Cycles of length n in an undirected and connected graph

Given an undirected and connected graph and a number n, count total number of cycles of length n in the graph. A cycle of length n simply means that the cycle contains n vertices and n edges. And we have to count all such cycles that exist. 

Example : 

Input :  n = 4

Output : Total cycles = 3
Explanation : Following 3 unique cycles 
   0 -> 1 -> 2 -> 3 -> 0
   0 -> 1 -> 4 -> 3 -> 0
   1 -> 2 -> 3 -> 4 -> 1
Note* : There are more cycles but
these 3 are unique as 0 -> 3 -> 2 -> 1
-> 0 and 0 -> 1 -> 2 -> 3 -> 0 are 
same cycles and hence will be counted as 1.

To solve this Problem, DFS(Depth First Search) can be effectively used. Using DFS we find every possible path of length (n-1) for a particular source (or starting point). Then we check if this path ends with the vertex it started with, if yes then we count this as the cycle of length n. Notice that we looked for path of length (n-1) because the n^th edge will be the closing edge of cycle.

Every possible path of length (n-1) can be searched using only V – (n – 1) vertices (where V is the total number of vertices). 
For above example, all the cycles of length 4 can be searched using only 5-(4-1) = 2 vertices. The reason behind this is quite simple, because we search for all possible path of length (n-1) = 3 using these 2 vertices which include the remaining 3 vertices. So, these 2 vertices cover the cycles of remaining 3 vertices as well, and using only 3 vertices we can’t form a cycle of length 4 anyways. 

One more thing to notice is that, every vertex finds 2 duplicate cycles for every cycle that it forms. For above example 0^th vertex finds two duplicate cycle namely 0 -> 3 -> 2 -> 1 -> 0 and 0 -> 1 -> 2 -> 3 -> 0. Hence the total count must be divided by 2 because every cycle is counted twice.

Implementation:

C++

// CPP Program to count cycles of length n
// in a given graph.
#include <bits/stdc++.h>
using namespace std;

// Number of vertices
const int V = 5;

void DFS(bool graph[][V], bool marked[], int n, int vert,
         int start, int& count)
{
    // mark the vertex vert as visited
    marked[vert] = true;

    // if the path of length (n-1) is found
    if (n == 0) {

        // mark vert as un-visited to make
        // it usable again.
        marked[vert] = false;

        // Check if vertex vert can end with
        // vertex start
        if (graph[vert][start] && graph[start][vert]) {
            count++;
            return;
        }
        else
            return;
    }

    // For searching every possible path of
    // length (n-1)
    for (int i = 0; i < V; i++)
        if (!marked[i] && graph[vert][i])

            // DFS for searching path by decreasing
            // length by 1
            DFS(graph, marked, n - 1, i, start, count);

    // marking vert as unvisited to make it
    // usable again.
    marked[vert] = false;
}

// Counts cycles of length N in an undirected
// and connected graph.
int countCycles(bool graph[][V], int n)
{
    // all vertex are marked un-visited initially.
    bool marked[V];
    memset(marked, 0, sizeof(marked));

    // Searching for cycle by using v-n+1 vertices
    int count = 0;
    for (int i = 0; i < V - (n - 1); i++) {
        DFS(graph, marked, n - 1, i, i, count);

        // ith vertex is marked as visited and
        // will not be visited again.
        marked[i] = true;
    }

    return count / 2;
}

int main()
{
    bool graph[][V] = { { 0, 1, 0, 1, 0 },
                        { 1, 0, 1, 0, 1 },
                        { 0, 1, 0, 1, 0 },
                        { 1, 0, 1, 0, 1 },
                        { 0, 1, 0, 1, 0 } };
    int n = 4;
    cout << "Total cycles of length " << n << " are "
         << countCycles(graph, n);
    return 0;
}

Java

// Java program to calculate cycles of
// length n in a given graph
public class Main {

    // Number of vertices
    public static final int V = 5;
    static int count = 0;

    static void DFS(int graph[][], boolean marked[],
                    int n, int vert, int start) {

        // mark the vertex vert as visited
        marked[vert] = true;

        // if the path of length (n-1) is found
        if (n == 0) {

            // mark vert as un-visited to
            // make it usable again
            marked[vert] = false;

            // Check if vertex vert end
            // with vertex start
            if (graph[vert][start] == 1) {
                count++;
                return;
            } else
                return;
        }

        // For searching every possible
        // path of length (n-1)
        for (int i = 0; i < V; i++)
            if (!marked[i] && graph[vert][i] == 1)

                // DFS for searching path by
                // decreasing length by 1
                DFS(graph, marked, n-1, i, start);

        // marking vert as unvisited to make it
        // usable again
        marked[vert] = false;
    }

    // Count cycles of length N in an
    // undirected and connected graph.
    static int countCycles(int graph[][], int n) {

        // all vertex are marked un-visited
        // initially.
        boolean marked[] = new boolean[V];

        // Searching for cycle by using
        // v-n+1 vertices
        for (int i = 0; i < V - (n - 1); i++) {
            DFS(graph, marked, n-1, i, i);

            // ith vertex is marked as visited
            // and will not be visited again
            marked[i] = true;
        }

        return count / 2;
    }

    // driver code
    public static void main(String[] args) {
        int graph[][] = {{0, 1, 0, 1, 0},
                        {1, 0, 1, 0, 1},
                        {0, 1, 0, 1, 0},
                        {1, 0, 1, 0, 1},
                        {0, 1, 0, 1, 0}};

        int n = 4;

        System.out.println("Total cycles of length "+
                          n + " are "+
                          countCycles(graph, n));
    }
}

// This code is contributed by nuclode

Python3

# Python Program to count
# cycles of length n
# in a given graph.

# Number of vertices
V = 5

def DFS(graph, marked, n, vert, start, count):

    # mark the vertex vert as visited
    marked[vert] = True

    # if the path of length (n-1) is found
    if n == 0:

        # mark vert as un-visited to make
        # it usable again.
        marked[vert] = False

        # Check if vertex vert can end with
        # vertex start
        if graph[vert][start] == 1:
            count = count + 1
            return count
        else:
            return count

    # For searching every possible path of
    # length (n-1)
    for i in range(V):
        if marked[i] == False and graph[vert][i] == 1:

            # DFS for searching path by decreasing
            # length by 1
            count = DFS(graph, marked, n-1, i, start, count)

    # marking vert as unvisited to make it
    # usable again.
    marked[vert] = False
    return count

# Counts cycles of length
# N in an undirected
# and connected graph.
def countCycles( graph, n):

    # all vertex are marked un-visited initially.
    marked = [False] * V

    # Searching for cycle by using v-n+1 vertices
    count = 0
    for i in range(V-(n-1)):
        count = DFS(graph, marked, n-1, i, i, count)

        # ith vertex is marked as visited and
        # will not be visited again.
        marked[i] = True

    return int(count/2)

# main :
graph = [[0, 1, 0, 1, 0],
         [1 ,0 ,1 ,0, 1],
         [0, 1, 0, 1, 0],
         [1, 0, 1, 0, 1],
         [0, 1, 0, 1, 0]]

n = 4
print("Total cycles of length ",n," are ",countCycles(graph, n))

# this code is contributed by Shivani Ghughtyal

C#

// C# program to calculate cycles of
// length n in a given graph
using System;

class GFG
{

    // Number of vertices
    public static int V = 5;
    static int count = 0;

    static void DFS(int [,]graph, bool []marked,
                    int n, int vert, int start)
    {

        // mark the vertex vert as visited
        marked[vert] = true;

        // if the path of length (n-1) is found
        if (n == 0)
        {

            // mark vert as un-visited to
            // make it usable again
            marked[vert] = false;

            // Check if vertex vert end
            // with vertex start
            if (graph[vert, start] == 1)
            {
                count++;
                return;
            }
            else
                return;
        }

        // For searching every possible
        // path of length (n-1)
        for (int i = 0; i < V; i++)
            if (!marked[i] && graph[vert, i] == 1)

                // DFS for searching path by
                // decreasing length by 1
                DFS(graph, marked, n - 1, i, start);

        // marking vert as unvisited to make it
        // usable again
        marked[vert] = false;
    }

    // Count cycles of length N in an
    // undirected and connected graph.
    static int countCycles(int [,]graph, int n)
    {

        // all vertex are marked un-visited
        // initially.
        bool []marked = new bool[V];

        // Searching for cycle by using
        // v-n+1 vertices
        for (int i = 0; i < V - (n - 1); i++)
        {
            DFS(graph, marked, n - 1, i, i);

            // ith vertex is marked as visited
            // and will not be visited again
            marked[i] = true;
        }

        return count / 2;
    }

    // Driver code
    public static void Main()
    {
        int [,]graph = {{0, 1, 0, 1, 0},
                        {1, 0, 1, 0, 1},
                        {0, 1, 0, 1, 0},
                        {1, 0, 1, 0, 1},
                        {0, 1, 0, 1, 0}};

        int n = 4;

        Console.WriteLine("Total cycles of length "+
                        n + " are "+
                        countCycles(graph, n));
    }
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>

// JavaScript program to calculate cycles of
// length n in a given graph

// Number of vertices
var V = 5;
var count = 0;

function DFS(graph, marked, n, vert, start)
{

    // mark the vertex vert as visited
    marked[vert] = true;

    // if the path of length (n-1) is found
    if (n == 0)
    {

        // mark vert as un-visited to
        // make it usable again
        marked[vert] = false;

        // Check if vertex vert end
        // with vertex start
        if (graph[vert][start] == 1)
        {
            count++;
            return;
        }
        else
            return;
    }

    // For searching every possible
    // path of length (n-1)
    for (var i = 0; i < V; i++)
        if (!marked[i] && graph[vert][i] == 1)

            // DFS for searching path by
            // decreasing length by 1
            DFS(graph, marked, n - 1, i, start);

    // marking vert as unvisited to make it
    // usable again
    marked[vert] = false;
}

// Count cycles of length N in an
// undirected and connected graph.
function countCycles(graph, n)
{

    // all vertex are marked un-visited
    // initially.
    var marked = Array(V).fill(false);

    // Searching for cycle by using
    // v-n+1 vertices
    for (var i = 0; i < V - (n - 1); i++)
    {
        DFS(graph, marked, n - 1, i, i);

        // ith vertex is marked as visited
        // and will not be visited again
        marked[i] = true;
    }

    return parseInt(count / 2);
}

// Driver code
var graph = [[0, 1, 0, 1, 0],
                [1, 0, 1, 0, 1],
                [0, 1, 0, 1, 0],
                [1, 0, 1, 0, 1],
                [0, 1, 0, 1, 0]];

var n = 4;

document.write("Total cycles of length "+
                n + " are "+
                countCycles(graph, n));

</script>

Output

Total cycles of length 4 are 3

Time Complexity: O(V*V)
Space Complexity: O(V)

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