Permutations

From Rosetta Code
Task
Permutations
You are encouraged to solve this task according to the task description, using any language you may know.
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
Task: Combinations Task: Permutations
With replacement
Task: Combinations with repetitions Task: Permutations with repetitions



360 Assembly[edit]

Translation of: Liberty BASIC
*        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 [email protected]
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) [email protected](j)
LA R3,A-1(R8) [email protected](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) [email protected](j)
LA R3,A-1(R6) [email protected](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
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[edit]

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.
Output:
1,  3,  2

2,  1,  3

2,  3,  1

3,  1,  2

3,  2,  1

Ada[edit]

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[edit]

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:

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;

Here is the implementation of the package:

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;

The procedure Print_Perms[edit]

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;
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[edit]

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
# -*- 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
File: test_permutations.a68
#!/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 (#,
## (PERMDATALIST permutation)VOID:(
printf((data list fmt, permutation, $l$))
# OD #))
 
)
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[edit]

Translation of: JavaScript

(Functional ES5 version)

on run
 
permutations({"aardvarks", "eat", "ants"})
 
end run
 
-- 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
 
 
 
-- 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)
set mf to mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to mf's lambda(item i of xs, i, xs)
end repeat
return lst
end map
 
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
set mf to mReturn(f)
 
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to mf's lambda(v, item i of xs, i, xs)
end repeat
return v
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)
set lng to length of xs
if lng > 0 then
if lng > 1 then
{item 1 of xs, items 2 thru -1 of xs}
else
{item 1 of xs, {}}
end if
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
 
 
Output:
{{"aardvarks", "eat", "ants"}, {"aardvarks", "ants", "eat"}, 
{"eat", "aardvarks", "ants"}, {"eat", "ants", "aardvarks"}, 
{"ants", "aardvarks", "eat"}, {"ants", "eat", "aardvarks"}}

AutoHotkey[edit]

from the forum topic http://www.autohotkey.com/forum/viewtopic.php?t=77959

#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
}
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[edit]

Alternate version to produce numerical permutations of combinations.

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
}
Output:
MsgBox % P(3)
---------------------------
permute.ahk
---------------------------
123
132
213
231
312
321
---------------------------
OK   
---------------------------
MsgBox % P("Hello",3)
---------------------------
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   
---------------------------
MsgBox % P("2`n3`n4`n5",2,3)
---------------------------
permute.ahk
---------------------------

2
22
23
24
25
3
32
33
34
35
4
42
43
44
45
5
52
53
54
55
---------------------------
OK   
---------------------------
MsgBox % P("11 a text ] u+z",3,0," ")
---------------------------
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[edit]

Recursive permutation generator.

 
@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
 
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[edit]

The procedure PROC_NextPermutation() will give the next lexicographic permutation of an integer array.

      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

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[edit]

  ( 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);

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[edit]

See lexicographic generation of permutations.

#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;
}

C++[edit]

The C++ standard library provides for this in the form of std::next_permutation and std::prev_permutation.

#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;
}
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#[edit]

A recursive Iterator. Runs under C#2 (VS2005), i.e. no `var`, no lambdas,...

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];
}
);
}
}
}
}
}

Usage:

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));
}
}
}
}

Clojure[edit]

Library function[edit]

In an REPL:

 
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))

Explicit[edit]

Replacing the call to the combinatorics library function by its real implementation.

 
(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]))
 
 

CoffeeScript[edit]

# 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...

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:
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

Common Lisp[edit]

(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)))
Output:
((A B Z) (A Z B) (B A Z) (B Z A) (Z A B) (Z B A))

Lexicographic next permutation:

(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))

D[edit]

Simple Eager version[edit]

Compile with -version=permutations1_main to see the output.

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);
}
}
Output:
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]

Fast Lazy Version[edit]

Compiled with -version=permutations2_main produces its output.

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;
}
}

Standard Version[edit]

void main() {
import std.stdio, std.algorithm;
 
auto items = [1, 2, 3];
do
items.writeln;
while (items.nextPermutation);
}

Delphi[edit]

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.
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[edit]

 
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
 
 
Output:
251
215
521
512
152
125

Elixir[edit]

Translation of: Erlang
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])
Output:
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

Erlang[edit]

Shortest form:

-module(permute).
-export([permute/1]).
 
permute([]) -> [[]];
permute(L) -> [[X|Y] || X<-L, Y<-permute(L--[X])].

Y-combinator (for shell):

F = fun(L) -> G = fun(_, []) -> [[]]; (F, L) -> [[X|Y] || X<-L, Y<-F(F, L--[X])] end, G(G, L) end.

More efficient zipper implementation:

-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]).

Demonstration (escript):

main(_) -> io:fwrite("~p~n", [permute:permute([1,2,3])]).
Output:
[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

Euphoria[edit]

Translation of: PureBasic
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
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#[edit]

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

Factor[edit]

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

Fortran[edit]

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
Output:
           1           2           3
           1           3           2
           2           1           3
           2           3           1
           3           1           2
           3           2           1

Alternate solution[edit]

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.

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

Fortran 77[edit]

Here is an alternate, iterative version in Fortran 77.

Translation of: Ada
      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

GAP[edit]

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.

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 ] ]

GAP has also built-in functions to get permutations

# All arrangements of 4 elements in 1 .. 4
Arrangements([1 .. 4], 4);
# All permutations of 1 .. 4
PermutationsList([1 .. 4]);

Here is an implementation using a function to compute next permutation in lexicographic order:

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 ] ]

Go[edit]

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))
}
Output:
[1 2 3]
[1 3 2]
[3 1 2]
[2 1 3]
[2 3 1]
[3 2 1]

Groovy[edit]

Solution:

def makePermutations = { l -> l.permutations() }

Test:

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 }
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[edit]

import Data.List (permutations)
 
main = mapM_ print (permutations [1,2,3])

A simple implementation, that assumes elements are unique and support equality:

import Data.List (delete)
 
permutations :: Eq a => [a] -> [[a]]
permutations [] = [[]]
permutations xs = [ x:ys | x <- xs, ys <- permutations (delete x xs)]

A slightly more efficient implementation that doesn't have the above restrictions:

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 ]

The above are all selection-based approaches. The following is an insertion-based approach:

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)

Icon and Unicon[edit]

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
Output:
->permute Aardvarks eat ants      
Aardvarks eat ants 
Aardvarks ants eat 
eat Aardvarks ants 
eat ants Aardvarks 
ants eat Aardvarks 
ants Aardvarks eat 
->

J[edit]

perms=: A.&i.~ !
Example use:
   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

Java[edit]

Using the code of Michael Gilleland.

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
Output:
1 2 3 
1 3 2 
2 1 3 
2 3 1 
3 1 2 
3 2 1 

optimized

Following needs: Utils.java

public class Permutations {
public static void main(String[] args) {
System.out.println(Utils.Permutations(Utils.mRange(1, 3)));
}
}
Output:
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

JavaScript[edit]

Imperative[edit]

Copy the following as an HTML file and load in a browser.

<html><head><title>Permutations</title></head>
<body><pre id="result"></pre>
<script type="text/javascript">
var d = document.getElementById('result');
 
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>

Functional[edit]

ES5[edit]

Translation of: Haskell

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

(function () {
 
// [a] -> [[a]]
function permutations(xs) {
return xs.length ? (concatMap(
function (x) {
return concatMap(
function (ys) {
return ([[x].concat(ys)]);
}, permutations(delete1(x, xs)))
}, xs)) : [[]]
}
 
 
// GENERIC LIBRARY FUNCTIONS
 
// concatMap :: (a -> [b]) -> [a] -> [b]
function concatMap(f, xs) {
return [].concat.apply([], xs.map(f));
}
 
// delete first instance of a in [a]
// delete1 :: a -> [a] -> [a]
function delete1(x, xs) {
return deleteBy(function (a, b) {
return a === b;
}, x, xs);
}
 
// deleteBy :: (a -> a -> Bool) -> a -> [a] -> [a]
function deleteBy(fnEq, x, xs) {
return xs.length ? fnEq(x, xs[0]) ? xs.slice(1) : [xs[0]]
.concat(deleteBy(fnEq, x, xs.slice(1))) : [];
}
 
 
return permutations(['Aardvarks', 'eat', 'ants'])
 
})();
Output:
[["Aardvarks", "eat", "ants"], ["Aardvarks", "ants", "eat"],
["eat", "Aardvarks", "ants"], ["eat", "ants", "Aardvarks"],
["ants", "Aardvarks", "eat"], ["ants", "eat", "Aardvarks"]]


ES6[edit]

(function (lst) {
'use strict';
 
let permutations = (xs) => xs.length ? (
flatMap((x) => flatMap((xs) => [[x].concat(xs)],
permutations(del(x, xs))), xs)
) : [[]],
 
flatMap = (f, xs) => [].concat.apply([], xs.map(f)),
 
del = (x, xs) => xs.length ? x === xs[0] ? (
xs.slice(1)
) : [xs[0]].concat(del(x, xs.slice(1))
) : [];
 
 
return permutations(lst);
 
})(["Aardvarks", "eat", "ants"]);


Output:
[["Aardvarks", "eat", "ants"], ["Aardvarks", "ants", "eat"],
["eat", "Aardvarks", "ants"], ["eat", "ants", "Aardvarks"],
["ants", "Aardvarks", "eat"], ["ants", "eat", "Aardvarks"]]

jq[edit]

"permutations" generates a stream of the permutations of the input array.

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

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

Example 3: 10!

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

Julia[edit]

Julia has native support for permutation creation and processing. permutations(v) creates an iterator over all permutations of v.

 
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()
 
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[edit]

Translation of: J
   perm:{:[1<x;,/(>:'(x,x)#1,x#0)[;0,'1+_f x-1];,!x]}
perm 2
(0 1
1 0)
 
`0:{1_,/" ",/:[email protected]@#r:("some";"random";"text")
some random text
some text random
random some text
random text some
text some random
text random some

Alternative:

 
perm:[email protected]@&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
 

LFE[edit]

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

REPL usage:

 
> (permute '(1 2 3))
((1 2 3) (1 3 2) (2 1 3) (2 3 1) (3 1 2) (3 2 1))
 

Liberty BASIC[edit]

Permuting numerical array (non-recursive):

Translation of: PowerBASIC
 
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
 
Output:
123
132
213
231
312
321

Permuting string (recursive):

 
n = 3
 
s$=""
for i = 1 to n
s$=s$;i
next
 
res$=permutation$("", s$)
 
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
 
 
Output:
123
132
213
231
312
321

Logtalk[edit]

:- 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.
Usage example:
| ?- 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

Lua[edit]

 
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)
 
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)
 
1 2 3 
1 3 2 
2 1 3 
2 3 1 
3 1 2 
3 2 1

Maple[edit]

 
> 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]]
 

Mathematica[edit]

Permutations[{1,2,3,4}]
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[edit]

perms([1,2,3,4])
Output:
4321
4312
4231
4213
4123
4132
3421
3412
3241
3214
3124
3142
2341
2314
2431
2413
2143
2134
1324
1342
1234
1243
1423
1432

Maxima[edit]

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] */

Builtin version[edit]

 
(%i1) permutations([1, 2, 3]);
(%o1) {[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]}
 

Modula-2[edit]

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.

NetRexx[edit]

/* NetRexx */
options replace format comments java crossref symbols nobinary
 
import java.util.List
import java.util.ArrayList
 
-- =============================================================================
/**
* Permutation Iterator
* <br />
* <br />
* 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: [email protected] 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
 
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[edit]

Translation of: C
# 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

Output:

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

OCaml[edit]

(* 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 *)

Permutations can also be defined on lists recursively:

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])

or permutations indexed independently:

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]

PARI/GP[edit]

vector(n!,k,numtoperm(n,k))

Pascal[edit]

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.

alternative[edit]

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.

{$IFDEF FPC}
{$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.
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

Perl[edit]

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
use ntheory qw/forperm/;
my @tasks = (qw/party sleep study/);
forperm {
print "@tasks[@_]\n";
} scalar(@tasks);
Output:
party sleep study
party study sleep
sleep party study
sleep study party
study party sleep
study sleep party

A simple recursive routine:

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);
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[edit]

Works with: rakudo version 2014-1-24

First, you can just use the built-in method on any list type.

.say for <a b c>.permutations
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 >.

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 ...^ !*;
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.

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' )
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:

.say for permutations(3);
Output:
0 1 2
0 2 1
1 0 2
1 2 0
2 0 1
2 1 0

Phix[edit]

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.

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

Example use:

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")
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[edit]

(load "@lib/simul.l")
 
(permute (1 2 3))
Output:
-> ((1 2 3) (1 3 2) (2 1 3) (2 3 1) (3 1 2) (3 2 1))

PowerBASIC[edit]

Works with: PowerBASIC version 10.00+
  #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
Output:
 1 2 3
 1 3 2
 2 1 3
 2 3 1
 3 1 2
 3 2 1

PowerShell[edit]

 
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')
 

Output:

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

Prolog[edit]

Works with SWI-Prolog and library clpfd,

:- use_module(library(clpfd)).
 
permut_clpfd(L, N) :-
length(L, N),
L ins 1..N,
all_different(L),
label(L).
Output:
?- 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.
 

A declarative way of fetching permutations:

% 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).
Output:
 ?- 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.

PureBasic[edit]

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.

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
Output:
1, 2, 3
1, 3, 2
2, 1, 3
2, 3, 1
3, 1, 2
3, 2, 1

Python[edit]

Standard library function[edit]

Works with: Python version 2.6+
import itertools
for values in itertools.permutations([1,2,3]):
print (values)
Output:
(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)

Recursive implementation[edit]

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.

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)

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:

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)

Iterative implementation[edit]

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.

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)

Qi[edit]

Translation of: Erlang
 
(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))))

R[edit]

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

Racket[edit]

 
#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))
 

REXX[edit]

using names[edit]

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.

/*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'; @[email protected]; 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

output   when the following was used for input:   3   3

123
132
213
231
312
321

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[edit]

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.

/*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

output   when using the default input:

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

Ring[edit]

 
list = [1, 2, 3, 4]
for perm = 1 to 24
for i = 1 to len(list)
see list[i] + " "
next
see nl
nextPermutation(list)
next
 
func nextPermutation a
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
 

Output:

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

Ruby[edit]

Works with: Ruby version 1.8.7+
p [1,2,3].permutation.to_a
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.

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

Run BASIC[edit]

Works with Run BASIC, Liberty BASIC and Just BASIC

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

SAS[edit]

/* 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;

Scala[edit]

There is a built-in function that works on any sequential collection. It could be used as follows given a List of symbols:

List('a, 'b, 'c).permutations foreach println
Output:
List('a, 'b, 'c)
List('a, 'c, 'b)
List('b, 'a, 'c)
List('b, 'c, 'a)
List('c, 'a, 'b)
List('c, 'b, 'a)

The following function returns all the unique permutation of a list:

def permutations[T](list: List[T]):List[List[T]] = {
    list match {
                case Nil => Nil
                case elem :: Nil => List(list)
                case head :: tail => list.distinct.foldLeft(List[List[T]]()) ((lst, elem)=> lst ++ ((permutations(list.diff(List(elem)))).map((l)=> (elem :: l))))
		}

Scheme[edit]

Translation of: Erlang
(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))))))
Translation of: OCaml
; 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

Completely recursive on lists:

(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)))

Seed7[edit]

$ 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;
Output:
1 2 3 
1 3 2 
2 1 3 
2 3 1 
3 1 2 
3 2 1 

Sidef[edit]

Built-in[edit]

[0,1,2].permutations { |p|
say p
}

Iterative[edit]

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)

Recursive[edit]

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])
Output:
[0, 1, 2]
[0, 2, 1]
[1, 0, 2]
[1, 2, 0]
[2, 0, 1]
[2, 1, 0]

Smalltalk[edit]

Works with: Squeak
Works with: Pharo
(1 to: 4) permutationsDo: [ :x | 
Transcript show: x printString; cr ].

Swift[edit]

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]]

Tcl[edit]

Library: Tcllib (Package: struct::list)
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
}

Testing with tclsh listPerms.tcl 3 produces this output:

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

Ursala[edit]

In practice there's no need to write this because it's in the standard library.

#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

test program:

#cast %nLL
 
test = permutations <1,2,3>
Output:
<
   <1,2,3>,
   <2,1,3>,
   <2,3,1>,
   <1,3,2>,
   <3,1,2>,
   <3,2,1>>

VBA[edit]

Translation of: Pascal
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
Next
 
count = 0
Last = False
 
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
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[edit]

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)
]

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[edit]

Using the solution from task Permutations by swapping#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),...)