通过Prolog随时建立牢固连接的组件 [英] Anytime strongly connected components via Prolog

问题描述

Markus Triska报告了一种算法，该算法可以确定强连接的组件(SCC).是否有一种解决方案，可以在不使用属性变量的情况下确定SCC，并且可以随时使用.这样某些顶点可以具有无限多个边?

Markus Triska has reported an algorithm to determine strongly connected components (SCC). Is there a solution which determines the SCC without making use of attributed variable and that can work anytime. So that some vertices can have infinitely many edges?

我之所以问是因为我想知道B-Prologs 他们随时渴望的制表是否可以复制. B-Prolog确定群集，它们是SCC的名称.但是，它还有一种制表模式，可以快速返回表格结果.

I am asking because I am wondering whether B-Prologs anytime tabling which they call eager can be replicated. B-Prolog determines clusters which is their name for SCC. But it has also a tabling mode where it returns tabled results eagerly.

推荐答案

我猜这算法符合要求，因为只需稍加修改，它就可以在根中实现无限度.当所有边都指向有限多个顶点时，这是SCC算法:

I guess this algorithm fits the bill, since with a little modification it would allow an infinite degree in the root. This is the SCC algorithm when all edges point to finitely many vertices:

% plan(+Atom, -Assoc)
plan(P, R) :-
  sccforpred(P, [], [], R, _).

% sccforpred(+Atom, +List, +Assoc, -Assoc, -List)
sccforpred(P, S, H, H, [P]) :-
  member(P, S), !.
sccforpred(P, _, H, H, []) :-
  member(Q-L, H), (Q = P; member(P, L)), !.
sccforpred(P, S, I, O2, R2) :-
  findall(Q, edge(P, Q), L),
  sccforlist(L, [P|S], I, O, R),
  (member(U, R), member(U, S) ->
      O2 = O, R2 = [P|R];
      O2 = [P-R|O], R2 = []).

% sccforlist(+List, +List, +Assoc, -Assoc, -List)
sccforlist([], _, H, H, []).
sccforlist([P|Q], S, I, O, R) :-
  sccforpred(P, S, I, H, R1),
  sccforlist(Q, S, H, O, R2),
  findall(U, (member(U, R1); member(U, R2)), K),
  sort(K, R).

使用此示例图:

我得到这个结果:

Jekejeke Prolog 4, Runtime Library 1.4.1 (August 20, 2019)

?- plan(21,X).
X = [21-[], 22-[22, 23], 26-[24, 25, 26], 27-[]]

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