Graph find all vertices that are part of a cycle

Question

I was wondering how I can find all the vertices in a graph that are part of a cycle. To determine this for one vertex, I could just start a DFS from that vertex, but this seems unefficient to me when I want to know for all vertices.

Thanks for your help!

score 4 · Accepted Answer · edited Dec 27 '19 at 17:26

There is a duplicate question in StackOverflow. I quote the highest voted answer from Craig Gidney here.

What you want to do is remove all of the Bridges (i.e. edges that disconnect a component when removed). The Wikipedia article gives a linear time algorithm for finding all of the bridges.

Once all the bridges are gone, each node is either isolated (has degree 0) or is part of a cycle. Throw out the lonely nodes, and what's left is the nodes you want.

David Zhou · Answer 2 · 2022-05-09T12:09:12.607

I just had an exam. There is the same question, we being ask to find all vertex that not part of cycle.

The function is like this (In C)

void HasCycle(int hasCycle[], int size, Graph g);

In my function, I use

An array to track visit status(Using hasCycle provided).

Then I will run DFS on the vertex has not being vsisited, and in each DFS search I kept the following

An array to track parent. For example, I visit 3 through edge 5->3, I will put parent[3] = 5.
An array to track backedge. The authentic way should be using a timeStamp, this can be found in CLRS, chapter discuss Graph DFS search. I used color. At begin of loop, I color the vertex and at the end, I decolor the vertex so that it can be visited for multiple times. And by this, each time I have a edge conenct two colored node, means there is a back edge.
A 2-D matrix, that track the status of edge. Idea is once DFS used the edge, it will be marked so won't do DFS again through that edge. In this way DFS can visit same node for several times but can't using same edge. At the end, I want to achieve, node can be visited several times but have to through different edge.

So that in the end. The time complexity will be O(V^2), for each DFS I run, it will versify the cycle in that connected component. And I will only run DFS if I havn't visit that node in previous DFS.

I am unable to proof the correctness of this algorithm. I did pass the autotest in the exam. So if there is some problem you spotted, just point it out.

score 0 · Answer 3 · answered Jun 20 '25 at 03:42

For a connected, directed graph, with starting node N0.

type State: { unvisited, visited, part_of_cycle }
reset(N0):
  for all nodes N in graph N0:
     state(N) = unvisited
identify_cycle_nodes_rec(N, T) is
  match state(N)
    case part_of_cycle:
      return
    case visited:
      state(N) = part_of_cycle
      for M in neighbors(N):
         visit(M, part_of_cycle)
    case unvisited:
      state(N) = T
      for M in neighbors(N):
         visit(M, T)
identify_cycle_nodes(N):
  reset(N)
  identify_cycle_nodes(N, visited)

The idea here is to traverse the graph, painting it with two successive colors. Our initial paint is "visited". We paint unvisited nodes with our currently chosen paint. If we encounter a node we have already painted "visited", we switch our paintbrush to the "part_of_cycle" color and paint the node again, visiting its neighbors with that brush. Whenever we encounter a node painted "part_of_cycle", we terminate the recursion. The "part_of_cycle" color will flow into all the nodes that are part of a cycle.

I used this in a compiler's optimizer to identify all the basic blocks which are part of a cycle.

Graph find all vertices that are part of a cycle

3 Answers3