Sorting algorithms/Permutation 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

ActionScript

<lang 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); }</lang>

AutoHotkey

ahk forum: discussion <lang AutoHotkey>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

}</lang>

C

<lang c>#include <stdlib.h>

include <stdio.h>
include <string.h>

typedef struct pi *Permutations;

/* Type of element on list to be sorted */ typedef const char *ElementType;

struct pi {

   short list_size;
   short *counts;
   ElementType *crntperm;

};

Permutations PermutationIterator( ElementType *list, short listSize) {

   int ix;
   Permutations p = (Permutations)malloc(sizeof(struct pi));
   p->list_size = listSize;
   p->counts = (short *)malloc( p->list_size * sizeof(short));
   p->crntperm = (ElementType *)malloc( p->list_size * sizeof(ElementType));

   for (ix=0; ix<p->list_size; ix++) {
       p->counts[ix] = ix;
       p->crntperm[ix] = list[ix];
   }
   return p;

}

void FreePermutations( Permutations p) {

   if (NULL == p) return;
   if (p->crntperm) free(p->crntperm);
   if (p->counts) free(p->counts);
   free(p);

}

define FREE_Permutations(pi) do {\

   FreePermutations(pi); pi=NULL; } while(0)

ElementType *FirstPermutation(Permutations p) {

   return p->crntperm;

}

ElementType *NextPermutation( Permutations p) {

   int ix, j;
   ElementType *crntp, t;

   crntp = p->crntperm;
   ix = 1;
   do {
       t = crntp[0];
       for(j=0; j<ix; j++) crntp[j] = crntp[j+1];
       crntp[ix] = t;
       if (p->counts[ix] > 0) break;
       ix += 1;
   } while (ix < p->list_size);
   if (ix == p->list_size) return NULL;

   p->counts[ix] -= 1;
   while(--ix) {
       p->counts[ix] = ix;
   }
   return crntp;

}

/* Checks to see if list is ordered */ int isInOrder(ElementType *letrList, int size ) {

   int j;
   ElementType *p0 = letrList, *p1 = letrList+1;
   for (j= 1; j<size; j++) {
       if ( strcmp( *p0, *p1) > 0) break;  /* compare strings */

// if ( *p0 > *p1) break; /* compare numeric values */

       p0++, p1++;
   }
   return ( j == size );

}

int main( ) {

   short size =5;
   ElementType *prm;
   ElementType mx[] = {"another", "sorted", "to_be", "list", "here's" };
   Permutations pi = PermutationIterator(mx, size);
   for ( prm = FirstPermutation(pi); prm; prm = NextPermutation(pi))
       if (isInOrder( prm, size) ) break;

   if (prm) {
       int j;
       printf("Sorted: ");
       for (j=0; j<size; j++) 
           printf("%s ",prm[j]);
       printf("\n");
   }

   FreePermutations( pi);
   return 0;

}</lang>

Clojure

<lang lisp> (use '[clojure.contrib.combinatorics :only (permutations)])

(defn permutation-sort [s]

 (first (filter (partial apply <=) (permutations s))))

(permutation-sort [2 3 5 3 5]) </lang>

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.

<lang 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)))</lang>

<lang lisp>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)</lang>

C++

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

<lang cpp>#include <algorithm>

template<typename ForwardIterator>

void permutation_sort(ForwardIterator begin, ForwardIterator end)

{

 while (std::next_permutation(begin, end))
 {
   // -- this block intentionally left empty --
 }

}</lang>

D

<lang d>module permsort ; import std.stdio ;

bool isSorted(T)(inout T[] a) { // test if a is already sorted

 if(a.length <= 1) return true ; // 1-elemented/empty array is defined as sorted
 for(int i = 1 ; i < a.length ; i++) if(a[i] < a[i-1]) return false ;
 return true ;

}

Permutator!(T) Perm(T)(T[] x) { return Permutator!(T)(x) ; } struct Permutator(T) { // permutation iterator

 T[] s ;
 alias int delegate(inout T[]) DG ;
 void swap(int i, int j) { T tmp = s[i] ; s[i] = s[j] ; s[j] = tmp ; }   
 int opApply(DG dg) { return perm(0, s.length, dg) ; }
 int perm(int breaked, int n, DG dg) {   
   if(breaked) return breaked ;
   else if(n <= 1) breaked = dg(s) ;
   else {
     for(int i = 0 ; i < n ; i++) {
       if((breaked = perm(breaked, n - 1, dg)) != 0) break ;
       if(0 == (n % 2)) swap(i, n-1) ; else swap(0, n-1) ;
     }
   }
   return breaked ;
 }

}

T[] permsort(T)(T[] s) {

 foreach( p ; Perm(s))
   if(isSorted(p)) 
     return p.dup ;
 assert(false, "Should not be here") ;

}

void main() {

 auto p = [2,7,4,3,5,1,0,9,8,6] ;
 
 writefln("%s", permsort(p)) ;
 writefln("%s", p) ;                       // sort is in place
 writefln("%s", permsort(["rosetta"])) ;   // test with one element
 writefln("%s", permsort(cast(int[])[])) ; // test empty array

}</lang>

E

Translation of: C++

<lang e>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)) {}

}</lang>

Haskell

<lang 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') }</lang>

Works with: GHC version 6.10

<lang haskell>import Data.List (permutations)

permutationSort l = head [p | p <- permutations l, sorted p]

sorted (e1 : e2 : r) = e1 <= e2 && sorted (e2 : r) sorted _ = True</lang>

Icon

Partly from here <lang icon>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</lang>

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. <lang ocaml>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)</lang>

J

A function to locate the permuation index, in the naive manner prescribed by the task: <lang j>ps =:(1+])^:((-.@-:/:~)@A.~)^:_ 0:</lang> Of course, this can be calculated much more directly (and efficiently): <lang j>ps =: A.@:/:</lang> Either way: <lang 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</lang>

PHP

<lang 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);</lang>

1,2,3,4,5,6,7,8,9,10

Prolog

<lang 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).</lang>

Python

Works with: Python version 2.6

<lang python>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()</lang>

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. <lang r>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))</lang>

Ruby

Works with: Ruby version 1.8.7+

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

 def permutationsort
   permutations = permutation
   begin
     perm = permutations.next
   end until perm.sorted?
   perm
 end
 
 def sorted?
   each_cons(2).all? {|a, b| a <= b}
 end

end</lang>

Scheme

<lang 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)))))</lang>

Tcl

using package struct::list from

Library: tcllib

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

<lang tcl>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

}</lang>

Ursala

Standard library functions to generate permutations and test for ordering by a given predicate are used. <lang Ursala>#import std

permsort "p" = ~&ihB+ ordered"p"*~+ permutations

cast %sL

example = permsort(lleq) <'pmf','oao','ejw','hhp','oqh','ock','dwj'></lang>

output:

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