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Task

Write a program that generates all permutations of n different objects. (Practically numerals!)

Related tasks

The number of samples of size k from n objects.

With combinations and permutations generation tasks.
	Order Unimportant	Order Important
Without replacement	${\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}$	$^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)$
Without replacement	Task: Combinations	Task: Permutations
With replacement	${\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}$	$n^{k}$
With replacement	Task: Combinations with repetitions	Task: Permutations with repetitions

360 Assembly

Translation of: Liberty BASIC

<lang 360asm>* Permutations 26/10/2015 PERMUTE CSECT

        USING  PERMUTE,R15        set base register
        LA     R9,TMP-A           n=hbound(a)
        SR     R10,R10            nn=0

LOOP LA R10,1(R10) nn=nn+1

        LA     R11,PG             pgi=@pg
        LA     R6,1               i=1

LOOPI1 CR R6,R9 do i=1 to n

        BH     ELOOPI1
        LA     R2,A-1(R6)         @a(i)
        MVC    0(1,R11),0(R2)     output a(i)
        LA     R11,1(R11)         pgi=pgi+1
        LA     R6,1(R6)           i=i+1
        B      LOOPI1

ELOOPI1 XPRNT PG,80

        LR     R6,R9              i=n

LOOPUIM BCTR R6,0 i=i-1

        LTR    R6,R6              until i=0
        BE     ELOOPUIM
        LA     R2,A-1(R6)         @a(i)
        LA     R3,A(R6)           @a(i+1)
        CLC    0(1,R2),0(R3)      or until a(i)<a(i+1)
        BNL    LOOPUIM

ELOOPUIM LR R7,R6 j=i

        LA     R7,1(R7)           j=i+1
        LR     R8,R9              k=n

LOOPWJ CR R7,R8 do while j<k

        BNL    ELOOPWJ
        LA     R2,A-1(R7)         r2=@a(j)
        LA     R3,A-1(R8)         r3=@a(k)
        MVC    TMP,0(R2)          tmp=a(j)
        MVC    0(1,R2),0(R3)      a(j)=a(k)
        MVC    0(1,R3),TMP        a(k)=tmp
        LA     R7,1(R7)           j=j+1
        BCTR   R8,0               k=k-1
        B      LOOPWJ

ELOOPWJ LTR R6,R6 if i>0

        BNP    ILE0
        LR     R7,R6              j=i
        LA     R7,1(R7)           j=i+1

LOOPWA LA R2,A-1(R7) @a(j)

        LA     R3,A-1(R6)         @a(i)
        CLC    0(1,R2),0(R3)      do while a(j)<a(i)
        BNL    AJGEAI
        LA     R7,1(R7)           j=j+1
        B      LOOPWA

AJGEAI LA R2,A-1(R7) r2=@a(j)

        LA     R3,A-1(R6)         r3=@a(i)
        MVC    TMP,0(R2)          tmp=a(j)
        MVC    0(1,R2),0(R3)      a(j)=a(i)
        MVC    0(1,R3),TMP        a(i)=tmp

ILE0 LTR R6,R6 until i<>0

        BNE    LOOP
        XR     R15,R15            set return code
        BR     R14                return to caller

A DC C'ABCD' <== input TMP DS C temp for swap PG DC CL80' ' buffer

        YREGS
        END    PERMUTE</lang>

Output:

ABCD
ABDC
ACBD
ACDB
ADBC
ADCB
BACD
BADC
BCAD
BCDA
BDAC
BDCA
CABD
CADB
CBAD
CBDA
CDAB
CDBA
DABC
DACB
DBAC
DBCA
DCAB
DCBA

ABAP

<lang ABAP>data: lv_flag type c,

     lv_number type i,
     lt_numbers type table of i.

append 1 to lt_numbers. append 2 to lt_numbers. append 3 to lt_numbers.

do.

 perform permute using lt_numbers changing lv_flag.
 if lv_flag = 'X'.
   exit.
 endif.
 loop at lt_numbers into lv_number.
   write (1) lv_number no-gap left-justified.
   if sy-tabix <> '3'.
     write ', '.
   endif.
 endloop.
 skip.

enddo.

" Permutation function - this is used to permute: " Can be used for an unbounded size set. form permute using iv_set like lt_numbers

            changing ev_last type c.
 data: lv_len     type i,
       lv_first   type i,
       lv_third   type i,
       lv_count   type i,
       lv_temp    type i,
       lv_temp_2  type i,
       lv_second  type i,
       lv_changed type c,
       lv_perm    type i.
 describe table iv_set lines lv_len.

 lv_perm = lv_len - 1.
 lv_changed = ' '.
 " Loop backwards through the table, attempting to find elements which
 " can be permuted. If we find one, break out of the table and set the
 " flag indicating a switch.
 do.
   if lv_perm <= 0.
     exit.
   endif.
   " Read the elements.
   read table iv_set index lv_perm into lv_first.
   add 1 to lv_perm.
   read table iv_set index lv_perm into lv_second.
   subtract 1 from lv_perm.
   if lv_first < lv_second.
     lv_changed = 'X'.
     exit.
   endif.
   subtract 1 from lv_perm.
 enddo.

 " Last permutation.
 if lv_changed <> 'X'.
   ev_last = 'X'.
   exit.
 endif.

 " Swap tail decresing to get a tail increasing.
 lv_count = lv_perm + 1.
 do.
   lv_first = lv_len + lv_perm - lv_count + 1.
   if lv_count >= lv_first.
     exit.
   endif.

   read table iv_set index lv_count into lv_temp.
   read table iv_set index lv_first into lv_temp_2.
   modify iv_set index lv_count from lv_temp_2.
   modify iv_set index lv_first from lv_temp.
   add 1 to lv_count.
 enddo.

 lv_count = lv_len - 1.
 do.
   if lv_count <= lv_perm.
     exit.
   endif.

   read table iv_set index lv_count into lv_first.
   read table iv_set index lv_perm into lv_second.
   read table iv_set index lv_len into lv_third.
   if ( lv_first < lv_third ) and ( lv_first > lv_second ).
     lv_len = lv_count.
   endif.

   subtract 1 from lv_count.
 enddo.

 read table iv_set index lv_perm into lv_temp.
 read table iv_set index lv_len into lv_temp_2.
 modify iv_set index lv_perm from lv_temp_2.
 modify iv_set index lv_len from lv_temp.

endform.</lang>

Output:

We split the task into two parts: The first part is to represent permutations, to initialize them and to go from one permutation to another one, until the last one has been reached. This can be used elsewhere, e.g., for the Topswaps [[1]] task. The second part is to read the N from the command line, and to actually print all permutations over 1 .. N.

The generic package Generic_Perm

When given N, this package defines the Element and Permutation types and exports procedures to set a permutation P to the first one, and to change P into the next one: <lang ada>generic

  N: positive;

package Generic_Perm is

  subtype Element is Positive range 1 .. N;
  type Permutation is array(Element) of Element;
  
  procedure Set_To_First(P: out Permutation; Is_Last: out Boolean);
  procedure Go_To_Next(P: in out Permutation; Is_Last: out Boolean);

end Generic_Perm;</lang>

Here is the implementation of the package: <lang ada>package body Generic_Perm is

  procedure Set_To_First(P: out Permutation; Is_Last: out Boolean) is
  begin
     for I in P'Range loop

P (I) := I;

     end loop;
     Is_Last := P'Length = 1; 
     -- if P has a single element, the fist permutation is the last one
  end Set_To_First;
  
  procedure Go_To_Next(P: in out Permutation; Is_Last: out Boolean) is
     
     procedure Swap (A, B : in out Integer) is
        C : Integer := A;
     begin
        A := B;
        B := C;
     end Swap;
     
     I, J, K : Element;      
  begin
     -- find longest tail decreasing sequence
     -- after the loop, this sequence is I+1 .. n,
     -- and the ith element will be exchanged later
     -- with some element of the tail
     Is_Last := True;
     I := N - 1;
     loop

if P (I) < P (I+1) then Is_Last := False; exit; end if;

-- next instruction will raise an exception if I = 1, so -- exit now (this is the last permutation) exit when I = 1; I := I - 1;

     end loop;
     
     -- if all the elements of the permutation are in
     -- decreasing order, this is the last one
     if Is_Last then

return;

     end if;
     
     -- sort the tail, i.e. reverse it, since it is in decreasing order
     J := I + 1;
     K := N;
     while J < K loop

Swap (P (J), P (K)); J := J + 1; K := K - 1;

     end loop;
     
     -- find lowest element in the tail greater than the ith element
     J := N;
     while P (J) > P (I) loop

J := J - 1;

     end loop;
     J := J + 1;
     
     -- exchange them
     -- this will give the next permutation in lexicographic order,
     -- since every element from ith to the last is minimum
     Swap (P (I), P (J));
  end Go_To_Next;

end Generic_Perm;</lang>

The procedure Print_Perms

<lang ada>with Ada.Text_IO, Ada.Command_Line, Generic_Perm;

procedure Print_Perms is

  package CML renames Ada.Command_Line;
  package TIO renames Ada.Text_IO;

begin

  declare
     package Perms is new Generic_Perm(Positive'Value(CML.Argument(1)));
     P : Perms.Permutation;
     Done : Boolean := False;
     
     procedure Print(P: Perms.Permutation) is
     begin
        for I in P'Range loop
           TIO.Put (Perms.Element'Image (P (I)));
        end loop;
        TIO.New_Line;
     end Print;         
  begin
     Perms.Set_To_First(P, Done);
     loop
        Print(P);
        exit when Done;
        Perms.Go_To_Next(P, Done);
     end loop;
  end;

exception

  when Constraint_Error 
    => TIO.Put_Line ("*** Error: enter one numerical argument n with n >= 1");

end Print_Perms;</lang>

Output:

>./print_perms 3
 1 2 3
 1 3 2
 2 1 3
 2 3 1
 3 1 2
 3 2 1
 3 2 1

ALGOL 68

Works with: ALGOL 68 version Revision 1 - one minor extension to language used - PRAGMA READ, similar to C's #include directive.

Works with: ALGOL 68G version Any - tested with release algol68g-2.6.

File: prelude_permutations.a68<lang algol68># -*- coding: utf-8 -*- #

COMMENT REQUIRED BY "prelude_permutations.a68"

 MODE PERMDATA = ~;

PROVIDES:

PERMDATA*=~* #
perm*=~ list* #

END COMMENT

MODE PERMDATALIST = REF[]PERMDATA; MODE PERMDATALISTYIELD = PROC(PERMDATALIST)VOID;

Generate permutations of the input data list of data list #

PROC perm gen permutations = (PERMDATALIST data list, PERMDATALISTYIELD yield)VOID: (

Warning: this routine does not correctly handle duplicate elements #

 IF LWB data list = UPB data list THEN
   yield(data list)
 ELSE
   FOR elem FROM LWB data list TO UPB data list DO
     PERMDATA first = data list[elem];
     data list[LWB data list+1:elem] := data list[:elem-1];
     data list[LWB data list] := first;
   # FOR PERMDATALIST next data list IN # perm gen permutations(data list[LWB data list+1:] # ) DO #,
   ##   (PERMDATALIST next)VOID:(
       yield(data list)
   # OD #));
     data list[:elem-1] := data list[LWB data list+1:elem];
     data list[elem] := first
   OD
 FI

);

SKIP</lang>File: test_permutations.a68<lang algol68>#!/usr/bin/a68g --script #

-*- coding: utf-8 -*- #

CO REQUIRED BY "prelude_permutations.a68" CO

 MODE PERMDATA = INT;

PROVIDES:#
PERM*=INT* #
perm *=int list *#

PR READ "prelude_permutations.a68" PR;

main:(

 FLEX[0]PERMDATA test case := (1, 22, 333, 44444);

 INT upb data list = UPB test case;
 FORMAT 
   data fmt := $g(0)$,
   data list fmt := $"("n(upb data list-1)(f(data fmt)", ")f(data fmt)")"$;

FOR DATALIST permutation IN # perm gen permutations(test case#) DO (#,
1. (PERMDATALIST permutation)VOID:(

   printf((data list fmt, permutation, $l$))

OD #))

)</lang>Output:

(1, 22, 333, 44444)
(1, 22, 44444, 333)
(1, 333, 22, 44444)
(1, 333, 44444, 22)
(1, 44444, 22, 333)
(1, 44444, 333, 22)
(22, 1, 333, 44444)
(22, 1, 44444, 333)
(22, 333, 1, 44444)
(22, 333, 44444, 1)
(22, 44444, 1, 333)
(22, 44444, 333, 1)
(333, 1, 22, 44444)
(333, 1, 44444, 22)
(333, 22, 1, 44444)
(333, 22, 44444, 1)
(333, 44444, 1, 22)
(333, 44444, 22, 1)
(44444, 1, 22, 333)
(44444, 1, 333, 22)
(44444, 22, 1, 333)
(44444, 22, 333, 1)
(44444, 333, 1, 22)
(44444, 333, 22, 1)

AppleScript

Translation of: JavaScript

(Functional ES5 version)

<lang AppleScript>-- permutations :: [a] -> a on permutations(xs)

   script firstElement
       on lambda(x)
           script tailElements
               on lambda(ys)
                   {{x} & ys}
               end lambda
           end script
           
           concatMap(tailElements, permutations(|delete|(x, xs)))
       end lambda
   end script
   
   if length of xs > 0 then
       concatMap(firstElement, xs)
   else
       {{}}
   end if

end permutations

-- TEST on run

   permutations({"aardvarks", "eat", "ants"})

end run

-- GENERIC LIBRARY FUNCTIONS

-- concatMap :: (a -> [b]) -> [a] -> [b] on concatMap(f, xs)

   script append
       on lambda(a, b)
           a & b
       end lambda
   end script
   
   foldl(append, {}, map(f, xs))

end concatMap

-- map :: (a -> b) -> [a] -> [b] on map(f, xs)

   tell mReturn(f)
       set lng to length of xs
       set lst to {}
       repeat with i from 1 to lng
           set end of lst to lambda(item i of xs, i, xs)
       end repeat
       return lst
   end tell

end map

-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)

   tell mReturn(f)
       set v to startValue
       set lng to length of xs
       repeat with i from 1 to lng
           set v to lambda(v, item i of xs, i, xs)
       end repeat
       return v
   end tell

end foldl

-- delete :: a -> [a] -> [a] on |delete|(x, xs)

   script Eq
       on lambda(a, b)
           a = b
       end lambda
   end script
   
   deleteBy(Eq, x, xs)

end |delete|

-- deleteBy :: (a -> a -> Bool) -> a -> [a] -> [a] on deleteBy(fnEq, x, xs)

   if length of xs > 0 then
       set {h, t} to uncons(xs)
       if lambda(x, h) of mReturn(fnEq) then
           t
       else
           {h} & deleteBy(fnEq, x, t)
       end if
   else
       {}
   end if

end deleteBy

-- uncons :: [a] -> Maybe (a, [a]) on uncons(xs)

   if length of xs > 0 then
       {item 1 of xs, rest of xs}
   else
       missing value
   end if

end uncons

-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)

   if class of f is script then
       f
   else
       script
           property lambda : f
       end script
   end if

end mReturn</lang>

Output:

{{"aardvarks", "eat", "ants"}, {"aardvarks", "ants", "eat"}, 
{"eat", "aardvarks", "ants"}, {"eat", "ants", "aardvarks"}, 
{"ants", "aardvarks", "eat"}, {"ants", "eat", "aardvarks"}}

Translation of: Pseudocode

(Fast recursive Heap's algorithm) <lang AppleScript> to DoPermutations(aList, n) --> Heaps's algorithm (Permutation by interchanging pairs) AppleScript by Jean.O.matiC if n = 1 then tell (a reference to Permlist) to copy aList to its end -- or: copy aList as text (for concatenated results) else repeat with i from 1 to n DoPermutations(aList, n - 1) if n mod 2 = 0 then -- n is even tell aList to set [item i, item n] to [item n, item i] -- swaps items i and n of aList else tell aList to set [item 1, item n] to [item n, item 1] -- swaps items 1 and n of aList end if set i to i + 1 end repeat end if return (a reference to Permlist) as list end DoPermutations

--> Example 1 (list of words) set [SourceList, Permlist] to [{"Good", "Johnny", "Be"}, {}] DoPermutations(SourceList, SourceList's length) --> result {{"Good", "Johnny", "Be"}, {"Johnny", "Good", "Be"}, {"Be", "Good", "Johnny"},¬

{"Good", "Be", "Johnny"}, {"Johnny", "Be", "Good"}, {"Be", "Johnny", "Good"}}

--> Example 2 (with concatenated results) set [SourceList, Permlist] to [{"X", "Y", "Z"}, {}] DoPermutations(SourceList, SourceList's length) --> result {"XYZ", "YXZ", "ZXY", "XZY", "YZX", "ZYX"}

--> Example 3 (Integers with concatenated results) set [SourceList, Permlist] to [{1, 2, 3}, {}] DoPermutations(SourceList, SourceList's length) --> result {"123", "213", "312", "132", "231", "321"}</lang>

AutoHotkey

from the forum topic http://www.autohotkey.com/forum/viewtopic.php?t=77959 <lang AutoHotkey>#NoEnv StringCaseSense On

o := str := "Hello"

Loop {

  str := perm_next(str)
  If !str
  {
     MsgBox % clipboard := o
     break
  }
  o.= "`n" . str

}

perm_Next(str){

  p := 0, sLen := StrLen(str)
  Loop % sLen
  {
     If A_Index=1
        continue
     t := SubStr(str, sLen+1-A_Index, 1)
     n := SubStr(str, sLen+2-A_Index, 1)
     If ( t < n )
     {
        p := sLen+1-A_Index, pC := SubStr(str, p, 1)
        break
     }
  }
  If !p
     return false
  Loop
  {
     t := SubStr(str, sLen+1-A_Index, 1)
     If ( t > pC )
     {
        n := sLen+1-A_Index, nC := SubStr(str, n, 1)
        break
     }
  }
  return SubStr(str, 1, p-1) . nC . Reverse(SubStr(str, p+1, n-p-1) . pC .  SubStr(str, n+1))

}

Reverse(s){

  Loop Parse, s
     o := A_LoopField o
  return o

}</lang>

Output:

Hello
Helol
Heoll
Hlelo
Hleol
Hlleo
Hlloe
Hloel
Hlole
Hoell
Holel
Holle
eHllo
eHlol
eHoll
elHlo
elHol
ellHo
elloH
eloHl
elolH
eoHll
eolHl
eollH
lHelo
lHeol
lHleo
lHloe
lHoel
lHole
leHlo
leHol
lelHo
leloH
leoHl
leolH
llHeo
llHoe
lleHo
lleoH
lloHe
lloeH
loHel
loHle
loeHl
loelH
lolHe
loleH
oHell
oHlel
oHlle
oeHll
oelHl
oellH
olHel
olHle
oleHl
olelH
ollHe
olleH

Alternate Version

Alternate version to produce numerical permutations of combinations. <lang ahk>P(n,k="",opt=0,delim="",str="") { ; generate all n choose k permutations lexicographically ;1..n = range, or delimited list, or string to parse ; to process with a different min index, pass a delimited list, e.g. "0`n1`n2" ;k = length of result ;opt 0 = no repetitions ;opt 1 = with repetitions ;opt 2 = run for 1..k ;opt 3 = run for 1..k with repetitions ;str = string to prepend (used internally) ;returns delimited string, error message, or (if k > n) a blank string i:=0 If !InStr(n,"`n") If n in 2,3,4,5,6,7,8,9 Loop, %n% n := A_Index = 1 ? A_Index : n "`n" A_Index Else Loop, Parse, n, %delim% n := A_Index = 1 ? A_LoopField : n "`n" A_LoopField If (k = "") RegExReplace(n,"`n","",k), k++ If k is not Digit Return "k must be a digit." If opt not in 0,1,2,3 Return "opt invalid." If k = 0 Return str Else Loop, Parse, n, `n If (!InStr(str,A_LoopField) || opt & 1) s .= (!i++ ? (opt & 2 ? str "`n" : "") : "`n" ) . P(n,k-1,opt,delim,str . A_LoopField . delim) Return s }</lang>

Output:

<lang ahk>MsgBox % P(3)</lang>

---------------------------
permute.ahk
---------------------------
123
132
213
231
312
321
---------------------------
OK   
---------------------------

<lang ahk>MsgBox % P("Hello",3)</lang>

---------------------------
permute.ahk
---------------------------
Hel
Hel
Heo
Hle
Hlo
Hle
Hlo
Hoe
Hol
Hol
eHl
eHl
eHo
elH
elo
elH
elo
eoH
eol
eol
lHe
lHo
leH
leo
loH
loe
lHe
lHo
leH
leo
loH
loe
oHe
oHl
oHl
oeH
oel
oel
olH
ole
olH
ole
---------------------------
OK   
---------------------------

<lang ahk>MsgBox % P("2`n3`n4`n5",2,3)</lang>

---------------------------
permute.ahk
---------------------------

2
22
23
24
25
3
32
33
34
35
4
42
43
44
45
5
52
53
54
55
---------------------------
OK   
---------------------------

<lang ahk>MsgBox % P("11 a text ] u+z",3,0," ")</lang>

---------------------------
permute.ahk
---------------------------
11 a text 
11 a ] 
11 a u+z 
11 text a 
11 text ] 
11 text u+z 
11 ] a 
11 ] text 
11 ] u+z 
11 u+z a 
11 u+z text 
11 u+z ] 
a 11 text 
a 11 ] 
a 11 u+z 
a text 11 
a text ] 
a text u+z 
a ] 11 
a ] text 
a ] u+z 
a u+z 11 
a u+z text 
a u+z ] 
text 11 a 
text 11 ] 
text 11 u+z 
text a 11 
text a ] 
text a u+z 
text ] 11 
text ] a 
text ] u+z 
text u+z 11 
text u+z a 
text u+z ] 
] 11 a 
] 11 text 
] 11 u+z 
] a 11 
] a text 
] a u+z 
] text 11 
] text a 
] text u+z 
] u+z 11 
] u+z a 
] u+z text 
u+z 11 a 
u+z 11 text 
u+z 11 ] 
u+z a 11 
u+z a text 
u+z a ] 
u+z text 11 
u+z text a 
u+z text ] 
u+z ] 11 
u+z ] a 
u+z ] text 
---------------------------
OK   
---------------------------

Batch File

Recursive permutation generator. <lang Batch File> @echo off setlocal enabledelayedexpansion set arr=ABCD set /a n=4

echo !arr!

call :permu %n% arr goto:eof

permu num &arr

setlocal if %1 equ 1 call echo(!%2! & exit /b set /a "num=%1-1,n2=num-1" set arr=!%2! for /L %%c in (0,1,!n2!) do (

  call:permu !num! arr 
  set /a  n1="num&1"
  if !n1! equ 0 (call:swapit !num! 0 arr) else (call:swapit !num! %%c arr)
  )
  call:permu !num! arr

endlocal & set %2=%arr% exit /b

swapit from to &arr

setlocal set arr=!%3! set temp1=!arr:~%~1,1! set temp2=!arr:~%~2,1! set arr=!arr:%temp1%=@! set arr=!arr:%temp2%=%temp1%! set arr=!arr:@=%temp2%!

echo %1 %2 !%~3! !arr!

endlocal & set %3=%arr% exit /b </lang>

Output:

ABCD
BACD
CABD
ACBD
BCAD
CBAD
DBAC
BDAC
ADBC
DABC
BADC
ABDC
ACDB
CADB
DACB
ADCB
CDAB
DCAB
DCBA
CDBA
BDCA
DBCA
CBDA
BCDA

BBC BASIC

The procedure PROC_NextPermutation() will give the next lexicographic permutation of an integer array. <lang bbcbasic> DIM List%(3)

     List%() = 1, 2, 3, 4
     FOR perm% = 1 TO 24
       FOR i% = 0 TO DIM(List%(),1)
         PRINT List%(i%);
       NEXT
       PRINT
       PROC_NextPermutation(List%())
     NEXT
     END
     
     DEF PROC_NextPermutation(A%())
     LOCAL first, last, elementcount, pos
     elementcount = DIM(A%(),1)
     IF elementcount < 1 THEN ENDPROC
     pos = elementcount-1
     WHILE A%(pos) >= A%(pos+1)
       pos -= 1
       IF pos < 0 THEN
         PROC_Permutation_Reverse(A%(), 0, elementcount)
         ENDPROC
       ENDIF
     ENDWHILE
     last = elementcount
     WHILE A%(last) <= A%(pos)
       last -= 1
     ENDWHILE
     SWAP A%(pos), A%(last)
     PROC_Permutation_Reverse(A%(), pos+1, elementcount)
     ENDPROC
     
     DEF PROC_Permutation_Reverse(A%(), first, last)
     WHILE first < last
       SWAP A%(first), A%(last)
       first += 1
       last -= 1
     ENDWHILE
     ENDPROC</lang>

Output:

         1         2         3         4
         1         2         4         3
         1         3         2         4
         1         3         4         2
         1         4         2         3
         1         4         3         2
         2         1         3         4
         2         1         4         3
         2         3         1         4
         2         3         4         1
         2         4         1         3
         2         4         3         1
         3         1         2         4
         3         1         4         2
         3         2         1         4
         3         2         4         1
         3         4         1         2
         3         4         2         1
         4         1         2         3
         4         1         3         2
         4         2         1         3
         4         2         3         1
         4         3         1         2
         4         3         2         1

Bracmat

<lang bracmat> ( perm

 =   prefix List result original A Z
   .   !arg:(?.)
     |   !arg:(?prefix.?List:?original)
       & :?result
       &   whl
         ' ( !List:%?A ?Z
           & !result perm$(!prefix !A.!Z):?result
           & !Z !A:~!original:?List
           )
       & !result
 )

& out$(perm$(.a 2 "]" u+z);</lang> Output:

  (a 2 ] u+z.)
  (a 2 u+z ].)
  (a ] u+z 2.)
  (a ] 2 u+z.)
  (a u+z 2 ].)
  (a u+z ] 2.)
  (2 ] u+z a.)
  (2 ] a u+z.)
  (2 u+z a ].)
  (2 u+z ] a.)
  (2 a ] u+z.)
  (2 a u+z ].)
  (] u+z a 2.)
  (] u+z 2 a.)
  (] a 2 u+z.)
  (] a u+z 2.)
  (] 2 u+z a.)
  (] 2 a u+z.)
  (u+z a 2 ].)
  (u+z a ] 2.)
  (u+z 2 ] a.)
  (u+z 2 a ].)
  (u+z ] a 2.)
  (u+z ] 2 a.)

C

Full non-recursive algorithm generating permutation in reversed lexicographical order. Taken from here: [2]. It's a little bit modified algorythm of Narayana Pandit [3]. It reveres a string and works until a and b are not equal. It searches a new permutation directly (first internal cycle) as we do it on a sheet of paper. You can use English alphabet instead of digits. Run it from console with a parametr like "program.o 12345" or "program.o abcd": <lang c> //Author Ivan Gavryushin @dcc0

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

//This reverse x. Переворачиваем х

char revstring(char * x) {

     int i = strlen(x);
     int k=0;
     char c;

while( i > k ) {

        i--;      
        c=x[k];
        x[k]=x[i];
        x[i]=c;
        k++;
            
       }

}

//This cut x

 char  subb (char * x, int i) {
              x[i]='\0';

} //This cut y

 char  subb2 (char * y, int i) {
              int k = 0;
              while (k != strlen(y)+1) {
                    y[k]=y[i];
                   i++;
                   k++;
             }  
 }

//It gets an argumet like 1234 or abcd. All symbols must be uniqe

       int main (int argc, char *argv[]) {
            if (argc < 2) {
            printf("Enter an argument. Example 1234");
            return 0;         
                     }           
            char b[strlen(argv[1])];
            char a[strlen(argv[1])];
            int ij=0;
            while (ij!=strlen(argv[1])) {
                  a[ij]=argv[1][ij];
                  b[ij]=argv[1][ij];
                  ij++;
                  }
                  a[ij]='\0';
                  b[ij]='\0';
                   revstring(a);
                   printf("%s\n", a);
            int i;
            int j;
            char c;
            
         while (strcmp (a, b) !=0 ) {
         i=1;

         while(a[i] > a[i-1]) {
         i++;
         }
         
         j=0;
         while(a[j] < a[i]) {
         j++;

}

     c=a[j];
     a[j]=a[i];
     a[i]=c;
   
     char x[strlen(a)+1];
     char y[strlen(a)+1];
     strcpy(x,a);
     strcpy(y,a);
     subb(x, i);
     
    revstring(x);
  
     subb2(y, i);
     sprintf(a, "%s%s", x,y);       
     printf("%s\n", a);
      

  }

} </lang>

C

See lexicographic generation of permutations. <lang c>#include <stdio.h>

include <stdlib.h>

/* print a list of ints */ int show(int *x, int len) { int i; for (i = 0; i < len; i++) printf("%d%c", x[i], i == len - 1 ? '\n' : ' '); return 1; }

/* next lexicographical permutation */ int next_lex_perm(int *a, int n) {

define swap(i, j) {t = a[i]; a[i] = a[j]; a[j] = t;}

int k, l, t;

/* 1. Find the largest index k such that a[k] < a[k + 1]. If no such index exists, the permutation is the last permutation. */ for (k = n - 1; k && a[k - 1] >= a[k]; k--); if (!k--) return 0;

/* 2. Find the largest index l such that a[k] < a[l]. Since k + 1 is such an index, l is well defined */ for (l = n - 1; a[l] <= a[k]; l--);

/* 3. Swap a[k] with a[l] */ swap(k, l);

/* 4. Reverse the sequence from a[k + 1] to the end */ for (k++, l = n - 1; l > k; l--, k++) swap(k, l); return 1;

undef swap

}

void perm1(int *x, int n, int callback(int *, int)) { do { if (callback) callback(x, n); } while (next_lex_perm(x, n)); }

/* Boothroyd method; exactly N! swaps, about as fast as it gets */ void boothroyd(int *x, int n, int nn, int callback(int *, int)) { int c = 0, i, t; while (1) { if (n > 2) boothroyd(x, n - 1, nn, callback); if (c >= n - 1) return;

i = (n & 1) ? 0 : c; c++; t = x[n - 1], x[n - 1] = x[i], x[i] = t; if (callback) callback(x, nn); } }

/* entry for Boothroyd method */ void perm2(int *x, int n, int callback(int*, int)) { if (callback) callback(x, n); boothroyd(x, n, n, callback); }

/* same as perm2, but flattened recursions into iterations */ void perm3(int *x, int n, int callback(int*, int)) { /* calloc isn't strictly necessary, int c[32] would suffice for most practical purposes */ int d, i, t, *c = calloc(n, sizeof(int));

/* curiously, with GCC 4.6.1 -O3, removing next line makes it ~25% slower */ if (callback) callback(x, n); for (d = 1; ; c[d]++) { while (d > 1) c[--d] = 0; while (c[d] >= d) if (++d >= n) goto done;

t = x[ i = (d & 1) ? c[d] : 0 ], x[i] = x[d], x[d] = t; if (callback) callback(x, n); } done: free(c); }

define N 4

int main() { int i, x[N]; for (i = 0; i < N; i++) x[i] = i + 1;

/* three different methods */ perm1(x, N, show); perm2(x, N, show); perm3(x, N, show);

return 0; } </lang>

C

See lexicographic generation of permutations. <lang c>#include <stdio.h>

include <stdlib.h>

/* print a list of ints */ int show(int *x, int len) { int i; for (i = 0; i < len; i++) printf("%d%c", x[i], i == len - 1 ? '\n' : ' '); return 1; }

/* next lexicographical permutation */ int next_lex_perm(int *a, int n) {

define swap(i, j) {t = a[i]; a[i] = a[j]; a[j] = t;}

int k, l, t;

/* 1. Find the largest index k such that a[k] < a[k + 1]. If no such index exists, the permutation is the last permutation. */ for (k = n - 1; k && a[k - 1] >= a[k]; k--); if (!k--) return 0;

/* 2. Find the largest index l such that a[k] < a[l]. Since k + 1 is such an index, l is well defined */ for (l = n - 1; a[l] <= a[k]; l--);

/* 3. Swap a[k] with a[l] */ swap(k, l);

/* 4. Reverse the sequence from a[k + 1] to the end */ for (k++, l = n - 1; l > k; l--, k++) swap(k, l); return 1;

undef swap

}

void perm1(int *x, int n, int callback(int *, int)) { do { if (callback) callback(x, n); } while (next_lex_perm(x, n)); }

/* Boothroyd method; exactly N! swaps, about as fast as it gets */ void boothroyd(int *x, int n, int nn, int callback(int *, int)) { int c = 0, i, t; while (1) { if (n > 2) boothroyd(x, n - 1, nn, callback); if (c >= n - 1) return;

i = (n & 1) ? 0 : c; c++; t = x[n - 1], x[n - 1] = x[i], x[i] = t; if (callback) callback(x, nn); } }

/* entry for Boothroyd method */ void perm2(int *x, int n, int callback(int*, int)) { if (callback) callback(x, n); boothroyd(x, n, n, callback); }

/* same as perm2, but flattened recursions into iterations */ void perm3(int *x, int n, int callback(int*, int)) { /* calloc isn't strictly necessary, int c[32] would suffice for most practical purposes */ int d, i, t, *c = calloc(n, sizeof(int));

/* curiously, with GCC 4.6.1 -O3, removing next line makes it ~25% slower */ if (callback) callback(x, n); for (d = 1; ; c[d]++) { while (d > 1) c[--d] = 0; while (c[d] >= d) if (++d >= n) goto done;

t = x[ i = (d & 1) ? c[d] : 0 ], x[i] = x[d], x[d] = t; if (callback) callback(x, n); } done: free(c); }

define N 4

int main() { int i, x[N]; for (i = 0; i < N; i++) x[i] = i + 1;

/* three different methods */ perm1(x, N, show); perm2(x, N, show); perm3(x, N, show);

return 0; } </lang>

C++

The C++ standard library provides for this in the form of std::next_permutation and std::prev_permutation. <lang cpp>#include <algorithm>

include <string>
include <vector>
include <iostream>

template<class T> void print(const std::vector<T> &vec) {

   for (typename std::vector<T>::const_iterator i = vec.begin(); i != vec.end(); ++i)
   {
       std::cout << *i;
       if ((i + 1) != vec.end())
           std::cout << ",";
   }
   std::cout << std::endl;

}

int main() {

   //Permutations for strings
   std::string example("Hello");
   std::sort(example.begin(), example.end());
   do {
       std::cout << example << '\n';
   } while (std::next_permutation(example.begin(), example.end()));

   // And for vectors
   std::vector<int> another;
   another.push_back(1234);
   another.push_back(4321);
   another.push_back(1234);
   another.push_back(9999);

   std::sort(another.begin(), another.end());
   do {
       print(another);
   } while (std::next_permutation(another.begin(), another.end()));

   return 0;

}</lang>

Output:

Hello
Helol
Heoll
Hlelo
Hleol
Hlleo
Hlloe
Hloel
Hlole
Hoell
Holel
Holle
eHllo
eHlol
eHoll
elHlo
elHol
ellHo
elloH
eloHl
elolH
eoHll
eolHl
eollH
lHelo
lHeol
lHleo
lHloe
lHoel
lHole
leHlo
leHol
lelHo
leloH
leoHl
leolH
llHeo
llHoe
lleHo
lleoH
lloHe
lloeH
loHel
loHle
loeHl
loelH
lolHe
loleH
oHell
oHlel
oHlle
oeHll
oelHl
oellH
olHel
olHle
oleHl
olelH
ollHe
olleH
1234,1234,4321,9999
1234,1234,9999,4321
1234,4321,1234,9999
1234,4321,9999,1234
1234,9999,1234,4321
1234,9999,4321,1234
4321,1234,1234,9999
4321,1234,9999,1234
4321,9999,1234,1234
9999,1234,1234,4321
9999,1234,4321,1234
9999,4321,1234,1234

C#

A recursive Iterator. Runs under C#2 (VS2005), i.e. no `var`, no lambdas,... <lang csharp>public class Permutations<T> {

   public static System.Collections.Generic.IEnumerable<T[]> AllFor(T[] array)
   {
       if (array == null || array.Length == 0)
       {
           yield return new T[0];
       }
       else
       {
           for (int pick = 0; pick < array.Length; ++pick)
           {
               T item = array[pick];
               int i = -1;
               T[] rest = System.Array.FindAll<T>(
                   array, delegate(T p) { return ++i != pick; }
               );
               foreach (T[] restPermuted in AllFor(rest))
               {
                   i = -1;
                   yield return System.Array.ConvertAll<T, T>(
                       array,
                       delegate(T p) {
                           return ++i == 0 ? item : restPermuted[i - 1];
                       }
                   );
               }
           }
       }
   }

}</lang> Usage: <lang csharp>namespace Permutations_On_RosettaCode {

   class Program
   {
       static void Main(string[] args)
       {
           string[] list = "a b c d".Split();
           foreach (string[] permutation in Permutations<string>.AllFor(list))
           {
               System.Console.WriteLine(string.Join(" ", permutation));
           }
       }
   }

}</lang>

Clojure

Library function

In an REPL:

<lang clojure> user=> (require 'clojure.contrib.combinatorics) nil user=> (clojure.contrib.combinatorics/permutations [1 2 3]) ((1 2 3) (1 3 2) (2 1 3) (2 3 1) (3 1 2) (3 2 1))</lang>

Explicit

Replacing the call to the combinatorics library function by its real implementation. <lang clojure> (defn- iter-perm [v]

 (let [len (count v),

j (loop [i (- len 2)] (cond (= i -1) nil (< (v i) (v (inc i))) i :else (recur (dec i))))]

   (when j
     (let [vj (v j),

l (loop [i (dec len)] (if (< vj (v i)) i (recur (dec i))))] (loop [v (assoc v j (v l) l vj), k (inc j), l (dec len)] (if (< k l) (recur (assoc v k (v l) l (v k)) (inc k) (dec l)) v))))))

(defn- vec-lex-permutations [v]

 (when v (cons v (lazy-seq (vec-lex-permutations (iter-perm v))))))

(defn lex-permutations

 "Fast lexicographic permutation generator for a sequence of numbers"
 [c]
 (lazy-seq
  (let [vec-sorted (vec (sort c))]
    (if (zero? (count vec-sorted))
      (list [])
      (vec-lex-permutations vec-sorted)))))

(defn permutations

 "All the permutations of items, lexicographic by index"
 [items]
 (let [v (vec items)]
   (map #(map v %) (lex-permutations (range (count v))))))

(println (permutations [1 2 3]))

</lang>

CoffeeScript

<lang coffeescript># Returns a copy of an array with the element at a specific position

removed from it.

arrayExcept = (arr, idx) -> res = arr[0..] res.splice idx, 1 res

The actual function which returns the permutations of an array-like
object (or a proper array).

permute = (arr) -> arr = Array::slice.call arr, 0 return [[]] if arr.length == 0

permutations = (for value,idx in arr [value].concat perm for perm in permute arrayExcept arr, idx)

# Flatten the array before returning it. [].concat permutations...</lang> This implementation utilises the fact that the permutations of an array could be defined recursively, with the fixed point being the permutations of an empty array.

Usage:

<lang coffeescript>coffee> console.log (permute "123").join "\n" 1,2,3 1,3,2 2,1,3 2,3,1 3,1,2 3,2,1</lang>

Common Lisp

<lang lisp>(defun permute (list)

 (if list
   (mapcan #'(lambda (x)

(mapcar #'(lambda (y) (cons x y)) (permute (remove x list)))) list)

   '(()))) ; else

(print (permute '(A B Z)))</lang>

Output:

((A B Z) (A Z B) (B A Z) (B Z A) (Z A B) (Z B A))

Lexicographic next permutation: <lang lisp>(defun next-perm (vec cmp) ; modify vector

 (declare (type (simple-array * (*)) vec))
 (macrolet ((el (i) `(aref vec ,i))
            (cmp (i j) `(funcall cmp (el ,i) (el ,j))))
   (loop with len = (1- (length vec))
      for i from (1- len) downto 0
      when (cmp i (1+ i)) do
        (loop for k from len downto i
           when (cmp i k) do
             (rotatef (el i) (el k))
             (setf k (1+ len))
             (loop while (< (incf i) (decf k)) do
                  (rotatef (el i) (el k)))
             (return-from next-perm vec)))))

test code

(loop for a = "1234" then (next-perm a #'char<) while a do

    (write-line a))</lang>

D

Simple Eager version

Compile with -version=permutations1_main to see the output. <lang d>T[][] permutations(T)(T[] items) pure nothrow {

   T[][] result;

   void perms(T[] s, T[] prefix=[]) nothrow {
       if (s.length)
           foreach (immutable i, immutable c; s)
              perms(s[0 .. i] ~ s[i+1 .. $], prefix ~ c);
       else
           result ~= prefix;
   }

   perms(items);
   return result;

}

version (permutations1_main) {

   void main() {
       import std.stdio;
       writefln("%(%s\n%)", [1, 2, 3].permutations);
   }

}</lang>

Output:

[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]

Fast Lazy Version

Compiled with -version=permutations2_main produces its output. <lang d>import std.algorithm, std.conv, std.traits;

struct Permutations(bool doCopy=true, T) if (isMutable!T) {

   private immutable size_t num;
   private T[] items;
   private uint[31] indexes;
   private ulong tot;

   this (T[] items) pure nothrow @safe @nogc
   in {
       static enum string L = indexes.length.text;
       assert(items.length >= 0 && items.length <= indexes.length,
              "Permutations: items.length must be >= 0 && < " ~ L);
   } body {
       static ulong factorial(in size_t n) pure nothrow @safe @nogc {
           ulong result = 1;
           foreach (immutable i; 2 .. n + 1)
               result *= i;
           return result;
       }

       this.num = items.length;
       this.items = items;
       foreach (immutable i; 0 .. cast(typeof(indexes[0]))this.num)
           this.indexes[i] = i;
       this.tot = factorial(this.num);
   }

   @property T[] front() pure nothrow @safe {
       static if (doCopy) {
           return items.dup;
       } else
           return items;
   }

   @property bool empty() const pure nothrow @safe @nogc {
       return tot == 0;
   }

   @property size_t length() const pure nothrow @safe @nogc {
       // Not cached to keep the function pure.
       typeof(return) result = 1;
       foreach (immutable x; 1 .. items.length + 1)
           result *= x;
       return result;
   }

   void popFront() pure nothrow @safe @nogc {
       tot--;
       if (tot > 0) {
           size_t j = num - 2;

           while (indexes[j] > indexes[j + 1])
               j--;
           size_t k = num - 1;
           while (indexes[j] > indexes[k])
               k--;
           swap(indexes[k], indexes[j]);
           swap(items[k], items[j]);

           size_t r = num - 1;
           size_t s = j + 1;
           while (r > s) {
               swap(indexes[s], indexes[r]);
               swap(items[s], items[r]);
               r--;
               s++;
           }
       }
   }

}

Permutations!(doCopy,T) permutations(bool doCopy=true, T)

                                   (T[] items)

pure nothrow if (isMutable!T) {

   return Permutations!(doCopy, T)(items);

}

version (permutations2_main) {

   void main() {
       import std.stdio, std.bigint;
       alias B = BigInt;
       foreach (p; [B(1), B(2), B(3)].permutations)
           assert((p[0] + 1) > 0);
       [1, 2, 3].permutations!false.writeln;
       [B(1), B(2), B(3)].permutations!false.writeln;
   }

}</lang>

Standard Version

<lang d>void main() {

   import std.stdio, std.algorithm;

   auto items = [1, 2, 3];
   do
       items.writeln;
   while (items.nextPermutation);

}</lang>

Delphi

<lang Delphi>program TestPermutations;

{$APPTYPE CONSOLE}

type

 TItem = Integer;                // declare ordinal type for array item
 TArray = array[0..3] of TItem;

const

 Source: TArray = (1, 2, 3, 4);

procedure Permutation(K: Integer; var A: TArray); var

 I, J: Integer;
 Tmp: TItem;

begin

 for I:= Low(A) + 1 to High(A) + 1 do begin
   J:= K mod I;
   Tmp:= A[J];
   A[J]:= A[I - 1];
   A[I - 1]:= Tmp;
   K:= K div I;
 end;

end;

var

 A: TArray;
 I, K, Count: Integer;
 S, S1, S2: ShortString;

begin

 Count:= 1;
 I:= Length(A);
 while I > 1 do begin
   Count:= Count * I;
   Dec(I);
 end;

 S:= ;
 for K:= 0 to Count - 1 do begin
   A:= Source;
   Permutation(K, A);
   S1:= ;
   for I:= Low(A) to High(A) do begin
     Str(A[I]:1, S2);
     S1:= S1 + S2;
   end;
   S:= S + '  ' + S1;
   if Length(S) > 40 then begin
     Writeln(S);
     S:= ;
   end;
 end;

 if Length(S) > 0 then Writeln(S);
 Readln;

end.</lang>

Output:

  4123  4213  4312  4321  4132  4231  3421
  3412  2413  1423  2431  1432  3142  3241
  2341  1342  2143  1243  3124  3214  2314
  1324  2134  1234

Eiffel

<lang Eiffel> class APPLICATION

create make

feature {NONE}

make do test := <<2, 5, 1>> permute (test, 1) end

test: ARRAY [INTEGER]

permute (a: ARRAY [INTEGER]; k: INTEGER) -- All permutations of 'a'. require count_positive: a.count > 0 k_valid_index: k > 0 local t: INTEGER do if k = a.count then across a as ar loop io.put_integer (ar.item) end io.new_line else across k |..| a.count as c loop t := a [k] a [k] := a [c.item] a [c.item] := t permute (a, k + 1) t := a [k] a [k] := a [c.item] a [c.item] := t end end end

end

</lang>

Output:

Elixir

Translation of: Erlang

<lang elixir>defmodule RC do

 def permute([]), do: [[]]
 def permute(list) do
   for x <- list, y <- permute(list -- [x]), do: [x|y]
 end

end

IO.inspect RC.permute([1, 2, 3])</lang>

Output:

[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

Erlang

Shortest form: <lang Erlang>-module(permute). -export([permute/1]).

permute([]) -> [[]]; permute(L) -> [[X|Y] || X<-L, Y<-permute(L--[X])].</lang> Y-combinator (for shell): <lang Erlang>F = fun(L) -> G = fun(_, []) -> [[]]; (F, L) -> [[X|Y] || X<-L, Y<-F(F, L--[X])] end, G(G, L) end.</lang> More efficient zipper implementation: <lang Erlang>-module(permute).

-export([permute/1]).

permute([]) -> [[]]; permute(L) -> zipper(L, [], []).

% Use zipper to pick up first element of permutation zipper([], _, Acc) -> lists:reverse(Acc); zipper([H|T], R, Acc) ->

 % place current member in front of all permutations
 % of rest of set - both sides of zipper
 prepend(H, permute(lists:reverse(R, T)),
   % pass zipper state for continuation
   T, [H|R], Acc).

prepend(_, [], T, R, Acc) -> zipper(T, R, Acc); % continue in zipper prepend(X, [H|T], ZT, ZR, Acc) -> prepend(X, T, ZT, ZR, [[X|H]|Acc]).</lang> Demonstration (escript): <lang Erlang>main(_) -> io:fwrite("~p~n", [permute:permute([1,2,3])]).</lang>

Output:

[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

Euphoria

Translation of: PureBasic

<lang euphoria>function reverse(sequence s, integer first, integer last)

   object x
   while first < last do
       x = s[first]
       s[first] = s[last]
       s[last] = x
       first += 1
       last -= 1
   end while
   return s

end function

function nextPermutation(sequence s)

   integer pos, last
   object x
   if length(s) < 1 then
       return 0
   end if
   
   pos = length(s)-1
   while compare(s[pos], s[pos+1]) >= 0 do
       pos -= 1
       if pos < 1 then
           return -1
       end if
   end while
   
   last = length(s)
   while compare(s[last], s[pos]) <= 0 do
       last -= 1
   end while
   x = s[pos]
   s[pos] = s[last]
   s[last] = x
   
   return reverse(s, pos+1, length(s))

end function

object s s = "abcd" puts(1, s & '\t') while 1 do

   s = nextPermutation(s)
   if atom(s) then
       exit
   end if
   puts(1, s & '\t')

end while</lang>

Output:

abcd    abdc    acbd    acdb    adbc    adcb    bacd    badc    bcad    bcda
bdac    bdca    cabd    cadb    cbad    cbda    cdab    cdba    dabc    dacb
dbac    dbca    dcab    dcba

F#

<lang fsharp> let rec insert left x right = seq {

   match right with
   | [] -> yield left @ [x]
   | head :: tail -> 
       yield left @ [x] @ right
       yield! insert (left @ [head]) x tail
   }

let rec perms permute =

   seq {
       match permute with
       | [] -> yield []
       | head :: tail -> yield! Seq.collect (insert [] head) (perms tail)
   }

[<EntryPoint>] let main argv =

   perms (Seq.toList argv)
   |> Seq.iter (fun x -> printf "%A\n" x)
   0

</lang>

>RosettaPermutations 1 2 3
["1"; "2"; "3"]
["2"; "1"; "3"]
["2"; "3"; "1"]
["1"; "3"; "2"]
["3"; "1"; "2"]
["3"; "2"; "1"]

Factor

The all-permutations word is part of factor's standard library. See http://docs.factorcode.org/content/word-all-permutations,math.combinatorics.html

Fortran

<lang fortran>program permutations

 implicit none
 integer, parameter :: value_min = 1
 integer, parameter :: value_max = 3
 integer, parameter :: position_min = value_min
 integer, parameter :: position_max = value_max
 integer, dimension (position_min : position_max) :: permutation

 call generate (position_min)

contains

 recursive subroutine generate (position)

   implicit none
   integer, intent (in) :: position
   integer :: value

   if (position > position_max) then
     write (*, *) permutation
   else
     do value = value_min, value_max
       if (.not. any (permutation (: position - 1) == value)) then
         permutation (position) = value
         call generate (position + 1)
       end if
     end do
   end if

 end subroutine generate

end program permutations</lang>

Output:

           1           2           3
           1           3           2
           2           1           3
           2           3           1
           3           1           2
           3           2           1

Alternate solution

Instead of looking up unused values, this program starts from [1, ..., n] and does only swaps, hence the array always represents a valid permutation. The values need to be "swapped back" after the recursive call.

<lang fortran>program allperm

   implicit none
   integer :: n, i
   integer, allocatable :: a(:)
   read *, n
   allocate(a(n))
   a = [ (i, i = 1, n) ]
   call perm(1)
   deallocate(a)

contains

   recursive subroutine perm(i)
       integer :: i, j, t
       if (i == n) then
           print *, a
       else
           do j = i, n
               t = a(i)
               a(i) = a(j)
               a(j) = t
               call perm(i + 1)
               t = a(i)
               a(i) = a(j)
               a(j) = t
           end do
       end if
   end subroutine

end program</lang>

Fortran 77

Here is an alternate, iterative version in Fortran 77.

Translation of: Ada

<lang fortran> program nptest

     integer n,i,a
     logical nextp
     external nextp
     parameter(n=4)
     dimension a(n)
     do i=1,n
     a(i)=i
     enddo
  10 print *,(a(i),i=1,n)
     if(nextp(n,a)) go to 10
     end
     
     function nextp(n,a)
     integer n,a,i,j,k,t
     logical nextp
     dimension a(n)
     i=n-1
  10 if(a(i).lt.a(i+1)) go to 20
     i=i-1
     if(i.eq.0) go to 20
     go to 10
  20 j=i+1
     k=n
  30 t=a(j)
     a(j)=a(k)
     a(k)=t
     j=j+1
     k=k-1
     if(j.lt.k) go to 30
     j=i
     if(j.ne.0) go to 40
     nextp=.false.
     return
  40 j=j+1
     if(a(j).lt.a(i)) go to 40
     t=a(i)
     a(i)=a(j)
     a(j)=t
     nextp=.true.
     end</lang>

GAP

GAP can handle permutations and groups. Here is a straightforward implementation : for each permutation p in S(n) (symmetric group), compute the images of 1 .. n by p. As an alternative, List(SymmetricGroup(n)) would yield the permutations as GAP Permutation objects, which would probably be more manageable in later computations. <lang gap>gap>List(SymmetricGroup(4), p -> Permuted([1 .. 4], p)); perms(4); [ [ 1, 2, 3, 4 ], [ 4, 2, 3, 1 ], [ 2, 4, 3, 1 ], [ 3, 2, 4, 1 ], [ 1, 4, 3, 2 ], [ 4, 1, 3, 2 ], [ 2, 1, 3, 4 ],

 [ 3, 1, 4, 2 ], [ 1, 3, 4, 2 ], [ 4, 3, 1, 2 ], [ 2, 3, 1, 4 ], [ 3, 4, 1, 2 ], [ 1, 2, 4, 3 ], [ 4, 2, 1, 3 ],
 [ 2, 4, 1, 3 ], [ 3, 2, 1, 4 ], [ 1, 4, 2, 3 ], [ 4, 1, 2, 3 ], [ 2, 1, 4, 3 ], [ 3, 1, 2, 4 ], [ 1, 3, 2, 4 ],
 [ 4, 3, 2, 1 ], [ 2, 3, 4, 1 ], [ 3, 4, 2, 1 ] ]</lang>

GAP has also built-in functions to get permutations <lang gap># All arrangements of 4 elements in 1 .. 4 Arrangements([1 .. 4], 4);

All permutations of 1 .. 4

PermutationsList([1 .. 4]);</lang> Here is an implementation using a function to compute next permutation in lexicographic order: <lang gap>NextPermutation := function(a)

  local i, j, k, n, t;
  n := Length(a);
  i := n - 1;
  while i > 0 and a[i] > a[i + 1] do
     i := i - 1;
  od;
  j := i + 1;
  k := n;
  while j < k do
     t := a[j];
     a[j] := a[k];
     a[k] := t;
     j := j + 1;
     k := k - 1;
  od;
  if i = 0 then
     return false;
  else
     j := i + 1;
     while a[j] < a[i] do
        j := j + 1;
     od;
     t := a[i];
     a[i] := a[j];
     a[j] := t;
     return true;
  fi;

end;

Permutations := function(n)

  local a, L;
  a := List([1 .. n], x -> x);
  L := [ ];
  repeat 
     Add(L, ShallowCopy(a));
  until not NextPermutation(a);
  return L;

end;

Permutations(3); [ [ 1, 2, 3 ], [ 1, 3, 2 ],

 [ 2, 1, 3 ], [ 2, 3, 1 ],
 [ 3, 1, 2 ], [ 3, 2, 1 ] ]</lang>

Go

<lang go>package main

import "fmt"

func main() {

   demoPerm(3)

}

func demoPerm(n int) {

   // create a set to permute.  for demo, use the integers 1..n.
   s := make([]int, n)
   for i := range s {
       s[i] = i + 1
   }
   // permute them, calling a function for each permutation.
   // for demo, function just prints the permutation.
   permute(s, func(p []int) { fmt.Println(p) })

}

// permute function. takes a set to permute and a function // to call for each generated permutation. func permute(s []int, emit func([]int)) {

   if len(s) == 0 {
       emit(s)
       return
   }
   // Steinhaus, implemented with a recursive closure.
   // arg is number of positions left to permute.
   // pass in len(s) to start generation.
   // on each call, weave element at pp through the elements 0..np-2,
   // then restore array to the way it was.
   var rc func(int)
   rc = func(np int) {
       if np == 1 {
           emit(s)
           return
       }
       np1 := np - 1
       pp := len(s) - np1
       // weave
       rc(np1)
       for i := pp; i > 0; i-- {
           s[i], s[i-1] = s[i-1], s[i]
           rc(np1)
       }
       // restore
       w := s[0]
       copy(s, s[1:pp+1])
       s[pp] = w
   }
   rc(len(s))

}</lang>

Output:

[1 2 3]
[1 3 2]
[3 1 2]
[2 1 3]
[2 3 1]
[3 2 1]

Groovy

Solution: <lang groovy>def makePermutations = { l -> l.permutations() }</lang> Test: <lang groovy>def list = ['Crosby', 'Stills', 'Nash', 'Young'] def permutations = makePermutations(list) assert permutations.size() == (1..<(list.size()+1)).inject(1) { prod, i -> prod*i } permutations.each { println it }</lang>

Output:

[Young, Crosby, Stills, Nash]
[Crosby, Stills, Young, Nash]
[Nash, Crosby, Young, Stills]
[Stills, Nash, Crosby, Young]
[Young, Stills, Crosby, Nash]
[Stills, Crosby, Nash, Young]
[Stills, Crosby, Young, Nash]
[Stills, Young, Nash, Crosby]
[Nash, Stills, Young, Crosby]
[Crosby, Young, Nash, Stills]
[Crosby, Nash, Young, Stills]
[Crosby, Nash, Stills, Young]
[Nash, Young, Stills, Crosby]
[Young, Nash, Stills, Crosby]
[Nash, Young, Crosby, Stills]
[Young, Stills, Nash, Crosby]
[Crosby, Stills, Nash, Young]
[Stills, Young, Crosby, Nash]
[Young, Nash, Crosby, Stills]
[Nash, Stills, Crosby, Young]
[Young, Crosby, Nash, Stills]
[Nash, Crosby, Stills, Young]
[Crosby, Young, Stills, Nash]
[Stills, Nash, Young, Crosby]

Haskell

<lang haskell>import Data.List (permutations)

main = mapM_ print (permutations [1,2,3])</lang>

A simple implementation, that assumes elements are unique and support equality: <lang haskell>import Data.List (delete)

permutations :: Eq a => [a] -> a permutations [] = [[]] permutations xs = [ x:ys | x <- xs, ys <- permutations (delete x xs)]</lang>

A slightly more efficient implementation that doesn't have the above restrictions: <lang haskell>permutations :: [a] -> a permutations [] = [[]] permutations xs = [ y:zs | (y,ys) <- select xs, zs <- permutations ys]

 where select []     = []
       select (x:xs) = (x,xs) : [ (y,x:ys) | (y,ys) <- select xs ]</lang>

The above are all selection-based approaches. The following is an insertion-based approach: <lang haskell>permutations :: [a] -> a permutations = foldr (concatMap . insertEverywhere) [[]]

 where insertEverywhere :: a -> [a] -> a
       insertEverywhere x [] = x
       insertEverywhere x l@(y:ys) = (x:l) : map (y:) (insertEverywhere x ys)</lang>

Icon and Unicon

<lang unicon>procedure main(A)

   every p := permute(A) do every writes((!p||" ")|"\n")

end

procedure permute(A)

   if *A <= 1 then return A
   suspend [(A[1]<->A[i := 1 to *A])] ||| permute(A[2:0])

end</lang>

Output:

->permute Aardvarks eat ants      
Aardvarks eat ants 
Aardvarks ants eat 
eat Aardvarks ants 
eat ants Aardvarks 
ants eat Aardvarks 
ants Aardvarks eat 
->

J

<lang j>perms=: A.&i.~ !</lang>

Example use:

<lang j> perms 2 0 1 1 0

  ({~ perms@#)&.;: 'some random text'

some random text some text random random some text random text some text some random text random some</lang>

Java

Using the code of Michael Gilleland. <lang java>public class PermutationGenerator {

   private int[] array;
   private int firstNum;
   private boolean firstReady = false;

   public PermutationGenerator(int n, int firstNum_) {
       if (n < 1) {
           throw new IllegalArgumentException("The n must be min. 1");
       }
       firstNum = firstNum_;
       array = new int[n];
       reset();
   }

   public void reset() {
       for (int i = 0; i < array.length; i++) {
           array[i] = i + firstNum;
       }
       firstReady = false;
   }

   public boolean hasMore() {
       boolean end = firstReady;
       for (int i = 1; i < array.length; i++) {
           end = end && array[i] < array[i-1];
       }
       return !end;
   }

   public int[] getNext() {

       if (!firstReady) {
           firstReady = true;
           return array;
       }

       int temp;
       int j = array.length - 2;
       int k = array.length - 1;

       // Find largest index j with a[j] < a[j+1]

       for (;array[j] > array[j+1]; j--);

       // Find index k such that a[k] is smallest integer
       // greater than a[j] to the right of a[j]

       for (;array[j] > array[k]; k--);

       // Interchange a[j] and a[k]

       temp = array[k];
       array[k] = array[j];
       array[j] = temp;

       // Put tail end of permutation after jth position in increasing order

       int r = array.length - 1;
       int s = j + 1;

       while (r > s) {
           temp = array[s];
           array[s++] = array[r];
           array[r--] = temp;
       }

       return array;
   } // getNext()

   // For testing of the PermutationGenerator class
   public static void main(String[] args) {
       PermutationGenerator pg = new PermutationGenerator(3, 1);

       while (pg.hasMore()) {
           int[] temp =  pg.getNext();
           for (int i = 0; i < temp.length; i++) {
               System.out.print(temp[i] + " ");
           }
           System.out.println();
       }
   }

} // class</lang>

Output:

optimized

Following needs: Utils.java <lang java>public class Permutations { public static void main(String[] args) { System.out.println(Utils.Permutations(Utils.mRange(1, 3))); } }</lang>

Output:

[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

JavaScript

ES5

Iteration

Copy the following as an HTML file and load in a browser. <lang javascript><html><head><title>Permutations</title></head>

<body>

function perm(list, ret) {

   if (list.length == 0) {
       var row = document.createTextNode(ret.join(' ') + '\n');
       d.appendChild(row);
       return;
   }
   for (var i = 0; i < list.length; i++) {
       var x = list.splice(i, 1);
       ret.push(x);
       perm(list, ret);
       ret.pop();
       list.splice(i, 0, x);
   }

}

perm([1, 2, 'A', 4], []); </script></body></html></lang>

Alternatively: 'Genuine' js code, assuming no duplicate.

<lang JavaScript> function perm(a) {

   if (a.length < 2) return [a];
   var c, d, b = [];
   for (c = 0; c < a.length; c++) {
       var e = a.splice(c, 1),
           f = perm(a);
       for (d = 0; d < f.length; d++) b.push([e].concat(f[d]));
       a.splice(c, 0, e[0])
   } return b

}

console.log(perm(['Aardvarks', 'eat', 'ants']).join("\n")); </lang>

Output:

<lang JavaScript>Aardvarks,eat,ants Aardvarks,ants,eat eat,Aardvarks,ants eat,ants,Aardvarks ants,Aardvarks,eat ants,eat,Aardvarks</lang>

Functional composition

Translation of: Haskell

(Simple version – assuming a unique list of objects comparable by the JS === operator)

<lang JavaScript>(function () {

   'use strict';

   // permutations :: [a] -> a
   var permutations = function (xs) {
       return xs.length ? concatMap(function (x) {
           return concatMap(function (ys) {
               return [[x].concat(ys)];
           }, permutations(delete_(x, xs)));
       }, xs) : [[]];
   };

   // GENERIC FUNCTIONS

   // concatMap :: (a -> [b]) -> [a] -> [b]
   var concatMap = function (f, xs) {
       return [].concat.apply([], xs.map(f));
   };

   // delete :: Eq a => a -> [a] -> [a]
   var delete_ = function (x, xs) {
       return deleteBy(function (a, b) {
           return a === b;
       }, x, xs);
   };

   // deleteBy :: (a -> a -> Bool) -> a -> [a] -> [a]
   var deleteBy = function (f, x, xs) {
       return xs.length > 0 ? f(x, xs[0]) ? xs.slice(1) : 
       [xs[0]].concat(deleteBy(f, x, xs.slice(1))) : [];
   };

   // TEST
   return permutations(['Aardvarks', 'eat', 'ants']);

})();</lang>

Output:

<lang JavaScript>[["Aardvarks", "eat", "ants"], ["Aardvarks", "ants", "eat"],

["eat", "Aardvarks", "ants"], ["eat", "ants", "Aardvarks"],

["ants", "Aardvarks", "eat"], ["ants", "eat", "Aardvarks"]]</lang>

ES6

<lang JavaScript>(() => {

   'use strict';

   // permutations :: [a] -> a
   const permutations = xs =>
       xs.length ? concatMap(x => concatMap(ys => [
               [x].concat(ys)
           ],
           permutations(delete_(x, xs))), xs) : [
           []
       ];

   // GENERIC FUNCTIONS

   // concatMap :: (a -> [b]) -> [a] -> [b]
   const concatMap = (f, xs) => [].concat.apply([], xs.map(f));

   // delete :: Eq a => a -> [a] -> [a]
   const delete_ = (x, xs) =>
       deleteBy((a, b) => a === b, x, xs);

   // deleteBy :: (a -> a -> Bool) -> a -> [a] -> [a]
   const deleteBy = (f, x, xs) =>
       xs.length > 0 ? (
           f(x, xs[0]) ? (
               xs.slice(1)
           ) : [xs[0]].concat(deleteBy(f, x, xs.slice(1)))
       ) : [];

   // TEST
   return permutations(['Aardvarks', 'eat', 'ants']);

})();</lang>

Output:

<lang JavaScript>[["Aardvarks", "eat", "ants"], ["Aardvarks", "ants", "eat"],

["eat", "Aardvarks", "ants"], ["eat", "ants", "Aardvarks"],

["ants", "Aardvarks", "eat"], ["ants", "eat", "Aardvarks"]]</lang>

jq

"permutations" generates a stream of the permutations of the input array. <lang jq>def permutations:

 if length == 0 then []
 else
   range(0;length) as $i
   | [.[$i]] + (del(.[$i])|permutations)
 end ;

</lang> Example 1: list them

[range(0;3)] | permutations
[0,1,2]
[0,2,1]
[1,0,2]
[1,2,0]
[2,0,1]
[2,1,0]

Example 2: count them

[[range(0;3)] | permutations] | length
6

Or more efficiently:

def count(s): reduce s as $i (0;.+1);

[range(0;3)] | count(permutations)
6

Example 3: 10!

[range(0;10)] | count(permutations)
3628800

Julia

Julia has native support for permutation creation and processing. permutations(v) creates an iterator over all permutations of v. <lang Julia> term = "RCode" i = 0 pcnt = factorial(length(term)) print("All the permutations of ", term, " (", pcnt, "):\n ") for p in permutations(split(term, ""))

   print(join(p), " ")
   i += 1
   i %= 12
   i != 0 || print("\n    ")

end println() </lang>

Output:

All the permutations of RCode (120):
    RCode RCoed RCdoe RCdeo RCeod RCedo RoCde RoCed RodCe RodeC RoeCd RoedC 
    RdCoe RdCeo RdoCe RdoeC RdeCo RdeoC ReCod ReCdo ReoCd ReodC RedCo RedoC 
    CRode CRoed CRdoe CRdeo CReod CRedo CoRde CoRed CodRe CodeR CoeRd CoedR 
    CdRoe CdReo CdoRe CdoeR CdeRo CdeoR CeRod CeRdo CeoRd CeodR CedRo CedoR 
    oRCde oRCed oRdCe oRdeC oReCd oRedC oCRde oCRed oCdRe oCdeR oCeRd oCedR 
    odRCe odReC odCRe odCeR odeRC odeCR oeRCd oeRdC oeCRd oeCdR oedRC oedCR 
    dRCoe dRCeo dRoCe dRoeC dReCo dReoC dCRoe dCReo dCoRe dCoeR dCeRo dCeoR 
    doRCe doReC doCRe doCeR doeRC doeCR deRCo deRoC deCRo deCoR deoRC deoCR 
    eRCod eRCdo eRoCd eRodC eRdCo eRdoC eCRod eCRdo eCoRd eCodR eCdRo eCdoR 
    eoRCd eoRdC eoCRd eoCdR eodRC eodCR edRCo edRoC edCRo edCoR edoRC edoCR

K

Translation of: J

<lang K> perm:{:[1<x;,/(>:'(x,x)#1,x#0)[;0,'1+_f x-1];,!x]}

  perm 2

(0 1

1 0)

  `0:{1_,/" ",/:x}'r@perm@#r:("some";"random";"text")

some random text some text random random some text random text some text some random text random some</lang>

Alternative: <lang K>

  perm:{x@m@&n=(#?:)'m:!n#n:#x}
  
  perm[!3]

(0 1 2

0 2 1
1 0 2
1 2 0
2 0 1
2 1 0)
 
  perm "abc"

("abc"

"acb"
"bac"
"bca"
"cab"
"cba")
  
  `0:{1_,/" ",/: $x}' perm `$" "\"some random text"

some random text some text random random some text random text some text some random text random some </lang>

LFE

<lang lisp> (defun permute

 (('())
   '(()))
 ((l)
   (lc ((<- x l)
        (<- y (permute (-- l `(,x)))))
       (cons x y))))

</lang> REPL usage: <lang lisp> > (permute '(1 2 3)) ((1 2 3) (1 3 2) (2 1 3) (2 3 1) (3 1 2) (3 2 1)) </lang>

Liberty BASIC

Permuting numerical array (non-recursive):

Translation of: PowerBASIC

<lang lb> n=3 dim a(n+1) '+1 needed due to bug in LB that checks loop condition

   '   until (i=0) or (a(i)<a(i+1))
   'before executing i=i-1 in loop body.

for i=1 to n: a(i)=i: next do

 for i=1 to n: print a(i);: next: print
 i=n
 do
   i=i-1
 loop until (i=0) or (a(i)<a(i+1))
 j=i+1
 k=n
 while j<k
   'swap a(j),a(k)
   tmp=a(j): a(j)=a(k): a(k)=tmp
   j=j+1
   k=k-1
 wend
 if i>0 then
   j=i+1
   while a(j)<a(i)
     j=j+1
   wend
   'swap a(i),a(j)
   tmp=a(j): a(j)=a(i): a(i)=tmp
 end if

loop until i=0 </lang>

Output:

Permuting string (recursive): <lang lb> n = 3

s$="" for i = 1 to n

   s$=s$;i

Function permutation$(pre$, post$)

   lgth = Len(post$)
   If lgth < 2 Then
       print pre$;post$
   Else
       For i = 1 To lgth
           tmp$=permutation$(pre$+Mid$(post$,i,1),Left$(post$,i-1)+Right$(post$,lgth-i))
       Next i
   End If

End Function

</lang>

Output:

Logtalk

<lang logtalk>:- object(list).

   :- public(permutation/2).

   permutation(List, Permutation) :-
       same_length(List, Permutation),
       permutation2(List, Permutation).

   permutation2([], []).
   permutation2(List, [Head| Tail]) :-
       select(Head, List, Remaining),
       permutation2(Remaining, Tail).

   same_length([], []).
   same_length([_| Tail1], [_| Tail2]) :-
       same_length(Tail1, Tail2).

   select(Head, [Head| Tail], Tail).
   select(Head, [Head2| Tail], [Head2| Tail2]) :-
       select(Head, Tail, Tail2).

- end_object.</lang>

Usage example:

<lang logtalk>| ?- forall(list::permutation([1, 2, 3], Permutation), (write(Permutation), nl)).

[1,2,3] [1,3,2] [2,1,3] [2,3,1] [3,1,2] [3,2,1] yes</lang>

Lua

<lang lua> local function permutation(a, n, cb) if n == 0 then cb(a) else for i = 1, n do a[i], a[n] = a[n], a[i] permutation(a, n - 1, cb) a[i], a[n] = a[n], a[i] end end end

--Usage local function callback(a) print('{'..table.concat(a, ', ')..'}') end permutation({1,2,3}, 3, callback) </lang>

Output:

{2, 3, 1}
{3, 2, 1}
{3, 1, 2}
{1, 3, 2}
{2, 1, 3}
{1, 2, 3}

-- Iterative version function ipermutations(a,b)

   if a==0 then return end
   local taken = {} local slots = {}
   for i=1,a do slots[i]=0 end
   for i=1,b do taken[i]=false end
   local index = 1
   while index > 0 do repeat
       repeat slots[index] = slots[index] + 1
       until slots[index] > b or not taken[slots[index]]
       if slots[index] > b then
           slots[index] = 0
           index = index - 1
           if index > 0 then
               taken[slots[index]] = false
           end
           break
       else
           taken[slots[index]] = true
       end
       if index == a then
           for i=1,a do io.write(slots[i]) io.write(" ") end
           io.write("\n")
           taken[slots[index]] = false
           break
       end
       index = index + 1
   until true end

end

ipermutations(3, 3) </lang>

Maple

<lang Maple> > combinat:-permute( 3 );

  [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

> combinat:-permute( [a,b,c] );

  [[a, b, c], [a, c, b], [b, a, c], [b, c, a], [c, a, b], [c, b, a]]

</lang>

Mathematica

<lang Mathematica>Permutations[{1,2,3,4}]</lang>

Output:

{{1, 2, 3, 4}, {1, 2, 4, 3}, {1, 3, 2, 4}, {1, 3, 4, 2}, {1, 4, 2, 3}, {1, 4, 3, 2}, {2, 1, 3, 4}, {2, 1, 4, 3}, {2, 3, 1, 4}, {2, 3, 
  4, 1}, {2, 4, 1, 3}, {2, 4, 3, 1}, {3, 1, 2, 4}, {3, 1, 4, 2}, {3, 2, 1, 4}, {3, 2, 4, 1}, {3, 4, 1, 2}, {3, 4, 2, 1}, {4, 1, 2, 
  3}, {4, 1, 3, 2}, {4, 2, 1, 3}, {4, 2, 3, 1}, {4, 3, 1, 2}, {4, 3, 2, 1}}

MATLAB / Octave

<lang MATLAB>perms([1,2,3,4])</lang>

Output:

Maxima

<lang maxima>next_permutation(v) := block([n, i, j, k, t],

  n: length(v), i: 0,
  for k: n - 1 thru 1 step -1 do (if v[k] < v[k + 1] then (i: k, return())),
  j: i + 1, k: n,
  while j < k do (t: v[j], v[j]: v[k], v[k]: t, j: j + 1, k: k - 1),
  if i = 0 then return(false),
  j: i + 1,
  while v[j] < v[i] do j: j + 1,
  t: v[j], v[j]: v[i], v[i]: t,
  true

)$

print_perm(n) := block([v: makelist(i, i, 1, n)],

  disp(v),
  while next_permutation(v) do disp(v)

)$

print_perm(3); /* [1, 2, 3]

  [1, 3, 2]
  [2, 1, 3]
  [2, 3, 1]
  [3, 1, 2]
  [3, 2, 1] */</lang>

Builtin version

<lang maxima> (%i1) permutations([1, 2, 3]); (%o1) {[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]} </lang>

Modula-2

Works with: ADW Modula-2 (1.6.291)

<lang Modula-2>MODULE Permute;

FROM Terminal IMPORT Read, Write, WriteLn;

FROM Terminal2 IMPORT WriteString;

CONST MAXIDX = 6; MINIDX = 1;

TYPE TInpCh = ['a'..'z']; TChr = SET OF TInpCh;

VAR n, nl: INTEGER; ch: CHAR; a: ARRAY[MINIDX..MAXIDX] OF CHAR; kt: TChr = TChr{'a'..'f'};

PROCEDURE output; VAR i: INTEGER; BEGIN FOR i := MINIDX TO n DO Write(a[i]) END; WriteString(" | "); END output;

PROCEDURE exchange(VAR x, y : CHAR); VAR z: CHAR; BEGIN z := x; x := y; y := z END exchange;

PROCEDURE permute(k: INTEGER); VAR i: INTEGER; BEGIN IF k = 1 THEN output; INC(nl); IF (nl MOD 8 = 1) THEN WriteLn END; ELSE permute(k-1); FOR i := MINIDX TO k-1 DO exchange(a[i], a[k]); permute(k-1); exchange(a[i], a[k]); END END END permute;

BEGIN n := 0; nl := 1; WriteString("Input {a,b,c,d,e,f} >"); REPEAT Read(ch); IF ch IN kt THEN INC(n); a[n] := ch; Write(ch) END UNTIL (ch <= " ") OR (n > MAXIDX);

WriteLn; IF n > 0 THEN permute(n) END; (*Wait*) END Permute.</lang>

NetRexx

<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary

import java.util.List import java.util.ArrayList

-- ============================================================================= /**

* Permutation Iterator
* 

* 

* Algorithm by E. W. Dijkstra, "A Discipline of Programming", Prentice-Hall, 1976, p.71
*/

class RPermutationIterator implements Iterator

 -- ---------------------------------------------------------------------------
 properties indirect
   perms = List
   permOrders = int[]
   maxN
   currentN
   first = boolean

 -- ---------------------------------------------------------------------------
 properties constant
   isTrue  = boolean (1 == 1)
   isFalse = boolean (1 \= 1)

 -- ---------------------------------------------------------------------------
 method RPermutationIterator(initial = List) public
   setUp(initial)
   return

 -- ---------------------------------------------------------------------------
 method RPermutationIterator(initial = Object[]) public
   init = ArrayList(initial.length)
   loop elmt over initial
     init.add(elmt)
     end elmt
   setUp(init)
   return

 -- ---------------------------------------------------------------------------
 method RPermutationIterator(initial = Rexx[]) public
   init = ArrayList(initial.length)
   loop elmt over initial
     init.add(elmt)
     end elmt
   setUp(init)
   return

 -- ---------------------------------------------------------------------------
 method setUp(initial = List) private
   setFirst(isTrue)
   setPerms(initial)
   setPermOrders(int[getPerms().size()])
   setMaxN(getPermOrders().length)
   setCurrentN(0)
   po = getPermOrders()
   loop i_ = 0 while i_ < po.length
     po[i_] = i_
     end i_
   return

 -- ---------------------------------------------------------------------------
 method hasNext() public returns boolean
   status = isTrue
   if getCurrentN() == factorial(getMaxN()) then status = isFalse
   setCurrentN(getCurrentN() + 1)    
   return status

 -- ---------------------------------------------------------------------------
 method next() public returns Object
   if isFirst() then setFirst(isFalse)
   else do
     po = getPermOrders()
     i_ = getMaxN() - 1
     loop while po[i_ - 1] >= po[i_]
       i_ = i_ - 1
       end

     j_ = getMaxN()
     loop while po[j_ - 1] <= po[i_ - 1]
       j_ = j_ - 1
       end

     swap(i_ - 1, j_ - 1)

     i_ = i_ + 1
     j_ = getMaxN()
     loop while i_ < j_
       swap(i_ - 1, j_ - 1)
       i_ = i_ + 1
       j_ = j_ - 1
       end
     end
   return reorder()

 -- ---------------------------------------------------------------------------
 method remove() public signals UnsupportedOperationException
   signal UnsupportedOperationException()

 -- ---------------------------------------------------------------------------
 method swap(i_, j_) private
   po = getPermOrders()
   save   = po[i_]
   po[i_] = po[j_]
   po[j_] = save
   return

 -- ---------------------------------------------------------------------------
 method reorder() private returns List
   result = ArrayList(getPerms().size())
   loop ix over getPermOrders()
     result.add(getPerms().get(ix))
     end ix
   return result

 -- ---------------------------------------------------------------------------
 /**
  * Calculate n factorial: {@code n! = 1 * 2 * 3 .. * n}
  * @param n
  * @return n!
  */
 method factorial(n) public static
   fact = 1
   if n > 1 then loop i = 1 while i <= n
     fact = fact * i
     end i
   return fact

 -- ---------------------------------------------------------------------------
 method main(args = String[]) public static
   thing02 = RPermutationIterator(['alpha', 'omega'])
   thing03 = RPermutationIterator([String 'one', 'two', 'three'])
   thing04 = RPermutationIterator(Arrays.asList([Integer(1), Integer(2), Integer(3), Integer(4)]))
   things = [thing02, thing03, thing04]
   loop thing over things
     N = thing.getMaxN()
     say 'Permutations:' N'! =' factorial(N)
     loop lineCount = 1 while thing.hasNext()
       prm = thing.next()
       say lineCount.right(8)':' prm.toString()
       end lineCount
     say 'Permutations:' N'! =' factorial(N)
     say
     end thing
   return

</lang>

Output:

Permutations: 2! = 2
       1: [alpha, omega]
       2: [omega, alpha]
Permutations: 2! = 2

Permutations: 3! = 6
       1: [one, two, three]
       2: [one, three, two]
       3: [two, one, three]
       4: [two, three, one]
       5: [three, one, two]
       6: [three, two, one]
Permutations: 3! = 6

Permutations: 4! = 24
       1: [1, 2, 3, 4]
       2: [1, 2, 4, 3]
       3: [1, 3, 2, 4]
       4: [1, 3, 4, 2]
       5: [1, 4, 2, 3]
       6: [1, 4, 3, 2]
       7: [2, 1, 3, 4]
       8: [2, 1, 4, 3]
       9: [2, 3, 1, 4]
      10: [2, 3, 4, 1]
      11: [2, 4, 1, 3]
      12: [2, 4, 3, 1]
      13: [3, 1, 2, 4]
      14: [3, 1, 4, 2]
      15: [3, 2, 1, 4]
      16: [3, 2, 4, 1]
      17: [3, 4, 1, 2]
      18: [3, 4, 2, 1]
      19: [4, 1, 2, 3]
      20: [4, 1, 3, 2]
      21: [4, 2, 1, 3]
      22: [4, 2, 3, 1]
      23: [4, 3, 1, 2]
      24: [4, 3, 2, 1]
Permutations: 4! = 24

Nim

Translation of: C

<lang nim># iterative Boothroyd method iterator permutations[T](ys: openarray[T]): seq[T] =

 var
   d = 1
   c = newSeq[int](ys.len)
   xs = newSeq[T](ys.len)

 for i, y in ys: xs[i] = y
 yield xs

 block outer:
   while true:
     while d > 1:
       dec d
       c[d] = 0
     while c[d] >= d:
       inc d
       if d >= ys.len: break outer

     let i = if (d and 1) == 1: c[d] else: 0
     swap xs[i], xs[d]
     yield xs
     inc c[d]

var x = @[1,2,3]

for i in permutations(x):

 echo i</lang>

Output:

@[1, 2, 3]
@[2, 1, 3]
@[3, 1, 2]
@[1, 3, 2]
@[2, 3, 1]
@[3, 2, 1]

OCaml

<lang ocaml>(* Iterative, though loops are implemented as auxiliary recursive functions.

  Translation of Ada version. *)

let next_perm p = let n = Array.length p in let i = let rec aux i = if (i < 0) || (p.(i) < p.(i+1)) then i else aux (i - 1) in aux (n - 2) in let rec aux j k = if j < k then let t = p.(j) in p.(j) <- p.(k); p.(k) <- t; aux (j + 1) (k - 1) else () in aux (i + 1) (n - 1); if i < 0 then false else let j = let rec aux j = if p.(j) > p.(i) then j else aux (j + 1) in aux (i + 1) in let t = p.(i) in p.(i) <- p.(j); p.(j) <- t; true;;

let print_perm p = let n = Array.length p in for i = 0 to n - 2 do print_int p.(i); print_string " " done; print_int p.(n - 1); print_newline ();;

let print_all_perm n = let p = Array.init n (function i -> i + 1) in print_perm p; while next_perm p do print_perm p done;;

print_all_perm 3;; (* 1 2 3

  1 3 2
  2 1 3
  2 3 1
  3 1 2
  3 2 1 *)</lang>

Permutations can also be defined on lists recursively: <lang OCaml>let rec permutations l =

  let n = List.length l in
  if n = 1 then [l] else
  let rec sub e = function
     | [] -> failwith "sub"
     | h :: t -> if h = e then t else h :: sub e t in
  let rec aux k =
     let e = List.nth l k in
     let subperms = permutations (sub e l) in
     let t = List.map (fun a -> e::a) subperms in
     if k < n-1 then List.rev_append t (aux (k+1)) else t in
  aux 0;;

let print l = List.iter (Printf.printf " %d") l; print_newline() in List.iter print (permutations [1;2;3;4])</lang> or permutations indexed independently: <lang OCaml>let rec pr_perm k n l =

  let a, b = let c = k/n in c, k-(n*c) in
  let e = List.nth l b in
  let rec sub e = function
     | [] -> failwith "sub"
     | h :: t -> if h = e then t else h :: sub e t in
  (Printf.printf " %d" e; if n > 1 then pr_perm a (n-1) (sub e l))

let show_perms l =

  let n = List.length l in 
  let rec fact n = if n < 3 then n else n * fact (n-1) in 
  for i = 0 to (fact n)-1 do
     pr_perm i n l;
     print_newline()
  done

let () = show_perms [1;2;3;4]</lang>

PARI/GP

<lang parigp>vector(n!,k,numtoperm(n,k))</lang>

Pascal

<lang pascal>program perm;

var p: array[1 .. 12] of integer; is_last: boolean; n: integer;

procedure next; var i, j, k, t: integer; begin is_last := true; i := n - 1; while i > 0 do begin if p[i] < p[i + 1] then begin is_last := false; break; end; i := i - 1; end;

if not is_last then begin j := i + 1; k := n; while j < k do begin t := p[j]; p[j] := p[k]; p[k] := t; j := j + 1; k := k - 1; end;

j := n; while p[j] > p[i] do j := j - 1; j := j + 1;

t := p[i]; p[i] := p[j]; p[j] := t; end; end;

procedure print; var i: integer; begin for i := 1 to n do write(p[i], ' '); writeln; end;

procedure init; var i: integer; begin n := 0; while (n < 1) or (n > 10) do begin write('Enter n (1 <= n <= 10): '); readln(n); end; for i := 1 to n do p[i] := i; end;

begin init; repeat print; next; until is_last; end.</lang>

alternative

a little bit more speed.I take n = 12. The above version takes more than 5 secs.My permlex takes 2.8s, but in the depth of my harddisk I found a version, creating all permutations using k places out of n.The cpu loves it! 1.33 s. But you have to use the integers [1..n] directly or as Index to your data. 1 to n are in lexicographic order. <lang pascal>{$IFDEF FPC}

 {$MODE DELPHI}

{$ELSE}

 {$APPTYPE CONSOLE}

{$ENDIF} uses

 sysutils;

type

 tPermfield  =  array[0..15] of Nativeint;

var

 permcnt: NativeUint;

procedure DoSomething(k: NativeInt;var x:tPermfield); var

 i:integer;
 kk:string;

begin

 kk:=;
 for i:=1 to k do kk:=kk+inttostr(x[i])+' ';
   writeln(kk);

end;

procedure PermKoutOfN(k,n: nativeInt); var

 x,y:tPermfield;
 i,yi,tmp:NativeInt;

begin

 //initialise
 permcnt:= 1;
 if k>n then
   k:=n;
 if k=n then
   k:=k-1;
 for i:=1 to n do x[i]:=i;
 for i:=1 to k do y[i]:=i;

// DoSomething(k,x);

 i := k;
 repeat
   yi:=y[i];
   if yi <n then
   begin
     inc(permcnt);
     inc(yi);
     y[i]:=yi;
     tmp:=x[i];x[i]:=x[yi];x[yi]:=tmp;
     i:=k;

// DoSomething(k,x);

   end
   else
   begin
     repeat
       tmp:=x[i];x[i]:=x[yi];x[yi]:=tmp;
       dec(yi);
     until yi<=i;
     y[i]:=yi;
     dec(i);
   end;
 until (i=0);

end;

var

 t1,t0 : TDateTime;

Begin

 permcnt:= 0;
 T0 := now;
 PermKoutOfN(12,12);
 T1 := now;
 writeln(permcnt);
 writeln(FormatDateTime('HH:NN:SS.zzz',T1-T0));

end.</lang>

Output:

{fpc 2.64/3.0 32Bit or 3.1 64 Bit i4330 3.5 Ghz same timings. //PermKoutOfN(12,12);

 
479001600 //= 12!
00:00:01.328

PHP

Full non-recursive algorithm generating permutation in reversed lexicographical order. Taken from here: [3]. It's a little bit modified algorythm of Narayana Pandit [4]. It reveres a string and works until a and b are not equal. It searches a new permutation directly (first internal cycle) as we do it on a sheet of paper. You can use English alphabet instead of digits.

<lang php> //Author Gavryushin Ivan @dcc0 <?php $b="0123"; $a=strrev($b);

while ($a !=$b) { $i=1; while($a[$i] > $a[$i-1]) { $i++; } $j=0; while($a[$j] < $a[$i]) {

$j++; } $c=$a[$j]; $a[$j]=$a[$i]; $a[$i]=$c; $a=strrev(substr($a, 0, $i)).substr($a, $i); print $a. "\n";

} ?> </lang> Output

Perl

There are many modules that can do permutations, or it can be fairly easily done by hand with an example below. In performance order for simple permutation of 10 scalars, a sampling of some solutions:

- 1.7s Algorithm::FastPermute permute iterator
- 1.7s Algorithm::Permute permute iterator
- 2.0s ntheory  forperm iterator
- 6.3s Algorithm::Combinatorics permutations iterator
- 9.1s the recursive sub below
- 21.1s Math::Combinatorics permutations iterator

Example:

Library: ntheory

<lang perl>use ntheory qw/forperm/; my @tasks = (qw/party sleep study/); forperm {

 print "@tasks[@_]\n";

} scalar(@tasks);</lang>

Output:

party sleep study
party study sleep
sleep party study
sleep study party
study party sleep
study sleep party

A simple recursive routine: <lang perl>sub permutation { my ($perm,@set) = @_; print "$perm\n" || return unless (@set); permutation($perm.$set[$_],@set[0..$_-1],@set[$_+1..$#set]) foreach (0..$#set); } my @input = (qw/a 2 c 4/); permutation(,@input);</lang>

Output:

a2c4
a24c
ac24
ac42
a42c
a4c2
2ac4
2a4c
2ca4
2c4a
24ac
24ca
ca24
ca42
c2a4
c24a
c4a2
c42a
4a2c
4ac2
42ac
42ca
4ca2
4c2a

Perl 6

Works with: rakudo version 2014-1-24

First, you can just use the built-in method on any list type. <lang Perl6>.say for <a b c>.permutations</lang>

Output:

a b c
a c b
b a c
b c a
c a b
c b a

Here is some generic code that works with any ordered type. To force lexicographic ordering, change after to gt. To force numeric order, replace it with >. <lang perl6>sub next_perm ( @a is copy ) {

   my $j = @a.end - 1;
   return Nil if --$j < 0 while @a[$j] after @a[$j+1];

   my $aj = @a[$j];
   my $k  = @a.end;
   $k-- while $aj after @a[$k];
   @a[ $j, $k ] .= reverse;

   my $r = @a.end;
   my $s = $j + 1;
   @a[ $r--, $s++ ] .= reverse while $r > $s;
   return $(@a);

}

.say for [<a b c>], &next_perm ...^ !*;</lang>

Output:

a b c
a c b
b a c
b c a
c a b
c b a

Here is another non-recursive implementation, which returns a lazy list. It also works with any type. <lang perl6>sub permute(+@items) {

  my @seq := 1..+@items;
  gather for (^[*] @seq) -> $n is copy {
     my @order;
     for @seq {
        unshift @order, $n mod $_;
        $n div= $_;
     }
     my @i-copy = @items;
     take map { |@i-copy.splice($_, 1) }, @order;
  }

} .say for permute( 'a'..'c' )</lang>

Output:

(a b c)
(a c b)
(b a c)
(b c a)
(c a b)
(c b a)

Finally, if you just want zero-based numbers, you can call the built-in function: <lang perl6>.say for permutations(3);</lang>

Output:

Phix

The distribution includes builtins\permute.e, which is reproduced below. This can be used to retrieve all possible permutations, in no particular order. The elements can be any type. It is just as fast to generate the (n!)th permutation as the first, so some applications may benefit by storing an integer key rather than duplicating all the elements of the given set. <lang Phix>global function permute(integer n, sequence set) -- -- return the nth permute of the given set. -- n should be an integer in the range 1 to factorial(length(set)) -- sequence res integer w

   n -= 1
   res = set
   for i=length(set) to 1 by -1 do 
       w = remainder(n,i)+1
       res[i] = set[w]
       set[w] = set[i]
       n = floor(n/i)
   end for
   return res

end function</lang> Example use: <lang Phix>function permutes(sequence set) sequence res = repeat(0,factorial(length(set)))

   for i=1 to length(res) do
       res[i] = permute(i,set)
   end for
   return res

end function ?permutes("abcd")</lang>

Output:

{"bcda","dcab","bdac","bcad","cdba","cadb","dabc","cabd","bdca","dacb","badc","bacd","cbda","cdab","dbac","cbad","dcba","acdb","adbc","acbd","dbca","adcb","abdc","abcd"}

PicoLisp

<lang PicoLisp>(load "@lib/simul.l")

(permute (1 2 3))</lang>

Output:

-> ((1 2 3) (1 3 2) (2 1 3) (2 3 1) (3 1 2) (3 2 1))

PowerBASIC

Works with: PowerBASIC version 10.00+

<lang ada> #COMPILE EXE

 #DIM ALL
 GLOBAL a, i, j, k, n  AS INTEGER
 GLOBAL d, ns, s AS STRING 'dynamic string
 FUNCTION PBMAIN () AS LONG
 ns = INPUTBOX$("   n =",, "3") 'input n
 n = VAL(ns)
 DIM a(1 TO n) AS INTEGER
 FOR i = 1 TO n: a(i)= i: NEXT
 DO
   s = " "
   FOR i = 1 TO n
     d = STR$(a(i))
     s = BUILD$(s, d) '  s & d concatenate
   NEXT
   ? s  'print and pause
   i = n
   DO
    DECR i
   LOOP UNTIL i = 0 OR a(i) < a(i+1)
   j = i+1
   k = n
   DO WHILE j < k
     SWAP a(j), a(k)
     INCR j
     DECR k
   LOOP
   IF i > 0 THEN
     j = i+1
     DO WHILE a(j) < a(i)
       INCR j
     LOOP
     SWAP a(i), a(j)
   END IF
 LOOP UNTIL i = 0
 END FUNCTION</lang>

Output:

PowerShell

<lang PowerShell> function permutation ($array) {

   function generate($n, $array, $A) {
       if($n -eq 1) {
           $array[$A] -join ' '
       }
       else{
           for( $i = 0; $i -lt ($n - 1); $i += 1) {
               generate ($n - 1) $array $A
               if($n % 2 -eq 0){
                   $i1, $i2 = $i, ($n-1)
                   $A[$i1], $A[$i2] = $A[$i2], $A[$i1]
               }
               else{
                   $i1, $i2 = 0, ($n-1)
                   $A[$i1], $A[$i2] = $A[$i2], $A[$i1]
               }
           }
           generate ($n - 1) $array $A
       }
   }
   $n = $array.Count
   if($n -gt 0) {
       (generate $n $array (0..($n-1)))
   } else {$array}

} permutation @('A','B','C') </lang> Output:

A B C
B A C
C A B
A C B
B C A
C B A

Prolog

Works with SWI-Prolog and library clpfd, <lang Prolog>:- use_module(library(clpfd)).

permut_clpfd(L, N) :-

   length(L, N),
   L ins 1..N,
   all_different(L),
   label(L).</lang>

Output:

<lang Prolog>?- permut_clpfd(L, 3), writeln(L), fail. [1,2,3] [1,3,2] [2,1,3] [2,3,1] [3,1,2] [3,2,1] false. </lang> A declarative way of fetching permutations: <lang Prolog>% permut_Prolog(P, L) % P is a permutation of L

permut_Prolog([], []). permut_Prolog([H | T], NL) :- select(H, NL, NL1), permut_Prolog(T, NL1).</lang>

Output:

<lang Prolog> ?- permut_Prolog(P, [ab, cd, ef]), writeln(P), fail. [ab,cd,ef] [ab,ef,cd] [cd,ab,ef] [cd,ef,ab] [ef,ab,cd] [ef,cd,ab] false.</lang>

PureBasic

The procedure nextPermutation() takes an array of integers as input and transforms its contents into the next lexicographic permutation of it's elements (i.e. integers). It returns #True if this is possible. It returns #False if there are no more lexicographic permutations left and arranges the elements into the lowest lexicographic permutation. It also returns #False if there is less than 2 elemetns to permute.

The integer elements could be the addresses of objects that are pointed at instead. In this case the addresses will be permuted without respect to what they are pointing to (i.e. strings, or structures) and the lexicographic order will be that of the addresses themselves. <lang PureBasic>Macro reverse(firstIndex, lastIndex)

 first = firstIndex
 last = lastIndex
 While first < last
   Swap cur(first), cur(last)
   first + 1
   last - 1
 Wend

EndMacro

Procedure nextPermutation(Array cur(1))

 Protected first, last, elementCount = ArraySize(cur())
 If elementCount < 1
   ProcedureReturn #False ;nothing to permute
 EndIf 
 
 ;Find the lowest position pos such that [pos] < [pos+1]
 Protected pos = elementCount - 1
 While cur(pos) >= cur(pos + 1)
   pos - 1
   If pos < 0
     reverse(0, elementCount)
     ProcedureReturn #False ;no higher lexicographic permutations left, return lowest one instead
   EndIf 
 Wend

 ;Swap [pos] with the highest positional value that is larger than [pos]
 last = elementCount
 While cur(last) <= cur(pos)
   last - 1
 Wend
 Swap cur(pos), cur(last)

 ;Reverse the order of the elements in the higher positions
 reverse(pos + 1, elementCount)
 ProcedureReturn #True ;next lexicographic permutation found

EndProcedure

Procedure display(Array a(1))

 Protected i, fin = ArraySize(a())
 For i = 0 To fin
   Print(Str(a(i)))
   If i = fin: Continue: EndIf
   Print(", ")
 Next
 PrintN("")

EndProcedure

If OpenConsole()

 Dim a(2)
 a(0) = 1: a(1) = 2: a(2) =  3
 display(a())
 While nextPermutation(a()): display(a()): Wend
 
 Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
 CloseConsole()

EndIf</lang>

Output:

1, 2, 3
1, 3, 2
2, 1, 3
2, 3, 1
3, 1, 2
3, 2, 1

Python

Standard library function

Works with: Python version 2.6+

<lang python>import itertools for values in itertools.permutations([1,2,3]):

   print (values)</lang>

Output:

(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)

Recursive implementation

The follwing functions start from a list [0 ... n-1] and exchange elements to always have a valid permutation. This is done recursively: first exchange a[0] with all the other elements, then a[1] with a[2] ... a[n-1], etc. thus yielding all permutations.

<lang python>def perm1(n):

   a = list(range(n))
   def sub(i):
       if i == n - 1:
           yield tuple(a)
       else:
           for k in range(i, n):
               a[i], a[k] = a[k], a[i]
               yield from sub(i + 1)
               a[i], a[k] = a[k], a[i]
   yield from sub(0)

def perm2(n):

   a = list(range(n))
   def sub(i):
       if i == n - 1:
           yield tuple(a)
       else:
           for k in range(i, n):
               a[i], a[k] = a[k], a[i]
               yield from sub(i + 1)
           x = a[i]
           for k in range(i + 1, n):
               a[k - 1] = a[k]
           a[n - 1] = x
   yield from sub(0)</lang>

These two solutions make use of a generator, and "yield from" introduced in PEP-380. They are slightly different: the latter produces permutations in lexicographic order, because the "remaining" part of a (that is, a[i+1:]) is always sorted, whereas the former always reverses the exchange just after the recursive call.

On three elements, the difference can be seen on the last two permutations:

<lang python>for u in perm1(3): print(u) (0, 1, 2) (0, 2, 1) (1, 0, 2) (1, 2, 0) (2, 1, 0) (2, 0, 1)

for u in perm2(3): print(u) (0, 1, 2) (0, 2, 1) (1, 0, 2) (1, 2, 0) (2, 0, 1) (2, 1, 0)</lang>

Iterative implementation

Given a permutation, one can easily compute the next permutation in some order, for example lexicographic order, here. Then to get all permutations, it's enough to start from [0, 1, ... n-1], and store the next permutation until [n-1, n-2, ... 0], which is the last in lexicographic order.

<lang python>def nextperm(a):

   n = len(a)
   i = n - 1
   while i > 0 and a[i - 1] > a[i]:
       i -= 1
   j = i
   k = n - 1
   while j < k:
       a[j], a[k] = a[k], a[j]
       j += 1
       k -= 1
   if i == 0:
       return False
   else:
       j = i
       while a[j] < a[i - 1]:
           j += 1
       a[i - 1], a[j] = a[j], a[i - 1]
       return True

def perm3(n):

   if type(n) is int:
       if n < 1:
           return []
       a = list(range(n))
   else:
       a = sorted(n)
   u = [tuple(a)]
   while nextperm(a):
       u.append(tuple(a))
   return u

for p in perm3(3): print(p) (0, 1, 2) (0, 2, 1) (1, 0, 2) (1, 2, 0) (2, 0, 1) (2, 1, 0)</lang>

Qi

Translation of: Erlang

<lang qi> (define insert

 L      0 E -> [E|L]
 [L|Ls] N E -> [L|(insert Ls (- N 1) E)])

(define seq

 Start Start -> [Start]
 Start End   -> [Start|(seq (+ Start 1) End)])

(define append-lists

 []    -> []
 [A|B] -> (append A (append-lists B)))

(define permutate

 []    -> [[]]
 [H|T] -> (append-lists (map (/. P
                                 (map (/. N
                                          (insert P N H))
                                      (seq 0 (length P))))
                             (permute T))))</lang>

R

<lang r>next.perm <- function(p) {

 n <- length(p)
 i <- n - 1
 r = T
 for (i in seq(n - 1, 1)) {
   if (p[i] < p[i + 1]) {
     r = F
     break
   }
 }

 j <- i + 1
 k <- n
 while (j < k) {
   x <- p[j]
   p[j] <- p[k]
   p[k] <- x
   j <- j + 1
   k <- k - 1
 }

 if(r) return(NULL)

 j <- n
 while (p[j] > p[i]) j <- j - 1
 j <- j + 1

 x <- p[i]
 p[i] <- p[j]
 p[j] <- x
 return(p)

}

print.perms <- function(n) {

 p <- 1:n
 while (!is.null(p)) {
   cat(p, "\n")
   p <- next.perm(p)
 }

}

print.perms(3)

1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1</lang>

Racket

lang racket

using a builtin

(permutations '(A B C))

-> '((A B C) (B A C) (A C B) (C A B) (B C A) (C B A))

a random simple version (which is actually pretty good for a simple version)

(define (perms l)

 (let loop ([l l] [tail '()])
   (if (null? l) (list tail)
       (append-map (λ(x) (loop (remq x l) (cons x tail))) l))))

(perms '(A B C))

-> '((C B A) (B C A) (C A B) (A C B) (B A C) (A B C))

</lang>

REXX

using names

This program could be simplified quite a bit if the "things" were just restricted to numbers (numerals),
but that would make it specific to numbers and not "things" or objects. <lang rexx>/*REXX program generates and displays all permutations of N different objects. */ parse arg things bunch inbetweenChars names

                      /* inbetweenChars  (optional)   defaults to a  [null].           */
                      /*          names  (optional)   defaults to digits (and letters).*/

call permSets things, bunch, inbetweenChars, names exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ p: return word(arg(1),1) /*P function (Pick first arg of many).*/ /*──────────────────────────────────────────────────────────────────────────────────────*/ permSets: procedure; parse arg x,y,between,uSyms /*X things taken Y at a time. */

         @.=;   sep=                            /*X  can't be  >  length(@0abcs).      */
         @abc  = 'abcdefghijklmnopqrstuvwxyz';     @abcU=@abc;         upper @abcU
         @abcS = @abcU || @abc;                    @0abcS=123456789 || @abcS

           do k=1  for x                        /*build a list of permutation symbols. */
           _=p(word(uSyms,k)  p(substr(@0abcS,k,1) k))      /*get or generate a symbol.*/
           if length(_)\==1  then sep='_'       /*if not 1st character,  then use sep. */
           $.k=_                                /*append the character to symbol list. */
           end   /*k*/

         if between==  then between=sep       /*use the appropriate separator chars. */
         call .permset 1                        /*start with the  first  permuation.   */
         return

.permset: procedure expose $. @. between x y; parse arg ?

         if ?>y then do; _=@.1;   do j=2  to y;  _=_ || between || @.j; end;  say _;  end
                else do q=1  for x              /*build the permutation recursively.   */
                         do k=1  for ?-1;  if @.k==$.q  then iterate q;  end  /*k*/
                     @.?=$.q;              call .permset ?+1
                     end    /*q*/
         return</lang>

output when the following was used for input: 3 3

output when the following was used for input: 4 4 --- A B C D

A---B---C---D
A---B---D---C
A---C---B---D
A---C---D---B
A---D---B---C
A---D---C---B
B---A---C---D
B---A---D---C
B---C---A---D
B---C---D---A
B---D---A---C
B---D---C---A
C---A---B---D
C---A---D---B
C---B---A---D
C---B---D---A
C---D---A---B
C---D---B---A
D---A---B---C
D---A---C---B
D---B---A---C
D---B---C---A
D---C---A---B
D---C---B---A

output when the following was used for input: 4 3 ~ aardvark gnu stegosaurus platypus

aardvark~gnu~stegosaurus
aardvark~gnu~platypus
aardvark~stegosaurus~gnu
aardvark~stegosaurus~platypus
aardvark~platypus~gnu
aardvark~platypus~stegosaurus
gnu~aardvark~stegosaurus
gnu~aardvark~platypus
gnu~stegosaurus~aardvark
gnu~stegosaurus~platypus
gnu~platypus~aardvark
gnu~platypus~stegosaurus
stegosaurus~aardvark~gnu
stegosaurus~aardvark~platypus
stegosaurus~gnu~aardvark
stegosaurus~gnu~platypus
stegosaurus~platypus~aardvark
stegosaurus~platypus~gnu
platypus~aardvark~gnu
platypus~aardvark~stegosaurus
platypus~gnu~aardvark
platypus~gnu~stegosaurus
platypus~stegosaurus~aardvark
platypus~stegosaurus~gnu

using numbers

This version is modeled after the Maxima program (as far as output).

It doesn't have the formatting capabilities of the REXX version 1, nor can it handle taking X items taken Y at-a-time. <lang rexx>/*REXX program displays permutations of N number of objects (1, 2, 3, ···). */ parse arg n .; if n== | n=="," then n=3 /*Not specified? Then use the default.*/

                                                /* [↓]  populate the first permutation.*/
        do pop=1  for n;            @.pop=pop  ;     end  /*pop  */;          call tell n
        do  while nPerm(n, 0);      call tell n;     end  /*while*/

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ nPerm: procedure expose @.; parse arg n,i; nm=n-1

        do k=nm  by -1  for nm;  kp=k+1; if @.k<@.kp  then do; i=k; leave; end; end /*k*/
        do j=i+1  while j<n;  parse value  @.j  @.n   with   @.n  @.j;   n=n-1; end /*j*/
        if i==0  then return 0
                                                    do m=i+1  while  @.m<@.i;  end  /*m*/
        parse value  @.m  @.i  with  @.i  @.m
        return 1

/*──────────────────────────────────────────────────────────────────────────────────────*/ tell: procedure expose @.; _=; do j=1 for arg(1); _=_ @.j; end; say _; return</lang> output when using the default input:

Ring

<lang ring> list = [1, 2, 3, 4] for perm = 1 to 24

   for i = 1 to len(list)
       see list[i] + " "
   next
   see nl
   nextPermutation(list)

    elementcount = len(a)
    if elementcount < 1 then return ok
    pos = elementcount-1
    while a[pos] >= a[pos+1] 
          pos -= 1
          if pos <= 0 permutationReverse(a, 1, elementcount)
             return ok
    end
    last = elementcount
    while a[last] <= a[pos]
          last -= 1
    end
    temp = a[pos]
    a[pos] = a[last]
    a[last] = temp
    permutationReverse(a, pos+1, elementcount)

func permutationReverse a, first, last
     while first < last
           temp = a[first]
           a[first] = a[last]
           a[last] = temp
           first += 1
           last -= 1
     end

</lang> Output:

Ruby

Works with: Ruby version 1.8.7+

<lang ruby>p [1,2,3].permutation.to_a</lang>

Output:

[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

However, this method will produce indistinct permutations if the array has indistinct elements. If you need to find all the permutations of an array of which many elements are the same, the method below will be more efficient. <lang ruby>class Array

 # Yields distinct permutations of _self_ to the block.
 # This method requires that all array elements be Comparable.
 def distinct_permutation  # :yields: _ary_
   # If no block, return an enumerator. Works with Ruby 1.8.7.
   block_given? or return enum_for(:distinct_permutation)

   copy = self.sort
   yield copy.dup
   return if size < 2

   while true
     # from: "The Art of Computer Programming" by Donald Knuth
     j = size - 2;
     j -= 1 while j > 0 && copy[j] >= copy[j+1]
     if copy[j] < copy[j+1]
       l = size - 1
       l -= 1 while copy[j] >= copy[l] 
       copy[j] , copy[l] = copy[l] , copy[j]
       copy[j+1..-1] = copy[j+1..-1].reverse
       yield copy.dup
     else
       break
     end
   end
 end

end

permutations = [] [1,1,2].distinct_permutation do |p| permutations << p end p permutations

=> [[1, 1, 2], [1, 2, 1], [2, 1, 1]]

if RUBY_VERSION >= "1.8.7"

 p [1,1,2].distinct_permutation.to_a
 # => [[1, 1, 2], [1, 2, 1], [2, 1, 1]]

end</lang>

Run BASIC

Works with Run BASIC, Liberty BASIC and Just BASIC <lang Runbasic>list$ = "h,e,l,l,o" ' supply list seperated with comma's

while word$(list$,d+1,",") <> "" 'Count how many in the list d = d + 1 wend

dim theList$(d) ' place list in array for i = 1 to d

 theList$(i) = word$(list$,i,",")

next i

for i = 1 to d ' print the Permutations

for j = 2 to d
  perm$ = ""
  for k = 1 to d
   perm$ = perm$ + theList$(k)
  next k
  if instr(perm2$,perm$+",") = 0 then print perm$ ' only list 1 time
  perm2$ 	 = perm2$ + perm$ + ","
  h$		 = theList$(j)
  theList$(j)	 = theList$(j - 1)
  theList$(j - 1) = h$
 next j

next i end</lang>Output:

hello
ehllo
elhlo
ellho
elloh
leloh
lleoh
lloeh
llohe
lolhe
lohle
lohel
olhel
ohlel
ohell
hoell
heoll
helol

SAS

<lang sas>/* Store permutations in a SAS dataset. Translation of Fortran 77 */ data perm;

 n=6;
 array a{6} p1-p6;
 do i=1 to n;
   a(i)=i;
 end;

L1:

 output;
 link L2;
 if next then goto L1;
 stop;

L2:

 next=0;
 i=n-1;

L10:

 if a(i)<a(i+1) then goto L20;
 i=i-1;
 if i=0 then goto L20;
 goto L10;
 L20:
 j=i+1;
 k=n;

L30:

 t=a(j);
 a(j)=a(k);
 a(k)=t;
 j=j+1;
 k=k-1;
 if j<k then goto L30;
 j=i;
 if j=0 then return;

L40:

 j=j+1;
 if a(j)<a(i) then goto L40;
 t=a(i);
 a(i)=a(j);
 a(j)=t;
 next=1;
 return;
 keep p1-p6;

run;</lang>

Scala

There is a built-in function in the Scala collections library, that is part of the language's standard library. The permutation function is available on any sequential collection. It could be used as follows given a list of numbers:

<lang scala>List(1, 2, 3).permutations.foreach(println)</lang>

Output:

 List(1, 2, 3)
 List(1, 3, 2)
 List(2, 1, 3)
 List(2, 3, 1)
 List(3, 1, 2)
 List(3, 2, 1)

The following function returns all the permutations of a list:

<lang scala> def permutations[T]: List[T] => Traversable[List[T]] = {

   case Nil => List(Nil)
   case xs => {
     for {
       (x, i) <- xs.zipWithIndex
       ys <- permutations(xs.take(i) ++ xs.drop(1 + i))
     } yield {
       x :: ys
     }
   }
 }</lang>

If you need the unique permutations, use distinct or toSet on either the result or on the input.

Scheme

Translation of: Erlang

<lang scheme>(define (insert l n e)

 (if (= 0 n)
     (cons e l)
     (cons (car l) 
           (insert (cdr l) (- n 1) e))))

(define (seq start end)

 (if (= start end)
     (list end)
     (cons start (seq (+ start 1) end))))

(define (permute l)

 (if (null? l)
     '(())
     (apply append (map (lambda (p)
                          (map (lambda (n)
                                 (insert p n (car l)))
                               (seq 0 (length p))))
                        (permute (cdr l))))))</lang>

Translation of: OCaml

<lang scheme>; translation of ocaml : mostly iterative, with auxiliary recursive functions for some loops (define (vector-swap! v i j) (let ((tmp (vector-ref v i))) (vector-set! v i (vector-ref v j)) (vector-set! v j tmp)))

(define (next-perm p) (let* ((n (vector-length p)) (i (let aux ((i (- n 2))) (if (or (< i 0) (< (vector-ref p i) (vector-ref p (+ i 1)))) i (aux (- i 1)))))) (let aux ((j (+ i 1)) (k (- n 1))) (if (< j k) (begin (vector-swap! p j k) (aux (+ j 1) (- k 1))))) (if (< i 0) #f (begin (vector-swap! p i (let aux ((j (+ i 1))) (if (> (vector-ref p j) (vector-ref p i)) j (aux (+ j 1))))) #t))))

(define (print-perm p) (let ((n (vector-length p))) (do ((i 0 (+ i 1))) ((= i n)) (display (vector-ref p i)) (display " ")) (newline)))

(define (print-all-perm n) (let ((p (make-vector n))) (do ((i 0 (+ i 1))) ((= i n)) (vector-set! p i i)) (print-perm p) (do ( ) ((not (next-perm p))) (print-perm p))))

(print-all-perm 3)

0 1 2
0 2 1
1 0 2
1 2 0
2 0 1
2 1 0

a more recursive implementation

(define (permute p i) (let ((n (vector-length p))) (if (= i (- n 1)) (print-perm p) (begin (do ((j i (+ j 1))) ((= j n)) (vector-swap! p i j) (permute p (+ i 1))) (do ((j (- n 1) (- j 1))) ((< j i)) (vector-swap! p i j))))))

(define (print-all-perm-rec n) (let ((p (make-vector n))) (do ((i 0 (+ i 1))) ((= i n)) (vector-set! p i i)) (permute p 0)))

(print-all-perm-rec 3)

0 1 2
0 2 1
1 0 2
1 2 0
2 0 1
2 1 0</lang>

Completely recursive on lists: <lang lisp>(define (perm s)

 (cond ((null? s) '())

((null? (cdr s)) (list s)) (else ;; extract each item in list in turn and perm the rest (let splice ((l '()) (m (car s)) (r (cdr s))) (append (map (lambda (x) (cons m x)) (perm (append l r))) (if (null? r) '() (splice (cons m l) (car r) (cdr r))))))))

(display (perm '(1 2 3)))</lang>

Seed7

<lang seed7>$ include "seed7_05.s7i";

const type: permutations is array array integer;

const func permutations: permutations (in array integer: items) is func

 result
   var permutations: permsList is 0 times 0 times 0;
 local
   const proc: perms (in array integer: sequence, in array integer: prefix) is func
     local
       var integer: element is 0;
       var integer: index is 0;
     begin
       if length(sequence) <> 0 then
         for element key index range sequence do
           perms(sequence[.. pred(index)] & sequence[succ(index) ..], prefix & [] (element));
         end for;
       else
         permsList &:= prefix;
       end if;
     end func;
 begin
   perms(items, 0 times 0);
 end func;

const proc: main is func

 local
   var array integer: perm is 0 times 0;
   var integer: element is 0;
 begin
   for perm range permutations([] (1, 2, 3)) do
     for element range perm do
       write(element <& " ");
     end for;
     writeln;
   end for;
 end func;</lang>

Output:

Sidef

Built-in

<lang ruby>[0,1,2].permutations { |p|

   say p

}</lang>

Iterative

<lang ruby>func forperm(callback, n) {

   var idx = @^n

   loop {
       callback([idx...])

       var p = n-1
       while (idx[p-1] > idx[p]) {--p}
       p == 0 && return()

       var d = p
       idx += idx.splice(p).reverse

       while (idx[p-1] > idx[d]) {++d}
       idx.swap(p-1, d)
   }

   return()

}

forperm({|p| say p }, 3)</lang>

Recursive

<lang ruby>func permutations(callback, set, perm=[]) {

   set.is_empty && callback(perm)
   for i in ^set {
       __FUNC__(callback, [
           set[(0 ..^ i)..., (i+1 ..^ set.len)...]
       ], [perm..., set[i]])
   }
   return()

}

permutations({|p| say p }, [0,1,2])</lang>

Output:

[0, 1, 2]
[0, 2, 1]
[1, 0, 2]
[1, 2, 0]
[2, 0, 1]
[2, 1, 0]

Smalltalk

Works with: Squeak

Works with: Pharo

<lang smalltalk>(1 to: 4) permutationsDo: [ :x | Transcript show: x printString; cr ].</lang>

Swift

<lang swift>func perms<T>(var ar: [T]) -> T {

 return heaps(&ar, ar.count)

}

func heaps<T>(inout ar: [T], n: Int) -> T {

 return n == 1 ? [ar] :
   Swift.reduce(0..<n, T()) {
     (var shuffles, i) in
     shuffles.extend(heaps(&ar, n - 1))
     swap(&ar[n % 2 == 0 ? i : 0], &ar[n - 1])
     return shuffles
 }

}

perms([1, 2, 3]) // [[1, 2, 3], [2, 1, 3], [3, 1, 2], [1, 3, 2], [2, 3, 1], [3, 2, 1]]</lang>

Tcl

Library: Tcllib (Package: struct::list)

<lang tcl>package require struct::list

Make the sequence of digits to be permuted

set n [lindex $argv 0] for {set i 1} {$i <= $n} {incr i} {lappend sequence $i}

Iterate over the permutations, printing as we go

struct::list foreachperm p $sequence {

   puts $p

}</lang> Testing with tclsh listPerms.tcl 3 produces this output:

Ursala

In practice there's no need to write this because it's in the standard library. <lang Ursala>#import std

permutations =

~&itB^?a( # are both the input argument list and its tail non-empty?

  @ahPfatPRD *= refer ^C(      # yes, recursively generate all permutations of the tail, and for each one
     ~&a,                        # insert the head at the first position
     ~&ar&& ~&arh2falrtPXPRD),   # if the rest is non-empty, recursively insert at all subsequent positions
  ~&aNC)                       # no, return the singleton list of the argument</lang>

test program: <lang Ursala>#cast %nLL

test = permutations <1,2,3></lang>

Output:

<
   <1,2,3>,
   <2,1,3>,
   <2,3,1>,
   <1,3,2>,
   <3,1,2>,
   <3,2,1>>

VBA

Translation of: Pascal

<lang VBA>Public Sub Permute(n As Integer, Optional printem As Boolean = True) 'Generate, count and print (if printem is not false) all permutations of first n integers

Dim P() As Integer Dim t As Integer, i As Integer, j As Integer, k As Integer Dim count As Long Dim Last As Boolean

If n <= 1 Then

 Debug.Print "Please give a number greater than 1"
 Exit Sub

End If

'Initialize ReDim P(n)

For i = 1 To n

 P(i) = i

Do While Not Last

  'print?
  If printem Then

     For t = 1 To n
       Debug.Print P(t);
     Next

     Debug.Print

  End If

count = count + 1

Last = True i = n - 1

  Do While i > 0

    If P(i) < P(i + 1) Then

      Last = False
      Exit Do

    End If

    i = i - 1
  Loop

 j = i + 1
 k = n
 
 While j < k
   ' Swap p(j) and p(k)
   t = P(j)
   P(j) = P(k)
   P(k) = t
   j = j + 1
   k = k - 1
 Wend
 
 j = n
 
 While P(j) > P(i)
   j = j - 1
 Wend
 
 j = j + 1
 'Swap p(i) and p(j)
 t = P(i)
 P(i) = P(j)
 P(j) = t

Loop 'While not last

Debug.Print "Number of permutations: "; count

End Sub</lang>

Sample dialogue:

permute 1
give a number greater than 1!
permute 2
 1  2 
 2  1 
Number of permutations:  2 
permute 4
 1  2  3  4 
 1  2  4  3 
 1  3  2  4 
 1  3  4  2 
 1  4  2  3 
 1  4  3  2 
 2  1  3  4 
 2  1  4  3 
 2  3  1  4 
 2  3  4  1 
 2  4  1  3 
 2  4  3  1 
 3  1  2  4 
 3  1  4  2 
 3  2  1  4 
 3  2  4  1 
 3  4  1  2 
 3  4  2  1 
 4  1  2  3 
 4  1  3  2 
 4  2  1  3 
 4  2  3  1 
 4  3  1  2 
 4  3  2  1 
Number of permutations:  24 
permute 10,False
Number of permutations:  3628800

XPL0

<lang XPL0>code ChOut=8, CrLf=9; def N=4; \number of objects (letters) char S0, S1(N);

proc Permute(D); \Display all permutations of letters in S0 int D; \depth of recursion int I, J; [if D=N then

       [for I:= 0 to N-1 do ChOut(0, S1(I));
       CrLf(0);
       return;
       ];

for I:= 0 to N-1 do

       [for J:= 0 to D-1 do    \check if object (letter) already used
               if S1(J) = S0(I) then J:=100;
       if J<100 then
               [S1(D):= S0(I); \object (letter) not used so append it
               Permute(D+1);   \recurse next level deeper
               ];
       ];

];

[S0:= "rose "; \N different objects (letters) Permute(0); \(space char avoids MSb termination) ]</lang>

Output:

rose
roes
rsoe
rseo
reos
reso
orse
ores
osre
oser
oers
oesr
sroe
sreo
sore
soer
sero
seor
eros
erso
eors
eosr
esro
esor

zkl

Using the solution from task Permutations by swapping#zkl: <lang zkl>zkl: Utils.Helpers.permute("rose").apply("concat") L("rose","roes","reos","eros","erso","reso","rseo","rsoe","sroe","sreo",...)

zkl: Utils.Helpers.permute("rose").len() 24

zkl: Utils.Helpers.permute(T(1,2,3,4)) L(L(1,2,3,4),L(1,2,4,3),L(1,4,2,3),L(4,1,2,3),L(4,1,3,2),L(1,4,3,2),L(1,3,4,2),L(1,3,2,4),...)</lang>