Find all the Anagrams!

Question

Despite having 17 questions tagged anagrams, we still don't have this question, so here it is.

Your Task

You must write a program or function that, when receiving a string, prints out all possible anagrams of it. For the purposes of this question, an anagram is a string that contains the same character as the original string, but is not an exact copy of the original string. An anagram does not have to be or contain actual words.

Input

You may accept the string, which may be of any length > 0, by any standard input method. It may contain any ASCII characters.

Output

You may output all of the possible anagrams of the inputted string in any standard way. You must not output the same string twice, or output a string equal to the input.

Other Rules

Standard Loopholes are disallowed

Scoring

This is code-golf, least bytes wins.

Answer 1

05AB1E, 3 bytes

œÙ¦

A function that leaves the stack with a list of the anagrams on top (and as its only item). As a full program prints a representation of that list.

Try it online!

How?

    - push input
œ   - pop and push a list of all permutations (input appears at the head)
 Ù  - pop and push a list of unique items (sorted by first appearance)
  ¦ - pop and push a dequeued list (removes the occurrence of the input)
    - As a full program: implicit print of the top of the stack

Answer 2

Ruby, 45 bytes

->x{(x.chars.permutation.map(&:join)-[x])|[]}

Try it online!

Despite having a built-in, the word "permutation" is really long :(

Answer 3

pyth, 8 4

-{.p

Online test.

  .pQ     # all permutations of the (implicit) input string
 {        # de-duplicate
-    Q    # subtract (implicit) input

Answer 4

Japt, 6 bytes

á â kU

Try it online!

Explanation

 á â kU
Uá â kU   // Ungolfed
          // Implicit: U = input string
Uá        // Take all permutations of U.
   â      // Remove duplicates.
     kU   // Remove U itself from the result.
          // Implicit: output resulting array, separated by commas

Answer 5

MATL, 7 bytes

tY@1&X~

Try it online!

Explanation

t     % Implicitly input a string, say of length n. Duplicate
Y@    % All permutations. May contain duplicates. Gives a 2D char array of 
      % size n!×n with each permutation in a row
1&X~  % Set symmetric difference, row-wise. Automatically removes duplicates.
      % This takes the n!×n char array and the input string (1×n char array)
      % and produces an m×n char array containing the rows that are present 
      % in exactly one of the two arrays
      % Implicitly display

Answer 6

Haskell, 48 40 bytes

import Data.List
a=tail.nub.permutations

Try it online!

Saved 8 bytes thanks to Leo's tail tip.

Answer 7

CJam, 8 bytes

l_e!\a-p

Try it online!

Explanation

l    e# Read string from input
_    e# Duplicate
e!   e# Unique permutations. Gives a list of strings
\    e# Swap
a    e# Wrap in a singleton array
-    e# Set difference. This removes the input string
p    e# Pretty print the list

Answer 8

Jelly, 4 bytes

Œ!QḊ

A monadic link taking a list of characters and returning a list of lists of characters - all distinct anagrams that are not equal to the input.

Try it online! (the footer forms a program that joins the list by newlines and prints to avoid the otherwise smashed representation).

How?

Œ!QḊ - Link: list of characters     e.g. "text"
Œ!   - all permutations of the list      ["text","tetx","txet","txte","ttex","ttxe","etxt","ettx","extt","extt","ettx","etxt","xtet","xtte","xett","xett","xtte","xtet","ttex","ttxe","tetx","text","txte","txet"]
  Q  - de-duplicate                      ["text","tetx","txet","txte","ttex","ttxe","etxt","ettx","extt","xtet","xtte","xett"]
   Ḋ - dequeue (the first one = input)          ["tetx","txet","txte","ttex","ttxe","etxt","ettx","extt","xtet","xtte","xett"]

Answer 9

Mathematica, 47 bytes

Drop[StringJoin/@Permutations[Characters@#],1]&

Answer 10

Python 3, 85 76 63 bytes

As a function, and returning strings as list of characters (thanks to @pizzapants184 for telling me it is allowed):

from itertools import*
lambda z:set(permutations(z))-{tuple(z)}

As a function:

from itertools import*
lambda z:map("".join,set(permutations(z))-{tuple(z)})

85 bytes as a full program:

from itertools import*
z=input()
print(*map("".join,set(permutations(z))-{tuple(z)}))

Could be reduced a bit if outputting strings as ('a', 'b', 'c') is allowed (I am not sure it is).

Answer 11

Java 8, 245 239 237 bytes

import java.util.*;s->{Set l=new HashSet();p("",s,l);l.remove(s);l.forEach(System.out::println);}void p(String p,String s,Set l){int n=s.length(),i=0;if(n<1)l.add(p);else for(;i<n;p(p+s.charAt(i),s.substring(0,i)+s.substring(++i,n),l));}

-6 bytes thanks to @OlivierGrégoire.

Typical verbose Java.. I see a lot of <10 byte answers, and here I am with 200+ bytes.. XD

Explanation:

Try it here.

import java.util.*;         // Required import for the Set and HashSet

s->{                        // Method (1) with String parameter and no return-type
  Set l=new HashSet();      //  Set to save all permutations in (without duplicates)
  p("",s);                  //  Determine all permutations, and save them in the Set
  l.remove(s);              //  Remove the input from the Set
  l.forEach(                //  Loop over the Set
    System.out::println);   //   And print all the items
}                           // End of method (1)

// This method will determine all permutations, and save them in the Set:
void p(String p,String s,Set l){
  int n=s.length(),         //  Length of the first input String
      i=0;                  //  And a temp index-integer
  if(n<1)                   //  If the length is 0:
    l.add(p);               //   Add the permutation input-String to the Set
  else                      //  Else:
    for(;i<n;               //   Loop over the first input-String
      p(                    //    And do a recursive-call with:
        p+s.charAt(i),      //     Permutation + char
        s.substring(0,i)+s.substring(++i,n),l)
                            //     Everything except this char
      );                    //   End of loop
}                           // End of method (2)

Answer 12

Perl 6,  39  38 bytes

*.comb.permutations».join.unique[1..*]

Try it

*.comb.permutations».join.unique.skip

Try it

Expanded

*               # WhateverCode lambda (this is the parameter)
.comb           # split into graphemes
.permutations\  # get all of the permutations
».join          # join each of them with a hyper method call
.unique         # make sure they are unique
.skip           # start after the first value (the input)

Answer 13

C++, 142 bytes

#include<algorithm>
void p(std::string s){auto b=s;sort(begin(s),end(s));do if(s!=b)puts(s.data());while(next_permutation(begin(s),end(s)));}

ungolfed

#include <algorithm>

void p(std::string s)
{
    auto b = s;                    // use auto to avoid std::string
    sort(begin(s), end(s));        // start at first permutation
    do
      if (s != b)                  // only print permutation different than given string
        puts(s.data());
    while (next_permutation(begin(s), end(s))); // move to next permutation
}

Answer 14

K (oK), 13 bytes

Solution:

1_?x@prm@!#x:

Try it online!

Explanation:

Evaluation is performed right-to-left.

1_?x@prm@!#x: / the solution
           x: / store input in variable x
          #   / count length of x, #"abc" => 3
         !    / range, !3 => 0 1 2
     prm@     / apply (@) function permutations (prm) to range
   x@         / apply (@) these pumuted indixes back to original input
  ?           / return distinct values
1_            / drop the first one (ie the original input)

Answer 15

JavaScript (ES6), 101 bytes

Adopted from a past answer of mine.

S=>(R=new Set,p=(s,m='')=>s[0]?s.map((_,i)=>p(a=[...s],m+a.splice(i,1))):R.add(m),_=p([...S]),[...R])

f=
S=>(R=new Set,p=(s,m='')=>s[0]?s.map((_,i)=>p(a=[...s],m+a.splice(i,1))):R.add(m),_=p([...S]),[...R])


console.log(
  f('ABC'),
  f('ABCD'),
  f('ABCC'),
  f('AABBC')
)

Answer 16

Perl 5, 89 + 2 (-F) = 91 bytes

$,=$_;$"=",";map{say if!$k{$_}++&&$,ne$_&&(join"",sort@F)eq join"",sort/./g}glob"{@F}"x@F

Try it online!