search all paths and the shortest path for a graph - Prolog
Asked Answered
M

2

1

I have a problem in my code with turbo prolog which searches all paths and the shortest path in a graph between 2 nodes. The problem that i have is to test if the node is in the list or not (exactly in the clause of member)

           1    ---- b ----   3
           ---       |        ---
        ---          |             -----
      a              |5                  d
        ---          |             -----
            ---      |         ---
             2  ---  |     ---   4
                  -- c  --

for example we have for b--->c 
([b,c],5) , ([b,a,c],3) and ([b,d,c],7) : possible paths.
([b,a,c],3) : the shortest path.

and this is my code :

DOMAINS
    list=Symbol *

PREDICATES
    distance(Symbol, Symbol)
    path1(Symbol, Symbol, list, integer)
    path(Symbol, Symbol,list, list, integer)
    distance(Symbol, list, integer)
    member(Symbol, list)
    shortest(Symbol, Symbol, list, integer)

CLAUSES
    distance(a, b, 1).
    distance(a, c, 2).
    distance(b, d, 3).
    distance(c, d, 4).
    distance(b, c, 5).
    distance(b, a, 1).
    distance(c, a, 2).
    distance(d, b, 3).
    distance(d, c, 4).
    distance(c, b, 5).

    member(X, [X|T]).
    member(X, [Y|T]) :- member(X, T).

    absent(X, L) :-
        member(X, L),
        !,
        fail.
    absent(_, _).

    /* find all paths */
    path1(X, Y, L, C) :- path(X, Y, L, I, C).
    path(X, X, [X], I, C) :- absent(X, I).
    path(X, Y, [X|R], I, C) :-
        distance(X, Z, A),
        absent(Z, I),
        path(Z, Y, R, [X|I], C1),
        C = C1 + A
        .

    /* to find the shortest path */
    shortest(X, Y, L, C) :-
        path(X, Y, L, C),
        path(X, Y, L1, C1),
        C < C1.
Middleoftheroader answered 1/4, 2010 at 22:22 Comment(1)
You did not state what your question is.Archimage
S
1

This shows the shortest path and it's weight:

edge(a,b,6).
edge(a,c,1).
edge(b,d,5).
edge(c,e,4).
edge(c,f,1).
edge(d,h,3).
edge(e,h,7).
edge(f,g,2).
edge(g,h,1).

path(X,Y,M,[Y]) :- edge(X,Y,M).
path(X,Y,P,[Z|T]) :- edge(X,Z,M),path(Z,Y,N,T),
            P is M+N.

pravilo(X,Y,Z) :-  assert(min(100)),assert(minpath([])),!,
                path(X,Y,K,PATH1),
                (min(Z),K<Z,
                retract(min(Z));assert(min(K))),
                minpath(Q),retract(minpath(Q)),
                assert(minpath([X|PATH1])),
                fail.

?- pravilo(a,h,X);
    write("Minimal Path:"),
    minpath(PATH),
    write(PATH),
    nl,
    write("Path weight:"),
    min(Z),
    write(Z).
Seicento answered 15/5, 2012 at 18:55 Comment(0)
F
0

Without knowing what the actual problem is, I can at least suggest that maybe shortest() and path() should take a maximum-length parameter that short-circuits the search.

Also, shortest() doesn't find the shortest path. It finds, for every possible pair of paths, the shortest of each pair.

Found answered 1/4, 2010 at 22:49 Comment(1)
ok i see could you propose a solution for this problem please thank you ! because i try but with no solution !!!Middleoftheroader

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