prolog - 在序言中实现 Dijkstra 的最佳结构图

Question

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

我找到了 this但这对我的工作来说太慢了。

Answer 1

该实现还不错:

?- 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，然后 this answer可能与您有关。
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 好吧，我必须说第一次使用属性变量进行编程并不太容易。

我正在使用我上面链接的@Mat 的答案中的建议，滥用属性变量来根据算法的要求获得对附加到数据的属性的恒定时间访问。我已经(盲目地)实现了 wikipedia algorithm ，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，我更喜欢你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 显然显示出改善

% 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])
....

但图表太小，无法做出任何明确的断言。如果此代码段减少了您的程序所需的时间，请告诉我们。

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。