Cycles of length n in an undirected and connected graph - GeeksforGeeks

02 Jul, 2021
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 nth 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 0th 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. 
C++
Java
Python3
C#
Javascript
// CPP Program to count cycles of length n// in a given graph.#include <bits/stdc++.h>using namespace std; // Number of verticesconst 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])        {            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 program to calculate cycles of// length n in a given graphpublic 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
# Python Program to count# cycles of length n# in a given graph.  # Number of verticesV = 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 = 4print("Total cycles of length ",n," are ",countCycles(graph, n)) # this code is contributed by Shivani Ghughtyal
// C# program to calculate cycles of// length n in a given graphusing 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 */
<script> // JavaScript program to calculate cycles of// length n in a given graph // Number of verticesvar 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 codevar 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
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