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It's become somewhat of a tradition in PPCG that some users temporarily change their names by an anagram (a new name formed by reordering the letters of the old).

Sometimes it gets difficult to find out who is who. I could use a program or function to tell if two phrases are anagrams of each other.

The challenge

The program of function should take two strings and produce a truthy result if they are anagrams of each other, and falsy otherwise.


  • Input will only contain letters (ASCII 65 to 90 and 97 to 122), digits (ASCII 48 to 57) or space (ASCII 32).
  • The anagram relation is independendent of case. So "Arm" and "RAM" are anagrams.
  • Spaces don't count either. So "keyboard" and "Barked Yo" are anagrams
  • All builtins allowed
  • Input format is flexible (two strings, an array of two strings, a string containing both phrases with a suitable separator ...)

Code golf. Fewest bytes wins.

Test cases


Lynn, Nyl N
Digital Trauma, Tau Digital Arm
Sp3000, P S 3000
Manage Trash So, Those anagrams


Calvins Hobbies, Helka Homba
Android, rains odd
In between days, bayed entwine
Code golf, cod elf got
share|improve this question
Related but different (only letters, no case, no spaces) – Luis Mendo Feb 25 at 16:09
This question's title is very perplexing to someone who's not had enough coffee. +1 :D – cat Feb 25 at 17:36
@DonMuesli I would argue that this is still a dupe. The slight changes are very trivial. – Mego Feb 25 at 17:59
Manage Trash So, Those anagrams. Nice. – mbomb007 Feb 25 at 20:13
So, the anagrams... – Leaky Nun May 21 at 17:23

27 Answers 27

up vote 11 down vote accepted

05AB1E, 9 8 bytes




l         # Lowercase the strings
 vy   }   # Map over the list, for each...
   ð-     #   remove spaces
     {    #   and sort
       Q  # Check equality

Try it online!

share|improve this answer
... but 2 bytes less! Well done! – Luis Mendo Feb 25 at 16:28
@DonMuesli Thanks! :) – Adnan Feb 25 at 16:29

Retina, 25


Try it Online! Additionally, you can run a modified multi-line version.

Delete letters from before the comma along with their matches after the comma. If we have no letters left then it was an anagram.

share|improve this answer
For Retina, if a positive number could be considered a failure, and zero be considered success, this could be three bytes shorter by using \w as the last stage. – FryAmTheEggman Feb 25 at 20:20
@dev-null "Input will only contain letters (ASCII 65 to 90 and 97 to 122), digits (ASCII 48 to 57) or space (ASCII 32)" – FryAmTheEggman Feb 26 at 2:36

Pyth, 11 10 bytes

Thanks to @FryAmTheEggman for teaching me the power of ;!


Try it here!

Takes a list of two strings as input.


qFmSr-d;0Q    # Q = input

  m      Q    # map Q with d as lambda variable
     -d;      # filter spaces out of the string
    r   0     # convert to lowercase
   S          # sort all characters in string
qF            # Unfold resulting list and check for equality
share|improve this answer

Python 2, 63 61 bytes

lambda*l:len({`sorted(s.lower())`[2::5].strip()for s in l})<2

An anonymous function that, in fact, takes n arguments and determines if all n of them are mutual palindromes! f("Lynn", "Nyl N") returns True.

This set comprehension trick is by xnor. It saved two bytes, but the old approach looked very neat:

exec"a=`sorted(input().lower())`[2::5].strip();a"*2;print a==aa
share|improve this answer
`sorted(input().lower())`.strip(" [',") is the same length :/ – Sp3000 Feb 25 at 16:43
The exec thing is clever but seems too complex. You can do better with lambda*l:len({`sorted(s.lower())`[2::5].strip()for s in l})<2. – xnor Feb 25 at 16:45
Thanks! I'm a bit disappointed – it looked very cool. Keeping it in the post anyway. – Lynn Feb 25 at 16:50

CJam, 11 12 14 bytes

3 2 bytes removed thanks to @FryAmTheEggman


Try it online!

{      }2*       e# do this twice
 l               e# read line as a string
  el             e# make lowercase
    S-           e# remove spaces from string
      $          e# sort
          =      e# compare strings
share|improve this answer
@FryAmTheEggman Thank you! – Luis Mendo Feb 25 at 16:56
@FryAmTheEggman Thanks again! I still have much to learn about CJam :-) – Luis Mendo Feb 25 at 16:59
Your code is secretly laughing. lel. – Cyoce Feb 27 at 21:12
Or is it a one? lel ==> 1e1 No one knows. It is a mystery. – zyabin101 Mar 2 at 12:23

Jelly, 12 bytes


Try it online!

How it works

ḟ€⁶O&95Ṣ€QLḂ  Main link. Input: A (list of strings)

  ⁶           Yield ' '.
ḟ€            Filter it from each string.
   O          Apply ordinal to all characters.
    &95       Take bitwise AND with 95 to make the ordinals case-insensitive.
       Ṣ€     Sort each list of ordinals.
         Q    Deduplicate the list.
          L   Get the length.
           Ḃ  Compute the length's parity (1 -> 1, 2 -> 0).

Alternate version, non-competing (9 bytes)

Jelly's uppercase atom had a bug, and Jelly still had no built-in to test lists for equality...


Try it online!

How it works

ḟ⁶ŒuṢµ€⁼/     Main link. Input: A (list of strings)

     µ€       Map the chain to the left over A.
 ⁶            Yield ' '.
ḟ             Filter it from the string.
  Œu          Cast to uppercase.
    Ṣ         Sort.
       ⁼/     Reduce by equality.
share|improve this answer

Javascript, 69 61 60 59 bytes

1 byte off thanks @ӍѲꝆΛҐӍΛПҒЦꝆ. 1 byte off with currying (pointed out by @apsillers)

n=>m=>(G=s=>[]+s.toLowerCase().split(/ */).sort())(n)==G(m)

        .split(/ */)

F=(n,m)=>document.body.innerHTML+=`<pre>f('${n}')('${m}') -> ${f(n)(m)}</pre>`

F('Luis Mendo','Don Muesli')
F('Calvins Hobbies','Helka Homba')
F('Android','rains odd')
F('In between days','bayed entwine')
F('Code golf','cod elf got')
F('Lynn','Nyl N')
F('Digital Trauma','Tau Digital Arm')
F('Sp3000','P S 3000')
F('Manage Trash So','Those anagrams')

share|improve this answer
Very nice, filtering out spaces and converting to an array at the same time! – Neil Feb 25 at 17:06
Very nice. You can save one byte using currying, which the community has decided is an acceptable form of arguments: n=>m=>... – apsillers Feb 25 at 17:23
Try n=>m=>(G=s=>[]+s.toLowerCase().split(/\S/).sort())(n)==G(m). Using split instead of match should save you a byte. – Mama Fun Roll Feb 26 at 3:42
@ӍѲꝆΛҐӍΛПҒЦꝆ. No, because suppose s='db cz'... Now s.match(/\S/g).sort() results in ['b','c','d','z']... and s.split(/\s/).sort() results in ['cz','db'] – removed Feb 26 at 12:59
@ӍѲꝆΛҐӍΛПҒЦꝆ. But... looking into your idea, I changed it a bit and saved one byte... thanks! – removed Feb 26 at 13:05

MATL, 11 bytes


EDIT (May 20, 2016) The code in the link uses Xz instead of Xv, owing to recent changes in the language.

Try it online!

2:"     ]       % do this twice
   j            % read input line as a string
    k           % convert to lowercase
     Xv         % remove spaces
       S        % sort
         X=     % are they equal?
share|improve this answer
Did you just change your name for that challenge? – DenkerAffe Feb 25 at 16:34
@DenkerAffe I had been thinking about it for some time. I just made it coincide with the challenge :-) – Luis Mendo Feb 25 at 16:36
Don Muesli lol. So you are the Lord of Muesli Luis!? Is this how you keep your healthy complexion? – rayryeng Feb 25 at 21:00
@rayryeng Heyyy! Good to see you here, Ray! Get back to golfing! – Luis Mendo Feb 25 at 21:22
I promise I will :) once this course ends... I see you are learning CJam now too. Very nice! – rayryeng Feb 25 at 21:32

Seriously, 11 9 bytes


Try It Online!

Everyone seems to be using the same algorithm. Here it is yet again.

2`    `n          Do it twice
  ,               Read a string
   ù              Make it lowercase
    S             Sort
     ô            Strip spaces.
        =         Check equality.

Edit: realized sorting does work correctly on strings, and sorts spaces to the front so strip() will work.

share|improve this answer

C, 165 bytes

#define d(x) int x(char*a,char*b){
#define n(x) for(qsort(x,strlen(x),1,(__compar_fn_t)q);*x<33;x++);
d(s)n(a)n(b)return strcasecmp(a,b);}

Readable and in working context,

#include <stdlib.h>
#include <string.h>
#include <stdio.h>

// start of comparison
int q(char *a, char *b){
     return ((*a)&0xdf)-((*b)&0xdf); // case-insensitive
int s(char *a, char *b){
    for(qsort(a,strlen(a),1,(__compar_fn_t)q); *a<33; a++) /**/;
    for(qsort(b,strlen(b),1,(__compar_fn_t)q); *b<33; b++) /**/;
    return strcasecmp(a,b);
// end of comparison

int main(int i, char **v){
    printf("'%s' '%s'", v[1], v[2]);
    printf("=> %d\n", s(v[1], v[2])); // 0 if equalish
    return 0;
share|improve this answer

zsh, 85 bytes

[ $(for x in $@;{tr -d \ <<<$x|tr A-Z a-z|fold -1|sort|paste -sd x}|uniq|wc -l) = 1 ]

Input as command line arguments, output as return code.

The for syntax makes this Bash-incompatible.

[               # test...
$(for x in $@;  # map over arguments
{tr -d \ <<<$x  # remove spaces
|tr A-Z a-z     # lowercase
|fold -1        # put each character on its own line
|sort           # sort lines
|paste -sd x    # remove all newlines except last
}|uniq          # take only unique lines
|wc -l          # how many lines remain?
) = 1 ]         # if only 1 line left, it must have been an anagram
share|improve this answer

Japt, 12 bytes

N®v ¬n ¬xÃä¥

Test it online!

How it works

        // Implicit: N = array of input strings
N®    Ã // Take N, and map each item Z to:
v ¬n    //  Take Z.toLowerCase(), split into chars, and sort.
¬x      //  Join and trim off whitespace.
ä¥      // Reduce each pair of items (that's exactly one pair) X and Y to X == Y.
share|improve this answer

GNU Sed, 33

Score includes +2 for -rn options to sed.

This is almost a direct port of @FryAmTheEggman's Retina answer:



share|improve this answer

Perl, 34 33 + 1 = 34 bytes


Requires the -n flag and the free -M5.010|-E:

$ perl -M5.010 -ne's/(.)(.*,.*)\1/$2/i?redo:say!/\w/' <<< 'hello, lloeh'

How it works:

                                   # '-n' make a implicit while loop around the code
 s/(.)(.*,.*)\1/$2/i               # Remove a letter that occurs on both sides of the comma.
                     redo:         # Redo is a glorified goto statement that goes to the top of the while loop
                          say!/\w/ # Check to see if any letter is left

Thanks to msh210 for suggesting using ternary operators to save one byte

share|improve this answer

Bash + GNU utilities, 51

f()(fold -1<<<${@^^}|sort)
f $1|diff -qBw - <(f $2)
  • Define a function f() which:
    • ${@^^} converts all parameters to upper case
    • fold -1 splits chars - one per line
    • sorts lines
  • call diff with -q to suppress full diff output and -Bw to ignore whitespace changes
share|improve this answer

Pyke (commit 30, noncompetitive), 9 bytes



F      )  -  for _ in eval_or_not(input())
 l1       -     ^.lower()
   dk:    -    ^.replace(" ", "")
      S   -   sorted(^)
        q - ^==^
share|improve this answer

Mathematica, 77 76 bytes

StringMatchQ[##,IgnoreCase->1>0]&@@(""<>Sort[Characters@#/." "->""]&/@{##})&

The first part is actually one of my answers to another question!

share|improve this answer

Pike, 54 112 109 109 96 bytes

#define a(x) sort((array)replace(lower_case(x)," ",""))
int s(mixed i){return a(i[0])==a(i[1]);}

mixed happens to be shorter than array(string).

s returns 1 if its arguments are anagrams.

share|improve this answer

Q, 25 Bytes


NOTE.- counting include function name f: to facilitate tests (as lambda we can decrement 2 Bytes)

Readable version

match over {ascending not null lower x} each x

{.. x ..} is an anonymous function with arg x
_x        lowers string x
&~^x      where not null x (space is considered null)
x@..      selects elements of x according to indexes .. 
<x        ascending indexes of x (not values). Ex <"cab" is 1 2 0
x@<x      ascending values of x (x at ascending indexes of x)
~         match (diad function). Ex "one"~"one" is true
f'..      applies function f for each argument ..
f/..      applies function f over elements of sequence (fold)


f("Lynn";"Nyl N")                       
f("Digital Trauma";"Tau Digital Arm")   
f("Sp3000";"P S 3000")                  
f("Manage Trash So";"Those anagrams")   
f("Calvins Hobbies";"Helka Homba")      
f("Android";"rains odd")                
f("In between days";"bayed entwine")    
f("Code golf";"cod elf got")    

generates (1b = true, 0b = false)


About Q

General-purpose language (APL derivative, specialized in data processing) developed by Free full functional evaluation version for Windows/Linux/MacOS.

share|improve this answer
What do you mean, other languages are not serious? :-P – Luis Mendo May 21 at 8:59
If the f is required for the code to evaluate properly, then it must be counted. Otherwise, just leave it off of your submission code, and only use it in examples to show how to assign the function. – Mego May 21 at 9:28
Of course, other languages are as serious as Q. I beg my poor english. But some languages sacrifices readability or are equiped with libraries ad-hoc for this type of contests. Q is a 'general purpose language', in spite of the fact that code is not very readable. – J. Sendra May 21 at 16:59

APL, 31 chars

{≡/{x[⍋x←('.'⎕R'\u0')⍵~' ']}¨⍵}

To be used so:

    {≡/{x[⍋x←('.'⎕R'\u0')⍵~' ']}¨⍵}'Sp3000' 'P S 3000' 

In English:

  • { ... }¨⍵: for each of the two elements of the argument
  • x←('.'⎕R'\u0')⍵~' ': transform to uppercase (using a regex...) the string without the spaces and assign the temporary result to x
  • x[⍋x]: sort x
  • ≡/: compare the two results of the sorting: if they match, return 1.
share|improve this answer
Is it possible to try it online? I tried with this but I don't really know how to use it – Luis Mendo 2 days ago
Sure. Here: definition after which you can just type f 'first avatar' 'second avatar' – lstefano 2 days ago
Thanks! Maybe add that to the answer? So that people can try – Luis Mendo 2 days ago
–9: ≡/{x[⍋x←0~⍨32|⎕UCS⍵]}¨ – Adám 2 days ago
@Adám: that won't work because ≡/{x[⍋x←0~⍨32|⎕UCS⍵]}¨'pp' '00' gives 1. – lstefano yesterday

PHP, 109 94 bytes

function f($x){return str_split((trim($x));}function g($x,$y){return array_diff(f($x),f($y));}

Blech, the two function/returns are killing me here.

Returns the difference between two string inputs as an array of characters. PHP considers [] falsy, satisfying the return requirements.

share|improve this answer
function($x,$y){$S=str_split;return array_diff($S(trim($x)),$S(trim($y)));} --> 75 bytes. Creates an anonymous function that returns the result. I've removed that long function and replaced the calls to str_split with an assignmed variable, to shorten it up. – Ismael Miguel Feb 25 at 16:52
Nice. I was tweaking it to reduce it to the one function, this is two steps ahead of that, well done. – ricdesi Feb 25 at 16:56

Ruby, 50 bytes

def f;gets.upcase.chars.sort.join.strip;end
p f==f

Writing f=->{...} and f[]==f[] is just as long. :(

share|improve this answer

PowerShell, 81 bytes

param([char[]]$a,[char[]]$b)-join($a-replace' '|sort)-eq-join($b-replace' '|sort)

A slight rewrite of my answer on the linked Anagram challenge.

Takes input as char-arrays, performs a -replace operation to remove spaces, sorts them (which sorts alphabetically, not by ASCII value), then -joins them back into a string. The -eq in PowerShell is by default case-insensitive, but here it must be performed on strings, as [char]'a' is not equal to [char]'A', hence the reason for -join.

share|improve this answer

Perl, 35 bytes

Include +1 for -p

Somewhat abusive since it depends on the program being given on the commandline.

perl -pe'<>=~s%\S%*_=s/$&//i?_:0%reg;$_=!//'

Then give the strings as 2 consecutive lines on STDIN

A very abusive solution is 30 bytes:

perl -ne'<>=~s%\w%1/!s/$&//i%reg;1/!//'

This crashes if the strings are not anagrams and therefore gives a false exit code from the point of view of the shell. It also gives garbage on STDERR for that case. If the strings are anagrams the program is silent and gives a "true" exit code

share|improve this answer

C#, 378 bytes

I need a handicap!!

using System;
using System.Linq;
using System.Text;

public class Program

    public static void Main()
        var l = "Hello World";

        var r = "Red Who Loll";

        var y = new Func<string,string>(s => new String(s.ToLower().Replace(" ","").OrderBy(v => v).ToArray()));
        var z = new Func<string,string,Func<string,string>,bool>((w,x,f) => f(w) == f(x));
        var o = z(l, r, y);

        Console.WriteLine("{0} & {1} are anagram: {2}",l, r, o);

                Console.WriteLine("C#, {0} bytes", Encoding.Unicode.GetByteCount(@"var y = new Func<string,string>(s => new String(s.ToLower().Replace("" "","""").OrderBy(v => v).ToArray()));
    var z = new Func<string,string,Func<string,string>,bool>((w,x,f) => f(w) == f(x));"));


share|improve this answer
Welcome to Programming Puzzles & Code Golf! General rule is to put your language together with your byte count in the headline of your post. You can do that by adding a leading # to the first line. Also for code-golf questions it is required to golf your program. For the start you should remove unecessary whitespaces and use single-character variable names. Also you can always use a function instead of a full program (unless explicitly forbidden) to save some more bytes. – DenkerAffe Feb 28 at 0:40
@DenkerAffe you ninja'd me :) – cat Feb 28 at 0:40
Why all the whitespace?? – CalculatorFeline Mar 6 at 20:32

D, 99 107 103 102 99 131 116 bytes

Edit: forgot about Dre case insensitivity imports lambdas

I don't like this language :D

import std.string,std.algorithm;(string[]a)=>sort(split(strip(a[0].toLower),""))==sort(split(strip(a[1].toLower),"")

To use it, assign it:

import std.string,std.algorithm;

void main () {
    auto x = (string[]a) => sort(split(strip(a[0].toLower),"")) == sort(split(strip(a[1].toLower),"")
share|improve this answer

Factor, 68 bytes

[ [ 32 swap remove >lower natural-sort ] map duplicates length 1 = ]

An anyonymous function. Call it like { "array" "of" "strings" } ~quotation~ call.

share|improve this answer

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