Hamiltonian Path Java
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PROGRAM TO SOLVE HAMILTONIAN CYCLE PROBLEM
class HamiltonianCycle{ final int V = 5; int path[]; /* A utility function to check if the vertex v can be added at index 'pos'in the Hamiltonian Cycle constructed so far (stored in 'path[]') */ boolean isSafe(int v, int graph[][], int path[], int pos) { /* Check if this vertex is an adjacent vertex of the previously added vertex. */ if (graph[path[pos - 1]][v] == 0) return false; /* Check if the vertex has already been included. This step can be optimized by creating an array of size V */ for (int i = 0; i < pos; i++) if (path[i] == v) return false; return true; } /* A recursive utility function to solve hamiltonian cycle problem */ boolean hamCycleUtil(int graph[][], int path[], int pos) { /* base case: If all vertices are included in Hamiltonian Cycle */ if (pos == V) { // And if there is an edge from the last included // vertex to the first vertex if (graph[path[pos - 1]][path[0]] == 1) return true; else return false; } // Try different vertices as a next candidate in // Hamiltonian Cycle. We don't try for 0 as we // included 0 as starting point in hamCycle() for (int v = 1; v < V; v++) { /* Check if this vertex can be added to Hamiltonian Cycle */ if (isSafe(v, graph, path, pos)) { path[pos] = v; /* recur to construct rest of the path */ if (hamCycleUtil(graph, path, pos + 1) == true) return true; /* If adding vertex v doesn't lead to a solution, then remove it */ path[pos] = -1; } } /* If no vertex can be added to Hamiltonian Cycle constructed so far, then return false */ return false; } /* This function solves the Hamiltonian Cycle problem using Backtracking. It mainly uses hamCycleUtil() to solve the problem. It returns false if there is no Hamiltonian Cycle possible, otherwise return true and prints the path. Please note that there may be more than one solutions, this function prints one of the feasible solutions. */ int hamCycle(int graph[][]) { path = new int[V]; for (int i = 0; i < V; i++) path[i] = -1; /* Let us put vertex 0 as the first vertex in the path. If there is a Hamiltonian Cycle, then the path can be started from any point of the cycle as the graph is undirected */ path[0] = 0; if (hamCycleUtil(graph, path, 1) == false) { System.out.println("\nSolution does not exist"); return 0; } printSolution(path); return 1; } /* A utility function to print solution */ void printSolution(int path[]) { System.out.println("Solution Exists: Following" + " is one Hamiltonian Cycle"); for (int i = 0; i < V; i++) System.out.print(" " + path[i] + " "); // Let us print the first vertex again to show the // complete cycle System.out.println(" " + path[0] + " "); } // driver program to test above function public static void main(String args[]) { HamiltonianCycle hamiltonian = new HamiltonianCycle(); /* Let us create the following graph (0)--(1)--(2) | / \ | | / \ | | / \ | (3)-------(4) */ int graph1[][] = {{0, 1, 0, 1, 0}, {1, 0, 1, 1, 1}, {0, 1, 0, 0, 1}, {1, 1, 0, 0, 1}, {0, 1, 1, 1, 0}, }; // Print the solution hamiltonian.hamCycle(graph1); /* Let us create the following graph (0)--(1)--(2) | / \ | | / \ | | / \ | (3) (4) */ int graph2[][] = {{0, 1, 0, 1, 0}, {1, 0, 1, 1, 1}, {0, 1, 0, 0, 1}, {1, 1, 0, 0, 0}, {0, 1, 1, 0, 0}, }; // Print the solution hamiltonian.hamCycle(graph2); }}
OUTPUT:
Solution Exists: Following is one Hamiltonian Cycle 0 1 2 4 3 0 Solution does not exist
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