In chess, a queen can move as far as as the board extends horizontally, vertically, or diagonally.

Given a NxN sized chessboard, print out how many possible positions N queens can be placed on the board and not be able to hit each other in 1 move.


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  • \$\begingroup\$ Do we need to handle 2 <= N <= 4 cases? If so how? \$\endgroup\$ – st0le Feb 2 '11 at 12:09
  • \$\begingroup\$ There is no solution for case: N = 2,3. The wikipedia has a excellent write up about this classic problem. It documents ver well about the solution number from N = 1 to N = 14. (I am still new to Code Golf. Not sure what is the best way to participate yet. :)) \$\endgroup\$ – Dongshengcn Jul 4 '11 at 0:01
  • \$\begingroup\$ A000170 \$\endgroup\$ – Peter Taylor Apr 12 '13 at 15:52

Here's a solution (originally from this blog entry) where I construct a logical description of the solution in conjunctive normal form which is then solved by Mathematica:

(* Define the variables: Q[i,j] indicates whether there is a 
   Queen in row i, column j *)
Qs = Array[Q, {8, 8}];

(* Define the logical constraints. *)
problem =
   (* Each row must have a queen. *)
   And @@ Map[(Or @@ #) &, Qs],
   (* for all i,j: Q[i,j] implies Not[...] *)
   And @@ Flatten[
     Qs /. Q[i_, j_] :>
       And @@ Map[Implies[Q[i, j], Not[#]] &, 
          Q[k_, l_] /;
           Not[(i == k) && (j == l)] && (
             (i == k) ||          (* same row *)
                 (j == l) ||          (* same column *)
             (i + j == k + l) ||  (* same / diagonal *)
             (i - j == k - l)),   (* same \ diagonal *)

(* Find the solution *)
solution = FindInstance[problem, Flatten[Qs], Booleans] ;

(* Display the solution *)
Qs /. First[solution] /. {True -> Q, False -> x} // MatrixForm

Here's the output:

x   x   x   x   Q   x   x   x
x   Q   x   x   x   x   x   x
x   x   x   Q   x   x   x   x
x   x   x   x   x   x   Q   x
x   x   Q   x   x   x   x   x
x   x   x   x   x   x   x   Q
x   x   x   x   x   Q   x   x
Q   x   x   x   x   x   x   x


I don't see a golf tag, so i'm assuming it's just a challenge.

Here's an implementation of the Algorithm mentioned on Wikipedia. It's not by me, it's at Rosetta Stone and can be found here

CommWikied this Answer.


Python 2, 190 185 chars

from itertools import*
print len(filter(lambda x:all(1^(y in(z,z+i-j,z-i+j))for i,y in enumerate(x)for j,z in enumerate(x[:i]+(1e9,)+x[i+1:])),permutations(range(1,n+1),n)))

I just assumed the code golf tag even though it wasn't there. N is read from stdin, the program calculates solutions up to n=10 in acceptable time.



  def x=0,y=0
  Set a=it.collect{it-x++} 
  Set b=it.collect{it+y++} 

Delivers a list of all queen solutions like this:

[ [4, 7, 3, 0, 6, 1, 5, 2], 
  [6, 2, 7, 1, 4, 0, 5, 3], 
  ... ]

For graphical representation add:

s.each { def size = it.size()
         it.each { (it-1).times { print "|_" }
                   print "|Q"
                   (size-it).times { print "|_" }
                   println "|"
         println ""

which looks like this:


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