Permutations: Difference between revisions
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end lexicalPermutations2 |
end lexicalPermutations2 |
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def alpha: ['ABCD'...]; |
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[ $alpha::length -> lexicalPermutations2 -> '$alpha($)...;' ] -> !OUT::write |
[ $alpha::length -> lexicalPermutations2 -> '$alpha($)...;' ] -> !OUT::write |
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</lang> |
</lang> |
||
Revision as of 07:41, 9 July 2022
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
11l
<lang 11l>V a = [1, 2, 3] L
print(a)
I !a.next_permutation()
L.break</lang>
- Output:
[1, 2, 3] [1, 3, 2] [2, 1, 3] [2, 3, 1] [3, 1, 2] [3, 2, 1]
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
AArch64 Assembly
<lang AArch64 Assembly> /* ARM assembly AARCH64 Raspberry PI 3B */ /* program permutation64.s */
/*******************************************/ /* Constantes file */ /*******************************************/ /* for this file see task include a file in language AArch64 assembly */ .include "../includeConstantesARM64.inc"
/*********************************/ /* Initialized data */ /*********************************/ .data
sMessResult: .asciz "Value : @\n" sMessCounter: .asciz "Permutations = @ \n" szCarriageReturn: .asciz "\n"
.align 4 TableNumber: .quad 1,2,3
.equ NBELEMENTS, (. - TableNumber) / 8
/*********************************/ /* UnInitialized data */ /*********************************/ .bss sZoneConv: .skip 24 /*********************************/ /* code section */ /*********************************/ .text .global main main: //entry of program
ldr x0,qAdrTableNumber //address number table mov x1,NBELEMENTS //number of élements mov x10,0 //counter bl heapIteratif mov x0,x10 //display counter ldr x1,qAdrsZoneConv // bl conversion10S //décimal conversion ldr x0,qAdrsMessCounter ldr x1,qAdrsZoneConv //insert conversion bl strInsertAtCharInc bl affichageMess //display message
100: //standard end of the program
mov x0,0 //return code mov x8,EXIT //request to exit program svc 0 //perform the system call
qAdrszCarriageReturn: .quad szCarriageReturn qAdrsMessResult: .quad sMessResult qAdrTableNumber: .quad TableNumber qAdrsMessCounter: .quad sMessCounter /******************************************************************/ /* permutation by heap iteratif (wikipedia) */ /******************************************************************/ /* x0 contains the address of table */ /* x1 contains the eléments number */ heapIteratif:
stp x2,lr,[sp,-16]! // save registers stp x3,x4,[sp,-16]! // save registers stp x5,x6,[sp,-16]! // save registers stp x7,fp,[sp,-16]! // save registers tst x1,1 // odd ? add x2,x1,1 csel x2,x2,x1,ne // the stack must be a multiple of 16 lsl x7,x2,3 // 8 bytes by count sub sp,sp,x7 mov fp,sp mov x3,#0 mov x4,#0 // index
1: // init area counter
str x4,[fp,x3,lsl 3] add x3,x3,#1 cmp x3,x1 blt 1b bl displayTable add x10,x10,#1 mov x3,#0 // index
2:
ldr x4,[fp,x3,lsl 3] // load count [i] cmp x4,x3 // compare with i bge 5f tst x3,#1 // even ? bne 3f ldr x5,[x0] // yes load value A[0] ldr x6,[x0,x3,lsl 3] // and swap with value A[i] str x6,[x0] str x5,[x0,x3,lsl 3] b 4f
3:
ldr x5,[x0,x4,lsl 3] // load value A[count[i]] ldr x6,[x0,x3,lsl 3] // and swap with value A[i] str x6,[x0,x4,lsl 3] str x5,[x0,x3,lsl 3]
4:
bl displayTable add x10,x10,1 add x4,x4,1 // increment count i str x4,[fp,x3,lsl 3] // and store on stack mov x3,0 // raz index b 2b // and loop
5:
mov x4,0 // raz count [i] str x4,[fp,x3,lsl 3] add x3,x3,1 // increment index cmp x3,x1 // end ? blt 2b // no -> loop add sp,sp,x7 // stack alignement
100:
ldp x7,fp,[sp],16 // restaur 2 registers ldp x5,x6,[sp],16 // restaur 2 registers ldp x3,x4,[sp],16 // restaur 2 registers ldp x2,lr,[sp],16 // restaur 2 registers ret // return to address lr x30
/******************************************************************/ /* Display table elements */ /******************************************************************/ /* x0 contains the address of table */ displayTable:
stp x1,lr,[sp,-16]! // save registers stp x2,x3,[sp,-16]! // save registers mov x2,x0 // table address mov x3,#0
1: // loop display table
ldr x0,[x2,x3,lsl 3] ldr x1,qAdrsZoneConv bl conversion10S // décimal conversion ldr x0,qAdrsMessResult ldr x1,qAdrsZoneConv // insert conversion bl strInsertAtCharInc bl affichageMess // display message add x3,x3,1 cmp x3,NBELEMENTS - 1 ble 1b ldr x0,qAdrszCarriageReturn bl affichageMess mov x0,x2
100:
ldp x2,x3,[sp],16 // restaur 2 registers ldp x1,lr,[sp],16 // restaur 2 registers ret // return to address lr x30
qAdrsZoneConv: .quad sZoneConv /********************************************************/ /* File Include fonctions */ /********************************************************/ /* for this file see task include a file in language AArch64 assembly */ .include "../includeARM64.inc"
</lang>
Value : +1 Value : +2 Value : +3 Value : +2 Value : +1 Value : +3 Value : +3 Value : +1 Value : +2 Value : +1 Value : +3 Value : +2 Value : +2 Value : +3 Value : +1 Value : +3 Value : +2 Value : +1 Permutations = +6
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
Action!
<lang Action!>PROC PrintArray(BYTE ARRAY a BYTE len)
BYTE i
FOR i=0 TO len-1
DO
PrintB(a(i))
OD
Print(" ")
RETURN
BYTE FUNC NextPermutation(BYTE ARRAY a BYTE len)
BYTE i,j,k,tmp
i=len-1 WHILE i>0 AND a(i-1)>a(i) DO i==-1 OD j=i k=len-1 WHILE j<k DO tmp=a(j) a(j)=a(k) a(k)=tmp j==+1 k==-1 OD IF i=0 THEN RETURN (0) FI
j=i WHILE a(j)<a(i-1) DO j==+1 OD tmp=a(i-1) a(i-1)=a(j) a(j)=tmp
RETURN (1)
PROC Main()
DEFINE len="5" BYTE ARRAY a(len) BYTE RMARGIN=$53,oldRMARGIN BYTE i
oldRMARGIN=RMARGIN RMARGIN=37 ;change right margin on the screen
FOR i=0 TO len-1 DO a(i)=i OD
DO PrintArray(a,len) UNTIL NextPermutation(a,len)=0 OD
RMARGIN=oldRMARGIN ;restore right margin on the screen
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
01234 01243 01324 01342 01423 01432 02134 02143 02314 02341 02413 02431 03124 03142 03214 03241 03412 03421 04123 04132 04213 04231 04312 04321 10234 10243 10324 10342 10423 10432 12034 12043 12304 12340 12403 12430 13024 13042 13204 13240 13402 13420 14023 14032 14203 14230 14302 14320 20134 20143 20314 20341 20413 20431 21034 21043 21304 21340 21403 21430 23014 23041 23104 23140 23401 23410 24013 24031 24103 24130 24301 24310 30124 30142 30214 30241 30412 30421 31024 31042 31204 31240 31402 31420 32014 32041 32104 32140 32401 32410 34012 34021 34102 34120 34201 34210 40123 40132 40213 40231 40312 40321 41023 41032 41203 41230 41302 41320 42013 42031 42103 42130 42301 42310 43012 43021 43102 43120 43201 43210
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
Aime
<lang aime>void f1(record r, ...) {
if (~r) {
for (text s in r) {
r.delete(s);
rcall(f1, -2, 0, -1, s);
r[s] = 0;
}
} else {
ocall(o_, -2, 1, -1, " ", ",");
o_newline();
}
}
main(...) {
record r;
ocall(r_put, -2, 1, -1, r, 0); f1(r);
0;
}</lang>
- Output:
aime permutations -a Aaa Bb C Aaa, Bb, C, Aaa, C, Bb, Bb, Aaa, C, Bb, C, Aaa, C, Aaa, Bb, C, Bb, Aaa,
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)
Amazing Hopper
<lang Amazing Hopper> /* hopper-JAMBO - a flavour of Amazing Hopper! */
- include <jambo.h>
Main
leng=0
Void(lista)
Set("la realidad","escapa","a los sentidos"), Apnd list(lista)
Length(lista), Move to(leng)
Toksep(" ")
Printnl( lista )
Set(1) Gosub(Permutar)
End-Return
Subrutines
Define( Permutar, pos )
If ( Sub(leng, pos) Isgeq(1) )
i=pos
Loop if( Less( i, leng ) )
Plusone(pos), Gosub(Permutar)
Set( pos ), Gosub(Rotate)
Printnl( lista )
++i
Back
Plusone(pos), Gosub(Permutar)
Set( pos ), Gosub(Rotate)
End If
Return
Define ( Rotate, pos )
c=0, [pos] Get(lista), Move to(c) [ Plusone(pos): leng ] Cget(lista) [ pos: Minusone(leng) ] Cput(lista) Set(c), [ leng ] Cput(lista)
Return </lang>
- Output:
la realidad escapa a los sentidos la realidad a los sentidos escapa escapa a los sentidos la realidad escapa la realidad a los sentidos a los sentidos la realidad escapa a los sentidos escapa la realidad
APL
For Dyalog APL(assumes index origin ⎕IO←1): <lang APL> ⍝ Builtin version, takes a vector: ⎕CY'dfns' perms←{↓⍵[pmat ≢⍵]} ⍝ pmat always gives lexicographically ordered permutations.
⍝ Recursive fast implementation, courtesy of dzaima from The APL Orchard: dpmat←{1=⍵:,⊂,0 ⋄ (⊃,/)¨(⍳⍵)⌽¨⊂(⊂(!⍵-1)⍴⍵-1),⍨∇⍵-1} perms2←{↓⍵[1+⍉↑dpmat ≢⍵]} </lang>
perms 'cat'
┌───┬───┬───┬───┬───┬───┐
│cat│cta│act│atc│tca│tac│
└───┴───┴───┴───┴───┴───┘
perms2 'cat'
┌───┬───┬───┬───┬───┬───┐
│cta│atc│tac│tca│act│cat│
└───┴───┴───┴───┴───┴───┘
AppleScript
Recursive
(Functional ES6 version)
Recursively, in terms of concatMap and delete: <lang AppleScript>----------------------- PERMUTATIONS -----------------------
-- permutations :: [a] -> a on permutations(xs)
script go
on |λ|(xs)
script h
on |λ|(x)
script ts
on |λ|(ys)
{{x} & ys}
end |λ|
end script
concatMap(ts, go's |λ|(|delete|(x, xs)))
end |λ|
end script
if {} ≠ xs then
concatMap(h, xs)
else
{{}}
end if
end |λ|
end script
go's |λ|(xs)
end permutations
TEST ---------------------------
on run
permutations({"aardvarks", "eat", "ants"})
end run
GENERIC 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 & |λ|(contents of item i of xs, i, xs))
end repeat
end tell
return lst
end concatMap
-- 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|
-- 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 |λ| : f
end script
end if
end mReturn
-- 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</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)
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
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>
Non-recursive
As a right fold (which turns out to be significantly faster than recurse + delete): <lang applescript>----------------------- PERMUTATIONS -----------------------
-- permutations :: [a] -> a on permutations(xs)
script go
on |λ|(x, a)
script
on |λ|(ys)
script infix
on |λ|(n)
if ys ≠ {} then
take(n, ys) & {x} & drop(n, ys)
else
{x}
end if
end |λ|
end script
map(infix, enumFromTo(0, (length of ys)))
end |λ|
end script
concatMap(result, a)
end |λ|
end script
foldr(go, {{}}, xs)
end permutations
TEST ---------------------------
on run
permutations({1, 2, 3})
--> {{1, 2, 3}, {2, 1, 3}, {2, 3, 1}, {1, 3, 2}, {3, 1, 2}, {3, 2, 1}}
end run
GENERIC --------------------------
-- concatMap :: (a -> [b]) -> [a] -> [b] on concatMap(f, xs)
set lng to length of xs
set acc to {}
tell mReturn(f)
repeat with i from 1 to lng
set acc to acc & |λ|(item i of xs, i, xs)
end repeat
end tell
return acc
end concatMap
-- drop :: Int -> [a] -> [a]
on drop(n, xs)
if n < length of xs then
items (1 + n) thru -1 of xs
else
{}
end if
end drop
-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if m ≤ n then
set lst to {}
repeat with i from m to n
set end of lst to i
end repeat
return lst
else
return {}
end if
end enumFromTo
-- foldr :: (a -> b -> b) -> b -> [a] -> b
on foldr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from lng to 1 by -1
set v to |λ|(item i of xs, v, i, xs)
end repeat
return v
end tell
end foldr
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- 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 |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- min :: Ord a => a -> a -> a
on min(x, y)
if y < x then
y
else
x
end if
end min
-- take :: Int -> [a] -> [a]
-- take :: Int -> String -> String
on take(n, xs)
if 0 < n then
items 1 thru min(n, length of xs) of xs
else
{}
end if
end take</lang>
- Output:
{{1, 2, 3}, {2, 1, 3}, {2, 3, 1}, {1, 3, 2}, {3, 1, 2}, {3, 2, 1}}
Recursive again
This is marginally faster even than the Pseudocode translation above and doesn't demarcate lists with square brackets, which don't officially exist in AppleScript. It now features tail call elimination and a rethink of way the results list is built which, on my machine, reduces the time taken to return the 362,880 permutations of a 9-item list from a minute to a second and a half. It'll even return the 3,628,800 permutations of a 10-item list without seizing up, in about 17 seconds. But make sure Script Editor doesn't attempt to display such large results or you'll need to force-quit it!
<lang applescript>-- AppleScript interpretation of "Improved version of Heap's method (recursive)" -- found in Robert Sedgewick's PDF document "Permutation Generation Methods" -- <https://www.cs.princeton.edu/~rs/talks/perms.pdf> -- Adapted to permute from right to left instead of vice versa and to eliminate tail calls.
on allPermutations(theList)
script o
property workList : missing value
property permutations : {}
property r : (count theList) -- The work list's rightmost …
property |r-1| : r - 1 -- … penultimate …
property |r-2| : r - 2 -- … and penpenultimate indices.
property p : 1 -- Index for the permutations list.
-- Recursive handler. Stores copies of workList after permuting items l thru r.
on prmt(l)
set evenCount to ((r - l) mod 2 = 1) -- Permuting an even number of the work list's items?
-- Tail call elimination repeat. Stops with the three rightmost items still to be permuted.
repeat until (l = |r-2|)
-- Recursively permute items (l + 1) thru r.
set |l+1| to l + 1
prmt(|l+1|)
-- And again after successive swaps of item l with others to its right.
if (evenCount) then
-- Permuting an even number of items. Swap item l with items from slots r to l + 1 in turn.
repeat with swapIndex from r to |l+1| + 1 by -1
tell item l of my workList
set item l of my workList to item swapIndex of my workList
set item swapIndex of my workList to it
end tell
prmt(|l+1|)
end repeat
set swapIndex to |l+1|
else
-- Permuting an odd number of items. Always swap item l with item r.
repeat (r - |l+1|) times
tell item l of my workList
set item l of my workList to item r of my workList
set item r of my workList to it
end tell
prmt(|l+1|)
end repeat
set swapIndex to r
end if
-- Do the last swap with the current l, then reset to repeat in lieu of a tail recursion.
tell item l of my workList
set item l of my workList to item swapIndex of my workList
set item swapIndex of my workList to it
end tell
set l to |l+1|
set evenCount to (not evenCount)
end repeat
-- Store the current state of the work list.
copy workList to item p of my permutations
-- And five more times with permutations of the rightmost three items.
-- (Written out here to save doing five more recursion branches.)
set {v1, v2, v3} to items |r-2| thru r of my workList
set item |r-1| of my workList to v3
set item r of my workList to v2
copy workList to item (p + 1) of my permutations
set item |r-2| of my workList to v2
set item r of my workList to v1
copy workList to item (p + 2) of my permutations
set item |r-1| of my workList to v1
set item r of my workList to v3
copy workList to item (p + 3) of my permutations
set item |r-2| of my workList to v3
set item r of my workList to v2
copy workList to item (p + 4) of my permutations
set item |r-1| of my workList to v2
set item r of my workList to v1
copy workList to item (p + 5) of my permutations
set p to p + 6
end prmt
end script
if (o's r < 3) then
-- Special-case fewer than three items.
copy theList to the beginning of o's permutations
if (o's r is 2) then set the end of o's permutations to theList's reverse
else
-- Otherwise set up to use the recursive handler.
copy theList to o's workList
-- "Growing" a long results list by appending each permutation to it
-- takes a disproportionately long time. Instead, build a list of the
-- appropriate factorial length beforehand, using concatenation.
set o's permutations to {missing value, missing value}
repeat with i from 3 to (count theList)
set temp to o's permutations
repeat (i - 1) times
set o's permutations to o's permutations & temp
end repeat
end repeat
o's prmt(1)
end if
return o's permutations
end allPermutations
return allPermutations({1, 2, "cat", "dog"})</lang>
- Output:
<lang applescript>{{1, 2, "cat", "dog"}, {1, 2, "dog", "cat"}, {1, "cat", "dog", 2}, {1, "cat", 2, "dog"}, {1, "dog", 2, "cat"}, {1, "dog", "cat", 2}, {2, "dog", "cat", 1}, {2, "dog", 1, "cat"}, {2, "cat", 1, "dog"}, {2, "cat", "dog", 1}, {2, 1, "dog", "cat"}, {2, 1, "cat", "dog"}, {"cat", 1, 2, "dog"}, {"cat", 1, "dog", 2}, {"cat", 2, "dog", 1}, {"cat", 2, 1, "dog"}, {"cat", "dog", 1, 2}, {"cat", "dog", 2, 1}, {"dog", "cat", 2, 1}, {"dog", "cat", 1, 2}, {"dog", 2, 1, "cat"}, {"dog", 2, "cat", 1}, {"dog", 1, "cat", 2}, {"dog", 1, 2, "cat"}}</lang>
ARM Assembly
<lang ARM Assembly> /* ARM assembly Raspberry PI */ /* program permutation.s */
/* REMARK 1 : this program use routines in a include file see task Include a file language arm assembly for the routine affichageMess conversion10 see at end of this program the instruction include */
/* for constantes see task include a file in arm assembly */ /************************************/ /* Constantes */ /************************************/ .include "../constantes.inc"
/*********************************/ /* Initialized data */ /*********************************/ .data
sMessResult: .asciz "Value : @ \n" sMessCounter: .asciz "Permutations = @ \n" szCarriageReturn: .asciz "\n"
.align 4 TableNumber: .int 1,2,3
.equ NBELEMENTS, (. - TableNumber) / 4
/*********************************/ /* UnInitialized data */ /*********************************/ .bss sZoneConv: .skip 24 /*********************************/ /* code section */ /*********************************/ .text .global main main: @ entry of program
ldr r0,iAdrTableNumber @ address number table mov r1,#NBELEMENTS @ number of élements mov r10,#0 @ counter bl heapIteratif mov r0,r10 @ display counter ldr r1,iAdrsZoneConv @ bl conversion10S @ décimal conversion ldr r0,iAdrsMessCounter ldr r1,iAdrsZoneConv @ insert conversion bl strInsertAtCharInc bl affichageMess @ display message
100: @ standard end of the program
mov r0, #0 @ return code mov r7, #EXIT @ request to exit program svc #0 @ perform the system call
iAdrszCarriageReturn: .int szCarriageReturn iAdrsMessResult: .int sMessResult iAdrTableNumber: .int TableNumber iAdrsMessCounter: .int sMessCounter /******************************************************************/ /* permutation by heap iteratif (wikipedia) */ /******************************************************************/ /* r0 contains the address of table */ /* r1 contains the eléments number */ heapIteratif:
push {r3-r9,lr} @ save registers
lsl r9,r1,#2 @ four bytes by count
sub sp,sp,r9
mov fp,sp
mov r3,#0
mov r4,#0 @ index
1: @ init area counter
str r4,[fp,r3,lsl #2] add r3,r3,#1 cmp r3,r1 blt 1b bl displayTable add r10,r10,#1 mov r3,#0 @ index
2:
ldr r4,[fp,r3,lsl #2] @ load count [i] cmp r4,r3 @ compare with i bge 5f tst r3,#1 @ even ? bne 3f ldr r5,[r0] @ yes load value A[0] ldr r6,[r0,r3,lsl #2] @ and swap with value A[i] str r6,[r0] str r5,[r0,r3,lsl #2] b 4f
3:
ldr r5,[r0,r4,lsl #2] @ load value A[count[i]] ldr r6,[r0,r3,lsl #2] @ and swap with value A[i] str r6,[r0,r4,lsl #2] str r5,[r0,r3,lsl #2]
4:
bl displayTable add r10,r10,#1 add r4,r4,#1 @ increment count i str r4,[fp,r3,lsl #2] @ and store on stack mov r3,#0 @ raz index b 2b @ and loop
5:
mov r4,#0 @ raz count [i] str r4,[fp,r3,lsl #2] add r3,r3,#1 @ increment index cmp r3,r1 @ end ? blt 2b @ no -> loop add sp,sp,r9 @ stack alignement
100:
pop {r3-r9,lr}
bx lr @ return
/******************************************************************/ /* Display table elements */ /******************************************************************/ /* r0 contains the address of table */ displayTable:
push {r0-r3,lr} @ save registers
mov r2,r0 @ table address
mov r3,#0
1: @ loop display table
ldr r0,[r2,r3,lsl #2] ldr r1,iAdrsZoneConv @ bl conversion10S @ décimal conversion ldr r0,iAdrsMessResult ldr r1,iAdrsZoneConv @ insert conversion bl strInsertAtCharInc bl affichageMess @ display message add r3,#1 cmp r3,#NBELEMENTS - 1 ble 1b ldr r0,iAdrszCarriageReturn bl affichageMess mov r0,r2
100:
pop {r0-r3,lr}
bx lr
iAdrsZoneConv: .int sZoneConv /***************************************************/ /* ROUTINES INCLUDE */ /***************************************************/ .include "../affichage.inc" </lang>
Value : +1 Value : +2 Value : +3 Value : +2 Value : +1 Value : +3 Value : +3 Value : +1 Value : +2 Value : +1 Value : +3 Value : +2 Value : +2 Value : +3 Value : +1 Value : +3 Value : +2 Value : +1 Permutations = +6
Arturo
<lang rebol>print permutate [1 2 3]</lang>
- Output:
[1 2 3] [1 3 2] [2 1 3] [2 3 1] [3 1 2] [3 2 1]
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 ---------------------------
AWK
<lang AWK>
- syntax: GAWK -f PERMUTATIONS.AWK [-v sep=x] [word]
- examples:
- REM all permutations on one line
- GAWK -f PERMUTATIONS.AWK
- REM all permutations on a separate line
- GAWK -f PERMUTATIONS.AWK -v sep="\n"
- REM use a different word
- GAWK -f PERMUTATIONS.AWK Gwen
- REM command used for RosettaCode output
- GAWK -f PERMUTATIONS.AWK -v sep="\n" Gwen
BEGIN {
sep = (sep == "") ? " " : substr(sep,1,1)
str = (ARGC == 1) ? "abc" : ARGV[1]
printf("%s%s",str,sep)
leng = length(str)
for (i=1; i<=leng; i++) {
arr[i-1] = substr(str,i,1)
}
ana_permute(0)
exit(0)
} function ana_permute(pos, i,j,str) {
if (leng - pos < 2) { return }
for (i=pos; i<leng-1; i++) {
ana_permute(pos+1)
ana_rotate(pos)
for (j=0; j<=leng-1; j++) {
printf("%s",arr[j])
}
printf(sep)
}
ana_permute(pos+1)
ana_rotate(pos)
} function ana_rotate(pos, c,i) {
c = arr[pos]
for (i=pos; i<leng-1; i++) {
arr[i] = arr[i+1]
}
arr[leng-1] = c
} </lang>
sample command:
GAWK -f PERMUTATIONS.AWK Gwen
- Output:
Gwen Gwne Genw Gewn Gnwe Gnew wenG weGn wnGe wneG wGen wGne enGw enwG eGwn eGnw ewnG ewGn nGwe nGew nweG nwGe neGw newG
BASIC256
<lang BASIC256>arraybase 1 n = 4 : cont = 0 dim a(n) dim c(n)
for j = 1 to n
a[j] = j
next j
do
for i = 1 to n
print a[i];
next
print " ";
i = n
cont += 1
if cont = 12 then
print
cont = 0
else
print " ";
end if
do
i -= 1
until (i = 0) or (a[i] < a[i+1])
j = i + 1
k = n
while j < k
tmp = a[j] : a[j] = a[k] : a[k] = tmp
j += 1
k -= 1
end while
if i > 0 then
j = i + 1
while a[j] < a[i]
j += 1
end while
tmp = a[j] : a[j] = a[i] : a[i] = tmp
end if
until i = 0 end</lang>
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
version 1
Non-recursive algorithm to generate all permutations. It prints objects in lexicographical order. <lang c>
- include <stdio.h>
int main (int argc, char *argv[]) { //here we check arguments if (argc < 2) {
printf("Enter an argument. Example 1234 or dcba:\n");
return 0;
} //it calculates an array's length
int x;
for (x = 0; argv[1][x] != '\0'; x++);
//buble sort the array int f, v, m; for(f=0; f < x; f++) {
for(v = x-1; v > f; v-- ) {
if (argv[1][v-1] > argv[1][v]) {
m=argv[1][v-1]; argv[1][v-1]=argv[1][v]; argv[1][v]=m;
} }
}
//it calculates a factorial to stop the algorithm
char a[x];
int k=0; int fact=k+1;
while (k!=x) {
a[k]=argv[1][k];
k++;
fact = k*fact;
}
a[k]='\0';
//Main part: here we permutate
int i, j;
int y=0;
char c;
while (y != fact) {
printf("%s\n", a);
i=x-2;
while(a[i] > a[i+1] ) i--;
j=x-1;
while(a[j] < a[i] ) j--;
c=a[j];
a[j]=a[i];
a[i]=c;
i++; for (j = x-1; j > i; i++, j--) {
c = a[i];
a[i] = a[j];
a[j] = c;
}
y++;
}
} </lang>
version 2
Non-recursive algorithm to generate all permutations. It prints them from right to left. <lang c>
- include <stdio.h>
int main() {
char a[] = "4321"; //array
int i, j;
int f=24; //factorial
char c; //buffer
while (f--) {
printf("%s\n", a);
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;
i--; for (j = 0; j < i; i--, j++) {
c = a[i];
a[i] = a[j];
a[j] = c;
}
}
}
</lang>
version 3
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>
version 4
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#
Recursive Linq
<lang csharp>public static class Extension {
public static IEnumerable<IEnumerable<T>> Permutations<T>(this IEnumerable<T> values) where T : IComparable<T>
{
if (values.Count() == 1)
return new[] { values };
return values.SelectMany(v => Permutations(values.Where(x => x.CompareTo(v) != 0)), (v, p) => p.Prepend(v));
}
}</lang> Usage <lang sharp>Enumerable.Range(0,5).Permutations()</lang> 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>
Recursive version
<lang csharp>using System;
class Permutations
{
static int n = 4; static int [] buf = new int [n]; static bool [] used = new bool [n];
static void Main()
{
for (int i = 0; i < n; i++) used [i] = false;
rec(0);
}
static void rec(int ind)
{
for (int i = 0; i < n; i++)
{
if (!used [i])
{
used [i] = true;
buf [ind] = i;
if (ind + 1 < n) rec(ind + 1);
else Console.WriteLine(string.Join(",", buf));
used [i] = false;
} } }
}</lang>
Alternate recursive version
<lang csharp> using System; class Permutations {
static int n = 4; static int [] buf = new int [n]; static int [] next = new int [n+1];
static void Main()
{
for (int i = 0; i < n; i++) next [i] = i + 1;
next[n] = 0;
rec(0);
}
static void rec(int ind)
{
for (int i = n; next[i] != n; i = next[i])
{
buf [ind] = next[i];
next[i]=next[next[i]];
if (ind < n - 1) rec(ind + 1);
else Console.WriteLine(string.Join(",", buf));
next[i] = buf [ind];
}
}
} </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
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>
Recursive implementation of Heap's algorithm: <lang lisp>(defun heap-permutations (seq)
(let ((permutations nil)) (labels ((permute (seq k)
(if (= k 1) (push seq permutations) (progn (permute seq (1- k)) (loop for i from 0 below (1- k) do (if (evenp k) (rotatef (elt seq i) (elt seq (1- k))) (rotatef (elt seq 0) (elt seq (1- k)))) (permute seq (1- k)))))))
(permute seq (length seq))
permutations)))</lang>
Crystal
<lang Ruby>puts [1, 2, 3].permutations</lang>
- Output:
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
Curry
<lang curry> insert :: a -> [a] -> [a] insert x xs = x : xs insert x (y:ys) = y : insert x ys
permutation :: [a] -> [a] permutation [] = [] permutation (x:xs) = insert x $ permutation xs </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
EDSAC order code
Uses two subroutines which respectively (1) Generate the first permutation in lexicographic order; (2) Return the next permutation in lexicographic order, or set a flag to indicate there are no more permutations. The algorithm for (2) is the same as in the Wikipedia article "Permutation". <lang edsac> [Permutations task for Rosetta Code.] [EDSAC program, Initial Orders 2.]
T51K P200F [G parameter: start address of subroutines] T47K P100F [M parameter: start address of main routine]
[====================== G parameter: Subroutines =====================]
E25K TG GK
[Constants used in the subroutines]
[0] AF [add to address to make A order for that address]
[1] SF [add to address to make S order for that address]
[2] UF [(1) add to address to make U order for that address]
[(2) subtract from S order to make T order, same address]
[3] OF [add to A order to make T order, same address]
[----------------------------------------------------------- Subroutine to initialize an array of n short (17-bit) words to 0, 1, 2, ..., n-1 (in the address field). Parameters: 4F = address of array; 5F = n = length of array. Workspace: 0F, 1F.]
[4] A3F [plant return link as usual]
T19@
A4F [address of array]
A2@ [make U order for that address]
T1F [store U order in 1F]
A5F [load n = number of elements (in address field)]
S2F [make n-1]
[Start of loop; works backwards, n-1 to 0]
[11] UF [store array element in 0F]
A1F [make order to store element in array]
T15@ [plant that order in code]
AF [pick up element fron 0F]
[15] UF [(planted) store element in array]
S2F [dec to next element]
E11@ [loop if still >= 0]
TF [clear acc. before return]
[19] ZF [overwritten by jump back to caller]
[------------------------------------------------------------------- Subroutine to get next permutation in lexicographic order. Uses same 4-step algorithm as Wikipedia article "Permutations", but notation in comments differs from that in Wikipedia. Parameters: 4F = address of array; 5F = n = length of array.
0F is returned as 0 for success, < 0 if passed-in
permutation is the last.
Workspace: 0F, 1F.]
[20] A3F [plant return link as usual]
T103@
[Step 1: Find the largest index k such that a{k} > a{k-1}.
If no such index exists, the passed-in permutation is the last.]
A4F [load address of a{0}]
A@ [make A order for a{0}]
U1F [store as test for end of loop]
A5F [make A order for a{n}]
U96@ [plant in code below]
S2F [make A order for a{n-1}]
T43@ [plant in code below]
A4F [load address of a{0}]
A5F [make address of a{n}]
A1@ [make S order for a{n}]
T44@ [plant in code below]
[Start of loop for comparing a{k} with a{k-1}]
[33] TF [clear acc]
A43@ [load A order for a{k}]
S2F [make A order for a{k-1}]
S1F [tested all yet?]
G102@ [if yes, jump to failed (no more permutations)]
A1F [restore accumulator after test]
T43@ [plant updated A order]
A44@ [dec address in S order]
S2F
T44@
[43] SF [(planted) load a{k-1}]
[44] AF [(planted) subtract a{k}]
E33@ [loop back if a{k-1} > a{k}]
[Step 2: Find the largest index j >= k such that a{j} > a{k-1}.
Such an index j exists, because j = k is an instance.]
TF [clear acc]
A4F [load address of a{0}]
A5F [make address of a{n}]
A1@ [make S order for a{n}]
T1F [save as test for end of loop]
A44@ [load S order for a{k}]
T64@ [plant in code below]
A43@ [load A order for a{k-1}]
T63@ [plant in code below]
[Start of loop]
[55] TF [clear acc]
A64@ [load S order for a{j} (initially j = k)]
U75@ [plant in code below]
A2F [inc address (in effect inc j)]
S1F [test for end of array]
E66@ [jump out if so]
A1F [restore acc after test]
T64@ [update S order]
[63] AF [(planted) load a{k-1}]
[64] SF [(planted) subtract a{j}]
G55@ [loop back if a{j} still > a{k-1}]
[66]
[Step 3: Swap a{k-1} and a{j}]
TF [clear acc]
A63@ [load A order for a{k-1}]
U77@ [plant in code below, 2 places]
U94@
A3@ [make T order for a{k-1}]
T80@ [plant in code below]
A75@ [load S order for a{j}]
S2@ [make T order for a{j}]
T78@ [plant in code below]
[75] SF [(planted) load -a{j}]
TF [park -a{j} in 0F]
[77] AF [(planted) load a{k-1}]
[78] TF [(planted) store a{j}]
SF [load a{j} by subtracting -a{j}]
[80] TF [(planted) store in a{k-1}]
[Step 4: Now a{k}, ..., a{n-1} are in decreasing order.
Change to increasing order by repeated swapping.]
[81] A96@ [counting down from a{n} (exclusive end of array)]
S2F [make A order for a{n-1}]
U96@ [plant in code]
A3@ [make T order for a{n-1}]
T99@ [plant]
A94@ [counting up from a{k-1} (exclusive)]
A2F [make A order for a{k}]
U94@ [plant]
A3@ [make T order for a{k}]
U97@ [plant]
S99@ [swapped all yet?]
E101@ [if yes, jump to exit from subroutine]
[Swapping two array elements, initially a{k} and a{n-1}]
TF [clear acc]
[94] AF [(planted) load 1st element]
TF [park in 0F]
[96] AF [(planted) load 2nd element]
[97] TF [(planted) copy to 1st element]
AF [load old 1st element]
[99] TF [(planted) copy to 2nd element]
E81@ [always loop back]
[101] TF [done, return 0 in location 0F]
[102] TF [return status to caller in 0F; also clears acc]
[103] ZF [(planted) jump back to caller]
[==================== M parameter: Main routine ==================] [Prints all 120 permutations of the letters in 'EDSAC'.]
E25K TM GK
[Constants used in the main routine]
[0] P900F [address of permutation array]
[1] P5F [number of elements in permutation (in address field)]
[Array of letters in 'EDSAC', in alphabetical order]
[2] AF CF DF EF SF
[7] O2@ [add to index to make O order for letter in array]
[8] P12F [permutations per printed line (in address field)]
[9] AF [add to address to make A order for that address]
[Teleprinter characters]
[10] K2048F [set letters mode]
[11] !F [space]
[12] @F [carriage return]
[13] &F [line feed]
[14] K4096F [null]
[Entry point, with acc = 0.]
[15] O10@ [set teleprinter to letters]
S8@ [intialize -ve count of permutations per line]
T7F [keep count in 7F]
A@ [pass address of permutation array in 4F]
T4F
A1@ [pass number of elements in 5F]
T5F
[22] A22@ [call subroutine to initialize permutation array]
G4G
[Loop: print current permutation, then get next (if any)]
[24] A4F [address]
A9@ [make A order]
T29@ [plant in code]
S5F [initialize -ve count of array elements]
[28] T6F [keep count in 6F]
[29] AF [(planted) load permutation element]
A7@ [make order to print letter from table]
T32@ [plant in code]
[32] OF [(planted) print letter from table]
A29@ [inc address in permutation array]
A2F
T29@
A6F [inc -ve count of array elements]
A2F
G28@ [loop till count becomes 0]
A7F [inc -ve count of perms per line]
A2F
E44@ [jump if end of line]
O11@ [else print a space]
G47@ [join common code]
[44] O12@ [print CR]
O13@ [print LF]
S8@
[47] T7F [update -ve count of permutations in line]
[48] A48@ [call subroutine for next permutation (if any)]
G20G
AF [test 0F: got a new permutation?]
E24@ [if so, loop to print it]
O14@ [no more, output null to flush teleprinter buffer]
ZF [halt program]
E15Z [define entry point]
PF [enter with acc = 0]
[end] </lang>
- Output:
ACDES ACDSE ACEDS ACESD ACSDE ACSED ADCES ADCSE ADECS ADESC ADSCE ADSEC AECDS AECSD AEDCS AEDSC AESCD AESDC ASCDE ASCED ASDCE ASDEC ASECD ASEDC CADES CADSE CAEDS CAESD CASDE CASED CDAES CDASE CDEAS CDESA CDSAE CDSEA CEADS CEASD CEDAS CEDSA CESAD CESDA CSADE CSAED CSDAE CSDEA CSEAD CSEDA DACES DACSE DAECS DAESC DASCE DASEC DCAES DCASE DCEAS DCESA DCSAE DCSEA DEACS DEASC DECAS DECSA DESAC DESCA DSACE DSAEC DSCAE DSCEA DSEAC DSECA EACDS EACSD EADCS EADSC EASCD EASDC ECADS ECASD ECDAS ECDSA ECSAD ECSDA EDACS EDASC EDCAS EDCSA EDSAC EDSCA ESACD ESADC ESCAD ESCDA ESDAC ESDCA SACDE SACED SADCE SADEC SAECD SAEDC SCADE SCAED SCDAE SCDEA SCEAD SCEDA SDACE SDAEC SDCAE SDCEA SDEAC SDECA SEACD SEADC SECAD SECDA SEDAC SEDCA
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"]
Translation of Haskell "insertion-based approach" (last version) <lang fsharp> let permutations xs =
let rec insert x = function
| [] -> x
| head :: tail -> (x :: (head :: tail)) :: (List.map (fun l -> head :: l) (insert x tail))
List.fold (fun s e -> List.collect (insert e) s) [[]] xs
</lang>
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 Speed Test
So ... what is the fastest algorithm?
Here below is the speed test for a couple of algorithms of permutation. We can add more algorithms into this frame-work. When they work in the same circumstance, we can see which is the fastest one.
<lang fortran> program testing_permutation_algorithms
implicit none integer :: nmax integer, dimension(:),allocatable :: ida logical :: mtc logical :: even integer :: i integer(8) :: ic integer :: clock_rate, clock_max, t1, t2 real(8) :: dt integer :: pos_min, pos_max
! ! ! Beginning: !
write(*,*) 'INPUT N:' read *, nmax write(*,*) 'N =', nmax allocate ( ida(1:nmax) )
! ! ! (1) Starting: !
do i = 1, nmax
ida(i) = i
enddo
!
ic = 0 call system_clock ( t1, clock_rate, clock_max )
!
mtc = .false.
!
do
call subnexper ( nmax, ida, mtc, even )
! ! 1) counting the number of permutatations !
ic = ic + 1
! ! 2) writing out the result: ! ! do i = 1, nmax ! write (100,"(i3,',')",advance = "no") ida(i) ! enddo ! write(100,*) ! ! repeat if not being finished yet, otherwise exit. !
if (mtc) then
cycle
else
exit
endif
!
enddo
!
call system_clock ( t2, clock_rate, clock_max ) dt = ( dble(t2) - dble(t1) )/ dble(clock_rate)
! ! Finishing (1) !
write(*,*) "1) subnexper:" write(*,*) 'Total permutations :', ic write(*,*) 'Total time elapsed :', dt
! ! ! (2) Starting: !
do i = 1, nmax
ida(i) = i
enddo
!
pos_min = 1 pos_max = nmax
!
ic = 0 call system_clock ( t1, clock_rate, clock_max )
!
call generate ( pos_min )
!
call system_clock ( t2, clock_rate, clock_max ) dt = ( dble(t2) - dble(t1) )/ dble(clock_rate)
! ! Finishing (2) !
write(*,*) "2) generate:" write(*,*) 'Total permutations :', ic write(*,*) 'Total time elapsed :', dt
! ! ! (3) Starting: !
do i = 1, nmax
ida(i) = i
enddo
!
ic = 0 call system_clock ( t1, clock_rate, clock_max )
!
i = 1 call perm ( i )
!
call system_clock ( t2, clock_rate, clock_max ) dt = ( dble(t2) - dble(t1) )/ dble(clock_rate)
! ! Finishing (3) !
write(*,*) "3) perm:" write(*,*) 'Total permutations :', ic write(*,*) 'Total time elapsed :', dt
! ! ! (4) Starting: !
do i = 1, nmax
ida(i) = i
enddo
!
ic = 0 call system_clock ( t1, clock_rate, clock_max )
!
do
! ! 1) counting the number of permutatations !
ic = ic + 1
! ! 2) writing out the result: ! ! do i = 1, nmax ! write (100,"(i3,',')",advance = "no") ida(i) ! enddo ! write(100,*) ! ! repeat if not being finished yet, otherwise exit. !
if ( nextp(nmax,ida) ) then
cycle
else
exit
endif
!
enddo
!
call system_clock ( t2, clock_rate, clock_max ) dt = ( dble(t2) - dble(t1) )/ dble(clock_rate)
! ! Finishing (4) !
write(*,*) "4) nextp:" write(*,*) 'Total permutations :', ic write(*,*) 'Total time elapsed :', dt
! ! ! What's else? ! ... ! !==
deallocate(ida)
!
stop
!==
contains
!== ! Modified version of SUBROUTINE NEXPER from the book of ! Albert Nijenhuis and Herbert S. Wilf, "Combinatorial ! Algorithms For Computers and Calculators", 2nd Ed, p.59. !
subroutine subnexper ( n, a, mtc, even )
implicit none
integer,intent(in) :: n
integer,dimension(n),intent(inout) :: a
logical,intent(inout) :: mtc, even
! ! local varialbes: !
integer,save :: nm3
integer :: ia, i, s, d, i1, l, j, m
!
if (mtc) goto 10
nm3 = n-3
do i = 1,n
a(i) = i
enddo
mtc = .true.
5 even = .true.
if ( n .eq. 1 ) goto 8
6 if ( a(n) .ne. 1 .or. a(1) .ne. 2+mod(n,2) ) return
if ( n .le. 3 ) goto 8
do i = 1,nm3
if( a(i+1) .ne. a(i)+1 ) return
enddo
8 mtc = .false.
return
10 if ( n .eq. 1 ) goto 27
if( .not. even ) goto 20
ia = a(1)
a(1) = a(2)
a(2) = ia
even = .false.
goto 6
20 s = 0
do i1 = 2,n
ia = a(i1)
i = i1-1
d = 0
do j = 1,i
if ( a(j) .gt. ia ) d = d+1
enddo
s = d+s
if ( d .ne. i*mod(s,2) ) goto 35
enddo
27 a(1) = 0
goto 8
35 m = mod(s+1,2)*(n+1)
do j = 1,i
if(isign(1,a(j)-ia) .eq. isign(1,a(j)-m)) cycle
m = a(j)
l = j
enddo
a(l) = ia
a(i1) = m
even = .true.
return
end subroutine
!===== ! ! http://rosettacode.org/wiki/Permutations#Fortran !
recursive subroutine generate (pos)
implicit none
integer,intent(in) :: pos
integer :: val
if (pos > pos_max) then
! ! 1) counting the number of permutatations !
ic = ic + 1
! ! 2) writing out the result: ! ! write (*,*) permutation !
else
do val = 1, nmax
if (.not. any (ida( : pos-1) == val)) then
ida(pos) = val
call generate (pos + 1)
endif
enddo
endif
end subroutine
!===== ! ! http://rosettacode.org/wiki/Permutations#Fortran !
recursive subroutine perm (i)
implicit none
integer,intent(inout) :: i
!
integer :: j, t, ip1
!
if (i == nmax) then
! ! 1) couting the number of permutatations !
ic = ic + 1
! ! 2) writing out the result: ! ! write (*,*) a !
else
ip1 = i+1
do j = i, nmax
t = ida(i)
ida(i) = ida(j)
ida(j) = t
call perm ( ip1 )
t = ida(i)
ida(i) = ida(j)
ida(j) = t
enddo
endif
return
end subroutine
!===== ! ! http://rosettacode.org/wiki/Permutations#Fortran !
function nextp ( n, a )
logical :: nextp
integer,intent(in) :: n
integer,dimension(n),intent(inout) :: a
! ! local variables: !
integer i,j,k,t
!
i = n-1
10 if ( a(i) .lt. a(i+1) ) goto 20
i = i-1
if ( i .eq. 0 ) goto 20
goto 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 ) goto 30
j = i
if (j .ne. 0 ) goto 40
!
nextp = .false.
!
return
!
40 j = j+1
if ( a(j) .lt. a(i) ) goto 40
t = a(i)
a(i) = a(j)
a(j) = t
!
nextp = .true.
!
return
end function
!===== ! ! What's else ? ! ...
!=====
end program</lang>
An example of performance:
1) Compiled with GNU fortran compiler:
gfortran -O3 testing_permutation_algorithms.f90 ; ./a.out
INPUT N:
10
N = 10 1) subnexper: Total permutations : 3628800 Total time elapsed : 4.9000000000000002E-002 2) generate: Total permutations : 3628800 Total time elapsed : 0.84299999999999997 3) perm: Total permutations : 3628800 Total time elapsed : 5.6000000000000001E-002 4) nextp: Total permutations : 3628800 Total time elapsed : 2.9999999999999999E-002
b) Compiled with Intel compiler:
ifort -O3 testing_permutation_algorithms.f90 ; ./a.out
INPUT N: 10
N = 10 1) subnexper: Total permutations : 3628800 Total time elapsed : 8.240000000000000E-002 2) generate: Total permutations : 3628800 Total time elapsed : 0.616200000000000 3) perm: Total permutations : 3628800 Total time elapsed : 5.760000000000000E-002 4) nextp: Total permutations : 3628800 Total time elapsed : 3.600000000000000E-002
So far, we have conclusion from the above performance: 1) subnexper is the 3rd fast with ifort and the 2nd with gfortran. 2) generate is the slowest one with not only ifort but gfortran. 3) perm is the 2nd fast one with ifort and the 3rd one with gfortran. 4) nextp is the fastest one with both ifort and gfortran (the winner in this test).
Note: It is worth mentioning that the performance of this test is dependent not only on algorithm, but also on computer where the test runs. Therefore we should run the test on our own computer and make conclusion by ourselves.
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>
Ratfor 77
See RATFOR.
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
Frink
Frink's array class has built-in methods permute[] and lexicographicPermute[] which permute the elements of an array in reflected Gray code order and lexicographic order respectively.
<lang frink>a = [1,2,3,4]
println[formatTable[a.lexicographicPermute[]]]</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
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>
GNU make
Recursive on unique elements
<lang make>
- delimiter should not occur inside elements
delimiter=;
- convert list to delimiter separated string
implode=$(subst $() $(),$(delimiter),$(strip $1))
- convert delimiter separated string to list
explode=$(strip $(subst $(delimiter), ,$1))
- enumerate all permutations and subpermutations
permutations0=$(if $1,$(foreach x,$1,$x $(addprefix $x$(delimiter),$(call permutations0,$(filter-out $x,$1)))),)
- remove subpermutations from permutations0 output
permutations=$(strip $(foreach x,$(call permutations0,$1),$(if $(filter $(words $1),$(words $(call explode,$x))),$(call implode,$(call explode,$x)),)))
delimiter_separated_output=$(call permutations,a b c d) $(info $(delimiter_separated_output)) </lang>
- Output:
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
Go
recursive
<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]
non-recursive, lexicographical order
<lang go>package main
import "fmt"
func main() {
var a = []int{1, 2, 3}
fmt.Println(a)
var n = len(a) - 1
var i, j int
for c := 1; c < 6; c++ { // 3! = 6:
i = n - 1
j = n
for a[i] > a[i+1] {
i--
}
for a[j] < a[i] {
j--
}
a[i], a[j] = a[j], a[i]
j = n
i += 1
for i < j {
a[i], a[j] = a[j], a[i]
i++
j--
}
fmt.Println(a)
}
}</lang>
- Output:
[1 2 3] [1 3 2] [2 1 3] [2 3 1] [3 1 2] [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>
A serialized version:
<lang haskell>import Data.Bifunctor (second)
permutations :: [a] -> a permutations =
let ins x xs n = uncurry (<>) $ second (x :) (splitAt n xs)
in foldr
( \x a ->
a >>= (fmap . ins x)
<*> (enumFromTo 0 . length)
)
[[]]
main :: IO () main = print $ permutations [1, 2, 3]</lang>
- Output:
[[1,2,3],[2,3,1],[3,1,2],[2,1,3],[1,3,2],[3,2,1]]
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 ->
IS-BASIC
<lang IS-BASIC>100 PROGRAM "Permutat.bas" 110 LET N=4 ! Number of elements 120 NUMERIC T(1 TO N) 130 FOR I=1 TO N 140 LET T(I)=I 150 NEXT 160 LET S=0 170 CALL PERM(N) 180 PRINT "Number of permutations:";S 190 END 200 DEF PERM(I) 210 NUMERIC J,X 220 IF I=1 THEN 230 FOR X=1 TO N 240 PRINT T(X); 250 NEXT 260 PRINT :LET S=S+1 270 ELSE 280 CALL PERM(I-1) 290 FOR J=1 TO I-1 300 LET C=T(J):LET T(J)=T(I):LET T(I)=C 310 CALL PERM(I-1) 320 LET C=T(J):LET T(J)=T(I):LET T(I)=C 330 NEXT 340 END IF 350 END DEF</lang>
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
Recursively, in terms of concatMap and delete: <lang JavaScript>(() => {
'use strict';
// permutations :: [a] -> a const permutations = xs => { const go = xs => xs.length ? ( concatMap( x => concatMap( ys => [[x].concat(ys)], go(delete_(x, xs))), xs ) ) : [[]]; return go(xs); };
// GENERIC FUNCTIONS ----------------------------------
// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = (f, xs) =>
xs.reduce((a, x) => a.concat(f(x)), []);
// delete :: Eq a => a -> [a] -> [a]
const delete_ = (x, xs) => {
const go = xs => {
return 0 < xs.length ? (
(x === xs[0]) ? (
xs.slice(1)
) : [xs[0]].concat(go(xs.slice(1)))
) : [];
}
return go(xs);
};
// TEST
return JSON.stringify(
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>
Or, without recursion, in terms of concatMap and reduce:
<lang javascript>(() => {
'use strict';
// permutations :: [a] -> a const permutations = xs => xs.reduceRight( (a, x) => concatMap( xs => enumFromTo(0, xs.length) .map(n => xs.slice(0, n) .concat(x) .concat(xs.slice(n)) ), a ), [[]] );
// GENERIC FUNCTIONS ----------------------------------
// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = (f, xs) =>
xs.reduce((a, x) => a.concat(f(x)), []);
// ft :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: 1 + n - m
}, (_, i) => m + i);
// showLog :: a -> IO ()
const showLog = (...args) =>
console.log(
args
.map(JSON.stringify)
.join(' -> ')
);
// TEST -----------------------------------------------
showLog(
permutations([1, 2, 3])
);
})();</lang>
- Output:
[[1,2,3],[2,1,3],[2,3,1],[1,3,2],[3,1,2],[3,2,1]]
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
<lang Julia> julia> perms(l) = isempty(l) ? [l] : [[x; y] for x in l for y in perms(setdiff(l, x))] </lang>
- Output:
<lang Julia> julia> perms([1,2,3]) 6-element Vector{Vector{Int64}}:
[1, 2, 3] [1, 3, 2] ⋮ [3, 1, 2] [3, 2, 1]
</lang>
Further support for permutation creation and processing is available in the Combinatorics.jl package. permutations(v) creates an iterator over all permutations of v. Julia 0.7 and 1.0+ require the line global i inside the for to update the i variable. <lang Julia> using Combinatorics
term = "RCode" i = 0 pcnt = factorial(length(term)) print("All the permutations of ", term, " (", pcnt, "):\n ") for p in permutations(split(term, ""))
global i
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
<lang>
- Generate all permutations of size t from an array a with possibly duplicated elements.
collect(Combinatorics.multiset_permutations([1,1,0,0,0],3)) </lang>
- Output:
7-element Array{Array{Int64,1},1}:
[1, 1, 0]
[1, 0, 1]
[1, 0, 0]
[0, 1, 1]
[0, 1, 0]
[0, 0, 1]
[0, 0, 0]
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]
Lambdatalk
<lang scheme> {def inject
{lambda {:x :a}
{if {A.empty? :a}
then {A.new {A.new :x}}
else {let { {:c {{lambda {:a :b} {A.cons {A.first :a} :b}} :a}}
{:d {inject :x {A.rest :a}}}
{:e {A.cons :x :a}}
} {A.cons :e {A.map :c :d}}}}}}
-> inject
{def permut
{lambda {:a}
{if {A.empty? :a}
then {A.new :a}
else {let { {:c {{lambda {:a :b} {inject {A.first :a} :b}} :a}}
{:d {permut {A.rest :a}}}
} {A.reduce A.concat {A.map :c :d}}}}}}
-> permut
{permut {A.new 1 2 3}} -> [[1,2,3],[2,1,3],[2,3,1],[1,3,2],[3,1,2],[3,2,1]]
{permut {A.new 1 2 3 4}} -> [[1,2,3,4],[2,1,3,4],[2,3,1,4],[2,3,4,1],[1,3,2,4],[3,1,2,4],[3,2,1,4],[3,2,4,1],[1,3,4,2],[3,1,4,2],[3,4,1,2],[3,4,2,1],[1,2,4,3],[2,1,4,3],[2,4,1,3],[2,4,3,1],[1,4,2,3],[4,1,2,3],[4,2,1,3],[4,2,3,1],[1,4,3,2],[4,1,3,2],[4,3,1,2],[4,3,2,1]] </lang>
And this is an illustration of the way lambdatalk builds an interface for javascript functions (the first one is given in this page):
<lang javascript> 1) permutations on sentences
{script
var S_perm = function(a) {
if (a.length < 2) return [a];
var b = [];
for (var c = 0; c < a.length; c++) {
var e = a.splice(c, 1), f = S_perm(a);
for (var d = 0; d < f.length; d++)
b.push( e.concat( f[d]) );
a.splice(c, 0, e[0])
}
return b
}
LAMBDATALK.DICT['S.perm'] = function() { // {S.perm 1 2 3}
return S_perm( arguments[0].trim()
.split(" ") )
.join(" ")
.replace(/\s/g,"{br}")
};
}
{S.perm 1 2 3} -> 1,2,3 1,3,2 2,1,3 2,3,1 3,1,2 3,2,1
{S.perm hello brave world} -> hello,brave,world hello,world,brave brave,hello,world brave,world,hello world,hello,brave world,brave,hello
2) permutations on words
{script
var W_perm = function(word) {
if (word.length === 1) return [word]
var results = [];
for (var i = 0; i < word.length; i++) {
var buti = W_perm( word.substring(0, i) + word.substring(i + 1) );
for (var j = 0; j < buti.length; j++)
results.push(word[i] + buti[j]);
}
return results;
};
LAMBDATALK.DICT['W.perm'] = function() { // {W.perm 123}
return W_perm( arguments[0].trim() ).join("{br}")
};
}
{W.perm 123} -> 123 132 213 231 312 321
</lang>
langur
This follows the Go language non-recursive example, but is not limited to integers, or even to numbers.
Prior to 0.10, multi-variable declaration/assignment would use parentheses around variable names and values. 0.10 also parses the increment section of a for loop as a multi-variable assignment, not as a list of assignments.
<lang langur>val .factorial = f if(.x < 2: 1; .x x self(.x - 1))
val .permute = f(.arr) {
if not isArray(.arr): throw "expected array"
val .limit = 10 if len(.arr) > .limit: throw $"permutation limit exceeded (currently \.limit;)"
var .elements = .arr var .ordinals = pseries len .elements
val .n = len(.ordinals) var .i, .j
for[.p=[.arr]] of .factorial(len .arr)-1 {
.i = .n - 1
.j = .n
while .ordinals[.i] > .ordinals[.i+1] {
.i -= 1
}
while .ordinals[.j] < .ordinals[.i] {
.j -= 1
}
.ordinals[.i], .ordinals[.j] = .ordinals[.j], .ordinals[.i]
.elements[.i], .elements[.j] = .elements[.j], .elements[.i]
.i += 1
for .j = .n; .i < .j ; .i, .j = .i+1, .j-1 {
.ordinals[.i], .ordinals[.j] = .ordinals[.j], .ordinals[.i]
.elements[.i], .elements[.j] = .elements[.j], .elements[.i]
}
.p = more .p, .elements
}
}
for .e in .permute([1, 3.14, 7]) {
writeln .e
}</lang>
- Output:
[1, 3.14, 7] [1, 7, 3.14] [3.14, 1, 7] [3.14, 7, 1] [7, 1, 3.14] [7, 3.14, 1]
LFE
<lang lisp> (defun permute
(('())
'(()))
((l)
(lc ((<- x l)
(<- y (permute (-- l `(,x)))))
(cons x y))))
</lang> REPL usage: <lang lisp> > (permute '(1 2 3)) ((1 2 3) (1 3 2) (2 1 3) (2 3 1) (3 1 2) (3 2 1)) </lang>
Liberty BASIC
Permuting numerical array (non-recursive):
<lang lb> n=3 dim a(n+1) '+1 needed due to bug in LB that checks loop condition
' until (i=0) or (a(i)<a(i+1)) 'before executing i=i-1 in loop body.
for i=1 to n: a(i)=i: next do
for i=1 to n: print a(i);: next: print
i=n
do
i=i-1
loop until (i=0) or (a(i)<a(i+1))
j=i+1
k=n
while j<k
'swap a(j),a(k)
tmp=a(j): a(j)=a(k): a(k)=tmp
j=j+1
k=k-1
wend
if i>0 then
j=i+1
while a(j)<a(i)
j=j+1
wend
'swap a(i),a(j)
tmp=a(j): a(j)=a(i): a(i)=tmp
end if
loop until i=0 </lang>
- Output:
123 132 213 231 312 321
Permuting string (recursive): <lang lb> n = 3
s$="" 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
Lobster
<lang Lobster> // Lobster implementation of the (very fast) Go example // http://rosettacode.org/wiki/Permutations#Go // implementing the plain changes (bell ringers) algorithm, using a recursive function // https://en.wikipedia.org/wiki/Steinhaus–Johnson–Trotter_algorithm
def permr(s, f):
if s.length == 0:
f(s)
return
def rc(np: int):
if np == 1:
f(s)
return
let np1 = np - 1
let pp = s.length - np1
rc(np1) // recurs prior swaps
var i = pp
while i > 0:
// swap s[i], s[i-1]
let t = s[i]
s[i] = s[i-1]
s[i-1] = t
rc(np1) // recurs swap
i -= 1
let w = s[0]
for(pp): s[_] = s[_+1]
s[pp] = w
rc(s.length)
// Heap's recursive method https://en.wikipedia.org/wiki/Heap%27s_algorithm
def permh(s, f):
def rc(k: int):
if k <= 1:
f(s)
else:
// Generate permutations with kth unaltered
// Initially k == length(s)
rc(k-1)
// Generate permutations for kth swapped with each k-1 initial
for(k-1) i:
// Swap choice dependent on parity of k (even or odd)
// zero-indexed, the kth is at k-1
if (k & 1) == 0:
let t = s[i]
s[i] = s[k-1]
s[k-1] = t
else:
let t = s[0]
s[0] = s[k-1]
s[k-1] = t
rc(k-1)
rc(s.length)
// iterative Boothroyd method
import std
def permi(xs, f):
var d = 1
let c = map(xs.length): 0
f(xs)
while true:
while d > 1:
d -= 1
c[d] = 0
while c[d] >= d:
d += 1
if d >= xs.length:
return
let i = if (d & 1) == 1: c[d] else: 0
let t = xs[i]
xs[i] = xs[d]
xs[d] = t
f(xs)
c[d] = c[d] + 1
// next lexicographical permutation // to get all permutations the initial input `a` must be in sorted order // returns false when input `a` is in reverse sorted order
def next_lex_perm(a):
def swap(i, j):
let t = a[i]
a[i] = a[j]
a[j] = t
let n = a.length
/* 1. Find the largest index k such that a[k] < a[k + 1]. If no such
index exists, the permutation is the last permutation. */
var k = n - 1
while k > 0 and a[k-1] >= a[k]: k--
if k == 0: return false
k -= 1
/* 2. Find the largest index l such that a[k] < a[l]. Since k + 1 is
such an index, l is well defined */
var l = n - 1
while 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 */
k += 1
l = n - 1
while l > k:
swap(k, l)
l -= 1
k += 1
return true
var se = [0, 1, 2, 3] //, 4, 5, 6, 7, 8, 9, 10]
print "Iterative lexicographical permuter"
print se while next_lex_perm(se): print se
print "Recursive plain changes iterator"
se = [0, 1, 2, 3]
permr(se): print(_)
print "Recursive Heap\'s iterator"
se = [0, 1, 2, 3]
permh(se): print(_)
print "Iterative Boothroyd iterator"
se = [0, 1, 2, 3]
permi(se): print(_) </lang>
- Output:
Iterative lexicographical permuter [0, 1, 2, 3] [0, 1, 3, 2] [0, 2, 1, 3] [0, 2, 3, 1] [0, 3, 1, 2] [0, 3, 2, 1] [1, 0, 2, 3] [1, 0, 3, 2] [1, 2, 0, 3] [1, 2, 3, 0] [1, 3, 0, 2] [1, 3, 2, 0] [2, 0, 1, 3] [2, 0, 3, 1] [2, 1, 0, 3] [2, 1, 3, 0] [2, 3, 0, 1] [2, 3, 1, 0] [3, 0, 1, 2] [3, 0, 2, 1] [3, 1, 0, 2] [3, 1, 2, 0] [3, 2, 0, 1] [3, 2, 1, 0] Recursive plain changes iterator [0, 1, 2, 3] [0, 1, 3, 2] [0, 3, 1, 2] [3, 0, 1, 2] [0, 2, 1, 3] [0, 2, 3, 1] [0, 3, 2, 1] [3, 0, 2, 1] [2, 0, 1, 3] [2, 0, 3, 1] [2, 3, 0, 1] [3, 2, 0, 1] [1, 0, 2, 3] [1, 0, 3, 2] [1, 3, 0, 2] [3, 1, 0, 2] [1, 2, 0, 3] [1, 2, 3, 0] [1, 3, 2, 0] [3, 1, 2, 0] [2, 1, 0, 3] [2, 1, 3, 0] [2, 3, 1, 0] [3, 2, 1, 0] Recursive Heap's iterator [0, 1, 2, 3] [1, 0, 2, 3] [2, 0, 1, 3] [0, 2, 1, 3] [1, 2, 0, 3] [2, 1, 0, 3] [3, 1, 0, 2] [1, 3, 0, 2] [0, 3, 1, 2] [3, 0, 1, 2] [1, 0, 3, 2] [0, 1, 3, 2] [0, 2, 3, 1] [2, 0, 3, 1] [3, 0, 2, 1] [0, 3, 2, 1] [2, 3, 0, 1] [3, 2, 0, 1] [3, 2, 1, 0] [2, 3, 1, 0] [1, 3, 2, 0] [3, 1, 2, 0] [2, 1, 3, 0] [1, 2, 3, 0] Iterative Boothroyd iterator [0, 1, 2, 3] [1, 0, 2, 3] [2, 0, 1, 3] [0, 2, 1, 3] [1, 2, 0, 3] [2, 1, 0, 3] [3, 1, 0, 2] [1, 3, 0, 2] [0, 3, 1, 2] [3, 0, 1, 2] [1, 0, 3, 2] [0, 1, 3, 2] [0, 2, 3, 1] [2, 0, 3, 1] [3, 0, 2, 1] [0, 3, 2, 1] [2, 3, 0, 1] [3, 2, 0, 1] [3, 2, 1, 0] [2, 3, 1, 0] [1, 3, 2, 0] [3, 1, 2, 0] [2, 1, 3, 0] [1, 2, 3, 0]
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
fast, iterative with coroutine to use as a generator
<lang lua>
- !/usr/bin/env luajit
-- Iterative version local function ipermgen(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
coroutine.yield(slots)
taken[slots[index]] = false
break
end
index = index + 1
until true end
end local function iperm(a) local co=coroutine.create(function() ipermgen(a,a) end) return function() local code,res=coroutine.resume(co) return res end end
local a=arg[1] and tonumber(arg[1]) or 3 for p in iperm(a) do print(table.concat(p, " ")) end </lang>
- Output:
> ./perm_iter_coroutine.lua 3 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1
M2000 Interpreter
All permutations in one module
<lang M2000 Interpreter> Module Checkit {
Global a$
Document a$
Module Permutations (s){
Module Level (n, s, h) {
If n=1 then {
while Len(s) {
m1=each(h)
while m1 {
Print Array$(m1);" ";
}
Print Array$(S)
ToClipBoard()
s=cdr(s)
}
} Else {
for i=1 to len(s) {
call Level n-1, cdr(s), cons(h, car(s))
s=cons(cdr(s), car(s))
}
}
Sub ToClipBoard()
local m=each(h)
Local b$=""
While m {
b$+=If$(Len(b$)<>0->" ","")+Array$(m)+" "
}
b$+=If$(Len(b$)<>0->" ","")+Array$(s,0)+" "+{
}
a$<=b$ ' assign to global need <=
End Sub
}
If len(s)=0 then Error
Head=(,)
Call Level Len(s), s, Head
}
Clear a$
Permutations (1,2,3,4)
Permutations (100, 200, 500)
Permutations ("A", "B", "C","D")
Permutations ("DOG", "CAT", "BAT")
ClipBoard a$
} Checkit </lang>
Step by step Generator
<lang M2000 Interpreter> Module StepByStep {
Function PermutationStep (a) {
c1=lambda (&f, a) ->{
=car(a)
f=true
}
m=len(a)
c=c1
while m>1 {
c1=lambda c2=c,p, m=(,) (&f, a) ->{
if len(m)=0 then m=a
=cons(car(m),c2(&f, cdr(m)))
if f then f=false:p++: m=cons(cdr(m), car(m)) : if p=len(m) then p=0 : m=(,):: f=true
}
c=c1
m--
}
=lambda c, a (&f) -> {
=c(&f, a)
}
}
k=false
StepA=PermutationStep((1,2,3,4))
while not k {
Print StepA(&k)
}
k=false
StepA=PermutationStep((100,200,300))
while not k {
Print StepA(&k)
}
k=false
StepA=PermutationStep(("A", "B", "C", "D"))
while not k {
Print StepA(&k)
}
k=false
StepA=PermutationStep(("DOG", "CAT", "BAT"))
while not k {
Print StepA(&k)
}
} StepByStep
</lang>
- Output:
1 2 3 4 1 2 4 3 1 3 4 2 1 3 2 4 1 4 2 3 1 4 3 2 2 3 4 1 2 3 1 4 2 4 1 3 2 4 3 1 2 1 3 4 2 1 4 3 3 4 1 2 3 4 2 1 3 1 2 4 3 1 4 2 3 2 4 1 3 2 1 4 4 1 2 3 4 1 3 2 4 2 3 1 4 2 1 3 4 3 1 2 4 3 2 1 100 200 500 100 500 200 200 500 100 200 100 500 500 100 200 500 200 100 A B C D A B D C A C D B A C B D A D B C A D C B B C D A B C A D B D A C B D C A B A C D B A D C C D A B C D B A C A B D C A D B C B D A C B A D D A B C D A C B D B C A D B A C D C A B D C B A DOG CAT BAT DOG BAT CAT CAT BAT DOG CAT DOG BAT BAT DOG CAT BAT CAT DOG
m4
A peculiarity of this implementation is my use of arithmetic rather than branching to compute Sedgewick’s ‘k’. (I use arithmetic similarly in my Ratfor 77 implementation.)
<lang m4>divert(-1)
- 1-based indexing of a string's characters.
define(`get',`substr(`$1',decr(`$2'),1)') define(`set',`substr(`$1',0,decr(`$2'))`'$3`'substr(`$1',`$2')') define(`swap', `pushdef(`_u',`get(`$1',`$2')')`'dnl pushdef(`_v',`get(`$1',`$3')')`'dnl set(set(`$1',`$2',_v),`$3',_u)`'dnl popdef(`_u',`_v')')
- $1-fold repetition of $2.
define(`repeat',`ifelse($1,0,`',`$2`'$0(decr($1),`$2')')')
- Heap's algorithm. Algorithm 2 in Robert Sedgewick, 1977. Permutation
- generation methods. ACM Comput. Surv. 9, 2 (June 1977), 137-164.
- This implementation permutes the characters in a string of length no
- more than 9. On longer strings, it may strain the resources of a
- very old implementation of m4.
define(`permutations',
`ifelse($2,`',`$1
$0(`$1',repeat(len(`$1'),1),2)',
`ifelse(eval(($3) <= len(`$1')),1,
`ifelse(eval(get($2,$3) < $3),1,
`swap(`$1',_$0($2,$3),$3)
$0(swap(`$1',_$0($2,$3),$3),set($2,$3,incr(get($2,$3))),2)',
`$0(`$1',set($2,$3,1),incr($3))')')')')
define(`_permutations',`eval((($2) % 2) + ((1 - (($2) % 2)) * get($1,$2)))')
divert`'dnl permutations(`123') permutations(`abcd')</lang>
- Output:
$ m4 permutations.m4
123 213 312 132 231 321 abcd bacd cabd acbd bcad cbad dbac bdac adbc dabc badc abdc acdb cadb dacb adcb cdab dcab dcba cdba bdca dbca cbda bcda
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>
An implementation based on Mathematica solution:
<lang maple>fold:=(f,a,v)->`if`(nops(v)=0,a,fold(f,f(a,op(1,v)),[op(2...,v)])): insert:=(v,a,n)->`if`(n>nops(v),[op(v),a],subsop(n=(a,v[n]),v)): perm:=s->fold((a,b)->map(u->seq(insert(u,b,k+1),k=0..nops(u)),a),[[]],s): perm([$1..3]);
[[3, 2, 1], [2, 3, 1], [2, 1, 3], [3, 1, 2], [1, 3, 2], [1, 2, 3]]</lang>
Mathematica/Wolfram Language
Note: The built-in version will have better performance.
Version from scratch
<lang Mathematica> (***Standard list functions:*) fold[f_, x_, {}] := x fold[f_, x_, {h_, t___}] := fold[f, f[x, h], {t}] insert[L_, x_, n_] := Join[L;; n - 1, {x}, Ln ;;]
(***Generate all permutations of a list S:*)
permutations[S_] :=
fold[Join @@ (Function[{L},
Table[insert[L, #2, k + 1], {k, 0, Length[L]}]] /@ #1) &, {{}},
S]
</lang>
- Output:
{{4, 3, 2, 1}, {3, 4, 2, 1}, {3, 2, 4, 1}, {3, 2, 1, 4}, {4, 2, 3,
1}, {2, 4, 3, 1}, {2, 3, 4, 1}, {2, 3, 1, 4}, {4, 2, 1, 3}, {2, 4,
1, 3}, {2, 1, 4, 3}, {2, 1, 3, 4}, {4, 3, 1, 2}, {3, 4, 1, 2}, {3,
1, 4, 2}, {3, 1, 2, 4}, {4, 1, 3, 2}, {1, 4, 3, 2}, {1, 3, 4,
2}, {1, 3, 2, 4}, {4, 1, 2, 3}, {1, 4, 2, 3}, {1, 2, 4, 3}, {1, 2,
3, 4}}
Built-in version
<lang Mathematica>Permutations[{1,2,3,4}]</lang>
- Output:
{{1, 2, 3, 4}, {1, 2, 4, 3}, {1, 3, 2, 4}, {1, 3, 4, 2}, {1, 4, 2, 3}, {1, 4, 3, 2}, {2, 1, 3, 4}, {2, 1, 4, 3}, {2, 3, 1, 4}, {2, 3,
4, 1}, {2, 4, 1, 3}, {2, 4, 3, 1}, {3, 1, 2, 4}, {3, 1, 4, 2}, {3, 2, 1, 4}, {3, 2, 4, 1}, {3, 4, 1, 2}, {3, 4, 2, 1}, {4, 1, 2,
3}, {4, 1, 3, 2}, {4, 2, 1, 3}, {4, 2, 3, 1}, {4, 3, 1, 2}, {4, 3, 2, 1}}
MATLAB / Octave
<lang MATLAB>perms([1,2,3,4])</lang>
- Output:
4321 4312 4231 4213 4123 4132 3421 3412 3241 3214 3124 3142 2341 2314 2431 2413 2143 2134 1324 1342 1234 1243 1423 1432
Maxima
<lang maxima>next_permutation(v) := block([n, i, j, k, t],
n: length(v), i: 0, for k: n - 1 thru 1 step -1 do (if v[k] < v[k + 1] then (i: k, return())), j: i + 1, k: n, while j < k do (t: v[j], v[j]: v[k], v[k]: t, j: j + 1, k: k - 1), if i = 0 then return(false), j: i + 1, while v[j] < v[i] do j: j + 1, t: v[j], v[j]: v[i], v[i]: t, true
)$
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>
Mercury
<lang mercury>
- - module permutations2.
- - interface.
- - import_module io.
- - pred main(io::di, io::uo) is det.
- - import_module list.
- - import_module set_ordlist.
- - import_module set.
- - import_module solutions.
%% permutationSet(List, Set) is true if List is a permutation of Set:
- - pred permutationSet(list(A)::out,set(A)::in) is nondet.
%% Two ways to compute all permutations of a given list (using backtracking):
- - func all_permutations1(list(int))=set_ordlist.set_ordlist(list(int)).
- - func all_permutations2(list(int))=set_ordlist.set_ordlist(list(int)).
- - implementation.
permutationSet([],set.init).
permutationSet([H|T], S) :- set.member(H,S), permutationSet(T,set.delete(S,H)).
all_permutations1(L) =
solutions_set(pred(X::out) is nondet:-permutationSet(X,set.from_list(L))).
%%Alternatively, using the imported list.perm predicate: all_permutations2(L) =
solutions_set(pred(X::out) is nondet:-perm(L,X)).
main(!IO) :-
print(all_permutations1([1,2,3,4]),!IO), nl(!IO), print(all_permutations2([1,2,3,4]),!IO).
</lang>
- Output:
>./permutations2 sol([[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]]) sol([[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]])
Microsoft Small Basic
<lang smallbasic>'Permutations - sb
n=4
printem = "True"
For i = 1 To n
p[i] = i
EndFor
count = 0
Last = "False"
While Last = "False"
If printem Then
For t = 1 To n
TextWindow.Write(p[t])
EndFor
TextWindow.WriteLine("")
EndIf
count = count + 1
Last = "True"
i = n - 1
While i > 0
If p[i] < p[i + 1] Then
Last = "False"
Goto exitwhile
EndIf
i = i - 1
EndWhile
exitwhile:
j = i + 1
k = n
While j < k
t = p[j]
p[j] = p[k]
p[k] = t
j = j + 1
k = k - 1
EndWhile
j = n
While p[j] > p[i]
j = j - 1
EndWhile
j = j + 1
t = p[i]
p[i] = p[j]
p[j] = t
EndWhile
TextWindow.WriteLine("Number of permutations: "+count) </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 Number of permutations: 24
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>
Modula-3
Simple version
This implementation merely prints out the orbit of the list (1, 2, ..., n) under the action of Sn. It shows off Modula-3's built-in Set type and uses the standard IntSeq library module.
<lang modula2>MODULE Permutations EXPORTS Main;
IMPORT IO, IntSeq;
CONST n = 3;
TYPE Domain = SET OF [ 1.. n ];
VAR
chosen: IntSeq.T;
values := Domain { };
PROCEDURE GeneratePermutations(VAR chosen: IntSeq.T; remaining: Domain) = (*
Recursively generates all the permutations of elements in the union of "chosen" and "values". Values in "chosen" have already been chosen; values in "remaining" can still be chosen. If "remaining" is empty, it prints the sequence and returns. Otherwise, it picks each element in "remaining", removes it, adds it to "chosen", recursively calls itself, then removes the last element of "chosen" and adds it back to "remaining".
- )
BEGIN
FOR i := 1 TO n DO
(* check if each element is in "remaining" *)
IF i IN remaining THEN
(* if so, remove from "remaining" and add to "chosen" *)
remaining := remaining - Domain { i };
chosen.addhi(i);
IF remaining # Domain { } THEN
(* still something to process? do it *)
GeneratePermutations(chosen, remaining);
ELSE
(* otherwise, print what we've chosen *)
FOR j := 0 TO chosen.size() - 2 DO
IO.PutInt(chosen.get(j)); IO.Put(", ");
END;
IO.PutInt(chosen.gethi());
IO.PutChar('\n');
END;
(* add "i" back to "remaining" and remove from "chosen" *)
remaining := remaining + Domain { i };
EVAL chosen.remhi();
END;
END;
END GeneratePermutations;
BEGIN
(* initial setup *)
chosen := NEW(IntSeq.T).init(n);
FOR i := 1 TO n DO values := values + Domain { i }; END;
GeneratePermutations(chosen, values);
END Permutations.</lang>
- Output:
For reasons of space, we show only the elements of S3, but we have tested it with higher.
1, 2, 3 1, 3, 2 2, 1, 3 2, 3, 1 3, 1, 2 3, 2, 1
Generic version
This version works on any type, and requires the library's Set and Sequence. As usual in Modula-3, the generic instance will need to be instantiated for whatever type you want to use, and you will also need to instantiate a set of, sequence of, and sequence of sequences of the domain elements. This will have to be taken care of by the m3makefile.
- interface
Suppose that D is the domain of elements to be permuted. This module requires a DomainSeq (Sequence of D), a DomainSet (Set of D), and a DomainSeqSeq (Sequence of Sequences of Domain).
<lang modula3>GENERIC INTERFACE GenericPermutations(DomainSeq, DomainSet, DomainSeqSeq);
(*
"Domain" is where the things to permute come from (unused in interface). "DomainSeq" is a "Sequence" of "Domain". "DomainSet" is a "Set" of "Domain". "DomainSeqSeq" is a "Sequence" of "DomainSeq".
- )
PROCEDURE GeneratePermutations(
READONLY chosen: DomainSeq.T; READONLY remaining: DomainSet.T; READONLY result: DomainSeqSeq.T
); (*
Recursively generates all the permutations of elements in the union of "chosen" and "remaining". Values in "chosen" have already been chosen; values in "remaining" can still be chosen. If "remaining" is empty, it adds the permutation to "result". Otherwise, it picks each element in "remaining", removes it, adds it to "chosen", recursively calls itself, then removes the last element of "chosen" and adds it back to "remaining". Although the parameters are modified, we can describe them as "READONLY" because we do not re-assign them.
- )
END GenericPermutations.</lang>
- implementation
In addition to the interface's specifications, this requires a generic Domain. Some implementations of a set are not safe to iterate over while modifying (e.g., a tree), so this copies the values and iterates over them.
<lang modula3>GENERIC MODULE GenericPermutations(Domain, DomainSeq, DomainSet, DomainSeqSeq);
(*
"Domain" is where the things to permute come from. "DomainSeq" is a "Sequence" of "Domain". "DomainSet" is a "Set" of "Domain". "DomainSeqSeq" is a "Sequence" of "DomainSeq".
- )
PROCEDURE GeneratePermutations(
READONLY chosen: DomainSeq.T; READONLY remaining: DomainSet.T; READONLY result: DomainSeqSeq.T
) =
(*
Recursively generates all the permutations of elements in the union of "chosen" and "remaining". Values in "chosen" have already been chosen; values in "remaining" can still be chosen. If "remaining" is empty, it adds the permutation to "result". Otherwise, it picks each element in "remaining", removes it, adds it to "chosen", recursively calls itself, then removes the last element of "chosen" and adds it back to "remaining".
- )
VAR
r: Domain.T; (* element added to permutation *)
iterator := remaining.iterate(); (* to iterate through remaining elements *)
values := NEW(DomainSeq.T).init(remaining.size()); (* used to store values for iteration *)
BEGIN
(* cannot safely modify a set while iterating, so we'll store the values *) WHILE iterator.next(r) DO values.addhi(r); END;
(* now loop through the stored values *) FOR i := 0 TO values.size() - 1 DO
(* remove from "remaining" and add to "chosen" *) r := values.get(i); EVAL remaining.delete(r); chosen.addhi(r);
(* if this is not the last remaining elements, call recursively *)
IF remaining.size() # 0 THEN
GeneratePermutations(chosen, remaining, result);
ELSE
(* we have a new permutation; add a copy to the set *)
VAR newPerm := NEW(DomainSeq.T).init(chosen.size());
BEGIN
FOR i := 0 TO chosen.size() - 1 DO
newPerm.addhi(chosen.get(i));
END;
result.addhi(newPerm);
END;
END;
(* move r back from chosen *) EVAL remaining.insert(chosen.remhi());
END;
END GeneratePermutations;
BEGIN END GenericPermutations.</lang>
- Sample Usage
Here the domain is Integer, but the interface doesn't require that, so we "merely" need IntSeq (a Sequence of Integer), IntSetTree (a set type I use, but you could use SetDef or SetList if you prefer; I've tested it and it works), IntSeqSeq (a Sequence of Sequences of Integer), and IntPermutations, which is GenericPermutations instantiated for Integer.
<lang modula3>MODULE GPermutations EXPORTS Main;
IMPORT IO, IntSeq, IntSetTree, IntSeqSeq, IntPermutations;
CONST
n = 7;
VAR
chosen: IntSeq.T; remaining: IntSetTree.T; result: IntSeqSeq.T;
PROCEDURE Factorial(n: CARDINAL): CARDINAL = VAR result := 1; BEGIN
FOR i := 2 TO n DO result := result * i; END; RETURN result;
END Factorial;
BEGIN
(* initial setup *) chosen := NEW(IntSeq.T).init(n); remaining := NEW(IntSetTree.T).init(); result := NEW(IntSeqSeq.T).init(Factorial(n)); FOR i := 1 TO n DO EVAL remaining.insert(i); END;
IntPermutations.GeneratePermutations(chosen, remaining, result);
IO.Put("Printing "); IO.PutInt(result.size());
IO.Put(" permutations of "); IO.PutInt(n); IO.Put(" elements \n");
FOR i := 0 TO result.size() - 1 DO
FOR j := 0 TO result.get(i).size() - 1 DO
IO.PutInt(result.get(i).get(j)); IO.PutChar(' ');
END;
IO.PutChar('\n');
END;
END GPermutations.</lang>
- Output:
(somewhat edited!)
Printing 5040 permutations of 7 elements 1 2 3 4 5 6 7 1 2 3 4 5 7 6 1 2 3 4 6 5 7 ... 7 6 5 4 2 3 1 7 6 5 4 3 1 2 7 6 5 4 3 2 1
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
Using the standard library
<lang nim>import algorithm var v = [1, 2, 3] # List has to start sorted echo v while v.nextPermutation():
echo v</lang>
- Output:
[1, 2, 3] [1, 3, 2] [2, 1, 3] [2, 3, 1] [3, 1, 2] [3, 2, 1]
Single yield iterator
<lang nim> iterator inplacePermutations[T](xs: var seq[T]): var seq[T] =
assert xs.len <= 24, "permutation of array longer than 24 is not supported"
let n = xs.len - 1
var
c: array[24, int8]
i: int = 0
for i in 0 .. n: c[i] = int8(i+1)
while true:
yield xs
if i >= n: break
c[i] -= 1
let j = if (i and 1) == 1: 0 else: int(c[i])
swap(xs[i+1], xs[j])
i = 0
while c[i] == 0:
let t = i+1
c[i] = int8(t)
i = t
</lang> verification <lang nim> import intsets from math import fac block:
# test all permutations of length from 0 to 9 for l in 0..9: # prepare data var xs = newSeq[int](l) for i in 0..<l: xs[i] = i var s = initIntSet()
for cs in inplacePermutations(xs):
# each permutation must be of length l
assert len(cs) == l
# each permutation must contain digits from 0 to l-1 exactly once
var ds = newSeq[bool](l)
for c in cs:
assert not ds[c]
ds[c] = true
# generate a unique number for each permutation
var h = 0
for e in cs:
h = l * h + e
assert not s.contains(h)
s.incl(h)
# check exactly l! unique number of permutations
assert len(s) == fac(l)
</lang>
Translation of C
<lang nim># iterative Boothroyd method iterator permutations[T](ys: openarray[T]): seq[T] =
var d = 1 c = newSeq[int](ys.len) xs = newSeq[T](ys.len)
for i, y in ys: xs[i] = y yield xs
block outer:
while true:
while d > 1:
dec d
c[d] = 0
while c[d] >= d:
inc d
if d >= ys.len: break outer
let i = if (d and 1) == 1: c[d] else: 0
swap xs[i], xs[d]
yield xs
inc c[d]
var x = @[1,2,3]
for i in permutations(x):
echo i</lang>
Output:
@[1, 2, 3] @[2, 1, 3] @[3, 1, 2] @[1, 3, 2] @[2, 3, 1] @[3, 2, 1]
Translation of Go
<lang nim># Nim implementation of the (very fast) Go example.
- http://rosettacode.org/wiki/Permutations#Go
- implementing a recursive https://en.wikipedia.org/wiki/Steinhaus–Johnson–Trotter_algorithm
import algorithm
proc perm(s: openArray[int]; emit: proc(emit: openArray[int])) =
var s = @s if s.len == 0: emit(s) return
proc rc(np: int) =
if np == 1:
emit(s)
return
var
np1 = np - 1
pp = s.len - np1
rc(np1) # Recurse prior swaps.
for i in countDown(pp, 1):
swap s[i], s[i-1]
rc(np1) # Recurse swap.
s.rotateLeft(0..pp, 1)
rc(s.len)
var se = @[0, 1, 2, 3] #, 4, 5, 6, 7, 8, 9, 10]
perm(se, proc(s: openArray[int])= echo s)</lang>
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>
ooRexx
Essentially derived fom the program shown under rexx. This program works also with Regina (and other REXX implementations?) <lang oorexx> /* REXX Compute bunch permutations of things elements */ Parse Arg bunch things If bunch='?' Then
Call help
If bunch== Then bunch=3 If datatype(bunch)<>'NUM' Then Call help 'bunch ('bunch') must be numeric' thing.= Select
When things= Then things=bunch
When datatype(things)='NUM' Then Nop
Otherwise Do
data=things
things=words(things)
Do i=1 To things
Parse Var data thing.i data
End
End
End
If things<bunch Then Call help 'things ('things') must be >= bunch ('bunch')'
perms =0 Call time 'R' Call permSets things, bunch Say perms 'Permutations' Say time('E') 'seconds' Exit
/*--------------------------------------------------------------------------------------*/ first_word: return word(Arg(1),1) /*--------------------------------------------------------------------------------------*/ permSets: Procedure Expose perms thing.
Parse Arg things,bunch aa.= sep= perm_elements='123456789ABCDEF' Do k=1 To things perm=first_word(first_word(substr(perm_elements,k,1) k)) dd.k=perm End Call .permSet 1 Return
.permSet: Procedure Expose dd. aa. things bunch perms thing.
Parse Arg iteration
If iteration>bunch Then do
perm= aa.1
Do j=2 For bunch-1
perm= perm aa.j
End
perms+=1
If thing.1<> Then Do
ol=
Do pi=1 To words(perm)
z=word(perm,pi)
If datatype(z)<>'NUM' Then
z=9+pos(z,'ABCDEF')
ol=ol thing.z
End
Say strip(ol)
End
Else
Say perm
End
Else Do
Do q=1 for things
Do k=1 for iteration-1
If aa.k==dd.q Then
iterate q
End
aa.iteration= dd.q
Call .permSet iteration+1
End
End
Return
help:
Parse Arg msg If msg<> Then Do Say 'ERROR:' msg Say End Say 'rexx perm -> Permutations of 1 2 3 ' Say 'rexx perm 2 -> Permutations of 1 2 ' Say 'rexx perm 2 4 -> Permutations of 1 2 3 4 in 2 positions' Say 'rexx perm 2 a b c d -> Permutations of a b c d in 2 positions' Exit</lang>
- Output:
H:\>rexx perm 2 U V W X U V U W U X V U V W V X W U W V W X X U X V X W 12 Permutations 0.006000 seconds H:\>rexx perm ? rexx perm -> Permutations of 1 2 3 rexx perm 2 -> Permutations of 1 2 rexx perm 2 4 -> Permutations of 1 2 3 4 in 2 positions rexx perm 2 a b c d -> Permutations of a b c d in 2 positions
OpenEdge/Progress
<lang OpenEdge/Progress> DEFINE VARIABLE charArray AS CHARACTER EXTENT 3 INITIAL ["A","B","C"]. DEFINE VARIABLE sizeofArray AS INTEGER.
sizeOfArray = EXTENT(charArray).
RUN GetPermutations(1).
PROCEDURE GetPermutations:
DEFINE INPUT PARAMETER n AS INTEGER.
DEFINE VARIABLE i AS INTEGER.
DEFINE VARIABLE j AS INTEGER.
DEFINE VARIABLE currentPermutation AS CHARACTER.
REPEAT i = n TO sizeOfArray:
RUN swapValues(i,n).
RUN GetPermutations(n + 1).
RUN swapValues(i,n).
END.
IF n = sizeOfArray THEN DO:
DO j = 1 TO EXTENT(charArray):
currentPermutation = currentPermutation + charArray[j].
END.
DISPLAY currentPermutation WITH FRAME A DOWN.
END.
END PROCEDURE.
PROCEDURE swapValues:
DEFINE INPUT PARAMETER a AS INTEGER. DEFINE INPUT PARAMETER b AS INTEGER. DEFINE VARIABLE temp AS CHARACTER. temp = charArray[a]. charArray[a] = charArray[b]. charArray[b] = temp.
END PROCEDURE. </lang>
- Output:
ABC ACB BAC BCA CAB CBA
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
Permutations from integers
A console application in Free Pascal, created with the Lazarus IDE. <lang pascal> program Permutations; (* Demonstrates four closely related ways of establishing a bijection between permutations of 0..(n-1) and integers 0..(n! - 1). Each integer in that range is represented by mixed-base digits d[0..n-1], where each d[j] satisfies 0 <= d[j] <=j. The integer represented by d[0..n-1] is
d[n-1]*(n-1)! + d[n-2]*(n-2)! + ... + d[1]*1! + d[0]*0!
where the last term can be omitted in practice because d[0] is always 0. See the section "Numbering permutations" in the Wikipedia article "Permutation" (NB their digit array d is 1-based).
- )
uses SysUtils, TypInfo; type TPermIntMapping = (map_I, map_J, map_K, map_L); type TPermutation = array of integer;
// Function to map an integer to a permutation. function IntToPerm( map : TPermIntMapping;
nrItems, z : integer) : TPermutation;
var
d, lookup : array of integer; x, y : integer; h, j, k, m : integer;
begin
SetLength( result, nrItems); SetLength( lookup, nrItems); SetLength( d, nrItems); m := nrItems - 1; // Convert z to digits d[0..m] (see comment at head of program). d[0] := 0; y := z; for j := 1 to m - 1 do begin x := y div (j + 1); d[j] := y - x*(j + 1); y := x; end; d[m] := y;
// Set up the permutation elements
case map of
map_I, map_L: for j := 0 to m do lookup[j] := j;
map_J, map_K: for j := 0 to m do lookup[j] := m - j;
end;
for j := m downto 0 do begin
k := d[j];
case map of
map_I: result[lookup[k]] := m - j;
map_J: result[j] := lookup[k];
map_K: result[lookup[k]] := j;
map_L: result[m - j] := lookup[k];
end;
// When lookup[k] has been used, it's removed from the lookup table
// and the elements above it are moved down one place.
for h := k to j - 1 do lookup[h] := lookup[h + 1];
end;
end;
// Function to map a permutation to an integer; inverse of the above. // Put in for completeness, not required for Rosetta Code task. function PermToInt( map : TPermIntMapping;
p : TPermutation) : integer;
var
m, i, j, k : integer; d : array of integer;
begin
m := High(p); // number of items in permutation is m + 1 SetLength( d, m + 1); for k := 0 to m do d[k] := 0; // initialize all digits to 0
// Looking for inversions
for i := 0 to m - 1 do begin
for j := i + 1 to m do begin
if p[j] < p[i] then begin
case map of
map_I : inc( d[m - p[j]]);
map_J : inc( d[j]);
map_K : inc( d[p[i]]);
map_L : inc( d[m - i]);
end;
end;
end;
end;
// Get result from its digits (see comment at head of program).
result := d[m];
for j := m downto 2 do result := result*j + d[j - 1];
end;
// Main routine to generate permutations of the integers 0..(n-1), // where n is passed as a command-line parameter, e.g. Permutations 4 var
n, n_fac, z, j : integer; nrErrors : integer; perm : TPermutation; map : TPermIntMapping; lineOut : string; pinfo : TypInfo.PTypeInfo;
begin
n := SysUtils.StrToInt( ParamStr(1));
n_fac := 1;
for j := 2 to n do n_fac := n_fac*j;
pinfo := System.TypeInfo( TPermIntMapping);
lineOut := 'integer';
for map := Low( TPermIntMapping) to High( TPermIntMapping) do begin
lineOut := lineOut + ' ' + TypInfo.GetEnumName( pinfo, ord(map));
end;
WriteLn( lineOut);
for z := 0 to n_fac - 1 do begin
lineOut := SysUtils.Format( '%7d', [z]);
for map := Low( TPermIntMapping) to High( TPermIntMapping) do begin
perm := IntToPerm( map, n, z);
// Check the inverse mapping (not required for Rosetta Code task)
Assert( z = PermToInt( map, perm));
lineOut := lineOut + ' ';
for j := 0 to n - 1 do
lineOut := lineOut + SysUtils.Format( '%d', [perm[j]]);
end;
WriteLn( lineOut);
end;
end. </lang>
- Output:
integer map_I map_J map_K map_L
0 0123 0123 0123 0123
1 0132 1023 1023 0132
2 0213 0213 0213 0213
3 0312 2013 1203 0231
4 0231 1203 2013 0312
5 0321 2103 2103 0321
6 1023 0132 0132 1023
7 1032 1032 1032 1032
8 2013 0312 0231 1203
9 3012 3012 1230 1230
10 2031 1302 2031 1302
11 3021 3102 2130 1320
12 1203 0231 0312 2013
13 1302 2031 1302 2031
14 2103 0321 0321 2103
15 3102 3021 1320 2130
16 2301 2301 2301 2301
17 3201 3201 2310 2310
18 1230 1230 3012 3012
19 1320 2130 3102 3021
20 2130 1320 3021 3102
21 3120 3120 3120 3120
22 2310 2310 3201 3201
23 3210 3210 3210 3210
Perl
A simple recursive implementation. <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 b c d/); permutation(,@input);</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
For better performance, use a module like ntheory or Algorithm::Permute.
<lang perl>use ntheory qw/forperm/; my @tasks = (qw/party sleep study/); forperm {
print "@tasks[@_]\n";
} @tasks;</lang>
- Output:
party sleep study party study sleep sleep party study sleep study party study party sleep study sleep party
Phix
with javascript_semantics requires("1.0.2") ?shorten(permutes("abcd"),"elements",5)
- Output:
{"abcd","abdc","acbd","acdb","adbc","...","dacb","dbac","dbca","dcab","dcba"," (24 elements)"}
The elements can be any type. There is also a permute() function which accepts an integer between 1 and factorial(length(s)) and returns the permutations in lexicographical position order. 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.
Phixmonti
<lang Phixmonti>include ..\Utilitys.pmt
def save
over over chain ps> swap 0 put >ps
enddef
def permute /# l l -- #/
len 2 > if
len for drop
pop swap rot swap 1 put swap permute
endfor
else
save rotate save rotate
endif
swap len if
pop rot rot 0 put
else
drop drop
endif
enddef
( ) >ps ( ) ( 1 2 3 4 ) permute ps> sort print</lang>
Picat
Picat has built-in support for permutations:
permutation(L): Generates all permutations for a list L.permutation(L,P): Generates (via backtracking) all permutations for a list L.
Recursion
Use findall/2 to find all permutations. See example below.
<lang Picat>permutation_rec1([X|Y],Z) :-
permutation_rec1(Y,W), select(X,Z,W).
permutation_rec1([],[]).
permutation_rec2([], []). permutation_rec2([X], [X]) :-!. permutation_rec2([T|H], X) :-
permutation_rec2(H, H1), append(L1, L2, H1), append(L1, [T], X1), append(X1, L2, X).</lang>
Constraint modelling
Constraint modelling only handles integers, and here generates all permutations of a list 1..N for a given N.
permutation_cp_list(L) permutes a list via permutation_cp2/1.
<lang Picat>import cp.
% Returns all permutations permutation_cp1(N) = solve_all(X) =>
X = new_list(N), X :: 1..N, all_different(X).
% Find next permutation on backtracking permutation_cp2(N,X) =>
X = new_list(N), X :: 1..N, all_different(X), solve(X).
% Use the cp approach on a list L. permutation_cp_list(L) = Perms =>
Perms = [ [L[I] : I in P] : P in permutation_cp1(L.len)].</lang>
Tests
Here is a test of the different approaches, including the two built-ins. <lang Picat>import util, cp. main =>
N = 3,
println(permutations=permutations(1..N)), % built in
println(permutation=findall(P,permutation([a,b,c],P))), % built-in
println(permutation_rec1=findall(P,permutation_rec1(1..N,P))),
println(permutation_rec2=findall(P,permutation_rec2(1..N,P))),
println(permutation_cp1=permutation_cp1(N)),
println(permutation_cp2=findall(P,permutation_cp2(N,P))),
println(permutation_cp_list=permutation_cp_list("abc")).</lang>
- Output:
permutations = [[1,2,3],[2,1,3],[2,3,1],[1,3,2],[3,1,2],[3,2,1]] permutation = [abc,acb,bac,bca,cab,cba] permutation_rec1 = [[1,2,3],[2,1,3],[2,3,1],[1,3,2],[3,1,2],[3,2,1]] permutation_rec2 = [[1,2,3],[2,1,3],[2,3,1],[1,3,2],[3,1,2],[3,2,1]] permutation_cp1 = [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]] permutation_cp2 = [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]] permutation_cp_list = [abc,acb,bac,bca,cab,cba]
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>
<lang Prolog> insert(X, L, [X|L]). insert(X, [Y|Ys], [Y|L2]) :- insert(X, Ys, L2).
permutation([], []). permutation([X|Xs], P) :- permutation(Xs, L), insert(X, L, P). </lang>
- Output:
?- permutation([a,b,c],X). X = [a, b, c] ; X = [b, a, c] ; X = [b, c, a] ; X = [a, c, b] ; X = [c, a, b] ; X = [c, b, a] ; false.
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>
Implementation using destructive list updates
<lang python> def permutations(xs):
ac = [[]]
for x in xs:
ac_new = []
for ts in ac:
for n in range(0,ts.__len__()+1):
new_ts = ts[:] #(shallow) copy of ts
new_ts.insert(n,x)
ac_new.append(new_ts)
ac=ac_new
return ac
print(permutations([1,2,3,4])) </lang>
Functional :: type-preserving
The itertools.permutations function is polymorphic in its inputs but not in its outputs – it discards the type of input lists and strings, coercing all inputs to tuples.
In this type-preserving variant, permutation is defined (without the need for mutating name-bindings) in terms of two universal abstractions: reduce and concatMap:
<lang python>Permutations of a list, string or tuple
from functools import (reduce) from itertools import (chain)
- permutations :: [a] -> a
def permutations(xs):
Type-preserving permutations of xs.
ps = reduce(
lambda a, x: concatMap(
lambda xs: (
xs[n:] + [x] + xs[0:n] for n in range(0, 1 + len(xs)))
)(a),
xs, [[]]
)
t = type(xs)
return ps if list == t else (
[.join(x) for x in ps] if str == t else [
t(x) for x in ps
]
)
- TEST ----------------------------------------------------
- main :: IO ()
def main():
Permutations of lists, strings and tuples.
print(
fTable(__doc__ + ':\n')(repr)(showList)(
permutations
)([
[1, 2, 3],
'abc',
(1, 2, 3),
])
)
- GENERIC -------------------------------------------------
- concatMap :: (a -> [b]) -> [a] -> [b]
def concatMap(f):
A concatenated list over which a function has been mapped.
The list monad can be derived by using a function f which
wraps its output in a list,
(using an empty list to represent computational failure).
return lambda xs: list(
chain.from_iterable(map(f, xs))
)
- FORMATTING ----------------------------------------------
- fTable :: String -> (a -> String) ->
- (b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
Heading -> x display function -> fx display function ->
f -> xs -> tabular string.
def go(xShow, fxShow, f, xs):
ys = [xShow(x) for x in xs]
w = max(map(len, ys))
return s + '\n' + '\n'.join(map(
lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),
xs, ys
))
return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
xShow, fxShow, f, xs
)
- showList :: [a] -> String
def showList(xs):
Stringification of a list.
return '[' + ','.join(showList(x) for x in xs) + ']' if (
isinstance(xs, list)
) else repr(xs)
- MAIN ---
if __name__ == '__main__':
main()</lang>
- Output:
[1, 2, 3] -> [[1,2,3],[2,3,1],[3,1,2],[2,1,3],[1,3,2],[3,2,1]]
'abc' -> ['abc','bca','cab','bac','acb','cba']
(1, 2, 3) -> [(1, 2, 3),(2, 3, 1),(3, 1, 2),(2, 1, 3),(1, 3, 2),(3, 2, 1)]
QBasic
<lang qbasic>SUB perms (n)
DIM a(0 TO n - 1), c(0 TO n - 1)
FOR j = 0 TO n - 1
a(j) = j + 1
PRINT a(j);
NEXT j
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 j
PRINT
c(i) = c(i) + 1
i = 0
ELSE
c(i) = 0
i = i + 1
END IF
WEND
END SUB
perms(4)</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>
Quackery
General Solution
The word perms solves a more general task; generate permutations of between a and b items (inclusive) from the specified nest.
<lang Quackery> [ stack ] is perms.min ( --> [ )
[ stack ] is perms.max ( --> [ )
forward is (perms)
[ over size
perms.min share > if
[ over temp take
swap nested join
temp put ]
over size
perms.max share < if
[ dup size times
[ 2dup i^ pluck
rot swap nested join
swap (perms) ] ]
2drop ] resolves (perms) ( [ [ --> )
[ perms.max put 1 - perms.min put [] temp put [] swap (perms) temp take perms.min release perms.max release ] is perms ( [ a b --> [ )
[ dup size dup perms ] is permutations ( [ --> [ )
' [ 1 2 3 ] permutations echo cr $ "quack" permutations 60 wrap$ $ "quack" 3 4 perms 46 wrap$</lang>
Output:
[ [ 1 2 3 ] [ 1 3 2 ] [ 2 1 3 ] [ 2 3 1 ] [ 3 1 2 ] [ 3 2 1 ] ] quack quakc qucak qucka qukac qukca qauck qaukc qacuk qacku qakuc qakcu qcuak qcuka qcauk qcaku qckua qckau qkuac qkuca qkauc qkacu qkcua qkcau uqack uqakc uqcak uqcka uqkac uqkca uaqck uaqkc uacqk uackq uakqc uakcq ucqak ucqka ucaqk ucakq uckqa uckaq ukqac ukqca ukaqc ukacq ukcqa ukcaq aquck aqukc aqcuk aqcku aqkuc aqkcu auqck auqkc aucqk auckq aukqc aukcq acquk acqku acuqk acukq ackqu ackuq akquc akqcu akuqc akucq akcqu akcuq cquak cquka cqauk cqaku cqkua cqkau cuqak cuqka cuaqk cuakq cukqa cukaq caquk caqku cauqk caukq cakqu cakuq ckqua ckqau ckuqa ckuaq ckaqu ckauq kquac kquca kqauc kqacu kqcua kqcau kuqac kuqca kuaqc kuacq kucqa kucaq kaquc kaqcu kauqc kaucq kacqu kacuq kcqua kcqau kcuqa kcuaq kcaqu kcauq qua quac quak quc quca quck quk quka qukc qau qauc qauk qac qacu qack qak qaku qakc qcu qcua qcuk qca qcau qcak qck qcku qcka qku qkua qkuc qka qkau qkac qkc qkcu qkca uqa uqac uqak uqc uqca uqck uqk uqka uqkc uaq uaqc uaqk uac uacq uack uak uakq uakc ucq ucqa ucqk uca ucaq ucak uck uckq ucka ukq ukqa ukqc uka ukaq ukac ukc ukcq ukca aqu aquc aquk aqc aqcu aqck aqk aqku aqkc auq auqc auqk auc aucq auck auk aukq aukc acq acqu acqk acu acuq acuk ack ackq acku akq akqu akqc aku akuq akuc akc akcq akcu cqu cqua cquk cqa cqau cqak cqk cqku cqka cuq cuqa cuqk cua cuaq cuak cuk cukq cuka caq caqu caqk cau cauq cauk cak cakq caku ckq ckqu ckqa cku ckuq ckua cka ckaq ckau kqu kqua kquc kqa kqau kqac kqc kqcu kqca kuq kuqa kuqc kua kuaq kuac kuc kucq kuca kaq kaqu kaqc kau kauq kauc kac kacq kacu kcq kcqu kcqa kcu kcuq kcua kca kcaq kcau
An Uncommon Ordering
Edit: I think this process is called "iterative deepening". Would love to have this confirmed or corrected.
The central idea is that given a list of the permutations of say 3 items, each permutation can be used to generate 4 of the permutations of 4 items, so for example, from [ 3 1 2 ] we can generate
[ 0 3 1 2 ][ 3 0 1 2 ][ 3 1 0 2 ][ 3 1 2 0 ]
by stuffing the 0 into each of the 4 possible positions that it could go.
The code start with a nest of all the permutations of 0 items [ [ ] ], and each time though the outer times loop (i.e. 4 times in the example) it takes each of the permutations generated so far (this is the witheach loop) and applies the central idea descried above (that is the inner times loop.)
Some aids to reading the code.
Quackery is a stack based language. If you are unfamiliar the with words swap, rot, dup, 2dup, dip, unrot or drop they can be skimmed over as "noise" to get a gist of the process.
[] creates an empty nest [ ].
times indicates that the word or nest following it is to be repeated a specified number of times. (The specified number is on the top of the stack, so 4 times [ ... ]repeats some arbitrary code 4 times.)
i returns the number of times a times loop has left to repeat. It counts down to zero.
i^ returns the number of times a times loop has been repeated. It counts up from zero.
size returns the number of items (words, numbers, nests) in a nest.
witheach indicates that the word or nest following it is to be repeated once for each item in a specified nest, with successive items from the nest available on the top of stack on each repetition.
999 ' [ 10 11 12 13 ] 3 stuff will return [ 10 11 12 999 13 ]by stuffing the number 999 into the 3rd position in the nest. (The start of a nest is the zeroth position, the end of this nest is the 5th position.)
nested join adds a nest to the end of a nest as its last item.
<lang Quackery> [ ' [ [ ] ] swap times
[ [] i rot witheach
[ dup size 1+ times
[ 2dup i^ stuff
dip rot nested join
unrot ] drop ] drop ] ] is perms ( n --> [ )
4 perms witheach [ echo cr ]</lang>
- Output:
[ 0 1 2 3 ] [ 1 0 2 3 ] [ 1 2 0 3 ] [ 1 2 3 0 ] [ 0 2 1 3 ] [ 2 0 1 3 ] [ 2 1 0 3 ] [ 2 1 3 0 ] [ 0 2 3 1 ] [ 2 0 3 1 ] [ 2 3 0 1 ] [ 2 3 1 0 ] [ 0 1 3 2 ] [ 1 0 3 2 ] [ 1 3 0 2 ] [ 1 3 2 0 ] [ 0 3 1 2 ] [ 3 0 1 2 ] [ 3 1 0 2 ] [ 3 1 2 0 ] [ 0 3 2 1 ] [ 3 0 2 1 ] [ 3 2 0 1 ] [ 3 2 1 0 ]
R
Iterative version
<lang r>next.perm <- function(a) {
n <- length(a)
i <- n
while (i > 1 && a[i - 1] >= a[i]) i <- i - 1
if (i == 1) {
NULL
} else {
j <- i
k <- n
while (j < k) {
s <- a[j]
a[j] <- a[k]
a[k] <- s
j <- j + 1
k <- k - 1
}
s <- a[i - 1]
j <- i
while (a[j] <= s) j <- j + 1
a[i - 1] <- a[j]
a[j] <- s
a
}
}
perm <- function(n) {
e <- NULL
a <- 1:n
repeat {
e <- cbind(e, a)
a <- next.perm(a)
if (is.null(a)) break
}
unname(e)
}</lang>
Example
<lang>> perm(3)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 2 3 3 [2,] 2 3 1 3 1 2 [3,] 3 2 3 1 2 1</lang>
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) 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))
- permutations in lexicographic order
(define (lperms s)
(cond [(empty? s) '()]
[(empty? (cdr s)) (list s)]
[else
(let splice ([l '()][m (car s)][r (cdr s)])
(append
(map (lambda (x) (cons m x)) (lperms (append l r)))
(if (empty? r) '()
(splice (append l (list m)) (car r) (cdr r)))))]))
(display (lperms '(A B C)))
- -> ((A B C) (A C B) (B A C) (B C A) (C A B) (C B A))
- permutations in lexicographical order using generators
(require racket/generator) (define (splice s)
(generator ()
(let outer-loop ([l '()][m (car s)][r (cdr s)])
(let ([permuter (lperm (append l r))])
(let inner-loop ([p (permuter)])
(when (not (void? p))
(let ([q (cons m p)])
(yield q)
(inner-loop (permuter))))))
(if (not (empty? r))
(outer-loop (append l (list m)) (car r) (cdr r))
(void)))))
(define (lperm s)
(generator ()
(cond [(empty? s) (yield '())]
[(empty? (cdr s)) (yield s)]
[else
(let ([splicer (splice s)])
(let loop ([q (splicer)])
(when (not (void? q))
(begin
(yield q)
(loop (splicer))))))])
(void)))
(let ([permuter (lperm '(A B C))])
(let next-perm ([p (permuter)])
(when (not (void? p))
(begin
(display p)
(next-perm (permuter))))))
- -> (A B C)(A C B)(B A C)(B C A)(C A B)(C B A)
</lang>
Raku
(formerly 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
RATFOR
For translation to FORTRAN 77 with the public domain ratfor77 preprocessor.
<lang ratfor># Heap’s algorithm for generating permutations. Algorithm 2 in
- Robert Sedgewick, 1977. Permutation generation methods. ACM
- Comput. Surv. 9, 2 (June 1977), 137-164.
define(n, 3) define(n_minus_1, 2)
implicit none
integer a(1:n)
integer c(1:n) integer i, k integer tmp
10000 format ('(', I1, n_minus_1(' ', I1), ')')
- Initialize the data to be permuted.
do i = 1, n {
a(i) = i
}
- What follows is a non-recursive Heap’s algorithm as presented by
- Sedgewick. Sedgewick neglects to fully initialize c, so I have
- corrected for that. Also I compute k without branching, by instead
- doing a little arithmetic.
do i = 1, n {
c(i) = 1
} i = 2 write (*, 10000) a while (i <= n) {
if (c(i) < i) {
k = mod (i, 2) + ((1 - mod (i, 2)) * c(i))
tmp = a(i)
a(i) = a(k)
a(k) = tmp
c(i) = c(i) + 1
i = 2
write (*, 10000) a
} else {
c(i) = 1
i = i + 1
}
}
end</lang>
Here is what the generated FORTRAN 77 code looks like: <lang fortran>C Output from Public domain Ratfor, version 1.0
implicit none
integer a(1: 3)
integer c(1: 3)
integer i, k
integer tmp
10000 format ('(', i1, 2(' ', i1), ')')
do23000 i = 1, 3
a(i) = i
23000 continue 23001 continue
do23002 i = 1, 3
c(i) = 1
23002 continue 23003 continue
i = 2
write (*, 10000) a
23004 if(i .le. 3)then
if(c(i) .lt. i)then
k = mod (i, 2) + ((1 - mod (i, 2)) * c(i))
tmp = a(i)
a(i) = a(k)
a(k) = tmp
c(i) = c(i) + 1
i = 2
write (*, 10000) a
else
c(i) = 1
i = i + 1
endif
goto 23004
endif
23005 continue
end</lang>
- Output:
$ ratfor77 permutations.r > permutations.f && f2c permutations.f && cc -o permutations permutations.c -lf2c && ./permutations
(1 2 3) (2 1 3) (3 1 2) (1 3 2) (2 3 1) (3 2 1)
REXX
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 pgm generates/displays all permutations of N different objects taken M at a time.*/
parse arg things bunch inbetweenChars names /*obtain optional arguments from the CL*/
if things== | things=="," then things= 3 /*Not specified? Then use the default.*/
if bunch== | bunch=="," then bunch= things /* " " " " " " */
/* ╔════════════════════════════════════════════════════════════════╗ */
/* ║ 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/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 permutation. */
return /* [↓] this is a recursive subroutine.*/
.permSet: procedure expose $. @. between x y; parse arg ?
if ?>y then do; _= @.1; do j=2 for y-1
_= _ || between || @.j
end /*j*/
say _
end
else do q=1 for x /*build the permutation recursively. */
do k=1 for ?-1; if @.k==$.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
Ring
<lang ring> load "stdlib.ring"
list = 1:4 lenList = len(list) permut = [] for perm = 1 to factorial(len(list))
for i = 1 to len(list)
add(permut,list[i])
next
perm(list)
next for n = 1 to len(permut)/lenList
for m = (n-1)*lenList+1 to n*lenList
see "" + permut[m]
if m < n*lenList
see ","
ok
next
see nl
next
func perm 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
permReverse(a, pos+1, elementcount)
func permReverse a, first, last
while first < last
temp = a[first]
a[first] = a[last]
a[last] = temp
first += 1
last -= 1
end
</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
Ring
<lang ring>
Another Solution
// Permutations -- Bert Mariani 2020-07-12 // Ask User for number of digits to permutate
? "Enter permutations number : " Give n n = number(n) x = 1:n // array ? "Permutations are : " count = 0
nPermutation(1,n) //===>>> START
? " " // ? = print
? "Exiting of the program... "
? "Enter to Exit : " Give m // To Exit CMD window
//====================== // Returns true only if uniq number on row
Func Place(k,i)
for j=1 to k-1
if x[j] = i // Two numbers in same row
return 0
ok
next
return 1
//====================== Func nPermutation(k, n)
for i = 1 to n
if( Place(k,i)) //===>>> Call
x[k] = i
if(k=n)
See nl
for i= 1 to n
See " "+ x[i]
next
See " "+ (count++)
else
nPermutation(k+1,n) //===>>> Call RECURSION
ok
ok
next
return
</lang>
Output:
Enter permutations number : 4 Permutations are : 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 Exiting of the program... Enter to Exit :
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]]
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
Uses Heap's algorithm. An in-place version is possible but is incompatible with Iterator.
<lang rust>pub fn permutations(size: usize) -> Permutations {
Permutations { idxs: (0..size).collect(), swaps: vec![0; size], i: 0 }
}
pub struct Permutations {
idxs: Vec<usize>, swaps: Vec<usize>, i: usize,
}
impl Iterator for Permutations {
type Item = Vec<usize>;
fn next(&mut self) -> Option<Self::Item> {
if self.i > 0 {
loop {
if self.i >= self.swaps.len() { return None; }
if self.swaps[self.i] < self.i { break; }
self.swaps[self.i] = 0;
self.i += 1;
}
self.idxs.swap(self.i, (self.i & 1) * self.swaps[self.i]);
self.swaps[self.i] += 1;
}
self.i = 1;
Some(self.idxs.clone())
}
}
fn main() {
let perms = permutations(3).collect::<Vec<_>>();
assert_eq!(perms, vec![
vec![0, 1, 2],
vec![1, 0, 2],
vec![2, 0, 1],
vec![0, 2, 1],
vec![1, 2, 0],
vec![2, 1, 0],
]);
}</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());
}
}
}
fn main() {
let mut queue = (1..4).collect::<VecDeque<_>>();
permute(&mut Vec::new(), &mut queue, &|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
Shen
<lang Shen> (define permute [] -> [] [X] -> X X -> (permute-helper [] X))
(define permute-helper _ [] -> [] Done [X|Rest] -> (append (prepend-all X (permute (append Done Rest))) (permute-helper [X|Done] Rest)) )
(define prepend-all _ [] -> [] X [Next|Rest] -> [[X|Next]|(prepend-all X Rest)] )
(set *maximum-print-sequence-size* 50)
(permute [a b c d]) </lang>
- Output:
[[a b c d] [a b d c] [a c b d] [a c d b] [a d c b] [a d b c] [b a c d] [b a d c] [b c a d] [b c d a] [b d c a] [b d a c] [c b a d] [c b d a] [c a b d] [c a d b] [c d a b] [c d b a] [d c b a] [d c a b] [d b c a] [d b a c] [d a b c] [d a c b]]
For lexical order, make a small change: <lang Shen> (define permute-helper _ [] -> [] Done [X|Rest] -> (append (prepend-all X (permute (append Done Rest))) (permute-helper (append Done [X]) Rest)) ) </lang>
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>
<lang smalltalk> ArrayedCollection extend [
permuteAndDo: aBlock
["Permute receiver in-place, and call aBlock.
Requires integer keys."
self permuteUpto: self size andDo: aBlock]
permuteUpto: n andDo: aBlock
[n = 0 ifTrue: [^aBlock value].
1 to: n do:
[:i |
self swap: i with: n.
self permuteUpto: n-1 andDo: aBlock.
self swap: i with: n]]
]
SequenceableCollection extend [
permutations
["Answer a ReadStream of permuted shallow copies of receiver."
| c |
c := MappedCollection
collection: self
map: self keys asArray.
^Generator on:
[:g |
c map permuteAndDo: [g yield: (c copyFrom: 1 to: c size)]]]
</lang>
Use example: <lang Smalltalk> st> 'Abc' permutations contents ('bcA' 'cbA' 'cAb' 'Acb' 'bAc' 'Abc' ) </lang>
Stata
Program to build a dataset containing all permutations of 1...n. Each permutation is stored as an observation.
For instance:
<lang stata>perm 4</lang>
Program
<lang stata>program perm local n=`1' local r=1 forv i=1/`n' { local r=`r'*`i' } clear qui set obs `r' forv i=1/`n' { gen p`i'=0 } mata: genperm() end
mata void genperm() { real scalar n, i, j, k, s, p real rowvector u st_view(a=., ., .) n = cols(a) u = 1..n p = 1 do { a[p++, .] = u for (i = n; i > 1; i--) { if (u[i-1] < u[i]) break } if (i > 1) { j = i k = n while (j < k) u[(j++, k--)] = u[(k, j)]
s = u[i-1] for (j = i; u[j] < s; j++) { } u[i-1] = u[j] u[j] = s } } while (i > 1) } end</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>
Tailspin
This solution seems to be the same as the Kotlin solution. Permutations flow independently without being collected until the end. <lang tailspin> templates permutations
when <=1> do [1] !
otherwise
def n: $;
templates expand
def p: $;
1..$n -> \(def k: $;
[$p(1..$k-1)..., $n, $p($k..last)...] !\) !
end expand
$n - 1 -> permutations -> expand !
end permutations
def alpha: ['ABCD'...]; [ $alpha::length -> permutations -> '$alpha($)...;' ] -> !OUT::write </lang>
- Output:
[DCBA, CDBA, CBDA, CBAD, DBCA, BDCA, BCDA, BCAD, DBAC, BDAC, BADC, BACD, DCAB, CDAB, CADB, CABD, DACB, ADCB, ACDB, ACBD, DABC, ADBC, ABDC, ABCD]
If we collect all the permutations of the next size down, we can output permutations in lexical order <lang tailspin> templates lexicalPermutations
when <=1> do [1] !
otherwise
def n: $;
def p: [ $n - 1 -> lexicalPermutations ];
1..$n -> \(def k: $;
$p... -> [ $k, $... -> \(when <$k..> do $+1! otherwise $!\)] !\) !
end lexicalPermutations
def alpha: ['ABCD'...]; [ $alpha::length -> lexicalPermutations -> '$alpha($)...;' ] -> !OUT::write </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]
That algorithm can also be written from the bottom up to produce an infinite stream of sets of larger and larger permutations, until we stop <lang tailspin> templates lexicalPermutations2
def N: $; 1 -> # when <[<[]($N)>]> do $... ! otherwise def tails: $; [1..$tails(1)::length+1 -> \( def first: $; $tails... -> [$first, $... -> \(when <$first..> do $+1! otherwise $!\)] ! \)] -> #
end lexicalPermutations2
def alpha: ['ABCD'...]; [ $alpha::length -> lexicalPermutations2 -> '$alpha($)...;' ] -> !OUT::write </lang>
- Output:
[ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, DCBA]
The solutions above create a lot of new arrays at various stages. We can also use mutable state and just emit a copy for each generated solution. <lang tailspin> templates perms
templates findPerms
when <$@perms::length..> do $@perms !
otherwise
def index: $;
$index..$@perms::length
-> \(
@perms([$, $index]): $@perms([$index, $])...;
$index + 1 -> findPerms !
\) !
@perms([last, $index..last-1]): $@perms($index..last)...;
end findPerms
@: [1..$];
1 -> findPerms !
end perms
def alpha: ['ABCD'...]; [4 -> perms -> '$alpha($)...;' ] -> !OUT::write </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]
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
True BASIC
<lang qbasic>SUB SWAP(vb1, vb2)
LET temp = vb1 LET vb1 = vb2 LET vb2 = temp
END SUB
LET n = 4 DIM a(4) DIM c(4)
FOR i = 1 TO n
LET a(i) = i
NEXT i PRINT
DO
FOR i = 1 TO n
PRINT a(i);
NEXT i
PRINT
LET i = n
DO
LET i = i - 1
LOOP UNTIL (i = 0) OR (a(i) < a(i + 1))
LET j = i + 1
LET k = n
DO WHILE j < k
CALL SWAP (a(j), a(k))
LET j = j + 1
LET k = k - 1
LOOP
IF i > 0 THEN
LET j = i + 1
DO WHILE a(j) < a(i)
LET j = j + 1
LOOP
CALL SWAP (a(i), a(j))
END IF
LOOP UNTIL i = 0 END</lang>
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 VB>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
VBScript
A recursive implementation. Arrays can contain anything, I stayed with with simple variables. (Elements could be arrays but then the printing routine should be recursive...) <lang vb> 'permutation ,recursive a=array("Hello",1,True,3.141592) cnt=0 perm a,0 wscript.echo vbcrlf &"Count " & cnt
sub print(a)
s="" for i=0 to ubound(a): s=s &" " & a(i): next: wscript.echo s : cnt=cnt+1 :
end sub sub swap(a,b) t=a: a=b :b=t: end sub
sub perm(byval a,i)
if i=ubound(a) then print a: exit sub
for j= i to ubound(a)
swap a(i),a(j)
perm a,i+1
swap a(i),a(j)
next
end sub </lang> Output
Hello 1 Verdadero 3.141592 Hello 1 3.141592 Verdadero Hello Verdadero 1 3.141592 Hello Verdadero 3.141592 1 Hello 3.141592 Verdadero 1 Hello 3.141592 1 Verdadero 1 Hello Verdadero 3.141592 1 Hello 3.141592 Verdadero 1 Verdadero Hello 3.141592 1 Verdadero 3.141592 Hello 1 3.141592 Verdadero Hello 1 3.141592 Hello Verdadero Verdadero 1 Hello 3.141592 Verdadero 1 3.141592 Hello Verdadero Hello 1 3.141592 Verdadero Hello 3.141592 1 Verdadero 3.141592 Hello 1 Verdadero 3.141592 1 Hello 3.141592 1 Verdadero Hello 3.141592 1 Hello Verdadero 3.141592 Verdadero 1 Hello 3.141592 Verdadero Hello 1 3.141592 Hello Verdadero 1 3.141592 Hello 1 Verdadero Count 24
Wren
Recursive
<lang ecmascript>var permute // recursive permute = Fn.new { |input|
if (input.count == 1) return [input]
var perms = []
var toInsert = input[0]
for (perm in permute.call(input[1..-1])) {
for (i in 0..perm.count) {
var newPerm = perm.toList
newPerm.insert(i, toInsert)
perms.add(newPerm)
}
}
return perms
}
var input = [1, 2, 3] var perms = permute.call(input) System.print("There are %(perms.count) permutations of %(input), namely:\n") perms.each { |perm| System.print(perm) }</lang>
- Output:
There are 6 permutations of [1, 2, 3], namely: [1, 2, 3] [2, 1, 3] [2, 3, 1] [1, 3, 2] [3, 1, 2] [3, 2, 1]
Iterative, lexicographical order
Output modified to follow the pattern of the recursive version. <lang ecmascript>import "/math" for Int
var input = [1, 2, 3] var perms = [input] var a = input.toList var n = a.count - 1 for (c in 1...Int.factorial(n+1)) {
var i = n - 1
var j = n
while (a[i] > a[i+1]) i = i - 1
while (a[j] < a[i]) j = j - 1
var t = a[i]
a[i] = a[j]
a[j] = t
j = n
i = i + 1
while (i < j) {
t = a[i]
a[i] = a[j]
a[j] = t
i = i + 1
j = j - 1
}
perms.add(a.toList)
} System.print("There are %(perms.count) permutations of %(input), namely:\n") perms.each { |perm| System.print(perm) }</lang>
- Output:
There are 6 permutations of [1, 2, 3], namely: [1, 2, 3] [1, 3, 2] [2, 1, 3] [2, 3, 1] [3, 1, 2] [3, 2, 1]
Library based
<lang ecmascript>import "./perm" for Perm
var a = [1, 2, 3] System.print(Perm.list(a)) // not lexicographic System.print() System.print(Perm.listLex(a)) // lexicographic</lang>
- Output:
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 2, 1], [3, 1, 2]] [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
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
Yabasic
<lang yabasic>n = 4 dim a(n), c(n)
for j = 1 to n : a(j) = j : next j
repeat
for i = 1 to n: print a(i);: next: print
i = n
repeat
i = i - 1
until (i = 0) or (a(i) < a(i+1))
j = i + 1
k = n
while j < 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
tmp = a(j) : a(j) = a(i) : a(i) = tmp
endif
until i = 0 end</lang>
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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