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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
| #OCaml | OCaml | let best_shuffle s =
let len = String.length s in
let r = String.copy s in
for i = 0 to pred len do
for j = 0 to pred len do
if i <> j && s.[i] <> r.[j] && s.[j] <> r.[i] then
begin
let tmp = r.[i] in
r.[i] <- r.[j];
r.[j] <- tmp;
end
done;
done;
(r)
let count_same s1 s2 =
let len1 = String.length s1
and len2 = String.length s2 in
let n = ref 0 in
for i = 0 to pred (min len1 len2) do
if s1.[i] = s2.[i] then incr n
done;
!n
let () =
let test s =
let s2 = best_shuffle s in
Printf.printf " '%s', '%s' -> %d\n" s s2 (count_same s s2);
in
test "tree";
test "abracadabra";
test "seesaw";
test "elk";
test "grrrrrr";
test "up";
test "a";
;; |
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.
| #Raku | Raku | # Raku is perfectly fine with NUL *characters* in strings:
my Str $s = 'nema' ~ 0.chr ~ 'problema!';
say $s;
# However, Raku makes a clear distinction between strings
# (i.e. sequences of characters), like your name, or …
my Str $str = "My God, it's full of chars!";
# … and sequences of bytes (called Bufs), for example a PNG image, or …
my Buf $buf = Buf.new(255, 0, 1, 2, 3);
say $buf;
# Strs can be encoded into Blobs …
my Blob $this = 'round-trip'.encode('ascii');
# … and Blobs can be decoded into Strs …
my Str $that = $this.decode('ascii');
# So it's all there. Nevertheless, let's solve this task explicitly
# in order to see some nice language features:
# We define a class …
class ByteStr {
# … that keeps an array of bytes, and we delegate some
# straight-forward stuff directly to this attribute:
# (Note: "has byte @.bytes" would be nicer, but that is
# not yet implemented in Rakudo.)
has Int @.bytes handles(< Bool elems gist push >);
# A handful of methods …
method clone() {
self.new(:@.bytes);
}
method substr(Int $pos, Int $length) {
self.new(:bytes(@.bytes[$pos .. $pos + $length - 1]));
}
method replace(*@substitutions) {
my %h = @substitutions;
@.bytes.=map: { %h{$_} // $_ }
}
}
# A couple of operators for our new type:
multi infix:<cmp>(ByteStr $x, ByteStr $y) { $x.bytes.join cmp $y.bytes.join }
multi infix:<~> (ByteStr $x, ByteStr $y) { ByteStr.new(:bytes(|$x.bytes, |$y.bytes)) }
# create some byte strings (destruction not needed due to garbage collection)
my ByteStr $b0 = ByteStr.new;
my ByteStr $b1 = ByteStr.new(:bytes( |'foo'.ords, 0, 10, |'bar'.ords ));
# assignment ($b1 and $b2 contain the same ByteStr object afterwards):
my ByteStr $b2 = $b1;
# comparing:
say 'b0 cmp b1 = ', $b0 cmp $b1;
say 'b1 cmp b2 = ', $b1 cmp $b2;
# cloning:
my $clone = $b1.clone;
$b1.replace('o'.ord => 0);
say 'b1 = ', $b1;
say 'b2 = ', $b2;
say 'clone = ', $clone;
# to check for (non-)emptiness we evaluate the ByteStr in boolean context:
say 'b0 is ', $b0 ?? 'not empty' !! 'empty';
say 'b1 is ', $b1 ?? 'not empty' !! 'empty';
# appending a byte:
$b1.push: 123;
say 'appended = ', $b1;
# extracting a substring:
my $sub = $b1.substr(2, 4);
say 'substr = ', $sub;
# replacing a byte:
$b2.replace(102 => 103);
say 'replaced = ', $b2;
# joining:
my ByteStr $b3 = $b1 ~ $sub;
say 'joined = ', $b3; |
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.
| #Brainf.2A.2A.2A | Brainf*** | +[ Start with n=1 to kick off the loop
[>>++<< Set up {n 0 2} for divmod magic
[->+>- Then
[>+>>]> do
[+[-<+>]>+>>] the
<<<<<<] magic
>>>+ Increment n % 2 so that 0s don't break things
>] Move into n / 2 and divmod that unless it's 0
-< Set up sentinel ‑1 then move into the first binary digit
[++++++++ ++++++++ ++++++++ Add 47 to get it to ASCII
++++++++ ++++++++ +++++++. and print it
[<]<] Get to a 0; the cell to the left is the next binary digit
>>[<+>-] Tape is {0 n}; make it {n 0}
>[>+] Get to the ‑1
<[[-]<] Zero the tape for the next iteration
++++++++++. Print a newline
[-]<+] Zero it then increment n and go again |
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.
| #PicoLisp | PicoLisp | (de brez (Img X Y DX DY)
(let SX
(cond
((=0 DX) 0)
((gt0 DX) 1)
(T (setq DX (- DX)) -1) )
(let SY
(cond
((=0 DY) 0)
((gt0 DY) 1)
(T (setq DY (- DY)) -1) )
(if (>= DX DY)
(let E (- (* 2 DY) DX)
(do DX
(set (nth Img Y X) 1)
(when (ge0 E)
(inc 'Y SY)
(dec 'E (* 2 DX)) )
(inc 'X SX)
(inc 'E (* 2 DY)) ) )
(let E (- (* 2 DX) DY)
(do DY
(set (nth Img Y X) 1)
(when (ge0 E)
(inc 'X SX)
(dec 'E (* 2 DY)) )
(inc 'Y SY)
(inc 'E (* 2 DX)) ) ) ) ) ) )
(let Img (make (do 90 (link (need 120 0)))) # Create image 120 x 90
(brez Img 10 10 100 30) # Draw five lines
(brez Img 10 10 100 50)
(brez Img 10 10 100 70)
(brez Img 10 10 60 70)
(brez Img 10 10 20 70)
(out "img.pbm" # Write to bitmap file
(prinl "P1")
(prinl 120 " " 90)
(mapc prinl 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
| #UNIX_Shell | UNIX Shell | if
echo 'Looking for file' # This is the evaluation block
test -e foobar.fil # The exit code from this statement determines whether the branch runs
then
echo 'The file exists' # This is the optional branch
echo 'I am going to delete it'
rm foobar.fil
fi |
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
| #Ursa | Ursa | decl boolean bool |
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
| #VBA | VBA | Dim a As Integer
Dim b As Boolean
Debug.Print b
a = b
Debug.Print a
b = True
Debug.Print b
a = b
Debug.Print a |
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)..
| #PureBasic | PureBasic | DataSection
Data.s "N", "north", "E", "east", "W", "west", "S", "south", "b", " by " ;abbreviations, expansions
Data.s "N NbE N-NE NEbN NE NEbE E-NE EbN E EbS E-SE SEbE SE SEbS S-SE SbE" ;dirs
Data.s "S SbW S-SW SWbS SW SWbW W-SW WbS W WbN W-NW NWbW NW NWbN N-NW NbW"
EndDataSection
;initialize data
NewMap dirSubst.s()
Define i, abbr.s, expansion.s
For i = 1 To 5
Read.s abbr
Read.s expansion
dirSubst(abbr) = expansion
Next
Dim dirs.s(32)
Define j, s.s
For j = 0 To 1
Read.s s.s
For i = 0 To 15
abbr.s = StringField(s, i + 1, " ")
dirs(j * 16 + i) = abbr
Next
Next
;expand abbreviated compass point and capitalize
Procedure.s abbr2compassPoint(abbr.s)
Shared dirSubst()
Protected i, compassPoint.s, key.s
For i = 1 To Len(abbr)
key.s = Mid(abbr, i, 1)
If FindMapElement(dirSubst(), key)
compassPoint + dirSubst(key)
Else
compassPoint + key
EndIf
Next
ProcedureReturn UCase(Left(compassPoint, 1)) + Mid(compassPoint, 2)
EndProcedure
Procedure.s angle2compass(angle.f)
Shared dirs()
Static segment.f = 360.0 / 32 ;width of each compass segment
Protected dir
;work out which segment contains the compass angle
dir = Int((Mod(angle, 360) / segment) + 0.5)
;convert to a named direction
ProcedureReturn abbr2compassPoint(dirs(dir))
EndProcedure
;box the compass
If OpenConsole()
Define i, heading.f, index
For i = 0 To 32
heading = i * 11.25
If i % 3 = 1
heading + 5.62
EndIf
If i % 3 = 2
heading - 5.62
EndIf
index = i % 32 + 1
PrintN(RSet(Str(index), 2) + " " + LSet(angle2compass(heading), 18) + RSet(StrF(heading, 2), 7))
Next
Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf |
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.
| #Julia | Julia | # Version 5.2
@show 1 & 2 # AND
@show 1 | 2 # OR
@show 1 ^ 2 # XOR -- for Julia 6.0 the operator is `⊻`
@show ~1 # NOT
@show 1 >>> 2 # SHIFT RIGHT (LOGICAL)
@show 1 >> 2 # SHIFT RIGHT (ARITMETIC)
@show 1 << 2 # SHIFT LEFT (ARITMETIC/LOGICAL)
A = BitArray([true, true, false, false, true])
@show A ror(A,1) ror(A,2) ror(A,5) # ROTATION RIGHT
@show rol(A,1) rol(A,2) rol(A,5) # ROTATION LEFT
|
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.
| #Pascal | Pascal | Interface
uses crt, { GetDir }
graph; { function GetPixel }
type { integer numbers }
{ from unit bitmaps XPERT software production Tamer Fakhoury }
_bit = $00000000..$00000001; {number 1 bit without sign = (0..1) }
_byte = $00000000..$000000FF; {number 1 byte without sign = (0..255)}
_word = $00000000..$0000FFFF; {number 2 bytes without sign = (0..65 535)}
_dWord = $00000000..$7FFFFFFF; {number 4 bytes without sign = (0..4 294 967 296)}
_longInt = $80000000..$7FFFFFFF; {number 4 bytes with sign
= (-2 147 483 648..2 147 483 648}
TbmpFileHeader =
record
ID: _word; { Must be 'BM' =19778=$424D for windows }
FileSize: _dWord; { Size of this file in bytes }
Reserved: _dWord; { ??? }
bmpDataOffset: _dword; { = 54 = $36 from begining of file to begining of bmp data }
end;
TbmpInfoHeader =
record
InfoHeaderSize: _dword; { Size of Info header
= 28h = 40 (decimal)
for windows }
Width,
Height: _longInt; { Width and Height of image in pixels }
Planes, { number of planes of bitmap }
BitsPerPixel: _word; { Bits can be 1, 4, 8, 24 or 32 }
Compression,
bmpDataSize: _dword; { in bytes rounded to the next 4 byte boundary }
XPixPerMeter, { horizontal resolution in pixels }
YPixPerMeter: _longInt; { vertical }
NumbColorsUsed,
NumbImportantColors: _dword; {= NumbColorUsed}
end; { TbmpHeader = Record ... }
T32Color =
record { 4 byte = 32 bit }
Blue: byte;
Green: byte;
Red: byte;
Alfa: byte
end;
var directory,
bmpFileName: string;
bmpFile: file; { untyped file }
bmpFileHeader: TbmpFileHeader;
bmpInfoHeader: TbmpInfoHeader;
color32: T32Color;
RowSizeInBytes: integer;
BytesPerPixel: integer;
const defaultBmpFileName = 'test';
DefaultDirectory = 'c:\bp\';
DefaultExtension = '.bmp';
bmpFileHeaderSize = 14;
{ compression specyfication }
bi_RGB = 0; { compression }
bi_RLE8 = 1;
bi_RLE4 = 2;
bi_BITFIELDS = 3;
bmp_OK = 0;
bmp_NotBMP = 1;
bmp_OpenError = 2;
bmp_ReadError = 3;
Procedure CreateBmpFile32(directory: string; FileName: string;
iWidth, iHeight: _LongInt);
{************************************************}
Implementation {-----------------------------}
{************************************************}
Procedure CreateBmpFile32(directory: string; FileName: string;
iWidth, iHeight: _LongInt);
var
x, y: integer;
begin
if directory = '' then
GetDir(0, directory);
if FileName = '' then
FileName: = DefaultBmpFileName;
{ create a new file on a disk in a given directory with given name }
Assign(bmpFile, directory + FileName + DefaultExtension);
ReWrite(bmpFile, 1);
{ fill the headers }
with bmpInfoHeader, bmpFileHeader do
begin
ID := 19778;
InfoheaderSize := 40;
width := iWidth;
height := iHeight;
BitsPerPixel := 32;
BytesPerPixel := BitsPerPixel div 8;
reserved := 0;
bmpDataOffset := InfoHeaderSize + bmpFileHeaderSize;
planes := 1;
compression := bi_RGB;
XPixPerMeter := 0;
YPixPerMeter := 0;
NumbColorsUsed := 0;
NumbImportantColors := 0;
RowSizeInBytes := (Width * BytesPerPixel); { only for >=8 bits per pixel }
bmpDataSize := height * RowSizeinBytes;
FileSize := InfoHeaderSize + bmpFileHeaderSize + bmpDataSize;
{ copy headers to disk file }
BlockWrite(bmpFile, bmpFileHeader, bmpFileHeaderSize);
BlockWrite(bmpFile, bmpInfoHeader, infoHeaderSize);
{ fill the pixel data area }
for y := (height - 1) downto 0 do
begin
for x := 0 to (width - 1) do
begin { Pixel(x,y) }
color32.Blue := 255;
color32.Green := 0;
color32.Red := 0;
color32.Alfa := 0;
BlockWrite(bmpFile, color32, 4);
end; { for x ... }
end; { for y ... }
Close(bmpFile);
end; { with bmpInfoHeader, bmpFileHeader }
end; { procedure }
|
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 | #Sidef | Sidef | say 15.of { .bell } |
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 | #Swift | Swift | public struct BellTriangle<T: BinaryInteger> {
@usableFromInline
var arr: [T]
@inlinable
public internal(set) subscript(row row: Int, col col: Int) -> T {
get { arr[row * (row - 1) / 2 + col] }
set { arr[row * (row - 1) / 2 + col] = newValue }
}
@inlinable
public init(n: Int) {
arr = Array(repeating: 0, count: n * (n + 1) / 2)
self[row: 1, col: 0] = 1
for i in 2...n {
self[row: i, col: 0] = self[row: i - 1, col: i - 2]
for j in 1..<i {
self[row: i, col: j] = self[row: i, col: j - 1] + self[row: i - 1, col: j - 1]
}
}
}
}
let tri = BellTriangle<Int>(n: 15)
print("First 15 Bell numbers:")
for i in 1...15 {
print("\(i): \(tri[row: i, col: 0])")
}
for i in 1...10 {
print(tri[row: i, col: 0], terminator: "")
for j in 1..<i {
print(", \(tri[row: i, col: j])", terminator: "")
}
print()
} |
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 .
| #Pascal | Pascal | program fibFirstdigit;
{$IFDEF FPC}{$MODE Delphi}{$ELSE}{$APPTYPE CONSOLE}{$ENDIF}
uses
sysutils;
type
tDigitCount = array[0..9] of LongInt;
var
s: Ansistring;
dgtCnt,
expectedCnt : tDigitCount;
procedure GetFirstDigitFibonacci(var dgtCnt:tDigitCount;n:LongInt=1000);
//summing up only the first 9 digits
//n = 1000 -> difference to first 9 digits complete fib < 100 == 2 digits
var
a,b,c : LongWord;//about 9.6 decimals
Begin
for a in dgtCnt do dgtCnt[a] := 0;
a := 0;b := 1;
while n > 0 do
Begin
c := a+b;
//overflow? round and divide by base 10
IF c < a then
Begin a := (a+5) div 10;b := (b+5) div 10;c := a+b;end;
a := b;b := c;
s := IntToStr(a);inc(dgtCnt[Ord(s[1])-Ord('0')]);
dec(n);
end;
end;
procedure InitExpected(var dgtCnt:tDigitCount;n:LongInt=1000);
var
i: integer;
begin
for i := 1 to 9 do
dgtCnt[i] := trunc(n*ln(1 + 1 / i)/ln(10));
end;
var
reldiff: double;
i,cnt: integer;
begin
cnt := 1000;
InitExpected(expectedCnt,cnt);
GetFirstDigitFibonacci(dgtCnt,cnt);
writeln('Digit count expected rel diff');
For i := 1 to 9 do
Begin
reldiff := 100*(expectedCnt[i]-dgtCnt[i])/expectedCnt[i];
writeln(i:5,dgtCnt[i]:7,expectedCnt[i]:10,reldiff:10:5,' %');
end;
end. |
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.
| #Maple | Maple | print(select(n->n[2]<>0,[seq([n,bernoulli(n,1)],n=0..60)])); |
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.
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | bernoulli[n_] := Module[{a = ConstantArray[0, n + 2]},
Do[
a[[m]] = 1/m;
If[m == 1 && a[[1]] != 0, Print[{m - 1, a[[1]]}]];
Do[
a[[j - 1]] = (j - 1)*(a[[j - 1]] - a[[j]]);
If[j == 2 && a[[1]] != 0, Print[{m - 1, a[[1]]}]];
, {j, m, 2, -1}];
, {m, 1, n + 1}];
]
bernoulli[60] |
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.
| #D | D | import std.stdio, std.array, std.range, std.traits;
/// Recursive.
bool binarySearch(R, T)(/*in*/ R data, in T x) pure nothrow @nogc
if (isRandomAccessRange!R && is(Unqual!T == Unqual!(ElementType!R))) {
if (data.empty)
return false;
immutable i = data.length / 2;
immutable mid = data[i];
if (mid > x)
return data[0 .. i].binarySearch(x);
if (mid < x)
return data[i + 1 .. $].binarySearch(x);
return true;
}
/// Iterative.
bool binarySearchIt(R, T)(/*in*/ R data, in T x) pure nothrow @nogc
if (isRandomAccessRange!R && is(Unqual!T == Unqual!(ElementType!R))) {
while (!data.empty) {
immutable i = data.length / 2;
immutable mid = data[i];
if (mid > x)
data = data[0 .. i];
else if (mid < x)
data = data[i + 1 .. $];
else
return true;
}
return false;
}
void main() {
/*const*/ auto items = [2, 4, 6, 8, 9].assumeSorted;
foreach (const x; [1, 8, 10, 9, 5, 2])
writefln("%2d %5s %5s %5s", x,
items.binarySearch(x),
items.binarySearchIt(x),
// Standard Binary Search:
!items.equalRange(x).empty);
} |
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
| #Pascal | Pascal | program BestShuffleDemo(output);
function BestShuffle(s: string): string;
var
tmp: char;
i, j: integer;
t: string;
begin
t := s;
for i := 1 to length(t) do
for j := 1 to length(t) do
if (i <> j) and (s[i] <> t[j]) and (s[j] <> t[i]) then
begin
tmp := t[i];
t[i] := t[j];
t[j] := tmp;
end;
BestShuffle := t;
end;
const
original: array[1..6] of string =
('abracadabra', 'seesaw', 'elk', 'grrrrrr', 'up', 'a');
var
shuffle: string;
i, j, score: integer;
begin
for i := low(original) to high(original) do
begin
shuffle := BestShuffle(original[i]);
score := 0;
for j := 1 to length(shuffle) do
if original[i][j] = shuffle[j] then
inc(score);
writeln(original[i], ', ', shuffle, ', (', score, ')');
end;
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.
| #Red | Red | Red []
s: copy "abc" ;; string creation
s: none ;; destruction
t: "Abc"
if t == "abc" [print "equal case"] ;; comparison , case sensitive
if t = "abc" [print "equal (case insensitive)"] ;; comparison , case insensitive
s: copy "" ;; copying string
if empty? s [print "string is empty "] ;; check if string is empty
append s #"a" ;; append byte
substr: copy/part at "1234" 3 2 ;; ~ substr ("1234" ,3,2) , red has 1 based indices !
?? substr
s: replace/all "abcabcabc" "bc" "x" ;; replace all "bc" by "x"
?? s
s: append "hello " "world" ;; join 2 strings
?? s
s: rejoin ["hello " "world" " !"] ;; join multiple strings
?? s
|
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.
| #Burlesque | Burlesque |
blsq ) {5 50 9000}{2B!}m[uN
101
110010
10001100101000
|
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.
| #PL.2FI | PL/I |
/* Draw a line from (x0, y0) to (x1, y1). 13 May 2010 */
/* Based on Rosetta code proforma. */
/* Declarations for image and selected color, for 4-bit colors. */
declare image(40,40) bit (4), color bit (4) static initial ('1000'b);
draw_line: procedure (xi, yi, xf, yf );
declare (xi, yi, xf, yf) fixed binary (31) nonassignable;
declare (x0, y0, x1, y1) fixed binary (31);
declare (deltax, deltay, x, y, ystep) fixed binary;
declare (error initial (0), delta_error) float;
declare steep bit (1);
x0 = xi; y = YI; y0 = yi; x1 = xf; y1 = yf;
steep = abs(y1 - y0) > abs (x1 - x0);
if steep then
do; call swap (x0, y0); call swap (x1, y1); end;
if x0 > x1 then
do; call swap (x0, x1); call swap (y0, y1); end;
deltax = x1 - x0; deltay = abs(y1 - y0);
delta_error = deltay/deltax;
if y0 < y1 then ystep = 1; else ystep = -1;
do x = x0 to x1;
if steep then image(y, x) = color; else image(x, y) = color;
if steep then put skip list (y, x); else put skip list (x, y);
error = error + delta_error;
if error >= 0.5 then do; y = y + ystep; error = error - 1; end;
end;
swap: procedure (a, b);
declare (a, b) fixed binary (31);
declare t fixed binary (31);
t = a; a = b; b = t;
end swap;
end draw_line;
|
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
| #VBScript | VBScript |
a = True
b = False
Randomize Timer
x = Int(Rnd * 2) <> 0
y = Int(Rnd * 2) = 0
MsgBox a & " " & b & " " & x & " " & y |
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
| #Vim_Script | Vim Script | if "foo"
echo "true"
else
echo "false"
endif |
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
| #Vlang | Vlang | // Boolean Value, in V
// Tectonics: v run boolean-value.v
module main
// V bool type, with values true or false are the V booleans.
// true and false are V keywords, and display as true/false
// Numeric values are not booleans in V, 0 is not boolean false
pub fn main() {
t := true
f := false
if t { println(t) }
// this code would fail to compile
// if 1 { println(t) }
if 0 == 1 { println("bad result") } else { println(f) }
} |
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)..
| #Python | Python | majors = 'north east south west'.split()
majors *= 2 # no need for modulo later
quarter1 = 'N,N by E,N-NE,NE by N,NE,NE by E,E-NE,E by N'.split(',')
quarter2 = [p.replace('NE','EN') for p in quarter1]
def degrees2compasspoint(d):
d = (d % 360) + 360/64
majorindex, minor = divmod(d, 90.)
majorindex = int(majorindex)
minorindex = int( (minor*4) // 45 )
p1, p2 = majors[majorindex: majorindex+2]
if p1 in {'north', 'south'}:
q = quarter1
else:
q = quarter2
return q[minorindex].replace('N', p1).replace('E', p2).capitalize()
if __name__ == '__main__':
for i in range(33):
d = i * 11.25
m = i % 3
if m == 1: d += 5.62
elif m == 2: d -= 5.62
n = i % 32 + 1
print( '%2i %-18s %7.2f°' % (n, degrees2compasspoint(d), 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.
| #Kotlin | Kotlin | /* for symmetry with Kotlin's other binary bitwise operators
we wrap Java's 'rotate' methods as infix functions */
infix fun Int.rol(distance: Int): Int = Integer.rotateLeft(this, distance)
infix fun Int.ror(distance: Int): Int = Integer.rotateRight(this, distance)
fun main(args: Array<String>) {
// inferred type of x and y is Int i.e. 32 bit signed integers
val x = 10
val y = 2
println("x = $x")
println("y = $y")
println("NOT x = ${x.inv()}")
println("x AND y = ${x and y}")
println("x OR y = ${x or y}")
println("x XOR y = ${x xor y}")
println("x SHL y = ${x shl y}")
println("x ASR y = ${x shr y}") // arithmetic shift right (sign bit filled)
println("x LSR y = ${x ushr y}") // logical shift right (zero filled)
println("x ROL y = ${x rol y}")
println("x ROR y = ${x ror y}")
} |
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.
| #Perl | Perl | #! /usr/bin/perl
use strict;
use Image::Imlib2;
# create the "canvas"
my $img = Image::Imlib2->new(200,200);
# fill with a plain RGB(A) color
$img->set_color(255, 0, 0, 255);
$img->fill_rectangle(0,0, 200, 200);
# set a pixel to green (at 40,40)
$img->set_color(0, 255, 0, 255);
$img->draw_point(40,40);
# "get" pixel rgb(a)
my ($red, $green, $blue, $alpha) = $img->query_pixel(40,40);
undef $img;
# another way of creating a canvas with a bg colour (or from
# an existing "raw" data)
my $col = pack("CCCC", 255, 255, 0, 0); # a, r, g, b
my $img = Image::Imlib2->new_using_data(200, 200, $col x (200 * 200));
exit 0; |
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 | #Visual_Basic_.NET | Visual Basic .NET | Imports System.Numerics
Imports System.Runtime.CompilerServices
Module Module1
<Extension()>
Sub Init(Of T)(array As T(), value As T)
If IsNothing(array) Then Return
For i = 0 To array.Length - 1
array(i) = value
Next
End Sub
Function BellTriangle(n As Integer) As BigInteger()()
Dim tri(n - 1)() As BigInteger
For i = 0 To n - 1
Dim temp(i - 1) As BigInteger
tri(i) = temp
tri(i).Init(0)
Next
tri(1)(0) = 1
For i = 2 To n - 1
tri(i)(0) = tri(i - 1)(i - 2)
For j = 1 To i - 1
tri(i)(j) = tri(i)(j - 1) + tri(i - 1)(j - 1)
Next
Next
Return tri
End Function
Sub Main()
Dim bt = BellTriangle(51)
Console.WriteLine("First fifteen Bell numbers:")
For i = 1 To 15
Console.WriteLine("{0,2}: {1}", i, bt(i)(0))
Next
Console.WriteLine("50: {0}", bt(50)(0))
Console.WriteLine()
Console.WriteLine("The first ten rows of Bell's triangle:")
For i = 1 To 10
Dim it = bt(i).GetEnumerator()
Console.Write("[")
If it.MoveNext() Then
Console.Write(it.Current)
End If
While it.MoveNext()
Console.Write(", ")
Console.Write(it.Current)
End While
Console.WriteLine("]")
Next
End Sub
End Module |
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 .
| #Perl | Perl | #!/usr/bin/perl
use strict ;
use warnings ;
use POSIX qw( log10 ) ;
my @fibonacci = ( 0 , 1 ) ;
while ( @fibonacci != 1000 ) {
push @fibonacci , $fibonacci[ -1 ] + $fibonacci[ -2 ] ;
}
my @actuals ;
my @expected ;
for my $i( 1..9 ) {
my $sum = 0 ;
map { $sum++ if $_ =~ /\A$i/ } @fibonacci ;
push @actuals , $sum / 1000 ;
push @expected , log10( 1 + 1/$i ) ;
}
print " Observed Expected\n" ;
for my $i( 1..9 ) {
print "$i : " ;
my $result = sprintf ( "%.2f" , 100 * $actuals[ $i - 1 ] ) ;
printf "%11s %%" , $result ;
$result = sprintf ( "%.2f" , 100 * $expected[ $i - 1 ] ) ;
printf "%15s %%\n" , $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.
| #Nim | Nim | import bignum
import strformat
const Lim = 60
#---------------------------------------------------------------------------------------------------
proc bernoulli(n: Natural): Rat =
## Compute a Bernoulli number using Akiyama–Tanigawa algorithm.
var a = newSeq[Rat](n + 1)
for m in 0..n:
a[m] = newRat(1, m + 1)
for j in countdown(m, 1):
a[j-1] = j * (a[j] - a[j-1])
result = a[0]
#———————————————————————————————————————————————————————————————————————————————————————————————————
type Info = tuple
n: int # Number index in Bernoulli sequence.
val: Rat # Bernoulli number.
var values: seq[Info] # List of values as Info tuples.
var maxLen = -1 # Maximum length.
# First step: compute the values and prepare for display.
for n in 0..Lim:
# Compute value.
if n != 1 and (n and 1) == 1: continue # Ignore odd "n" except 1.
let b = bernoulli(n)
# Check numerator length.
let len = ($b.num).len
if len > maxLen: maxLen = len
# Store information for next step.
values.add((n, b))
# Second step: display the values with '/' aligned.
for (n, b) in values:
let s = fmt"{($b.num).alignString(maxLen, '>')} / {b.denom}"
echo fmt"{n:2}: {s}" |
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.
| #Delphi | Delphi | /** Returns null if the value is not found. */
def binarySearch(collection, value) {
var low := 0
var high := collection.size() - 1
while (low <= high) {
def mid := (low + high) // 2
def comparison := value.op__cmp(collection[mid])
if (comparison.belowZero()) { high := mid - 1 } \
else if (comparison.aboveZero()) { low := mid + 1 } \
else if (comparison.isZero()) { return mid } \
else { throw("You expect me to binary search with a partial order?") }
}
return null
} |
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
| #Perl | Perl | use strict;
use warnings;
use List::Util qw(shuffle);
use Algorithm::Permute;
best_shuffle($_) for qw(abracadabra seesaw elk grrrrrr up a);
sub best_shuffle {
my ($original_word) = @_;
my $best_word = $original_word;
my $best_score = length $best_word;
my @shuffled = shuffle split //, $original_word;
my $iterator = Algorithm::Permute->new(\@shuffled);
while( my @array = $iterator->next ) {
my $word = join '', @array;
# For each letter which is the same in the two words,
# there will be a \x00 in the "^" of the two words.
# The tr operator is then used to count the "\x00"s.
my $score = ($original_word ^ $word) =~ tr/\x00//;
next if $score >= $best_score;
($best_word, $best_score) = ($word, $score);
last if $score == 0;
}
print "$original_word, $best_word, $best_score\n";
}
|
http://rosettacode.org/wiki/Binary_strings | Binary strings | Many languages have powerful and useful (binary safe) string manipulation functions, while others don't, making it harder for these languages to accomplish some tasks.
This task is about creating functions to handle binary strings (strings made of arbitrary bytes, i.e. byte strings according to Wikipedia) for those languages that don't have built-in support for them.
If your language of choice does have this built-in support, show a possible alternative implementation for the functions or abilities already provided by the language.
In particular the functions you need to create are:
String creation and destruction (when needed and if there's no garbage collection or similar mechanism)
String assignment
String comparison
String cloning and copying
Check if a string is empty
Append a byte to a string
Extract a substring from a string
Replace every occurrence of a byte (or a string) in a string with another string
Join strings
Possible contexts of use: compression algorithms (like LZW compression), L-systems (manipulation of symbols), many more.
| #REXX | REXX | /*REXX program demonstrates methods (code examples) to use and express binary strings.*/
dingsta= '11110101'b /*four versions, bit string assignment.*/
dingsta= "11110101"b /*this is the same assignment as above.*/
dingsta= '11110101'B /* " " " " " " " */
dingsta= '1111 0101'B /* " " " " " " */
dingsta2= dingsta /*clone one string to another (a copy).*/
other= '1001 0101 1111 0111'b /*another binary (or bit) string. */
if dingsta= other then say 'they are equal' /*compare the two (binary) strings. */
if other== '' then say "OTHER is empty." /*see if the OTHER string is empty.*/
otherA= other || '$' /*append a dollar sign ($) to OTHER. */
otherB= other'$' /*same as above, but with less fuss. */
guts= substr(c2b(other), 10, 3) /*obtain the 10th through 12th bits.*/
new= changeStr('A' , other, "Z") /*change the upper letter A ──► Z. */
tt= changeStr('~~', other, ";") /*change two tildes ──► one semicolon.*/
joined= dingsta || dingsta2 /*join two strings together (concat). */
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
c2b: return x2b( c2x( arg(1) ) ) /*return the string as a binary string.*/ |
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.
| #C | C | #define _CRT_SECURE_NO_WARNINGS // turn off panic warnings
#define _CRT_NONSTDC_NO_DEPRECATE // enable old-gold POSIX names in MSVS
#include <stdio.h>
#include <stdlib.h>
char* bin2str(unsigned value, char* buffer)
{
// This algorithm is not the fastest one, but is relativelly simple.
//
// A faster algorithm would be conversion octets to strings by a lookup table.
// There is only 2**8 == 256 octets, therefore we would need only 2048 bytes
// for the lookup table. Conversion of a 64-bit integers would need 8 lookups
// instead 64 and/or/shifts of bits etc. Even more... lookups may be implemented
// with XLAT or similar CPU instruction... and AVX/SSE gives chance for SIMD.
const unsigned N_DIGITS = sizeof(unsigned) * 8;
unsigned mask = 1 << (N_DIGITS - 1);
char* ptr = buffer;
for (int i = 0; i < N_DIGITS; i++)
{
*ptr++ = '0' + !!(value & mask);
mask >>= 1;
}
*ptr = '\0';
// Remove leading zeros.
//
for (ptr = buffer; *ptr == '0'; ptr++)
;
return ptr;
}
char* bin2strNaive(unsigned value, char* buffer)
{
// This variation of the solution doesn't use bits shifting etc.
unsigned n, m, p;
n = 0;
p = 1; // p = 2 ** n
while (p <= value / 2)
{
n = n + 1;
p = p * 2;
}
m = 0;
while (n > 0)
{
buffer[m] = '0' + value / p;
value = value % p;
m = m + 1;
n = n - 1;
p = p / 2;
}
buffer[m + 1] = '\0';
return buffer;
}
int main(int argc, char* argv[])
{
const unsigned NUMBERS[] = { 5, 50, 9000 };
const int RADIX = 2;
char buffer[(sizeof(unsigned)*8 + 1)];
// Function itoa is an POSIX function, but it is not in C standard library.
// There is no big surprise that Microsoft deprecate itoa because POSIX is
// "Portable Operating System Interface for UNIX". Thus it is not a good
// idea to use _itoa instead itoa: we lost compatibility with POSIX;
// we gain nothing in MS Windows (itoa-without-underscore is not better
// than _itoa-with-underscore). The same holds for kbhit() and _kbhit() etc.
//
for (int i = 0; i < sizeof(NUMBERS) / sizeof(unsigned); i++)
{
unsigned value = NUMBERS[i];
itoa(value, buffer, RADIX);
printf("itoa: %u decimal = %s binary\n", value, buffer);
}
// Yeep, we can use a homemade bin2str function. Notice that C is very very
// efficient (as "hi level assembler") when bit manipulation is needed.
//
for (int i = 0; i < sizeof(NUMBERS) / sizeof(unsigned); i++)
{
unsigned value = NUMBERS[i];
printf("bin2str: %u decimal = %s binary\n", value, bin2str(value, buffer));
}
// Another implementation - see above.
//
for (int i = 0; i < sizeof(NUMBERS) / sizeof(unsigned); i++)
{
unsigned value = NUMBERS[i];
printf("bin2strNaive: %u decimal = %s binary\n", value, bin2strNaive(value, buffer));
}
return EXIT_SUCCESS;
}
|
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.
| #Prolog | Prolog |
:- use_module(bitmap).
:- use_module(bitmapIO).
:- use_module(library(clpfd)).
% ends when X1 = X2 and Y1 = Y2
draw_recursive_line(NPict,Pict,Color,X,X,_DX,_DY,Y,Y,_E,_Sx,_Sy):-
set_pixel0(NPict,Pict,[X,Y],Color).
draw_recursive_line(NPict,Pict,Color,X,X2,DX,DY,Y,Y2,E,Sx,Sy):-
set_pixel0(TPict,Pict,[X,Y],Color),
E2 #= 2*E,
% because we can't accumulate error we set Ey or Ex to 1 or 0
% depending on whether we need to add dY or dX to the error term
( E2 >= DY ->
Ey = 1, NX #= X + Sx;
Ey = 0, NX = X),
( E2 =< DX ->
Ex = 1, NY #= Y + Sy;
Ex = 0, NY = Y),
NE #= E + DX*Ex + DY*Ey,
draw_recursive_line(NPict,TPict,Color,NX,X2,DX,DY,NY,Y2,NE,Sx,Sy).
draw_line(NPict,Pict,Color,X1,Y1,X2,Y2):-
DeltaY #= Y2-Y1,
DeltaX #= X2-X1,
( DeltaY < 0 -> Sy = -1; Sy = 1),
( DeltaX < 0 -> Sx = -1; Sx = 1),
DX #= abs(DeltaX),
DY #= -1*abs(DeltaY),
E #= DY+DX,
draw_recursive_line(NPict,Pict,Color,X1,X2,DX,DY,Y1,Y2,E,Sx,Sy).
init:-
new_bitmap(B,[100,100],[255,255,255]),
draw_line(NB,B,[0,0,0],2,2,10,90),
write_ppm_p6('line.ppm',NB).
|
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
| #WDTE | WDTE | let io => import 'io';
let ex => switch 'this is irrelevant for this example' {
false => 'This is, obviously, not returned.';
'a string' => 'This is also not returned.';
true => 'This is returned.';
};
ex -- io.writeln io.stdout; |
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
| #Wren | Wren | var embed = true
System.printAll([embed, ", ", !embed, ", ", "Is Wren embeddable? " + embed.toString]) |
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)..
| #QBasic | QBasic | DECLARE FUNCTION compasspoint$ (h!)
DIM SHARED POINT$(32)
FOR i = 1 TO 32
READ d$: POINT$(i) = d$
NEXT i
FOR i = 0 TO 32
heading = i * 11.25
IF (i MOD 3) = 1 THEN
heading = heading + 5.62
ELSE
IF (i MOD 3) = 2 THEN heading = heading - 5.62
END IF
ind = i MOD 32 + 1
PRINT ind, compasspoint$(heading), heading
NEXT i
END
DATA "North ", "North by east ", "North-northeast ", "Northeast by north"
DATA "Northeast ", "Northeast by east ", "East-northeast ", "East by north "
DATA "East ", "East by south ", "East-southeast ", "Southeast by east "
DATA "Southeast ", "Southeast by south", "South-southeast ", "South by east "
DATA "South ", "South by west ", "South-southwest ", "Southwest by south"
DATA "Southwest ", "Southwest by west ", "West-southwest ", "West by south "
DATA "West ", "West by north ", "West-northwest ", "Northwest by west "
DATA "Northwest ", "Northwest by north", "North-northwest ", "North by west "
FUNCTION compasspoint$ (h)
x = h / 11.25 + 1.5
IF (x >= 33!) THEN x = x - 32!
compasspoint$ = POINT$(INT(x))
END FUNCTION |
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.
| #LFE | LFE | (defun bitwise (a b)
(lists:map
(lambda (x) (io:format "~p~n" `(,x)))
`(,(band a b)
,(bor a b)
,(bxor a b)
,(bnot a)
,(bsl a b)
,(bsr a b)))
'ok)
(defun dec->bin (x)
(integer_to_list x 2))
(defun describe (func arg1 arg2 result)
(io:format "(~s ~s ~s): ~s~n"
(list func (dec->bin arg1) (dec->bin arg2) (dec->bin result))))
(defun bitwise
((a b 'binary)
(describe "band" a b (band a b))
(describe "bor" a b (bor a b))
(describe "bxor" a b (bxor a b))
(describe "bnot" a b (bnot a))
(describe "bsl" a b (bsl a b))
(describe "bsr" a b (bsr a b))))
|
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.
| #Phix | Phix | with javascript_semantics
-- Some colour constants:
constant black = #000000,
-- blue = #0000FF,
-- green = #00FF00,
-- red = #FF0000,
white = #FFFFFF
-- Create new image filled with some colour
function new_image(integer width, integer height, integer fill_colour=black)
return repeat(repeat(fill_colour,height),width)
end function
-- Usage example:
sequence image = new_image(800,600)
-- Set pixel color:
image[400][300] = white
-- Get pixel color
integer colour = image[400][300] -- Now colour is #FFFFFF
|
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 | #Vlang | Vlang | import math.big
fn bell_triangle(n int) [][]big.Integer {
mut tri := [][]big.Integer{len: n}
for i in 0..n {
tri[i] = []big.Integer{len: i}
for j in 0..i {
tri[i][j] = big.zero_int
}
}
tri[1][0] = big.integer_from_u64(1)
for i in 2..n {
tri[i][0] = tri[i-1][i-2]
for j := 1; j < i; j++ {
tri[i][j] = tri[i][j-1] + tri[i-1][j-1]
}
}
return tri
}
fn main() {
bt := bell_triangle(51)
println("First fifteen and fiftieth Bell numbers:")
for i := 1; i <= 15; i++ {
println("${i:2}: ${bt[i][0]}")
}
println("50: ${bt[50][0]}")
println("\nThe first ten rows of Bell's triangle:")
for i := 1; i <= 10; i++ {
println(bt[i])
}
} |
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 | #Wren | Wren | import "/big" for BigInt
import "/fmt" for Fmt
var bellTriangle = Fn.new { |n|
var tri = List.filled(n, null)
for (i in 0...n) {
tri[i] = List.filled(i, null)
for (j in 0...i) tri[i][j] = BigInt.zero
}
tri[1][0] = BigInt.one
for (i in 2...n) {
tri[i][0] = tri[i-1][i-2]
for (j in 1...i) {
tri[i][j] = tri[i][j-1] + tri[i-1][j-1]
}
}
return tri
}
var bt = bellTriangle.call(51)
System.print("First fifteen and fiftieth Bell numbers:")
for (i in 1..15) Fmt.print("$2d: $,i", i, bt[i][0])
Fmt.print("$2d: $,i", 50, bt[50][0])
System.print("\nThe first ten rows of Bell's triangle:")
for (i in 1..10) Fmt.print("$,7i", bt[i]) |
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 .
| #Phix | Phix | with javascript_semantics
function benford(sequence s, string title)
sequence f = repeat(0,9)
for i=1 to length(s) do
integer fdx = sprint(s[i])[1]-'0'
f[fdx] += 1
end for
sequence res = {title,"Digit Observed% Predicted%"}
for i=1 to length(f) do
atom o = f[i]/length(s)*100,
e = log10(1+1/i)*100
res = append(res,sprintf(" %d %9.3f %8.3f", {i, o, e}))
end for
return res
end function
function fib(integer lim)
atom a=0, b=1
sequence res = repeat(0,lim)
for i=1 to lim do
{res[i], a, b} = {b, b, b+a}
end for
return res
end function
sequence res = {benford(fib(1000),"First 1000 Fibonacci numbers"),
benford(get_primes(-10000),"First 10000 Prime numbers"),
benford(sq_power(3,tagset(500)),"First 500 powers of three")}
papply(true,printf,{1,{"%-40s%-40s%-40s\n"},columnize(res)})
|
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.
| #PARI.2FGP | PARI/GP | for(n=0,60,t=bernfrac(n);if(t,print(n" "t))) |
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.
| #E | E | /** Returns null if the value is not found. */
def binarySearch(collection, value) {
var low := 0
var high := collection.size() - 1
while (low <= high) {
def mid := (low + high) // 2
def comparison := value.op__cmp(collection[mid])
if (comparison.belowZero()) { high := mid - 1 } \
else if (comparison.aboveZero()) { low := mid + 1 } \
else if (comparison.isZero()) { return mid } \
else { throw("You expect me to binary search with a partial order?") }
}
return null
} |
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
| #Phix | Phix | with javascript_semantics
constant tests = {"abracadabra", "seesaw", "elk", "grrrrrr", "up", "a"}
for test=1 to length(tests) do
string s = tests[test],
t = shuffle(s)
for i=1 to length(t) do
for j=1 to length(t) do
integer {ti,tj} = {t[i],t[j]}
if i!=j and ti!=s[j] and tj!=s[i] then
t[i] = tj
t[j] = ti
exit
end if
end for
end for
printf(1,"%s -> %s (%d)\n",{s,t,sum(sq_eq(t,s))})
end for
|
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.
| #Ring | Ring | # string creation
x = "hello world"
# string destruction
x = NULL
# string assignment with a null byte
x = "a"+char(0)+"b"
see len(x) # ==> 3
# string comparison
if x = "hello"
See "equal"
else
See "not equal"
ok
y = 'bc'
if strcmp(x,y) < 0
See x + " is lexicographically less than " + y
ok
# string cloning
xx = x
See x = xx # true, same length and content
# check if empty
if x = NULL
See "is empty"
ok
# append a byte
x += char(7)
# substring
x = "hello"
x[1] = "H"
See x + nl
# join strings
a = "hel"
b = "lo w"
c = "orld"
See a + b + 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.
| #Ruby | Ruby | # string creation
x = "hello world"
# string destruction
x = nil
# string assignment with a null byte
x = "a\0b"
x.length # ==> 3
# string comparison
if x == "hello"
puts "equal"
else
puts "not equal"
end
y = 'bc'
if x < y
puts "#{x} is lexicographically less than #{y}"
end
# string cloning
xx = x.dup
x == xx # true, same length and content
x.equal?(xx) # false, different objects
# check if empty
if x.empty?
puts "is empty"
end
# append a byte
p x << "\07"
# substring
p xx = x[0..-2]
x[1,2] = "XYZ"
p x
# replace bytes
p y = "hello world".tr("l", "L")
# join strings
a = "hel"
b = "lo w"
c = "orld"
p d = a + b + c |
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.
| #C.23 | C# | using System;
class Program
{
static void Main()
{
foreach (var number in new[] { 5, 50, 9000 })
{
Console.WriteLine(Convert.ToString(number, 2));
}
}
} |
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.
| #PureBasic | PureBasic | Procedure BresenhamLine(x0 ,y0 ,x1 ,y1)
If Abs(y1 - y0) > Abs(x1 - x0);
steep =#True
Swap x0, y0
Swap x1, y1
EndIf
If x0 > x1
Swap x0, x1
Swap y0, y1
EndIf
deltax = x1 - x0
deltay = Abs(y1 - y0)
error = deltax / 2
y = y0
If y0 < y1
ystep = 1
Else
ystep = -1
EndIf
For x = x0 To x1
If steep
Plot(y,x)
Else
Plot(x,y)
EndIf
error - deltay
If error < 0
y + ystep
error + deltax
EndIf
Next
EndProcedure
#Window1 = 0
#Image1 = 0
#ImgGadget = 0
#width = 300
#height = 300
Define.i Event
Define.f Angle
If OpenWindow(#Window1, 0, 0, #width, #height, "Bresenham's Line PureBasic Example", #PB_Window_SystemMenu|#PB_Window_ScreenCentered)
If CreateImage(#Image1, #width, #height)
ImageGadget(#ImgGadget, 0, 0, #width, #height, ImageID(#Image1))
StartDrawing(ImageOutput(#Image1))
FillArea(0,0,-1,$FFFFFF) :FrontColor(0)
While Angle < 2*#PI
BresenhamLine(150,150,150+Cos(Angle)*120,150+Sin(Angle)*120)
Angle + #PI/60
Wend
StopDrawing()
SetGadgetState(#ImgGadget, ImageID(#Image1))
Repeat
Event = WaitWindowEvent()
Until Event = #PB_Event_CloseWindow
EndIf
EndIf |
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #XLISP | XLISP | <xsl:if test="true() or false()">
True and false are returned by built-in XPath functions.
</xsl:if>
<xsl:if test="@myAttribute='true'">
Node attributes set to "true" or "false" are just strings. Use string comparison to convert them to booleans.
</xsl:if>
<xsl:if test="@myAttribute or not($nodeSet)">
Test an attribute for its presence (empty or not), or whether a node set is empty.
</xsl:if> |
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
| #XPL0 | XPL0 | <xsl:if test="true() or false()">
True and false are returned by built-in XPath functions.
</xsl:if>
<xsl:if test="@myAttribute='true'">
Node attributes set to "true" or "false" are just strings. Use string comparison to convert them to booleans.
</xsl:if>
<xsl:if test="@myAttribute or not($nodeSet)">
Test an attribute for its presence (empty or not), or whether a node set is empty.
</xsl:if> |
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)..
| #Racket | Racket | #lang racket
;;; Generate the headings and boxes
(define (i->heading/box i)
(values (let ((heading (* i #e11.25)))
(case (modulo i 3)
((1) (+ heading #e5.62))
((2) (- heading #e5.62))
(else heading)))
(add1 (modulo i 32))))
(define-values (headings-list box-list)
(for/lists (h-lst i-lst) ((i (in-range 0 (add1 32))))
(i->heading/box i)))
(define box-names
(list "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" "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"))
(define (heading->box h)
(let* ((n-boxes (length box-names))
(box-width (/ 360 n-boxes)))
(add1 (modulo (ceiling (- (/ h box-width) 1/2)) n-boxes))))
;;; displays a single row of the table, can also be used for titles
(define (display-row b a p)
(printf "~a | ~a | ~a~%"
(~a b #:width 2 #:align 'right)
(~a a #:width 6 #:align 'right)
(~a p)))
;;; display the table
(display-row "#" "Angle" "Point")
(displayln "---+--------+-------------------")
(for ((heading headings-list))
(let* ((box-number (heading->box heading)))
;; by coincidence, default value of the second argument to
;; real->decimal-string "decimal-digits" is 2,... just what we want!
(display-row box-number
(real->decimal-string heading)
(list-ref box-names (sub1 box-number)))))
(module+ test
(require rackunit)
;;; unit test heading->box (the business end of the implementation)
(check-= (heading->box 354.38) 1 0)
(check-= (heading->box 5.62) 1 0)
(check-= (heading->box 5.63) 2 0)
(check-= (heading->box 16.87) 2 0)) |
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.
| #Liberty_BASIC | Liberty BASIC |
' bitwise operations on byte-sized variables
v =int( 256 *rnd( 1))
s = 1
print "Shift ="; s; " place."
print
print "Number as dec. = "; v; " & as 8-bits byte = ", dec2Bin$( v)
print "NOT as dec. = "; bitInvert( v), dec2Bin$( bitInvert( v))
print "Shifted left as dec. = "; shiftLeft( v, s), dec2Bin$( shiftLeft( v, s))
print "Shifted right as dec. = "; shiftRight( v, s), dec2Bin$( shiftRight( v, s))
print "Rotated left as dec. = "; rotateLeft( v, s), dec2Bin$( rotateLeft( v, s))
print "Rotated right as dec. = "; rotateRight( v, s), dec2Bin$( rotateRight( v, s))
end
function shiftLeft( b, n)
shiftLeft =( b *2^n) and 255
end function
function shiftRight( b, n)
shiftRight =int(b /2^n)
end function
function rotateLeft( b, n)
rotateLeft = (( 2^n *b) mod 256) or ( b >127)
end function
function rotateRight( b, n)
rotateRight = (128*( b and 1)) or int( b /2)
end function
function bitInvert( b)
bitInvert =b xor 255
end function
function dec2Bin$( num) ' Given an integer decimal 0 <--> 255, returns binary equivalent as a string
n =num
dec2Bin$ =""
while ( num >0)
dec2Bin$ =str$( num mod 2) +dec2Bin$
num =int( num /2)
wend
dec2Bin$ =right$( "00000000" +dec2Bin$, 8)
end function
|
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.
| #PHP | PHP | class Bitmap {
public $data;
public $w;
public $h;
public function __construct($w = 16, $h = 16){
$white = array_fill(0, $w, array(255,255,255));
$this->data = array_fill(0, $h, $white);
$this->w = $w;
$this->h = $h;
}
//Fills a rectangle, or the whole image with black by default
public function fill($x = 0, $y = 0, $w = null, $h = null, $color = array(0,0,0)){
if (is_null($w)) $w = $this->w;
if (is_null($h)) $h = $this->h;
$w += $x;
$h += $y;
for ($i = $y; $i < $h; $i++){
for ($j = $x; $j < $w; $j++){
$this->setPixel($j, $i, $color);
}
}
}
public function setPixel($x, $y, $color = array(0,0,0)){
if ($x >= $this->w) return false;
if ($x < 0) return false;
if ($y >= $this->h) return false;
if ($y < 0) return false;
$this->data[$y][$x] = $color;
}
public function getPixel($x, $y){
return $this->data[$y][$x];
}
}
$b = new Bitmap(16,16);
$b->fill();
$b->fill(2, 2, 18, 18, array(240,240,240));
$b->setPixel(0, 15, array(255,0,0));
print_r($b->getPixel(3,3)); //(240,240,240) |
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 | #zkl | zkl | fcn bellTriangleW(start=1,wantRow=False){ // --> iterator
Walker.zero().tweak('wrap(row){
row.insert(0,row[-1]);
foreach i in ([1..row.len()-1]){ row[i]+=row[i-1] }
wantRow and row or row[-1]
}.fp(List(start))).push(start,start);
} |
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 .
| #Picat | Picat | go =>
N = 1000,
Fib = [fib(I) : I in 1..N],
check_benford(Fib),
nl.
% Check a list of numbers for Benford's law
check_benford(L) =>
Len = L.len,
println(len=Len),
Count = [F[1].to_integer() : Num in L, F=Num.to_string()].occurrences(),
P = new_map([I=D/Len : I=D in Count]),
println("Benford (percent):"),
foreach(D in 1..9)
B = benford(D)*100,
PI = P.get(D,0)*100,
Diff = abs(PI - B),
printf("%d: count=%5d observed: %0.2f%% Benford: %0.2f%% diff=%0.3f\n", D,Count.get(D,0),PI,B,Diff)
end,
nl.
benford(D) = log10(1+1/D).
% create an occurrences map of a list
occurrences(List) = Map =>
Map = new_map(),
foreach(E in List)
Map.put(E, cond(Map.has_key(E),Map.get(E)+1,1))
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 .
| #PicoLisp | PicoLisp |
(scl 4)
(load "@lib/misc.l")
(load "@lib/math.l")
(setq LOG10E 0.4343)
(de fibo (N)
(let (A 0 B 1 C NIL)
(make
(link B)
(do (dec N)
(setq C (+ A B) A B B C)
(link B)))))
(setq Actual
(let (
Digits (sort (mapcar '((N) (format (car (chop N)))) (fibo 1000)))
Count 0
)
(make
(for (Ds Digits Ds (cdr Ds))
(inc 'Count)
(when (n== (car Ds) (cadr Ds))
(link Count)
(setq Count 0))))))
(setq Expected
(mapcar
'((D) (*/ (log (+ 1. (/ 1. D))) LOG10E 1.))
(range 1 9)))
(prinl "Digit\tActual\tExpected")
(let (As Actual Bs Expected)
(for D 9
(prinl D "\t" (format (pop 'As) 3) "\t" (round (pop 'Bs) 3))))
(bye)
|
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.
| #Pascal | Pascal |
(* Taken from the 'Ada 99' project, https://marquisdegeek.com/code_ada99 *)
program BernoulliForAda99;
uses BigDecimalMath; {library for arbitary high precision BCD numbers}
type
Fraction = object
private
numerator, denominator: BigDecimal;
public
procedure assign(n, d: Int64);
procedure subtract(rhs: Fraction);
procedure multiply(value: Int64);
procedure reduce();
procedure writeOutput();
end;
function gcd(a, b: BigDecimal):BigDecimal;
begin
if (b = 0) then begin
gcd := a;
end
else begin
gcd := gcd(b, a mod b);
end;
end;
procedure Fraction.writeOutput();
var sign : char;
begin
sign := ' ';
if (numerator<0) then sign := '-';
if (denominator<0) then sign := '-';
write(sign + BigDecimalToStr(abs(numerator)):45);
write(' / ');
write(BigDecimalToStr(abs(denominator)));
end;
procedure Fraction.assign(n, d: Int64);
begin
numerator := n;
denominator := d;
end;
procedure Fraction.subtract(rhs: Fraction);
begin
numerator := numerator * rhs.denominator;
numerator := numerator - (rhs.numerator * denominator);
denominator := denominator * rhs.denominator;
end;
procedure Fraction.multiply(value: Int64);
var
temp :BigDecimal;
begin
temp := value;
numerator := numerator * temp;
end;
procedure Fraction.reduce();
var gcdResult: BigDecimal;
begin
gcdResult := gcd(numerator, denominator);
begin
numerator := numerator div gcdResult; (* div is Int64 division *)
denominator := denominator div gcdResult; (* could also use round(d/r) *)
end;
end;
function calculateBernoulli(n: Int64) : Fraction;
var
m, j: Int64;
results: array of Fraction;
begin
setlength(results, 60) ; {largest value 60}
for m:= 0 to n do
begin
results[m].assign(1, m+1);
for j:= m downto 1 do
begin
results[j-1].subtract(results[j]);
results[j-1].multiply(j);
results[j-1].reduce();
end;
end;
calculateBernoulli := results[0];
end;
(* Main program starts here *)
var
b: Int64;
result: Fraction;
begin
writeln('Calculating Bernoulli numbers...');
writeln('B( 0) : 1 / 1');
for b:= 1 to 60 do
begin
if (b<3) or ((b mod 2) = 0) then begin
result := calculateBernoulli(b);
write('B(',b:2,')');
write(' : ');
result.writeOutput();
writeln;
end;
end;
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.
| #EasyLang | EasyLang | func bin_search val . a[] res .
low = 0
high = len a[] - 1
res = -1
while low <= high and res = -1
mid = (low + high) div 2
if a[mid] > val
high = mid - 1
elif a[mid] < val
low = mid + 1
else
res = mid
.
.
.
a[] = [ 2 4 6 8 9 ]
call bin_search 8 a[] r
print r |
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
| #PHP | PHP | foreach (split(' ', 'abracadabra seesaw pop grrrrrr up a') as $w)
echo bestShuffle($w) . '<br>';
function bestShuffle($s1) {
$s2 = str_shuffle($s1);
for ($i = 0; $i < strlen($s2); $i++) {
if ($s2[$i] != $s1[$i]) continue;
for ($j = 0; $j < strlen($s2); $j++)
if ($i != $j && $s2[$i] != $s1[$j] && $s2[$j] != $s1[$i]) {
$t = $s2[$i];
$s2[$i] = $s2[$j];
$s2[$j] = $t;
break;
}
}
return "$s1 $s2 " . countSame($s1, $s2);
}
function countSame($s1, $s2) {
$cnt = 0;
for ($i = 0; $i < strlen($s2); $i++)
if ($s1[$i] == $s2[$i])
$cnt++;
return "($cnt)";
} |
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.
| #Run_BASIC | Run BASIC | ' Create string
s$ = "Hello, world"
' String destruction
s$ = ""
' String comparison
If s$ = "Hello, world" then print "Equal String"
' Copying string
a$ = s$
' Check If empty
If s$ = "" then print "String is MT"
' Append a byte
s$ = s$ + Chr$(65)
' Extract a substring
a$ = Mid$(s$, 1, 5) ' bytes 1 -> 5
'substitute string "world" with "universe"
a$ = "Hello, world"
for i = 1 to len(a$)
if mid$(a$,i,5)="world" then
a$=left$(a$,i-1)+"universe"+mid$(a$,i+5)
end if
next
print a$
'join strings
s$ = "See " + "you " + "later."
print s$ |
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.
| #Rust | Rust | use std::str;
fn main() {
// Create new string
let string = String::from("Hello world!");
println!("{}", string);
assert_eq!(string, "Hello world!", "Incorrect string text");
// Create and assign value to string
let mut assigned_str = String::new();
assert_eq!(assigned_str, "", "Incorrect string creation");
assigned_str += "Text has been assigned!";
println!("{}", assigned_str);
assert_eq!(assigned_str, "Text has been assigned!","Incorrect string text");
// String comparison, compared lexicographically byte-wise same as the asserts above
if string == "Hello world!" && assigned_str == "Text has been assigned!" {
println!("Strings are equal");
}
// Cloning -> string can still be used after cloning
let clone_str = string.clone();
println!("String is:{} and Clone string is: {}", string, clone_str);
assert_eq!(clone_str, string, "Incorrect string creation");
// Copying, string won't be usable anymore, accessing it will cause compiler failure
let copy_str = string;
println!("String copied now: {}", copy_str);
// Check if string is empty
let empty_str = String::new();
assert!(empty_str.is_empty(), "Error, string should be empty");
// Append byte, Rust strings are a stream of UTF-8 bytes
let byte_vec = [65]; // contains A
let byte_str = str::from_utf8(&byte_vec).unwrap();
assert_eq!(byte_str, "A", "Incorrect byte append");
// Substrings can be accessed through slices
let test_str = "Blah String";
let mut sub_str = &test_str[0..11];
assert_eq!(sub_str, "Blah String", "Error in slicing");
sub_str = &test_str[1..5];
assert_eq!(sub_str, "lah ", "Error in slicing");
sub_str = &test_str[3..];
assert_eq!(sub_str, "h String", "Error in slicing");
sub_str = &test_str[..2];
assert_eq!(sub_str, "Bl", "Error in slicing");
// String replace, note string is immutable
let org_str = "Hello";
assert_eq!(org_str.replace("l", "a"), "Heaao", "Error in replacement");
assert_eq!(org_str.replace("ll", "r"), "Hero", "Error in replacement");
// Joining strings requires a `String` and an &str or a two `String`s one of which needs an & for coercion
let str1 = "Hi";
let str2 = " There";
let fin_str = str1.to_string() + str2;
assert_eq!(fin_str, "Hi There", "Error in concatenation");
// Joining strings requires a `String` and an &str or two `Strings`s, one of which needs an & for coercion
let str1 = "Hi";
let str2 = " There";
let fin_str = str1.to_string() + str2;
assert_eq!(fin_str, "Hi There", "Error in concatenation");
// Splits -- note Rust supports passing patterns to splits
let f_str = "Pooja and Sundar are up in Tumkur";
let split_str: Vec<_> = f_str.split(' ').collect();
assert_eq!(split_str, ["Pooja", "and", "Sundar", "are", "up", "in", "Tumkur"], "Error in string split");
} |
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.
| #C.2B.2B | C++ | #include <bitset>
#include <iostream>
#include <limits>
#include <string>
void print_bin(unsigned int n) {
std::string str = "0";
if (n > 0) {
str = std::bitset<std::numeric_limits<unsigned int>::digits>(n).to_string();
str = str.substr(str.find('1')); // remove leading zeros
}
std::cout << str << '\n';
}
int main() {
print_bin(0);
print_bin(5);
print_bin(50);
print_bin(9000);
}
|
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.
| #Python | Python | def line(self, x0, y0, x1, y1):
"Bresenham's line algorithm"
dx = abs(x1 - x0)
dy = abs(y1 - y0)
x, y = x0, y0
sx = -1 if x0 > x1 else 1
sy = -1 if y0 > y1 else 1
if dx > dy:
err = dx / 2.0
while x != x1:
self.set(x, y)
err -= dy
if err < 0:
y += sy
err += dx
x += sx
else:
err = dy / 2.0
while y != y1:
self.set(x, y)
err -= dx
if err < 0:
x += sx
err += dy
y += sy
self.set(x, y)
Bitmap.line = line
bitmap = Bitmap(17,17)
for points in ((1,8,8,16),(8,16,16,8),(16,8,8,1),(8,1,1,8)):
bitmap.line(*points)
bitmap.chardisplay()
'''
The origin, 0,0; is the lower left, with x increasing to the right,
and Y increasing upwards.
The chardisplay above produces the following output :
+-----------------+
| @ |
| @ @ |
| @ @ |
| @ @ |
| @ @ |
| @ @ |
| @ @ |
| @ @ |
| @ @|
| @ @ |
| @ @ |
| @ @@ |
| @ @ |
| @ @ |
| @ @ |
| @ |
| |
+-----------------+
''' |
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
| #XSLT | XSLT | <xsl:if test="true() or false()">
True and false are returned by built-in XPath functions.
</xsl:if>
<xsl:if test="@myAttribute='true'">
Node attributes set to "true" or "false" are just strings. Use string comparison to convert them to booleans.
</xsl:if>
<xsl:if test="@myAttribute or not($nodeSet)">
Test an attribute for its presence (empty or not), or whether a node set is empty.
</xsl:if> |
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
| #Z80_Assembly | Z80 Assembly | BIT 0,A
jr nz,true
;;;;;;;;;;;;;;;;
AND 1
jr nz,true
;;;;;;;;;;;;;;;;
RRCA
jr c,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
| #zkl | zkl | a:=True;
b:=False;
True.dir(); |
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)..
| #Raku | Raku | sub point (Int $index) {
my $ix = $index % 32;
if $ix +& 1
{ "&point(($ix + 1) +& 28) by &point(((2 - ($ix +& 2)) * 4) + $ix +& 24)" }
elsif $ix +& 2
{ "&point(($ix + 2) +& 24)-&point(($ix +| 4) +& 28)" }
elsif $ix +& 4
{ "&point(($ix + 8) +& 16)&point(($ix +| 8) +& 24)" }
else
{ <north east south west>[$ix div 8]; }
}
sub test-angle ($ix) { $ix * 11.25 + (0, 5.62, -5.62)[ $ix % 3 ] }
sub angle-to-point(\𝜽) { floor 𝜽 / 360 * 32 + 0.5 }
for 0 .. 32 -> $ix {
my \𝜽 = test-angle($ix);
printf " %2d %6.2f° %s\n",
$ix % 32 + 1,
𝜽,
tc point angle-to-point 𝜽;
} |
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.
| #Lingo | Lingo | put bitAND(2,7)
put bitOR(2,7)
put bitXOR(2,7)
put bitNOT(7) |
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.
| #PicoLisp | PicoLisp | # Create an empty image of 120 x 90 pixels
(setq *Ppm (make (do 90 (link (need 120)))))
# Fill an image with a given color
(de ppmFill (Ppm R G B)
(for Y Ppm
(map
'((X) (set X (list R G B)))
Y ) ) )
# Set pixel with a color
(de ppmSetPixel (Ppm X Y R G B)
(set (nth Ppm Y X) (list R G B)) )
# Get the color of a pixel
(de ppmGetPixel (Ppm X Y)
(get Ppm Y X) ) |
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.
| #PL.2FI | PL/I |
/* Declaration for an image, suitable for BMP files. */
declare image(0:500, 0:500) bit (24) aligned;
image = '000000000000000011111111'b;
/* Sets the entire image to red. */
image(10,40) = '111111110000000000000000'b;
/* Sets one pixel to blue. */
declare color bit (24) aligned;
color = image(20,50); /* Obtain the color of a pixel */
/* To allocate an image of size (x,y) */
allocate_image: procedure (image, x, y);
declare image (*, *) controlled bit (24) aligned;
declare (x, y) fixed binary (31);
allocate image (0:x, 0:y);
end allocate_image;
/* To use the above procedure, it's necessary to define */
/* the image in the calling program thus, for BMP images: */
declare image(*,*) controlled bit (24) aligned;
|
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 .
| #PL.2FI | PL/I |
(fofl, size, subrg):
Benford: procedure options(main); /* 20 October 2013 */
declare sc(1000) char(1), f(1000) float (16);
declare d fixed (1);
call Fibonacci(f);
call digits(sc, f);
put skip list ('digit expected obtained');
do d= 1 upthru 9;
put skip edit (d, log10(1 + 1/d), tally(sc, trim(d))/1000)
(f(3), 2 f(13,8) );
end;
Fibonacci: procedure (f);
declare f(*) float (16);
declare i fixed binary;
f(1), f(2) = 1;
do i = 3 to 1000;
f(i) = f(i-1) + f(i-2);
end;
end Fibonacci;
digits: procedure (sc, f);
declare sc(*) char(1), f(*) float (16);
sc = substr(trim(f), 1, 1);
end digits;
tally: procedure (sc, d) returns (fixed binary);
declare sc(*) char(1), d char(1);
declare (i, t) fixed binary;
t = 0;
do i = 1 to 1000;
if sc(i) = d then t = t + 1;
end;
return (t);
end tally;
end Benford;
|
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 .
| #PL.2FpgSQL | PL/pgSQL |
WITH recursive
constant(val) AS
(
SELECT 1000.
)
,
fib(a,b) AS
(
SELECT CAST(0 AS NUMERIC), CAST(1 AS NUMERIC)
UNION ALL
SELECT b,a+b
FROM fib
)
,
benford(first_digit, probability_real, probability_theoretical) AS
(
SELECT *,
CAST(log(1. + 1./CAST(first_digit AS INT)) AS NUMERIC(5,4)) probability_theoretical
FROM (
SELECT first_digit, CAST(COUNT(1)/(SELECT val FROM constant) AS NUMERIC(5,4)) probability_real FROM
(
SELECT SUBSTRING(CAST(a AS VARCHAR(100)),1,1) first_digit
FROM fib
WHERE SUBSTRING(CAST(a AS VARCHAR(100)),1,1) <> '0'
LIMIT (SELECT val FROM constant)
) t
GROUP BY first_digit
) f
ORDER BY first_digit ASC
)
SELECT *
FROM benford CROSS JOIN
(SELECT CAST(corr(probability_theoretical,probability_real) AS NUMERIC(5,4)) correlation
FROM benford) c
|
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.
| #Perl | Perl | #!perl
use strict;
use warnings;
use List::Util qw(max);
use Math::BigRat;
my $one = Math::BigRat->new(1);
sub bernoulli_print {
my @a;
for my $m ( 0 .. 60 ) {
push @a, $one / ($m + 1);
for my $j ( reverse 1 .. $m ) {
# This line:
( $a[$j-1] -= $a[$j] ) *= $j;
# is a faster version of the following line:
# $a[$j-1] = $j * ($a[$j-1] - $a[$j]);
# since it avoids unnecessary object creation.
}
next unless $a[0];
printf "B(%2d) = %44s/%s\n", $m, $a[0]->parts;
}
}
bernoulli_print();
|
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.
| #Eiffel | Eiffel | class
APPLICATION
create
make
feature {NONE} -- Initialization
make
local
a: ARRAY [INTEGER]
keys: ARRAY [INTEGER]
do
a := <<0, 1, 4, 5, 6, 7, 8, 9,
12, 26, 45, 67, 78, 90,
98, 123, 211, 234, 456,
769, 865, 2345, 3215,
14345, 24324>>
keys := <<0, 42, 45, 24324, 99999>>
across keys as k loop
if has_binary (a, k.item) then
print ("The array has an element " + k.item.out)
else
print ("The array has NOT an element " + k.item.out)
end
print ("%N")
end
end
feature -- Search
has_binary (a: ARRAY [INTEGER]; key: INTEGER): BOOLEAN
-- Does `a[a.lower..a.upper]' include an element `key'?
require
is_sorted (a, a.lower, a.upper)
local
i: INTEGER
do
i := where_binary (a, key)
if a.lower <= i and i <= a.upper then
Result := True
else
Result := False
end
end
where_binary (a: ARRAY [INTEGER]; key: INTEGER): INTEGER
-- The index of an element `key' within `a[a.lower..a.upper]' if it exists.
-- Otherwise an integer outside `[a.lower..a.upper]'
require
is_sorted (a, a.lower, a.upper)
do
Result := where_binary_range (a, key, a.lower, a.upper)
end
where_binary_range (a: ARRAY [INTEGER]; key: INTEGER; low, high: INTEGER): INTEGER
-- The index of an element `key' within `a[low..high]' if it exists.
-- Otherwise an integer outside `[low..high]'
note
source: "http://arxiv.org/abs/1211.4470"
require
is_sorted (a, low, high)
local
i, j, mid: INTEGER
do
if low > high then
Result := low - 1
else
from
i := low
j := high
mid := low
Result := low - 1
invariant
low <= i and i <= mid + 1
low <= mid and mid <= j and j <= high
i <= j
has (a, key, i, j) = has (a, key, low, high)
until
i >= j
loop
mid := i + (j - i) // 2
if a [mid] < key then
i := mid + 1
else
j := mid
end
variant
j - i
end
if a [i] = key then
Result := i
end
end
ensure
low <= Result and Result <= high implies a [Result] = key
Result < low or Result > high implies not has (a, key, low, high)
end
feature -- Implementation
is_sorted (a: ARRAY [INTEGER]; low, high: INTEGER): BOOLEAN
-- Is `a[low..high]' sorted in nondecreasing order?
require
a.lower <= low
high <= a.upper
do
Result := across low |..| (high - 1) as i all a [i.item] <= a [i.item + 1] end
end
has (a: ARRAY [INTEGER]; key: INTEGER; low, high: INTEGER): BOOLEAN
-- Is there an element `key' in `a[low..high]'?
require
a.lower <= low
high <= a.upper
do
Result := across low |..| high as i some a [i.item] = key end
end
end |
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
| #Picat | Picat | import cp.
go =>
Words = ["abracadabra",
"seesaw",
"elk",
"grrrrrr",
"up",
"a",
"shuffle",
"aaaaaaa"
],
foreach(Word in Words)
best_shuffle(Word,Best,_Score),
printf("%s, %s, (%d)\n", Word,Best,diff_word(Word, Best))
end,
nl.
best_shuffle(Word,Best,Score) =>
WordAlpha = Word.map(ord), % convert to integers
WordAlphaNoDups = WordAlpha.remove_dups(),
% occurrences of each character in the word
Occurrences = occurrences(WordAlpha),
Len = Word.length,
% Decision variables
WordC = new_list(Len),
WordC :: WordAlphaNoDups,
%
% The constraints
%
% Ensure that the shuffled word has the same
% occurrences for each character
foreach(V in WordAlphaNoDups)
count(V, WordC,#=, Occurrences.get(V))
end,
% The score is the number of characters
% in the same position as the origin word
% (to be minimized).
Score #= sum([WordC[I] #= WordAlpha[I] : I in 1..Len]),
if var(Score) then
% We don't have a score yet: minimize Score
solve([$min(Score),split], WordC)
else
% Get a solution for the given Score
solve([split], WordC)
end,
% convert back to alpha
Best = WordC.map(chr).
diff_word(W1,W2) = Diff =>
Diff = sum([1 : I in 1..W1.length, W1[I]==W2[I]]).
occurrences(L) = Occ =>
Occ = new_map(),
foreach(E in L)
Occ.put(E, Occ.get(E,0) + 1)
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.
| #Seed7 | Seed7 | var string: stri is "asdf"; # variable declaration
const string: stri is "jkl"; # constant declaration
|
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.
| #Smalltalk | Smalltalk | s := "abc" # create a string (immutable if its a literal constant in the program)
s := #[16r01 16r02 16r00 16r03] asString # strings can contain any value, even nulls
s := String new:3. # a mutable string
v := s # assignment
s = t # same contents
s < t # less
s <= t # less or equal
s = '' # equal empty string
s isEmpty # ditto
s size # string length
t := s copy # a copy
t := s copyFrom:2 to:3 # a substring
t := s copyReplaceFrom:2 to:3 with:'**' # a copy with some replacements
s replaceFrom:2 to:3 with:'**' # inplace replace (must be mutable)
s replaceAll:$x with:$y # destructive replace of characters
s copyReplaceAll:$x with:$y # non-destructive replace
s replaceString:s1 withString:s2 # substring replace
s3 := s1 , s2 # concatenation of strings
s2 := s1 , $x # append a character
s2 := s1 , 123 asCharacter # append an arbitrary byte
s := 'Hello / 今日は' # they support unicode (at least up to 16rFFFF, some more) |
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.
| #Ceylon | Ceylon | shared void run() {
void printBinary(Integer integer) =>
print(Integer.format(integer, 2));
printBinary(5);
printBinary(50);
printBinary(9k);
} |
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.
| #Racket | Racket |
#lang racket
(require racket/draw)
(define (draw-line dc x0 y0 x1 y1)
(define dx (abs (- x1 x0)))
(define dy (abs (- y1 y0)))
(define sx (if (> x0 x1) -1 1))
(define sy (if (> y0 y1) -1 1))
(cond
[(> dx dy)
(let loop ([x x0] [y y0] [err (/ dx 2.0)])
(unless (= x x1)
(send dc draw-point x y)
(define newerr (- err dy))
(if (< newerr 0)
(loop (+ x sx) (+ y sy) (+ newerr dx))
(loop (+ x sx) y newerr))))]
[else
(let loop ([x x0] [y y0] [err (/ dy 2.0)])
(unless (= y y1)
(send dc draw-point x y)
(define newerr (- err dy))
(if (< newerr 0)
(loop (+ x sx) (+ y sy) newerr)
(loop x (+ y sy) (+ newerr dy)))))]))
(define bm (make-object bitmap% 17 17))
(define dc (new bitmap-dc% [bitmap bm]))
(send dc set-smoothing 'unsmoothed)
(send dc set-pen "red" 1 'solid)
(for ([points '((1 8 8 16) (8 16 16 8) (16 8 8 1) (8 1 1 8))])
(apply draw-line (cons dc points)))
bm
|
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
| #zonnon | zonnon |
var
a,b,c: boolean;
begin
a := false;
b := true;
c := 1 > 2;
...
|
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)..
| #Red | Red | Red []
d: charset [#"N" #"E" #"S" #"W"] ;; main directions
hm: #() ;; hm = hashmap - key= heading, value = [box , compass point]
compass-points: [N NbE NNE NEbN NE NEbE ENE EbN E EbS ESE SEbE SE SEbS SSE
SbE S SbW SSW SWbS SW SWbW WSW WbS W WbN WNW NWbW NW NWbN NNW NbW N ]
expand: func [cp repl][ ;; expand compass point to words
parse cp [ copy a thru d ahead 2 d insert "-" ] ;; insert "-" after first direction, if followed by 2 more
foreach [src dst ] repl [ replace/all cp to-string src to-string dst ] ;; N -> north ...
uppercase/part cp 1 ;; convert first letter to uppercase
]
print-line: does [ print [pad/left hm/:heading/1 3 pad hm/:heading/2 20 heading ] ]
forall compass-points [
i: (index? compass-points) - 1 ;; so i = 0..33
heading: i * 11.25 + either 1 = rem: i % 3 [ 5.62] ;; rem = remainder
[ either rem = 2 [-5.62] [0.0] ]
hm/:heading: reduce [ (i % 32 + 1 ) expand to-string compass-points/1 [N north b " by " S south E east W west] ]
print-line heading
] |
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.
| #LiveCode | LiveCode | put "and:" && (255 bitand 2) & comma into bitops
put " or:" && (255 bitor 2) & comma after bitops
put " xor:" && (255 bitxor 2) & comma after bitops
put " not:" && (bitnot 255) after bitops
put bitops
-- Ouput
and: 2, or: 255, xor: 253, not: 4294967040 |
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.
| #Processing | Processing | PGraphics bitmap = createGraphics(100,100); // Create the bitmap
bitmap.beginDraw();
bitmap.background(255, 0, 0); // Fill bitmap with red rgb color
bitmap.endDraw();
image(bitmap, 0, 0); // Place bitmap on screen.
color b = color(0, 0, 255); // Define a blue rgb color
set(50, 50, b); // Set blue colored pixel in the middle of the screen
color c = get(50, 50); // Get the color of same pixel
if(b == c) print("Color changed correctly"); // Verify
|
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 .
| #PowerShell | PowerShell |
$url = "https://oeis.org/A000045/b000045.txt"
$file = "$env:TEMP\FibonacciNumbers.txt"
(New-Object System.Net.WebClient).DownloadFile($url, $file)
$benford = Get-Content -Path $file |
Select-Object -Skip 1 -First 1000 |
ForEach-Object {(($_ -split " ")[1].ToString().ToCharArray())[0]} |
Group-Object |
Select-Object -Property @{Name="Digit" ; Expression={[int]($_.Name)}},
Count,
@{Name="Actual" ; Expression={$_.Count/1000}},
@{Name="Expected"; Expression={[double]("{0:f5}" -f [Math]::Log10(1 + 1 / $_.Name))}}
$benford | Sort-Object -Property Digit | Format-Table -AutoSize
Remove-Item -Path $file -Force -ErrorAction SilentlyContinue
|
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.
| #Phix | Phix | with javascript_semantics
include builtins/mpfr.e
procedure bernoulli(mpq rop, integer n)
sequence a = mpq_inits(n+1)
for m=1 to n+1 do
mpq_set_si(a[m], 1, m)
for j=m-1 to 1 by -1 do
mpq_sub(a[j], a[j+1], a[j])
mpq_set_si(rop, j, 1)
mpq_mul(a[j], a[j], rop)
end for
end for
mpq_set(rop, a[1])
a = mpq_free(a)
end procedure
mpq rop = mpq_init()
mpz n = mpz_init(),
d = mpz_init()
for i=0 to 60 do
bernoulli(rop, i)
if mpq_cmp_si(rop, 0, 1) then
mpq_get_num(n, rop)
mpq_get_den(d, rop)
string ns = mpz_get_str(n),
ds = mpz_get_str(d)
printf(1,"B(%2d) = %44s / %s\n", {i,ns,ds})
end if
end for
{n,d} = mpz_free({n,d})
rop = mpq_free(rop)
|
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.
| #Elixir | Elixir | defmodule Binary do
def search(list, value), do: search(List.to_tuple(list), value, 0, length(list)-1)
def search(_tuple, _value, low, high) when high < low, do: :not_found
def search(tuple, value, low, high) do
mid = div(low + high, 2)
midval = elem(tuple, mid)
cond do
value < midval -> search(tuple, value, low, mid-1)
value > midval -> search(tuple, value, mid+1, high)
value == midval -> mid
end
end
end
list = [0,1,4,5,6,7,8,9,12,26,45,67,78,90,98,123,211,234,456,769,865,2345,3215,14345,24324]
Enum.each([0,42,45,24324,99999], fn val ->
case Binary.search(list, val) do
:not_found -> IO.puts "#{val} not found in list"
index -> IO.puts "found #{val} at index #{index}"
end
end) |
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
| #PicoLisp | PicoLisp | (de bestShuffle (Str)
(let Lst NIL
(for C (setq Str (chop Str))
(if (assoc C Lst)
(con @ (cons C (cdr @)))
(push 'Lst (cons C)) ) )
(setq Lst (apply conc (flip (by length sort Lst))))
(let Res
(mapcar
'((C)
(prog1 (or (find <> Lst (circ C)) C)
(setq Lst (delete @ Lst)) ) )
Str )
(prinl Str " " Res " (" (cnt = Str Res) ")") ) ) ) |
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
| #PL.2FI | PL/I | shuffle: procedure options (main); /* 14/1/2011 */
declare (s, saves) character (20) varying, c character (1);
declare t(length(s)) bit (1);
declare (i, k, moves initial (0)) fixed binary;
get edit (s) (L);
put skip list (s);
saves = s;
t = '0'b;
do i = 1 to length (s);
if t(i) then iterate; /* This character has already been moved. */
c = substr(s, i, 1);
k = search (s, c, i+1);
if k > 0 then
do;
substr(s, i, 1) = substr(s, k, 1);
substr(s, k, 1) = c;
t(k), t(i) = '1'b;
end;
end;
do k = length(s) to 2 by -1;
if ^t(k) then /* this character wasn't moved. */
all: do;
c = substr(s, k, 1);
do i = k-1 to 1 by -1;
if c ^= substr(s, i, 1) then
if substr(saves, i, 1) ^= c then
do;
substr(s, k, 1) = substr(s, i, 1);
substr(s, i, 1) = c;
t(k) = '1'b;
leave all;
end;
end;
end;
end;
moves = length(s) - sum(t);
put skip edit (s, trim(moves))(a, x(1));
search: procedure (s, c, k) returns (fixed binary);
declare s character (*) varying;
declare c character (1);
declare k fixed binary;
declare i fixed binary;
do i = k to length(s);
if ^t(i) then if c ^= substr(s, i, 1) then return (i);
end;
return (0); /* No eligible character. */
end search;
end 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.
| #Tcl | Tcl | # string creation
set x "hello world"
# string destruction
unset x
# string assignment with a null byte
set x a\0b
string length $x ;# ==> 3
# string comparison
if {$x eq "hello"} {puts equal} else {puts "not equal"}
set y bc
if {$x < $y} {puts "$x is lexicographically less than $y"}
# string copying; cloning happens automatically behind the scenes
set xx $x
# check if empty
if {$x eq ""} {puts "is empty"}
if {[string length $x] == 0} {puts "is empty"}
# append a byte
append x \07
# substring
set xx [string range $x 0 end-1]
# replace bytes
set y [string map {l L} "hello world"]
# join strings
set a "hel"
set b "lo w"
set c "orld"
set d $a$b$c |
http://rosettacode.org/wiki/Binary_digits | Binary digits | Task
Create and display the sequence of binary digits for a given non-negative integer.
The decimal value 5 should produce an output of 101
The decimal value 50 should produce an output of 110010
The decimal value 9000 should produce an output of 10001100101000
The results can be achieved using built-in radix functions within the language (if these are available), or alternatively a user defined function can be used.
The output produced should consist just of the binary digits of each number followed by a newline.
There should be no other whitespace, radix or sign markers in the produced output, and leading zeros should not appear in the results.
| #Clojure | Clojure | (Integer/toBinaryString 5)
(Integer/toBinaryString 50)
(Integer/toBinaryString 9000) |
http://rosettacode.org/wiki/Bitmap/Bresenham%27s_line_algorithm | Bitmap/Bresenham's line algorithm | Task
Using the data storage type defined on the Bitmap page for raster graphics images,
draw a line given two points with Bresenham's line algorithm.
| #Raku | Raku | class Pixel { has UInt ($.R, $.G, $.B) }
class Bitmap {
has UInt ($.width, $.height);
has Pixel @!data;
method fill(Pixel $p) {
@!data = $p.clone xx ($!width*$!height)
}
method pixel(
$i where ^$!width,
$j where ^$!height
--> Pixel
) is rw { @!data[$i + $j * $!width] }
method set-pixel ($i, $j, Pixel $p) {
self.pixel($i, $j) = $p.clone;
}
method get-pixel ($i, $j) returns Pixel {
self.pixel($i, $j);
}
}
sub line(Bitmap $bitmap, $x0 is copy, $x1 is copy, $y0 is copy, $y1 is copy) {
my $steep = abs($y1 - $y0) > abs($x1 - $x0);
if $steep {
($x0, $y0) = ($y0, $x0);
($x1, $y1) = ($y1, $x1);
}
if $x0 > $x1 {
($x0, $x1) = ($x1, $x0);
($y0, $y1) = ($y1, $y0);
}
my $Δx = $x1 - $x0;
my $Δy = abs($y1 - $y0);
my $error = 0;
my $Δerror = $Δy / $Δx;
my $y-step = $y0 < $y1 ?? 1 !! -1;
my $y = $y0;
for $x0 .. $x1 -> $x {
my $pix = Pixel.new(R => 100, G => 200, B => 0);
if $steep {
$bitmap.set-pixel($y, $x, $pix);
} else {
$bitmap.set-pixel($x, $y, $pix);
}
$error += $Δerror;
if $error >= 0.5 {
$y += $y-step;
$error -= 1.0;
}
}
} |
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)..
| #REXX | REXX | /*REXX program "boxes the compass" [from degree (º) headings ───► a 32 point set]. */
parse arg $ /*allow º headings to be specified.*/
if $='' then $= 0 16.87 16.88 33.75 50.62 50.63 67.5 84.37 84.38 101.25 118.12 118.13 ,
135 151.87 151.88 168.75 185.62 185.63 202.5 219.37 219.38 236.25 ,
253.12 253.13 270 286.87 286.88 303.75 320.62 320.63 337.5 354.37 354.38
/* [↑] use default, they're in degrees*/
@pts= 'n nbe n-ne nebn ne nebe e-ne ebn e ebs e-se sebe se sebs s-se sbe',
"s sbw s-sw swbs sw swbw w-sw wbs w wbn w-nw nwbw nw nwbn n-nw nbw"
#= words(@pts) + 1 /*#: used for integer ÷ remainder (//)*/
dirs= 'north south east west' /*define cardinal compass directions. */
/* [↓] choose a glyph for degree (°).*/
if 4=='f4'x then degSym= "a1"x /*is this system an EBCDIC system? */
else degSym= "a7"x /*'f8'x is the degree symbol: ° vs º */
/*──────────────────────────── f8 vs a7*/
say right(degSym'heading', 30) center("compass heading", 20)
say right( '════════', 30) copies( "═", 20)
pad= ' ' /*used to interject a blank for output.*/
do j=1 for words($); x= word($, j) /*obtain one of the degree headings. */
say right( format(x, , 2)degSym, 30-1) pad boxHeading(x)
end /*j*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
boxHeading: y= arg(1)//360; if y<0 then y=360-y /*normalize heading within unit circle.*/
z= word(@pts, trunc(max(1, (y/11.25 +1.5) // #))) /*convert degrees─►heading*/
do k=1 for words(dirs); d= word(dirs, k)
z= changestr( left(d, 1), z, d)
end /*k*/ /* [↑] old, haystack, new*/
return changestr('b', z, " by ") /*expand "b" ───► " by ".*/ |
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.
| #LLVM | LLVM | ; ModuleID = 'test.o'
;e means little endian
;p: { pointer size : pointer abi : preferred alignment for pointers }
;i same for integers
;v is for vectors
;f for floats
;a for aggregate types
;s for stack objects
;n: {size:size:size...}, best integer sizes
target datalayout = "e-p:32:32:32-i1:8:8-i8:8:8-i16:16:16-i32:32:32-i64:64:64-f32:32:32-f64:64:64-v64:64:64-v128:128:128-a0:0:64-f80:32:32-n8:16:32"
;this was compiled with mingw32; thus it must be linked to an ABI compatible c library
target triple = "i386-mingw32"
@.str = private constant [13 x i8] c"a and b: %d\0A\00", align 1 ; <[13 x i8]*> [#uses=1]
@.str1 = private constant [12 x i8] c"a or b: %d\0A\00", align 1 ; <[12 x i8]*> [#uses=1]
@.str2 = private constant [13 x i8] c"a xor b: %d\0A\00", align 1 ; <[13 x i8]*> [#uses=1]
@.str3 = private constant [11 x i8] c"not a: %d\0A\00", align 1 ; <[11 x i8]*> [#uses=1]
@.str4 = private constant [12 x i8] c"a << n: %d\0A\00", align 1 ; <[12 x i8]*> [#uses=1]
@.str5 = private constant [12 x i8] c"a >> n: %d\0A\00", align 1 ; <[12 x i8]*> [#uses=1]
@.str6 = private constant [12 x i8] c"c >> b: %d\0A\00", align 1 ; <[12 x i8]*> [#uses=1]
;A function that will do many bitwise opreations to two integer arguments, %a and %b
define void @bitwise(i32 %a, i32 %b) nounwind {
;entry block
entry:
;Register to store (a & b)
%0 = and i32 %b, %a ; <i32> [#uses=1]
;print the results
%1 = tail call i32 (i8*, ...)* @printf(i8* getelementptr inbounds ([13 x i8]* @.str, i32 0, i32 0), i32 %0) nounwind ; <i32> [#uses=0]
;Register to store (a | b)
%2 = or i32 %b, %a ; <i32> [#uses=1]
;print the results
%3 = tail call i32 (i8*, ...)* @printf(i8* getelementptr inbounds ([12 x i8]* @.str1, i32 0, i32 0), i32 %2) nounwind ; <i32> [#uses=0]
;Register to store (a ^ b)
%4 = xor i32 %b, %a ; <i32> [#uses=1]
;print the results
%5 = tail call i32 (i8*, ...)* @printf(i8* getelementptr inbounds ([13 x i8]* @.str2, i32 0, i32 0), i32 %4) nounwind ; <i32> [#uses=0]
;Register to store (~a)
%not = xor i32 %a, -1 ; <i32> [#uses=1]
;print the results
%6 = tail call i32 (i8*, ...)* @printf(i8* getelementptr inbounds ([11 x i8]* @.str3, i32 0, i32 0), i32 %not) nounwind ; <i32> [#uses=0]
;Register to store (a << b)
%7 = shl i32 %a, %b ; <i32> [#uses=1]
;print the results
%8 = tail call i32 (i8*, ...)* @printf(i8* getelementptr inbounds ([12 x i8]* @.str4, i32 0, i32 0), i32 %7) nounwind ; <i32> [#uses=0]
;Register to store (a >> b) (a is signed)
%9 = ashr i32 %a, %b ; <i32> [#uses=1]
;print the results
%10 = tail call i32 (i8*, ...)* @printf(i8* getelementptr inbounds ([12 x i8]* @.str5, i32 0, i32 0), i32 %9) nounwind ; <i32> [#uses=0]
;Register to store (c >> b), where c is unsiged (eg. logical right shift)
%11 = lshr i32 %a, %b ; <i32> [#uses=1]
;print the results
%12 = tail call i32 (i8*, ...)* @printf(i8* getelementptr inbounds ([12 x i8]* @.str6, i32 0, i32 0), i32 %11) nounwind ; <i32> [#uses=0]
;terminator instruction
ret void
}
;Declare external fuctions
declare i32 @printf(i8* nocapture, ...) nounwind |
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.
| #Prolog | Prolog |
:- module(bitmap, [
new_bitmap/3,
fill_bitmap/3,
get_pixel0/3,
set_pixel0/4 ]).
:- use_module(library(lists)).
%-----------------------------------------------------------------------------%
% Convenience Predicates
replicate(Term,Times,L):-
length(L,Times),
maplist(=(Term),L).
replace0(N,OL,E,NL):-
nth0(N,OL,_,TL),
nth0(N,NL,E,TL).
%-----------------------------------------------------------------------------%
% Bitmap Utilities
%
% The Bitmap structure is a list with pixels kept in row major order:
% [dimensions-[X,Y],pixels-[[n11,n12...],[n21,n22...]]]
% In this code what exactly an RGB value is doesn't matter however
% in other bitmap tasks it is assumed to be a list [R,G,B] where
% each is an int between 0 and 255, in code:
rgb_pixel(RGB):-
length(RGB,3),
maplist(integer,RGB),
maplist(between(0,255),RGB).
%new_bitmap(Bitmap,Dimensions,RGB)
new_bitmap([[X,Y],Pixels],[X,Y],RGB) :-
replicate(RGB,X,Row),
replicate(Row,Y,Pixels).
%fill_bitmap(New_Bitmap,Bitmap,RGB)
fill_bitmap(New_Bitmap,[[X,Y],_],RGB) :-
new_bitmap(New_Bitmap,[X,Y],RGB).
%here get and set use 0 based indexing
%get_pixel0(Bitmap,Coordinates,RGB)
get_pixel0([[_DimX,_DimY],Pixels],[X,Y],RGB) :-
nth0(Y,Pixels,Row),
nth0(X,Row,RGB).
%set_pixel0(New Bitmap, Bitmap, Coordinates, RGB)
set_pixel0([[DimX,DimY],New_Pixels],[[DimX,DimY],Pixels],[X,Y],RGB) :-
nth0(Y,Pixels,Row),
replace0(X,Row,RGB,New_Row),
replace0(Y,Pixels,New_Row,New_Pixels).
|
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 .
| #Prolog | Prolog | %_________________________________________________________________
% Does the Fibonacci sequence follow Benford's law?
%~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Fibonacci sequence generator
fib(C, [P,S], C, N) :- N is P + S.
fib(C, [P,S], Cv, V) :- succ(C, Cn), N is P + S, !, fib(Cn, [S,N], Cv, V).
fib(0, 0).
fib(1, 1).
fib(C, N) :- fib(2, [0,1], C, N). % Generate from 3rd sequence on
% The benford law calculated
benford(D, Val) :- Val is log10(1+1/D).
% Retrieves the first characters of the first 1000 fibonacci numbers
% (excluding zero)
firstchar(V) :-
fib(C,N), N =\= 0, atom_chars(N, [Ch|_]), number_chars(V, [Ch]),
(C>999-> !; true).
% Increment the n'th list item (1 based), result -> third argument.
incNth(1, [Dh|Dt], [Ch|Dt]) :- !, succ(Dh, Ch).
incNth(H, [Dh|Dt], [Dh|Ct]) :- succ(Hn, H), !, incNth(Hn, Dt, Ct).
% Calculate the frequency of the all the list items
freq([], D, D).
freq([H|T], D, C) :- incNth(H, D, L), !, freq(T, L, C).
freq([H|T], Freq) :-
length([H|T], Len), min_list([H|T], Min), max_list([H|T], Max),
findall(0, between(Min,Max,_), In),
freq([H|T], In, F), % Frequency stored in F
findall(N, (member(V, F), N is V/Len), Freq). % Normalise F->Freq
% Output the results
writeHdr :-
format('~t~w~15| - ~t~w\n', ['Benford', 'Measured']).
writeData(Benford, Freq) :-
format('~t~2f%~15| - ~t~2f%\n', [Benford*100, Freq*100]).
go :- % main goal
findall(B, (between(1,9,N), benford(N,B)), Benford),
findall(C, firstchar(C), Fc), freq(Fc, Freq),
writeHdr, maplist(writeData, Benford, Freq). |
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.
| #PicoLisp | PicoLisp | (load "@lib/frac.l")
(de fact (N)
(cache '(NIL) N
(if (=0 N) 1 (apply * (range 1 N))) ) )
(de binomial (N K)
(frac
(/
(fact N)
(* (fact (- N K)) (fact K)) )
1 ) )
(de A (N M)
(let Sum (0 . 1)
(for X M
(setq Sum
(f+
Sum
(f*
(binomial (+ N 3) (- N (* X 6)))
(berno (- N (* X 6)) ) ) ) ) )
Sum ) )
(de berno (N)
(cache '(NIL) N
(cond
((=0 N) (1 . 1))
((= 1 N) (-1 . 2))
((bit? 1 N) (0 . 1))
(T
(case (% N 6)
(0
(f/
(f-
(frac (+ N 3) 3)
(A N (/ N 6)) )
(binomial (+ N 3) N) ) )
(2
(f/
(f-
(frac (+ N 3) 3)
(A N (/ (- N 2) 6)) )
(binomial (+ N 3) N) ) )
(4
(f/
(f-
(f* (-1 . 1) (frac (+ N 3) 6))
(A N (/ (- N 4) 6)) )
(binomial (+ N 3) N) ) ) ) ) ) ) )
(de berno-brute (N)
(cache '(NIL) N
(let Sum (0 . 1)
(cond
((=0 N) (1 . 1))
((= 1 N) (-1 . 2))
((bit? 1 N) (0 . 1))
(T
(for (X 0 (> N X) (inc X))
(setq Sum
(f+
Sum
(f* (binomial (inc N) X) (berno-brute X)) ) ) )
(f/ (f* (-1 . 1) Sum) (binomial (inc N) N)) ) ) ) ) )
(for (N 0 (> 62 N) (inc N))
(if (or (= N 1) (not (bit? 1 N)))
(tab (2 4 -60) N " => " (sym (berno N))) ) )
(for (N 0 (> 400 N) (inc N))
(test (berno N) (berno-brute N)) )
(bye) |
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.
| #Emacs_Lisp | Emacs Lisp |
(defun binary-search (value array)
(let ((low 0)
(high (1- (length array))))
(cl-do () ((< high low) nil)
(let ((middle (floor (+ low high) 2)))
(cond ((> (aref array middle) value)
(setf high (1- middle)))
((< (aref array middle) value)
(setf low (1+ middle)))
(t (cl-return middle))))))) |
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
| #PowerShell | PowerShell | # Calculate best possible shuffle score for a given string
# (Split out into separate function so we can use it separately in our output)
function Get-BestScore ( [string]$String )
{
# Convert to array of characters, group identical characters,
# sort by frequecy, get size of first group
$MostRepeats = $String.ToCharArray() |
Group |
Sort Count -Descending |
Select -First 1 -ExpandProperty Count
# Return count of most repeated character minus all other characters (math simplified)
return [math]::Max( 0, 2 * $MostRepeats - $String.Length )
}
function Get-BestShuffle ( [string]$String )
{
# Convert to arrays of characters, one for comparison, one for manipulation
$S1 = $String.ToCharArray()
$S2 = $String.ToCharArray()
# Calculate best possible score as our goal
$BestScore = Get-BestScore $String
# Unshuffled string has score equal to number of characters
$Length = $String.Length
$Score = $Length
# While still striving for perfection...
While ( $Score -gt $BestScore )
{
# For each character
ForEach ( $i in 0..($Length-1) )
{
# If the shuffled character still matches the original character...
If ( $S1[$i] -eq $S2[$i] )
{
# Swap it with a random character
# (Random character $j may be the same as or may even be
# character $i. The minor impact on speed was traded for
# a simple solution to guarantee randomness.)
$j = Get-Random -Maximum $Length
$S2[$i], $S2[$j] = $S2[$j], $S2[$i]
}
}
# Count the number of indexes where the two arrays match
$Score = ( 0..($Length-1) ).Where({ $S1[$_] -eq $S2[$_] }).Count
}
# Put it back into a string
$Shuffle = ( [string[]]$S2 -join '' )
return $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.
| #VBA | VBA | The Option Compare instruction is used at module level to declare the default comparison method to use when string data is compared.
The default text comparison method is Binary.
|
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