Detecting and printing cycle in undirected graph

Question

I have question similar - same - as this one. So I want to know how to not only detect cycle but also print out vertices which this cycle contains. I tried ways which were mentioned in the question above, but I must have done something wrong, why it doesn't work for my. Also my program checks just if one specific vertex makes cycle. My code is here:

#include<iostream>
#include <list>
using namespace std;


class Graph
{
    int V;    
    list<int> *adj;    
public:
    Graph(int V);   
    void addEdge(int v, int w);   
    bool Graph::isCyclicUtil(int v, bool visited[], int *cycleVertices, int parent, int index);
};

Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}

void Graph::addEdge(int v, int w)
{
    adj[v].push_back(w); 
    adj[w].push_back(v);
}


bool Graph::isCyclicUtil(int v, bool visited[], int *cycleVertices, int parent, int index)
{

    visited[v] = true;


    list<int>::iterator i;
    for (i = adj[v].begin(); i != adj[v].end(); ++i)
    {

        if (!visited[*i])
        {
            if (isCyclicUtil(*i, visited, cycleVertices, v, index)) {
                if (index <= 1 || cycleVertices[0] != cycleVertices[index - 1])
                    cycleVertices[index++] = *i;
                return true;
            }
        }

        else if (*i != parent) {
            cycleVertices[index++] = *i;
            return true;
        }
    }
    return false;
}


int main()
{
    bool *visited = new bool[5];
    for (int i = 0; i < 5; i++)
        visited[i] = false;
    int cycleVertices[5];
    for (int i = 0; i < 5; i++)
        cycleVertices[i] = -1;
    Graph g1(5);
    g1.addEdge(1, 0);
    g1.addEdge(0, 2);
    g1.addEdge(2, 1);
    g1.addEdge(0, 3);
    g1.addEdge(3, 4);
    g1.isCyclicUtil(4, visited, cycleVertices, -1, 0) ? cout << "Graph contains cycle\n" :
        cout << "Graph doesn't contain cycle\n";
    int x = 0;
    while (cycleVertices[x] != -1)
        cout << cycleVertices[x++] << " ";
    return 0;
}

score 1 · Accepted Answer · edited May 23 '17 at 11:53

I've found a solution. I've tried j_random_hacker's solution in this post and it didn't work. But problem was with indexing in cycleVertices in my code. Variable index was always same. So I've added a new attribute index in the class Graph and now it works. So here is the edited code:

#include<iostream>
#include <list>

#define FINISHED -1
#define NOCYCLE -2
using namespace std;


class Graph
{
    int V;   
    int index;
    list<int> *adj;   
public:
    Graph(int V);   
    void addEdge(int v, int w);  
    void set_index();
    int Graph::isCyclicUtil(int v, bool visited[], int *cycleVertices, int parent);
};

Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
    this->index = 0;
}

void Graph::set_index()
{
    this->index = 0;
}

void Graph::addEdge(int v, int w)
{
    adj[v].push_back(w); 
    adj[w].push_back(v); 
}

int Graph::isCyclicUtil(int v, bool visited[], int *cycleVertices, int parent)
{
    visited[v] = true;

    list<int>::iterator i;
    for (i = adj[v].begin(); i != adj[v].end(); ++i)
    {
        if (!visited[*i])
        {
            int result = isCyclicUtil(*i, visited, cycleVertices, v);
            if (result == FINISHED)
                return FINISHED;
            else if (result != NOCYCLE) {
                cycleVertices[index++] = v;
                if (result == v)
                    return FINISHED;
                else
                    return result;
            }
        }

        else if (*i != parent) {
            return *i;
        }
    }
    return NOCYCLE;
}


int main()
{
    bool *visited = new bool[4];
    for (int i = 0; i < 4; i++)
        visited[i] = false;
    int cycleVertices[4];
    for (int i = 0; i < 4; i++)
        cycleVertices[i] = -1;
    Graph g1(4);
    g1.addEdge(0, 1);
    g1.addEdge(1, 2);
    g1.addEdge(2, 3);
    g1.addEdge(3, 0);
    g1.isCyclicUtil(3, visited, cycleVertices, -1) ? cout << "Graph contains cycle\n" :
        cout << "Graph doesn't contain cycle\n";
    int x = 0;
    while (cycleVertices[x] != -1)
        cout << cycleVertices[x++] << " ";
    return 0;
}

score 0 · Answer 2 · answered Apr 17 '21 at 22:29

Here is my DFS solution in Python

#program to print all nodes included in the cycle in the given undirected graph 

import collections

edges = [[1, 2], [2, 3], [1, 3], [2, 4], [4, 5], [5, 6], [4, 6]]
n = 6

parent = [0] * (n + 1)
color = [0] * (n + 1)
mark = [0] * (n + 1)
cycleno = 0

graph = collections.defaultdict(set)

for i, j in edges:
    graph[i].add(j)
    graph[j].add(i)

def dfs(u, v):
    global cycleno 

    if color[u] == 2: #node explore complete
        return 
    elif color[u] == 1: #cycle found
        cycleno += 1
        cur = v 
        mark[cur] = cycleno 
        while cur != u:
            cur = parent[cur]
            mark[cur] = cycleno 
    else:
        parent[u] = v 
        color[u] = 1
        for nei in graph[u]:
            if nei == parent[u]:    continue 
            dfs(nei, u)
        color[u] = 2 #exploration for this node completed
    

dfs(1, 0)
print(mark)

Output: [0, 1, 1, 1, 2, 2, 2] which means nodes 1, 2, 3 and 4, 5, 6 belongs to the cycles respectively.

Reference: https://www.tutorialspoint.com/print-all-the-cycles-in-an-undirected-graph-in-cplusplus

Detecting and printing cycle in undirected graph

2 Answers2