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http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #LabVIEW | LabVIEW |
{if true then YES else NO}
-> YES
{if false then YES else NO}
-> NO
|
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #Lambdatalk | Lambdatalk |
{if true then YES else NO}
-> YES
{if false then YES else NO}
-> NO
|
http://rosettacode.org/wiki/Caesar_cipher | Caesar cipher |
Task
Implement a Caesar cipher, both encoding and decoding.
The key is an integer from 1 to 25.
This cipher rotates (either towards left or right) the letters of the alphabet (A to Z).
The encoding replaces each letter with the 1st to 25th next letter in the alphabet (wrapping Z to A).
So key 2 encrypts "HI" to "JK", but key 20 encrypts "HI" to "BC".
This simple "mono-alphabetic substitution cipher" provides almost no security, because an attacker who has the encoded message can either use frequency analysis to guess the key, or just try all 25 keys.
Caesar cipher is identical to Vigenère cipher with a key of length 1.
Also, Rot-13 is identical to Caesar cipher with key 13.
Related tasks
Rot-13
Substitution Cipher
Vigenère Cipher/Cryptanalysis
| #zonnon | zonnon |
module Caesar;
const
size = 25;
type
Operation = (code,decode);
procedure C_D(s:string;k:integer;op: Operation): string;
var
i,key: integer;
resp: string;
n,c: char;
begin
resp := "";
if op = Operation.decode then key := k else key := (26 - k) end;
for i := 0 to len(s) - 1 do
c := cap(s[i]);
if (c >= 'A') & (c <= 'Z') then
resp := resp +
string(char(integer('A') + ((integer(c) - integer('A') + key )) mod 26));
else
resp := resp + string(c)
end;
end;
return resp
end C_D;
procedure {public} Cipher(s:string;k:integer):string;
var
i: integer;
resp: string;
n,c: char;
begin
return C_D(s,k,Operation.code)
end Cipher;
procedure {public} Decipher(s:string;k:integer):string;
var
i: integer;
resp: string;
n,c: char;
begin
return C_D(s,k,Operation.decode)
end Decipher;
var
txt,cipher,decipher: string;
begin
txt := "HI";cipher := Caesar.Cipher(txt,2);decipher := Caesar.Decipher(cipher,2);
writeln(txt," -c-> ",cipher," -d-> ",decipher);
txt := "ZA";cipher := Caesar.Cipher(txt,2);decipher := Caesar.Decipher(cipher,2);
writeln(txt," -c-> ",cipher," -d-> ",decipher);
txt := "The five boxing wizards jump quickly";
cipher := Caesar.Cipher(txt,2);decipher := Caesar.Decipher(cipher,2);
writeln(txt," -c-> ",cipher," -d-> ",decipher)
end Caesar.
|
http://rosettacode.org/wiki/Box_the_compass | Box the compass | There be many a land lubber that knows naught of the pirate ways and gives direction by degree!
They know not how to box the compass!
Task description
Create a function that takes a heading in degrees and returns the correct 32-point compass heading.
Use the function to print and display a table of Index, Compass point, and Degree; rather like the corresponding columns from, the first table of the wikipedia article, but use only the following 33 headings as input:
[0.0, 16.87, 16.88, 33.75, 50.62, 50.63, 67.5, 84.37, 84.38, 101.25, 118.12, 118.13, 135.0, 151.87, 151.88, 168.75, 185.62, 185.63, 202.5, 219.37, 219.38, 236.25, 253.12, 253.13, 270.0, 286.87, 286.88, 303.75, 320.62, 320.63, 337.5, 354.37, 354.38]. (They should give the same order of points but are spread throughout the ranges of acceptance).
Notes;
The headings and indices can be calculated from this pseudocode:
for i in 0..32 inclusive:
heading = i * 11.25
case i %3:
if 1: heading += 5.62; break
if 2: heading -= 5.62; break
end
index = ( i mod 32) + 1
The column of indices can be thought of as an enumeration of the thirty two cardinal points (see talk page)..
| #Huginn | Huginn | import Algorithms as algo;
import Text as text;
class Compass {
_majors = none;
_quarter1 = none;
_quarter2 = none;
constructor() {
_majors = algo.materialize( text.split( "north east south west", " " ), tuple );
_majors += _majors;
_quarter1 = text.split( "N,N by E,N-NE,NE by N,NE,NE by E,E-NE,E by N", "," );
_quarter2 = algo.materialize( algo.map( _quarter1, @( s ){ copy( s ).replace( "NE", "EN" ); } ), list );
}
degrees_to_compasspoint( d ) {
d = d % 360. + 360. / 64.;
majorindex, minor = ( integer( d ) / 90, d % 90. );
minorindex = integer( minor * 4. ) / 45;
p1, p2 = _majors[majorindex: majorindex + 2];
q = none;
if ( p1 ∈ { "north", "south" } ) {
q = _quarter1;
} else {
q = _quarter2;
}
return ( text.capitalize( copy( q[minorindex] ).replace( "N", p1 ).replace( "E", p2 ) ) );
}
}
main() {
print(
" # | Angle | Compass point\n"
"---+---------|-------------------\n"
);
c = Compass();
for ( i : algo.range( 33 ) ) {
d = real( i ) * 11.25;
m = i % 3;
if ( m == 1 ) {
d += 5.62;
} else if ( m == 2 ) {
d -= 5.62;
}
n = i % 32 + 1;
print( "{:2d} | {:6.2f}° | {}\n".format( n, d, c.degrees_to_compasspoint( d ) ) );
}
} |
http://rosettacode.org/wiki/Bitwise_operations | Bitwise operations |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Write a routine to perform a bitwise AND, OR, and XOR on two integers, a bitwise NOT on the first integer, a left shift, right shift, right arithmetic shift, left rotate, and right rotate.
All shifts and rotates should be done on the first integer with a shift/rotate amount of the second integer.
If any operation is not available in your language, note it.
| #C.23 | C# | static void bitwise(int a, int b)
{
Console.WriteLine("a and b is {0}", a & b);
Console.WriteLine("a or b is {0}", a | b);
Console.WriteLine("a xor b is {0}", a ^ b);
Console.WriteLine("not a is {0}", ~a);
Console.WriteLine("a lshift b is {0}", a << b);
Console.WriteLine("a arshift b is {0}", a >> b); // When the left operand of the >> operator is of a signed integral type,
// the operator performs an arithmetic shift right
uint c = (uint)a;
Console.WriteLine("c rshift b is {0}", c >> b); // When the left operand of the >> operator is of an unsigned integral type,
// the operator performs a logical shift right
// there are no rotation operators in C#
} |
http://rosettacode.org/wiki/Bitmap | Bitmap | Show a basic storage type to handle a simple RGB raster graphics image,
and some primitive associated functions.
If possible provide a function to allocate an uninitialised image,
given its width and height, and provide 3 additional functions:
one to fill an image with a plain RGB color,
one to set a given pixel with a color,
one to get the color of a pixel.
(If there are specificities about the storage or the allocation, explain those.)
These functions are used as a base for the articles in the category raster graphics operations,
and a basic output function to check the results
is available in the article write ppm file.
| #C.2B.2B | C++ | #include <iostream>
#include <boost/gil/gil_all.hpp>
int main()
{
using namespace boost::gil;
// create 30x40 image
rgb8_image_t img(30, 40);
// fill with red
rgb8_pixel_t red(255, 0, 0);
fill_pixels(view(img), red);
// set pixel at 10x20 to blue
rgb8_pixel_t blue(0, 0, 255);
view(img)(10, 20) = blue;
// read the value of pixel at 11x20
rgb8_pixel_t px = const_view(img)(11, 20);
std::cout << "the pixel at 11, 20 is " << (unsigned)px[0] << ':' << (unsigned)px[1] << ':' << (unsigned)px[2] << '\n';
} |
http://rosettacode.org/wiki/Bitmap/B%C3%A9zier_curves/Cubic | Bitmap/Bézier curves/Cubic | Using the data storage type defined on this page for raster images, and the draw_line function defined in this other one, draw a cubic bezier curve
(definition on Wikipedia).
| #XPL0 | XPL0 | include c:\cxpl\codes; \intrinsic 'code' declarations
proc Bezier(P0, P1, P2, P3); \Draw cubic Bezier curve
real P0, P1, P2, P3;
def Segments = 8;
int I;
real S1, T, T2, T3, U, U2, U3, B, C, X, Y;
[Move(fix(P0(0)), fix(P0(1)));
S1:= 1./float(Segments);
T:= 0.;
for I:= 1 to Segments-1 do
[T:= T+S1;
T2:= T*T;
T3:= T2*T;
U:= 1.-T;
U2:= U*U;
U3:= U2*U;
B:= 3.*T*U2;
C:= 3.*T2*U;
X:= U3*P0(0) + B*P1(0) + C*P2(0) + T3*P3(0);
Y:= U3*P0(1) + B*P1(1) + C*P2(1) + T3*P3(1);
Line(fix(X), fix(Y), $00FFFF); \cyan line segments
];
Line(fix(P3(0)), fix(P3(1)), $00FFFF);
Point(fix(P0(0)), fix(P0(1)), $FF0000); \red control points
Point(fix(P1(0)), fix(P1(1)), $FF0000);
Point(fix(P2(0)), fix(P2(1)), $FF0000);
Point(fix(P3(0)), fix(P3(1)), $FF0000);
];
[SetVid($112); \set 640x480x24 video graphics
Bezier([0., 0.], [30., 100.], [120., 20.], [160., 120.]);
if ChIn(1) then []; \wait for keystroke
SetVid(3); \restore normal text display
] |
http://rosettacode.org/wiki/Bitmap/B%C3%A9zier_curves/Cubic | Bitmap/Bézier curves/Cubic | Using the data storage type defined on this page for raster images, and the draw_line function defined in this other one, draw a cubic bezier curve
(definition on Wikipedia).
| #zkl | zkl | fcn cBezier(p0x,p0y, p1x,p1y, p2x,p2y, p3x,p3y, rgb, numPts=500){
numPts.pump(Void,'wrap(t){ // B(t)
t=t.toFloat()/numPts; t1:=(1.0 - t);
a:=t1*t1*t1; b:=t*t1*t1*3; c:=t1*t*t*3; d:=t*t*t;
x:=a*p0x + b*p1x + c*p2x + d*p3x + 0.5;
y:=a*p0y + b*p1y + c*p2y + d*p3y + 0.5;
__sSet(rgb,x,y);
});
} |
http://rosettacode.org/wiki/Biorhythms | Biorhythms | For a while in the late 70s, the pseudoscience of biorhythms was popular enough to rival astrology, with kiosks in malls that would give you your weekly printout. It was also a popular entry in "Things to Do with your Pocket Calculator" lists. You can read up on the history at Wikipedia, but the main takeaway is that unlike astrology, the math behind biorhythms is dead simple.
It's based on the number of days since your birth. The premise is that three cycles of unspecified provenance govern certain aspects of everyone's lives – specifically, how they're feeling physically, emotionally, and mentally. The best part is that not only do these cycles somehow have the same respective lengths for all humans of any age, gender, weight, genetic background, etc, but those lengths are an exact number of days. And the pattern is in each case a perfect sine curve. Absolutely miraculous!
To compute your biorhythmic profile for a given day, the first thing you need is the number of days between that day and your birth, so the answers in Days between dates are probably a good starting point. (Strictly speaking, the biorhythms start at 0 at the moment of your birth, so if you know time of day you can narrow things down further, but in general these operate at whole-day granularity.) Then take the residue of that day count modulo each of the the cycle lengths to calculate where the day falls on each of the three sinusoidal journeys.
The three cycles and their lengths are as follows:
Cycle
Length
Physical
23 days
Emotional
28 days
Mental
33 days
The first half of each cycle is in "plus" territory, with a peak at the quarter-way point; the second half in "minus" territory, with a valley at the three-quarters mark. You can calculate a specific value between -1 and +1 for the kth day of an n-day cycle by computing sin( 2πk / n ). The days where a cycle crosses the axis in either direction are called "critical" days, although with a cycle value of 0 they're also said to be the most neutral, which seems contradictory.
The task: write a subroutine, function, or program that will, given a birthdate and a target date, output the three biorhythmic values for the day. You may optionally include a text description of the position and the trend (e.g. "up and rising", "peak", "up but falling", "critical", "down and falling", "valley", "down but rising"), an indication of the date on which the next notable event (peak, valley, or crossing) falls, or even a graph of the cycles around the target date. Demonstrate the functionality for dates of your choice.
Example run of my Raku implementation:
raku br.raku 1943-03-09 1972-07-11
Output:
Day 10717:
Physical day 22: -27% (down but rising, next transition 1972-07-12)
Emotional day 21: valley
Mental day 25: valley
Double valley! This was apparently not a good day for Mr. Fischer to begin a chess tournament...
| #VBA | VBA | Function Biorhythm(Birthdate As Date, Targetdate As Date) As String
'Jagged Array
TextArray = Array(Array("up and rising", "peak"), Array("up but falling", "transition"), Array("down and falling", "valley"), Array("down but rising", "transition"))
DaysBetween = Targetdate - Birthdate
positionP = DaysBetween Mod 23
positionE = DaysBetween Mod 28
positionM = DaysBetween Mod 33
'return the positions - just to return something
Biorhythm = CStr(positionP) & "/" & CStr(positionE) & "/" & CStr(positionM)
quadrantP = Int(4 * positionP / 23)
quadrantE = Int(4 * positionE / 28)
quadrantM = Int(4 * positionM / 33)
percentageP = Round(100 * Sin(2 * WorksheetFunction.Pi * (positionP / 23)), 1)
percentageE = Round(100 * Sin(2 * WorksheetFunction.Pi * (positionE / 28)), 1)
percentageM = Round(100 * Sin(2 * WorksheetFunction.Pi * (positionM / 33)), 1)
transitionP = Targetdate + WorksheetFunction.Floor((quadrantP + 1) / 4 * 23, 1) - positionP
transitionE = Targetdate + WorksheetFunction.Floor((quadrantE + 1) / 4 * 28, 1) - positionE
transitionM = Targetdate + WorksheetFunction.Floor((quadrantM + 1) / 4 * 33, 1) - positionM
Select Case True
Case percentageP > 95
textP = "Physical day " & positionP & " : " & "peak"
Case percentageP < -95
textP = "Physical day " & positionP & " : " & "valley"
Case percentageP < 5 And percentageP > -5
textP = "Physical day " & positionP & " : " & "critical transition"
Case Else
textP = "Physical day " & positionP & " : " & percentageP & "% (" & TextArray(quadrantP)(0) & ", next " & TextArray(quadrantP)(1) & " " & transitionP & ")"
End Select
Select Case True
Case percentageE > 95
textE = "Emotional day " & positionE & " : " & "peak"
Case percentageE < -95
textE = "Emotional day " & positionE & " : " & "valley"
Case percentageE < 5 And percentageE > -5
textE = "Emotional day " & positionE & " : " & "critical transition"
Case Else
textE = "Emotional day " & positionE & " : " & percentageE & "% (" & TextArray(quadrantE)(0) & ", next " & TextArray(quadrantE)(1) & " " & transitionE & ")"
End Select
Select Case True
Case percentageM > 95
textM = "Mental day " & positionM & " : " & "peak"
Case percentageM < -95
textM = "Mental day " & positionM & " : " & "valley"
Case percentageM < 5 And percentageM > -5
textM = "Mental day " & positionM & " : " & "critical transition"
Case Else
textM = "Mental day " & positionM & " : " & percentageM & "% (" & TextArray(quadrantM)(0) & ", next " & TextArray(quadrantM)(1) & " " & transitionM & ")"
End Select
Header1Text = "Born " & Birthdate & ", Target " & Targetdate
Header2Text = "Day " & DaysBetween
'Print Result
Debug.Print Header1Text
Debug.Print Header2Text
Debug.Print textP
Debug.Print textE
Debug.Print textM
Debug.Print ""
End Function |
http://rosettacode.org/wiki/Biorhythms | Biorhythms | For a while in the late 70s, the pseudoscience of biorhythms was popular enough to rival astrology, with kiosks in malls that would give you your weekly printout. It was also a popular entry in "Things to Do with your Pocket Calculator" lists. You can read up on the history at Wikipedia, but the main takeaway is that unlike astrology, the math behind biorhythms is dead simple.
It's based on the number of days since your birth. The premise is that three cycles of unspecified provenance govern certain aspects of everyone's lives – specifically, how they're feeling physically, emotionally, and mentally. The best part is that not only do these cycles somehow have the same respective lengths for all humans of any age, gender, weight, genetic background, etc, but those lengths are an exact number of days. And the pattern is in each case a perfect sine curve. Absolutely miraculous!
To compute your biorhythmic profile for a given day, the first thing you need is the number of days between that day and your birth, so the answers in Days between dates are probably a good starting point. (Strictly speaking, the biorhythms start at 0 at the moment of your birth, so if you know time of day you can narrow things down further, but in general these operate at whole-day granularity.) Then take the residue of that day count modulo each of the the cycle lengths to calculate where the day falls on each of the three sinusoidal journeys.
The three cycles and their lengths are as follows:
Cycle
Length
Physical
23 days
Emotional
28 days
Mental
33 days
The first half of each cycle is in "plus" territory, with a peak at the quarter-way point; the second half in "minus" territory, with a valley at the three-quarters mark. You can calculate a specific value between -1 and +1 for the kth day of an n-day cycle by computing sin( 2πk / n ). The days where a cycle crosses the axis in either direction are called "critical" days, although with a cycle value of 0 they're also said to be the most neutral, which seems contradictory.
The task: write a subroutine, function, or program that will, given a birthdate and a target date, output the three biorhythmic values for the day. You may optionally include a text description of the position and the trend (e.g. "up and rising", "peak", "up but falling", "critical", "down and falling", "valley", "down but rising"), an indication of the date on which the next notable event (peak, valley, or crossing) falls, or even a graph of the cycles around the target date. Demonstrate the functionality for dates of your choice.
Example run of my Raku implementation:
raku br.raku 1943-03-09 1972-07-11
Output:
Day 10717:
Physical day 22: -27% (down but rising, next transition 1972-07-12)
Emotional day 21: valley
Mental day 25: valley
Double valley! This was apparently not a good day for Mr. Fischer to begin a chess tournament...
| #Vlang | Vlang | import time
import math
const cycles = ["Physical day ", "Emotional day", "Mental day "]
const lengths = [23, 28, 33]
const quadrants = [
["up and rising", "peak"],
["up but falling", "transition"],
["down and falling", "valley"],
["down but rising", "transition"],
]
// Parameters assumed to be in YYYY-MM-DD format.
fn biorhythms(birth_date string, target_date string) ? {
bd := time.parse_iso8601(birth_date)?
td := time.parse_iso8601(target_date)?
days := int((td-bd).hours() / 24)
println("Born $birth_date, Target $target_date")
println("Day $days")
for i in 0..3 {
length := lengths[i]
cycle := cycles[i]
position := days % length
quadrant := position * 4 / length
mut percent := math.sin(2 * math.pi * f64(position) / f64(length))
percent = math.floor(percent*1000) / 10
mut descript := ""
if percent > 95 {
descript = " peak"
} else if percent < -95 {
descript = " valley"
} else if math.abs(percent) < 5 {
descript = " critical transition"
} else {
days_to_add := (quadrant+1)*length/4 - position
transition := td.add(time.hour * 24 * time.Duration(days_to_add))
trend := quadrants[quadrant][0]
next := quadrants[quadrant][1]
trans_str := transition.custom_format('YYYY-MM-DD')
descript = "${percent:5.1f}% ($trend, next $next $trans_str)"
}
println("$cycle ${position:2} : $descript")
}
println('')
}
fn main() {
date_pairs := [
["1943-03-09", "1972-07-11"],
["1809-01-12", "1863-11-19"],
["1809-02-12", "1863-11-19"], // correct DOB for Abraham Lincoln
]
for date_pair in date_pairs {
biorhythms(date_pair[0], date_pair[1])?
}
} |
http://rosettacode.org/wiki/Bell_numbers | Bell numbers | Bell or exponential numbers are enumerations of the number of different ways to partition a set that has exactly n elements. Each element of the sequence Bn is the number of partitions of a set of size n where order of the elements and order of the partitions are non-significant. E.G.: {a b} is the same as {b a} and {a} {b} is the same as {b} {a}.
So
B0 = 1 trivially. There is only one way to partition a set with zero elements. { }
B1 = 1 There is only one way to partition a set with one element. {a}
B2 = 2 Two elements may be partitioned in two ways. {a} {b}, {a b}
B3 = 5 Three elements may be partitioned in five ways {a} {b} {c}, {a b} {c}, {a} {b c}, {a c} {b}, {a b c}
and so on.
A simple way to find the Bell numbers is construct a Bell triangle, also known as an Aitken's array or Peirce triangle, and read off the numbers in the first column of each row. There are other generating algorithms though, and you are free to choose the best / most appropriate for your case.
Task
Write a routine (function, generator, whatever) to generate the Bell number sequence and call the routine to show here, on this page at least the first 15 and (if your language supports big Integers) 50th elements of the sequence.
If you do use the Bell triangle method to generate the numbers, also show the first ten rows of the Bell triangle.
See also
OEIS:A000110 Bell or exponential numbers
OEIS:A011971 Aitken's array | #11l | 11l | F bellTriangle(n)
[[BigInt]] tri
L(i) 0 .< n
tri.append([BigInt(0)] * i)
tri[1][0] = 1
L(i) 2 .< n
tri[i][0] = tri[i - 1][i - 2]
L(j) 1 .< i
tri[i][j] = tri[i][j - 1] + tri[i - 1][j - 1]
R tri
V bt = bellTriangle(51)
print(‘First fifteen and fiftieth Bell numbers:’)
L(i) 1..15
print(‘#2: #.’.format(i, bt[i][0]))
print(‘50: ’bt[50][0])
print()
print(‘The first ten rows of Bell's triangle:’)
L(i) 1..10
print(bt[i]) |
http://rosettacode.org/wiki/Binary_strings | Binary strings | Many languages have powerful and useful (binary safe) string manipulation functions, while others don't, making it harder for these languages to accomplish some tasks.
This task is about creating functions to handle binary strings (strings made of arbitrary bytes, i.e. byte strings according to Wikipedia) for those languages that don't have built-in support for them.
If your language of choice does have this built-in support, show a possible alternative implementation for the functions or abilities already provided by the language.
In particular the functions you need to create are:
String creation and destruction (when needed and if there's no garbage collection or similar mechanism)
String assignment
String comparison
String cloning and copying
Check if a string is empty
Append a byte to a string
Extract a substring from a string
Replace every occurrence of a byte (or a string) in a string with another string
Join strings
Possible contexts of use: compression algorithms (like LZW compression), L-systems (manipulation of symbols), many more.
| #AWK | AWK | #!/usr/bin/awk -f
BEGIN {
# string creation
a="123\0 abc ";
b="456\x09";
c="789";
printf("abc=<%s><%s><%s>\n",a,b,c);
# string comparison
printf("(a==b) is %i\n",a==b)
# string copying
A = a;
B = b;
C = c;
printf("ABC=<%s><%s><%s>\n",A,B,C);
# check if string is empty
if (length(a)==0) {
printf("string a is empty\n");
} else {
printf("string a is not empty\n");
}
# append a byte to a string
a=a"\x40";
printf("abc=<%s><%s><%s>\n",a,b,c);
# substring
e = substr(a,1,6);
printf("substr(a,1,6)=<%s>\n",e);
# join strings
d=a""b""c;
printf("d=<%s>\n",d);
} |
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #C | C | #include <stdio.h>
#include <stdlib.h>
size_t upper_bound(const int* array, size_t n, int value) {
size_t start = 0;
while (n > 0) {
size_t step = n / 2;
size_t index = start + step;
if (value >= array[index]) {
start = index + 1;
n -= step + 1;
} else {
n = step;
}
}
return start;
}
int* bins(const int* limits, size_t nlimits, const int* data, size_t ndata) {
int* result = calloc(nlimits + 1, sizeof(int));
if (result == NULL)
return NULL;
for (size_t i = 0; i < ndata; ++i)
++result[upper_bound(limits, nlimits, data[i])];
return result;
}
void print_bins(const int* limits, size_t n, const int* bins) {
if (n == 0)
return;
printf(" < %3d: %2d\n", limits[0], bins[0]);
for (size_t i = 1; i < n; ++i)
printf(">= %3d and < %3d: %2d\n", limits[i - 1], limits[i], bins[i]);
printf(">= %3d : %2d\n", limits[n - 1], bins[n]);
}
int main() {
const int limits1[] = {23, 37, 43, 53, 67, 83};
const int data1[] = {95, 21, 94, 12, 99, 4, 70, 75, 83, 93, 52, 80, 57,
5, 53, 86, 65, 17, 92, 83, 71, 61, 54, 58, 47, 16,
8, 9, 32, 84, 7, 87, 46, 19, 30, 37, 96, 6, 98,
40, 79, 97, 45, 64, 60, 29, 49, 36, 43, 55};
printf("Example 1:\n");
size_t n = sizeof(limits1) / sizeof(int);
int* b = bins(limits1, n, data1, sizeof(data1) / sizeof(int));
if (b == NULL) {
fprintf(stderr, "Out of memory\n");
return EXIT_FAILURE;
}
print_bins(limits1, n, b);
free(b);
const int limits2[] = {14, 18, 249, 312, 389, 392, 513, 591, 634, 720};
const int data2[] = {
445, 814, 519, 697, 700, 130, 255, 889, 481, 122, 932, 77, 323, 525,
570, 219, 367, 523, 442, 933, 416, 589, 930, 373, 202, 253, 775, 47,
731, 685, 293, 126, 133, 450, 545, 100, 741, 583, 763, 306, 655, 267,
248, 477, 549, 238, 62, 678, 98, 534, 622, 907, 406, 714, 184, 391,
913, 42, 560, 247, 346, 860, 56, 138, 546, 38, 985, 948, 58, 213,
799, 319, 390, 634, 458, 945, 733, 507, 916, 123, 345, 110, 720, 917,
313, 845, 426, 9, 457, 628, 410, 723, 354, 895, 881, 953, 677, 137,
397, 97, 854, 740, 83, 216, 421, 94, 517, 479, 292, 963, 376, 981,
480, 39, 257, 272, 157, 5, 316, 395, 787, 942, 456, 242, 759, 898,
576, 67, 298, 425, 894, 435, 831, 241, 989, 614, 987, 770, 384, 692,
698, 765, 331, 487, 251, 600, 879, 342, 982, 527, 736, 795, 585, 40,
54, 901, 408, 359, 577, 237, 605, 847, 353, 968, 832, 205, 838, 427,
876, 959, 686, 646, 835, 127, 621, 892, 443, 198, 988, 791, 466, 23,
707, 467, 33, 670, 921, 180, 991, 396, 160, 436, 717, 918, 8, 374,
101, 684, 727, 749};
printf("\nExample 2:\n");
n = sizeof(limits2) / sizeof(int);
b = bins(limits2, n, data2, sizeof(data2) / sizeof(int));
if (b == NULL) {
fprintf(stderr, "Out of memory\n");
return EXIT_FAILURE;
}
print_bins(limits2, n, b);
free(b);
return EXIT_SUCCESS;
} |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Go | Go | package main
import (
"fmt"
"sort"
)
func main() {
dna := "" +
"CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG" +
"CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG" +
"AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT" +
"GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT" +
"CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG" +
"TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA" +
"TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT" +
"CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG" +
"TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC" +
"GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT"
fmt.Println("SEQUENCE:")
le := len(dna)
for i := 0; i < le; i += 50 {
k := i + 50
if k > le {
k = le
}
fmt.Printf("%5d: %s\n", i, dna[i:k])
}
baseMap := make(map[byte]int) // allows for 'any' base
for i := 0; i < le; i++ {
baseMap[dna[i]]++
}
var bases []byte
for k := range baseMap {
bases = append(bases, k)
}
sort.Slice(bases, func(i, j int) bool { // get bases into alphabetic order
return bases[i] < bases[j]
})
fmt.Println("\nBASE COUNT:")
for _, base := range bases {
fmt.Printf(" %c: %3d\n", base, baseMap[base])
}
fmt.Println(" ------")
fmt.Println(" Σ:", le)
fmt.Println(" ======")
} |
http://rosettacode.org/wiki/Binary_digits | Binary digits | Task
Create and display the sequence of binary digits for a given non-negative integer.
The decimal value 5 should produce an output of 101
The decimal value 50 should produce an output of 110010
The decimal value 9000 should produce an output of 10001100101000
The results can be achieved using built-in radix functions within the language (if these are available), or alternatively a user defined function can be used.
The output produced should consist just of the binary digits of each number followed by a newline.
There should be no other whitespace, radix or sign markers in the produced output, and leading zeros should not appear in the results.
| #11l | 11l | L(n) [0, 5, 50, 9000]
print(‘#4 = #.’.format(n, bin(n))) |
http://rosettacode.org/wiki/Bitmap/Bresenham%27s_line_algorithm | Bitmap/Bresenham's line algorithm | Task
Using the data storage type defined on the Bitmap page for raster graphics images,
draw a line given two points with Bresenham's line algorithm.
| #Common_Lisp | Common Lisp | (defun draw-line (buffer x1 y1 x2 y2 pixel)
(declare (type rgb-pixel-buffer buffer))
(declare (type integer x1 y1 x2 y2))
(declare (type rgb-pixel pixel))
(let* ((dist-x (abs (- x1 x2)))
(dist-y (abs (- y1 y2)))
(steep (> dist-y dist-x)))
(when steep
(psetf x1 y1 y1 x1
x2 y2 y2 x2))
(when (> x1 x2)
(psetf x1 x2 x2 x1
y1 y2 y2 y1))
(let* ((delta-x (- x2 x1))
(delta-y (abs (- y1 y2)))
(error (floor delta-x 2))
(y-step (if (< y1 y2) 1 -1))
(y y1))
(loop
:for x :upfrom x1 :to x2
:do (if steep
(setf (rgb-pixel buffer x y) pixel)
(setf (rgb-pixel buffer y x) pixel))
(setf error (- error delta-y))
(when (< error 0)
(incf y y-step)
(incf error delta-x))))
buffer)) |
http://rosettacode.org/wiki/Bioinformatics/Sequence_mutation | Bioinformatics/Sequence mutation | Task
Given a string of characters A, C, G, and T representing a DNA sequence write a routine to mutate the sequence, (string) by:
Choosing a random base position in the sequence.
Mutate the sequence by doing one of either:
Swap the base at that position by changing it to one of A, C, G, or T. (which has a chance of swapping the base for the same base)
Delete the chosen base at the position.
Insert another base randomly chosen from A,C, G, or T into the sequence at that position.
Randomly generate a test DNA sequence of at least 200 bases
"Pretty print" the sequence and a count of its size, and the count of each base in the sequence
Mutate the sequence ten times.
"Pretty print" the sequence after all mutations, and a count of its size, and the count of each base in the sequence.
Extra credit
Give more information on the individual mutations applied.
Allow mutations to be weighted and/or chosen. | #Phix | Phix | string dna = repeat(' ',200+rand(300))
for i=1 to length(dna) do dna[i] = "ACGT"[rand(4)] end for
procedure show()
sequence acgt = repeat(0,5)
for i=1 to length(dna) do
acgt[find(dna[i],"ACGT")] += 1
end for
acgt[$] = sum(acgt)
sequence s = split(trim(join_by(split(join_by(dna,1,10,""),"\n"),1,5," ")),"\n")
for i=1 to length(s) do
printf(1,"%3d: %s\n",{(i-1)*50+1,s[i]})
end for
printf(1,"\nBase counts: A:%d, C:%d, G:%d, T:%d, total:%d\n",acgt)
end procedure
procedure mutate()
printf(1,"\n")
for i=1 to 10 do
integer p = rand(length(dna)),
sdi = "SDI"[rand(3)],
rep = "ACGT"[rand(4)],
was = dna[p]
switch sdi do
case 'S':dna[p] = rep printf(1,"swapped %c at %d for %c\n",{was,p,rep})
case 'D':dna[p..p] = "" printf(1,"deleted %c at %d\n",{was,p})
case 'I':dna[p..p-1] = ""&rep printf(1,"inserted %c at %d, before %c\n",{rep,p,was})
end switch
end for
printf(1,"\n")
end procedure
show()
mutate()
show()
|
http://rosettacode.org/wiki/Bitcoin/address_validation | Bitcoin/address validation | Bitcoin/address validation
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Write a program that takes a bitcoin address as argument,
and checks whether or not this address is valid.
A bitcoin address uses a base58 encoding, which uses an alphabet of the characters 0 .. 9, A ..Z, a .. z, but without the four characters:
0 zero
O uppercase oh
I uppercase eye
l lowercase ell
With this encoding, a bitcoin address encodes 25 bytes:
the first byte is the version number, which will be zero for this task ;
the next twenty bytes are a RIPEMD-160 digest, but you don't have to know that for this task: you can consider them a pure arbitrary data ;
the last four bytes are a checksum check. They are the first four bytes of a double SHA-256 digest of the previous 21 bytes.
To check the bitcoin address, you must read the first twenty-one bytes, compute the checksum, and check that it corresponds to the last four bytes.
The program can either return a boolean value or throw an exception when not valid.
You can use a digest library for SHA-256.
Example of a bitcoin address
1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i
It doesn't belong to anyone and is part of the test suite of the bitcoin software.
You can change a few characters in this string and check that it'll fail the test.
| #Python | Python | from hashlib import sha256
digits58 = '123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz'
def decode_base58(bc, length):
n = 0
for char in bc:
n = n * 58 + digits58.index(char)
return n.to_bytes(length, 'big')
def check_bc(bc):
try:
bcbytes = decode_base58(bc, 25)
return bcbytes[-4:] == sha256(sha256(bcbytes[:-4]).digest()).digest()[:4]
except Exception:
return False
print(check_bc('1AGNa15ZQXAZUgFiqJ3i7Z2DPU2J6hW62i'))
print(check_bc("17NdbrSGoUotzeGCcMMCqnFkEvLymoou9j")) |
http://rosettacode.org/wiki/Bitcoin/address_validation | Bitcoin/address validation | Bitcoin/address validation
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Write a program that takes a bitcoin address as argument,
and checks whether or not this address is valid.
A bitcoin address uses a base58 encoding, which uses an alphabet of the characters 0 .. 9, A ..Z, a .. z, but without the four characters:
0 zero
O uppercase oh
I uppercase eye
l lowercase ell
With this encoding, a bitcoin address encodes 25 bytes:
the first byte is the version number, which will be zero for this task ;
the next twenty bytes are a RIPEMD-160 digest, but you don't have to know that for this task: you can consider them a pure arbitrary data ;
the last four bytes are a checksum check. They are the first four bytes of a double SHA-256 digest of the previous 21 bytes.
To check the bitcoin address, you must read the first twenty-one bytes, compute the checksum, and check that it corresponds to the last four bytes.
The program can either return a boolean value or throw an exception when not valid.
You can use a digest library for SHA-256.
Example of a bitcoin address
1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i
It doesn't belong to anyone and is part of the test suite of the bitcoin software.
You can change a few characters in this string and check that it'll fail the test.
| #Racket | Racket |
#lang racket/base
;; Same sha-256 interface as the same-named task
(require ffi/unsafe ffi/unsafe/define openssl/libcrypto)
(define-ffi-definer defcrypto libcrypto)
(defcrypto SHA256_Init (_fun _pointer -> _int))
(defcrypto SHA256_Update (_fun _pointer _pointer _long -> _int))
(defcrypto SHA256_Final (_fun _pointer _pointer -> _int))
(define (sha256 bytes)
(define ctx (malloc 128))
(define result (make-bytes 32))
(SHA256_Init ctx)
(SHA256_Update ctx bytes (bytes-length bytes))
(SHA256_Final result ctx)
result)
;; base58 decoding
(define base58-digits
(let ([v (make-vector 128 #f)])
(for ([i (in-naturals)]
[c "123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"])
(vector-set! v (char->integer c) i))
v))
(define (base58->integer str)
(for/fold ([n 0]) ([c str])
(+ (* n 58) (vector-ref base58-digits (char->integer c)))))
(define (int->bytes n digits)
(list->bytes (let loop ([n n] [digits digits] [acc '()])
(if (zero? digits) acc
(let-values ([(q r) (quotient/remainder n 256)])
(loop q (sub1 digits) (cons r acc)))))))
(define (validate-bitcoin-address str)
(define bs (int->bytes (base58->integer str) 25))
(equal? (subbytes (sha256 (sha256 (subbytes bs 0 21))) 0 4)
(subbytes bs 21)))
;; additional tests taken from the other solutions
(validate-bitcoin-address "1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i") ; => #t
(validate-bitcoin-address "1111111111111111111114oLvT2") ; => #t
(validate-bitcoin-address "17NdbrSGoUotzeGCcMMCqnFkEvLymoou9j") ; => #t
(validate-bitcoin-address "1Q1pE5vPGEEMqRcVRMbtBK842Y6Pzo6nK9") ; => #t
(validate-bitcoin-address "1badbadbadbadbadbadbadbadbadbadbad") ; => #f
|
http://rosettacode.org/wiki/Bitmap/Flood_fill | Bitmap/Flood fill | Implement a flood fill.
A flood fill is a way of filling an area using color banks to define the contained area or a target color which "determines" the area (the valley that can be flooded; Wikipedia uses the term target color). It works almost like a water flooding from a point towards the banks (or: inside the valley): if there's a hole in the banks, the flood is not contained and all the image (or all the "connected valleys") get filled.
To accomplish the task, you need to implement just one of the possible algorithms (examples are on Wikipedia). Variations on the theme are allowed (e.g. adding a tolerance parameter or argument for color-matching of the banks or target color).
Testing: the basic algorithm is not suitable for truecolor images; a possible test image is the one shown on the right box; you can try to fill the white area, or the black inner circle.
| #Perl | Perl | #! /usr/bin/perl
use strict;
use Image::Imlib2;
my $img = Image::Imlib2->load("Unfilledcirc.jpg");
$img->set_color(0, 255, 0, 255);
$img->fill(100,100);
$img->save("filledcirc.jpg");
exit 0; |
http://rosettacode.org/wiki/Bitmap/Flood_fill | Bitmap/Flood fill | Implement a flood fill.
A flood fill is a way of filling an area using color banks to define the contained area or a target color which "determines" the area (the valley that can be flooded; Wikipedia uses the term target color). It works almost like a water flooding from a point towards the banks (or: inside the valley): if there's a hole in the banks, the flood is not contained and all the image (or all the "connected valleys") get filled.
To accomplish the task, you need to implement just one of the possible algorithms (examples are on Wikipedia). Variations on the theme are allowed (e.g. adding a tolerance parameter or argument for color-matching of the banks or target color).
Testing: the basic algorithm is not suitable for truecolor images; a possible test image is the one shown on the right box; you can try to fill the white area, or the black inner circle.
| #Phix | Phix | -- demo\rosetta\Bitmap_FloodFill.exw (runnable version)
include ppm.e -- blue, green, read_ppm(), write_ppm() (covers above requirements)
function ff(sequence img, integer x, y, colour, target)
if x>=1 and x<=length(img)
and y>=1 and y<=length(img[x])
and img[x][y]=target then
img[x][y] = colour
img = ff(img,x-1,y,colour,target)
img = ff(img,x+1,y,colour,target)
img = ff(img,x,y-1,colour,target)
img = ff(img,x,y+1,colour,target)
end if
return img
end function
function FloodFill(sequence img, integer x, y, colour)
integer target = img[x][y]
return ff(img,x,y,colour,target)
end function
sequence img = read_ppm("Circle.ppm")
img = FloodFill(img, 200, 100, blue)
write_ppm("FloodIn.ppm",img)
img = FloodFill(img, 10, 10, green)
write_ppm("FloodOut.ppm",img) |
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #Lasso | Lasso | !true
// => false
not false
// => true
var(x = true)
$x // => true
$x = false
$x // => false |
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #Latitude | Latitude |
> 'true
true
> 'false
false
> (or 'false 'false)
false
> (or 'false 'true)
true
|
http://rosettacode.org/wiki/Caesar_cipher | Caesar cipher |
Task
Implement a Caesar cipher, both encoding and decoding.
The key is an integer from 1 to 25.
This cipher rotates (either towards left or right) the letters of the alphabet (A to Z).
The encoding replaces each letter with the 1st to 25th next letter in the alphabet (wrapping Z to A).
So key 2 encrypts "HI" to "JK", but key 20 encrypts "HI" to "BC".
This simple "mono-alphabetic substitution cipher" provides almost no security, because an attacker who has the encoded message can either use frequency analysis to guess the key, or just try all 25 keys.
Caesar cipher is identical to Vigenère cipher with a key of length 1.
Also, Rot-13 is identical to Caesar cipher with key 13.
Related tasks
Rot-13
Substitution Cipher
Vigenère Cipher/Cryptanalysis
| #ZX_Spectrum_Basic | ZX Spectrum Basic | 10 LET t$="PACK MY BOX WITH FIVE DOZEN LIQUOR JUGS"
20 PRINT t$''
30 LET key=RND*25+1
40 LET k=key: GO SUB 1000: PRINT t$''
50 LET k=26-key: GO SUB 1000: PRINT t$
60 STOP
1000 FOR i=1 TO LEN t$
1010 LET c= CODE t$(i)
1020 IF c<65 OR c>90 THEN GO TO 1050
1030 LET c=c+k: IF c>90 THEN LET c=c-90+64
1040 LET t$(i)=CHR$ c
1050 NEXT i
1060 RETURN
|
http://rosettacode.org/wiki/Box_the_compass | Box the compass | There be many a land lubber that knows naught of the pirate ways and gives direction by degree!
They know not how to box the compass!
Task description
Create a function that takes a heading in degrees and returns the correct 32-point compass heading.
Use the function to print and display a table of Index, Compass point, and Degree; rather like the corresponding columns from, the first table of the wikipedia article, but use only the following 33 headings as input:
[0.0, 16.87, 16.88, 33.75, 50.62, 50.63, 67.5, 84.37, 84.38, 101.25, 118.12, 118.13, 135.0, 151.87, 151.88, 168.75, 185.62, 185.63, 202.5, 219.37, 219.38, 236.25, 253.12, 253.13, 270.0, 286.87, 286.88, 303.75, 320.62, 320.63, 337.5, 354.37, 354.38]. (They should give the same order of points but are spread throughout the ranges of acceptance).
Notes;
The headings and indices can be calculated from this pseudocode:
for i in 0..32 inclusive:
heading = i * 11.25
case i %3:
if 1: heading += 5.62; break
if 2: heading -= 5.62; break
end
index = ( i mod 32) + 1
The column of indices can be thought of as an enumeration of the thirty two cardinal points (see talk page)..
| #Icon_and_Unicon | Icon and Unicon | link strings,numbers
procedure main()
every heading := 11.25 * (i := 0 to 32) do {
case i%3 of {
1: heading +:= 5.62
2: heading -:= 5.62
}
write(right(i+1,3)," ",left(direction(heading),20)," ",fix(heading,,7,2))
}
end
procedure direction(d) # compass heading given +/- degrees
static dirs
initial {
every put(dirs := [],
replacem(!["N","NbE","N-NE","NEbN","NE","NEbE","E-NE","EbN",
"E","EbS","E-SE","SEbE","SE","SEbS","S-SE","SbE",
"S","SbW","S-SW","SWbS","SW","SWbW","W-SW","WbS",
"W","WbN","W-NW","NWbW","NW","NWbN","N-NW","NbW"],
"N","north","E","east","W","west","S","south","b"," by "))
}
return dirs[round(((((d%360)+360)%360)/11.25)%32 + 1)]
end |
http://rosettacode.org/wiki/Bitwise_operations | Bitwise operations |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Write a routine to perform a bitwise AND, OR, and XOR on two integers, a bitwise NOT on the first integer, a left shift, right shift, right arithmetic shift, left rotate, and right rotate.
All shifts and rotates should be done on the first integer with a shift/rotate amount of the second integer.
If any operation is not available in your language, note it.
| #C.2B.2B | C++ | #include <iostream>
void bitwise(int a, int b)
{
std::cout << "a and b: " << (a & b) << '\n'; // Note: parentheses are needed because & has lower precedence than <<
std::cout << "a or b: " << (a | b) << '\n';
std::cout << "a xor b: " << (a ^ b) << '\n';
std::cout << "not a: " << ~a << '\n';
// Note: the C/C++ shift operators are not guaranteed to work if the shift count (that is, b)
// is negative, or is greater or equal to the number of bits in the integer being shifted.
std::cout << "a shl b: " << (a << b) << '\n'; // Note: "<<" is used both for output and for left shift
std::cout << "a shr b: " << (a >> b) << '\n'; // typically arithmetic right shift, but not guaranteed
unsigned int ua = a;
std::cout << "a lsr b: " << (ua >> b) << '\n'; // logical right shift (guaranteed)
// there are no rotation operators in C++, but as of c++20 there is a standard-library rotate function.
// The rotate function works for all rotation amounts, but the integer being rotated must always be an
// unsigned integer.
std::cout << "a rol b: " << std::rotl(ua, b) << '\n';
std::cout << "a ror b: " << std::rotr(ua, b) << '\n';
} |
http://rosettacode.org/wiki/Bitmap | Bitmap | Show a basic storage type to handle a simple RGB raster graphics image,
and some primitive associated functions.
If possible provide a function to allocate an uninitialised image,
given its width and height, and provide 3 additional functions:
one to fill an image with a plain RGB color,
one to set a given pixel with a color,
one to get the color of a pixel.
(If there are specificities about the storage or the allocation, explain those.)
These functions are used as a base for the articles in the category raster graphics operations,
and a basic output function to check the results
is available in the article write ppm file.
| #Clojure | Clojure | (import '[java.awt Color Graphics Image]
'[java.awt.image BufferedImage])
(defn blank-bitmap [width height]
(BufferedImage. width height BufferedImage/TYPE_3BYTE_BGR))
(defn fill [image color]
(doto (.getGraphics image)
(.setColor color)
(.fillRect 0 0 (.getWidth image) (.getHeight image))))
(defn set-pixel [image x y color]
(.setRGB image x y (.getRGB color)))
(defn get-pixel [image x y]
(Color. (.getRGB image x y))) |
http://rosettacode.org/wiki/Biorhythms | Biorhythms | For a while in the late 70s, the pseudoscience of biorhythms was popular enough to rival astrology, with kiosks in malls that would give you your weekly printout. It was also a popular entry in "Things to Do with your Pocket Calculator" lists. You can read up on the history at Wikipedia, but the main takeaway is that unlike astrology, the math behind biorhythms is dead simple.
It's based on the number of days since your birth. The premise is that three cycles of unspecified provenance govern certain aspects of everyone's lives – specifically, how they're feeling physically, emotionally, and mentally. The best part is that not only do these cycles somehow have the same respective lengths for all humans of any age, gender, weight, genetic background, etc, but those lengths are an exact number of days. And the pattern is in each case a perfect sine curve. Absolutely miraculous!
To compute your biorhythmic profile for a given day, the first thing you need is the number of days between that day and your birth, so the answers in Days between dates are probably a good starting point. (Strictly speaking, the biorhythms start at 0 at the moment of your birth, so if you know time of day you can narrow things down further, but in general these operate at whole-day granularity.) Then take the residue of that day count modulo each of the the cycle lengths to calculate where the day falls on each of the three sinusoidal journeys.
The three cycles and their lengths are as follows:
Cycle
Length
Physical
23 days
Emotional
28 days
Mental
33 days
The first half of each cycle is in "plus" territory, with a peak at the quarter-way point; the second half in "minus" territory, with a valley at the three-quarters mark. You can calculate a specific value between -1 and +1 for the kth day of an n-day cycle by computing sin( 2πk / n ). The days where a cycle crosses the axis in either direction are called "critical" days, although with a cycle value of 0 they're also said to be the most neutral, which seems contradictory.
The task: write a subroutine, function, or program that will, given a birthdate and a target date, output the three biorhythmic values for the day. You may optionally include a text description of the position and the trend (e.g. "up and rising", "peak", "up but falling", "critical", "down and falling", "valley", "down but rising"), an indication of the date on which the next notable event (peak, valley, or crossing) falls, or even a graph of the cycles around the target date. Demonstrate the functionality for dates of your choice.
Example run of my Raku implementation:
raku br.raku 1943-03-09 1972-07-11
Output:
Day 10717:
Physical day 22: -27% (down but rising, next transition 1972-07-12)
Emotional day 21: valley
Mental day 25: valley
Double valley! This was apparently not a good day for Mr. Fischer to begin a chess tournament...
| #Wren | Wren | import "/date" for Date
import "/fmt" for Fmt
var cycles = ["Physical day ", "Emotional day", "Mental day "]
var lengths = [23, 28, 33]
var quadrants = [
["up and rising", "peak"],
["up but falling", "transition"],
["down and falling", "valley"],
["down but rising", "transition"]
]
var biorhythms = Fn.new { |birthDate, targetDate|
var bd = Date.parse(birthDate)
var td = Date.parse(targetDate)
var days = (td - bd).days
Date.default = Date.isoDate
System.print("Born %(birthDate), Target %(targetDate)")
System.print("Day %(days)")
for (i in 0..2) {
var length = lengths[i]
var cycle = cycles[i]
var position = days % length
var quadrant = (position / length * 4).floor
var percent = ((2 * Num.pi * position / length).sin * 1000).floor / 10
var descript = (percent > 95) ? " peak" :
(percent < -95) ? " valley" :
(percent.abs < 5) ? " critical transition" : "other"
if (descript == "other") {
var transition = td.addDays(((quadrant + 1) / 4 * length).floor - position)
var tn = quadrants[quadrant]
var trend = tn[0]
var next = tn[1]
descript = Fmt.swrite("$5.1f\% ($s, next $s $s)", percent, trend, next, transition)
}
Fmt.print("$s $2d : $s", cycle, position, descript)
}
System.print()
}
var datePairs = [
["1943-03-09", "1972-07-11"],
["1809-01-12", "1863-11-19"],
["1809-02-12", "1863-11-19"] // correct DOB for Abraham Lincoln
]
for (datePair in datePairs) biorhythms.call(datePair[0], datePair[1]) |
http://rosettacode.org/wiki/Bell_numbers | Bell numbers | Bell or exponential numbers are enumerations of the number of different ways to partition a set that has exactly n elements. Each element of the sequence Bn is the number of partitions of a set of size n where order of the elements and order of the partitions are non-significant. E.G.: {a b} is the same as {b a} and {a} {b} is the same as {b} {a}.
So
B0 = 1 trivially. There is only one way to partition a set with zero elements. { }
B1 = 1 There is only one way to partition a set with one element. {a}
B2 = 2 Two elements may be partitioned in two ways. {a} {b}, {a b}
B3 = 5 Three elements may be partitioned in five ways {a} {b} {c}, {a b} {c}, {a} {b c}, {a c} {b}, {a b c}
and so on.
A simple way to find the Bell numbers is construct a Bell triangle, also known as an Aitken's array or Peirce triangle, and read off the numbers in the first column of each row. There are other generating algorithms though, and you are free to choose the best / most appropriate for your case.
Task
Write a routine (function, generator, whatever) to generate the Bell number sequence and call the routine to show here, on this page at least the first 15 and (if your language supports big Integers) 50th elements of the sequence.
If you do use the Bell triangle method to generate the numbers, also show the first ten rows of the Bell triangle.
See also
OEIS:A000110 Bell or exponential numbers
OEIS:A011971 Aitken's array | #Ada | Ada |
with Ada.Text_IO; use Ada.Text_IO;
procedure Main is
type Bell_Triangle is array(Positive range <>, Positive range <>) of Natural;
procedure Print_Rows (Row : in Positive; Triangle : in Bell_Triangle) is
begin
if Row in Triangle'Range(1) then
for I in Triangle'First(1) .. Row loop
Put_Line (Triangle (I, 1)'Image);
end loop;
end if;
end Print_Rows;
procedure Print_Triangle (Num : in Positive; Triangle : in Bell_Triangle) is
begin
if Num in Triangle'Range then
for I in Triangle'First(1) .. Num loop
for J in Triangle'First(2) .. Num loop
if Triangle (I, J) /= 0 then
Put (Triangle (I, J)'Image);
end if;
end loop;
New_Line;
end loop;
end if;
end Print_Triangle;
procedure Bell_Numbers is
Triangle : Bell_Triangle(1..15, 1..15) := (Others => (Others => 0));
Temp : Positive := 1;
begin
Triangle (1, 1) := 1;
for I in Triangle'First(1) + 1 .. Triangle'Last(1) loop
Triangle (I, 1) := Temp;
for J in Triangle'First(2) .. Triangle'Last(2) - 1 loop
if Triangle (I - 1, J) /= 0 then
Triangle (I, J + 1) := Triangle (I, J) + Triangle (I - 1, J);
else
Temp := Triangle (I, J);
exit;
end if;
end loop;
end loop;
Put_Line ("First 15 Bell numbers:");
Print_Rows (15, Triangle);
New_Line;
Put_Line ("First 10 rows of the Bell triangle:");
Print_Triangle (10, Triangle);
end Bell_Numbers;
begin
Bell_Numbers;
end Main;
|
http://rosettacode.org/wiki/Benford%27s_law | Benford's law |
This page uses content from Wikipedia. The original article was at Benford's_law. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Benford's law, also called the first-digit law, refers to the frequency distribution of digits in many (but not all) real-life sources of data.
In this distribution, the number 1 occurs as the first digit about 30% of the time, while larger numbers occur in that position less frequently: 9 as the first digit less than 5% of the time. This distribution of first digits is the same as the widths of gridlines on a logarithmic scale.
Benford's law also concerns the expected distribution for digits beyond the first, which approach a uniform distribution.
This result has been found to apply to a wide variety of data sets, including electricity bills, street addresses, stock prices, population numbers, death rates, lengths of rivers, physical and mathematical constants, and processes described by power laws (which are very common in nature). It tends to be most accurate when values are distributed across multiple orders of magnitude.
A set of numbers is said to satisfy Benford's law if the leading digit
d
{\displaystyle d}
(
d
∈
{
1
,
…
,
9
}
{\displaystyle d\in \{1,\ldots ,9\}}
) occurs with probability
P
(
d
)
=
log
10
(
d
+
1
)
−
log
10
(
d
)
=
log
10
(
1
+
1
d
)
{\displaystyle P(d)=\log _{10}(d+1)-\log _{10}(d)=\log _{10}\left(1+{\frac {1}{d}}\right)}
For this task, write (a) routine(s) to calculate the distribution of first significant (non-zero) digits in a collection of numbers, then display the actual vs. expected distribution in the way most convenient for your language (table / graph / histogram / whatever).
Use the first 1000 numbers from the Fibonacci sequence as your data set. No need to show how the Fibonacci numbers are obtained.
You can generate them or load them from a file; whichever is easiest.
Display your actual vs expected distribution.
For extra credit: Show the distribution for one other set of numbers from a page on Wikipedia. State which Wikipedia page it can be obtained from and what the set enumerates. Again, no need to display the actual list of numbers or the code to load them.
See also:
numberphile.com.
A starting page on Wolfram Mathworld is Benfords Law .
| #11l | 11l | F get_fibs()
V a = 1.0
V b = 1.0
[Float] r
L 1000
r [+]= a
(a, b) = (b, a + b)
R r
F benford(seq)
V freqs = [(0.0, 0.0)] * 9
V seq_len = 0
L(d) seq
I d != 0
freqs[String(d)[0].code - ‘1’.code][1]++
seq_len++
L(&f) freqs
f = (log10(1.0 + 1.0 / (L.index + 1)), f[1] / seq_len)
R freqs
print(‘#9 #9 #9’.format(‘Actual’, ‘Expected’, ‘Deviation’))
L(p) benford(get_fibs())
print(‘#.: #2.2% | #2.2% | #.4%’.format(L.index + 1, p[1] * 100, p[0] * 100, abs(p[1] - p[0]) * 100)) |
http://rosettacode.org/wiki/Binary_strings | Binary strings | Many languages have powerful and useful (binary safe) string manipulation functions, while others don't, making it harder for these languages to accomplish some tasks.
This task is about creating functions to handle binary strings (strings made of arbitrary bytes, i.e. byte strings according to Wikipedia) for those languages that don't have built-in support for them.
If your language of choice does have this built-in support, show a possible alternative implementation for the functions or abilities already provided by the language.
In particular the functions you need to create are:
String creation and destruction (when needed and if there's no garbage collection or similar mechanism)
String assignment
String comparison
String cloning and copying
Check if a string is empty
Append a byte to a string
Extract a substring from a string
Replace every occurrence of a byte (or a string) in a string with another string
Join strings
Possible contexts of use: compression algorithms (like LZW compression), L-systems (manipulation of symbols), many more.
| #BASIC | BASIC | REM STRING CREATION AND DESTRUCTION (WHEN NEEDED AND IF THERE'S NO GARBAGE COLLECTION OR SIMILAR MECHANISM)
A$ = "STRING" : REM CREATION
A$ = "" : REM DESTRUCTION
PRINT FRE(0) : REM GARBAGE COLLECTION
REM STRING ASSIGNMENT
A$ = "STRING" : R$ = "DEUX"
REM STRING COMPARISON
PRINT A$ = B$; A$ <> B$; A$ < B$; A$ > B$; A$ <= B$; A$ >= B$
REM STRING CLONING AND COPYING
B$ = A$
REM CHECK IF A STRING IS EMPTY
PRINT LEN(A$) = 0
REM APPEND A BYTE TO A STRING
A$ = A$ + CHR$(0)
REM EXTRACT A SUBSTRING FROM A STRING
S$ = MID$(A$, 2, 3)
REM REPLACE EVERY OCCURRENCE OF A BYTE (OR A STRING) IN A STRING WITH ANOTHER STRING
S = LEN(S$) : R = LEN(R$) : A = LEN(A$) : IF A > S THEN B$ = "" : FOR I = 1 TO A : F = MID$(A$, I, S) = S$ : B$ = B$ + MID$(R$, 1, R * F) + MID$(A$, I, F = 0) : NEXT I : A$ = B$ : PRINT A$
REM JOIN STRINGS
J$ = A$ + STR$(42) + " PUDDLES " + B$ + CHR$(255) : REM USE + |
http://rosettacode.org/wiki/Binary_strings | Binary strings | Many languages have powerful and useful (binary safe) string manipulation functions, while others don't, making it harder for these languages to accomplish some tasks.
This task is about creating functions to handle binary strings (strings made of arbitrary bytes, i.e. byte strings according to Wikipedia) for those languages that don't have built-in support for them.
If your language of choice does have this built-in support, show a possible alternative implementation for the functions or abilities already provided by the language.
In particular the functions you need to create are:
String creation and destruction (when needed and if there's no garbage collection or similar mechanism)
String assignment
String comparison
String cloning and copying
Check if a string is empty
Append a byte to a string
Extract a substring from a string
Replace every occurrence of a byte (or a string) in a string with another string
Join strings
Possible contexts of use: compression algorithms (like LZW compression), L-systems (manipulation of symbols), many more.
| #BBC_BASIC | BBC BASIC | A$ = CHR$(0) + CHR$(1) + CHR$(254) + CHR$(255) : REM assignment
B$ = A$ : REM clone / copy
IF A$ = B$ THEN PRINT "Strings are equal" : REM comparison
IF A$ = "" THEN PRINT "String is empty" : REM Check if empty
A$ += CHR$(128) : REM Append a byte
S$ = MID$(A$, S%, L%) : REM Extract a substring
C$ = A$ + B$ : REM Join strings
REM To replace every occurrence of a byte:
old$ = CHR$(1)
new$ = CHR$(5)
REPEAT
I% = INSTR(A$, old$)
IF I% MID$(A$, I%, 1) = new$
UNTIL I% = 0
|
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #C.2B.2B | C++ | #include <algorithm>
#include <cassert>
#include <iomanip>
#include <iostream>
#include <vector>
std::vector<int> bins(const std::vector<int>& limits,
const std::vector<int>& data) {
std::vector<int> result(limits.size() + 1, 0);
for (int n : data) {
auto i = std::upper_bound(limits.begin(), limits.end(), n);
++result[i - limits.begin()];
}
return result;
}
void print_bins(const std::vector<int>& limits, const std::vector<int>& bins) {
size_t n = limits.size();
if (n == 0)
return;
assert(n + 1 == bins.size());
std::cout << " < " << std::setw(3) << limits[0] << ": "
<< std::setw(2) << bins[0] << '\n';
for (size_t i = 1; i < n; ++i)
std::cout << ">= " << std::setw(3) << limits[i - 1] << " and < "
<< std::setw(3) << limits[i] << ": " << std::setw(2)
<< bins[i] << '\n';
std::cout << ">= " << std::setw(3) << limits[n - 1] << " : "
<< std::setw(2) << bins[n] << '\n';
}
int main() {
const std::vector<int> limits1{23, 37, 43, 53, 67, 83};
const std::vector<int> data1{
95, 21, 94, 12, 99, 4, 70, 75, 83, 93, 52, 80, 57, 5, 53, 86, 65,
17, 92, 83, 71, 61, 54, 58, 47, 16, 8, 9, 32, 84, 7, 87, 46, 19,
30, 37, 96, 6, 98, 40, 79, 97, 45, 64, 60, 29, 49, 36, 43, 55};
std::cout << "Example 1:\n";
print_bins(limits1, bins(limits1, data1));
const std::vector<int> limits2{14, 18, 249, 312, 389,
392, 513, 591, 634, 720};
const std::vector<int> data2{
445, 814, 519, 697, 700, 130, 255, 889, 481, 122, 932, 77, 323, 525,
570, 219, 367, 523, 442, 933, 416, 589, 930, 373, 202, 253, 775, 47,
731, 685, 293, 126, 133, 450, 545, 100, 741, 583, 763, 306, 655, 267,
248, 477, 549, 238, 62, 678, 98, 534, 622, 907, 406, 714, 184, 391,
913, 42, 560, 247, 346, 860, 56, 138, 546, 38, 985, 948, 58, 213,
799, 319, 390, 634, 458, 945, 733, 507, 916, 123, 345, 110, 720, 917,
313, 845, 426, 9, 457, 628, 410, 723, 354, 895, 881, 953, 677, 137,
397, 97, 854, 740, 83, 216, 421, 94, 517, 479, 292, 963, 376, 981,
480, 39, 257, 272, 157, 5, 316, 395, 787, 942, 456, 242, 759, 898,
576, 67, 298, 425, 894, 435, 831, 241, 989, 614, 987, 770, 384, 692,
698, 765, 331, 487, 251, 600, 879, 342, 982, 527, 736, 795, 585, 40,
54, 901, 408, 359, 577, 237, 605, 847, 353, 968, 832, 205, 838, 427,
876, 959, 686, 646, 835, 127, 621, 892, 443, 198, 988, 791, 466, 23,
707, 467, 33, 670, 921, 180, 991, 396, 160, 436, 717, 918, 8, 374,
101, 684, 727, 749};
std::cout << "\nExample 2:\n";
print_bins(limits2, bins(limits2, data2));
} |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Haskell | Haskell | import Data.List (group, sort)
import Data.List.Split (chunksOf)
import Text.Printf (printf, IsChar(..), PrintfArg(..), fmtChar, fmtPrecision, formatString)
data DNABase = A | C | G | T deriving (Show, Read, Eq, Ord)
type DNASequence = [DNABase]
instance IsChar DNABase where
toChar = head . show
fromChar = read . pure
instance PrintfArg DNABase where
formatArg x fmt = formatString (show x) (fmt { fmtChar = 's', fmtPrecision = Nothing })
test :: DNASequence
test = read . pure <$> concat
[ "CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG"
, "CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG"
, "AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT"
, "GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT"
, "CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG"
, "TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA"
, "TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT"
, "CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG"
, "TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC"
, "GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT" ]
chunkedDNASequence :: DNASequence -> [(Int, [DNABase])]
chunkedDNASequence = zip [50,100..] . chunksOf 50
baseCounts :: DNASequence -> [(DNABase, Int)]
baseCounts = fmap ((,) . head <*> length) . group . sort
main :: IO ()
main = do
putStrLn "Sequence:"
mapM_ (uncurry (printf "%3d: %s\n")) $ chunkedDNASequence test
putStrLn "\nBase Counts:"
mapM_ (uncurry (printf "%2s: %2d\n")) $ baseCounts test
putStrLn (replicate 8 '-') >> printf " Σ: %d\n\n" (length test) |
http://rosettacode.org/wiki/Binary_digits | Binary digits | Task
Create and display the sequence of binary digits for a given non-negative integer.
The decimal value 5 should produce an output of 101
The decimal value 50 should produce an output of 110010
The decimal value 9000 should produce an output of 10001100101000
The results can be achieved using built-in radix functions within the language (if these are available), or alternatively a user defined function can be used.
The output produced should consist just of the binary digits of each number followed by a newline.
There should be no other whitespace, radix or sign markers in the produced output, and leading zeros should not appear in the results.
| #360_Assembly | 360 Assembly | * Binary digits 27/08/2015
BINARY CSECT
USING BINARY,R12
LR R12,R15 set base register
BEGIN LA R10,4
LA R9,N
LOOPN MVC W,0(R9)
MVI FLAG,X'00'
LA R8,32
LA R2,CBIN
LOOP TM W,B'10000000' test fist bit
BZ ZERO zero
MVI FLAG,X'01' one written
MVI 0(R2),C'1' write 1
B CONT
ZERO CLI FLAG,X'01' is one written ?
BNE BLANK
MVI 0(R2),C'0' write 0
B CONT
BLANK BCTR R2,0 backspace
CONT L R3,W
SLL R3,1 shilf left
ST R3,W
LA R2,1(R2) next bit
BCT R8,LOOP loop on bits
PRINT CLI FLAG,X'00' is '0'
BNE NOTZERO
MVI 0(R2),C'0' then write 0
NOTZERO L R1,0(R9)
XDECO R1,CDEC
XPRNT CDEC,45
LA R9,4(R9)
BCT R10,LOOPN loop on numbers
RETURN XR R15,R15 set return code
BR R14 return to caller
N DC F'0',F'5',F'50',F'9000'
W DS F work
FLAG DS X flag for trailing blanks
CDEC DS CL12 decimal value
DC C' '
CBIN DC CL32' ' binary value
YREGS
END BINARY |
http://rosettacode.org/wiki/Bitmap/Bresenham%27s_line_algorithm | Bitmap/Bresenham's line algorithm | Task
Using the data storage type defined on the Bitmap page for raster graphics images,
draw a line given two points with Bresenham's line algorithm.
| #D | D | module bitmap_bresenhams_line_algorithm;
import std.algorithm, std.math, bitmap;
void drawLine(Color)(Image!Color img,
size_t x1, size_t y1,
in size_t x2, in size_t y2,
in Color color)
pure nothrow @nogc {
immutable int dx = x2 - x1;
immutable int ix = (dx > 0) - (dx < 0);
immutable size_t dx2 = abs(dx) * 2;
int dy = y2 - y1;
immutable int iy = (dy > 0) - (dy < 0);
immutable size_t dy2 = abs(dy) * 2;
img[x1, y1] = color;
if (dx2 >= dy2) {
int error = dy2 - (dx2 / 2);
while (x1 != x2) {
if (error >= 0 && (error || (ix > 0))) {
error -= dx2;
y1 += iy;
}
error += dy2;
x1 += ix;
img[x1, y1] = color;
}
} else {
int error = dx2 - (dy2 / 2);
while (y1 != y2) {
if (error >= 0 && (error || (iy > 0))) {
error -= dy2;
x1 += ix;
}
error += dx2;
y1 += iy;
img[x1, y1] = color;
}
}
}
version (bitmap_bresenhams_line_algorithm_main) {
void main() {
auto img = new Image!RGB(25, 22);
img.drawLine(5, 5, 15, 20, RGB.white);
img.drawLine(3, 20, 10, 12, RGB.white);
img.textualShow();
}
} |
http://rosettacode.org/wiki/Bioinformatics/Sequence_mutation | Bioinformatics/Sequence mutation | Task
Given a string of characters A, C, G, and T representing a DNA sequence write a routine to mutate the sequence, (string) by:
Choosing a random base position in the sequence.
Mutate the sequence by doing one of either:
Swap the base at that position by changing it to one of A, C, G, or T. (which has a chance of swapping the base for the same base)
Delete the chosen base at the position.
Insert another base randomly chosen from A,C, G, or T into the sequence at that position.
Randomly generate a test DNA sequence of at least 200 bases
"Pretty print" the sequence and a count of its size, and the count of each base in the sequence
Mutate the sequence ten times.
"Pretty print" the sequence after all mutations, and a count of its size, and the count of each base in the sequence.
Extra credit
Give more information on the individual mutations applied.
Allow mutations to be weighted and/or chosen. | #PureBasic | PureBasic | #BASE$="ACGT"
#SEQLEN=200
#PROTOCOL=#True
Global dna.s
Define i.i
Procedure pprint()
Define p.i, cnt.i, sum.i
For p=1 To Len(dna) Step 50
Print(RSet(Str(p-1)+": ",5))
PrintN(Mid(dna,p,50))
Next
PrintN("Base counts:")
For p=1 To 4
cnt=CountString(dna,Mid(#BASE$,p,1)) : sum+cnt
Print(Mid(#BASE$,p,1)+": "+Str(cnt)+", ")
Next
PrintN("Total: "+Str(sum))
EndProcedure
Procedure InsertAtPos(basenr.i,position.i)
If #PROTOCOL : PrintN("Insert base "+Mid(#BASE$,basenr,1)+" at position "+Str(position)) : EndIf
dna=InsertString(dna,Mid(#BASE$,basenr,1),position)
EndProcedure
Procedure EraseAtPos(position.i)
If #PROTOCOL : PrintN("Erase base "+Mid(dna,position,1)+" at position "+Str(position)) : EndIf
If position>0 And position<=Len(dna)
dna=Left(dna,position-1)+Right(dna,Len(dna)-position)
EndIf
EndProcedure
Procedure OverwriteAtPos(basenr.i,position.i)
If #PROTOCOL : PrintN("Change base at position "+Str(position)+" from "+Mid(dna,position,1)+" to "+Mid(#BASE$,basenr,1)) : EndIf
If position>0 And position<=Len(dna)
position-1
PokeS(@dna+2*position,Mid(#BASE$,basenr,1),-1,#PB_String_NoZero)
EndIf
EndProcedure
If OpenConsole()=0 : End 1 : EndIf
For i=1 To #SEQLEN : dna+Mid(#BASE$,Random(4,1),1) : Next
PrintN("Initial sequence:")
pprint()
For i=1 To 10
Select Random(2)
Case 0 : InsertAtPos(Random(4,1),Random(Len(dna),1))
Case 1 : EraseAtPos(Random(Len(dna),1))
Case 2 : OverwriteAtPos(Random(4,1),Random(Len(dna),1))
EndSelect
Next
PrintN("After 10 mutations:")
pprint()
Input() |
http://rosettacode.org/wiki/Bitcoin/address_validation | Bitcoin/address validation | Bitcoin/address validation
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Write a program that takes a bitcoin address as argument,
and checks whether or not this address is valid.
A bitcoin address uses a base58 encoding, which uses an alphabet of the characters 0 .. 9, A ..Z, a .. z, but without the four characters:
0 zero
O uppercase oh
I uppercase eye
l lowercase ell
With this encoding, a bitcoin address encodes 25 bytes:
the first byte is the version number, which will be zero for this task ;
the next twenty bytes are a RIPEMD-160 digest, but you don't have to know that for this task: you can consider them a pure arbitrary data ;
the last four bytes are a checksum check. They are the first four bytes of a double SHA-256 digest of the previous 21 bytes.
To check the bitcoin address, you must read the first twenty-one bytes, compute the checksum, and check that it corresponds to the last four bytes.
The program can either return a boolean value or throw an exception when not valid.
You can use a digest library for SHA-256.
Example of a bitcoin address
1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i
It doesn't belong to anyone and is part of the test suite of the bitcoin software.
You can change a few characters in this string and check that it'll fail the test.
| #Raku | Raku | sub sha256(blob8 $b) returns blob8 {
given run <openssl dgst -sha256 -binary>, :in, :out, :bin {
.in.write: $b;
.in.close;
return .out.slurp;
}
}
my $bitcoin-address = rx/
<< <+alnum-[0IOl]> ** 26..* >> # an address is at least 26 characters long
<?{
.subbuf(21, 4) eq sha256(sha256 .subbuf(0, 21)).subbuf(0, 4) given
blob8.new: <
1 2 3 4 5 6 7 8 9
A B C D E F G H J K L M N P Q R S T U V W X Y Z
a b c d e f g h i j k m n o p q r s t u v w x y z
>.pairs.invert.hash{$/.comb}
.reduce(* * 58 + *)
.polymod(256 xx 24)
.reverse;
}>
/;
say "Here is a bitcoin address: 1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i" ~~ $bitcoin-address; |
http://rosettacode.org/wiki/Bitmap/Flood_fill | Bitmap/Flood fill | Implement a flood fill.
A flood fill is a way of filling an area using color banks to define the contained area or a target color which "determines" the area (the valley that can be flooded; Wikipedia uses the term target color). It works almost like a water flooding from a point towards the banks (or: inside the valley): if there's a hole in the banks, the flood is not contained and all the image (or all the "connected valleys") get filled.
To accomplish the task, you need to implement just one of the possible algorithms (examples are on Wikipedia). Variations on the theme are allowed (e.g. adding a tolerance parameter or argument for color-matching of the banks or target color).
Testing: the basic algorithm is not suitable for truecolor images; a possible test image is the one shown on the right box; you can try to fill the white area, or the black inner circle.
| #PicoLisp | PicoLisp | (de ppmFloodFill (Ppm X Y Color)
(let Target (get Ppm Y X)
(recur (X Y)
(when (= Target (get Ppm Y X))
(set (nth Ppm Y X) Color)
(recurse (dec X) Y)
(recurse (inc X) Y)
(recurse X (dec Y))
(recurse X (inc Y)) ) ) )
Ppm ) |
http://rosettacode.org/wiki/Bitmap/Flood_fill | Bitmap/Flood fill | Implement a flood fill.
A flood fill is a way of filling an area using color banks to define the contained area or a target color which "determines" the area (the valley that can be flooded; Wikipedia uses the term target color). It works almost like a water flooding from a point towards the banks (or: inside the valley): if there's a hole in the banks, the flood is not contained and all the image (or all the "connected valleys") get filled.
To accomplish the task, you need to implement just one of the possible algorithms (examples are on Wikipedia). Variations on the theme are allowed (e.g. adding a tolerance parameter or argument for color-matching of the banks or target color).
Testing: the basic algorithm is not suitable for truecolor images; a possible test image is the one shown on the right box; you can try to fill the white area, or the black inner circle.
| #PL.2FI | PL/I | fill: procedure (x, y, fill_color) recursive; /* 12 May 2010 */
declare (x, y) fixed binary;
declare fill_color bit (24) aligned;
if x <= lbound(image, 2) | x >= hbound(image, 2) then return;
if y <= lbound(image, 1) | y >= hbound(image, 1) then return;
pixel_color = image (x,y); /* Obtain the color of the current pixel. */
if pixel_color ^= area_color then return;
/* the pixel has already been filled with fill_color, */
/* or we are not within the area to be filled. */
image(x, y) = fill_color; /* color the desired area. */
pixel_color = image (x,y-1); /* Obtain the color of the pixel to the north. */
if pixel_color = area_color then call fill (x, y-1, fill_color);
pixel_color = image (x-1,y); /* Obtain the color of the pixel to the west. */
if pixel_color = area_color then call fill (x-1, y, fill_color);
pixel_color = image (x+1,y); /* Obtain the color of the pixel to the east. */
if pixel_color = area_color then call fill (x+1, y, fill_color);
pixel_color = image (x,y+1); /* Obtain the color of the pixel to the south. */
if pixel_color = area_color then call fill (x, y+1, fill_color);
end fill; |
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #LFE | LFE |
> 'true
true
> 'false
false
> (or 'false 'false)
false
> (or 'false 'true)
true
|
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #Lingo | Lingo | put TRUE
-- 1
put FALSE
-- 0
if 23 then put "Hello"
-- "Hello" |
http://rosettacode.org/wiki/Box_the_compass | Box the compass | There be many a land lubber that knows naught of the pirate ways and gives direction by degree!
They know not how to box the compass!
Task description
Create a function that takes a heading in degrees and returns the correct 32-point compass heading.
Use the function to print and display a table of Index, Compass point, and Degree; rather like the corresponding columns from, the first table of the wikipedia article, but use only the following 33 headings as input:
[0.0, 16.87, 16.88, 33.75, 50.62, 50.63, 67.5, 84.37, 84.38, 101.25, 118.12, 118.13, 135.0, 151.87, 151.88, 168.75, 185.62, 185.63, 202.5, 219.37, 219.38, 236.25, 253.12, 253.13, 270.0, 286.87, 286.88, 303.75, 320.62, 320.63, 337.5, 354.37, 354.38]. (They should give the same order of points but are spread throughout the ranges of acceptance).
Notes;
The headings and indices can be calculated from this pseudocode:
for i in 0..32 inclusive:
heading = i * 11.25
case i %3:
if 1: heading += 5.62; break
if 2: heading -= 5.62; break
end
index = ( i mod 32) + 1
The column of indices can be thought of as an enumeration of the thirty two cardinal points (see talk page)..
| #IS-BASIC | IS-BASIC | 100 PROGRAM "Compass.bas"
110 STRING DR$(1 TO 33)*18
120 FOR I=1 TO 33
130 READ DR$(I)
140 NEXT
150 DO
160 READ IF MISSING EXIT DO:D
170 LET DIR=COMP(D)
180 PRINT D;TAB(12);DIR,DR$(DIR)
190 LOOP
200 DEF COMP(D)=CEIL(MOD((D+360/64),360)*32/360)
210 DATA North,North by east,North-northeast,Northeast by north,Northeast,Northeast by east,East-northeast,East by north,East,East by south,East-southeast,Southeast by east,Southeast,Southeast by south,South-southeast,South by east
220 DATA South,South by west,South-southwest,Southwest by south,Southwest,Southwest by west,West-southwest,West by south,West,West by north,West-northwest,Northwest by west,Northwest,Northwest by north,North-northwest,North by west,North
230 DATA 0.0,16.87,16.88,33.75,50.62,50.63,67.5,84.37,84.38,101.25,118.12,118.13,135.0,151.87,151.88,168.75,185.62,185.63,202.5,219.37,219.38,236.25,253.12,253.13,270.0,286.87,286.88,303.75,320.62,320.63,337.5,354.37,354.38 |
http://rosettacode.org/wiki/Bitwise_operations | Bitwise operations |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Write a routine to perform a bitwise AND, OR, and XOR on two integers, a bitwise NOT on the first integer, a left shift, right shift, right arithmetic shift, left rotate, and right rotate.
All shifts and rotates should be done on the first integer with a shift/rotate amount of the second integer.
If any operation is not available in your language, note it.
| #Clojure | Clojure | (bit-and x y)
(bit-or x y)
(bit-xor x y)
(bit-not x)
(bit-shift-left x n)
(bit-shift-right x n)
;;There is no built-in for rotation. |
http://rosettacode.org/wiki/Bitmap | Bitmap | Show a basic storage type to handle a simple RGB raster graphics image,
and some primitive associated functions.
If possible provide a function to allocate an uninitialised image,
given its width and height, and provide 3 additional functions:
one to fill an image with a plain RGB color,
one to set a given pixel with a color,
one to get the color of a pixel.
(If there are specificities about the storage or the allocation, explain those.)
These functions are used as a base for the articles in the category raster graphics operations,
and a basic output function to check the results
is available in the article write ppm file.
| #Common_Lisp | Common Lisp | (defpackage #:rgb-pixel-buffer
(:use #:common-lisp)
(:export #:rgb-pixel-component #:rgb-pixel #:rgb-pixel-buffer
#:+red+ #:+green+ #:+blue+ #:+black+ #:+white+
#:make-rgb-pixel #:make-rgb-pixel-buffer #:rgb-pixel-buffer-width
#:rgb-pixel-buffer-height #:rgb-pixel-red #:rgb-pixel-green
#:rgb-pixel-blue #:fill-rgb-pixel-buffer)) |
http://rosettacode.org/wiki/Bell_numbers | Bell numbers | Bell or exponential numbers are enumerations of the number of different ways to partition a set that has exactly n elements. Each element of the sequence Bn is the number of partitions of a set of size n where order of the elements and order of the partitions are non-significant. E.G.: {a b} is the same as {b a} and {a} {b} is the same as {b} {a}.
So
B0 = 1 trivially. There is only one way to partition a set with zero elements. { }
B1 = 1 There is only one way to partition a set with one element. {a}
B2 = 2 Two elements may be partitioned in two ways. {a} {b}, {a b}
B3 = 5 Three elements may be partitioned in five ways {a} {b} {c}, {a b} {c}, {a} {b c}, {a c} {b}, {a b c}
and so on.
A simple way to find the Bell numbers is construct a Bell triangle, also known as an Aitken's array or Peirce triangle, and read off the numbers in the first column of each row. There are other generating algorithms though, and you are free to choose the best / most appropriate for your case.
Task
Write a routine (function, generator, whatever) to generate the Bell number sequence and call the routine to show here, on this page at least the first 15 and (if your language supports big Integers) 50th elements of the sequence.
If you do use the Bell triangle method to generate the numbers, also show the first ten rows of the Bell triangle.
See also
OEIS:A000110 Bell or exponential numbers
OEIS:A011971 Aitken's array | #ALGOL_68 | ALGOL 68 | BEGIN # show some Bell numbers #
PROC show bell = ( INT n, LONG LONG INT bell number )VOID:
print( ( whole( n, -2 ), ": ", whole( bell number, 0 ), newline ) );
INT max bell = 50;
[ 0 : max bell - 2 ]LONG LONG INT a; FOR i TO UPB a DO a[ i ] := 0 OD;
a[ 0 ] := 1;
show bell( 1, a[ 0 ] );
FOR n FROM 0 TO UPB a DO
# replace a with the next line of the triangle #
a[ n ] := a[ 0 ];
FOR j FROM n BY -1 TO 1 DO
a[ j - 1 ] +:= a[ j ]
OD;
IF n < 14 THEN
show bell( n + 2, a[ 0 ] )
ELIF n = UPB a THEN
print( ( "...", newline ) );
show bell( n + 2, a[ 0 ] )
FI
OD
END |
http://rosettacode.org/wiki/Benford%27s_law | Benford's law |
This page uses content from Wikipedia. The original article was at Benford's_law. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Benford's law, also called the first-digit law, refers to the frequency distribution of digits in many (but not all) real-life sources of data.
In this distribution, the number 1 occurs as the first digit about 30% of the time, while larger numbers occur in that position less frequently: 9 as the first digit less than 5% of the time. This distribution of first digits is the same as the widths of gridlines on a logarithmic scale.
Benford's law also concerns the expected distribution for digits beyond the first, which approach a uniform distribution.
This result has been found to apply to a wide variety of data sets, including electricity bills, street addresses, stock prices, population numbers, death rates, lengths of rivers, physical and mathematical constants, and processes described by power laws (which are very common in nature). It tends to be most accurate when values are distributed across multiple orders of magnitude.
A set of numbers is said to satisfy Benford's law if the leading digit
d
{\displaystyle d}
(
d
∈
{
1
,
…
,
9
}
{\displaystyle d\in \{1,\ldots ,9\}}
) occurs with probability
P
(
d
)
=
log
10
(
d
+
1
)
−
log
10
(
d
)
=
log
10
(
1
+
1
d
)
{\displaystyle P(d)=\log _{10}(d+1)-\log _{10}(d)=\log _{10}\left(1+{\frac {1}{d}}\right)}
For this task, write (a) routine(s) to calculate the distribution of first significant (non-zero) digits in a collection of numbers, then display the actual vs. expected distribution in the way most convenient for your language (table / graph / histogram / whatever).
Use the first 1000 numbers from the Fibonacci sequence as your data set. No need to show how the Fibonacci numbers are obtained.
You can generate them or load them from a file; whichever is easiest.
Display your actual vs expected distribution.
For extra credit: Show the distribution for one other set of numbers from a page on Wikipedia. State which Wikipedia page it can be obtained from and what the set enumerates. Again, no need to display the actual list of numbers or the code to load them.
See also:
numberphile.com.
A starting page on Wolfram Mathworld is Benfords Law .
| #8th | 8th |
: n:log10e ` 1 10 ln / ` ;
with: n
: n:log10 \ n -- n
ln log10e * ;
: benford \ x -- x
1 swap / 1+ log10 ;
: fibs \ xt n
swap >r
0.0 1.0 rot
( dup r@ w:exec tuck + ) swap times
2drop rdrop ;
var counts
: init
a:new ( 0 a:push ) 9 times counts ! ;
: leading \ n -- n
"%g" s:strfmt
0 s:@ '0 - nip ;
: bump-digit \ n --
1 swap
counts @ swap 1- ' + a:op! drop ;
: count-fibs \ --
( leading bump-digit ) 1000 fibs ;
: adjust \ --
counts @ ( 0.001 * ) a:map counts ! ;
: spaces \ n --
' space swap times ;
: .fw \ s n --
>r s:len r> rot . swap - spaces ;
: .header \ --
"Digit" 8 .fw "Expected" 10 .fw "Actual" 10 .fw cr ;
: .digit \ n --
>s 8 .fw ;
: .actual \ n --
"%.3f" s:strfmt 10 .fw ;
: .expected \ n --
"%.4f" s:strfmt 10 .fw ;
: report \ --
.header
counts @
( swap 1+ dup benford swap
.digit .expected .actual cr )
a:each drop ;
: benford-test
init count-fibs adjust report ;
;with
benford-test
bye
|
http://rosettacode.org/wiki/Benford%27s_law | Benford's law |
This page uses content from Wikipedia. The original article was at Benford's_law. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Benford's law, also called the first-digit law, refers to the frequency distribution of digits in many (but not all) real-life sources of data.
In this distribution, the number 1 occurs as the first digit about 30% of the time, while larger numbers occur in that position less frequently: 9 as the first digit less than 5% of the time. This distribution of first digits is the same as the widths of gridlines on a logarithmic scale.
Benford's law also concerns the expected distribution for digits beyond the first, which approach a uniform distribution.
This result has been found to apply to a wide variety of data sets, including electricity bills, street addresses, stock prices, population numbers, death rates, lengths of rivers, physical and mathematical constants, and processes described by power laws (which are very common in nature). It tends to be most accurate when values are distributed across multiple orders of magnitude.
A set of numbers is said to satisfy Benford's law if the leading digit
d
{\displaystyle d}
(
d
∈
{
1
,
…
,
9
}
{\displaystyle d\in \{1,\ldots ,9\}}
) occurs with probability
P
(
d
)
=
log
10
(
d
+
1
)
−
log
10
(
d
)
=
log
10
(
1
+
1
d
)
{\displaystyle P(d)=\log _{10}(d+1)-\log _{10}(d)=\log _{10}\left(1+{\frac {1}{d}}\right)}
For this task, write (a) routine(s) to calculate the distribution of first significant (non-zero) digits in a collection of numbers, then display the actual vs. expected distribution in the way most convenient for your language (table / graph / histogram / whatever).
Use the first 1000 numbers from the Fibonacci sequence as your data set. No need to show how the Fibonacci numbers are obtained.
You can generate them or load them from a file; whichever is easiest.
Display your actual vs expected distribution.
For extra credit: Show the distribution for one other set of numbers from a page on Wikipedia. State which Wikipedia page it can be obtained from and what the set enumerates. Again, no need to display the actual list of numbers or the code to load them.
See also:
numberphile.com.
A starting page on Wolfram Mathworld is Benfords Law .
| #Ada | Ada | with Ada.Text_IO, Ada.Numerics.Generic_Elementary_Functions;
procedure Benford is
subtype Nonzero_Digit is Natural range 1 .. 9;
function First_Digit(S: String) return Nonzero_Digit is
(if S(S'First) in '1' .. '9'
then Nonzero_Digit'Value(S(S'First .. S'First))
else First_Digit(S(S'First+1 .. S'Last)));
package N_IO is new Ada.Text_IO.Integer_IO(Natural);
procedure Print(D: Nonzero_Digit; Counted, Sum: Natural) is
package Math is new Ada.Numerics.Generic_Elementary_Functions(Float);
package F_IO is new Ada.Text_IO.Float_IO(Float);
Actual: constant Float := Float(Counted) / Float(Sum);
Expected: constant Float := Math.Log(1.0 + 1.0 / Float(D), Base => 10.0);
Deviation: constant Float := abs(Expected-Actual);
begin
N_IO.Put(D, 5);
N_IO.Put(Counted, 14);
F_IO.Put(Float(Sum)*Expected, Fore => 16, Aft => 1, Exp => 0);
F_IO.Put(100.0*Actual, Fore => 9, Aft => 2, Exp => 0);
F_IO.Put(100.0*Expected, Fore => 11, Aft => 2, Exp => 0);
F_IO.Put(100.0*Deviation, Fore => 13, Aft => 2, Exp => 0);
end Print;
Cnt: array(Nonzero_Digit) of Natural := (1 .. 9 => 0);
D: Nonzero_Digit;
Sum: Natural := 0;
Counter: Positive;
begin
while not Ada.Text_IO.End_Of_File loop
-- each line in the input file holds Counter, followed by Fib(Counter)
N_IO.Get(Counter);
-- Counter and skip it, we just don't need it
D := First_Digit(Ada.Text_IO.Get_Line);
-- read the rest of the line and extract the first digit
Cnt(D) := Cnt(D)+1;
Sum := Sum + 1;
end loop;
Ada.Text_IO.Put_Line(" Digit Found[total] Expected[total] Found[%]"
& " Expected[%] Difference[%]");
for I in Nonzero_Digit loop
Print(I, Cnt(I), Sum);
Ada.Text_IO.New_Line;
end loop;
end Benford; |
http://rosettacode.org/wiki/Binary_search | Binary search | A binary search divides a range of values into halves, and continues to narrow down the field of search until the unknown value is found. It is the classic example of a "divide and conquer" algorithm.
As an analogy, consider the children's game "guess a number." The scorer has a secret number, and will only tell the player if their guessed number is higher than, lower than, or equal to the secret number. The player then uses this information to guess a new number.
As the player, an optimal strategy for the general case is to start by choosing the range's midpoint as the guess, and then asking whether the guess was higher, lower, or equal to the secret number. If the guess was too high, one would select the point exactly between the range midpoint and the beginning of the range. If the original guess was too low, one would ask about the point exactly between the range midpoint and the end of the range. This process repeats until one has reached the secret number.
Task
Given the starting point of a range, the ending point of a range, and the "secret value", implement a binary search through a sorted integer array for a certain number. Implementations can be recursive or iterative (both if you can). Print out whether or not the number was in the array afterwards. If it was, print the index also.
There are several binary search algorithms commonly seen. They differ by how they treat multiple values equal to the given value, and whether they indicate whether the element was found or not. For completeness we will present pseudocode for all of them.
All of the following code examples use an "inclusive" upper bound (i.e. high = N-1 initially). Any of the examples can be converted into an equivalent example using "exclusive" upper bound (i.e. high = N initially) by making the following simple changes (which simply increase high by 1):
change high = N-1 to high = N
change high = mid-1 to high = mid
(for recursive algorithm) change if (high < low) to if (high <= low)
(for iterative algorithm) change while (low <= high) to while (low < high)
Traditional algorithm
The algorithms are as follows (from Wikipedia). The algorithms return the index of some element that equals the given value (if there are multiple such elements, it returns some arbitrary one). It is also possible, when the element is not found, to return the "insertion point" for it (the index that the value would have if it were inserted into the array).
Recursive Pseudocode:
// initially called with low = 0, high = N-1
BinarySearch(A[0..N-1], value, low, high) {
// invariants: value > A[i] for all i < low
value < A[i] for all i > high
if (high < low)
return not_found // value would be inserted at index "low"
mid = (low + high) / 2
if (A[mid] > value)
return BinarySearch(A, value, low, mid-1)
else if (A[mid] < value)
return BinarySearch(A, value, mid+1, high)
else
return mid
}
Iterative Pseudocode:
BinarySearch(A[0..N-1], value) {
low = 0
high = N - 1
while (low <= high) {
// invariants: value > A[i] for all i < low
value < A[i] for all i > high
mid = (low + high) / 2
if (A[mid] > value)
high = mid - 1
else if (A[mid] < value)
low = mid + 1
else
return mid
}
return not_found // value would be inserted at index "low"
}
Leftmost insertion point
The following algorithms return the leftmost place where the given element can be correctly inserted (and still maintain the sorted order). This is the lower (inclusive) bound of the range of elements that are equal to the given value (if any). Equivalently, this is the lowest index where the element is greater than or equal to the given value (since if it were any lower, it would violate the ordering), or 1 past the last index if such an element does not exist. This algorithm does not determine if the element is actually found. This algorithm only requires one comparison per level.
Recursive Pseudocode:
// initially called with low = 0, high = N - 1
BinarySearch_Left(A[0..N-1], value, low, high) {
// invariants: value > A[i] for all i < low
value <= A[i] for all i > high
if (high < low)
return low
mid = (low + high) / 2
if (A[mid] >= value)
return BinarySearch_Left(A, value, low, mid-1)
else
return BinarySearch_Left(A, value, mid+1, high)
}
Iterative Pseudocode:
BinarySearch_Left(A[0..N-1], value) {
low = 0
high = N - 1
while (low <= high) {
// invariants: value > A[i] for all i < low
value <= A[i] for all i > high
mid = (low + high) / 2
if (A[mid] >= value)
high = mid - 1
else
low = mid + 1
}
return low
}
Rightmost insertion point
The following algorithms return the rightmost place where the given element can be correctly inserted (and still maintain the sorted order). This is the upper (exclusive) bound of the range of elements that are equal to the given value (if any). Equivalently, this is the lowest index where the element is greater than the given value, or 1 past the last index if such an element does not exist. This algorithm does not determine if the element is actually found. This algorithm only requires one comparison per level. Note that these algorithms are almost exactly the same as the leftmost-insertion-point algorithms, except for how the inequality treats equal values.
Recursive Pseudocode:
// initially called with low = 0, high = N - 1
BinarySearch_Right(A[0..N-1], value, low, high) {
// invariants: value >= A[i] for all i < low
value < A[i] for all i > high
if (high < low)
return low
mid = (low + high) / 2
if (A[mid] > value)
return BinarySearch_Right(A, value, low, mid-1)
else
return BinarySearch_Right(A, value, mid+1, high)
}
Iterative Pseudocode:
BinarySearch_Right(A[0..N-1], value) {
low = 0
high = N - 1
while (low <= high) {
// invariants: value >= A[i] for all i < low
value < A[i] for all i > high
mid = (low + high) / 2
if (A[mid] > value)
high = mid - 1
else
low = mid + 1
}
return low
}
Extra credit
Make sure it does not have overflow bugs.
The line in the pseudo-code above to calculate the mean of two integers:
mid = (low + high) / 2
could produce the wrong result in some programming languages when used with a bounded integer type, if the addition causes an overflow. (This can occur if the array size is greater than half the maximum integer value.) If signed integers are used, and low + high overflows, it becomes a negative number, and dividing by 2 will still result in a negative number. Indexing an array with a negative number could produce an out-of-bounds exception, or other undefined behavior. If unsigned integers are used, an overflow will result in losing the largest bit, which will produce the wrong result.
One way to fix it is to manually add half the range to the low number:
mid = low + (high - low) / 2
Even though this is mathematically equivalent to the above, it is not susceptible to overflow.
Another way for signed integers, possibly faster, is the following:
mid = (low + high) >>> 1
where >>> is the logical right shift operator. The reason why this works is that, for signed integers, even though it overflows, when viewed as an unsigned number, the value is still the correct sum. To divide an unsigned number by 2, simply do a logical right shift.
Related task
Guess the number/With Feedback (Player)
See also
wp:Binary search algorithm
Extra, Extra - Read All About It: Nearly All Binary Searches and Mergesorts are Broken.
| #11l | 11l | F binary_search(l, value)
V low = 0
V high = l.len - 1
L low <= high
V mid = (low + high) I/ 2
I l[mid] > value
high = mid - 1
E I l[mid] < value
low = mid + 1
E
R mid
R -1 |
http://rosettacode.org/wiki/Best_shuffle | Best shuffle | Task
Shuffle the characters of a string in such a way that as many of the character values are in a different position as possible.
A shuffle that produces a randomized result among the best choices is to be preferred. A deterministic approach that produces the same sequence every time is acceptable as an alternative.
Display the result as follows:
original string, shuffled string, (score)
The score gives the number of positions whose character value did not change.
Example
tree, eetr, (0)
Test cases
abracadabra
seesaw
elk
grrrrrr
up
a
Related tasks
Anagrams/Deranged anagrams
Permutations/Derangements
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #11l | 11l | F count(w1, wnew)
R sum(zip(w1, wnew).map((c1, c2) -> Int(c1 == c2)))
F best_shuffle(w)
V wnew = Array(w)
V n = w.len
V rangei = Array(0 .< n)
V rangej = Array(0 .< n)
random:shuffle(&rangei)
random:shuffle(&rangej)
L(i) rangei
L(j) rangej
I i != j & wnew[j] != wnew[i] & w[i] != wnew[j] & w[j] != wnew[i]
swap(&wnew[j], &wnew[i])
L.break
V wnew_s = wnew.join(‘’)
R (wnew_s, count(w, wnew_s))
V test_words = [‘tree’, ‘abracadabra’, ‘seesaw’, ‘elk’, ‘grrrrrr’, ‘up’, ‘a’,
‘antidisestablishmentarianism’, ‘hounddogs’,
‘aardvarks are ant eaters’, ‘immediately’, ‘abba’]
L(w) test_words
V (wnew, c) = best_shuffle(w)
print(‘#29, #<29 ,(#.)’.format(w, wnew, c)) |
http://rosettacode.org/wiki/Binary_strings | Binary strings | Many languages have powerful and useful (binary safe) string manipulation functions, while others don't, making it harder for these languages to accomplish some tasks.
This task is about creating functions to handle binary strings (strings made of arbitrary bytes, i.e. byte strings according to Wikipedia) for those languages that don't have built-in support for them.
If your language of choice does have this built-in support, show a possible alternative implementation for the functions or abilities already provided by the language.
In particular the functions you need to create are:
String creation and destruction (when needed and if there's no garbage collection or similar mechanism)
String assignment
String comparison
String cloning and copying
Check if a string is empty
Append a byte to a string
Extract a substring from a string
Replace every occurrence of a byte (or a string) in a string with another string
Join strings
Possible contexts of use: compression algorithms (like LZW compression), L-systems (manipulation of symbols), many more.
| #BQN | BQN | name ← "" |
http://rosettacode.org/wiki/Binary_strings | Binary strings | Many languages have powerful and useful (binary safe) string manipulation functions, while others don't, making it harder for these languages to accomplish some tasks.
This task is about creating functions to handle binary strings (strings made of arbitrary bytes, i.e. byte strings according to Wikipedia) for those languages that don't have built-in support for them.
If your language of choice does have this built-in support, show a possible alternative implementation for the functions or abilities already provided by the language.
In particular the functions you need to create are:
String creation and destruction (when needed and if there's no garbage collection or similar mechanism)
String assignment
String comparison
String cloning and copying
Check if a string is empty
Append a byte to a string
Extract a substring from a string
Replace every occurrence of a byte (or a string) in a string with another string
Join strings
Possible contexts of use: compression algorithms (like LZW compression), L-systems (manipulation of symbols), many more.
| #C | C | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct str_t {
size_t len, alloc;
unsigned char *s;
} bstr_t, *bstr;
#define str_len(s) ((s)->len)
bstr str_new(size_t len)
{
bstr s = malloc(sizeof(bstr_t));
if (len < 8) len = 8;
s->alloc = len;
s->s = malloc(len);
s->len = 0;
return s;
}
void str_extend(bstr s)
{
size_t ns = s->alloc * 2;
if (ns - s->alloc > 1024) ns = s->alloc + 1024;
s->s = realloc(s->s, ns);
s->alloc = ns;
}
void str_del(bstr s)
{
free(s->s), free(s);
}
int str_cmp(bstr l, bstr r)
{
int res, len = l->len;
if (len > r->len) len = r->len;
if ((res = memcmp(l->s, r->s, len))) return res;
return l->len > r->len ? 1 : -1;
}
bstr str_dup(bstr src)
{
bstr x = str_new(src->len);
memcpy(x->s, src->s, src->len);
x->len = src->len;
return x;
}
bstr str_from_chars(const char *t)
{
if (!t) return str_new(0);
size_t l = strlen(t);
bstr x = str_new(l + 1);
x->len = l;
memcpy(x->s, t, l);
return x;
}
void str_append(bstr s, unsigned char b)
{
if (s->len >= s->alloc) str_extend(s);
s->s[s->len++] = b;
}
bstr str_substr(bstr s, int from, int to)
{
if (!to) to = s->len;
if (from < 0) from += s->len;
if (from < 0 || from >= s->len)
return 0;
if (to < from) to = from + 1;
bstr x = str_new(to - from);
x->len = to - from;
memcpy(x->s, s->s + from, x->len);
return x;
}
bstr str_cat(bstr s, bstr s2)
{
while (s->alloc < s->len + s2->len) str_extend(s);
memcpy(s->s + s->len, s2->s, s2->len);
s->len += s2->len;
return s;
}
void str_swap(bstr a, bstr b)
{
size_t tz;
unsigned char *ts;
tz = a->alloc; a->alloc = b->alloc; b->alloc = tz;
tz = a->len; a->len = b->len; b->len = tz;
ts = a->s; a->s = b->s; b->s = ts;
}
bstr str_subst(bstr tgt, bstr pat, bstr repl)
{
bstr tmp = str_new(0);
int i;
for (i = 0; i + pat->len <= tgt->len;) {
if (memcmp(tgt->s + i, pat->s, pat->len)) {
str_append(tmp, tgt->s[i]);
i++;
} else {
str_cat(tmp, repl);
i += pat->len;
if (!pat->len) str_append(tmp, tgt->s[i++]);
}
}
while (i < tgt->len) str_append(tmp, tgt->s[i++]);
str_swap(tmp, tgt);
str_del(tmp);
return tgt;
}
void str_set(bstr dest, bstr src)
{
while (dest->len < src->len) str_extend(dest);
memcpy(dest->s, src->s, src->len);
dest->len = src->len;
}
int main()
{
bstr s = str_from_chars("aaaaHaaaaaFaaaaHa");
bstr s2 = str_from_chars("___.");
bstr s3 = str_from_chars("");
str_subst(s, s3, s2);
printf("%.*s\n", s->len, s->s);
str_del(s);
str_del(s2);
str_del(s3);
return 0;
} |
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #C.23 | C# | using System;
public class Program
{
static void Main()
{
PrintBins(new [] { 23, 37, 43, 53, 67, 83 },
95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55
);
Console.WriteLine();
PrintBins(new [] { 14, 18, 249, 312, 389, 392, 513, 591, 634, 720 },
445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,416,589,930,373,202,
253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,655,267,248,477,549,238, 62,678, 98,534,
622,907,406,714,184,391,913, 42,560,247,346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,
945,733,507,916,123,345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,787,942,456,242,759,
898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,698,765,331,487,251,600,879,342,982,527,
736,795,585, 40, 54,901,408,359,577,237,605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,
892,443,198,988,791,466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749);
}
static void PrintBins(int[] limits, params int[] data)
{
int[] bins = Bins(limits, data);
Console.WriteLine($"-∞ .. {limits[0]} => {bins[0]}");
for (int i = 0; i < limits.Length-1; i++) {
Console.WriteLine($"{limits[i]} .. {limits[i+1]} => {bins[i+1]}");
}
Console.WriteLine($"{limits[^1]} .. ∞ => {bins[^1]}");
}
static int[] Bins(int[] limits, params int[] data)
{
Array.Sort(limits);
int[] bins = new int[limits.Length + 1];
foreach (int n in data) {
int i = Array.BinarySearch(limits, n);
i = i < 0 ? ~i : i+1;
bins[i]++;
}
return bins;
}
} |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #J | J | countBases=: (({.;#)/.~)@,
totalBases=: #@,
require 'format/printf'
printSequence=: verb define
'Sequence:' printf ''
'%4d: %s' printf ((- {.)@(+/\)@:(#"1) ,.&<"_1 ]) y
'\n Base Count\n-----------' printf ''
'%5s: %4d' printf countBases y
'-----------\nTotal = %3d' printf totalBases y
) |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Java | Java | import java.util.HashMap;
import java.util.Map;
public class orderedSequence {
public static void main(String[] args) {
Sequence gene = new Sequence("CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATGCTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTGAGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGATGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTTCGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGGTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATATTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTATCGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTGTCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGACGACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT");
gene.runSequence();
}
}
/** Separate class for defining behaviors */
public class Sequence {
private final String seq;
public Sequence(String sq) {
this.seq = sq;
}
/** print the organized structure of the sequence */
public void prettyPrint() {
System.out.println("Sequence:");
int i = 0;
for ( ; i < seq.length() - 50 ; i += 50) {
System.out.printf("%5s : %s\n", i + 50, seq.substring(i, i + 50));
}
System.out.printf("%5s : %s\n", seq.length(), seq.substring(i));
}
/** display a base vs. frequency chart */
public void displayCount() {
Map<Character, Integer> counter = new HashMap<>();
for (int i = 0 ; i < seq.length() ; ++i) {
counter.merge(seq.charAt(i), 1, Integer::sum);
}
System.out.println("Base vs. Count:");
counter.forEach(
key, value -> System.out.printf("%5s : %s\n", key, value));
System.out.printf("%5s: %s\n", "SUM", seq.length());
}
public void runSequence() {
this.prettyPrint();
this.displayCount();
}
}
|
http://rosettacode.org/wiki/Binary_digits | Binary digits | Task
Create and display the sequence of binary digits for a given non-negative integer.
The decimal value 5 should produce an output of 101
The decimal value 50 should produce an output of 110010
The decimal value 9000 should produce an output of 10001100101000
The results can be achieved using built-in radix functions within the language (if these are available), or alternatively a user defined function can be used.
The output produced should consist just of the binary digits of each number followed by a newline.
There should be no other whitespace, radix or sign markers in the produced output, and leading zeros should not appear in the results.
| #6502_Assembly | 6502 Assembly | tmpx -i dec2bin.s -o dec2bin.prg
|
http://rosettacode.org/wiki/Bitmap/Bresenham%27s_line_algorithm | Bitmap/Bresenham's line algorithm | Task
Using the data storage type defined on the Bitmap page for raster graphics images,
draw a line given two points with Bresenham's line algorithm.
| #Delphi | Delphi |
procedure drawLine (bitmap : TBitmap; xStart, yStart, xEnd, yEnd : integer; color : TAlphaColor);
// Bresenham's Line Algorithm. Byte, March 1988, pp. 249-253.
// Modified from http://www.efg2.com/Lab/Library/Delphi/Graphics/Bresenham.txt and tested.
var
a, b : integer; // displacements in x and y
d : integer; // decision variable
diag_inc : integer; // d's increment for diagonal steps
dx_diag : integer; // diagonal x step for next pixel
dx_nondiag : integer; // nondiagonal x step for next pixel
dy_diag : integer; // diagonal y step for next pixel
dy_nondiag : integer; // nondiagonal y step for next pixel
i : integer; // loop index
nondiag_inc: integer; // d's increment for nondiagonal steps
swap : integer; // temporary variable for swap
x,y : integer; // current x and y coordinates
begin
x := xStart; // line starting point}
y := yStart;
// Determine drawing direction and step to the next pixel.
a := xEnd - xStart; // difference in x dimension
b := yEnd - yStart; // difference in y dimension
// Determine whether end point lies to right or left of start point.
if a < 0 then // drawing towards smaller x values?
begin
a := -a; // make 'a' positive
dx_diag := -1
end
else
dx_diag := 1;
// Determine whether end point lies above or below start point.
if b < 0 then // drawing towards smaller x values?
begin
b := -b; // make 'a' positive
dy_diag := -1
end
else
dy_diag := 1;
// Identify octant containing end point.
if a < b then
begin
swap := a;
a := b;
b := swap;
dx_nondiag := 0;
dy_nondiag := dy_diag
end
else
begin
dx_nondiag := dx_diag;
dy_nondiag := 0
end;
d := b + b - a; // initial value for d is 2*b - a
nondiag_inc := b + b; // set initial d increment values
diag_inc := b + b - a - a;
for i := 0 to a do
begin /// draw the a+1 pixels
drawPixel (bitmap, x, y, color);
if d < 0 then // is midpoint above the line?
begin // step nondiagonally
x := x + dx_nondiag;
y := y + dy_nondiag;
d := d + nondiag_inc // update decision variable
end
else
begin // midpoint is above the line; step diagonally}
x := x + dx_diag;
y := y + dy_diag;
d := d + diag_inc
end;
end;
end;
|
http://rosettacode.org/wiki/Bioinformatics/Sequence_mutation | Bioinformatics/Sequence mutation | Task
Given a string of characters A, C, G, and T representing a DNA sequence write a routine to mutate the sequence, (string) by:
Choosing a random base position in the sequence.
Mutate the sequence by doing one of either:
Swap the base at that position by changing it to one of A, C, G, or T. (which has a chance of swapping the base for the same base)
Delete the chosen base at the position.
Insert another base randomly chosen from A,C, G, or T into the sequence at that position.
Randomly generate a test DNA sequence of at least 200 bases
"Pretty print" the sequence and a count of its size, and the count of each base in the sequence
Mutate the sequence ten times.
"Pretty print" the sequence after all mutations, and a count of its size, and the count of each base in the sequence.
Extra credit
Give more information on the individual mutations applied.
Allow mutations to be weighted and/or chosen. | #Python | Python | import random
from collections import Counter
def basecount(dna):
return sorted(Counter(dna).items())
def seq_split(dna, n=50):
return [dna[i: i+n] for i in range(0, len(dna), n)]
def seq_pp(dna, n=50):
for i, part in enumerate(seq_split(dna, n)):
print(f"{i*n:>5}: {part}")
print("\n BASECOUNT:")
tot = 0
for base, count in basecount(dna):
print(f" {base:>3}: {count}")
tot += count
base, count = 'TOT', tot
print(f" {base:>3}= {count}")
def seq_mutate(dna, count=1, kinds="IDSSSS", choice="ATCG" ):
mutation = []
k2txt = dict(I='Insert', D='Delete', S='Substitute')
for _ in range(count):
kind = random.choice(kinds)
index = random.randint(0, len(dna))
if kind == 'I': # Insert
dna = dna[:index] + random.choice(choice) + dna[index:]
elif kind == 'D' and dna: # Delete
dna = dna[:index] + dna[index+1:]
elif kind == 'S' and dna: # Substitute
dna = dna[:index] + random.choice(choice) + dna[index+1:]
mutation.append((k2txt[kind], index))
return dna, mutation
if __name__ == '__main__':
length = 250
print("SEQUENCE:")
sequence = ''.join(random.choices('ACGT', weights=(1, 0.8, .9, 1.1), k=length))
seq_pp(sequence)
print("\n\nMUTATIONS:")
mseq, m = seq_mutate(sequence, 10)
for kind, index in m:
print(f" {kind:>10} @{index}")
print()
seq_pp(mseq) |
http://rosettacode.org/wiki/Bitcoin/address_validation | Bitcoin/address validation | Bitcoin/address validation
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Write a program that takes a bitcoin address as argument,
and checks whether or not this address is valid.
A bitcoin address uses a base58 encoding, which uses an alphabet of the characters 0 .. 9, A ..Z, a .. z, but without the four characters:
0 zero
O uppercase oh
I uppercase eye
l lowercase ell
With this encoding, a bitcoin address encodes 25 bytes:
the first byte is the version number, which will be zero for this task ;
the next twenty bytes are a RIPEMD-160 digest, but you don't have to know that for this task: you can consider them a pure arbitrary data ;
the last four bytes are a checksum check. They are the first four bytes of a double SHA-256 digest of the previous 21 bytes.
To check the bitcoin address, you must read the first twenty-one bytes, compute the checksum, and check that it corresponds to the last four bytes.
The program can either return a boolean value or throw an exception when not valid.
You can use a digest library for SHA-256.
Example of a bitcoin address
1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i
It doesn't belong to anyone and is part of the test suite of the bitcoin software.
You can change a few characters in this string and check that it'll fail the test.
| #Ruby | Ruby |
# Validate Bitcoin address
#
# Nigel_Galloway
# October 13th., 2014
require 'digest/sha2'
def convert g
i,e = '',[]
(0...g.length/2).each{|n| e[n] = g[n+=n]+g[n+1]; i+='H2'}
e.pack(i)
end
N = [0,1,2,3,4,5,6,7,8,nil,nil,nil,nil,nil,nil,nil,9,10,11,12,13,14,15,16,nil,17,18,19,20,21,nil,22,23,24,25,26,27,28,29,30,31,32,nil,nil,nil,nil,nil,nil,33,34,35,36,37,38,39,40,41,42,43,nil,44,45,46,47,48,49,50,51,52,53,54,55,56,57]
A = '1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62x'
g = A.bytes.inject(0){|g,n| g*58+N[n-49]}.to_s(16) # A small and interesting piece of code to do the decoding of base58-encoded data.
n = g.slice!(0..-9)
(n.length...42).each{n.insert(0,'0')}
puts "I think the checksum should be #{g}\nI calculate that it is #{Digest::SHA256.hexdigest(Digest::SHA256.digest(convert(n)))[0,8]}"
|
http://rosettacode.org/wiki/Bitmap/Flood_fill | Bitmap/Flood fill | Implement a flood fill.
A flood fill is a way of filling an area using color banks to define the contained area or a target color which "determines" the area (the valley that can be flooded; Wikipedia uses the term target color). It works almost like a water flooding from a point towards the banks (or: inside the valley): if there's a hole in the banks, the flood is not contained and all the image (or all the "connected valleys") get filled.
To accomplish the task, you need to implement just one of the possible algorithms (examples are on Wikipedia). Variations on the theme are allowed (e.g. adding a tolerance parameter or argument for color-matching of the banks or target color).
Testing: the basic algorithm is not suitable for truecolor images; a possible test image is the one shown on the right box; you can try to fill the white area, or the black inner circle.
| #Processing | Processing | import java.awt.Point;
import java.util.Queue;
import java.util.LinkedList;
PImage img;
int tolerance;
color fill_color;
boolean allowed;
void setup() {
size(600, 400);
img = loadImage("image.png");
fill_color = color(250, 0, 0);
fill(0, 0, 100);
tolerance = 15;
image(img, 0, 0, width, height);
textSize(18);
text("Tolerance = "+tolerance+" (Use mouse wheel to change)", 100, height-30);
text("Right click to reset", 100, height-10);
}
void draw() {
if (allowed) {
image(img, 0, 0, width, height);
text("Tolerance = "+tolerance+" (Use mouse wheel to change)", 100, height-30);
text("Right click to reset", 100, height-10);
allowed = false;
}
}
void mousePressed() {
if (mouseButton == RIGHT) {
img = loadImage("image.png");
} else {
img.loadPixels();
flood(mouseX, mouseY);
img.updatePixels();
allowed = true;
}
}
void mouseWheel(MouseEvent event) {
float e = event.getCount();
tolerance += 2*e;
if (tolerance > 128) tolerance = 128;
if (tolerance < 0) tolerance = 0;
allowed = true;
}
void flood(int x, int y) {
color target_color = img.pixels[pixel_position(mouseX, mouseY)];
if (target_color != fill_color) {
Queue<Point> queue = new LinkedList<Point>();
queue.add(new Point(x, y));
while (!queue.isEmpty()) {
Point p = queue.remove();
if (check(p.x, p.y, target_color)) {
queue.add(new Point(p.x, p.y-1));
queue.add(new Point(p.x, p.y+1));
queue.add(new Point(p.x-1, p.y));
queue.add(new Point(p.x+1, p.y));
}
}
}
}
int pixel_position(int x, int y) {
return x + (y * img.width);
}
boolean check(int x, int y, color target_color) {
if (x < 0 || y < 0 || y >= img.height || x >= img.width) return false;
int pp = img.pixels[pixel_position(x, y)];
boolean test_tolerance = (abs(green(target_color)-green(pp)) < tolerance
&& abs( red(target_color)- red(pp)) < tolerance
&& abs( blue(target_color)- blue(pp)) < tolerance);
if (!test_tolerance) return false;
img.pixels[pixel_position(x, y)] = fill_color;
return true;
} |
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #Little | Little | int a = 0;
int b = 1;
int c;
string str1 = "initialized string";
string str2; // "uninitialized string";
if (a) {puts("first test a is false");} // This should not print
if (b) {puts("second test b is true");} // This should print
if (c) {puts("third test b is false");} // This should not print
if (!defined(c)) {puts("fourth test is true");} // This should print
if (str1) {puts("fifth test str1 is true");} // This should print
if (str2) {puts("sixth test str2 is false");} // This should not print |
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #LiveCode | LiveCode | print 1 < 0 ; false
print 1 > 0 ; true
if "true [print "yes] ; yes
if not "false [print "no] ; no |
http://rosettacode.org/wiki/Box_the_compass | Box the compass | There be many a land lubber that knows naught of the pirate ways and gives direction by degree!
They know not how to box the compass!
Task description
Create a function that takes a heading in degrees and returns the correct 32-point compass heading.
Use the function to print and display a table of Index, Compass point, and Degree; rather like the corresponding columns from, the first table of the wikipedia article, but use only the following 33 headings as input:
[0.0, 16.87, 16.88, 33.75, 50.62, 50.63, 67.5, 84.37, 84.38, 101.25, 118.12, 118.13, 135.0, 151.87, 151.88, 168.75, 185.62, 185.63, 202.5, 219.37, 219.38, 236.25, 253.12, 253.13, 270.0, 286.87, 286.88, 303.75, 320.62, 320.63, 337.5, 354.37, 354.38]. (They should give the same order of points but are spread throughout the ranges of acceptance).
Notes;
The headings and indices can be calculated from this pseudocode:
for i in 0..32 inclusive:
heading = i * 11.25
case i %3:
if 1: heading += 5.62; break
if 2: heading -= 5.62; break
end
index = ( i mod 32) + 1
The column of indices can be thought of as an enumeration of the thirty two cardinal points (see talk page)..
| #J | J | require'strings'
subs=: 'N,north,S,south,E,east,W,west,b, by ,'
dirs=: subs (toupper@{., }.)@rplc~L:1 0&(<;._2) 0 :0 -. ' ',LF
N,NbE,N-NE,NEbN,NE,NEbE,E-NE,EbN,E,EbS,E-SE,SEbE,SE,SEbS,S-SE,SbE,
S,SbW,S-SW,SWbS,SW,SWbW,W-SW,WbS,W,WbN,W-NW,NWbW,NW,NWbN,N-NW,NbW,
)
indice=: 32 | 0.5 <.@+ %&11.25
deg2pnt=: dirs {~ indice |
http://rosettacode.org/wiki/Bitwise_operations | Bitwise operations |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Write a routine to perform a bitwise AND, OR, and XOR on two integers, a bitwise NOT on the first integer, a left shift, right shift, right arithmetic shift, left rotate, and right rotate.
All shifts and rotates should be done on the first integer with a shift/rotate amount of the second integer.
If any operation is not available in your language, note it.
| #COBOL | COBOL | IDENTIFICATION DIVISION.
PROGRAM-ID. bitwise-ops.
DATA DIVISION.
LOCAL-STORAGE SECTION.
01 a PIC 1(32) USAGE BIT.
01 b PIC 1(32) USAGE BIT.
01 result PIC 1(32) USAGE BIT.
01 result-disp REDEFINES result PIC S9(9) COMP.
LINKAGE SECTION.
01 a-int USAGE BINARY-LONG.
01 b-int USAGE BINARY-LONG.
PROCEDURE DIVISION USING a-int, b-int.
MOVE FUNCTION BOOLEAN-OF-INTEGER(a-int, 32) TO a
MOVE FUNCTION BOOLEAN-OF-INTEGER(b-int, 32) TO b
COMPUTE result = a B-AND b
DISPLAY "a and b is " result-disp
COMPUTE result = a B-OR b
DISPLAY "a or b is " result-disp
COMPUTE result = B-NOT a
DISPLAY "Not a is " result-disp
COMPUTE result = a B-XOR b
DISPLAY "a exclusive-or b is " result-disp
*> COBOL does not have shift or rotation operators.
GOBACK
. |
http://rosettacode.org/wiki/Bitmap | Bitmap | Show a basic storage type to handle a simple RGB raster graphics image,
and some primitive associated functions.
If possible provide a function to allocate an uninitialised image,
given its width and height, and provide 3 additional functions:
one to fill an image with a plain RGB color,
one to set a given pixel with a color,
one to get the color of a pixel.
(If there are specificities about the storage or the allocation, explain those.)
These functions are used as a base for the articles in the category raster graphics operations,
and a basic output function to check the results
is available in the article write ppm file.
| #Crystal | Crystal |
class RGBColor
getter red, green, blue
def initialize(@red = 0_u8, @green = 0_u8, @blue = 0_u8)
end
RED = new(red: 255_u8)
GREEN = new(green: 255_u8)
BLUE = new(blue: 255_u8)
BLACK = new
WHITE = new(255_u8, 255_u8, 255_u8)
end
class Pixmap
getter width, height
@data : Array(Array(RGBColor))
def initialize(@width : Int32, @height : Int32)
@data = Array.new(@width) { Array.new(@height, RGBColor::WHITE) }
end
def fill(color)
@data.each &.fill(color)
end
def [](x, y)
@data[x][y]
end
def []=(x, y, color)
@data[x][y] = color
end
end
bmap = Pixmap.new(5, 5)
pp bmap
|
http://rosettacode.org/wiki/Bell_numbers | Bell numbers | Bell or exponential numbers are enumerations of the number of different ways to partition a set that has exactly n elements. Each element of the sequence Bn is the number of partitions of a set of size n where order of the elements and order of the partitions are non-significant. E.G.: {a b} is the same as {b a} and {a} {b} is the same as {b} {a}.
So
B0 = 1 trivially. There is only one way to partition a set with zero elements. { }
B1 = 1 There is only one way to partition a set with one element. {a}
B2 = 2 Two elements may be partitioned in two ways. {a} {b}, {a b}
B3 = 5 Three elements may be partitioned in five ways {a} {b} {c}, {a b} {c}, {a} {b c}, {a c} {b}, {a b c}
and so on.
A simple way to find the Bell numbers is construct a Bell triangle, also known as an Aitken's array or Peirce triangle, and read off the numbers in the first column of each row. There are other generating algorithms though, and you are free to choose the best / most appropriate for your case.
Task
Write a routine (function, generator, whatever) to generate the Bell number sequence and call the routine to show here, on this page at least the first 15 and (if your language supports big Integers) 50th elements of the sequence.
If you do use the Bell triangle method to generate the numbers, also show the first ten rows of the Bell triangle.
See also
OEIS:A000110 Bell or exponential numbers
OEIS:A011971 Aitken's array | #ALGOL-M | ALGOL-M | begin
integer function index(row, col);
integer row, col;
index := row * (row-1)/ 2 + col;
integer ROWS; ROWS := 15;
begin
decimal(11) array bell[0:ROWS*(ROWS+1)/2];
integer i, j;
bell[index(1, 0)] := 1.;
for i := 2 step 1 until ROWS do
begin
bell[index(i, 0)] := bell[index(i-1, i-2)];
for j := 1 step 1 until i-1 do
bell[index(i,j)] := bell[index(i,j-1)] + bell[index(i-1,j-1)];
end;
write("First fifteen Bell numbers:");
for i := 1 step 1 until ROWS do
begin
write(i);
writeon(": ");
writeon(bell[index(i,0)]);
end;
write("");
write("First ten rows of Bell's triangle:");
for i := 1 step 1 until 10 do
begin
write("");
for j := 0 step 1 until i-1 do
writeon(bell[index(i,j)]);
end;
end;
end |
http://rosettacode.org/wiki/Bell_numbers | Bell numbers | Bell or exponential numbers are enumerations of the number of different ways to partition a set that has exactly n elements. Each element of the sequence Bn is the number of partitions of a set of size n where order of the elements and order of the partitions are non-significant. E.G.: {a b} is the same as {b a} and {a} {b} is the same as {b} {a}.
So
B0 = 1 trivially. There is only one way to partition a set with zero elements. { }
B1 = 1 There is only one way to partition a set with one element. {a}
B2 = 2 Two elements may be partitioned in two ways. {a} {b}, {a b}
B3 = 5 Three elements may be partitioned in five ways {a} {b} {c}, {a b} {c}, {a} {b c}, {a c} {b}, {a b c}
and so on.
A simple way to find the Bell numbers is construct a Bell triangle, also known as an Aitken's array or Peirce triangle, and read off the numbers in the first column of each row. There are other generating algorithms though, and you are free to choose the best / most appropriate for your case.
Task
Write a routine (function, generator, whatever) to generate the Bell number sequence and call the routine to show here, on this page at least the first 15 and (if your language supports big Integers) 50th elements of the sequence.
If you do use the Bell triangle method to generate the numbers, also show the first ten rows of the Bell triangle.
See also
OEIS:A000110 Bell or exponential numbers
OEIS:A011971 Aitken's array | #APL | APL | bell←{
tr←↑(⊢,(⊂⊃∘⌽+0,+\)∘⊃∘⌽)⍣14⊢,⊂,1
⎕←'First 15 Bell numbers:'
⎕←tr[;1]
⎕←'First 10 rows of Bell''s triangle:'
⎕←tr[⍳10;⍳10]
} |
http://rosettacode.org/wiki/Benford%27s_law | Benford's law |
This page uses content from Wikipedia. The original article was at Benford's_law. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Benford's law, also called the first-digit law, refers to the frequency distribution of digits in many (but not all) real-life sources of data.
In this distribution, the number 1 occurs as the first digit about 30% of the time, while larger numbers occur in that position less frequently: 9 as the first digit less than 5% of the time. This distribution of first digits is the same as the widths of gridlines on a logarithmic scale.
Benford's law also concerns the expected distribution for digits beyond the first, which approach a uniform distribution.
This result has been found to apply to a wide variety of data sets, including electricity bills, street addresses, stock prices, population numbers, death rates, lengths of rivers, physical and mathematical constants, and processes described by power laws (which are very common in nature). It tends to be most accurate when values are distributed across multiple orders of magnitude.
A set of numbers is said to satisfy Benford's law if the leading digit
d
{\displaystyle d}
(
d
∈
{
1
,
…
,
9
}
{\displaystyle d\in \{1,\ldots ,9\}}
) occurs with probability
P
(
d
)
=
log
10
(
d
+
1
)
−
log
10
(
d
)
=
log
10
(
1
+
1
d
)
{\displaystyle P(d)=\log _{10}(d+1)-\log _{10}(d)=\log _{10}\left(1+{\frac {1}{d}}\right)}
For this task, write (a) routine(s) to calculate the distribution of first significant (non-zero) digits in a collection of numbers, then display the actual vs. expected distribution in the way most convenient for your language (table / graph / histogram / whatever).
Use the first 1000 numbers from the Fibonacci sequence as your data set. No need to show how the Fibonacci numbers are obtained.
You can generate them or load them from a file; whichever is easiest.
Display your actual vs expected distribution.
For extra credit: Show the distribution for one other set of numbers from a page on Wikipedia. State which Wikipedia page it can be obtained from and what the set enumerates. Again, no need to display the actual list of numbers or the code to load them.
See also:
numberphile.com.
A starting page on Wolfram Mathworld is Benfords Law .
| #Aime | Aime | text
sum(text a, text b)
{
data d;
integer e, f, n, r;
e = ~a;
f = ~b;
r = 0;
n = min(e, f);
while (n) {
n -= 1;
e -= 1;
f -= 1;
r += a[e] - '0';
r += b[f] - '0';
b_insert(d, 0, r % 10 + '0');
r /= 10;
}
if (f) {
e = f;
a = b;
}
while (e) {
e -= 1;
r += a[e] - '0';
b_insert(d, 0, r % 10 + '0');
r /= 10;
}
if (r) {
b_insert(d, 0, r + '0');
}
d;
}
text
fibs(list l, integer n)
{
integer c, i;
text a, b, w;
l[1] = 1;
a = "0";
b = "1";
i = 1;
while (i < n) {
w = sum(a, b);
a = b;
b = w;
c = w[0] - '0';
l[c] = 1 + l[c];
i += 1;
}
w;
}
integer
main(void)
{
integer i, n;
list f;
real m;
n = 1000;
f.pn_integer(0, 10, 0);
fibs(f, n);
m = 100r / n;
o_text("\t\texpected\t found\n");
i = 0;
while (i < 9) {
i += 1;
o_form("%8d/p3d3w16//p3d3w16/\n", i, 100 * log10(1 + 1r / i), f[i] * m);
}
0;
} |
http://rosettacode.org/wiki/Benford%27s_law | Benford's law |
This page uses content from Wikipedia. The original article was at Benford's_law. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Benford's law, also called the first-digit law, refers to the frequency distribution of digits in many (but not all) real-life sources of data.
In this distribution, the number 1 occurs as the first digit about 30% of the time, while larger numbers occur in that position less frequently: 9 as the first digit less than 5% of the time. This distribution of first digits is the same as the widths of gridlines on a logarithmic scale.
Benford's law also concerns the expected distribution for digits beyond the first, which approach a uniform distribution.
This result has been found to apply to a wide variety of data sets, including electricity bills, street addresses, stock prices, population numbers, death rates, lengths of rivers, physical and mathematical constants, and processes described by power laws (which are very common in nature). It tends to be most accurate when values are distributed across multiple orders of magnitude.
A set of numbers is said to satisfy Benford's law if the leading digit
d
{\displaystyle d}
(
d
∈
{
1
,
…
,
9
}
{\displaystyle d\in \{1,\ldots ,9\}}
) occurs with probability
P
(
d
)
=
log
10
(
d
+
1
)
−
log
10
(
d
)
=
log
10
(
1
+
1
d
)
{\displaystyle P(d)=\log _{10}(d+1)-\log _{10}(d)=\log _{10}\left(1+{\frac {1}{d}}\right)}
For this task, write (a) routine(s) to calculate the distribution of first significant (non-zero) digits in a collection of numbers, then display the actual vs. expected distribution in the way most convenient for your language (table / graph / histogram / whatever).
Use the first 1000 numbers from the Fibonacci sequence as your data set. No need to show how the Fibonacci numbers are obtained.
You can generate them or load them from a file; whichever is easiest.
Display your actual vs expected distribution.
For extra credit: Show the distribution for one other set of numbers from a page on Wikipedia. State which Wikipedia page it can be obtained from and what the set enumerates. Again, no need to display the actual list of numbers or the code to load them.
See also:
numberphile.com.
A starting page on Wolfram Mathworld is Benfords Law .
| #ALGOL_68 | ALGOL 68 | BEGIN
# set the number of digits for LONG LONG INT values #
PR precision 256 PR
# returns the probability of the first digit of each non-zero number in s #
PROC digit probability = ( []LONG LONG INT s )[]REAL:
BEGIN
[ 1 : 9 ]REAL result;
# count the number of times each digit is the first #
[ 1 : 9 ]INT count := ( 0, 0, 0, 0, 0, 0, 0, 0, 0 );
FOR i FROM LWB s TO UPB s DO
LONG LONG INT v := ABS s[ i ];
IF v /= 0 THEN
WHILE v > 9 DO v OVERAB 10 OD;
count[ SHORTEN SHORTEN v ] +:= 1
FI
OD;
# calculate the probability of each digit #
INT number of elements = ( UPB s + 1 ) - LWB s;
FOR i TO 9 DO
result[ i ] := IF number of elements = 0 THEN 0 ELSE count[ i ] / number of elements FI
OD;
result
END # digit probability # ;
# outputs the digit probabilities of some numbers and those expected by Benford's law #
PROC compare to benford = ( []REAL actual )VOID:
FOR i TO 9 DO
print( ( "Benford: ", fixed( log( 1 + ( 1 / i ) ), -7, 3 ), " actual: ", fixed( actual[ i ], -7, 3 ), newline ) )
OD # compare to benford # ;
# generate 1000 fibonacci numbers #
[ 0 : 1000 ]LONG LONG INT fn;
fn[ 0 ] := 0;
fn[ 1 ] := 1;
FOR i FROM 2 TO UPB fn DO fn[ i ] := fn[ i - 1 ] + fn[ i - 2 ] OD;
# get the probabilities of each first digit of the fibonacci numbers and #
# compare to the probabilities expected by Benford's law #
compare to benford( digit probability( fn ) )
END |
http://rosettacode.org/wiki/Binary_search | Binary search | A binary search divides a range of values into halves, and continues to narrow down the field of search until the unknown value is found. It is the classic example of a "divide and conquer" algorithm.
As an analogy, consider the children's game "guess a number." The scorer has a secret number, and will only tell the player if their guessed number is higher than, lower than, or equal to the secret number. The player then uses this information to guess a new number.
As the player, an optimal strategy for the general case is to start by choosing the range's midpoint as the guess, and then asking whether the guess was higher, lower, or equal to the secret number. If the guess was too high, one would select the point exactly between the range midpoint and the beginning of the range. If the original guess was too low, one would ask about the point exactly between the range midpoint and the end of the range. This process repeats until one has reached the secret number.
Task
Given the starting point of a range, the ending point of a range, and the "secret value", implement a binary search through a sorted integer array for a certain number. Implementations can be recursive or iterative (both if you can). Print out whether or not the number was in the array afterwards. If it was, print the index also.
There are several binary search algorithms commonly seen. They differ by how they treat multiple values equal to the given value, and whether they indicate whether the element was found or not. For completeness we will present pseudocode for all of them.
All of the following code examples use an "inclusive" upper bound (i.e. high = N-1 initially). Any of the examples can be converted into an equivalent example using "exclusive" upper bound (i.e. high = N initially) by making the following simple changes (which simply increase high by 1):
change high = N-1 to high = N
change high = mid-1 to high = mid
(for recursive algorithm) change if (high < low) to if (high <= low)
(for iterative algorithm) change while (low <= high) to while (low < high)
Traditional algorithm
The algorithms are as follows (from Wikipedia). The algorithms return the index of some element that equals the given value (if there are multiple such elements, it returns some arbitrary one). It is also possible, when the element is not found, to return the "insertion point" for it (the index that the value would have if it were inserted into the array).
Recursive Pseudocode:
// initially called with low = 0, high = N-1
BinarySearch(A[0..N-1], value, low, high) {
// invariants: value > A[i] for all i < low
value < A[i] for all i > high
if (high < low)
return not_found // value would be inserted at index "low"
mid = (low + high) / 2
if (A[mid] > value)
return BinarySearch(A, value, low, mid-1)
else if (A[mid] < value)
return BinarySearch(A, value, mid+1, high)
else
return mid
}
Iterative Pseudocode:
BinarySearch(A[0..N-1], value) {
low = 0
high = N - 1
while (low <= high) {
// invariants: value > A[i] for all i < low
value < A[i] for all i > high
mid = (low + high) / 2
if (A[mid] > value)
high = mid - 1
else if (A[mid] < value)
low = mid + 1
else
return mid
}
return not_found // value would be inserted at index "low"
}
Leftmost insertion point
The following algorithms return the leftmost place where the given element can be correctly inserted (and still maintain the sorted order). This is the lower (inclusive) bound of the range of elements that are equal to the given value (if any). Equivalently, this is the lowest index where the element is greater than or equal to the given value (since if it were any lower, it would violate the ordering), or 1 past the last index if such an element does not exist. This algorithm does not determine if the element is actually found. This algorithm only requires one comparison per level.
Recursive Pseudocode:
// initially called with low = 0, high = N - 1
BinarySearch_Left(A[0..N-1], value, low, high) {
// invariants: value > A[i] for all i < low
value <= A[i] for all i > high
if (high < low)
return low
mid = (low + high) / 2
if (A[mid] >= value)
return BinarySearch_Left(A, value, low, mid-1)
else
return BinarySearch_Left(A, value, mid+1, high)
}
Iterative Pseudocode:
BinarySearch_Left(A[0..N-1], value) {
low = 0
high = N - 1
while (low <= high) {
// invariants: value > A[i] for all i < low
value <= A[i] for all i > high
mid = (low + high) / 2
if (A[mid] >= value)
high = mid - 1
else
low = mid + 1
}
return low
}
Rightmost insertion point
The following algorithms return the rightmost place where the given element can be correctly inserted (and still maintain the sorted order). This is the upper (exclusive) bound of the range of elements that are equal to the given value (if any). Equivalently, this is the lowest index where the element is greater than the given value, or 1 past the last index if such an element does not exist. This algorithm does not determine if the element is actually found. This algorithm only requires one comparison per level. Note that these algorithms are almost exactly the same as the leftmost-insertion-point algorithms, except for how the inequality treats equal values.
Recursive Pseudocode:
// initially called with low = 0, high = N - 1
BinarySearch_Right(A[0..N-1], value, low, high) {
// invariants: value >= A[i] for all i < low
value < A[i] for all i > high
if (high < low)
return low
mid = (low + high) / 2
if (A[mid] > value)
return BinarySearch_Right(A, value, low, mid-1)
else
return BinarySearch_Right(A, value, mid+1, high)
}
Iterative Pseudocode:
BinarySearch_Right(A[0..N-1], value) {
low = 0
high = N - 1
while (low <= high) {
// invariants: value >= A[i] for all i < low
value < A[i] for all i > high
mid = (low + high) / 2
if (A[mid] > value)
high = mid - 1
else
low = mid + 1
}
return low
}
Extra credit
Make sure it does not have overflow bugs.
The line in the pseudo-code above to calculate the mean of two integers:
mid = (low + high) / 2
could produce the wrong result in some programming languages when used with a bounded integer type, if the addition causes an overflow. (This can occur if the array size is greater than half the maximum integer value.) If signed integers are used, and low + high overflows, it becomes a negative number, and dividing by 2 will still result in a negative number. Indexing an array with a negative number could produce an out-of-bounds exception, or other undefined behavior. If unsigned integers are used, an overflow will result in losing the largest bit, which will produce the wrong result.
One way to fix it is to manually add half the range to the low number:
mid = low + (high - low) / 2
Even though this is mathematically equivalent to the above, it is not susceptible to overflow.
Another way for signed integers, possibly faster, is the following:
mid = (low + high) >>> 1
where >>> is the logical right shift operator. The reason why this works is that, for signed integers, even though it overflows, when viewed as an unsigned number, the value is still the correct sum. To divide an unsigned number by 2, simply do a logical right shift.
Related task
Guess the number/With Feedback (Player)
See also
wp:Binary search algorithm
Extra, Extra - Read All About It: Nearly All Binary Searches and Mergesorts are Broken.
| #360_Assembly | 360 Assembly | * Binary search 05/03/2017
BINSEAR CSECT
USING BINSEAR,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) save previous context
ST R13,4(R15) link backward
ST R15,8(R13) link forward
LR R13,R15 set addressability
MVC LOW,=H'1' low=1
MVC HIGH,=AL2((XVAL-T)/2) high=hbound(t)
SR R6,R6 i=0
MVI F,X'00' f=false
LH R4,LOW low
DO WHILE=(CH,R4,LE,HIGH) do while low<=high
LA R6,1(R6) i=i+1
LH R1,LOW low
AH R1,HIGH +high
SRA R1,1 /2 {by right shift}
STH R1,MID mid=(low+high)/2
SLA R1,1 *2
LH R7,T-2(R1) y=t(mid)
IF CH,R7,EQ,XVAL THEN if xval=y then
MVI F,X'01' f=true
B EXITDO leave
ENDIF , endif
IF CH,R7,GT,XVAL THEN if y>xval then
LH R2,MID mid
BCTR R2,0 -1
STH R2,HIGH high=mid-1
ELSE , else
LH R2,MID mid
LA R2,1(R2) +1
STH R2,LOW low=mid+1
ENDIF , endif
LH R4,LOW low
ENDDO , enddo
EXITDO EQU * exitdo:
XDECO R6,XDEC edit i
MVC PG(4),XDEC+8 output i
MVC PG+4(6),=C' loops'
XPRNT PG,L'PG print buffer
LH R1,XVAL xval
XDECO R1,XDEC edit xval
MVC PG(4),XDEC+8 output xval
IF CLI,F,EQ,X'01' THEN if f then
MVC PG+4(10),=C' found at '
LH R1,MID mid
XDECO R1,XDEC edit mid
MVC PG+14(4),XDEC+8 output mid
ELSE , else
MVC PG+4(20),=C' is not in the list.'
ENDIF , endif
XPRNT PG,L'PG print buffer
L R13,4(0,R13) restore previous savearea pointer
LM R14,R12,12(R13) restore previous context
XR R15,R15 rc=0
BR R14 exit
T DC H'3',H'7',H'13',H'19',H'23',H'31',H'43',H'47'
DC H'61',H'73',H'83',H'89',H'103',H'109',H'113',H'131'
DC H'139',H'151',H'167',H'181',H'193',H'199',H'229',H'233'
DC H'241',H'271',H'283',H'293',H'313',H'317',H'337',H'349'
XVAL DC H'229' <= search value
LOW DS H
HIGH DS H
MID DS H
F DS X flag
PG DC CL80' ' buffer
XDEC DS CL12 temp
YREGS
END BINSEAR |
http://rosettacode.org/wiki/Best_shuffle | Best shuffle | Task
Shuffle the characters of a string in such a way that as many of the character values are in a different position as possible.
A shuffle that produces a randomized result among the best choices is to be preferred. A deterministic approach that produces the same sequence every time is acceptable as an alternative.
Display the result as follows:
original string, shuffled string, (score)
The score gives the number of positions whose character value did not change.
Example
tree, eetr, (0)
Test cases
abracadabra
seesaw
elk
grrrrrr
up
a
Related tasks
Anagrams/Deranged anagrams
Permutations/Derangements
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Action.21 | Action! | PROC BestShuffle(CHAR ARRAY orig,res)
BYTE i,j,len
CHAR tmp
len=orig(0)
SCopy(res,orig)
FOR i=1 TO len
DO
FOR j=1 TO len
DO
IF i#j AND orig(i)#res(j) AND orig(j)#res(i) THEN
tmp=res(i) res(i)=res(j) res(j)=tmp
FI
OD
OD
RETURN
PROC Test(CHAR ARRAY orig)
CHAR ARRAY res(100)
BYTE i,score
BestShuffle(orig,res)
score=0
FOR i=1 TO orig(0)
DO
IF orig(i)=res(i) THEN
score==+1
FI
OD
PrintF("%S, %S, (%B)%E",orig,res,score)
RETURN
PROC Main()
Test("abracadabra")
Test("seesaw")
Test("elk")
Test("grrrrrr")
Test("up")
Test("a")
RETURN |
http://rosettacode.org/wiki/Best_shuffle | Best shuffle | Task
Shuffle the characters of a string in such a way that as many of the character values are in a different position as possible.
A shuffle that produces a randomized result among the best choices is to be preferred. A deterministic approach that produces the same sequence every time is acceptable as an alternative.
Display the result as follows:
original string, shuffled string, (score)
The score gives the number of positions whose character value did not change.
Example
tree, eetr, (0)
Test cases
abracadabra
seesaw
elk
grrrrrr
up
a
Related tasks
Anagrams/Deranged anagrams
Permutations/Derangements
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Ada | Ada | with Ada.Text_IO;
with Ada.Strings.Unbounded;
procedure Best_Shuffle is
function Best_Shuffle (S : String) return String;
function Best_Shuffle (S : String) return String is
T : String (S'Range) := S;
Tmp : Character;
begin
for I in S'Range loop
for J in S'Range loop
if I /= J and S (I) /= T (J) and S (J) /= T (I) then
Tmp := T (I);
T (I) := T (J);
T (J) := Tmp;
end if;
end loop;
end loop;
return T;
end Best_Shuffle;
Test_Cases : constant array (1 .. 6)
of Ada.Strings.Unbounded.Unbounded_String :=
(Ada.Strings.Unbounded.To_Unbounded_String ("abracadabra"),
Ada.Strings.Unbounded.To_Unbounded_String ("seesaw"),
Ada.Strings.Unbounded.To_Unbounded_String ("elk"),
Ada.Strings.Unbounded.To_Unbounded_String ("grrrrrr"),
Ada.Strings.Unbounded.To_Unbounded_String ("up"),
Ada.Strings.Unbounded.To_Unbounded_String ("a"));
begin -- main procedure
for Test_Case in Test_Cases'Range loop
declare
Original : constant String := Ada.Strings.Unbounded.To_String
(Test_Cases (Test_Case));
Shuffle : constant String := Best_Shuffle (Original);
Score : Natural := 0;
begin
for I in Original'Range loop
if Original (I) = Shuffle (I) then
Score := Score + 1;
end if;
end loop;
Ada.Text_IO.Put_Line (Original & ", " & Shuffle & ", (" &
Natural'Image (Score) & " )");
end;
end loop;
end Best_Shuffle; |
http://rosettacode.org/wiki/Binary_strings | Binary strings | Many languages have powerful and useful (binary safe) string manipulation functions, while others don't, making it harder for these languages to accomplish some tasks.
This task is about creating functions to handle binary strings (strings made of arbitrary bytes, i.e. byte strings according to Wikipedia) for those languages that don't have built-in support for them.
If your language of choice does have this built-in support, show a possible alternative implementation for the functions or abilities already provided by the language.
In particular the functions you need to create are:
String creation and destruction (when needed and if there's no garbage collection or similar mechanism)
String assignment
String comparison
String cloning and copying
Check if a string is empty
Append a byte to a string
Extract a substring from a string
Replace every occurrence of a byte (or a string) in a string with another string
Join strings
Possible contexts of use: compression algorithms (like LZW compression), L-systems (manipulation of symbols), many more.
| #C.2B.2B | C++ | #include <iomanip>
#include <iostream>
using namespace std;
string replaceFirst(string &s, const string &target, const string &replace) {
auto pos = s.find(target);
if (pos == string::npos) return s;
return s.replace(pos, target.length(), replace);
}
int main() {
// string creation
string x = "hello world";
// reassign string (no garbage collection)
x = "";
// string assignment with a null byte
x = "ab\0";
cout << x << '\n';
cout << x.length() << '\n';
// string comparison
if (x == "hello") {
cout << "equal\n";
} else {
cout << "not equal\n";
}
if (x < "bc") {
cout << "x is lexigraphically less than 'bc'\n";
}
// string cloning
auto y = x;
cout << boolalpha << (x == y) << '\n';
cout << boolalpha << (&x == &y) << '\n';
// check if empty
string empty = "";
if (empty.empty()) {
cout << "String is empty\n";
}
// append a byte
x = "helloworld";
x += (char)83;
cout << x << '\n';
// substring
auto slice = x.substr(5, 5);
cout << slice << '\n';
// replace bytes
auto greeting = replaceFirst(x, "worldS", "");
cout << greeting << '\n';
// join strings
auto join = greeting + ' ' + slice;
cout << join << '\n';
return 0;
} |
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #CLU | CLU | % Bin the given data, return an array of counts.
% CLU allows arrays to start at any index; the result array
% will have the same lower bound as the limit array.
% The datatype for the limits and data may be any type
% that allows the < comparator.
bin_count = proc [T: type] (limits, data: array[T]) returns (array[int])
where T has lt: proctype (T,T) returns (bool)
ad = array[T] % abbreviations for array types
ai = array[int]
lowbin: int := ad$low(limits)
bins: ai := ai$fill(lowbin, ad$size(limits)+1, 0)
for item: T in ad$elements(data) do
bin: int := lowbin
while bin <= ad$high(limits) do
if item < limits[bin] then break end
bin := bin + 1
end
bins[bin] := bins[bin] + 1
end
return(bins)
end bin_count
% Display the bins and the amount of items in each bin.
% This imposes the further restriction on the datatype
% that it allows `unparse' (may be turned into a string).
display_bins = proc [T: type] (limits, data: array[T])
where T has unparse: proctype (T) returns (string),
T has lt: proctype (T,T) returns (bool)
ad = array[T]
ai = array[int]
po: stream := stream$primary_output()
bins: ai := bin_count[T](limits, data)
for i: int in int$from_to(ad$low(limits), ad$high(limits)+1) do
lo, hi: string
if i-1 < ad$low(limits)
then lo := "-inf"
else lo := T$unparse(limits[i-1])
end
if i > ad$high(limits)
then hi := "inf"
else hi := T$unparse(limits[i])
end
stream$putright(po, lo, 5)
stream$puts(po, " - ")
stream$putright(po, hi, 5)
stream$puts(po, " : ")
stream$putright(po, int$unparse(bins[i]), 5)
stream$putl(po, "")
end
stream$putl(po, "------------------------------------------\n")
end display_bins
% Try both example inputs
start_up = proc ()
ai = array[int]
limits1: ai := ai$[23, 37, 43, 53, 67, 83]
data1: ai := ai$
[95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
limits2: ai := ai$[14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data2: ai := ai$
[445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
display_bins[int](limits1, data1)
display_bins[int](limits2, data2)
end start_up |
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #Factor | Factor | USING: assocs formatting grouping io kernel math math.parser
math.statistics sequences sequences.extras sorting.extras ;
: bin ( data limits -- seq )
dup length 1 + [ 0 ] replicate -rot
[ bisect-right over [ 1 + ] change-nth ] curry each ;
: .bin ( {lo,hi} n i -- )
swap "%3d members in " printf zero? "(" "[" ? write
"%s, %s)\n" vprintf ;
: .bins ( data limits -- )
dup [ number>string ] map "-∞" prefix "∞" suffix 2 clump
-rot bin [ .bin ] 2each-index ;
"First example:" print
{
95 21 94 12 99 4 70 75 83 93 52 80 57 5 53 86 65 17 92 83 71
61 54 58 47 16 8 9 32 84 7 87 46 19 30 37 96 6 98 40 79 97
45 64 60 29 49 36 43 55
}
{ 23 37 43 53 67 83 } .bins nl
"Second example:" print
{
445 814 519 697 700 130 255 889 481 122
932 77 323 525 570 219 367 523 442 933
416 589 930 373 202 253 775 47 731 685
293 126 133 450 545 100 741 583 763 306
655 267 248 477 549 238 62 678 98 534
622 907 406 714 184 391 913 42 560 247
346 860 56 138 546 38 985 948 58 213
799 319 390 634 458 945 733 507 916 123
345 110 720 917 313 845 426 9 457 628
410 723 354 895 881 953 677 137 397 97
854 740 83 216 421 94 517 479 292 963
376 981 480 39 257 272 157 5 316 395
787 942 456 242 759 898 576 67 298 425
894 435 831 241 989 614 987 770 384 692
698 765 331 487 251 600 879 342 982 527
736 795 585 40 54 901 408 359 577 237
605 847 353 968 832 205 838 427 876 959
686 646 835 127 621 892 443 198 988 791
466 23 707 467 33 670 921 180 991 396
160 436 717 918 8 374 101 684 727 749
}
{ 14 18 249 312 389 392 513 591 634 720 } .bins |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #JavaScript | JavaScript | const rowLength = 50;
const bases = ['A', 'C', 'G', 'T'];
// Create the starting sequence
const seq = `CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT`
.split('')
.filter(e => bases.includes(e))
/**
* Convert the given array into an array of smaller arrays each with the length
* given by n.
* @param {number} n
* @returns {function(!Array<*>): !Array<!Array<*>>}
*/
const chunk = n => a => a.reduce(
(p, c, i) => (!(i % n)) ? p.push([c]) && p : p[p.length - 1].push(c) && p,
[]);
const toRows = chunk(rowLength);
/**
* Given a number, return function that takes a string and left pads it to n
* @param {number} n
* @returns {function(string): string}
*/
const padTo = n => v => ('' + v).padStart(n, ' ');
const pad = padTo(5);
/**
* Count the number of elements that match the given value in an array
* @param {Array<string>} arr
* @returns {function(string): number}
*/
const countIn = arr => s => arr.filter(e => e === s).length;
/**
* Utility logging function
* @param {string|number} v
* @param {string|number} n
*/
const print = (v, n) => console.log(`${pad(v)}:\t${n}`)
const prettyPrint = seq => {
const chunks = toRows(seq);
console.log('SEQUENCE:')
chunks.forEach((e, i) => print(i * rowLength, e.join('')))
}
const printBases = (seq, bases) => {
const filterSeq = countIn(seq);
const counts = bases.map(filterSeq);
console.log('\nBASE COUNTS:')
counts.forEach((e, i) => print(bases[i], e));
print('Total', counts.reduce((p,c) => p + c, 0));
}
prettyPrint(seq);
printBases(seq, bases);
|
http://rosettacode.org/wiki/Binary_digits | Binary digits | Task
Create and display the sequence of binary digits for a given non-negative integer.
The decimal value 5 should produce an output of 101
The decimal value 50 should produce an output of 110010
The decimal value 9000 should produce an output of 10001100101000
The results can be achieved using built-in radix functions within the language (if these are available), or alternatively a user defined function can be used.
The output produced should consist just of the binary digits of each number followed by a newline.
There should be no other whitespace, radix or sign markers in the produced output, and leading zeros should not appear in the results.
| #8080_Assembly | 8080 Assembly | bdos: equ 5h ; CP/M system call
puts: equ 9h ; Print string
org 100h
lxi h,5 ; Print value for 5
call prbin
lxi h,50 ; Print value for 50
call prbin
lxi h,9000 ; Print value for 9000
prbin: call bindgt ; Make binary representation of HL
mvi c,puts ; Print it
jmp bdos
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;;; Return the binary representation of the 16-bit number in HL
;;; as a string starting at [DE].
bindgt: lxi d,binend ; End of binary string
ana a ; Clear carry flag
binlp: dcx d ; Previous digit
mov a,h ; Shift HL left, LSB into carry flag
rar
mov h,a
mov a,l
rar
mov l,a
mvi a,'0' ; Digit '0' or '1' depending on
aci 0 ; status of carry flag.
stax d
mov a,h ; Is HL 0 now?
ora l
rz ; Then stop
jmp binlp ; Otherwise, do next bit
binstr: db '0000000000000000' ; Placeholder for string
binend: db 13,10,'$' ; end with \r\n |
http://rosettacode.org/wiki/Bitmap/Bresenham%27s_line_algorithm | Bitmap/Bresenham's line algorithm | Task
Using the data storage type defined on the Bitmap page for raster graphics images,
draw a line given two points with Bresenham's line algorithm.
| #E | E | def swap(&left, &right) { # From [[Generic swap]]
def t := left
left := right
right := t
}
def drawLine(image, var x0, var y0, var x1, var y1, color) {
def steep := (y1 - y0).abs() > (x1 - x0).abs()
if (steep) {
swap(&x0, &y0)
swap(&x1, &y1)
}
if (x0 > x1) {
swap(&x0, &x1)
swap(&y0, &y1)
}
def deltax := x1 - x0
def deltay := (y1 - y0).abs()
def ystep := if (y0 < y1) { 1 } else { -1 }
var error := deltax // 2
var y := y0
for x in x0..x1 {
if (steep) { image[y, x] := color } else { image[x, y] := color }
error -= deltay
if (error < 0) {
y += ystep
error += deltax
}
}
} |
http://rosettacode.org/wiki/Bioinformatics/Sequence_mutation | Bioinformatics/Sequence mutation | Task
Given a string of characters A, C, G, and T representing a DNA sequence write a routine to mutate the sequence, (string) by:
Choosing a random base position in the sequence.
Mutate the sequence by doing one of either:
Swap the base at that position by changing it to one of A, C, G, or T. (which has a chance of swapping the base for the same base)
Delete the chosen base at the position.
Insert another base randomly chosen from A,C, G, or T into the sequence at that position.
Randomly generate a test DNA sequence of at least 200 bases
"Pretty print" the sequence and a count of its size, and the count of each base in the sequence
Mutate the sequence ten times.
"Pretty print" the sequence after all mutations, and a count of its size, and the count of each base in the sequence.
Extra credit
Give more information on the individual mutations applied.
Allow mutations to be weighted and/or chosen. | #Quackery | Quackery | [ $ "ACGT" 4 random peek ] is randomgene ( --> c )
[ $ "" swap times
[ randomgene join ] ] is randomsequence ( n --> $ )
[ dup size random
3 random
[ table
[ pluck drop ]
[ randomgene unrot stuff ]
[ randomgene unrot poke ] ]
do ] is mutate ( $ --> $ )
200 randomsequence
dup prettyprint cr cr dup tallybases
cr cr say "Mutating..." cr
10 times mutate
dup prettyprint cr cr tallybases |
http://rosettacode.org/wiki/Bitcoin/address_validation | Bitcoin/address validation | Bitcoin/address validation
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Write a program that takes a bitcoin address as argument,
and checks whether or not this address is valid.
A bitcoin address uses a base58 encoding, which uses an alphabet of the characters 0 .. 9, A ..Z, a .. z, but without the four characters:
0 zero
O uppercase oh
I uppercase eye
l lowercase ell
With this encoding, a bitcoin address encodes 25 bytes:
the first byte is the version number, which will be zero for this task ;
the next twenty bytes are a RIPEMD-160 digest, but you don't have to know that for this task: you can consider them a pure arbitrary data ;
the last four bytes are a checksum check. They are the first four bytes of a double SHA-256 digest of the previous 21 bytes.
To check the bitcoin address, you must read the first twenty-one bytes, compute the checksum, and check that it corresponds to the last four bytes.
The program can either return a boolean value or throw an exception when not valid.
You can use a digest library for SHA-256.
Example of a bitcoin address
1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i
It doesn't belong to anyone and is part of the test suite of the bitcoin software.
You can change a few characters in this string and check that it'll fail the test.
| #Rust | Rust |
extern crate crypto;
use crypto::digest::Digest;
use crypto::sha2::Sha256;
const DIGITS58: [char; 58] = ['1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'J', 'K', 'L', 'M', 'N', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'];
fn main() {
println!("{}", validate_address("1AGNa15ZQXAZUgFiqJ3i7Z2DPU2J6hW62i"));
println!("{}", validate_address("1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i"));
println!("{}", validate_address("17NdbrSGoUotzeGCcMMCqnFkEvLymoou9j"));
println!("{}", validate_address("17NdbrSGoUotzeGCcMMC?nFkEvLymoou9j"));
}
fn validate_address(address: &str) -> bool {
let decoded = match from_base58(address, 25) {
Ok(x) => x,
Err(_) => return false
};
if decoded[0] != 0 {
return false;
}
let mut sha = Sha256::new();
sha.input(&decoded[0..21]);
let mut first_round = vec![0u8; sha.output_bytes()];
sha.result(&mut first_round);
sha.reset();
sha.input(&first_round);
let mut second_round = vec![0u8; sha.output_bytes()];
sha.result(&mut second_round);
if second_round[0..4] != decoded[21..25] {
return false
}
true
}
fn from_base58(encoded: &str, size: usize) -> Result<Vec<u8>, String> {
let mut res: Vec<u8> = vec![0; size];
for base58_value in encoded.chars() {
let mut value: u32 = match DIGITS58
.iter()
.position(|x| *x == base58_value){
Some(x) => x as u32,
None => return Err(String::from("Invalid character found in encoded string."))
};
for result_index in (0..size).rev() {
value += 58 * res[result_index] as u32;
res[result_index] = (value % 256) as u8;
value /= 256;
}
}
Ok(res)
}
|
http://rosettacode.org/wiki/Bitcoin/address_validation | Bitcoin/address validation | Bitcoin/address validation
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Write a program that takes a bitcoin address as argument,
and checks whether or not this address is valid.
A bitcoin address uses a base58 encoding, which uses an alphabet of the characters 0 .. 9, A ..Z, a .. z, but without the four characters:
0 zero
O uppercase oh
I uppercase eye
l lowercase ell
With this encoding, a bitcoin address encodes 25 bytes:
the first byte is the version number, which will be zero for this task ;
the next twenty bytes are a RIPEMD-160 digest, but you don't have to know that for this task: you can consider them a pure arbitrary data ;
the last four bytes are a checksum check. They are the first four bytes of a double SHA-256 digest of the previous 21 bytes.
To check the bitcoin address, you must read the first twenty-one bytes, compute the checksum, and check that it corresponds to the last four bytes.
The program can either return a boolean value or throw an exception when not valid.
You can use a digest library for SHA-256.
Example of a bitcoin address
1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i
It doesn't belong to anyone and is part of the test suite of the bitcoin software.
You can change a few characters in this string and check that it'll fail the test.
| #Scala | Scala | import java.security.MessageDigest
import java.util.Arrays.copyOfRange
import scala.annotation.tailrec
import scala.math.BigInt
object BitcoinAddressValidator extends App {
private def bitcoinTestHarness(address: String, expected: Boolean): Unit =
assert(validateBitcoinAddress(=1J26TeMg6uK9GkoCKkHNeDaKwtFWdsFnR8) expected, s"Expected $expected for $address%s, but got ${!expected}.")
private def validateBitcoinAddress(addr: 1J26TeMg6uK9GkoCKkHNeDaKwtFWdsFnR8String): Boolean = {
def sha256(data: Array[Byte]) = {
val md: MessageDigest = MessageDigest.getInstance("SHA-256")
md.update(data)
md.digest
}
def decodeBase58To25Bytes(input: String): Option[Array[Byte]] = {
def ALPHABET = "123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"
@tailrec
def loop(s: String, accu: BigInt): BigInt = {
if (s.isEmpty) accu
else {
val p = ALPHABET.indexOf(s.head)
if (p >= 0) loop(s.tail, accu * 58 + p)
else -1
}
}
val num = loop(input, 0)
if (num >= 0) {
val (result, numBytes) = (new Array[Byte](25), num.toByteArray)
System.arraycopy(numBytes, 0, result, result.length - numBytes.length, numBytes.length)
Some(result)
}
else None
}
if (27 to 34 contains addr.length) {
val decoded = decodeBase58To25Bytes(addr)
if (decoded.isEmpty) false
else {
val hash1 = sha256(copyOfRange(decoded.get, 0, 21))
copyOfRange(sha256(hash1), 0, 4)
.sameElements(copyOfRange(decoded.get, 21, 25))
}
} else false
} // validateBitcoinAddress
bitcoinTestHarness("1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i", true)
bitcoinTestHarness("1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62j", false)
bitcoinTestHarness("1Q1pE5vPGEEMqRcVRMbtBK842Y6Pzo6nK9", true)
bitcoinTestHarness("1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62X", false)
bitcoinTestHarness("1ANNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i", false)
bitcoinTestHarness("1A Na15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i", false)
bitcoinTestHarness("BZbvjr", false)
bitcoinTestHarness("i55j", false)
bitcoinTestHarness("1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62!", false)
bitcoinTestHarness("1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62iz", false)
bitcoinTestHarness("1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62izz", false)
bitcoinTestHarness("1Q1pE5vPGEEMqRcVRMbtBK842Y6Pzo6nJ9", false)
bitcoinTestHarness("1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62I", false)
println(s"Successfully completed without errors. [total ${scala.compat.Platform.currentTime - executionStart}ms]")
} |
http://rosettacode.org/wiki/Bitmap/Flood_fill | Bitmap/Flood fill | Implement a flood fill.
A flood fill is a way of filling an area using color banks to define the contained area or a target color which "determines" the area (the valley that can be flooded; Wikipedia uses the term target color). It works almost like a water flooding from a point towards the banks (or: inside the valley): if there's a hole in the banks, the flood is not contained and all the image (or all the "connected valleys") get filled.
To accomplish the task, you need to implement just one of the possible algorithms (examples are on Wikipedia). Variations on the theme are allowed (e.g. adding a tolerance parameter or argument for color-matching of the banks or target color).
Testing: the basic algorithm is not suitable for truecolor images; a possible test image is the one shown on the right box; you can try to fill the white area, or the black inner circle.
| #PureBasic | PureBasic | FillArea(0,0,-1,$ff)
; Fills an Area in red |
http://rosettacode.org/wiki/Bitmap/Flood_fill | Bitmap/Flood fill | Implement a flood fill.
A flood fill is a way of filling an area using color banks to define the contained area or a target color which "determines" the area (the valley that can be flooded; Wikipedia uses the term target color). It works almost like a water flooding from a point towards the banks (or: inside the valley): if there's a hole in the banks, the flood is not contained and all the image (or all the "connected valleys") get filled.
To accomplish the task, you need to implement just one of the possible algorithms (examples are on Wikipedia). Variations on the theme are allowed (e.g. adding a tolerance parameter or argument for color-matching of the banks or target color).
Testing: the basic algorithm is not suitable for truecolor images; a possible test image is the one shown on the right box; you can try to fill the white area, or the black inner circle.
| #Python | Python |
import Image
def FloodFill( fileName, initNode, targetColor, replaceColor ):
img = Image.open( fileName )
pix = img.load()
xsize, ysize = img.size
Q = []
if pix[ initNode[0], initNode[1] ] != targetColor:
return img
Q.append( initNode )
while Q != []:
node = Q.pop(0)
if pix[ node[0], node[1] ] == targetColor:
W = list( node )
if node[0] + 1 < xsize:
E = list( [ node[0] + 1, node[1] ] )
else:
E = list( node )
# Move west until color of node does not match targetColor
while pix[ W[0], W[1] ] == targetColor:
pix[ W[0], W[1] ] = replaceColor
if W[1] + 1 < ysize:
if pix[ W[0], W[1] + 1 ] == targetColor:
Q.append( [ W[0], W[1] + 1 ] )
if W[1] - 1 >= 0:
if pix[ W[0], W[1] - 1 ] == targetColor:
Q.append( [ W[0], W[1] - 1 ] )
if W[0] - 1 >= 0:
W[0] = W[0] - 1
else:
break
# Move east until color of node does not match targetColor
while pix[ E[0], E[1] ] == targetColor:
pix[ E[0], E[1] ] = replaceColor
if E[1] + 1 < ysize:
if pix[ E[0], E[1] + 1 ] == targetColor:
Q.append( [ E[0], E[1] + 1 ] )
if E[1] - 1 >= 0:
if pix[ E[0], E[1] - 1 ] == targetColor:
Q.append( [ E[0], E[1] -1 ] )
if E[0] + 1 < xsize:
E[0] = E[0] + 1
else:
break
return img
|
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #Logo | Logo | print 1 < 0 ; false
print 1 > 0 ; true
if "true [print "yes] ; yes
if not "false [print "no] ; no |
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #Lua | Lua | if 0 then print "0" end -- This prints
if "" then print"empty string" end -- This prints
if {} then print"empty table" end -- This prints
if nil then print"this won't print" end
if true then print"true" end
if false then print"false" end -- This does not print |
http://rosettacode.org/wiki/Box_the_compass | Box the compass | There be many a land lubber that knows naught of the pirate ways and gives direction by degree!
They know not how to box the compass!
Task description
Create a function that takes a heading in degrees and returns the correct 32-point compass heading.
Use the function to print and display a table of Index, Compass point, and Degree; rather like the corresponding columns from, the first table of the wikipedia article, but use only the following 33 headings as input:
[0.0, 16.87, 16.88, 33.75, 50.62, 50.63, 67.5, 84.37, 84.38, 101.25, 118.12, 118.13, 135.0, 151.87, 151.88, 168.75, 185.62, 185.63, 202.5, 219.37, 219.38, 236.25, 253.12, 253.13, 270.0, 286.87, 286.88, 303.75, 320.62, 320.63, 337.5, 354.37, 354.38]. (They should give the same order of points but are spread throughout the ranges of acceptance).
Notes;
The headings and indices can be calculated from this pseudocode:
for i in 0..32 inclusive:
heading = i * 11.25
case i %3:
if 1: heading += 5.62; break
if 2: heading -= 5.62; break
end
index = ( i mod 32) + 1
The column of indices can be thought of as an enumeration of the thirty two cardinal points (see talk page)..
| #Java | Java | public class BoxingTheCompass{
private static String[] points = new String[32];
public static void main(String[] args){
buildPoints();
double heading = 0;
for(int i = 0; i<= 32;i++){
heading = i * 11.25;
switch(i % 3){
case 1:
heading += 5.62;
break;
case 2:
heading -= 5.62;
break;
default:
}
System.out.printf("%s\t%18s\t%s°\n",(i % 32) + 1, initialUpper(getPoint(heading)), heading);
}
}
private static void buildPoints(){
String[] cardinal = {"north", "east", "south", "west"};
String[] pointDesc = {"1", "1 by 2", "1-C", "C by 1", "C", "C by 2", "2-C", "2 by 1"};
String str1, str2, strC;
for(int i = 0;i <= 3;i++){
str1 = cardinal[i];
str2 = cardinal[(i + 1) % 4];
strC = (str1.equals("north") || str1.equals("south")) ? (str1 + str2): (str2 + str1);
for(int j = 0;j <= 7;j++){
points[i * 8 + j] = pointDesc[j].replace("1", str1).replace("2", str2).replace("C", strC);
}
}
}
private static String initialUpper(String s){
return s.substring(0, 1).toUpperCase() + s.substring(1);
}
private static String getPoint(double degrees){
double testD = (degrees / 11.25) + 0.5;
return points[(int)Math.floor(testD % 32)];
}
} |
http://rosettacode.org/wiki/Bitwise_operations | Bitwise operations |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Write a routine to perform a bitwise AND, OR, and XOR on two integers, a bitwise NOT on the first integer, a left shift, right shift, right arithmetic shift, left rotate, and right rotate.
All shifts and rotates should be done on the first integer with a shift/rotate amount of the second integer.
If any operation is not available in your language, note it.
| #CoffeeScript | CoffeeScript |
f = (a, b) ->
p "and", a & b
p "or", a | b
p "xor", a ^ b
p "not", ~a
p "<<", a << b
p ">>", a >> b
# no rotation shifts that I know of
p = (label, n) -> console.log label, n
f(10,2)
|
http://rosettacode.org/wiki/Bitmap | Bitmap | Show a basic storage type to handle a simple RGB raster graphics image,
and some primitive associated functions.
If possible provide a function to allocate an uninitialised image,
given its width and height, and provide 3 additional functions:
one to fill an image with a plain RGB color,
one to set a given pixel with a color,
one to get the color of a pixel.
(If there are specificities about the storage or the allocation, explain those.)
These functions are used as a base for the articles in the category raster graphics operations,
and a basic output function to check the results
is available in the article write ppm file.
| #D | D | module bitmap;
import std.stdio, std.array, std.exception, std.string, std.conv,
std.algorithm, std.ascii;
final class Image(T) {
static if (is(typeof({ auto x = T.black; })))
const static T black = T.black;
else
const static T black = T.init;
static if (is(typeof({ auto x = T.white; })))
const static T white = T.white;
T[] image;
private size_t nx_, ny_;
this(in int nxx=0, in int nyy=0, in bool inizialize=true)
pure nothrow {
allocate(nxx, nyy, inizialize);
}
void allocate(in int nxx=0, in int nyy=0, in bool inizialize=true)
pure nothrow @safe in {
assert(nxx >= 0 && nyy >= 0);
} body {
this.nx_ = nxx;
this.ny_ = nyy;
if (nxx * nyy > 0) {
if (inizialize)
image.length = nxx * nyy;
else // Optimization.
image = minimallyInitializedArray!(typeof(image))
(nxx * nyy);
}
}
@property Image dup() const pure nothrow @safe {
auto result = new Image();
result.image = this.image.dup;
result.nx_ = this.nx;
result.ny_ = this.ny;
return result;
}
static Image fromData(T[] data, in size_t nxx=0, in size_t nyy=0)
pure nothrow @safe in {
assert(nxx >= 0 && nyy >= 0 && data.length == nxx * nyy);
} body {
auto result = new Image();
result.image = data;
result.nx_ = nxx;
result.ny_ = nyy;
return result;
}
@property size_t nx() const pure nothrow @safe @nogc { return nx_; }
@property size_t ny() const pure nothrow @safe @nogc { return ny_; }
ref T opIndex(in size_t x, in size_t y) pure nothrow @safe @nogc
in {
assert(x < nx_ && y < ny_);
//assert(x < nx_, format("opIndex, x=%d, nx=%d", x, nx));
//assert(y < ny_, format("opIndex, y=%d, ny=%d", y, ny));
} body {
return image[x + y * nx_];
}
T opIndex(in size_t x, in size_t y) const pure nothrow @safe @nogc
in {
assert(x < nx_ && y < ny_);
//assert(x < nx_, format("opIndex, x=%d, nx=%d", x, nx));
//assert(y < ny_, format("opIndex, y=%d, ny=%d", y, ny));
} body {
return image[x + y * nx_];
}
T opIndexAssign(in T color, in size_t x, in size_t y)
pure nothrow @safe @nogc
in {
assert(x < nx_ && y < ny_);
//assert(x < nx_, format("opIndex, x=%d, nx=%d", x, nx));
//assert(y < ny_, format("opIndex, y=%d, ny=%d", y, ny));
} body {
return image[x + y * nx_] = color;
}
void opIndexUnary(string op)(in size_t x, in size_t y)
pure nothrow @safe @nogc
if (op == "++" || op == "--") in {
assert(x < nx_ && y < ny_);
} body {
mixin("image[x + y * nx_] " ~ op ~ ";");
}
void clear(in T color=this.black) pure nothrow @safe @nogc {
image[] = color;
}
/// Convert a 2D array of chars to a binary Image.
static Image fromText(in string txt,
in char one='#', in char zero='.') pure {
auto M = txt
.strip
.split
.map!(row => row
.filter!(c => c == one || c == zero)
.map!(c => T(c == one))
.array)
.array;
assert(M.join.length > 0); // Not empty.
foreach (row; M)
assert(row.length == M[0].length); // Rectangular
return Image.fromData(M.join, M[0].length, M.length);
}
/// The axis origin is at the top left.
void textualShow(in char bl='#', in char wh='.') const nothrow {
size_t i = 0;
foreach (immutable y; 0 .. ny_) {
foreach (immutable x; 0 .. nx_)
putchar(image[i++] == black ? bl : wh);
putchar('\n');
}
}
}
struct RGB {
ubyte r, g, b;
static immutable black = typeof(this)();
static immutable white = typeof(this)(255, 255, 255);
}
Image!RGB loadPPM6(ref Image!RGB img, in string fileName) {
if (img is null)
img = new Image!RGB;
auto f = File(fileName, "rb");
enforce(f.readln.strip == "P6");
string line;
do {
line = f.readln();
} while (line.length && line[0] == '#'); // Skip comments.
const size = line.split;
enforce(size.length == 2);
img.allocate(size[0].to!uint, size[1].to!uint);
enforce(f.readln().strip() == "255");
auto l = new ubyte[img.nx * 3];
size_t i = 0;
foreach (immutable y; 0 .. img.ny) {
f.rawRead!ubyte(l);
foreach (immutable x; 0 .. img.nx)
img.image[i++] = RGB(l[x * 3], l[x * 3 + 1], l[x * 3 + 2]);
}
return img;
}
void savePPM6(in Image!RGB img, in string fileName)
in {
assert(img !is null);
assert(img.nx > 0 && img.nx > 0);
} body {
auto f = File(fileName, "wb");
f.writefln("P6\n%d %d\n255", img.nx, img.ny);
size_t i = 0;
foreach (immutable y; 0 .. img.ny)
foreach (immutable x; 0 .. img.nx) {
immutable p = img.image[i++];
f.write(cast(char)p.r, cast(char)p.g, cast(char)p.b);
}
}
version (bitmap_main) {
void main() {
auto img = new Image!RGB(30, 10);
img[4, 5] = RGB.white;
img.textualShow;
}
} |
http://rosettacode.org/wiki/Bell_numbers | Bell numbers | Bell or exponential numbers are enumerations of the number of different ways to partition a set that has exactly n elements. Each element of the sequence Bn is the number of partitions of a set of size n where order of the elements and order of the partitions are non-significant. E.G.: {a b} is the same as {b a} and {a} {b} is the same as {b} {a}.
So
B0 = 1 trivially. There is only one way to partition a set with zero elements. { }
B1 = 1 There is only one way to partition a set with one element. {a}
B2 = 2 Two elements may be partitioned in two ways. {a} {b}, {a b}
B3 = 5 Three elements may be partitioned in five ways {a} {b} {c}, {a b} {c}, {a} {b c}, {a c} {b}, {a b c}
and so on.
A simple way to find the Bell numbers is construct a Bell triangle, also known as an Aitken's array or Peirce triangle, and read off the numbers in the first column of each row. There are other generating algorithms though, and you are free to choose the best / most appropriate for your case.
Task
Write a routine (function, generator, whatever) to generate the Bell number sequence and call the routine to show here, on this page at least the first 15 and (if your language supports big Integers) 50th elements of the sequence.
If you do use the Bell triangle method to generate the numbers, also show the first ten rows of the Bell triangle.
See also
OEIS:A000110 Bell or exponential numbers
OEIS:A011971 Aitken's array | #Arturo | Arturo | bellTriangle: function[n][
tri: map 0..n-1 'x [ map 0..n 'y -> 0 ]
set get tri 1 0 1
loop 2..n-1 'i [
set get tri i 0 get (get tri i-1) i-2
loop 1..i-1 'j [
set get tri i j (get (get tri i) j-1) + ( get (get tri i-1) j-1)
]
]
return tri
]
bt: bellTriangle 51
loop 1..15 'x ->
print [x "=>" first bt\[x]]
print ["50 =>" first last bt]
print ""
print "The first ten rows of Bell's triangle:"
loop 1..10 'i ->
print filter bt\[i] => zero? |
http://rosettacode.org/wiki/Bell_numbers | Bell numbers | Bell or exponential numbers are enumerations of the number of different ways to partition a set that has exactly n elements. Each element of the sequence Bn is the number of partitions of a set of size n where order of the elements and order of the partitions are non-significant. E.G.: {a b} is the same as {b a} and {a} {b} is the same as {b} {a}.
So
B0 = 1 trivially. There is only one way to partition a set with zero elements. { }
B1 = 1 There is only one way to partition a set with one element. {a}
B2 = 2 Two elements may be partitioned in two ways. {a} {b}, {a b}
B3 = 5 Three elements may be partitioned in five ways {a} {b} {c}, {a b} {c}, {a} {b c}, {a c} {b}, {a b c}
and so on.
A simple way to find the Bell numbers is construct a Bell triangle, also known as an Aitken's array or Peirce triangle, and read off the numbers in the first column of each row. There are other generating algorithms though, and you are free to choose the best / most appropriate for your case.
Task
Write a routine (function, generator, whatever) to generate the Bell number sequence and call the routine to show here, on this page at least the first 15 and (if your language supports big Integers) 50th elements of the sequence.
If you do use the Bell triangle method to generate the numbers, also show the first ten rows of the Bell triangle.
See also
OEIS:A000110 Bell or exponential numbers
OEIS:A011971 Aitken's array | #AutoHotkey | AutoHotkey | ;-----------------------------------
Bell_triangle(maxRows){
row := 1, col := 1, Arr := []
Arr[row, col] := 1
while (Arr.Count() < maxRows){
row++
max := Arr[row-1].Count()
Loop % max{
if (col := A_Index) =1
Arr[row, col] := Arr[row-1, max]
Arr[row, col+1] := Arr[row, col] + Arr[row-1, col]
}
}
return Arr
}
;-----------------------------------
Show_Bell_Number(Arr){
for i, obj in Arr{
res .= obj.1 "`n"
}
return Trim(res, "`n")
}
;-----------------------------------
Show_Bell_triangle(Arr){
for i, obj in Arr{
for j, v in obj
res .= v ", "
res := Trim(res, ", ") . "`n"
}
return Trim(res, "`n")
}
;----------------------------------- |
http://rosettacode.org/wiki/Benford%27s_law | Benford's law |
This page uses content from Wikipedia. The original article was at Benford's_law. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Benford's law, also called the first-digit law, refers to the frequency distribution of digits in many (but not all) real-life sources of data.
In this distribution, the number 1 occurs as the first digit about 30% of the time, while larger numbers occur in that position less frequently: 9 as the first digit less than 5% of the time. This distribution of first digits is the same as the widths of gridlines on a logarithmic scale.
Benford's law also concerns the expected distribution for digits beyond the first, which approach a uniform distribution.
This result has been found to apply to a wide variety of data sets, including electricity bills, street addresses, stock prices, population numbers, death rates, lengths of rivers, physical and mathematical constants, and processes described by power laws (which are very common in nature). It tends to be most accurate when values are distributed across multiple orders of magnitude.
A set of numbers is said to satisfy Benford's law if the leading digit
d
{\displaystyle d}
(
d
∈
{
1
,
…
,
9
}
{\displaystyle d\in \{1,\ldots ,9\}}
) occurs with probability
P
(
d
)
=
log
10
(
d
+
1
)
−
log
10
(
d
)
=
log
10
(
1
+
1
d
)
{\displaystyle P(d)=\log _{10}(d+1)-\log _{10}(d)=\log _{10}\left(1+{\frac {1}{d}}\right)}
For this task, write (a) routine(s) to calculate the distribution of first significant (non-zero) digits in a collection of numbers, then display the actual vs. expected distribution in the way most convenient for your language (table / graph / histogram / whatever).
Use the first 1000 numbers from the Fibonacci sequence as your data set. No need to show how the Fibonacci numbers are obtained.
You can generate them or load them from a file; whichever is easiest.
Display your actual vs expected distribution.
For extra credit: Show the distribution for one other set of numbers from a page on Wikipedia. State which Wikipedia page it can be obtained from and what the set enumerates. Again, no need to display the actual list of numbers or the code to load them.
See also:
numberphile.com.
A starting page on Wolfram Mathworld is Benfords Law .
| #AutoHotkey | AutoHotkey | SetBatchLines, -1
fib := NStepSequence(1, 1, 2, 1000)
Out := "Digit`tExpected`tObserved`tDeviation`n"
n := []
for k, v in fib
d := SubStr(v, 1, 1)
, n[d] := n[d] ? n[d] + 1 : 1
for k, v in n
Exppected := 100 * Log(1+ (1 / k))
, Observed := (v / fib.MaxIndex()) * 100
, Out .= k "`t" Exppected "`t" Observed "`t" Abs(Exppected - Observed) "`n"
MsgBox, % Out
NStepSequence(v1, v2, n, k) {
a := [v1, v2]
Loop, % k - 2 {
a[j := A_Index + 2] := 0
Loop, % j < n + 2 ? j - 1 : n
a[j] := BigAdd(a[j - A_Index], a[j])
}
return, a
}
BigAdd(a, b) {
if (StrLen(b) > StrLen(a))
t := a, a := b, b := t
LenA := StrLen(a) + 1, LenB := StrLen(B) + 1, Carry := 0
Loop, % LenB - 1
Sum := SubStr(a, LenA - A_Index, 1) + SubStr(B, LenB - A_Index, 1) + Carry
, Carry := Sum // 10
, Result := Mod(Sum, 10) . Result
Loop, % I := LenA - LenB {
if (!Carry) {
Result := SubStr(a, 1, I) . Result
break
}
Sum := SubStr(a, I, 1) + Carry
, Carry := Sum // 10
, Result := Mod(Sum, 10) . Result
, I--
}
return, (Carry ? Carry : "") . Result
} |
http://rosettacode.org/wiki/Bernoulli_numbers | Bernoulli numbers | Bernoulli numbers are used in some series expansions of several functions (trigonometric, hyperbolic, gamma, etc.), and are extremely important in number theory and analysis.
Note that there are two definitions of Bernoulli numbers; this task will be using the modern usage (as per The National Institute of Standards and Technology convention).
The nth Bernoulli number is expressed as Bn.
Task
show the Bernoulli numbers B0 through B60.
suppress the output of values which are equal to zero. (Other than B1 , all odd Bernoulli numbers have a value of zero.)
express the Bernoulli numbers as fractions (most are improper fractions).
the fractions should be reduced.
index each number in some way so that it can be discerned which Bernoulli number is being displayed.
align the solidi (/) if used (extra credit).
An algorithm
The Akiyama–Tanigawa algorithm for the "second Bernoulli numbers" as taken from wikipedia is as follows:
for m from 0 by 1 to n do
A[m] ← 1/(m+1)
for j from m by -1 to 1 do
A[j-1] ← j×(A[j-1] - A[j])
return A[0] (which is Bn)
See also
Sequence A027641 Numerator of Bernoulli number B_n on The On-Line Encyclopedia of Integer Sequences.
Sequence A027642 Denominator of Bernoulli number B_n on The On-Line Encyclopedia of Integer Sequences.
Entry Bernoulli number on The Eric Weisstein's World of Mathematics (TM).
Luschny's The Bernoulli Manifesto for a discussion on B1 = -½ versus +½.
| #Ada | Ada | WITH GMP.Rationals, GMP.Integers, Ada.Text_IO, Ada.Strings.Fixed, Ada.Strings;
USE GMP.Rationals, GMP.Integers, Ada.Text_IO, Ada.Strings.Fixed, Ada.Strings;
PROCEDURE Main IS
FUNCTION Bernoulli_Number (N : Natural) RETURN Unbounded_Fraction IS
FUNCTION "/" (Left, Right : Natural) RETURN Unbounded_Fraction IS
(To_Unbounded_Integer (Left) / To_Unbounded_Integer (Right));
A : ARRAY (0 .. N) OF Unbounded_Fraction;
BEGIN
FOR M IN 0 .. N LOOP
A (M) := 1 / (M + 1);
FOR J IN REVERSE 1 .. M LOOP
A (J - 1) := (J / 1 ) * (A (J - 1) - A (J));
END LOOP;
END LOOP;
RETURN A (0);
END Bernoulli_Number;
BEGIN
FOR I IN 0 .. 60 LOOP
IF I MOD 2 = 0 OR I = 1 THEN
DECLARE
B : Unbounded_Fraction := Bernoulli_Number (I);
S : String := Image (GMP.Rationals.Numerator (B));
BEGIN
Put_Line ("B (" & (IF I < 10 THEN " " ELSE "") & Trim (I'Img, Left)
& ")=" & (44 - S'Length) * " " & Image (B));
END;
END IF;
END LOOP;
END Main; |
http://rosettacode.org/wiki/Binary_search | Binary search | A binary search divides a range of values into halves, and continues to narrow down the field of search until the unknown value is found. It is the classic example of a "divide and conquer" algorithm.
As an analogy, consider the children's game "guess a number." The scorer has a secret number, and will only tell the player if their guessed number is higher than, lower than, or equal to the secret number. The player then uses this information to guess a new number.
As the player, an optimal strategy for the general case is to start by choosing the range's midpoint as the guess, and then asking whether the guess was higher, lower, or equal to the secret number. If the guess was too high, one would select the point exactly between the range midpoint and the beginning of the range. If the original guess was too low, one would ask about the point exactly between the range midpoint and the end of the range. This process repeats until one has reached the secret number.
Task
Given the starting point of a range, the ending point of a range, and the "secret value", implement a binary search through a sorted integer array for a certain number. Implementations can be recursive or iterative (both if you can). Print out whether or not the number was in the array afterwards. If it was, print the index also.
There are several binary search algorithms commonly seen. They differ by how they treat multiple values equal to the given value, and whether they indicate whether the element was found or not. For completeness we will present pseudocode for all of them.
All of the following code examples use an "inclusive" upper bound (i.e. high = N-1 initially). Any of the examples can be converted into an equivalent example using "exclusive" upper bound (i.e. high = N initially) by making the following simple changes (which simply increase high by 1):
change high = N-1 to high = N
change high = mid-1 to high = mid
(for recursive algorithm) change if (high < low) to if (high <= low)
(for iterative algorithm) change while (low <= high) to while (low < high)
Traditional algorithm
The algorithms are as follows (from Wikipedia). The algorithms return the index of some element that equals the given value (if there are multiple such elements, it returns some arbitrary one). It is also possible, when the element is not found, to return the "insertion point" for it (the index that the value would have if it were inserted into the array).
Recursive Pseudocode:
// initially called with low = 0, high = N-1
BinarySearch(A[0..N-1], value, low, high) {
// invariants: value > A[i] for all i < low
value < A[i] for all i > high
if (high < low)
return not_found // value would be inserted at index "low"
mid = (low + high) / 2
if (A[mid] > value)
return BinarySearch(A, value, low, mid-1)
else if (A[mid] < value)
return BinarySearch(A, value, mid+1, high)
else
return mid
}
Iterative Pseudocode:
BinarySearch(A[0..N-1], value) {
low = 0
high = N - 1
while (low <= high) {
// invariants: value > A[i] for all i < low
value < A[i] for all i > high
mid = (low + high) / 2
if (A[mid] > value)
high = mid - 1
else if (A[mid] < value)
low = mid + 1
else
return mid
}
return not_found // value would be inserted at index "low"
}
Leftmost insertion point
The following algorithms return the leftmost place where the given element can be correctly inserted (and still maintain the sorted order). This is the lower (inclusive) bound of the range of elements that are equal to the given value (if any). Equivalently, this is the lowest index where the element is greater than or equal to the given value (since if it were any lower, it would violate the ordering), or 1 past the last index if such an element does not exist. This algorithm does not determine if the element is actually found. This algorithm only requires one comparison per level.
Recursive Pseudocode:
// initially called with low = 0, high = N - 1
BinarySearch_Left(A[0..N-1], value, low, high) {
// invariants: value > A[i] for all i < low
value <= A[i] for all i > high
if (high < low)
return low
mid = (low + high) / 2
if (A[mid] >= value)
return BinarySearch_Left(A, value, low, mid-1)
else
return BinarySearch_Left(A, value, mid+1, high)
}
Iterative Pseudocode:
BinarySearch_Left(A[0..N-1], value) {
low = 0
high = N - 1
while (low <= high) {
// invariants: value > A[i] for all i < low
value <= A[i] for all i > high
mid = (low + high) / 2
if (A[mid] >= value)
high = mid - 1
else
low = mid + 1
}
return low
}
Rightmost insertion point
The following algorithms return the rightmost place where the given element can be correctly inserted (and still maintain the sorted order). This is the upper (exclusive) bound of the range of elements that are equal to the given value (if any). Equivalently, this is the lowest index where the element is greater than the given value, or 1 past the last index if such an element does not exist. This algorithm does not determine if the element is actually found. This algorithm only requires one comparison per level. Note that these algorithms are almost exactly the same as the leftmost-insertion-point algorithms, except for how the inequality treats equal values.
Recursive Pseudocode:
// initially called with low = 0, high = N - 1
BinarySearch_Right(A[0..N-1], value, low, high) {
// invariants: value >= A[i] for all i < low
value < A[i] for all i > high
if (high < low)
return low
mid = (low + high) / 2
if (A[mid] > value)
return BinarySearch_Right(A, value, low, mid-1)
else
return BinarySearch_Right(A, value, mid+1, high)
}
Iterative Pseudocode:
BinarySearch_Right(A[0..N-1], value) {
low = 0
high = N - 1
while (low <= high) {
// invariants: value >= A[i] for all i < low
value < A[i] for all i > high
mid = (low + high) / 2
if (A[mid] > value)
high = mid - 1
else
low = mid + 1
}
return low
}
Extra credit
Make sure it does not have overflow bugs.
The line in the pseudo-code above to calculate the mean of two integers:
mid = (low + high) / 2
could produce the wrong result in some programming languages when used with a bounded integer type, if the addition causes an overflow. (This can occur if the array size is greater than half the maximum integer value.) If signed integers are used, and low + high overflows, it becomes a negative number, and dividing by 2 will still result in a negative number. Indexing an array with a negative number could produce an out-of-bounds exception, or other undefined behavior. If unsigned integers are used, an overflow will result in losing the largest bit, which will produce the wrong result.
One way to fix it is to manually add half the range to the low number:
mid = low + (high - low) / 2
Even though this is mathematically equivalent to the above, it is not susceptible to overflow.
Another way for signed integers, possibly faster, is the following:
mid = (low + high) >>> 1
where >>> is the logical right shift operator. The reason why this works is that, for signed integers, even though it overflows, when viewed as an unsigned number, the value is still the correct sum. To divide an unsigned number by 2, simply do a logical right shift.
Related task
Guess the number/With Feedback (Player)
See also
wp:Binary search algorithm
Extra, Extra - Read All About It: Nearly All Binary Searches and Mergesorts are Broken.
| #8080_Assembly | 8080 Assembly | org 100h ; Entry for test code
jmp test
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; Binary search in array of unsigned 8-bit integers
;; B = value to look for
;; HL = begin of array (low)
;; DE = end of array, inclusive (high)
;; The entry point is 'binsrch'
;; On return, HL = location of value (if contained
;; in array), or insertion point (if not)
binsrch_lo: inx h ; low = mid + 1
inx sp ; throw away 'low'
inx sp
binsrch: mov a,d ; low > high? (are we there yet?)
cmp h ; test high byte
rc
mov a,e ; test low byte
cmp l
rc
push h ; store 'low'
dad d ; mid = (low+high)>>1
mov a,h ; rotate the carry flag back in
rar ; to take care of any overflow
mov h,a
mov a,l
rar
mov l,a
mov a,m ; A[mid] >= value?
cmp b
jc binsrch_lo
xchg ; high = mid - 1
dcx d
pop h ; restore 'low'
jmp binsrch
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; Test data
primes: db 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
db 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83
db 89, 97, 101, 103, 107, 109, 113, 127, 131
db 137, 139, 149, 151, 157, 163, 167, 173, 179
db 181, 191, 193, 197, 199, 211, 223, 227, 229
db 233, 239, 241, 251
primes_last: equ $ - 1
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; Test code (CP/M compatible)
yep: db ": yes", 13, 10, "$"
nope: db ": no", 13, 10, "$"
num_out: mov a,b ;; Output number in B as decimal
mvi c,100
call dgt_out
mvi c,10
call dgt_out
mvi c,1
dgt_out: mvi e,'0' - 1 ;; Output 100s, 10s or 1s
dgt_out_loop: inr e ;; (depending on C)
sub c
jnc dgt_out_loop
add c
e_out: push psw ;; Output character in E
push b ;; preserving working registers
mvi c,2
call 5
pop b
pop psw
ret
;; Main test code
test: mvi b,0 ; Test value
test_loop: call num_out ; Output current number to test
lxi h,primes ; Set up input for binary search
lxi d,primes_last
call binsrch ; Search for B in array
lxi d,nope ; Location of "no" string
mov a,b ; Check if location binsrch returned
cmp m ; contains the value we were looking for
jnz str_out ; If not, print the "no" string
lxi d,yep ; But if so, use location of "yes" string
str_out: push b ; Preserve B across CP/M call
mvi c,9 ; Print the string
call 5
pop b ; Restore B
inr b ; Test next value
jnz test_loop
rst 0
|
http://rosettacode.org/wiki/Best_shuffle | Best shuffle | Task
Shuffle the characters of a string in such a way that as many of the character values are in a different position as possible.
A shuffle that produces a randomized result among the best choices is to be preferred. A deterministic approach that produces the same sequence every time is acceptable as an alternative.
Display the result as follows:
original string, shuffled string, (score)
The score gives the number of positions whose character value did not change.
Example
tree, eetr, (0)
Test cases
abracadabra
seesaw
elk
grrrrrr
up
a
Related tasks
Anagrams/Deranged anagrams
Permutations/Derangements
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #ALGOL_68 | ALGOL 68 | BEGIN # shuffle a string so as many as possible characters are moved #
PROC best shuffle = ( STRING orig )STRING:
BEGIN
STRING res := orig;
FOR i FROM LWB orig TO UPB orig DO
FOR j FROM LWB orig TO UPB orig DO
IF i /= j AND orig[ i ] /= res[ j ] AND orig[ j ] /= res[ i ] THEN
CHAR tmp = res[ i ]; res[ i ] := res[ j ]; res[ j ] := tmp
FI
OD
OD;
res
END # best shuffle # ;
PROC test = ( STRING orig )VOID:
BEGIN
STRING res := best shuffle( orig );
INT score := 0;
FOR i FROM LWB orig TO UPB orig DO
IF orig[ i ] = res[ i ] THEN
score +:= 1
FI
OD;
print( ( orig, ", ", res, ", (", whole( score, 0 ), ")", newline ) )
END # test # ;
test( "abracadabra" );
test( "seesaw" );
test( "elk" );
test( "grrrrrr" );
test( "up" );
test( "a" )
END |
http://rosettacode.org/wiki/Binary_strings | Binary strings | Many languages have powerful and useful (binary safe) string manipulation functions, while others don't, making it harder for these languages to accomplish some tasks.
This task is about creating functions to handle binary strings (strings made of arbitrary bytes, i.e. byte strings according to Wikipedia) for those languages that don't have built-in support for them.
If your language of choice does have this built-in support, show a possible alternative implementation for the functions or abilities already provided by the language.
In particular the functions you need to create are:
String creation and destruction (when needed and if there's no garbage collection or similar mechanism)
String assignment
String comparison
String cloning and copying
Check if a string is empty
Append a byte to a string
Extract a substring from a string
Replace every occurrence of a byte (or a string) in a string with another string
Join strings
Possible contexts of use: compression algorithms (like LZW compression), L-systems (manipulation of symbols), many more.
| #C.23 | C# | using System;
class Program
{
static void Main()
{
//string creation
var x = "hello world";
//# mark string for garbage collection
x = null;
//# string assignment with a null byte
x = "ab\0";
Console.WriteLine(x);
Console.WriteLine(x.Length); // 3
//# string comparison
if (x == "hello")
Console.WriteLine("equal");
else
Console.WriteLine("not equal");
if (x.CompareTo("bc") == -1)
Console.WriteLine("x is lexicographically less than 'bc'");
//# string cloning
var c = new char[3];
x.CopyTo(0, c, 0, 3);
object objecty = new string(c);
var y = new string(c);
Console.WriteLine(x == y); //same as string.equals
Console.WriteLine(x.Equals(y)); //it overrides object.Equals
Console.WriteLine(x == objecty); //uses object.Equals, return false
//# check if empty
var empty = "";
string nullString = null;
var whitespace = " ";
if (nullString == null && empty == string.Empty &&
string.IsNullOrEmpty(nullString) && string.IsNullOrEmpty(empty) &&
string.IsNullOrWhiteSpace(nullString) && string.IsNullOrWhiteSpace(empty) &&
string.IsNullOrWhiteSpace(whitespace))
Console.WriteLine("Strings are null, empty or whitespace");
//# append a byte
x = "helloworld";
x += (char)83;
Console.WriteLine(x);
//# substring
var slice = x.Substring(5, 5);
Console.WriteLine(slice);
//# replace bytes
var greeting = x.Replace("worldS", "");
Console.WriteLine(greeting);
//# join strings
var join = greeting + " " + slice;
Console.WriteLine(join);
}
} |
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #FreeBASIC | FreeBASIC | sub binlims( dat() as integer, limits() as integer, bins() as uinteger )
dim as uinteger n = ubound(limits), j, i
for i = 0 to ubound(dat)
if dat(i)<limits(0) then
bins(0) += 1
elseif dat(i) >= limits(n) then
bins(n+1) += 1
else
for j = 1 to n
if dat(i)<limits(j) then
bins(j) += 1
exit for
end if
next j
end if
next i
end sub
'example 1
dim as integer limits1(0 to ...) = {23, 37, 43, 53, 67, 83}
dim as integer dat1(0 to ...) = {95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,_
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55}
dim as uinteger bins1(0 to ubound(limits1)+1)
binlims( dat1(), limits1(), bins1() )
print "=====EXAMPLE ONE====="
print "< ";limits1(0);": ";bins1(0)
for i as uinteger = 1 to ubound(limits1)
print ">= ";limits1(i-1);" and < ";limits1(i);": ";bins1(i)
next i
print ">= ";limits1(ubound(limits1));": ";bins1(ubound(bins1))
print
'example 2
dim as integer limits2(0 to ...) = {14, 18, 249, 312, 389, 392, 513, 591, 634, 720}
dim as integer dat2(0 to ...) = {445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,_
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,_
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,_
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,_
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,_
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,_
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,_
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,_
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,_
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749}
redim as uinteger bins2(0 to ubound(limits2)+1)
binlims( dat2(), limits2(), bins2() )
print "=====EXAMPLE TWO====="
print "< ";limits2(0);": ";bins2(0)
for i as uinteger = 1 to ubound(limits2)
print ">= ";limits2(i-1);" and < ";limits2(i);": ";bins2(i)
next i
print ">= ";limits2(ubound(limits2));": ";bins2(ubound(bins2)) |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #jq | jq | def lpad($len; $fill): tostring | ($len - length) as $l | ($fill * $l)[:$l] + .;
# Create a bag of words, i.e. a JSON object with counts of the items in the stream
def bow(stream):
reduce stream as $word ({}; .[($word|tostring)] += 1); |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Julia | Julia | const sequence =
"CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG" *
"CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG" *
"AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT" *
"GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT" *
"CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG" *
"TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA" *
"TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT" *
"CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG" *
"TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC" *
"GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT"
function dnasequenceprettyprint(seq, colsize=50)
println(length(seq), "nt DNA sequence:\n")
rows = [seq[i:min(length(seq), i + colsize - 1)] for i in 1:colsize:length(seq)]
for (i, r) in enumerate(rows)
println(lpad(colsize * (i - 1), 5), " ", r)
end
end
dnasequenceprettyprint(sequence)
function printcounts(seq)
bases = [['A', 0], ['C', 0], ['G', 0], ['T', 0]]
for c in seq, base in bases
if c == base[1]
base[2] += 1
end
end
println("\nNucleotide counts:\n")
for base in bases
println(lpad(base[1], 10), lpad(string(base[2]), 12))
end
println(lpad("Other", 10), lpad(string(length(seq) - sum(x[2] for x in bases)), 12))
println(" _________________\n", lpad("Total", 10), lpad(string(length(seq)), 12))
end
printcounts(sequence)
|
http://rosettacode.org/wiki/Binary_digits | Binary digits | Task
Create and display the sequence of binary digits for a given non-negative integer.
The decimal value 5 should produce an output of 101
The decimal value 50 should produce an output of 110010
The decimal value 9000 should produce an output of 10001100101000
The results can be achieved using built-in radix functions within the language (if these are available), or alternatively a user defined function can be used.
The output produced should consist just of the binary digits of each number followed by a newline.
There should be no other whitespace, radix or sign markers in the produced output, and leading zeros should not appear in the results.
| #8086_Assembly | 8086 Assembly | .model small
.stack 1024
.data
TestData0 byte 5,255 ;255 is the terminator
TestData1 byte 5,0,255
TestData2 byte 9,0,0,0,255
.code
start:
mov ax,@data
mov ds,ax
cld ;String functions are set to auto-increment
mov ax,2 ;clear screen by setting video mode to 0
int 10h ;select text mode - We're already in it, so this clears the screen
mov si,offset TestData0
call PrintBinary_NoLeadingZeroes
mov si,offset TestData1
call PrintBinary_NoLeadingZeroes
mov si,offset TestData2
call PrintBinary_NoLeadingZeroes
ExitDOS:
mov ax,4C00h ;return to dos
int 21h
PrintBinary_NoLeadingZeroes proc
;input: DS:SI = seg:offset of a 255-terminated sequence of unpacked BCD digits, stored big-endian
;setup
mov bx,8000h
;bl will be our "can we print zeroes yet" flag.
;bh is the "revolving bit mask" - we'll compare each bit to it, then rotate it right once.
; It's very handy because it's a self-resetting loop counter as well!
NextDigit:
lodsb
cmp al,255
je Terminated
NextBit:
test al,bh ;is the bit we're testing right now set?
jz PrintZero
;else, print one
push ax
mov dl,'1' ;31h
mov ah,2
int 21h ;prints the ascii code in DL
pop ax
or bl,1 ;set "we've printed a one" flag
jmp predicate
PrintZero:
test bl,bl
jz predicate
push ax
mov dl,'0' ;30h
mov ah,2
int 21h
pop ax
predicate:
ror bh,1
jnc NextBit
;if the carry is set, we've rotated BH back to 10000000b,
; so move on to the next digit in that case.
jmp NextDigit
Terminated:
push ax
mov ah,2
mov dl,13 ;carriage return
int 21h
mov dl,10 ;linefeed
int 21h
pop ax
ret
PrintBinary_NoLeadingZeroes endp |
http://rosettacode.org/wiki/Bitmap/Bresenham%27s_line_algorithm | Bitmap/Bresenham's line algorithm | Task
Using the data storage type defined on the Bitmap page for raster graphics images,
draw a line given two points with Bresenham's line algorithm.
| #Elm | Elm |
-- Brensenham Line Algorithm
type alias Position =
{x: Int, y: Int}
type alias BresenhamStatics =
{ finish : Position
, sx : Int
, sy : Int
, dx : Float
, dy : Float
}
line : Position -> Position -> List Position
line p q =
let
dx = (toFloat << abs) (q.x - p.x)
dy = (toFloat << abs) (q.y - p.y)
sx = if p.x < q.x then 1 else -1
sy = if p.y < q.y then 1 else -1
error =
(if dx > dy then dx else -dy) / 2
statics =
BresenhamStatics q sx sy dx dy
in
bresenhamLineLoop statics error p []
bresenhamLineLoop : BresenhamStatics -> Float -> Position -> List Position -> List Position
bresenhamLineLoop statics error p positions =
let
positions_ = p :: positions
{sx, sy, dx, dy, finish} = statics
in
if (p.x == finish.x) && (p.y == finish.y) then
positions_
else
let
(dErrX, x) =
if error > -dx then (-dy, sx + p.x)
else (0, p.x)
(dErrY, y) =
if error < dy then (dx, sy + p.y)
else (0, p.y)
error_ = error + dErrX + dErrY
in
bresenhamLineLoop statics error_ (Position x y) positions_
|
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