Permutations: Difference between revisions
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=={{header|Kotlin}}==
Translation of C# recursive 'insert' solution in Wikipedia article on Permutations:
<lang scala>// version 1.1.
fun <T> permute(input: List<T>): List<List<T>> {
Line 2,848:
}
}
return perms
}
| |||
Revision as of 22:57, 10 May 2017
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
<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:
1, 3, 2 2, 1, 3 2, 3, 1 3, 1, 2 3, 2, 1
Ada
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
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 (#,
- (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
(Functional ES6 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)
set lst to {}
set lng to length of xs
tell mReturn(f)
repeat with i from 1 to lng
set lst to (lst & lambda(contents of item i of xs, i, xs))
end repeat
end tell
return lst
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
-- delete :: a -> [a] -> [a] on |delete|(x, xs)
if length of xs > 0 then
set {h, t} to uncons(xs)
if x = h then
t
else
{h} & |delete|(x, t)
end if
else
{}
end if
end |delete|
-- 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"}}
(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 (value of Permlist) {{"Good", "Johnny", "Be"}, {"Johnny", "Good", "Be"}, {"Be", "Good", "Johnny"}, ¬
{"Good", "Be", "Johnny"}, {"Johnny", "Be", "Good"}, {"Be", "Johnny", "Good"}}
--> Example 2 (characters with concatenated results) set [SourceList, Permlist] to [{"X", "Y", "Z"}, {}] DoPermutations(SourceList, SourceList's length) --> result (value of Permlist) {"XYZ", "YXZ", "ZXY", "XZY", "YZX", "ZYX"}
--> Example 3 (Integers) set [SourceList, Permlist] to [{1, 2, 3}, {}] DoPermutations(SourceList, SourceList's length) --> result (value of Permlist) {{1, 2, 3}, {2, 1, 3}, {3, 1, 2}, {1, 3, 2}, {2, 3, 1}, {3, 2, 1}}
--> Example 4 (Integers with concatenated results) set [SourceList, Permlist] to [{1, 2, 3}, {}] DoPermutations(SourceList, SourceList's length) --> result (value of Permlist) {"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. Author: I. Gavryushin aka dcc0. Run it from console with a parametr like "program.o 12345" or "program.o abcd": <lang c>
- 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:
251 215 521 512 152 125
Elixir
<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
<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.
<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>
FreeBASIC
<lang freebasic>' version 07-04-2017 ' compile with: fbc -s console
' Heap's algorithm non-recursive Sub perms(n As Long)
Dim As ULong i, j, count = 1 Dim As ULong a(0 To n -1), c(0 To n -1)
For j = 0 To n -1
a(j) = j +1
Print a(j);
Next
Print " ";
i = 0
While i < n
If c(i) < i Then
If (i And 1) = 0 Then
Swap a(0), a(i)
Else
Swap a(c(i)), a(i)
End If
For j = 0 To n -1
Print a(j);
Next
count += 1
If count = 12 Then
Print
count = 0
Else
Print " ";
End If
c(i) += 1
i = 0
Else
c(i) = 0
i += 1
End If
Wend
End Sub
' ------=< MAIN >=------
perms(4)
' empty keyboard buffer While Inkey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
1234 2134 3124 1324 2314 3214 4213 2413 1423 4123 2143 1243 1342 3142 4132 1432 3412 4312 4321 3421 2431 4231 3241 2341
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>
Glee
<lang glee>$$ n !! k dyadic: Permutations for k out of n elements (in this case k = n) $$ #s monadic: number of elements in s $$ ,, monadic: expose with space-lf separators $$ s[n] index n of s
'Hello' 123 7.9 '•'=>s; s[s# !! (s#)],,</lang>
Result: <lang glee>Hello 123 7.9 • Hello 123 • 7.9 Hello 7.9 123 • Hello 7.9 • 123 Hello • 123 7.9 Hello • 7.9 123 123 Hello 7.9 • 123 Hello • 7.9 123 7.9 Hello • 123 7.9 • Hello 123 • Hello 7.9 123 • 7.9 Hello 7.9 Hello 123 • 7.9 Hello • 123 7.9 123 Hello • 7.9 123 • Hello 7.9 • Hello 123 7.9 • 123 Hello • Hello 123 7.9 • Hello 7.9 123 • 123 Hello 7.9 • 123 7.9 Hello • 7.9 Hello 123 • 7.9 123 Hello</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:
1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1
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>
<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></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
(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);
// delete_ :: Eq a => a -> [a] -> [a]
const delete_ = (x, xs) =>
xs.length > 0 ? (
(x === xs[0]) ? (
xs.slice(1)
) : [xs[0]].concat(delete_(x, xs.slice(1)))
) : [];
// range :: Int -> Int -> [Int]
const range = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);
// 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
<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>
Kotlin
Translation of C# recursive 'insert' solution in Wikipedia article on Permutations: <lang scala>// version 1.1.2
fun <T> permute(input: List<T>): List<List<T>> {
if (input.size == 1) return listOf(input)
val perms = mutableListOf<List<T>>()
val toInsert = input[0]
for (perm in permute(input.drop(1))) {
for (i in 0..perm.size) {
val newPerm = perm.toMutableList()
newPerm.add(i, toInsert)
perms.add(newPerm)
}
}
return perms
}
fun main(args: Array<String>) {
val input = listOf('a', 'b', 'c', 'd')
val perms = permute(input)
println("There are ${perms.size} permutations of $input, namely:\n")
for (perm in perms) println(perm)
}</lang>
- Output:
There are 24 permutations of [a, b, c, d], namely: [a, b, c, d] [b, a, c, d] [b, c, a, d] [b, c, d, a] [a, c, b, d] [c, a, b, d] [c, b, a, d] [c, b, d, a] [a, c, d, b] [c, a, d, b] [c, d, a, b] [c, d, b, a] [a, b, d, c] [b, a, d, c] [b, d, a, c] [b, d, c, a] [a, d, b, c] [d, a, b, c] [d, b, a, c] [d, b, c, a] [a, d, c, b] [d, a, c, b] [d, c, a, b] [d, c, b, a]
LFE
<lang lisp> (defun permute
(('())
'(()))
((l)
(lc ((<- x l)
(<- y (permute (-- l `(,x)))))
(cons x y))))
</lang> REPL usage: <lang lisp> > (permute '(1 2 3)) ((1 2 3) (1 3 2) (2 1 3) (2 3 1) (3 1 2) (3 2 1)) </lang>
Liberty BASIC
Permuting numerical array (non-recursive):
<lang lb> n=3 dim a(n+1) '+1 needed due to bug in LB that checks loop condition
' until (i=0) or (a(i)<a(i+1)) 'before executing i=i-1 in loop body.
for i=1 to n: a(i)=i: next do
for i=1 to n: print a(i);: next: print
i=n
do
i=i-1
loop until (i=0) or (a(i)<a(i+1))
j=i+1
k=n
while j<k
'swap a(j),a(k)
tmp=a(j): a(j)=a(k): a(k)=tmp
j=j+1
k=k-1
wend
if i>0 then
j=i+1
while a(j)<a(i)
j=j+1
wend
'swap a(i),a(j)
tmp=a(j): a(j)=a(i): a(i)=tmp
end if
loop until i=0 </lang>
- Output:
123 132 213 231 312 321
Permuting string (recursive): <lang lb> n = 3
s$="" 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
</lang>
- Output:
123 132 213 231 312 321
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}
<lang lua>
-- 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>
1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1
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:
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
<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
<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
<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. Author: I. Gavryushin aka dcc0.
<lang php>
<?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 $a. "\n";
} ?> </lang> Output
3210 2310 3120 1320 2130 1230 3201 2301 3021 0321 2031 0231 3102 1302 3012 0312 1032 0132 2103 1203 2013 0213 1023 0123
Perl
There are many modules that can do permutations, or it can be fairly easily done by hand with an example below. In performance order for simple permutation of 10 scalars, a sampling of some solutions:
- 1.7s Algorithm::FastPermute permute iterator - 1.7s Algorithm::Permute permute iterator - 2.0s ntheory forperm iterator - 6.3s Algorithm::Combinatorics permutations iterator - 9.1s the recursive sub below - 21.1s Math::Combinatorics permutations iterator
Example:
<lang perl>use ntheory qw/forperm/; my @tasks = (qw/party sleep study/); forperm {
print "@tasks[@_]\n";
} scalar(@tasks);</lang>
- Output:
party sleep study party study sleep sleep party study sleep study party study party sleep study sleep party
A simple recursive routine: <lang perl>sub permutation { my ($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
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:
0 1 2 0 2 1 1 0 2 1 2 0 2 0 1 2 1 0
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
<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$(s, d) ' s & d concatenate
NEXT
? s 'print and pause
i = n
DO
DECR i
LOOP UNTIL i = 0 OR a(i) < a(i+1)
j = i+1
k = n
DO WHILE j < k
SWAP a(j), a(k)
INCR j
DECR k
LOOP
IF i > 0 THEN
j = i+1
DO WHILE a(j) < a(i)
INCR j
LOOP
SWAP a(i), a(j)
END IF
LOOP UNTIL i = 0
END FUNCTION</lang>
- Output:
1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1
PowerShell
<lang PowerShell> function permutation ($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
<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
<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>
Here is another recursive version.
<lang r># list of the vectors by inserting x in s at position 0...end. linsert <- function(x,s) lapply(0:length(s), function(k) append(s,x,k))
- list of all permutations of 1:n
perm <- function(n){
if (n == 1) list(1)
else unlist(lapply(perm(n-1), function(s) linsert(n,s)),
recursive = F)}
- permutations of a vector s
permutation <- function(s) unique(lapply(perm(length(s)), function(i) s[i])) </lang>
Output: <lang r>> permutation(letters[1:3]) 1 [1] "c" "b" "a"
2 [1] "b" "c" "a"
3 [1] "b" "a" "c"
4 [1] "c" "a" "b"
5 [1] "a" "c" "b"
6 [1] "a" "b" "c"</lang>
Racket
<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
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
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:
1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1
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)
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
</lang> 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
<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
Rust
Iterative
Using the QuickPerm algorithm. This solution assumes you want seperate copies of each permutation. An in-place version is possible, but incompatible with the Iterator trait.
<lang rust>struct QuickPerm<T> {
idxs: Vec<usize>, elems: Vec<T>, idx: usize,
}
impl<T: Clone> QuickPerm<T> {
fn new(elems: Vec<T>) -> Self {
QuickPerm { idxs: (0..elems.len()+1).collect(), elems: elems, idx: 1 }
}
}
impl<T: Clone> Iterator for QuickPerm<T> {
type Item = Vec<T>;
fn next(&mut self) -> Option<Self::Item> {
if self.idx == self.elems.len() { return None; }
self.idxs[self.idx] -= 1;
let other = if self.idx % 2 == 1 { self.idxs[self.idx] } else { 0 };
self.elems.swap(self.idx, other);
self.idx = 1;
while self.idxs[self.idx] == 0 {
self.idxs[self.idx] = self.idx;
self.idx += 1;
}
Some(self.elems.clone())
}
}
fn main() {
for perm in QuickPerm::new(vec![1,2,3]) {
println!("{:?}", perm);
}
}</lang>
Recursive
<lang rust>use std::collections::VecDeque; fn permute<T, F: Fn(&[T])>(used: &mut Vec<T>, unused: &mut VecDeque<T>, action: &F) {
if unused.is_empty() {
action(used);
} else {
for _ in 0..unused.len() {
used.push(unused.pop_front().unwrap());
permute(used, unused, action);
unused.push_back(used.pop().unwrap());
}
}
}
// Same as the vec! macro, but for VecDeques as well macro_rules! vec_deque {
($($item:expr),*) => {{
let mut deque = ::std::collections::VecDeque::new();
$(deque.push_back($item);)*
deque
}}
}
fn main() {
permute(&mut Vec::new(), &mut vec_deque![1,2,3], &|perm| println!("{:?}", perm));
}</lang>
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
<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>
<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:
1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1
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
<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
<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:
1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1
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
<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
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</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>
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