N-queens problem: Difference between revisions

{{trans|Nim}}

 

 

F underAttack(col, queens)

print(‘ ’, end' ‘’)

print(Char(code' ‘a’.code + row)‘’col, end' ‘’)

 

{{out}}

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>

 

=={{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

 

=={{header|APL}}==

{{works with|Dyalog APL}}

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) ...)`:

=={{header|Arturo}}==

 

 

queens: function [n, i, a, b, c][

]

print ""

 

{{out}}

=={{header|AWK}}==

Inspired by Raymond Hettinger's Python solution, but builds the vector incrementally.

#!/usr/bin/gawk -f

# Solve the Eight Queens Puzzle

 

 

{{out}}

<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 sharp|C#}}==

{{works with|C sharp|C#|7}}

<!-- By Martin Freedman, 13/02/2018 -->

using static System.Linq.Enumerable;

using static System.Console;

public static IEnumerable<T> ToSingleton<T>(this T item) { yield return item; }

}

Output

<pre>8-queens has 92 solutions

===Hettinger Algorithm===

Compare this to the Hettinger solution used in the first Python answer. The logic is similar but the diagonal calculation is different and more expensive computationally (Both suffer from being unable to eliminate permutation prefixes that are invalid e.g. 0 1 ...)

using System.Collections.Generic;

using static System.Linq.Enumerable;

public static IEnumerable<T> ToSingleton<T>(this T item) { yield return item; }

}

=== Amb solution===

This uses the second version of the [https://rosettacode.org/wiki/Amb#C.23 Amb C# class] in the Amb challenge. Really that is not McCarthy's Amb (Ambiguous function) and here it is used just as a simple general interface by lambdas to a standalone backtrack algorithm. Due to the specification of the Amb challenge, this, ironically (given the notion of ambiguous functions), only produces one solution not 92. It is trivial to update Amb (might be better called a backtracker rather than Amb too) but here it is just used to show how easy it is to go from a generate and prune Linq solution to a backtrack solution. The Linq filters becoming "amb" requirements.

{{works with|C sharp|C#|7.1}}

<!-- By Martin Freedman, 9/02/2018 -->

using static System.Console;

 

}

}

 

=={{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)) [+ -]))

===Backtracking as Tree processing===

Each state of the board can be represented as a sequence of the row coordinate for a queen, the column being the index in the sequence (coordinates starting at 0). Each state can have 'children' states if it is legal (no conflict) and has less than n queens. A child state is the result of adding a new queen on the next column, there are as many children states as rows as we are trying all of them. A depth first traversal of this virtual tree of states gives us the solutions when we filter out the illegal states and the incomplete states. The sequence of states is lazy so we could read only one result and not have to compute the other states.

 

(defn n-queens [n]

(let[children #(map (partial conj %) (range n))

no-conflict?)

children []))))

 

=={{header|CLU}}==

{{trans|C}}

rep = null

own hist: array[int] := array[int]$[]

stream$putl(po, "No. " || int$unparse(count) || "\n-------\n" || s)

end

{{out}}

<pre style='height:50ex'>No. 1

 

=={{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 * *

{{libheader| System.SysUtils}}

{{Trans|Go}}

program N_queens_problem;

 

writeln(i, ' ', x[i]);

readln;

 

=={{header|Draco}}==

{{trans|C}}

word count;

 

count := 0;

solve(hist, 0)

{{out}}

<pre>No. 1

.

.

{{out}}

<pre>Solution 1

 

=={{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>

 

=={{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}}

=={{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>

 

===Alternative Version===

%%%For 8X8 chessboard with N queens.

-module(queens).

(Row /= Column + N) andalso (Row /= Column - N) andalso

safe(Row, Columns, (N+1)).

 

=={{header|ERRE}}==

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:

10 724

11 2680

 

=={{header|Factor}}==

{{works with|Factor|0.98}}

IN: queens

 

[

[ 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===

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

go to 60

end function

 

===Fortran 2008 in a Lisp-like fashion===

 

Part of the intent here is to show that Fortran can do quite a few things people would not think it could, if it is given adequate library support.

 

use, intrinsic :: iso_fortran_env, only: output_unit

end subroutine check_garbage

 

{{out}}$ ./example__n_queens 1 2 3 4

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

=={{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|Frink}}==

This example uses Frink's built-in <CODE>array.permute[]</CODE> method to generate possible permutations of the board efficiently.

{

for q = 0 to length[board] - 1

for b = array[1 to 8].permute[]

if solution[b]

 

=={{header|Fōrmulæ}}==

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}}==

===Niklaus Wirth algorithm (Wikipedia)===

// 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

}

}

 

{{out}}

 

=== Refactored Niklaus Wirth algorithm (clearer/Go friendly solution) ===

* N-Queens Problem

*

trycol(0)

printresults()

{{out}}

<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:

 

=={{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}}

 

--------------------- N QUEENS PROBLEM -------------------

 

main :: IO ()

{{Out}}

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

 

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

import System.Environment

 

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.

"use strict";

 

// MAIN ---

return main();

{{Out}}

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

This section presents a function for finding a single solution using

the formulae for explicit solutions at [[WP:Eight_queens_puzzle|Eight Queens Puzzle]].

def q: "♛";

def init(k): reduce range(0;k) as $i ([]; . + ["."]);

(""; reduce $row[] as $x (.; . + $x) + "\n");

 

{{out}}

$ jq -M -n -r -f n-queens-single-solution.jq

.....♛..

.......♛

♛.......

..♛.....

====Generate-and-test counter====

{{ works with|jq|1.4}}

'''Part 1: Generic functions'''

def permutations(n):

# Given a single array, generate a stream by inserting n at different positions:

end;

 

'''Part 2: n-queens'''

def sums:

. as $board

 

count( permutations(n) | select(allowable) );

'''Example''':

{{out}}

92

=={{header|Julia}}==

 

# EightQueensPuzzle

 

println()

end

<pre>

n = 1

=={{header|Kotlin}}==

{{trans|FreeBASIC}}

 

var count = 0

println()

}

 

{{out}}

=={{header|Liberty BASIC}}==

Program uses permutation generator (stores all permutations) and solves tasks 4x4 to 9x9. It prints all the solutions.

'N queens

'>10 would not work due to way permutations used

End Function

 

 

=={{header|Locomotive Basic}}==

Uses the heuristic from the Wikipedia article to get one solution.

 

20 while n<4:input "How many queens (N>=4)";n:wend

30 dim q(n),e(n),o(n)

500 for i=1 to n

510 if o(i)=1 or o(i)=3 then o(i)=-1 else if o(i)=0 then o(i)=1:o(i+1)=3:return

 

[[File:Queens Puzzle, Locomotive Basic.png]]

 

=={{header|Logo}}==

if :files = 0 [make "solutions :solutions+1 show :tried stop]

localmake "safe (bitand :files :diag1 :diag2)

end

 

 

=={{header|Lua}}==

 

-- We'll use nil to indicate no queen is present.

else

print(string.format("No solution for %d queens.\n", N))

 

=={{header|M2000 Interpreter}}==

{{trans|VBA}}

Module N_queens {

Const l = 15 'number of queens

}

N_queens

 

=={{header|m4}}==

It finds one solution of the Eight Queens problem.

 

 

The following macro find one solution to the eight-queens problem:

divert`'dnl

dnl

 

{{out}}

{{trans|Python}}

 

local a,u,v,m,aux;

a:=[$1..n];

end:

 

 

{{out}}

=={{header|Mathematica}}/{{header|Wolfram Language}}==

This code recurses through the possibilities, using the "safe" method to check if the current set is allowed. The recursive method has the advantage that finding all possibilities is about as hard (code-wise, not computation-wise) as finding just one.

With[{l = Length@q},

Length@Union@q == Length@Union[q + Range@l] ==

If[Length[q] == n, {q},

Cases[nQueen[Append[q, #], n] & /@ Range[n],

 

This returns a list of valid permutations by giving the queen's column number for each row. It can be displayed in a list of chess-board tables like this:

Grid[Normal@

SparseArray[MapIndexed[{#, First@#2} -> "Q" &, #], {n, n}, "."],

Frame -> All] & /@ nQueen[n]

{{out}}

<pre>{. . . Q . ., . . . . Q ., . Q . . . ., . . Q . . .}

This solution uses Permutations and subsets, also prints out a board representation.

 

Do[per[[q]]=Partition[Riffle[Reverse[Range[n]],per[[q]]],2],{q,1,Length[per]}];(* Riffled in the reverse of [range n] partitioned into pairs*)

Do[w=Subsets[per[[t]],{2}];(* This is a full subset of the previous set of pairs taken 2 at a time *)

If[tot==0,g=Grid[Table[" ",{n},{n}],Alignment->Center,Frame->All,Spacings->{1.2,1}];(* If no clashing diagonals setup an array and print the permutation and the grid*)

Do[g[[1,per[[t,w,1]],per[[t,w,2]]]]="Q",{w,1,n}];

 

Alternative Solution using Linear Programming:

 

dispSol[sol_] := sol /. {1 -> "Q" , 0 -> "-"} // Grid

 

 

solveNqueens[8] // dispSol

<pre>- - - - Q - - -

- Q - - - - - -

=={{header|MATLAB}}==

This solution is inspired by Raymond Hettinger's permutations based solution which was made in Python: https://code.activestate.com/recipes/576647/

solutions=[[]];

v = 1:n;

end

s

{{out}}

<pre>

 

=={{header|Maxima}}==

 

queens(n) := block([a, i, j, m, p, q, r, s, u, v, w, y, z],

[1, 6, 8, 3, 7, 4, 2, 5],

...]] */

 

=={{header|MiniZinc}}==

array [1..n] of var 1..n: q; % queen in column i is in row q[i]

 

satisfy;

output [ if fix(q[j]) == i then "Q" else "." endif ++

 

This solution appears in the official MiniZinc tutorial documentation, and is generalized.

=={{header|Modula-2}}==

{{trans|C}}

FROM InOut IMPORT Write, WriteCard, WriteString, WriteLn;

 

count := 0;

Solve(N, 0);

 

{{out}}

 

=={{header|MUMPS}}==

Set sol=0

For row(1)=1:1:4 Do try(2) ; Not 8, the other 4 are symmetric...

Quit

Do Queens

<div style="overflow:scroll; height:400px;">

+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+

8 | |##| Q|##| |##| |##| | |##| Q|##| |##| |##| | |##| | Q| |##| |##|

1 |##| |##| | Q| |##| |

+--+--+--+--+--+--+--+--+

 

=={{header|Nim}}==

 

proc underAttack(col: int; queens: seq[int]): bool =

if row > 0: stdout.write ' '

stdout.write chr(ord('a') + row), col

 

{{out}}

{{trans|Java}}

 

class NQueens {

b : static : Int[];

}

}

 

=={{header|OCaml}}==

{{libheader|FaCiLe}}

 

Copyright 2004 CENA. All rights reserved.

This code is distributed under the terms of the GNU LGPL *)

then raise (Failure "Usage: queens <nb of queens>");

Gc.set ({(Gc.get ()) with Gc.space_overhead = 500}); (* May help except with an underRAMed system *)

===A stand-alone OCaml solution===

 

let show board =

else 8 in

 

{{out}}

<pre>$ ocaml queens.ml 6

=={{header|Oz}}==

A pretty naive solution, using constraint programming:

fun {Queens N}

proc {$ Board}

in

{Length Solutions} = 92 %% assert

 

There is a more concise and much more efficient [http://www.mozart-oz.org/documentation/fdt/node25.html#section.scripts.queens solution] in the Mozart documentation.

 

=={{header|Pascal}}==

 

const l=16;

14 365596

15 2279184

 

===Alternative===

Solution found</pre>

{$IFDEF FPC}

{$MODE DELPHI}

end;

WriteLn('Fertig');

{{out}}

<pre>

 

=={{header|PDP-11 Assembly}}==

; "eight queens problem" benchmark test

 

 

scr: ;display RAM

 

=={{header|Perl}}==

 

sub try_column {

 

#print for @solutions; # un-comment to see all solutions

{{out}}

<pre>total 14200 solutions</pre>

 

=={{header|Phix}}==

<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>

<span style="color: #000080;font-style:italic;">--

<span style="color: #000000;">n_queens</span><span style="color: #0000FF;">(</span><span style="color: #000000;">N</span><span style="color: #0000FF;">,</span><span style="color: #000000;">N</span><span style="color: #0000FF;"><</span><span style="color: #000000;">5</span><span style="color: #0000FF;">)</span>

<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>

{{out}}

<pre>

 

=={{header|PHP}}==

<html>

<head>

</body>

</html>

 

<h2>Solution with recursion</h2>

 

<html>

<body>

</body>

</html>

 

=={{header|Picat}}==

===0/1 encoding a N x N matrix===

Using constraint modelling using an 0/1 encoding of an N x N matrix. It is the probably the fastest approach when using SAT and MIP solvers.

% import mip.

 

sum([Q[I,J] : I in 1..N]) #= 1

end,

 

===Constraint programming===

This is the "standard" model using constraint programming (in contract to the SAT 0/1 approach). Instead of an NxN matrix, this encoding uses a single list representing the columns. The three <code>all_different/1</code> then ensures that the rows, and the two diagonals are distinct.

 

queens(N, Q) =>

all_different([$Q[I]-I : I in 1..N]),

all_different([$Q[I]+I : I in 1..N]),

 

==="Naive" approach===

This approach might be called "naive" (in the constraint programming context) since it doesn't use the (general) faster <code>all_different/1</code> constraint.

Q = new_list(N),

Q :: 1..N,

Q[I] - I #!= Q[J] - J

end,

 

 

=={{header|PicoLisp}}==

===Calling 'permute'===

 

(de queens (N)

(length (uniq (mapcar - L R))) )

(inc 'Cnt) ) )

===Permuting inline===

This alternative version does not first pre-generate all permutations with

'permute', but creates them recursively. Also, it directly checks for

duplicates, instead of calling 'uniq' and 'length'. This is much faster.

(let (R (range 1 N) L (copy R) X L Cnt 0)

(recur (X) # Permute

(mapcar - L R) )

(inc 'Cnt) ) ) )

{{out}} for both cases:

<pre>: (queens 8)

=={{header|PL/I}}==

This code compiles with PL/I compilers ranging from the ancient IBM MVT PL/I F compiler of the 1960s, the IBM PL/I Optimizing compiler, thru the IBM PL/I compiler for MVS and VM, to the z/OS Enterprise PL/I v4.60 compiler;spanning 50 years of PL/I compilers. It only outputs the number of solutions found for a given N instead of printing out each individual chess board solution to avoid filling up spool space for large values of N. It's trivial to add a print-out of the individual solutions.

NQUEENS: PROC OPTIONS (MAIN);

DCL A(35) BIN FIXED(31) EXTERNAL;

END QUEEN;

 

=={{header|PowerBASIC}}==

=== Recursive version ===

{{trans|Stata}}

#DIM ALL

 

PRINT m

END IF

 

=== Iterative version ===

{{trans|Stata}}

#DIM ALL

 

GOTO 3

END IF

 

=={{header|PowerShell}}==

{{works with|PowerShell|2}}

function PlaceQueen ( [ref]$Board, $Row, $N )

{

}

}

Get-NQueensBoard 8

''

''

Get-NQueensBoard 14

{{out}}

<pre>

=={{header|Processing}}==

{{trans|Java}}

int n = 8;

int[] b = new int[n];

b[0] = -1;

}

 

==={{header|Processing Python mode}}===

This solution, originally by Raymond Hettinger for demonstrating the power of the itertools module, generates all solutions.

 

 

n = 8

global i

i = (i + 1) % len(solutions)

 

=={{header|Prolog}}==

 

Solution #1:

solution([X/Y|Others]) :-

member(Item,Rest).

 

Solution #2:

permutation([1,2,3,4,5,6,7,8], Queens),

safe(Queens).

Y-Y1=\=Xdist,

Dist1 is Xdist + 1,

 

Solution #3:

sol(Ylist,[1,2,3,4,5,6,7,8],

[1,2,3,4,5,6,7,8],

del(Item,[First|List],[First|List1]) :-

 

[http://ideone.com/Y6olN Output]:

===Alternative version===

Uses non-ISO predicates between/3 and select/3 (available in SWI Prolog and GNU Prolog).

 

 

 

 

[http://ideone.com/3bbIx0 Runs in: time: 0.02 memory: 68352]

 

Uses backtracking- a highly efficient mechanism in Prolog to find all solutions.

{{works with|SWI Prolog|version 6.2.6 by Jan Wielemaker, University of Amsterdam}}

% q(Row) represents a queen, allocated one per row. No rows ever clash.

% The columns are chosen iteratively from available columns held in a

length(Boards, Len), writef('%w solutions:\n', [Len]), % Output solutions

member(R,Boards), reverse(R,Board), writef(' - %w\n', [Board]), fail.

{{out}}

<pre>?- queens.

===Short version===

SWI-Prolog 7.2.3

maplist(plus, X, N, Z1), maplist(plus, X, Z2, N), is_set(Z1), is_set(Z2).

 

queens(N, Qs) :-

{{out}}

<pre>

 

===SWISH Prolog version===

% John Devou: 26-Nov-2021

% Short solution to use on https://swish.swi-prolog.org/.

not((member((U,V),Qs), (V =:= C; R-U =:= abs(C-V)))).

q(N,X):- q(N,N,_,X).

 

===CLP(FD): Constraint Logic Programming over Finite Domains Version===

</ul>

<br/>

 

% DOC: http://www.pathwayslms.com/swipltuts/clpfd/clpfd.html

main :-

print_nqueens_all(8).

{{out}}

<pre>

From the Pure (programming language) Wikipedia page

 

n-queens.pure

Tectonics:

 

compiling || (puts("queens 4: " + str(queens 4)) $$

 

{{out}}

=={{header|PureBasic}}==

A recursive approach is taken. A queen is placed in an unused column for each new row. An array keeps track if a queen has already been placed in a given column so that no duplicate columns result. That handles the Rook attacks. Bishop attacks are handled by checking the diagonal alignments of each new placement against the previously placed queens and if an attack is possible the solution backtracks. The solutions are kept track of in a global variable and the routine <tt>queens(n)</tt> is called with the required number of queens specified.

 

Procedure showBoard(Array queenCol(1))

Input()

CloseConsole()

Sample output showing the last solution (all are actually displayed) for 1 - 12 queens:

<pre style="height:40ex;overflow:scroll"> Solution 1

This solution, originally by [http://code.activestate.com/recipes/576647/ Raymond Hettinger] for demonstrating the power of the itertools module, generates all solutions. On a regular 8x8 board only 40,320 possible queen positions are examined.

 

 

n = 8

if n == len(set(vec[i]+i for i in cols)) \

== len(set(vec[i]-i for i in cols)):

 

The output is presented in vector form (each number represents the column position of a queen on consecutive rows).

The vector can be pretty printed by substituting a call to <code>board</code> instead of <code>print</code>, with the same argument, and where board is pre-defined as:

 

Raymond's description is:

On a regular 8x8 board only 15,720 possible queen positions are examined.

{{works with|Python|2.6, 3.x}}

BOARD_SIZE = 8

 

return solutions

 

 

===Python: Simple Backtracking Solution===

to a generator expression) produces a backtracking solution. On a regular 8x8 board only 15,720 possible queen positions are examined.

{{works with|Python|2.6, 3.x}}

 

def under_attack(col, queens):

answers = solve(BOARD_SIZE)

first_answer = next(answers)

 

===Python: Simple Backtracking Solution (Niklaus Wirth Algorithm)===

The following program is a translation of Niklaus Wirth's solution into the Python programming language, but does without the index arithmetic used in the original and uses simple lists instead, which means that the array ''x'' for recording the solution can be omitted. A generator replaces the procedure (see [https://www.inf.ethz.ch/personal/wirth/AD.pdf Algorithms and Data Structures], pages 114 to 118). On a regular 8x8 board only 15,720 possible queen positions are examined.

if i < n:

for j in range(n):

 

for solution in queens(8, 0, [], [], []):

The algorithm can be slightly improved by using sets instead of lists (cf. backtracking on permutations). But this makes the algorithm a bit harder to read, since the list x has to be added to record the solution. On a regular 8x8 board only 5,508 possible queen positions are examined. However, since these two solutions are intended for educational purposes, they are neither resource-friendly nor optimized for speed. The next program (backtracking on permutations) shows a much faster solution that also uses less space on the stack.

if a: # a is not empty

for j in a:

 

for solution in queens([], 0, set(range(8)), set(), set()):

 

===Python: backtracking on permutations===

The solutions are returned as a generator, using the "yield from" functionality of Python 3.3, described in [https://www.python.org/dev/peps/pep-0380/ PEP-380].

 

a = list(range(n))

up = [True]*(2*n - 1)

#Count solutions for n=8:

sum(1 for p in queens(8))

 

The preceding function does not enumerate solutions in lexicographic order, see [[Permutations#Recursive implementation]] for an explanation. The following does, but is almost 50% slower, because the exchange is always made (otherwise the loop to shift the array a by one place would not work). On a regular 8x8 board only 5,508 possible queen positions are examined.

However, it may be interesting to look at the first solution in lexicographic order: for growing n, and apart from a +1 offset, it gets closer and closer to the sequence [http://oeis.org/A065188 A065188] at OEIS. The first n for which the first solutions differ is n=26.

 

a = list(range(n))

up = [True]*(2*n - 1)

 

#Compare to A065188

 

===Python: fold/reduce===

Expressed in terms of nested folds, allowing for graphic display of results, and listing the number of solutions found for boards of various sizes. On a regular 8x8 board only 15,720 possible queen positions are examined.

{{Works with|Python|3.7}}

 

from functools import reduce

# MAIN ---

if __name__ == '__main__':

{{Out}}

<pre>10 solutions for a 5 * 5 board:

{{works with|QBasic}}

{{trans|QB64}}

CLS

COLOR 15

continue1:

NEXT icol

 

 

=={{header|QB64}}==

DIM SHARED QUEENS AS INTEGER

PRINT "# of queens:";: INPUT QUEENS

NEXT icol

END SUB

 

=={{header|R}}==

This solution uses recursive backtracking.

 

a <- seq(n)

u <- rep(T, 2 * n - 1)

aux(1)

m

 

Show the first solution found for size 8 as a permutation matrix.

 

a <- queens(8)

 

{{out}}

Count solutions for board size 4 to 12.

 

 

{{out}}

Backtracking algorithm; returns one solution

 

#lang racket

 

(nqueens 8)

; => (list (Q 3 7) (Q 1 6) (Q 6 5) (Q 2 4) (Q 5 3) (Q 7 2) (Q 4 1) (Q 0 0))

 

Show result with "How to Design Programs" GUI.

(require htdp/show-queen)

 

 

(show-nqueens 8)

 

[[image:Racket-nqueens.png]]

Computes all solutions.

 

#lang racket

 

'() qss-so-far)))

(lazy-filter valid? all-possible-solutions))

 

Taking the first solution does not compute the other solutions:

 

(car (nqueens 8))

;; => (list (Q 7 3) (Q 6 1) (Q 5 6) (Q 4 2) (Q 3 5) (Q 2 7) (Q 1 4) (Q 0 0))

 

Computing all solutions is also possible:

 

(define (force-and-print qs)

(define forced (force qs))

;(list (Q 7 3) (Q 6 6) (Q 5 4) (Q 4 1) (Q 3 5) (Q 2 0) (Q 1 2) (Q 0 7))

;(list (Q 7 4) (Q 6 6) (Q 5 1) (Q 4 5) (Q 3 2) (Q 2 0) (Q 1 3) (Q 0 7))

 

Logic borrowed from the Ruby example

#lang racket

(define (remove x lst)

(define (print-queens n)

(for ([x (queens n)]) (displayln (string-join x))))

 

=={{header|Raku}}==

Neither pretty nor efficient, a simple backtracking solution

 

sub collision(@field, $row) {

for ^$row -> $i {

}

}

{{out}}

<pre>[0 4 7 5 2 6 1 3]</pre>

=={{header|Rascal}}==

 

 

public set[list[int]] Nqueens(int n){

result += vector;}

return result;

 

=={{header|REXX}}==

 

About half of the REXX code involves presentation (and colorization achieved through dithering) of the chessboard and queens.

parse arg N . /*obtain optional argument from the CL.*/

if N=='' | N=="," then N= 8 /*Not specified: Then use the default.*/

say pad _ || bar /*show a single rank of the board.*/

end /*rank*/ /*80 cols can view a 19x19 board.*/

{{out|output|text=&nbsp; when using the default of an &nbsp; '''8'''<small>x</small>'''8''' &nbsp; chessboard:}}

<pre>

 

=={{header|Ring}}==

 

// Bert Mariani 2020-07-17

//================================

 

Output:

<pre>

 

=={{header|Ring}}==

load "stdlib.ring"

load "guilib.ring"

 

###============================================================

[https://www.mediafire.com/file/53bxu7kpuc4tlx5/Images.zip/file Necessary images]

 

=={{header|Ruby}}==

This implements the heuristics found on the wikipedia page to return just one solution

# puzzle).

# 2. Write a list of the even numbers from 2 to n in order.

end

 

{{out}}

===Alternate solution===

If there is not specification, it outputs all solutions.

attr_reader :count

puts @frame

end

 

'''Example:'''

puzzle = Queen.new(n)

puts " #{n} Queen : #{puzzle.count}"

puzzle = Queen.new(n, false) # do not display

puts " #{n} Queen : #{puzzle.count}"

 

{{out}}

 

=={{header|Run BASIC}}==

input "How many queens (N>=4)";n

if n < 4 then

end if

end if

<pre>abcdefgh

* 8

 

=={{header|Rust}}==

 

fn try(mut board: &mut [[bool; N]; N], row: usize, mut count: &mut i64) {

try (&mut board, 0, &mut count);

println!("Found {} solutions", count)

 

===Using Iterators===

Solution to the puzzle using an iterator that yields the 92 solutions for 8 queens.

use std::iter::IntoIterator;

 

str

}

 

=={{header|SAS}}==

data queens;

array a{8} p1-p8;

put n m;

keep p1-p8;

 

=={{header|Scala}}==

Lazily generates permutations with an <code>Iterator</code>.

 

object NQueens {

 

}

}

 

<pre>

This is a simple breadth-first technique to retrieve all solutions.

 

(import (scheme base)

(scheme write)

(pretty-print (n-queens 5) 5)

(pretty-print (n-queens 8) 8)

 

{{out}}

string(Board_size)+"x"+string(Board_size)+" board.");

//Time elapsed

 

{{out}}

 

=={{header|Seed7}}==

 

var array integer: board is 8 times 0;

end if;

end while;

 

=={{header|Sidef}}==

{{trans|Raku}}

 

func collision(field, row) {

for n in (1..15) {

say "#{'%2d' % n}: #{N_queens_solution(n) || 'No solution'}"

{{out}}

<pre>

 

=={{header|SNOBOL4}}==

* N queens problem

* Set N to the desired number. The program prints out all solution boards.

PRTLOOP B LEN(NP1) . OUTPUT = :S(PRTLOOP)F(RETURN)

END

 

=={{header|Sparkling}}==

This is somewhat a transliteration of the "shortened" C++ code above.

 

pos.foreach(function (_, i) {

stdout.printf(" %c", 'a' + i);

};

 

 

=={{header|SQL}}==

This implementation, which solves the problem for n=8, makes use of Common Table Expressions and has been tested with SQLite (>=3.8.3) and Postgres (please note the related comment in the code). It might be compatible with other SQL dialects as well. A gist with the SQL file and a Python script that runs it using SQLite is available on Github: https://gist.github.com/adewes/5e5397b693eb50e67f07

 

WITH RECURSIVE

positions(i) as (

SELECT board,n_queens FROM solutions WHERE n_queens = 8;

 

 

=={{header|SQL PL}}==

{{works with|Db2 LUW}} version 9.7 or higher.

With SQL PL:

-- A column of a matrix.

CREATE TYPE INTEGER_ARRAY AS INTEGER ARRAY[]@

 

 

Output:

<pre>

=={{header|Standard ML}}==

This implementation uses failure continuations for backtracking.

(*

* val threat : (int * int) -> (int * int) -> bool

(* NONE *)

queens(2);

 

=={{header|Stata}}==

Adapted from the Fortran 77 program, to illustrate the '''[http://www.stata.com/help.cgi?m2_goto goto]''' statement in Stata.

 

real matrix queens(real scalar n) {

real scalar i, j, k, p, q

rows(a)

92

 

It's also possible to save the solutions to a Stata dataset:

 

mata: a=queens(8)

getmata (a*)=a

 

=== Recursive version ===

The recursive solution is adapted from one of the Python programs.

 

real matrix queens_rec(real scalar n) {

real rowvector a, u, v

}

}

 

The iterative and the recursive programs are equivalent:

 

 

=={{header|Swift}}==

Port of the optimized C code above

let maxn = 31

 

}

 

 

=={{header|SystemVerilog}}==

Create a random board configuration, with the 8-queens as a constraint

 

parameter SIZE_LOG2 = 3;

 

endprogram

 

=={{header|Tailspin}}==

A functional-ish solution utilising tailspin's data flows

templates queens

def n: $;

'For 3 queens there are $:[3 -> queens] -> $::length; solutions

' -> !OUT::write

{{out}}

<pre>

 

A solution using state to find one solution if any exist

templates queens

def n: $;

'A solution to the 3 queens problem is $:3 -> queens;

' -> !OUT::write

{{out}}

<pre>

 

{{works with|Tcl|8.5}}

 

proc unsafe {y} {

}

 

{{out}}

<pre>$ tclsh8.5 8queens.tcl 6

The total number of solutions for 8 queens is displayed at the end of the run. The code could be adapted to display a selected solution or multiple solutions. This code runs anywhere you can get bash to run.

 

# variable declaration

work

out

 

=={{header|Ursala}}==

n is a number greater than 3. Multiple solutions may be reported

but reflections and rotations thereof are omitted.

#import nat

 

-<&l^|*DlrTS/~& ~&iiDlSzyCK9hlPNNXXtCS,

^jrX/~& @rZK20lrpblPOlrEkPK13lhPK2 ~&i&& nleq$-&lh+-,

The output shows one solution on each line.

A solution is reported as a sequence of <math>n</math> numbers

=={{header|VBA}}==

{{trans|BBC BASIC}}

Sub n_queens()

Const l = 15 'number of queens

Debug.Print n, m 'number of queens, number of solutions

Next n

{{out}}

<pre>

{{trans|BBC BASIC}}

To have the solutions printed (raw format) uncomment the ad hoc statement.

const l=15

dim a(),s(),u(): redim a(l),s(l),u(4*l-2)

Loop Until i=0

wscript.echo n &":"& m

{{out}}

<pre>

{{works with|Visual Basic|VB6 Standard}}

{{trans|BBC BASIC}}

Sub n_queens()

Const l = 15 'number of queens

Debug.Print n, m 'number of queens, number of solutions

Next n

{{out}}

<pre>

=={{header|Visual Basic .NET}}==

{{trans|BBC BASIC}}

Module Mod_n_queens

Sub n_queens()

Next n

End Sub 'n_queens

{{out}}

<pre>

 

=={{header|Wart}}==

prn "step: " queens # show progress

if (len.queens = n)

def (diagonal_match curr other)

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

 

=={{header|Wren}}==

{{trans|Kotlin}}

Very slow for the larger boards.

var c = []

var f = []

if (count > 0) System.print(" First is %(f)")

System.print()

 

{{out}}

 

Copied from http://www.cs.bu.edu/~hwxi/Xanadu/Examples/

int abs(i: int) {

if (i >= 0) return i; else return -i;

int main () {

return queen (8);

 

=={{header|XPL0}}==

[[File:NQueensXPL0.GIF|right]]

int R, C; \row and column of board

char B(N,N); \board

C:= 0; \start at left column

Try;

 

=={{header|XSLT}}==

83162574

84136275

 

You can view the results directly in your browser (Chrome/FF/IE/Opera/Safari) here: [[http://stamm-wilbrandt.de/en/xsl-list/n-queens/8-queens.xsl.xml]]

 

Here is stylesheet 8-queens.xsl.xml which produces the (simple) output by having itself as input: [[http://stamm-wilbrandt.de/en/xsl-list/n-queens/8-queens.xsl.xml]]

<!-- 8-queens.xsl disguised as XML file for the browsers -->

 

 

</xsl:stylesheet>

 

=={{header|Yabasic}}==

DOCU Place N Queens on an NxN chess board

DOCU such that they don't threaten each other.

wend

end sub

 

=={{header|Zig}}==

Outputs all 92 solutions.

const std = @import("std");

const stdout = std.io.getStdOut().outStream();

}

}

{{Out}}

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=={{header|zkl}}==

Modified from a Haskell version (if I remember correctly)

{ r,c:=q; (r==x or c==y or r+c==x+y or r-c==x-y) }

fcn isSafe(r,c,qs) // queen safe at (r,c)?, qs=( (r,c),(r,c)..) solution so far

if (row == N) return(qs);

return(qs.apply(self.fcn.fp(N,row+1)).flatten());

println(queens.len()," solution(s):");

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