## Backtracking | Set 6 (Hamiltonian Cycle)

Hamiltonian Path in an undirected graph is a path that visits each vertex exactly once. A Hamiltonian cycle (or Hamiltonian circuit) is a Hamiltonian Path such that there is an edge (in graph) from the last vertex to the first vertex of the Hamiltonian Path. Determine whether a given graph contains Hamiltonian Cycle or not. If it contains, then print the path. Following are the input and output of the required function.

*Input:*

A 2D array graph[V][V] where V is the number of vertices in graph and graph[V][V] is adjacency matrix representation of the graph. A value graph[i][j] is 1 if there is a direct edge from i to j, otherwise graph[i][j] is 0.

*Output:*

An array path[V] that should contain the Hamiltonian Path. path[i] should represent the ith vertex in the Hamiltonian Path. The code should also return false if there is no Hamiltonian Cycle in the graph.

For example, a Hamiltonian Cycle in the following graph is {0, 1, 2, 4, 3, 0}. There are more Hamiltonian Cycles in the graph like {0, 3, 4, 2, 1, 0}

(0)--(1)--(2) | / \ | | / \ | | / \ | (3)-------(4)

And the following graph doesn’t contain any Hamiltonian Cycle.

(0)--(1)--(2) | / \ | | / \ | | / \ | (3) (4)

**Naive Algorithm**

Generate all possible configurations of vertices and print a configuration that satisfies the given constraints. There will be n! (n factorial) configurations.

while there are untried conflagrations { generate the next configuration if ( there are edges between two consecutive vertices of this configuration and there is an edge from the last vertex to the first ). { print this configuration; break; } }

**Backtracking Algorithm**

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**

Following is C/C++ implementation of the Backtracking solution.

// Program to print Hamiltonian cycle #include<stdio.h> // Number of vertices in the graph #define V 5 void printSolution(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[]') */ bool isSafe(int v, bool graph[V][V], 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 */ bool hamCycleUtil(bool graph[V][V], 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. */ bool hamCycle(bool graph[V][V]) { int *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 ) { printf("\nSolution does not exist"); return false; } printSolution(path); return true; } /* A utility function to print solution */ void printSolution(int path[]) { printf ("Solution Exists:" " Following is one Hamiltonian Cycle \n"); for (int i = 0; i < V; i++) printf(" %d ", path[i]); // Let us print the first vertex again to show the complete cycle printf(" %d ", path[0]); printf("\n"); } // driver program to test above function int main() { /* Let us create the following graph (0)--(1)--(2) | / \ | | / \ | | / \ | (3)-------(4) */ bool graph1[V][V] = {{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 hamCycle(graph1); /* Let us create the following graph (0)--(1)--(2) | / \ | | / \ | | / \ | (3) (4) */ bool graph2[V][V] = {{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 hamCycle(graph2); return 0; }

Output:

Solution Exists: Following is one Hamiltonian Cycle 0 1 2 4 3 0 Solution does not exist

Note that the above code always prints cycle starting from 0. Starting point should not matter as cycle can be started from any point. If you want to change the starting point, you should make two changes to above code.

Change “path[0] = 0;” to “path[0] = s;” where s is your new starting point. Also change loop “for (int v = 1; v < V; v++)” in hamCycleUtil() to “for (int v = 0; v < V; v++)”.

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### Related Topics:

- Channel Assignment Problem
- Some interesting shortest path questions | Set 1
- Graph Coloring | Set 2 (Greedy Algorithm)
- Graph Coloring | Set 1 (Introduction and Applications)
- Johnson’s algorithm for All-pairs shortest paths
- Travelling Salesman Problem | Set 2 (Approximate using MST)
- Travelling Salesman Problem | Set 1 (Naive and Dynamic Programming)
- Detect cycle in an undirected graph

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