Depth limited search in prolog (vanilla meta-interpreter)
Asked Answered
M

3

6

I need to modify the vanilla meta-interpreter in order to make a search with limited depth. I'm using the following code for testing my sollution:

value(wire1,1).
connected(wire2, wire1).
connected(wire3, wire2).
connected(wire4, wire3).
connected(wire5, wire4).
connected(wire6, wire5).
connected(wire7, wire6).
connected(wire8, wire7).
connected(wire9, wire8).
value(W,X):-connected(W,V), value(V,X).

And the target is that something like:

solve(value(w9,X), 3).     /*depth =3, it should return false*/
solve(value(w9,X), 20).    /*depth=20 is enought for returning X=1*/

By the way my code is

solve(true,_):-!.
solve((A,B),D) :-!, solve(A,D), solve(B,D).
solve(A,D) :- clause(A, B),solve(B,D2),D=D2+1,D>0).

But it don't work property. Can you help me? Thanks a lot in advance

Mendoza answered 23/10, 2011 at 19:44 Comment(0)
G
4

An interesting page on metaprogramming came from a good developer: Markus Triska. Here (A Couple of Meta-interpreters in Prolog) you find both theory and practice. For instance:

... Another group of extensions aims to improve the incomplete default computation strategy. We start from an MI that limits the depth of the search tree:

        mi_limit(Goal, Max) :-
                mi_limit(Goal, Max, _).

        mi_limit(true, N, N).
        mi_limit((A,B), N0, N) :-
                mi_limit(A, N0, N1),
                mi_limit(B, N1, N).
        mi_limit(g(G), N0, N) :-
                N0 > 0,
                mi_clause(G, Body),
                N1 is N0 - 1,
                mi_limit(Body, N1, N).
Gasper answered 23/10, 2011 at 20:24 Comment(0)
C
3

You were almost there. Only the last clause needs a slight rearranging:

solve(A, D) :- clause(A, B), D1 is D - 1, D1 > 0, solve(B, D1).

?- solve(value(wire9, X), 9).       ===> false.
?- solve(value(wire9, X), 10).      ===> X = 1.
Catarinacatarrh answered 24/10, 2011 at 15:1 Comment(1)
You could reorder the goals in the body such that the depth-test comes first: solve(A, D) :- D > 1, clause(A, B), D1 is D - 1, solve(B, D1).Croquet
A
-1
dls(X,X,[X],L):-
    L >0 goal(X).
dls(X,Y,[A|p],L):-
    L > 0 ,goal(Y) ,
    move(X,Y),
    L1 is L - 1 ,
    dls(Z,Y ,P,L1).
Ashtray answered 1/5, 2015 at 21:59 Comment(0)

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