2 SAT問題解決報告

2 SATはP問題で多項式時間内に解決できる.ここでは方法の1つをまとめます.
テストデータのフォーマットは次のとおりです.
1000000 -383037 -585864 575603 -494722 -628477 805085 -502855 685868 923336 -606921 -653863 424389 333542 -656925 -809061 874253
すなわち範囲は1~100000,000であり,「-」号は取逆を示す.例えば-383037-585864表示~383037 U~58864575603-494722表示575603 U~494722
上記のすべてのclauseを満たす取法があるか否かを判断する.
解法は図論における強い連通子図を用いた.具体的には,まず有向図を構築する.各数値とその逆は個別のノードです.各エントリについて、AUBのように、~A->Bと~B->Aの2つのエッジが図に追加される.すなわち、Aが負であれば、Bは正である.逆にBが負であればAは正である.この2つのエッジはAUBと等価であり,いずれもAとBの値を制約している.有向図を構築すると強連通サブ図が構築される.1つの強い連通サブマップにおける変数は互いに制約され,同時に成立しなければならない.
したがって、一対の変数Aと~Aが同一の強連通サブマップにあると、この2 SATは満たされない.変数Aと~Aは同時に成り立たないからです.

public class TwoSAT {

	public static void main(String[] args)
	{
		In in = new In(args[0]);
		int N = in.readInt();
		int V = 2 * N;
		Digraph digraph = new Digraph(V);
		while (!in.isEmpty())
		{
			int v = in.readInt();
			int w = in.readInt();
			int negv = 2 * Math.abs(v) - 1, posv = 2 * Math.abs(v) - 2;
			int negw = 2 * Math.abs(w) - 1, posw = 2 * Math.abs(w) - 2;
			digraph.addEdge(v > 0 ? negv : posv, w > 0 ? posw : negw);
			digraph.addEdge(w > 0 ? negw : posw, v > 0 ? posv : negv);
		}
		StdOut.println("input done.");
		KosarajuSharirSCC scc = new KosarajuSharirSCC(digraph);
		boolean satisfied = true;
		for (int i = 0; i < N; ++i)
		{
			if (scc.stronglyConnected(2 * i, 2 * i + 1))
			{
				satisfied = false;
				break;
			}
		}
		if (satisfied)
		{
			StdOut.println(true);
		} else
		{
			StdOut.println(false);
		}
	}

}

ここではAlgos 4における強い連通サブマップを求める方法を用い,過程は逆を求め,dfsを2回加えた.

/*************************************************************************
 *  Compilation:  javac KosarajuSharirSCC.java
 *  Execution:    java KosarajuSharirSCC filename.txt
 *  Dependencies: Digraph.java TransitiveClosure.java StdOut.java In.java
 *  Data files:   http://algs4.cs.princeton.edu/42directed/tinyDG.txt
 *
 *  Compute the strongly-connected components of a digraph using the
 *  Kosaraju-Sharir algorithm.
 *
 *  Runs in O(E + V) time.
 *
 *  % java KosarajuSCC tinyDG.txt
 *  5 components
 *  1 
 *  0 2 3 4 5 
 *  9 10 11 12 
 *  6 
 *  7 8 
 *
 *  % java KosarajuSharirSCC mediumDG.txt 
 *  10 components
 *  21 
 *  2 5 6 8 9 11 12 13 15 16 18 19 22 23 25 26 28 29 30 31 32 33 34 35 37 38 39 40 42 43 44 46 47 48 49 
 *  14 
 *  3 4 17 20 24 27 36 
 *  41 
 *  7 
 *  45 
 *  1 
 *  0 
 *  10 
 *
 *************************************************************************/

public class KosarajuSharirSCC {
    private boolean[] marked;     // marked[v] = has vertex v been visited?
    private int[] id;             // id[v] = id of strong component containing v
    private int count;            // number of strongly-connected components


    public KosarajuSharirSCC(Digraph G) {

        // compute reverse postorder of reverse graph
        DepthFirstOrder dfs = new DepthFirstOrder(G.reverse());

        // run DFS on G, using reverse postorder to guide calculation
        marked = new boolean[G.V()];
        id = new int[G.V()];
        for (int v : dfs.reversePost()) {
            if (!marked[v]) {
                dfs(G, v);
                count++;
            }
        }

        // check that id[] gives strong components
        assert check(G);
    }

    // DFS on graph G
    private void dfs(Digraph G, int v) { 
        marked[v] = true;
        id[v] = count;
        for (int w : G.adj(v)) {
            if (!marked[w]) dfs(G, w);
        }
    }

    // return the number of strongly connected components
    public int count() { return count; }

    // are v and w strongly connected?
    public boolean stronglyConnected(int v, int w) {
        return id[v] == id[w];
    }

    // id of strong component containing v
    public int id(int v) {
        return id[v];
    }

    // does the id[] array contain the strongly connected components?
    private boolean check(Digraph G) {
        TransitiveClosure tc = new TransitiveClosure(G);
        for (int v = 0; v < G.V(); v++) {
            for (int w = 0; w < G.V(); w++) {
                if (stronglyConnected(v, w) != (tc.reachable(v, w) && tc.reachable(w, v)))
                    return false;
            }
        }
        return true;
    }

    public static void main(String[] args) {
        In in = new In(args[0]);
        Digraph G = new Digraph(in);
        KosarajuSharirSCC scc = new KosarajuSharirSCC(G);

        // number of connected components
        int M = scc.count();
        StdOut.println(M + " components");

        // compute list of vertices in each strong component
        Queue<Integer>[] components = (Queue<Integer>[]) new Queue[M];
        for (int i = 0; i < M; i++) {
            components[i] = new Queue<Integer>();
        }
        for (int v = 0; v < G.V(); v++) {
            components[scc.id(v)].enqueue(v);
        }

        // print results
        for (int i = 0; i < M; i++) {
            for (int v : components[i]) {
                StdOut.print(v + " ");
            }
            StdOut.println();
        }

    }

}