Check if a string is entirely made of the same substring

This is taken from this question (with permission ofcourse). I'll quote:

Create a function which takes a string, and it should return true or false based on whether the input consists of only a repeated character sequence. The length of given string is always greater than 1 and the character sequence must have at least one repetition.

Some examples:

'aa' //true
'aaa' //true
'abcabcabc' //true
'aba' //false
'ababa' //false
'weqweqweqweqweqw' // false


Specifically, the check for a string strictly composed of repeating substrings (Update) can output any true or false representation, but no error output please. Strictly alphhanumeric strings. Otherwise standard code golf rules. Shortest answer in bytes for each language wins.

• Hm, I was going to close this challenge as a dupe of that one, but I noticed that the other one scores on character count. So maybe we should close the other one (it also has an accepted answer) as a dupe of this one instead. – Erik the Outgolfer Apr 24 at 15:14
• – Erik the Outgolfer Apr 24 at 15:36

Brachylog, 4 3 bytes

ġ=Ṁ


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Explanation

ġ=Ṁ    Implicit input, say "abcabc"
ġ      Split into chunks of equal lengths (except maybe the last one): ["abc","abc"]
=     Apply the constraint that all of the chunks are equal,
Ṁ    and that there are multiple of them.


The program prints true. if the constraints can be satisfied, and false. if not.

• I was just struggling through trying to get something like ~j↙ or =Ṁc working before I noticed you posted this an hour ago – Unrelated String Apr 24 at 21:27
• Oh, yeah, this could be one byte shorter: ġ=Ṁ – Unrelated String Apr 24 at 21:36
• (Ṁ is a variable constrained to be a list of two or more elements) – Unrelated String Apr 24 at 21:39
• @UnrelatedString Great, thanks! I didn't think to check the variables wiki page. – Zgarb Apr 25 at 9:12
• A lot of great answers, and the LUA answer has a special place in my heart. Arnauld's answer is particularly sweet since the original question that I based this on (not the dupe) is actually tagged Javascript. Mainly selecting this one just because it does appear to be the overall shortest for all languages and, as this is my first question, I get a badge. – ouflak May 1 at 6:40

JavaScript (ES6), 22 bytes

Returns a Boolean value.

s=>/^(.*)\1+$/.test(s)  Try it online! Without a regular expression, 33 29 bytes Returns either null (falsy) or an object (truthy). s=>(s+s).slice(1,-1).match(s)  Try it online! NB: Technically, $$\s\$$ is converted to a regular expression for match(), so the above title is a lie. grep, 19 grep -qxE '(.+)\1+'  Test while read; do <<<"$REPLY" grep -qxE '(.+)\1+' && t="true" || t="false"
echo "$REPLY:$t"
done < infile


Output:

aa: true
aaa: true
abcabcabc: true
aba: false
ababa: false
weqweqweqweqweqw: false


Japt, 6 bytes

²é ¤øU


Saved one byte thanks to @Shaggy

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        Implicit input, stored in variable 'U'
²       U+U, "abcabc" -> "abcabcabcabc"
é      Rotate 1 char to the right "abcabcabcabc" -> "cabcabcabcab"
¤    Remove first two chars, "cabcabcabcab" -> "bcabcabcab"
øU  Check if U is in the above

• Nice one :) You can replace the p<space> with ² to save a byte. – Shaggy Apr 24 at 16:25

Java, 25 24 bytes

-1 byte thanks to Olivier Grégoire!

s->s.matches("(.+)\\1+")


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It's just 1 byte longer than the python answer aaaaa I'm tied now :)

• You can remove the final $ as the matches method is an exact match, not a substring match by default. – Olivier Grégoire Apr 24 at 23:18 • I forgot matches adds its own $ to the regex. Thanks! – Benjamin Urquhart Apr 24 at 23:34

Excel, 26 bytes

=FIND(A1,A1&A1,2)<=LEN(A1)


Inputs from A1, outputs to whatever cell you put this formula.

• You could save 4 bytes if you defined a single-letter range name (e.g. A) and set that as your input. – i_saw_drones Apr 24 at 18:54
• @i_saw_drones - I think that is disallowed by standard I/O rules: here's a link to the meta answer that would apply to that method; it's currently at -36 votes. – Sophia Lechner Apr 24 at 20:17
• Apologies I hadn't seen that post, although thinking about it, isn't A1 also a "variable" since it contains the input value? :) – i_saw_drones Apr 24 at 23:05
• I would feel that way if I were doing anything special with the fact that it's A1 specifically, like if I relied somehow on its ROW(_) being 1. As is, though, it's just the most natural way of providing an Excel function with an arbitrary input. – Sophia Lechner Apr 25 at 17:03

Retina 0.8.2, 9 bytes

^(.+)\1+$ Try it online! Link includes test cases. Jelly, 5 4 bytes I see now that the optimal way is to follow xnor's method! Ḋ;Ṗw  A monadic Link that accepts a list of characters and outputs an integer - the shortest possible length of a repeating slice or zero if none exists. Note that zero is falsey while non-zero numbers are truthy in Jelly. Try it online! How? Ḋ;Ṗw - Link: list of characters, S e.g. "abcabcabc" or "abababa" Ḋ - dequeue S "bcabcabc" "bababa" Ṗ - pop from S "abcabcab" "ababab" ; - concatenate "bcabcabcabcabcab" "bababaababab" w - first index of sublist 3 ^---here! 0 (not found)  R, 28 bytes grepl("(.+)\\1+$",scan(,''))


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Simple Regex version. R is (sometimes) very similar to Python, so this is similar to TFeld's Python 2 regex answer, albeit shorter!

Question (if anyone knows the answer)

I am still confused why this works, as the substring can be any length and will always work, and still works when I add a letter to the front of a valid string, like "cABABABABAB". If I personally read the regex, I see (.+), which captures any group of any length. And then \\1+$ which repeats the captured group any number of times until the end. So why doesn't it capture just "AB" and find that it is repeated until the end of the string, especially since there is no restriction specified as to where the substring can start? • Interesting, this seems to be a bug in R's regex engine. Adding the option perl=TRUE makes it match cABABAB, as you'd expect. Running grep -E '(.*)\1+$' in bash also matches cABABAB, even though grep -E uses ERE, the same regex flavor R is supposed to support. – Grimy Apr 25 at 15:43
• My guess is that this is an incorrectly applied optimization. Changing .+ at the start of a pattern to ^.+ is an important optimization, but if the .+ is inside capturing parens it stops being valid. – Grimy Apr 25 at 15:50

Perl 5-p, 14 bytes

$_=/^(.*)\1+$/


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Python 2, 24 bytes

lambda s:s in(s*2)[1:-1]


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Shamelessly stolen from xnor's answer to the original question.

More intuitive version:

Python 2, 5955 53 bytes

lambda s:s in[len(s)/i*s[:i]for i in range(1,len(s))]


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Boring regex version:

Python 2, 44 bytes

lambda s:re.match(r'(.+)\1+$',s)>0 import re  Try it online! Wolfram Language (Mathematica), 24 23 bytes StringMatchQ[x__..~~x_]  Try it online! StringMatchQ[ (*a function that checks if its input (string) matches:*) x__.. (*a sequence of one or more characters, repeated one or more times*) ~~x_] (*and one more time*)  Pyke, 4 bytes +tO{  Try it here! + - input+input t - ^[1:] O - ^[:-1] { - input in ^  J, 26 25 15 14 bytes Using xnor method +./@E.}:@}.@,~  Try it online! original (two different approaches) J, 25 bytes 1<1#.(#%#\)=<\+/@E.&:>"{]  Try it online! J, 26 bytes 1<1#.-@#$$[:(-:##{.)<$$"{]  Try it online! PowerShell, 23 24 bytes +1 byte to fully match rules "$args"-match"^(.+)\1+$"  Try it online! Pretty boring. Based on the other Regex answers. Luckily PowerShell doesn't use \ as an escape character! • it returns true for aabcabc – mazzy Apr 25 at 3:41 • @mazzy just fixed! – Gabriel Mills Apr 25 at 11:38 C# (Visual C# Interactive Compiler), 70 bytes xnor's shameless adaptation (46 bytes) s=>(s+s).Substring(1,s.Length*2-2).Contains(s)  My non Regex Solution: s=>s.Select((x,y)=>y).Count(z=>s.Replace(s.Substring(0,z+1),"")=="")>1  Explanation: Replace every possible substring that starts at index 0 with an empty string. If the result is an empty string, the string is entirely made of that substring. Since this includes evaluating the entire string with itself, the amount of expected results must be greater than 1. Example: abcabc Possible substrings starting at index 0: 'a', 'ab', 'abc', 'abca', 'abcab', 'abcabc'  If we replace them with empty strings Substring Result 'a' => 'bcbc' 'ab' => 'cc' 'abc' => '' 'abca' => 'bc' 'abcab' => 'c' 'abcabc' => ''  Since there is a substring other than 'abcabc' that returns an empty string, the string is entirely made of another substring ('abc') Try it online! Python 3, 626056 54 bytes -4 bytes thanx to ArBo lambda s:s in(len(s)//l*s[:l]for l in range(1,len(s)))  1. Iterate over all possible prefixes in the string. 2. Try to build the string out of the prefix. 3. Return whether this succeeds with any prefix at all. Try it online! • Nice answer! The f= can be dropped; anonymous functions are generally allowed. Also, by switching to Python 2 and checking membership of a list instead of the any construct, you can get to 55 bytes – ArBo Apr 26 at 8:09 • Nice catch with the list membership, thanx! I won't switch to Python 2, as this is like switching the language, which is obviously not the point here ;) Also, is there a convenient way to test an anonymous function in TIO, keeping the byte-count? – movatica Apr 26 at 14:13 • @movatica In the header, put f =  (\ is the line continuation character in python) – Artemis Fowl Apr 26 at 15:33 • Annoyingly, \ is also an escape character. Here, without code formatting, is what you should put in the header: f = \ – Artemis Fowl Apr 26 at 15:35 Japt, 10 bytes Returns a positive number if truthy and 0 if falsey. If you want a bool output just add -¡ flag å+ k@rXÃÊÉ  å+ k@rXÃÊÉ Full program. Implicit input U. e.g: U = "abcabcabc" å+ Take all prefixes U = ["a","ab","abc","abca","abcab","abcabc","abcabca","abcabcab","abcabcabc"] k@ Filter U by: rXÃ Values that return false (empty string) when replacing each prefix in U e.g: ["bcbcbc","ccc","","bcabc","cabc","abc","bc","c",""] take ↑ and ↑ U = ["abc","abcabcabc"] ÊÉ Get U length and subtract 1. Then return the result  Try it online! Husk, 6 bytes Ṡ€ȯhtD  Try it online! I feel like this is one byte more than optimal, but I couldn't find an arrangement that made the explicit composition ȯ unnecessary. Explanation Ṡ€ Find the argument in the result of applying the following function to the argument ȯhtD Duplicate the argument, then remove the first and last elements.  • €htD¹ avoids the ȯ. – Zgarb Apr 24 at 19:30 • That's fantastic! I had thought about λ€htD¹ but I didn't realize that lambdas would be added implicitly – Sophia Lechner Apr 24 at 20:10 Mathematica 11.x, 74 bytes {}!=StringCases[#,StartOfString~~x__/;(x!=#&&StringReplace[#,x->""]=="")]&  where, throughout, # represents the input string, and StringCases[#,<pattern>]  finds substrings of the input string matching the pattern StartOfString~~x__/;(x!=#&&StringReplace[#,x->""]=="")  This pattern requires matches, x, must start at the start of the string and must satisfy the condition that (1) the match is not the whole input string and (2) if we replace occurrences of the match in the input string with the empty string we obtain the empty string. Finally, comparing the list of matches to the empty list, {}!=  is True if the list of matches is nonempty and False if the list of matches is empty. Test cases: {}!=StringCases[#,StartOfString~~x__/;(x!=#&&StringReplace[#,x->""]=="")]&["aa"] (* True *) {}!=StringCases[#,StartOfString~~x__/;(x!=#&&StringReplace[#,x->""]=="")]&["aaa"] (* True *) {}!=StringCases[#,StartOfString~~x__/;(x!=#&&StringReplace[#,x->""]=="")]&["abcabc"] (* True *)  and {}!=StringCases[#,StartOfString~~x__/;(x!=#&&StringReplace[#,x->""]=="")]&["aba"] (* False *) {}!=StringCases[#,StartOfString~~x__/;(x!=#&&StringReplace[#,x->""]=="")]&["ababa"] (* False *) {}!=StringCases[#,StartOfString~~x__/;(x!=#&&StringReplace[#,x->""]=="")]&["weqweqweqweqweqw"] (* False *)  Python 3, 84 bytes import textwrap lambda s:any(len(set(textwrap.wrap(s,l)))<2 for l in range(1,len(s)))  Uses textwrap.wrap (thanks to this answer) to split the string into pieces of length n to test each possible length of repeating substring. The split pieces are then compared to each other by adding them to a set. If all of the pieces are equal, and the set is of length 1, then the string must be a repeating string. I used <2 instead of ==1 because it saves a byte, and the length of the input string was guaranteed to be greater than zero. If there is no n for which repeating substrings of length n make up the entire string, then return false for the whole function. 05AB1E, 5 bytes xnor's method from the previous question appears to be optimal in 05AB1E as well. «¦¨så  Try it online! or as a Test Suite Explanation « # append input to input ¦¨ # remove the first and last character of the resulting string så # check if the input is in this string  • Of course.. I was about to make a 05AB1E answer when I saw none were there. Colleague asked me some questions and talked about his vacation. I look back at the screen: one new answer. Tada, beat again XD – Kevin Cruijssen Apr 25 at 6:55 • @KevinCruijssen: That's typical. Has happened to me a bunch of times as well ;) – Emigna Apr 25 at 6:56 Clean, 73 bytes Doesn't use regex. import StdEnv,Data.List$s=or[isPrefixOf s(cycle t)\\t<-tl(tails s)|t>[]]


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Defines $:: [Char] -> Bool. Checks if the given string is a prefix of the repetition of any sub-string taken from the end. C++ (gcc), 36 bytes #define f(x)(x+x).find(x,1)<x.size()  Try it online! Another port of xnor's solution. Uses a macro to expand the argument into the expression. The argument is assumed to be of type std::string. QlikView Variable, 27 bytes This should be defined as a variable, which then allows you to pass parameters, e.g. $1 as your input value.

It returns 0 or -1 (equivalent to QlikView's TRUE() function).

=substringcount($1&$1,$1)>2  Swift, 196 bytes func r(s:String)->Bool{guard let k=s.dropFirst().firstIndex(where:{$0==s.first}) else{return false};let v=s[...k].dropLast();var w=v;while s.hasPrefix(w) && s.count>=(w+v).count{w+=v};return s==w}


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• I don't use Swift, but I'm sure that extra whitespace can be removed – Benjamin Urquhart Apr 24 at 19:56
• 193 bytes using @benjamin's suggestion. – Artemis Fowl Apr 26 at 15:39
• @ArtemisFowl or even 123 bytes – Roman Podymov Apr 26 at 16:59

Icon, 46 bytes

procedure f(s);return find(s,(s||s)[2:-1]);end


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Another port of xnor's solution.

K (oK), 29 bytes

{0<+/(1=#?:)'(0N,'1_!#x)#\:x}


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Red, 72 bytes

func[s][repeat i length? s[parse s[copy t i skip some t end(return 1)]]]


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Returns 1 for True

T-SQL, 47 bytes

Using @Xnor's method:

DECLARE @ varchar(max)='ababab'

PRINT sign(charindex(@,left(@+@,len(@)*2-1),2))


Keeping old answer as it contains some nice golfing(67 bytes):

DECLARE @y varchar(max)='abababa'

,@ INT=0WHILE
replace(@y,left(@y,@),'')>''SET
@+=1PRINT @/len(@y)^1


Explanation: This script is repeatingly trying to replace the input '@y' with the first '@' characters of the input '@y' with nothing, while increasing '@'.

if you replace 'ab' in 'ababab' with nothing you have an empty string

Eventually the result will be empty. If this happens when the loop variable is equal to the length of the varchar, the criteria is false/0 because '@'=len(@y) (there was no repeating varchar).

iif(@=len(@y),0,1)


can be golfed into this

@/len(@y)^1


because the length of '@y' can not be 0 and '@' will never exceed the length @y.

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