N-queens problem: Difference between revisions

[[File:chess_queen.jpg|400px||right]]

 

You can extend the problem to solve the puzzle with a board of size &nbsp; <big>'''N'''x'''N'''</big>.

 

 

<br><br>

 

 

=={{header|360 Assembly}}==

Translated from the Fortran 77 solution.<br>

For maximum compatibility, this program uses only the basic instruction set (S/360).

MACRO

&LAB XDECO &REG,&TARGET

U DC (4*LL-2)H'0' stack

REGS make sure to include copybook jcl

{{out}}

<pre>

</pre>

 

 

 

 

=={{header|ABAP}}==

TYPES: BEGIN OF gty_matrix,

1 TYPE c,

SKIP 1.

ENDFORM. " SHOW_MATRIX

 

=={{header|Ada}}==

 

procedure Queens is

end loop;

Put_Line (" A B C D E F G H");

{{out}}

<pre>

This one only counts solutions, though it's easy to do something else with each one (instead of the <code>M := M + 1;</code> line).

 

use Ada.Text_IO;

 

Put_Line (Long_Integer'Image (Queens (N)));

end loop;

 

=={{header|ALGOL 68}}==

 

{{works with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release [http://sourceforge.net/projects/algol68/files/algol68toc/algol68toc-1.8.8d/algol68toc-1.8-8d.fc9.i386.rpm/download 1.8.8d.fc9.i386]}}

dim = 8; # dim X dim chess board #

[ofs:dim+ofs-1]INT b;

FI

OD

 

More or less copied from the "DFS" lesson on tryapl.org .

⍝Solution

accm←{⍺,((⍴⍵)=⍴⊃⍺)↑⊂⍵}

⍝Example

printqueens 6

{{out}}

<pre>

 

=={{header|AppleScript}}==

 

property Grid_Size : 8

return newRows

 

=={{header|Arc}}==

This program prints out all possible solutions:

(if (< len.queens n)

(let row (if queens (+ 1 queens.0.0) 0)

(def diagonal-match (curr other)

(is (abs (- curr.0 other.0))

{{out}}

The output is one solution per line, each solution in the form `((row col) (row col) (row col) ...)`:

((3 2) (2 0) (1 3) (0 1))

((3 1) (2 3) (1 0) (0 2))</pre>

 

=={{header|ATS}}==

(* ****** ****** *)

//

 

(* end of [queens.dats] *)

 

=={{header|AutoHotkey}}==

=== Output to formatted Message box ===

{{trans|C}}

; Post: http://www.autohotkey.com/forum/viewtopic.php?p=353059#353059

; Timestamp: 05/may/2010

Output .= "|`n" , yyy++

}

=== Includes a solution browser GUI ===

This implementation supports N = 4..12 queens, and will find ALL solutions

The screenshot shows the first solution of 10 possible solutions for N = 5 queens.

 

Number: ; main entrance for different # of queens

SI := 1

 

GuiClose:

[[image:N-Queens_SolutionBrowserGUI.png]]

 

[[Image:queens9_bbc.gif|right]]

[[Image:queens10_bbc.gif|right]]

Cell% = 32

VDU 23,22,Size%*Cell%;Size%*Cell%;Cell%,Cell%,16,128+8,5

ENDWHILE

UNTIL i% = 0

 

=={{header|BCPL}}==

 

GET "libhdr.h"

RESULTIS 0

}

The following is a re-implementation of the algorithm given above but

using the MC package that allows machine independent runtime generation

It runs about 25 times faster that the version given above.

 

GET "libhdr.h"

GET "mc.h"

i, n, all)

}

 

=={{header|Befunge}}==

This algorithm works with any board size from 4 upwards.

 

"| Q"$$$>:01p:2%!00g0>>^<<!:-1\<1>00p::2%-:40p2/50p2*1+

!77**48*+31p\:1\g,::2\g:,\3\g,,^g>0g++40g%40g\-\40g\`*-

 

{{out}}

 

=={{header|Bracmat}}==

= board M L x y S R row line

. :?board

| out$(found !solutions solutions)

)

{{out}} (tail):

<pre>

 

=={{header|C}}==

#include <stdlib.h>

 

int hist[n];

solve(n, 0, hist);

 

#include <stdlib.h>

#include <stdint.h>

printf("\nSolutions: %d\n", count);

return 0;

Take that and unwrap the recursion, plus some heavy optimizations, and we have a very fast and very unreadable solution:

#include <stdlib.h>

 

printf("\nSolutions: %d\n", count);

return 0;

 

A slightly cleaned up version of the code above where some optimizations were redundant. This version is also further optimized, and runs about 15% faster than the one above on modern compilers:

 

#define MAXN 31

 

printf("Number of solution for %d is %d\n",n,nqueens(n));

}

 

=={{header|C++}}==

// uses C++11 threads to parallelize the computation; also uses backtracking

// Outputs all solutions for any table size

return 0;

}

{{out}}Output for N = 4:

<pre> a b c d

3 #

4 # </pre>

// A straight-forward brute-force C++ version with formatted output,

// eschewing obfuscation and C-isms, producing ALL solutions, which

std::cout << queens( N ) << "\n";

}

{{out}} for N=4:

<pre>

=== Alternate version ===

Windows-only

#include <windows.h>

#include <iostream>

}

//--------------------------------------------------------------------------------------------------

{{out}}

<pre>

 

Version using Heuristics - explained here: [http://en.wikipedia.org/wiki/8_queens_puzzle#Solution_construction Solution_construction]

#include <windows.h>

#include <iostream>

}

//--------------------------------------------------------------------------------------------------

 

=={{header|Clojure}}==

This produces all solutions by essentially a backtracking algorithm. The heart is the ''extends?'' function, which takes a partial solution for the first ''k<size'' columns and sees if the solution can be extended by adding a queen at row ''n'' of column ''k+1''. The ''extend'' function takes a list of all partial solutions for ''k'' columns and produces a list of all partial solutions for ''k+1'' columns. The final list ''solutions'' is calculated by starting with the list of 0-column solutions (obviously this is the list ''[ [] ]'', and iterates ''extend'' for ''size'' times.

 

(defn extends? [v n]

(println s))

 

===Short Version===

(:require [clojure.math.combinatorics :as combo]

 

(defn queens [n]

(filter (fn [x] (every? #(apply distinct? (map-indexed % x)) [+ -]))

 

=={{header|CoffeeScript}}==

# Unlike traditional N-Queens solutions that use recursion, this

# program attempts to more closely model the "human" algorithm.

 

nqueens(8)

 

=={{header|Common Lisp}}==

(if (zerop n)

(list nil)

 

(defun print-queens (n)

 

=== Alternate solution ===

Translation of Fortran 77

(let ((a (make-array n))

(s (make-array n))

> (loop for n from 1 to 14 collect (cons n (queens1 n)))

((1 . 1) (2 . 0) (3 . 0) (4 . 2) (5 . 10) (6 . 4) (7 . 40) (8 . 92) (9 . 352)

 

As in Fortran, the iterative function above is equivalent to the recursive function below:

 

(let ((a (make-array n))

(u (make-array (+ n n -1) :initial-element t))

(rotatef (aref a i) (aref a k))))))))

(sub 0))

 

=={{header|Curry}}==

Three different ways of attacking the same problem. All copied from [http://web.cecs.pdx.edu/~antoy/flp/patterns/ A Catalog of Design Patterns in FLP]

-- 8-queens implementation with the Constrained Constructor pattern

-- Sergio Antoy

 

main = extend []

 

Another approach from the same source.

 

-- N-queens puzzle implemented with "Distinct Choices" pattern

-- Sergio Antoy

store = free

-- end

 

Yet another approach, also from the same source.

 

-- 8-queens implementation with both the Constrained Constructor

-- and the Fused Generate and Test patterns.

 

main = extend []

Mainly [http://www-ps.informatik.uni-kiel.de/~pakcs/webpakcs/main.cgi?queens webpakcs], uses constraint-solver.

import Findall

 

 

-- oneSolution = unpack $ queens 8

 

=={{header|D}}==

===Short Version===

This high-level version uses the second solution of the Permutations task.

import std.stdio, std.algorithm, std.range, permutations2;

 

n.iota.map!(i => p[i] - i).array.sort().uniq.count == n)

.count.writeln;

{{out}}

<pre>92</pre>

This version shows all the solutions.

{{trans|C}}

__gshared int[side] board;

 

y--;

}

{{out}}

<pre>

===Fast Version===

{{trans|C}}

in {

assert(nn > 0 && nn <= 27,

immutable uint side = (args.length >= 2) ? args[1].to!uint : 8;

writefln("N-queens(%d) = %d solutions.", side, side.nQueens);

{{out}}

<pre>N-queens(8) = 92 solutions.</pre>

 

=={{header|Dart}}==

Return true if queen placement q[n] does not conflict with

other queens q[0] through q[n-1]

void main() {

enumerate(4);

{{out}}

<pre>* Q * *

* Q * *

</pre>

 

 

else

.

.

.

.

.

{{out}}

 

 

 

=={{header|EchoLisp}}==

;; square num is i + j*N

(define-syntax-rule (sq i j) (+ i (* j N)))

(define (task up-to-n)

(for ((i up-to-n)) (writeln i ' ♕ (q-count i) 'solutions)))

{{out}}

<pre>

</pre>

 

 

=={{header|Eiffel}}==

class

QUEENS

end

end

{{out}}

<pre>

=={{header|Elixir}}==

{{trans|Ruby}}

def queen(n, display \\ true) do

solve(n, [], [], [], display)

Enum.each(7..12, fn n ->

IO.puts " #{n} Queen : #{RC.queen(n, false)}" # no display

 

{{out}}

11 Queen : 2680

12 Queen : 14200

</pre>

 

=={{header|Erlang}}==

Instead of spawning a new process to search for each possible solution I backtrack.

-module( n_queens ).

 

Board = solve( N ),

display( Board ).

{{out}}

<pre>

. . Q . . . . .

</pre>

 

=={{header|ERRE}}==

!------------------------------------------------

! QUEENS.R : solve queens problem on a NxN board

END IF

END PROGRAM

Note: The program prints solutions one per line. This version works well for the PC and the C-64. For PC only you can omit the % integer-type specificator with a <code>!$INTEGER</code> pragma directive.

 

=={{header|F Sharp}}==

let rec iterate f value = seq {

yield value

 

printNumberOfSolutions()

 

The output:

 

| | | |X| | | | | |

| |X| | | | | | | |

10 724

11 2680

 

=={{header|Factor}}==

IN: queens

 

:: safe? ( board q -- ? )

[let q board nth :> x

x swap

[ board nth ] keep

 

: solution? ( board -- ? )

 

: queens ( n -- l )

 

: .queens ( n -- )

[

[ 1 + "%d " printf ] each nl

 

=={{header|Forth}}==

variable nodes

 

solutions @ . ." solutions, " nodes @ . ." nodes" ;

 

 

=={{header|Fortran}}==

 

Using a back tracking method to find one solution

implicit none

 

write(*, "(a)") line

end subroutine

{{out}} for 8, 16 and 32 queens

<pre style="height:40ex;overflow:scroll">n = 8

 

===Alternate Fortran 77 solution===

C See the 2nd program for Scheme on the "Permutations" page for the

C main idea.

C 17 95815104

C 18 666090624

 

!functions. The same algorithm is much easier to follow in Fortran 90, using the RECURSIVE keyword.

!Like previously, the program only counts solutions. It's pretty straightforward to adapt it to print

print *, n, m

end do

 

===Alternate Fortran 95 solution with OpenMP===

column three.

 

 

With some versions of GCC the function OMP_GET_WTIME is not known, which seems to be a bug. Then it's enough to comment out the two calls, and the program won't display timings.

 

use omp_lib

implicit none

integer, parameter :: long = selected_int_kind(17)

integer, parameter :: l = 18

integer :: n, i, j, a(l*l, 2), k, p, q

integer(long) :: s, b(l*l)

real(kind(1d0)) :: t1, t2

do n = 6, l

k = 0

go to 60

end function

=={{header|FreeBASIC}}==

Get slower for N > 14

' compile with: fbc -s console

Dim Shared As ULong count, c()

Print : Print "hit any key to end program"

Sleep

{{out}}

<pre> 1 3 5 2 4

=== Alternate version : recursive ===

 

u() As Integer, v() As Integer, ByRef m As LongInt)

 

aux(n, 1, a(), u(), v(), m)

Print m

 

=== Alternate version : iterative ===

 

Dim m As LongInt = 0

 

u(p) = 1 : v(q) = 1

Goto L3

 

=={{header|GAP}}==

Translation of Fortran 77. See also alternate Python implementation. One function to return the number of solutions, another to return the list of permutations.

 

local a, up, down, m, sub;

a := [1 .. n];

[ 0, 0, 0, 0, 0, 0, 1, 0 ],

[ 0, 1, 0, 0, 0, 0, 0, 0 ],

 

=={{header|Go}}==

// WP page. Well, it happened to be the example program the day I completed

// the task. It seems from the WP history that there has been some churn

// Data Structures = Programs."

package main

import "fmt"

var (

i int

x [9]int

)

func try(i int) {

for j := 1; ; j++ {

}

}

func main() {

for i := 1; i <= 8; i++ {

}

}

{{out}}

<pre>

7 2

8 4

</pre>

 

[[N-queens_problem/dlx_go|dlx packge]].

 

 

import (

}

return nil

{{out}}

<pre>

===Distinct Solutions===

This solver starts with the N! distinct solutions to the N-Rooks problem and then keeps only the candidates in which all Queens are mutually diagonal-safe.

def k = [a.size(), b.size()].min()

def i = (0..<k).find { a[it] != b[it] }

// each permutation is an N-Rooks solution

orderedPermutations((0..<n)).findAll (diagonalSafe)

 

===Unique Solutions===

Unique solutions are equivalence classes of distinct solutions, factoring out all reflections and rotations of a given solution. See the [[WP:Eight_queens_puzzle|Wikipedia page]] for more details.

public static final diag = { list ->

final n = list.size()

}

qus

 

===Test and Results===

This script tests both distinct and unique solution lists.

def qds = queensDistinctSolutions(n)

def qus = queensUniqueSolutions(qds)

else { println "first:${qus[0]}"; println "last:${qus[-1]}" }

println()

 

Interpreting the Results:

 

In other words, this:

 

Results:

 

=={{header|Haskell}}==

import Data.List

 

 

-- prints all the solutions for 6 queens

 

If you just want one solution, simply take the <code>head</code> of the result of <code>queens n</code>; since Haskell is lazy, it will only do as much work as needed to find one solution and stop.

 

===Alternative version===

import Data.List ((\\))

 

where

f qs _ = [q:qs | q <- [1..n] \\ qs, q `notDiag` qs]

 

===Using permutations===

This version uses permutations to generate unique horizontal and vertical position for each queen. Thus, we only need to check diagonals. However, it is less efficient than the previous version because it does not prune out prefixes that are found to be unsuitable.

 

-- checks if queens are on the same diagonal

 

-- 8 is for "8 queens"

 

===In terms of foldr===

A back-tracking variant using the Prelude's plain '''foldr''':

{{Trans|JavaScript}}

 

queenPuzzle :: Int -> Int -> [[Int]]

| nRows <= 0 = [[]]

| otherwise =

 

safe :: Int -> Int -> [Int] -> Bool

safe iRow iCol qs =

(not . or) $

 

-- 10 columns of solutions for the 7*7 board:

showSolutions :: Int -> Int -> [String]

showSolutions nCols nSize =

where

boardLines rows =

 

chunksOf :: Int -> [a] -> [[a]]

where

 

main :: IO ()

{{Out}}

<pre>......♛ ......♛ ......♛ ......♛ .....♛. .....♛. .....♛. .....♛. .....♛. .....♛.

 

===Breadth-first search and Depth-first search===

import System.Environment

 

-- | Breadth-first search

bfs :: Int -> [(State, Thread)] -> (State, Thread)

bfs n sts | (not.empty) goal = head goal

where

 

-- | Depth-first search

| infeasible n st = [emptySt]

| otherwise = do x <- [1..n]

 

main = do

let n = read narg :: Int

print (bfs n [emptySt])

 

{{Out}}

 

=={{header|Heron}}==

inherits {

Heron.Windows.Console;

}

}

 

=={{header|Icon}} and {{header|Unicon}}==

Here's a solution to the <tt>n = 8</tt> case:

write(q(1), " ", q(2), " ", q(3), " ", q(4), " ", q(5), " ", q(6), " ", q(7), " ", q(8))

end

every 0 = row[r := 1 to 8] = ddiag[r + c - 1] = udiag[8 + r - c] do # test if free

suspend row[r] <- ddiag[r + c - 1] <- udiag[8 + r - c] <- r # place and yield

 

Notes:

* As the calls to q() are evaluated in main, each one will suspend a possible row, thereby allowing the next q(n) in main to be evaluated. If any of the q() fails to yield a row for the nth queen (or runs out of solutions) the previous, suspended calls to q() are backtracked progressively. If the final q(8) yields a row, the write() will be called with the row positions of each queen. Note that even the final q(8) will be suspended along with the other 7 calls to q(). Unless the write() is driven to produce more solutions (see next point) the suspended procedures will be closed at the "end of statement" ie after the write has "succeeded".

* If you want to derive all possible solutions, main() can be embellished with the '''every''' keyword:

procedure main()

every write(q(1), " ", q(2), " ", q(3), " ", q(4), " ", q(5), " ", q(6), " ", q(7), " ", q(8))

end

This drives the backtracking to find more solutions.

 

The comment explains how to modify the program to produce <i>all</i>

solutions for a given <tt>N</tt>.

 

procedure main(args)

write()

return # Comment out to see all possible solutions

 

A sample run for <tt>N = 6</tt>:

 

=={{header|IS-BASIC}}==

110 TEXT 80

120 DO

450 LET T(I)=1

460 NEXT

 

=={{header|J}}==

This is one of several J solutions shown and explained on this [[J:Essays/N%20Queens%20Problem|J wiki page]]

 

comb2 =: (, #: I.@,@(</)&i.)~ NB. all size 2 combinations of integers 0 to y

mask =: [ */@:~:&(|@-/) {

 

Note that the Roger Hui's approach (used here) matches the description attributed to Raymond Hettinger (in the Python implementation). (Both were posted years ago: 1981 for Hui's version which was used here, and 2009 for Hettinger's.) However they do use different diagonal queen clash elimination approaches -see [http://rosettacode.org/wiki/N-queens_problem#Roger_Hui_.281981.29_Algorithm C# Roger Hui Algorithm] for a comparison of the two approaches.

Example use:

 

 

92 distinct solutions for an 8 by 8 board.

 

 

One of the solutions. Position indicates row number, the integer indicates column number (0..7) for each queen -- though of course you could just as validly think of that the other way around.

 

=={{header|Java}}==

 

private static int[] b = new int[8];

}

}

 

=={{header|Javascript}}==

===ES5===

Algorithm uses recursive Backtracking. Checks for correct position on subfields, whichs saves a lot position checks. Needs 15.720 position checks for a 8x8 field.

if (rows <= 0) {

return [[]];

}

 

 

===ES6===

Translating the ES5 version, and adding a function to display columns of solutions.

 

 

// queenPuzzle :: Int -> Int -> [[Int]]

 

// safe : Int -> Int -> [Int] -> Bool

 

 

 

 

 

 

 

 

};

 

// enumFromTo :: Int -> Int -> [Int]

}, (_, i) => m + i);