Rep-string
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.
The task is to:
- 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.
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
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
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$1 );</lang>
- Output:
1001110011 : 10011 1110111011 : 1110 0010010010 : 001 1010101010 : 1010 10 1111111111 : 11111 1111 111 11 1 0100101101 is not a rep-string 0100100 : 010 101 is not a rep-string 11 : 1 00 : 0 1 is not a rep-string
C
<lang c>
- include <stdio.h>
- include <string.h>
int repstr(char *str) {
if (!str) return 0;
size_t sl = strlen(str) / 2;
while (sl > 0) {
if (strstr(str, str + sl) == str)
return sl;
--sl;
}
return 0;
}
int main(void) {
char *strs[] = { "1001110011", "1110111011", "0010010010", "1111111111",
"0100101101", "0100100", "101", "11", "00", "1" };
size_t strslen = sizeof(strs) / sizeof(strs[0]);
size_t i;
for (i = 0; i < strslen; ++i) {
int n = repstr(strs[i]);
if (n)
printf("\"%s\" = rep-string \"%.*s\"\n", strs[i], n, strs[i]);
else
printf("\"%s\" = not a rep-string\n", strs[i]);
}
return 0;
} </lang>
- Output:
"1001110011" = rep-string "10011" "1110111011" = rep-string "1110" "0010010010" = rep-string "001" "1111111111" = rep-string "11111" "0100101101" = not a rep-string "0100100" = rep-string "010" "101" = not a rep-string "11" = rep-string "1" "00" = rep-string "0" "1" = not a rep-string
C++
<lang cpp>#include <string>
- include <vector>
- include <boost/regex.hpp>
bool is_repstring( const std::string & teststring , std::string & repunit ) {
std::string regex( "^(.+)\\1+(.*)$" ) ;
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!
D
Two different algorithms. The second is from the Perl 6 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)
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)
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 (maximumBy, inits)
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 = do
mapM_ processIO examples
where
examples = ["1001110011", "1110111011", "0010010010",
"1010101010", "1111111111", "0100101101", "0100100",
"101", "11", "00", "1"]
process = maybe "Not a rep string" id . 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
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
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
list_reps 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, list_reps indexes its input using the chr2ind built-in. <lang Julia> function list_reps{T<:String}(r::T)
n = length(r)
replst = T[]
for m in 1:n>>1
s = r[1:chr2ind(r,m)]
(s^(div(n,m)+1))[1:chr2ind(r,n)] == r || continue
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 = list_reps(r)
rlen = length(replst)
print(@sprintf(" %s ", r))
if rlen == 0
println("is not a rep-string.")
else
println("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.
ββββββββ is a rep-string of ββ, ββββ.
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. ${test[0..r]})" "is not a rep-string"}") }) }</lang>
- Output:
1001110011 has repetition of length 5 (i.e. 10011)1110111011 has repetition of length 4 (i.e. 1110) 0010010010 has repetition of length 3 (i.e. 001) 1010101010 has repetition of length 4 (i.e. 1010) 1111111111 has repetition of length 5 (i.e. 11111) 0100101101 is not a rep-string 0100100 has repetition of length 3 (i.e. 010) 101 is not a rep-string 11 has repetition of length 1 (i.e. 1) 00 has repetition of length 1 (i.e. 0)
1 is not a rep-string
Nim
<lang nim>import strutils
proc isRepeated(text): int =
for x in countdown(text.len div 2, 0): if text.startsWith(text[x..text.high]): return x
const matchstr = """1001110011 1110111011 0010010010 1010101010 1111111111 0100101101 0100100 101 11 00 1"""
for line in matchstr.split():
let ln = isRepeated(line) echo "'", line, "' has a repetition length of ", ln, " i.e ", (if ln > 0: "'" & line[0 .. <ln] & "'" else: "*not* a rep-string")</lang>
- Output:
'1001110011' has a repetition length of 5 i.e '10011' '1110111011' has a repetition length of 4 i.e '1110' '0010010010' has a repetition length of 3 i.e '001' '1010101010' has a repetition length of 4 i.e '1010' '1111111111' has a repetition length of 5 i.e '11111' '0100101101' has a repetition length of 0 i.e *not* a rep-string '0100100' has a repetition length of 3 i.e '010' '101' has a repetition length of 0 i.e *not* a rep-string '11' has a repetition length of 1 i.e '1' '00' has a repetition length of 1 i.e '0' '1' has a repetition length of 0 i.e *not* a rep-string
Objeck
<lang objeck>class RepString {
function : Main(args : String[]) ~ Nil {
strings := ["1001110011", "1110111011", "0010010010", "1111111111",
"0100101101", "0100100", "101", "11", "00", "1"];
each(i : strings) {
string := strings[i];
repstring := RepString(string);
if(repstring->Size() > 0) {
"\"{$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)
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' => 'ππ'), .substr($rep*2);
}
else {
say "$_ (no repeat)";
}
}</lang>
- Output:
10011πππππ 1110ππππ11 001πππ0010 1010ππππ10 11111πππππ 0100101101 (no repeat) 010πππ0 101 (no repeat) 1π 0π 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' => 'ππ'), .substr($rep*2);
}
else {
say "$_ (no repeat)";
}
}</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$+ "1" +#CRLF$
Define.i : 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>
Python: Regexp
This version, inspired by the Perl 6 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
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
/* [β] 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*/ /* [β] the "result" may be negatory.*/
exit /*stick a fork in it, we're all done. */ /*ββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββ*/ 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($,15) "[length" j']'
end /*j*/ /* [β] we have found a good repString.*/
return ' (no repetitions)' /*failure to find repString.*/</lang>
output when using the default binary strings for input:
1001110011 rep string= 10011 [length 5]
1110111011 rep string= 1110 [length 4]
0010010010 rep string= 001 [length 3]
1010101010 rep string= 10 [length 2]
1111111111 rep string= 1 [length 1]
0100101101 (no repetitions)
0100100 rep string= 010 [length 3]
101 (no repetitions)
11 rep string= 1 [length 1]
00 rep string= 0 [length 1]
1 (no repetitions)
45 ***error*** string isn't a binary string.
Ruby
<lang ruby>ar = %w(1001110011
1110111011
0010010010
1010101010
1111111111
0100101101
0100100
101
11
00
1)
ar.each do |str|
rep_pos = (str.size/2).downto(1).find{|pos| str.start_with? str[pos..-1]}
puts str, rep_pos ? " "*rep_pos + str[0, rep_pos] : "(no repetition)", ""
end</lang>
- Output (as Perl):
1001110011
10011
1110111011
1110
0010010010
001
1010101010
1010
1111111111
11111
0100101101
(no repetition)
0100100
010
101
(no repetition)
11
1
00
0
1
(no repetition)
Scala
<lang Scala>object RepString extends App {
def repsOf(s: String) = s.trim match {
case s if s.length < 2 => Nil
case s => (1 to (s.length/2)).map(s take _)
.filter(_ * s.length take s.length equals s)
}
val tests = Array(
"1001110011",
"1110111011",
"0010010010",
"1010101010",
"1111111111",
"0100101101",
"0100100",
"101",
"11",
"00",
"1"
)
def printReps(s: String) = repsOf(s) match {
case Nil => s+": NO"
case r => s+": YES ("+r.mkString(", ")+")"
}
val todo = if (args.length > 0) args else tests
todo.map(printReps).foreach(println)
}</lang>
- Output:
1001110011: YES (10011) 1110111011: YES (1110) 0010010010: YES (001) 1010101010: YES (10, 1010) 1111111111: YES (1, 11, 111, 1111, 11111) 0100101101: NO 0100100: YES (010) 101: NO 11: YES (1) 00: YES (0) 1: NO
Seed7
<lang seed7>$ 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', 'ππ'),
n.substr(i*2));
} else {
say "#{n} (no repeat)";
}
}</lang>
- Output:
10011πππππ 1110ππππ11 001πππ0010 1010ππππ10 11111πππππ 0100101101 (no repeat) 010πππ0 101 (no repeat) 1π 0π 1 (no repeat)
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
zkl
<lang zkl>fcn repString(s){
foreach n in ([s.len()/2+1..1,-1]){
s[0,n]:Utils.Helpers.cycle(_).aggregate(String).walk(s.len()) :
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:=repString(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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