Java Program to Give an Implementation of the Traditional Chinese Postman Problem

This is a java program to implement chinese Postman Problem. In graph theory, a branch of mathematics, the Chinese postman problem (CPP), postman tour or route inspection problem is to find a shortest closed path or circuit that visits every edge of a (connected) undirected graph. When the graph has an Eulerian circuit (a closed walk that covers every edge once), that circuit is an optimal solution. Otherwise, the optimization problem is to find the fewest number of edges to add to the graph so that the resulting multigraph does have an Eulerian circuit.

Here is the source code of the Java Program to Give an Implementation of the Traditional Chinese Postman Problem. The Java program is successfully compiled and run on a Windows system. The program output is also shown below.

package com.maixuanviet.graph;
 
import java.util.Vector;
 
public class ChinesePostmanProblem
{
    int            N;                // number of vertices
    int            delta[];          // deltas of vertices
    int            neg[], pos[];     // unbalanced vertices
    int            arcs[][];         // adjacency matrix, counts arcs between
                                      // vertices
    Vector<String> label[][];        // vectors of labels of arcs (for each
                                      // vertex
    // pair)
    int            f[][];            // repeated arcs in CPT
    float          c[][];            // costs of cheapest arcs or paths
    String         cheapestLabel[][]; // labels of cheapest arcs
    boolean        defined[][];      // whether path cost is defined between
                                      // vertices
    int            path[][];         // spanning tree of the graph
    float          basicCost;        // total cost of traversing each arc once
 
    void solve()
    {
        leastCostPaths();
        checkValid();
        findUnbalanced();
        findFeasible();
        while (improvements())
            ;
    }
 
    // allocate array memory, and instantiate graph object
    @SuppressWarnings("unchecked")
    ChinesePostmanProblem(int vertices)
    {
        if ((N = vertices) <= 0)
            throw new Error("Graph is empty");
        delta = new int[N];
        defined = new boolean[N][N];
        label = new Vector[N][N];
        c = new float[N][N];
        f = new int[N][N];
        arcs = new int[N][N];
        cheapestLabel = new String[N][N];
        path = new int[N][N];
        basicCost = 0;
    }
 
    ChinesePostmanProblem addArc(String lab, int u, int v, float cost)
    {
        if (!defined[u][v])
            label[u][v] = new Vector<String>();
        label[u][v].addElement(lab);
        basicCost += cost;
        if (!defined[u][v] || c[u][v] > cost)
        {
            c[u][v] = cost;
            cheapestLabel[u][v] = lab;
            defined[u][v] = true;
            path[u][v] = v;
        }
        arcs[u][v]++;
        delta[u]++;
        delta[v]--;
        return this;
    }
 
    void leastCostPaths()
    {
        for (int k = 0; k < N; k++)
            for (int i = 0; i < N; i++)
                if (defined[i][k])
                    for (int j = 0; j < N; j++)
                        if (defined[k][j]
                                && (!defined[i][j] || c[i][j] > c[i][k]
                                        + c[k][j]))
                        {
                            path[i][j] = path[i][k];
                            c[i][j] = c[i][k] + c[k][j];
                            defined[i][j] = true;
                            if (i == j && c[i][j] < 0)
                                return; // stop on negative cycle
                        }
    }
 
    void checkValid()
    {
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < N; j++)
                if (!defined[i][j])
                    throw new Error("Graph is not strongly connected");
            if (c[i][i] < 0)
                throw new Error("Graph has a negative cycle");
        }
    }
 
    float cost()
    {
        return basicCost + phi();
    }
 
    float phi()
    {
        float phi = 0;
        for (int i = 0; i < N; i++)
            for (int j = 0; j < N; j++)
                phi += c[i][j] * f[i][j];
        return phi;
    }
 
    void findUnbalanced()
    {
        int nn = 0, np = 0; // number of vertices of negative/positive delta
        for (int i = 0; i < N; i++)
            if (delta[i] < 0)
                nn++;
            else if (delta[i] > 0)
                np++;
        neg = new int[nn];
        pos = new int[np];
        nn = np = 0;
        for (int i = 0; i < N; i++)
            // initialise sets
            if (delta[i] < 0)
                neg[nn++] = i;
            else if (delta[i] > 0)
                pos[np++] = i;
    }
 
    void findFeasible()
    {   // delete next 3 lines to be faster, but non-reentrant
        int delta[] = new int[N];
        for (int i = 0; i < N; i++)
            delta[i] = this.delta[i];
        for (int u = 0; u < neg.length; u++)
        {
            int i = neg[u];
            for (int v = 0; v < pos.length; v++)
            {
                int j = pos[v];
                f[i][j] = -delta[i] < delta[j] ? -delta[i] : delta[j];
                delta[i] += f[i][j];
                delta[j] -= f[i][j];
            }
        }
    }
 
    boolean improvements()
    {
        ChinesePostmanProblem residual = new ChinesePostmanProblem(N);
        for (int u = 0; u < neg.length; u++)
        {
            int i = neg[u];
            for (int v = 0; v < pos.length; v++)
            {
                int j = pos[v];
                residual.addArc(null, i, j, c[i][j]);
                if (f[i][j] != 0)
                    residual.addArc(null, j, i, -c[i][j]);
            }
        }
        residual.leastCostPaths(); // find a negative cycle
        for (int i = 0; i < N; i++)
            if (residual.c[i][i] < 0) // cancel the cycle (if any)
            {
                int k = 0, u, v;
                boolean kunset = true;
                u = i;
                do // find k to cancel
                {
                    v = residual.path[u][i];
                    if (residual.c[u][v] < 0 && (kunset || k > f[v][u]))
                    {
                        k = f[v][u];
                        kunset = false;
                    }
                }
                while ((u = v) != i);
                u = i;
                do // cancel k along the cycle
                {
                    v = residual.path[u][i];
                    if (residual.c[u][v] < 0)
                        f[v][u] -= k;
                    else
                        f[u][v] += k;
                }
                while ((u = v) != i);
                return true; // have another go
            }
        return false; // no improvements found
    }
 
    static final int NONE = -1; // anything < 0
 
    int findPath(int from, int f[][]) // find a path between unbalanced vertices
    {
        for (int i = 0; i < N; i++)
            if (f[from][i] > 0)
                return i;
        return NONE;
    }
 
    void printCPT(int startVertex)
    {
        int v = startVertex;
        // delete next 7 lines to be faster, but non-reentrant
        int arcs[][] = new int[N][N];
        int f[][] = new int[N][N];
        for (int i = 0; i < N; i++)
            for (int j = 0; j < N; j++)
            {
                arcs[i][j] = this.arcs[i][j];
                f[i][j] = this.f[i][j];
            }
        while (true)
        {
            int u = v;
            if ((v = findPath(u, f)) != NONE)
            {
                f[u][v]--; // remove path
                for (int p; u != v; u = p) // break down path into its arcs
                {
                    p = path[u][v];
                    System.out.println("Take arc " + cheapestLabel[u][p]
                            + " from " + u + " to " + p);
                }
            }
            else
            {
                int bridgeVertex = path[u][startVertex];
                if (arcs[u][bridgeVertex] == 0)
                    break; // finished if bridge already used
                v = bridgeVertex;
                for (int i = 0; i < N; i++)
                    // find an unused arc, using bridge last
                    if (i != bridgeVertex && arcs[u][i] > 0)
                    {
                        v = i;
                        break;
                    }
                arcs[u][v]--; // decrement count of parallel arcs
                System.out.println("Take arc "
                        + label[u][v].elementAt(arcs[u][v]) + " from " + u
                        + " to " + v); // use each arc label in turn
            }
        }
    }
 
    static public void main(String args[])
    {
        // create a graph of four vertices
        ChinesePostmanProblem G = new ChinesePostmanProblem(4);
        // add the arcs for the example graph
        G.addArc("a", 0, 1, 1).addArc("b", 0, 2, 1).addArc("c", 1, 2, 1)
                .addArc("d", 1, 3, 1).addArc("e", 2, 3, 1).addArc("f", 3, 0, 1);
        G.solve(); // find the CPT
        G.printCPT(0); // print it, starting from vertex 0
        System.out.println("Cost = " + G.cost());
        OpenCPP.test();
    }
 
    // Print arcs and f
    void debugarcf()
    {
        for (int i = 0; i < N; i++)
        {
            System.out.print("f[" + i + "]= ");
            for (int j = 0; j < N; j++)
                System.out.print(f[i][j] + " ");
            System.out.print("  arcs[" + i + "]= ");
            for (int j = 0; j < N; j++)
                System.out.print(arcs[i][j] + " ");
            System.out.println();
        }
    }
 
    // Print out most of the matrices: defined, path and f
    void debug()
    {
        for (int i = 0; i < N; i++)
        {
            System.out.print(i + " ");
            for (int j = 0; j < N; j++)
                System.out
                        .print(j + ":" + (defined[i][j] ? "T" : "F") + " "
                                + c[i][j] + " p=" + path[i][j] + " f="
                                + f[i][j] + "; ");
            System.out.println();
        }
    }
 
    // Print out non zero f elements, and phi
    void debugf()
    {
        float sum = 0;
        for (int i = 0; i < N; i++)
        {
            boolean any = false;
            for (int j = 0; j < N; j++)
                if (f[i][j] != 0)
                {
                    any = true;
                    System.out.print("f(" + i + "," + j + ":" + label[i][j]
                            + ")=" + f[i][j] + "@" + c[i][j] + "  ");
                    sum += f[i][j] * c[i][j];
                }
            if (any)
                System.out.println();
        }
        System.out.println("-->phi=" + sum);
    }
 
    // Print out cost matrix.
    void debugc()
    {
        for (int i = 0; i < N; i++)
        {
            boolean any = false;
            for (int j = 0; j < N; j++)
                if (c[i][j] != 0)
                {
                    any = true;
                    System.out.print("c(" + i + "," + j + ":" + label[i][j]
                            + ")=" + c[i][j] + "  ");
                }
            if (any)
                System.out.println();
        }
    }
}
 
class OpenCPP
{
    class Arc
    {
        String lab;
        int    u, v;
        float  cost;
 
        Arc(String lab, int u, int v, float cost)
        {
            this.lab = lab;
            this.u = u;
            this.v = v;
            this.cost = cost;
        }
    }
 
    Vector<Arc> arcs = new Vector<Arc>();
    int         N;
 
    OpenCPP(int vertices)
    {
        N = vertices;
    }
 
    OpenCPP addArc(String lab, int u, int v, float cost)
    {
        if (cost < 0)
            throw new Error("Graph has negative costs");
        arcs.addElement(new Arc(lab, u, v, cost));
        return this;
    }
 
    float printCPT(int startVertex)
    {
        ChinesePostmanProblem bestGraph = null, g;
        float bestCost = 0, cost;
        int i = 0;
        do
        {
            g = new ChinesePostmanProblem(N + 1);
            for (int j = 0; j < arcs.size(); j++)
            {
                Arc it = arcs.elementAt(j);
                g.addArc(it.lab, it.u, it.v, it.cost);
            }
            cost = g.basicCost;
            g.findUnbalanced(); // initialise g.neg on original graph
            g.addArc("'virtual start'", N, startVertex, cost);
            g.addArc("'virtual end'",
            // graph is Eulerian if neg.length=0
                    g.neg.length == 0 ? startVertex : g.neg[i], N, cost);
            g.solve();
            if (bestGraph == null || bestCost > g.cost())
            {
                bestCost = g.cost();
                bestGraph = g;
            }
        }
        while (++i < g.neg.length);
        System.out.println("Open CPT from " + startVertex
                + " (ignore virtual arcs)");
        bestGraph.printCPT(N);
        return cost + bestGraph.phi();
    }
 
    static void test()
    {
        OpenCPP G = new OpenCPP(4); // create a graph of four vertices
        // add the arcs for the example graph
        G.addArc("a", 0, 1, 1).addArc("b", 0, 2, 1).addArc("c", 1, 2, 1)
                .addArc("d", 1, 3, 1).addArc("e", 2, 3, 1).addArc("f", 3, 0, 1);
        int besti = 0;
        float bestCost = 0;
        for (int i = 0; i < 4; i++)
        {
            System.out.println("Solve from " + i);
            float c = G.printCPT(i);
            System.out.println("Cost = " + c);
            if (i == 0 || c < bestCost)
            {
                bestCost = c;
                besti = i;
            }
        }
        G.printCPT(besti);
        System.out.println("Cost = " + bestCost);
    }
}

Output:

$ javac ChinesePostmanProblem.java
$ java ChinesePostmanProblem
 
Take arc b from 0 to 2
Take arc e from 2 to 3
Take arc f from 3 to 0
Take arc a from 0 to 1
Take arc c from 1 to 2
Take arc e from 2 to 3
Take arc f from 3 to 0
Take arc a from 0 to 1
Take arc d from 1 to 3
Take arc f from 3 to 0
Cost = 10.0
Solve from 0
Open CPT from 0 (ignore virtual arcs)
Take arc 'virtual start' from 4 to 0
Take arc a from 0 to 1
Take arc d from 1 to 3
Take arc f from 3 to 0
Take arc a from 0 to 1
Take arc c from 1 to 2
Take arc e from 2 to 3
Take arc f from 3 to 0
Take arc b from 0 to 2
Take arc 'virtual end' from 2 to 4
Cost = 8.0
Solve from 1
Open CPT from 1 (ignore virtual arcs)
Take arc 'virtual start' from 4 to 1
Take arc d from 1 to 3
Take arc f from 3 to 0
Take arc a from 0 to 1
Take arc c from 1 to 2
Take arc e from 2 to 3
Take arc f from 3 to 0
Take arc b from 0 to 2
Take arc 'virtual end' from 2 to 4
Cost = 7.0
Solve from 2
Open CPT from 2 (ignore virtual arcs)
Take arc 'virtual start' from 4 to 2
Take arc e from 2 to 3
Take arc f from 3 to 0
Take arc a from 0 to 1
Take arc d from 1 to 3
Take arc f from 3 to 0
Take arc a from 0 to 1
Take arc c from 1 to 2
Take arc e from 2 to 3
Take arc f from 3 to 0
Take arc b from 0 to 2
Take arc 'virtual end' from 2 to 4
Cost = 10.0
Solve from 3
Open CPT from 3 (ignore virtual arcs)
Take arc 'virtual start' from 4 to 3
Take arc f from 3 to 0
Take arc a from 0 to 1
Take arc d from 1 to 3
Take arc f from 3 to 0
Take arc a from 0 to 1
Take arc c from 1 to 2
Take arc e from 2 to 3
Take arc f from 3 to 0
Take arc b from 0 to 2
Take arc 'virtual end' from 2 to 4
Cost = 9.0
Open CPT from 1 (ignore virtual arcs)
Take arc 'virtual start' from 4 to 1
Take arc d from 1 to 3
Take arc f from 3 to 0
Take arc a from 0 to 1
Take arc c from 1 to 2
Take arc e from 2 to 3
Take arc f from 3 to 0
Take arc b from 0 to 2
Take arc 'virtual end' from 2 to 4
Cost = 7.0