Permutations: Difference between revisions
(→{{header|Common Lisp}}: CL is about not repeating yourself) |
m (adding sas) |
||
| Line 1,998: | Line 1,998: | ||
# => [[1, 1, 2], [1, 2, 1], [2, 1, 1]] |
# => [[1, 1, 2], [1, 2, 1], [2, 1, 1]] |
||
end</lang> |
end</lang> |
||
=={{header|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> |
|||
=={{header|Scala}}== |
=={{header|Scala}}== |
||
Revision as of 15:59, 15 November 2011
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>
ALGOL 68
File: Template_Permutations.a68 <lang algol68># Document prelude template usage: TEMPLATE(
INT upb values := 4; MODE VALUE = INT; FORMAT value fmt := $g(0)$
); #
MODE
VALVALUES = [upb values]VALUE, VALUES = REF VALVALUES, YIELDVALUES = PROC(VALUES)VOID;
FORMAT
values fmt := $"("n(upb values-1)(f(value fmt)", ")f(value fmt)")"$;
- Generate permutations of the input values of valueues #
PROC gen values permutations = (VALUES values, YIELDVALUES yield)VOID: (
- Warning: this routine does not correctly handle duplicate elements #
IF LWB values = UPB values THEN
yield(values)
ELSE
FOR elem FROM LWB values TO UPB values DO
VALUE first = values[elem];
values[LWB values+1:elem] := values[:elem-1];
values[LWB values] := first;
# FOR VALUES next values IN # gen values permutations(values[LWB values+1:] # ) DO #,
## (VALUES next)VOID:(
yield(values)
# OD #));
values[:elem-1] := values[LWB values+1:elem];
values[elem] := first
OD
FI
);
- Define some additional utility OPerators #
PRIO P = 7; # OP to calculate number of permutations # OP P = (INT n, k)INT: ( # n! OVER (n-k)! #
# ( n>k | n * ((n-1) P k) | n ); # INT out := k; FOR i FROM k+1 TO n DO out *:= i OD; out
);
- Define an operator for doing iterations over permutations #
PRIO DOPERM = 1; OP (VALUES, YIELDVALUES)VOID DOPERM = gen values permutations;
- Return an a matrix of permutations #
OP PERM = (VALUES in values)[, ]VALUE: (
[(UPB in values-LWB in values+1) P 1, LWB in values:UPB in values]VALUE out; INT elem := LWB out;
- FOR VALUES values IN # in values DOPERM (
- (VALUES values)VOID:(
out[elem, ] := values; elem +:= 1
- OD #));
out
);</lang>File: test_Permutations.a68 <lang algol68>#!/usr/local/bin/a68g --script #
PR READ "Template_Permutations.a68" PR # n.b. READ is nonstandard #
- USING( #
INT upb values := 4; MODE VALUE = INT; # user defined # FORMAT value fmt := $g(0)$
- ) #;
main:(
VALVALUES test case := (1, 22, 333, 44444);
print(("Number of permutations: ", UPB test case P 1, new line));
COMMENT # Use the generator: #
# FOR ARRAY values IN # test case DOPERM (
## (ARRAY values)VOID:(
printf((values fmt, values, $l$))
# OD #));
END COMMENT
- or simply the operator: #
printf(($f(values fmt)l$, PERM test case))
)</lang> Output:
Number of permutations: +24 (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)
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
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%(), first, last)
WHILE first < last
SWAP A%(first), A%(last)
first += 1
last -= 1
ENDWHILE
ENDPROC
</lang>
C
See lexicographic generation of permutations. <lang c>#include <stdio.h>
- include <stdlib.h>
/* print a list of ints */ int show(int *x, int len) { int i; for (i = 0; i < len; i++) printf("%d%c", x[i], i == len - 1 ? '\n' : ' '); return 1; }
/* next lexicographical permutation */ int next_lex_perm(int *a, int n) {
- define swap(i, j) {t = a[i]; a[i] = a[j]; a[j] = t;}
int k, l, t;
/* 1. Find the largest index k such that a[k] < a[k + 1]. If no such index exists, the permutation is the last permutation. */ for (k = n - 1; k && a[k - 1] >= a[k]; k--); if (!k--) return 0;
/* 2. Find the largest index l such that a[k] < a[l]. Since k + 1 is such an index, l is well defined */ for (l = n - 1; a[l] <= a[k]; l--);
/* 3. Swap a[k] with a[l] */ swap(k, l);
/* 4. Reverse the sequence from a[k + 1] to the end */ for (k++, l = n - 1; l > k; l--, k++) swap(k, l); return 1;
- undef swap
}
void perm1(int *x, int n, int callback(int *, int)) { do { if (callback) callback(x, n); } while (next_lex_perm(x, n)); }
/* Boothroyd method; exactly N! swaps, about as fast as it gets */ void boothroyd(int *x, int n, int nn, int callback(int *, int)) { int c = 0, i, t; while (1) { if (n > 2) boothroyd(x, n - 1, nn, callback); if (c >= n - 1) return;
i = (n & 1) ? 0 : c; c++; t = x[n - 1], x[n - 1] = x[i], x[i] = t; if (callback) callback(x, nn); } }
/* entry for Boothroyd method */ void perm2(int *x, int n, int callback(int*, int)) { if (callback) callback(x, n); boothroyd(x, n, n, callback); }
/* same as perm2, but flattened recursions into iterations */ void perm3(int *x, int n, int callback(int*, int)) { /* calloc isn't strictly necessary, int c[32] would suffice for most practical purposes */ int d, i, t, *c = calloc(n, sizeof(int));
/* curiously, with GCC 4.6.1 -O3, removing next line makes it ~25% slower */ if (callback) callback(x, n); for (d = 1; ; c[d]++) { while (d > 1) c[--d] = 0; while (c[d] >= d) if (++d >= n) goto done;
t = x[ i = (d & 1) ? c[d] : 0 ], x[i] = x[d], x[d] = t; if (callback) callback(x, n); } done: free(c); }
- define N 4
int main() { int i, x[N]; for (i = 0; i < N; i++) x[i] = i + 1;
/* three different methods */ perm1(x, N, show); perm2(x, N, show); perm3(x, N, show);
return 0; }</lang>
C++
The C++ standard library provides for this in the form of std::next_permutation and std::prev_permutation.
<lang cpp>#include <algorithm>
- include <string>
- include <vector>
- include <iostream>
template<class T> void print(const std::vector<T> &vec) {
for (typename std::vector<T>::const_iterator i = vec.begin(); i != vec.end(); ++i)
{
std::cout << *i;
if ((i + 1) != vec.end())
std::cout << ",";
}
std::cout << std::endl;
}
int main() {
//Permutations for strings
std::string example("Hello");
std::sort(example.begin(), example.end());
do {
std::cout << example << '\n';
} while (std::next_permutation(example.begin(), example.end()));
// And for vectors std::vector<int> another; another.push_back(1234); another.push_back(4321); another.push_back(1234); another.push_back(9999);
std::sort(another.begin(), another.end());
do {
print(another);
} while (std::next_permutation(another.begin(), another.end()));
return 0;
}</lang> Output:
Hello Helol Heoll Hlelo Hleol Hlleo Hlloe Hloel Hlole Hoell Holel Holle eHllo eHlol eHoll elHlo elHol ellHo elloH eloHl elolH eoHll eolHl eollH lHelo lHeol lHleo lHloe lHoel lHole leHlo leHol lelHo leloH leoHl leolH llHeo llHoe lleHo lleoH lloHe lloeH loHel loHle loeHl loelH lolHe loleH oHell oHlel oHlle oeHll oelHl oellH olHel olHle oleHl olelH ollHe olleH 1234,1234,4321,9999 1234,1234,9999,4321 1234,4321,1234,9999 1234,4321,9999,1234 1234,9999,1234,4321 1234,9999,4321,1234 4321,1234,1234,9999 4321,1234,9999,1234 4321,9999,1234,1234 9999,1234,1234,4321 9999,1234,4321,1234 9999,4321,1234,1234
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>
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<lang>((A B Z) (A Z B) (B A Z) (B Z A) (Z A B) (Z B A))</lang>
Lexicographic next permutation: <lang lisp> (defun next-perm (vec cmp) ; modify vector
(declare (type (simple-array * (*)) vec))
(macrolet ((el (i) `(aref vec ,i))
(cmp (i j) `(funcall cmp (el ,i) (el ,j))))
(loop with len = (1- (length vec))
for i from (1- len) downto 0
when (cmp i (1+ i)) do
(loop for k from len downto i
when (cmp i k) do
(rotatef (el i) (el k))
(setf k (1+ len))
(loop while (< (incf i) (decf k)) do
(rotatef (el i) (el k)))
(return-from next-perm vec)))))
- test code
(loop for a = "1234" then (next-perm a #'char<) while a do
(write-line a))
</lang>
D
<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, map/2, seq/2, split/2]).
%% insert element E into list L immediately after the item at position N %% (N=0 prepends E to the list) insert( L, N, E ) ->
{H, T} = split( N, L ),
append( [H, [E], T] ).
% Return a list of all the permutations of the given list permute( [] ) -> [[]]; permute( [H|T] ) ->
append( map( fun(P) ->
map( fun(N) ->
insert(P, N, H)
end,
seq( 0, length(P) ))
end,
permute( T ))).
</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
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>
<lang haskell>import Data.List (delete)
permutations [] = [[]] permutations xs = [ x:ys | x <- xs, ys <- permutations (delete x xs)] </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>
Mathematica
<lang Mathematica>Permutations[{1,2,3,4}]</lang>
Output:
{{1, 2, 3, 4}, {1, 2, 4, 3}, {1, 3, 2, 4}, {1, 3, 4, 2}, {1, 4, 2, 3}, {1, 4, 3, 2}, {2, 1, 3, 4}, {2, 1, 4, 3}, {2, 3, 1, 4}, {2, 3,
4, 1}, {2, 4, 1, 3}, {2, 4, 3, 1}, {3, 1, 2, 4}, {3, 1, 4, 2}, {3, 2, 1, 4}, {3, 2, 4, 1}, {3, 4, 1, 2}, {3, 4, 2, 1}, {4, 1, 2,
3}, {4, 1, 3, 2}, {4, 2, 1, 3}, {4, 2, 3, 1}, {4, 3, 1, 2}, {4, 3, 2, 1}}
Maxima
<lang maxima>next_permutation(v) := block([n, i, j, k],
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>
OCaml
<lang ocaml>(* Iterative, though loops are implemented as auxiliary recursive functions.
Translation of Ada version. *)
let next_perm p = let n = Array.length p in let i = let rec aux i = if (i < 0) || (p.(i) < p.(i+1)) then i else aux (i - 1) in aux (n - 2) in let rec aux j k = if j < k then let t = p.(j) in p.(j) <- p.(k); p.(k) <- t; aux (j + 1) (k - 1) else () in aux (i + 1) (n - 1); if i < 0 then false else let j = let rec aux j = if p.(j) > p.(i) then j else aux (j + 1) in aux (i + 1) in let t = p.(i) in p.(i) <- p.(j); p.(j) <- t; true;;
let print_perm p = let n = Array.length p in for i = 0 to n - 2 do print_int p.(i); print_string " " done; print_int p.(n - 1); print_newline ();;
let print_all_perm n = let p = Array.init n (function i -> i + 1) in print_perm p; while next_perm p do print_perm p done;;
print_all_perm 3;; (* 1 2 3
1 3 2 2 1 3 2 3 1 3 1 2 3 2 1 *)</lang>
Permutations can also be defined on lists recursively: <lang OCaml>let rec permutations l =
let n = List.length l in
if n = 1 then [l] else
let rec sub e = function
| [] -> failwith "sub"
| h :: t -> if h = e then t else h :: sub e t in
let rec aux k =
let e = List.nth l k in
let subperms = permutations (sub e l) in
let t = List.map (fun a -> e::a) subperms in
if k < n-1 then List.rev_append t (aux (k+1)) else t in
aux 0;;
let print l = List.iter (Printf.printf " %d") l; print_newline() in List.iter print (permutations [1;2;3;4])</lang> or permutations indexed independently: <lang OCaml>let rec pr_perm k n l =
let a, b = let c = k/n in c, k-(n*c) in
let e = List.nth l b in
let rec sub e = function
| [] -> failwith "sub"
| h :: t -> if h = e then t else h :: sub e t in
(Printf.printf " %d" e; if n > 1 then pr_perm a (n-1) (sub e l))
let show_perms l =
let n = List.length l in
let rec fact n = if n < 3 then n else n * fact (n-1) in
for i = 0 to (fact n)-1 do
pr_perm i n l;
print_newline()
done
let () = show_perms [1;2;3;4]</lang>
PARI/GP
<lang>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>
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)
Qi
<lang qi> (define insert
L 0 E -> [E|L] [L|Ls] N E -> [L|(insert Ls (- N 1) E)])
(define seq
Start Start -> [Start] Start End -> [Start|(seq (+ Start 1) End)])
(define append-lists
[] -> [] [A|B] -> (append A (append-lists B)))
(define permutate
[] -> [[]]
[H|T] -> (append-lists (map (/. P
(map (/. N
(insert P N H))
(seq 0 (length P))))
(permute T))))
</lang>
R
<lang r>next.perm <- function(p) { n <- length(p) i <- n - 1 r = 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]]
However, this method will produce indistinct permutations if the array has indistinct elements. If you need to find all the permutations of an array of which many elements are the same, the method below will be more efficient.
<lang ruby>class Array
# Yields distinct permutations of _self_ to the block. # This method requires that all array elements be Comparable. def distinct_permutation # :yields: _ary_ # If no block, return an enumerator. Works with Ruby 1.8.7. block_given? or return enum_for(:distinct_permutation)
copy = self.sort yield copy.dup return if size < 2
while true
# from: "The Art of Computer Programming" by Donald Knuth
j = size - 2;
j -= 1 while j > 0 && copy[j] >= copy[j+1]
if copy[j] < copy[j+1]
l = size - 1
l -= 1 while copy[j] >= copy[l]
copy[j] , copy[l] = copy[l] , copy[j]
copy[j+1..-1] = copy[j+1..-1].reverse
yield copy.dup
else
break
end
end
end
end
permutations = [] [1,1,2].distinct_permutation do |p| permutations << p end p permutations
- => [[1, 1, 2], [1, 2, 1], [2, 1, 1]]
if RUBY_VERSION >= "1.8.7"
p [1,1,2].distinct_permutation.to_a # => [[1, 1, 2], [1, 2, 1], [2, 1, 1]]
end</lang>
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 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> (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
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>>
VBA
Based on the Pascal version above.
<lang VBA> Public Sub Permute(n As Integer, Optional printem As Boolean = True) 'generate, count and print (if printem is not false) all permutations of first n integers
Dim P() As Integer Dim count As Long dim Last as boolean Dim t, i, j, k As Integer
If n <= 1 Then
Debug.Print "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
If Not Last Then
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
End If 'not last
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
- Programming Tasks
- Discrete math
- ABAP
- Ada
- ALGOL 68
- AutoHotkey
- BBC BASIC
- C
- C++
- Clojure
- CoffeeScript
- Common Lisp
- D
- Delphi
- Erlang
- Euphoria
- Factor
- Fortran
- GAP
- Go
- Haskell
- Icon
- Unicon
- J
- Java
- JavaScript
- Logtalk
- Mathematica
- Maxima
- OCaml
- PARI/GP
- Pascal
- Perl
- Perl 6
- PicoLisp
- Prolog
- PureBasic
- Python
- Qi
- R
- REXX
- Ruby
- SAS
- Scala
- Scheme
- Seed7
- Smalltalk
- Tcl
- Tcllib
- Ursala
- VBA