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Using Java write a program that implements a backtracking algorithm that solves

ID: 3711320 • Letter: U

Question

Using Java write a program that implements a backtracking algorithm that solves the Hamilton CircuitsProblem . Your program should conform to the following specifications.

Give the pre condition and post condition

Prompt the user for the number of vertices in the graph.

Prompt the user to enter the adjacency matrix of the graph one row at a time.

Print the adjacency matrix of the graph.

Print the first solution and ask the user if they want the rest of the solutions.

If the user indicates they want the rest of the solutions, print them without any additional prompts.

Explanation / Answer

Backtracking algorithm that solves the Hamilton Circuits Problem:

Create an empty path array and add vertex 0 to it. Add other vertices, starting from the vertex 1. Before adding a vertex, check for whether it is adjacent to the previously added vertex and not already added. If we find such a vertex, we add the vertex as part of the solution. If we do not find a vertex then we return false.

Implementation of Backtracking solution

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 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(" Solution 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);

    }

}

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