/* Article: 5653 of comp.lang.prolog Newsgroups: comp.lang.prolog Path: ecrc!Germany.EU.net!mcsun!ub4b!news.cs.kuleuven.ac.be!bimbart From: bimbart@cs.kuleuven.ac.be (Bart Demoen) Subject: boolean constraint solvers Message-ID: <1992Oct19.093131.11399@cs.kuleuven.ac.be> Sender: news@cs.kuleuven.ac.be Nntp-Posting-Host: hera.cs.kuleuven.ac.be Organization: Dept. Computerwetenschappen K.U.Leuven Date: Mon, 19 Oct 1992 09:31:31 GMT Lines: 120 ?- calc_constr(N,C,L) . % with N instantiated to a positive integer generates in the variable C a datastructure that can be interpreted as a boolean expression (and in fact is so by SICStus Prolog's bool:sat) and in L the list of variables involved in this boolean expression; so ?- calc_constr(N,C,L) , bool:sat(C) , bool:labeling(L) . % with N instantiated to a positive integer shows the instantiations of L for which the boolean expression is true e.g. | ?- calc_constr(3,C,L) , bool:sat(C) , bool:labeling(L) . % C = omitted L = [1,0,1,0,1,0,1,0,1] ? ; no it is related to a puzzle which I can describe if people are interested SICStus Prolog can solve this puzzle up to N = 9 on my machine; it then fails because of lack of memory (my machine has relatively little: for N=9 SICStus needs 14 Mb - and about 50 secs runtime + 20 secs for gc on Sparc 1) I am interested in hearing about boolean constraint solvers that can deal with the expression generated by the program below, for large N and in reasonable time and space; say N in the range 10 to 20: the number of solutions for different N varies wildly; there is exactly one solution for N = 10,12,13,15,20 but for N=18 or 19 there are several thousand, so perhaps it is best to restrict attention to N with only one solution - unless that is unfair to your solver in case you have to adapt the expression for your own boolean solver, in the expression generated, ~ means negation, + means disjunction, * means conjunction and somewhere in the program, 1 means true Thanks Bart Demoen */ % test(N,L) :- calc_constr(N,C,L) , bool:sat(C) , bool:labeling(L) . test(N,L) :- calc_constr(N,C,L) , solve_bool(C,1). testbl(N,L) :- calc_constr(N,C,L) , solve_bool(C,1), labeling. testul(N,L) :- calc_constr(N,C,L) , solve_bool(C,1), label_bool(L). calc_constr(N,C,L) :- M is N * N , functor(B,b,M) , B =.. [_|L] , cc(N,N,N,B,C,1) . cc(0,M,N,B,C,T) :- ! , NewM is M - 1 , cc(N,NewM,N,B,C,T) . cc(_,0,_,B,C,C) :- ! . cc(I,J,N,B,C,T) :- neighbours(I,J,N,B,C,S) , NewI is I - 1 , cc(NewI,J,N,B,S,T) . neighbours(I,J,N,B,C,S) :- add(I,J,N,B,L,R1) , add(I-1,J,N,B,R1,R2) , add(I+1,J,N,B,R2,R3) , add(I,J-1,N,B,R3,R4) , add(I,J+1,N,B,R4,[]) , % L is the list of neighbours of (I,J) % including (I,J) odd(L,C,S) . add(I,J,N,B,S,S) :- I =:= 0 , ! . add(I,J,N,B,S,S) :- J =:= 0 , ! . add(I,J,N,B,S,S) :- I > N , ! . add(I,J,N,B,S,S) :- J > N , ! . add(I,J,N,B,[X|S],S) :- A is (I-1) * N + J , arg(A,B,X) . % odd/2 generates the constraint that an odd number of elements of its first % argument must be 1, the rest must be 0 odd(L,C*S,S):- exors(L,C). exors([X],X). exors([X|L],X#R):- L=[_|_], exors(L,R). /* % did this by enumeration, because there are only 4 possibilities odd([A], A * S,S) :- ! . odd([A,B,C], ((A * ~~(B) * ~~(C)) + (A * B * C) + ( ~~(A) * B * ~~(C)) + ( ~~(A) * ~~(B) * C)) * S,S) :- ! . odd([A,B,C,D], ((A * ~~(B) * ~~(C) * ~~(D)) + (A * B * C * ~~(D)) + (A * B * ~~(C) * D) + (A * ~~(B) * C * D) + ( ~~(A) * B * ~~(C) * ~~(D)) + ( ~~(A) * B * C * D) + ( ~~(A) * ~~(B) * C * ~~(D)) + ( ~~(A) * ~~(B) * ~~(C) * D)) * S,S ) :- ! . odd([A,B,C,D,E],((A * ~~(B) * ~~(C) * ~~(D) * ~~(E)) + (A * B * C * ~~(D) * ~~(E)) + (A * B * ~~(C) * D * ~~(E)) + (A * ~~(B) * C * D * ~~(E)) + (A * B * ~~(C) * ~~(D) * E) + (A * ~~(B) * C * ~~(D) * E) + (A * ~~(B) * ~~(C) * D * E) + (A * B * C * D * E) + ( ~~(A) * B * ~~(C) * ~~(D) * ~~(E)) + ( ~~(A) * B * ~~(C) * D * E) + ( ~~(A) * B * C * ~~(D) * E) + ( ~~(A) * B * C * D * ~~(E)) + ( ~~(A) * ~~(B) * C * ~~(D) * ~~(E)) + ( ~~(A) * ~~(B) * C * D * E) + ( ~~(A) * ~~(B) * ~~(C) * D * ~~(E)) + ( ~~(A) * ~~(B) * ~~(C) * ~~(D) * E)) * S,S ) :- ! . */