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/*
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 ) :- ! .
*/