Code-Golf: Permutations

Question

Write a function that takes as input a set of integers (can be a list, array or any other container with distinct numbers), and outputs the list of all its permutations.

Python (95 chars):

p=lambda s:s and sum(map(lambda e:map(lambda p:[e]+p,p(filter(lambda x:x!=e,s))),s),[]) or [[]]

It'd be nice to be beaten in the same language, but implementations in other languages are more than welcome!

Answer 1

Python, 52

Input is a set. Output is a list of lists.

f=lambda a:[p+[x]for x in a for p in f(a-{x})]or[[]]

This is shorter than the answer that does all the work with a builtin.

Answer 2

Python - 76 chars

Longer than gnibbler's, but implements things from scratch.

p=lambda x:x and[[a]+b for a in x for b in p([c for c in x if c!=a])]or[[]]

Answer 3

J, 11 characters

(i.@!@#A.[)

Usage:

   (i.@!@#A.[) 1 3 5
1 3 5
1 5 3
3 1 5
3 5 1
5 1 3
5 3 1

Explanation:

i.@!@# uses three verbs to return a list from 0 to (!n)-1 where n is the number of items in the given list.

[ returns the list itself. In the example shown that gives 0 1 2 3 4 5 A. 1 3 5.

A. returns one possible permutation of the second list for each item in the first list (kind of - the proper explanation is given here).

Answer 4

Python - 55 chars

from itertools import*
p=lambda x:list(permutations(x))

Answer 5

Haskell, _⁴⁴ 43

p[]=[[]]
p l=[e:r|e<-l,r<-p$filter(/=e)l]

Essentially the same as ugoren's solution, but Haskell is better at list comprehensions!

---

Of course, it can also do

30

import Data.List
p=permutations

---

More efficient approach, that doesn't require an equality comparison:

92

import Data.List
p[]=[[]]
p l=(\(l,(e:r))->map(e:)$p(l++r))=<<(init$zip(inits l)(tails l))

As a consequece, this one also works when there are duplicate elements in the list.

Answer 6

C, 270 243 239 characters

#define S t=*a;*a=a[i];a[i]=t;
#define R o=p(n,r-1,a+1,o,r-2,0)
int*p(n,r,a,o,i,t)int*a,*o;{if(!r)for(;n;--n)*o++=*--a;else{R;for(;i;--i){S R;S}}return o;}
P(n,a)int*a;{int N=1,i=n;for(;i;N*=i--);return p(n,n,a,malloc(N*n*8),n-1,0)-N*n;}

The function P(n,a) returns a pointer to the n! permutations of a, packed one after another in one giant array.

Answer 7

Prolog (SWI), 33 bytes

X-Y:-bagof(A,permutation(X,A),Y).

Try it online!

-2 from steffan

Answer 8

in Q (48)

g:{$[x=1;y;raze .z.s[x-1;y]{x,/:y except x}\:y]}

Sample usage:

q)g[3;1 2 3]
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1

Answer 9

Ruby - 23 chars

f=->x{p *x.permutation}

for example f[[1,2,3]] outputs this.

but using [].permutation feels like cheating, so:

Ruby - 59 chars

f=->a{a.size<2?[a]:a.flat_map{|x|f[(a-x=[x])].map{|y|x+y}}}

tested with

100.times.all?{arr=(1..99).to_a.sample(rand(5)); arr.permutation.to_a==f[arr]}
=> true

Answer 10

Scala 30:

def p(s:Seq[_])=s.permutations 

Scala 195, quick'n'dirty, without permutations from library:

def c(x:Int,t:List[_]):List[_]={val l=t.size
val o=x%l
if(l>1){val r=c(x/l,t.tail)
r.take(o):::(t.head::r.drop(o))}else
t}
def p(y:List[_])=(0 to(1 to y.size).product).foreach(z=>println(c(z,y)))

val y=List(0,1,2,3)
p(y)

Scala 293, full grown, type safe iterator:

class P[A](val l:Seq[A])extends Iterator[Seq[A]]{
var c=0
val s=(1 to l.size).product
def g(c:Int,t:List[A]):List[A]={
val n=t.size
val o=c%n
if(n>1){val r=g(c/n,t.tail)
r.take(o):::(t.head::r.drop(o))
}else
t}
def hasNext=c!=s
def next={c+=1
g(c-1,l.toList)}
}
for(e<-new P("golf"))println(e)

Answer 11

Python - 58 chars

Slightly shorter than ugoren's, by taking a set as input:

p=lambda x:x and[[y]+l for y in x for l in p(x-{y})]or[[]]

Answer 12

K, 30 bytes

{x@v@&((#x;1)~^=:)'v:!(#x)##x}

No builtins!

Answer 13

JS - 154 146 chars

function f(x){var a=[],m;(m=x.length)>1?f(x.slice(1)).map(function(y){for(l=m;l--;a.push(y.slice(0,l).concat(x[0],y.slice(l))));}):a=[x];return a}

Test : f([1,2,3,4,5]).map(function(a){return a.join('')}).join('\n') returns this.

Answer 14

R

Since we are talking about permutations let me show at least one solution in R:

library(gtools);v=c(3,4,5);permutations(length(v),length(v),v)

Answer 15

Perl 188

No library routines, no recursion

sub p{$l=(@_=sort split'',shift)-1;while(print@_){$k=$j=$l;--$k while($_[$k-1]cmp$_[$k])>=0;$k||last;--$j while($_[$k-1]cmp$_[$j])>=0;@_[$j,$k-1]=@_[$k-1,$j];@_[$k..$l]=reverse@_[$k..$l]}}

Answer 16

Python - 50 chars

import itertools
list(itertools.permutations("123"))

Answer 17

Pyth, 4 bytes

L.pb

Yeah, Pyth was created after this challenge was posted and all. This is still really cool. :D

Live demo.

Reading from stdin is a byte shorter:

.pQ

Answer 18

JavaScript 143 136 134 123

function p(s,a="",c="",i,z=[]){a+=c,i=s.length
!i?z.push(a):0
for(;i--;s.splice(i,0,c))p(s,a,c=s.splice(i,1),0,z);return z}

var perms = p([1,2,3]);

document.getElementById('output').innerHTML = perms.join("\n");

<pre id="output"></pre>

Answer 19

05AB1E - 2 1 bytes

œ

The input must be an array/list.

Explanation:

œ //Takes all the permutations of the elements in the top of the stack (the input is a list, so it would work)

Saved a byte thanks to Erik the Outgolfer

Answer 20

Brachylog, 2 bytes

pᵘ

Try it online!

 ᵘ    Find every unique
p     permutation of the input.

Answer 21

Vyxal, 2 bytes

ṖU

Try it Online!

Ṗ  # All permutations of input 
 U # Filtered by unique

Answer 22

JavaScript (ES6), 91 78 bytes

-5 thanks to @Neil

f=a=>a.length<2?[a]:a.flatMap((x,i)=>f(a.filter((_,j)=>i-j)).map(t=>[x,...t]))

Pretty normal recursive approach, and my first time using copyWithin :p

Okay, copyWithin modifies the array, so that doesn't work. Used a different approach, which now saves more bytes.

Answer 23

Curry, 41 26 bytes

15 bytes saved by alephalpha

Tested in PAKCS

p[]=[]
p(x++a:y)=a:p(x++y)

Try it online!

Uses non-determinism to output all the permutations. This same code is used in my answer here for which I originally wrote it.

Answer 24

Knight, 180 bytes

;=i=jF;W=pP E++"=x"=i+1i" p";=s"";W>i-=j+1jT=s+++++++s";=y"j"F W>i-=y"j"+1y"j"T"E+s';=z=bF;W>i-=z+1zT;=aF W>i-=a+1aT|?z a=b|b?E+"y"zE+"y"a|b;=rF;=s"";W>z=r+1r=s++sE+"x"E+"y"r" "Os'

Try it online!

This was a pain. Nested evals and stuff. Since recursion doesn't really work, I had to literally compose a string of nested while loops, and then eval it. Extremely inefficient, \$O(n^{n+1})\$ where \$n\$ is the length of the input. My browser started to freeze up with \$n=6\$.

-2 bytes thanks to Aiden Chow.

Answer 25

K (ngn/k), 15 bytes

{x@?<'+!a#a:#x}

Try it online!

Answer 26

Perl 5, 66 bytes

sub{$"=",";grep{my%t;@t{@$_}++;@_==%t}map{[/\d+/g]}glob"{@_}-"x@_}

Try it online!

Relies on the problem definition of distinct integers. Also horribly wasteful--it creates n**n candidates only to discard the ones that duplicate members within them.

Answer 27

Python, 53 bytes

from itertools import*;lambda x:list(permutations(x))

Answer 28

Jelly, 2 bytes

Œ!

Try it online!

Yay for builtins!

Answer 29

K (oK), 3 bytes

Solution

prm

Try it online!

Explanation:

It's a 3 byte built-in shortcut to the following built-in 47 byte function:

{[x]{[x]$[x;,/x ,''o'x ^/:x;,x]}@$[-8>@x;!x;x]}

... which can be shortened to 23 bytes if we know we're getting a list of ints as input:

{$[x;,/x,''o'x^/:x;,x]} / golfed built in
{                     } / lambda function with implicit input x
 $[ ;             ;  ]  / if[condition;true;false]
   x                    / if x is not null...
             x^/:x      / x except (^) each-right (/:) x; create length-x combinations
           o'           / call self (o) with each of these
       x,''             / x concatenated with each-each of these results (this is kinda magic to me)
     ,/                 / flatten list
                    ,x  / otherwise enlist x (enlisted empty list)

Answer 30

Axiom, 160 bytes

p(a)==(#a=0=>[[]];r:=[[a.1]];r:=delete(r,1);n:=#a;m:=factorial n;m>1.E7=>r;b:=permutations n;for j in 1..m repeat(x:=b.j;r:=concat([a.(x.i)for i in 1..n],r));r)

ungolfed

--Permutation of a
pmt(a)==
     #a=0=>[[]]
     r:=[[a.1]]; r:=delete(r,1) -- r has the type List List typeof(a)
     n:=#a
     m:=factorial n
     m>1.E7=>r
     b:=permutations(n)         --one built in for permutation indices 
     for j in 1..m repeat
        x:=b.j
        r:=concat([a.(x.i) for i in 1..n],r)
     r

All this call one library function that give permutation on index (only integers as permutation as permutations on [1], permutations on [1,2], permutations on[1,2,3] etc).So it is enough get these set of indices and build the lists; One has to note that this seems to be compiled good for every List of type X

(4) -> p([1,2,3])
   Compiling function p with type List PositiveInteger -> List List
      PositiveInteger
   (4)  [[1,2,3],[1,3,2],[3,1,2],[2,1,3],[2,3,1],[3,2,1]]
                                          Type: List List PositiveInteger
(5) -> p([x^2,y*x,y^2])
   Compiling function p with type List Polynomial Integer -> List List
      Polynomial Integer
   (5)
      2      2    2  2        2  2            2  2        2  2    2      2
   [[x ,x y,y ],[x ,y ,x y],[y ,x ,x y],[x y,x ,y ],[x y,y ,x ],[y ,x y,x ]]
                                       Type: List List Polynomial Integer
(6) -> p([sin(x),log(y)])
   Compiling function p with type List Expression Integer -> List List
      Expression Integer
   (6)  [[sin(x),log(y)],[log(y),sin(x)]]
                                       Type: List List Expression Integer
(7) -> m:=p("abc")::List List Character
   Compiling function p with type String -> Any
   (7)  [[a,b,c],[a,c,b],[c,a,b],[b,a,c],[b,c,a],[c,b,a]]
                                                Type: List List Character
(8) -> [concat(map(x+->x::String, m.j))  for j in 1..#m]
   (8)  ["abc","acb","cab","bac","bca","cba"]
                                                        Type: List String