Sorting algorithms/Permutation sort

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Task
Sorting algorithms/Permutation sort
You are encouraged to solve this task according to the task description, using any language you may know.

Sorting Algorithm
This is a sorting algorithm. It may be applied to a set of data in order to sort it.

For other sorting algorithms, see Category:Sorting Algorithms, or:
O(n logn) Sorts
Heapsort | Mergesort | Quicksort
O(n log2n) Sorts
Shell Sort
O(n2) Sorts
Bubble sort | Cocktail sort | Comb sort | Gnome sort | Insertion sort | Selection sort | Strand sort
Other Sorts
Bead sort | Bogosort | Counting sort | Pancake sort | Permutation sort | Radix sort | Sleep sort | Stooge sort
Permutation sort, which proceeds by generating the possible permutations of the input array/list until discovering the sorted one.

Pseudocode:

while not InOrder(list) do
    nextPermutation(list)
done

Contents

[edit] ActionScript

//recursively builds the permutations of permutable, appended to front, and returns the first sorted permutation it encounters
function permutations(front:Array, permutable:Array):Array {
//If permutable has length 1, there is only one possible permutation. Check whether it's sorted
if (permutable.length==1)
return isSorted(front.concat(permutable));
else
//There are multiple possible permutations. Generate them.
var i:uint=0,tmp:Array=null;
do
{
tmp=permutations(front.concat([permutable[i]]),remove(permutable,i));
i++;
}while (i< permutable.length && tmp == null);
//If tmp != null, it contains the sorted permutation. If it does not contain the sorted permutation, return null. Either way, return tmp.
return tmp;
}
//returns the array if it's sorted, or null otherwise
function isSorted(data:Array):Array {
for (var i:uint = 1; i < data.length; i++)
if (data[i]<data[i-1])
return null;
return data;
}
//returns a copy of array with the i'th element removed
function remove(array:Array, i:uint):Array {
return array.filter(function(item,index,array){return(index !=i)}) ;
}
//wrapper around the permutation function to provide a more logical interface
function permutationSort(array:Array):Array {
return permutations([],array);
}

[edit] AutoHotkey

ahk forum: discussion

MsgBox % PermSort("")
MsgBox % PermSort("xxx")
MsgBox % PermSort("3,2,1")
MsgBox % PermSort("dog,000000,xx,cat,pile,abcde,1,cat")
 
PermSort(var) { ; SORT COMMA SEPARATED LIST
Local i, sorted
StringSplit a, var, `, ; make array, size = a0
 
v0 := a0 ; auxiliary array for permutations
Loop %v0%
v%A_Index% := A_Index
 
While unSorted("a","v") ; until sorted
NextPerm("v") ; try new permutations
 
Loop % a0 ; construct string from sorted array
i := v%A_Index%, sorted .= "," . a%i%
Return SubStr(sorted,2) ; drop leading comma
}
 
unSorted(a,v) {
Loop % %a%0-1 {
i := %v%%A_Index%, j := A_Index+1, j := %v%%j%
If (%a%%i% > %a%%j%)
Return 1
}
}
 
NextPerm(v) { ; the lexicographically next LARGER permutation of v1..v%v0%
Local i, i1, j, t
i := %v%0, i1 := i-1
While %v%%i1% >= %v%%i% {
--i, --i1
IfLess i1,1, Return 1 ; Signal the end
}
j := %v%0
While %v%%j% <= %v%%i1%
--j
t := %v%%i1%, %v%%i1% := %v%%j%, %v%%j% := t, j := %v%0
While i < j
t := %v%%i%, %v%%i% := %v%%j%, %v%%j% := t, ++i, --j
}

[edit] BBC BASIC

      DIM test(9)
test() = 4, 65, 2, 31, 0, 99, 2, 83, 782, 1
 
perms% = 0
WHILE NOT FNsorted(test())
perms% += 1
PROCnextperm(test())
ENDWHILE
PRINT ;perms% " permutations required to sort "; DIM(test(),1)+1 " items."
END
 
DEF PROCnextperm(a())
LOCAL last%, maxindex%, p%
maxindex% = DIM(a(),1)
IF maxindex% < 1 THEN ENDPROC
p% = maxindex%-1
WHILE a(p%) >= a(p%+1)
p% -= 1
IF p% < 0 THEN
PROCreverse(a(), 0, maxindex%)
ENDPROC
ENDIF
ENDWHILE
last% = maxindex%
WHILE a(last%) <= a(p%)
last% -= 1
ENDWHILE
SWAP a(p%), a(last%)
PROCreverse(a(), p%+1, maxindex%)
ENDPROC
 
DEF PROCreverse(a(), first%, last%)
WHILE first% < last%
SWAP a(first%), a(last%)
first% += 1
last% -= 1
ENDWHILE
ENDPROC
 
DEF FNsorted(d())
LOCAL I%
FOR I% = 1 TO DIM(d(),1)
IF d(I%) < d(I%-1) THEN = FALSE
NEXT
= TRUE

Output:

980559 permutations required to sort 10 items.

[edit] C

Just keep generating next lexicographic permutation until the last one; it's sorted by definition.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
typedef int(*cmp_func)(const void*, const void*);
 
void perm_sort(void *a, int n, size_t msize, cmp_func _cmp)
{
char *p, *q, *tmp = malloc(msize);
# define A(i) ((char *)a + msize * (i))
# define swap(a, b) {\
memcpy(tmp, a, msize);\
memcpy(a, b, msize);\
memcpy(b, tmp, msize); }

while (1) {
/* find largest k such that a[k - 1] < a[k] */
for (p = A(n - 1); (void*)p > a; p = q)
if (_cmp(q = p - msize, p) > 0)
break;
 
if ((void*)p <= a) break;
 
/* find largest l such that a[l] > a[k - 1] */
for (p = A(n - 1); p > q; p-= msize)
if (_cmp(q, p) > 0) break;
 
swap(p, q); /* swap a[k - 1], a[l] */
/* flip a[k] through a[end] */
for (q += msize, p = A(n - 1); q < p; q += msize, p -= msize)
swap(p, q);
}
free(tmp);
}
 
int scmp(const void *a, const void *b) { return strcmp(*(const char *const *)a, *(const char *const *)b); }
 
int main()
{
int i;
const char *strs[] = { "spqr", "abc", "giant squid", "stuff", "def" };
perm_sort(strs, 5, sizeof(*strs), scmp);
 
for (i = 0; i < 5; i++)
printf("%s\n", strs[i]);
return 0;
}

[edit] C#

 
public static class PermutationSorter
{
public static void Sort<T>(List<T> list) where T : IComparable
{
PermutationSort(list, 0);
}
public static bool PermutationSort<T>(List<T> list, int i) where T : IComparable
{
int j;
if (issorted(list, i))
{
return true;
}
for (j = i + 1; j < list.Count; j++)
{
T temp = list[i];
list[i] = list[j];
list[j] = temp;
if (PermutationSort(list, i + 1))
{
return true;
}
temp = list[i];
list[i] = list[j];
list[j] = temp;
}
return false;
}
public static bool issorted<T>(List<T> list, int i) where T : IComparable
{
for (int j = list.Count-1; j > 0; j--)
{
if(list[j].CompareTo(list[j-1])<0)
{
return false;
}
}
return true;
}
}
 

[edit] Clojure

 
(use '[clojure.contrib.combinatorics :only (permutations)])
 
(defn permutation-sort [s]
(first (filter (partial apply <=) (permutations s))))
 
(permutation-sort [2 3 5 3 5])
 

[edit] CoffeeScript

# This code takes a ridiculously inefficient algorithm and rather futilely
# optimizes one part of it. Permutations are computed lazily.
 
sorted_copy = (a) ->
# This returns a sorted copy of an array by lazily generating
# permutations of indexes and stopping when the indexes yield
# a sorted array.
indexes = [0...a.length]
ans = find_matching_permutation indexes, (permuted_indexes) ->
new_array = (a[i] for i in permuted_indexes)
console.log permuted_indexes, new_array
in_order(new_array)
(a[i] for i in ans)
 
in_order = (a) ->
# return true iff array a is in increasing order.
return true if a.length <= 1
for i in [0...a.length-1]
return false if a[i] > a[i+1]
true
 
get_factorials = (n) ->
# return an array of the first n+1 factorials, starting with 0!
ans = [1]
f = 1
for i in [1..n]
f *= i
ans.push f
ans
 
permutation = (a, i, factorials) ->
# Return the i-th permutation of an array by
# using remainders of factorials to determine
# elements.
while a.length > 0
f = factorials[a.length-1]
n = Math.floor(i / f)
i = i % f
elem = a[n]
a = a[0...n].concat(a[n+1...])
elem
# The above loop gets treated like
# an array expression, so it returns
# all the elements.
 
find_matching_permutation = (a, f_match) ->
factorials = get_factorials(a.length)
for i in [0...factorials[a.length]]
permuted_array = permutation(a, i, factorials)
if f_match permuted_array
return permuted_array
null
 
 
do ->
a = ['c', 'b', 'a', 'd']
console.log 'input:', a
ans = sorted_copy a
console.log 'DONE!'
console.log 'sorted copy:', ans
 

output

 
> coffee permute_sort.coffee
input: [ 'c', 'b', 'a', 'd' ]
[ 0, 1, 2, 3 ] [ 'c', 'b', 'a', 'd' ]
[ 0, 1, 3, 2 ] [ 'c', 'b', 'd', 'a' ]
[ 0, 2, 1, 3 ] [ 'c', 'a', 'b', 'd' ]
[ 0, 2, 3, 1 ] [ 'c', 'a', 'd', 'b' ]
[ 0, 3, 1, 2 ] [ 'c', 'd', 'b', 'a' ]
[ 0, 3, 2, 1 ] [ 'c', 'd', 'a', 'b' ]
[ 1, 0, 2, 3 ] [ 'b', 'c', 'a', 'd' ]
[ 1, 0, 3, 2 ] [ 'b', 'c', 'd', 'a' ]
[ 1, 2, 0, 3 ] [ 'b', 'a', 'c', 'd' ]
[ 1, 2, 3, 0 ] [ 'b', 'a', 'd', 'c' ]
[ 1, 3, 0, 2 ] [ 'b', 'd', 'c', 'a' ]
[ 1, 3, 2, 0 ] [ 'b', 'd', 'a', 'c' ]
[ 2, 0, 1, 3 ] [ 'a', 'c', 'b', 'd' ]
[ 2, 0, 3, 1 ] [ 'a', 'c', 'd', 'b' ]
[ 2, 1, 0, 3 ] [ 'a', 'b', 'c', 'd' ]
DONE!
sorted copy: [ 'a', 'b', 'c', 'd' ]
 

[edit] Common Lisp

Too bad sorted? vector code has to be copypasta'd. Could use map nil but that would in turn make it into spaghetti code.

The nth-permutation function is some classic algorithm from Wikipedia.

(defun factorial (n)
(loop for result = 1 then (* i result)
for i from 2 to n
finally (return result)))
 
(defun nth-permutation (k sequence)
(if (zerop (length sequence))
(coerce () (type-of sequence))
(let ((seq (etypecase sequence
(vector (copy-seq sequence))
(sequence (coerce sequence 'vector)))))
(loop for j from 2 to (length seq)
do (setq k (truncate (/ k (1- j))))
do (rotatef (aref seq (mod k j))
(aref seq (1- j)))
finally (return (coerce seq (type-of sequence)))))))
 
(defun sortedp (fn sequence)
(etypecase sequence
(list (loop for previous = #1='#:foo then i
for i in sequence
always (or (eq previous #1#)
(funcall fn i previous))))
;; copypasta
(vector (loop for previous = #1# then i
for i across sequence
always (or (eq previous #1#)
(funcall fn i previous))))))
 
(defun permutation-sort (fn sequence)
(loop for i below (factorial (length sequence))
for permutation = (nth-permutation i sequence)
when (sortedp fn permutation)
do (return permutation)))
CL-USER> (time (permutation-sort #'> '(8 3 10 6 1 9 7 2 5 4)))
Evaluation took:
5.292 seconds of real time
5.204325 seconds of total run time (5.176323 user, 0.028002 system)
[ Run times consist of 0.160 seconds GC time, and 5.045 seconds non-GC time. ]
98.34% CPU
12,337,938,025 processor cycles
611,094,240 bytes consed
 
(1 2 3 4 5 6 7 8 9 10)

[edit] C++

Since next_permutation already returns whether the resulting sequence is sorted, the code is quite simple:

#include <algorithm>
 
template<typename ForwardIterator>
void permutation_sort(ForwardIterator begin, ForwardIterator end)
{
while (std::next_permutation(begin, end))
{
// -- this block intentionally left empty --
}
}

[edit] D

[edit] Basic Version

This uses the second (lazy) permutations from the Permutations Task.

import std.stdio, std.algorithm, permutations2;
 
void permutationSort(T)(T[] items) pure nothrow @safe {
foreach (const perm; items.permutations!false)
if (perm.isSorted) {
items[] = perm[];
break;
}
}
 
void main() {
auto data = [2, 7, 4, 3, 5, 1, 0, 9, 8, 6, -1];
data.permutationSort;
data.writeln;
}
Output:
[-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Run-time about 1.34 seconds with dmd.

[edit] Alternative Version

Translation of: C++
import std.stdio, std.algorithm;
 
void permutationSort(T)(T[] items) pure nothrow {
while (items.nextPermutation) {}
}
 
void main() {
auto data = [2, 7, 4, 3, 5, 1, 0, 9, 8, 6, -1];
data.permutationSort;
data.writeln;
}

The output is the same. Run-time about 1.8 seconds with dmd (the C++ entry with G++ takes about 0.4 seconds).

[edit] E

Translation of: C++
def swap(container, ixA, ixB) {
def temp := container[ixA]
container[ixA] := container[ixB]
container[ixB] := temp
}
 
/** Reverse order of elements of 'sequence' whose indexes are in the interval [ixLow, ixHigh] */
def reverseRange(sequence, var ixLow, var ixHigh) {
while (ixLow < ixHigh) {
swap(sequence, ixLow, ixHigh)
ixLow += 1
ixHigh -= 1
}
}
 
/** Algorithm from <http://marknelson.us/2002/03/01/next-permutation>, allegedly from a version of the C++ STL */
def nextPermutation(sequence) {
def last := sequence.size() - 1
var i := last
while (true) {
var ii := i
i -= 1
if (sequence[i] < sequence[ii]) {
var j := last + 1
while (!(sequence[i] < sequence[j -= 1])) {} # buried side effect
swap(sequence, i, j)
reverseRange(sequence, ii, last)
return true
}
if (i == 0) {
reverseRange(sequence, 0, last)
return false
}
}
}
 
/** Note: Worst case on sorted list */
def permutationSort(flexList) {
while (nextPermutation(flexList)) {}
}

[edit] Go

Not following the pseudocode, it seemed simpler to just test sorted at the bottom of a recursive permutation generator.

package main
 
import "fmt"
 
var a = []int{170, 45, 75, -90, -802, 24, 2, 66}
 
// in place permutation sort of slice a
func main() {
fmt.Println("before:", a)
if len(a) > 1 && !recurse(len(a) - 1) {
// recurse should never return false from the top level.
// if it does, it means some code somewhere is busted,
// either the the permutation generation code or the
// sortedness testing code.
panic("sorted permutation not found!")
}
fmt.Println("after: ", a)
}
 
// recursive permutation generator
func recurse(last int) bool {
if last <= 0 {
// bottom of recursion. test if sorted.
for i := len(a) - 1; a[i] >= a[i-1]; i-- {
if i == 1 {
return true
}
}
return false
}
for i := 0; i <= last; i++ {
a[i], a[last] = a[last], a[i]
if recurse(last - 1) {
return true
}
a[i], a[last] = a[last], a[i]
}
return false
}

[edit] Groovy

Permutation sort is an astonishingly inefficient sort algorithm. To even begin to make it tractable, we need to be able to create enumerated permutations on the fly, rather than relying on Groovy's List.permutations() method. For a list of length N there are N! permutations. In this solution, makePermutation creates the Ith permutation to order based on a recursive construction of a unique indexed permutation. The sort method then checks to see if that permutation is sorted, and stops when it is.

I believe that this method of constructing permutations results in a stable sort, but I have not actually proven that assertion.

def factorial = { (it > 1) ? (2..it).inject(1) { i, j -> i*j } : 1 }
 
def makePermutation;
makePermutation = { list, i ->
def n = list.size()
if (n < 2) return list
def fact = factorial(n-1)
assert i < fact*n
 
def index = i.intdiv(fact)
[list[index]] + makePermutation(list[0..<index] + list[(index+1)..<n], i % fact)
}
 
def sorted = { a -> (1..<(a.size())).every { a[it-1] <= a[it] } }
 
def permutationSort = { a ->
def n = a.size()
def fact = factorial(n)
def permuteA = makePermutation.curry(a)
def pIndex = (0..<fact).find { print "."; sorted(permuteA(it)) }
permuteA(pIndex)
}

Test:

println permutationSort([7,0,12,-45,-1])
println ()
println permutationSort([10, 10.0, 10.00, 1])
println permutationSort([10, 10.00, 10.0, 1])
println permutationSort([10.0, 10, 10.00, 1])
println permutationSort([10.0, 10.00, 10, 1])
println permutationSort([10.00, 10, 10.0, 1])
println permutationSort([10.00, 10.0, 10, 1])

The examples with distinct integer and decimal values that compare as equal are there to demonstrate, but not to prove, that the sort is stable.

Output:

.............................................................................................[-45, -1, 0, 7, 12]

...................[1, 10, 10.0, 10.00]
...................[1, 10, 10.00, 10.0]
...................[1, 10.0, 10, 10.00]
...................[1, 10.0, 10.00, 10]
...................[1, 10.00, 10, 10.0]
...................[1, 10.00, 10.0, 10]

[edit] Haskell

import Control.Monad
 
permutationSort l = head [p | p <- permute l, sorted p]
 
sorted (e1 : e2 : r) = e1 <= e2 && sorted (e2 : r)
sorted _ = True
 
permute = foldM (flip insert) []
 
insert e [] = return [e]
insert e l@(h : t) = return (e : l) `mplus`
do { t' <- insert e t ; return (h : t') }
Works with: GHC version 6.10
import Data.List (permutations)
 
permutationSort l = head [p | p <- permutations l, sorted p]
 
sorted (e1 : e2 : r) = e1 <= e2 && sorted (e2 : r)
sorted _ = True

[edit] Icon and Unicon

Partly from here

procedure do_permute(l, i, n)
if i >= n then
return l
else
suspend l[i to n] <-> l[i] & do_permute(l, i+1, n)
end
 
procedure permute(l)
suspend do_permute(l, 1, *l)
end
 
procedure sorted(l)
local i
if (i := 2 to *l & l[i] >= l[i-1]) then return &fail else return 1
end
 
procedure main()
local l
l := [6,3,4,5,1]
|( l := permute(l) & sorted(l)) \1 & every writes(" ",!l)
end

[edit] NetRexx

Uses the permutation iterator RPermutationIterator at Permutations to generate the permutations.

/* NetRexx */
options replace format comments java crossref symbols nobinary
 
import java.util.List
import java.util.ArrayList
 
numeric digits 20
 
class RSortingPermutationsort public
 
properties private static
iterations
maxIterations
 
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method permutationSort(vlist = List) public static returns List
perm = RPermutationIterator(vlist)
iterations = 0
maxIterations = RPermutationIterator.factorial(vlist.size())
loop while perm.hasNext()
iterations = iterations + 1
pl = List perm.next()
if isSorted(pl) then leave
else pl = null
end
return pl
 
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method isSorted(ss = List) private static returns boolean
status = isTrue
loop ix = 1 while ix < ss.size()
vleft = Rexx ss.get(ix - 1)
vright = Rexx ss.get(ix)
if vleft.datatype('N') & vright.datatype('N')
then vtest = vleft > vright -- For numeric types we must use regular comparison.
else vtest = vleft >> vright -- For non-numeric/mixed types we must do strict comparison.
if vtest then do
status = isFalse
leave ix
end
end ix
return status
 
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method runSample(arg) private static
placesList = -
"UK London, US New York, US Boston, US Washington" -
"UK Washington, US Birmingham, UK Birmingham, UK Boston"
anotherList = 'Alpha, Beta, Gamma, Beta'
reversed = '7, 6, 5, 4, 3, 2, 1'
unsorted = '734, 3, 1, 24, 324, -1024, -666, -1, 0, 324, 99999999'
lists = [makeList(placesList), makeList(anotherList), makeList(reversed), makeList(unsorted)]
loop il = 0 while il < lists.length
vlist = lists[il]
say vlist
runtime = System.nanoTime()
rlist = permutationSort(vlist)
runtime = System.nanoTime() - runtime
if rlist \= null then say rlist
else say 'sort failed'
say 'This permutation sort of' vlist.size() 'elements took' iterations 'passes (of' maxIterations') to complete. \-'
say 'Elapsed time:' (runtime / 10 ** 9)'s.'
say
end il
return
 
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method makeList(in) public static returns List
lst = ArrayList()
loop while in > ''
parse in val ',' in
lst.add(val.strip())
end
return lst
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method main(args = String[]) public static
runSample(Rexx(args))
return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method isTrue() public static returns boolean
return (1 == 1)
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method isFalse() public static returns boolean
return (1 == 0)
 
Output:
[UK  London, US  New York, US  Boston, US  Washington UK  Washington, US  Birmingham, UK  Birmingham, UK  Boston]
[UK  Birmingham, UK  Boston, UK  London, US  Birmingham, US  Boston, US  New York, US  Washington UK  Washington]
This permutation sort of 7 elements took 4221 passes (of 5040) to complete. Elapsed time: 0.361959s.

[Alpha, Beta, Gamma, Beta]
[Alpha, Beta, Beta, Gamma]
This permutation sort of 4 elements took 2 passes (of 24) to complete. Elapsed time: 0.000113s.

[7, 6, 5, 4, 3, 2, 1]
[1, 2, 3, 4, 5, 6, 7]
This permutation sort of 7 elements took 5040 passes (of 5040) to complete. Elapsed time: 0.267956s.

[734, 3, 1, 24, 324, -1024, -666, -1, 0, 324, 99999999]
[-1024, -666, -1, 0, 1, 3, 24, 324, 324, 734, 99999999]
This permutation sort of 11 elements took 20186793 passes (of 39916800) to complete. Elapsed time: 141.461863s.

[edit] OCaml

Like the Haskell version, except not evaluated lazily. So it always computes all the permutations, before searching through them for a sorted one; which is more expensive than necessary; unlike the Haskell version, which stops generating at the first sorted permutation.

let rec sorted = function
| e1 :: e2 :: r -> e1 <= e2 && sorted (e2 :: r)
| _ -> true
 
let rec insert e = function
| [] -> [[e]]
| h :: t as l -> (e :: l) :: List.map (fun t' -> h :: t') (insert e t)
 
let permute xs = List.fold_right (fun h z -> List.concat (List.map (insert h) z))
xs [[]]
 
let permutation_sort l = List.find sorted (permute l)

[edit] J

Generally, this task should be accomplished in J using /:~. Here we take an approach that's more comparable with the other examples on this page.

A function to locate the permuation index, in the naive manner prescribed by the task:

ps =:(1+])^:((-.@-:/:~)@A.~)^:_ 0:

Of course, this can be calculated much more directly (and efficiently):

ps =: A.@:/:

Either way:

   list =: 2 7 4 3 5 1 0 9 8 6
 
ps list
2380483
 
2380483 A. list
0 1 2 3 4 5 6 7 8 9
 
(A.~ps) list
0 1 2 3 4 5 6 7 8 9

[edit] Java

import java.util.List;
import java.util.ArrayList;
import java.util.Arrays;
 
public class PermutationSort
{
public static void main(String[] args)
{
int[] a={3,2,1,8,9,4,6};
System.out.println("Unsorted: " + Arrays.toString(a));
a=pSort(a);
System.out.println("Sorted: " + Arrays.toString(a));
}
public static int[] pSort(int[] a)
{
List<int[]> list=new ArrayList<int[]>();
permute(a,a.length,list);
for(int[] x : list)
if(isSorted(x))
return x;
return a;
}
private static void permute(int[] a, int n, List<int[]> list)
{
if (n == 1)
{
int[] b=new int[a.length];
System.arraycopy(a, 0, b, 0, a.length);
list.add(b);
return;
}
for (int i = 0; i < n; i++)
{
swap(a, i, n-1);
permute(a, n-1, list);
swap(a, i, n-1);
}
}
private static boolean isSorted(int[] a)
{
for(int i=1;i<a.length;i++)
if(a[i-1]>a[i])
return false;
return true;
}
private static void swap(int[] arr,int i, int j)
{
int temp=arr[i];
arr[i]=arr[j];
arr[j]=temp;
}
}

Output:

Unsorted: [3, 2, 1, 8, 9, 4, 6]
Sorted: [1, 2, 3, 4, 6, 8, 9]

[edit] Mathematica

Here is a one-line solution. A custom order relation can be defined for the OrderedQ[] function.

PermutationSort[x_List] := NestWhile[RandomSample, x, Not[OrderedQ[#]] &]

[edit] MATLAB / Octave

function list = permutationSort(list)
 
permutations = perms(1:numel(list)); %Generate all permutations of the item indicies
 
%Test every permutation of the indicies of the original list
for i = (1:size(permutations,1))
if issorted( list(permutations(i,:)) )
list = list(permutations(i,:));
return %Once the correct permutation of the original list is found break out of the program
end
end
 
end

Sample Usage:

>> permutationSort([4 3 1 5 6 2])
 
ans =
 
1 2 3 4 5 6

[edit] Nimrod

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 outter:
while true:
while d > 1:
dec d
c[d] = 0
while c[d] >= d:
inc d
if d >= ys.len: break outter
 
let i = if (d and 1) == 1: c[d] else: 0
swap xs[i], xs[d]
yield xs
inc c[d]
 
proc isSorted[T](s: openarray[T]): bool =
var last = low(T)
for c in s:
if c < last:
return false
last = c
return true
 
proc permSort[T](a: openarray[T]): seq[T] =
for p in a.permutations:
if p.isSorted:
return p
 
var a = @[4, 65, 2, -31, 0, 99, 2, 83, 782]
echo a.permSort

Output:

@[-31, 0, 2, 2, 4, 65, 83, 99, 782]

[edit] PARI/GP

permutationSort(v)={
my(u);
for(k=1,(#v)!,
u=vecextract(v, numtoperm(#v,k));
for(i=2,#u,
if(u[i]<u[i-1], next(2))
);
return(u)
)
};

[edit] Perl

Pass a list in by reference, and sort in situ.

sub psort {
my ($x, $d) = @_;
 
unless ($d //= $#$x) {
$x->[$_] < $x->[$_ - 1] and return for 1 .. $#$x;
return 1
}
 
for (0 .. $d) {
unshift @$x, splice @$x, $d, 1;
next if $x->[$d] < $x->[$d - 1];
return 1 if psort($x, $d - 1);
}
}
 
my @a = map+(int rand 100), 0 .. 10;
print "Before:\t@a\n";
psort(\@a);
print "After:\t@a\n"
Sample output:
Before: 94 15 42 35 55 24 96 14 61 94 43
After:  14 15 24 35 42 43 55 61 94 94 96

[edit] Perl 6

# Lexicographic permuter from "Permutations" task.
sub next_perm ( @a ) {
my $j = @a.end - 1;
$j-- while $j >= 1 and [>] @a[ $j, $j+1 ];
 
my $aj = @a[$j];
my $k = @a.end;
$k-- while [>] $aj, @a[$k];
 
@a[ $j, $k ] .= reverse;
 
my Int $r = @a.end;
my Int $s = $j + 1;
while $r > $s {
@a[ $r, $s ] .= reverse;
$r--;
$s++;
}
}
 
sub permutation_sort ( @a ) {
my @n = @a.keys;
my $perm_count = [*] 1 .. +@n; # Factorial
for ^$perm_count {
my @permuted_a = @a[ @n ];
return @permuted_a if [le] @permuted_a;
next_perm(@n);
}
}
 
my @data = < c b e d a >; # Halfway between abcde and edcba
say 'Input = ' ~ @data;
say 'Output = ' ~ @data.&permutation_sort;
 
Output:
Input  = c b e d a
Output = a b c d e

[edit] PHP

function inOrder($arr){
for($i=0;$i<count($arr);$i++){
if(isset($arr[$i+1])){
if($arr[$i] > $arr[$i+1]){
return false;
}
}
}
return true;
}
 
function permute($items, $perms = array( )) {
if (empty($items)) {
if(inOrder($perms)){
return $perms;
}
} else {
for ($i = count($items) - 1; $i >= 0; --$i) {
$newitems = $items;
$newperms = $perms;
list($foo) = array_splice($newitems, $i, 1);
array_unshift($newperms, $foo);
$res = permute($newitems, $newperms);
if($res){
return $res;
}
}
}
}
 
$arr = array( 8, 3, 10, 6, 1, 9, 7, 2, 5, 4);
$arr = permute($arr);
echo implode(',',$arr);
1,2,3,4,5,6,7,8,9,10

[edit] PicoLisp

(de permutationSort (Lst)
(let L Lst
(recur (L) # Permute
(if (cdr L)
(do (length L)
(T (recurse (cdr L)) Lst)
(rot L)
NIL )
(apply <= Lst) ) ) ) )

Output:

: (permutationSort (make (do 9 (link (rand 1 999)))))
-> (82 120 160 168 205 226 408 708 719)

: (permutationSort (make (do 9 (link (rand 1 999)))))
-> (108 212 330 471 667 716 739 769 938)

: (permutationSort (make (do 9 (link (rand 1 999)))))
-> (118 253 355 395 429 548 890 900 983)

[edit] PowerShell

Function PermutationSort( [Object[]] $indata, $index = 0, $k = 0 )
{
$data = $indata.Clone()
$datal = $data.length - 1
if( $datal -gt 0 ) {
for( $j = $index; $j -lt $datal; $j++ )
{
$sorted = ( PermutationSort $data ( $index + 1 ) $j )[0]
if( -not $sorted )
{
$temp = $data[ $index ]
$data[ $index ] = $data[ $j + 1 ]
$data[ $j + 1 ] = $temp
}
}
if( $index -lt ( $datal - 1 ) )
{
PermutationSort $data ( $index + 1 ) $j
} else {
$sorted = $true
for( $i = 0; ( $i -lt $datal ) -and $sorted; $i++ )
{
$sorted = ( $data[ $i ] -le $data[ $i + 1 ] )
}
$sorted
$data
}
}
}
 
0..4 | ForEach-Object { $a = $_; 0..4 | Where-Object { -not ( $_ -match "$a" ) } |
ForEach-Object { $b = $_; 0..4 | Where-Object { -not ( $_ -match "$a|$b" ) } |
ForEach-Object { $c = $_; 0..4 | Where-Object { -not ( $_ -match "$a|$b|$c" ) } |
ForEach-Object { $d = $_; 0..4 | Where-Object { -not ( $_ -match "$a|$b|$c|$d" ) } |
ForEach-Object { $e=$_; "$( PermutationSort ( $a, $b, $c, $d, $e ) )" }
}
}
}
}
$l = 8; PermutationSort ( 1..$l | ForEach-Object { $Rand = New-Object Random }{ $Rand.Next( 0, $l - 1 ) } )

[edit] Prolog

permutation_sort(L,S) :- permutation(L,S), sorted(S).
 
sorted([]).
sorted([_]).
sorted([X,Y|ZS]) :- X =< Y, sorted([Y|ZS]).
 
permutation([],[]).
permutation([X|XS],YS) :- permutation(XS,ZS), select(X,YS,ZS).

[edit] 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 < 2
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(9)
a(0) = 8: a(1) = 3: a(2) = 10: a(3) = 6: a(4) = 1: a(5) = 9: a(6) = 7: a(7) = -4: a(8) = 5: a(9) = 3
display(a())
While nextPermutation(a()): Wend
display(a())
 
Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf

Sample output:

8, 3, 10, 6, 1, 9, 7, -4, 5, 3
-4, 1, 3, 3, 5, 6, 7, 8, 9, 10

[edit] Python

Works with: Python version 2.6
from itertools import permutations
 
in_order = lambda s: all(x <= s[i+1] for i,x in enumerate(s[:-1]))
perm_sort = lambda s: (p for p in permutations(s) if in_order(p)).next()

[edit] R

Library: e1071

Warning: This function keeps all the possible permutations in memory at once, which becomes silly when x has 10 or more elements.

permutationsort <- function(x)
{
if(!require(e1071) stop("the package e1071 is required")
is.sorted <- function(x) all(diff(x) >= 0)
 
perms <- permutations(length(x))
i <- 1
while(!is.sorted(x))
{
x <- x[perms[i,]]
i <- i + 1
}
x
}
permutationsort(c(1, 10, 9, 7, 3, 0))

[edit] Racket

 
#lang racket
(define (sort l)
(for/first ([p (in-permutations l)] #:when (apply <= p)) p))
(sort '(6 1 5 2 4 3)) ; => '(1 2 3 4 5 6)
 

[edit] REXX

/*REXX program sorts an array using the  permutation-sort  method.      */
call gen@ /*generate the array elements. */
call show@ 'before sort' /*show the before array elements.*/
call permsets items /*generate items! permutations.*/
call permSort items /*invoke the permutation sort. */
call show@ ' after sort' /*show after array elements*/
say; say 'Permutation sort took' ? "permutations to find the sorted list."
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────GEN@ subroutine─────────────────────*/
gen@: @.= /*assign default value. */
@.1 = '---Four_horsemen_of_the_Apocalypse---'
@.2 = '====================================='
@.3 = 'Famine───black_horse'
@.4 = 'Death───pale_horse'
@.5 = 'Pestilence_[Slaughter]───red_horse'
@.6 = 'Conquest_[War]───white_horse'
list= /*[↓] find # of entries in array.*/
do items=1 while @.items\==''; @@.items=@.items; end /*items*/
items=items-1 /*adjust items slightly. */
return
/*──────────────────────────────────INORDER subroutine──────────────────*/
inOrder: parse arg q /*see if list Q is in order. */
_=word(q,1); do j=2 to words(q); x=word(q,j)
if x<_ then return 0 /*Out of order? Then not sorted.*/
_=x
end /*j*/
do k=1 for items; _=word(#.?,k); @.k=@@._; end /*k*/ /*here it is*/
return 1 /*they're all in order finally. */
/*──────────────────────────────────PERMSETS subroutine─────────────────*/
permsets: procedure expose !. # #.; parse arg n,#.; #=0
do f=1 for n;  !.f=f; end /*f*/; call .permAdd /*populate 1st perm*/
do while .permNext(n,0); call .permAdd; end /*while ···*/
return #
.permNext: procedure expose !.; parse arg n,i; nm=n-1
do k=nm by -1 for nm; kp=k+1
if !.k<!.kp then do; i=k; leave; end
end /*k*/
do j=i+1 while j<n; parse value !.j !.n with !.n !.j; n=n-1; end
if i==0 then return 0; do j=i+1 while !.j<!.i; end /*j*/
parse value  !.j  !.i with  !.i  !.j
return 1
.permAdd: #=#+1; do j=1 for N; #.#=#.# !.j; end /*j*/; return
/*──────────────────────────────────PERMSORT subroutine─────────────────*/
permSort: do ?=1 until inOrder(aList) /*look for the sorted permutation*/
aList=; do m=1 for items; _=word(#.?,m); aList=aList @._; end /*m*/
end /*?*/
return
/*──────────────────────────────────SHOW@ subroutine────────────────────*/
show@: widthH=length(items) /*maximum width of any line. */
do j=1 for items; say 'element' right(j,widthH) arg(1)":" @.j; end /*j*/
say copies('─', 79) /*show a nice separator line. */
return

output

element 1 before sort: ---Four_horsemen_of_the_Apocalypse---
element 2 before sort: =====================================
element 3 before sort: Famine───black_horse
element 4 before sort: Death───pale_horse
element 5 before sort: Pestilence_[Slaughter]───red_horse
element 6 before sort: Conquest_[War]───white_horse
───────────────────────────────────────────────────────────────────────────────
element 1  after sort: ---Four_horsemen_of_the_Apocalypse---
element 2  after sort: =====================================
element 3  after sort: Conquest_[War]───white_horse
element 4  after sort: Death───pale_horse
element 5  after sort: Famine───black_horse
element 6  after sort: Pestilence_[Slaughter]───red_horse
───────────────────────────────────────────────────────────────────────────────

Permuation sort took 21 "sorts".

[edit] Ruby

Works with: Ruby version 1.8.7+

The Array class has a permutation method that, with no arguments, returns an enumerable object.

class Array
def permutationsort
permutation.each{|perm| return perm if perm.sorted?}
end
 
def sorted?
each_cons(2).all? {|a, b| a <= b}
end
end

[edit] Scheme

(define (insertions e list)
(if (null? list)
(cons (cons e list) list)
(cons (cons e list)
(map (lambda (tail) (cons (car list) tail))
(insertions e (cdr list))))))
 
(define (permutations list)
(if (null? list)
(cons list list)
(apply append (map (lambda (permutation)
(insertions (car list) permutation))
(permutations (cdr list))))))
 
(define (sorted? list)
(cond ((null? list) #t)
((null? (cdr list)) #t)
((<= (car list) (cadr list)) (sorted? (cdr list)))
(else #f)))
 
(define (permutation-sort list)
(let loop ((permutations (permutations list)))
(if (sorted? (car permutations))
(car permutations)
(loop (cdr permutations)))))

[edit] Tcl

Library: Tcllib (Package: struct::list)

The firstperm procedure actually returns the lexicographically first permutation of the input list. However, to meet the letter of the problem, let's loop:

package require Tcl 8.5
package require struct::list
 
proc inorder {list} {::tcl::mathop::<= {*}$list}
 
proc permutationsort {list} {
while { ! [inorder $list]} {
set list [struct::list nextperm $list]
}
return $list
}

[edit] Ursala

Standard library functions to generate permutations and test for ordering by a given predicate are used.

#import std
 
permsort "p" = ~&ihB+ ordered"p"*~+ permutations
 
#cast %sL
 
example = permsort(lleq) <'pmf','oao','ejw','hhp','oqh','ock','dwj'>

output:

<'dwj','ejw','hhp','oao','ock','oqh','pmf'>
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