 # Hamilton Circuit using Backtracking Algorithm

Hamilton Circuit is a uni-directed path with a feedback mechanism. The Hamiltonian path can be defined as a singular directed path which visits each of the nodes in the graph exactly once.

Challenging task: In a given series of numbers in a list, check if a Hamiltonian path can be formed among them and also check if it can form a Hamilton circuit i.e the path comes back to its origin after passing through every node in the given matrix. If such a path exists, print the path else, exit false -1.

Solution: 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)
```

The simplest solution for the above problem is to find all possible configurations and check for the required condition to be satisfied. If this approach is considered, the program has to be iterated n! times which would, in turn, delay the results for large samples.

The above-discussed problem can be tackled using Backtracking algorithm:

Here, we create an empty array which is considered to be vertex 0. Now, add vertices one by one and check for possible solutions. While adding a new vertex care is taken to add a vertex which could possibly move us towards achieving results faster. If we complete adding all vertices we print the path or if the vertex cannot be added any further, then exit the program with false -1.

## C/C++ implementation using backtracking algorithm:

```/* 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 ] == 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;
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);
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 implementation using backtracking algorithm:

```/* 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] == 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;
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 + " ");
}

// 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);
}
}

```

## Python implementation using backtracking algorithm:

```# 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 ] == 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

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, "\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();

```

Output:

```Solution Exists: Following is one Hamiltonian Cycle
0  1  2  4  3  0

Solution does not exist
```

POST A NEW COMMENT

• Input (stdin)

Output (stdout)

Input (stdin)

Expected Output

Compiler Message

Input (stdin)

`2    3`

`5`
`5`
`5`