Find all chordless cycles in an undirected graph

Question

How to find all chordless cycles in an undirected graph?

For example, given the graph

0 --- 1
|     | \
|     |  \
4 --- 3 - 2

the algorithm should return 1-2-3 and 0-1-3-4, but never 0-1-2-3-4.

---

(Note: [1] This question is not the same as small cycle finding in a planar graph because the graph is not necessarily planar. [2] I have read the paper Generating all cycles, chordless cycles, and Hamiltonian cycles with the principle of exclusion but I don't understand what they're doing :). [3] I have tried CYPATH but the program only gives the count, algorithm EnumChordlessPath in readme.txt has significant typos, and the C code is a mess. [4] I am not trying to find an arbitrary set of fundametal cycles. Cycle basis can have chords.)

Answer 1

@aioobe has a point. Just find all the cycles and then exclude the non-chordless ones. This may be too inefficient, but the search space can be pruned along the way to reduce the inefficiencies. Here is a general algorithm:

void printChordlessCycles( ChordlessCycle path) {

  System.out.println( path.toString() );
  for( Node n : path.lastNode().neighbors() ) {

    if( path.canAdd( n) ) {

      path.add( n);
      printChordlessCycles( path);
      path.remove( n);
    }
  }
}

Graph g = loadGraph(...);
ChordlessCycle p = new ChordlessCycle();
for( Node n : g.getNodes()) {
  p.add(n);
  printChordlessCycles( p);
  p.remove( n);
}

class ChordlessCycle {
   private CountedSet<Node> connected_nodes;
   private List<Node> path;

   ...

   public void add( Node n) {
     for( Node neighbor : n.getNeighbors() ) {
       connected_nodes.increment( neighbor);
     }
     path.add( n);
   }

   public void remove( Node n) {
     for( Node neighbor : n.getNeighbors() ) {
       connected_nodes.decrement( neighbor);
     }
     path.remove( n);
   }

   public boolean canAdd( Node n) {
     return (connected_nodes.getCount( n) == 0);
   }
}

Answer 2

Assign numbers to nodes from 1 to n.

1. Pick the node number 1. Call it 'A'.

2. Enumerate pairs of links coming out of 'A'.

3. Pick one. Let's call the adjacent nodes 'B' and 'C' with B less than C.

4. If B and C are connected, then output the cycle ABC, return to step 3 and pick a different pair.

5. If B and C are not connected:

   • Enumerate all nodes connected to B. Suppose it's connected to D, E, and F. Create a list of vectors CABD, CABE, CABF. For each of these:
   • if the last node is connected to any internal node except C and B, discard the vector
   • if the last node is connected to C, output and discard
   • if it's not connected to either, create a new list of vectors, appending all nodes to which the last node is connected.

   Repeat until you run out of vectors.

6. Repeat steps 3-5 with all pairs.

7. Remove node 1 and all links that lead to it. Pick the next node and go back to step 2.

Edit: and you can do away with one nested loop.

This seems to work at the first sight, there may be bugs, but you should get the idea:

void chordless_cycles(int* adjacency, int dim)
{
    for(int i=0; i<dim-2; i++)
    {
        for(int j=i+1; j<dim-1; j++)
        {
            if(!adjacency[i+j*dim])
                continue;
            list<vector<int> > candidates;
            for(int k=j+1; k<dim; k++)
            {
                if(!adjacency[i+k*dim])
                    continue;
                if(adjacency[j+k*dim])
                {
                    cout << i+1 << " " << j+1 << " " << k+1 << endl;
                    continue;
                }
                vector<int> v;
                v.resize(3);
                v[0]=j;
                v[1]=i;
                v[2]=k;
                candidates.push_back(v);
            }
            while(!candidates.empty())
            {
                vector<int> v = candidates.front();
                candidates.pop_front();
                int k = v.back();
                for(int m=i+1; m<dim; m++)
                {
                    if(find(v.begin(), v.end(), m) != v.end())
                        continue;
                    if(!adjacency[m+k*dim])
                        continue;
                    bool chord = false;
                    int n;
                    for(n=1; n<v.size()-1; n++)
                        if(adjacency[m+v[n]*dim])
                            chord = true;
                    if(chord)
                        continue;
                    if(adjacency[m+j*dim])
                    {
                        for(n=0; n<v.size(); n++)
                            cout<<v[n]+1<<" ";
                        cout<<m+1<<endl;
                        continue;
                    }
                    vector<int> w = v;
                    w.push_back(m);
                    candidates.push_back(w);
                }
            }
        }
    }
}

Answer 3

How about this. First, reduce the problem to finding all chordless cycles that pass through a given vertex A. Once you've found all of those, you can remove A from the graph, and repeat with another point until there's nothing left.

And how to find all the chordless cycles that pass through vertex A? Reduce this to finding all chordless paths from B to A, given a list of permitted vertices, and search either breadth-first or depth-first. Note that when iterating over the vertices reachable (in one step) from B, when you choose one of them you must remove all of the others from the list of permitted vertices (take special care when B=A, so as not to eliminate three-edge paths).

Answer 4

Just a thought:

Let's say you are enumerating cycles on your example graph and you are starting from node 0.

If you do a breadth-first search for each given edge, e.g. 0 - 1, you reach a fork at 1. Then the cycles that reach 0 again first are chordless, and the rest are not and can be eliminated... at least I think this is the case.

Could you use an approach like this? Or is there a counterexample?

Answer 5

Find all cycles.

Definition of a chordless cycle is a set of points in which a subset cycle of those points don't exist. So, once you have all cycles problem is simply to eliminate cycles which do have a subset cycle.

For efficiency, for each cycle you find, loop through all existing cycles and verify that it is not a subset of another cycle or vice versa, and if so, eliminate the larger cycle.

Beyond that, only difficulty is figuring out how to write an algorithm that determines if a set is a subset of another.