* The model is based on a directed graph G = (V,E). * We have one binary variable x[e] for each edge e in E. That variable * is set to 1 if edge e is selected in the optimal tour and 0 otherwise. *
** The model contains only two explicit constraints: *
for each v in V: sum(u in V : u != v) x[uv] == 1 for each v in V: sum(u in V : u != v) x[vu] == 1* These state that node u should have exactly one outgoing and exactly * one incoming edge in a tour. * *
* The above constraints ensures that the selected edges form tours. However, * it allows multiple tours, also known as subtours. So we need a constraint * that requires that there is only one tour (which then necessarily hits * all the nodes). This constraint is known as a subtour elimination constraint * and is *
sum(e in S) x[e] <= |S|-1 for each subtour S
* * * Since there are exponentially many subtours in a graph, this constraint is * not stated explicitly. Instead we check for any solution that the optimizer * finds, whether it satisfies the subtour elimination constraint. If it does * then we accept the solution. Otherwise we reject the solution and augment the * model by the violated subtour eliminiation constraint. *
* * This lazy addition of constraints is implemented using a preintsol callback * that rejects any solution that violates a subtour elimination constraint and * injects a violated subtour elimination constraint in case the solution * candidate came from an integral node. *
** An important thing to note about this strategy is that dual reductions have * to be disabled. Since the optimizer does not see the whole model (subtour * elimination constraints are only generated on the fly), dual reductions may * cut off the optimal solution. *
*/ public final class TravelingSalesPerson { /** Number of nodes in the instance. */ private final int nodes; /** X coordinate of nodes. */ private final double[] nodeX; /** Y coordinate of nodes. */ private final double[] nodeY; /** Variables the edges. */ private Variable[][] x; /** * Construct a new random instance with random seed 0. * * @param nodes The number of nodes in the instance. */ public TravelingSalesPerson(int nodes) { this(nodes, 0); } /** * Construct a new random instance. * * @param nodes The number of nodes in the instance. * @param seed Random number seed. */ public TravelingSalesPerson(int nodes, int seed) { this.nodes = nodes; nodeX = new double[nodes]; nodeY = new double[nodes]; Random rand = new Random(seed); for (int i = 0; i < nodes; ++i) { nodeX[i] = 4 * rand.nextDouble(); nodeY[i] = 4 * rand.nextDouble(); } } /** * Get the distance between two nodes. * * @param u First node. * @param v Second node. * @return The distance betweenu and v. The distance
* is symmetric.
*/
public double distance(int u, int v) {
return Math.sqrt((nodeX[u] - nodeX[v]) * (nodeX[u] - nodeX[v]) + (nodeY[u] - nodeY[v]) * (nodeY[u] - nodeY[v]));
}
/**
* Find the tour rooted at 0 in a solution. As a side effect, the tour is
* printed to the console.
*
* @param sol The current solution.
* @param from Stores the tour. from[u] yields the predecessor of
* u in the tour. If from[u] is negative
* then u is not in the tour. This parameter can be
* null.
* @return The length of the tour.
*/
private int findTour(double[] sol, int[] from) {
if (from == null)
from = new int[nodes];
Arrays.fill(from, -1);
int node = 0;
int used = 0;
System.out.print("0");
while (node != 0 || used == 0) {
// Find the edge leaving node
Variable edge = null;
for (int i = 0; i < nodes; ++i) {
if (i != node && x[node][i].getValue(sol) > 0,5) {
System.out.printf(" -> %d", i);
edge = x[node][i];
from[i] = node;
node = i;
++used;
break;
}
}
if (edge == null)
break;
}
System.out.println();
return used;
}
/**
* Integer solution check callback.
*/
private final class PreIntsolCallback implements XpressProblem.CallbackAPI.PreIntsolCallback {
@Override
public void preIntsol(XpressProblem prob, int soltype, IntHolder p_reject, DoubleHolder p_cutoff) {
System.out.println("Checking candidate solution ...");
// Get current solution and check whether it is feasible
double[] sol = prob.getLpSolX();
int[] from = new int[nodes];
int used = findTour(sol, from);
System.out.print("Solution is ");
if (used < nodes) {
// The tour given by the current solution does not pass through
// all the nodes and is thus infeasible.
// If soltype is non-zero then we reject by setting
// p_reject.value=1.
// If instead soltype is zero then the solution came from an
// integral node. In this case we can reject by adding a cut
// that cuts off that solution. Note that we must NOT reject
// in that case because that would result in just dropping
// the node, no matter whether we add cuts or not.
System.out.println("infeasible (" + used + " edges)");
if (soltype != 0) {
p_reject.value = 1;
} else {
// The tour is too short. Get the edges on the tour and
// add a subtour elimination constraint
LinExpression subtour = LinExpression.create();
for (int u = 0; u < nodes; ++u) {
if (from[u] >= 0)
subtour.addTerm(x[from[u]][u]);
}
// We add the constraint. The solver must translate the
// constraint from the original space into the presolved
// space. This may fail (in theory). In that case the
// addCut() function will return non-zero.
if (prob.addCut(1, subtour.leq(used - 1)) != 0)
throw new RuntimeException("failed to presolve subtour elimination constraint");
}
} else {
System.out.println("feasible");
}
}
}
/** Create a feasible tour and add this as initial MIP solution. */
private void createInitialTour(XpressProblem prob) {
Variable[] variable = new Variable[nodes];
double[] value = new double[nodes];
// Create a tour that just visits each node in order.
for (int i = 0; i < nodes; ++i) {
variable[i] = x[i][(i + 1) % nodes];
value[i] = 1;
}
prob.addMipSol(value, variable, "init");
}
/**
* Solve the TSP represented by this instance.
*/
public void solve() {
try (XpressProblem prob = new XpressProblem(null)) {
// Create variables. We create one variable for each edge in
// the (complete) graph. That is, we create variables from each
// node u to all other nodes v. We even create a variable for
// the self-loop from u to u, but that variable is fixed to 0.
// x[u][v] gives the variable that represents edge uv.
// All variables are binary.
x = prob.addVariables(nodes, nodes)
.withType(ColumnType.Binary)
.withName((i,j) -> String.format("x_%d_%d", i, j))
.withUB((i,j) -> (i == j) ? 0.0 : 1)
.toArray();
// Objective. All variables are in the objective and their
// respective coefficient is the distance between the two nodes.
prob.setObjective(sum(nodes,
u -> sum(nodes,
v -> x[u][v].mul(distance(u, v)))),
ObjSense.MINIMIZE);
// Constraint: In the graph that is induced by the selected
// edges, each node should have exactly one outgoing
// and exactly one incoming edge.
// These are the only constraints we add explicitly.
// Subtour elimination constraints are added
// dynamically via a callback.
prob.addConstraints(nodes,
u -> sum(range(0, nodes)
.filter(v -> v != u)
.mapToObj(v -> x[u][v]))
.eq(1));
prob.addConstraints(nodes,
u -> sum(range(0, nodes)
.filter(v -> v != u)
.mapToObj(v -> x[v][u]))
.eq(1));
// Create a starting solution.
// This is optional but having a feasible solution available right
// from the beginning can improve optimizer performance.
createInitialTour(prob);
// Write out the model in case we want to look at it.
prob.writeProb("travelingsalesperson.lp", "l");
// We don't have all constraints explicitly in the matrix, hence
// we must disable dual reductions. Otherwise MIP presolve may
// cut off the optimal solution.
prob.controls().setMIPDualReductions(0);
// Add a callback that rejects solutions that do not satisfy
// the subtour constraints.
prob.callbacks.addPreIntsolCallback(new PreIntsolCallback());
// Add a message listener to display log information.
prob.addMessageListener(DefaultMessageListener::console);
prob.optimize();
if (prob.attributes().getSolStatus() != SolStatus.OPTIMAL)
throw new RuntimeException("failed to solve");
double[] sol = prob.getSolution();
// Print the optimal tour.
System.out.println("Tour with length " + prob.attributes().getMIPBestObjVal());
findTour(sol, null);
x = null; // We are done with the variables
}
}
public static void main(String[] args) {
new TravelingSalesPerson(10).solve();
}
}