Permutations
You are encouraged to solve this task according to the task description, using any language you may know.
Write a program which generates the all permutations of n different objects. (Practically numerals!)
- C.f.
ABAP
<lang ABAP> data: lv_flag type c,
lv_number type i,
lt_numbers type table of i.
append 1 to lt_numbers. append 2 to lt_numbers. append 3 to lt_numbers.
do.
perform permute using lt_numbers changing lv_flag.
if lv_flag = 'X'.
exit.
endif.
loop at lt_numbers into lv_number.
write (1) lv_number no-gap left-justified.
if sy-tabix <> '3'.
write ', '.
endif.
endloop.
skip.
enddo.
" Permutation function - this is used to permute: " Can be used for an unbounded size set. form permute using iv_set like lt_numbers
changing ev_last type c.
data: lv_len type i,
lv_first type i,
lv_third type i,
lv_count type i,
lv_temp type i,
lv_temp_2 type i,
lv_second type i,
lv_changed type c,
lv_perm type i.
describe table iv_set lines lv_len.
lv_perm = lv_len - 1.
lv_changed = ' '.
" Loop backwards through the table, attempting to find elements which
" can be permuted. If we find one, break out of the table and set the
" flag indicating a switch.
do.
if lv_perm <= 0.
exit.
endif.
" Read the elements.
read table iv_set index lv_perm into lv_first.
add 1 to lv_perm.
read table iv_set index lv_perm into lv_second.
subtract 1 from lv_perm.
if lv_first < lv_second.
lv_changed = 'X'.
exit.
endif.
subtract 1 from lv_perm.
enddo.
" Last permutation. if lv_changed <> 'X'. ev_last = 'X'. exit. endif.
" Swap tail decresing to get a tail increasing.
lv_count = lv_perm + 1.
do.
lv_first = lv_len + lv_perm - lv_count + 1.
if lv_count >= lv_first.
exit.
endif.
read table iv_set index lv_count into lv_temp. read table iv_set index lv_first into lv_temp_2. modify iv_set index lv_count from lv_temp_2. modify iv_set index lv_first from lv_temp. add 1 to lv_count. enddo.
lv_count = lv_len - 1.
do.
if lv_count <= lv_perm.
exit.
endif.
read table iv_set index lv_count into lv_first.
read table iv_set index lv_perm into lv_second.
read table iv_set index lv_len into lv_third.
if ( lv_first < lv_third ) and ( lv_first > lv_second ).
lv_len = lv_count.
endif.
subtract 1 from lv_count. enddo.
read table iv_set index lv_perm into lv_temp. read table iv_set index lv_len into lv_temp_2. modify iv_set index lv_perm from lv_temp_2. modify iv_set index lv_len from lv_temp.
endform.</lang>
Output:
1, 3, 2 2, 1, 3 2, 3, 1 3, 1, 2 3, 2, 1
Ada
<lang ada>-- perm.adb -- print all permutations of 1 .. n -- where n is given as a command line argument -- to compile with gnat : gnatmake perm.adb -- to call : perm n with ada.text_io, ada.command_line;
procedure perm is
use ada.text_io, ada.command_line; n : integer;
begin
if argument_count /= 1
then
put_line (command_name & " n (with n >= 1)");
return;
else
n := integer'value (argument (1));
end if;
declare
subtype element is integer range 1 .. n;
type permutation is array (element'range) of element;
p : permutation;
is_last : boolean := false;
-- compute next permutation in lexicographic order
-- iterative algorithm :
-- find longest tail-decreasing sequence in p
-- the elements from this tail cannot be permuted to get a new permutation, so
-- reverse this tail, to start from an increaing sequence, and
-- exchange the element x preceding the tail, with the minimum value in the tail,
-- that is also greater than x
procedure next is
i, j, k, t : 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
t := p (j);
p (j) := p (k);
p (k) := t;
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
t := p (i);
p (i) := p (j);
p (j) := t;
end next;
procedure print is
begin
for i in element'range loop
put (integer'image (p (i)));
end loop;
new_line;
end print;
-- initialize the permutation
procedure init is
begin
for i in element'range loop
p (i) := i;
end loop;
end init;
begin
init;
loop
print;
next;
exit when is_last;
end loop;
end;
end perm;</lang>
BBC BASIC
The procedure PROC_NextPermutation() will give the next lexicographic permutation of an integer array. <lang BBC BASIC>
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%(), firstindex, lastindex)
LOCAL first, last
first = firstindex
last = lastindex
WHILE first < last
SWAP A%(first), A%(last)
first += 1
last -= 1
ENDWHILE
ENDPROC
</lang>
C
Iterative method. Given a permutation, find the next in lexicographic order. Iterate until the last one. Here the main program shows letters and is thus limited to permutations of 26 objects. The function computing next permutation is not limited. <lang c>#include <stdio.h>
- include <stdlib.h>
/* Find next permutation in lexicographic order. Return -1 if already the last, otherwise 0. A permutation is an array[n] of value between 0 and n-1. */ int nextperm(int n, int *perm) { int i,j,k,t;
t = 0; for(i=n-2; i>=0; i--) { if(perm[i] < perm[i+1]) { t = 1; break; } }
/* last permutation if decreasing sequence */ if(!t) { return -1; }
/* swap tail decreasing to get a tail increasing */ for(j=i+1; j<n+i-j; j++) { t = perm[j]; perm[j] = perm[n+i-j]; perm[n+i-j] = t; }
/* find min acceptable value in tail */ k = n-1; for(j=n-2; j>i; j--) { if((perm[j] < perm[k]) && (perm[j] > perm[i])) { k = j; } }
/* swap with ith element (head) */ t = perm[i]; perm[i] = perm[k]; perm[k] = t;
return 0; }
int main(int argc, char **argv) { int i, n, *perm;
if(argc != 2) { printf("perm n (1 <= n <= 26\n"); return 1; } n = strtol(argv[1], NULL, 0);
perm = (int *)calloc(n, sizeof(int)); for(i=0; i<n; i++) { perm[i] = i; } do { for(i=0; i<n; i++) { printf("%c", perm[i] + 'A'); } printf("\n"); } while(!nextperm(n, perm)); free(perm); return 0; }</lang>
Sample result :
perm 3
ABC ACB BAC BCA CAB CBA
C++
The STL 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(int argc, char **argv) {
//Permutations for strings
std::string example("Hello");
while (std::next_permutation(example.begin(), example.end()))
std::cout << example << std::endl;
// 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());
while (std::next_permutation(another.begin(), another.end()))
print(another);
/* and so on...*/ return 0;
}</lang> Output:
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 1234,1234,9999,4321 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
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>
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>
D
<lang d>import std.stdio: writeln;
T[][] permutations(T)(T[] items) {
T[][] result;
void perms(T[] s, T[] prefix=[]) {
if (s.length)
foreach (i, c; s)
perms(s[0 .. i] ~ s[i+1 .. $], prefix ~ c);
else
result ~= prefix;
}
perms(items); return result;
}
void main() {
foreach (p; permutations([1, 2, 3]))
writeln(p);
}</lang> Output:
[1, 2, 3] [1, 3, 2] [2, 1, 3] [2, 3, 1] [3, 1, 2] [3, 2, 1]
Delphi
<lang Delphi>program TestPermutations;
{$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
Erlang
<lang Erlang>module(permute). -export([permute/1]). -import(lists, [append/1, append/2, delete/2, map/2]).
permute([]) ->
[[]];
permute(L) ->
append(map(fun(X) ->
map(fun(P) -> append([X],P) end, permute(delete(X, L)))
end, L)).</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]]
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: <lang> 1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1</lang>
Here is an alternate, iterative version in Fortran 77. Based on Ada version. <lang fortran> program nptest
integer n,i,a
logical nextp
external nextp
parameter(n=4)
dimension a(n)
do i=1,n
a(i)=i
enddo
10 print *,(a(i),i=1,n)
if(nextp(n,a)) go to 10
end
function nextp(n,a)
integer n,a,i,j,k,t
logical nextp
dimension a(n)
i=n-1
10 if(a(i).lt.a(i+1)) go to 20
i=i-1
if(i.eq.0) go to 20
go to 10
20 j=i+1
k=n
30 t=a(j)
a(j)=a(k)
a(k)=t
j=j+1
k=k-1
if(j.lt.k) go to 30
j=i
if(j.ne.0) go to 40
nextp=.false.
return
40 j=j+1
if(a(j).lt.a(i)) go to 40
t=a(i)
a(i)=a(j)
a(j)=t
nextp=.true.
end</lang>
GAP
GAP can handle permutations and groups. Here is a straightforward implementation : for each permutation p in S(n) (symmetric group), compute the images of 1...n by p. As an alternative, List(SymmetricGroup(n)) would yield the permutations as GAP Permutation objects, which would probably be more manageable in later computations. <lang gap>gap>perms := n -> List(SymmetricGroup(n), p -> List([1..n], x -> x^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>
Go
<lang go>package main
import "fmt"
func main() {
demoPerm(3)
}
func demoPerm(n int) {
// create a set to permute. for demo, use the integers 1..n.
s := make([]int, n)
for i := range s {
s[i] = i + 1
}
// permute them, calling a function for each permutation.
// for demo, function just prints the permutation.
permute(s, func(p []int) { fmt.Println(p) })
}
// permute function. takes a set to permute and a function // to call for each generated permutation. func permute(s []int, emit func([]int)) {
if len(s) == 0 {
emit(s)
return
}
// Steinhaus, implemented with a recursive closure.
// arg is number of positions left to permute.
// pass in len(s) to start generation.
// on each call, weave element at pp through the elements 0..np-2,
// then restore array to the way it was.
var rc func(int)
rc = func(np int) {
if np == 1 {
emit(s)
return
}
np1 := np - 1
pp := len(s) - np1
// weave
rc(np1)
for i := pp; i > 0; i-- {
s[i], s[i-1] = s[i-1], s[i]
rc(np1)
}
// restore
w := s[0]
copy(s, s[1:pp+1])
s[pp] = w
}
rc(len(s))
}</lang> Output:
[1 2 3] [1 3 2] [3 1 2] [2 1 3] [2 3 1] [3 2 1]
Haskell
<lang haskell>import Data.List (permutations)
main = mapM_ print (permutations [1,2,3])</lang>
Icon and Unicon
<lang unicon>procedure main(A)
every p := permute(A) do every writes((!p||" ")|"\n")
end
procedure permute(A)
if *A <= 1 then return A suspend [(A[1]<->A[i := 1 to *A])] ||| permute(A[2:0])
end</lang> A sample run:
->permute Aardvarks eat ants Aardvarks eat ants Aardvarks ants eat eat Aardvarks ants eat ants Aardvarks ants eat Aardvarks ants Aardvarks eat ->
J
<lang j>perms=: A.&i.~ !</lang>
Example use:
<lang j> perms 2 0 1 1 0
({~ perms@#)&.;: 'some random text'
some random text some text random random some text random text some text some random text random some</lang>
Java
Using the code of Michael Gilleland. <lang java>public class PermutationGenerator {
private int[] array; private int firstNum; private boolean firstReady = false;
public PermutationGenerator(int n, int firstNum_) {
if (n < 1) {
throw new IllegalArgumentException("The n must be min. 1");
}
firstNum = firstNum_;
array = new int[n];
reset();
}
public void reset() {
for (int i = 0; i < array.length; i++) {
array[i] = i + firstNum;
}
firstReady = false;
}
public boolean hasMore() {
boolean end = firstReady;
for (int i = 1; i < array.length; i++) {
end = end && array[i] < array[i-1];
}
return !end;
}
public int[] getNext() {
if (!firstReady) {
firstReady = true;
return array;
}
int temp;
int j = array.length - 2;
int k = array.length - 1;
// Find largest index j with a[j] < a[j+1]
for (;array[j] > array[j+1]; j--);
// Find index k such that a[k] is smallest integer
// greater than a[j] to the right of a[j]
for (;array[j] > array[k]; k--);
// Interchange a[j] and a[k]
temp = array[k];
array[k] = array[j];
array[j] = temp;
// Put tail end of permutation after jth position in increasing order
int r = array.length - 1;
int s = j + 1;
while (r > s) {
temp = array[s];
array[s++] = array[r];
array[r--] = temp;
}
return array; } // getNext()
// For testing of the PermutationGenerator class
public static void main(String[] args) {
PermutationGenerator pg = new PermutationGenerator(3, 1);
while (pg.hasMore()) {
int[] temp = pg.getNext();
for (int i = 0; i < temp.length; i++) {
System.out.print(temp[i] + " ");
}
System.out.println();
}
}
} // class</lang>
If I tested the program for n=3 with beginning 1, I got this 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
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>
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>
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>
PARI/GP
<lang>vector(n!,k,numtoperm(n,k))</lang>
Perl
<lang perl6># quick and dirty recursion sub permutation(){ my ($perm,@set) = @_; print "$perm\n" || return unless (@set); &permutation($perm.$set[$_],@set[0..$_-1],@set[$_+1..$#set]) foreach (0..$#set); } @input = (a,2,c,4); &permutation(,@input); </lang>
Output:
a2c4 a24c ac24 ac42 a42c a4c2 2ac4 2a4c 2ca4 2c4a 24ac 24ca ca24 ca42 c2a4 c24a c4a2 c42a 4a2c 4ac2 42ac 42ca 4ca2 4c2a
Perl 6
This is 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
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))
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>
Example of 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> Example of output : <lang Prolog> ?- permut_Prolog(P, [ab, cd, ef]), writeln(P), fail. [ab,cd,ef] [ab,ef,cd] [cd,ab,ef] [cd,ef,ab] [ef,ab,cd] [ef,cd,ab] false. </lang>
PureBasic
The procedure nextPermutation() takes an array of integers as input and transforms its contents into the next lexicographic permutation of it's elements (i.e. integers). It returns #True if this is possible. It returns #False if there are no more lexicographic permutations left and arranges the elements into the lowest lexicographic permutation. It also returns #False if there is less than 2 elemetns to permute.
The integer elements could be the addresses of objects that are pointed at instead. In this case the addresses will be permuted without respect to what they are pointing to (i.e. strings, or structures) and the lexicographic order will be that of the addresses themselves. <lang PureBasic>Macro reverse(firstIndex, lastIndex)
first = firstIndex last = lastIndex While first < last Swap cur(first), cur(last) first + 1 last - 1 Wend
EndMacro
Procedure nextPermutation(Array cur(1))
Protected first, last, elementCount = ArraySize(cur())
If elementCount < 1
ProcedureReturn #False ;nothing to permute
EndIf
;Find the lowest position pos such that [pos] < [pos+1]
Protected pos = elementCount - 1
While cur(pos) >= cur(pos + 1)
pos - 1
If pos < 0
reverse(0, elementCount)
ProcedureReturn #False ;no higher lexicographic permutations left, return lowest one instead
EndIf
Wend
;Swap [pos] with the highest positional value that is larger than [pos] last = elementCount While cur(last) <= cur(pos) last - 1 Wend Swap cur(pos), cur(last)
;Reverse the order of the elements in the higher positions reverse(pos + 1, elementCount) ProcedureReturn #True ;next lexicographic permutation found
EndProcedure
Procedure display(Array a(1))
Protected i, fin = ArraySize(a())
For i = 0 To fin
Print(Str(a(i)))
If i = fin: Continue: EndIf
Print(", ")
Next
PrintN("")
EndProcedure
If OpenConsole()
Dim a(2) a(0) = 1: a(1) = 2: a(2) = 3 display(a()) While nextPermutation(a()): display(a()): Wend Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input() CloseConsole()
EndIf</lang> Sample output:
1, 2, 3 1, 3, 2 2, 1, 3 2, 3, 1 3, 1, 2 3, 2, 1
Python
<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)
R
<lang r>next.perm <- function(p) { n <- length(p) i <- n - 1 r = TRUE for(i in (n-1):1) { if(p[i] < p[i+1]) { r = FALSE break } }
j <- i + 1 k <- n while(j < k) { x <- p[j] p[j] <- p[k] p[k] <- x j <- j + 1 k <- k - 1 }
if(r) return(NULL)
j <- n while(p[j] > p[i]) j <- j - 1 j <- j + 1
x <- p[i] p[i] <- p[j] p[j] <- x return(p) }
print.perms <- function(n) { p <- 1:n while(!is.null(p)) { cat(p,"\n") p <- next.perm(p) } }
print.perms(3)
- 1 2 3
- 1 3 2
- 2 1 3
- 2 3 1
- 3 1 2
- 3 2 1</lang>
REXX
<lang rexx> /*REXX program to find the missing permutation. */
/*inbetweenChars & names are optional.*/
parse arg things bunch inbetweenChars names
/*inbetweenChars defaults to a [null]. */
/* names defaults to digits (and letters). */
call permSets things,bunch,inbetweenChars,names exit
/*──────────────────────────────────────────────────────────────────────*/
permSets: procedure; parse arg x,y,between,usyms /*X things Y at a time.*/
/*X can't be > length(@0abcs). */
@abc='abcdefghijklmnopqrstuvwxyz' @abcu=@abc; upper @abcu @abcs=@abcu||@abc @0abcs=123456789||@abcs @.= sep=
do k=1 for x /*build list of symbols. */ _=p(word(usyms,k) p(substr(@0abcs,k,1) k)) /*get or generate a symbol. */ if length(_)\==1 then sep='_' /*if not 1char, then use sep*/ $.k=_ /*append to the sumbol list.*/ end
if between== then between=sep /*use appropriate seperator.*/
list='$. @. between x y' call permset(1) exit
/*────────────────────────────────PERMSET subroutine────────────────────*/ permset: procedure expose (list); parse arg ?
if ?>y then do
_=@.1
do j=2 to y
_=_||between||@.j
end
say _
end
else do q=1 for x /*construction permutation recursively*/
do k=1 for ?-1
if @.k==$.q then iterate q
end
@.?=$.q
call permset(?+1)
end
return
/*────────────────────────────────P subroutine (Pick one)───────────────*/
p: return word(arg(1),1)
</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
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]]
Scala
There is a built-in function that works on any sequential collection. It could be used as follows given a List of symbols: <lang>List('a, 'b, 'c).permutations foreach println</lang> Output:
List('a, 'b, 'c)
List('a, 'c, 'b)
List('b, 'a, 'c)
List('b, 'c, 'a)
List('c, 'a, 'b)
List('c, 'b, 'a)
Scheme
<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>
Smalltalk
<lang smalltalk>(1 to: 4) permutationsDo: [ :x | Transcript show: x printString; cr ].</lang>
Tcl
<lang tcl>package require struct::list
- Make the sequence of digits to be permuted
set n [lindex $argv 0] for {set i 1} {$i <= $n} {incr i} {lappend sequence $i}
- Iterate over the permutations, printing as we go
struct::list foreachperm p $sequence {
puts $p
}</lang>
Testing with tclsh listPerms.tcl 3 produces this output:
1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1
Ursala
In practice there's no need to write this because it's in the standard library.
<lang Ursala>#import std
permutations =
~&itB^?a( # are both the input argument list and its tail non-empty?
@ahPfatPRD *= refer ^C( # yes, recursively generate all permutations of the tail, and for each one
~&a, # insert the head at the first position
~&ar&& ~&arh2falrtPXPRD), # if the rest is non-empty, recursively insert at all subsequent positions
~&aNC) # no, return the singleton list of the argument</lang>
test program: <lang Ursala>#cast %nLL
test = permutations <1,2,3></lang> output:
< <1,2,3>, <2,1,3>, <2,3,1>, <1,3,2>, <3,1,2>, <3,2,1>>