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http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #Python | Python | from bisect import bisect_right
def bin_it(limits: list, data: list) -> list:
"Bin data according to (ascending) limits."
bins = [0] * (len(limits) + 1) # adds under/over range bins too
for d in data:
bins[bisect_right(limits, d)] += 1
return bins
def bin_print(limits: list, bins: list) -> list:
print(f" < {limits[0]:3} := {bins[0]:3}")
for lo, hi, count in zip(limits, limits[1:], bins[1:]):
print(f">= {lo:3} .. < {hi:3} := {count:3}")
print(f">= {limits[-1]:3} := {bins[-1]:3}")
if __name__ == "__main__":
print("RC FIRST EXAMPLE\n")
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
bins = bin_it(limits, data)
bin_print(limits, bins)
print("\nRC SECOND EXAMPLE\n")
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
bins = bin_it(limits, data)
bin_print(limits, bins) |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Racket | Racket | #lang racket
(define (fold-sequence seq kons #:finalise (finalise (λ x (apply values x))) . k0s)
(define (recur seq . ks)
(if (null? seq)
(call-with-values (λ () (apply finalise ks)) (λ vs (apply values vs)))
(call-with-values (λ () (apply kons (car seq) ks)) (λ ks+ (apply recur (cdr seq) ks+)))))
(apply recur (if (string? seq) (string->list (regexp-replace* #px"[^ACGT]" seq "")) seq) k0s))
(define (sequence->pretty-printed-string seq)
(define (fmt idx cs-rev) (format "~a: ~a" (~a idx #:width 3 #:align 'right) (list->string (reverse cs-rev))))
(fold-sequence
seq
(λ (b n start-idx lns-rev cs-rev)
(if (zero? (modulo n 50))
(values (+ n 1) n (if (pair? cs-rev) (cons (fmt start-idx cs-rev) lns-rev) lns-rev) (cons b null))
(values (+ n 1) start-idx lns-rev (cons b cs-rev))))
0 0 null null
#:finalise (λ (n idx lns-rev cs-rev)
(string-join (reverse (if (null? cs-rev) lns-rev (cons (fmt idx cs-rev) lns-rev))) "\n"))))
(define (count-bases b as cs gs ts n)
(values (+ as (if (eq? b #\A) 1 0))
(+ cs (if (eq? b #\C) 1 0))
(+ gs (if (eq? b #\T) 1 0))
(+ ts (if (eq? b #\G) 1 0))
(add1 n)))
(define (bioinformatics-Base_count s)
(define-values (as cs gs ts n) (fold-sequence s count-bases 0 0 0 0 0))
(printf "SEQUENCE:~%~%~a~%~%" (sequence->pretty-printed-string s))
(printf "BASE COUNT:~%-----------~%~%~a~%~%"
(string-join (map (λ (c n) (format " ~a :~a" c (~a #:width 4 #:align 'right n)))
'(A T C G)
(list as ts cs gs)) "\n"))
(newline)
(printf "TOTAL: ~a~%" n))
(module+
main
(define the-string
#<<EOS
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
EOS
)
(bioinformatics-Base_count the-string)) |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Raku | Raku | my $dna = join '', lines q:to/END/;
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
END
put pretty($dna, 80);
put "\nTotal bases: ", +my $bases = $dna.comb.Bag;
put $bases.sort(~*.key).join: "\n";
sub pretty ($string, $wrap = 50) {
$string.comb($wrap).map( { sprintf "%8d: %s", $++ * $wrap, $_ } ).join: "\n"
} |
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.
| #ALGOL_W | ALGOL W | begin
% prints an integer in binary - the number must be greater than zero %
procedure printBinaryDigits( integer value n ) ;
begin
if n not = 0 then begin
printBinaryDigits( n div 2 );
writeon( if n rem 2 = 1 then "1" else "0" )
end
end binaryDigits ;
% prints an integer in binary - the number must not be negative %
procedure printBinary( integer value n ) ;
begin
if n = 0 then writeon( "0" )
else printBinaryDigits( n )
end printBinary ;
% test the printBinaryDigits procedure %
for i := 5, 50, 9000 do begin
write();
printBinary( i );
end
end. |
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.
| #Go | Go | package raster
// Line draws line by Bresenham's algorithm.
func (b *Bitmap) Line(x0, y0, x1, y1 int, p Pixel) {
// implemented straight from WP pseudocode
dx := x1 - x0
if dx < 0 {
dx = -dx
}
dy := y1 - y0
if dy < 0 {
dy = -dy
}
var sx, sy int
if x0 < x1 {
sx = 1
} else {
sx = -1
}
if y0 < y1 {
sy = 1
} else {
sy = -1
}
err := dx - dy
for {
b.SetPx(x0, y0, p)
if x0 == x1 && y0 == y1 {
break
}
e2 := 2 * err
if e2 > -dy {
err -= dy
x0 += sx
}
if e2 < dx {
err += dx
y0 += sy
}
}
}
func (b *Bitmap) LineRgb(x0, y0, x1, y1 int, c Rgb) {
b.Line(x0, y0, x1, y1, c.Pixel())
} |
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
| #ooRexx | ooRexx | my $x = 0.0;
my $true_or_false = $x ? 'true' : 'false'; # false |
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #Order | Order | my $x = 0.0;
my $true_or_false = $x ? 'true' : 'false'; # false |
http://rosettacode.org/wiki/Box_the_compass | Box the compass | There be many a land lubber that knows naught of the pirate ways and gives direction by degree!
They know not how to box the compass!
Task description
Create a function that takes a heading in degrees and returns the correct 32-point compass heading.
Use the function to print and display a table of Index, Compass point, and Degree; rather like the corresponding columns from, the first table of the wikipedia article, but use only the following 33 headings as input:
[0.0, 16.87, 16.88, 33.75, 50.62, 50.63, 67.5, 84.37, 84.38, 101.25, 118.12, 118.13, 135.0, 151.87, 151.88, 168.75, 185.62, 185.63, 202.5, 219.37, 219.38, 236.25, 253.12, 253.13, 270.0, 286.87, 286.88, 303.75, 320.62, 320.63, 337.5, 354.37, 354.38]. (They should give the same order of points but are spread throughout the ranges of acceptance).
Notes;
The headings and indices can be calculated from this pseudocode:
for i in 0..32 inclusive:
heading = i * 11.25
case i %3:
if 1: heading += 5.62; break
if 2: heading -= 5.62; break
end
index = ( i mod 32) + 1
The column of indices can be thought of as an enumeration of the thirty two cardinal points (see talk page)..
| #M2000_Interpreter | M2000 Interpreter |
Module CheckIt {
Locale 1033 'change decimal point char to dot.
Form 80,50 ' set console to 80 characters by 50 lines
\\ Function heading() get a positive double as degrees and return the compass index (1 for North)
Function heading(d) {
d1=d div 11.25
if d1 mod 3= 1 then d+=5.62 :d1=d div 11.25
=d1 mod 32 +1
}
Dim wind$(1 to 32)
wind$(1)="North", "North by east", "North-northeast", "Northeast by north", "Northeast"
wind$(6)="Northeast by east", "East-northeast", "East by north", "East", "East by south", "East-southeast"
wind$(12)="Southeast by east", "Southeast", "Southeast by south", "South-southeast", "South by east", "South"
wind$(18)="South by west", "South-southwest", "Southwest by south", "Southwest", "Southwest by west", "West-southwest"
wind$(24)="West by south", "West", "West by north", "West-northwest", "Northwest by west", "Northwest", "Northwest by north"
wind$(31)="North-northwest", "North by west"
oldvalue=-2
newvalue=2
Print " angle | box | compass point"
Print "-------+-----+---------------------"
For i=0 to 360 step 0.005
newvalue=heading(i)
if (newvalue mod 3) =2 then i+=5.62: newvalue=heading(i)
if oldvalue<>newvalue then Print format$("{0:2:-6}°| {1::-2} | {2}",i, newvalue, wind$(newvalue)) : oldvalue=newvalue : refresh
Next i
}
CheckIt
|
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.
| #Factor | Factor | "a=" "b=" [ write readln string>number ] bi@
{
[ bitand "a AND b: " write . ]
[ bitor "a OR b: " write . ]
[ bitxor "a XOR b: " write . ]
[ drop bitnot "NOT a: " write . ]
[ abs shift "a asl b: " write . ]
[ neg shift "a asr b: " write . ]
} 2cleave |
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.
| #Fortran | Fortran | Screenres 320, 240, 8
Dim Shared As Integer w, h
Screeninfo w, h
Const As Ubyte cyan = 3
Const As Ubyte red = 4
Sub rellenar(c As Integer)
Line (0,0) - (w/3, h/3), red, BF
End Sub
Sub establecePixel(x As Integer, y As Integer, c As Integer)
Pset (x,y), cyan
End Sub
rellenar(12)
establecePixel(10,10, cyan)
Locate 12
Print "pixel 10,10 es " & Point(10,10)
Print "pixel 20,20 es " & Point(20,10)
Bsave "FreeBASIC_bitmap.bmp", 0
Sleep |
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 | #Haskell | Haskell | bellTri :: [[Integer]]
bellTri =
let f xs = (last xs, xs)
in map snd (iterate (f . uncurry (scanl (+))) (1, [1]))
bell :: [Integer]
bell = map head bellTri
main :: IO ()
main = do
putStrLn "First 10 rows of Bell's Triangle:"
mapM_ print (take 10 bellTri)
putStrLn "\nFirst 15 Bell numbers:"
mapM_ print (take 15 bell)
putStrLn "\n50th Bell number:"
print (bell !! 49) |
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 | #J | J |
bell=: ([: +/\ (,~ {:))&.>@:{:
,. bell^:(<5) <1
+--------------+
|1 |
+--------------+
|1 2 |
+--------------+
|2 3 5 |
+--------------+
|5 7 10 15 |
+--------------+
|15 20 27 37 52|
+--------------+
{.&> bell^:(<15) <1
1 1 2 5 15 52 203 877 4140 21147 115975 678570 4213597 27644437 190899322
{:>bell^:49<1x
185724268771078270438257767181908917499221852770
|
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 .
| #F.23 | F# | open System
let fibonacci = Seq.unfold (fun (x, y) -> Some(x, (y, x + y))) (0I,1I)
let fibFirstNumbers nth =
fibonacci |> Seq.take nth |> Seq.map (fun n -> n.ToString().[0] |> string |> Int32.Parse)
let fibFirstNumbersFrequency nth =
let firstNumbers = fibFirstNumbers nth |> Seq.toList
let counts = firstNumbers |> List.countBy id |> List.sort |> List.filter (fun (k, _) -> k <> 0)
let total = firstNumbers |> List.length |> float
counts |> List.map (fun (_, v) -> float v/total)
let benfordLaw d = log10(1.0 + (1.0/float d))
let benfordLawFigures = [1..9] |> List.map benfordLaw
let run () =
printfn "Frequency of the first digit (1 through 9) in the Fibonacci sequence:"
fibFirstNumbersFrequency 1000 |> List.iter (fun f -> printf $"{f:N5} ")
printfn "\nBenford's law for 1 through 9:"
benfordLawFigures |> List.iter (fun f -> printf $"{f:N5} ")
|
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.
| #EchoLisp | EchoLisp |
(lib 'bigint) ;; lerge numbers
(lib 'gloops) ;; classes
(define-class Rational null ((a :initform #0) (b :initform #1)))
(define-method tostring (Rational) (lambda (r) (format "%50d / %d" r.a r.b)))
(define-method normalize (Rational) (lambda (r) ;; divide a and b by gcd
(let ((g (gcd r.a r.b)))
(set! r.a (/ r.a g)) (set! r.b (/ r.b g))
(when (< r.b 0) (set! r.a ( - r.a)) (set! r.b (- r.b))) ;; denominator > 0
r)))
(define-method initialize (Rational) (lambda (r) (normalize r)))
(define-method add (Rational) (lambda (r n) ;; + Rational any number
(normalize (Rational (+ (* (+ #0 n) r.b) r.a) r.b))))
(define-method add (Rational Rational) (lambda (r q) ;;; + Rational Rational
(normalize (Rational (+ (* r.a q.b) (* r.b q.a)) (* r.b q.b)))))
(define-method sub (Rational Rational) (lambda (r q)
(normalize (Rational (- (* r.a q.b) (* r.b q.a)) (* r.b q.b)))))
(define-method mul (Rational Rational) (lambda (r q)
(normalize (Rational (* r.a q.a) (* r.b q.b)))))
(define-method mul (Rational) (lambda (r n)
(normalize (Rational (* r.a (+ #0 n)) r.b ))))
(define-method div (Rational Rational) (lambda (r q)
(normalize (Rational (* r.a q.b) (* r.b q.a)))))
|
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.
| #AutoHotkey | AutoHotkey | array := "1,2,4,6,8,9"
StringSplit, A, array, `, ; creates associative array
MsgBox % x := BinarySearch(A, 4, 1, A0) ; Recursive
MsgBox % A%x%
MsgBox % x := BinarySearchI(A, A0, 4) ; Iterative
MsgBox % A%x%
BinarySearch(A, value, low, high) { ; A0 contains length of array
If (high < low) ; A1, A2, A3...An are array elements
Return not_found
mid := Floor((low + high) / 2)
If (A%mid% > value) ; A%mid% is automatically global since no such locals are present
Return BinarySearch(A, value, low, mid - 1)
Else If (A%mid% < value)
Return BinarySearch(A, value, mid + 1, high)
Else
Return mid
}
BinarySearchI(A, lengthA, value) {
low := 0
high := lengthA - 1
While (low <= high) {
mid := Floor((low + high) / 2) ; round to lower integer
If (A%mid% > value)
high := mid - 1
Else If (A%mid% < value)
low := mid + 1
Else
Return mid
}
Return not_found
} |
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
| #D | D | import std.stdio, std.random, std.algorithm, std.conv, std.range,
std.traits, std.typecons;
auto bestShuffle(S)(in S orig) @safe if (isSomeString!S) {
static if (isNarrowString!S)
immutable o = orig.dtext;
else
alias o = orig;
auto s = o.dup;
s.randomShuffle;
foreach (immutable i, ref ci; s) {
if (ci != o[i])
continue;
foreach (immutable j, ref cj; s)
if (ci != cj && ci != o[j] && cj != o[i]) {
swap(ci, cj);
break;
}
}
return tuple(s, s.zip(o).count!q{ a[0] == a[1] });
} unittest {
assert("abracadabra".bestShuffle[1] == 0);
assert("immediately".bestShuffle[1] == 0);
assert("grrrrrr".bestShuffle[1] == 5);
assert("seesaw".bestShuffle[1] == 0);
assert("pop".bestShuffle[1] == 1);
assert("up".bestShuffle[1] == 0);
assert("a".bestShuffle[1] == 1);
assert("".bestShuffle[1] == 0);
}
void main(in string[] args) @safe {
if (args.length > 1) {
immutable entry = args.dropOne.join(' ');
const res = entry.bestShuffle;
writefln("%s : %s (%d)", entry, res[]);
}
} |
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.
| #J | J | name=: '' |
http://rosettacode.org/wiki/Binary_strings | Binary strings | Many languages have powerful and useful (binary safe) string manipulation functions, while others don't, making it harder for these languages to accomplish some tasks.
This task is about creating functions to handle binary strings (strings made of arbitrary bytes, i.e. byte strings according to Wikipedia) for those languages that don't have built-in support for them.
If your language of choice does have this built-in support, show a possible alternative implementation for the functions or abilities already provided by the language.
In particular the functions you need to create are:
String creation and destruction (when needed and if there's no garbage collection or similar mechanism)
String assignment
String comparison
String cloning and copying
Check if a string is empty
Append a byte to a string
Extract a substring from a string
Replace every occurrence of a byte (or a string) in a string with another string
Join strings
Possible contexts of use: compression algorithms (like LZW compression), L-systems (manipulation of symbols), many more.
| #Java | Java | import java.io.ByteArrayOutputStream;
import java.nio.charset.StandardCharsets;
import java.util.Arrays;
public class MutableByteString {
private byte[] bytes;
private int length;
public MutableByteString(byte... bytes) {
setInternal(bytes);
}
public int length() {
return length;
}
public boolean isEmpty() {
return length == 0;
}
public byte get(int index) {
return bytes[check(index)];
}
public void set(byte[] bytes) {
setInternal(bytes);
}
public void set(int index, byte b) {
bytes[check(index)] = b;
}
public void append(byte b) {
if (length >= bytes.length) {
int len = 2 * bytes.length;
if (len < 0)
len = Integer.MAX_VALUE;
bytes = Arrays.copyOf(bytes, len);
}
bytes[length] = b;
length++;
}
public MutableByteString substring(int from, int to) {
return new MutableByteString(Arrays.copyOfRange(bytes, from, to));
}
public void replace(byte[] from, byte[] to) {
ByteArrayOutputStream copy = new ByteArrayOutputStream();
if (from.length == 0) {
for (byte b : bytes) {
copy.write(to, 0, to.length);
copy.write(b);
}
copy.write(to, 0, to.length);
} else {
for (int i = 0; i < length; i++) {
if (regionEquals(i, from)) {
copy.write(to, 0, to.length);
i += from.length - 1;
} else {
copy.write(bytes[i]);
}
}
}
set(copy.toByteArray());
}
public boolean regionEquals(int offset, MutableByteString other, int otherOffset, int len) {
if (Math.max(offset, otherOffset) + len < 0)
return false;
if (offset + len > length || otherOffset + len > other.length())
return false;
for (int i = 0; i < len; i++) {
if (bytes[offset + i] != other.get(otherOffset + i))
return false;
}
return true;
}
public String toHexString() {
char[] hex = new char[2 * length];
for (int i = 0; i < length; i++) {
hex[2 * i] = "0123456789abcdef".charAt(bytes[i] >> 4 & 0x0F);
hex[2 * i + 1] = "0123456789abcdef".charAt(bytes[i] & 0x0F);
}
return new String(hex);
}
public String toStringUtf8() {
return new String(bytes, 0, length, StandardCharsets.UTF_8);
}
private void setInternal(byte[] bytes) {
this.bytes = bytes.clone();
this.length = bytes.length;
}
private boolean regionEquals(int offset, byte[] other) {
int len = other.length;
if (offset < 0 || offset + len < 0)
return false;
if (offset + len > length)
return false;
for (int i = 0; i < len; i++) {
if (bytes[offset + i] != other[i])
return false;
}
return true;
}
private int check(int index) {
if (index < 0 || index >= length)
throw new IndexOutOfBoundsException(String.valueOf(index));
return index;
}
} |
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #R | R | limits1 <- c(23, 37, 43, 53, 67, 83)
data1 <- c(95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16,8,9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55)
limits2 <- c(14, 18, 249, 312, 389, 392, 513, 591, 634, 720)
data2 <- c(445,814,519,697,700,130,255,889,481,122,932,77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775,47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238,62,678,98,534,622,907,406,714,184,391,913,42,560,247,
346,860,56,138,546,38,985,948,58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426,9,457,628,410,723,354,895,881,953,677,137,397,97,
854,740,83,216,421,94,517,479,292,963,376,981,480,39,257,272,157,5,316,395,
787,942,456,242,759,898,576,67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585,40,54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466,23,707,467,33,670,921,180,991,396,160,436,717,918,8,374,101,684,727,749)
createBin <- function(limits, data) sapply(0:length(limits), function(x) sum(findInterval(data, limits) == x))
#Contains some unicode magic so that we can get the infinity symbol and <= to print nicely.
#Half of the battle here is making sure that we're not being thrown by R being 1-indexed.
#The other half is avoiding the mathematical sin of saying that anything can be >=infinity.
printBin <- function(limits, bin)
{
invisible(sapply(0:length(limits), function(x) cat("Bin", x, "covers the range:",
if(x == 0) "-\U221E < x <" else paste(limits[x], "\U2264 x <"),
if(x == length(limits)) "\U221E" else limits[x + 1],
"and contains", bin[x + 1], "elements.\n")))
}
#Showing off a one-line solution. Admittedly, calling the massive anonymous function "one-line" is generous.
oneLine <- function(limits, data)
{
invisible(sapply(0:length(limits), function(x) cat("Bin", x, "covers the range:",
if(x == 0) "-\U221E < x <" else paste(limits[x], "\U2264 x <"),
if(x == length(limits)) "\U221E" else limits[x + 1],
"and contains", sum(findInterval(data, limits) == x),
"elements.\n")))
}
createBin(limits1, data1)
printBin(limits1, createBin(limits1, data1))
createBin(limits2, data2)
printBin(limits2, createBin(limits2, data2))
oneLine(limits2, c(data1, data2))#Not needed. |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #REXX | REXX | /*REXX program finds the number of each base in a DNA string (along with a total). */
parse arg dna .
if dna=='' | dna=="," then dna= 'cgtaaaaaattacaacgtcctttggctatctcttaaactcctgctaaatg' ,
'ctcgtgctttccaattatgtaagcgttccgagacggggtggtcgattctg' ,
'aggacaaaggtcaagatggagcgcatcgaacgcaataaggatcatttgat' ,
'gggacgtttcgtcgacaaagtcttgtttcgagagtaacggctaccgtctt' ,
'cgattctgcttataacactatgttcttatgaaatggatgttctgagttgg' ,
'tcagtcccaatgtgcggggtttcttttagtacgtcgggagtggtattata' ,
'tttaatttttctatatagcgatctgtatttaagcaattcatttaggttat' ,
'cgccgcgatgctcggttcggaccgccaagcatctggctccactgctagtg' ,
'tcctaaatttgaatggcaaacacaaataagatttagcaattcgtgtagac' ,
'gaccggggacttgcatgatgggagcagctttgttaaactacgaacgtaat'
dna= space(dna, 0); upper dna /*elide blanks from DNA; uppercase it. */
say '────────length of the DNA string: ' length(dna)
@.= 0 /*initialize the count for all bases. */
w= 1 /*the maximum width of a base count. */
$= /*a placeholder for the names of bases.*/
do j=1 for length(dna) /*traipse through the DNA string. */
_= substr(dna, j, 1) /*obtain a base name from the DNA str. */
if pos(_, $)==0 then $= $ || _ /*if not found before, add it to list. */
@._= @._ + 1 /*bump the count of this base. */
w= max(w, length(@._) ) /*compute the maximum width number. */
end /*j*/
say
do k=0 for 255; z= d2c(k) /*traipse through all possibilities. */
if pos(z, $)==0 then iterate /*Was this base found? No, then skip. */
say ' base ' z " has a basecount of: " right(@.z, w)
@.tot= @.tot + @.z /*add to a grand total to verify count.*/
end /*k*/ /*stick a fork in it, we're all done. */
say
say '────────total for all basecounts:' right(@.tot, w+1) |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Ring | Ring |
dna = "" +
"CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG" +
"CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG" +
"AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT" +
"GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT" +
"CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG" +
"TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA" +
"TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT" +
"CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG" +
"TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC" +
"GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT"
dnaBase = [:A=0, :C=0, :G=0, :T=0]
lenDna = len(dna)
for n = 1 to lenDna
dnaStr = substr(dna,n,1)
switch dnaStr
on "A"
strA = dnaBase["A"]
strA++
dnaBase["A"] = strA
on "C"
strC = dnaBase["C"]
strC++
dnaBase["C"] = strC
on "G"
strG = dnaBase["G"]
strG++
dnaBase["G"] = strG
on "T"
strT = dnaBase["T"]
strT++
dnaBase["T"] = strT
off
next
? "A : " + dnaBase["A"]
? "T : " + dnaBase["T"]
? "C : " + dnaBase["C"]
? "G : " + dnaBase["G"]
|
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.
| #AppleScript | AppleScript | ---------------------- BINARY STRING -----------------------
-- showBin :: Int -> String
on showBin(n)
script binaryChar
on |λ|(n)
text item (n + 1) of "01"
end |λ|
end script
showIntAtBase(2, binaryChar, n, "")
end showBin
--------------------------- TEST ---------------------------
on run
script
on |λ|(n)
intercalate(" -> ", {n as string, showBin(n)})
end |λ|
end script
return unlines(map(result, {5, 50, 9000}))
end run
-------------------- GENERIC FUNCTIONS ---------------------
-- showIntAtBase :: Int -> (Int -> Char) -> Int -> String -> String
on showIntAtBase(base, toChr, n, rs)
script showIt
on |λ|(nd_, r)
set {n, d} to nd_
set r_ to toChr's |λ|(d) & r
if n > 0 then
|λ|(quotRem(n, base), r_)
else
r_
end if
end |λ|
end script
if base ≤ 1 then
"error: showIntAtBase applied to unsupported base: " & base as string
else if n < 0 then
"error: showIntAtBase applied to negative number: " & base as string
else
showIt's |λ|(quotRem(n, base), rs)
end if
end showIntAtBase
-- quotRem :: Integral a => a -> a -> (a, a)
on quotRem(m, n)
{m div n, m mod n}
end quotRem
-------------------- GENERICS FOR TEST ---------------------
-- intercalate :: Text -> [Text] -> Text
on intercalate(strText, lstText)
set {dlm, my text item delimiters} to {my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- unlines :: [String] -> String
on unlines(xs)
intercalate(linefeed, xs)
end unlines |
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.
| #Haskell | Haskell | module Bitmap.Line(line) where
import Bitmap
import Control.Monad
import Control.Monad.ST
import qualified Data.STRef
var = Data.STRef.newSTRef
get = Data.STRef.readSTRef
mutate = Data.STRef.modifySTRef
line :: Color c => Image s c -> Pixel -> Pixel -> c -> ST s ()
line i (Pixel (xa, ya)) (Pixel (xb, yb)) c = do
yV <- var y1
errorV <- var $ deltax `div` 2
forM_ [x1 .. x2] (\x -> do
y <- get yV
setPix i (Pixel $ if steep then (y, x) else (x, y)) c
mutate errorV $ subtract deltay
error <- get errorV
when (error < 0) (do
mutate yV (+ ystep)
mutate errorV (+ deltax)))
where steep = abs (yb - ya) > abs (xb - xa)
(xa', ya', xb', yb') = if steep
then (ya, xa, yb, xb)
else (xa, ya, xb, yb)
(x1, y1, x2, y2) = if xa' > xb'
then (xb', yb', xa', ya')
else (xa', ya', xb', yb')
deltax = x2 - x1
deltay = abs $ y2 - y1
ystep = if y1 < y2 then 1 else -1 |
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
| #Oz | Oz | my $x = 0.0;
my $true_or_false = $x ? 'true' : 'false'; # false |
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #PARI.2FGP | PARI/GP | my $x = 0.0;
my $true_or_false = $x ? 'true' : 'false'; # false |
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)..
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | Map[List[Part[#,1], dirs[[Part[#,1]]], ToString@Part[#,2]<>"°"]&,
Map[{Floor[Mod[ #+5.625 , 360]/11.25]+1,#}&,input] ]//TableForm |
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.
| #FALSE | FALSE | 10 3
\$@$@$@$@\ { 3 copies }
"a & b = "&."
a | b = "|."
~a = "%~."
" |
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.
| #FreeBASIC | FreeBASIC | Screenres 320, 240, 8
Dim Shared As Integer w, h
Screeninfo w, h
Const As Ubyte cyan = 3
Const As Ubyte red = 4
Sub rellenar(c As Integer)
Line (0,0) - (w/3, h/3), red, BF
End Sub
Sub establecePixel(x As Integer, y As Integer, c As Integer)
Pset (x,y), cyan
End Sub
rellenar(12)
establecePixel(10,10, cyan)
Locate 12
Print "pixel 10,10 es " & Point(10,10)
Print "pixel 20,20 es " & Point(20,10)
Bsave "FreeBASIC_bitmap.bmp", 0
Sleep |
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 | #Java | Java | import java.util.ArrayList;
import java.util.List;
public class Bell {
private static class BellTriangle {
private List<Integer> arr;
BellTriangle(int n) {
int length = n * (n + 1) / 2;
arr = new ArrayList<>(length);
for (int i = 0; i < length; ++i) {
arr.add(0);
}
set(1, 0, 1);
for (int i = 2; i <= n; ++i) {
set(i, 0, get(i - 1, i - 2));
for (int j = 1; j < i; ++j) {
int value = get(i, j - 1) + get(i - 1, j - 1);
set(i, j, value);
}
}
}
private int index(int row, int col) {
if (row > 0 && col >= 0 && col < row) {
return row * (row - 1) / 2 + col;
} else {
throw new IllegalArgumentException();
}
}
public int get(int row, int col) {
int i = index(row, col);
return arr.get(i);
}
public void set(int row, int col, int value) {
int i = index(row, col);
arr.set(i, value);
}
}
public static void main(String[] args) {
final int rows = 15;
BellTriangle bt = new BellTriangle(rows);
System.out.println("First fifteen Bell numbers:");
for (int i = 0; i < rows; ++i) {
System.out.printf("%2d: %d\n", i + 1, bt.get(i + 1, 0));
}
for (int i = 1; i <= 10; ++i) {
System.out.print(bt.get(i, 0));
for (int j = 1; j < i; ++j) {
System.out.printf(", %d", bt.get(i, j));
}
System.out.println();
}
}
} |
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 .
| #Factor | Factor | USING: assocs compiler.tree.propagation.call-effect formatting
kernel math math.functions math.statistics math.text.utils
sequences ;
IN: rosetta-code.benfords-law
: expected ( n -- x ) recip 1 + log10 ;
: next-fib ( vec -- vec' )
[ last2 ] keep [ + ] dip [ push ] keep ;
: data ( -- seq ) V{ 1 1 } clone 998 [ next-fib ] times ;
: 1st-digit ( n -- m ) 1 digit-groups last ;
: leading ( -- seq ) data [ 1st-digit ] map ;
: .header ( -- )
"Digit" "Expected" "Actual" "%-10s%-10s%-10s\n" printf ;
: digit-report ( digit digit-count -- digit expected actual )
dupd [ expected ] dip 1000 /f ;
: .digit-report ( digit digit-count -- )
digit-report "%-10d%-10.4f%-10.4f\n" printf ;
: main ( -- )
.header leading histogram [ .digit-report ] assoc-each ;
MAIN: main |
http://rosettacode.org/wiki/Benford%27s_law | Benford's law |
This page uses content from Wikipedia. The original article was at Benford's_law. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)
Benford's law, also called the first-digit law, refers to the frequency distribution of digits in many (but not all) real-life sources of data.
In this distribution, the number 1 occurs as the first digit about 30% of the time, while larger numbers occur in that position less frequently: 9 as the first digit less than 5% of the time. This distribution of first digits is the same as the widths of gridlines on a logarithmic scale.
Benford's law also concerns the expected distribution for digits beyond the first, which approach a uniform distribution.
This result has been found to apply to a wide variety of data sets, including electricity bills, street addresses, stock prices, population numbers, death rates, lengths of rivers, physical and mathematical constants, and processes described by power laws (which are very common in nature). It tends to be most accurate when values are distributed across multiple orders of magnitude.
A set of numbers is said to satisfy Benford's law if the leading digit
d
{\displaystyle d}
(
d
∈
{
1
,
…
,
9
}
{\displaystyle d\in \{1,\ldots ,9\}}
) occurs with probability
P
(
d
)
=
log
10
(
d
+
1
)
−
log
10
(
d
)
=
log
10
(
1
+
1
d
)
{\displaystyle P(d)=\log _{10}(d+1)-\log _{10}(d)=\log _{10}\left(1+{\frac {1}{d}}\right)}
For this task, write (a) routine(s) to calculate the distribution of first significant (non-zero) digits in a collection of numbers, then display the actual vs. expected distribution in the way most convenient for your language (table / graph / histogram / whatever).
Use the first 1000 numbers from the Fibonacci sequence as your data set. No need to show how the Fibonacci numbers are obtained.
You can generate them or load them from a file; whichever is easiest.
Display your actual vs expected distribution.
For extra credit: Show the distribution for one other set of numbers from a page on Wikipedia. State which Wikipedia page it can be obtained from and what the set enumerates. Again, no need to display the actual list of numbers or the code to load them.
See also:
numberphile.com.
A starting page on Wolfram Mathworld is Benfords Law .
| #Forth | Forth | : 3drop drop 2drop ;
: f2drop fdrop fdrop ;
: int-array create cells allot does> swap cells + ;
: 1st-fib 0e 1e ;
: next-fib ftuck f+ ;
: 1st-digit ( fp -- n )
pad 6 represent 3drop pad c@ [char] 0 - ;
10 int-array counts
: tally
0 counts 10 cells erase
1st-fib
1000 0 DO
1 fdup 1st-digit counts +!
next-fib
LOOP f2drop ;
: benford ( d -- fp )
s>f 1/f 1e f+ flog ;
: tab 9 emit ;
: heading ( -- )
cr ." Leading digital distribution of the first 1,000 Fibonacci numbers:"
cr ." Digit" tab ." Actual" tab ." Expected" ;
: .fixed ( n -- ) \ print count as decimal fraction
s>d <# # # # [char] . hold #s #> type space ;
: report ( -- )
precision 3 set-precision
heading
10 1 DO
cr i 3 .r
tab i counts @ .fixed
tab i benford f.
LOOP
set-precision ;
: compute-benford tally report ; |
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.
| #Elixir | Elixir | defmodule Bernoulli do
defmodule Rational do
import Kernel, except: [div: 2]
defstruct numerator: 0, denominator: 1
def new(numerator, denominator\\1) do
sign = if numerator * denominator < 0, do: -1, else: 1
{numerator, denominator} = {abs(numerator), abs(denominator)}
gcd = gcd(numerator, denominator)
%Rational{numerator: sign * Kernel.div(numerator, gcd),
denominator: Kernel.div(denominator, gcd)}
end
def sub(a, b) do
new(a.numerator * b.denominator - b.numerator * a.denominator,
a.denominator * b.denominator)
end
def mul(a, b) when is_integer(a) do
new(a * b.numerator, b.denominator)
end
defp gcd(a,0), do: a
defp gcd(a,b), do: gcd(b, rem(a,b))
end
def numbers(n) do
Stream.transform(0..n, {}, fn m,acc ->
acc = Tuple.append(acc, Rational.new(1,m+1))
if m>0 do
new =
Enum.reduce(m..1, acc, fn j,ar ->
put_elem(ar, j-1, Rational.mul(j, Rational.sub(elem(ar,j-1), elem(ar,j))))
end)
{[elem(new,0)], new}
else
{[elem(acc,0)], acc}
end
end) |> Enum.to_list
end
def task(n \\ 61) do
b_nums = numbers(n)
width = Enum.map(b_nums, fn b -> b.numerator |> to_string |> String.length end)
|> Enum.max
format = 'B(~2w) = ~#{width}w / ~w~n'
Enum.with_index(b_nums)
|> Enum.each(fn {b,i} ->
if b.numerator != 0, do: :io.fwrite format, [i, b.numerator, b.denominator]
end)
end
end
Bernoulli.task |
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.
| #F.23 | F# |
open MathNet.Numerics
open System
open System.Collections.Generic
let calculateBernoulli n =
let ℚ(x) = BigRational.FromInt x
let A = Array.init<BigRational> (n+1) (fun x -> ℚ(x+1))
for m in [1..n] do
A.[m] <- ℚ(1) / (ℚ(m) + ℚ(1))
for j in [m..(-1)..1] do
A.[j-1] <- ℚ(j) * (A.[j-1] - A.[j])
A.[0]
[<EntryPoint>]
let main argv =
for n in [0..60] do
let bernoulliNumber = calculateBernoulli n
match bernoulliNumber.Numerator.IsZero with
| false ->
let formatedString = String.Format("B({0, 2}) = {1, 44} / {2}", n, bernoulliNumber.Numerator, bernoulliNumber.Denominator)
printfn "%s" formatedString
| true ->
printf ""
0
|
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.
| #AWK | AWK | function binary_search(array, value, left, right, middle) {
if (right < left) return 0
middle = int((right + left) / 2)
if (value == array[middle]) return 1
if (value < array[middle])
return binary_search(array, value, left, middle - 1)
return binary_search(array, value, middle + 1, right)
} |
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
| #Delphi | Delphi |
program Best_shuffle;
{$APPTYPE CONSOLE}
uses
System.SysUtils,
System.Generics.Collections;
type
TShuffledString = record
private
original: string;
Shuffled: TStringBuilder;
ignoredChars: Integer;
procedure DetectIgnores;
procedure Shuffle;
procedure Swap(pos1, pos2: Integer);
function TrySwap(pos1, pos2: Integer): Boolean;
function GetShuffled: string;
public
class operator Implicit(convert: string): TShuffledString;
constructor Create(Word: string);
procedure Free;
property Ignored: integer read ignoredChars;
property ToString: string read GetShuffled;
end;
{ TShuffledString }
procedure TShuffledString.Swap(pos1, pos2: Integer);
var
temp: char;
begin
temp := shuffled[pos1];
shuffled[pos1] := shuffled[pos2];
shuffled[pos2] := temp;
end;
function TShuffledString.TrySwap(pos1, pos2: Integer): Boolean;
begin
if (original[pos1] = shuffled[pos2]) or (original[pos2] = shuffled[pos1]) then
Exit(false)
else
Exit(true);
end;
procedure TShuffledString.Shuffle;
var
length, swaps: Integer;
used: TList<Integer>;
i, j, k: Integer;
begin
Randomize;
length := original.Length;
used := TList<Integer>.create();
for i := 0 to length - 1 do
begin
swaps := 0;
while used.Count <= (length - i) do
begin
j := i + Random(length - 1 - i);
if (original[i] <> original[j]) and TrySwap(i, j) and (not used.Contains(j)) then
begin
Swap(i, j);
Inc(swaps);
break;
end
else
used.Add(j);
end;
if swaps = 0 then
begin
for k := i downto 0 do
begin
if TrySwap(i, k) then
Swap(i, k);
end;
end;
used.Clear();
end;
used.Free;
end;
constructor TShuffledString.Create(Word: string);
begin
original := Word;
shuffled := TStringBuilder.create(Word);
Shuffle();
DetectIgnores();
end;
procedure TShuffledString.DetectIgnores;
var
ignores, i: Integer;
begin
ignores := 0;
for i := 0 to original.Length - 1 do
begin
if original[i] = shuffled[i] then
Inc(ignores);
end;
ignoredChars := ignores;
end;
procedure TShuffledString.Free;
begin
Shuffled.Free;
end;
function TShuffledString.GetShuffled: string;
begin
result := shuffled.ToString();
end;
class operator TShuffledString.Implicit(convert: string): TShuffledString;
begin
result := TShuffledString.Create(convert);
end;
var
words: array of string;
Word: TShuffledString;
w: string;
begin
words := ['abracadabra', 'seesaw', 'elk', 'grrrrrr', 'up', 'a'];
for w in words do
begin
Word := w;
writeln(format('%s, %s, (%d)', [Word.Original, Word.ToString, Word.Ignored]));
Word.Free;
end;
Readln;
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.
| #JavaScript | JavaScript | //String creation
var str='';
//or
str2=new String();
//String assignment
str="Hello";
//or
str2=', Hey there'; //can use " or '
str=str+str2;//concantenates
//string deletion
delete str2;//this will return true or false, true when it has been successfully deleted, it shouldn't/won't work when the variable has been declared with the keyword 'var', you don't have to initialize variables with the var keyword in JavaScript, but when you do, you cannot 'delete' them. However JavaScript garbage collects, so when the function returns, the variable declared on the function is erased.
//String comparison
str!=="Hello"; //!== not equal-> returns true there's also !===
str=="Hello, Hey there"; //returns true
//compares 'byte' by 'byte'
"Character Z">"Character A"; //returns true, when using > or < operators it converts the string to an array and evalues the first index that is higher than another. (using unicode values) in this case 'Z' char code is 90 (decimal) and 'A' char code is 65, therefore making one string "larger" than the other.
//String cloning and copying
string=str;//Strings are immutable therefore when you assign a string to a variable another one is created. So for two variables to have the 'same' string you have to add that string to an object, and get/set the string from that object
//Check if a string is empty
Boolean(''); //returns false
''[0]; //returns undefined
''.charCodeAt(); //returns NaN
''==0; //returns true
''===0; //returns false
''==false; //returns true
//Append byte to String
str+="\x40";//using + operator before the equal sign on a string makes it equal to str=str+"\x40"
//Extract a substring from a string
//str is "Hello, Hey there@"
str.substr(3); //returns "lo, Hey there@"
str.substr(-5); //returns "here@" negative values just go to the end
str.substr(7,9); //returns "Hey there" index of 7 + 9 characters after the 7
str.substring(3); //same as substr
str.substring(-5); //negative values don't work on substring same as substr(0)
str.substring(7,9); //returns "He" that is, whatever is between index 7 and index 9, same as substring(9,7)
//Replace every occurence of x byte with another string
str3="url,url,url,url,url";
str3.replace(/,/g,'\n') //Regex ,returns the same string with the , replaced by \n
str4=str3.replace(/./g,function(index){//it also supports callback functions, the function will be called when a match has been found..
return index==','?'\n':index;//returns replacement
})
//Join Strings
[str," ",str3].join(" "/*this is the character that will glue the strings*/)//we can join an array of strings
str3+str4;
str.concat('\n',str4); //concantenate them |
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #Racket | Racket | #lang racket
(define (find-bin-index limits v)
(let inner ((l 0) (r (vector-length limits)))
(let ((m (quotient (+ l r) 2)))
(if (< v (vector-ref limits m))
(if (= m l) l (inner l m))
(if (= m (sub1 r)) r (inner m r))))))
(define ((bin-given-limits! limits) data (bins (make-vector (add1 (vector-length limits)) 0)))
(for ((d data))
(let ((i (find-bin-index limits d)))
(vector-set! bins i (add1 (vector-ref bins i)))))
bins)
(define (report-bins-given-limits limits data)
(for ((b ((bin-given-limits! limits) data))
(ge (in-sequences (in-value -Inf.0) limits))
(lt (in-sequences limits (in-value +Inf.0))))
(printf "~a <= v < ~a : ~a~%" ge lt b)))
(define (Bin-given-limits)
(report-bins-given-limits
#[23 37 43 53 67 83]
(list 95 21 94 12 99 4 70 75 83 93 52 80 57 5 53 86 65 17 92 83 71 61 54 58 47
16 8 9 32 84 7 87 46 19 30 37 96 6 98 40 79 97 45 64 60 29 49 36 43 55))
(newline)
(report-bins-given-limits
#[14 18 249 312 389 392 513 591 634 720]
(list 445 814 519 697 700 130 255 889 481 122 932 77 323 525 570 219 367 523 442 933
416 589 930 373 202 253 775 47 731 685 293 126 133 450 545 100 741 583 763 306
655 267 248 477 549 238 62 678 98 534 622 907 406 714 184 391 913 42 560 247
346 860 56 138 546 38 985 948 58 213 799 319 390 634 458 945 733 507 916 123
345 110 720 917 313 845 426 9 457 628 410 723 354 895 881 953 677 137 397 97
854 740 83 216 421 94 517 479 292 963 376 981 480 39 257 272 157 5 316 395
787 942 456 242 759 898 576 67 298 425 894 435 831 241 989 614 987 770 384 692
698 765 331 487 251 600 879 342 982 527 736 795 585 40 54 901 408 359 577 237
605 847 353 968 832 205 838 427 876 959 686 646 835 127 621 892 443 198 988 791
466 23 707 467 33 670 921 180 991 396 160 436 717 918 8 374 101 684 727 749)))
(module+ main
(Bin-given-limits)) |
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #Raku | Raku |
sub bin_it ( @limits, @data ) {
my @ranges = ( -Inf, |@limits, Inf ).rotor( 2 => -1 ).map: { .[0] ..^ .[1] };
my @binned = @data.classify(-> $d { @ranges.grep(-> $r { $d ~~ $r }) });
my %counts = @binned.map: { .key => .value.elems };
return @ranges.map: { $_ => ( %counts{$_} // 0 ) };
}
sub bin_format ( @bins ) {
return @bins.map: { .key.gist.fmt('%9s => ') ~ .value.fmt('%2d') };
}
my @tests =
{
limits => (23, 37, 43, 53, 67, 83),
data => (95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,16,8,9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55),
},
{
limits => (14, 18, 249, 312, 389, 392, 513, 591, 634, 720),
data => (
445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749
),
},
;
for @tests -> ( :@limits, :@data ) {
my @bins = bin_it( @limits, @data );
.say for bin_format(@bins);
say '';
}
|
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Ruby | Ruby | dna = <<DNA_STR
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
DNA_STR
chunk_size = 60
dna = dna.delete("\n")
size = dna.size
0.step(size, chunk_size) do |pos|
puts "#{pos.to_s.ljust(6)} #{dna[pos, chunk_size]}"
end
puts dna.chars.tally.sort.map{|ar| ar.join(" : ") }
puts "Total : #{dna.size}"
|
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Rust | Rust |
use std::collections::HashMap;
fn main() {
let dna = "CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG\
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG\
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT\
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT\
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG\
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA\
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT\
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG\
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC\
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT";
let mut base_count = HashMap::new();
let mut total_count = 0;
print!("Sequence:");
for base in dna.chars() {
if total_count % 50 == 0 {
print!("\n{:3}: ", total_count);
}
print!("{}", base);
total_count += 1;
let count = base_count.entry(base).or_insert(0); // Return current count for base or insert 0
*count += 1;
}
println!("\n");
println!("Base count:");
println!("-----------");
let mut base_count: Vec<_> = base_count.iter().collect(); // HashMaps can't be sorted, so collect into Vec
base_count.sort_by_key(|bc| bc.0); // Sort bases alphabetically
for (base, count) in base_count.iter() {
println!(" {}: {:3}", base, count);
}
println!();
println!("Total: {}", total_count);
}
|
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.
| #ARM_Assembly | ARM Assembly |
/* ARM assembly Raspberry PI */
/* program binarydigit.s */
/* Constantes */
.equ STDOUT, 1
.equ WRITE, 4
.equ EXIT, 1
/* Initialized data */
.data
sMessAffBin: .ascii "The decimal value "
sZoneDec: .space 12,' '
.ascii " should produce an output of "
sZoneBin: .space 36,' '
.asciz "\n"
/* code section */
.text
.global main
main: /* entry of program */
push {fp,lr} /* save des 2 registres */
mov r0,#5
ldr r1,iAdrsZoneDec
bl conversion10S @ decimal conversion
bl conversion2 @ binary conversion and display résult
mov r0,#50
ldr r1,iAdrsZoneDec
bl conversion10S
bl conversion2
mov r0,#-1
ldr r1,iAdrsZoneDec
bl conversion10S
bl conversion2
mov r0,#1
ldr r1,iAdrsZoneDec
bl conversion10S
bl conversion2
100: /* standard end of the program */
mov r0, #0 @ return code
pop {fp,lr} @restaur 2 registers
mov r7, #EXIT @ request to exit program
swi 0 @ perform the system call
iAdrsZoneDec: .int sZoneDec
/******************************************************************/
/* register conversion in binary */
/******************************************************************/
/* r0 contains the register */
conversion2:
push {r0,lr} /* save registers */
push {r1-r5} /* save others registers */
ldr r1,iAdrsZoneBin @ address reception area
clz r2,r0 @ number of left zeros bits
rsb r2,#32 @ number of significant bits
mov r4,#' ' @ space
add r3,r2,#1 @ position counter in reception area
1:
strb r4,[r1,r3] @ space in other location of reception area
add r3,#1
cmp r3,#32 @ end of area ?
ble 1b @ no! loop
mov r3,r2 @ position counter of the written character
2: @ loop
lsrs r0,#1 @ shift right one bit with flags
movcc r4,#48 @ carry clear => character 0
movcs r4,#49 @ carry set => character 1
strb r4,[r1,r3] @ character in reception area at position counter
sub r3,r3,#1 @
subs r2,r2,#1 @ 0 bits ?
bgt 2b @ no! loop
ldr r0,iAdrsZoneMessBin
bl affichageMess
100:
pop {r1-r5} /* restaur others registers */
pop {r0,lr}
bx lr
iAdrsZoneBin: .int sZoneBin
iAdrsZoneMessBin: .int sMessAffBin
/******************************************************************/
/* display text with size calculation */
/******************************************************************/
/* r0 contains the address of the message */
affichageMess:
push {fp,lr} /* save registres */
push {r0,r1,r2,r7} /* save others registres */
mov r2,#0 /* counter length */
1: /* loop length calculation */
ldrb r1,[r0,r2] /* read octet start position + index */
cmp r1,#0 /* if 0 its over */
addne r2,r2,#1 /* else add 1 in the length */
bne 1b /* and loop */
/* so here r2 contains the length of the message */
mov r1,r0 /* address message in r1 */
mov r0,#STDOUT /* code to write to the standard output Linux */
mov r7, #WRITE /* code call system "write" */
swi #0 /* call systeme */
pop {r0,r1,r2,r7} /* restaur others registres */
pop {fp,lr} /* restaur des 2 registres */
bx lr /* return */
/***************************************************/
/* conversion registre en décimal signé */
/***************************************************/
/* r0 contient le registre */
/* r1 contient l adresse de la zone de conversion */
conversion10S:
push {fp,lr} /* save des 2 registres frame et retour */
push {r0-r5} /* save autres registres */
mov r2,r1 /* debut zone stockage */
mov r5,#'+' /* par defaut le signe est + */
cmp r0,#0 /* nombre négatif ? */
movlt r5,#'-' /* oui le signe est - */
mvnlt r0,r0 /* et inversion en valeur positive */
addlt r0,#1
mov r4,#10 /* longueur de la zone */
1: /* debut de boucle de conversion */
bl divisionpar10 /* division */
add r1,#48 /* ajout de 48 au reste pour conversion ascii */
strb r1,[r2,r4] /* stockage du byte en début de zone r5 + la position r4 */
sub r4,r4,#1 /* position précedente */
cmp r0,#0
bne 1b /* boucle si quotient different de zéro */
strb r5,[r2,r4] /* stockage du signe à la position courante */
subs r4,r4,#1 /* position précedente */
blt 100f /* si r4 < 0 fin */
/* sinon il faut completer le debut de la zone avec des blancs */
mov r3,#' ' /* caractere espace */
2:
strb r3,[r2,r4] /* stockage du byte */
subs r4,r4,#1 /* position précedente */
bge 2b /* boucle si r4 plus grand ou egal a zero */
100: /* fin standard de la fonction */
pop {r0-r5} /*restaur des autres registres */
pop {fp,lr} /* restaur des 2 registres frame et retour */
bx lr
/***************************************************/
/* division par 10 signé */
/* Thanks to http://thinkingeek.com/arm-assembler-raspberry-pi/*
/* and http://www.hackersdelight.org/ */
/***************************************************/
/* r0 contient le dividende */
/* r0 retourne le quotient */
/* r1 retourne le reste */
divisionpar10:
/* r0 contains the argument to be divided by 10 */
push {r2-r4} /* save others registers */
mov r4,r0
ldr r3, .Ls_magic_number_10 /* r1 <- magic_number */
smull r1, r2, r3, r0 /* r1 <- Lower32Bits(r1*r0). r2 <- Upper32Bits(r1*r0) */
mov r2, r2, ASR #2 /* r2 <- r2 >> 2 */
mov r1, r0, LSR #31 /* r1 <- r0 >> 31 */
add r0, r2, r1 /* r0 <- r2 + r1 */
add r2,r0,r0, lsl #2 /* r2 <- r0 * 5 */
sub r1,r4,r2, lsl #1 /* r1 <- r4 - (r2 * 2) = r4 - (r0 * 10) */
pop {r2-r4}
bx lr /* leave function */
.align 4
.Ls_magic_number_10: .word 0x66666667
|
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.
| #J | J | thru=: <./ + -~ i.@+ _1 ^ > NB. integers from x through y
NB.*getBresenhamLine v Returns points for a line given start and end points
NB. y is: y0 x0 ,: y1 x1
getBresenhamLine=: monad define
steep=. ([: </ |@-~/) y
points=. |."1^:steep y
slope=. %~/ -~/ points
ypts=. thru/ {."1 points
xpts=. ({: + 0.5 <.@:+ slope * ypts - {.) {.points
|."1^:steep ypts,.xpts
)
NB.*drawLines v Draws lines (x) on image (y)
NB. x is: 2-item list (start and end points) ; (color)
drawLines=: (1&{:: ;~ [: ; [: <@getBresenhamLine"2 (0&{::))@[ setPixels ] |
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
| #Pascal | Pascal | my $x = 0.0;
my $true_or_false = $x ? 'true' : 'false'; # false |
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #Perl | Perl | my $x = 0.0;
my $true_or_false = $x ? 'true' : 'false'; # false |
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)..
| #MATLAB_.2F_Octave | MATLAB / Octave | function b = compassbox(d)
b = ceil(mod(d+360/64,360)*32/360);
end; |
http://rosettacode.org/wiki/Bitwise_operations | Bitwise operations |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Write a routine to perform a bitwise AND, OR, and XOR on two integers, a bitwise NOT on the first integer, a left shift, right shift, right arithmetic shift, left rotate, and right rotate.
All shifts and rotates should be done on the first integer with a shift/rotate amount of the second integer.
If any operation is not available in your language, note it.
| #Forth | Forth | : arshift 0 ?do 2/ loop ; \ 2/ is an arithmetic shift right by one bit (2* shifts left one bit)
: bitwise ( a b -- )
cr ." a = " over . ." b = " dup .
cr ." a and b = " 2dup and .
cr ." a or b = " 2dup or .
cr ." a xor b = " 2dup xor .
cr ." not a = " over invert .
cr ." a shl b = " 2dup lshift .
cr ." a shr b = " 2dup rshift .
cr ." a ashr b = " 2dup arshift .
2drop ; |
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.
| #Go | Go | package main
import (
"bytes"
"fmt"
"image"
"image/color"
"image/draw"
"image/png"
)
func main() {
// A rectangle from 0,0 to 300,240.
r := image.Rect(0, 0, 300, 240)
// Create an image
im := image.NewNRGBA(r)
// set some color variables for convience
var (
red = color.RGBA{0xff, 0x00, 0x00, 0xff}
blue = color.RGBA{0x00, 0x00, 0xff, 0xff}
)
// Fill with a uniform color
draw.Draw(im, r, &image.Uniform{red}, image.ZP, draw.Src)
// Set individual pixels
im.Set(10, 20, blue)
im.Set(20, 30, color.Black)
im.Set(30, 40, color.RGBA{0x10, 0x20, 0x30, 0xff})
// Get the values of specific pixels as color.Color types.
// The color will be in the color.Model of the image (in this
// case color.NRGBA) but color models can convert their values
// to other models.
c1 := im.At(0, 0)
c2 := im.At(10, 20)
// or directly as RGB components (scaled values)
redc, greenc, bluec, _ := c1.RGBA()
redc, greenc, bluec, _ = im.At(30, 40).RGBA()
// Images can be read and writen in various formats
var buf bytes.Buffer
err := png.Encode(&buf, im)
if err != nil {
fmt.Println(err)
}
fmt.Println("Image size:", im.Bounds().Dx(), "×", im.Bounds().Dy())
fmt.Println(buf.Len(), "bytes when encoded as PNG.")
fmt.Printf("Pixel at %7v is %v\n", image.Pt(0, 0), c1)
fmt.Printf("Pixel at %7v is %#v\n", image.Pt(10, 20), c2) // %#v shows type details
fmt.Printf("Pixel at %7v has R=%d, G=%d, B=%d\n",
image.Pt(30, 40), redc, greenc, bluec)
} |
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 | #jq | jq | # nth Bell number
def bell:
. as $n
| if $n < 0 then "non-negative integer expected"
elif $n < 2 then 1
else
reduce range(1; $n) as $i ([1];
reduce range(1; $i) as $j (.;
.[$i - $j] as $x
| .[$i - $j - 1] += $x )
| .[$i] = .[0] + .[$i - 1] )
| .[$n - 1]
end;
# The task
range(1;51) | 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 | #Julia | Julia | """
bellnum(n)
Compute the ``n``th Bell number.
"""
function bellnum(n::Integer)
if n < 0
throw(DomainError(n))
elseif n < 2
return 1
end
list = Vector{BigInt}(undef, n)
list[1] = 1
for i = 2:n
for j = 1:i - 2
list[i - j - 1] += list[i - j]
end
list[i] = list[1] + list[i - 1]
end
return list[n]
end
for i in 1:50
println(bellnum(i))
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 .
| #Fortran | Fortran | -*- mode: compilation; default-directory: "/tmp/" -*-
Compilation started at Sat May 18 01:13:00
a=./f && make $a && $a
f95 -Wall -ffree-form f.F -o f
0.301030010 0.176091254 0.124938756 9.69100147E-02 7.91812614E-02 6.69467747E-02 5.79919666E-02 5.11525236E-02 4.57575098E-02 THE LAW
0.300999999 0.177000001 0.125000000 9.60000008E-02 7.99999982E-02 6.70000017E-02 5.70000000E-02 5.29999994E-02 4.50000018E-02 LEADING FIBONACCI DIGIT
Compilation finished at Sat May 18 01:13:00 |
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 .
| #FreeBASIC | FreeBASIC | ' version 27-10-2016
' compile with: fbc -s console
#Define max 1000 ' total number of Fibonacci numbers
#Define max_sieve 15485863 ' should give 1,000,000
#Include Once "gmp.bi" ' uses the GMP libary
Dim As ZString Ptr z_str
Dim As ULong n, d
ReDim As ULong digit(1 To 9)
Dim As Double expect, found
Dim As mpz_ptr fib1, fib2
fib1 = Allocate(Len(__mpz_struct)) : Mpz_init_set_ui(fib1, 0)
fib2 = Allocate(Len(__mpz_struct)) : Mpz_init_set_ui(fib2, 1)
digit(1) = 1 ' fib2
For n = 2 To max
Swap fib1, fib2 ' fib1 = 1, fib2 = 0
mpz_add(fib2, fib1, fib2) ' fib1 = 1, fib2 = 1 (fib1 + fib2)
z_str = mpz_get_str(0, 10, fib2)
d = Val(Left(*z_str, 1)) ' strip the 1 digit on the left off
digit(d) = digit(d) +1
Next
mpz_clear(fib1) : DeAllocate(fib1)
mpz_clear(fib2) : DeAllocate(fib2)
Print
Print "First 1000 Fibonacci numbers"
Print "nr: total found expected difference"
For d = 1 To 9
n = digit(d)
found = n / 10
expect = (Log(1 + 1 / d) / Log(10)) * 100
Print Using " ## ##### ###.## % ###.## % ##.### %"; _
d; n ; found; expect; expect - found
Next
ReDim digit(1 To 9)
ReDim As UByte sieve(max_sieve)
'For d = 4 To max_sieve Step 2
' sieve(d) = 1
'Next
Print : Print "start sieve"
For d = 3 To sqr(max_sieve)
If sieve(d) = 0 Then
For n = d * d To max_sieve Step d * 2
sieve(n) = 1
Next
End If
Next
digit(2) = 1 ' 2
Print "start collecting first digits"
For n = 3 To max_sieve Step 2
If sieve(n) = 0 Then
d = Val(Left(Trim(Str(n)), 1))
digit(d) = digit(d) +1
End If
Next
Dim As ulong total
For n = 1 To 9
total = total + digit(n)
Next
Print
Print "First";total; " primes"
Print "nr: total found expected difference"
For d = 1 To 9
n = digit(d)
found = n / total * 100
expect = (Log(1 + 1 / d) / Log(10)) * 100
Print Using " ## ######## ###.## % ###.## % ###.### %"; _
d; n ; found; expect; expect - found
Next
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
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.
| #Factor | Factor | IN: scratchpad
[
0 1 1 "%2d : %d / %d\n" printf
1 -1 2 "%2d : %d / %d\n" printf
30 iota [
1 + 2 * dup bernoulli [ numerator ] [ denominator ] bi
"%2d : %d / %d\n" printf
] each
] time
0 : 1 / 1
1 : -1 / 2
2 : 1 / 6
4 : -1 / 30
6 : 1 / 42
8 : -1 / 30
10 : 5 / 66
12 : -691 / 2730
14 : 7 / 6
16 : -3617 / 510
18 : 43867 / 798
20 : -174611 / 330
22 : 854513 / 138
24 : -236364091 / 2730
26 : 8553103 / 6
28 : -23749461029 / 870
30 : 8615841276005 / 14322
32 : -7709321041217 / 510
34 : 2577687858367 / 6
36 : -26315271553053477373 / 1919190
38 : 2929993913841559 / 6
40 : -261082718496449122051 / 13530
42 : 1520097643918070802691 / 1806
44 : -27833269579301024235023 / 690
46 : 596451111593912163277961 / 282
48 : -5609403368997817686249127547 / 46410
50 : 495057205241079648212477525 / 66
52 : -801165718135489957347924991853 / 1590
54 : 29149963634884862421418123812691 / 798
56 : -2479392929313226753685415739663229 / 870
58 : 84483613348880041862046775994036021 / 354
60 : -1215233140483755572040304994079820246041491 / 56786730
Running time: 0.00489444 seconds |
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.
| #Axe | Axe | Lbl BSEARCH
0→L
r₃-1→H
While L≤H
(L+H)/2→M
If {L+M}>r₂
M-1→H
ElseIf {L+M}<r₂
M+1→L
Else
M
Return
End
End
-1
Return |
http://rosettacode.org/wiki/Best_shuffle | Best shuffle | Task
Shuffle the characters of a string in such a way that as many of the character values are in a different position as possible.
A shuffle that produces a randomized result among the best choices is to be preferred. A deterministic approach that produces the same sequence every time is acceptable as an alternative.
Display the result as follows:
original string, shuffled string, (score)
The score gives the number of positions whose character value did not change.
Example
tree, eetr, (0)
Test cases
abracadabra
seesaw
elk
grrrrrr
up
a
Related tasks
Anagrams/Deranged anagrams
Permutations/Derangements
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Elena | Elena | import system'routines;
import extensions;
import extensions'text;
extension op
{
get Shuffled()
{
var original := self.toArray();
var shuffled := self.toArray();
for (int i := 0, i < original.Length, i += 1) {
for (int j := 0, j < original.Length, j += 1) {
if (i != j && original[i] != shuffled[j] && original[j] != shuffled[i])
{
shuffled.exchange(i,j)
}
}
};
^ shuffled.summarize(new StringWriter()).toString()
}
score(originalText)
{
var shuffled := self.toArray();
var original := originalText.toArray();
int score := 0;
for (int i := 0, i < original.Length, i += 1) {
if (original[i] == shuffled[i]) { score += 1 }
};
^ score
}
}
public program()
{
new string[]{"abracadabra", "seesaw", "grrrrrr", "pop", "up", "a"}.forEach:(s)
{
var shuffled_s := s.Shuffled;
console.printLine("The best shuffle of ",s," is ",shuffled_s,"(",shuffled_s.score(s),")")
};
console.readChar()
} |
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
| #Erlang | Erlang |
-module( best_shuffle ).
-export( [sameness/2, string/1, task/0] ).
sameness( String1, String2 ) -> lists:sum( [1 || {X, X} <- lists:zip(String1, String2)] ).
string( String ) ->
{"", String, Acc} = lists:foldl( fun different/2, {lists:reverse(String), String, []}, String ),
lists:reverse( Acc ).
task() ->
Strings = ["abracadabra", "seesaw", "elk", "grrrrrr", "up", "a"],
Shuffleds = [string(X) || X <- Strings],
[io:fwrite("~p ~p ~p~n", [X, Y, sameness(X,Y)]) || {X, Y} <- lists:zip(Strings, Shuffleds)].
different( Character, {[Character], Original, Acc} ) ->
try_to_save_last( Character, Original, Acc );
different( Character, {[Character | T]=Not_useds, Original, Acc} ) ->
Different_or_same = different_or_same( [X || X <- T, X =/= Character], Character ),
{lists:delete(Different_or_same, Not_useds), Original, [Different_or_same | Acc]};
different( _Character1, {[Character2 | T], Original, Acc} ) ->
{T, Original, [Character2 | Acc]}.
different_or_same( [Different | _T], _Character ) -> Different;
different_or_same( [], Character ) -> Character.
try_to_save_last( Character, Original_string, Acc ) ->
Fun = fun ({X, Y}) -> (X =:= Y) orelse (X =:= Character) end,
New_acc = try_to_save_last( lists:splitwith(Fun, lists:zip(lists:reverse(Original_string), [Character | Acc])), [Character | Acc] ),
{"", Original_string, New_acc}.
try_to_save_last( {_Not_split, []}, Acc ) -> Acc;
try_to_save_last( {Last_reversed_zip, First_reversed_zip}, _Acc ) ->
{_Last_reversed_original, [Last_character_acc | Last_part_acc]} = lists:unzip( Last_reversed_zip ),
{_First_reversed_original, [Character_acc | First_part_acc]} = lists:unzip( First_reversed_zip ),
[Character_acc | Last_part_acc] ++ [Last_character_acc | First_part_acc].
|
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.
| #jq | jq | # If the input is a valid representation of a binary string
# then pass it along:
def check_binary:
. as $a
| reduce .[] as $x
($a;
if $x | (type == "number" and . == floor
and 0 <= . and . <= 255) then $a
else error("\(.) is an invalid representation of a byte")
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.
| #Julia | Julia |
# String assignment. Creation and garbage collection are automatic.
a = "123\x00 abc " # strings can contain bytes that are not printable in the local font
b = "456" * '\x09'
c = "789"
println(a)
println(b)
println(c)
# string comparison
println("(a == b) is $(a == b)")
# String copying.
A = a
B = b
C = c
println(A)
println(B)
println(C)
# check if string is empty
if length(a) == 0
println("string a is empty")
else
println("string a is not empty")
end
# append a byte (actually this is a Char in Julia, and may also be up to 32 bit Unicode) to a string
a= a * '\x64'
println(a)
# extract a substring from string
e = a[1:6]
println(e)
# repeat strings with ^
b4 = b ^ 4
println(b4)
# Replace every occurrence of a string in another string with third string
r = replace(b4, "456" => "xyz")
println(r)
# join strings with *
d = a * b * c
println(d)
|
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #REXX | REXX | /*REXX program counts how many numbers of a set that fall in the range of each bin. */
lims= 23 37 43 53 67 83 /* ◄■■■■■■1st set of bin limits & data.*/
data= 95 21 94 12 99 4 70 75 83 93 52 80 57 5 53 86 65 17 92 83 71 61 54 58 47 ,
16 8 9 32 84 7 87 46 19 30 37 96 6 98 40 79 97 45 64 60 29 49 36 43 55
call lims lims; call bins data
call show 'the 1st set of bin counts for the specified data:'
say; say; say
lims= 14 18 249 312 389 392 513 591 634 720 /* ◄■■■■■■2nd set of bin limits & data.*/
data= 445 814 519 697 700 130 255 889 481 122 932 77 323 525 570 219 367 523 442 933 ,
416 589 930 373 202 253 775 47 731 685 293 126 133 450 545 100 741 583 763 306 ,
655 267 248 477 549 238 62 678 98 534 622 907 406 714 184 391 913 42 560 247 ,
346 860 56 138 546 38 985 948 58 213 799 319 390 634 458 945 733 507 916 123 ,
345 110 720 917 313 845 426 9 457 628 410 723 354 895 881 953 677 137 397 97 ,
854 740 83 216 421 94 517 479 292 963 376 981 480 39 257 272 157 5 316 395 ,
787 942 456 242 759 898 576 67 298 425 894 435 831 241 989 614 987 770 384 692 ,
698 765 331 487 251 600 879 342 982 527 736 795 585 40 54 901 408 359 577 237 ,
605 847 353 968 832 205 838 427 876 959 686 646 835 127 621 892 443 198 988 791 ,
466 23 707 467 33 670 921 180 991 396 160 436 717 918 8 374 101 684 727 749
call lims lims; call bins data
call show 'the 2nd set of bin counts for the specified data:'
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
bins: parse arg nums; !.= 0; datum= words(nums); wc= length(datum) /*max width count.*/
do j=1 for datum; x= word(nums, j)
do k=0 for # /*find the bin that this number is in. */
if x < @.k then do; !.k= !.k + 1; iterate j; end /*bump a bin count*/
end /*k*/
!.k= !.k + 1 /*number is > the highest bin specified*/
end /*j*/; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
lims: parse arg limList; #= words(limList); wb= 0 /*max width binLim*/
do j=1 for #; _= j - 1; @._= word(limList, j); wb= max(wb, length(@._) )
end /*j*/; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
show: parse arg t; say center(t, 51 ); $= left('', 9) /*$: for indentation*/
say center('', 51, "═") /*show title separator.*/
jp= # - 1; ge= '≥'; le='<'; eq= ' count='
do j=0 for #; jm= j - 1; bin= right(@.j, wb)
if j==0 then say $ left('', length(ge) +3+wb+length(..) )le bin eq right(!.j, wc)
else say $ ge right(@.jm, wb) .. le bin eq right(!.j, wc)
if j==jp then say $ ge right(@.jp,wb) left('', 3+length(..)+wb) eq right(!.#, wc)
end /*j*/; return |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Swift | Swift | import Foundation
let dna = """
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
"""
print("input:\n\(dna)\n")
let counts =
dna.replacingOccurrences(of: "\n", with: "").reduce(into: [:], { $0[$1, default: 0] += 1 })
print("Counts: \(counts)")
print("Total: \(counts.values.reduce(0, +))") |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Tcl | Tcl | namespace path ::tcl::mathop
proc process {data {width 50}} {
set len [string length $data]
set addrwidth [string length [* [/ $len $width] $width]]
for {set i 0} {$i < $len} {incr i $width} {
puts "[format %${addrwidth}u $i] [string range $data $i $i+[- $width 1]]"
}
puts "\nBase count:"
foreach base {A C G T} {
puts "$base [regexp -all $base $data]"
}
puts "Total $len"
}
set test [string cat \
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG \
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG \
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT \
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT \
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG \
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA \
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT \
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG \
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC \
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT]
process $test 50 |
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.
| #Arturo | Arturo | print as.binary 5
print as.binary 50
print as.binary 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.
| #Java | Java | import java.awt.Color;
import java.awt.Dimension;
import java.awt.Graphics;
import javax.swing.JFrame;
import javax.swing.JPanel;
import javax.swing.SwingUtilities;
import javax.swing.WindowConstants;
public class Bresenham {
public static void main(String[] args) {
SwingUtilities.invokeLater(Bresenham::run);
}
private static void run() {
JFrame f = new JFrame();
f.setDefaultCloseOperation(WindowConstants.DISPOSE_ON_CLOSE);
f.setTitle("Bresenham");
f.getContentPane().add(new BresenhamPanel());
f.pack();
f.setLocationRelativeTo(null);
f.setVisible(true);
}
}
class BresenhamPanel extends JPanel {
private final int pixelSize = 10;
BresenhamPanel() {
setPreferredSize(new Dimension(600, 500));
setBackground(Color.WHITE);
}
@Override
public void paintComponent(Graphics g) {
super.paintComponent(g);
int w = (getWidth() - 1) / pixelSize;
int h = (getHeight() - 1) / pixelSize;
int maxX = (w - 1) / 2;
int maxY = (h - 1) / 2;
int x1 = -maxX, x2 = maxX * -2 / 3, x3 = maxX * 2 / 3, x4 = maxX;
int y1 = -maxY, y2 = maxY * -2 / 3, y3 = maxY * 2 / 3, y4 = maxY;
drawLine(g, 0, 0, x3, y1); // NNE
drawLine(g, 0, 0, x4, y2); // ENE
drawLine(g, 0, 0, x4, y3); // ESE
drawLine(g, 0, 0, x3, y4); // SSE
drawLine(g, 0, 0, x2, y4); // SSW
drawLine(g, 0, 0, x1, y3); // WSW
drawLine(g, 0, 0, x1, y2); // WNW
drawLine(g, 0, 0, x2, y1); // NNW
}
private void plot(Graphics g, int x, int y) {
int w = (getWidth() - 1) / pixelSize;
int h = (getHeight() - 1) / pixelSize;
int maxX = (w - 1) / 2;
int maxY = (h - 1) / 2;
int borderX = getWidth() - ((2 * maxX + 1) * pixelSize + 1);
int borderY = getHeight() - ((2 * maxY + 1) * pixelSize + 1);
int left = (x + maxX) * pixelSize + borderX / 2;
int top = (y + maxY) * pixelSize + borderY / 2;
g.setColor(Color.black);
g.drawOval(left, top, pixelSize, pixelSize);
}
private void drawLine(Graphics g, int x1, int y1, int x2, int y2) {
// delta of exact value and rounded value of the dependent variable
int d = 0;
int dx = Math.abs(x2 - x1);
int dy = Math.abs(y2 - y1);
int dx2 = 2 * dx; // slope scaling factors to
int dy2 = 2 * dy; // avoid floating point
int ix = x1 < x2 ? 1 : -1; // increment direction
int iy = y1 < y2 ? 1 : -1;
int x = x1;
int y = y1;
if (dx >= dy) {
while (true) {
plot(g, x, y);
if (x == x2)
break;
x += ix;
d += dy2;
if (d > dx) {
y += iy;
d -= dx2;
}
}
} else {
while (true) {
plot(g, x, y);
if (y == y2)
break;
y += iy;
d += dx2;
if (d > dy) {
x += ix;
d -= dy2;
}
}
}
}
} |
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
| #Phix | Phix | with javascript_semantics
for i=1 to 3 do
integer c = (i==2), -- fine
d = (c==1), -- oops
e = (c==true), -- fine
f = equal(c,1) -- fine, ditto equal(c,true)
printf(1,"%d==2:%5t(%d) ==1:%5t, eq1:%5t, ==true:%5t\n",
{i, c, c, d, e, f})
end for
--
-- output on desktop/Phix: 1==2:false(0) ==1:false, eq1:false, ==true:false
-- 2==2: true(1) ==1: true, eq1: true, ==true: true
-- 3==2:false(0) ==1:false, eq1:false, ==true:false
--
-- output on pwa/p2js: 1==2:false(0) ==1:false, eq1:false, ==true:false
-- 2==2: true(1) ==1:false, eq1: true, ==true: true
-- 3==2:false(0) ==1:false, eq1:false, ==true:false
--
|
http://rosettacode.org/wiki/Boolean_values | Boolean values | Task
Show how to represent the boolean states "true" and "false" in a language.
If other objects represent "true" or "false" in conditionals, note it.
Related tasks
Logical operations
| #PHP | PHP | go ?=>
member(N,1..5),
println(N),
fail, % or false/0 to get other solutions
nl.
go => true. |
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)..
| #Modula-2 | Modula-2 | MODULE BoxTheCompass;
FROM FormatString IMPORT FormatString;
FROM RealStr IMPORT RealToStr;
FROM Terminal IMPORT WriteString,WriteLn,Write,ReadChar;
PROCEDURE expand(cp : ARRAY OF CHAR);
VAR i : INTEGER = 0;
BEGIN
WHILE cp[i] # 0C DO
IF i=0 THEN
CASE cp[i] OF
'N': WriteString("North") |
'E': WriteString("East") |
'S': WriteString("South") |
'W': WriteString("West") |
'b': WriteString(" by ")
ELSE
WriteString("-");
END;
ELSE
CASE cp[i] OF
'N': WriteString("north") |
'E': WriteString("east") |
'S': WriteString("south") |
'W': WriteString("west") |
'b': WriteString(" by ")
ELSE
WriteString("-");
END;
END;
INC(i)
END;
END expand;
PROCEDURE FormatReal(r : REAL);
VAR
buf : ARRAY[0..63] OF CHAR;
u,v : INTEGER;
w : REAL;
BEGIN
u := TRUNC(r);
w := r - FLOAT(u);
v := TRUNC(100.0 * w);
FormatString("%6i.%'02i", buf, u, v);
WriteString(buf);
END FormatReal;
VAR
cp : ARRAY[0..31] OF ARRAY[0..4] OF CHAR = {
"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"
};
buf : ARRAY[0..63] OF CHAR;
i,index : INTEGER;
heading : REAL;
BEGIN
WriteString("Index Degrees Compass point");
WriteLn;
WriteString("----- ------- -------------");
WriteLn;
FOR i:=0 TO 32 DO
index := i MOD 32;
heading := FLOAT(i) * 11.25;
CASE i MOD 3 OF
1: heading := heading + 5.62; |
2: heading := heading - 5.62;
ELSE
(* empty *)
END;
FormatString("%2i ", buf, index+1);
WriteString(buf);
FormatReal(heading);
WriteString(" ");
expand(cp[index]);
WriteLn
END;
ReadChar
END BoxTheCompass. |
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.
| #Fortran | Fortran | integer :: i, j = -1, k = 42
logical :: a
i = bit_size(j) ! returns the number of bits in the given INTEGER variable
! bitwise boolean operations on integers
i = iand(k, j) ! returns bitwise AND of K and J
i = ior(k, j) ! returns bitwise OR of K and J
i = ieor(k, j) ! returns bitwise EXCLUSIVE OR of K and J
i = not(j) ! returns bitwise NOT of J
! single-bit integer/logical operations (bit positions are zero-based)
a = btest(i, 4) ! returns logical .TRUE. if bit position 4 of I is 1, .FALSE. if 0
i = ibclr(k, 8) ! returns value of K with 8th bit position "cleared" (set to 0)
i = ibset(k, 13) ! returns value of K with 13th bit position "set" (set to 1)
! multi-bit integer operations
i = ishft(k, j) ! returns value of K shifted by J bit positions, with ZERO fill
! (right shift if J < 0 and left shift if J > 0).
i = ishftc(k, j) ! returns value of K shifted CIRCULARLY by J bit positions
! (right circular shift if J < 0 and left if J > 0)
i = ishftc(k, j, 20) ! returns value as before except that ONLY the 20 lowest order
! (rightmost) bits are circularly shifted
i = ibits(k, 7, 8) ! extracts 8 contiguous bits from K starting at position 7 and
! returns them as the rightmost bits of an otherwise
! zero-filled integer. For non-negative K this is
! arithmetically equivalent to: MOD((K / 2**7), 2**8) |
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.
| #Haskell | Haskell | module Bitmap(module Bitmap) where
import Control.Monad
import Control.Monad.ST
import Data.Array.ST
newtype Pixel = Pixel (Int, Int) deriving Eq
instance Ord Pixel where
compare (Pixel (x1, y1)) (Pixel (x2, y2)) =
case compare y1 y2 of
EQ -> compare x1 x2
v -> v
instance Ix Pixel where
{- This instance differs from the one for (Int, Int) in that
the ordering of indices is
(0,0), (1,0), (2,0), (0,1), (1,1), (2,1)
instead of
(0,0), (0,1), (1,0), (1,1), (2,0), (2,1). -}
range (Pixel (xa, ya), Pixel (xz, yz)) =
[Pixel (x, y) | y <- [ya .. yz], x <- [xa .. xz]]
index (Pixel (xa, ya), Pixel (xz, _)) (Pixel (xi, yi)) =
(yi - ya)*(xz - xa + 1) + (xi - xa)
inRange (Pixel (xa, ya), Pixel (xz, yz)) (Pixel (xi, yi)) =
not $ xi < xa || xi > xz || yi < ya || yi > yz
rangeSize (Pixel (xa, ya), Pixel (xz, yz)) =
(xz - xa + 1) * (yz - ya + 1)
instance Show Pixel where
show (Pixel p) = show p
class Ord c => Color c where
luminance :: c -> Int
-- The Int should be in the range [0 .. 255].
black, white :: c
toNetpbm :: [c] -> String
fromNetpbm :: [Int] -> [c]
netpbmMagicNumber, netpbmMaxval :: c -> String
{- The argument to these two functions is ignored; the
parameter is only for typechecking. -}
newtype Color c => Image s c = Image (STArray s Pixel c)
image :: Color c => Int -> Int -> c -> ST s (Image s c)
{- Creates a new image with the given width and height, filled
with the given color. -}
image w h = liftM Image .
newArray (Pixel (0, 0), Pixel (w - 1, h - 1))
listImage :: Color c => Int -> Int -> [c] -> ST s (Image s c)
{- Creates a new image with the given width and height, with
each pixel set to the corresponding element of the given list. -}
listImage w h = liftM Image .
newListArray (Pixel (0, 0), Pixel (w - 1, h - 1))
dimensions :: Color c => Image s c -> ST s (Int, Int)
dimensions (Image i) = do
(_, Pixel (x, y)) <- getBounds i
return (x + 1, y + 1)
getPix :: Color c => Image s c -> Pixel -> ST s c
getPix (Image i) = readArray i
getPixels :: Color c => Image s c -> ST s [c]
getPixels (Image i) = getElems i
setPix :: Color c => Image s c -> Pixel -> c -> ST s ()
setPix (Image i) = writeArray i
fill :: Color c => Image s c -> c -> ST s ()
fill (Image i) c = getBounds i >>= mapM_ f . range
where f p = writeArray i p c
mapImage :: (Color c, Color c') =>
(c -> c') -> Image s c -> ST s (Image s c')
mapImage f (Image i) = liftM Image $ mapArray f 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 | #Kotlin | Kotlin | class BellTriangle(n: Int) {
private val arr: Array<Int>
init {
val length = n * (n + 1) / 2
arr = Array(length) { 0 }
set(1, 0, 1)
for (i in 2..n) {
set(i, 0, get(i - 1, i - 2))
for (j in 1 until i) {
val value = get(i, j - 1) + get(i - 1, j - 1)
set(i, j, value)
}
}
}
private fun index(row: Int, col: Int): Int {
require(row > 0)
require(col >= 0)
require(col < row)
return row * (row - 1) / 2 + col
}
operator fun get(row: Int, col: Int): Int {
val i = index(row, col)
return arr[i]
}
private operator fun set(row: Int, col: Int, value: Int) {
val i = index(row, col)
arr[i] = value
}
}
fun main() {
val rows = 15
val bt = BellTriangle(rows)
println("First fifteen Bell numbers:")
for (i in 1..rows) {
println("%2d: %d".format(i, bt[i, 0]))
}
for (i in 1..10) {
print("${bt[i, 0]}")
for (j in 1 until i) {
print(", ${bt[i, j]}")
}
println()
}
} |
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 .
| #F.C5.8Drmul.C3.A6 | Fōrmulæ | package main
import (
"fmt"
"math"
)
func Fib1000() []float64 {
a, b, r := 0., 1., [1000]float64{}
for i := range r {
r[i], a, b = b, b, b+a
}
return r[:]
}
func main() {
show(Fib1000(), "First 1000 Fibonacci numbers")
}
func show(c []float64, title string) {
var f [9]int
for _, v := range c {
f[fmt.Sprintf("%g", v)[0]-'1']++
}
fmt.Println(title)
fmt.Println("Digit Observed Predicted")
for i, n := range f {
fmt.Printf(" %d %9.3f %8.3f\n", i+1, float64(n)/float64(len(c)),
math.Log10(1+1/float64(i+1)))
}
} |
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.
| #Fermat | Fermat | Func Bern(m) = Sigma<k=0,m>[Sigma<v=0,k>[(-1)^v*Bin(k,v)*(v+1)^m/(k+1)]].;
for i=0, 60 do b:=Bern(i); if b<>0 then !!(i,b) fi od; |
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.
| #BASIC | BASIC | FUNCTION binary_search ( array() AS Integer, value AS Integer, lo AS Integer, hi AS Integer) AS Integer
DIM middle AS Integer
IF hi < lo THEN
binary_search = 0
ELSE
middle = (hi + lo) / 2
SELECT CASE value
CASE IS < array(middle)
binary_search = binary_search(array(), value, lo, middle-1)
CASE IS > array(middle)
binary_search = binary_search(array(), value, middle+1, hi)
CASE ELSE
binary_search = middle
END SELECT
END IF
END FUNCTION |
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
| #FreeBASIC | FreeBASIC |
Dim As String*11 lista(6) => {"abracadabra","seesaw","pop","grrrrrr","up","a"}
Function bestShuffle(s1 As String) As String
Dim As String s2 = s1
Dim As Integer i, j, i1, j1
For i = 1 To Len(s2)
For j = 1 To Len(s2)
If (i <> j) And (Mid(s2,i,1) <> Mid(s1,j,1)) And (Mid(s2,j,1) <> Mid(s1,i,1)) Then
If j < i Then i1 = j : j1 = i Else i1 = i : j1 = j
s2 = Left(s2,i1-1) + Mid(s2,j1,1) + Mid(s2,i1+1,(j1-i1)-1) + Mid(s2,i1,1) + Mid(s2,j1+1)
End If
Next j
Next i
bestShuffle = s2
End Function
Dim As String palabra, bs
Dim As Integer puntos
For b As Integer = 0 To Ubound(lista)-1
palabra = lista(b)
bs = bestShuffle(palabra)
puntos = 0
For i As Integer = 1 To Len(palabra)
If Mid(palabra,i,1) = Mid(bs,i,1) Then puntos += 1
Next i
Print palabra; " ==> "; bs; " (puntuaci¢n:"; puntos; ")"
Next b
Sleep
|
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.
| #Kotlin | Kotlin | class ByteString(private val bytes: ByteArray) : Comparable<ByteString> {
val length get() = bytes.size
fun isEmpty() = bytes.isEmpty()
operator fun plus(other: ByteString): ByteString = ByteString(bytes + other.bytes)
operator fun plus(byte: Byte) = ByteString(bytes + byte)
operator fun get(index: Int): Byte {
require (index in 0 until length)
return bytes[index]
}
fun toByteArray() = bytes
fun copy() = ByteString(bytes.copyOf())
override fun compareTo(other: ByteString) = this.toString().compareTo(other.toString())
override fun equals(other: Any?): Boolean {
if (other == null || other !is ByteString) return false
return compareTo(other) == 0
}
override fun hashCode() = this.toString().hashCode()
fun substring(startIndex: Int) = ByteString(bytes.sliceArray(startIndex until length))
fun substring(startIndex: Int, endIndex: Int) =
ByteString(bytes.sliceArray(startIndex until endIndex))
fun replace(oldByte: Byte, newByte: Byte): ByteString {
val ba = ByteArray(length) { if (bytes[it] == oldByte) newByte else bytes[it] }
return ByteString(ba)
}
fun replace(oldValue: ByteString, newValue: ByteString) =
this.toString().replace(oldValue.toString(), newValue.toString()).toByteString()
override fun toString(): String {
val chars = CharArray(length)
for (i in 0 until length) {
chars[i] = when (bytes[i]) {
in 0..127 -> bytes[i].toChar()
else -> (256 + bytes[i]).toChar()
}
}
return chars.joinToString("")
}
}
fun String.toByteString(): ByteString {
val bytes = ByteArray(this.length)
for (i in 0 until this.length) {
bytes[i] = when (this[i].toInt()) {
in 0..127 -> this[i].toByte()
in 128..255 -> (this[i] - 256).toByte()
else -> '?'.toByte() // say
}
}
return ByteString(bytes)
}
/* property to be used as an abbreviation for String.toByteString() */
val String.bs get() = this.toByteString()
fun main(args: Array<String>) {
val ba = byteArrayOf(65, 66, 67)
val ba2 = byteArrayOf(68, 69, 70)
val bs = ByteString(ba)
val bs2 = ByteString(ba2)
val bs3 = bs + bs2
val bs4 = "GHI£€".toByteString()
println("The length of $bs is ${bs.length}")
println("$bs + $bs2 = $bs3")
println("$bs + D = ${bs + 68}")
println("$bs == ABC is ${bs == bs.copy()}")
println("$bs != ABC is ${bs != bs.copy()}")
println("$bs >= $bs2 is ${bs > bs2}")
println("$bs <= $bs2 is ${bs < bs2}")
println("$bs is ${if (bs.isEmpty()) "empty" else "not empty"}")
println("ABC[1] = ${bs[1].toChar()}")
println("ABC as a byte array is ${bs.toByteArray().contentToString()}")
println("ABCDEF(1..5) = ${bs3.substring(1)}")
println("ABCDEF(2..4) = ${bs3.substring(2,5)}")
println("ABCDEF with C replaced by G is ${bs3.replace(67, 71)}")
println("ABCDEF with CD replaced by GH is ${bs3.replace("CD".bs, "GH".bs)}")
println("GHI£€ as a ByteString is $bs4")
} |
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #Ring | Ring |
limit = [0, 23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
data = sort(data)
dn = list(len(limit))
see "Example 1:" + nl + nl
limits(limit,data,dn)
see nl
limit = [0, 14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
data = sort(data)
dn = list(len(limit))
see "Example 2:" + nl + nl
limits(limit,data,dn)
func limits(limit,data,dn)
for n = 1 to len(data)
for m = 1 to len(limit)-1
if data[n] >= limit[m] and data[n] < limit[m+1]
dn[m] += 1
ok
next
if data[n] >= limit[len(limit)]
dn[len(limit)] += 1
ok
next
for n = 1 to len(limit)-1
see ">= " + limit[n] + " and < " + limit[n+1] + " := " + dn[n] + nl
next
see ">= " + limit[n] + " := " + dn[n] + nl
|
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #Ruby | Ruby | Test = Struct.new(:limits, :data)
tests = Test.new( [23, 37, 43, 53, 67, 83],
[95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]),
Test.new( [14, 18, 249, 312, 389, 392, 513, 591, 634, 720],
[445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749])
def bin(limits, data)
data.map{|d| limits.bsearch{|limit| limit > d} }.tally
end
def present_bins(limits, bins)
ranges = ([nil]+limits+[nil]).each_cons(2).map{|low, high| Range.new(low, high, true) }
ranges.each{|range| puts "#{range.to_s.ljust(12)} #{bins[range.end].to_i}"}
end
tests.each do |test|
present_bins(test.limits, bin(test.limits, test.data))
puts
end
|
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Vlang | Vlang | fn main() {
dna := "" +
"CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG" +
"CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG" +
"AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT" +
"GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT" +
"CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG" +
"TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA" +
"TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT" +
"CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG" +
"TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC" +
"GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT"
println("SEQUENCE:")
le := dna.len
for i := 0; i < le; i += 50 {
mut k := i + 50
if k > le {
k = le
}
println("${i:5}: ${dna[i..k]}")
}
mut base_map := map[byte]int{} // allows for 'any' base
for i in 0..le {
base_map[dna[i]]++
}
mut bases := base_map.keys()
bases.sort()
println("\nBASE COUNT:")
for base in bases {
println(" $base: ${base_map[base]:3}")
}
println(" ------")
println(" Σ: $le")
println(" ======")
} |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #Wren | Wren | import "/fmt" for Fmt
import "/sort" for Sort
import "/trait" for Stepped
var dna = "CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG" +
"CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG" +
"AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT" +
"GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT" +
"CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG" +
"TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA" +
"TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT" +
"CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG" +
"TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC" +
"GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT"
System.print("SEQUENCE:")
var le = dna.count
for (i in Stepped.new(0...le, 50)) {
var k = i + 50
if (k > le) k = le
System.print("%(Fmt.d(5, i)): %(dna[i...k])")
}
var baseMap = {} // allows for 'any' base
for (i in 0...le) {
var d = dna[i]
var v = baseMap[d]
baseMap[d] = !v ? 1 : v + 1
}
var bases = baseMap.keys.toList
Sort.quick(bases)
System.print("\nBASE COUNT:")
for (base in bases) {
System.print(" %(base): %(Fmt.d(3, baseMap[base]))")
}
System.print(" ------")
System.print(" Σ: %(le)")
System.print(" ======") |
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.
| #AutoHotkey | AutoHotkey | MsgBox % NumberToBinary(5) ;101
MsgBox % NumberToBinary(50) ;110010
MsgBox % NumberToBinary(9000) ;10001100101000
NumberToBinary(InputNumber)
{
While, InputNumber
Result := (InputNumber & 1) . Result, InputNumber >>= 1
Return, Result
} |
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.
| #JavaScript | JavaScript | function bline(x0, y0, x1, y1) {
var dx = Math.abs(x1 - x0), sx = x0 < x1 ? 1 : -1;
var dy = Math.abs(y1 - y0), sy = y0 < y1 ? 1 : -1;
var err = (dx>dy ? dx : -dy)/2;
while (true) {
setPixel(x0,y0);
if (x0 === x1 && y0 === y1) break;
var e2 = err;
if (e2 > -dx) { err -= dy; x0 += sx; }
if (e2 < dy) { err += dx; y0 += sy; }
}
} |
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
| #Picat | Picat | go ?=>
member(N,1..5),
println(N),
fail, % or false/0 to get other solutions
nl.
go => 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
| #PicoLisp | PicoLisp |
> 0;
(3) Result: 0
> false;
(4) Result: 0
> 0;
(6) Result: 0
> !true;
(7) Result: 0
> true;
(8) Result: 1
> 1;
(9) Result: 1
>
|
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)..
| #MUMPS | MUMPS | BOXING(DEGREE)
;This takes in a degree heading, nominally from 0 to 360, and returns the compass point name.
QUIT:((DEGREE<0)||(DEGREE>360)) "land lubber can't read a compass"
NEW DIRS,UNP,UNPACK,SEP,DIR,DOS,D,DS,I,F
SET DIRS="N^NbE^N-NE^NEbN^NE^NEbE^E-NE^EbN^E^EbS^E-SE^SEbE^SE^SEbS^S-SE^SbE^"
SET DIRS=DIRS_"S^SbW^S-SW^SWbS^SW^SWbW^W-SW^WbS^W^WbN^W-NW^NWbW^NW^NWbN^N-NW^NbW"
SET UNP="NESWb"
SET UNPACK="north^east^south^west^ by "
SET SEP=360/$LENGTH(DIRS,"^")
SET DIR=(DEGREE/SEP)+1.5
SET DIR=$SELECT((DIR>33):DIR-32,1:DIR)
SET DOS=$NUMBER(DIR-.5,0)
SET D=$PIECE(DIRS,"^",DIR)
SET DS=""
FOR I=1:1:$LENGTH(D) DO
. SET F=$FIND(UNP,$EXTRACT(D,I)) SET DS=DS_$SELECT((F>0):$PIECE(UNPACK,"^",F-1),1:$E(D,I))
KILL DIRS,UNP,UNPACK,SEP,DIR,D,I,F
QUIT DOS_"^"_DS
BOXWRITE
NEW POINTS,UP,LO,DIR,P,X
SET POINTS="0.0,16.87,16.88,33.75,50.62,50.63,67.5,84.37,84.38,101.25,118.12,118.13,135.0,151.87,"
SET POINTS=POINTS_"151.88,168.75,185.62,185.63,202.5,219.37,219.38,236.25,253.12,253.13,270.0,286.87,"
SET POINTS=POINTS_"286.88,303.75,320.62,320.63,337.5,354.37,354.38"
SET UP="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
SET LO="abcdefghijklmnopqrstuvwxyz"
FOR P=1:1:$LENGTH(POINTS,",") DO
. SET X=$$BOXING($PIECE(POINTS,",",P))
. ;Capitalize the initial letter of the direction
. SET DIR=$PIECE(X,"^",2)
. SET DIR=$TRANSLATE($EXTRACT(DIR,1),LO,UP)_$EXTRACT(DIR,2,$LENGTH(DIR))
. WRITE $PIECE(X,"^"),?5,DIR,?40,$JUSTIFY($PIECE(POINTS,",",P),10,2),!
KILL POINTS,UP,LO,DIR,P,X
QUIT |
http://rosettacode.org/wiki/Bitwise_operations | Bitwise operations |
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Task
Write a routine to perform a bitwise AND, OR, and XOR on two integers, a bitwise NOT on the first integer, a left shift, right shift, right arithmetic shift, left rotate, and right rotate.
All shifts and rotates should be done on the first integer with a shift/rotate amount of the second integer.
If any operation is not available in your language, note it.
| #Free_Pascal | Free Pascal | program Bitwise;
{$mode objfpc}
var
// Pascal uses a native int type as a default literal type
// Make sure the operants work on an exact type.
x:shortint = 2;
y:ShortInt = 3;
begin
Writeln('2 and 3 = ', x and y);
Writeln('2 or 3 = ', x or y);
Writeln('2 xor 3 = ', x xor y);
Writeln('not 2 = ', not x);
Writeln('2 shl 3 = ', x >> y);
Writeln('2 shr 3 = ', x << y);
writeln('2 rol 3 = ', rolbyte(x,y));
writeln('2 ror 3 = ', rorbyte(x,y));
writeln('2 sar 3 = ', sarshortint(x,y));
Readln;
end. |
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.
| #Icon_and_Unicon | Icon and Unicon | procedure makebitmap(width,height)
return open("bitmap", "g", "canvas=hidden",
"size="||width||","||height)
end
procedure fillimage(w,color)
Fg(w,color)
FillRectangle(w)
end
procedure setpixel(w,x,y,color)
Fg(w,color)
DrawPixel(x,y)
end
procedure getpixel(w,x,y)
return Pixel(w,x,y)
end |
http://rosettacode.org/wiki/Bell_numbers | Bell numbers | Bell or exponential numbers are enumerations of the number of different ways to partition a set that has exactly n elements. Each element of the sequence Bn is the number of partitions of a set of size n where order of the elements and order of the partitions are non-significant. E.G.: {a b} is the same as {b a} and {a} {b} is the same as {b} {a}.
So
B0 = 1 trivially. There is only one way to partition a set with zero elements. { }
B1 = 1 There is only one way to partition a set with one element. {a}
B2 = 2 Two elements may be partitioned in two ways. {a} {b}, {a b}
B3 = 5 Three elements may be partitioned in five ways {a} {b} {c}, {a b} {c}, {a} {b c}, {a c} {b}, {a b c}
and so on.
A simple way to find the Bell numbers is construct a Bell triangle, also known as an Aitken's array or Peirce triangle, and read off the numbers in the first column of each row. There are other generating algorithms though, and you are free to choose the best / most appropriate for your case.
Task
Write a routine (function, generator, whatever) to generate the Bell number sequence and call the routine to show here, on this page at least the first 15 and (if your language supports big Integers) 50th elements of the sequence.
If you do use the Bell triangle method to generate the numbers, also show the first ten rows of the Bell triangle.
See also
OEIS:A000110 Bell or exponential numbers
OEIS:A011971 Aitken's array | #Little_Man_Computer | Little Man Computer |
// Little Man Computer, for Rosetta Code.
// Calculate Bell numbers, using a 1-dimensional array and addition.
//
// After the calculation of B_n (n > 0), the array contains n elements,
// of which B_n is the first. Example to show calculation of B_(n+1):
// After calc. of B_3 = 5, array holds: 5, 3, 2
// Extend array by copying B_3 to high end: 5, 3, 2, 5
// Replace 2 by 5 + 2 = 7: 5, 3, 7, 5
// Replace 3 by 7 + 3 = 10: 5, 10, 7, 5
// Replace first 5 by 10 + 5 = 15: 15, 10, 7, 5
// First element of array is now B_4 = 15.
// Initialize; B_0 := 1
LDA c1
STA Bell
LDA c0
STA index
BRA print // skip increment of index
// Increment index of Bell number
inc_ix LDA index
ADD c1
STA index
// Here acc = index; print index and Bell number
print OUT
LDA colon
OTC // non-standard instruction; cosmetic only
LDA Bell
OUT
LDA index
BRZ inc_ix // if index = 0, skip rest and loop back
SUB c7 // reached maximum index yet?
BRZ done // if so, jump to exit
// Manufacture some instructions
LDA lda_0
ADD index
STA lda_ix
SUB c200 // convert LDA to STA with same address
STA sta_ix
// Copy latest Bell number to end of array
lda_0 LDA Bell // load Bell number
sta_ix STA 0 // address was filled in above
// Manufacture more instructions
LDA lda_ix // load LDA instruction
loop SUB c401 // convert to ADD with address 1 less
STA add_ix_1
ADD c200 // convert to STA
STA sta_ix_1
// Execute instructions; zero addresses were filled in above
lda_ix LDA 0 // load element of array
add_ix_1 ADD 0 // add to element below
sta_ix_1 STA 0 // update element below
LDA sta_ix_1 // load previous STA instruction
SUB sta_Bell // does it refer to first element of array?
BRZ inc_ix // yes, loop to inc index and print
LDA lda_ix // no, repeat with addresses 1 less
SUB c1
STA lda_ix
BRA loop
// Here when done
done HLT
// Constants
colon DAT 58
c0 DAT 0
c1 DAT 1
c7 DAT 7 // maximum index
c200 DAT 200
c401 DAT 401
sta_Bell STA Bell // not executed; used for comparison
// Variables
index DAT
Bell DAT
// Rest of array goes here
// end
|
http://rosettacode.org/wiki/Bell_numbers | Bell numbers | Bell or exponential numbers are enumerations of the number of different ways to partition a set that has exactly n elements. Each element of the sequence Bn is the number of partitions of a set of size n where order of the elements and order of the partitions are non-significant. E.G.: {a b} is the same as {b a} and {a} {b} is the same as {b} {a}.
So
B0 = 1 trivially. There is only one way to partition a set with zero elements. { }
B1 = 1 There is only one way to partition a set with one element. {a}
B2 = 2 Two elements may be partitioned in two ways. {a} {b}, {a b}
B3 = 5 Three elements may be partitioned in five ways {a} {b} {c}, {a b} {c}, {a} {b c}, {a c} {b}, {a b c}
and so on.
A simple way to find the Bell numbers is construct a Bell triangle, also known as an Aitken's array or Peirce triangle, and read off the numbers in the first column of each row. There are other generating algorithms though, and you are free to choose the best / most appropriate for your case.
Task
Write a routine (function, generator, whatever) to generate the Bell number sequence and call the routine to show here, on this page at least the first 15 and (if your language supports big Integers) 50th elements of the sequence.
If you do use the Bell triangle method to generate the numbers, also show the first ten rows of the Bell triangle.
See also
OEIS:A000110 Bell or exponential numbers
OEIS:A011971 Aitken's array | #Lua | Lua | -- Bell numbers in Lua
-- db 6/11/2020 (to replace missing original)
local function bellTriangle(n)
local tri = { {1} }
for i = 2, n do
tri[i] = { tri[i-1][i-1] }
for j = 2, i do
tri[i][j] = tri[i][j-1] + tri[i-1][j-1]
end
end
return tri
end
local N = 25 -- in lieu of 50, practical limit with double precision floats
local tri = bellTriangle(N)
print("First 15 and "..N.."th Bell numbers:")
for i = 1, 15 do
print(i, tri[i][1])
end
print(N, tri[N][1])
print()
print("First 10 rows of Bell triangle:")
for i = 1, 10 do
print("[ " .. table.concat(tri[i],", ") .. " ]")
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 .
| #Go | Go | package main
import (
"fmt"
"math"
)
func Fib1000() []float64 {
a, b, r := 0., 1., [1000]float64{}
for i := range r {
r[i], a, b = b, b, b+a
}
return r[:]
}
func main() {
show(Fib1000(), "First 1000 Fibonacci numbers")
}
func show(c []float64, title string) {
var f [9]int
for _, v := range c {
f[fmt.Sprintf("%g", v)[0]-'1']++
}
fmt.Println(title)
fmt.Println("Digit Observed Predicted")
for i, n := range f {
fmt.Printf(" %d %9.3f %8.3f\n", i+1, float64(n)/float64(len(c)),
math.Log10(1+1/float64(i+1)))
}
} |
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.
| #FreeBASIC | FreeBASIC | ' version 08-10-2016
' compile with: fbc -s console
' uses gmp
#Include Once "gmp.bi"
#Define max 60
Dim As Long n
Dim As ZString Ptr gmp_str :gmp_str = Allocate(1000) ' 1000 char
Dim Shared As Mpq_ptr tmp, big_j
tmp = Allocate(Len(__mpq_struct)) :Mpq_init(tmp)
big_j = Allocate(Len(__mpq_struct)) :Mpq_init(big_j)
Dim Shared As Mpq_ptr a(max), b(max)
For n = 0 To max
A(n) = Allocate(Len(__mpq_struct)) :Mpq_init(A(n))
B(n) = Allocate(Len(__mpq_struct)) :Mpq_init(B(n))
Next
Function Bernoulli(n As Integer) As Mpq_ptr
Dim As Long m, j
For m = 0 To n
Mpq_set_ui(A(m), 1, m + 1)
For j = m To 1 Step - 1
Mpq_sub(tmp, A(j - 1), A(j))
Mpq_set_ui(big_j, j, 1) 'big_j = j
Mpq_mul(A(j - 1), big_j, tmp)
Next
Next
Return A(0)
End Function
' ------=< MAIN >=------
For n = 0 To max
Mpq_set(B(n), Bernoulli(n))
Mpq_get_str(gmp_str, 10, B(n))
If *gmp_str <> "0" Then
If *gmp_str = "1" Then *gmp_str = "1/1"
Print Using "B(##) = "; n;
Print Space(45 - InStr(*gmp_str, "/")); *gmp_str
End If
Next
' empty keyboard buffer
While Inkey <> "" :Wend
Print :Print "hit any key to end program"
Sleep
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.
| #BASIC256 | BASIC256 | function binarySearchR(array, valor, lb, ub)
if ub < lb then
return false
else
mitad = floor((lb + ub) / 2)
if valor < array[mitad] then return binarySearchR(array, valor, lb, mitad-1)
if valor > array[mitad] then return binarySearchR(array, valor, mitad+1, ub)
if valor = array[mitad] then return mitad
end if
end function |
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
| #Free_Pascal | Free Pascal |
Program BestShuffle;
Const
arr : array[1..6] Of string = ('abracadabra','seesaw','elk','grrrrrr','up','a');
Function Shuffle(inp: String): STRING;
Var x,ReplacementDigit : longint;
ch : char;
Begin
If length(inp) > 1 Then
Begin
Randomize;
For x := 1 To length(inp) Do
Begin
Repeat
ReplacementDigit := random(length(inp))+1;
Until (ReplacementDigit <> x);
ch := inp[x];
inp[x] := inp[ReplacementDigit];
inp[ReplacementDigit] := ch;
End;
End;
shuffle := inp;
End;
Function score(OrgString,ShuString : String) : integer;
Var i : integer;
Begin
score := 0;
For i := 1 To length(OrgString) Do
If OrgString[i] = ShuString[i] Then inc(score);
End;
Var i : integer;
shuffled : string;
Begin
For i := low(arr) To high(arr) Do
Begin
shuffled := shuffle(arr[i]);
writeln(arr[i],' , ',shuffled,' , (',score(arr[i],shuffled),')');
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.
| #Liberty_BASIC | Liberty BASIC |
'string creation
s$ = "Hello, world!"
'string destruction - not needed because of garbage collection
s$ = ""
'string comparison
s$ = "Hello, world!"
If s$ = "Hello, world!" then print "Equal Strings"
'string copying
a$ = s$
'check If empty
If s$ = "" then print "Empty String"
'append a byte
s$ = s$ + Chr$(33)
'extract a substring
a$ = Mid$(s$, 1, 5)
'replace bytes
a$ = "Hello, world!"
for i = 1 to len(a$)
if mid$(a$,i,1)="l" then
a$=left$(a$,i-1)+"L"+mid$(a$,i+1)
end if
next
print a$
'join strings
s$ = "Good" + "bye" + " for now."
|
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #Rust | Rust | fn make_bins(limits: &Vec<usize>, data: &Vec<usize>) -> Vec<Vec<usize>> {
let mut bins: Vec<Vec<usize>> = Vec::with_capacity(limits.len() + 1);
for _ in 0..=limits.len() {bins.push(Vec::new());}
limits.iter().enumerate().for_each(|(idx, limit)| {
data.iter().for_each(|elem| {
if idx == 0 && elem < limit { bins[0].push(*elem); } // smaller than the smallest limit
else if idx == limits.len()-1 && elem >= limit { bins[limits.len()].push(*elem); } // larger than the largest limit
else if elem < limit && elem >= &limits[idx-1] { bins[idx].push(*elem); } // otherwise
});
});
bins
}
fn print_bins(limits: &Vec<usize>, bins: &Vec<Vec<usize>>) {
for (idx, bin) in bins.iter().enumerate() {
if idx == 0 {
println!(" < {:3} := {:3}", limits[idx], bin.len());
} else if idx == limits.len() {
println!(">= {:3} := {:3}", limits[idx-1], bin.len());
}else {
println!(">= {:3} .. < {:3} := {:3}", limits[idx-1], limits[idx], bin.len());
}
};
}
fn main() {
let limits1 = vec![23, 37, 43, 53, 67, 83];
let data1 = vec![95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55];
let limits2 = vec![14, 18, 249, 312, 389, 392, 513, 591, 634, 720];
let data2 = vec![
445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749
];
// Why are we calling it RC anyways???
println!("RC FIRST EXAMPLE");
let bins1 = make_bins(&limits1, &data1);
print_bins(&limits1, &bins1);
println!("\nRC SECOND EXAMPLE");
let bins2 = make_bins(&limits2, &data2);
print_bins(&limits2, &bins2);
}
|
http://rosettacode.org/wiki/Bin_given_limits | Bin given limits | You are given a list of n ascending, unique numbers which are to form limits
for n+1 bins which count how many of a large set of input numbers fall in the
range of each bin.
(Assuming zero-based indexing)
bin[0] counts how many inputs are < limit[0]
bin[1] counts how many inputs are >= limit[0] and < limit[1]
..
bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]
bin[n] counts how many inputs are >= limit[n-1]
Task
The task is to create a function that given the ascending limits and a stream/
list of numbers, will return the bins; together with another function that
given the same list of limits and the binning will print the limit of each bin
together with the count of items that fell in the range.
Assume the numbers to bin are too large to practically sort.
Task examples
Part 1: Bin using the following limits the given input data
limits = [23, 37, 43, 53, 67, 83]
data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,
16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]
Part 2: Bin using the following limits the given input data
limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]
data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,
416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,
655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,
346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,
345,110,720,917,313,845,426, 9,457,628,410,723,354,895,881,953,677,137,397, 97,
854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157, 5,316,395,
787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,
698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,
605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,
466, 23,707,467, 33,670,921,180,991,396,160,436,717,918, 8,374,101,684,727,749]
Show output here, on this page.
| #Tcl | Tcl | namespace path {::tcl::mathop ::tcl::mathfunc}
# Not necessary but useful helper
proc lincr {_list index} {
upvar $_list list
lset list $index [+ [lindex $list $index] 1]
}
proc distribute_bins {binlims data} {
set bins [lrepeat [+ [llength $binlims] 1] 0]
foreach val $data {
lincr bins [+ [lsearch -exact -integer -sorted -bisect $binlims $val] 1]
}
return $bins
}
proc print_bins {binlims bins} {
set binlims [list -∞ {*}$binlims ∞]
for {set i 0} {$i < [llength $bins]} {incr i} {
puts "[lindex $binlims $i]..[lindex $binlims [+ $i 1]]: [lindex $bins $i]"
}
}
set binlims {23 37 43 53 67 83}
set data {95 21 94 12 99 4 70 75 83 93 52 80 57 5 53 86 65 17 92 83 71 61 54 58 47
16 8 9 32 84 7 87 46 19 30 37 96 6 98 40 79 97 45 64 60 29 49 36 43 55}
print_bins $binlims [distribute_bins $binlims $data]
puts ""
set binlims {14 18 249 312 389 392 513 591 634 720}
set data {445 814 519 697 700 130 255 889 481 122 932 77 323 525 570 219 367 523 442 933
416 589 930 373 202 253 775 47 731 685 293 126 133 450 545 100 741 583 763 306
655 267 248 477 549 238 62 678 98 534 622 907 406 714 184 391 913 42 560 247
346 860 56 138 546 38 985 948 58 213 799 319 390 634 458 945 733 507 916 123
345 110 720 917 313 845 426 9 457 628 410 723 354 895 881 953 677 137 397 97
854 740 83 216 421 94 517 479 292 963 376 981 480 39 257 272 157 5 316 395
787 942 456 242 759 898 576 67 298 425 894 435 831 241 989 614 987 770 384 692
698 765 331 487 251 600 879 342 982 527 736 795 585 40 54 901 408 359 577 237
605 847 353 968 832 205 838 427 876 959 686 646 835 127 621 892 443 198 988 791
466 23 707 467 33 670 921 180 991 396 160 436 717 918 8 374 101 684 727 749}
print_bins $binlims [distribute_bins $binlims $data] |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #XPL0 | XPL0 | char Bases;
int Counts(256), Cnt, I, Ch;
[Bases:= "
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAATx ";
for I:= 0 to 255 do Counts(I):= 0;
Format(5, 0);
Cnt:= 0;
I:= 0;
loop [repeat Ch:= Bases(I);
I:= I+1;
if Ch = ^x then quit;
Counts(Ch):= Counts(Ch)+1;
ChOut(0, Ch);
until Ch = \LF\$0A;
RlOut(0, float(Cnt)); Text(0, ": ");
Cnt:= Cnt + 50;
];
CrLf(0); CrLf(0);
Text(0, "Base counts A: "); IntOut(0, Counts(^A));
Text(0, " C: "); IntOut(0, Counts(^C));
Text(0, " G: "); IntOut(0, Counts(^G));
Text(0, " T: "); IntOut(0, Counts(^T));
Text(0, "
Total: "); IntOut(0, Cnt); CrLf(0);
] |
http://rosettacode.org/wiki/Bioinformatics/base_count | Bioinformatics/base count | Given this string representing ordered DNA bases:
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT
Task
"Pretty print" the sequence followed by a summary of the counts of each of the bases: (A, C, G, and T) in the sequence
print the total count of each base in the string.
Other tasks related to string operations:
Metrics
Array length
String length
Copy a string
Empty string (assignment)
Counting
Word frequency
Letter frequency
Jewels and stones
I before E except after C
Bioinformatics/base count
Count occurrences of a substring
Count how many vowels and consonants occur in a string
Remove/replace
XXXX redacted
Conjugate a Latin verb
Remove vowels from a string
String interpolation (included)
Strip block comments
Strip comments from a string
Strip a set of characters from a string
Strip whitespace from a string -- top and tail
Strip control codes and extended characters from a string
Anagrams/Derangements/shuffling
Word wheel
ABC problem
Sattolo cycle
Knuth shuffle
Ordered words
Superpermutation minimisation
Textonyms (using a phone text pad)
Anagrams
Anagrams/Deranged anagrams
Permutations/Derangements
Find/Search/Determine
ABC words
Odd words
Word ladder
Semordnilap
Word search
Wordiff (game)
String matching
Tea cup rim text
Alternade words
Changeable words
State name puzzle
String comparison
Unique characters
Unique characters in each string
Extract file extension
Levenshtein distance
Palindrome detection
Common list elements
Longest common suffix
Longest common prefix
Compare a list of strings
Longest common substring
Find common directory path
Words from neighbour ones
Change e letters to i in words
Non-continuous subsequences
Longest common subsequence
Longest palindromic substrings
Longest increasing subsequence
Words containing "the" substring
Sum of the digits of n is substring of n
Determine if a string is numeric
Determine if a string is collapsible
Determine if a string is squeezable
Determine if a string has all unique characters
Determine if a string has all the same characters
Longest substrings without repeating characters
Find words which contains all the vowels
Find words which contains most consonants
Find words which contains more than 3 vowels
Find words which first and last three letters are equals
Find words which odd letters are consonants and even letters are vowels or vice_versa
Formatting
Substring
Rep-string
Word wrap
String case
Align columns
Literals/String
Repeat a string
Brace expansion
Brace expansion using ranges
Reverse a string
Phrase reversals
Comma quibbling
Special characters
String concatenation
Substring/Top and tail
Commatizing numbers
Reverse words in a string
Suffixation of decimal numbers
Long literals, with continuations
Numerical and alphabetical suffixes
Abbreviations, easy
Abbreviations, simple
Abbreviations, automatic
Song lyrics/poems/Mad Libs/phrases
Mad Libs
Magic 8-ball
99 Bottles of Beer
The Name Game (a song)
The Old lady swallowed a fly
The Twelve Days of Christmas
Tokenize
Text between
Tokenize a string
Word break problem
Tokenize a string with escaping
Split a character string based on change of character
Sequences
Show ASCII table
De Bruijn sequences
Self-referential sequences
Generate lower case ASCII alphabet
| #zkl | zkl | bases:=
#<<<"
CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG
CTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG
AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT
GGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT
CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG
TCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA
TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT
CGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG
TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC
GACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT" - " \n";
#<<<
[0..*,50].zipWith(fcn(n,bases){ println("%6d: %s".fmt(n,bases.concat())) },
bases.walker().walk.fp(50)).pump(Void); // .pump forces the iterator
println("\nBase Counts: ", bases.counts().pump(String,Void.Read,"%s: %d ".fmt));
println("Total: ",bases.len()); |
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.
| #AutoIt | AutoIt |
ConsoleWrite(IntToBin(50) & @CRLF)
Func IntToBin($iInt)
$Stack = ObjCreate("System.Collections.Stack")
Local $b = -1, $r = ""
While $iInt <> 0
$b = Mod($iInt, 2)
$iInt = INT($iInt/2)
$Stack.Push ($b)
WEnd
For $i = 1 TO $Stack.Count
$r &= $Stack.Pop
Next
Return $r
EndFunc ;==>IntToBin
|
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.
| #Julia | Julia | function drawline!(img::Matrix{T}, x0::Int, y0::Int, x1::Int, y1::Int, col::T) where T
δx = abs(x1 - x0)
δy = abs(y1 - y0)
δe = abs(δy / δx)
er = 0.0
y = y0
for x in x0:x1
img[x, y] = col
er += δe
if er > 0.5
y += 1
er -= 1.0
end
end
return img
end
using Images
img = fill(Gray(255.0), 5, 5);
println("\nImage:")
display(img); println()
drawline!(img, 1, 1, 5, 5, Gray(0.0));
println("\nModified image:")
display(img); println() |
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
| #Pike | Pike |
> 0;
(3) Result: 0
> false;
(4) Result: 0
> 0;
(6) Result: 0
> !true;
(7) Result: 0
> true;
(8) Result: 1
> 1;
(9) Result: 1
>
|
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
| #PL.2FI | PL/I | Declare x bit(1);
x='1'b; /* True */
x='0'b; /* False */ |
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)..
| #NetRexx | NetRexx | /* NetRexx */
options replace format comments java crossref savelog symbols nobinary utf8
class RCBoxTheCompass
properties public constant
_FULL = '_FULL'
properties indirect
headings = Rexx
rosepoints = Rexx
/* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ */
method RCBoxTheCompass() public
setHeadings(makeHeadings)
setRosepoints(makeRosepoints)
return
/* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ */
method main(args = String[]) public static
box = RCBoxTheCompass()
cp = box.getCompassPoints
loop r_ = 1 to cp[0]
say cp[r_]
end r_
say
hx = box.getHeadingsReport(box.getHeadings)
loop r_ = 1 to hx[0]
say hx[r_]
end r_
say
return
/* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ */
method getDecimalAngle(degrees, minutes, seconds) public static returns Rexx
degrees = degrees * 10 % 10
minutes = minutes * 10 % 10
angle = degrees + (minutes / 60 + (seconds / 60 / 60))
return angle
/* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ */
method getDegreesMinutesSeconds(angle) public static returns Rexx
degrees = angle * 100 % 100
minutes = ((angle - degrees) * 60) * 100 % 100
seconds = ((((angle - degrees) * 60) - minutes) * 60) * 100 % 100
return degrees minutes seconds
/* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ */
method getHeadingsReport(bearings) public returns Rexx
r_ = 0
table = ''
r_ = r_ + 1; table[0] = r_; table[r_] = 'Idx' -
'Abbr' -
'Compass Point'.left(20) -
'Degrees'.right(10)
loop h_ = 1 to bearings[0]
heading = bearings[h_]
index = getRosepointIndex(heading)
parse getRosepoint(index) p_abbrev p_full
r_ = r_ + 1; table[0] = r_; table[r_] = index.right(3) -
p_abbrev.left(4) -
p_full.left(20) -
heading.format(6, 3).right(10)
end h_
return table
/* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ */
method getRosepointIndex(heading) public returns Rexx
one_pt = 360 / 32
hn = heading // 360
hi = hn % one_pt
if hn > hi * one_pt + one_pt / 2 then do
hi = hi + 1 -- greater than max range for this point; bump to next point
end
idx = hi // 32 + 1 -- add one to get index into rosepoints indexed string
return idx
/* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ */
method getRosepoint(index) public returns Rexx
rp = getRosepoints
return rp[index] rp[index, _FULL]
/* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ */
method makeRosepoints() private returns Rexx
p_ = 0
rp = ''
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'N'; rp[p_, _FULL] = 'North'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'NbE'; rp[p_, _FULL] = 'North by East'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'NNE'; rp[p_, _FULL] = 'North-Northeast'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'NEbn'; rp[p_, _FULL] = 'Northeast by North'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'NE'; rp[p_, _FULL] = 'Northeast'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'NEbE'; rp[p_, _FULL] = 'Northeast by East'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'ENE'; rp[p_, _FULL] = 'East-Northeast'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'EbN'; rp[p_, _FULL] = 'East by North'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'E'; rp[p_, _FULL] = 'East'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'EbS'; rp[p_, _FULL] = 'East by South'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'ESE'; rp[p_, _FULL] = 'East-Southeast'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'SEbE'; rp[p_, _FULL] = 'Southeast by East'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'SE'; rp[p_, _FULL] = 'Southeast'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'SEbS'; rp[p_, _FULL] = 'Southeast by South'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'SSE'; rp[p_, _FULL] = 'South-Southeast'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'SbE'; rp[p_, _FULL] = 'South by East'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'S'; rp[p_, _FULL] = 'South'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'SbW'; rp[p_, _FULL] = 'South by West'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'SSW'; rp[p_, _FULL] = 'South-Southwest'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'SWbS'; rp[p_, _FULL] = 'Southwest by South'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'SW'; rp[p_, _FULL] = 'Southwest'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'SWbW'; rp[p_, _FULL] = 'Southwest by West'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'WSW'; rp[p_, _FULL] = 'Southwest'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'WbS'; rp[p_, _FULL] = 'West by South'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'W'; rp[p_, _FULL] = 'West'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'WbN'; rp[p_, _FULL] = 'West by North'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'WNW'; rp[p_, _FULL] = 'West-Northwest'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'NWbW'; rp[p_, _FULL] = 'Northwest by West'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'NW'; rp[p_, _FULL] = 'Northwest'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'NWbN'; rp[p_, _FULL] = 'Northwest by North'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'NNW'; rp[p_, _FULL] = 'North-Northwest'
p_ = p_ + 1; rp[0] = p_; rp[p_] = 'NbW'; rp[p_, _FULL] = 'North by West'
return rp
/* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ */
method makeHeadings() private returns Rexx
hdg = ''
hdg[0] = 0
points = 32
loop i_ = 0 to points
heading = i_ * 360 / points
select case i_ // 3
when 1 then heading_h = heading + 5.62
when 2 then heading_h = heading - 5.62
otherwise heading_h = heading
end
ix = hdg[0] + 1; hdg[0] = ix; hdg[ix] = heading_h
end i_
return hdg
/* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ */
method getCompassPoints() public returns Rexx
r_ = 0
table = ''
r_ = r_ + 1; table[0] = r_; table[r_] = 'Idx' -
'Abbr' -
'Compass Point'.left(20) -
'Low (Deg.)'.right(10) -
'Mid (Deg.)'.right(10) -
'Hi (Deg.)'.right(10)
-- one point of the compass is 360 / 32 (11.25) degrees
-- using functions to calculate this just for fun
one_pt = 360 / 32
parse getDegreesMinutesSeconds(one_pt / 2) ad am as .
points = 32
loop h_ = 0 to points - 1
heading = h_ * 360 / points
hmin = heading - getDecimalAngle(ad, am, as)
hmax = heading + getDecimalAngle(ad, am, as)
if hmin < 0 then do
hmin = hmin + 360
end
if hmax >= 360 then do
hmax = hmax - 360
end
index = getRosepointIndex(heading)
parse getRosepoint(index) p_abbrev p_full
r_ = r_ + 1; table[0] = r_; table[r_] = index.right(3) -
p_abbrev.left(4) -
p_full.left(20) -
hmin.format(6, 3).right(10) -
heading.format(6, 3).right(10) -
hmax.format(6, 3).right(10)
end h_
return table
|
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