如何将强连接的组件缩减为一个顶点？ [英] How to reduce a strongly connected component to one vertex?

问题描述

来自

我在任何地方都找不到的是我上面介绍的那种细节。诸如好吧，这很容易，您先做那件事，然后再做其他事情之类的评论就没有具体细节了。

From https://algs4.cs.princeton.edu/42digraph/

43. Reachable vertex in a digraph. Design a linear-time algorithm to determine whether a digraph has a vertex that is reachable from every other vertex.

Kosaraju-Sharir algorithm gives us the strongly connected components. Java code for that can be seen here. Reducing each SCC to a single vertex, a vertex that has outdegree zero is reachable from every other.

Problem is, everyone seems to be talking about reducing a SCC without providing details. What is an efficient algorithm to do so?

解决方案

Following is a Java solution to my own question. For the graph representation, it uses edu.princeton.cs:algs4:1.0.3 from https://github.com/kevin-wayne/algs4. There appears to be general algorithms for graph contraction, as outlined in this paper; however, for my purposes, the following is sufficient.

/**
 * 43. <b>Reachable vertex.</b>
 * <p>
 * DAG: Design a linear-time algorithm to determine whether a DAG has a vertex that is reachable from every other
 * vertex, and if so, find one.
 * Digraph: Design a linear-time algorithm to determine whether a digraph has a vertex that is reachable from every
 * other vertex, and if so, find one.
 * <p>
 * Answer:
 * DAG: Consider an edge (u, v) ∈ E. Since the graph is acyclic, u is not reachable from v.
 * Thus u cannot be the solution to the problem. From this it follows that only a vertex of
 * outdegree zero can be a solution. Furthermore, there has to be exactly one vertex with outdegree zero,
 * or the problem has no solution. This is because if there were multiple vertices with outdegree zero,
 * they wouldn't be reachable from each other.
 * <p>
 * Digraph: Reduce the graph to it's Kernel DAG, then find a vertex of outdegree zero.
 */
public class Scc {
    private final Digraph g;
    private final Stack<Integer> s = new Stack<>();
    private final boolean marked[];
    private final Digraph r;
    private final int[] scc;
    private final Digraph kernelDag;

    public Scc(Digraph g) {
        this.g = g;
        this.r = g.reverse();
        marked = new boolean[g.V()];
        scc = new int[g.V()];
        Arrays.fill(scc, -1);

        for (int v = 0; v < r.V(); v++) {
            if (!marked[v]) visit(v);
        }

        int i = 0;
        while (!s.isEmpty()) {
            int v = s.pop();

            if (scc[v] == -1) visit(v, i++);
        }
        Set<Integer> vPrime = new HashSet<>();
        Set<Map.Entry<Integer, Integer>> ePrime = new HashSet<>();

        for (int v = 0; v < scc.length; v++) {
            vPrime.add(scc[v]);
            for (int w : g.adj(v)) {
                // no self-loops, no parallel edges
                if (scc[v] != scc[w]) {
                    ePrime.add(new SimpleImmutableEntry<>(scc[v], scc[w]));
                }
            }
        }
        kernelDag = new Digraph(vPrime.size());
        for (Map.Entry<Integer, Integer> e : ePrime) kernelDag.addEdge(e.getKey(), e.getValue());
    }

    public int reachableFromAllOther() {
        for (int v = 0; v < kernelDag.V(); v++) {
            if (kernelDag.outdegree(v) == 0) return v;
        }
        return -1;
    }

    // reverse postorder
    private void visit(int v) {
        marked[v] = true;

        for (int w : r.adj(v)) {
            if (!marked[w]) visit(w);
        }
        s.push(v);
    }

    private void visit(int v, int i) {
        scc[v] = i;

        for (int w : g.adj(v)) {
            if (scc[w] == -1) visit(w, i);
        }
    }
}

Running it on the graph below produces the strongly-connected components as shown. Vertex 0 in the reduced DAG is reachable from every other vertex.

What I couldn't find anywhere is the kind of detail that I presented above. Comments like "well, this is easy, you do that, then you do something else" are thrown around without concrete details.

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