Sorting algorithms/Permutation sort: Difference between revisions

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do { t' <- insert e t ; return (h : t') }
do { t' <- insert e t ; return (h : t') }
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=={{header|Icon}}==
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 e, o
o := l[1]
every e := !l do if e > o then return &fail else o := e
return 1
end
procedure main()
local l, m
l := [6,3,4,5,1]
every m := permute(l) & sorted(m) do
every writes(" ",!m)
end


=={{header|OCaml}}==
=={{header|OCaml}}==

Revision as of 13:58, 10 December 2008

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

Permutation sort.

Pseudocode:

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

C++

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

<cpp>

  1. include <algorithm>

template<typename ForwardIterator>

void permutation_sort(ForwardIterator begin, ForwardIterator end)

{

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

} </cpp>

D

<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

}</d>

Haskell

import Control.Monad

permutationSort l = head $ do p <- permute l
                              if sorted p then return p else mzero

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

permute []           = return []
permute (h:t)        = do { t' <- permute t ; insert h t' }

insert e []          = return [e]
insert e l@(h : t)   = return (e : l) `mplus`
                       do { t' <- insert e t ; return (h : t') }

Icon

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 e, o
   o := l[1]
   every e := !l do if e > o then return &fail else o := e
   return 1
end

procedure main()
   local l, m
   l := [6,3,4,5,1]
   every m := permute(l) & sorted(m) do
       every writes(" ",!m)
end

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. <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 rec permute = function

| []     -> [[]]
| h :: t -> List.concat (List.map (fun t' -> insert h t') (permute t))

let permutation_sort l = List.find sorted (permute l)</ocaml>

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).

Scheme

<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)))))

</scheme>