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问题很简单。如何在 SWI prolog 中构造我的 Graph 以实现 Dijkstra 算法？

我找到了这个，但它对我的工作来说太慢了。

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1 回答 1

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这个实现还不错：

?- time(dijkstra(penzance, Ss)).
% 3,778 inferences, 0,003 CPU in 0,003 seconds (99% CPU, 1102647 Lips)
Ss = [s(aberdeen, 682, [penzance, exeter, bristol, birmingham, manchester, carlisle, edinburgh|...]), s(aberystwyth, 352, [penzance, exeter, bristol, swansea, aberystwyth]), s(birmingham, 274, [penzance, exeter, bristol, birmingham]), s(brighton, 287, [penzance, exeter, portsmouth, brighton]), s(bristol, 188, [penzance, exeter, bristol]), s(cambridge, 339, [penzance, exeter|...]), s(cardiff, 322, [penzance|...]), s(carlisle, 474, [...|...]), s(..., ..., ...)|...].

SWI-Prolog 提供属性变量，那么这个答案可能与您相关。我希望我今天晚些时候会发布一个使用属性变量的 dijkstra/2 实现。

编辑得好，我必须说第一次用属性变量编程并不容易。

我正在使用上面链接的@Mat 的答案中的建议，滥用属性变量以根据算法的要求获得对附加到数据的属性的持续时间访问。我（盲目地）实现了维基百科算法，这是我的努力：

/*  File:    dijkstra_av.pl
    Author:  Carlo,,,
    Created: Aug  3 2012
    Purpose: learn graph programming with attribute variables
*/

:- module(dijkstra_av, [dijkstra_av/3]).

dijkstra_av(Graph, Start, Solution) :-
    setof(X, Y^D^(member(d(X,Y,D), Graph)
             ;member(d(Y,X,D), Graph)), Xs),
    length(Xs, L),
    length(Vs, L),
    aggregate_all(sum(D), member(d(_, _, D), Graph), Infinity),
    catch((algo(Graph, Infinity, Xs, Vs, Start, Solution),
           throw(sol(Solution))
          ), sol(Solution), true).

algo(Graph, Infinity, Xs, Vs, Start, Solution) :-
    pairs_keys_values(Ps, Xs, Vs),
    maplist(init_adjs(Ps), Graph),
    maplist(init_dist(Infinity), Ps),
    ord_memberchk(Start-Sv, Ps),
    put_attr(Sv, dist, 0),
    time(main_loop(Vs)),
    maplist(solution(Start), Vs, Solution).

solution(Start, V, s(N, D, [Start|P])) :-
    get_attr(V, name, N),
    get_attr(V, dist, D),
    rpath(V, [], P).

rpath(V, X, P) :-
    get_attr(V, name, N),
    (   get_attr(V, previous, Q)
    ->  rpath(Q, [N|X], P)
    ;   P = X
    ).

init_dist(Infinity, N-V) :-
    put_attr(V, name, N),
    put_attr(V, dist, Infinity).

init_adjs(Ps, d(X, Y, D)) :-
    ord_memberchk(X-Xv, Ps),
    ord_memberchk(Y-Yv, Ps),
    adj_add(Xv, Yv, D),
    adj_add(Yv, Xv, D).

adj_add(X, Y, D) :-
    (   get_attr(X, adjs, L)
    ->  put_attr(X, adjs, [Y-D|L])
    ;   put_attr(X, adjs, [Y-D])
    ).

main_loop([]).
main_loop([Q|Qs]) :-
    smallest_distance(Qs, Q, U, Qn),
    put_attr(U, assigned, true),
    get_attr(U, adjs, As),
    update_neighbours(As, U),
    main_loop(Qn).

smallest_distance([A|Qs], C, M, [T|Qn]) :-
    get_attr(A, dist, Av),
    get_attr(C, dist, Cv),
    (   Av < Cv
    ->  (N,T) = (A,C)
    ;   (N,T) = (C,A)
    ),
    !, smallest_distance(Qs, N, M, Qn).
smallest_distance([], U, U, []).

update_neighbours([V-Duv|Vs], U) :-
    (   get_attr(V, assigned, true)
    ->  true
    ;   get_attr(U, dist, Du),
        get_attr(V, dist, Dv),
        Alt is Du + Duv,
        (   Alt < Dv
        ->  put_attr(V, dist, Alt),
        put_attr(V, previous, U)
        ;   true
        )
    ),
    update_neighbours(Vs, U).
update_neighbours([], _).

:- begin_tests(dijkstra_av).

test(1) :-
    nl,
    time(dijkstra_av([d(a,b,1),d(b,c,1),d(c,d,1),d(a,d,2)], a, L)),
    maplist(writeln, L).

test(2) :-
    open('salesman.pl', read, F),
    readf(F, L),
    close(F),
    nl,
    dijkstra_av(L, penzance, R),
    maplist(writeln, R).

readf(F, [d(X,Y,D)|R]) :-
    read(F, dist(X,Y,D)), !, readf(F, R).
readf(_, []).

:- end_tests(dijkstra_av).

说实话，我更喜欢您在问题中链接的代码。有一个明显的优化点，smallest_distance/4 现在使用哑线性扫描，使用 rbtree 运行时应该更好。但是必须小心处理属性变量。

time/1 显然显示出改进

% 2,278 inferences, 0,003 CPU in 0,003 seconds (97% CPU, 747050 Lips)
s(aberdeen,682,[penzance,exeter,bristol,birmingham,manchester,carlisle,edinburgh,aberdeen])
....

但该图太小，无法做出任何明确的断言。让我们知道此代码段是否会减少您的程序所需的时间。

文件salesman.pl包含 dist/3 事实，它是从问题中的链接中逐字提取的。

于 2012-08-03T08:13:24.223 回答