Print all the cycles in an undirected graph

Given an undirected graph, print all the vertices that form cycles in it.

Pre-requisite: Detect Cycle in a directed graph using colors

In the above diagram, the cycles have been marked with dark green color. The output for the above will be

1st cycle: 3 5 4 6
2nd cycle: 11 12 13

Approach: Using the graph coloring method, mark all the vertex of the different cycles with unique numbers. Once the graph traversal is completed, push all the similar marked numbers to an adjacency list and print the adjacency list accordingly. Given below is the algorithm:

  • Insert the edges into an adjacency list.
  • Call the DFS function which uses the coloring method to mark the vertex.
  • Whenever there is a partially visited vertex, backtrack till the current vertex is reached and mark all of them with cycle numbers. Once all the vertexes are marked, increase the cycle number.
  • Once Dfs is completed, iterate for the edges and push the same marked number edges to another adjacency list.
  • Iterate in the another adjacency list and print the vertex cycle-number wise.

Below is the implementation of the above approach:





// C++ program to print all the cycles
// in an undirected graph
#include <bits/stdc++.h>
using namespace std;
const int N = 100000;
// variables to be used
// in both functions
vector<int> graph[N];
vector<int> cycles[N];
// Function to mark the vertex with
// different colors for different cycles
void dfs_cycle(int u, int p, int color[],
               int mark[], int par[], int& cyclenumber)
    // already (completely) visited vertex.
    if (color[u] == 2) {
    // seen vertex, but was not completely visited -> cycle detected.
    // backtrack based on parents to find the complete cycle.
    if (color[u] == 1) {
        int cur = p;
        mark[cur] = cyclenumber;
        // backtrack the vertex which are
        // in the current cycle thats found
        while (cur != u) {
            cur = par[cur];
            mark[cur] = cyclenumber;
    par[u] = p;
    // partially visited.
    color[u] = 1;
    // simple dfs on graph
    for (int v : graph[u]) {
        // if it has not been visited previously
        if (v == par[u]) {
        dfs_cycle(v, u, color, mark, par, cyclenumber);
    // completely visited.
    color[u] = 2;
// add the edges to the graph
void addEdge(int u, int v)
// Function to print the cycles
void printCycles(int edges, int mark[], int& cyclenumber)
    // push the edges that into the
    // cycle adjacency list
    for (int i = 1; i <= edges; i++) {
        if (mark[i] != 0)
    // print all the vertex with same cycle
    for (int i = 1; i <= cyclenumber; i++) {
        // Print the i-th cycle
        cout << "Cycle Number " << i << ": ";
        for (int x : cycles[i])
            cout << x << " ";
        cout << endl;
// Driver Code
int main()
    // add edges
    addEdge(1, 2);
    addEdge(2, 3);
    addEdge(3, 4);
    addEdge(4, 6);
    addEdge(4, 7);
    addEdge(5, 6);
    addEdge(3, 5);
    addEdge(7, 8);
    addEdge(6, 10);
    addEdge(5, 9);
    addEdge(10, 11);
    addEdge(11, 12);
    addEdge(11, 13);
    addEdge(12, 13);
    // arrays required to color the
    // graph, store the parent of node
    int color[N];
    int par[N];
    // mark with unique numbers
    int mark[N];
    // store the numbers of cycle
    int cyclenumber = 0;
    int edges = 13;
    // call DFS to mark the cycles
    dfs_cycle(1, 0, color, mark, par, cyclenumber);
    // function to print the cycles
    printCycles(edges, mark, cyclenumber);



Cycle Number 1: 3 4 5 6 
Cycle Number 2: 11 12 13 

Time Complexity: O(N + M), where N is number of vertex and M is the number of edges.
Auxiliary Space: O(N + M)

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Striver(underscore)79 at Codechef and codeforces D

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