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# Wolf, Goat and Cabbage #
### What is this repository for? ###
This code uses a mixed integer programming (MIP) model to solve a [logic puzzle](https://en.wikipedia.org/wiki/Wolf,_goat_and_cabbage_problem) posted in which a farmer, for unspecified reasons, buys a wolf, a goat and a cabbage. On the way home, the farmer encounters a river that must be crossed by boat. The boat has only enough room for the farmer and one purchase at a time. If the wolf and goat are left alone together, the wolf will eat the goat. Similarly, if the goat and cabbage are left alone together, the goat will eat the cabbage. (The wolf and the cabbage apparently get along amicably.) The puzzle is to find a sequence of trips across the river that result in the farmer and all three purchases ending up intact on the far side of the river.
The MIP model solves the puzzle with the objective of minimizing the number of river crossings required to achieve the desired terminal state.
### Details ###
The code was developed using a beta version of CPLEX 22.1 but will run with any version. CPLEX is overkill for a problem this size. Any MIP solver would do.
The model formulation is discussed in my [blog post]().
### License ###
The code here is copyrighted by Paul A. Rubin (yours truly) and licensed under a [Creative Commons Attribution 3.0 Unported License](http://creativecommons.org/licenses/by/3.0/deed.en_US).
package wolfgoatcabbage;
import ilog.concert.IloException;
import ilog.concert.IloNumVar;
import ilog.cplex.IloCplex;
/**
* Model provides an optimization model to solve the logic puzzle.
*
* The model captures trips between the "near" bank (starting point) and
* "far" bank (destination). At time 0, the farmer is on the near bank with
* three items (wolf, goat, cabbage). By the ending time (TMAX) the farmer
* must be on the far bank with all three items.
*
* @author Paul A. Rubin (rubin@msu.edu)
*/
public final class Model implements AutoCloseable {
// Item indices.
private static final int WOLF = 0;
private static final int GOAT = 1;
private static final int CABBAGE = 2;
private static final int ITEMCOUNT = 3;
// Item names.
private static final String[] ITEMS =
new String[] {"WOLF", "GOAT", "CABBAGE"};
// Formatting information.
private static final int WIDTH = 20; // field width in formatted lines
// Time period limit (1 greater than maximum trips allowed).
private static final int TMAX = 17;
// Cutoff for rounding integers.
private static final double HALF = 0.5;
// CPLEX objects.
private final IloCplex cplex; // the model instance
private final IloNumVar[][] inventoryNear; // item inventories on near bank
private final IloNumVar[][] inventoryFar; // item inventories on far bank
private final IloNumVar[][] nearToFar; // items carried near to far
private final IloNumVar[][] farToNear; // items carried far to near
private final IloNumVar[] inProgress; // 1 if in progress, 0 if done
/**
* Constructor.
* @throws IloException if the model instance cannot be created
*/
public Model() throws IloException {
// Instantiate the model.
cplex = new IloCplex();
// Create the variables.
inventoryNear = new IloNumVar[TMAX][ITEMCOUNT];
inventoryFar = new IloNumVar[TMAX][ITEMCOUNT];
nearToFar = new IloNumVar[TMAX][ITEMCOUNT];
farToNear = new IloNumVar[TMAX][ITEMCOUNT];
inProgress = new IloNumVar[TMAX];
for (int t = 0; t < TMAX; t++) {
inProgress[t] = cplex.intVar(0, 1, "inProgress_" + t);
for (int i = 0; i < ITEMCOUNT; i++) {
inventoryNear[t][i] = cplex.numVar(0, 1, "nearInv_" + i + "_" + t);
inventoryFar[t][i] = cplex.numVar(0, 1, "farInv_" + i + "_" + t);
nearToFar[t][i] = cplex.intVar(0, 1, "nearToFar_" + i + "_" + t);
farToNear[t][i] = cplex.intVar(0, 1, "farToNear_" + i + "_" + t);
}
}
// The objective is to minimize the number of trips, or equivalently the
// number of time periods during which movement occurs.
cplex.addMinimize(cplex.sum(inProgress));
for (int i = 0; i < ITEMCOUNT; i++) {
// We being with all inventory on the near bank.
inventoryNear[0][i].setLB(1);
inventoryFar[0][i].setUB(0);
// No movement occurs at time 0.
nearToFar[0][i].setUB(0);
farToNear[0][i].setUB(0);
}
// In odd numbered time periods, movement is from the near bank to the
// far bank. In even numbered periods, movement is from the far bank
// to the near bank. When movement occurs, at most one item moves.
for (int t = 1; t < TMAX; t++) {
if (t % 2 == 0) {
// Even period; flow is far to near.
cplex.addEq(cplex.sum(nearToFar[t]), 0);
cplex.addLe(cplex.sum(farToNear[t]), 1);
} else {
// Odd period; flow is near to far.
cplex.addEq(cplex.sum(farToNear[t]), 0);
cplex.addLe(cplex.sum(nearToFar[t]), 1);
}
}
// In each period, flows add to destination inventory and deplete
// origin inventory.
for (int t = 1; t < TMAX; t++) {
for (int i = 0; i < ITEMCOUNT; i++) {
cplex.addEq(inventoryNear[t][i],
cplex.sum(inventoryNear[t - 1][i],
cplex.diff(farToNear[t][i], nearToFar[t][i])));
cplex.addEq(inventoryFar[t][i],
cplex.sum(inventoryFar[t - 1][i],
cplex.diff(nearToFar[t][i], farToNear[t][i])));
}
}
// In an odd numbered period (where the farmer ends up on the far bank),
// neither wolf and goat nor goat and cabbage can be on the near bank.
for (int t = 1; t < TMAX; t += 2) {
cplex.addLe(cplex.sum(inventoryNear[t][WOLF], inventoryNear[t][GOAT]), 1);
cplex.addLe(cplex.sum(inventoryNear[t][GOAT], inventoryNear[t][CABBAGE]),
1);
}
// In an even numbered period (where the farmer ends up on the near bank),
// neither wolf and goat nor goat and cabbage can be on the far bank
// UNLESS the problem is completed (in which case the farmer remains on
// the far bank).
for (int t = 2; t < TMAX; t += 2) {
cplex.addLe(cplex.sum(inventoryFar[t][WOLF], inventoryFar[t][GOAT]),
cplex.diff(2, inProgress[t]));
cplex.addLe(cplex.sum(inventoryFar[t][GOAT], inventoryFar[t][CABBAGE]),
cplex.diff(2, inProgress[t]));
}
// The transfer remains in progress until nothing is left on the near bank.
for (int t = 1; t < TMAX; t++) {
cplex.addLe(cplex.sum(inventoryNear[t - 1]),
cplex.prod(ITEMCOUNT, inProgress[t]));
}
// Suppress solver output.
cplex.setOut(null);
}
/**
* Solves the model.
* @return the optimal objective value
* @throws IloException if CPLEX blows up
*/
public double solve() throws IloException {
cplex.solve();
return cplex.getObjValue();
}
/**
* Closes the model.
*/
@Override
public void close() {
cplex.close();
}
/**
* Gets the solution to the puzzle.
* @return a string displaying the solution
* @throws IloException if the solution does not exist
*/
public String getSolution() throws IloException {
StringBuilder sb = new StringBuilder();
sb.append("""
The farmer arrives at the left bank at time 0.
""");
double[] running = cplex.getValues(inProgress);
double[] left, right, x;
boolean lr;
for (int t = 1; t < TMAX; t++) {
if (running[t] > HALF) {
if (t % 2 == 0) {
left = cplex.getValues(inventoryNear[t - 1]);
right = cplex.getValues(inventoryFar[t]);
x = cplex.getValues(farToNear[t]);
lr = false;
} else {
left = cplex.getValues(inventoryNear[t]);
right = cplex.getValues(inventoryFar[t - 1]);
x = cplex.getValues(nearToFar[t]);
lr = true;
}
sb.append("Time ").append(t).append(":\t")
.append(format(what(left), what(x), what(right), lr))
.append("\n");
} else {
return sb.toString();
}
}
return sb.toString();
}
/**
* Gets the items (if any) indicated by a vector of 0-1 values.
* @param x the vector of variable values
* @return the symbols of the indicated items (if any); "nothing" if none are
* indicated
*/
private String what(final double[] x) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < ITEMCOUNT; i++) {
if (x[i] > HALF) {
if (!sb.isEmpty()) {
sb.append(", ");
}
sb.append(ITEMS[i]);
}
}
if (sb.isEmpty()) {
sb.append("nothing");
}
return sb.toString();
}
/**
* Merges three strings into a formatted string describing a trip.
* @param left the left bank inventory
* @param center the contents of the boat
* @param right the right bank inventory
* @param lr true if the trip is left to right, false if right to left
* @return a formatted string displaying the trip
*/
private String format(final String left, final String center,
final String right, final boolean lr) {
StringBuilder sb = new StringBuilder();
int n = left.length();
sb.append(String.format("%" + WIDTH + "s", left));
String m;
if (lr) {
m = "-- " + center + " ->";
} else {
m = "<- " + center + " --";
}
n = m.length();
int p0 = n + (WIDTH - n) / 2; // width including left padding
m = String.format("%" + p0 + "s", m);
m = String.format("%-" + WIDTH + "s", m);
sb.append(" ").append(m).append(" ");
n = right.length();
sb.append(String.format("%-" + WIDTH + "s", right));
return sb.toString();
}
}
package wolfgoatcabbage;
import ilog.concert.IloException;
/**
* Solves the Wolf-Goat-Cabbage logic puzzle.
* @author Paul A. Rubin (rubin@msu.edu)
*/
public final class WolfGoatCabbage {
/**
* Dummy constructor.
*/
private WolfGoatCabbage() { }
/**
* Solves the puzzle.
* @param args the command line arguments (ignored)
*/
public static void main(final String[] args) {
try (Model model = new Model()) {
double z = model.solve();
System.out.println("Optimal number of trips = " + z + ".");
System.out.println("Solution:");
System.out.println(model.getSolution());
} catch (IloException ex) {
System.out.println("CPLEX exception:\n" + ex.getMessage());
}
}
}
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