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{{task|Sorting Algorithms}} {{Sorting Algorithm}} {{omit from|GUISS}}
;Task: Implement a 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'''
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);
}
AutoHotkey
ahk forum: [http://www.autohotkey.com/forum/post-276680.html#276680 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
}
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
{{out}}
980559 permutations required to sort 10 items.
C
Just keep generating [[wp:Permutation#Systematic_generation_of_all_permutations|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;
}
C#
<lang C sharp|C#>
public static class PermutationSorter
{
public static void Sort
## C++
Since <tt>next_permutation</tt> already returns whether the resulting sequence is sorted, the code is quite simple:
```cpp
#include <algorithm>
template<typename ForwardIterator>
void permutation_sort(ForwardIterator begin, ForwardIterator end)
{
while (std::next_permutation(begin, end))
{
// -- this block intentionally left empty --
}
}
Clojure
(use '[clojure.contrib.combinatorics :only (permutations)])
(defn permutation-sort [s]
(first (filter (partial apply <=) (permutations s))))
(permutation-sort [2 3 5 3 5])
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
{{out}}
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' ]
## Common Lisp
Too bad <code>sorted?</code> vector code has to be copypasta'd. Could use <tt>map nil</tt> but that would in turn make it into spaghetti code.
The <code>nth-permutation</code> function is some classic algorithm from Wikipedia.
```lisp
(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)
Crystal
def sorted?(items : Array)
prev = items[0]
items.each do |item|
if item < prev
return false
end
prev = item
end
return true
end
def permutation_sort(items : Array)
items.each_permutation do |permutation|
if sorted?(permutation)
return permutation
end
end
end
D
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 @nogc {
foreach (const perm; items.permutations!false)
if (perm.isSorted)
break;
}
void main() {
auto data = [2, 7, 4, 3, 5, 1, 0, 9, 8, 6, -1];
data.permutationSort;
data.writeln;
}
{{out}}
[-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
The run-time is about 0.52 seconds with ldc2.
Alternative Version
{{trans|C++}}
import std.stdio, std.algorithm;
void permutationSort(T)(T[] items) pure nothrow @safe @nogc {
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 1.04 seconds with ldc2 (the C++ entry with G++ takes about 0.4 seconds).
E
{{trans|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)) {}
}
EchoLisp
;; This efficient sort method uses the list library for permutations
(lib 'list)
(define (in-order L)
(cond
((empty? L) #t)
((empty? (rest L)) #t)
(else (and ( < (first L) (second L)) (in-order (rest L))))))
(define L (shuffle (iota 6)))
→ (1 5 4 2 0 3)
(for ((p (in-permutations (length L ))))
#:when (in-order (list-permute L p))
(writeln (list-permute L p)) #:break #t)
→ (0 1 2 3 4 5)
Elixir
defmodule Sort do
def permutation_sort([]), do: []
def permutation_sort(list) do
Enum.find(permutation(list), fn [h|t] -> in_order?(t, h) end)
end
defp permutation([]), do: [[]]
defp permutation(list) do
for x <- list, y <- permutation(list -- [x]), do: [x|y]
end
defp in_order?([], _), do: true
defp in_order?([h|_], pre) when h<pre, do: false
defp in_order?([h|t], _), do: in_order?(t, h)
end
IO.inspect list = for _ <- 1..9, do: :rand.uniform(20)
IO.inspect Sort.permutation_sort(list)
{{out}}
[18, 2, 19, 10, 17, 10, 14, 8, 3]
[2, 3, 8, 10, 10, 14, 17, 18, 19]
Factor
USING: grouping io math.combinatorics math.order prettyprint ;
IN: rosetta-code.permutation-sort
: permutation-sort ( seq -- seq' )
[ [ before=? ] monotonic? ] find-permutation ;
{ 10 2 6 8 1 4 3 } permutation-sort .
"apple" permutation-sort print
{{out}}
{ 1 2 3 4 6 8 10 }
aelpp
FreeBASIC
' version 07-04-2017
' compile with: fbc -s console
' Heap's algorithm non-recursive
Function permutation_sort(a() As ULong) As ULong
Dim As ULong i, j, count
Dim As ULong lb = LBound(a), ub = UBound(a)
Dim As ULong n = ub - lb +1
Dim As ULong c(lb To ub)
While i < n
If c(i) < i Then
If (i And 1) = 0 Then
Swap a(0), a(i)
Else
Swap a(c(i)), a(i)
End If
count += 1
For j = lb To ub -1
If a(j) > a(j +1) Then j = 99
Next
If j < 99 Then Return count
c(i) += 1
i = 0
Else
c(i) = 0
i += 1
End If
Wend
End Function
' ------=< MAIN >=------
Dim As ULong k, p, arr(0 To 9)
Randomize Timer
Print "unsorted array"
For k = LBound(arr) To UBound(arr)
arr(k) = Rnd * 1000
Print arr(k) & IIf(k = UBound(arr), "", ", ");
Next
Print : Print
p = permutation_sort(arr())
Print "sorted array"
For k = LBound(arr) To UBound(arr)
Print arr(k) & IIf(k = UBound(arr), "", ", ");
Next
Print : Print
Print "sorted array in "; p; " permutations"
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
{{out}}
unsorted array
81, 476, 915, 357, 934, 683, 413, 450, 2, 407
sorted array
2, 81, 357, 407, 413, 450, 476, 683, 915, 934
sorted array in 1939104 permutations
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
}
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.
{{out}}
.............................................................................................[-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]
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|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
=={{header|Icon}} and {{header|Unicon}}== Partly from [http://infohost.nmt.edu/tcc/help/lang/icon/backtrack.html 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
J
{{eff note|J|/:~}} 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):
Either way:
```j
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
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;
}
}
{{out}}
Unsorted: [3, 2, 1, 8, 9, 4, 6]
Sorted: [1, 2, 3, 4, 6, 8, 9]
jq
'''Infrastructure''': The following function generates a stream of permutations of an arbitrary JSON array:
def permutations:
if length == 0 then []
else
. as $in
| range(0;length) as $i
| ($in|del(.[$i])|permutations)
| [$in[$i]] + .
end ;
Next is a generic function for checking whether the input array is non-decreasing. If your jq has until/2 then its definition here can be removed.
def sorted:
def until(cond; next):
def _until: if cond then . else (next|_until) end;
_until;
length as $length
| if $length <= 1 then true
else . as $in
| 1 | until( . == $length or $in[.-1] > $in[.] ; .+1) == $length
end;
'''Permutation-sort''':
The first permutation-sort solution presented here works with jq 1.4 but is slower than the subsequent solution, which uses the "foreach" construct introduced after the release of jq 1.4. "foreach" allows a stream generator to be interrupted.
{{works with|jq|1.4}}
def permutation_sort_slow:
reduce permutations as $p (null; if . then . elif ($p | sorted) then $p else . end);
{{works with|jq|with foreach}}
def permutation_sort:
# emit the first item in stream that satisfies the condition
def first(stream; cond):
label $out
| foreach stream as $item
( [false, null];
if .[0] then break $out else [($item | cond), $item] end;
if .[0] then .[1] else empty end );
first(permutations; sorted);
'''Example''':
["too", true, 1, 0, {"a":1}, {"a":0} ] | permutation_sort
{{out}}
$ jq -c -n -f Permutation_sort.jq
[true,0,1,"too",{"a":0},{"a":1}]
Julia
# v0.6
using Combinatorics
function permsort(x::Array)
for perm in permutations(x)
if issorted(perm)
return perm
end
end
end
x = randn(10)
@show x permsort(x)
{{out}}
x = [-0.799206, -2.52542, 0.677947, -1.85139, 0.744764, 1.5327, 0.808935, -0.876105, -0.234308, 0.874579]
permsort(x) = [-2.52542, -1.85139, -0.876105, -0.799206, -0.234308, 0.677947, 0.744764, 0.808935, 0.874579, 1.5327]
Kotlin
// version 1.1.2
fun <T : Comparable<T>> isSorted(list: List<T>): Boolean {
val size = list.size
if (size < 2) return true
for (i in 1 until size) {
if (list[i] < list[i - 1]) return false
}
return true
}
fun <T : Comparable<T>> permute(input: List<T>): List<List<T>> {
if (input.size == 1) return listOf(input)
val perms = mutableListOf<List<T>>()
val toInsert = input[0]
for (perm in permute(input.drop(1))) {
for (i in 0..perm.size) {
val newPerm = perm.toMutableList()
newPerm.add(i, toInsert)
perms.add(newPerm)
}
}
return perms
}
fun <T : Comparable<T>> permutationSort(input: List<T>): List<T> {
if (input.size == 1) return input
val toInsert = input[0]
for (perm in permute(input.drop(1))) {
for (i in 0..perm.size) {
val newPerm = perm.toMutableList()
newPerm.add(i, toInsert)
if (isSorted(newPerm)) return newPerm
}
}
return input
}
fun main(args: Array<String>) {
val input = listOf('d', 'b', 'e', 'a', 'f', 'c')
println("Before sorting : $input")
val output = permutationSort(input)
println("After sorting : $output")
println()
val input2 = listOf("first", "second", "third", "fourth", "fifth", "sixth")
println("Before sorting : $input2")
val output2 = permutationSort(input2)
println("After sorting : $output2")
}
{{out}}
Before sorting : [d, b, e, a, f, c]
After sorting : [a, b, c, d, e, f]
Before sorting : [first, second, third, fourth, fifth, sixth]
After sorting : [fifth, first, fourth, second, sixth, third]
Lua
-- Return an iterator to produce every permutation of list
function permute (list)
local function perm (list, n)
if n == 0 then coroutine.yield(list) end
for i = 1, n do
list[i], list[n] = list[n], list[i]
perm(list, n - 1)
list[i], list[n] = list[n], list[i]
end
end
return coroutine.wrap(function() perm(list, #list) end)
end
-- Return true if table t is in ascending order or false if not
function inOrder (t)
for pos = 2, #t do
if t[pos] < t[pos - 1] then
return false
end
end
return true
end
-- Main procedure
local list = {2,3,1} --\ Written to match task pseudocode,
local nextPermutation = permute(list) --\ more idiomatic would be:
while not inOrder(list) do --\
list = nextPermutation() --/ for p in permute(list) do
end --/ stuffWith(p)
print(unpack(list)) --/ end
{{out}}
1 2 3
Maple
arr := Array([17,0,-1,72,0]):
len := numelems(arr):
P := Iterator:-Permute(len):
for p in P do
lst:= convert(arr[sort(convert(p,list),output=permutation)],list):
if (ListTools:-Sorted(lst)) then
print(lst):
break:
end if:
end do:
{{Out|Output}}
[-1,0,0,17,72]
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[#]] &]
=={{header|MATLAB}} / {{header|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
MAXScript
fn inOrder arr =
(
if arr.count < 2 then return true
else
(
local i = 1
while i < arr.count do
(
if arr[i+1] < arr[i] do return false
i += 1
)
return true
)
)
fn permutations arr =
(
if arr.count <= 1 then return arr
else
(
for i = 1 to arr.count do
(
local rest = for r in 1 to arr.count where r != i collect arr[r]
local permRest = permutations rest
local new = join #(arr[i]) permRest
if inOrder new do return new
)
)
)
Output:
a = for i in 1 to 9 collect random 1 20
#(10, 20, 17, 15, 17, 15, 3, 11, 15)
permutations a
#(3, 10, 11, 15, 15, 15, 17, 17, 20)
Warning: This algorithm is very inefficient and Max will crash very quickly with bigger arrays.
NetRexx
Uses the permutation iterator '''RPermutationIterator''' at [[Permutations#NetRexx|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)
{{out}}
[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.
Nim
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
{{out}}
@[-31, 0, 2, 2, 4, 65, 83, 99, 782]
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)
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)
)
};
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"
{{out}}
Before: 94 15 42 35 55 24 96 14 61 94 43
After: 14 15 24 35 42 43 55 61 94 94 96
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;
{{out}}
Input = c b e d a
Output = a b c d e
Phix
function inOrder(sequence s)
for i=2 to length(s) do
if s[i]<s[i-1] then return 0 end if
end for
return 1
end function
function permutationSort(sequence s)
for n=1 to factorial(length(s)) do
sequence perm = permute(n,s)
if inOrder(perm) then return perm end if
end for
?9/0 -- should never happen
end function
?permutationSort({"dog",0,15.545,{"cat","pile","abcde",1},"cat"})
{{out}}
{0,15.545,"cat","dog",{"cat","pile","abcde",1}}
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
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) ) ) ) )
{{out}}
: (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)
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, $l - 1 ) } )
Prolog
permutation_sort(L,S) :- permutation(L,S), sorted(S).
sorted([]).
sorted([_]).
sorted([X,Y|ZS]) :- X =< Y, sorted([Y|ZS]).
permutation([],[]).
permutation([X|XS],YS) :- permutation(XS,ZS), select(X,YS,ZS).
PureBasic
Macro reverse(firstIndex, lastIndex)
first = firstIndex
last = lastIndex
While first < last
Swap cur(first), cur(last)
first + 1
last - 1
Wend
EndMacro
Procedure nextPermutation(Array cur(1))
Protected first, last, elementCount = ArraySize(cur())
If elementCount < 2
ProcedureReturn #False ;nothing to permute
EndIf
;Find the lowest position pos such that [pos] < [pos+1]
Protected pos = elementCount - 1
While cur(pos) >= cur(pos + 1)
pos - 1
If pos < 0
reverse(0, elementCount)
ProcedureReturn #False ;no higher lexicographic permutations left, return lowest one instead
EndIf
Wend
;Swap [pos] with the highest positional value that is larger than [pos]
last = elementCount
While cur(last) <= cur(pos)
last - 1
Wend
Swap cur(pos), cur(last)
;Reverse the order of the elements in the higher positions
reverse(pos + 1, elementCount)
ProcedureReturn #True ;next lexicographic permutation found
EndProcedure
Procedure display(Array a(1))
Protected i, fin = ArraySize(a())
For i = 0 To fin
Print(Str(a(i)))
If i = fin: Continue: EndIf
Print(", ")
Next
PrintN("")
EndProcedure
If OpenConsole()
Dim a(9)
a(0) = 8: a(1) = 3: a(2) = 10: a(3) = 6: a(4) = 1: a(5) = 9: a(6) = 7: a(7) = -4: a(8) = 5: a(9) = 3
display(a())
While nextPermutation(a()): Wend
display(a())
Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf
{{out}}
8, 3, 10, 6, 1, 9, 7, -4, 5, 3
-4, 1, 3, 3, 5, 6, 7, 8, 9, 10
Python
{{works with|Python|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()
R
{{libheader|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))
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)
REXX
/*REXX program sorts and displays an array using the permutation-sort method. */
call gen /*generate the array elements. */
call show 'before sort' /*show the before array elements. */
say copies('░', 75) /*show separator line between displays.*/
call pSort L /*invoke the permutation sort. */
call show ' after sort' /*show the after array elements. */
say; say 'Permutation sort took ' ? " permutations to find the sorted list."
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
.pAdd: #=#+1; do j=1 for N; #.#=#.# !.j; end; return /*add a permutation.*/
show: do j=1 for L; say @e right(j,wL) arg(1)":" translate(@.j,,'_'); end; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
gen: @.=; @.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'
@e=right('element', 15) /*literal used for the display.*/
do L=1 while @.L\==''; @@.L=@.L; end; L=L-1; wL=length(L); return
/*──────────────────────────────────────────────────────────────────────────────────────*/
isOrd: parse arg q /*see if Q list is in order. */
_=word(q, 1); do j=2 to words(q); x=word(q, j); if x<_ then return 0; _=x
end /*j*/ /* [↑] Out of order? ¬sorted*/
do k=1 for #; _=word(#.?, k); @.k=@@._; end /*k*/; return 1 /*in order.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
.pNext: 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*/ /* [↓] swap two array elements*/
do j=i+1 while j<n; parse value !.j !.n with !.n !.j; n=n-1; end /*j*/
if i==0 then return 0 /*0: indicates no more perms. */
do j=i+1 while !.j<!.i; end /*j*/ /*search perm for a lower value*/
parse value !.j !.i with !.i !.j /*swap two values in !. array.*/
return 1
/*──────────────────────────────────────────────────────────────────────────────────────*/
pSort: parse arg n,#.; #=0 /*generate L items (!) permutations.*/
do f=1 for n; !.f=f; end /*f*/
call .pAdd; do while .pNext(n, 0); call .pAdd; end /*while*/
do ?=1 until isOrd($); $= /*find perm.*/
do m=1 for #; _=word(#.?, m); $=$ @._; end /*m*/ /*build list*/
end /*?*/
return
{{out|output|text= when using the default (internal) inputs:}}
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
Permutation sort took 21 permutations to find the sorted list.
Ring
# Project : Sorting algorithms/Permutation sort
a = [4, 65, 2, 31, 0, 99, 2, 83, 782]
result = []
permute(a,1)
for n = 1 to len(result)
num = 0
for m = 1 to len(result[n]) - 1
if result[n][m] <= result[n][m+1]
num = num + 1
ok
next
if num = len(result[n]) - 1
nr = n
exit
ok
next
see "" + nr + " permutations required to sort " + len(a) + " items." + nl
func permute(a,k)
if k = len(a)
add(result,a)
else
for i = k to len(a)
temp=a[k]
a[k]=a[i]
a[i]=temp
permute(a,k+1)
temp=a[k]
a[k]=a[i]
a[i]=temp
next
ok
return a
Output:
169329 permutations required to sort 9 items.
Ruby
{{works with|Ruby|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
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)))))
Sidef
{{trans|Perl}}
func psort(x, d=x.end) {
if (d.is_zero) {
for i in (1 .. x.end) {
(x[i] < x[i-1]) && return false;
}
return true;
}
(d+1).times {
x.prepend(x.splice(d, 1)...);
x[d] < x[d-1] && next;
psort(x, d-1) && return true;
}
return false;
}
var a = 10.of { 100.irand };
say "Before:\t#{a}";
psort(a);
say "After:\t#{a}";
{{out}}
Before: 60 98 85 85 37 0 62 96 95 2
After: 0 2 37 60 62 85 85 95 96 98
Tcl
{{tcllib|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
}
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'>
{{out}}
<'dwj','ejw','hhp','oao','ock','oqh','pmf'>
zkl
Performance is horrid
rns:=T(4, 65, 2, 31, 0, 99, 2, 83, 782, 1);
fcn psort(list){ len:=list.len(); cnt:=Ref(0);
foreach ns in (Utils.Helpers.permuteW(list)){ // lasy permutations
cnt.set(1);
ns.reduce('wrap(p,n){ if(p>n)return(Void.Stop); cnt.inc(); n });
if(cnt.value==len) return(ns);
}
}(rns).println();
{{out}}
L(0,1,2,2,4,31,65,83,99,782)