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http://rosettacode.org/wiki/File_size | File size | Verify the size of a file called input.txt for a file in the current working directory, and another one in the file system root.
| #8086_Assembly | 8086 Assembly | putch: equ 2 ; Print character
puts: equ 9 ; Print $-terminated string
setdta: equ 1Ah ; Set DTA
stat: equ 4Eh ; Get file info
cpu 8086
bits 16
org 100h
section .text
mov si,curf ; Print file size for 'INPUT.TXT'
call pfsize ; (in current directory),
mov si,rootf ; Then for '\INPUT.TXT' in root directory
;;; Print file name and size for file in DS:SI
pfsize: mov ah,setdta ; Set disc transfer area pointer
mov dx,dta
int 21h
call puts0 ; Print the filename in SI
mov ah,puts ; Print colon and space
mov dx,colspc
int 21h
mov ah,stat ; Find file info
xor cx,cx ; We want a normal file
mov dx,si ; Filename is in SI
int 21h
jnc .ok ; Carry clear = found
mov ah,puts ; Carry set = not found = print 'not found'
mov dx,nofile
int 21h
ret
.ok: les bp,[dta+26] ; 32-bit file size in bytes at DTA+26
mov di,es ; DI:BP = 32-bit file size
mov bx,numbuf ; ASCII number buffer
mov cx,10 ; Divisor (10)
.dgt: xor dx,dx ; 32-bit division (to get digits)
mov ax,di ; can be done with chained DIVs
div cx
mov di,ax
mov ax,bp
div cx
mov bp,ax
add dl,'0' ; DX is now remainder, i.e. digit
dec bx ; Move digit pointer backwards,
mov [bx],dl ; Store ASCII digit,
or ax,di ; If the new divisor is not zero,
jnz .dgt ; then there is another digit.
mov ah,puts ; If so, the number is done,
mov dx,bx ; and we can print it.
int 21h
ret
;;; Print 0-terminated string in SI
puts0: push si ; Save SI register
mov ah,putch ; Print char syscall
.loop: lodsb ; Load character from SI
test al,al ; If zero,
jz .out ; then stop.
mov dl,al ; Tell DOS to print character
int 21h
jmp .loop ; go get another.
.out: pop si ; Restore SI register
ret
section .data
rootf: db '\' ; \INPUT.TXT (for root) and
curf: db 'INPUT.TXT',0 ; INPUT.TXT (for current directory)
nofile: db 'Not found.',13,10,'$' ; "Not found" message
db '0000000000' ; Number output buffer
numbuf: db ' bytes',13,10,'$'
colspc: db ': $' ; Colon and space
section .bss
dta: resb 512 ; Disc transfer area |
http://rosettacode.org/wiki/File_modification_time | File modification time | Task
Get and set the modification time of a file.
| #ALGOL_68 | ALGOL 68 | PROC get output = (STRING cmd) VOID:
IF STRING sh cmd = " " + cmd + " ; 2>&1";
STRING output;
execve output ("/bin/sh", ("sh", "-c", sh cmd), "", output) >= 0
THEN print (output) FI;
get output ("rm -rf WTC_1"); CO Ensure file doesn't exist CO
get output ("touch WTC_1"); CO Create file CO
get output ("ls -l --time-style=full-iso WTC_1"); CO Display its last modified time CO
get output ("touch -t 200109111246.40 WTC_1"); CO Change its last modified time CO
get output ("ls -l --time-style=full-iso WTC_1") CO Verify it changed CO
|
http://rosettacode.org/wiki/File_modification_time | File modification time | Task
Get and set the modification time of a file.
| #AutoHotkey | AutoHotkey | FileGetTime, OutputVar, output.txt
MsgBox % OutputVar
FileSetTime, 20080101, output.txt
FileGetTime, OutputVar, output.txt
MsgBox % OutputVar |
http://rosettacode.org/wiki/File_size_distribution | File size distribution | Task
Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.
My suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.
Don't forget that empty files may exist, to serve as a marker.
Is your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?
| #Factor | Factor | USING: accessors assocs formatting io io.directories.search
io.files.types io.pathnames kernel math math.functions
math.statistics namespaces sequences ;
: classify ( m -- n ) [ 0 ] [ log10 >integer 1 + ] if-zero ;
: file-size-histogram ( path -- assoc )
recursive-directory-entries
[ type>> +directory+ = ] reject
[ size>> classify ] map histogram ;
current-directory get file-size-histogram dup
[ "Count of files < 10^%d bytes: %4d\n" printf ] assoc-each
nl values sum "Total files: %d\n" printf |
http://rosettacode.org/wiki/File_size_distribution | File size distribution | Task
Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.
My suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.
Don't forget that empty files may exist, to serve as a marker.
Is your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?
| #Go | Go | package main
import (
"fmt"
"log"
"math"
"os"
"path/filepath"
)
func commatize(n int64) string {
s := fmt.Sprintf("%d", n)
if n < 0 {
s = s[1:]
}
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
if n >= 0 {
return s
}
return "-" + s
}
func fileSizeDistribution(root string) {
var sizes [12]int
files := 0
directories := 0
totalSize := int64(0)
walkFunc := func(path string, info os.FileInfo, err error) error {
if err != nil {
return err
}
files++
if info.IsDir() {
directories++
}
size := info.Size()
if size == 0 {
sizes[0]++
return nil
}
totalSize += size
logSize := math.Log10(float64(size))
index := int(math.Floor(logSize))
sizes[index+1]++
return nil
}
err := filepath.Walk(root, walkFunc)
if err != nil {
log.Fatal(err)
}
fmt.Printf("File size distribution for '%s' :-\n\n", root)
for i := 0; i < len(sizes); i++ {
if i == 0 {
fmt.Print(" ")
} else {
fmt.Print("+ ")
}
fmt.Printf("Files less than 10 ^ %-2d bytes : %5d\n", i, sizes[i])
}
fmt.Println(" -----")
fmt.Printf("= Total number of files : %5d\n", files)
fmt.Printf(" including directories : %5d\n", directories)
c := commatize(totalSize)
fmt.Println("\n Total size of files :", c, "bytes")
}
func main() {
fileSizeDistribution("./")
} |
http://rosettacode.org/wiki/Find_common_directory_path | Find common directory path | Create a routine that, given a set of strings representing directory paths and a single character directory separator, will return a string representing that part of the directory tree that is common to all the directories.
Test your routine using the forward slash '/' character as the directory separator and the following three strings as input paths:
'/home/user1/tmp/coverage/test'
'/home/user1/tmp/covert/operator'
'/home/user1/tmp/coven/members'
Note: The resultant path should be the valid directory '/home/user1/tmp' and not the longest common string '/home/user1/tmp/cove'.
If your language has a routine that performs this function (even if it does not have a changeable separator character), then mention it as part of the task.
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
| #BBC_BASIC | BBC BASIC | DIM path$(3)
path$(1) = "/home/user1/tmp/coverage/test"
path$(2) = "/home/user1/tmp/covert/operator"
path$(3) = "/home/user1/tmp/coven/members"
PRINT FNcommonpath(path$(), "/")
END
DEF FNcommonpath(p$(), s$)
LOCAL I%, J%, O%
REPEAT
O% = I%
I% = INSTR(p$(1), s$, I%+1)
FOR J% = 2 TO DIM(p$(), 1)
IF LEFT$(p$(1), I%) <> LEFT$(p$(J%), I%) EXIT REPEAT
NEXT J%
UNTIL I% = 0
= LEFT$(p$(1), O%-1) |
http://rosettacode.org/wiki/Find_common_directory_path | Find common directory path | Create a routine that, given a set of strings representing directory paths and a single character directory separator, will return a string representing that part of the directory tree that is common to all the directories.
Test your routine using the forward slash '/' character as the directory separator and the following three strings as input paths:
'/home/user1/tmp/coverage/test'
'/home/user1/tmp/covert/operator'
'/home/user1/tmp/coven/members'
Note: The resultant path should be the valid directory '/home/user1/tmp' and not the longest common string '/home/user1/tmp/cove'.
If your language has a routine that performs this function (even if it does not have a changeable separator character), then mention it as part of the task.
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
| #C | C | #include <stdio.h>
int common_len(const char *const *names, int n, char sep)
{
int i, pos;
for (pos = 0; ; pos++) {
for (i = 0; i < n; i++) {
if (names[i][pos] != '\0' &&
names[i][pos] == names[0][pos])
continue;
/* backtrack */
while (pos > 0 && names[0][--pos] != sep);
return pos;
}
}
return 0;
}
int main()
{
const char *names[] = {
"/home/user1/tmp/coverage/test",
"/home/user1/tmp/covert/operator",
"/home/user1/tmp/coven/members",
};
int len = common_len(names, sizeof(names) / sizeof(const char*), '/');
if (!len) printf("No common path\n");
else printf("Common path: %.*s\n", len, names[0]);
return 0;
} |
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #Ada | Ada | with Ada.Integer_Text_Io; use Ada.Integer_Text_Io;
with Ada.Text_Io; use Ada.Text_Io;
procedure Array_Selection is
type Array_Type is array (Positive range <>) of Integer;
Null_Array : Array_Type(1..0);
function Evens (Item : Array_Type) return Array_Type is
begin
if Item'Length > 0 then
if Item(Item'First) mod 2 = 0 then
return Item(Item'First) & Evens(Item((Item'First + 1)..Item'Last));
else
return Evens(Item((Item'First + 1)..Item'Last));
end if;
else
return Null_Array;
end if;
end Evens;
procedure Print(Item : Array_Type) is
begin
for I in Item'range loop
Put(Item(I));
New_Line;
end loop;
end Print;
Foo : Array_Type := (1,2,3,4,5,6,7,8,9,10);
begin
Print(Evens(Foo));
end Array_Selection; |
http://rosettacode.org/wiki/Find_if_a_point_is_within_a_triangle | Find if a point is within a triangle | Find if a point is within a triangle.
Task
Assume points are on a plane defined by (x, y) real number coordinates.
Given a point P(x, y) and a triangle formed by points A, B, and C, determine if P is within triangle ABC.
You may use any algorithm.
Bonus: explain why the algorithm you chose works.
Related tasks
Determine_if_two_triangles_overlap
Also see
Discussion of several methods. [[1]]
Determine if a point is in a polygon [[2]]
Triangle based coordinate systems [[3]]
Wolfram entry [[4]]
| #jq | jq | def sum_of_squares(stream): reduce stream as $x (0; . + $x * $x);
def distanceSquared(P1; P2): sum_of_squares(P1[0]-P2[0], P1[1]-P2[1]);
# Emit {x1,y1, ...} for the input triangle
def xy:
{ x1: .[0][0],
y1: .[0][1],
x2: .[1][0],
y2: .[1][1],
x3: .[2][0],
y3: .[2][1] };
def EPS: 0.001;
def EPS_SQUARE: EPS * EPS; |
http://rosettacode.org/wiki/Find_if_a_point_is_within_a_triangle | Find if a point is within a triangle | Find if a point is within a triangle.
Task
Assume points are on a plane defined by (x, y) real number coordinates.
Given a point P(x, y) and a triangle formed by points A, B, and C, determine if P is within triangle ABC.
You may use any algorithm.
Bonus: explain why the algorithm you chose works.
Related tasks
Determine_if_two_triangles_overlap
Also see
Discussion of several methods. [[1]]
Determine if a point is in a polygon [[2]]
Triangle based coordinate systems [[3]]
Wolfram entry [[4]]
| #Julia | Julia | Point(x, y) = [x, y]
Triangle(a, b, c) = [a, b, c]
LEzero(x) = x < 0 || isapprox(x, 0, atol=0.00000001)
GEzero(x) = x > 0 || isapprox(x, 0, atol=0.00000001)
""" Determine which side of plane cut by line (p2, p3) p1 is on """
side(p1, p2, p3) = (p1[1] - p3[1]) * (p2[2] - p3[2]) - (p2[1] - p3[1]) * (p1[2] - p3[2])
"""
Determine if point is within triangle formed by points p1, p2, p3.
If so, the point will be on the same side of each of the half planes
defined by vectors p1p2, p2p3, and p3p1. Each z is positive if outside,
negative if inside such a plane. All should be positive or all negative
if point is within the triangle.
"""
function iswithin(point, p1, p2, p3)
z1 = side(point, p1, p2)
z2 = side(point, p2, p3)
z3 = side(point, p3, p1)
notanyneg = GEzero(z1) && GEzero(z2) && GEzero(z3)
notanypos = LEzero(z1) && LEzero(z2) && LEzero(z3)
return notanyneg || notanypos
end
const POINTS = [Point(0 // 1, 0 // 1), Point(0 // 1, 1 // 1), Point(3 // 1, 1 // 1),
Point(1 // 10 + (3 // 7) * (100 // 8 - 1 // 10), 1 // 9 + (3 // 7) * (100 // 3 - 1 // 9)),
Point(3 // 2, 12 // 5), Point(51 // 100, -31 // 100), Point(-19 // 50, 6 // 5),
Point(1 // 10, 1 // 9), Point(25 / 2, 100 // 3), Point(25, 100 // 9),
Point(-25 // 2, 50 // 3)
]
const TRI = [
Triangle(POINTS[5], POINTS[6], POINTS[7]),
Triangle(POINTS[8], POINTS[9], POINTS[10]),
Triangle(POINTS[8], POINTS[9], POINTS[11])
]
for tri in TRI
pstring(pt) = "[$(Float32(pt[1])), $(Float32(pt[2]))]"
println("\nUsing triangle [", join([pstring(x) for x in tri], ", "), "]:")
a, b, c = tri[1], tri[2], tri[3]
for p in POINTS[1:4]
isornot = iswithin(p, a, b, c) ? "is" : "is not"
println("Point $(pstring(p)) $isornot within the triangle.")
end
end
|
http://rosettacode.org/wiki/Flatten_a_list | Flatten a list | Task
Write a function to flatten the nesting in an arbitrary list of values.
Your program should work on the equivalent of this list:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
Where the correct result would be the list:
[1, 2, 3, 4, 5, 6, 7, 8]
Related task
Tree traversal
| #TI-89_BASIC | TI-89 BASIC | [[1] 2 [[3 4] 5] [[[]]] [[[6]]] 7 8 []] flatten |
http://rosettacode.org/wiki/Find_limit_of_recursion | Find limit of recursion | Find limit of recursion is part of Short Circuit's Console Program Basics selection.
Task
Find the limit of recursion.
| #Elixir | Elixir | (defun my-recurse (n)
(my-recurse (1+ n)))
(my-recurse 1)
=>
enters debugger at (my-recurse 595),
per the default max-lisp-eval-depth 600 in Emacs 24.1 |
http://rosettacode.org/wiki/Find_limit_of_recursion | Find limit of recursion | Find limit of recursion is part of Short Circuit's Console Program Basics selection.
Task
Find the limit of recursion.
| #Emacs_Lisp | Emacs Lisp | (defun my-recurse (n)
(my-recurse (1+ n)))
(my-recurse 1)
=>
enters debugger at (my-recurse 595),
per the default max-lisp-eval-depth 600 in Emacs 24.1 |
http://rosettacode.org/wiki/Find_palindromic_numbers_in_both_binary_and_ternary_bases | Find palindromic numbers in both binary and ternary bases | Find palindromic numbers in both binary and ternary bases
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Find and show (in decimal) the first six numbers (non-negative integers) that are palindromes in both:
base 2
base 3
Display 0 (zero) as the first number found, even though some other definitions ignore it.
Optionally, show the decimal number found in its binary and ternary form.
Show all output here.
It's permissible to assume the first two numbers and simply list them.
See also
Sequence A60792, numbers that are palindromic in bases 2 and 3 on The On-Line Encyclopedia of Integer Sequences.
| #JavaScript | JavaScript | (() => {
'use strict';
// GENERIC FUNCTIONS
// range :: Int -> Int -> [Int]
const range = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);
// compose :: (b -> c) -> (a -> b) -> (a -> c)
const compose = (f, g) => x => f(g(x));
// listApply :: [(a -> b)] -> [a] -> [b]
const listApply = (fs, xs) =>
[].concat.apply([], fs.map(f =>
[].concat.apply([], xs.map(x => [f(x)]))));
// pure :: a -> [a]
const pure = x => [x];
// curry :: Function -> Function
const curry = (f, ...args) => {
const go = xs => xs.length >= f.length ? (f.apply(null, xs)) :
function () {
return go(xs.concat([].slice.apply(arguments)));
};
return go([].slice.call(args, 1));
};
// transpose :: [[a]] -> [[a]]
const transpose = xs =>
xs[0].map((_, iCol) => xs.map(row => row[iCol]));
// reverse :: [a] -> [a]
const reverse = xs =>
typeof xs === 'string' ? (
xs.split('')
.reverse()
.join('')
) : xs.slice(0)
.reverse();
// take :: Int -> [a] -> [a]
const take = (n, xs) => xs.slice(0, n);
// drop :: Int -> [a] -> [a]
const drop = (n, xs) => xs.slice(n);
// maximum :: [a] -> a
const maximum = xs =>
xs.reduce((a, x) => (x > a || a === undefined ? x : a), undefined);
// quotRem :: Integral a => a -> a -> (a, a)
const quotRem = (m, n) => [Math.floor(m / n), m % n];
// length :: [a] -> Int
const length = xs => xs.length;
// justifyLeft :: Int -> Char -> Text -> Text
const justifyLeft = (n, cFiller, strText) =>
n > strText.length ? (
(strText + cFiller.repeat(n))
.substr(0, n)
) : strText;
// unwords :: [String] -> String
const unwords = xs => xs.join(' ');
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
// BASES AND PALINDROMES
// show, showBinary, showTernary :: Int -> String
const show = n => n.toString(10);
const showBinary = n => n.toString(2);
const showTernary = n => n.toString(3);
// readBase3 :: String -> Int
const readBase3 = s => parseInt(s, 3);
// base3Palindrome :: Int -> String
const base3Palindrome = n => {
const s = showTernary(n);
return s + '1' + reverse(s);
};
// isBinPal :: Int -> Bool
const isBinPal = n => {
const
s = showBinary(n),
[q, r] = quotRem(s.length, 2);
return (r !== 0) && drop(q + 1, s) === reverse(take(q, s));
};
// solutions :: [Int]
const solutions = [0, 1].concat(range(1, 10E5)
.map(compose(readBase3, base3Palindrome))
.filter(isBinPal));
// TABULATION
// cols :: [[Int]]
const cols = transpose(
[
['Decimal', 'Ternary', 'Binary']
].concat(
solutions.map(
compose(
xs => listApply([show, showTernary, showBinary], xs),
pure
)
)
)
);
return unlines(
transpose(cols.map(col => col.map(
curry(justifyLeft)(maximum(col.map(length)) + 1, ' ')
)))
.map(unwords));
})(); |
http://rosettacode.org/wiki/Find_largest_left_truncatable_prime_in_a_given_base | Find largest left truncatable prime in a given base | A truncatable prime is one where all non-empty substrings that finish at the end of the number (right-substrings) are also primes when understood as numbers in a particular base. The largest such prime in a given (integer) base is therefore computable, provided the base is larger than 2.
Let's consider what happens in base 10. Obviously the right most digit must be prime, so in base 10 candidates are 2,3,5,7. Putting a digit in the range 1 to base-1 in front of each candidate must result in a prime. So 2 and 5, like the whale and the petunias in The Hitchhiker's Guide to the Galaxy, come into existence only to be extinguished before they have time to realize it, because 2 and 5 preceded by any digit in the range 1 to base-1 is not prime. Some numbers formed by preceding 3 or 7 by a digit in the range 1 to base-1 are prime. So 13,17,23,37,43,47,53,67,73,83,97 are candidates. Again, putting a digit in the range 1 to base-1 in front of each candidate must be a prime. Repeating until there are no larger candidates finds the largest left truncatable prime.
Let's work base 3 by hand:
0 and 1 are not prime so the last digit must be 2. 123 = 510 which is prime, 223 = 810 which is not so 123 is the only candidate. 1123 = 1410 which is not prime, 2123 = 2310 which is, so 2123 is the only candidate. 12123 = 5010 which is not prime, 22123 = 7710 which also is not prime. So there are no more candidates, therefore 23 is the largest left truncatable prime in base 3.
The task is to reconstruct as much, and possibly more, of the table in the OEIS as you are able.
Related Tasks:
Miller-Rabin primality test
| #zkl | zkl | var [const] BN=Import("zklBigNum"); // libGMP
fcn largest_lefty_prime(base){
primes,p:=List(),BN(1); while(p.nextPrime()<base){ primes.append(p.copy()) }
b,biggest := BN(1),0;
while(primes){
b*=base; // base,base^2,base^3... gets big
ps:=List();
foreach p,n in (primes,[1..base-1]){
if((z:=(p + b*n)).probablyPrime()){
ps.append(z);
if(z>biggest) biggest=z;
}
}
primes=ps; // the number of lists is small
}
biggest
}
foreach n in ([3..17]){ println("%2d %s".fmt(n,largest_lefty_prime(n))) } |
http://rosettacode.org/wiki/FizzBuzz | FizzBuzz | Task
Write a program that prints the integers from 1 to 100 (inclusive).
But:
for multiples of three, print Fizz (instead of the number)
for multiples of five, print Buzz (instead of the number)
for multiples of both three and five, print FizzBuzz (instead of the number)
The FizzBuzz problem was presented as the lowest level of comprehension required to illustrate adequacy.
Also see
(a blog) dont-overthink-fizzbuzz
(a blog) fizzbuzz-the-programmers-stairway-to-heaven
| #Emojicode | Emojicode | 🏁🍇
🔂 i 🆕⏩ 1 101 1 ❗ 🍇
↪️ i 🚮 15 🙌 0 🍇
😀 🔤FizzBuzz🔤 ❗
🍉
🙅↪️ i 🚮 3 🙌 0 🍇
😀 🔤Fizz🔤 ❗
🍉
🙅↪️ i 🚮 5 🙌 0 🍇
😀 🔤Buzz🔤 ❗
🍉🙅🍇
😀 🔤🧲i🧲🔤 ❗
🍉
🍉
🍉 |
http://rosettacode.org/wiki/File_size | File size | Verify the size of a file called input.txt for a file in the current working directory, and another one in the file system root.
| #Action.21 | Action! | INCLUDE "D2:IO.ACT" ;from the Action! Tool Kit
PROC Dir(CHAR ARRAY filter)
BYTE dev=[1]
CHAR ARRAY line(255)
Close(dev)
Open(dev,filter,6)
DO
InputSD(dev,line)
PrintE(line)
IF line(0)=0 THEN
EXIT
FI
OD
Close(dev)
RETURN
CARD FUNC FileSize(CHAR ARRAY src,dst)
DEFINE BUF_LEN="100"
BYTE dev=[1]
BYTE ARRAY buff(BUF_LEN)
CARD len,size
size=0
Close(dev)
Open(dev,src,4)
DO
len=Bget(dev,buff,BUF_LEN)
size==+len
UNTIL len#BUF_LEN
OD
Close(dev)
RETURN (size)
PROC Main()
CHAR ARRAY filter="D:*.*", fname="D:INPUT.TXT"
CARD size
Put(125) PutE() ;clear screen
PrintF("Dir ""%S""%E",filter)
Dir(filter)
size=FileSize(fname)
PrintF("Size of ""%S"" is %U bytes%E",fname,size)
RETURN |
http://rosettacode.org/wiki/File_size | File size | Verify the size of a file called input.txt for a file in the current working directory, and another one in the file system root.
| #Ada | Ada | with Ada.Directories; use Ada.Directories;
with Ada.Text_IO; use Ada.Text_IO;
procedure Test_File_Size is
begin
Put_Line (File_Size'Image (Size ("input.txt")) & " bytes");
Put_Line (File_Size'Image (Size ("/input.txt")) & " bytes");
end Test_File_Size; |
http://rosettacode.org/wiki/File_modification_time | File modification time | Task
Get and set the modification time of a file.
| #AWK | AWK | @load "filefuncs"
BEGIN {
name = "input.txt"
# display time
stat(name, fd)
printf("%s\t%s\n", name, strftime("%a %b %e %H:%M:%S %Z %Y", fd["mtime"]) )
# change time
cmd = "touch -t 201409082359.59 " name
system(cmd)
close(cmd)
} |
http://rosettacode.org/wiki/File_modification_time | File modification time | Task
Get and set the modification time of a file.
| #Batch_File | Batch File | for %%f in (file.txt) do echo.%%~tf |
http://rosettacode.org/wiki/File_size_distribution | File size distribution | Task
Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.
My suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.
Don't forget that empty files may exist, to serve as a marker.
Is your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?
| #Haskell | Haskell | {-# LANGUAGE LambdaCase #-}
import Control.Concurrent (forkIO, setNumCapabilities)
import Control.Concurrent.Chan (Chan, newChan, readChan,
writeChan, writeList2Chan)
import Control.Exception (IOException, catch)
import Control.Monad (filterM, forever, join,
replicateM, replicateM_, (>=>))
import Control.Parallel.Strategies (parTraversable, rseq, using,
withStrategy)
import Data.Char (isDigit)
import Data.List (find, sort)
import qualified Data.Map.Strict as Map
import GHC.Conc (getNumProcessors)
import System.Directory (doesDirectoryExist, doesFileExist,
listDirectory,
pathIsSymbolicLink)
import System.Environment (getArgs)
import System.FilePath.Posix ((</>))
import System.IO (FilePath, IOMode (ReadMode),
hFileSize, hPutStrLn, stderr,
withFile)
import Text.Printf (hPrintf, printf)
data Item = File FilePath Integer | Folder FilePath deriving (Show)
type FGKey = (Integer, Integer)
type FrequencyGroup = (FGKey, Integer)
type FrequencyGroups = Map.Map FGKey Integer
newFrequencyGroups :: FrequencyGroups
newFrequencyGroups = Map.empty
fileSizes :: [Item] -> [Integer]
fileSizes = foldr f [] where f (File _ n) acc = n:acc
f _ acc = acc
folders :: [Item] -> [FilePath]
folders = foldr f [] where f (Folder p) acc = p:acc
f _ acc = acc
totalBytes :: [Item] -> Integer
totalBytes = sum . fileSizes
counts :: [Item] -> (Integer, Integer)
counts = foldr (\x (a, b) -> case x of File _ _ -> (succ a, b)
Folder _ -> (a, succ b)) (0, 0)
-- |Creates 'FrequencyGroups' from the provided size and data set.
frequencyGroups :: Int -- ^ Desired number of frequency groups.
-> [Integer] -- ^ List of collected file sizes. Must be sorted.
-> FrequencyGroups -- ^ Returns a 'FrequencyGroups' for the file sizes.
frequencyGroups _ [] = newFrequencyGroups
frequencyGroups totalGroups xs
| length xs == 1 = Map.singleton (head xs, head xs) 1
| otherwise = foldr placeGroups newFrequencyGroups xs `using` parTraversable rseq
where
range = maximum xs - minimum xs
groupSize = succ $ ceiling $ realToFrac range / realToFrac totalGroups
groups = takeWhile (<=groupSize + maximum xs) $ iterate (+groupSize) 0
groupMinMax = zip groups (pred <$> tail groups)
findGroup n = find (\(low, high) -> n >= low && n <= high)
incrementCount (Just n) = Just (succ n) -- Update count for range.
incrementCount Nothing = Just 1 -- Insert new range with initial count.
placeGroups n fgMap = case findGroup n groupMinMax of
Just k -> Map.alter incrementCount k fgMap
Nothing -> fgMap -- Should never happen.
expandGroups :: Int -- ^ Desired number of frequency groups.
-> [Integer] -- ^ List of collected file sizes.
-> Integer -- ^ Computed frequency group limit.
-> FrequencyGroups -- ^ Expanded 'FrequencyGroups'
expandGroups gsize fileSizes groupThreshold
| groupThreshold > 0 = loop 15 $ frequencyGroups gsize sortedFileSizes
| otherwise = frequencyGroups gsize sortedFileSizes
where
sortedFileSizes = sort fileSizes
loop 0 gs = gs -- break out in case we can't go below threshold
loop n gs | all (<= groupThreshold) $ Map.elems gs = gs
| otherwise = loop (pred n) (expand gs)
expand :: FrequencyGroups -> FrequencyGroups
expand = foldr f . withStrategy (parTraversable rseq) <*>
Map.mapWithKey groupsFromGroup . Map.filter (> groupThreshold)
where
f :: Maybe (FGKey, FrequencyGroups) -- ^ expanded frequency group
-> FrequencyGroups -- ^ accumulator
-> FrequencyGroups -- ^ merged accumulator
f (Just (k, fg)) acc = Map.union (Map.delete k acc) fg
f Nothing acc = acc
groupsFromGroup
:: FGKey -- ^ Group Key
-> Integer -- ^ Count
-> Maybe (FGKey, FrequencyGroups) -- ^ Returns expanded 'FrequencyGroups' with base key it replaces.
groupsFromGroup (min, max) count
| length range > 1 = Just ((min, max), frequencyGroups gsize range)
| otherwise = Nothing
where
range = filter (\n -> n >= min && n <= max) sortedFileSizes
displaySize :: Integer -> String
displaySize n
| n <= 2^10 = printf "%8dB " n
| n >= 2^10 && n <= 2^20 = display (2^10) "KB"
| n >= 2^20 && n <= 2^30 = display (2^20) "MB"
| n >= 2^30 && n <= 2^40 = display (2^30) "GB"
| n >= 2^40 && n <= 2^50 = display (2^40) "TB"
| otherwise = "Too large!"
where
display :: Double -> String -> String
display b = printf "%7.2f%s " (realToFrac n / b)
displayFrequency :: Integer -> FrequencyGroup -> IO ()
displayFrequency filesCount ((min, max), count) = do
printf "%s <-> %s" (displaySize min) (displaySize max)
printf "= %-10d %6.3f%%: %-5s\n" count percentage bars
where
percentage :: Double
percentage = (realToFrac count / realToFrac filesCount) * 100
size = round percentage
bars | size == 0 = "▍"
| otherwise = replicate size '█'
folderWorker :: Chan FilePath -> Chan [Item] -> IO ()
folderWorker folderChan resultItemsChan =
forever (readChan folderChan >>= collectItems >>= writeChan resultItemsChan)
collectItems :: FilePath -> IO [Item]
collectItems folderPath = catch tryCollect $ \e -> do
hPrintf stderr "Skipping: %s\n" $ show (e :: IOException)
pure []
where
tryCollect = (fmap (folderPath </>) <$> listDirectory folderPath) >>=
mapM (\p -> doesDirectoryExist p >>=
\case True -> pure $ Folder p
False -> File p <$> withFile p ReadMode hFileSize)
parallelItemCollector :: FilePath -> IO [Item]
parallelItemCollector folder = do
wCount <- getNumProcessors
setNumCapabilities wCount
printf "Using %d worker threads\n" wCount
folderChan <- newChan
resultItemsChan <- newChan
replicateM_ wCount (forkIO $ folderWorker folderChan resultItemsChan)
loop folderChan resultItemsChan [Folder folder]
where
loop :: Chan FilePath -> Chan [Item] -> [Item] -> IO [Item]
loop folderChan resultItemsChan xs = do
regularFolders <- filterM (pathIsSymbolicLink >=> (pure . not)) $ folders xs
if null regularFolders then pure []
else do
writeList2Chan folderChan regularFolders
childItems <- replicateM (length regularFolders) (readChan resultItemsChan)
result <- mapM (loop folderChan resultItemsChan) childItems
pure (join childItems <> join result)
parseArgs :: [String] -> Either String (FilePath, Int)
parseArgs (x:y:xs)
| all isDigit y = Right (x, read y)
| otherwise = Left "Invalid frequency group size"
parseArgs (x:xs) = Right (x, 4)
parseArgs _ = Right (".", 4)
main :: IO ()
main = parseArgs <$> getArgs >>= \case
Left errorMessage -> hPutStrLn stderr errorMessage
Right (path, groupSize) -> do
items <- parallelItemCollector path
let (fileCount, folderCount) = counts items
printf "Total files: %d\nTotal folders: %d\n" fileCount folderCount
printf "Total size: %s\n" $ displaySize $ totalBytes items
printf "\nDistribution:\n\n%9s <-> %9s %7s\n" "From" "To" "Count"
putStrLn $ replicate 46 '-'
let results = expandGroups groupSize (fileSizes items) (groupThreshold fileCount)
mapM_ (displayFrequency fileCount) $ Map.assocs results
where
groupThreshold = round . (*0.25) . realToFrac |
http://rosettacode.org/wiki/Find_common_directory_path | Find common directory path | Create a routine that, given a set of strings representing directory paths and a single character directory separator, will return a string representing that part of the directory tree that is common to all the directories.
Test your routine using the forward slash '/' character as the directory separator and the following three strings as input paths:
'/home/user1/tmp/coverage/test'
'/home/user1/tmp/covert/operator'
'/home/user1/tmp/coven/members'
Note: The resultant path should be the valid directory '/home/user1/tmp' and not the longest common string '/home/user1/tmp/cove'.
If your language has a routine that performs this function (even if it does not have a changeable separator character), then mention it as part of the task.
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
| #C.23 | C# |
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace RosettaCodeTasks
{
class Program
{
static void Main ( string[ ] args )
{
FindCommonDirectoryPath.Test ( );
}
}
class FindCommonDirectoryPath
{
public static void Test ( )
{
Console.WriteLine ( "Find Common Directory Path" );
Console.WriteLine ( );
List<string> PathSet1 = new List<string> ( );
PathSet1.Add ( "/home/user1/tmp/coverage/test" );
PathSet1.Add ( "/home/user1/tmp/covert/operator" );
PathSet1.Add ( "/home/user1/tmp/coven/members" );
Console.WriteLine("Path Set 1 (All Absolute Paths):");
foreach ( string path in PathSet1 )
{
Console.WriteLine ( path );
}
Console.WriteLine ( "Path Set 1 Common Path: {0}", FindCommonPath ( "/", PathSet1 ) );
}
public static string FindCommonPath ( string Separator, List<string> Paths )
{
string CommonPath = String.Empty;
List<string> SeparatedPath = Paths
.First ( str => str.Length == Paths.Max ( st2 => st2.Length ) )
.Split ( new string[ ] { Separator }, StringSplitOptions.RemoveEmptyEntries )
.ToList ( );
foreach ( string PathSegment in SeparatedPath.AsEnumerable ( ) )
{
if ( CommonPath.Length == 0 && Paths.All ( str => str.StartsWith ( PathSegment ) ) )
{
CommonPath = PathSegment;
}
else if ( Paths.All ( str => str.StartsWith ( CommonPath + Separator + PathSegment ) ) )
{
CommonPath += Separator + PathSegment;
}
else
{
break;
}
}
return CommonPath;
}
}
}
|
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #Aime | Aime | integer
even(integer e)
{
return !(e & 1);
}
list
filter(list l, integer (*f)(integer))
{
integer i;
list v;
i = 0;
while (i < l_length(l)) {
integer e;
e = l_q_integer(l, i);
if (f(e)) {
lb_p_integer(v, e);
}
i += 1;
}
return v;
}
integer
main(void)
{
integer i;
list l;
i = 0;
while (i < 10) {
lb_p_integer(l, i);
i += 1;
}
l = filter(l, even);
i = 0;
while (i < l_length(l)) {
o_space(1);
o_integer(l_q_integer(l, i));
i += 1;
}
o_byte('\n');
return 0;
} |
http://rosettacode.org/wiki/Find_if_a_point_is_within_a_triangle | Find if a point is within a triangle | Find if a point is within a triangle.
Task
Assume points are on a plane defined by (x, y) real number coordinates.
Given a point P(x, y) and a triangle formed by points A, B, and C, determine if P is within triangle ABC.
You may use any algorithm.
Bonus: explain why the algorithm you chose works.
Related tasks
Determine_if_two_triangles_overlap
Also see
Discussion of several methods. [[1]]
Determine if a point is in a polygon [[2]]
Triangle based coordinate systems [[3]]
Wolfram entry [[4]]
| #Kotlin | Kotlin | import kotlin.math.max
import kotlin.math.min
private const val EPS = 0.001
private const val EPS_SQUARE = EPS * EPS
private fun test(t: Triangle, p: Point) {
println(t)
println("Point $p is within triangle ? ${t.within(p)}")
}
fun main() {
var p1 = Point(1.5, 2.4)
var p2 = Point(5.1, -3.1)
var p3 = Point(-3.8, 1.2)
var tri = Triangle(p1, p2, p3)
test(tri, Point(0.0, 0.0))
test(tri, Point(0.0, 1.0))
test(tri, Point(3.0, 1.0))
println()
p1 = Point(1.0 / 10, 1.0 / 9)
p2 = Point(100.0 / 8, 100.0 / 3)
p3 = Point(100.0 / 4, 100.0 / 9)
tri = Triangle(p1, p2, p3)
val pt = Point(p1.x + 3.0 / 7 * (p2.x - p1.x), p1.y + 3.0 / 7 * (p2.y - p1.y))
test(tri, pt)
println()
p3 = Point(-100.0 / 8, 100.0 / 6)
tri = Triangle(p1, p2, p3)
test(tri, pt)
}
class Point(val x: Double, val y: Double) {
override fun toString(): String {
return "($x, $y)"
}
}
class Triangle(private val p1: Point, private val p2: Point, private val p3: Point) {
private fun pointInTriangleBoundingBox(p: Point): Boolean {
val xMin = min(p1.x, min(p2.x, p3.x)) - EPS
val xMax = max(p1.x, max(p2.x, p3.x)) + EPS
val yMin = min(p1.y, min(p2.y, p3.y)) - EPS
val yMax = max(p1.y, max(p2.y, p3.y)) + EPS
return !(p.x < xMin || xMax < p.x || p.y < yMin || yMax < p.y)
}
private fun nativePointInTriangle(p: Point): Boolean {
val checkSide1 = side(p1, p2, p) >= 0
val checkSide2 = side(p2, p3, p) >= 0
val checkSide3 = side(p3, p1, p) >= 0
return checkSide1 && checkSide2 && checkSide3
}
private fun distanceSquarePointToSegment(p1: Point, p2: Point, p: Point): Double {
val p1P2SquareLength = (p2.x - p1.x) * (p2.x - p1.x) + (p2.y - p1.y) * (p2.y - p1.y)
val dotProduct = ((p.x - p1.x) * (p2.x - p1.x) + (p.y - p1.y) * (p2.y - p1.y)) / p1P2SquareLength
if (dotProduct < 0) {
return (p.x - p1.x) * (p.x - p1.x) + (p.y - p1.y) * (p.y - p1.y)
}
if (dotProduct <= 1) {
val pP1SquareLength = (p1.x - p.x) * (p1.x - p.x) + (p1.y - p.y) * (p1.y - p.y)
return pP1SquareLength - dotProduct * dotProduct * p1P2SquareLength
}
return (p.x - p2.x) * (p.x - p2.x) + (p.y - p2.y) * (p.y - p2.y)
}
private fun accuratePointInTriangle(p: Point): Boolean {
if (!pointInTriangleBoundingBox(p)) {
return false
}
if (nativePointInTriangle(p)) {
return true
}
if (distanceSquarePointToSegment(p1, p2, p) <= EPS_SQUARE) {
return true
}
return if (distanceSquarePointToSegment(p2, p3, p) <= EPS_SQUARE) {
true
} else distanceSquarePointToSegment(p3, p1, p) <= EPS_SQUARE
}
fun within(p: Point): Boolean {
return accuratePointInTriangle(p)
}
override fun toString(): String {
return "Triangle[$p1, $p2, $p3]"
}
companion object {
private fun side(p1: Point, p2: Point, p: Point): Double {
return (p2.y - p1.y) * (p.x - p1.x) + (-p2.x + p1.x) * (p.y - p1.y)
}
}
} |
http://rosettacode.org/wiki/Flatten_a_list | Flatten a list | Task
Write a function to flatten the nesting in an arbitrary list of values.
Your program should work on the equivalent of this list:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
Where the correct result would be the list:
[1, 2, 3, 4, 5, 6, 7, 8]
Related task
Tree traversal
| #Trith | Trith | [[1] 2 [[3 4] 5] [[[]]] [[[6]]] 7 8 []] flatten |
http://rosettacode.org/wiki/Flatten_a_list | Flatten a list | Task
Write a function to flatten the nesting in an arbitrary list of values.
Your program should work on the equivalent of this list:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
Where the correct result would be the list:
[1, 2, 3, 4, 5, 6, 7, 8]
Related task
Tree traversal
| #True_BASIC | True BASIC | LET sstring$ = "[[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8 []]"
FOR sicount = 1 TO LEN(sstring$)
IF POS("[] ,",(sstring$)[sicount:sicount+1-1]) = 0 THEN
LET sflatter$ = sflatter$ & scomma$ & (sstring$)[sicount:sicount+1-1]
LET scomma$ = ", "
END IF
NEXT sicount
PRINT "["; sflatter$; "]"
END |
http://rosettacode.org/wiki/Find_limit_of_recursion | Find limit of recursion | Find limit of recursion is part of Short Circuit's Console Program Basics selection.
Task
Find the limit of recursion.
| #Erlang | Erlang | let rec recurse n =
recurse (n+1)
recurse 0 |
http://rosettacode.org/wiki/Find_limit_of_recursion | Find limit of recursion | Find limit of recursion is part of Short Circuit's Console Program Basics selection.
Task
Find the limit of recursion.
| #F.23 | F# | let rec recurse n =
recurse (n+1)
recurse 0 |
http://rosettacode.org/wiki/Find_palindromic_numbers_in_both_binary_and_ternary_bases | Find palindromic numbers in both binary and ternary bases | Find palindromic numbers in both binary and ternary bases
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Find and show (in decimal) the first six numbers (non-negative integers) that are palindromes in both:
base 2
base 3
Display 0 (zero) as the first number found, even though some other definitions ignore it.
Optionally, show the decimal number found in its binary and ternary form.
Show all output here.
It's permissible to assume the first two numbers and simply list them.
See also
Sequence A60792, numbers that are palindromic in bases 2 and 3 on The On-Line Encyclopedia of Integer Sequences.
| #Julia | Julia | ispalindrome(n, bas) = (s = string(n, base=bas); s == reverse(s))
prin3online(n) = println(lpad(n, 15), lpad(string(n, base=2), 40), lpad(string(n, base=3), 30))
reversebase3(n) = (x = 0; while n != 0 x = 3x + (n %3); n = div(n, 3); end; x)
function printpalindromes(N)
lo, hi, pow2, pow3, count, i = 0, 1, 1, 1, 1, 0
println(lpad("Number", 15), lpad("Base 2", 40), lpad("Base 3", 30))
prin3online(0)
while true
for j in lo:hi-1
i = j
n = (3 * j + 1) * pow3 + reversebase3(j)
if ispalindrome(n, 2)
prin3online(n)
count += 1
if count >= N
return
end
end
end
if i == pow3
pow3 *= 3
else
pow2 *= 4
end
while true
while pow2 <= pow3
pow2 *= 4
end
lo2 = div(div(pow2, pow3) - 1, 3)
hi2 = div(div(pow2 * 2, pow3), 3) + 1
lo3 = div(pow3, 3)
hi3 = pow3
if lo2 >= hi3
pow3 *= 3
elseif lo3 >= hi2
pow2 *= 4
else
lo = max(lo2, lo3)
hi = min(hi2, hi3)
break
end
end
end
end
printpalindromes(6)
|
http://rosettacode.org/wiki/Find_palindromic_numbers_in_both_binary_and_ternary_bases | Find palindromic numbers in both binary and ternary bases | Find palindromic numbers in both binary and ternary bases
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Find and show (in decimal) the first six numbers (non-negative integers) that are palindromes in both:
base 2
base 3
Display 0 (zero) as the first number found, even though some other definitions ignore it.
Optionally, show the decimal number found in its binary and ternary form.
Show all output here.
It's permissible to assume the first two numbers and simply list them.
See also
Sequence A60792, numbers that are palindromic in bases 2 and 3 on The On-Line Encyclopedia of Integer Sequences.
| #Kotlin | Kotlin | // version 1.0.5-2
/** converts decimal 'n' to its ternary equivalent */
fun Long.toTernaryString(): String = when {
this < 0L -> throw IllegalArgumentException("negative numbers not allowed")
this == 0L -> "0"
else -> {
var result = ""
var n = this
while (n > 0) {
result += n % 3
n /= 3
}
result.reversed()
}
}
/** wraps java.lang.Long.toBinaryString in a Kotlin extension function */
fun Long.toBinaryString(): String = java.lang.Long.toBinaryString(this)
/** check if a binary or ternary numeric string 's' is palindromic */
fun isPalindromic(s: String): Boolean = (s == s.reversed())
/** print a number which is both a binary and ternary palindrome in all three bases */
fun printPalindrome(n: Long) {
println("Decimal : $n")
println("Binary : ${n.toBinaryString()}")
println("Ternary : ${n.toTernaryString()}")
println()
}
/** create a ternary palindrome whose left part is the ternary equivalent of 'n' and return its decimal equivalent */
fun createPalindrome3(n: Long): Long {
val ternary = n.toTernaryString()
var power3 = 1L
var sum = 0L
val length = ternary.length
for (i in 0 until length) { // right part of palindrome is mirror image of left part
if (ternary[i] > '0') sum += (ternary[i].toInt() - 48) * power3
power3 *= 3L
}
sum += power3 // middle digit must be 1
power3 *= 3L
sum += n * power3 // value of left part is simply 'n' multiplied by appropriate power of 3
return sum
}
fun main(args: Array<String>) {
var i = 1L
var p3: Long
var count = 2
var binStr: String
println("The first 6 numbers which are palindromic in both binary and ternary are:\n")
// we can assume the first two palindromic numbers as per the task description
printPalindrome(0L) // 0 is a palindrome in all 3 bases
printPalindrome(1L) // 1 is a palindrome in all 3 bases
do {
p3 = createPalindrome3(i)
if (p3 % 2 > 0L) { // cannot be even as binary equivalent would end in zero
binStr = p3.toBinaryString()
if (binStr.length % 2 == 1) { // binary palindrome must have an odd number of digits
if (isPalindromic(binStr)) {
printPalindrome(p3)
count++
}
}
}
i++
}
while (count < 6)
} |
http://rosettacode.org/wiki/FizzBuzz | FizzBuzz | Task
Write a program that prints the integers from 1 to 100 (inclusive).
But:
for multiples of three, print Fizz (instead of the number)
for multiples of five, print Buzz (instead of the number)
for multiples of both three and five, print FizzBuzz (instead of the number)
The FizzBuzz problem was presented as the lowest level of comprehension required to illustrate adequacy.
Also see
(a blog) dont-overthink-fizzbuzz
(a blog) fizzbuzz-the-programmers-stairway-to-heaven
| #Erlang | Erlang | -spec fizzbuzz() -> Result :: string().
fizzbuzz() ->
F = fun(N) when N rem 15 == 0 -> "FizzBuzz";
(N) when N rem 3 == 0 -> "Fizz";
(N) when N rem 5 == 0 -> "Buzz";
(N) -> integer_to_list(N)
end,
lists:flatten([[F(N)] ++ ["\n"] || N <- lists:seq(1,100)]). |
http://rosettacode.org/wiki/File_size | File size | Verify the size of a file called input.txt for a file in the current working directory, and another one in the file system root.
| #Aime | Aime | o_(stat("input.txt", ST_SIZE), "\n");
o_("/Cygwin.ico".stat(ST_SIZE), "\n"); |
http://rosettacode.org/wiki/File_size | File size | Verify the size of a file called input.txt for a file in the current working directory, and another one in the file system root.
| #ALGOL_68 | ALGOL 68 | PROC set = (REF FILE file, INT page, line, character)VOID: ~ |
http://rosettacode.org/wiki/File_extension_is_in_extensions_list | File extension is in extensions list | File extension is in extensions list
You are encouraged to solve this task according to the task description, using any language you may know.
Filename extensions are a rudimentary but commonly used way of identifying files types.
Task
Given an arbitrary filename and a list of extensions, tell whether the filename has one of those extensions.
Notes:
The check should be case insensitive.
The extension must occur at the very end of the filename, and be immediately preceded by a dot (.).
You may assume that none of the given extensions are the empty string, and none of them contain a dot. Other than that they may be arbitrary strings.
Extra credit:
Allow extensions to contain dots. This way, users of your function/program have full control over what they consider as the extension in cases like:
archive.tar.gz
Please state clearly whether or not your solution does this.
Test cases
The following test cases all assume this list of extensions: zip, rar, 7z, gz, archive, A##
Filename
Result
MyData.a##
true
MyData.tar.Gz
true
MyData.gzip
false
MyData.7z.backup
false
MyData...
false
MyData
false
If your solution does the extra credit requirement, add tar.bz2 to the list of extensions, and check the following additional test cases:
Filename
Result
MyData_v1.0.tar.bz2
true
MyData_v1.0.bz2
false
Motivation
Checking if a file is in a certain category of file formats with known extensions (e.g. archive files, or image files) is a common problem in practice, and may be approached differently from extracting and outputting an arbitrary extension (see e.g. FileNameExtensionFilter in Java).
It also requires less assumptions about the format of an extension, because the calling code can decide what extensions are valid.
For these reasons, this task exists in addition to the Extract file extension task.
Related tasks
Extract file extension
String matching
| #11l | 11l | F is_ext(file_name, extensions)
R any(extensions.map(e -> @file_name.lowercase().ends_with(‘.’e.lowercase())))
F test(file_names, extensions)
L(file_name) file_names
print(file_name.ljust(max(file_names.map(f_n -> f_n.len)))‘ ’String(is_ext(file_name, extensions)))
test([‘MyData.a##’, ‘MyData.tar.Gz’, ‘MyData.gzip’, ‘MyData.7z.backup’, ‘MyData...’, ‘MyData’], [‘zip’, ‘rar’, ‘7z’, ‘gz’, ‘archive’, ‘A##’, ‘tar.bz2’])
test([‘MyData_v1.0.tar.bz2’, ‘MyData_v1.0.bz2’], [‘tar.bz2’]) |
http://rosettacode.org/wiki/File_modification_time | File modification time | Task
Get and set the modification time of a file.
| #BBC_BASIC | BBC BASIC | DIM ft{dwLowDateTime%, dwHighDateTime%}
DIM st{wYear{l&,h&}, wMonth{l&,h&}, wDayOfWeek{l&,h&}, \
\ wDay{l&,h&}, wHour{l&,h&}, wMinute{l&,h&}, \
\ wSecond{l&,h&}, wMilliseconds{l&,h&} }
REM File is assumed to exist:
file$ = @tmp$ + "rosetta.tmp"
REM Get and display the modification time:
file% = OPENIN(file$)
SYS "GetFileTime", @hfile%(file%), 0, 0, ft{}
CLOSE #file%
SYS "FileTimeToSystemTime", ft{}, st{}
date$ = STRING$(16, CHR$0)
time$ = STRING$(16, CHR$0)
SYS "GetDateFormat", 0, 0, st{}, 0, date$, LEN(date$) TO N%
date$ = LEFT$(date$, N%-1)
SYS "GetTimeFormat", 0, 0, st{}, 0, time$, LEN(time$) TO N%
time$ = LEFT$(time$, N%-1)
PRINT date$ " " time$
REM Set the modification time to the current time:
SYS "GetSystemTime", st{}
SYS "SystemTimeToFileTime", st{}, ft{}
file% = OPENUP(file$)
SYS "SetFileTime", @hfile%(file%), 0, 0, ft{}
CLOSE #file%
|
http://rosettacode.org/wiki/File_modification_time | File modification time | Task
Get and set the modification time of a file.
| #C | C | #include <sys/stat.h>
#include <stdio.h>
#include <time.h>
#include <utime.h>
const char *filename = "input.txt";
int main() {
struct stat foo;
time_t mtime;
struct utimbuf new_times;
if (stat(filename, &foo) < 0) {
perror(filename);
return 1;
}
mtime = foo.st_mtime; /* seconds since the epoch */
new_times.actime = foo.st_atime; /* keep atime unchanged */
new_times.modtime = time(NULL); /* set mtime to current time */
if (utime(filename, &new_times) < 0) {
perror(filename);
return 1;
}
return 0;
} |
http://rosettacode.org/wiki/File_size_distribution | File size distribution | Task
Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.
My suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.
Don't forget that empty files may exist, to serve as a marker.
Is your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?
| #J | J | ((10x^~.),.#/.~) <.10 ^.1>. /:~;{:|:dirtree '~'
1 2
10 8
100 37
1000 49
10000 20
100000 9
1000000 4
10000000 4 |
http://rosettacode.org/wiki/File_size_distribution | File size distribution | Task
Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.
My suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.
Don't forget that empty files may exist, to serve as a marker.
Is your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?
| #Julia | Julia | using Humanize
function sizelist(path::AbstractString)
rst = Vector{Int}(0)
for (root, dirs, files) in walkdir(path)
files = joinpath.(root, files)
tmp = collect(filesize(f) for f in files if !islink(f))
append!(rst, tmp)
end
return rst
end
byclass(y, classes) = Dict{eltype(classes),Int}(c => count(c[1] .≤ y .< c[2]) for c in classes)
function main(path::AbstractString)
s = sizelist(path)
cls = append!([(0, 1)], collect((10 ^ (i-1), 10 ^ i) for i in 1:9))
f = byclass(s, cls)
println("filesizes: ")
for c in cls
@printf(" - between %8s and %8s bytes: %3i\n", datasize(c[1]), datasize(c[2]), f[c])
end
println("\n-> total: $(datasize(sum(s))) bytes and $(length(s)) files")
end
main(".") |
http://rosettacode.org/wiki/File_size_distribution | File size distribution | Task
Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.
My suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.
Don't forget that empty files may exist, to serve as a marker.
Is your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?
| #Kotlin | Kotlin | // version 1.2.10
import java.io.File
import kotlin.math.log10
import kotlin.math.floor
fun fileSizeDistribution(path: String) {
val sizes = IntArray(12)
val p = File(path)
val files = p.walk()
var accessible = 0
var notAccessible = 0
var totalSize = 0L
for (file in files) {
try {
if (file.isFile()) {
val len = file.length()
accessible++
if (len == 0L) {
sizes[0]++
continue
}
totalSize += len
val logLen = log10(len.toDouble())
val index = floor(logLen).toInt()
sizes[index + 1]++
}
}
catch (se: SecurityException) {
notAccessible++
}
}
println("File size distribution for '$path' :-\n")
for (i in 0 until sizes.size) {
print(if (i == 0) " " else "+ ")
print("Files less than 10 ^ ${"%-2d".format(i)} bytes : ")
println("%5d".format(sizes[i]))
}
println(" -----")
println("= Number of accessible files : ${"%5d".format(accessible)}")
println("\n Total size in bytes : $totalSize")
println("\n Number of inaccessible files : ${"%5d".format(notAccessible)}")
}
fun main(args: Array<String>) {
fileSizeDistribution("./") // current directory
} |
http://rosettacode.org/wiki/Find_common_directory_path | Find common directory path | Create a routine that, given a set of strings representing directory paths and a single character directory separator, will return a string representing that part of the directory tree that is common to all the directories.
Test your routine using the forward slash '/' character as the directory separator and the following three strings as input paths:
'/home/user1/tmp/coverage/test'
'/home/user1/tmp/covert/operator'
'/home/user1/tmp/coven/members'
Note: The resultant path should be the valid directory '/home/user1/tmp' and not the longest common string '/home/user1/tmp/cove'.
If your language has a routine that performs this function (even if it does not have a changeable separator character), then mention it as part of the task.
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
| #C.2B.2B | C++ | #include <algorithm>
#include <iostream>
#include <string>
#include <vector>
std::string longestPath( const std::vector<std::string> & , char ) ;
int main( ) {
std::string dirs[ ] = {
"/home/user1/tmp/coverage/test" ,
"/home/user1/tmp/covert/operator" ,
"/home/user1/tmp/coven/members" } ;
std::vector<std::string> myDirs ( dirs , dirs + 3 ) ;
std::cout << "The longest common path of the given directories is "
<< longestPath( myDirs , '/' ) << "!\n" ;
return 0 ;
}
std::string longestPath( const std::vector<std::string> & dirs , char separator ) {
std::vector<std::string>::const_iterator vsi = dirs.begin( ) ;
int maxCharactersCommon = vsi->length( ) ;
std::string compareString = *vsi ;
for ( vsi = dirs.begin( ) + 1 ; vsi != dirs.end( ) ; vsi++ ) {
std::pair<std::string::const_iterator , std::string::const_iterator> p =
std::mismatch( compareString.begin( ) , compareString.end( ) , vsi->begin( ) ) ;
if (( p.first - compareString.begin( ) ) < maxCharactersCommon )
maxCharactersCommon = p.first - compareString.begin( ) ;
}
std::string::size_type found = compareString.rfind( separator , maxCharactersCommon ) ;
return compareString.substr( 0 , found ) ;
} |
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #ALGOL_68 | ALGOL 68 | MODE TYPE = INT;
PROC select = ([]TYPE from, PROC(TYPE)BOOL where)[]TYPE:
BEGIN
FLEX[0]TYPE result;
FOR key FROM LWB from TO UPB from DO
IF where(from[key]) THEN
[UPB result+1]TYPE new result;
new result[:UPB result] := result;
new result[UPB new result] := from[key];
result := new result
FI
OD;
result
END;
[]TYPE from values = (1,2,3,4,5,6,7,8,9,10);
PROC where even = (TYPE value)BOOL: NOT ODD value;
print((select(from values, where even), new line));
# Or as a simple one line query #
print((select((1,4,9,16,25,36,49,64,81,100), (TYPE x)BOOL: NOT ODD x ), new line)) |
http://rosettacode.org/wiki/Find_if_a_point_is_within_a_triangle | Find if a point is within a triangle | Find if a point is within a triangle.
Task
Assume points are on a plane defined by (x, y) real number coordinates.
Given a point P(x, y) and a triangle formed by points A, B, and C, determine if P is within triangle ABC.
You may use any algorithm.
Bonus: explain why the algorithm you chose works.
Related tasks
Determine_if_two_triangles_overlap
Also see
Discussion of several methods. [[1]]
Determine if a point is in a polygon [[2]]
Triangle based coordinate systems [[3]]
Wolfram entry [[4]]
| #Lua | Lua | EPS = 0.001
EPS_SQUARE = EPS * EPS
function side(x1, y1, x2, y2, x, y)
return (y2 - y1) * (x - x1) + (-x2 + x1) * (y - y1)
end
function naivePointInTriangle(x1, y1, x2, y2, x3, y3, x, y)
local checkSide1 = side(x1, y1, x2, y2, x, y) >= 0
local checkSide2 = side(x2, y2, x3, y3, x, y) >= 0
local checkSide3 = side(x3, y3, x1, y1, x, y) >= 0
return checkSide1 and checkSide2 and checkSide3
end
function pointInTriangleBoundingBox(x1, y1, x2, y2, x3, y3, x, y)
local xMin = math.min(x1, x2, x3) - EPS
local xMax = math.max(x1, x2, x3) + EPS
local yMin = math.min(y1, y2, y3) - EPS
local yMax = math.max(y1, y2, y3) + EPS
return not (x < xMin or xMax < x or y < yMin or yMax < y)
end
function distanceSquarePointToSegment(x1, y1, x2, y2, x, y)
local p1_p2_squareLength = (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1)
local dotProduct = ((x - x1) * (x2 - x1) + (y - y1) * (y2 - y1)) / p1_p2_squareLength
if dotProduct < 0 then
return (x - x1) * (x - x1) + (y - y1) * (y - y1)
end
if dotProduct <= 1 then
local p_p1_squareLength = (x1 - x) * (x1 - x) + (y1 - y) * (y1 - y)
return p_p1_squareLength - dotProduct * dotProduct * p1_p2_squareLength
end
return (x - x2) * (x - x2) + (y - y2) * (y - y2)
end
function accuratePointInTriangle(x1, y1, x2, y2, x3, y3, x, y)
if not pointInTriangleBoundingBox(x1, y1, x2, y2, x3, y3, x, y) then
return false
end
if naivePointInTriangle(x1, y1, x2, y2, x3, y3, x, y) then
return true
end
if distanceSquarePointToSegment(x1, y1, x2, y2, x, y) <= EPS_SQUARE then
return true
end
if distanceSquarePointToSegment(x2, y2, x3, y3, x, y) <= EPS_SQUARE then
return true
end
if distanceSquarePointToSegment(x3, y3, x1, y1, x, y) <= EPS_SQUARE then
return true
end
return false
end
function printPoint(x, y)
io.write('('..x..", "..y..')')
end
function printTriangle(x1, y1, x2, y2, x3, y3)
io.write("Triangle is [")
printPoint(x1, y1)
io.write(", ")
printPoint(x2, y2)
io.write(", ")
printPoint(x3, y3)
print("]")
end
function test(x1, y1, x2, y2, x3, y3, x, y)
printTriangle(x1, y1, x2, y2, x3, y3)
io.write("Point ")
printPoint(x, y)
print(" is within triangle? " .. tostring(accuratePointInTriangle(x1, y1, x2, y2, x3, y3, x, y)))
end
test(1.5, 2.4, 5.1, -3.1, -3.8, 1.2, 0, 0)
test(1.5, 2.4, 5.1, -3.1, -3.8, 1.2, 0, 1)
test(1.5, 2.4, 5.1, -3.1, -3.8, 1.2, 3, 1)
print()
test(0.1, 0.1111111111111111, 12.5, 33.333333333333336, 25, 11.11111111111111, 5.414285714285714, 14.349206349206348)
print()
test(0.1, 0.1111111111111111, 12.5, 33.333333333333336, -12.5, 16.666666666666668, 5.414285714285714, 14.349206349206348)
print() |
http://rosettacode.org/wiki/Flatten_a_list | Flatten a list | Task
Write a function to flatten the nesting in an arbitrary list of values.
Your program should work on the equivalent of this list:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
Where the correct result would be the list:
[1, 2, 3, 4, 5, 6, 7, 8]
Related task
Tree traversal
| #TXR | TXR | @(bind foo ((1) 2 ((3 4) 5) ((())) (((6))) 7 8 ()))
@(bind bar foo)
@(flatten bar) |
http://rosettacode.org/wiki/Flatten_a_list | Flatten a list | Task
Write a function to flatten the nesting in an arbitrary list of values.
Your program should work on the equivalent of this list:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
Where the correct result would be the list:
[1, 2, 3, 4, 5, 6, 7, 8]
Related task
Tree traversal
| #VBScript | VBScript |
class flattener
dim separator
sub class_initialize
separator = ","
end sub
private function makeflat( a )
dim i
dim res
for i = lbound( a ) to ubound( a )
if isarray( a( i ) ) then
res = res & makeflat( a( i ) )
else
res = res & a( i ) & separator
end if
next
makeflat = res
end function
public function flatten( a )
dim res
res = makeflat( a )
res = left( res, len( res ) - len(separator))
res = split( res, separator )
flatten = res
end function
public property let itemSeparator( c )
separator = c
end property
end class
|
http://rosettacode.org/wiki/Find_limit_of_recursion | Find limit of recursion | Find limit of recursion is part of Short Circuit's Console Program Basics selection.
Task
Find the limit of recursion.
| #Factor | Factor | : recurse ( n -- n ) 1 + recurse ;
0 recurse |
http://rosettacode.org/wiki/Find_limit_of_recursion | Find limit of recursion | Find limit of recursion is part of Short Circuit's Console Program Basics selection.
Task
Find the limit of recursion.
| #Fermat | Fermat |
Func Sisyphus(n)=!!n;Sisyphus(n+1).
Sisyphus(0)
|
http://rosettacode.org/wiki/Find_palindromic_numbers_in_both_binary_and_ternary_bases | Find palindromic numbers in both binary and ternary bases | Find palindromic numbers in both binary and ternary bases
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Find and show (in decimal) the first six numbers (non-negative integers) that are palindromes in both:
base 2
base 3
Display 0 (zero) as the first number found, even though some other definitions ignore it.
Optionally, show the decimal number found in its binary and ternary form.
Show all output here.
It's permissible to assume the first two numbers and simply list them.
See also
Sequence A60792, numbers that are palindromic in bases 2 and 3 on The On-Line Encyclopedia of Integer Sequences.
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | palindromify3[n_] :=
Block[{digits},
If[Divisible[n, 3], {},
digits = IntegerDigits[n, 3];
FromDigits[#, 3] & /@
{Join[Reverse[digits], digits], Join[Reverse[Rest[digits]], {First[digits]}, Rest[digits]]}
]
];
base2PalindromeQ[n_] := IntegerDigits[n, 2] === Reverse[IntegerDigits[n, 2]];
Select[Flatten[palindromify3 /@ Range[1000000]], base2PalindromeQ] |
http://rosettacode.org/wiki/Find_palindromic_numbers_in_both_binary_and_ternary_bases | Find palindromic numbers in both binary and ternary bases | Find palindromic numbers in both binary and ternary bases
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Find and show (in decimal) the first six numbers (non-negative integers) that are palindromes in both:
base 2
base 3
Display 0 (zero) as the first number found, even though some other definitions ignore it.
Optionally, show the decimal number found in its binary and ternary form.
Show all output here.
It's permissible to assume the first two numbers and simply list them.
See also
Sequence A60792, numbers that are palindromic in bases 2 and 3 on The On-Line Encyclopedia of Integer Sequences.
| #Nim | Nim | import bitops, strformat, times
#---------------------------------------------------------------------------------------------------
func isPal2(k: uint64; digitCount: Natural): bool =
## Return true if the "digitCount" + 1 bits of "k" form a palindromic number.
for i in 0..digitCount:
if k.testBit(i) != k.testBit(digitCount - i):
return false
result = true
#---------------------------------------------------------------------------------------------------
func reverseNumber(k: uint64): uint64 =
## Return the reverse number of "n".
var p = k
while p > 0:
result += 2 * result + p mod 3
p = p div 3
#---------------------------------------------------------------------------------------------------
func toBase2(n: uint64): string =
## Return the string representation of "n" in base 2.
var n = n
while true:
result.add(chr(ord('0') + (n and 1)))
n = n shr 1
if n == 0: break
#---------------------------------------------------------------------------------------------------
func toBase3(n: uint64): string =
## Return the string representation of "n" in base 3.
var n = n
while true:
result.add(chr(ord('0') + n mod 3))
n = n div 3
if n == 0: break
#---------------------------------------------------------------------------------------------------
proc print(n: uint64) =
## Print the value in bases 10, 2 and 3.
echo &"{n:>18} {n.toBase2():^59} {n.toBase3():^41}"
#---------------------------------------------------------------------------------------------------
proc findPal23() =
## Find the seven first palindromic numbers in binary and ternary bases.
var p3 = 1u64
var countPal = 1
print(0)
for p in 0..31:
while (3 * p3 + 1) * p3 < 1u64 shl (2 * p):
p3 *= 3
let bound = 1u64 shl (2 * p) div (3 * p3)
for k in max(p3 div 3, bound) .. min(2 * bound, p3 - 1):
let n = (3 * k + 1) * p3 + reverseNumber(k)
if isPal2(n, 2 * p):
print(n)
inc countPal
if countPal == 7:
return
#———————————————————————————————————————————————————————————————————————————————————————————————————
let t0 = cpuTime()
findPal23()
echo fmt"\nTime: {cpuTime() - t0:.2f}s" |
http://rosettacode.org/wiki/FizzBuzz | FizzBuzz | Task
Write a program that prints the integers from 1 to 100 (inclusive).
But:
for multiples of three, print Fizz (instead of the number)
for multiples of five, print Buzz (instead of the number)
for multiples of both three and five, print FizzBuzz (instead of the number)
The FizzBuzz problem was presented as the lowest level of comprehension required to illustrate adequacy.
Also see
(a blog) dont-overthink-fizzbuzz
(a blog) fizzbuzz-the-programmers-stairway-to-heaven
| #ERRE | ERRE |
PROGRAM FIZZ_BUZZ
!
! for rosettacode.org
!
BEGIN
FOR A=1 TO 100 DO
IF A MOD 15=0 THEN
PRINT("FizzBuzz")
ELSIF A MOD 3=0 THEN
PRINT("Fizz")
ELSIF A MOD 5=0 THEN
PRINT("Buzz")
ELSE
PRINT(A)
END IF
END FOR
END PROGRAM
|
http://rosettacode.org/wiki/File_size | File size | Verify the size of a file called input.txt for a file in the current working directory, and another one in the file system root.
| #Arturo | Arturo | print volume "input.txt"
print volume "/input.txt" |
http://rosettacode.org/wiki/File_size | File size | Verify the size of a file called input.txt for a file in the current working directory, and another one in the file system root.
| #AutoHotkey | AutoHotkey | FileGetSize, FileSize, input.txt ; Retrieve the size in bytes.
MsgBox, Size of input.txt is %FileSize% bytes
FileGetSize, FileSize, \input.txt, K ; Retrieve the size in Kbytes.
MsgBox, Size of \input.txt is %FileSize% Kbytes |
http://rosettacode.org/wiki/File_extension_is_in_extensions_list | File extension is in extensions list | File extension is in extensions list
You are encouraged to solve this task according to the task description, using any language you may know.
Filename extensions are a rudimentary but commonly used way of identifying files types.
Task
Given an arbitrary filename and a list of extensions, tell whether the filename has one of those extensions.
Notes:
The check should be case insensitive.
The extension must occur at the very end of the filename, and be immediately preceded by a dot (.).
You may assume that none of the given extensions are the empty string, and none of them contain a dot. Other than that they may be arbitrary strings.
Extra credit:
Allow extensions to contain dots. This way, users of your function/program have full control over what they consider as the extension in cases like:
archive.tar.gz
Please state clearly whether or not your solution does this.
Test cases
The following test cases all assume this list of extensions: zip, rar, 7z, gz, archive, A##
Filename
Result
MyData.a##
true
MyData.tar.Gz
true
MyData.gzip
false
MyData.7z.backup
false
MyData...
false
MyData
false
If your solution does the extra credit requirement, add tar.bz2 to the list of extensions, and check the following additional test cases:
Filename
Result
MyData_v1.0.tar.bz2
true
MyData_v1.0.bz2
false
Motivation
Checking if a file is in a certain category of file formats with known extensions (e.g. archive files, or image files) is a common problem in practice, and may be approached differently from extracting and outputting an arbitrary extension (see e.g. FileNameExtensionFilter in Java).
It also requires less assumptions about the format of an extension, because the calling code can decide what extensions are valid.
For these reasons, this task exists in addition to the Extract file extension task.
Related tasks
Extract file extension
String matching
| #Action.21 | Action! | DEFINE PTR="CARD"
CHAR FUNC ToLower(CHAR c)
IF c>='A AND c<='Z THEN
c==+'a-'A
FI
RETURN (c)
BYTE FUNC CheckExt(CHAR ARRAY file,ext)
BYTE i,j
CHAR c1,c2
i=file(0) j=ext(0)
IF i<j THEN RETURN (0) FI
WHILE j>0
DO
c1=ToLower(file(i))
c2=ToLower(ext(j))
IF c1#c2 THEN RETURN (0) FI
i==-1
j==-1
OD
IF file(i)#'. THEN
RETURN (0)
FI
RETURN (1)
BYTE FUNC Check(CHAR ARRAY file PTR ARRAY exts BYTE count)
BYTE i
FOR i=0 TO count-1
DO
IF CheckExt(file,exts(i)) THEN
RETURN (1)
FI
OD
RETURN (0)
PROC Main()
PTR ARRAY exts(7),files(8)
BYTE i
exts(0)="zip"
exts(1)="rar"
exts(2)="7z"
exts(3)="gz"
exts(4)="archive"
exts(5)="A##"
exts(6)="tar.bz2"
files(0)="MyData.a##"
files(1)="MyData.tar.Gz"
files(2)="MyData.gzip"
files(3)="MyData.7z.backup"
files(4)="MyData..."
files(5)="MyData"
files(6)="MyData_v1.0.tar.bz2"
files(7)="MyData_v1.0.bz2"
FOR i=0 to 7
DO
Print(files(i)) Print(" -> ")
IF Check(files(i),exts,7) THEN
PrintE("true")
ELSE
PrintE("false")
FI
OD
RETURN |
http://rosettacode.org/wiki/File_extension_is_in_extensions_list | File extension is in extensions list | File extension is in extensions list
You are encouraged to solve this task according to the task description, using any language you may know.
Filename extensions are a rudimentary but commonly used way of identifying files types.
Task
Given an arbitrary filename and a list of extensions, tell whether the filename has one of those extensions.
Notes:
The check should be case insensitive.
The extension must occur at the very end of the filename, and be immediately preceded by a dot (.).
You may assume that none of the given extensions are the empty string, and none of them contain a dot. Other than that they may be arbitrary strings.
Extra credit:
Allow extensions to contain dots. This way, users of your function/program have full control over what they consider as the extension in cases like:
archive.tar.gz
Please state clearly whether or not your solution does this.
Test cases
The following test cases all assume this list of extensions: zip, rar, 7z, gz, archive, A##
Filename
Result
MyData.a##
true
MyData.tar.Gz
true
MyData.gzip
false
MyData.7z.backup
false
MyData...
false
MyData
false
If your solution does the extra credit requirement, add tar.bz2 to the list of extensions, and check the following additional test cases:
Filename
Result
MyData_v1.0.tar.bz2
true
MyData_v1.0.bz2
false
Motivation
Checking if a file is in a certain category of file formats with known extensions (e.g. archive files, or image files) is a common problem in practice, and may be approached differently from extracting and outputting an arbitrary extension (see e.g. FileNameExtensionFilter in Java).
It also requires less assumptions about the format of an extension, because the calling code can decide what extensions are valid.
For these reasons, this task exists in addition to the Extract file extension task.
Related tasks
Extract file extension
String matching
| #Ada | Ada | with Ada.Text_IO; use Ada.Text_IO;
with Ada.Strings.Fixed.Equal_Case_Insensitive; use Ada.Strings.Fixed;
with Ada.Strings.Bounded;
procedure Main is
package B_String is new Ada.Strings.Bounded.Generic_Bounded_Length (30);
use B_String;
function is_equal (left, right : String) return Boolean renames
Ada.Strings.Fixed.Equal_Case_Insensitive;
type extension_list is array (Positive range <>) of Bounded_String;
Ext_List : extension_list :=
(To_Bounded_String ("zip"), To_Bounded_String ("rar"),
To_Bounded_String ("7z"), To_Bounded_String ("gz"),
To_Bounded_String ("archive"), To_Bounded_String ("A##"),
To_Bounded_String ("tar.bz2"));
type filename_list is array (Positive range <>) of Bounded_String;
fnames : filename_list :=
(To_Bounded_String ("MyData.a##"), To_Bounded_String ("MyData.tar.Gz"),
To_Bounded_String ("MyData.gzip"), To_Bounded_String ("MyData..."),
To_Bounded_String ("Mydata"), To_Bounded_String ("MyData_V1.0.tar.bz2"),
To_Bounded_String ("MyData_v1.0.bz2"));
Valid_Extension : Boolean;
begin
for name of fnames loop
Valid_Extension := False;
Put (To_String (name));
for ext of Ext_List loop
declare
S : String := "." & To_String (ext);
T : String := Tail (Source => To_String (name), Count => S'Length);
begin
if is_equal (S, T) then
Valid_Extension := True;
end if;
end;
end loop;
Set_Col (22);
Put_Line (": " & Valid_Extension'Image);
end loop;
end Main; |
http://rosettacode.org/wiki/File_modification_time | File modification time | Task
Get and set the modification time of a file.
| #C.23 | C# | using System;
using System.IO;
Console.WriteLine(File.GetLastWriteTime("file.txt"));
File.SetLastWriteTime("file.txt", DateTime.Now); |
http://rosettacode.org/wiki/File_modification_time | File modification time | Task
Get and set the modification time of a file.
| #C.2B.2B | C++ | #include <boost/filesystem/operations.hpp>
#include <ctime>
#include <iostream>
int main( int argc , char *argv[ ] ) {
if ( argc != 2 ) {
std::cerr << "Error! Syntax: moditime <filename>!\n" ;
return 1 ;
}
boost::filesystem::path p( argv[ 1 ] ) ;
if ( boost::filesystem::exists( p ) ) {
std::time_t t = boost::filesystem::last_write_time( p ) ;
std::cout << "On " << std::ctime( &t ) << " the file " << argv[ 1 ]
<< " was modified the last time!\n" ;
std::cout << "Setting the modification time to now:\n" ;
std::time_t n = std::time( 0 ) ;
boost::filesystem::last_write_time( p , n ) ;
t = boost::filesystem::last_write_time( p ) ;
std::cout << "Now the modification time is " << std::ctime( &t ) << std::endl ;
return 0 ;
} else {
std::cout << "Could not find file " << argv[ 1 ] << '\n' ;
return 2 ;
}
} |
http://rosettacode.org/wiki/File_size_distribution | File size distribution | Task
Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.
My suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.
Don't forget that empty files may exist, to serve as a marker.
Is your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | SetDirectory[NotebookDirectory[]];
Histogram[FileByteCount /@ Select[FileNames[__], DirectoryQ /* Not], {"Log", 15}, {"Log", "Count"}] |
http://rosettacode.org/wiki/File_size_distribution | File size distribution | Task
Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.
My suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.
Don't forget that empty files may exist, to serve as a marker.
Is your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?
| #Nim | Nim | import math, os, strformat
const
MaxPower = 10
Powers = [1, 10, 100]
func powerWithUnit(idx: int): string =
## Return a string representing value 10^idx with a unit.
if idx < 0:
"0B"
elif idx < 3:
fmt"{Powers[idx]}B"
elif idx < 6:
fmt"{Powers[idx - 3]}kB"
elif idx < 9:
fmt"{Powers[idx - 6]}MB"
else:
fmt"{Powers[idx - 9]}GB"
# Retrieve the directory path.
var dirpath: string
if paramCount() == 0:
dirpath = getCurrentDir()
else:
dirpath = paramStr(1)
if not dirExists(dirpath):
raise newException(ValueError, "wrong directory path: " & dirpath)
# Distribute sizes.
var counts: array[-1..MaxPower, Natural]
for path in dirpath.walkDirRec():
if not path.fileExists():
continue # Not a regular file.
let size = getFileSize(path)
let index = if size == 0: -1 else: log10(size.float).toInt
inc counts[index]
# Display distribution.
let total = sum(counts)
echo "File size distribution for directory: ", dirpath
echo ""
for idx, count in counts:
let rangeString = fmt"[{powerWithUnit(idx)}..{powerWithUnit(idx + 1)}[:"
echo fmt"Size in {rangeString: 14} {count:>7} {100 * count / total:5.2f}%"
echo ""
echo "Total number of files: ", sum(counts) |
http://rosettacode.org/wiki/File_size_distribution | File size distribution | Task
Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.
My suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.
Don't forget that empty files may exist, to serve as a marker.
Is your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?
| #Perl | Perl | use File::Find;
use List::Util qw(max);
my %fsize;
$dir = shift || '.';
find(\&fsize, $dir);
$max = max($max,$fsize{$_}) for keys %fsize;
$total += $size while (undef,$size) = each %fsize;
print "File size distribution in bytes for directory: $dir\n";
for (0 .. max(keys %fsize)) {
printf "# files @ %4sb %8s: %s\n", $_ ? '10e'.($_-1) : 0, $fsize{$_} // 0,
histogram( $max, $fsize{$_} // 0, 80);
}
print "$total total files.\n";
sub histogram {
my($max, $value, $width) = @_;
my @blocks = qw<| ▏ ▎ ▍ ▌ ▋ ▊ ▉ █>;
my $scaled = int $value * $width / $max;
my $end = $scaled % 8;
my $bar = int $scaled / 8;
my $B = $blocks[8] x ($bar * 8) . ($end ? $blocks[$end] : '');
}
sub fsize { $fsize{ log10( (lstat($_))[7] ) }++ }
sub log10 { my($s) = @_; $s ? int log($s)/log(10) : 0 } |
http://rosettacode.org/wiki/Find_common_directory_path | Find common directory path | Create a routine that, given a set of strings representing directory paths and a single character directory separator, will return a string representing that part of the directory tree that is common to all the directories.
Test your routine using the forward slash '/' character as the directory separator and the following three strings as input paths:
'/home/user1/tmp/coverage/test'
'/home/user1/tmp/covert/operator'
'/home/user1/tmp/coven/members'
Note: The resultant path should be the valid directory '/home/user1/tmp' and not the longest common string '/home/user1/tmp/cove'.
If your language has a routine that performs this function (even if it does not have a changeable separator character), then mention it as part of the task.
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
| #Clojure | Clojure | (use '[clojure.string :only [join,split]])
(defn common-prefix [sep paths]
(let [parts-per-path (map #(split % (re-pattern sep)) paths)
parts-per-position (apply map vector parts-per-path)]
(join sep
(for [parts parts-per-position :while (apply = parts)]
(first parts)))))
(println
(common-prefix "/"
["/home/user1/tmp/coverage/test"
"/home/user1/tmp/covert/operator"
"/home/user1/tmp/coven/members"])) |
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #ALGOL_W | ALGOL W | begin
% sets the elements of out to the elements of in that return true from applying the where procedure to them %
% the bounds of in must be 1 :: inUb - out must be at least as big as in and the number of matching %
% elements is returned in outUb - in and out can be the same array %
procedure select ( integer array in ( * ); integer value inUb
; integer array out ( * ); integer result outUb
; logical procedure where % ( integer value n ) %
) ;
begin
outUb := 0;
for i := 1 until inUb do begin
if where( in( i ) ) then begin
outUb := outUb + 1;
out( outUb ) := in( i )
end f_where_in_i
end for_i
end select ;
% test the select procedure %
logical procedure isEven ( integer value n ) ; not odd( n );
integer array t, out ( 1 :: 10 );
integer outUb;
for i := 1 until 10 do t( i ) := i;
select( t, 10, out, outUb, isEven );
for i := 1 until outUb do writeon( i_w := 3, s_w := 0, out( i ) );
write()
end.
|
http://rosettacode.org/wiki/Find_if_a_point_is_within_a_triangle | Find if a point is within a triangle | Find if a point is within a triangle.
Task
Assume points are on a plane defined by (x, y) real number coordinates.
Given a point P(x, y) and a triangle formed by points A, B, and C, determine if P is within triangle ABC.
You may use any algorithm.
Bonus: explain why the algorithm you chose works.
Related tasks
Determine_if_two_triangles_overlap
Also see
Discussion of several methods. [[1]]
Determine if a point is in a polygon [[2]]
Triangle based coordinate systems [[3]]
Wolfram entry [[4]]
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | RegionMember[Polygon[{{1, 2}, {3, 1}, {2, 4}}], {2, 2}] |
http://rosettacode.org/wiki/Find_if_a_point_is_within_a_triangle | Find if a point is within a triangle | Find if a point is within a triangle.
Task
Assume points are on a plane defined by (x, y) real number coordinates.
Given a point P(x, y) and a triangle formed by points A, B, and C, determine if P is within triangle ABC.
You may use any algorithm.
Bonus: explain why the algorithm you chose works.
Related tasks
Determine_if_two_triangles_overlap
Also see
Discussion of several methods. [[1]]
Determine if a point is in a polygon [[2]]
Triangle based coordinate systems [[3]]
Wolfram entry [[4]]
| #Nim | Nim | import strformat
const
Eps = 0.001
Eps2 = Eps * Eps
type
Point = tuple[x, y: float]
Triangle = object
p1, p2, p3: Point
func initTriangle(p1, p2, p3: Point): Triangle =
Triangle(p1: p1, p2: p2, p3: p3)
func side(p1, p2, p: Point): float =
(p2.y - p1.y) * (p.x - p1.x) + (-p2.x + p1.x) * (p.y - p1.y)
func distanceSquarePointToSegment(p1, p2, p: Point): float =
let p1P2SquareLength = (p2.x - p1.x) * (p2.x - p1.x) + (p2.y - p1.y) * (p2.y - p1.y)
let dotProduct = ((p.x - p1.x) * (p2.x - p1.x) + (p.y - p1.y) * (p2.y - p1.y)) / p1P2SquareLength
if dotProduct < 0:
return (p.x - p1.x) * (p.x - p1.x) + (p.y - p1.y) * (p.y - p1.y)
if dotProduct <= 1:
let pP1SquareLength = (p1.x - p.x) * (p1.x - p.x) + (p1.y - p.y) * (p1.y - p.y)
return pP1SquareLength - dotProduct * dotProduct * p1P2SquareLength
result = (p.x - p2.x) * (p.x - p2.x) + (p.y - p2.y) * (p.y - p2.y)
func pointInTriangleBoundingBox(t: Triangle; p: Point): bool =
let xMin = min(t.p1.x, min(t.p2.x, t.p3.x)) - EPS
let xMax = max(t.p1.x, max(t.p2.x, t.p3.x)) + EPS
let yMin = min(t.p1.y, min(t.p2.y, t.p3.y)) - EPS
let yMax = max(t.p1.y, max(t.p2.y, t.p3.y)) + EPS
result = p.x in xMin..xMax and p.y in yMin..yMax
func nativePointInTriangle(t: Triangle; p: Point): bool =
let checkSide1 = side(t.p1, t.p2, p) >= 0
let checkSide2 = side(t.p2, t.p3, p) >= 0
let checkSide3 = side(t.p3, t.p1, p) >= 0
result = checkSide1 and checkSide2 and checkSide3
func accuratePointInTriangle(t: Triangle; p: Point): bool =
if not t.pointInTriangleBoundingBox(p):
return false
if t.nativePointInTriangle(p):
return true
if distanceSquarePointToSegment(t.p1, t.p2, p) <= Eps2 or
distanceSquarePointToSegment(t.p3, t.p1, p) <= Eps2:
return true
func `$`(p: Point): string = &"({p.x}, {p.y})"
func `$`(t: Triangle): string = &"Triangle[{t.p1}, {t.p2}, {t.p3}]"
func contains(t: Triangle; p: Point): bool = t.accuratePointInTriangle(p)
when isMainModule:
proc test(t: Triangle; p: Point) =
echo t
echo &"Point {p} is within triangle ? {p in t}"
var p1: Point = (1.5, 2.4)
var p2: Point = (5.1, -3.1)
var p3: Point = (-3.8, 1.2)
var tri = initTriangle(p1, p2, p3)
test(tri, (0.0, 0.0))
test(tri, (0.0, 1.0))
test(tri, (3.0, 1.0))
echo()
p1 = (1 / 10, 1 / 9)
p2 = (100 / 8, 100 / 3)
p3 = (100 / 4, 100 / 9)
tri = initTriangle(p1, p2, p3)
let pt = (p1.x + 3.0 / 7 * (p2.x - p1.x), p1.y + 3.0 / 7 * (p2.y - p1.y))
test(tri, pt)
echo()
p3 = (-100 / 8, 100 / 6)
tri = initTriangle(p1, p2, p3)
test(tri, pt) |
http://rosettacode.org/wiki/Flatten_a_list | Flatten a list | Task
Write a function to flatten the nesting in an arbitrary list of values.
Your program should work on the equivalent of this list:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
Where the correct result would be the list:
[1, 2, 3, 4, 5, 6, 7, 8]
Related task
Tree traversal
| #Wart | Wart | def (flatten seq acc)
if no.seq
acc
~list?.seq
(cons seq acc)
:else
(flatten car.seq (flatten cdr.seq acc)) |
http://rosettacode.org/wiki/Find_limit_of_recursion | Find limit of recursion | Find limit of recursion is part of Short Circuit's Console Program Basics selection.
Task
Find the limit of recursion.
| #Forth | Forth | : munge ( n -- n' ) 1+ recurse ;
: test 0 ['] munge catch if ." Recursion limit at depth " . then ;
test \ Default gforth: Recursion limit at depth 3817 |
http://rosettacode.org/wiki/Find_limit_of_recursion | Find limit of recursion | Find limit of recursion is part of Short Circuit's Console Program Basics selection.
Task
Find the limit of recursion.
| #Fortran | Fortran | program recursion_depth
implicit none
call recurse (1)
contains
recursive subroutine recurse (i)
implicit none
integer, intent (in) :: i
write (*, '(i0)') i
call recurse (i + 1)
end subroutine recurse
end program recursion_depth |
http://rosettacode.org/wiki/Find_palindromic_numbers_in_both_binary_and_ternary_bases | Find palindromic numbers in both binary and ternary bases | Find palindromic numbers in both binary and ternary bases
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Find and show (in decimal) the first six numbers (non-negative integers) that are palindromes in both:
base 2
base 3
Display 0 (zero) as the first number found, even though some other definitions ignore it.
Optionally, show the decimal number found in its binary and ternary form.
Show all output here.
It's permissible to assume the first two numbers and simply list them.
See also
Sequence A60792, numbers that are palindromic in bases 2 and 3 on The On-Line Encyclopedia of Integer Sequences.
| #PARI.2FGP | PARI/GP | check(n)={ \\ Check for 2n+1-digit palindromes in base 3
my(N=3^n);
forstep(i=N+1,2*N,[1,2],
my(base2,base3=digits(i,3),k);
base3=concat(Vecrev(base3[2..n+1]), base3);
k=subst(Pol(base3),'x,3);
base2=binary(k);
if(base2==Vecrev(base2), print1(", "k))
)
};
print1("0, 1"); for(i=1,11,check(i)) |
http://rosettacode.org/wiki/FizzBuzz | FizzBuzz | Task
Write a program that prints the integers from 1 to 100 (inclusive).
But:
for multiples of three, print Fizz (instead of the number)
for multiples of five, print Buzz (instead of the number)
for multiples of both three and five, print FizzBuzz (instead of the number)
The FizzBuzz problem was presented as the lowest level of comprehension required to illustrate adequacy.
Also see
(a blog) dont-overthink-fizzbuzz
(a blog) fizzbuzz-the-programmers-stairway-to-heaven
| #Euphoria | Euphoria | include std/utils.e
function fb( atom n )
sequence fb
if remainder( n, 15 ) = 0 then
fb = "FizzBuzz"
elsif remainder( n, 5 ) = 0 then
fb = "Fizz"
elsif remainder( n, 3 ) = 0 then
fb = "Buzz"
else
fb = sprintf( "%d", n )
end if
return fb
end function
function fb2( atom n )
return iif( remainder(n, 15) = 0, "FizzBuzz",
iif( remainder( n, 5 ) = 0, "Fizz",
iif( remainder( n, 3) = 0, "Buzz", sprintf( "%d", n ) ) ) )
end function
for i = 1 to 30 do
printf( 1, "%s ", { fb( i ) } )
end for
puts( 1, "\n" )
for i = 1 to 30 do
printf( 1, "%s ", { fb2( i ) } )
end for
puts( 1, "\n" ) |
http://rosettacode.org/wiki/File_size | File size | Verify the size of a file called input.txt for a file in the current working directory, and another one in the file system root.
| #AWK | AWK | @load "filefuncs"
function filesize(name ,fd) {
if ( stat(name, fd) == -1)
return -1 # doesn't exist
else
return fd["size"]
}
BEGIN {
print filesize("input.txt")
print filesize("/input.txt")
} |
http://rosettacode.org/wiki/File_size | File size | Verify the size of a file called input.txt for a file in the current working directory, and another one in the file system root.
| #Axe | Axe | If GetCalc("appvINPUT")→I
Disp {I-2}ʳ▶Dec,i
Else
Disp "NOT FOUND",i
End |
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #11l | 11l | V file_contents = File(‘input.txt’).read()
File(‘output.txt’, ‘w’).write(file_contents) |
http://rosettacode.org/wiki/File_extension_is_in_extensions_list | File extension is in extensions list | File extension is in extensions list
You are encouraged to solve this task according to the task description, using any language you may know.
Filename extensions are a rudimentary but commonly used way of identifying files types.
Task
Given an arbitrary filename and a list of extensions, tell whether the filename has one of those extensions.
Notes:
The check should be case insensitive.
The extension must occur at the very end of the filename, and be immediately preceded by a dot (.).
You may assume that none of the given extensions are the empty string, and none of them contain a dot. Other than that they may be arbitrary strings.
Extra credit:
Allow extensions to contain dots. This way, users of your function/program have full control over what they consider as the extension in cases like:
archive.tar.gz
Please state clearly whether or not your solution does this.
Test cases
The following test cases all assume this list of extensions: zip, rar, 7z, gz, archive, A##
Filename
Result
MyData.a##
true
MyData.tar.Gz
true
MyData.gzip
false
MyData.7z.backup
false
MyData...
false
MyData
false
If your solution does the extra credit requirement, add tar.bz2 to the list of extensions, and check the following additional test cases:
Filename
Result
MyData_v1.0.tar.bz2
true
MyData_v1.0.bz2
false
Motivation
Checking if a file is in a certain category of file formats with known extensions (e.g. archive files, or image files) is a common problem in practice, and may be approached differently from extracting and outputting an arbitrary extension (see e.g. FileNameExtensionFilter in Java).
It also requires less assumptions about the format of an extension, because the calling code can decide what extensions are valid.
For these reasons, this task exists in addition to the Extract file extension task.
Related tasks
Extract file extension
String matching
| #ALGOL_68 | ALGOL 68 | # returns the length of str #
OP LENGTH = ( STRING str )INT: ( UPB str - LWB str ) + 1;
# returns TRUE if str ends with ending FALSE otherwise #
PRIO ENDSWITH = 9;
OP ENDSWITH = ( STRING str, STRING ending )BOOL:
IF INT str length = LENGTH str;
INT ending length = LENGTH ending;
ending length > str length
THEN
# the ending is longer than the string #
FALSE
ELSE
# the string is at least as long as the ending #
str[ ( str length - ending length ) + 1 : AT 1 ] = ending
FI # ENDSWITH # ;
# returns str cnverted to upper case #
OP TOUPPER = ( STRING str )STRING:
BEGIN
STRING result := str;
FOR s pos FROM LWB result TO UPB result DO
result[ s pos ] := to upper( result[ s pos ] )
OD;
result
END # TOUPPER # ;
# tests whether file name has one of the extensions and returns #
# the index of the extension in extensions or LWB extensions - 1 #
# if it does not end with one of the extensions #
# the tests are not case-sensitive #
PROC has extension in list = ( STRING file name, []STRING extensions )INT:
BEGIN
INT extension number := LWB extensions - 1;
STRING upper name = TOUPPER file name;
FOR pos FROM LWB extensions TO UPB extensions WHILE extension number < LWB extensions DO
IF upper name ENDSWITH ( "." + TOUPPER extensions[ pos ] )
THEN
# found the extension #
extension number := pos
FI
OD;
extension number
END # has extension # ;
# test the has extension in list procedure #
PROC test has extension in list = ( STRING file name, []STRING extensions, BOOL expected result )VOID:
IF INT extension number = has extension in list( file name, extensions );
extension number < LWB extensions
THEN
# the file does not have one of the extensions #
print( ( file name
, " does not have an extension in the list "
, IF expected result THEN "NOT AS EXPECTED" ELSE "" FI
, newline
)
)
ELSE
# the file does have one of the extensions #
print( ( file name
, " has extension """
, extensions[ extension number ]
, """ "
, IF NOT expected result THEN "NOT AS EXPECTED" ELSE "" FI
, newline
)
)
FI # test has extension in list # ;
# the extensions for the task #
[]STRING task extensions = ( "zip", "rar", "7z", "gz", "archive", "A##" );
# test the file names in the standard task #
test has extension in list( "MyData.a##", task extensions, TRUE );
test has extension in list( "MyData.tar.Gz", task extensions, TRUE );
test has extension in list( "MyData.gzip", task extensions, FALSE );
test has extension in list( "MyData.7z.backup", task extensions, FALSE );
test has extension in list( "MyData...", task extensions, FALSE );
test has extension in list( "MyData", task extensions, FALSE );
# the extensions for the extra credit #
[]STRING ec extensions = ( "zip", "rar", "7z", "gz", "archive", "A##", "tar.bz2" );
# test the file names in the extra credit #
test has extension in list( "MyData_v1.0.tar.bz2", ec extensions, TRUE );
test has extension in list( "MyData_v1.0.bz2", ec extensions, FALSE ) |
http://rosettacode.org/wiki/File_modification_time | File modification time | Task
Get and set the modification time of a file.
| #Clojure | Clojure | (import '(java.io File)
'(java.util Date))
(Date. (.lastModified (File. "output.txt")))
(Date. (.lastModified (File. "docs")))
(.setLastModified (File. "output.txt")
(.lastModified (File. "docs"))) |
http://rosettacode.org/wiki/File_modification_time | File modification time | Task
Get and set the modification time of a file.
| #Common_Lisp | Common Lisp | (file-write-date "input.txt") |
http://rosettacode.org/wiki/File_size_distribution | File size distribution | Task
Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.
My suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.
Don't forget that empty files may exist, to serve as a marker.
Is your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?
| #Phix | Phix | without js -- file i/o
sequence sizes = {1},
res = {0}
atom t1 = time()+1
function store_res(string filepath, sequence dir_entry)
if not find('d', dir_entry[D_ATTRIBUTES]) then
atom size = dir_entry[D_SIZE]
integer sdx = 1
while size>sizes[sdx] do
if sdx=length(sizes) then
sizes &= sizes[$]*iff(mod(length(sizes),3)?10:10.24)
res &= 0
end if
sdx += 1
end while
res[sdx] += 1
if time()>t1 then
printf(1,"%,d files found\r",sum(res))
t1 = time()+1
end if
end if
return 0 -- keep going
end function
integer exit_code = walk_dir(".", store_res, true)
printf(1,"%,d files found\n",sum(res))
integer w = max(res)
--include builtins/pfile.e
for i=1 to length(res) do
integer ri = res[i]
string s = file_size_k(sizes[i], 5),
p = repeat('*',floor(60*ri/w))
printf(1,"files < %s: %s%,d\n",{s,p,ri})
end for
|
http://rosettacode.org/wiki/Find_common_directory_path | Find common directory path | Create a routine that, given a set of strings representing directory paths and a single character directory separator, will return a string representing that part of the directory tree that is common to all the directories.
Test your routine using the forward slash '/' character as the directory separator and the following three strings as input paths:
'/home/user1/tmp/coverage/test'
'/home/user1/tmp/covert/operator'
'/home/user1/tmp/coven/members'
Note: The resultant path should be the valid directory '/home/user1/tmp' and not the longest common string '/home/user1/tmp/cove'.
If your language has a routine that performs this function (even if it does not have a changeable separator character), then mention it as part of the task.
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
| #Common_Lisp | Common Lisp |
(defun common-directory-path (&rest paths)
(do* ((pathnames (mapcar #'(lambda (path) (cdr (pathname-directory (pathname path)))) paths)) ; convert strings to lists of subdirectories
(rem pathnames (cdr rem))
(pos (length (first rem))) ) ; position of first mismatched element
((null (cdr rem)) (make-pathname :directory (cons :absolute (subseq (first pathnames) 0 pos)))) ; take the common sublists and convert back to a pathname
(setq pos (min pos (mismatch (first rem) (second rem) :test #'string-equal))) )) ; compare two paths
|
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #AmigaE | AmigaE | PROC main()
DEF l : PTR TO LONG, r : PTR TO LONG, x
l := [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
r := List(ListLen(l))
SelectList({x}, l, r, `Mod(x,2)=0)
ForAll({x}, r, `WriteF('\d\n', x))
ENDPROC |
http://rosettacode.org/wiki/Find_if_a_point_is_within_a_triangle | Find if a point is within a triangle | Find if a point is within a triangle.
Task
Assume points are on a plane defined by (x, y) real number coordinates.
Given a point P(x, y) and a triangle formed by points A, B, and C, determine if P is within triangle ABC.
You may use any algorithm.
Bonus: explain why the algorithm you chose works.
Related tasks
Determine_if_two_triangles_overlap
Also see
Discussion of several methods. [[1]]
Determine if a point is in a polygon [[2]]
Triangle based coordinate systems [[3]]
Wolfram entry [[4]]
| #Perl | Perl | # 20201123 added Perl programming solution
use strict;
use warnings;
use List::AllUtils qw(min max natatime);
use constant EPSILON => 0.001;
use constant EPSILON_SQUARE => EPSILON*EPSILON;
sub side {
my ($x1, $y1, $x2, $y2, $x, $y) = @_;
return ($y2 - $y1)*($x - $x1) + (-$x2 + $x1)*($y - $y1);
}
sub naivePointInTriangle {
my ($x1, $y1, $x2, $y2, $x3, $y3, $x, $y) = @_;
my $checkSide1 = side($x1, $y1, $x2, $y2, $x, $y) >= 0 ;
my $checkSide2 = side($x2, $y2, $x3, $y3, $x, $y) >= 0 ;
my $checkSide3 = side($x3, $y3, $x1, $y1, $x, $y) >= 0 ;
return $checkSide1 && $checkSide2 && $checkSide3 || 0 ;
}
sub pointInTriangleBoundingBox {
my ($x1, $y1, $x2, $y2, $x3, $y3, $x, $y) = @_;
my $xMin = min($x1, min($x2, $x3)) - EPSILON;
my $xMax = max($x1, max($x2, $x3)) + EPSILON;
my $yMin = min($y1, min($y2, $y3)) - EPSILON;
my $yMax = max($y1, max($y2, $y3)) + EPSILON;
( $x < $xMin || $xMax < $x || $y < $yMin || $yMax < $y ) ? 0 : 1
}
sub distanceSquarePointToSegment {
my ($x1, $y1, $x2, $y2, $x, $y) = @_;
my $p1_p2_squareLength = ($x2 - $x1)**2 + ($y2 - $y1)**2;
my $dotProduct = ($x-$x1)*($x2-$x1)+($y-$y1)*($y2-$y1) ;
if ( $dotProduct < 0 ) {
return ($x - $x1)**2 + ($y - $y1)**2;
} elsif ( $dotProduct <= $p1_p2_squareLength ) {
my $p_p1_squareLength = ($x1 - $x)**2 + ($y1 - $y)**2;
return $p_p1_squareLength - $dotProduct**2 / $p1_p2_squareLength;
} else {
return ($x - $x2)**2 + ($y - $y2)**2;
}
}
sub accuratePointInTriangle {
my ($x1, $y1, $x2, $y2, $x3, $y3, $x, $y) = @_;
return 0 unless pointInTriangleBoundingBox($x1,$y1,$x2,$y2,$x3,$y3,$x,$y);
return 1 if ( naivePointInTriangle($x1, $y1, $x2, $y2, $x3, $y3, $x, $y)
or distanceSquarePointToSegment($x1, $y1, $x2, $y2, $x, $y) <= EPSILON_SQUARE
or distanceSquarePointToSegment($x2, $y2, $x3, $y3, $x, $y) <= EPSILON_SQUARE
or distanceSquarePointToSegment($x3, $y3, $x1, $y1, $x, $y) <= EPSILON_SQUARE);
return 0
}
my @DATA = (1.5, 2.4, 5.1, -3.1, -3.8, 0.5);
for my $point ( [0,0] , [0,1] ,[3,1] ) {
print "Point (", join(',',@$point), ") is within triangle ";
my $iter = natatime 2, @DATA;
while ( my @vertex = $iter->()) { print '(',join(',',@vertex),') ' }
print ': ',naivePointInTriangle (@DATA, @$point) ? 'True' : 'False', "\n" ;
} |
http://rosettacode.org/wiki/Flatten_a_list | Flatten a list | Task
Write a function to flatten the nesting in an arbitrary list of values.
Your program should work on the equivalent of this list:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
Where the correct result would be the list:
[1, 2, 3, 4, 5, 6, 7, 8]
Related task
Tree traversal
| #WDTE | WDTE | let a => import 'arrays';
let s => import 'stream';
let flatten array =>
a.stream array
-> s.flatMap (@ f v => v {
reflect 'Array' => a.stream v -> s.flatMap f;
})
-> s.collect
; |
http://rosettacode.org/wiki/Find_limit_of_recursion | Find limit of recursion | Find limit of recursion is part of Short Circuit's Console Program Basics selection.
Task
Find the limit of recursion.
| #GAP | GAP | f := function(n)
return f(n+1);
end;
# Now loop until an error occurs
f(0);
# Error message :
# Entering break read-eval-print loop ...
# you can 'quit;' to quit to outer loop, or
# you may 'return;' to continue
n;
# 4998
# quit "brk mode" and return to GAP
quit; |
http://rosettacode.org/wiki/Find_limit_of_recursion | Find limit of recursion | Find limit of recursion is part of Short Circuit's Console Program Basics selection.
Task
Find the limit of recursion.
| #gnuplot | gnuplot | # Put this in a file foo.gnuplot and run as
# gnuplot foo.gnuplot
# probe by 1 up to 1000, then by 1% increases
if (! exists("try")) { try=0 }
try=(try<1000 ? try+1 : try*1.01)
recurse(n) = (n > 0 ? recurse(n-1) : 'ok')
print "try recurse ", try
print recurse(try)
reread |
http://rosettacode.org/wiki/Find_palindromic_numbers_in_both_binary_and_ternary_bases | Find palindromic numbers in both binary and ternary bases | Find palindromic numbers in both binary and ternary bases
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Find and show (in decimal) the first six numbers (non-negative integers) that are palindromes in both:
base 2
base 3
Display 0 (zero) as the first number found, even though some other definitions ignore it.
Optionally, show the decimal number found in its binary and ternary form.
Show all output here.
It's permissible to assume the first two numbers and simply list them.
See also
Sequence A60792, numbers that are palindromic in bases 2 and 3 on The On-Line Encyclopedia of Integer Sequences.
| #Perl | Perl | use ntheory qw/fromdigits todigitstring/;
print "0 0 0\n"; # Hard code the 0 result
for (0..2e5) {
# Generate middle-1-palindrome in base 3.
my $pal = todigitstring($_, 3);
my $b3 = $pal . "1" . reverse($pal);
# Convert base 3 number to base 2
my $b2 = todigitstring(fromdigits($b3, 3), 2);
# Print results (including base 10) if base-2 palindrome
print fromdigits($b2,2)," $b3 $b2\n" if $b2 eq reverse($b2);
} |
http://rosettacode.org/wiki/FizzBuzz | FizzBuzz | Task
Write a program that prints the integers from 1 to 100 (inclusive).
But:
for multiples of three, print Fizz (instead of the number)
for multiples of five, print Buzz (instead of the number)
for multiples of both three and five, print FizzBuzz (instead of the number)
The FizzBuzz problem was presented as the lowest level of comprehension required to illustrate adequacy.
Also see
(a blog) dont-overthink-fizzbuzz
(a blog) fizzbuzz-the-programmers-stairway-to-heaven
| #F.23 | F# | let fizzbuzz n =
match n%3 = 0, n%5 = 0 with
| true, false -> "fizz"
| false, true -> "buzz"
| true, true -> "fizzbuzz"
| _ -> string n
let printFizzbuzz() =
[1..100] |> List.iter (fizzbuzz >> printfn "%s") |
http://rosettacode.org/wiki/File_size | File size | Verify the size of a file called input.txt for a file in the current working directory, and another one in the file system root.
| #BaCon | BaCon | ' file size
' Return the entire message, FILELEN returns a NUMBER
FUNCTION printlen$(STRING name$)
IF FILEEXISTS(name$) THEN
RETURN name$ & ": " & STR$(FILELEN(name$))
ELSE
RETURN "file " & name$ & " not found"
END IF
END FUNCTION
PRINT printlen$("input.txt")
PRINT printlen$("/input.txt") |
http://rosettacode.org/wiki/File_size | File size | Verify the size of a file called input.txt for a file in the current working directory, and another one in the file system root.
| #Batch_File | Batch File |
@echo off
if not exist "%~1" exit /b 1 & rem If file doesn't exist exit with error code of 1.
for /f %%i in (%~1) do echo %~zi
pause>nul
|
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #AArch64_Assembly | AArch64 Assembly |
/* ARM assembly AARCH64 Raspberry PI 3B */
/* program readwrtFile64.s */
/*******************************************/
/* Constantes file */
/*******************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeConstantesARM64.inc"
.equ TAILLEBUF, 1000
/*********************************/
/* Initialized data */
/*********************************/
.data
szMessErreur: .asciz "Error open input file.\n"
szMessErreur4: .asciz "Error open output file.\n"
szMessErreur1: .asciz "Error close file.\n"
szMessErreur2: .asciz "Error read file.\n"
szMessErreur3: .asciz "Error write output file.\n"
/*************************************************/
szMessCodeErr: .asciz "Error code décimal : @ \n"
szNameFileInput: .asciz "input.txt"
szNameFileOutput: .asciz "output.txt"
/*******************************************/
/* UnInitialized data */
/*******************************************/
.bss
sBuffer: .skip TAILLEBUF
sZoneConv: .skip 24
/**********************************************/
/* -- Code section */
/**********************************************/
.text
.global main
main: // entry of program
mov x0,AT_FDCWD
ldr x1,qAdrszNameFileInput // file name
mov x2,#O_RDWR // flags
mov x3,#0 // mode
mov x8,#OPEN // call system OPEN
svc #0
cmp x0,0 // open error ?
ble erreur
mov x19,x0 // save File Descriptor
ldr x1,qAdrsBuffer // buffer address
mov x2,TAILLEBUF // buffer size
mov x8,READ // call system READ
svc 0
cmp x0,0 // read error ?
ble erreur2
mov x20,x0 // length read characters
// close imput file
mov x0,x19 // Fd
mov x8,CLOSE // call system CLOSE
svc 0
cmp x0,0 // close error ?
blt erreur1
// create output file
mov x0,AT_FDCWD
ldr x1,qAdrszNameFileOutput // file name
mov x2,O_CREAT|O_RDWR // flags
ldr x3,qFicMask1 // Mode
mov x8,OPEN // call system open file
svc 0
cmp x0,#0 // create error ?
ble erreur4
mov x19,x0 // file descriptor
ldr x1,qAdrsBuffer
mov x2,x20 // length to write
mov x8, #WRITE // select system call 'write'
svc #0 // perform the system call
cmp x0,#0 // error write ?
blt erreur3
// close output file
mov x0,x19 // Fd fichier
mov x8, #CLOSE // call system CLOSE
svc #0
cmp x0,#0 // error close ?
blt erreur1
mov x0,#0 // return code OK
b 100f
erreur:
ldr x1,qAdrszMessErreur
bl displayError
mov x0,#1 // error return code
b 100f
erreur1:
ldr x1,qAdrszMessErreur1
bl displayError
mov x0,#1 // error return code
b 100f
erreur2:
ldr x1,qAdrszMessErreur2
bl displayError
mov x0,#1 // error return code
b 100f
erreur3:
ldr x1,qAdrszMessErreur3
bl displayError
mov x0,#1 // error return code
b 100f
erreur4:
ldr x1,qAdrszMessErreur4
bl displayError
mov x0,#1 // error return code
b 100f
100: // end program
mov x8,EXIT
svc 0
qAdrszNameFileInput: .quad szNameFileInput
qAdrszNameFileOutput: .quad szNameFileOutput
qAdrszMessErreur: .quad szMessErreur
qAdrszMessErreur1: .quad szMessErreur1
qAdrszMessErreur2: .quad szMessErreur2
qAdrszMessErreur3: .quad szMessErreur3
qAdrszMessErreur4: .quad szMessErreur4
qAdrsBuffer: .quad sBuffer
qFicMask1: .quad 0644
/******************************************************************/
/* display error message */
/******************************************************************/
/* x0 contains error code */
/* x1 contains address error message */
displayError:
stp x2,lr,[sp,-16]! // save registers
mov x2,x0 // save error code
mov x0,x1 // display message error
bl affichageMess
mov x0,x2
ldr x1,qAdrsZoneConv // conversion error code
bl conversion10S // decimal conversion
ldr x0,qAdrszMessCodeErr
ldr x1,qAdrsZoneConv
bl strInsertAtCharInc // insert result at @ character
bl affichageMess // display message final
ldp x2,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
qAdrsZoneConv: .quad sZoneConv
qAdrszMessCodeErr: .quad szMessCodeErr
/********************************************************/
/* File Include fonctions */
/********************************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeARM64.inc"
|
http://rosettacode.org/wiki/File_input/output | File input/output | File input/output is part of Short Circuit's Console Program Basics selection.
Task
Create a file called "output.txt", and place in it the contents of the file "input.txt", via an intermediate variable.
In other words, your program will demonstrate:
how to read from a file into a variable
how to write a variable's contents into a file
Oneliners that skip the intermediate variable are of secondary interest — operating systems have copy commands for that.
| #ACL2 | ACL2 | :set-state-ok t
(defun read-channel (channel limit state)
(mv-let (ch state)
(read-char$ channel state)
(if (or (null ch)
(zp limit))
(let ((state (close-input-channel channel state)))
(mv nil state))
(mv-let (so-far state)
(read-channel channel (1- limit) state)
(mv (cons ch so-far) state)))))
(defun read-from-file (filename limit state)
(mv-let (channel state)
(open-input-channel filename :character state)
(mv-let (contents state)
(read-channel channel limit state)
(mv (coerce contents 'string) state))))
(defun write-channel (channel cs state)
(if (endp cs)
(close-output-channel channel state)
(let ((state (write-byte$ (char-code (first cs))
channel state)))
(let ((state (write-channel channel
(rest cs)
state)))
state))))
(defun write-to-file (filename str state)
(mv-let (channel state)
(open-output-channel filename :byte state)
(write-channel channel (coerce str 'list) state)))
(defun copy-file (in out state)
(mv-let (contents state)
(read-from-file in (expt 2 40) state)
(write-to-file out contents state))) |
http://rosettacode.org/wiki/File_extension_is_in_extensions_list | File extension is in extensions list | File extension is in extensions list
You are encouraged to solve this task according to the task description, using any language you may know.
Filename extensions are a rudimentary but commonly used way of identifying files types.
Task
Given an arbitrary filename and a list of extensions, tell whether the filename has one of those extensions.
Notes:
The check should be case insensitive.
The extension must occur at the very end of the filename, and be immediately preceded by a dot (.).
You may assume that none of the given extensions are the empty string, and none of them contain a dot. Other than that they may be arbitrary strings.
Extra credit:
Allow extensions to contain dots. This way, users of your function/program have full control over what they consider as the extension in cases like:
archive.tar.gz
Please state clearly whether or not your solution does this.
Test cases
The following test cases all assume this list of extensions: zip, rar, 7z, gz, archive, A##
Filename
Result
MyData.a##
true
MyData.tar.Gz
true
MyData.gzip
false
MyData.7z.backup
false
MyData...
false
MyData
false
If your solution does the extra credit requirement, add tar.bz2 to the list of extensions, and check the following additional test cases:
Filename
Result
MyData_v1.0.tar.bz2
true
MyData_v1.0.bz2
false
Motivation
Checking if a file is in a certain category of file formats with known extensions (e.g. archive files, or image files) is a common problem in practice, and may be approached differently from extracting and outputting an arbitrary extension (see e.g. FileNameExtensionFilter in Java).
It also requires less assumptions about the format of an extension, because the calling code can decide what extensions are valid.
For these reasons, this task exists in addition to the Extract file extension task.
Related tasks
Extract file extension
String matching
| #Arturo | Arturo | fileExtensions: map ["zip" "rar" "7z" "gz" "archive" "A##"] => ["." ++ lower]
hasExtension?: function [file][
in? extract.extension lower file
fileExtensions
]
files: ["MyData.a##" "MyData.tar.Gz" "MyData.gzip" "MyData.7z.backup" "MyData..." "MyData"]
loop files 'file ->
print [file "=> hasExtension?:" hasExtension? file] |
http://rosettacode.org/wiki/File_extension_is_in_extensions_list | File extension is in extensions list | File extension is in extensions list
You are encouraged to solve this task according to the task description, using any language you may know.
Filename extensions are a rudimentary but commonly used way of identifying files types.
Task
Given an arbitrary filename and a list of extensions, tell whether the filename has one of those extensions.
Notes:
The check should be case insensitive.
The extension must occur at the very end of the filename, and be immediately preceded by a dot (.).
You may assume that none of the given extensions are the empty string, and none of them contain a dot. Other than that they may be arbitrary strings.
Extra credit:
Allow extensions to contain dots. This way, users of your function/program have full control over what they consider as the extension in cases like:
archive.tar.gz
Please state clearly whether or not your solution does this.
Test cases
The following test cases all assume this list of extensions: zip, rar, 7z, gz, archive, A##
Filename
Result
MyData.a##
true
MyData.tar.Gz
true
MyData.gzip
false
MyData.7z.backup
false
MyData...
false
MyData
false
If your solution does the extra credit requirement, add tar.bz2 to the list of extensions, and check the following additional test cases:
Filename
Result
MyData_v1.0.tar.bz2
true
MyData_v1.0.bz2
false
Motivation
Checking if a file is in a certain category of file formats with known extensions (e.g. archive files, or image files) is a common problem in practice, and may be approached differently from extracting and outputting an arbitrary extension (see e.g. FileNameExtensionFilter in Java).
It also requires less assumptions about the format of an extension, because the calling code can decide what extensions are valid.
For these reasons, this task exists in addition to the Extract file extension task.
Related tasks
Extract file extension
String matching
| #AWK | AWK |
# syntax: GAWK -f FILE_EXTENSION_IS_IN_EXTENSIONS_LIST.AWK
BEGIN {
n = split("zip,rar,7z,gz,archive,A##,tar.bz2", arr, ",")
for (i=1; i<=n; i++) {
ext_arr[tolower(arr[i])] = ""
}
filenames = "MyData.a##,MyData.tar.Gz,MyData.gzip,MyData.7z.backup,MyData...,MyData,MyData_v1.0.tar.bz2,MyData_v1.0.bz2"
n = split(filenames, fn_arr, ",")
for (i=1; i<=n; i++) {
ext_found = ""
for (ext in ext_arr) {
if (tolower(fn_arr[i]) ~ (".*\\." ext "$")) {
ext_found = ext
break
}
}
ans = (ext_found == "") ? "is not in list" : ("is in list: " ext_found)
printf("%s extension %s\n", fn_arr[i], ans)
}
exit(0)
} |
http://rosettacode.org/wiki/File_modification_time | File modification time | Task
Get and set the modification time of a file.
| #D | D | import std.stdio;
import std.file: getTimes, setTimes, SysTime;
void main() {
auto fname = "unixdict.txt";
SysTime fileAccessTime, fileModificationTime;
getTimes(fname, fileAccessTime, fileModificationTime);
writeln(fileAccessTime, "\n", fileModificationTime);
setTimes(fname, fileAccessTime, fileModificationTime);
} |
http://rosettacode.org/wiki/File_modification_time | File modification time | Task
Get and set the modification time of a file.
| #Delphi | Delphi | function GetModifiedDate(const aFilename: string): TDateTime;
var
hFile: Integer;
iDosTime: Integer;
begin
hFile := FileOpen(aFilename, fmOpenRead);
iDosTime := FileGetDate(hFile);
FileClose(hFile);
if (hFile = -1) or (iDosTime = -1) then raise Exception.Create('Cannot read file: ' + sFilename);
Result := FileDateToDateTime(iDosTime);
end;
procedure ChangeModifiedDate(const aFilename: string; aDateTime: TDateTime);
begin
FileSetDate(aFileName, DateTimeToFileDate(aDateTime));
end; |
http://rosettacode.org/wiki/File_size_distribution | File size distribution | Task
Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.
My suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.
Don't forget that empty files may exist, to serve as a marker.
Is your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?
| #Python | Python | import sys, os
from collections import Counter
def dodir(path):
global h
for name in os.listdir(path):
p = os.path.join(path, name)
if os.path.islink(p):
pass
elif os.path.isfile(p):
h[os.stat(p).st_size] += 1
elif os.path.isdir(p):
dodir(p)
else:
pass
def main(arg):
global h
h = Counter()
for dir in arg:
dodir(dir)
s = n = 0
for k, v in sorted(h.items()):
print("Size %d -> %d file(s)" % (k, v))
n += v
s += k * v
print("Total %d bytes for %d files" % (s, n))
main(sys.argv[1:]) |
http://rosettacode.org/wiki/File_size_distribution | File size distribution | Task
Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.
My suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.
Don't forget that empty files may exist, to serve as a marker.
Is your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?
| #Racket | Racket | #lang racket
(define (file-size-distribution (d (current-directory)) #:size-group-function (sgf values))
(for/fold ((rv (hash)) (Σ 0) (n 0)) ((f (in-directory d)) #:when (file-exists? f))
(define sz (file-size f))
(values (hash-update rv (sgf sz) add1 0) (+ Σ sz) (add1 n))))
(define (log10-or-so x) (if (zero? x) #f (round (/ (log x) (log 10)))))
(define number-maybe-<
(match-lambda** [(#f #f) #f]
[(#f _) #t]
[(_ #f) #f]
[(a b) (< a b)]))
(define ...s? (match-lambda** [(one 1) one] [(one n) (string-append one "s")]))
(define ((report-fsd f) fsd Σ n)
(for/list ((k (in-list (sort (hash-keys fsd) number-maybe-<))))
(printf "~a(size): ~a -> ~a ~a~%"
(object-name f)
k
(hash-ref fsd k) (...s? "file" (hash-ref fsd k))))
(printf "Total: ~a ~a in ~a ~a~%" Σ (...s? "byte" Σ) n (...s? "file" n)))
(module+ test
(call-with-values (λ () (file-size-distribution #:size-group-function log10-or-so))
(report-fsd log10-or-so))) |
http://rosettacode.org/wiki/Fibonacci_word/fractal | Fibonacci word/fractal |
The Fibonacci word may be represented as a fractal as described here:
(Clicking on the above website (hal.archives-ouvertes.fr) will leave a cookie.)
For F_wordm start with F_wordCharn=1
Draw a segment forward
If current F_wordChar is 0
Turn left if n is even
Turn right if n is odd
next n and iterate until end of F_word
Task
Create and display a fractal similar to Fig 1.
(Clicking on the above website (hal.archives-ouvertes.fr) will leave a cookie.)
| #AutoHotkey | AutoHotkey | #NoEnv
SetBatchLines, -1
p := 0.3 ; Segment length (pixels)
F_Word := 30
SysGet, Mon, MonitorWorkArea
W := FibWord(F_Word)
d := 1
x1 := 0
y1 := MonBottom
Width := A_ScreenWidth
Height := A_ScreenHeight
If (!pToken := Gdip_Startup()) {
MsgBox, 48, Gdiplus Error!, Gdiplus failed to start. Please ensure you have Gdiplus on your system.
ExitApp
}
OnExit, Shutdown
Gui, 1: -Caption +E0x80000 +LastFound +AlwaysOnTop +ToolWindow +OwnDialogs
Gui, 1: Show, NA
hwnd1 := WinExist()
hbm := CreateDIBSection(Width, Height)
hdc := CreateCompatibleDC()
obm := SelectObject(hdc, hbm)
G := Gdip_GraphicsFromHDC(hdc)
Gdip_SetSmoothingMode(G, 4)
pPen := Gdip_CreatePen(0xffff0000, 1)
Loop, Parse, W
{
if (d = 0)
x2 := x1 + p, y2 := y1
else if (d = 1 || d = -3)
x2 := x1, y2 := y1 - p
else if (d = 2 || d = -2)
x2 := x1 - p, y2 := y1
else if (d = 3 || d = -1)
x2 := x1, y2 := y1 + p
Gdip_DrawLine(G, pPen, x1, y1, x2, y2)
if (!Mod(A_Index, 1500))
UpdateLayeredWindow(hwnd1, hdc, 0, 0, Width, Height)
if (A_LoopField = 0) {
if (!Mod(A_Index, 2))
d += 1
else
d -= 1
}
x1 := x2, y1 := y2, d := Mod(d, 4)
}
Gdip_DeletePen(pPen)
UpdateLayeredWindow(hwnd1, hdc, 0, 0, Width, Height)
SelectObject(hdc, obm)
DeleteObject(hbm)
DeleteDC(hdc)
Gdip_DeleteGraphics(G)
return
FibWord(n, FW1=1, FW2=0) {
Loop, % n - 2
FW3 := FW2 FW1, FW1 := FW2, FW2 := FW3
return FW3
}
Esc::
Shutdown:
Gdip_DeletePen(pPen)
SelectObject(hdc, obm)
DeleteObject(hbm)
DeleteDC(hdc)
Gdip_DeleteGraphics(G)
Gdip_Shutdown(pToken)
ExitApp |
http://rosettacode.org/wiki/Find_common_directory_path | Find common directory path | Create a routine that, given a set of strings representing directory paths and a single character directory separator, will return a string representing that part of the directory tree that is common to all the directories.
Test your routine using the forward slash '/' character as the directory separator and the following three strings as input paths:
'/home/user1/tmp/coverage/test'
'/home/user1/tmp/covert/operator'
'/home/user1/tmp/coven/members'
Note: The resultant path should be the valid directory '/home/user1/tmp' and not the longest common string '/home/user1/tmp/cove'.
If your language has a routine that performs this function (even if it does not have a changeable separator character), then mention it as part of the task.
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.string, std.algorithm, std.path, std.array;
string commonDirPath(in string[] paths, in string sep = "/") pure {
if (paths.empty)
return null;
return paths.map!(p => p.split(sep)).reduce!commonPrefix.join(sep);
}
void main() {
immutable paths = ["/home/user1/tmp/coverage/test",
"/home/user1/tmp/covert/operator",
"/home/user1/tmp/coven/members"];
writeln(`The common path is: "`, paths.commonDirPath, '"');
} |
http://rosettacode.org/wiki/Filter | Filter | Task
Select certain elements from an Array into a new Array in a generic way.
To demonstrate, select all even numbers from an Array.
As an option, give a second solution which filters destructively,
by modifying the original Array rather than creating a new Array.
| #AntLang | AntLang | x:range[100]
{1- x mod 2}hfilter x |
http://rosettacode.org/wiki/Find_if_a_point_is_within_a_triangle | Find if a point is within a triangle | Find if a point is within a triangle.
Task
Assume points are on a plane defined by (x, y) real number coordinates.
Given a point P(x, y) and a triangle formed by points A, B, and C, determine if P is within triangle ABC.
You may use any algorithm.
Bonus: explain why the algorithm you chose works.
Related tasks
Determine_if_two_triangles_overlap
Also see
Discussion of several methods. [[1]]
Determine if a point is in a polygon [[2]]
Triangle based coordinate systems [[3]]
Wolfram entry [[4]]
| #Phix | Phix | with javascript_semantics
constant p0 = {0,0},
p1 = {0,1},
p2 = {3,1},
triangle = {{3/2, 12/5}, {51/10, -31/10}, {-19/5, 1/2}}
function inside(sequence p)
return sort(convex_hull({p}&triangle))==sort(deep_copy(triangle))
end function
printf(1,"Point %v is with triangle %v?:%t\n",{p0,triangle,inside(p0)})
printf(1,"Point %v is with triangle %v?:%t\n",{p1,triangle,inside(p1)})
printf(1,"Point %v is with triangle %v?:%t\n",{p2,triangle,inside(p2)})
|
http://rosettacode.org/wiki/Flatten_a_list | Flatten a list | Task
Write a function to flatten the nesting in an arbitrary list of values.
Your program should work on the equivalent of this list:
[[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
Where the correct result would be the list:
[1, 2, 3, 4, 5, 6, 7, 8]
Related task
Tree traversal
| #Wren | Wren | import "/seq" for Lst
var a = [[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]
System.print(Lst.flatten(a)) |
http://rosettacode.org/wiki/Find_limit_of_recursion | Find limit of recursion | Find limit of recursion is part of Short Circuit's Console Program Basics selection.
Task
Find the limit of recursion.
| #Go | Go | package main
import (
"flag"
"fmt"
"runtime/debug"
)
func main() {
stack := flag.Int("stack", 0, "maximum per goroutine stack size or 0 for the default")
flag.Parse()
if *stack > 0 {
debug.SetMaxStack(*stack)
}
r(1)
}
func r(l int) {
if l%1000 == 0 {
fmt.Println(l)
}
r(l + 1)
} |
http://rosettacode.org/wiki/Find_limit_of_recursion | Find limit of recursion | Find limit of recursion is part of Short Circuit's Console Program Basics selection.
Task
Find the limit of recursion.
| #Gri | Gri | `Recurse'
{
show .depth.
.depth. = {rpn .depth. 1 +}
Recurse
}
.depth. = 1
Recurse |
http://rosettacode.org/wiki/Find_palindromic_numbers_in_both_binary_and_ternary_bases | Find palindromic numbers in both binary and ternary bases | Find palindromic numbers in both binary and ternary bases
You are encouraged to solve this task according to the task description, using any language you may know.
Task
Find and show (in decimal) the first six numbers (non-negative integers) that are palindromes in both:
base 2
base 3
Display 0 (zero) as the first number found, even though some other definitions ignore it.
Optionally, show the decimal number found in its binary and ternary form.
Show all output here.
It's permissible to assume the first two numbers and simply list them.
See also
Sequence A60792, numbers that are palindromic in bases 2 and 3 on The On-Line Encyclopedia of Integer Sequences.
| #Phix | Phix | with javascript_semantics
-- widths and limits for 32/64 bit running (see output below):
constant {dsize,w3,w2,limit} = iff(machine_bits()=32?{12,23,37,6}
:{18,37,59,7}),
-- [atoms on 32-bit have only 53 bits of precision, but 7th ^^^^ requires 59]
dfmt = sprintf("%%%dd",dsize), -- ie "%12d" or "%18d"
esc = #1B
function center(string s, integer l)
l = max(0,floor((l-length(s))/2))
string space = repeat(' ',l)
s = space & s & space
return s
end function
integer count = 1
procedure show(atom n, string p2, p3)
if count=1 then
printf(1," %s %s %s\n",{pad_head("decimal",dsize),center("ternary",w3),center(" binary",w2)})
end if
string ns = sprintf(dfmt,n)
printf(1,"%2d: %s %s %s\n",{count, ns, center(p3,w3), center(p2,w2)})
count += 1
end procedure
procedure progress64(string e, p2, p3)
e = pad_head(e,dsize)
printf(1,"--: %s %s %s\r",{e, center(p3,w3), center(p2,w2)})
end procedure
function to_base(atom i, integer base)
string s = ""
while i>0 do
s = append(s,remainder(i,base)+'0')
i = floor(i/base)
end while
s = reverse(s)
if s="" then
s = "0"
end if
return s
end function
function from_base(string s, integer base)
atom res = 0
for i=1 to length(s) do
res = res*base+s[i]-'0'
end for
return res
end function
function sn(string s, integer f, base)
-- helper function, return s mirrored (if f!=0)
-- and as a (decimal) number (if base!=0)
-- all returns from next_palindrome() get fed through here.
if f then
s[f+2..$] = reverse(s[1..f])
end if
atom n = iff(base?from_base(s,base):0)
return {s,n}
end function
function next_palindrome(integer base, object s)
--
-- base is 2 or 3
-- s is not usually a palindrome, but derived from one in <5-base>
--
-- all done with very obvious string manipulations, plus a few
-- less obvious optimisations (odd length, middle 1 in base 3).
--
-- example: next_palindrome(2,"10001000100") -> "10001010001"
--
if not string(s) then s = to_base(s,base) end if
integer l = length(s),
f = floor(l/2),
m = f+1, c
if mod(l,2) then -- optimisation: palindromes must be odd-length
-- 1) is a plain mirror greater? (as in the example just given)
{string r} = sn(s,f,0)
-- optimisation: base 3 palindromes have '1' in the middle
if base=3 and r[m]!='1' then r[m] = '1' end if
if r>s then return sn(r,0,base) end if
-- 2) can we (just) increment the middle digit?
c = s[m]-'0'+1
if base=2 or c=1 then
if c<base then
s[m] = c+'0'
return sn(s,f,base)
end if
s[m] = '0'
elsif base=3 then
s[m] = '1'
end if
-- 3) can we increment left half (or is it all <base-1>s?)
for i=f to 1 by -1 do
if s[i]<base-1+'0' then
s[i] += 1
return sn(s,f,base)
else
s[i] = '0'
end if
end for
l += 2 -- (stay odd)
else
l += 1 -- (even->odd)
end if
-- 4) well then, next palindrome is longer, 1000..0001-style
s = sprintf("1%s1",{repeat('0',l-2)})
-- optimisation: base 3 palindromes have '1' in the middle
if base=3 then
m = (l+1)/2
s[m] = '1'
end if
return sn(s,0,base)
end function
string p2 = "0", p3 = "0" -- palindromes as strings in base 2 and 3
atom n2 = 0, n3 = 0, -- decimal equivalents of the above.
t0 = time(),
t1 = time()+1
while count<=limit do
if n2=n3 then
show(n2,p2,p3)
{p2,n2} = next_palindrome(2,p2)
{p3,n3} = next_palindrome(3,p3)
elsif n2<n3 then
{p2,n2} = next_palindrome(2,n3-1)
elsif n2>n3 then
{p3,n3} = next_palindrome(3,n2-1)
end if
if time()>t1 and platform()!=JS then
progress64(elapsed_short(time()-t0),p2,p3)
t1 = time()+1
if find(get_key(),{'q','Q',esc}) then exit end if
end if
end while
?elapsed(time()-t0)
|
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