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Finding all cycles in a directed graph

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How do you find the number of cycles in a graph?

Here for finding the number of cycles in the undirected graph, the time complexity is the same as DFS traversal, i.e., O(V+E), where V is the number of vertices and E is the number of edges in the graph. Space complexity is O(V) as extra color[ ] and par[ ] is used here.

Can DFS be used to find cycles?

DFS is used to find cycles in directed graphs, because it works. The vertex is started. The subgraph reachable from that vertex is visited. This includes tracing all untraced edges that are reachable from that vertex, and visiting all reachable unvisited vertexes.

Can DFS detect cycle in directed graph?

Using a Depth First Search (DFS) traversal algorithm we can detect cycles in a directed graph. If there is any self-loop in any node, it will be considered as a cycle, otherwise, when the child node has another edge to connect its parent, it will also a cycle.

Can a directed graph have a cycle?

Directed acyclic graphs (DAGs) are specific names given to acyclic graphs. We can determine if a graph has a cycle by doing DFS and see if we re-explore a vertex that's on our current exploration path.


I found this page in my search and since cycles are not same as strongly connected components, I kept on searching and finally, I found an efficient algorithm which lists all (elementary) cycles of a directed graph. It is from Donald B. Johnson and the paper can be found in the following link:

http://www.cs.tufts.edu/comp/150GA/homeworks/hw1/Johnson%2075.PDF

A java implementation can be found in:

http://normalisiert.de/code/java/elementaryCycles.zip

A Mathematica demonstration of Johnson's algorithm can be found here, implementation can be downloaded from the right ("Download author code").

Note: Actually, there are many algorithms for this problem. Some of them are listed in this article:

http://dx.doi.org/10.1137/0205007

According to the article, Johnson's algorithm is the fastest one.


Depth first search with backtracking should work here. Keep an array of boolean values to keep track of whether you visited a node before. If you run out of new nodes to go to (without hitting a node you have already been), then just backtrack and try a different branch.

The DFS is easy to implement if you have an adjacency list to represent the graph. For example adj[A] = {B,C} indicates that B and C are the children of A.

For example, pseudo-code below. "start" is the node you start from.

dfs(adj,node,visited):  
  if (visited[node]):  
    if (node == start):  
      "found a path"  
    return;  
  visited[node]=YES;  
  for child in adj[node]:  
    dfs(adj,child,visited)
  visited[node]=NO;

Call the above function with the start node:

visited = {}
dfs(adj,start,visited)

First of all - you do not really want to try find literally all cycles because if there is 1 then there is an infinite number of those. For example A-B-A, A-B-A-B-A etc. Or it may be possible to join together 2 cycles into an 8-like cycle etc., etc... The meaningful approach is to look for all so called simple cycles - those that do not cross themselves except in the start/end point. Then if you wish you can generate combinations of simple cycles.

One of the baseline algorithms for finding all simple cycles in a directed graph is this: Do a depth-first traversal of all simple paths (those that do not cross themselves) in the graph. Every time when the current node has a successor on the stack a simple cycle is discovered. It consists of the elements on the stack starting with the identified successor and ending with the top of the stack. Depth first traversal of all simple paths is similar to depth first search but you do not mark/record visited nodes other than those currently on the stack as stop points.

The brute force algorithm above is terribly inefficient and in addition to that generates multiple copies of the cycles. It is however the starting point of multiple practical algorithms which apply various enhancements in order to improve performance and avoid cycle duplication. I was surprised to find out some time ago that these algorithms are not readily available in textbooks and on the web. So I did some research and implemented 4 such algorithms and 1 algorithm for cycles in undirected graphs in an open source Java library here : http://code.google.com/p/niographs/ .

BTW, since I mentioned undirected graphs : The algorithm for those is different. Build a spanning tree and then every edge which is not part of the tree forms a simple cycle together with some edges in the tree. The cycles found this way form a so called cycle base. All simple cycles can then be found by combining 2 or more distinct base cycles. For more details see e.g. this : http://dspace.mit.edu/bitstream/handle/1721.1/68106/FTL_R_1982_07.pdf .


The simplest choice I found to solve this problem was using the python lib called networkx.

It implements the Johnson's algorithm mentioned in the best answer of this question but it makes quite simple to execute.

In short you need the following:

import networkx as nx
import matplotlib.pyplot as plt

# Create Directed Graph
G=nx.DiGraph()

# Add a list of nodes:
G.add_nodes_from(["a","b","c","d","e"])

# Add a list of edges:
G.add_edges_from([("a","b"),("b","c"), ("c","a"), ("b","d"), ("d","e"), ("e","a")])

#Return a list of cycles described as a list o nodes
list(nx.simple_cycles(G))

Answer: [['a', 'b', 'd', 'e'], ['a', 'b', 'c']]

enter image description here


To clarify:

  1. Strongly Connected Components will find all subgraphs that have at least one cycle in them, not all possible cycles in the graph. e.g. if you take all strongly connected components and collapse/group/merge each one of them into one node (i.e. a node per component), you'll get a tree with no cycles (a DAG actually). Each component (which is basically a subgraph with at least one cycle in it) can contain many more possible cycles internally, so SCC will NOT find all possible cycles, it will find all possible groups that have at least one cycle, and if you group them, then the graph will not have cycles.

  2. to find all simple cycles in a graph, as others mentioned, Johnson's algorithm is a candidate.