_0)^(x) f(s, phi_(n-1)) dif s $ 类似的，计算： $ partialDer(phi_n, x_0) = -f(x_0, phi_(n-1)) + integral_(x_0)^(x) partialDer(f, y) partialDer(phi_(n-1), x_0) dif s $ 接下来的计算是完全类似的 ] #remark[][ 这里我们不对关于 $x$ 的光滑性有很高要求，当然是因为微分方程的解相当于 $x$ 的积分，当然会提高光滑性。 ] #corollary[][ $ cases( y' = f(x, y, lambda), y(x_0) = y_0 ) $ 其中 $f$ 连续，对 $y, lambda$ 是 $C^k$ 的（$k$ 可能为无穷），且对于任意 $x_0, y_0, lambda$ 的解都存在唯一，则设 $phi(x, x_0, y_0, lambda)$ 是上述方程的解，它将是关于 $y_0, lambda$ 是 $C^k$ 的，关于 $x, x_0$ 是 $C^1$ 的 ] #proof[ $C^1$ 刚刚已经证明，同时也可以将 $y_0$ 吸收，化为： $ cases( y' = f(x, y, lambda), y(0) = 0 ) $ 化为积分方程： $ y = integral_(0)^(x) f(s, y, lambda) dif s $ 可以直接求导： $ partialDer(y, lambda) = integral_(0)^(x) partialDer(f, y) (s, y, lambda) partialDer(y, lambda) + partialDer(f, lambda) (s, y, lambda) dif s $ 设 $p = partialDer(y, lambda)$，将有： $ p' = partialDer(f, y) (x, y, lambda) p + partialDer(f, lambda) (x, y, lambda) $ 这是关于 $p$ 的线性微分方程，而右侧关于 $lambda$ 是 $C^(k-1)$，关于 $y$ 是 $C^infinity$ 的，不断利用定理即可 ] #remark[][ 上面的定理换成关于 $y, lambda$ 解析也对，既然 Picard 序列中每一项的解析，而解析函数的一致收敛极限也是解析的 ] #corollary[][ 在上面的定理中将条件换成 $f$ 对所有参数都 $C^k$，则 $phi$ 就是 $C^k$ 的 ] #proof[ 此时不妨将所有系数吸收进参数，只需研究方程： $ cases( der(y, x) = f(x, y, x_0, y_0, lambda), y(0) = 0 ) $ 前面已经证明关于 $x_0, y_0, lambda$ 都是 $C^k$ 的，只需考虑关于 $x$ 的，然而： $ partialDer(phi, x) = f(x, phi, x_0, y_0, lambda) $ 前面已经证明 $phi$ 是 $C^1$ 的，上式表明 $phi$ 将是 $C^2$ 的，继而归纳可得 $phi$ 是 $C^k$ 的 ] #corollary[解对初值和参数的光滑依赖性最终版本][ 在上面的定理中将条件换成 $f$ 对所有参数都 $C^(k-1)$，且对 $y, lambda$ 是 $C^k$ 的，则 $phi$ 就是 $C^k$ 的 ] #proof[ 已经证明了 $k = 1$ 时情景，同时 $phi in C^(k-1)$ 也已经成立，同样可以直接看到关于 $x$ 是 $C^k$ 的，并且注意到： $ partialDer(phi, x) = f(x, phi, lambda)\ partialDer(partialDer(phi, x), x_0) = partialDer(f, y) partialDer(phi, x_0) $ 令 $Y = partialDer(phi, x_0)$ ，将有微分方程： $ Y' = partialDer(f, y) Y $ 上式右端关于所有参数都是 $C^(k-1)$ 的，因此 $Y$ 关于 $x_0$ 也是 $C^(k-1)$，这就证明了对 $x_0$ 的可微性\ ] #example[][ $ cases( y' = y + mu (x^2 + y^2), y(0) = 1 ) $ 试求 $partialDer(phi, mu)|_(mu = 0)$ 事实上，容易看出 $f$ 是解析函数，继而它的解都解析，我们直接求偏导并交换顺序： $ partialDer(partialDer(phi, x), mu) = partialDer(phi, mu) + (x^2 + phi^2) + 2 mu phi partialDer(phi, mu) $ 设 $u = partialDer(phi, mu)$，它满足微分方程： $ u' = u + (x^2 + phi^2) + 2 phi u mu = (1 + 2 phi mu) u+ x^2 + phi^2 $ 这是关于 $u$ 的一阶线性微分方程\ 同时，注意到 $mu = 0$ 时方程是好解的，因此： $ phi(x, 0) = e^x $ 代入得： $ u' = u + x^2 + e^(2 x) $ 解出 $u(x, 0) = e^(2 x) - x - 1$ ] == 局部变换本章的内容是从理论上研究微分方程。 #definition[自治系统][ 自治系统是指形如： $ cases( der(x, t) = f(x), x(t_0) = x_0 ) $ ] 局部变换是希望将一个自治系统的解在局部变换为另一个常微分方程的解。 #definition[][ - 称 $x_0$ 是常点，如果 $f(x_0) != 0$ - 称 $x_0$ 是奇点/平衡点，如果 $f(x_0) = 0$ ] #theorem[][ 设 $U$ 是微分同胚，$x = U(y)$，将微分方程： $ der(x, t) = f(x) $ 变成 $ der(y, t) = g(y) $ 则： $ g(y) = Inv((U')) f(x) $<local-transform> 这里 $U$ 指微分同胚的雅可比矩阵 ] #proof[ $ der(U y, t) = f(x)\ U' der(y, t) = f(x)\ U' g(y) = f(x) $ ] #definition[][ 设两个微分方程： $ der(x, t) = f(x)\ der(y, t) = g(y) $ 之间，存在 $C^k$ 的微分同胚使得满足@local-transform ，则称两个方程 $C^k$ 等价。\ 类似的，若在 $x_0$ 处的某个邻域存在 $C^k$ 的微分同胚使得满足@local-transform ，则称两个方程在 $x_0$ 处局部 $C^k$ 等价。 ] #theorem[常点的 $C^k$ 分类/拉直定理][ 设微分方程 $der(x, t) = f(x)$ 满足 $f(0) != 0$，则方程在 $0$ 处局部 $C^k$ 等价于： $ der(y, t) = Y(y), Y(y) = vec(1, 0, dots.v, 0) $ ] #proof[ #let vc = $vec(1, 0, dots.v, 0)$ 设 $f_i (x)$ 是分量，不妨设 $f_1 (0) != 0$\ 显然第二个方程的解就是： $ psi = y_0 + t vc $ 设 $phi(t, x_0)$ 是原方程的解，令: $ U(0) = 0\ U(y) = phi(y_1, 0, y_2, dots, y_(n)) $ 为了验证微分同胚，计算： $ abs(der(U, y)) = abs(partialDer(phi, t) der(y_1, y) + partialDer(phi, x_0) der((0, y_2, ..., y_n)^T, y))\ = abs(f(phi) der(y_1, y) + partialDer(phi, x_0) der((0, y_2, ..., y_n)^T, y)) $ 此外，将有： $ U(psi(t, y)) = U(t+y_1, y_2, ..., y_n) = phi(t+y_1, 0, y_2, ..., y_n) $ 求导再令 $t = 0$ 可得： $ U'(y) Y(y) = f(phi(y_1, 0, y_2, ... y_n)) = f(U(y)) $ 证毕 ] 说明常点的局部等价分类是非常简单的，接下来我们考虑奇点处的局部等价分类 #theorem[][ 线性微分方程 $x' = A x, x' = B x$ 在 $0$ 处局部 $C^k$ 等价当且仅当 $A, B$ 相似 ] #proof[ 任取微分同胚 $U$ 在 $0$ 处泰勒展开计算即可 ] 对于一般的方程，我们当然希望通过泰勒展开将其化为线性方程。然而方程 $x' = A x + o(x)$ 当然不总是与 $x' = A x$ 等价，至少我们需要 $A$ 非退化。在什么条件下可以将其化为线性方程是上个世纪常微分方程研究的重要课题之一 #theorem[][ 设 $f(x) = A x + o(x)$ 是 $C^infinity$ 的，而 $A$ 的特征根满足非共振条件： $ sum_(i = 1)^n m_i lambda_i != 0, (m_i) in ZZ^n - {0} $ 则方程 $x' = f(x)$ 在 $0$ 处局部 $C^infinity$ 等价于 $x' = A x$ ] = 线性微分方程 == 一般线性微分方程方便起见，这里约定对向量/矩阵的求导/积分都是逐元素进行的 #definition[][ 设 $x in RR^n, t in RR$\ 称微分方程： $ der(x, t) = A(t) x + B(t) $<linear-equation> 为一般线性微分方程，如果 $A, B$ 都关于 $t$ 连续 ] #proposition[][ - 一般线性微分方程在任何点处的解都存在唯一 - 每个解总是在大范围存在 ] #proof[ - 就是 @peano - 就是 @linear_all_range ] 一般而言，当 $A(t)$ 不是常数时，方程是无法写出解的。我们的目标是研究解空间的性质。 == 齐次线性微分方程 #definition[][ 在 @linear-equation 中，若 $B(t) = 0$，则称之为齐次微分方程 ] #proposition[][ - 齐次线性方程的解构成线性空间 - 齐次线性方程的解要么恒为零，要么恒不为零 - 齐次线性方程的若干解线性相关当且仅当在存在某点，它们在该点线性相关 ]<homogeneous-linear> #proof[ - 简单验证即可 - 注意到 $x = 0$ 是平凡解，结合解的唯一性立得 - 注意到若干解的线性组合还是解，利用上一条性质立得 ] #theorem[][ $n$ 维齐次线性微分方程的解空间恰为 $n$ 维线性空间。换言之，若可以找到 $n$ 个线性无关的解，则它们生成的线性空间恰为解空间，这称为方程的通解 ] #proof[ 令 $e_i$ 是 $RR^n$ 中一组标准基，令 $x_i (t)$ 是方程： $ cases( der(x, t) = A(t) x, x(t_0) = e_i ) $ 的一个解，断言它们就是原方程解的基 - 首先，它们线性无关，既然它们在 $t_0$ 处线性无关，利用 @homogeneous-linear 即可 - 其次，任取原方程的初值为 $x(t_0) = x_0$，令： $ x(t) = sum_i x_0^i x_i (t) $ 容易看出 $x(t)$ 也是符合该初值的解，由唯一性这就是唯一一个解证毕 ] #definition[][ 若矩阵 $X$ 是齐次线性微分方程的解，也即： $ der(X, t) = A(t) X $ 则称之为矩阵解。显然矩阵是解当且仅当每一列都是原方程的一个解，继而该矩阵的列秩至多为 $n$，恰为 $n$ 时称之为基本解矩阵或者基解阵 ] #proposition[][ 矩阵解 $X$ 是基础解矩阵当且仅当在某个点上有 $det(X) != 0$ ] #proof[ 就是 @homogeneous-linear ] #theorem()[][ 设 $Phi(t)$ 是基解阵，则 $X(t)$ 是基解阵当且仅当存在可逆矩阵 $C$ 使得： $ Phi(t) C = X(t) $ ] #proof[ 注意到求导是线性的，因此： $ der(Phi(t) C, t) = der(Phi(t), t) C = A(t) Phi(t) C $ 因此 $Phi(t) C$ 确实是解，计算行列式可得它是基础解矩阵另一方面，设 $X(t)$ 是基础解矩阵，任取 $t_0$ 并设有： $ Phi(t_0) C = X(t_0) $ 注意到 $ Phi(t_0), X(t_0)$ 都是可逆矩阵，$C$ 当然存在\ 此时 $X(t) - Phi(t) C$ 是原方程有零点的解，就是零 ] #proposition()[Liouville][ 设 $x_i$ 是 $n$ 个解，设 $det(x_1, x_2, ..., x_n) := W(t)$，则： $ W'(t) = tr(A(t)) W(t) $ ] #proof[ 容易发现： $ W'(t) = sum_i det(x_1, x_2, ..., x'_i, ..., x_n) = sum_i det(x_1, x_2, ..., A x_i, ..., x_n) $ #lemmaLinear[][ $ sum_i det(x_1, x_2, ..., A x_i, ..., x_n) = tr(A) det(x_1, x_2, ..., x_n) $ ] ] == 非齐次线性微分方程 #proposition[][ 设线性微分方程： $ x' = A(t) x + B(t) $ 则： - 任意两个解的差是对应齐次线性微分方程 $x' = A(t) x$ 的解 - 设 $gamma(t)$ 是一个特解（任意一个解），$X$ 是对应齐次微分方程的解空间，则原方程的所有解为 $gamma(t) + X$ ] #proposition[常数变易法][ 设 $Phi(t)$ 是 $x' = A(t) x$ 的基本解矩阵，则 $x' = A(t) x + f(t)$ 的解可以通过常数变易： $ x = Phi(t) C(t)\ Phi'(t) C(t) + Phi(t) C'(t) = A(t) Phi(t) C(t) + f(t)\ Phi(t) C'(t) = f(t)\ C'(t) = Inv(Phi(t)) f(t)\ C(t) = integral_(t_0)^(t) Inv(Phi(s)) f(s) dif s $ 这给出了可行的 $C$ ]<constant-variation> 上述命题表明，解一般线性微分方程的困难根本上来源于求齐次线性微分方程基础解矩阵的困难 == 常系数线性微分方程为了叙述方便，我们先给出矩阵幂级数，指数等定义 #definition[][ - 本节中定义矩阵的模为 $sum_(i, j) abs(a_(i j))$ 或 $max_(abs(x) = 1) abs(A x)$，它们都满足 $abs(A B) <= abs(A) abs(B)$ - 形式定义： $ e^A = sum_(n = 0)^infinity A^n/n! $ 注意到矩阵绝对收敛只需要每个分量绝对收敛，同时 $ sum_(n = 0)^infinity abs(A^n/n!) <= sum_(n = 0)^infinity abs(A)^n/n! = e^(abs(A)) < +infinity $ 满足性质： - 若 $A B = B A$ 则 $e^(A + B) = e^A e^B$ - $det(e^A) = e^(tr(A)) > 0$（证明需要若当标准型） ] #theorem[][ 给定常系数微分方程 $y' = A y + f(x)$ ，则 $e^(A x)$ 是方程的基础解矩阵 ] 以上结论非常漂亮，结合之前的理论我们可以求解出方程的通解。唯一的问题是按照定义求出 $e^(A x)$ 并不容易。 #example[][ 由约当标准型，可设： $ A = P D P^(-1)\ D = sum_(d=1)^(d_max) sum_(lambda in Lambda) sum_s lambda I + J_(d s) $ 其中 $d$ 是约当块维度，$s$ 代表不同的块。这些块都在不同的位置上，乘积为零，因此： $ D^n = (sum_(d) sum_(lambda in Lambda) sum_s lambda I + J_(d s))^n\ = sum_(d) sum_(lambda in Lambda) sum_s (lambda I + J_(d s))^n\ = sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^n C_n^i lambda^(n-i) J_(d s)^i\ = sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(min {d, n}) C_n^i lambda^(n-i) J_(d s)^i\ $ 注意到 $n > d$ 时上式的求和项数已与 $n$ 无关，$n < d$ 仅有有限个，因此可以求出 $D^n$ 的通式，利用： $ e^(A x) = sum_(k=0)^infinity A^k/k! x^k = P (sum_(k=0)^infinity D^k/k! x^k) Inv(P)\ = P (sum_(k=0)^infinity 1/k! (sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(min {d, k}) C_k^i lambda^(k-i) J_(d s)^i x^k)) Inv(P)\ = P (sum_(k=0)^infinity 1/k! (sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(min {d_max, k}) C_k^i lambda^(k-i) J_(d s)^i x^k)) Inv(P)\ = P (sum_(k=0)^d_max 1/k! (sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(n) C_k^i lambda^(k-i) J_(d s)^i x^k) + sum_(k=d_max + 1)^infinity 1/k! (sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(d_max) C_k^i lambda^(k-i) J_(d s)^i x^k)) Inv(P)\ = P (sum_(k=0)^d_max 1/k! (sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(k) C_k^i lambda^(k-i) J_(d s)^i x^k) + sum_(d) sum_(lambda in Lambda) sum_s sum_(i=0)^(d_max) J_(d s)^i sum_(k=d_max + 1)^infinity 1/k! C_k^i lambda^(k-i) x^k ) Inv(P)\ $ 上式第二项可以求得，第一项仅有有限项当然，这里的计算是极其麻烦的。有些时候，我们可以用技巧大大简化，例如： - 若 $A$ 可对角化，则 $d_(max) = 0$ 上式简化为： $ P (sum_(lambda in Lambda) sum_s sum_(k=0)^infinity 1/k! lambda^(k) x^k J_(0 s) ) Inv(P) = P e^(D x) Inv(P) $ - 若可求得 $A$ 的零化多项式，可以尝试降次 - 若 $A$ 仅有一个特征值 $lambda$ 则有： $ e^(A x) = e^(lambda x) e^((A - lambda) x) $ 后者第二项是幂零矩阵，可以直接求出 ] #example[][ 大部分时候如果只为了求方程的解，没必要完整算出 $e^(A x)$ ，既然若设： $ e^(A x) = P e^(D x) Inv(P) $ 则 $e^(A x) P = P e^(D x)$ 也是基础解矩阵，它会更加好求，既然 $D$ 是对角/约当矩阵，$e^(D x)$ 是容易求的，而 $P$ 恰由特征向量/循环子空间的基构成。 ] #example[][ 有些形式较好的方程并不用使用上面的一般方法求解，例如： $ cases( x' = y + z, y' = x + y, z' = x + z ) $ 三式相加，立得 $(x + y + z)' = x' + y' + z' = 2 (x + y + z), x+ y + z = C_1 e^(2t)$，代回得 $x' = e^(2 t) - x$ 解出即可 ] == 高阶线性微分方程 #theorem[][ 给定高阶线性微分方程： $ x^((n)) + a_(n-1)(t) x^((n-1)) + dots + a_1(t) x' + a_0(t) x = f(t) $ 可做换元，令： $ x_0 = x\ x_1 = x'\ ...\ x_n = x^((n)) $ 方程变为多元线性微分方程： $ vec(x_0, x_1, dots.v, x_(n-1), x_n)' = mat(0, 1, ...,0, 0; 0, 0, ...,0, 0; dots.v, dots.v, ..., dots.v, dots.v; 0, 0, ..., 0, 1; -a_0 (t), -a_1 (t), ..., -a_(n-1) (t), 1) vec(x_0, x_1, dots.v, x_(n-1), x_n) + vec(0, 0, dots.v, 0, f(t)) $ 因此，可以利用线性方程组的理论知道： - 方程的解是齐次线性方程的解空间 $n$ 维线性空间加上一个特解 - 给定初值 $x(0), x'(0), ..., x^((n))(0)$ 则方程的解唯一存在，且解是大范围的 ] #lemma[Wronskian][ 假设 $x_1 (t), x_2 (t), ..., x_n (t)$ 都 $n-1$ 阶可导，则称： $ Det(x_1, x_2, ..., x_n; x'_1, x'_2, ..., x'_n; dots.v, dots.v, ..., dots.v; x_1^((n-1)), x_2^((n-1)), ..., x_n^((n-1))) = W(t) $ 为朗斯基行列式。若这些函数线性相关，则朗斯基行列式恒为零。反之结论一般是不成立的，但若这些函数是齐次线性微分方程的解，则反命题也成立。 ] #proof[ 设 $sum_i c_i x_i (t) = 0$，对于任意 $t$ 反复求导可得： $ mat(x_1, x_2, ..., x_n; x'_1, x'_2, ..., x'_n; dots.v, dots.v, ..., dots.v; x_1^((n-1)), x_2^((n-1)), ..., x_n^((n-1)))vec(c_1, c_2, dots.v, c_n) = 0 $ 线性相关性相当于上面的齐次线性方程有非零解，因此行列式为零，证毕。至于反命题，一个简单的反例是 $x^2, x abs(x)$，而都是同一个齐次方程的解的情形就是 @homogeneous-linear ] #example[][ 假设我们已求出齐次方程的解，根据 @constant-variation 当然可以通过常数变易求出线性方程的解。实际这么做时，我们要遇到对： $ sum_i c_i (t) x_i (t) $ 求 $n$ 阶导的问题，由于我们只需要一个解即可，不妨添加条件： $ sum_i c'_i (t) x_i (t) = 0\ sum_i c'_i (t) x'_i (t) = 0\ ...\ sum_i c'_i (t) x_i^((n-1)) (t) = 0 $ 这样求导的形式就很简单： $ (sum_i c_i (t) x_i (t)) ' = sum_i c_i (t) x'_i (t) \ (sum_i c_i (t) x_i (t)) '' = sum_i c_i (t) x''_i (t) \ ...\ (sum_i c_i (t) x_i (t)) ^((n-1)) = sum_i c_i (t) x^((n-1))_i (t)\ (sum_i c_i (t) x_i (t)) ^((n)) = sum_i c_i (t) x^((n))_i (t) + sum_i c'_i (t) x^((n-1))_i (t) $ 最终方程变成： $ sum_i a_i (sum_i c_i (t) x_i (t)) ^((i)) = f(t)\ a_n sum_j c'_i (t) x^((n-1))_i (t) + sum_i a_i (sum_j c_j (t) x^((i))_j (t)) = f(t)\ a_n sum_j c'_i (t) x^((n-1))_i (t) = f(t) $ 这是因为中间凑出了齐次方程的解，因此可以直接消去。加上之前的假设方程变成关于 $c'_i$ 的线性方程组，求解即可。 ] #lemma[][ 对于二阶齐次线性微分方程： $ x'' + a(t) x' + b(t) x = 0 $ 设有两无关解 $x_1, x_2$ 不难验证若令： $ W = Det(x_1, x_2; x'_1, x'_2) = x_1 x'_2 - x'_1 x_2 $ 则有： $ W' &= x'_1 x'_2 + x_1 x''_2 - x''_1 x_2 - x'_1 x'_2 = x_1 x''_2 - x''_1 x_2 \ &= - x_1 (a(t) x'_2 + x_2) + x_2 (a(t) x'_1 + x_1) \ &= - a(t) (x_1 x'_2 - x'_1 x_2) \ &= - a(t) W $ 可以直接解出 $W$，此时假设 $x_1$ 已知，方程变成了关于 $x_2$ 的一阶线性微分方程，是可解的。这表明对于二阶齐次线性微分方程，若已知一个解，另一个解可以通过一阶线性微分方程求出。 ] #theorem[][ 对于二阶常系数齐次线性微分方程： $ x^((n)) + a_(n-1)(t) x^((n-1)) + dots + a_1(t) x' + a_0(t) x = 0 $ 设： $ p(lambda) = lambda^n + a_(n-1) lambda^(n-1) + dots + a_1 lambda + a_0 $ 有 $s$ 个不同根 $lambda_i$ ，重数分别为 $m_1, m_2, ..., m_s$，则： $ x^k e^(lambda_i x) forall k <= m_i, forall i = 1, 2, ..., s $ 恰构成一组基础解 ] #proof[ 设 $L(x) = x^((n)) + a_(n-1)(t) x^((n-1)) + dots + a_1(t) x' + a_0(t) x$，则有： $ L(e^(lambda x)) = p(lambda) e^(lambda x) $ 因此当然 $p(lambda) = 0$ 时 $e^(lambda x)$ 是解。进一步，两边对 $lambda$ 求导得： $ L(x e^(lambda x)) = (lambda p(lambda) + p'(lambda)) e^(lambda x) $ 若 $lambda$ 是至少二重根，上式也是零，反复进行即得结论。它们的线性无关性是显然的 ] = 幂级数解法 == 一般幂级数本章中 $y$ 允许多元函数 #lemma[][ 设微分方程： $ cases( der(y, x) = f(x, y), y(x_0) = y_0 ) $ 其中 $f$ 在 $x_0$ 附近解析，则它的解存在唯一，且是解析函数。 ] #proof[ 前面 Picard 序列的证明中给出了这个推论 ] 理论上来说，幂级数展开并比对系数可以将一般的微分方程化为代数方程。然而一般的情形仍然难以计算，最常见的情形是对线性方程做展开。 #example[][ - $der(y, x) = y - x$，令 $y = sum_i a_i x^i$，计算得： $ sum_(i >= 1) i a_i x^(i-1) = sum_i a_i x^i - x $ 有： $ a_1 = a_0\ 2 a_2 = a_1 - 1\ (i+1) a_(i+1) = a_i\ $ 可以递推解得 $a_i$ - $y'' - 2 x y' + 4 y = 0$，令 $y = sum_i a_i x^i$，计算得： $ sum_i (i+1)(i+2)a_(i+2)x^i - 2 sum_i i a_(i) x^i - 4 sum_i a_i x^i = 0 $ 得到一般的递推关系： $ (i+1)(i+2)a_(i+2) = 2 i a_i + 4 a_i\ (i+1) a_(i+2) = 2 a_i $ - $y'' = x y$，计算得： $ sum_i (i+1)(i+2)a_(i+2)x^i = sum_i a_(i-1) x^i $ 有： $ a_2 = 0\ (i+1)(i+2)a_(i+2) = a_(i-1) $ 可以解得： $ a_(3 k + 2) = 0\ a_(3 k) = ((3k - 3)!!!)/((3k) !) a_0 $ ] #remark[][ 对于形如： $ u(x) der(y, x) = v(x, y) $ 的微分方程，如果 $u(x) > 0$，将其除掉即可得到解的解析性。但若 $u(x)$ 有零点情形未必。例如： $ cases( x^2 der(y, x) = y - x, y(0) = 0 ) $ 若其有解析解，比对系数发现一定有 $a_n = n!$，但是这个幂级数不收敛，因此是荒谬的。下节的目标便是处理这种方程。 ] == 广义幂级数 #definition[广义幂级数][ 称： $ sum_(n=0)^(+infinity) a_n x^(n + alpha), alpha in RR $ 为广义幂级数。 ] #theorem[][ 设二阶微分方程： $ y'' + p(x) y' + q(x) y = 0 $ 其中 $p, q$ 可能以 $0$ 为奇点，但 $x p, x^2 q$ 都在 $0$ 处解析且不全为零，则它在 $0$ 附近有广义幂级数解 ] #proof[ 方程等价于： $ x^2 y'' + x (sum_i a_i x^i) y' + (sum_i b_i x^i) y = 0 $ 设 $y = sum_(n=0)^(+infinity) c_n x^(n + alpha)$，代入得： $ x^(alpha)(sum_(n=2)^(+infinity) c_n (n+alpha)(n+alpha-1) x^(n) \ + (sum_(n=1)^(+infinity) c_n (n+alpha) x^(n))(sum_(i=0)^infinity a_i x^i) \ + (sum_(n=0)^(+infinity) c_n x^(n))(sum_(i=0)^infinity a_i x^i)) = 0 $ ] #example[贝塞尔方程][ 方程： $ y'' + 1/x y' + (x^2 - n^2) / x^2 y = 0 $ 称为贝塞尔方程，由上面的定理它在 $0$ 附近有广义幂级数解，并且计算可得 $n$ 是正整数时解是整函数。 ] = 微分方程定性理论：边值问题 == Sturm 比较定理本节我们的研究的是形如： $ y'' + p(x) y' + q(x) y = 0 $<obj-def> 其中 $p, q$ 是某个区间 $J$ 上的连续函数 #lemma[][ @obj-def 的非零解都是简单零点（导数非零），进一步都是孤立零点 ] #proof[ 如若不然，设 $x_0$ 处函数值和导数值均为零，由存在唯一性这将导致解恒为零，矛盾！ ] #lemma[][ 设 $f, g$ 是 @obj-def 的两个非零解，且都有零点： - $f, g$ 线性相关当且仅当有相同的零点集 - $f, g$ 线性无关当且仅当零点相间，也即每两个相邻零点构成的开区间内有对方的零点 ] #proof[ - - 若 $f, g$ 相关则 $lambda f + mu g = 0$，不难发现 $lambda, mu$ 非零，因此零点集相同 - 若零点集相同，考虑 Wronskian 行列式： $ W(x) = Det(f, g; f', g') $ 由条件可知存在一个共同零点，则在该点处 $W(x) = 0$，之前的结论表明 $f, g$ 线性相关 - 设 $x_1, x_2$ 是 $f$ 的相邻零点，不妨假设在 $(x_1, x_2)$ 上有 $f > 0$ - 设 $f, g$ 无关，则： $ W(x) = Det(f, g; f', g') $ 定号，进而 $W(x_1) W(x_2) > 0$，而： $ W(x_1) = -g(x_1) f'(x_1)\ W(x_2) = -g(x_2) f'(x_2) $ 之前证明了 $f'(x_1), f'(x_2) != 0$，不难发现一定有 $f'(x_1) > 0, f'(x_1) < 0$，上式表明 $g(x_1), g(x_2)$ 异号，当然就有介质定理。同时其间只能有一个零点，否则若有两个可以反过来找到 $f$ 的零点，与 $x_1, x_2$ 相邻矛盾！ - 反之，若零点相间结论由前一条结论知零点集不同，当然无关 ] #remark[][ 上面的引理中需要留意零点的存在性，例如 $f, g$ 无关而 $f$ 仅有一个零点，此时 $g$ 的零点个数可能是 $0, 1, 2$，与定理都不矛盾 ] #theorem[比较定理][ 设有两个微分方程： $ y'' + p(x) y' + q(x) y = 0 $<eq-1> $ y'' + p(x) y' + r(x) y = 0 $<eq-2> 且满足： $ r(x) >= q(x) $ 设 $f, g$ 分别是@eq-1 和@eq-2 的两个非零解，$x_1, x_2$ 是 $f$ 的两个相邻零点，则 $g$ 在 $[x_1, x_2]$ 上有零点。 ]<compare-two> #proof[ 不妨假设 $f$ 在 $(x_1, x_2)$ 上恒正，有： $ f'(x_1) > 0, f'(x_2) < 0 $ 假设 $g(x)$ 在 $[x_1, x_2]$ 上无零点，不妨设其恒正。令： $ W(x) = Det(f, g;f', g') = f g' - f' g\ W'(x) = Det(f, g;f'', g'') = Det(f, g;-p(x)f' - q(x) f, -p(x) g' - r(x) g)\ =Det(f, g;-p(x)f', -p(x) g' - (r(x) - p(x)) g)\ = - p(x) W(x) - f g (r(x) - p(x)) $ 注意到一定有： $ f g(r(x) - p(x)) >= 0 $ 由一阶方程的比较定理，有： $ B e^(-p(x))<= W(x) <= A e^(-p(x)) $ 其中 $A e^(-p(x))$ 是： $ cases( W'(x) = -p(x) W(x), W(x_1) = - g(x_1) f'(x_1) ) $ 的解，而 $B e^(-p(x))$ 是： $ cases( W'(x) = -p(x) W(x), W(x_1) = - g(x_2) f'(x_2) ) $ 的解，然而不难发现 $- g(x_1) f'(x_1) < 0, - g(x_2) f'(x_2) > 0$ 导致 $A < 0, B > 0$，这是荒谬的！ ] #remark[][ 上面的定理实际上是说 $y$ 前面的系数表示解振荡的频率，系数越大振荡越快。因此有如下对于振动的研究： ] #definition[][ 设@obj-def 的有至少两个零点的非零解为振动解，有无穷多个零点的非零解为无穷解 ] #example[][ - 考虑方程： $ y'' + p(x) y' + r(x) = 0 $<eq-3> 其中 $r(x) <= 0$，注意到它可以对： $ y'' + p(x) y' = 0 $<eq-4> 利用比较定理。假如@eq-3 有振动解，则由 @compare-two 可知@eq-4 的任意一个解都有零点，然而@eq-4 有解 $y = 1$ 无零点，矛盾！因此@eq-3 不可能有振动解 - 考虑方程： $ y'' + q(x) y = 0 $ 其中 $q(x) >= m > 0$，则它任意非零解无限振动，且相邻零点间距离不超过 $pi / sqrt(m)$\ 首先证明对于任何 $a$, $[a, a + pi / sqrt(m)]$ 的区间都有方程的解即可。考虑方程： $ y'' + m y = 0 $ 后面的方程有解： $ y = sin (sqrt(m) (x - a)) $ 以 $a, a + pi / sqrt(m)$ 为零点，利用 @compare-two 立得结论。当然这样的区间有无穷多，因此方程的解有无穷多个零点。\ 注意这个结论不能加强到 $q(x) > 0$，例如： $ y'' + 1/(4 x^2) y = 0, x in [1, +infinity] $ 这是欧拉方程，可以解得： $ y = sqrt(x) (c_1 + c_2 ln x) $ 当然至多只有一个零点 - 考虑微分方程： $ y'' + q(x) y = 0 $ 且存在非负整数 $n$ 使得： $ n^2 < q(x) < (n+1)^2 $ 则方程的任意非零解不是 $2 pi$ 周期的。\ 如若不然，设 $f$ 是非零的 $2 pi$ 周期解，代入方程不难发现 $q(x) f$ 也是以 $2 pi$ 周期的，而 $f$ 的零点孤立，由 $q$ 的连续性知它应该以 $2 pi$ 为周期，因此有最大最小值。设： $ m^2 <= q(x) <= M^2 $ 仿照上面的例子，利用 @compare-two 可以导出 $f$ 的两个相邻零点间的距离在 $pi/M, pi/m$ 之间\ 设 $f$ 在一个周期上有 $2n$ 个零点，零点分别为： $ t_0 < t_1 < ... < t_(2 n) = t_0 + 2 pi $ 对距离求和，可得： $ 2n/M pi <= 2 pi <= 2n/m pi\ m <= n <= M $ 然而由条件 $m, M$ 夹在两个连续整数之间，与 $n$ 是整数矛盾！ - 考虑微分方程： $ y'' + q(y) = f(x) $ 且存在非负整数 $n$ 使得： $ n^2 < q'(y) < (n+1)^2 $ 则方程至多有一个 $2 pi$ 周期解，否则设 $y_1, y_2$ 是两个 $2 pi$ 周期解，容易得到： $ (y_1 - y_2)' + q(y_1) - q(y_2) = 0 $ 令 $p(x) = (q(y_1) - q(y_2))/(y_1 - y_2)$，则 $y_1 - y_2$ 成为微分方程： $ y'' + p(x) y = 0 $ 的 $2 pi$ 周期解，然而上面的结论表明这样的方程没有 $2 pi$ 周期解，矛盾！ ] == 边值问题 #definition[][ 设有微分方程： $ cases( (p(x) y')' + (lambda r(x) + q(x)) y = 0, k y(a) + l y'(a) = M y(b) + N y'(b) = 0 ) $ 其中 $p, q, r in C[a, b], p, r > 0, (K, L) != 0, (M, N) != 0$ 该微分方程非平凡解的存在性称为二阶微分方程的边值问题。若对 $lambda = lambda_0$ 方程有非零解 $phi$，则称 $lambda_0$ 为一个特征值，$phi$ 为对应的特征函数。 ]<boundary-value> #remark[][ 这个叫法是因为若设： $ A y = ((p y')' + q y)/(-r) $ 则它是线性算子，而边值问题实际上是求解 $A y = lambda y$ 的问题 ] #example[][ - 考虑方程： $ cases( y'' + lambda y = 0, y'(0) = y'(l) = 0 ) $ - 当 $lambda = -a^2 < 0$ 时，方程的通解为： $ y = linearCombinationC(e^(a x), e^(-a x)) $ 计算发现若要符合边值，将有： $ cases( C_1 = C_2, C_1 = -C_2 ) $ 导出零解 - 当 $lambda = 0$ 时方程的解是线性函数，因此每个常函数都是特征函数 - 当 $lambda = a^2 > 0$，通解为： $ linearCombinationC(cos a x, sin a x) $ 计算得： $ cases( C_2 = 0, - C_1 sin a l + C_2 cos a l = 0 ) $ 当且仅当 $a$ 为 $(n pi)/l$ 时有非零解综上，特征值为 $(n^2 pi^2)/l^2, forall n in NN$ ]<sin-cos-example> @boundary-value 中的形式当然可以再简化，通过一些简单的线性变换，可设 $a = 0, b = 1$，进一步由 $p > 0$，设： $ t = 1/(c_0) integral_(0)^(x) 1/(p(s)) dif s \ tilde(y) (t) = y(x(t)) $ $x(t)$ 存在是因为 $t$ 关于 $x$ 单调上升，有反函数。不难计算得： $ p(x) der(y, x) = 1/(c_0) der(tilde(y), t)\ der(p(x) der(y, x) , x) = 1/(c_0^2 p(x)) tilde(y)'' $ 总之，方程化为了 $ cases( y'' + (lambda c_0^2 p(x(t))r(x(t)) + c_0^2 p(x(t)) q(x(t))) y = 0, K y(0) + L /(c_0 p(a)) y'(0) = M y(1) + N /(c_0 p(b)) y'(1) = 0) $ #lemma[][ @boundary-value 中的问题可以化归为以下标准形式： $ cases( y'' + (lambda r(x) + q(x)) y = 0, y(0) cos alpha - y'(0) sin alpha = y(1) cos beta - y'(1) sin beta = 0 ) $<standard-form-boundary-value> 其中 $r, q$ 连续，$r > 0, alpha, beta in [0, pi)$ ] #theorem[Sturm-Liouville][ @standard-form-boundary-value 给出的边值问题有可数多个特征值，且所有特征值形成链： $ lambda_0 < lambda_1 < ... < lambda_n < ... $ 其中 $lambda_n -> +infinity$\ 设 $phi_n (x)$ 是 $lambda_n$ 对应的特征函数，则 $phi_n$ 恰有 $n$ 个零点 ]<sturm-liouville-theorem> #proof[ 令 $y = phi(x, lambda)$ 是满足： $ y(0) = sin alpha, y'(0) = cos alpha $ 的解。@standard-form-boundary-value 中第一个边值条件已经满足，只需找到 $lambda$ 满足第二个方程。显然 $y$ 不是零解，$y, y'$ 不同时为零，因此可做极坐标变换： $ cases( phi = rho cos theta, phi' = rho sin theta ) $ 此时，第二个边值条件等价于 $theta(1) = beta + 2 k pi$\ 计算： $ der(theta, x) = der(arctan phi/phi', x) = (phi'^2 - phi phi'' )/(phi^2 + phi'^2) = (phi'^2 + (lambda r + q) phi^2 )/(phi^2 + phi'^2)\ = cos^2 theta + (lambda r + q) sin^2 theta $<eq-order1> 结合 $theta(0, lambda) = 0$，这是关于 $theta$ 的一阶微分方程，接下来的讨论都是关于这个方程的。 - 上式对 $lambda$ 求偏导得到的是关于 $lambda$ 的一阶线性微分方程，可以解得： $ partialDer(theta, lambda) = integral_(0)^(x) e^(integral_(t)^(x) E(s, lambda) dif x) r(t) sin^2 theta dif t $ （其中 $E(s, lambda) = (lambda r(s) + q(x) - 1) sin(2 theta)$）由条件知被积函数恒正，从而 $theta$ 关于 $lambda$ 严格递增 - 观察方程可以发现 $theta = 0$ 时 $theta' = 1$，且当 $lambda$ 充分小时，只要 $theta$ 不太小，就有 $theta' < 0$，我们猜测并证明以下结论： - $theta(x) > 0$\ 如若不然，假设 $x_0$ 是最小的零点，之前的论述表明 $theta'(x_0) = 1 > 0$，而 $theta(0) = alpha > 0$，利用介质定理将可构造出更小的零点，矛盾！ - $ lim_(lambda -> +infinity) theta(x) = +infinity, forall x_0 $ 设 $h = min r > 0$，将有： $ theta' >= (lambda h + q) sin^2 theta + cos^2 theta $ 取 $lambda$ 充分大，可设 $lambda h + q > n + 1$，将有： $ theta' >= 1 + n sin^2 theta $ 由比较定理，只需证明 $theta' = 1 + n sin^2 theta$ 的解充分大即可。事实上： $ x = integral_(alpha)^(theta) 1/(1 + n sin^2 t) dif t $ 观察不难发现 $n$ 充分大时 $theta$ 也应该充分大 - $ lim_(lambda -> -infinity) theta(x) = 0, forall x_0 $ 任取 $epsilon > 0$，当 $lambda$ 充分小时可以证明，$theta$ 落在线段 $(0, alpha) -> (1, epsilon)$ 时，一定有 $theta' < 0$，因此 $theta$ 被限制在该折线和 $y = 0$ 之间，当然意味着 $theta(1) < epsilon$，证毕以上论断表明，$theta(1, lambda) = beta + 2 k pi$ 对于每个 $k in NN$ 都恰有一解，且每个解都逐个增大。此外，特征函数为： $ phi(x) = rho(x) sin (theta_(lambda_k) (x)) $ 由介质定理，至少： $ theta_(lambda_k) (x) = j pi $ 对于 $j = 1, 2, ..., k$ 都有一个解 $x_k$，并且注意到 $forall x with theta(x) = i pi, i in ZZ$，都有： $ phi'(x) = cos^2 theta(x) + (lambda r(x) + q(x)) sin^2 theta(x) = 1 $ 换言之： $ theta_(lambda_k) (x) = j pi $ 只能有一个解（否则两个相邻解的导数值必然反号）\ 同时，对于 $j > k$， $theta_(lambda_k) (x) = j pi$ 将无解，否则由于 $theta(0), theta(1) < j pi$，至少产生两解，与上面的论述矛盾！ ] == 特征函数系的正交性 #lemma[][ 在 @sturm-liouville-theorem 中，每个特征值对应的特征函数彼此相关（特征空间只有一维） ] #proof[ 假设 $phi, psi$ 是 $lambda_n$ 的特征函数，将有： $ Det(phi(0), phi'(0);psi(0), psi'(0)) = 0 $ 由 Wronskian 行列式的结论，这意味着 $phi, psi$ 线性相关 ] #theorem[][ 在 @sturm-liouville-theorem 中，每个特征值对应的特征函数彼此正交，也即： $ integral_(0)^(1) r(x) phi_m phi_n dif x = 0, m != n $ ] #proof[ 有两个方程： $ cases( phi''_n + (lambda_n r + q)phi_n = 0, phi''_m + (lambda_m r + q)phi_m = 0 ) $ 两式分别乘以 $phi_m, phi_n$ 并相减，得到： $ phi''_n phi_m - phi''_m phi_n + (lambda_n - lambda_m) r phi_m phi_n = 0 $ 只需计算： $ &integral_(0)^(1) phi''_n phi_m - phi''_m phi_n dif x \ &= integral_(0)^(1) (phi'_n phi_m - phi_n phi'_m)' dif x\ &= (phi'_n phi_m - phi_n phi'_m)|_0^1\ &= 0 $ 证毕 ] #remark[][ 有了正交性，对于任意的以上形式的二阶线性微分方程，我们都可以考虑在特征函数系上做傅里叶展开。事实上，@sin-cos-example 说明了通常 $sin, cos$ 产生的傅里叶展开是本章定理的一种特殊情况。 ] = 一阶偏微分方程 == 首次积分 #let yv = $bold(y)$ #let xv = $bold(x)$ #definition[][ 在微分方程： $ der(yv, x) = f(x, yv), f in C^1 $ 中，称 $H(xv, yv)$ 为首次积分，若 $H$ 不是常数且在方程的任意解曲线上取常值 ] #example[][ - 考虑方程： $ cases( der(x,t) = -y, der(y, t) = x ) $ 则 $x^2 + y^2$ 就是一个首次积分，既然设 $x, y$ 是一族解，将有： $ (x^2 + y^2)' = 2 x x' + 2 y y' = 0 $ - 一般的，方程： $ cases( der(x,t) = f(y), der(y, t) = g(x) ) $ 则 $integral f(y) - integral g(x)$ 就是一个首次积分，既然： $ (integral f(y) - integral g(x))' = f(y)y' - g(x)x' = 0 $ - 考虑方程： $ cases( der(x,t) = y - x(x^2 + y^2 - 1), der(y, t) = -x - y(x^2 + y^2 - 1) ) $ 注意到： $ (x^2 + y^2)' = 2 x' x + 2 y' y = -x^2 (x^2 + y^2 - 1) - y^2 (x^2 + y^2 - 1) \ = (x^2 + y^2) - (x^2 + y^2)^2 $ 可以解出 $x^2 + y^2$ 进而产生一个首次积分\ 同时： $ x y' - y x' = -x^2 - y^2\ (y/x)' = -1 - (y/x)^2 $ 也可以解出 $y/x$ 产生一个首次积分。事实上，可以发现方程有唯一的平衡点 $(0, 0)$ 和孤立的有界闭解曲线（称为极限环）。对于形如： $ cases( x' = f(x, y), y' = g(x, y) ) $ 其中 $f, g$ 是不超过 $n$ 次的多项式，这个系统称为 $n$ 次系统。$n$ 次系统极限环个数的上界及极限环的分布情况是希尔伯特第十六问题的重要部分，至今二次系统的情况都仍未解决，目前最多举出了四个极限环的例子，并且证明了每个二次系统的极限环个数都是有限的。 ] #lemma[][ $H(x, yv)$ 是 $yv' = f(x, yv)$ 的首次积分当且仅当： $ partialDer(H, x) + partialDer(H, yv) dot der(yv, x) = partialDer(H, x) + partialDer(H, yv) dot f(x, yv) = 0 $ ] #proof[ 由定义显然可得 ] 验证首次积分是非常简单的，然而找到一个首次积分是极其困难的。 #theorem[][ 首次积分可以将原方程降维。具体而言，假设首次积分形如： $ phi(yv) $ 由于 $phi(yv) !=0$，若还有 $der(phi(yv), yv) != 0$，便可在局部找到隐函数消去若干变量 ] #definition[][ 设 $phi_i$ 是 $n$ 个首次积分，如果： $ abs(partialDer((phi_1, phi_2, ..., phi_n), (yv_1, yv_2, ..., yv_n))) != 0 $ 则称 $phi_1, phi_2, ..., phi_n$ 为相互独立的首次积分 ] #theorem[][ $n$ 维自治系统至多 $n$ 个独立的首次积分，且在局部恰有 $n$ 个独立的首次积分。设 $phi_i$ 是 $n$ 个独立的首次积分，则任何首次积分在局部都形如 $h(phi_1, ..., phi_n), H in C^1$ ]<first-integral> #proof[ 第一部分我们不证明，可以参考教材对于第二个结论，设 $c = vec(c_1, dots.v, c_n)$，在任何一个点 $(x_0, c)$ 由存在唯一性可以找到解 $y = phi(x, c)$，显然有： $ partialDer(phi, c)|_(x = x_0) = id $ 因此由隐函数定理，可以反解出： $ c = psi(x, y) $ 并有： $ der(y, c) = partialDer(phi, c)\ der(c, y) = Inv((partialDer(phi, c))) $ 这个 $psi$ 当然就是局部的首次积分，恰有 $n$ 个独立的分量（既然在 $x_0$ 处上式表明偏导是 $id$）也即 $n$ 个首次积分。第三部分是类似的，也不证明。 ] == 一阶线性齐次偏微分方程 #definition[一阶线性齐次偏微分方程][ 称： $ A(xv) der(u, xv) = f(xv) $ 是一阶线性齐次偏微分方程 ] #definition[特征方程][ 称： $ der(x_i, A_i (xv)) = der(x_j, A_j (xv)) $ 为齐次一阶线性偏微分方程的特征方程，它可以看成关于某个分量的 $n-1$ 阶常微分方程 ] #theorem[][ 设特征方程有 $n-1$ 个独立的首次积分 $phi_i$ ，则原偏微分方程的通解恰为： $ Phi(phi_1, phi_2, ..., phi_(n-1)) $ 其中 $Phi$ 是任何 $C^1$ 函数 ] #proof[ 设 $A_1 (xv) != 0$，特征方程实际上形如： $ der(x_i, x_1) = (A_i (x))/(A_1 (x)) $ 不难发现 $u$ 是偏微分方程的解当且仅当 $u$ 是常数或者是特征方程的一个首次积分，由 @first-integral 立得结论。 ] #example[][ - 考虑方程： $ (x + y) partialDer(u, x) - (x - y) partialDer(u, y) = 0 $ 其特征方程为： $ der(y, x) = (y - x)/(y + x) $ 可以解出一个首次积分： $ (x^2 + y^2)e^(2 arctan y/x) $ 因此原方程的通解为： $ phi((x^2 + y^2)e^(2 arctan y/x) ) $ - $ sqrt(x) partialDer(f, x) + sqrt(y) partialDer(f, y) + z partialDer(f, z) = 0 $ 且 $f(x, y, 1) = x y$，求 $f$\ 类似的可以解出： $ f = phi(sqrt(x) - sqrt(y), sqrt(y) - ln z)\ $ 代入可得： $ x y = f(x, y, 1) = phi(sqrt(x) - sqrt(y), sqrt(y) ) $ 做变量替换： $ cases( u = sqrt(x) - sqrt(y), v = sqrt(y) ) $ 可反解出 $phi$ ] == 一阶拟线性偏微分方程 #definition[一阶拟线性偏微分方程][ 称： $ A(xv, u) der(u, xv) = B(xv, u) $ 是一阶拟线性偏微分方程 ] #theorem[][ #let sumf = sumf.with(lower: $1$, upper: $n$) 对于一阶拟线性偏微分方程，考虑微分方程： $ sumf() A_i (xv, u) der(Phi, x_i) + B(xv, u) der(Phi, u) = 0 $<eq-t> 这是关于 $xv, u$ 的一阶线性齐次偏微分方程，设其解为： $ Phi(xv, u) $ 则 $Phi(xv, u) = 0$ 对应的 $u$ 就是原方程的解。反之，任意原方程的解 $u = phi(xv)$ 都有： $ Phi = phi(xv) - u $ 是@eq-t 的解 ] #theorem[][ 若称： $ der(x_i, A_i (xv)) = der(x_j, A_j (xv)) = der(u, B (xv, u)) $ 为拟线性方程的特征方程，且有 $n$ 个独立的首次积分，则原方程的通解恰为： $ Phi(phi_1, phi_2, ..., phi_n) = 0 $ ]

https://github.com/jgm/typst-hs

https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/compute/calc-12.typ

typst

Other

// Test the `quo` function. #test(calc.quo(1, 1), 1) #test(calc.quo(5, 3), 1) #test(calc.quo(5, -3), -1) #test(calc.quo(22.5, 10), 2) #test(calc.quo(9, 4.5), 2)

https://github.com/herbhuang/utdallas-thesis-template-typst

https://raw.githubusercontent.com/herbhuang/utdallas-thesis-template-typst/main/layout/proposal_template.typ

typst

MIT License

#import "/layout/titlepage.typ": * #import "/layout/transparency_ai_tools.typ": transparency_ai_tools as transparency_ai_tools_layout #import "/utils/print_page_break.typ": * // The project function defines how your document looks. // It takes your content and some metadata and formats it. // Go ahead and customize it to your liking! #let proposal( title: "", titleGerman: "", degree: "", program: "", supervisor: "", advisors: (), author: "", startDate: datetime, submissionDate: datetime, transparency_ai_tools: "", is_print: false, body, ) = { titlepage( title: title, titleGerman: titleGerman, degree: degree, program: program, supervisor: supervisor, advisors: advisors, author: author, startDate: startDate, submissionDate: submissionDate ) print_page_break(print: is_print) // Set the document's basic properties. set page( margin: (left: 30mm, right: 30mm, top: 40mm, bottom: 40mm), numbering: "1", number-align: center, ) // Save heading and body font families in variables. let body-font = "New Computer Modern" let sans-font = "New Computer Modern Sans" // Set body font family. set text( font: body-font, size: 12pt, lang: "en" ) show math.equation: set text(weight: 400) // --- Headings --- show heading: set block(below: 0.85em, above: 1.75em) show heading: set text(font: body-font) set heading(numbering: "1.1") // --- Paragraphs --- let firstParagraphIndent = 1.45em show heading: it => { it h(firstParagraphIndent) } set par(leading: 1em, justify: true, first-line-indent: 2em) // --- Citation Style --- set cite(style: "alphanumeric") // --- Figures --- show figure: set text(size: 0.85em) body pagebreak() bibliography("/thesis.bib") pagebreak() transparency_ai_tools_layout(transparency_ai_tools) }

https://github.com/fenjalien/metro

https://raw.githubusercontent.com/fenjalien/metro/main/tests/num/rounding/round-pad/test.typ

typst

Apache License 2.0

#import "/src/lib.typ": * #set page(width: auto, height: auto, margin: 1cm) #metro-setup(round-mode: "figures", round-precision: 4) #num(12.3) #num(12.3, round-pad: false)

https://github.com/ymgyt/techbook

https://raw.githubusercontent.com/ymgyt/techbook/master/programmings/js/typescript/specification/literal_type.md

markdown

# literal type * primitive型の特定の値だけを代入にする型を表現できる。 ```typescript // xにはなんでもassigneできる let x: number; x = 1; ``` ```typescript // xには1だけを代入できる let x: 1 x = 1; // compile error // x = 2; ``` ## literal typeが利用できるprimitive型 * bool * number * string ```typescript let status: 1 | 2 | 3 = 1; ```

https://github.com/bamboovir/typst-resume-template

https://raw.githubusercontent.com/bamboovir/typst-resume-template/main/README.md

markdown

MIT License

# Typst Resume Template A simple resume template for [typst.app](https://typst.app/). Aesthetic style inspired by the following project: - [Awesome-CV](https://github.com/posquit0/Awesome-CV) - [LaTeX Resume](https://github.com/billryan/resume) This is not a perfect clone, the main purpose of this project is to explore and experiment with Typst's typography features. ## [Sample](./resume.pdf) ![awesome-sample](./assets/image/awesome-sample.png) ![latex-sample](./assets/image/latex-sample.png) ## Declaration If you want to see a more realistic example rendered using this template, check [this](https://github.com/bamboovir/typst-resume-template/blob/main/huiming-sun-sde-resume.pdf). This is the resume I built in 2022, may be somewhat out of date and not actively maintained, and is not intended to be an accurate description of any of my current experiences but is intended solely to demonstrate the aesthetics of this template. You are free to take my .typ template and modify it to create your own resume. **Please don't use my resume for anything else without my permission, though!** ## Development Environment - Install [Typst](https://github.com/typst/typst) - Install [Just](https://github.com/casey/just) ## Build Resume ```bash just build ``` ## Interactive Development Resume ```bash just dev ``` ## Containerized Build ```bash just containerized-build ``` ## GitHub Action for resume build automation - [Resume Build CI Pipeline](https://github.com/bamboovir/typst-resume-template/actions/workflows/build-resume.yml) ## Credit [**Typst**](https://github.com/typst/typst) is a new markup-based typesetting system that is designed to be as powerful as LaTeX while being much easier to learn and use. [**FontAwesome**](https://fontawesome.com/) is the Internet's icon library and toolkit, used by millions of designers, developers, and content creators. [**Roboto**](https://github.com/google/roboto) is the default font on Android and ChromeOS, and the recommended font for Google’s visual language, Material Design. [**Source Sans Pro**](https://github.com/adobe-fonts/source-sans-pro) is a set of OpenType fonts that have been designed to work well in user interface (UI) environments.

https://github.com/jgm/typst-hs

https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/compiler/recursion-02.typ

typst

Other

// Test capturing with named function. #let f = 10 #let f() = f #test(type(f()), "function")

https://github.com/ParaN3xus/tex2typ

https://raw.githubusercontent.com/ParaN3xus/tex2typ/main/README.md

markdown

MIT License

# tex2typ A tool to rebuild [Typst](https://typst.app/) mathematical formulas from [KaTeX](https://katex.org/) syntax tree. ## Features - Convert LaTeX mathematical formulas to Typst mathematical formulas. - Extremely useful for building Typst formula datasets. ## Differences from [MiTeX](https://github.com/mitex-rs/mitex) - Focuses on ensuring a generally similar visual effect rather than an identical one, aiming to make the formulas look like they were written by a human. - The generated formulas do not rely on any special Typst environments or packages and can be compiled directly with the standard Typst. - We didn't provide any Typst package. ## TODO - [ ] Improve the handling of spaces in TeX formulas. - [ ] Refactor and optimize the code logic to reduce redundancy. - [x] Fix the issue with incorrect delimiter passing when reconstructing functions like `cases` and `vec`. ## Credits This project makes use of the following open-source projects: - [KaTeX](https://github.com/KaTeX/KaTeX): Fast math typesetting for the web. - [mitex](https://github.com/mitex-rs/mitex): LaTeX support for Typst, powered by Rust and WASM. - [im2markup](https://github.com/harvardnlp/im2markup/): Neural model for converting Image-to-Markup. Thanks to the developers and contributors of these projects for their hard work and dedication. ## LICENSE MIT

https://github.com/typst/packages

https://raw.githubusercontent.com/typst/packages/main/packages/preview/outrageous/0.1.0/examples/basic.typ

typst

Apache License 2.0

#import "../outrageous.typ" #set heading(numbering: "1.") #set outline(indent: auto) #page(height: auto, width: 15cm, margin: 1cm)[ #show outline.entry: outrageous.show-entry #outline() ] #page(height: auto, width: 15cm, margin: 1cm)[ #show outline.entry: outrageous.show-entry.with( // the typst preset retains the normal Typst appearance ..outrageous.presets.typst, // we only override a few things: // level-1 entries are italic, all others keep their font style font-style: ("italic", auto), // no fill for level-1 entries, a thin gray line for all deeper levels fill: (none, line(length: 100%, stroke: gray + .5pt)), ) #outline() ] = Introduction #lorem(400) #lorem(400) == What is this About? #lorem(400) #lorem(400) == Why am I Here? #lorem(400) #lorem(400) = The Backstory #lorem(400) #lorem(400) == How it all Started #lorem(400) #lorem(400) === Early Beginnings #lorem(400) #lorem(400) === First Settlements #lorem(400) #lorem(400) = The Consequences #lorem(400) #lorem(400) = Happy Ending #lorem(400) #lorem(400)

https://github.com/HezelTm/TypstDocuments

https://raw.githubusercontent.com/HezelTm/TypstDocuments/main/README.md

markdown

MIT License

# TypstDocuments Repository to store all Typst Documents

https://github.com/lee-flower/Sn-nCr

https://raw.githubusercontent.com/lee-flower/Sn-nCr/main/main.typ

typst

Apache License 2.0

#import "template.typ": * #show: project.with( title: "等差数列的r阶前n项和与组合数的联系", authors: ( ( name: "江林锐", organization: [梅州市梅县区高级中学], email: "<EMAIL>" ), ), abstract: "本文的目的在于：用数学归纳法证明了等差数列的r阶前n项和与组合数之间的联系。", keywords: ( "等差数列r阶前n项和", "组合数", "数学归纳法" ), ) = 定义 == 等差数列${a_n}$的r阶前n项和设数列${a_n}$为等差数列，其通项公式为$a_n = n$，现定义$S_n^($r$)$为数列${a_n}$的*r 阶前 n 项和*，其中 $r in NN$. 而且对于$forall r >= 1$，都有： $ display(S_n^($r$) = sum_(i=1)^(n) S_i^($r-1$)) $ 另外，我们规定：$S_n^($0$) = a_n$. == 排列、排列数及其计算公式关于*排列*及*排列数*的定义，详见@pailie . 现给出*排列数*的计算公式： $ A_n^m = n(n-1)(n-2) dots.h.c (n-m+1) = n!/(n-m)! $ 其中，$n, m in NN$且$m <= n$. == 组合、组合数及其计算公式关于*组合*及*组合数*的定义，详见@zuhe . 现给出*组合数*的计算公式： $ binom(n,m) = (A_n^m)/(A_m^m) = n!/(m!(n-m)!) $ 其中，$n, m in NN$且$m <= n$. === 组合数的一个推论与一个性质现给出用于本文证明的*组合数*的一个推论和一个性质： *推论1*：对于任意的$n in NN$，有$binom(n,n)=1$. *性质1*：对于任意的$n, m in NN_(+)$且$m <= n$，有$binom(n,m) + binom(n,m-1) = binom(n+1,m)$. 对于上述*组合数*的推论和性质的证明，详见@xingzhi . = 猜想 == 举例由定义*1.1*可知，当$r=1$时，$S_(n)^($1$)$即$S_(n) = sum_(i=1)^(n) a_i = 1/2 n (n + 1)$；又如，当$r=2$时，$S_(n)^($2$) = sum_(i=1)^(n) S_i = 1/6 n(n+1)(n+2)$. #figure( image("f1.svg", width:76%), caption: [ 绿点、紫点、蓝点分别表示数列${a_n}$、${S_n}$及${S_n^($2$)}$ ] ) == 猜想等差数列${a_n}$的r阶前n项和的表达式据*2.1*所举例子，可得如下猜想： #set math.equation(numbering: "(1)") $ S_n^($r$) = 1/(r+1)! product_(i=0)^(r) (n+i) = (n+r)!/((r+1)!(n-1)!) = binom(n+r,r+1) $<chaixiang> = 证明现对所做猜想给出证明： *证*. (1) 当$r=0$或$r=1$时，显然有 @chaixiang 成立. (2) 假设当$r=k (k in NN_(+))$时，@chaixiang 成立，即 $ S_n^($k$) = binom(n+k,k+1) $ 则据定义*1.1*，有， $ S_n^($k+1$) & = sum_(i=1)^(n) S_i^($k$) \ & = S_1^($k$) + S_2^($k$) + S_3^($k$) + dots.h.c + S_n^($k$) \ & = binom(1+k,k+1) + binom(2+k,k+1) + binom(3+k,k+1) + dots.h.c + binom(n+k,k+1) $<tuidao3> 由*组合数*的*推论1*可知，@tuidao3 等价于 $ S_n^($k+1$) = binom(2+k,k+2) + binom(2+k,k+1) + binom(3+k,k+1) + dots.h.c + binom(n+k,k+1) $<tuidao4> 又由*组合数*的*性质1*可知，@tuidao4 等价于 $ S_n^($k+1$) & = binom(3+k,k+2) + binom(3+k,k+1) + dots.h.c + binom(n+k,k+1) \ & = binom(4+k,k+2) + binom(4+k,k+1) + dots.h.c + binom(n+k,k+1) \ & dots.h \ & = binom(n+k,k+2) + binom(n+k,k+1) \ & = binom(n+k+1,k+2) = binom(n+(k+1),(k+1)+1) $ 即当$n=k+1$时， @chaixiang 也成立.\ 由(1)(2)可知， @chaixiang 对任何$r in NN$都成立. *证毕* 故猜想成立. = 附录 == 排列及排列数<pailie> *排列*，一般地，从$n$个不同的元素中取出$m(m<=n)$个元素，按照一定的顺序排成一列，叫做从$n$个元素中取出$m$个元素的一个排列. 特别地，当$m=n$时，这个排列被称作_全排列_. *排列数*，排列数指的是从$n$个不同元素中任取$m(m <= n)$个元素排成一列（考虑元素先后出现次序）称此为一个排列，此种排列的总数即为排列数，即叫做从$n$个不同元素中取出$m$个元素的排列数，记作$A_n^m$. == 组合及组合数<zuhe> *组合*，一般地，从n个不同的元素中，任取$m(m<=n)$个元素为一组，叫作$n$个不同元素中取出$m$个元素的一个组合. *组合数*，从$n$个不同元素中取出$m(m<=n)$个元素的所有组合的个数，叫做从$n$个不同元素中取出$m$个元素的组合数，记作$binom(n,m)$. === 组合数的一个推论与一个性质的证明<xingzhi> 现给出对*推论1*和*性质1*的证明. ==== 推论1的证明 #set math.equation(numbering: none) 设$n,m in NN$且$m <= n$，则由*组合数*的计算公式$ binom(n,m) = A_m^n / A_m^m $知，\ 当$m=n$时，上式为$ binom(n,n) = A_n^n / A_n^n = 1 $. 故*推论1*得证. ==== 性质1的证明设$n,m in NN_(+)$且$m <= n$，则 $ binom(n,m) + binom(n,m-1) & = n!/(m!(n-m)!) + n!/((m-1)!(n-m+1)!) \ & = n!(1/(m(m-1)!(n-m)!) + 1/((n-m+1)(m-1)!(n-m)!)) \ & = n!/((m-1)!(n-m)!) (1/m + 1/(n-m+1)) \ & = n!/((m-1)!(n-m)!) dot.op (n-m+1+m)/(m(n-m+1)) \ & = (n!(n+1))/(m(m-1)!(n-m+1)(n-m)!) \ & = ((n+1)!)/(m!(n-m+1)!) \ & = binom(n+1,m) $ 故*性质1*得证. == 一个有趣的数当$n=4,r=20$时，$S_n^($r$) = S_4^($20$) = binom(24,5) = 2024$，恰好是本文的写作年份. 在此，本文作者也祝大家2024新年快乐（虽然已经过去快4个月了）！本文写于*2024*年*3*月*31*日. == 致谢在此感谢*钟运辉*老师、*李嘉才*同学、*王富平*同学在本文写作过程中给予的帮助.

https://github.com/7sDream/fonts-and-layout-zhCN

https://raw.githubusercontent.com/7sDream/fonts-and-layout-zhCN/master/chapters/02-concepts/dimension/dim-3.typ

typst

Other

#import "/lib/draw.typ": * #import "/lib/glossary.typ": tr #let start = (0, 0) #let end = (250, 220) #let base = (100, 70) #let up = 150 #let down = 35 #let width = 74 #let lt = (base.at(0), base.at(1) + up) #let rb = (base.at(0) + width, base.at(1) - down) #let bbox-lt = (93, 178) #let bbox-width = 74 #let bbox-height = 111 #let bbox-ct = (bbox-lt.at(0) + bbox-width / 2, bbox-lt.at(1)) #let line-color = gray.darken(30%) #let graph = with-unit((ux, uy) => { // mesh(start, end, (50, 50), stroke: 1 * ux + gray) rect( lt, end: rb, stroke: 1.4 * ux + line-color, ) let line-stroke = 1 * ux + line-color segment( (0, base.at(1)), (end.at(0), base.at(1)), stroke: line-stroke, ) let arrow-y = base.at(1) - down - 20 let arrow-length = 12 arrow( (base.at(0) + width - arrow-length, arrow-y), (base.at(0) + width, arrow-y), stroke: line-stroke, head-scale: 3, ) arrow( (base.at(0) + arrow-length, arrow-y), (base.at(0), arrow-y), stroke: line-stroke, head-scale: 3, ) txt(tr[horizontal advance], (base.at(0) + width / 2, arrow-y), size: 12 * ux) txt([#tr[baseline]], (0, base.at(1)), size: 12 * ux, anchor: "lb", dy: 2) rect( bbox-lt, width: bbox-width, height: bbox-height, stroke: 1.2 * ux + line-color, ) txt(text(fill: line-color)[#tr[outline]#tr[bounding box]], bbox-ct, size: 12 * ux, anchor: "cb", dy: 2) txt(text(font: ("Fresca",))[b], base, size: 150 * ux, anchor: "lb", dx: -5, dy: -2) }) #canvas(end, start: start, width: 50%, graph)

https://github.com/VisualFP/docs

https://raw.githubusercontent.com/VisualFP/docs/main/SA/design_concept/content/poc/ui.typ

typst

#import "../../../acronyms.typ": * = User Interface <ui> This section describes the features of the #ac("PoC") application #ac("UI"), the high-level implementation, and how functional reactive programming could be applied to VisualFP. == Features The #ac("UI") for the #ac("PoC") application includes two main components, as shown in @ui-empty-editor: A sidebar with pre-defined value blocks and the function editor. #figure( image("../../static/ui_empty_editor.png"), caption: "Undefined function value in the VisualFP UI" ) <ui-empty-editor> The #ac("PoC") allows the construction of a value, the "userDefinedFunction", which starts with a generic type hole. Starting with a generic function type allows more flexible testing. In a completed application version, the user can define the function name and type when creating it. #grid( columns: (60%, 40%), gutter: 5pt, [#figure( image("../../static/ui_editor_drag_lambda.png"), caption: "Dragging lambda block into value definition" ) <ui-dragging-lambda>], [#figure( image("../../static/ui_editor_dropped_lambda.png"), caption: "Updated function definition including a lambda block" ) <ui-dropped-lambda>] ) @ui-dragging-lambda and @ui-dropped-lambda show how a lambda block is inserted into the value definition. To build the value definition, the user drags the lambda block from the sidebar into the type hole. The drop event then triggers the application to insert the lambda block into the function definition and infer the types of the new function definition. This process can be repeated with suiting value blocks until no type hole is left. As the #ac("PoC") is intended to test the concept, only a reset button exists to return to the initial empty definition. In a full version, this would be replaced with the possibility to remove specific blocks from the definition. Finally, the user-built function definitions can be viewed as Haskell code by clicking the "View Haskell" button. @ui-view-haskell shows the Haskell code for the `mapAdd5` function. #figure( image("../../static/ui_view_haskell.png"), caption: "Haskell defintion of mapAdd5 function in VisualFP" ) <ui-view-haskell> == Implementation The #ac("UI") implementation consists of an Electron.js app hosting a Threepenny #ac("UI"). The Electron app is packaged with an executable of the Threepenny #ac("UI") and all #ac("UI") related static files, i.e. #ac("CSS") & JavaScript files. When starting the Threepenny #ac("UI"), the Electron app passes a usable port for the local web server and the file path of the static #ac("UI") files to the Threepenny #ac("UI"). The function editor is the most significant part of the Threepenny #ac("UI") and has two primary responsibilities: - Rendering of function value blocks - Reacting to value block drop events The rendering part creates an #ac("HTML") representation of each block in the value definition and annotates it with #ac("CSS") classes according to its block type. Reacting to the drop events is a bit more complicated. The block values in the application's sidebar carry their names as data transfer data. When the user drops a block value into a type hole, the data transfer data is included in the event data. Unfortunately, the drop events cannot be registered when creating the type hole elements in the rendering part. So, to register the drop event listeners, the IDs of type holes need to be collected upfront. With these IDs, the #ac("HTML") elements added to the #ac("DOM") can be loaded, and the event handlers registered. The drop event handlers always do the same, regardless of the block value that was dropped: 1. Replace the type hole with the dropped value 2. Infer the updated function definition 3. Clear all elements from the function editor 4. Render the inferred function definition == Functional Reactive Programming Threepenny includes an #ac("FRP") library, which follows the concepts described by <NAME> and <NAME>. #ac("FRP") has two main concepts: Events and Behaviors. An Event is defined as a list of occurrences in time. A Behavior represents a value that changes over time. @frp_elliott_hudak While the first intention was to build the #ac("PoC") with an #ac("FRP") architecture, it became clear over time that Threepenny's #ac("FRP") library is not yet ready for more complex use-cases like VisualFP's function editor. The main problem is that no function allows it to merge multiple events. Implementing the #ac("FRP") architecture through Threepenny could be considered again once the #ac("FRP") library is replaced by reactive-banana #footnote("https://github.com/HeinrichApfelmus/reactive-banana"). The author of Threepenny, <NAME>, plans to do that in a future release @threepenny-frp-replacement. Generally, there is no reason why VisualFP couldn't be implemented using #ac("FRP"). In such an implementation, there would be three kinds of events: - "Reset Editor" button is clicked - "View Haskell" button is clicked - A block value is dropped into a type hole. This event combines all events from every type hole in the function definition. The value definition of the user-defined function is a behavior that changes every time a block value is dropped into the value definition. When the value definition changes, the elements displayed in the function editor must also be updated.

https://github.com/justmejulian/typst-documentation-template

https://raw.githubusercontent.com/justmejulian/typst-documentation-template/main/README.md

markdown

# ZHAW typst-documentation-template Typst documentation template for a ZHAW thesis, based on [ZHAW guidelines](https://gpmpublic.zhaw.ch/GPMDocProdDPublic/Vorgabedokumente_ZHAW/Z_RL_Richtlinie_Corporate_Design_Markengrundelemente.pdf). Learn more about Typst [here](https://github.com/typst/typst). This template is based on the [School of TUM thesis-template-typst](https://github.com/ls1intum/thesis-template-typst). ## Installation Install typst. ```bash brew install typst ``` Once you have installed Typst, you can use it like this: ```bash # Creates `main.pdf` in working directory. typst compile main.typ # Watches source files and recompiles on changes. typst watch main.typ ``` I recommend using the skim pdf viewer, which can be installed via brew. Skim automatically reloads the pdf when it changes. ```bash brew install --cask skim ``` ## neovim todo: add how to setup typst-lsp / Treesitter. ## Todo Github action to build https://github.com/marketplace/actions/github-action-for-typst

https://github.com/jgm/typst-hs

https://raw.githubusercontent.com/jgm/typst-hs/main/test/typ/visualize/image-08.typ

typst

Other

// Error: 8-18 failed to parse svg: found closing tag 'g' instead of 'style' in line 4 #image("test/assets/files/bad.svg")

https://github.com/yaoyuanArtemis/resume

https://raw.githubusercontent.com/yaoyuanArtemis/resume/main/README-zh.md

markdown

Do What The F*ck You Want To Public License

# typst-cv-miku 这是一个简单、优雅、学术风格的 [typst](https://typst.app/) 个人简历（CV）模板。支持中英文（以及更多）。你可以在 [这里](https://typst.app/project/rbxGsQC-tEkDq0mnNIuxkv) 查看在线演示。 ## 示例 ![cv_1](./assets/cv_1.webp) ![cv_2](./assets/cv_2.webp) ## 使用说明 1. 阅读 [typst](https://typst.app/docs/) 文档。 2. 安装此模板需要的字体： - [kpfonts](https://ctan.org/pkg/kpfonts) - [Source Han Sans](https://github.com/adobe-fonts/source-han-sans) - [Source Han Serif](https://source.typekit.com/source-han-serif/cn/) 3. 根据需要修改 `.typ` 文件. 你可能需要了解 typst 的一些基本语法。 ## 此外 Typst 目前在 Emoji 输出上有一些 [bugs](https://github.com/typst/typst/issues/144)，所以暂时用 SVG 替代，你可以在 [twemoji utils](https://twemoji.godi.se/) 找到更多。小图标来自 Material Icons (Community). ## License Licensed under [WTFPL](http://www.wtfpl.net/).

https://github.com/protohaven/printed_materials

https://raw.githubusercontent.com/protohaven/printed_materials/main/meta-environments/env-branding.typ

typst

// Branding #let color = ( tablegrey: rgb(95%,95%,95%), lightgrey: rgb(65%,65%,65%), midgrey: rgb(50%,50%,50%), darkgrey: rgb(38%,38%,38%), warning: rgb("#900000"), accent: rgb("#6EC7E2"), link: blue, ) #let font = ( title: "Noto Sans", sans: "Noto Sans", link: "Fira Mono", )

https://github.com/Toniolo-Marco/git-for-dummies

https://raw.githubusercontent.com/Toniolo-Marco/git-for-dummies/main/book/roles-duties.typ

typst

= Ruoli e mansioni #let wgc = "Working group coordinator" #let gl = "Group leader" #let gm = "GitHub maintainer" Uno degli aspetti principali del corso è che il progetto viene strutturato per assomigliare il più possibile alla realtà lavorativa, questo si riflette in una struttura gerarchica ben definita dove ognuno ha un ruolo e dei compiti associati, nel 2023 la struttura era: - #wgc - #gl - #gm - Tester - Reporter - Member == #wgc il #wgc è il responsabile di tutta la parte comune del progetto, è lui che ha l'ultima parola su come interpretare le specifiche fornite dal prof, gestire le riunioni, accettare o rifiutare le proposte (anche se è comune indirre delle votazioni), fissare le scadenze e scrivere i report sullo stato del progetto dopo ogni riunione. Viene eletto a maggioranza dai membri presenti durante una delle lezioni, il professore vi dirà in anticipo quando sarà il giorno, solitamente poco dopo la presentazione del progetto. Il primo compito del #wgc è quello di scegliere i suoi collaboratori, ovvero i *#gl* (solitamente 2), i *Tester* (solitamente 2/3) e i *Reporter* (solitamente 2), fatto questo il assieme ai #gm, deve creare l'organizzazione e dare ad essi tutti i permessi così che possano gestirla. == #gl Ogni gruppo ha un responsabile, questo va scelto tra i membri e comunicato al docente entro la data stabilita assieme al nome del gruppo e all'elenco dei partecipanti. Il suo compito è quello di partecipare alle riunione generali con #wgc e di rappresentare gli interessi del gruppo, votando le varie sulle questioni. == #gm Si consiglia di scegliere qualcuno che ha dimestichezza con git e GitHub, meglio se ha la possibilità di hostare delle applicazioni e dimestichezza con Docker. Si occuperà di gestire il repository, l'organizzazione su GitHub, effettuare il setup dell'inviter (o in alternativa aggiungere a mano tutte le persone), creare le action e cosa più importante gestire le issue, le milestones e le pull requests che verranno create, lavorerà in stretto contatto con i principali sviluppatori del *Common Crate* e coi testers. È compito suo accertarsi che una pr non rompa tutto il codice già presente o in caso contrario, che ci sia un valido motivo, deve controllare che il codice rispetti gli standard decisi e che sia ben documentato. == Tester Solitamente 2/3 persone, si occupano di scrivere i test che il Common Crate deve superare quando si implementa una nuova feature, oltre a scrivere i test che le applicazioni dei gruppi devono superare per verificare che stiano usando il Common Crate nel modo corretto, forzando gli standard decisi dalle specifiche. L'idea è che ad ogni pull request vengano eseguiti i test e in base all'esito, si procede con la revisione manuale. == Reporter Sono coloro che partecipano ad ogni riunione col compito di stilare un resoconto degli eventi, utile per tracciare i progressi e la direzione del progetto, basandosi su esso il #wgc, stilerà il suo report da inviare al docente. Hanno inoltre il compito di scrivere le specifiche man mano che vengono corrette e definite. == Member Sono tutti i membri dei vari gruppi, il loro compito è partecipare all'implementazione del Common Crate, proporre feature e aprire pull requests con i cambiamenti proposti. Il #wgc potrebbe apportare delle modifiche, come decidere di avere più o meno persone per ruolo, oppure crearne di nuovi, quello che possiamo dirvi è che noi ci siamo trovati bene con questa struttura.

https://github.com/Mufanc/hnuslides-typst

https://raw.githubusercontent.com/Mufanc/hnuslides-typst/master/configs.typ

typst

#let slide = ( width: 640pt, height: 360pt, margin: (x: 3em, top: 2em, bottom: 0em), )

https://github.com/DrGo/typst-tips

https://raw.githubusercontent.com/DrGo/typst-tips/main/refs/samples/typst-uwthesis-master/README.md

markdown

# typst-uwthesis This is a [typst](https://typst.app/) template that should (almost) satisfy the [University of Waterloo's thesis formatting requirements](https://uwaterloo.ca/graduate-studies-postdoctoral-affairs/current-students/thesis/thesis-formatting). The resulted document is similar to, but not exactly the same as the [LaTeX template](https://uwaterloo.atlassian.net/wiki/spaces/ISTKB/pages/2666037269/LaTeX+Software+for+Thesis+and+Document+Preparation+and+the+Overleaf+Cloud+Service). I wrote this template for my thesis proposal. At this moment, `project`, `appendix` and `gls` are only documented using the code itself. ## Limitations 1. Equations are not labeled in `section.equ` format. We would either need to "hack" using `set equ` or wait for typst to implement this. 2. Some duplicated headings appear in PDF's outline (not in TOC). This [issue](https://github.com/typst/typst/pull/1566) has been addressed by typst's developer. If you compile from GitHub's `main` branch, instead of using the pre-compiled package, the problem will disappear. We expect the problem will also disappear in the next release of typst. 3. Limited reverse link from bibliography. This needs to be addressed by typst. 3. No reverse link from list of abbreviations. I might find a way to hack this later. ## License Placeholders in the example file might contain copyrighted example texts obtained from UW's IST. By the time you complete you thesis, all copyrighted texts will have been removed. Currently, there should be no legal problem for a UW student to use this template. We have plan to release the template under an Apache license similar to [simple-typst-thesis's](https://github.com/zagoli/simple-typst-thesis) after getting rid of the potentially copyrighted texts. If you need this template soon, feel free to replace the text by non-copyrighted material and submit a pull request.

https://github.com/davystrong/umbra

https://raw.githubusercontent.com/davystrong/umbra/main/src/lib.typ

typst

MIT License

#let version = version((0, 1, 0)) #import "shadow-path.typ": shadow-path

https://github.com/Maso03/Bachelor

https://raw.githubusercontent.com/Maso03/Bachelor/main/Bachelorarbeit/chapters/VR.typ

typst

MIT License

== Virtual Reality (VR) Virtual Reality (VR) ist eine Technologie, die es Benutzern ermöglicht, in eine computergenerierte, dreidimensionale Umgebung einzutauchen. Diese Umgebung kann mit Hilfe von VR-Headsets und anderen Peripheriegeräten erlebt werden. VR findet Anwendung in Bereichen wie Gaming, Ausbildung, Medizin und Architektur. Durch die Schaffung immersiver Erlebnisse kann VR das Lernen und die Interaktion mit digitalen Inhalten erheblich verbessern.

https://github.com/PA055/5839B-Notebook

https://raw.githubusercontent.com/PA055/5839B-Notebook/main/appendix.typ

typst

#import "./packages.typ": notebookinator #import notebookinator: * #import themes.radial.components #import "./utils.typ": get-page-number #import "./glossary.typ" #create-appendix-entry(title: "Glossary")[ #components.glossary() ]

https://github.com/Enter-tainer/typstyle

https://raw.githubusercontent.com/Enter-tainer/typstyle/master/tests/assets/unit/code/param-len.typ

typst

Apache License 2.0

#let get-page-dim-writer() = locate(w_loc => {}) #let get-page-dim-writer(a) = locate(w_loc => {}) #let get-page-dim-writer(a, b) = locate(w_loc => {}) #let get-page-dim-writer(a, b, c) = locate(w_loc => {}) #let get-page-dim-writer( a, b, c) = locate(w_loc => {})

https://github.com/VaranTavers/vspct

https://raw.githubusercontent.com/VaranTavers/vspct/main/vspct.typ

typst

MIT License

// Varan's simple Pseudocode for Typst #let lang = "en" #let pkeywords = ( en: ( if_pre: "if", if_post: "", while_pre: "while", while_post: "", do_start: "repeat", do_end_pre: "until", do_end_post: "", end_pre: "end", end_post: "", else_pre: "else", return_pre: "return", for_pre: "for", for_post: "", in_pre: "In", out_pre: "Out", algorithm_pre: "Algorithm", algorithm_post: "", ), hu: ( if_pre: "Ha", if_post: "akkor", while_pre: "Amíg", while_post: "végezd el", do_start: "Ismételd", do_end_pre: "Ameddig", do_end_post: "", end_pre: "", end_post: "vége", else_pre: "különben", return_pre: "térít", for_pre: "Minden", for_post: "végezd el", in_pre: "Be", out_pre: "Ki", algorithm_pre: "Algoritmus", algorithm_post: "", ) ) #let gets = sym.arrow.l.long #let algCounter = counter("alg") #let pseudo(body, caption: "") = { block( breakable: false, { set align(left) strong[ #pkeywords.at(lang).at("algorithm_pre") #algCounter.display()#pkeywords.at(lang).at("algorithm_post") ] ": " + caption algCounter.step() block( breakable: false, stroke: (top: black, bottom: black), above: 0.4em, inset: (top: 0.5em, left: 1em, right: 1em, bottom:0.5em), par( leading: 0.5em, body ) ) } ) } #let pblock(fname_pre, fname_post, f, body) = { text(weight: "bold", fname_pre)+" "+f+" "+text(weight: "bold",fname_post) block( above: 0.7em, below: 0.5em, inset: (top: 0em, left: 1em), par( leading: 0.5em, body ) ) text(weight: "bold", pkeywords.at(lang).at("end_pre") + " " + fname_pre + " " + pkeywords.at(lang).at("end_post")) } #let pfor(iterator, start, end, cumul: "", body) = { text(weight: "bold", pkeywords.at(lang).at("for_pre"))+" "+iterator+" "+gets+" "+start+", "+end if cumul != "" { ", " + cumul } " " text(weight: "bold", pkeywords.at(lang).at("for_post")) block( above: 0.7em, below: 0.5em, inset: (top: 0em, left: 1em), par( leading: 0.5em, body ) ) text(weight: "bold", pkeywords.at(lang).at("end_pre") + " " + pkeywords.at(lang).at("for_pre") + " " + pkeywords.at(lang).at("end_post")) } #let prepeat(f, body) = { text(weight: "bold", pkeywords.at(lang).at("do_start")) block( above: 0.7em, below: 0.5em, inset: (top: 0em, left: 1em), par( leading: 0.5em, body ) ) text(weight: "bold", pkeywords.at(lang).at("do_end_pre")+ " " + f + " " + pkeywords.at(lang).at("do_end_post")) } #let pwhile(f, body) = pblock(pkeywords.at(lang).at("while_pre"), pkeywords.at(lang).at("while_post"), f, body) #let pif(f, body) = pblock(pkeywords.at(lang).at("if_pre"), pkeywords.at(lang).at("if_post"), f, body) #let pifelse(f, body1, body2) = { text(weight: "bold", pkeywords.at(lang).at("if_pre"))+" "+f+text(weight: "bold",pkeywords.at(lang).at("if_post")) block( above: 0.7em, below: 0.5em, inset: (top: 0em, left: 1em), par( leading: 0.5em, body1 ) ) text(weight: "bold", pkeywords.at(lang).at("else_pre")) block( above: 0.7em, below: 0.5em, inset: (top: 0em, left: 1em), par( leading: 0.5em, body2 ) ) text(weight: "bold", pkeywords.at(lang).at("end_pre")+" "+pkeywords.at(lang).at("if_pre")) + " " + pkeywords.at(lang).at("end_post") } #let preturn(body) = { text(weight: "bold", pkeywords.at(lang).at("return_pre")) + " " + body } #let pin(body) = { text(weight: "bold", pkeywords.at(lang).at("in_pre")) + " " + body } #let pout(body) = { text(weight: "bold", pkeywords.at(lang).at("out_pre")) + " " + body } #let pref(label) = pkeywords.at(lang).at("algorithm_pre") + " " + locate(loc => { algCounter.at( query(label, loc).first().location() ).at(0) }) + " " + pkeywords.at(lang).at("algorithm_post") /* #show figure: it => if rev [ #set align(left) #strong[ #it.supplement #it.counter.display(it.numbering) ]: #it.caption #it.body ] else [ #set align(center) #it.body #emph[ #it.supplement #it.counter.display(it.numbering) ]: #it.caption ] */

https://github.com/LDemetrios/Typst4k

https://raw.githubusercontent.com/LDemetrios/Typst4k/master/src/test/resources/suite/foundations/panic.typ

typst

--- panic --- // Test panic. // Error: 2-9 panicked #panic() --- panic-with-int --- // Test panic. // Error: 2-12 panicked with: 123 #panic(123) --- panic-with-str --- // Test panic. // Error: 2-24 panicked with: "this is wrong" #panic("this is wrong")

https://github.com/VZkxr/Typst

https://raw.githubusercontent.com/VZkxr/Typst/master/Tests/document.typ

typst

Sea $\{X_n\}_{n in NN}$ una cadena de Markov con espacio de estados $EE=\{0,1,2,3,4\}$, distribución inicial $pi^0 = (1,0,0,0,0)$ y matriz de probabilidades de transición $ PP= mat( 0 , 1 , 0 , 0 , 0; frac(1, 4), 0, frac(3, 4), 0, 0; 0, frac(1, 2), 0, frac(1, 2), 0; 0, 0, frac(3, 4), 0, frac(1, 4); 0, 0, 0, 1, 0; ) $ Calcula la distribución estacionaria de $X_n_{n in NN}$ para $n$ par, es decir, utilizando $PP$. #let ofi = [Office] #let rem = [_Remote_] #let lea = [*On leave*] #table( columns: 6 * (1fr,), table.header( [Team member], [Monday], [Tuesday], [Wednesday], [Thursday], [Friday] ), [<NAME>], table.cell(colspan: 2, ofi), table.cell(colspan: 2, rem), ofi, [<NAME>], table.cell(colspan: 5, lea), [<NAME>], rem, table.cell(colspan: 2, ofi), rem, ofi, ) #table( columns: 5 * (1fr,), table.header( [], table.cell(colspan:2, [Blue chip]), [Fresh IPO], [Penny st'k], ), table.cell( rowspan: 4, align: horizon, [ USD/day ], ), [0.20], [104], [5], [3.17], [108], [4], [1.59], [84], [1], [0.26], [98], [15], [0.01], [195], [4], [7], [ USD/hr ], [57], [2], [3], [6.7] ) #table( columns: (auto, 1fr, auto, auto), table.header( table.cell(colspan:4, align: center, [*Índice temático*]), ), table.cell(rowspan: 2,[]), table.cell(rowspan: 2, align: center + horizon, [*Tema*]), table.cell(colspan: 2, align: center, [*Horas de curso*]), [*Teorías*], [*Prácticas*], [*1*], [*Conjuntos de números*], [*1*], [], [*2*], [*La recta real*], [*1*], [*1*], [*3*], [*Potencias*], [*1*], [*1*], [*4*], [*Raíces*], [*1*], [*1*], [*5*], [*Porcentajes*], [*1*], [*1*], table.cell(colspan: 2, align: right, [*Subtotal*]), [*5*], [*4*], table.cell(colspan: 2, align: right, [*Total*]), table.cell(colspan: 2, align: center, [*9 hrs*]) ) #pagebreak() #table( columns: (auto, 1fr), table.header( table.cell(colspan:2, align: center, [*Contenido temático*]), ), table.cell(rowspan: 2,[]), table.cell(rowspan: 2, align: center + horizon, [*Temas y subtemas*]), [*1*], [*Conjuntos de números* \ #v(.07cm) 1.1 #h(.4cm)Números naturales \ #v(.07cm) 1.2 #h(.4cm)Números enteros \ #v(.07cm) 1.3 #h(.4cm)Números racionales \ #v(.07cm) 1.4 #h(.4cm)Números irracionales \ #v(.15cm)], [*2*], [*La recta real* \ #v(.07cm) 2.1 #h(.4cm)Propiedades de los signos \ #v(.07cm) 2.2 #h(.4cm)Suma y resta \ #v(.07cm) 2.3 #h(.4cm)Multiplicación y división \ #v(.07cm) 2.4 #h(.4cm)Jerarquía de operaciones \ #v(.07cm) 2.5 #h(.4cm)Operaciones con fracciones \ #v(.15cm)], [*3*], [*Potencias* \ #v(.07cm) 3.1 #h(.4cm)Noción intuitiva de la potencia \ #v(.07cm) 3.2 #h(.4cm)Operaciones con potencias \ #v(.15cm)], [*4*], [*Raíces* \ #v(.07cm) 4.1 #h(.4cm)Raíces exactas \ #v(.07cm) 4.2 #h(.4cm)Raíces no exactas \ #v(.15cm)], [*5*], [*Porcentajes* \ #v(.07cm) 5.1 #h(.4cm)Representaciones \ #v(.07cm) 5.2 #h(.4cm)Conversiones \ #v(.07cm) 5.3 #h(.4cm)Problemas de aplicación \ #v(.15cm)] ) #pagebreak() #table( columns: (1fr, 1fr), table.header(align(center)[*Estrategias didácticas*], align(center)[*Evaluación de aprendizaje*]), [Exposición #h(1fr) (X)], [Exámenes parciales #h(1fr) (X)], [Trabajo en equipo #h(1fr) (#h(.2cm))], [Examen final #h(1fr) (X)], [Lecturas #h(1fr) (#h(.2cm))], [Trabajos y tareas #h(1fr) (#h(.2cm))], [Trabajo de investigación #h(1fr) (#h(.2cm))], [Presentación del tema #h(1fr) (#h(.2cm))], [Prácticas (taller o laboratorio) #h(1fr) (#h(.2cm))], [Participación en clase #h(1fr) (#h(.2cm))], [Prácticas de campo #h(1fr) (#h(.2cm))],[Asistencia #h(1fr) (#h(.2cm))], [Aprendizaje por proyectos #h(1fr) (#h(.2cm))],[Rúbricas #h(1fr) (#h(.2cm))], [Aprendizaje basado en problemas #h(1fr) (#h(.2cm))],[Portafolios #h(1fr) (#h(.2cm))], [Casos de enseñanza #h(1fr) (#h(.2cm))],[Listas de cotejo #h(1fr) (#h(.2cm))], [Otras (especificar) #h(1fr) (#h(.2cm))],[Otras (especificar) #h(1fr) (#h(.2cm))] )

https://github.com/typst/packages

https://raw.githubusercontent.com/typst/packages/main/packages/preview/cetz/0.1.2/src/aabb.typ

typst

Apache License 2.0

#import "vector.typ" /// Compute an axis aligned bounding box (aabb) /// for a list of vectors. /// /// An aabb dictionary has the following keys: /// - low (vector) Min. bounds vector /// - high (vector) Max. bounds vector /// /// - pts (array): List of vectors/points /// - init (dictionary): Initial aabb /// -> dictionary AABB object #let aabb(pts, init: none) = { let bounds = init if type(pts) == array { for (i, pt) in pts.enumerate() { if bounds == none and i == 0 { bounds = (low: pt, high: pt) } else { let (x, y, z) = pt let (lo-x, lo-y, lo-z) = bounds.low bounds.low = (calc.min(lo-x, x), calc.min(lo-y, y), calc.min(lo-z, z)) let (hi-x, hi-y, hi-z) = bounds.high bounds.high = (calc.max(hi-x, x), calc.max(hi-y, y), calc.max(hi-z, z)) } } return bounds } else if type(pts) == dictionary { if init == none { return pts } else { return aabb((pts.low, pts.high,), init: bounds) } } panic("Expected array of vectors or bbox dictionary, got: " + repr(pts)) } /// Get the mid-point of an aabb as vector. /// /// - bounds (AABB): AABB /// -> vector #let mid(bounds) = { return vector.scale(vector.add(bounds.low, bounds.high), .5) } /// Get the size of an aabb as vector /// this is a vector from the aabb's low to high. /// /// - bounds (AABB): AABB /// -> vector #let size(bounds) = { return vector.sub(bounds.high, bounds.low) }

https://github.com/thomasschuiki/thomasschuiki

https://raw.githubusercontent.com/thomasschuiki/thomasschuiki/main/cv/fontawesome.typ

typst

//! typst-fontawesome //! //! https://github.com/duskmoon314/typst-fontawesome // Implementation of `fa-icon` #import "lib-impl.typ": fa-icon // Generated icons #import "lib-gen.typ": *

https://github.com/Vaskozlov/Lectures

https://raw.githubusercontent.com/Vaskozlov/Lectures/main/Алгебра (многочлены).typ

typst

= Многочлены == Производная многочлена $ f(x) = product_(i = 1)^(n)(x - x_i) $ $ f'(x) = sum_(i = 1)^(n) frac(f(x), x - x_i) = f(x) sum_(i = 1)^(n) frac(1, x - x_i) $ $ f'(x) = f(x) sum_(i = 1)^(n) frac(1, x - x_i) => frac(f'(x), f(x)) = sum_(i = 1)^(n)frac(1, x - x_i) $ Если многочлен f(x) имеет корень кратности n, то это значит, что + $f(alpha) = 0$ + $f^((i))(alpha) = 0 "для" forall i in [1,n - 1]$ === Пример (связь производной с функцией и ее корнями) $ f(x) = x^2 - 3x + 2 = (x - 2)(x - 1) $ $ f'(x) = f(x) dot frac(1, x - 1) + frac(1, x - 2) = f(x) dot frac(2x - 3, x ^ 2 - 3x + 2) = f(x) dot frac(2x - 3, f(x)) = 2x - 3 $ $ f'(x) = 2x - 3 $ === Пример (задачи на нахождение суммы) == Простейшие дроби Дробь $frac(f, g) in K[t]$ - простейшая, если $g = p^k$, где $p in K[t]$ – неприводимый многочлен и $"deg"(f) < "deg"(g)$. == Связь разложения на простейшие дроби с интерполяцией === Алгоритм, когда все корни первой степени $ g(x) = (x - a_1) dots (x - a_n) $ $ frac(f(x), g(x)) = sum_(i = 1)^(n) frac(f(a_i), g'(a_i) dot (x - a_i)) $ #pagebreak() == Алгоритм интерполяции Лагранжа $ f(x) = product_(i = 1)^(n)(x - x_i) $ $ L(X) = sum_(i = 0)^(n)y_i dot l_(i)(x) = sum_(i = 0)^(n) y_i dot frac(f(x), f'(x_i) dot (x - x_i)) $ $ l_i(x) = product_(j = 0, j != i)frac(x - x_j, x_i - x_j) $ === Пример #columns(2)[ #align(right)[ $ l_0(x) = frac((x - x_1)(x - x_2)(x - x_3), (x_0 - x_1)(x_0 - x_2)(x_0 - x_3)) = \ = frac((x - 2)(x - 3)(x - 5), -30) $ $ l_1(x) = frac((x - x_0)(x - x_2)(x - x_3), (x_1 - x_0)(x_1 - x_2)(x_1 - x_3)) = \ = frac((x - 0)(x - 3)(x - 5), 6) $ ] #colbreak() #table( columns: 3, [*i*], [*x*], [*y*], [0], [0], [0], [1], [2], [1], [2], [3], [3], [3], [5], [2]) ] $ L(x) = y_0 dot l_0(x) + y_1 dot l_1(x) + y_2 dot l_2(x) + y_3 dot l_3(x) = 0 dot l_0(x) + 1 dot l_1(x) + 3 dot l_2(x) + 2 dot l_3(x) $ == Алгоритм интерполяции по Ньютону $ N = a_0 + a_1 dot (x - x_0) + a_2 dot (x - x_0)(x - x_1) + dots $ Чтобы найти многочлен по точкам, нужно постепенно подставлять значения x, тогда если подставляем $x_i$, то начиная с i будут нули. === Пример #columns(2)[ #align(right)[ $ 2 = a_0 + a_1(1 - 1) + a_2(1 - 1)(2 - 1) + dots $ $ 3 = a_0 + a_1(2 - 1) + a_2 (2 - 1)(2 - 2) + dots $ ] #colbreak() #table( columns: 3, [*i*], [*x*], [*y*], [0], [1], [3], [1], [2], [-10], [2], [3], [5]) ] === Упрощение алгоритма интерполяции Ньюетона при $Delta x = "const"$ Пусть $h = Delta x$, тогда построим табличку, где $Delta^k y_i = Delta^(k - 1)y_(i + 1) - Delta^(k - 1)y_i$, где i - строчка в таблице. #table( columns: 5, align: (center, center, center , center, center), [*i*], [*x*], [*y = $Delta^0$y*], [*$Delta^1$y*], [*$Delta^2$y*], [0], [1], [3], [-13], [28], [1], [2], [-10], [15], [], [2], [3], [5], [], [] ) Тогда коэффициент $a_k = frac(Delta^k y_0, k! dot h^k)$

https://github.com/jiamingluuu/mata35-notes

https://raw.githubusercontent.com/jiamingluuu/mata35-notes/main/diff-eqn.typ

typst

#set text(size: 13pt) #set math.equation(numbering: "(1)") #set rect(width: 100%, radius: 8pt, fill: rgb("#f2f2f2"), stroke: 1pt, inset: 12pt) = Differential Equations == Introduction *Definition.* _Differential equation (DE)_ is an equation involving functions and their derivatives. The study of differntial equations plays a significant role in the modern study of physics, engineering, economics, and etc. To dive deep in the mathematically rigorous dicussion of differnetial equation, it is inevitable to introduce a diverse branches of mathematics, for instance, analysis and numerical methods. In light of the limitation to the person who made the note, we are only going to talk about the most fundamental part of differneitlal equation. == First Order Differential Equations *Definition.* The _order_ of a differential equation is defined by the highest degree of derivative of the differential equation has. *Definition.* let $y: RR -> RR$ be a function with variable $t$, and $f: RR^2 -> RR$. A _first order differential equatoin_ has the the form $ (dif y)/(dif t) = f(y, t) $ *Example.* _Acceleration_ describes the rate of change of velocity of an object. Suppose a ball is dropped from air and undergoes a free fall, that is, only gravity is acting on the ball, its acceleration $a(t) = (dif v)/(dif t)$ is characterized by a first order DE $ (dif v)/(dif t) = g. $ Where $g$ is the _gravitational constant_ There is a special class of DE we are interested to study: separable DE. *Definition.* A _separable differential equation_ can be written as $ (dif y)/(dif t) = M(y) N(t). $ To solve a separable, we follows the following strategy in general: 1. Re-write the given DE into the form given in (3). 2. Separable varable, so each side of equation only contains one type of variable $ 1/(M(y))(dif y)/(dif t) = N(t). $ 3. Then integrate both sides of the equation at the same time. $ integral 1/(M(y)) (dif y)/(dif t) dif t = integral N(t) dif t. $ *Example* (Newton's Law of Cooling). Let $T(t)$ be the temperature of an object at time $t$, and $T_s$ be the temperature of surrounding environment. The rate of change of temperature of $T(t)$ is described as $ (dif T)/ (dif t) = k(T(t) - T_s), $ where both $k, T_s$ are a constant in real number. To solve (6), we are going to follow the strategy stated before: $ 1/(T(t) - T_s) (dif T)/(dif t) &= k\ integral 1/(T(t) - T_s) (dif T)/cancel(dif t) cancel(dif t) &= integral k dif t\ integral 1/(T(t) - T_s) dif T &= integral k dif t\ ln abs(T(t) - T_s) &= k t + c\ T(t) - T_s &= e^(k t + c)\ T(t) - T_s = e^(k t) times e^c\ T(t) = A e^(k t) + T_s\ $ #rect[ _Remark._ Notice that it is irrigorous to treat the derivative $(dif T)/(dif t)$ as a fraction and cancel it with the term $dif t$. The definition of derivative involving limit, and is to find the flactuation of the original function in the infidecimal change in $t$, that is $ (dif T)/(dif t) equiv lim_(h -> 0) (T(t + h) - T(t))/(h), $ which is a unity becuase the limit is not gaurantee to be congervent. Whereas when we write the whole term $ 1/(T(t) - T_s) (dif T)/(dif t) dif t, $ it is a 1-form of differential form, which is a smooth section of the co-tangent bundle on a manifold. Under no circumstance should those two notion to be inter-changibly used. ] However, what if the DE is not separable? For instace, how can we solve for a differenial equation has the form $ A(x)(dif y)/(dif x) + B(x) y = C(x), $ where function $A, B, C$ functions over real. In this senario, we need introduce some prior knowledge. *Definition.* Let $f: RR^n -> RR$ be a differentiable function over $RR$ with respect to variable $x_1, x_2, ..., x_n$. The _total derivative_ $dif f$ can be written as $ dif f = sum^n_(i = 1) (diff f)/(diff x_i) dif x_i $ *Example.* The total derivative of $f(x, y) = x^2 + y^2$ is given by $ dif f &= (diff f)/(diff x)dif x + (diff f)/(diff y)dif y\ &= diff/(diff x)(x^2 + y^2)dif x + diff/(diff y)(x^2 + y^2)dif y\ &= 2x dif x + 2y dif y\ $ *Definition.* A differential form $alpha$ is _exact_ if there exists some differential form $beta$ such that $ dif beta = alpha $ *Proposition.* Let $P, Q: RR^2 -> RR$ be multi-variable functions with respect to variable $x$ and $y$. Then the differential form $ P dif x + Q dif y $ is exact if and only if $ (diff P)/ (diff y) = (diff y)/ (diff x) $ #rect[ _Remark._ The notion of exactness is telling us, if a differential form $alpha$ is exact, then it can be obtained by computing the total derivative of another differential form $beta$. Furthermore, the proposition above provides us an easy, swift way of verifying if a given differential form is exact. The intuition behinds the proposition is the following: if equation (12) were exact, then $P$ and $Q$ are the derivative of same function but with respect to different variable. That is $ P = (diff f)/(diff x), quad Q = (diff f)/(diff y). $ Therefore, $ (diff P)/(diff y) = (diff^2 f)/(diff y diff x) "and" (diff Q)/(diff x) = (diff^2 f)/(diff x diff y), $ which are essentially equal. ] In light the the introductory of differential form and exactness, we can take their advantages in the discussion of solving DEs. Given a first order differential equation with the form: $ A (dif y)/(dif x) + B y = C, $ where $A, B, C: RR -> RR$ are functions over real with respect to variable $x$, we can re-write it by using differential forms $ underbrace(A dif y + B y dif x, alpha) = C dif x. $ Hence, if $alpha$ were exact, then by equation (11) there exists some other form $beta$ such that $ dif beta = alpha. $ It implies that (17) is equivalent to $ dif beta = C dif x. $ And by integrating both sides of (19), we can obtain the solution of our DE (16) in implicit form: $ integral dif beta &= integral C dif x\ beta &= integral C dif x $ *Example.* Solve the following differential equation $ (4 + t^2) (dif y)/(dif t) + 2 t y = 4t. $ Let's firstly write it using differential form and check if it is exact: $ underbrace((4 + t^2) dif y, P) + underbrace(2t y dif t, Q) = 4t dif t. $ $ (diff P)/(diff t) &= diff/(diff t) (4 + t^2)\ &= 2t\ (diff Q)/ (diff y) &= diff/(diff y)(2t y)\ &= 2t. $ So we can see the left part of (22) is an exact differential form, which is implying that $ P = (diff f)/(diff y), quad Q = (diff f)/(diff y). $ Hence $ f &= integral P dif y\ &= integral 4 + t^2 dif y\ &= 4y + t^2 y + h(t) $ And we can solve the unknown function $h(t)$ by computing the derivative of $f$ with respect to $t$: $ (diff f)/(diff t) &= diff/(diff t)(4y + t^2 y + h(t))\ &= 2t y + h'(t)\ $ Therefore, by using the fact that $ (diff f)/(diff t) &= Q\ 2t y + h'(t) &= 2t y\ h'(t) &= 0 $ And it follows that $h(t) = c$, where $c$ is an aribitrary constant in $RR$. Hence, we have the final answer: $ f = 2y + t^2 y + c $ However, what if the given DE is not exact? Then we need to develop some trick to modify the DE and change to exact. *Definition.* Given a DE of the from $ A dif y + B y dif x = C $ A _integrating factor_ is an auxiliary function $I(x)$ such that when we multiply it to the both sides of the eqaution, making $ I A dif y + I B dif x $ to be an exact differential form. *Proposition.* If a DE has the form $ A dif y + B y dif x = C, $ where $A, B, C: RR -> RR$ are function with respect to variable $x$, and $A$ is non-zero over its domain. Then the DE has an integrating factor $ I(x) = e^(integral P(x) dif x), $ where $P(x) = B(x)/A(x)$. *Example.* Solve the following DE $ x dif y + 2y dif x = 4x^3 dif x $ As you can verify, this DE is not exact, so we are going to find the integrating factor $ I(x) &= e^(integral P(x) dif x)\ &= e^(integral 2/x dif x)\ &= e^(2ln(x))\ &= x^2 $ By multiplying both sides by the integrating factor, we have $ x dif y + 2y dif x &= 4x^3 dif x\ dif y + (2y)/x dif x &= 4x^4 dif x\ x^2 (dif y + (2y)/x dif x) &= x^2 dot 4x^4 dif x\ x^2 dif y + 2y x dif x &= 4x^6 dif x\ d(x^2 y) &= 4x^6 dif x\ integral d(x^2 y) &= integral 4x^6 dif x\ x^2 y &= 4/7 x^7 + c $

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#show terms: it => { let title = label("Mon Exercice de Traitement du Signal") let duration = label("1h30") let difficulty = label("Facile") let solution = label("0") let figures = label("") let points = label("5") let bonus = label("0") let author = label("Moi") let references = label("") let language = label("Français") let material = label("") let name = label("exo4") } = Exercice 1 Considérez un signal sinusoïdal x(t)=$A⋅sin(2π*f*t+ϕ)$, où AA est l'amplitude, ff est la fréquence en Hz et ϕϕ est la phase en radians. Écrivez une fonction en Python pour calculer la valeur efficace (RMS) de ce signal sur une période donnée. Testez votre fonction avec $A=3$, $f=50$Hz et $ϕ=π*4$ sur une période de T=$1/f$. = Solution Solution de l'exercice: Pour calculer la valeur efficace (RMS) d'un signal sinusoïdal $x(t)=A⋅sin(2*π*f*t+ϕ)$ nous pouvons utiliser la formule suivante : $"RMS"=sqrt(1/T*integral_(0)^T x^2 dif x)$ où $A$ est l'amplitude, $f$ est la fréquence en Hz, $ϕ$ est la phase en radians, et $T$ est la période du signal ($T=1/f$). Dans notre cas, avec $A=3$, $f=50$ Hz et $ϕ=4π$ la formule devient : $"RMS"=sqrt(1/T*integral_(0)^T (3⋅sin(2π⋅50⋅t+4π))² dif x)$ Après résolution numérique de cette intégrale sur une période T, nous obtenons la valeur efficace du signal.

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#import "font.typ": * #import "utils.typ": * #show heading : it => { set align(center) set text(font:heiti, size: font_size.sanhao) it par(leading: 1.5em)[#text(size:0.0em)[#h(0.0em)]] } #set page(footer: [ #set align(center) #set text(size: 10pt, baseline: -3pt) #counter(page).display( "I", ) ] ) // 图目录 #heading(level: 1, outlined: false)[图目录] #v(2em) #show outline: it => { set heading(numbering: "1.1") set text(font: songti, size: font_size.xiaosi) set par(leading: 1em ) it } #outline( title: none, target: figure.where(kind: image), indent : true, ) #pagebreak() #heading(level: 1, outlined: false)[表目录] #v(2em) #show outline: it => { set text(font: songti, size: font_size.xiaosi) set par(leading: 1em ) it } #outline( title: none, target: figure.where(kind: table), indent : true, ) #pagebreak() // 目录 #heading(level: 1, outlined: false)[目录] #v(2em) #{ set align(right) set text(font: songti, size: font_size.xiaosi) set par(first-line-indent: 0pt) [摘要 ] + [.] * 144 + [ I] set par(leading: 1em) [Abstract ] + [.] * 137 + [ II] set par(leading: 1em) [图目录 ] + [.] * 137 + [ III] set par(leading: 1em) [表目录 ] + [.] * 137 + [ IV] set par(leading: 1em) [目录 ] + [.] * 143 + [ V] } #show outline: it => { set text(font: songti, size: font_size.xiaosi) set par(leading: 1em ) it } #outline( title: none, indent : true, )