which included commits to RCS files with non-trunk default branches. git-svn-id: https://yap.svn.sf.net/svnroot/yap/trunk@5 b08c6af1-5177-4d33-ba66-4b1c6b8b522a
		
			
				
	
	
		
			151 lines
		
	
	
		
			4.1 KiB
		
	
	
	
		
			Plaintext
		
	
	
	
	
	
			
		
		
	
	
			151 lines
		
	
	
		
			4.1 KiB
		
	
	
	
		
			Plaintext
		
	
	
	
	
	
| /*
 | |
| 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 ) :- ! .
 | |
| 
 | |
| */
 | |
| 
 | |
| 
 | |
| 
 | |
| 
 | |
| 
 | |
| 
 | |
| 
 | |
| 
 |