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.

edited May 1 at 7:40

asked Apr 24 at 14:55

ouflak

3201416

3

$begingroup$
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.
$endgroup$
– Erik the Outgolfer
Apr 24 at 15:14

$begingroup$
Let us continue this discussion in chat.
$endgroup$
– Erik the Outgolfer
Apr 24 at 15:36

add a comment |

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

edited May 1 at 7:40

asked Apr 24 at 14:55

ouflak

3201416

3

$begingroup$
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.
$endgroup$
– Erik the Outgolfer
Apr 24 at 15:14

$begingroup$
Let us continue this discussion in chat.
$endgroup$
– Erik the Outgolfer
Apr 24 at 15:36

add a comment |

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

edited May 1 at 7:40

asked Apr 24 at 14:55

ouflak

3201416

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

code-golf decision-problem

edited May 1 at 7:40

asked Apr 24 at 14:55

ouflak

3201416

edited May 1 at 7:40

asked Apr 24 at 14:55

ouflak

3201416

edited May 1 at 7:40

asked Apr 24 at 14:55

ouflak

3201416

asked Apr 24 at 14:55

ouflak

3201416

asked Apr 24 at 14:55

ouflak

3201416

3

$begingroup$
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.
$endgroup$
– Erik the Outgolfer
Apr 24 at 15:14

$begingroup$
Let us continue this discussion in chat.
$endgroup$
– Erik the Outgolfer
Apr 24 at 15:36

add a comment |

3

$begingroup$
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.
$endgroup$
– Erik the Outgolfer
Apr 24 at 15:14

$begingroup$
Let us continue this discussion in chat.
$endgroup$
– Erik the Outgolfer
Apr 24 at 15:36

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

Let us continue this discussion in chat.

– Erik the Outgolfer
Apr 24 at 15:36

add a comment |

40 Answers
40

active

oldest

votes

Brachylog, 4 3 bytes

ġ=Ṁ

Try it online!

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.

edited Apr 25 at 0:22

Unrelated String

2,190313

answered Apr 24 at 19:28

Zgarb

27.2k462233

$begingroup$
I was just struggling through trying to get something like ~j↙ or =Ṁc working before I noticed you posted this an hour ago
$endgroup$
– Unrelated String
Apr 24 at 21:27

4

$begingroup$
Oh, yeah, this could be one byte shorter: ġ=Ṁ
$endgroup$
– Unrelated String
Apr 24 at 21:36

$begingroup$
(Ṁ is a variable constrained to be a list of two or more elements)
$endgroup$
– Unrelated String
Apr 24 at 21:39

1

$begingroup$
@UnrelatedString Great, thanks! I didn't think to check the variables wiki page.
$endgroup$
– Zgarb
Apr 25 at 9:12

1

$begingroup$
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.
$endgroup$
– ouflak
May 1 at 6:40

|
show 4 more comments

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.

edited Apr 24 at 16:34

answered Apr 24 at 15:01

Arnauld

85.1k7100349

add a comment |

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

edited Apr 24 at 16:02

answered Apr 24 at 15:56

Thor

2,38621219

add a comment |

Japt, 6 bytes

²é ¤øU

Saved one byte thanks to @Shaggy

Try it online!

        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

edited Apr 24 at 16:28

answered Apr 24 at 16:14

Embodiment of Ignorance

3,874128

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

add a comment |

Java, 25 24 bytes

-1 byte thanks to Olivier Grégoire!

Boring regex answer

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

Try it online!

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

edited Apr 24 at 23:35

answered Apr 24 at 19:49

Benjamin Urquhart

1,036116

3

$begingroup$
You can remove the final $ as the matches method is an exact match, not a substring match by default.
$endgroup$
– Olivier Grégoire
Apr 24 at 23:18

$begingroup$
I forgot matches adds its own $ to the regex. Thanks!
$endgroup$
– Benjamin Urquhart
Apr 24 at 23:34

add a comment |

Excel, 26 bytes

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

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

answered Apr 24 at 16:51

Sophia Lechner

1,03018

$begingroup$
You could save 4 bytes if you defined a single-letter range name (e.g. A) and set that as your input.
$endgroup$
– i_saw_drones
Apr 24 at 18:54

$begingroup$
@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.
$endgroup$
– Sophia Lechner
Apr 24 at 20:17

$begingroup$
Apologies I hadn't seen that post, although thinking about it, isn't A1 also a "variable" since it contains the input value? :)
$endgroup$
– i_saw_drones
Apr 24 at 23:05

1

$begingroup$
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.
$endgroup$
– Sophia Lechner
Apr 25 at 17:03

add a comment |

Retina 0.8.2, 9 bytes

^(.+)1+$

Try it online! Link includes test cases.

answered Apr 24 at 15:13

Neil

84.6k845183

add a comment |

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)

edited Apr 24 at 19:07

answered Apr 24 at 16:18

Jonathan Allan

55.9k538178

add a comment |

R, 28 bytes

grepl("(.+)\1+$",scan(,''))

Try it online!

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?

edited Apr 24 at 19:55

answered Apr 24 at 19:46

Sumner18

68018

1

$begingroup$
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.
$endgroup$
– Grimy
Apr 25 at 15:43

1

$begingroup$
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.
$endgroup$
– Grimy
Apr 25 at 15:50

add a comment |

Perl 5 `-p`, 14 bytes

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

Try it online!

answered Apr 24 at 16:32

Xcali

5,890523

add a comment |

Python 2, 24 bytes

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

Try it online!

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

More intuitive version:

Python 2, 59 55 53 bytes

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

Try it online!

Boring regex version:

Python 2, 44 bytes

lambda s:re.match(r'(.+)1+$',s)>0

import re

Try it online!

edited Apr 24 at 18:37

answered Apr 24 at 16:28

TFeld

16.9k31452

add a comment |

Pyke, 4 bytes

+tO{

Try it here!

+    -    input+input

 t   -   ^[1:]

  O  -  ^[:-1]

   { - input in ^

answered Apr 24 at 17:44

Blue

24.1k73891

add a comment |

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!

edited Apr 24 at 21:33

answered Apr 24 at 17:13

Jonah

3,3781019

add a comment |

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*)

edited Apr 25 at 9:47

answered Apr 25 at 3:48

attinat

1,05717

add a comment |

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!

edited Apr 25 at 11:36

answered Apr 24 at 20:13

Gabriel Mills

673213

$begingroup$
it returns true for aabcabc
$endgroup$
– mazzy
Apr 25 at 3:41

1

$begingroup$
@mazzy just fixed!
$endgroup$
– Gabriel Mills
Apr 25 at 11:38

add a comment |

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!

edited Apr 26 at 12:37

answered Apr 25 at 9:09

Innat3

3014

add a comment |

Python 3, 62 60 56 54 bytes

-4 bytes thanx to ArBo

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

Iterate over all possible prefixes in the string.

Try to build the string out of the prefix.

Return whether this succeeds with any prefix at all.

Try it online!

edited Apr 28 at 0:50

answered Apr 25 at 19:11

movatica

1786

1

$begingroup$
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
$endgroup$
– ArBo
Apr 26 at 8:09

1

$begingroup$
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?
$endgroup$
– movatica
Apr 26 at 14:13

1

$begingroup$
@movatica In the header, put `f = ` ( is the line continuation character in python)
$endgroup$
– Artemis Fowl
Apr 26 at 15:33

$begingroup$
Annoyingly, is also an escape character. Here, without code formatting, is what you should put in the header: f =
$endgroup$
– Artemis Fowl
Apr 26 at 15:35

add a comment |

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!

edited Apr 24 at 15:33

answered Apr 24 at 15:18

Luis felipe De jesus Munoz

6,30121874

add a comment |

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.

answered Apr 24 at 17:16

Sophia Lechner

1,03018

2

$begingroup$
€htD¹ avoids the ȯ.
$endgroup$
– Zgarb
Apr 24 at 19:30

$begingroup$
That's fantastic! I had thought about λ€htD¹ but I didn't realize that lambdas would be added implicitly
$endgroup$
– Sophia Lechner
Apr 24 at 20:10

add a comment |

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  *)

answered Apr 24 at 17:40

Eric Towers

66637

add a comment |

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.

answered Apr 24 at 20:20

Delya Erricson

8113

add a comment |

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

answered Apr 25 at 6:33

Emigna

49.5k534150

1

$begingroup$
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
$endgroup$
– Kevin Cruijssen
Apr 25 at 6:55

$begingroup$
@KevinCruijssen: That's typical. Has happened to me a bunch of times as well ;)
$endgroup$
– Emigna
Apr 25 at 6:56

add a comment |

Clean, 73 bytes

Doesn't use regex.

import StdEnv,Data.List

$s=or[isPrefixOf s(cycle t)\t<-tl(tails s)|t>]

Try it online!

Defines $ :: [Char] -> Bool.

Checks if the given string is a prefix of the repetition of any sub-string taken from the end.

answered Apr 26 at 0:43

Οurous

7,47611136

add a comment |

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.

edited Apr 26 at 15:35

answered Apr 25 at 23:42

jxh

32128

add a comment |

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

answered Apr 24 at 19:19

i_saw_drones

2835

add a comment |

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}

Try it online!

answered Apr 24 at 19:32

onnoweb

1713

$begingroup$
I don't use Swift, but I'm sure that extra whitespace can be removed
$endgroup$
– Benjamin Urquhart
Apr 24 at 19:56

$begingroup$
193 bytes using @benjamin's suggestion.
$endgroup$
– Artemis Fowl
Apr 26 at 15:39

$begingroup$
@ArtemisFowl or even 123 bytes
$endgroup$
– Roman Podymov
Apr 26 at 16:59

add a comment |

Icon, 46 bytes

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

Try it online!

Another port of xnor's solution.

edited Apr 25 at 7:41

answered Apr 25 at 7:13

Galen Ivanov

8,33211237

add a comment |

K (oK), 29 bytes

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

Try it online!

answered Apr 25 at 8:52

Galen Ivanov

8,33211237

add a comment |

Red, 72 bytes

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

Try it online!

Returns 1 for True

answered Apr 25 at 10:36

Galen Ivanov

8,33211237

add a comment |

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.

Try it online

edited Apr 25 at 10:47

answered Apr 24 at 17:04

t-clausen.dk

2,214514

add a comment |

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40 Answers
40

active

oldest

votes

40 Answers
40

active

oldest

votes

Brachylog, 4 3 bytes

ġ=Ṁ

Try it online!

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.

edited Apr 25 at 0:22

Unrelated String

2,190313

answered Apr 24 at 19:28

Zgarb

27.2k462233

$begingroup$
I was just struggling through trying to get something like ~j↙ or =Ṁc working before I noticed you posted this an hour ago
$endgroup$
– Unrelated String
Apr 24 at 21:27

4

$begingroup$
Oh, yeah, this could be one byte shorter: ġ=Ṁ
$endgroup$
– Unrelated String
Apr 24 at 21:36

$begingroup$
(Ṁ is a variable constrained to be a list of two or more elements)
$endgroup$
– Unrelated String
Apr 24 at 21:39

1

$begingroup$
@UnrelatedString Great, thanks! I didn't think to check the variables wiki page.
$endgroup$
– Zgarb
Apr 25 at 9:12

1

$begingroup$
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.
$endgroup$
– ouflak
May 1 at 6:40

|
show 4 more comments

Brachylog, 4 3 bytes

ġ=Ṁ

Try it online!

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.

edited Apr 25 at 0:22

Unrelated String

2,190313

answered Apr 24 at 19:28

Zgarb

27.2k462233

$begingroup$
I was just struggling through trying to get something like ~j↙ or =Ṁc working before I noticed you posted this an hour ago
$endgroup$
– Unrelated String
Apr 24 at 21:27

4

$begingroup$
Oh, yeah, this could be one byte shorter: ġ=Ṁ
$endgroup$
– Unrelated String
Apr 24 at 21:36

$begingroup$
(Ṁ is a variable constrained to be a list of two or more elements)
$endgroup$
– Unrelated String
Apr 24 at 21:39

1

$begingroup$
@UnrelatedString Great, thanks! I didn't think to check the variables wiki page.
$endgroup$
– Zgarb
Apr 25 at 9:12

1

$begingroup$
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.
$endgroup$
– ouflak
May 1 at 6:40

|
show 4 more comments

Brachylog, 4 3 bytes

ġ=Ṁ

Try it online!

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.

edited Apr 25 at 0:22

Unrelated String

2,190313

answered Apr 24 at 19:28

Zgarb

27.2k462233

Brachylog, 4 3 bytes

ġ=Ṁ

Try it online!

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.

edited Apr 25 at 0:22

Unrelated String

2,190313

answered Apr 24 at 19:28

Zgarb

27.2k462233

edited Apr 25 at 0:22

Unrelated String

2,190313

edited Apr 25 at 0:22

Unrelated String

2,190313

edited Apr 25 at 0:22

Unrelated String

2,190313

answered Apr 24 at 19:28

Zgarb

27.2k462233

answered Apr 24 at 19:28

Zgarb

27.2k462233

answered Apr 24 at 19:28

Zgarb

27.2k462233

$begingroup$
I was just struggling through trying to get something like ~j↙ or =Ṁc working before I noticed you posted this an hour ago
$endgroup$
– Unrelated String
Apr 24 at 21:27

4

$begingroup$
Oh, yeah, this could be one byte shorter: ġ=Ṁ
$endgroup$
– Unrelated String
Apr 24 at 21:36

$begingroup$
(Ṁ is a variable constrained to be a list of two or more elements)
$endgroup$
– Unrelated String
Apr 24 at 21:39

1

$begingroup$
@UnrelatedString Great, thanks! I didn't think to check the variables wiki page.
$endgroup$
– Zgarb
Apr 25 at 9:12

1

$begingroup$
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.
$endgroup$
– ouflak
May 1 at 6:40

|
show 4 more comments

$begingroup$
I was just struggling through trying to get something like ~j↙ or =Ṁc working before I noticed you posted this an hour ago
$endgroup$
– Unrelated String
Apr 24 at 21:27

4

$begingroup$
Oh, yeah, this could be one byte shorter: ġ=Ṁ
$endgroup$
– Unrelated String
Apr 24 at 21:36

$begingroup$
(Ṁ is a variable constrained to be a list of two or more elements)
$endgroup$
– Unrelated String
Apr 24 at 21:39

1

$begingroup$
@UnrelatedString Great, thanks! I didn't think to check the variables wiki page.
$endgroup$
– Zgarb
Apr 25 at 9:12

1

$begingroup$
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.
$endgroup$
– ouflak
May 1 at 6:40

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

|
show 4 more comments

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.

edited Apr 24 at 16:34

answered Apr 24 at 15:01

Arnauld

85.1k7100349

add a comment |

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.

edited Apr 24 at 16:34

answered Apr 24 at 15:01

Arnauld

85.1k7100349

add a comment |

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.

edited Apr 24 at 16:34

answered Apr 24 at 15:01

Arnauld

85.1k7100349

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.

edited Apr 24 at 16:34

answered Apr 24 at 15:01

Arnauld

85.1k7100349

edited Apr 24 at 16:34

answered Apr 24 at 15:01

Arnauld

85.1k7100349

answered Apr 24 at 15:01

Arnauld

85.1k7100349

answered Apr 24 at 15:01

Arnauld

85.1k7100349

add a comment |

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

edited Apr 24 at 16:02

answered Apr 24 at 15:56

Thor

2,38621219

add a comment |

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

edited Apr 24 at 16:02

answered Apr 24 at 15:56

Thor

2,38621219

add a comment |

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

edited Apr 24 at 16:02

answered Apr 24 at 15:56

Thor

2,38621219

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

edited Apr 24 at 16:02

answered Apr 24 at 15:56

Thor

2,38621219

edited Apr 24 at 16:02

answered Apr 24 at 15:56

Thor

2,38621219

answered Apr 24 at 15:56

Thor

2,38621219

answered Apr 24 at 15:56

Thor

2,38621219

add a comment |

Japt, 6 bytes

²é ¤øU

Saved one byte thanks to @Shaggy

Try it online!

        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

edited Apr 24 at 16:28

answered Apr 24 at 16:14

Embodiment of Ignorance

3,874128

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

add a comment |

Japt, 6 bytes

²é ¤øU

Saved one byte thanks to @Shaggy

Try it online!

        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

edited Apr 24 at 16:28

answered Apr 24 at 16:14

Embodiment of Ignorance

3,874128

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

add a comment |

Japt, 6 bytes

²é ¤øU

Saved one byte thanks to @Shaggy

Try it online!

        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

edited Apr 24 at 16:28

answered Apr 24 at 16:14

Embodiment of Ignorance

3,874128

Japt, 6 bytes

²é ¤øU

Saved one byte thanks to @Shaggy

Try it online!

        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

edited Apr 24 at 16:28

answered Apr 24 at 16:14

Embodiment of Ignorance

3,874128

edited Apr 24 at 16:28

answered Apr 24 at 16:14

Embodiment of Ignorance

3,874128

answered Apr 24 at 16:14

Embodiment of Ignorance

3,874128

answered Apr 24 at 16:14

Embodiment of Ignorance

3,874128

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

add a comment |

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

Nice one :) You can replace the p<space> with ² to save a byte.

– Shaggy
Apr 24 at 16:25

add a comment |

Java, 25 24 bytes

-1 byte thanks to Olivier Grégoire!

Boring regex answer

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

Try it online!

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

edited Apr 24 at 23:35

answered Apr 24 at 19:49

Benjamin Urquhart

1,036116

3

$begingroup$
You can remove the final $ as the matches method is an exact match, not a substring match by default.
$endgroup$
– Olivier Grégoire
Apr 24 at 23:18

$begingroup$
I forgot matches adds its own $ to the regex. Thanks!
$endgroup$
– Benjamin Urquhart
Apr 24 at 23:34

add a comment |

Java, 25 24 bytes

-1 byte thanks to Olivier Grégoire!

Boring regex answer

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

Try it online!

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

edited Apr 24 at 23:35

answered Apr 24 at 19:49

Benjamin Urquhart

1,036116

3

$begingroup$
You can remove the final $ as the matches method is an exact match, not a substring match by default.
$endgroup$
– Olivier Grégoire
Apr 24 at 23:18

$begingroup$
I forgot matches adds its own $ to the regex. Thanks!
$endgroup$
– Benjamin Urquhart
Apr 24 at 23:34

add a comment |

Java, 25 24 bytes

-1 byte thanks to Olivier Grégoire!

Boring regex answer

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

Try it online!

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

edited Apr 24 at 23:35

answered Apr 24 at 19:49

Benjamin Urquhart

1,036116

Java, 25 24 bytes

-1 byte thanks to Olivier Grégoire!

Boring regex answer

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

Try it online!

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

edited Apr 24 at 23:35

answered Apr 24 at 19:49

Benjamin Urquhart

1,036116

edited Apr 24 at 23:35

answered Apr 24 at 19:49

Benjamin Urquhart

1,036116

answered Apr 24 at 19:49

Benjamin Urquhart

1,036116

answered Apr 24 at 19:49

Benjamin Urquhart

1,036116

3

$begingroup$
You can remove the final $ as the matches method is an exact match, not a substring match by default.
$endgroup$
– Olivier Grégoire
Apr 24 at 23:18

$begingroup$
I forgot matches adds its own $ to the regex. Thanks!
$endgroup$
– Benjamin Urquhart
Apr 24 at 23:34

add a comment |

3

$begingroup$
You can remove the final $ as the matches method is an exact match, not a substring match by default.
$endgroup$
– Olivier Grégoire
Apr 24 at 23:18

$begingroup$
I forgot matches adds its own $ to the regex. Thanks!
$endgroup$
– Benjamin Urquhart
Apr 24 at 23:34

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

add a comment |

Excel, 26 bytes

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

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

answered Apr 24 at 16:51

Sophia Lechner

1,03018

$begingroup$
You could save 4 bytes if you defined a single-letter range name (e.g. A) and set that as your input.
$endgroup$
– i_saw_drones
Apr 24 at 18:54

$begingroup$
@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.
$endgroup$
– Sophia Lechner
Apr 24 at 20:17

$begingroup$
Apologies I hadn't seen that post, although thinking about it, isn't A1 also a "variable" since it contains the input value? :)
$endgroup$
– i_saw_drones
Apr 24 at 23:05

1

$begingroup$
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.
$endgroup$
– Sophia Lechner
Apr 25 at 17:03

add a comment |

Excel, 26 bytes

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

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

answered Apr 24 at 16:51

Sophia Lechner

1,03018

$begingroup$
You could save 4 bytes if you defined a single-letter range name (e.g. A) and set that as your input.
$endgroup$
– i_saw_drones
Apr 24 at 18:54

$begingroup$
@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.
$endgroup$
– Sophia Lechner
Apr 24 at 20:17

$begingroup$
Apologies I hadn't seen that post, although thinking about it, isn't A1 also a "variable" since it contains the input value? :)
$endgroup$
– i_saw_drones
Apr 24 at 23:05

1

$begingroup$
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.
$endgroup$
– Sophia Lechner
Apr 25 at 17:03

add a comment |

Excel, 26 bytes

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

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

answered Apr 24 at 16:51

Sophia Lechner

1,03018

Excel, 26 bytes

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

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

answered Apr 24 at 16:51

Sophia Lechner

1,03018

answered Apr 24 at 16:51

Sophia Lechner

1,03018

answered Apr 24 at 16:51

Sophia Lechner

1,03018

answered Apr 24 at 16:51

Sophia Lechner

1,03018

$begingroup$
You could save 4 bytes if you defined a single-letter range name (e.g. A) and set that as your input.
$endgroup$
– i_saw_drones
Apr 24 at 18:54

$begingroup$
@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.
$endgroup$
– Sophia Lechner
Apr 24 at 20:17

$begingroup$
Apologies I hadn't seen that post, although thinking about it, isn't A1 also a "variable" since it contains the input value? :)
$endgroup$
– i_saw_drones
Apr 24 at 23:05

1

$begingroup$
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.
$endgroup$
– Sophia Lechner
Apr 25 at 17:03

add a comment |

$begingroup$
You could save 4 bytes if you defined a single-letter range name (e.g. A) and set that as your input.
$endgroup$
– i_saw_drones
Apr 24 at 18:54

$begingroup$
@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.
$endgroup$
– Sophia Lechner
Apr 24 at 20:17

$begingroup$
Apologies I hadn't seen that post, although thinking about it, isn't A1 also a "variable" since it contains the input value? :)
$endgroup$
– i_saw_drones
Apr 24 at 23:05

1

$begingroup$
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.
$endgroup$
– Sophia Lechner
Apr 25 at 17:03

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

add a comment |

Retina 0.8.2, 9 bytes

^(.+)1+$

Try it online! Link includes test cases.

answered Apr 24 at 15:13

Neil

84.6k845183

add a comment |

Retina 0.8.2, 9 bytes

^(.+)1+$

Try it online! Link includes test cases.

answered Apr 24 at 15:13

Neil

84.6k845183

add a comment |

Retina 0.8.2, 9 bytes

^(.+)1+$

Try it online! Link includes test cases.

answered Apr 24 at 15:13

Neil

84.6k845183

Retina 0.8.2, 9 bytes

^(.+)1+$

Try it online! Link includes test cases.

answered Apr 24 at 15:13

Neil

84.6k845183

answered Apr 24 at 15:13

Neil

84.6k845183

answered Apr 24 at 15:13

Neil

84.6k845183

answered Apr 24 at 15:13

Neil

84.6k845183

add a comment |

Jelly, 5 4 bytes

I see now that the optimal way is to follow xnor's method!

Ḋ;Ṗw

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)

edited Apr 24 at 19:07

answered Apr 24 at 16:18

Jonathan Allan

55.9k538178

add a comment |

Jelly, 5 4 bytes

I see now that the optimal way is to follow xnor's method!

Ḋ;Ṗw

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)

edited Apr 24 at 19:07

answered Apr 24 at 16:18

Jonathan Allan

55.9k538178

add a comment |

Jelly, 5 4 bytes

I see now that the optimal way is to follow xnor's method!

Ḋ;Ṗw

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)

edited Apr 24 at 19:07

answered Apr 24 at 16:18

Jonathan Allan

55.9k538178

Jelly, 5 4 bytes

I see now that the optimal way is to follow xnor's method!

Ḋ;Ṗw

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)

edited Apr 24 at 19:07

answered Apr 24 at 16:18

Jonathan Allan

55.9k538178

edited Apr 24 at 19:07

answered Apr 24 at 16:18

Jonathan Allan

55.9k538178

answered Apr 24 at 16:18

Jonathan Allan

55.9k538178

answered Apr 24 at 16:18

Jonathan Allan

55.9k538178

add a comment |

R, 28 bytes

grepl("(.+)\1+$",scan(,''))

Try it online!

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)

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?

edited Apr 24 at 19:55

answered Apr 24 at 19:46

Sumner18

68018

1

$begingroup$
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.
$endgroup$
– Grimy
Apr 25 at 15:43

1

$begingroup$
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.
$endgroup$
– Grimy
Apr 25 at 15:50

add a comment |

R, 28 bytes

grepl("(.+)\1+$",scan(,''))

Try it online!

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)

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?

edited Apr 24 at 19:55

answered Apr 24 at 19:46

Sumner18

68018

1

$begingroup$
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.
$endgroup$
– Grimy
Apr 25 at 15:43

1

$begingroup$
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.
$endgroup$
– Grimy
Apr 25 at 15:50

add a comment |

R, 28 bytes

grepl("(.+)\1+$",scan(,''))

Try it online!

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)

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?

edited Apr 24 at 19:55

answered Apr 24 at 19:46

Sumner18

68018

R, 28 bytes

grepl("(.+)\1+$",scan(,''))

Try it online!

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)

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?

edited Apr 24 at 19:55

answered Apr 24 at 19:46

Sumner18

68018

edited Apr 24 at 19:55

answered Apr 24 at 19:46

Sumner18

68018

answered Apr 24 at 19:46

Sumner18

68018

answered Apr 24 at 19:46

Sumner18

68018

1

$begingroup$
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.
$endgroup$
– Grimy
Apr 25 at 15:43

1

$begingroup$
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.
$endgroup$
– Grimy
Apr 25 at 15:50

add a comment |

1

$begingroup$
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.
$endgroup$
– Grimy
Apr 25 at 15:43

1

$begingroup$
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.
$endgroup$
– Grimy
Apr 25 at 15:50

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

add a comment |

Perl 5 `-p`, 14 bytes

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

Try it online!

answered Apr 24 at 16:32

Xcali

5,890523

add a comment |

Perl 5 `-p`, 14 bytes

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

Try it online!

answered Apr 24 at 16:32

Xcali

5,890523

add a comment |

Perl 5 `-p`, 14 bytes

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

Try it online!

answered Apr 24 at 16:32

Xcali

5,890523

Perl 5 `-p`, 14 bytes

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

Try it online!

answered Apr 24 at 16:32

Xcali

5,890523

answered Apr 24 at 16:32

Xcali

5,890523

answered Apr 24 at 16:32

Xcali

5,890523

answered Apr 24 at 16:32

Xcali

5,890523

add a comment |

Python 2, 24 bytes

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

Try it online!

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

More intuitive version:

Python 2, 59 55 53 bytes

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

Try it online!

Boring regex version:

Python 2, 44 bytes

lambda s:re.match(r'(.+)1+$',s)>0

import re

Try it online!

edited Apr 24 at 18:37

answered Apr 24 at 16:28

TFeld

16.9k31452

add a comment |

Python 2, 24 bytes

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

Try it online!

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

More intuitive version:

Python 2, 59 55 53 bytes

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

Try it online!

Boring regex version:

Python 2, 44 bytes

lambda s:re.match(r'(.+)1+$',s)>0

import re

Try it online!

edited Apr 24 at 18:37

answered Apr 24 at 16:28

TFeld

16.9k31452

add a comment |

Python 2, 24 bytes

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

Try it online!

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

More intuitive version:

Python 2, 59 55 53 bytes

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

Try it online!

Boring regex version:

Python 2, 44 bytes

lambda s:re.match(r'(.+)1+$',s)>0

import re

Try it online!

edited Apr 24 at 18:37

answered Apr 24 at 16:28

TFeld

16.9k31452

Python 2, 24 bytes

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

Try it online!

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

More intuitive version:

Python 2, 59 55 53 bytes

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

Try it online!

Boring regex version:

Python 2, 44 bytes

lambda s:re.match(r'(.+)1+$',s)>0

import re

Try it online!

edited Apr 24 at 18:37

answered Apr 24 at 16:28

TFeld

16.9k31452

edited Apr 24 at 18:37

answered Apr 24 at 16:28

TFeld

16.9k31452

answered Apr 24 at 16:28

TFeld

16.9k31452

answered Apr 24 at 16:28

TFeld

16.9k31452

add a comment |

Pyke, 4 bytes

+tO{

Try it here!

+    -    input+input

 t   -   ^[1:]

  O  -  ^[:-1]

   { - input in ^

answered Apr 24 at 17:44

Blue

24.1k73891

add a comment |

Pyke, 4 bytes

+tO{

Try it here!

+    -    input+input

 t   -   ^[1:]

  O  -  ^[:-1]

   { - input in ^

answered Apr 24 at 17:44

Blue

24.1k73891

add a comment |

Pyke, 4 bytes

+tO{

Try it here!

+    -    input+input

 t   -   ^[1:]

  O  -  ^[:-1]

   { - input in ^

answered Apr 24 at 17:44

Blue

24.1k73891

Pyke, 4 bytes

+tO{

Try it here!

+    -    input+input

 t   -   ^[1:]

  O  -  ^[:-1]

   { - input in ^

answered Apr 24 at 17:44

Blue

24.1k73891

answered Apr 24 at 17:44

Blue

24.1k73891

answered Apr 24 at 17:44

Blue

24.1k73891

answered Apr 24 at 17:44

Blue

24.1k73891

add a comment |

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!

edited Apr 24 at 21:33

answered Apr 24 at 17:13

Jonah

3,3781019

add a comment |

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!

edited Apr 24 at 21:33

answered Apr 24 at 17:13

Jonah

3,3781019

add a comment |

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!

edited Apr 24 at 21:33

answered Apr 24 at 17:13

Jonah

3,3781019

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!

edited Apr 24 at 21:33

answered Apr 24 at 17:13

Jonah

3,3781019

edited Apr 24 at 21:33

answered Apr 24 at 17:13

Jonah

3,3781019

answered Apr 24 at 17:13

Jonah

3,3781019

answered Apr 24 at 17:13

Jonah

3,3781019

add a comment |

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*)

edited Apr 25 at 9:47

answered Apr 25 at 3:48

attinat

1,05717

add a comment |

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*)

edited Apr 25 at 9:47

answered Apr 25 at 3:48

attinat

1,05717

add a comment |

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*)

edited Apr 25 at 9:47

answered Apr 25 at 3:48

attinat

1,05717

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*)

edited Apr 25 at 9:47

answered Apr 25 at 3:48

attinat

1,05717

edited Apr 25 at 9:47

answered Apr 25 at 3:48

attinat

1,05717

answered Apr 25 at 3:48

attinat

1,05717

answered Apr 25 at 3:48

attinat

1,05717

add a comment |

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!

edited Apr 25 at 11:36

answered Apr 24 at 20:13

Gabriel Mills

673213

$begingroup$
it returns true for aabcabc
$endgroup$
– mazzy
Apr 25 at 3:41

1

$begingroup$
@mazzy just fixed!
$endgroup$
– Gabriel Mills
Apr 25 at 11:38

add a comment |

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!

edited Apr 25 at 11:36

answered Apr 24 at 20:13

Gabriel Mills

673213

$begingroup$
it returns true for aabcabc
$endgroup$
– mazzy
Apr 25 at 3:41

1

$begingroup$
@mazzy just fixed!
$endgroup$
– Gabriel Mills
Apr 25 at 11:38

add a comment |

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!

edited Apr 25 at 11:36

answered Apr 24 at 20:13

Gabriel Mills

673213

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!

edited Apr 25 at 11:36

answered Apr 24 at 20:13

Gabriel Mills

673213

edited Apr 25 at 11:36

answered Apr 24 at 20:13

Gabriel Mills

673213

answered Apr 24 at 20:13

Gabriel Mills

673213

answered Apr 24 at 20:13

Gabriel Mills

673213

$begingroup$
it returns true for aabcabc
$endgroup$
– mazzy
Apr 25 at 3:41

1

$begingroup$
@mazzy just fixed!
$endgroup$
– Gabriel Mills
Apr 25 at 11:38

add a comment |

$begingroup$
it returns true for aabcabc
$endgroup$
– mazzy
Apr 25 at 3:41

1

$begingroup$
@mazzy just fixed!
$endgroup$
– Gabriel Mills
Apr 25 at 11:38

it returns true for aabcabc

– mazzy
Apr 25 at 3:41

@mazzy just fixed!

– Gabriel Mills
Apr 25 at 11:38

add a comment |

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:

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!

edited Apr 26 at 12:37

answered Apr 25 at 9:09

Innat3

3014

add a comment |

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:

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!

edited Apr 26 at 12:37

answered Apr 25 at 9:09

Innat3

3014

add a comment |

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:

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!

edited Apr 26 at 12:37

answered Apr 25 at 9:09

Innat3

3014

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:

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!

edited Apr 26 at 12:37

answered Apr 25 at 9:09

Innat3

3014

edited Apr 26 at 12:37

answered Apr 25 at 9:09

Innat3

3014

answered Apr 25 at 9:09

Innat3

3014

answered Apr 25 at 9:09

Innat3

3014

add a comment |

Python 3, 62 60 56 54 bytes

-4 bytes thanx to ArBo

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

Iterate over all possible prefixes in the string.

Try to build the string out of the prefix.

Return whether this succeeds with any prefix at all.

Try it online!

edited Apr 28 at 0:50

answered Apr 25 at 19:11

movatica

1786

1

$begingroup$
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
$endgroup$
– ArBo
Apr 26 at 8:09

1

$begingroup$
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?
$endgroup$
– movatica
Apr 26 at 14:13

1

$begingroup$
@movatica In the header, put `f = ` ( is the line continuation character in python)
$endgroup$
– Artemis Fowl
Apr 26 at 15:33

$begingroup$
Annoyingly, is also an escape character. Here, without code formatting, is what you should put in the header: f =
$endgroup$
– Artemis Fowl
Apr 26 at 15:35

add a comment |

Python 3, 62 60 56 54 bytes

-4 bytes thanx to ArBo

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

Iterate over all possible prefixes in the string.

Try to build the string out of the prefix.

Return whether this succeeds with any prefix at all.

Try it online!

edited Apr 28 at 0:50

answered Apr 25 at 19:11

movatica

1786

1

$begingroup$
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
$endgroup$
– ArBo
Apr 26 at 8:09

1

$begingroup$
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?
$endgroup$
– movatica
Apr 26 at 14:13

1

$begingroup$
@movatica In the header, put `f = ` ( is the line continuation character in python)
$endgroup$
– Artemis Fowl
Apr 26 at 15:33

$begingroup$
Annoyingly, is also an escape character. Here, without code formatting, is what you should put in the header: f =
$endgroup$
– Artemis Fowl
Apr 26 at 15:35

add a comment |

Python 3, 62 60 56 54 bytes

-4 bytes thanx to ArBo

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

Iterate over all possible prefixes in the string.

Try to build the string out of the prefix.

Return whether this succeeds with any prefix at all.

Try it online!

edited Apr 28 at 0:50

answered Apr 25 at 19:11

movatica

1786

Python 3, 62 60 56 54 bytes

-4 bytes thanx to ArBo

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

Iterate over all possible prefixes in the string.

Try to build the string out of the prefix.

Return whether this succeeds with any prefix at all.

Try it online!

edited Apr 28 at 0:50

answered Apr 25 at 19:11

movatica

1786

edited Apr 28 at 0:50

answered Apr 25 at 19:11

movatica

1786

answered Apr 25 at 19:11

movatica

1786

answered Apr 25 at 19:11

movatica

1786

1

$begingroup$
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
$endgroup$
– ArBo
Apr 26 at 8:09

1

$begingroup$
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?
$endgroup$
– movatica
Apr 26 at 14:13

1

$begingroup$
@movatica In the header, put `f = ` ( is the line continuation character in python)
$endgroup$
– Artemis Fowl
Apr 26 at 15:33

$begingroup$
Annoyingly, is also an escape character. Here, without code formatting, is what you should put in the header: f =
$endgroup$
– Artemis Fowl
Apr 26 at 15:35

add a comment |

1

$begingroup$
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
$endgroup$
– ArBo
Apr 26 at 8:09

1

$begingroup$
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?
$endgroup$
– movatica
Apr 26 at 14:13

1

$begingroup$
@movatica In the header, put `f = ` ( is the line continuation character in python)
$endgroup$
– Artemis Fowl
Apr 26 at 15:33

$begingroup$
Annoyingly, is also an escape character. Here, without code formatting, is what you should put in the header: f =
$endgroup$
– Artemis Fowl
Apr 26 at 15:35

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

add a comment |

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!

edited Apr 24 at 15:33

answered Apr 24 at 15:18

Luis felipe De jesus Munoz

6,30121874

add a comment |

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!

edited Apr 24 at 15:33

answered Apr 24 at 15:18

Luis felipe De jesus Munoz

6,30121874

add a comment |

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!

edited Apr 24 at 15:33

answered Apr 24 at 15:18

Luis felipe De jesus Munoz

6,30121874

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!

edited Apr 24 at 15:33

answered Apr 24 at 15:18

Luis felipe De jesus Munoz

6,30121874

edited Apr 24 at 15:33

answered Apr 24 at 15:18

Luis felipe De jesus Munoz

6,30121874

answered Apr 24 at 15:18

Luis felipe De jesus Munoz

6,30121874

answered Apr 24 at 15:18

Luis felipe De jesus Munoz

6,30121874

add a comment |

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.

answered Apr 24 at 17:16

Sophia Lechner

1,03018

2

$begingroup$
€htD¹ avoids the ȯ.
$endgroup$
– Zgarb
Apr 24 at 19:30

$begingroup$
That's fantastic! I had thought about λ€htD¹ but I didn't realize that lambdas would be added implicitly
$endgroup$
– Sophia Lechner
Apr 24 at 20:10

add a comment |

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.

answered Apr 24 at 17:16

Sophia Lechner

1,03018

2

$begingroup$
€htD¹ avoids the ȯ.
$endgroup$
– Zgarb
Apr 24 at 19:30

$begingroup$
That's fantastic! I had thought about λ€htD¹ but I didn't realize that lambdas would be added implicitly
$endgroup$
– Sophia Lechner
Apr 24 at 20:10

add a comment |

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.

answered Apr 24 at 17:16

Sophia Lechner

1,03018

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.

answered Apr 24 at 17:16

Sophia Lechner

1,03018

answered Apr 24 at 17:16

Sophia Lechner

1,03018

answered Apr 24 at 17:16

Sophia Lechner

1,03018

answered Apr 24 at 17:16

Sophia Lechner

1,03018

2

$begingroup$
€htD¹ avoids the ȯ.
$endgroup$
– Zgarb
Apr 24 at 19:30

$begingroup$
That's fantastic! I had thought about λ€htD¹ but I didn't realize that lambdas would be added implicitly
$endgroup$
– Sophia Lechner
Apr 24 at 20:10

add a comment |

2

$begingroup$
€htD¹ avoids the ȯ.
$endgroup$
– Zgarb
Apr 24 at 19:30

$begingroup$
That's fantastic! I had thought about λ€htD¹ but I didn't realize that lambdas would be added implicitly
$endgroup$
– Sophia Lechner
Apr 24 at 20:10

€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

add a comment |

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->""]=="")

{}!=

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  *)

answered Apr 24 at 17:40

Eric Towers

66637

add a comment |

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->""]=="")

{}!=

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  *)

answered Apr 24 at 17:40

Eric Towers

66637

add a comment |

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->""]=="")

{}!=

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  *)

answered Apr 24 at 17:40

Eric Towers

66637

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->""]=="")

{}!=

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  *)

answered Apr 24 at 17:40

Eric Towers

66637

answered Apr 24 at 17:40

Eric Towers

66637

answered Apr 24 at 17:40

Eric Towers

66637

answered Apr 24 at 17:40

Eric Towers

66637

add a comment |

Python 3, 84 bytes

import textwrap

lambda s:any(len(set(textwrap.wrap(s,l)))<2 for l in range(1,len(s)))

If there is no n for which repeating substrings of length n make up the entire string, then return false for the whole function.

answered Apr 24 at 20:20

Delya Erricson

8113

add a comment |

Python 3, 84 bytes

import textwrap

lambda s:any(len(set(textwrap.wrap(s,l)))<2 for l in range(1,len(s)))

If there is no n for which repeating substrings of length n make up the entire string, then return false for the whole function.

answered Apr 24 at 20:20

Delya Erricson

8113

add a comment |

Python 3, 84 bytes

import textwrap

lambda s:any(len(set(textwrap.wrap(s,l)))<2 for l in range(1,len(s)))

If there is no n for which repeating substrings of length n make up the entire string, then return false for the whole function.

answered Apr 24 at 20:20

Delya Erricson

8113

Python 3, 84 bytes

import textwrap

lambda s:any(len(set(textwrap.wrap(s,l)))<2 for l in range(1,len(s)))

If there is no n for which repeating substrings of length n make up the entire string, then return false for the whole function.

answered Apr 24 at 20:20

Delya Erricson

8113

answered Apr 24 at 20:20

Delya Erricson

8113

answered Apr 24 at 20:20

Delya Erricson

8113

answered Apr 24 at 20:20

Delya Erricson

8113

add a comment |

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

answered Apr 25 at 6:33

Emigna

49.5k534150

1

$begingroup$
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
$endgroup$
– Kevin Cruijssen
Apr 25 at 6:55

$begingroup$
@KevinCruijssen: That's typical. Has happened to me a bunch of times as well ;)
$endgroup$
– Emigna
Apr 25 at 6:56

add a comment |

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

answered Apr 25 at 6:33

Emigna

49.5k534150

1

$begingroup$
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
$endgroup$
– Kevin Cruijssen
Apr 25 at 6:55

$begingroup$
@KevinCruijssen: That's typical. Has happened to me a bunch of times as well ;)
$endgroup$
– Emigna
Apr 25 at 6:56

add a comment |

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

answered Apr 25 at 6:33

Emigna

49.5k534150

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

answered Apr 25 at 6:33

Emigna

49.5k534150

answered Apr 25 at 6:33

Emigna

49.5k534150

answered Apr 25 at 6:33

Emigna

49.5k534150

answered Apr 25 at 6:33

Emigna

49.5k534150

1

$begingroup$
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
$endgroup$
– Kevin Cruijssen
Apr 25 at 6:55

$begingroup$
@KevinCruijssen: That's typical. Has happened to me a bunch of times as well ;)
$endgroup$
– Emigna
Apr 25 at 6:56

add a comment |

1

$begingroup$
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
$endgroup$
– Kevin Cruijssen
Apr 25 at 6:55

$begingroup$
@KevinCruijssen: That's typical. Has happened to me a bunch of times as well ;)
$endgroup$
– Emigna
Apr 25 at 6:56

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

add a comment |

Clean, 73 bytes

Doesn't use regex.

import StdEnv,Data.List

$s=or[isPrefixOf s(cycle t)\t<-tl(tails s)|t>]

Try it online!

Defines $ :: [Char] -> Bool.

Checks if the given string is a prefix of the repetition of any sub-string taken from the end.

answered Apr 26 at 0:43

Οurous

7,47611136

add a comment |

Clean, 73 bytes

Doesn't use regex.

import StdEnv,Data.List

$s=or[isPrefixOf s(cycle t)\t<-tl(tails s)|t>]

Try it online!

Defines $ :: [Char] -> Bool.

Checks if the given string is a prefix of the repetition of any sub-string taken from the end.

answered Apr 26 at 0:43

Οurous

7,47611136

add a comment |

Clean, 73 bytes

Doesn't use regex.

import StdEnv,Data.List

$s=or[isPrefixOf s(cycle t)\t<-tl(tails s)|t>]

Try it online!

Defines $ :: [Char] -> Bool.

Checks if the given string is a prefix of the repetition of any sub-string taken from the end.

answered Apr 26 at 0:43

Οurous

7,47611136

Clean, 73 bytes

Doesn't use regex.

import StdEnv,Data.List

$s=or[isPrefixOf s(cycle t)\t<-tl(tails s)|t>]

Try it online!

Defines $ :: [Char] -> Bool.

Checks if the given string is a prefix of the repetition of any sub-string taken from the end.

answered Apr 26 at 0:43

Οurous

7,47611136

answered Apr 26 at 0:43

Οurous

7,47611136

answered Apr 26 at 0:43

Οurous

7,47611136

answered Apr 26 at 0:43

Οurous

7,47611136

add a comment |

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.

edited Apr 26 at 15:35

answered Apr 25 at 23:42

jxh

32128

add a comment |

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.

edited Apr 26 at 15:35

answered Apr 25 at 23:42

jxh

32128

add a comment |

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.

edited Apr 26 at 15:35

answered Apr 25 at 23:42

jxh

32128

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.

edited Apr 26 at 15:35

answered Apr 25 at 23:42

jxh

32128

edited Apr 26 at 15:35

answered Apr 25 at 23:42

jxh

32128

answered Apr 25 at 23:42

jxh

32128

answered Apr 25 at 23:42

jxh

32128

add a comment |

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

answered Apr 24 at 19:19

i_saw_drones

2835

add a comment |

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

answered Apr 24 at 19:19

i_saw_drones

2835

add a comment |

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

answered Apr 24 at 19:19

i_saw_drones

2835

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

answered Apr 24 at 19:19

i_saw_drones

2835

answered Apr 24 at 19:19

i_saw_drones

2835

answered Apr 24 at 19:19

i_saw_drones

2835

answered Apr 24 at 19:19

i_saw_drones

2835

add a comment |

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}

Try it online!

answered Apr 24 at 19:32

onnoweb

1713

$begingroup$
I don't use Swift, but I'm sure that extra whitespace can be removed
$endgroup$
– Benjamin Urquhart
Apr 24 at 19:56

$begingroup$
193 bytes using @benjamin's suggestion.
$endgroup$
– Artemis Fowl
Apr 26 at 15:39

$begingroup$
@ArtemisFowl or even 123 bytes
$endgroup$
– Roman Podymov
Apr 26 at 16:59

add a comment |

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}

Try it online!

answered Apr 24 at 19:32

onnoweb

1713

$begingroup$
I don't use Swift, but I'm sure that extra whitespace can be removed
$endgroup$
– Benjamin Urquhart
Apr 24 at 19:56

$begingroup$
193 bytes using @benjamin's suggestion.
$endgroup$
– Artemis Fowl
Apr 26 at 15:39

$begingroup$
@ArtemisFowl or even 123 bytes
$endgroup$
– Roman Podymov
Apr 26 at 16:59

add a comment |

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}

Try it online!

answered Apr 24 at 19:32

onnoweb

1713

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}

Try it online!

answered Apr 24 at 19:32

onnoweb

1713

answered Apr 24 at 19:32

onnoweb

1713

answered Apr 24 at 19:32

onnoweb

1713

answered Apr 24 at 19:32

onnoweb

1713

$begingroup$
I don't use Swift, but I'm sure that extra whitespace can be removed
$endgroup$
– Benjamin Urquhart
Apr 24 at 19:56

$begingroup$
193 bytes using @benjamin's suggestion.
$endgroup$
– Artemis Fowl
Apr 26 at 15:39

$begingroup$
@ArtemisFowl or even 123 bytes
$endgroup$
– Roman Podymov
Apr 26 at 16:59

add a comment |

$begingroup$
I don't use Swift, but I'm sure that extra whitespace can be removed
$endgroup$
– Benjamin Urquhart
Apr 24 at 19:56

$begingroup$
193 bytes using @benjamin's suggestion.
$endgroup$
– Artemis Fowl
Apr 26 at 15:39

$begingroup$
@ArtemisFowl or even 123 bytes
$endgroup$
– Roman Podymov
Apr 26 at 16:59

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

add a comment |

Icon, 46 bytes

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

Try it online!

Another port of xnor's solution.

edited Apr 25 at 7:41

answered Apr 25 at 7:13

Galen Ivanov

8,33211237

add a comment |

Icon, 46 bytes

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

Try it online!

Another port of xnor's solution.

edited Apr 25 at 7:41

answered Apr 25 at 7:13

Galen Ivanov

8,33211237

add a comment |

Icon, 46 bytes

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

Try it online!

Another port of xnor's solution.

edited Apr 25 at 7:41

answered Apr 25 at 7:13

Galen Ivanov

8,33211237

Icon, 46 bytes

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

Try it online!

Another port of xnor's solution.

edited Apr 25 at 7:41

answered Apr 25 at 7:13

Galen Ivanov

8,33211237

edited Apr 25 at 7:41

answered Apr 25 at 7:13

Galen Ivanov

8,33211237

answered Apr 25 at 7:13

Galen Ivanov

8,33211237

answered Apr 25 at 7:13

Galen Ivanov

8,33211237

add a comment |

K (oK), 29 bytes

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

Try it online!

answered Apr 25 at 8:52

Galen Ivanov

8,33211237

add a comment |

K (oK), 29 bytes

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

Try it online!

answered Apr 25 at 8:52

Galen Ivanov

8,33211237

add a comment |

K (oK), 29 bytes

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

Try it online!

answered Apr 25 at 8:52

Galen Ivanov

8,33211237

K (oK), 29 bytes

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

Try it online!

answered Apr 25 at 8:52

Galen Ivanov

8,33211237

answered Apr 25 at 8:52

Galen Ivanov

8,33211237

answered Apr 25 at 8:52

Galen Ivanov

8,33211237

answered Apr 25 at 8:52

Galen Ivanov

8,33211237

add a comment |

Red, 72 bytes

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

Try it online!

Returns 1 for True

answered Apr 25 at 10:36

Galen Ivanov

8,33211237

add a comment |

Red, 72 bytes

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

Try it online!

Returns 1 for True

answered Apr 25 at 10:36

Galen Ivanov

8,33211237

add a comment |

Red, 72 bytes

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

Try it online!

Returns 1 for True

answered Apr 25 at 10:36

Galen Ivanov

8,33211237

Red, 72 bytes

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

Try it online!

Returns 1 for True

answered Apr 25 at 10:36

Galen Ivanov

8,33211237

answered Apr 25 at 10:36

Galen Ivanov

8,33211237

answered Apr 25 at 10:36

Galen Ivanov

8,33211237

answered Apr 25 at 10:36

Galen Ivanov

8,33211237

add a comment |

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.

Try it online

edited Apr 25 at 10:47

answered Apr 24 at 17:04

t-clausen.dk

2,214514

add a comment |

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.

Try it online

edited Apr 25 at 10:47

answered Apr 24 at 17:04

t-clausen.dk

2,214514

add a comment |

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.

Try it online

edited Apr 25 at 10:47

answered Apr 24 at 17:04

t-clausen.dk

2,214514

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.

Try it online

edited Apr 25 at 10:47

answered Apr 24 at 17:04

t-clausen.dk

2,214514

edited Apr 25 at 10:47

answered Apr 24 at 17:04

t-clausen.dk

2,214514

answered Apr 24 at 17:04

t-clausen.dk

2,214514

answered Apr 24 at 17:04

t-clausen.dk

2,214514

add a comment |

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

draft discarded

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

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This page is only for reference, If you need detailed information, please check here

Check if a string is entirely made of the same substring

40 Answers 40

Brachylog, 4 3 bytes

Explanation

JavaScript (ES6), 22 bytes

Without a regular expression, 33 29 bytes

grep, 19

Test

Japt, 6 bytes

Java, 25 24 bytes

Excel, 26 bytes

Retina 0.8.2, 9 bytes

Jelly, 5 4 bytes

How?

R, 28 bytes

Question (if anyone knows the answer)

Perl 5 -p, 14 bytes

Python 2, 24 bytes

Python 2, 59 55 53 bytes

Python 2, 44 bytes

Pyke, 4 bytes

J, 26 25 15 14 bytes

original (two different approaches)

J, 25 bytes

J, 26 bytes

Wolfram Language (Mathematica), 24 23 bytes

PowerShell, 23 24 bytes

C# (Visual C# Interactive Compiler), 70 bytes

Python 3, 62 60 56 54 bytes

Japt, 10 bytes

Husk, 6 bytes

Explanation

Mathematica 11.x, 74 bytes

Python 3, 84 bytes

05AB1E, 5 bytes

Clean, 73 bytes

C++ (gcc), 36 bytes

QlikView Variable, 27 bytes

Swift, 196 bytes

Icon, 46 bytes

K (oK), 29 bytes

Red, 72 bytes

T-SQL, 47 bytes

Your Answer

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40 Answers 40

40 Answers 40

Brachylog, 4 3 bytes

Explanation

Brachylog, 4 3 bytes

Explanation

Brachylog, 4 3 bytes

Explanation

Brachylog, 4 3 bytes

Explanation

JavaScript (ES6), 22 bytes

Without a regular expression, 33 29 bytes

JavaScript (ES6), 22 bytes

Without a regular expression, 33 29 bytes

JavaScript (ES6), 22 bytes

Without a regular expression, 33 29 bytes

JavaScript (ES6), 22 bytes

Without a regular expression, 33 29 bytes

grep, 19

Test

grep, 19

Test

grep, 19

Test

grep, 19

Test

Japt, 6 bytes

Japt, 6 bytes

Japt, 6 bytes

Japt, 6 bytes

Java, 25 24 bytes

Java, 25 24 bytes

Java, 25 24 bytes

40 Answers
40

Perl 5 `-p`, 14 bytes

J, ^{26 25 15} 14 bytes

40 Answers
40

40 Answers
40

Perl 5 `-p`, 14 bytes

Perl 5 `-p`, 14 bytes

Perl 5 `-p`, 14 bytes

Perl 5 `-p`, 14 bytes

J, ^{26 25 15} 14 bytes

J, ^{26 25 15} 14 bytes

J, ^{26 25 15} 14 bytes

J, ^{26 25 15} 14 bytes