Rep-string: Difference between revisions
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1: no rep string</pre> |
1: no rep string</pre> |
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=={{header|APL}}== |
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{{works with|Dyalog APL}} |
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This function returns a list of all possible repeated substrings. |
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It returns the empty list if there are none. |
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<lang APL>rep ← ⊢ (⊢(/⍨)(⊂⊣)≡¨(≢⊣)⍴¨⊢) ⍳∘(⌊0.5×≢)↑¨⊂</lang> |
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{{out}} |
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<pre> rep '1001110011' |
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10011 |
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rep '1110111011' |
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1110 |
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rep '0010010010' |
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001 |
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rep '1010101010' |
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10 1010 |
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rep '1111111111' |
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1 11 111 1111 11111 |
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rep '0100101101' |
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rep '0100100' |
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010 |
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rep '101' |
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rep '11' |
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1 |
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rep '00' |
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0 |
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rep ,'1' |
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</pre> |
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=={{header|AppleScript}}== |
=={{header|AppleScript}}== |
Revision as of 21:10, 9 May 2021
You are encouraged to solve this task according to the task description, using any language you may know.
Given a series of ones and zeroes in a string, define a repeated string or rep-string as a string which is created by repeating a substring of the first N characters of the string truncated on the right to the length of the input string, and in which the substring appears repeated at least twice in the original.
For example, the string 10011001100 is a rep-string as the leftmost four characters of 1001 are repeated three times and truncated on the right to give the original string.
Note that the requirement for having the repeat occur two or more times means that the repeating unit is never longer than half the length of the input string.
- Task
- Write a function/subroutine/method/... that takes a string and returns an indication of if it is a rep-string and the repeated string. (Either the string that is repeated, or the number of repeated characters would suffice).
- There may be multiple sub-strings that make a string a rep-string - in that case an indication of all, or the longest, or the shortest would suffice.
- Use the function to indicate the repeating substring if any, in the following:
-
1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1
- Show your output on this page.
- Metrics
- 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
11l
<lang 11l>F reps(text)
R (1 .< 1 + text.len I/ 2).filter(x -> @text.starts_with(@text[x..])).map(x -> @text[0 .< x])
V matchstr = ‘1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1’
L(line) matchstr.split("\n")
print(‘'#.' has reps #.’.format(line, reps(line)))</lang>
- Output:
'1001110011' has reps [10011] '1110111011' has reps [1110] '0010010010' has reps [001] '1010101010' has reps [10, 1010] '1111111111' has reps [1, 11, 111, 1111, 11111] '0100101101' has reps [] '0100100' has reps [010] '101' has reps [] '11' has reps [1] '00' has reps [0] '1' has reps []
Ada
<lang Ada>with Ada.Command_Line, Ada.Text_IO, Ada.Strings.Fixed;
procedure Rep_String is
function Find_Largest_Rep_String(S:String) return String is L: Natural := S'Length; begin for I in reverse 1 .. L/2 loop
declare use Ada.Strings.Fixed; T: String := S(S'First .. S'First + I-1); -- the first I characters of S U: String := (1+(L/I)) * T; -- repeat T so often that U'Length >= L begin -- compare first L characers of U with S if U(U'First .. U'First + S'Length -1) = S then return T; -- T is a rep-string end if; end;
end loop; return ""; -- no rep string; end Find_Largest_Rep_String; X: String := Ada.Command_Line.Argument(1); Y: String := Find_Largest_Rep_String(X);
begin
if Y="" then Ada.Text_IO.Put_Line("No rep-string for """ & X & """"); else Ada.Text_IO.Put_Line("Longest rep-string for """& X &""": """& Y &""""); end if;
end Rep_String;</lang>
- Output:
> ./rep_string 1001110011 Longest rep-string for "1001110011": "10011" > ./rep_string 1110111011 Longest rep-string for "1110111011": "1110" > ./rep_string 0010010010 Longest rep-string for "0010010010": "001" > ./rep_string 1010101010 Longest rep-string for "1010101010": "1010" > ./rep_string 1111111111 Longest rep-string for "1111111111": "11111" > ./rep_string 0100101101 No rep-string for "0100101101" > ./rep_string 0100100 Longest rep-string for "0100100": "010" > ./rep_string 101 No rep-string for "101" > ./rep_string 11 Longest rep-string for "11": "1" > ./rep_string 00 Longest rep-string for "00": "0" > ./rep_string 1 No rep-string for "1"
ALGOL 68
<lang algol68># procedure to find the longest rep-string in a given string #
- the input string is not validated to contain only "0" and "1" characters #
PROC longest rep string = ( STRING input )STRING: BEGIN
STRING result := "";
# ensure the string we are working on has a lower-bound of 1 # STRING str = input[ AT 1 ];
# work backwards from half the input string looking for a rep-string # FOR string length FROM UPB str OVER 2 BY -1 TO 1 WHILE STRING left substring = str[ 1 : string length ]; # if the left substgring repeated a sufficient number of times # # (truncated on the right) is equal to the original string, then # # we have found the longest rep-string # STRING repeated string = ( left substring * ( ( UPB str OVER string length ) + 1 ) )[ 1 : UPB str ]; IF str = repeated string THEN # found a rep-string # result := left substring; FALSE ELSE # not a rep-string, keep looking # TRUE FI DO SKIP OD; result
END; # longest rep string #
- test the longest rep string procedure #
main: (
[]STRING tests = ( "1001110011" , "1110111011" , "0010010010" , "1010101010" , "1111111111" , "0100101101" , "0100100" , "101" , "11" , "00" , "1" );
FOR test number FROM LWB tests TO UPB tests DO STRING rep string = longest rep string( tests[ test number ] ); print( ( tests[ test number ] , ": " , IF rep string = "" THEN "no rep string" ELSE "longest rep string: """ + rep string + """" FI , newline ) ) OD
)</lang>
- Output:
1001110011: longest rep string: "10011" 1110111011: longest rep string: "1110" 0010010010: longest rep string: "001" 1010101010: longest rep string: "1010" 1111111111: longest rep string: "11111" 0100101101: no rep string 0100100: longest rep string: "010" 101: no rep string 11: longest rep string: "1" 00: longest rep string: "0" 1: no rep string
APL
This function returns a list of all possible repeated substrings. It returns the empty list if there are none. <lang APL>rep ← ⊢ (⊢(/⍨)(⊂⊣)≡¨(≢⊣)⍴¨⊢) ⍳∘(⌊0.5×≢)↑¨⊂</lang>
- Output:
rep '1001110011' 10011 rep '1110111011' 1110 rep '0010010010' 001 rep '1010101010' 10 1010 rep '1111111111' 1 11 111 1111 11111 rep '0100101101' rep '0100100' 010 rep '101' rep '11' 1 rep '00' 0 rep ,'1'
AppleScript
<lang AppleScript>------------------------REP-CYCLES-------------------------
-- repCycles :: String -> [String] on repCycles(xs)
set n to length of xs script isCycle on |λ|(cs) xs = takeCycle(n, cs) end |λ| end script filter(isCycle, tail(inits(take(quot(n, 2), xs))))
end repCycles
-- cycleReport :: String -> [String] on cycleReport(xs)
set reps to repCycles(xs) if isNull(reps) then {xs, "(n/a)"} else {xs, item -1 of reps} end if
end cycleReport
TEST----------------------------
on run
set samples to {"1001110011", "1110111011", "0010010010", ¬ "1010101010", "1111111111", "0100101101", "0100100", ¬ "101", "11", "00", "1"} unlines(cons("Longest cycle:" & linefeed, ¬ map(intercalate(" -> "), ¬ map(cycleReport, samples))))
end run
GENERIC FUNCTIONS---------------------
-- concat :: a -> [a] | [String] -> String on concat(xs)
if length of xs > 0 and class of (item 1 of xs) is string then set acc to "" else set acc to {} end if repeat with i from 1 to length of xs set acc to acc & item i of xs end repeat acc
end concat
-- cons :: a -> [a] -> [a] on cons(x, xs)
{x} & xs
end cons
-- filter :: (a -> Bool) -> [a] -> [a] on filter(f, xs)
tell mReturn(f) set lst to {} set lng to length of xs repeat with i from 1 to lng set v to item i of xs if |λ|(v, i, xs) then set end of lst to v end repeat return lst end tell
end filter
-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)
tell mReturn(f) set v to startValue set lng to length of xs repeat with i from 1 to lng set v to |λ|(v, item i of xs, i, xs) end repeat return v end tell
end foldl
-- inits :: [a] -> a -- inits :: String -> [String] on inits(xs)
script elemInit on |λ|(_, i, xs) items 1 thru i of xs end |λ| end script script charInit on |λ|(_, i, xs) text 1 thru i of xs end |λ| end script if class of xs is string then {""} & map(charInit, xs) else {{}} & map(elemInit, xs) end if
end inits
-- intercalate :: Text -> [Text] -> Text on intercalate(strText)
script on |λ|(xs) set {dlm, my text item delimiters} to {my text item delimiters, strText} set strJoined to xs as text set my text item delimiters to dlm return strJoined end |λ| end script
end intercalate
-- map :: (a -> b) -> [a] -> [b] on map(f, xs)
tell mReturn(f) set lng to length of xs set lst to {} repeat with i from 1 to lng set end of lst to |λ|(item i of xs, i, xs) end repeat return lst end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then f else script property |λ| : f end script end if
end mReturn
-- isNull :: [a] -> Bool on isNull(xs)
xs = {}
end isNull
-- min :: Ord a => a -> a -> a on min(x, y)
if y < x then y else x end if
end min
-- quot :: Integral a => a -> a -> a on quot(n, m)
n div m
end quot
-- replicate :: Int -> a -> [a] on replicate(n, a)
set out to {} if n < 1 then return out set dbl to {a} repeat while (n > 1) if (n mod 2) > 0 then set out to out & dbl set n to (n div 2) set dbl to (dbl & dbl) end repeat return out & dbl
end replicate
-- tail :: [a] -> [a] on tail(xs)
if length of xs > 1 then items 2 thru -1 of xs else {} end if
end tail
-- take :: Int -> [a] -> [a] on take(n, xs)
if class of xs is string then if n > 0 then text 1 thru min(n, length of xs) of xs else "" end if else if n > 0 then items 1 thru min(n, length of xs) of xs else {} end if end if
end take
-- takeCycle :: Int -> [a] -> [a] on takeCycle(n, xs)
set lng to length of xs if lng ≥ n then set cycle to xs else set cycle to concat(replicate((n div lng) + 1, xs)) end if if class of xs is string then items 1 thru n of cycle as string else items 1 thru n of cycle end if
end takeCycle
-- unlines :: [String] -> String on unlines(xs)
|λ|(xs) of intercalate(linefeed)
end unlines</lang>
- Output:
Longest cycle: 1001110011 -> 10011 1110111011 -> 1110 0010010010 -> 001 1010101010 -> 1010 1111111111 -> 11111 0100101101 -> (n/a) 0100100 -> 010 101 -> (n/a) 11 -> 1 00 -> 0 1 -> (n/a)
Arturo
<lang rebol>repeated?: function [text][
loop ((size text)/2)..0 'x [ if prefix? text slice text x (size text)-1 [ (x>0)? -> return slice text 0 x-1 -> return false ] ] return false
]
strings: {
1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1
}
loop split.lines strings 'str [
rep: repeated? str if? false = rep -> print [str "-> *not* a rep-string"] else -> print [str "->" rep "( length:" size rep ")"]
]</lang>
- Output:
1001110011 -> 10011 ( length: 5 ) 1110111011 -> 1110 ( length: 4 ) 0010010010 -> 001 ( length: 3 ) 1010101010 -> 1010 ( length: 4 ) 1111111111 -> 11111 ( length: 5 ) 0100101101 -> *not* a rep-string 0100100 -> 010 ( length: 3 ) 101 -> *not* a rep-string 11 -> 1 ( length: 1 ) 00 -> 0 ( length: 1 ) 1 -> *not* a rep-string
AutoHotkey
<lang AutoHotkey>In := ["1001110011", "1110111011", "0010010010", "1010101010"
, "1111111111", "0100101101", "0100100", "101", "11", "00", "1"]
for k, v in In Out .= RepString(v) "`t" v "`n" MsgBox, % Out
RepString(s) { Loop, % StrLen(s) // 2 { i := A_Index Loop, Parse, s { pos := Mod(A_Index, i) if (A_LoopField != SubStr(s, !pos ? i : pos, 1)) continue, 2 } return SubStr(s, 1, i) } return "N/A" }</lang>
- Output:
10011 1001110011 1110 1110111011 001 0010010010 10 1010101010 1 1111111111 N/A 0100101101 010 0100100 N/A 101 1 11 0 00 N/A 1
BaCon
<lang freebasic>all$ = "1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1"
FOR word$ IN all$
FOR x = LEN(word$)/2 DOWNTO 1
ex$ = EXPLODE$(word$, x)
FOR st$ IN UNIQ$(ex$) IF NOT(REGEX(HEAD$(ex$, 1), "^" & st$)) THEN CONTINUE 2 NEXT
PRINT "Repeating string: ", word$, " -> ", HEAD$(ex$, 1) CONTINUE 2 NEXT
PRINT "Not a repeating string: ", word$
NEXT</lang>
- Output:
Repeating string: 1001110011 -> 10011 Repeating string: 1110111011 -> 1110 Repeating string: 0010010010 -> 001 Repeating string: 1010101010 -> 1010 Repeating string: 1111111111 -> 11111 Not a repeating string: 0100101101 Repeating string: 0100100 -> 010 Not a repeating string: 101 Repeating string: 11 -> 1 Repeating string: 00 -> 0 Not a repeating string: 1
Bracmat
<lang bracmat>( ( rep-string
= reps L x y . ( reps = x y z . !arg:(?x.?y) & ( @(!y:!x ?z)&reps$(!x.!z) | @(!x:!y ?) ) ) & ( :?L & @( !arg : %?x !x ( ?y & reps$(!x.!y) & !x !L:?L & ~ ) ) | !L: & out$(str$(!arg " is not a rep-string")) | out$(!arg ":" !L) ) )
& rep-string$1001110011 & rep-string$1110111011 & rep-string$0010010010 & rep-string$1010101010 & rep-string$1111111111 & rep-string$0100101101 & rep-string$0100100 & rep-string$101 & rep-string$11 & rep-string$00 & rep-string$1 );</lang>
- Output:
1001110011 : 10011 1110111011 : 1110 0010010010 : 001 1010101010 : 1010 10 1111111111 : 11111 1111 111 11 1 0100101101 is not a rep-string 0100100 : 010 101 is not a rep-string 11 : 1 00 : 0 1 is not a rep-string
C
Longest substring
<lang c>
- include <stdio.h>
- include <string.h>
int repstr(char *str) {
if (!str) return 0;
size_t sl = strlen(str) / 2; while (sl > 0) { if (strstr(str, str + sl) == str) return sl; --sl; }
return 0;
}
int main(void) {
char *strs[] = { "1001110011", "1110111011", "0010010010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1" };
size_t strslen = sizeof(strs) / sizeof(strs[0]); size_t i; for (i = 0; i < strslen; ++i) { int n = repstr(strs[i]); if (n) printf("\"%s\" = rep-string \"%.*s\"\n", strs[i], n, strs[i]); else printf("\"%s\" = not a rep-string\n", strs[i]); }
return 0;
} </lang>
- Output:
"1001110011" = rep-string "10011" "1110111011" = rep-string "1110" "0010010010" = rep-string "001" "1111111111" = rep-string "11111" "0100101101" = not a rep-string "0100100" = rep-string "010" "101" = not a rep-string "11" = rep-string "1" "00" = rep-string "0" "1" = not a rep-string
shortest substring
<lang c> // strstr : Returns a pointer to the first occurrence of str2 in str1, or a null pointer if str2 is not part of str1. // size_t is an unsigned integer typ // lokks for the shortest substring int repstr(char *str) {
if (!str) return 0; // if empty input size_t sl = 1; size_t sl_max = strlen(str) ; while (sl < sl_max) { if (strstr(str, str + sl) == str) // How it works ???? It checks the whole string str return sl; ++sl; } return 0;
} </lang>
C++
<lang cpp>#include <string>
- include <vector>
- include <boost/regex.hpp>
bool is_repstring( const std::string & teststring , std::string & repunit ) {
std::string regex( "^(.+)\\1+(.*)$" ) ; boost::regex e ( regex ) ; boost::smatch what ; if ( boost::regex_match( teststring , what , e , boost::match_extra ) ) { std::string firstbracket( what[1 ] ) ; std::string secondbracket( what[ 2 ] ) ; if ( firstbracket.length( ) >= secondbracket.length( ) &&
firstbracket.find( secondbracket ) != std::string::npos ) { repunit = firstbracket ;
} } return !repunit.empty( ) ;
}
int main( ) {
std::vector<std::string> teststrings { "1001110011" , "1110111011" , "0010010010" , "1010101010" , "1111111111" , "0100101101" , "0100100" , "101" , "11" , "00" , "1" } ; std::string theRep ; for ( std::string myString : teststrings ) { if ( is_repstring( myString , theRep ) ) {
std::cout << myString << " is a rep string! Here is a repeating string:\n" ; std::cout << theRep << " " ;
} else {
std::cout << myString << " is no rep string!" ;
} theRep.clear( ) ; std::cout << std::endl ; } return 0 ;
}</lang>
- Output:
1001110011 is a rep string! Here is a repeating string: 10011 1110111011 is a rep string! Here is a repeating string: 1110 0010010010 is a rep string! Here is a repeating string: 001 1010101010 is a rep string! Here is a repeating string: 1010 1111111111 is a rep string! Here is a repeating string: 11111 0100101101 is no rep string! 0100100 is a rep string! Here is a repeating string: 010 101 is no rep string! 11 is a rep string! Here is a repeating string: 1 00 is a rep string! Here is a repeating string: 0 1 is no rep string!
Clojure
<lang lisp>(defn rep-string [s]
(let [len (count s) first-half (subs s 0 (/ len 2)) test-group (take-while seq (iterate butlast first-half)) test-reptd (map (comp #(take len %) cycle) test-group)] (some #(= (seq s) %) test-reptd)))</lang>
- Output:
<lang lisp> (def test-strings ["1001110011"
"1110111011" "0010010010" "1010101010" "1111111111" "0100101101" "0100100" "101" "11" "00" "1"])
(map (juxt identity rep-string) test-strings) </lang>
(["1001110011" true] ["1110111011" true] ["0010010010" true] ["1010101010" true] ["1111111111" true] ["0100101101" nil] ["0100100" true] ["101" nil] ["11" true] ["00" true] ["1" nil])
Common Lisp
<lang lisp> (ql:quickload :alexandria) (defun rep-stringv (a-str &optional (max-rotation (floor (/ (length a-str) 2))))
;; Exit condition if no repetition found. (cond ((< max-rotation 1) "Not a repeating string") ;; Two checks: ;; 1. Truncated string must be equal to rotation by repetion size. ;; 2. Remaining chars (rest-str) are identical to starting chars (beg-str) ((let* ((trunc (* max-rotation (truncate (length a-str) max-rotation))) (truncated-str (subseq a-str 0 trunc)) (rest-str (subseq a-str trunc)) (beg-str (subseq a-str 0 (rem (length a-str) max-rotation)))) (and (string= beg-str rest-str) (string= (alexandria:rotate (copy-seq truncated-str) max-rotation) truncated-str))) ;; If both checks pass, return the repeting string. (subseq a-str 0 max-rotation)) ;; Recurse function reducing length of rotation. (t (rep-stringv a-str (1- max-rotation)))))
</lang>
- Output:
<lang lisp> (setf test-strings '("1001110011"
"1110111011" "0010010010" "1010101010" "1111111111" "0100101101" "0100100" "101" "11" "00" "1" ))
(loop for item in test-strings
collecting (cons item (rep-stringv item)))
</lang>
(("1001110011" . "10011") ("1110111011" . "1110") ("0010010010" . "001") ("1010101010" . "1010") ("1111111111" . "11111") ("0100101101" . "Not a repeating string") ("0100100" . "010") ("101" . "Not a repeating string") ("11" . "1") ("00" . "0") ("1" . "Not a repeating string"))
Crystal
<lang ruby>def rep(s : String) : Int32
x = s.size // 2
while x > 0 return x if s.starts_with? s[x..] x -= 1 end
0
end
def main
%w( 1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1 ).each do |s| n = rep s puts n > 0 ? "\"#{s}\" #{n} rep-string \"#{s[..(n - 1)]}\"" : "\"#{s}\" not a rep-string" end
end
main </lang>
- Output:
"1001110011" 5 rep-string "10011" "1110111011" 4 rep-string "1110" "0010010010" 3 rep-string "001" "1010101010" 4 rep-string "1010" "1111111111" 5 rep-string "11111" "0100101101" not a rep-string "0100100" 3 rep-string "010" "101" not a rep-string "11" 1 rep-string "1" "00" 1 rep-string "0" "1" not a rep-string
D
Two different algorithms. The second is from the Raku entry. <lang d>import std.stdio, std.string, std.conv, std.range, std.algorithm,
std.ascii, std.typecons;
Nullable!(size_t, 0) repString1(in string s) pure nothrow @safe @nogc in {
//assert(s.all!isASCII); assert(s.representation.all!isASCII);
} body {
immutable sr = s.representation; foreach_reverse (immutable n; 1 .. sr.length / 2 + 1) if (sr.take(n).cycle.take(sr.length).equal(sr)) return typeof(return)(n); return typeof(return)();
}
Nullable!(size_t, 0) repString2(in string s) pure @safe /*@nogc*/ in {
assert(s.countchars("01") == s.length);
} body {
immutable bits = s.to!ulong(2);
foreach_reverse (immutable left; 1 .. s.length / 2 + 1) { immutable right = s.length - left; if ((bits ^ (bits >> left)) == ((bits >> right) << right)) return typeof(return)(left); } return typeof(return)();
}
void main() {
immutable words = "1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1".split;
foreach (immutable w; words) { immutable r1 = w.repString1; //assert(r1 == w.repString2); immutable r2 = w.repString2; assert((r1.isNull && r2.isNull) || r1 == r2); if (r1.isNull) writeln(w, " (no repeat)"); else writefln("%(%s %)", w.chunks(r1)); }
}</lang>
- Output:
10011 10011 1110 1110 11 001 001 001 0 1010 1010 10 11111 11111 0100101101 (no repeat) 010 010 0 101 (no repeat) 1 1 0 0 1 (no repeat)
Delphi
<lang Delphi> program Rep_string;
{$APPTYPE CONSOLE}
uses
System.SysUtils;
const
m = '1001110011'#10 + '1110111011'#10 + '0010010010'#10 + '1010101010'#10 + '1111111111'#10 + '0100101101'#10 + '0100100'#10 + '101'#10 + '11'#10 + '00'#10 + '1';
function Rep(s: string; var sub:string): Integer; var
x: Integer;
begin
for x := s.Length div 2 downto 1 do begin sub := s.Substring(x); if s.StartsWith(sub) then exit(x); end; sub := ; Result := 0;
end;
begin
for var s in m.Split([#10]) do begin var sub := ; var n := rep(s,sub); if n > 0 then writeln(format('"%s" %d rep-string "%s"', [s, n, sub])) else writeln(format('"%s" not a rep-string', [s])); end; {$IFNDEF UNIX}readln;{$ENDIF}
end.</lang>
Dyalect
<lang dyalect>func rep(s) {
var x = s.len() / 2 while x > 0 { if s.startsWith(s.sub(x)) { return x } x -= 1 } return 0
}
let m = [
"1001110011", "1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1"
]
for s in m {
if (rep(s) is n) && n > 0 { print("\(s) \(n) rep-string \(s.sub(n))") } else { print("\(s) not a rep-string") }
}</lang>
- Output:
1001110011 5 rep-string 10011 1110111011 4 rep-string 111011 0010010010 3 rep-string 0010010 1010101010 4 rep-string 101010 1111111111 5 rep-string 11111 0100101101 not a rep-string 0100100 3 rep-string 0100 101 not a rep-string 11 1 rep-string 1 00 1 rep-string 0 1 not a rep-string
EchoLisp
<lang scheme> (lib 'list) ;; list-rotate
- a list is a rep-list if equal? to itself after a rotation of lam units
- lam <= list length / 2
- truncate to a multiple of lam before rotating
- try cycles in decreasing lam order (longest wins)
(define (cyclic? cyclic)
(define len (length cyclic)) (define trunc null)
(if (> len 1) (for ((lam (in-range (quotient len 2) 0 -1))) (set! trunc (take cyclic (- len (modulo len lam)))) #:break (equal? trunc (list-rotate trunc lam)) => (list->string (take cyclic lam)) 'no-rep ) 'too-short-no-rep))
</lang>
- Output:
<lang scheme> (define strings '["1001110011" "1110111011" "0010010010" "1010101010"
"1111111111" "0100101101" "0100100" "101" "11" "00" "1"])
(define (task strings) (for-each (lambda (s) (writeln s (cyclic? (string->list s)))) strings))
(task strings)
"1001110011" "10011" "1110111011" "1110" "0010010010" "001" "1010101010" "1010" "1111111111" "11111" "0100101101" no-rep "0100100" "010" "101" no-rep "11" "1" "00" "0" "1" too-short-no-rep </lang>
Elixir
<lang elixir>defmodule Rep_string do
def find(""), do: IO.puts "String was empty (no repetition)" def find(str) do IO.puts str rep_pos = Enum.find(div(String.length(str),2)..1, fn pos -> String.starts_with?(str, String.slice(str, pos..-1)) end) if rep_pos && rep_pos>0 do IO.puts String.duplicate(" ", rep_pos) <> String.slice(str, 0, rep_pos) else IO.puts "(no repetition)" end IO.puts "" end
end
strs = ~w(1001110011
1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1)
Enum.each(strs, fn str -> Rep_string.find(str) end)</lang>
- Output:
1001110011 10011 1110111011 1110 0010010010 001 1010101010 1010 1111111111 11111 0100101101 (no repetition) 0100100 010 101 (no repetition) 11 1 00 0 1 (no repetition)
F#
<lang fsharp>let isPrefix p (s : string) = s.StartsWith(p) let getPrefix n (s : string) = s.Substring(0,n)
let repPrefixOf str =
let rec isRepeatedPrefix p s = if isPrefix p s then isRepeatedPrefix p (s.Substring (p.Length)) else isPrefix s p
let rec getLongestRepeatedPrefix n = if n = 0 then None elif isRepeatedPrefix (getPrefix n str) str then Some(getPrefix n str) else getLongestRepeatedPrefix (n-1)
getLongestRepeatedPrefix (str.Length/2)
[<EntryPoint>] let main argv =
printfn "Testing for rep-string (and showing the longest repeated prefix in case):" [ "1001110011" "1110111011" "0010010010" "1010101010" "1111111111" "0100101101" "0100100" "101" "11" "00" "1" ] |> List.map (fun s -> match repPrefixOf s with | None -> s + ": NO" | Some(p) -> s + ": YES ("+ p + ")") |> List.iter (printfn "%s") 0</lang>
- Output:
Testing for rep-string (and showing the longest repeated prefix in case): 1001110011: YES (10011) 1110111011: YES (1110) 0010010010: YES (001) 1010101010: YES (1010) 1111111111: YES (11111) 0100101101: NO 0100100: YES (010) 101: NO 11: YES (1) 00: YES (0) 1: NO
Factor
<lang factor>USING: formatting grouping kernel math math.ranges qw sequences ; IN: rosetta-code.rep-string
- (find-rep-string) ( str -- str )
dup dup length 2/ [1,b] [ <groups> [ head? ] monotonic? ] with find nip dup [ head ] [ 2drop "N/A" ] if ;
- find-rep-string ( str -- str )
dup length 1 <= [ drop "N/A" ] [ (find-rep-string) ] if ;
qw{ 1001110011 1110111011 0010010010 1010101010 1111111111
0100101101 0100100 101 11 00 1 }
"Shortest cycle:\n\n" printf [ dup find-rep-string "%-10s -> %s\n" printf ] each</lang>
- Output:
Shortest cycle: 1001110011 -> 10011 1110111011 -> 1110 0010010010 -> 001 1010101010 -> 10 1111111111 -> 1 0100101101 -> N/A 0100100 -> 010 101 -> N/A 11 -> 1 00 -> 0 1 -> N/A
Forth
<lang forth>: rep-string ( caddr1 u1 -- caddr2 u2 ) \ u2=0: not a rep-string
2dup dup >r r@ 2/ /string begin 2over 2over string-prefix? 0= over r@ < and while -1 /string repeat r> swap - >r 2drop r> ;
- test ( caddr u -- )
2dup type ." has " rep-string ?dup 0= if drop ." no " else type ." as " then ." repeating substring" cr ;
- tests
s" 1001110011" test s" 1110111011" test s" 0010010010" test s" 1010101010" test s" 1111111111" test s" 0100101101" test s" 0100100" test s" 101" test s" 11" test s" 00" test s" 1" test ;</lang>
- Output:
<lang forth>cr tests 1001110011 has 10011 as repeating substring 1110111011 has 1110 as repeating substring 0010010010 has 001 as repeating substring 1010101010 has 1010 as repeating substring 1111111111 has 11111 as repeating substring 0100101101 has no repeating substring 0100100 has 010 as repeating substring 101 has no repeating substring 11 has 1 as repeating substring 00 has 0 as repeating substring 1 has no repeating substring
ok</lang>
Go
<lang go>package main
import (
"fmt" "strings"
)
func rep(s string) int {
for x := len(s) / 2; x > 0; x-- { if strings.HasPrefix(s, s[x:]) { return x } } return 0
}
const m = ` 1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1`
func main() {
for _, s := range strings.Fields(m) { if n := rep(s); n > 0 { fmt.Printf("%q %d rep-string %q\n", s, n, s[:n]) } else { fmt.Printf("%q not a rep-string\n", s) } }
}</lang>
- Output:
"1001110011" 5 rep-string "10011" "1110111011" 4 rep-string "1110" "0010010010" 3 rep-string "001" "1010101010" 4 rep-string "1010" "1111111111" 5 rep-string "11111" "0100101101" not a rep-string "0100100" 3 rep-string "010" "101" not a rep-string "11" 1 rep-string "1" "00" 1 rep-string "0" "1" not a rep-string
Haskell
<lang haskell>import Data.List (inits, maximumBy) import Data.Maybe (fromMaybe)
repstring :: String -> Maybe String -- empty strings are not rep strings repstring [] = Nothing -- strings with only one character are not rep strings repstring [_] = Nothing repstring xs
| any (`notElem` "01") xs = Nothing | otherwise = longest xs where -- length of the original string lxs = length xs -- half that length lq2 = lxs `quot` 2 -- make a string of same length using repetitions of a part -- of the original string, and also return the substring used subrepeat x = (x, take lxs $ concat $ repeat x) -- check if a repeated string matches the original string sndValid (_, ys) = ys == xs -- make all possible strings out of repetitions of parts of -- the original string, which have max. length lq2 possible = map subrepeat . take lq2 . tail . inits -- filter only valid possibilities, and return the substrings -- used for building them valid = map fst . filter sndValid . possible -- see which string is longer compLength a b = compare (length a) (length b) -- get the longest substring that, repeated, builds a string -- that matches the original string longest ys = case valid ys of [] -> Nothing zs -> Just $ maximumBy compLength zs
main :: IO () main =
mapM_ processIO examples where examples = [ "1001110011", "1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1" ] process = fromMaybe "Not a rep string" . repstring processIO xs = do putStr (xs <> ": ") putStrLn $ process xs</lang>
- Output:
1001110011: 10011 1110111011: 1110 0010010010: 001 1010101010: 1010 1111111111: 11111 0100101101: Not a rep string 0100100: 010 101: Not a rep string 11: 1 00: 0 1: Not a rep string
Or, alternatively: <lang haskell>import Data.Bool (bool) import Data.List (inits, intercalate, transpose)
REP-CYCLES ----------------------
repCycles :: String -> [String] repCycles cs =
filter ((cs ==) . take n . cycle) ((tail . inits) $ take (quot n 2) cs) where n = length cs
TEST -------------------------
main :: IO () main =
putStrLn $ fTable "Longest cycles:\n" id ((flip bool "n/a" . last) <*> null) repCycles [ "1001110011", "1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1" ]
GENERIC ------------------------
fTable ::
String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> String
fTable s xShow fxShow f xs =
let rjust n c = drop . length <*> (replicate n c <>) w = maximum (length . xShow <$> xs) in unlines $ s : fmap ( ((<>) . rjust w ' ' . xShow) <*> ((" -> " <>) . fxShow . f) ) xs</lang>
- Output:
Longest cycles: 1001110011 -> 10011 1110111011 -> 1110 0010010010 -> 001 1010101010 -> 1010 1111111111 -> 11111 0100101101 -> n/a 0100100 -> 010 101 -> n/a 11 -> 1 00 -> 0 1 -> n/a
Icon and Unicon
The following works in both languages.
<lang unicon>procedure main(A)
every write(s := !A,": ",(repString(s) | "Not a rep string!")\1)
end
procedure repString(s)
rs := s[1+:*s/2] while (*rs > 0) & (s ~== lrepl(rs,*s,rs)) do rs := rs[1:-1] return (*rs > 0, rs)
end
procedure lrepl(s1,n,s2) # The standard left() procedure won't work.
while *s1 < n do s1 ||:= s2 return s1[1+:n]
end</lang>
- Output:
->rs 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 1 1110111011: 1110 0010010010: 001 1010101010: 1010 1111111111: 11111 0100101101: Not a rep string! 0100100: 010 101: Not a rep string! 11: 1 1: Not a rep string! ->
J
Here's a test:
<lang j>replengths=: >:@i.@<.@-:@# rep=: $@] $ $
isRepStr=: +./@((] -: rep)"0 1~ replengths)</lang>
Example use:
<lang j> isRepStr '1001110011' 1
Tests=: noun define
1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1 )
isRepStr;._2 Tests NB. run all tests
1 1 1 1 1 0 1 0 1 1 0</lang>
We could also report the lengths of the repeated prefix, though this seems more arbitrary:
<lang j>nRepStr=: 0 -.~ (([ * ] -: rep)"0 1~ replengths)</lang>
With the above examples:
<lang j> ":@nRepStr;._2 Tests 5 4 3 2 4 1 2 3 4 5
3
1 1</lang>
Here, the "non-str-rep" cases are indicated by an empty list of prefix lengths.
Java
<lang java>public class RepString {
static final String[] input = {"1001110011", "1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1", "0100101"};
public static void main(String[] args) { for (String s : input) System.out.printf("%s : %s%n", s, repString(s)); }
static String repString(String s) { int len = s.length(); outer: for (int part = len / 2; part > 0; part--) { int tail = len % part; if (tail > 0 && !s.substring(0, tail).equals(s.substring(len - tail))) continue; for (int j = 0; j < len / part - 1; j++) { int a = j * part; int b = (j + 1) * part; int c = (j + 2) * part; if (!s.substring(a, b).equals(s.substring(b, c))) continue outer; } return s.substring(0, part); } return "none"; }
}</lang>
- Output:
1001110011 : 10011 1110111011 : 1110 0010010010 : 001 1010101010 : 1010 1111111111 : 11111 0100101101 : none 0100100 : 010 101 : none 11 : 1 00 : 0 1 : none 0100101 : none
JavaScript
ES6
<lang javascript>(() => {
'use strict';
const main = () => {
// REP-CYCLES -------------------------------------
// repCycles :: String -> [String] const repCycles = s => { const n = s.length; return filter( x => s === take(n, cycle(x)).join(), tail(inits(take(quot(n, 2), s))) ); };
// TEST ------------------------------------------- console.log(fTable( 'Longest cycles:\n', str, xs => 0 < xs.length ? concat(last(xs)) : '(none)', repCycles, [ '1001110011', '1110111011', '0010010010', '1010101010', '1111111111', '0100101101', '0100100', '101', '11', '00', '1' ] )); };
// GENERIC FUNCTIONS ----------------------------------
// concat :: a -> [a] // concat :: [String] -> String const concat = xs => 0 < xs.length ? (() => { const unit = 'string' !== typeof xs[0] ? ( [] ) : ; return unit.concat.apply(unit, xs); })() : [];
// cycle :: [a] -> Generator [a] function* cycle(xs) { const lng = xs.length; let i = 0; while (true) { yield(xs[i]) i = (1 + i) % lng; } }
// filter :: (a -> Bool) -> [a] -> [a] const filter = (f, xs) => xs.filter(f);
// fTable :: String -> (a -> String) -> // (b -> String) -> (a -> b) -> [a] -> String const fTable = (s, xShow, fxShow, f, xs) => { // Heading -> x display function -> // fx display function -> // f -> values -> tabular string const ys = xs.map(xShow), w = Math.max(...ys.map(length)); return s + '\n' + zipWith( (a, b) => a.padStart(w, ' ') + ' -> ' + b, ys, xs.map(x => fxShow(f(x))) ).join('\n'); };
// inits([1, 2, 3]) -> [[], [1], [1, 2], [1, 2, 3] // inits('abc') -> ["", "a", "ab", "abc"]
// inits :: [a] -> a // inits :: String -> [String] const inits = xs => [ [] ] .concat(('string' === typeof xs ? xs.split() : xs) .map((_, i, lst) => lst.slice(0, 1 + i)));
// last :: [a] -> a const last = xs => 0 < xs.length ? xs.slice(-1)[0] : undefined;
// Returns Infinity over objects without finite length. // This enables zip and zipWith to choose the shorter // argument when one is non-finite, like cycle, repeat etc
// length :: [a] -> Int const length = xs => (Array.isArray(xs) || 'string' === typeof xs) ? ( xs.length ) : Infinity;
// quot :: Int -> Int -> Int const quot = (n, m) => Math.floor(n / m);
// str :: a -> String const str = x => x.toString();
// tail :: [a] -> [a] const tail = xs => 0 < xs.length ? xs.slice(1) : [];
// take :: Int -> [a] -> [a] // take :: Int -> String -> String const take = (n, xs) => 'GeneratorFunction' !== xs.constructor.constructor.name ? ( xs.slice(0, n) ) : [].concat.apply([], Array.from({ length: n }, () => { const x = xs.next(); return x.done ? [] : [x.value]; }));
// unlines :: [String] -> String const unlines = xs => xs.join('\n');
// Use of `take` and `length` here allows zipping with non-finite lists // i.e. generators like cycle, repeat, iterate.
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c] const zipWith = (f, xs, ys) => { const lng = Math.min(length(xs), length(ys)), as = take(lng, xs), bs = take(lng, ys); return Array.from({ length: lng }, (_, i) => f(as[i], bs[i], i)); };
// MAIN --- return main();
})();</lang>
- Output:
Longest cycles: 1001110011 -> 10011 1110111011 -> 1110 0010010010 -> 001 1010101010 -> 1010 1111111111 -> 11111 0100101101 -> (none) 0100100 -> 010 101 -> (none) 11 -> 1 00 -> 0 1 -> (none)
jq
For each test string, a JSON object giving details about the prefixes that satisfy the requirement is presented; if the string is not a rep-string, the empty array ([]) is shown. <lang jq>def is_rep_string:
# if self is a rep-string then return [n, prefix] # where n is the number of full occurrences of prefix def _check(prefix; n; sofar): length as $length | if length <= (sofar|length) then [n, prefix] else (sofar+prefix) as $sofar | if startswith($sofar) then _check(prefix; n+1; $sofar) elif ($sofar|length) > $length and startswith($sofar[0:$length]) then [n, prefix] else [0, prefix] end end ;
[range (1; length/2 + 1) as $i | .[0:$i] as $prefix | _check($prefix; 1; $prefix) | select( .[0] > 1 ) ] ;
</lang> Example: <lang jq>def test:
( "1001110011", "1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1" ) | { (.) : is_rep_string }
test</lang>
- Output:
<lang sh> $ jq -n -c -f rep-string.jq
{"1001110011":2,"10011"} {"1110111011":2,"1110"} {"0010010010":3,"001"} {"1010101010":[[5,"10"],[2,"1010"]]} {"1111111111":[[10,"1"],[5,"11"],[3,"111"],[2,"1111"],[2,"11111"]]} {"0100101101":[]} {"0100100":2,"010"} {"101":[]} {"11":2,"1"} {"00":2,"0"} {"1":[]}</lang>
Julia
repstring returns a list of all of the substrings of its input that are the repeating units of a rep-string. If the input is not a valid rep-string, it returns an empty list. Julia indexes strings, including those that contain multi-byte characters, at the byte level. Because of this characteristic, repstring indexes its input using the chr2ind built-in.
<lang julia>function repstring(r::AbstractString)
n = length(r) replst = String[] for m in 1:n÷2 s = r[1:chr2ind(r, m)] if (s ^ cld(n, m))[1:chr2ind(r, n)] != r continue end push!(replst, s) end return replst
end
tests = ["1001110011", "1110111011", "0010010010", "1010101010", "1111111111",
"0100101101", "0100100", "101", "11", "00", "1", "\u2200\u2203\u2200\u2203\u2200\u2203\u2200\u2203"]
for r in tests
replst = repstring(r) if isempty(replst) println("$r is not a rep-string.") else println("$r is a rep-string of ", join(replst, ", "), ".") end
end</lang>
- Output:
1001110011 is a rep-string of 10011. 1110111011 is a rep-string of 1110. 0010010010 is a rep-string of 001. 1010101010 is a rep-string of 10, 1010. 1111111111 is a rep-string of 1, 11, 111, 1111, 11111. 0100101101 is not a rep-string. 0100100 is a rep-string of 010. 101 is not a rep-string. 11 is a rep-string of 1. 00 is a rep-string of 0. 1 is not a rep-string. ∀∃∀∃∀∃∀∃ is a rep-string of ∀∃, ∀∃∀∃.
Kotlin
<lang scala>// version 1.0.6
fun repString(s: String): MutableList<String> {
val reps = mutableListOf<String>() if (s.length < 2) return reps for (c in s) if (c != '0' && c != '1') throw IllegalArgumentException("Not a binary string") for (len in 1..s.length / 2) { val t = s.take(len) val n = s.length / len val r = s.length % len val u = t.repeat(n) + t.take(r) if (u == s) reps.add(t) } return reps
}
fun main(args: Array<String>) {
val strings = listOf( "1001110011", "1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1" ) println("The (longest) rep-strings are:\n") for (s in strings) { val reps = repString(s) val size = reps.size println("${s.padStart(10)} -> ${if (size > 0) reps[size - 1] else "Not a rep-string"}") }
}</lang>
- Output:
The (longest) rep-strings are: 1001110011 -> 10011 1110111011 -> 1110 0010010010 -> 001 1010101010 -> 1010 1111111111 -> 11111 0100101101 -> Not a rep-string 0100100 -> 010 101 -> Not a rep-string 11 -> 1 00 -> 0 1 -> Not a rep-string
LFE
The heavy lifting:
<lang lisp> (defun get-reps (text)
(lists:filtermap (lambda (x) (case (get-rep text (lists:split x text)) ('() 'false) (x `#(true ,x)))) (lists:seq 1 (div (length text) 2))))
(defun get-rep
((text `#(,head ,tail)) (case (string:str text tail) (1 head) (_ '()))))
</lang>
Displaying the results:
<lang lisp> (defun report
((`#(,text ())) (io:format "~p has no repeating characters.~n" `(,text))) ((`#(,text (,head . ,_))) (io:format "~p repeats ~p every ~p character(s).~n" `(,text ,head ,(length head)))) ((data) (lists:map #'report/1 (lists:zip data (lists:map #'get-reps/1 data))) 'ok))
</lang>
Running the code:
> (set data '("1001110011" "1110111011" "0010010010" "1010101010" "1111111111" "0100101101" "0100100" "101" "11" "00" "1")) > (report data) "1001110011" repeats "10011" every 5 character(s). "1110111011" repeats "1110" every 4 character(s). "0010010010" repeats "001" every 3 character(s). "1010101010" repeats "10" every 2 character(s). "1111111111" repeats "1" every 1 character(s). "0100101101" has no repeating characters. "0100100" repeats "010" every 3 character(s). "101" has no repeating characters. "11" repeats "1" every 1 character(s). "00" repeats "0" every 1 character(s). "1" has no repeating characters. ok
Maple
The built-in Period
command in the StringTools
package computes the length of the longest repeated prefix.
<lang Maple>repstr? := proc( s :: string )
local per := StringTools:-Period( s ); if 2 * per <= length( s ) then true, s[ 1 .. per ] else false, "" end if
end proc:</lang> For the given set of test strings, we can generate the following output. <lang Maple> > Test := ["1001110011", "1110111011", "0010010010", "1010101010", "1111111111", \
"0100101101", "0100100", "101", "11", "00", "1"]:
> for s in Test do > printf( "%*s\t%5s %s\n", 3 + max(map(length,Test)), s, repstr?( s ) ) > end do:
1001110011 true 10011 1110111011 true 1110 0010010010 true 001 1010101010 true 10 1111111111 true 1 0100101101 false 0100100 true 010 101 false 11 true 1 00 true 0 1 false
</lang>
Mathematica
Mathematica is based on pattern-based matching, so this is very easily implemented: <lang Mathematica>RepStringQ[strin_String]:=StringCases[strin,StartOfString~~Repeated[x__,{2,\[Infinity]}]~~y___~~EndOfString/;StringMatchQ[x,StartOfString~~y~~___]:>x, Overlaps -> All]</lang> Trying it out for the test-strings: <lang>str={"1001110011","1110111011","0010010010","1010101010","1111111111","0100101101","0100100","101","11","00","1"}; {#,RepStringQ[#]}&/@str//Grid</lang>
- Output:
1001110011 {10011} 1110111011 {1110} 0010010010 {001} 1010101010 {1010,10,10} 1111111111 {11111,1111,111,11,11,1,1} 0100101101 {} 0100100 {010} 101 {} 11 {1} 00 {0} 1 {}
It outputs all the possibilities for a rep-string, if there is no rep-string it will show an empty list {}.
NetRexx
<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary
/* REXX ***************************************************************
- 11.05.2013 Walter Pachl
- /
runSample(arg) return
/**
* Test for rep-strings * @param s_str a string to check for rep-strings * @return Rexx string: boolean indication of reps, length, repeated value */
method repstring(s_str) public static
s_str_n = s_str.length() rep_str = Loop lx = s_str.length() % 2 to 1 By -1 If s_str.substr(lx + 1, lx) = s_str.left(lx) Then Leave lx End lx If lx > 0 Then Do label reps rep_str = s_str.left(lx) Loop ix = 1 By 1 If s_str.substr(ix * lx + 1, lx) <> rep_str Then Leave ix End ix If rep_str.copies(s_str_n).left(s_str.length()) <> s_str Then rep_str = End reps Return (rep_str.length() > 0) rep_str.length() rep_str
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method runSample(arg) public static
parse arg samples if samples = then - samples = - '1001110011' - '1110111011' - '0010010010' - '1010101010' - '1111111111' - '0100101101' - '0100100' - '101' - '11' - '00' - '1'
loop w_ = 1 to samples.words() in_str = samples.word(w_) parse repstring(in_str) is_rep_str rep_str_len rep_str
sq = 'in_str' tstrlen = sq.length().max(20) sq=sq.right(tstrlen) if is_rep_str then Say sq 'has a repetition length of' rep_str_len "i.e. '"rep_str"'" else Say sq 'is not a repeated string' end w_ return
</lang>
- Output:
'1001110011' has a repetition length of 5 i.e. '10011' '1110111011' has a repetition length of 4 i.e. '1110' '0010010010' has a repetition length of 3 i.e. '001' '1010101010' has a repetition length of 4 i.e. '1010' '1111111111' has a repetition length of 5 i.e. '11111' '0100101101' is not a repeated string '0100100' has a repetition length of 3 i.e. '010' '101' is not a repeated string '11' has a repetition length of 1 i.e. '1' '00' has a repetition length of 1 i.e. '0' '1' is not a repeated string
NGS
<lang NGS>tests = [ '1001110011' '1110111011' '0010010010' '1010101010' '1111111111' '0100101101' '0100100' '101' '11' '00' '1' ]
F is_repeated(s:Str) (s.len()/2..0).first(F(x) s.starts_with(s[x..null]))
{ tests.each(F(test) { local r = is_repeated(test) echo("${test} ${if r "has repetition of length ${r} (i.e. ${test[0..r]})" "is not a rep-string"}") }) }</lang>
- Output:
1001110011 has repetition of length 5 (i.e. 10011)1110111011 has repetition of length 4 (i.e. 1110) 0010010010 has repetition of length 3 (i.e. 001) 1010101010 has repetition of length 4 (i.e. 1010) 1111111111 has repetition of length 5 (i.e. 11111) 0100101101 is not a rep-string 0100100 has repetition of length 3 (i.e. 010) 101 is not a rep-string 11 has repetition of length 1 (i.e. 1) 00 has repetition of length 1 (i.e. 0)
1 is not a rep-string
Nim
<lang nim>import strutils
proc isRepeated(text: string): int =
for x in countdown(text.len div 2, 0): if text.startsWith(text[x..text.high]): return x
const matchstr = """1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1"""
for line in matchstr.split():
let ln = isRepeated(line) echo "'", line, "' has a repetition length of ", ln, " i.e ", (if ln > 0: "'" & line[0 ..< ln] & "'" else: "*not* a rep-string")</lang>
- Output:
'1001110011' has a repetition length of 5 i.e '10011' '1110111011' has a repetition length of 4 i.e '1110' '0010010010' has a repetition length of 3 i.e '001' '1010101010' has a repetition length of 4 i.e '1010' '1111111111' has a repetition length of 5 i.e '11111' '0100101101' has a repetition length of 0 i.e *not* a rep-string '0100100' has a repetition length of 3 i.e '010' '101' has a repetition length of 0 i.e *not* a rep-string '11' has a repetition length of 1 i.e '1' '00' has a repetition length of 1 i.e '0' '1' has a repetition length of 0 i.e *not* a rep-string
Objeck
<lang objeck>class RepString {
function : Main(args : String[]) ~ Nil { strings := ["1001110011", "1110111011", "0010010010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1"]; each(i : strings) { string := strings[i]; repstring := RepString(string); if(repstring->Size() > 0) { "\"{$string}\" = rep-string \"{$repstring}\""->PrintLine(); } else { "\"{$string}\" = not a rep-string"->PrintLine(); }; }; }
function : RepString(string : String) ~ String { offset := string->Size() / 2;
while(offset > 0) { left := string->SubString(offset); right := string->SubString(left->Size(),left->Size()); if(left->Equals(right)) { if(ValidateMatch(left, string)) { return left; } else { return ""; }; }; offset--; }; return ""; }
function : ValidateMatch(left : String, string : String) ~ Bool { parts := string->Size() / left->Size(); tail := string->Size() % left->Size() <> 0;
for(i := 1; i < parts; i+=1;) { offset := i * left->Size(); right := string->SubString(offset, left->Size()); if(<>left->Equals(right)) { return false; }; };
if(tail) { offset := parts * left->Size(); right := string->SubString(offset, string->Size() - offset); each(i : right) { if(left->Get(i) <> right->Get(i)) { return false; }; }; }; return true; }
}</lang>
Output:
"1001110011" = rep-string "10011" "1110111011" = rep-string "1110" "0010010010" = rep-string "001" "1111111111" = rep-string "11111" "0100101101" = not a rep-string "0100100" = rep-string "010" "101" = not a rep-string "11" = rep-string "1" "00" = rep-string "0" "1" = not a rep-string
Oforth
Returns null if no rep string.
<lang oforth>: repString(s) | sz i |
s size dup ->sz 2 / 1 -1 step: i [ s left(sz i - ) s right(sz i -) == ifTrue: [ s left(i) return ] ] null ;</lang>
- Output:
["1001110011", "1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1"] map(#repString) . [10011, 1110, 001, 1010, 11111, null, 010, null, 1, 0, null] ok
PARI/GP
<lang parigp>rep(v)=for(i=1,#v\2,for(j=i+1,#v,if(v[j]!=v[j-i],next(2)));return(i));0; v=["1001110011","1110111011","0010010010","1010101010","1111111111","0100101101","0100100","101","11","00","1"]; for(i=1,#v,print(v[i]" "rep(Vec(v[i]))))</lang>
- Output:
1001110011 5 1110111011 4 0010010010 3 1010101010 2 1111111111 1 0100101101 0 0100100 3 101 0 11 1 00 1 1 0
Perl
<lang perl>foreach (qw(1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1)) {
print "$_\n"; if (/^(.+)\1+(.*$)(?(?{ substr($1, 0, length $2) eq $2 })|(?!))/) { print ' ' x length $1, "$1\n\n"; } else { print " (no repeat)\n\n"; }
}</lang>
- Output:
1001110011 10011 1110111011 1110 0010010010 001 1010101010 1010 1111111111 11111 0100101101 (no repeat) 0100100 010 101 (no repeat) 11 1 00 0 1 (no repeat)
Phix
Shows all possible repeated sub-strings, as Julia, but in the output style of Perl/Elixir <lang Phix>function list_reps(string r) sequence replist = {} integer n = length(r)
for m=1 to floor(n/2) do string s = r[1..m] if join(repeat(s,floor(n/m)+1),"")[1..n]=r then replist = append(replist,s) end if end for return replist
end function
constant tests = {"1001110011",
"1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1"}
for i=1 to length(tests) do
printf(1,"%s\n",{tests[i]}) sequence replist = list_reps(tests[i]) if length(replist)=0 then printf(1,"not a rep-string.\n") else for j=1 to length(replist) do string rj = replist[j], pad = repeat(' ',length(rj)) printf(1,"%s%s\n",{pad,rj}) end for end if printf(1,"\n")
end for</lang>
- Output:
1001110011 10011 1110111011 1110 0010010010 001 1010101010 10 1010 1111111111 1 11 111 1111 11111 0100101101 not a rep-string. 0100100 010 101 not a rep-string. 11 1 00 0 1 not a rep-string.
Phixmonti
<lang Phixmonti>include ..\Utilitys.pmt
def repstr /# s n -- s #/
"" swap for drop over chain endfor nip
enddef
def repString /# s -- s #/
len dup var sz 2 / 1 swap 2 tolist for var i 1 i slice var chunk chunk sz i / 1 + repstr 1 sz slice nip over == if chunk exitfor endif endfor len sz == sz 1 == or if ": No repeat string" chain else ": " swap chain chain endif
enddef
( "1001110011" "1110111011" "0010010010" "1010101010" "1111111111" "0100101101" "0100100" "101" "11" "00" "1" )
len for
get repString print nl
endfor</lang>
- Output:
1001110011: 10011 1110111011: 1110 0010010010: 001 1010101010: 10 1111111111: 1 0100101101: No repeat string 0100100: 010 101: No repeat string 11: 1 00: 0 1: No repeat string === Press any key to exit ===
PicoLisp
<lang PicoLisp>(de repString (Str)
(let Lst (chop Str) (for (N (/ (length Lst) 2) (gt0 N) (dec N)) (T (use (Lst X) (let H (cut N 'Lst) (loop (setq X (cut N 'Lst)) (NIL (head X H)) (NIL Lst T) ) ) ) N ) ) ) )</lang>
Test: <lang PicoLisp>(test 5 (repString "1001110011")) (test 4 (repString "1110111011")) (test 3 (repString "0010010010")) (test 4 (repString "1010101010")) (test 5 (repString "1111111111")) (test NIL (repString "0100101101")) (test 3 (repString "0100100")) (test NIL (repString "101")) (test 1 (repString "11")) (test 1 (repString "00")) (test NIL (repString "1")) (test NIL (repString "0100101"))</lang>
PL/I
<lang PL/I>rep: procedure options (main); /* 5 May 2015 */
declare s bit (10) varying; declare (i, k) fixed binary;
main_loop:
do s = '1001110011'b, '1110111011'b, '0010010010'b, '1010101010'b, '1111111111'b, '0100101101'b, '0100100'b, '101'b, '11'b, '00'b, '1'b; k = length(s); do i = k/2 to 1 by -1; if substr(s, 1, i) = substr(s, i+1, i) then do; put skip edit (s, ' is a rep-string containing ', substr(s, 1, i) ) (a); iterate main_loop; end; end; put skip edit (s, ' is not a rep-string') (a); end;
end rep;</lang>
- Output:
1001110011 is a rep-string containing 10011 1110111011 is a rep-string containing 1110 0010010010 is a rep-string containing 001 1010101010 is a rep-string containing 1010 1111111111 is a rep-string containing 11111 0100101101 is a rep-string containing 010 0100100 is a rep-string containing 010 101 is not a rep-string 11 is a rep-string containing 1 00 is a rep-string containing 0 1 is not a rep-string
Prolog
Using SWI-Prolog 7 library(func), for some functional syntax.
<lang Prolog>:- use_module(library(func)).
%% Implementation logic:
test_for_repstring(String, (String, Result, Reps)) :-
( setof(Rep, repstring(String, Rep), Reps) -> Result = 'no repstring' ; Result = 'repstrings', Reps = [] ).
repstring(Codes, R) :-
RepLength = between(1) of (_//2) of length $ Codes, length(R, RepLength), phrase( (rep(R), prefix(~,R)), Codes).
rep(X) --> X, X. rep(X) --> X, rep(X).
%% Demonstration output:
test_strings([`1001110011`, `1110111011`, `0010010010`, `1010101010`,
`1111111111`, `0100101101`, `0100100`, `101`, `11`, `00`, `1`]).
report_repstring((S,Result,Reps)):-
format('~s -- ~w: ', [S, Result]), foreach(member(R, Reps), format('~s, ', [R])), nl.
report_repstrings :-
Results = maplist(test_for_repstring) $ test_strings(~), maplist(report_repstring, Results).</lang>
Output
?- report_repstrings. 1001110011 -- repstrings: 10011, 1110111011 -- repstrings: 1110, 0010010010 -- repstrings: 001, 1010101010 -- repstrings: 10, 1010, 1111111111 -- repstrings: 1, 11, 111, 1111, 11111, 0100101101 -- no repstring: 0100100 -- repstrings: 010, 101 -- no repstring: 11 -- repstrings: 1, 00 -- repstrings: 0, 1 -- no repstring: true.
PureBasic
<lang purebasic>a$="1001110011"+#CRLF$+"1110111011"+#CRLF$+"0010010010"+#CRLF$+"1010101010"+#CRLF$+"1111111111"+#CRLF$+
"0100101101"+#CRLF$+"0100100" +#CRLF$+"101" +#CRLF$+"11" +#CRLF$+"00" +#CRLF$+ "1" +#CRLF$
OpenConsole()
Procedure isRepStr(s1$,s2$)
If Int(Len(s1$)/Len(s2$))>=2 : ProcedureReturn isRepStr(s1$,s2$+s2$) : EndIf If Len(s1$)>Len(s2$) : ProcedureReturn isRepStr(s1$,s2$+Left(s2$,Len(s1$)%Len(s2$))) : EndIf If s1$=s2$ : ProcedureReturn #True : Else : ProcedureReturn #False : EndIf
EndProcedure
For k=1 To CountString(a$,#CRLF$)
s1$=StringField(a$,k,#CRLF$) : s2$=Left(s1$,Len(s1$)/2) While Len(s2$) r=isRepStr(s1$,s2$) If Not r : s2$=Left(s2$,Len(s2$)-1) : Else : Break : EndIf Wend If Len(s2$) And r : PrintN(LSet(s1$,15,Chr(32))+#TAB$+"longest sequence: "+s2$) : EndIf If Not Len(s2$) : PrintN(LSet(s1$,15,Chr(32))+#TAB$+"found nothing.") : EndIf
Next Input()</lang>
- Output:
1001110011 longest sequence: 10011 1110111011 longest sequence: 1110 0010010010 longest sequence: 001 1010101010 longest sequence: 1010 1111111111 longest sequence: 11111 0100101101 found nothing. 0100100 longest sequence: 010 101 found nothing. 11 longest sequence: 1 00 longest sequence: 0 1 found nothing.
Python
Python: Procedural
<lang python>def is_repeated(text):
'check if the first part of the string is repeated throughout the string' for x in range(len(text)//2, 0, -1): if text.startswith(text[x:]): return x return 0
matchstr = """\ 1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1 """ for line in matchstr.split():
ln = is_repeated(line) print('%r has a repetition length of %i i.e. %s' % (line, ln, repr(line[:ln]) if ln else '*not* a rep-string'))</lang>
- Output:
'1001110011' has a repetition length of 5 i.e. '10011' '1110111011' has a repetition length of 4 i.e. '1110' '0010010010' has a repetition length of 3 i.e. '001' '1010101010' has a repetition length of 4 i.e. '1010' '1111111111' has a repetition length of 5 i.e. '11111' '0100101101' has a repetition length of 0 i.e. *not* a rep-string '0100100' has a repetition length of 3 i.e. '010' '101' has a repetition length of 0 i.e. *not* a rep-string '11' has a repetition length of 1 i.e. '1' '00' has a repetition length of 1 i.e. '0' '1' has a repetition length of 0 i.e. *not* a rep-string
Python: Functional
This returns all the possible repeated substrings <lang python>>>> def reps(text):
return [text[:x] for x in range(1, 1 + len(text) // 2) if text.startswith(text[x:])]
>>> matchstr = """\ 1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1 """ >>> print('\n'.join('%r has reps %r' % (line, reps(line)) for line in matchstr.split())) '1001110011' has reps ['10011'] '1110111011' has reps ['1110'] '0010010010' has reps ['001'] '1010101010' has reps ['10', '1010'] '1111111111' has reps ['1', '11', '111', '1111', '11111'] '0100101101' has reps [] '0100100' has reps ['010'] '101' has reps [] '11' has reps ['1'] '00' has reps ['0'] '1' has reps [] >>> </lang>
And we could also express this in terms of itertools.cycle
<lang python>Rep-strings
from itertools import (accumulate, chain, cycle, islice)
- repCycles :: String -> [String]
def repCycles(s):
Repeated sequences of characters in s. n = len(s) cs = list(s)
return [ x for x in tail(inits(take(n // 2)(s))) if cs == take(n)(cycle(x)) ]
- TEST ----------------------------------------------------
- main :: IO ()
def main():
Tests - longest cycle (if any) in each string. print( fTable('Longest cycles:\n')(repr)( lambda xs: .join(xs[-1]) if xs else '(none)' )(repCycles)([ '1001110011', '1110111011', '0010010010', '1010101010', '1111111111', '0100101101', '0100100', '101', '11', '00', '1', ]) )
- GENERIC -------------------------------------------------
- inits :: [a] -> a
def inits(xs):
all initial segments of xs, shortest first. return accumulate(chain([[]], xs), lambda a, x: a + [x])
- tail :: [a] -> [a]
- tail :: Gen [a] -> [a]
def tail(xs):
The elements following the head of a (non-empty) list or generator stream. if isinstance(xs, list): return xs[1:] else: list(islice(xs, 1)) # First item dropped. return xs
- take :: Int -> [a] -> [a]
- take :: Int -> String -> String
def take(n):
The prefix of xs of length n, or xs itself if n > length xs. return lambda xs: ( xs[0:n] if isinstance(xs, (list, tuple)) else list(islice(xs, n)) )
- OUTPUT FORMATTING ---------------------------------------
- fTable :: String -> (a -> String) ->
- (b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
Heading -> x display function -> fx display function -> f -> value list -> tabular string. def go(xShow, fxShow, f, xs): ys = [xShow(x) for x in xs] w = max(map(len, ys)) return s + '\n' + '\n'.join(map( lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)), xs, ys )) return lambda xShow: lambda fxShow: lambda f: lambda xs: go( xShow, fxShow, f, xs )
- MAIN ---
if __name__ == '__main__':
main()</lang>
- Output:
Longest cycles: '1001110011' -> 10011 '1110111011' -> 1110 '0010010010' -> 001 '1010101010' -> 1010 '1111111111' -> 11111 '0100101101' -> (none) '0100100' -> 010 '101' -> (none) '11' -> 1 '00' -> 0 '1' -> (none)
Python: Regexp
This version, inspired by the Raku entry uses the regexp substitute where what the match is substituted with is returned by a function. <lang python>import re
matchstr = """\ 1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1"""
def _checker(matchobj):
g0, (g1, g2, g3, g4) = matchobj.group(0), matchobj.groups() if not g4 and g1 and g1.startswith(g3): return '%r repeats %r' % (g0, g1) return '%r is not a rep-string' % (g0,)
def checkit(txt):
print(re.sub(r'(.+)(\1+)(.*)|(.*)', _checker, txt))
checkit(matchstr)</lang>
- Output:
'1001110011' repeats '10011' '1110111011' repeats '1110' '0010010010' repeats '001' '1010101010' repeats '1010' '1111111111' repeats '11111' '0100101101' is not a rep-string '0100100' repeats '010' '101' is not a rep-string '11' repeats '1' '00' repeats '0' '1' is not a rep-string
Python: find
See David Zhang's solution to the same question posed on Stack Overflow.
Racket
<lang Racket>#lang racket
(define (rep-string str)
(define len (string-length str)) (for/or ([n (in-range 1 len)]) (and (let loop ([from n]) (or (>= from len) (let ([m (min (- len from) n)]) (and (equal? (substring str from (+ from m)) (substring str 0 m)) (loop (+ n from)))))) (<= n (quotient len 2)) (substring str 0 n))))
(for ([str '("1001110011"
"1110111011" "0010010010" "1010101010" "1111111111" "0100101101" "0100100" "101" "11" "00" "1")]) (printf "~a => ~a\n" str (or (rep-string str) "not a rep-string")))</lang>
- Output:
1001110011 => 10011 1110111011 => 1110 0010010010 => 001 1010101010 => 10 1111111111 => 1 0100101101 => not a rep-string 0100100 => 010 101 => not a rep-string 11 => 1 00 => 0 1 => not a rep-string
Raku
(formerly Perl 6) <lang perl6>for <1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1> {
if /^ (.+) $0+: (.*$) <?{ $0.substr(0,$1.chars) eq $1 }> / {
my $rep = $0.chars; say .substr(0,$rep), .substr($rep,$rep).trans('01' => '𝟘𝟙'), .substr($rep*2);
} else {
say "$_ (no repeat)";
}
}</lang>
- Output:
10011𝟙𝟘𝟘𝟙𝟙 1110𝟙𝟙𝟙𝟘11 001𝟘𝟘𝟙0010 1010𝟙𝟘𝟙𝟘10 11111𝟙𝟙𝟙𝟙𝟙 0100101101 (no repeat) 010𝟘𝟙𝟘0 101 (no repeat) 1𝟙 0𝟘 1 (no repeat)
Here's a technique that relies on the fact that XORing the shifted binary number should set all the lower bits to 0 if there are repeats. (The cool thing is that shift will automatically throw away the bits on the right that you want thrown away.) This produces the same output as above. <lang perl6>sub repstr(Str $s) {
my $bits = :2($s); for reverse 1 .. $s.chars div 2 -> $left {
my $right = $s.chars - $left; return $left if $bits +^ ($bits +> $left) == $bits +> $right +< $right;
}
}
for '1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1'.words {
if repstr $_ -> $rep {
say .substr(0,$rep), .substr($rep,$rep).trans('01' => '𝟘𝟙'), .substr($rep*2);
} else {
say "$_ (no repeat)";
}
}</lang>
REXX
version 1
<lang rexx>/* REXX ***************************************************************
- 11.05.2013 Walter Pachl
- 14.05.2013 Walter Pachl extend to show additional rep-strings
- /
Call repstring '1001110011' Call repstring '1110111011' Call repstring '0010010010' Call repstring '1010101010' Call repstring '1111111111' Call repstring '0100101101' Call repstring '0100100' Call repstring '101' Call repstring '11' Call repstring '00' Call repstring '1' Exit
repstring: Parse Arg s sq='s' n=length(s) Do l=length(s)%2 to 1 By -1
If substr(s,l+1,l)=left(s,l) Then Leave End
If l>0 Then Do
rep_str=left(s,l) Do i=1 By 1 If substr(s,i*l+1,l)<>rep_str Then Leave End If left(copies(rep_str,n),length(s))=s Then Do Call show_rep rep_str /* show result */ Do i=length(rep_str)-1 To 1 By -1 /* look for shorter rep_str-s */ rep_str=left(s,i) If left(copies(rep_str,n),length(s))=s Then Call show_rep rep_str End End Else Call show_norep End
Else
Call show_norep
Return
show_rep:
Parse Arg rs Say right(sq,12) 'has a repetition length of' length(rs) 'i.e.' 'rs' Return
show_norep:
Say right(sq,12) 'is not a repeated string' Return</lang>
- Output:
'1001110011' has a repetition length of 5 i.e. '10011' '1110111011' has a repetition length of 4 i.e. '1110' '0010010010' has a repetition length of 3 i.e. '001' '1010101010' has a repetition length of 4 i.e. '1010' '1010101010' has a repetition length of 2 i.e. '10' '1111111111' has a repetition length of 5 i.e. '11111' '1111111111' has a repetition length of 4 i.e. '1111' '1111111111' has a repetition length of 3 i.e. '111' '1111111111' has a repetition length of 2 i.e. '11' '1111111111' has a repetition length of 1 i.e. '1' '0100101101' is not a repeated string '0100100' has a repetition length of 3 i.e. '010' '101' is not a repeated string '11' has a repetition length of 1 i.e. '1' '00' has a repetition length of 1 i.e. '0' '1' is not a repeated string
version 2
A check was added to validate if the strings are binary strings. The binary strings can be of any length. <lang rexx>/*REXX pgm determines if a string is a repString, it returns minimum length repString.*/ parse arg s /*get optional strings from the C.L. */ if s= then s=1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1 45
/* [↑] S not specified? Use defaults*/ do k=1 for words(s); _=word(s,k); w=length(_) /*process binary strings. */ say right(_,max(25,w)) repString(_) /*show repString & result.*/ end /*k*/ /* [↑] the "result" may be negatory.*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ repString: procedure; parse arg x; L=length(x); @rep=' rep string='
if \datatype(x,'B') then return " ***error*** string isn't a binary string." h=L%2 do j=1 for L-1 while j<=h; $=left(x,j); $$=copies($,L) if left($$,L)==x then return @rep left($,15) "[length" j']' end /*j*/ /* [↑] we have found a good repString.*/ return ' (no repetitions)' /*failure to find repString.*/</lang>
output when using the default binary strings for input:
1001110011 rep string= 10011 [length 5] 1110111011 rep string= 1110 [length 4] 0010010010 rep string= 001 [length 3] 1010101010 rep string= 10 [length 2] 1111111111 rep string= 1 [length 1] 0100101101 (no repetitions) 0100100 rep string= 010 [length 3] 101 (no repetitions) 11 rep string= 1 [length 1] 00 rep string= 0 [length 1] 1 (no repetitions) 45 ***error*** string isn't a binary string.
Ring
<lang ring>
- Project : Rep-string
test = ["1001110011",
"1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1"]
for n = 1 to len(test)
strend = "" for m=1 to len(test[n]) strbegin = substr(test[n], 1, m) strcut = right(test[n], len(test[n]) - m) nr = substr(strcut, strbegin) if nr=1 and len(test[n]) > 1 strend = strbegin ok next if strend = "" see "" + test[n] + " -> (none)" + nl else see "" + test[n] + " -> " + strend + nl ok
next </lang> Output:
1001110011 -> 10011 1110111011 -> 1110 0010010010 -> 001 1010101010 -> 1010 1111111111 -> 11111 0100101101 -> 010 0100100 -> 010 101 -> (none) 11 -> 1 00 -> 0 1 -> (none)
Ruby
<lang ruby>ar = %w(1001110011
1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1)
ar.each do |str|
rep_pos = (str.size/2).downto(1).find{|pos| str.start_with? str[pos..-1]} puts str, rep_pos ? " "*rep_pos + str[0, rep_pos] : "(no repetition)", ""
end</lang>
- Output (as Perl):
1001110011 10011 1110111011 1110 0010010010 001 1010101010 1010 1111111111 11111 0100101101 (no repetition) 0100100 010 101 (no repetition) 11 1 00 0 1 (no repetition)
Rust
<lang Rust>fn main() {
let strings = vec![ String::from("1001110011"), String::from("1110111011"), String::from("0010010010"), String::from("1010101010"), String::from("1111111111"), String::from("0100101101"), String::from("0100100"), String::from("101"), String::from("11"), String::from("00"), String::from("1"), ]; for string in strings { match rep_string(&string) { Some(rep_string) => println!( "Longuest rep-string for '{}' is '{}' ({} chars)", string, rep_string, rep_string.len(), ), None => println!("No rep-string found for '{}'", string), }; }
}
fn rep_string(string: &str) -> Option<&str> {
let index = string.len() / 2;
for split_index in (1..=index).rev() { let mut is_rep_string = true; let (first, last) = string.split_at(split_index);
let inter = last.chars().collect::<Vec<char>>(); let mut iter = inter.chunks_exact(split_index); for chunk in iter.by_ref() { if first != chunk.iter().collect::<String>() { is_rep_string = false; break; } } let rmnd = iter.remainder().iter().collect::<String>();
// Check that the remainder starts with the rep-string if !first.starts_with(rmnd.as_str()) { is_rep_string = false; }
if is_rep_string { return Some(first); } } None
}
- [cfg(test)]
mod tests {
use super::rep_string; use std::collections::HashMap;
#[test] fn test_rep_string() { let mut results = HashMap::new(); results.insert(String::from("1001110011"), Some("10011")); results.insert(String::from("1110111011"), Some("1110")); results.insert(String::from("0010010010"), Some("001")); results.insert(String::from("1010101010"), Some("1010")); results.insert(String::from("1111111111"), Some("11111")); results.insert(String::from("0100101101"), None); results.insert(String::from("0100100"), Some("010")); results.insert(String::from("101"), None); results.insert(String::from("11"), Some("1")); results.insert(String::from("00"), Some("0")); results.insert(String::from("1"), None);
for (input, expected) in results { assert_eq!(expected, rep_string(&input)); } }
}</lang>
Scala
<lang Scala>object RepString extends App {
def repsOf(s: String) = s.trim match { case s if s.length < 2 => Nil case s => (1 to (s.length/2)).map(s take _) .filter(_ * s.length take s.length equals s) }
val tests = Array( "1001110011", "1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1" ) def printReps(s: String) = repsOf(s) match { case Nil => s+": NO" case r => s+": YES ("+r.mkString(", ")+")" } val todo = if (args.length > 0) args else tests todo.map(printReps).foreach(println)
}</lang>
- Output:
1001110011: YES (10011) 1110111011: YES (1110) 0010010010: YES (001) 1010101010: YES (10, 1010) 1111111111: YES (1, 11, 111, 1111, 11111) 0100101101: NO 0100100: YES (010) 101: NO 11: YES (1) 00: YES (0) 1: NO
Seed7
<lang seed7>$ include "seed7_05.s7i";
const func integer: repeatLength (in string: text) is func
result var integer: length is 0; local var integer: pos is 0; begin for pos range succ(length(text) div 2) downto 1 until length <> 0 do if startsWith(text, text[pos ..]) then length := pred(pos); end if; end for; end func;
const proc: main is func
local var string: line is ""; var integer: length is 0; begin for line range [] ("1001110011", "1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1") do length := repeatLength(line); if length = 0 then writeln("No rep-string for " <& literal(line)); else writeln("Longest rep-string for " <& literal(line) <& ": " <& literal(line[.. length])); end if; end for; end func;</lang>
- Output:
Longest rep-string for "1001110011": "10011" Longest rep-string for "1110111011": "1110" Longest rep-string for "0010010010": "001" Longest rep-string for "1010101010": "1010" Longest rep-string for "1111111111": "11111" No rep-string for "0100101101" Longest rep-string for "0100100": "010" No rep-string for "101" Longest rep-string for "11": "1" Longest rep-string for "00": "0" No rep-string for "1"
Sidef
<lang ruby>var arr = <1001110011 1110111011
0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1>;
arr.each { |n| if (var m = /^(.+)\1+(.*$)(?(?{ substr($1, 0, length $2) eq $2 })|(?!))/.match(n)) { var i = m[0].len; say (n.substr(0, i), n.substr(i, i).tr('01', '𝟘𝟙'), n.substr(i*2)); } else { say "#{n} (no repeat)"; }
}</lang>
- Output:
10011𝟙𝟘𝟘𝟙𝟙 1110𝟙𝟙𝟙𝟘11 001𝟘𝟘𝟙0010 1010𝟙𝟘𝟙𝟘10 11111𝟙𝟙𝟙𝟙𝟙 0100101101 (no repeat) 010𝟘𝟙𝟘0 101 (no repeat) 1𝟙 0𝟘 1 (no repeat)
Snobol4
<lang Snobol4>* Rep-string strings = "1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1 " pat1 = (len(1) $ fc breakx(*fc)) $ x *x (arbno(*x) (rpos(0) | rem $ y *?(x ? y))) getstring strings ? (break(" ") . rs len(1)) = :f(end) rs ? pat1 :s(yes) output = rs " is not a rep-string -> n/a" :(getstring) yes output = rs " has shortest rep-string value of -> " x :(getstring) end </lang>
- Output:
1001110011 has shortest rep-string value of -> 10011 1110111011 has shortest rep-string value of -> 1110 0010010010 has shortest rep-string value of -> 001 1010101010 has shortest rep-string value of -> 10 1111111111 has shortest rep-string value of -> 1 0100101101 is not a rep-string -> n/a 0100100 has shortest rep-string value of -> 010 101 is not a rep-string -> n/a 11 has shortest rep-string value of -> 1 00 has shortest rep-string value of -> 0 1 is not a rep-string -> n/a
Swift
<lang swift>import Foundation
func repString(_ input: String) -> [String] {
return (1..<(1 + input.count / 2)).compactMap({x -> String? in let i = input.index(input.startIndex, offsetBy: x) return input.hasPrefix(input[i...]) ? String(input.prefix(x)) : nil })
}
let testCases = """
1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1 """.components(separatedBy: "\n")
for testCase in testCases {
print("\(testCase) has reps: \(repString(testCase))")
}</lang>
- Output:
1001110011 has reps: ["10011"] 1110111011 has reps: ["1110"] 0010010010 has reps: ["001"] 1010101010 has reps: ["10", "1010"] 1111111111 has reps: ["1", "11", "111", "1111", "11111"] 0100101101 has reps: [] 0100100 has reps: ["010"] 101 has reps: [] 11 has reps: ["1"] 00 has reps: ["0"] 1 has reps: []
Tcl
<lang tcl>proc repstring {text} {
set len [string length $text] for {set i [expr {$len/2}]} {$i > 0} {incr i -1} {
set sub [string range $text 0 [expr {$i-1}]] set eq [string repeat $sub [expr {int(ceil($len/double($i)))}]] if {[string equal -length $len $text $eq]} { return $sub }
} error "no repetition"
}</lang> Demonstrating: <lang tcl>foreach sample {
"1001110011" "1110111011" "0010010010" "1010101010" "1111111111" "0100101101" "0100100" "101" "11" "00" "1"
} {
if {[catch {
set rep [repstring $sample] puts [format "\"%s\" has repetition (length: %d) of \"%s\"" \ $sample [string length $rep] $rep]
}]} {
puts [format "\"%s\" is not a repeated string" $sample]
}
}</lang>
- Output:
"1001110011" has repetition (length: 5) of "10011" "1110111011" has repetition (length: 4) of "1110" "0010010010" has repetition (length: 3) of "001" "1010101010" has repetition (length: 4) of "1010" "1111111111" has repetition (length: 5) of "11111" "0100101101" is not a repeated string "0100100" has repetition (length: 3) of "010" "101" is not a repeated string "11" has repetition (length: 1) of "1" "00" has repetition (length: 1) of "0" "1" is not a repeated string
UNIX Shell
<lang bash>is_repeated() {
local str=$1 len rep part for (( len = ${#str} / 2; len > 0; len-- )); do part=${str:0:len} rep="" while (( ${#rep} < ${#str} )); do rep+=$part done if [[ ${rep:0:${#str}} == $str ]] && (( $len < ${#str} )); then echo "$part" return 0 fi done return 1
}
while read test; do
if part=$( is_repeated "$test" ); then echo "$test is composed of $part repeated" else echo "$test is not a repeated string" fi
done <<END_TESTS 1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1 END_TESTS</lang>
- Output:
1001110011 is composed of 10011 repeated 1110111011 is composed of 1110 repeated 0010010010 is composed of 001 repeated 1010101010 is composed of 1010 repeated 1111111111 is composed of 11111 repeated 0100101101 is not a repeated string 0100100 is composed of 010 repeated 101 is not a repeated string 11 is composed of 1 repeated 00 is composed of 0 repeated 1 is not a repeated string
VBScript
<lang vb> Function rep_string(s) max_len = Int(Len(s)/2) tmp = "" If max_len = 0 Then rep_string = "No Repeating String" Exit Function End If For i = 1 To max_len If InStr(i+1,s,tmp & Mid(s,i,1))Then tmp = tmp & Mid(s,i,1) Else Exit For End If Next Do While Len(tmp) > 0 If Mid(s,Len(tmp)+1,Len(tmp)) = tmp Then rep_string = tmp Exit Do Else tmp = Mid(tmp,1,Len(tmp)-1) End If Loop If Len(tmp) > 0 Then rep_string = tmp Else rep_string = "No Repeating String" End If End Function
'testing the function arr = Array("1001110011","1110111011","0010010010","1010101010",_ "1111111111","0100101101","0100100","101","11","00","1")
For n = 0 To UBound(arr) WScript.StdOut.Write arr(n) & ": " & rep_string(arr(n)) WScript.StdOut.WriteLine Next </lang>
- Output:
1001110011: 10011 1110111011: 1110 0010010010: 001 1010101010: 1010 1111111111: 11111 0100101101: 010 0100100: 010 101: No Repeating String 11: 1 00: 0 1: No Repeating String
Wren
<lang ecmascript>import "/fmt" for Fmt
var repString = Fn.new { |s|
var reps = [] if (s.count < 2) return reps for (c in s) if (c != "0" && c != "1") Fiber.abort("Argument is not a binary string.") var size = s.count for (len in 1..(size/2).floor) { var t = s[0...len] var n = (size/len).floor var r = size % len var u = t * n + t[0...r] if (u == s) reps.add(t) } return reps
}
var strings = [
"1001110011", "1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1"
] System.print("The (longest) rep-strings are:\n") for (s in strings) {
var reps = repString.call(s) var t = (reps.count > 0) ? reps[-1] : "Not a rep-string" Fmt.print("$10s -> $s", s, t)
}</lang>
- Output:
The (longest) rep-strings are: 1001110011 -> 10011 1110111011 -> 1110 0010010010 -> 001 1010101010 -> 1010 1111111111 -> 11111 0100101101 -> Not a rep-string 0100100 -> 010 101 -> Not a rep-string 11 -> 1 00 -> 0 1 -> Not a rep-string
Yabasic
<lang Yabasic>data "1001110011", "1110111011", "0010010010", "1010101010", "1111111111", "0100101101", "0100100", "101", "11", "00", "1", ""
sub rep$(c$, n)
local r$, i for i = 1 to n r$ = r$ + c$ next return r$
end sub
do
read p$ : if p$ = "" break b$ = "" : l = len(p$) : m = int(l / 2)
for i = m to 1 step -1 t$ = left$(p$, i) s$ = rep$(t$, l / i + 1) if p$ = left$(s$, l) b$ = t$ : break next
if b$ = "" then print p$, " is not a repeated string" else print p$, " is composed of ", b$, " repeated" end if
loop</lang>
zkl
<lang zkl>fcn repString(s){
foreach n in ([s.len()/2+1..1,-1]){ Walker.cycle(s[0,n]).pump(s.len(),String) : if(_==s and n*2<=s.len()) return(n); } return(False)
}</lang>
<lang zkl>fcn repString(s){
foreach n in ([s.len()/2..0,-1]){ if(s.matches(s[n,*] + "*") and n*2<=s.len()) return(n); } return(False)
}</lang> <lang zkl>words:=("1001110011 1110111011 0010010010 1010101010 "
"1111111111 0100101101 0100100 101 11 00 1").split(" ");
foreach w in (words){
if(not n:=repString2(w)) "No repeat in ".println(w); else [0..*,n].tweak('wrap(z){ if(s:=w[z,n]) s else Void.Stop }) .walk().concat(" ").println();
}</lang>
- Output:
10011 10011 1110 1110 11 001 001 001 0 1010 1010 10 11111 11111 No repeat in 0100101101 010 010 0 No repeat in 101 1 1 0 0 No repeat in 1
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