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| (Een tussenliggende versie door een andere gebruiker niet weergegeven) |
| Regel 1: |
Regel 1: |
| =DT: oplossing voor het probleem K-Queens=
| | #REDIRECT [[Declaratieve_Talen/Oplossing_N-queens]] |
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| (zie niet direct nog optimalisaties, nog iemand ideeën? Merel)
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| % beginmethode: Opl zal van de vorm [1,2,3,4,5,6,..] zijn
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| nqueens(N, Opl) :-
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| nqueens(N, 1, [], Opl).
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| | |
| % C is current kolum, Accum van vorm [(1,x), (2,y), ..]
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| nqueens(N, C, Accum, Opl) :-
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| (
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| check_row_constraint(Accum) ->
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| fail
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| ;
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| check_diagonal_constraint(N, Accum) ->
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| fail
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| ;
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| C > N ->
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| findall(B, member((_,B), Accum), Opl)
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| ;
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| get_next_choice(N, C, Choice),
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| C1 is C + 1,
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| nqueens(N, C1, [Choice|Accum], Opl)
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| ).
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| | |
| % constraintchecking
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| check_row_constraint(Accum) :-
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| member((A, B), Accum),
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| member((C, D), Accum),
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| A \== C,
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| B == D, !.
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| % diagonaal rechts naar boven check
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| check_diagonal_constraint(N, Accum) :-
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| member((A, B), Accum),
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| between(1, N, D),
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| D \== 0,
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| F is A + D,
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| G is B + D,
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| member((F, G), Accum).
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| | |
| % diagonaal links naar boven check
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| check_diagonal_constraint(N, Accum) :-
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| member((A, B), Accum),
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| between(1, N, D),
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| D \== 0,
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| F is A - D,
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| G is B + D,
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| member((F, G), Accum).
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| % volgende keuze
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| get_next_choice(N, C, Choice) :-
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| between(1, N, A),
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| Choice = (C, A).
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