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

Undirected Graph

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: 5 6 10 9
3rd 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 store them into a vector. Once all the vertices of this cycle are stored into the vector, store this vector into our global cycle answer vector and increase the cycle number.
• Once DFS is completed, iterate into our global cycle answer vector and print the vertex cycle-number wise.

Below is the implementation of the above approach: 

C++

// 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<vector<int>> cycles;

// Function to mark the vertex with
// different colors for different cycles
void dfs_cycle(int u, int p, int color[], int par[], int& cyclenumber)
{

    // already (completely) visited vertex.
    if (color[u] == 2) {
        return;
    }

    // seen vertex, but was not completely visited -> cycle detected.
    // backtrack based on parents to find the complete cycle.
    if (color[u] == 1) {
        vector<int> v;
        cyclenumber++;

        int cur = p;
          v.push_back(cur);

        // backtrack the vertex which are
        // in the current cycle thats found
        while (cur != u) {
            cur = par[cur];
              v.push_back(cur);
        }
          cycles.push_back(v);
        return;
    }
    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]) {
            continue;
        }
        dfs_cycle(v, u, color, par, cyclenumber);
    }

    // completely visited.
    color[u] = 2;
}

// add the edges to the graph
void addEdge(int u, int v)
{
    graph[u].push_back(v);
    graph[v].push_back(u);
}

// Function to print the cycles
void printCycles(int& cyclenumber)
{

    // print all the vertex with same cycle
    for (int i = 0; i < cyclenumber; i++) {
        // Print the i-th cycle
        cout << "Cycle Number " << i + 1 << ": ";
        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, 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];

    // store the numbers of cycle
    int cyclenumber = 0;
    int edges = 6;

    // call DFS to mark the cycles
    dfs_cycle(1, 0, color, par, cyclenumber);

    // function to print the cycles
    printCycles(cyclenumber);
}

Java

// Java program to print all the cycles
// in an undirected graph
import java.util.*;

class GFG
{

    static final int N = 100000;

    // variables to be used
    // in both functions
    @SuppressWarnings("unchecked")
    static Vector<Integer>[] graph = new Vector[N];
    @SuppressWarnings("unchecked")
    static Vector<Integer>[] cycles = new Vector[N];
    static int cyclenumber;

    // Function to mark the vertex with
    // different colors for different cycles
    static void dfs_cycle(int u, int p, int[] color,int[] par)
    {

        // already (completely) visited vertex.
        if (color[u] == 2)
        {
            return;
        }

        // seen vertex, but was not completely visited -> cycle detected.
        // backtrack based on parents to find the complete cycle.
        if (color[u] == 1)
        {


              Vector<Integer> v = new Vector<Integer>();
            int cur = p;
              v.add(cur);

            // backtrack the vertex which are
            // in the current cycle thats found
            while (cur != u)
            {
                cur = par[cur];
                v.add(cur);
            }
              cycles[cyclenumber] = v;
            cyclenumber++;
            return;
        }
        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])
            {
                continue;
            }
            dfs_cycle(v, u, color, par);
        }

        // completely visited.
        color[u] = 2;
    }

    // add the edges to the graph
    static void addEdge(int u, int v)
    {
        graph[u].add(v);
        graph[v].add(u);
    }

    // Function to print the cycles
    static void printCycles()
    {

        // print all the vertex with same cycle
        for (int i = 0; i < cyclenumber; i++)
        {
            // Print the i-th cycle
            System.out.printf("Cycle Number %d: ", i + 1);
            for (int x : cycles[i])
                System.out.printf("%d ", x);
            System.out.println();
        }
    }

    // Driver Code
    public static void main(String[] args)
    {

        for (int i = 0; i < N; i++)
        {
            graph[i] = new Vector<>();
            cycles[i] = new Vector<>();
        }

        // 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, 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 = new int[N];
        int[] par = new int[N];

        // mark with unique numbers
        int[] mark = new int[N];

        // store the numbers of cycle
        cyclenumber = 0;

        // call DFS to mark the cycles
        dfs_cycle(1, 0, color, par);

        // function to print the cycles
        printCycles();
    }
}

Python3

# Python3 program to print all the cycles
# in an undirected graph
N = 100000

# variables to be used
# in both functions
graph = [[] for i in range(N)]
cycles = [[] for i in range(N)]


# Function to mark the vertex with
# different colors for different cycles
def dfs_cycle(u, p, color: list,
              par: list):
    global cyclenumber

    # already (completely) visited vertex.
    if color[u] == 2:
        return

    # seen vertex, but was not
    # completely visited -> cycle detected.
    # backtrack based on parents to
    # find the complete cycle.
    if color[u] == 1:
        v = []
        cur = p
        v.append(cur)

        # backtrack the vertex which are
        # in the current cycle thats found
        while cur != u:
            cur = par[cur]
            v.append(cur)
        cycles[cyclenumber] = v
        cyclenumber += 1

        return

    par[u] = p

    # partially visited.
    color[u] = 1

    # simple dfs on graph
    for v in graph[u]:

        # if it has not been visited previously
        if v == par[u]:
            continue
        dfs_cycle(v, u, color, par)

    # completely visited.
    color[u] = 2

# add the edges to the graph
def addEdge(u, v):
    graph[u].append(v)
    graph[v].append(u)

# Function to print the cycles
def printCycles():

    # print all the vertex with same cycle
    for i in range(0, cyclenumber):

        # Print the i-th cycle
        print("Cycle Number %d:" % (i+1), end = " ")
        for x in cycles[i]:
            print(x, end = " ")
        print()

# Driver Code
if __name__ == "__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, 9)
    addEdge(10, 11)
    addEdge(11, 12)
    addEdge(11, 13)
    addEdge(12, 13)

    # arrays required to color the
    # graph, store the parent of node
    color = [0] * N
    par = [0] * N

    # store the numbers of cycle
    cyclenumber = 0

    # call DFS to mark the cycles
    dfs_cycle(1, 0, color, par)

    # function to print the cycles
    printCycles()

C#

// C# program to print all
// the cycles in an undirected
// graph
using System;
using System.Collections.Generic;
class GFG{

static readonly int N = 100000;

// variables to be used
// in both functions
static List<int>[] graph =
       new List<int>[N];
static List<int>[] cycles =
       new List<int>[N];
static int cyclenumber;

// Function to mark the vertex with
// different colors for different cycles
static void dfs_cycle(int u, int p,
                      int[] color,
                      int[] par)
{
  // already (completely)
  // visited vertex.
  if (color[u] == 2)
  {
    return;
  }

  // seen vertex, but was not
  // completely visited -> cycle
  // detected. backtrack based on
  // parents to find the complete
  // cycle.
  if (color[u] == 1)
  {

    List<int> v = new List<int>();
    int cur = p;
    v.Add(cur);

    // backtrack the vertex which
    // are in the current cycle
    // thats found
    while (cur != u)
    {
      cur = par[cur];
      v.Add(cur);
    }
    cycles[cyclenumber] = v;
    cyclenumber++;
    return;
  }
  par[u] = p;

  // partially visited.
  color[u] = 1;

  // simple dfs on graph
  foreach (int v in graph[u])
  {
    // if it has not been
    // visited previously
    if (v == par[u])
    {
      continue;
    }
    dfs_cycle(v, u, color, par);
  }

  // completely visited.
  color[u] = 2;
}

// add the edges to the
// graph
static void addEdge(int u,
                    int v)
{
  graph[u].Add(v);
  graph[v].Add(u);
}

// Function to print the cycles
static void printCycles()
{
  // print all the vertex with
  // same cycle
  for (int i = 0;
           i < cyclenumber; i++)
  {
    // Print the i-th cycle
    Console.Write("Cycle Number " + (i+1) + ":");
    foreach (int x in cycles[i])
      Console.Write(" " + x);
    Console.WriteLine();
  }
}

// Driver Code
public static void Main(String[] args)
{
  for (int i = 0; i < N; i++)
  {
    graph[i] = new List<int>();
    cycles[i] = new List<int>();
  }

  // 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, 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 = new int[N];
  int[] par = new int[N];

  // store the numbers of cycle
  cyclenumber = 0;

  // call DFS to mark
  // the cycles
  dfs_cycle(1, 0, color,
            par);

  // function to print the cycles
  printCycles();
}
}

Javascript

<script>

// JavaScript program to print all
// the cycles in an undirected
// graph

var N = 100000;

// variables to be used
// in both functions
var graph = Array.from(Array(N), ()=>Array());

var cycles = Array.from(Array(N), ()=>Array());

var cyclenumber = 0;

// Function to mark the vertex with
// different colors for different cycles
function dfs_cycle(u, p, color, par)
{
  // already (completely)
  // visited vertex.
  if (color[u] == 2)
  {
    return;
  }

  // seen vertex, but was not
  // completely visited -> cycle
  // detected. backtrack based on
  // parents to find the complete
  // cycle.
  if (color[u] == 1)
  {

    var v = [];
    var cur = p;
    v.push(cur);

    // backtrack the vertex which
    // are in the current cycle
    // thats found
    while (cur != u)
    {
      cur = par[cur];
      v.push(vur);
    }
    cycles[cyclenumber] = v;
    cyclenumber++;
    return;
  }
  par[u] = p;

  // partially visited.
  color[u] = 1;

  // simple dfs on graph
  for(var v of graph[u])
  {
    // if it has not been
    // visited previously
    if (v == par[u])
    {
      continue;
    }
    dfs_cycle(v, u, color,
              par);
  }

  // completely visited.
  color[u] = 2;
}

// add the edges to the
// graph
function addEdge(u, v)
{
  graph[u].push(v);
  graph[v].push(u);
}

// Function to print the cycles
function printCycles()
{
  // print all the vertex with
  // same cycle
  for (var i = 0;
           i < cyclenumber; i++)
  {
    // Print the i-th cycle
    document.write("Cycle Number " + (i+1) + ":");
    for(var x of cycles[i])
      document.write(" " + x);
    document.write("<br>");
  }
}

// Driver Code
// 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, 9);
addEdge(10, 11);
addEdge(11, 12);
addEdge(11, 13);
addEdge(12, 13);
// arrays required to color
// the graph, store the parent
// of node
var color = Array(N).fill(0);
var par = Array(N).fill(0);

// store the numbers of cycle
cyclenumber = 0;
// call DFS to mark
// the cycles
dfs_cycle(1, 0, color,
          par);
// function to print the cycles
printCycles();

</script>

Output

Cycle Number 1: 5 6 4 3 
Cycle Number 2: 10 9 5 6 
Cycle Number 3: 13 12 11 

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