// (c) 2023-2025 Fair Isaac Corporation
import static com.dashoptimization.objects.Utils.sum;
import static java.util.stream.IntStream.range;
import java.util.Arrays;
import java.util.Random;
import com.dashoptimization.ColumnType;
import com.dashoptimization.DefaultMessageListener;
import com.dashoptimization.DoubleHolder;
import com.dashoptimization.IntHolder;
import com.dashoptimization.XPRSenumerations.ObjSense;
import com.dashoptimization.XPRSenumerations.SolStatus;
import com.dashoptimization.objects.LinExpression;
import com.dashoptimization.objects.Variable;
import com.dashoptimization.objects.XpressProblem;
/**
* Example for solving a MIP using lazily separated cuts/constraints.
*
* We solve a random instance of the symmetric TSP using lazily separated
* cuts/constraints.
*
* <p>
* 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.
* </p>
* <p>
* The model contains only two explicit constraints:
* <pre>
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
</pre>
* These state that node u should have exactly one outgoing and exactly
* one incoming edge in a tour.
* </p>
* <p>
* 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
* <pre>
sum(e in S) x[e] <= |S|-1 for each subtour S
* </pre>
*
* 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.
* </p>
* <p>
* 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.
* </p>
* <p>
* 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.
* </p>
*/
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] = 4.0 * rand.nextDouble();
nodeY[i] = 4.0 * rand.nextDouble();
}
}
/**
* Get the distance between two nodes.
*
* @param u First node.
* @param v Second node.
* @return The distance between <code>u</code> and <code>v</code>. 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. <code>from[u]</code> yields the predecessor of
* <code>u</code> in the tour. If <code>from[u]</code> is negative
* then <code>u</code> is not in the tour. This parameter can be
* <code>null</code>.
* @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.getCallbackSolution();
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] = 1.0;
}
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 : 1.0)
.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();
}
}
|
/***********************************************************************
Xpress Optimizer Examples
=========================
file TSP.java
````````````
Solve a MIP using cuts/constraints that are lazily separated.
We take a random instance of the symmetric TSP and solve that using
lazily separated constraints.
(c) 2021-2025 Fair Isaac Corporation
***********************************************************************/
import com.dashoptimization.XPRSprob;
import com.dashoptimization.DefaultMessageListener;
import com.dashoptimization.AbstractPreIntsolListener;
import static com.dashoptimization.XPRSenumerations.ObjSense;
import static com.dashoptimization.XPRSprob.RowInfo;
import java.util.Arrays;
import java.util.Random;
/** Example for solving a MIP using lazily separated cuts/constraints.
*
* We solve a random instance of the symmetric TSP using lazily separated
* cuts/constraints.
*
* <p>
* The model is based on a 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.
* </p>
* <p>
* The model contains only one explicit constraint:
* <pre>
for each v in V: sum(u in V : u != v) x[uv] == 2
</pre>
* This states that from all the edges incident to a node u, exactly two
* must be selected in the optimal tour.
* </p>
* <p>
* 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 subtour elimination constraint
* and is
* <pre>
sum(e in S) x[e] <= |S|-1 for each subtour S
</pre>
* 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.
* </p>
* <p>
* 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.
* </p>
* <p>
* 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.
* </p>
*/
public final class TSP {
/** Number of nodes in the instance. */
private final int nodes;
/** Number of edges in the instance. */
private final int edges;
/** X coordinate of nodes. */
private final double[] nodeX;
/** Y coordinate of nodes. */
private final double[] nodeY;
/** Variable indices for the edges. */
private final int[][] x;
/** Construct a new random instance with random seed 0.
* @param nodes The number of nodes in the instance.
*/
public TSP(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 TSP(int nodes, int seed) {
this.nodes = nodes;
edges = (nodes * (nodes - 1)) / 2;
nodeX = new double[nodes];
nodeY = new double[nodes];
Random rand = new Random(seed);
for (int i = 0; i < nodes; ++i) {
nodeX[i] = 4.0 * rand.nextDouble();
nodeY[i] = 4.0 * rand.nextDouble();
}
x = new int[nodes][];
for (int i = 0; i < nodes; ++i)
x[i] = new int[nodes];
}
/** Get the distance between two nodes.
* @param u First node.
* @param v Second node.
* @return The distance between <code>u</code> and <code>v</code>.
* 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. <code>from[u]</code> yields the
* predecessor of <code>u</code> in the tour. If
* <code>from[u]</code> is negative then <code>u</code>
* is not in the tour.
* This parameter can be <code>null</code>.
* @return The length of the tour.
*/
private int findTour(double[] sol, int[] from) {
if (from == null)
from = new int[nodes];
Arrays.fill(from, -1);
boolean[] inTour = new boolean[edges]; // Marks edges on the subtour
int u = 0;
int used = 0;
System.out.print("0");
do {
for (int v = 0; v < nodes; ++v) {
if (u == v) // no self-loops
continue;
else if (from[v] != -1) // node already on tour
continue;
else if (sol[x[u][v]] < 0.5) // edge not selected in solution
continue;
else if (inTour[x[u][v]]) // edge already on tour
continue;
else {
System.out.print(" -> " + v);
inTour[x[u][v]] = true;
from[v] = u;
used += 1;
u = v;
break;
}
}
} while (u != 0);
System.out.println();
return used;
}
/** Integer solution check callback.
*/
private final class PreIntsolListener extends AbstractPreIntsolListener {
@Override
public Double preIntsolEvent(XPRSprob prob, int soltype, double cutoff) {
System.out.println("Checking candidate solution ...");
// Get current solution and check whether it is feasible
double[] sol = prob.getCallbackSolution();
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 returning null.
// 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 return
// null 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) {
return null;
}
else {
// The tour is too short. Get the edges on the tour and
// add a subtour elimination constraint
int[] ind = new int[used];
double[] val = new double[used];
for (int u = 0, next = 0; u < nodes; ++u) {
if (from[u] >= 0)
ind[next++] = x[u][from[u]];
}
Arrays.fill(val, 1);
// Since we created the constraint in the original space,
// we must crush it to the presolved space before we can
// add it.
RowInfo r = prob.presolveRow(ind, val, 'L', used - 1);
if (r != null)
prob.addCut(1, r);
else
throw new RuntimeException("failed to presolve subtour elimination constraint");
}
return cutoff; // Do not return null as that would drop the node
}
else {
System.out.println("feasible");
return cutoff;
}
}
}
/** Create a feasible tour and add this as initial MIP solution. */
private void createInitialTour(XPRSprob prob) {
int[] ind = new int[nodes];
double[] val = new double[nodes];
// Create a tour that just visits each node in order.
for (int i = 0; i < nodes; ++i) {
ind[i] = x[i][(i + 1) % nodes];
val[i] = 1.0;
}
prob.addMipSol(val, ind);
}
/** Solve the TSP represented by this instance.
*/
public void solve() {
try (XPRSprob prob = new XPRSprob(null)) {
// Create variables. We create one variable for each edge in
// the (complete) graph. x[u][v] gives the index of the variable
// that represents edge uv. Since edges are undirected, x[v][u]
// gives the same variable.
// All variables are binary.
for (int i = 0; i < nodes; ++i) {
for (int j = i + 1; j < nodes; ++j) {
x[j][i] = x[i][j] = prob.binVar("x_" + i + "_" + j);
}
}
// Objective. All variables are in the objective and their
// respective coefficient is the distance between the two nodes.
int[] objInd = new int[edges];
double[] objVal = new double[edges];
for (int i = 0, nz = 0; i < nodes; ++i) {
for (int j = i + 1; j < nodes; ++j) {
objInd[nz] = x[i][j];
objVal[nz] = distance(i, j);
++nz;
}
}
prob.setObjective(objInd, objVal, ObjSense.MINIMIZE);
// Constraint: In the graph that is induced by the selected
// edges, each node should have degree 2.
// This is the only constraint we add explicitly.
// Subtour elimination constraints are added
// dynamically via a callback.
for (int u = 0; u < nodes; ++u) {
int[] ind = new int[nodes - 1];
double[] val = new double[nodes - 1];
for (int v = 0, next = 0; v < nodes; ++v) {
if (u != v)
ind[next++] = x[u][v];
}
Arrays.fill(val, 1.0);
prob.addRow(ind, val, 'E', 2);
}
/* Note: with Java 1.8 this could also be written differently:
IntStream.range(0, nodes).forEach(
u -> prob.addRow(IntStream.range(0, nodes).filter(v -> v != u).map(v -> x[u][v]).toArray(),
DoubleStream.generate(() -> 1).limit(nodes - 1).toArray(),
'E', 2));
*/
// 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("tsp.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.addPreIntsolListener(new PreIntsolListener());
// Add a message listener to display log information.
prob.addMessageListener(DefaultMessageListener::console);
prob.mipOptimize();
double[] sol = prob.getSolution();
// Print the optimal tour.
System.out.println("Tour with length " + prob.attributes().getMIPBestObjVal());
findTour(sol, null);
}
}
public static void main(String[] args) {
new TSP(10).solve();
}
}
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