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algorithmprologclpfdn-queens

Solving N-Queens Problem... How far can we go?


The N-Queens Problem:

This problem states that given a chess board of size N by N, find the different permutations in which N queens can be placed on the board without any one threatening each other.

My question is:
What is the maximum value of N for which a program can calculate the answer in reasonable amount of time? Or what is the largest N we have seen so far?

Here is my program in CLPFD(Prolog):

generate([],_).
generate([H|T],N) :-
   H in 1..N ,
   generate(T,N).

lenlist(L,N) :-
   lenlist(L,0,N).

lenlist([],N,N).
lenlist([_|T],P,N) :-
   P1 is P+1,
   lenlist(T,P1,N).

queens(N,L) :-
   generate(L,N),lenlist(L,N),
   safe(L),
   !,
   labeling([ffc],L).

notattack(X,Xs) :-
   notattack(X,Xs,1).

notattack(X,[],N).
notattack(X,[Y|Ys],N) :-
   X #\= Y,
   X #\= Y - N,
   X #\= Y + N,
   N1 is N + 1,
   notattack(X,Ys,N1).

safe([]).
safe([F|T]) :-
   notattack(F,T),
   safe(T).

This program works just fine, but the the time it takes keeps on increasing with N. Here is a sample execution:

?- queens(4,L).

L = [2, 4, 1, 3] ;

L = [3, 1, 4, 2] ;

No

This means you place the 4 queens at Row 2 in Column1, Row 4 in Column 2, Row 1 in 3 and Row 3 in 4.(In a 4 By 4 chess board)

Now lets see how this program performs(Time taken in calculating the first permutation):
For N=4,5.....10 Computes within a second
For N=11-30 Takes between -1-3 seconds
For N=40..50 Still calculates within a minute
At N=60 It goes out of Global stack(Search space being enormous).

This was a past Homework problem. (The original problem was just to code N-Queens)

I am also interested in seeing alternate implementations in other languages(which performs better than my implementation) or If there is room for improvement in my algorithm/program


Solution

  • a short solution presented by raymond hettinger at pycon: easy ai in python

    #!/usr/bin/env python
    from itertools import permutations
    n = 12
    cols = range(n)
    for vec in permutations(cols):
        if (n == len(set(vec[i] + i for i in cols))
              == len(set(vec[i] - i for i in cols))):
            print vec
    

    computing all permutations is not scalable, though (O(n!))