### 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 are implementations of the Backtracking solution.

#### C/C++

```/* C/C++ program for solution of Hamiltonian Cycle problem
using backtracking */
#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
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;

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;
}
```

#### Java

```/* Java program for solution of Hamiltonian Cycle problem
using backtracking */
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
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;

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);
}
}
// This code is contributed by Abhishek Shankhadhar
```

#### Python

```# Python program for solution of
# hamiltonian cycle problem

class Graph():
def __init__(self, vertices):
self.graph = [[0 for column in range(vertices)]\
for row in range(vertices)]
self.V = vertices

''' Check if this vertex is an adjacent vertex
of the previously added vertex and is not
included in the path earlier '''
def isSafe(self, v, pos, path):
# Check if current vertex and last vertex
if self.graph[ path[pos-1] ][v] == 0:
return False

# Check if current vertex not already in path
for vertex in path:
if vertex == v:
return False

return True

# A recursive utility function to solve
# hamiltonian cycle problem
def hamCycleUtil(self, path, pos):

# base case: if all vertices are
# included in the path
if pos == self.V:
# Last vertex must be adjacent to the
# first vertex in path to make a cyle
if self.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 v in range(1,self.V):

if self.isSafe(v, pos, path) == True:

path[pos] = v

if self.hamCycleUtil(path, pos+1) == True:
return True

# Remove current vertex if it doesn't
path[pos] = -1

return False

def hamCycle(self):
path = [-1] * self.V

''' 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 self.hamCycleUtil(path,1) == False:
print "Solution does not exist\n"
return False

self.printSolution(path)
return True

def printSolution(self, path):
print "Solution Exists: Following is one Hamiltonian Cycle"
for vertex in path:
print vertex,
print path[0], "\n"

# Driver Code

''' Let us create the following graph
(0)--(1)--(2)
|   / \   |
|  /   \  |
| /     \ |
(3)-------(4)    '''
g1 = Graph(5)
g1.graph = [ [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
g1.hamCycle();

''' Let us create the following graph
(0)--(1)--(2)
|   / \   |
|  /   \  |
| /     \ |
(3)       (4)    '''
g2 = Graph(5)
g2.graph = [ [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
g2.hamCycle();

# This code is contributed by Divyanshu Mehta
```
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++)”.