AssignmentProblemToLP.java AssignmentProblemToLP.java Below is the syntax highlighted version of AssignmentProblemToLP.java from §6.5 Reductions. /******************************************************************************
* Compilation: javac AssignmentProblemToLP.java
* Execution: java AassignmentProblemToLP n
* Dependencies: LinearProgramming.java
*
* Solve an n-by-n assignment problem (maximum weighted bipartite matching)
* by reducing it to linear programming.
*
* Warning: in practice, use AssignmentProblem.java which runs in N^3 log B
* time instead of this version.
*
*
******************************************************************************/
public class AssignmentProblemToLP {
private int n; // number of rows and columns
private double[][] weight; // the n-by-n cost matrix
private double[] px; // px[i] = dual variable for row i
private double[] py; // py[j] = dual variable for col j
private int[] xy; // xy[i] = j means i-j is a match
private int[] yx; // yx[j] = i means i-j is a match
public AssignmentProblemToLP(double[][] weight) {
n = weight.length;
this.weight = new double[n][n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
this.weight[i][j] = weight[i][j];
double[] c = new double[n*n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
c[i*n+j] = weight[i][j]; // or vice versa?
double[] b = new double[2*n];
for (int i = 0; i < 2*n; i++)
b[i] = 1.0;
// constraint matrix
double[][] A = new double[2*n][n*n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
A[i][i*n+j] = 1.0;
}
}
for (int j = 0; j < n; j++) {
for (int i = 0; i < n; i++) {
A[j+n][i*n+j] = 1.0;
}
}
LinearProgramming lp = new LinearProgramming(A, b, c);
// dual variables
double[] y = lp.dual();
px = new double[n];
for (int i = 0; i < n; i++)
px[i] = y[i];
py = new double[n];
for (int i = 0; i < n; i++)
py[i] = y[i+n];
// primal variables
double[] x = lp.primal();
xy = new int[n];
yx = new int[n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (x[i*n+j] == 1.0) {
xy[i] = j;
yx[j] = i;
}
}
}
assert check();
}
// reduced cost of i-j
private double reducedCost(int i, int j) {
return weight[i][j] - px[i] - py[j];
}
// total weight of min weight perfect matching
public double weight() {
double total = 0.0;
for (int i = 0; i < n; i++)
total += weight[i][xy[i]];
return total;
}
public int sol(int i) {
return xy[i];
}
// check that dual variables are feasible
private boolean isDualFeasible() {
double EPSILON = 1E-10;
// check that all edges have >= 0 reduced cost
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (reducedCost(i, j) > EPSILON) {
StdOut.println("Dual variables are not feasible");
return false;
}
}
}
return true;
}
// check that primal and dual variables are complementary slack
private boolean isComplementarySlack() {
double EPSILON = 1E-10;
// check that all matched edges have 0-reduced cost
for (int i = 0; i < n; i++) {
if (reducedCost(i, xy[i]) < -EPSILON) {
StdOut.println("Primal and dual variables are not complementary slack");
StdOut.println("Reduced cost of " + i + "-" + xy[i] + " = " + reducedCost(i, xy[i]));
return false;
}
}
return true;
}
// check that primal variables are a perfect matching
private boolean isPerfectMatching() {
// check that xy[] is a perfect matching
boolean[] perm = new boolean[n];
for (int i = 0; i < n; i++) {
if (perm[xy[i]]) {
StdOut.println("Not a perfect matching");
return false;
}
perm[xy[i]] = true;
}
// check that xy[] and yx[] are inverses
for (int j = 0; j < n; j++) {
if (xy[yx[j]] != j) {
StdOut.println("xy[] and yx[] are not inverses");
return false;
}
}
for (int i = 0; i < n; i++) {
if (yx[xy[i]] != i) {
StdOut.println("xy[] and yx[] are not inverses");
return false;
}
}
return true;
}
// check optimality conditions
private boolean check() {
return isPerfectMatching() && isDualFeasible() && isComplementarySlack();
}
public static void main(String[] args) {
int n = Integer.parseInt(args[0]);
double[][] weight = new double[n][n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
weight[i][j] = StdRandom.uniform();
AssignmentProblemToLP assignment = new AssignmentProblemToLP(weight);
StdOut.println("weight = " + assignment.weight());
for (int i = 0; i < n; i++)
StdOut.println(i + "-" + assignment.sol(i));
}
}
Copyright © 2000–2017, Robert Sedgewick and Kevin Wayne. Last updated: Fri Oct 20 12:50:46 EDT 2017.
