Find all chordless cycles in an undirected graph

kennytm picture kennytm · Oct 26, 2010 · Viewed 9.1k times · Source

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

Eugene Smith picture Eugene Smith · Oct 27, 2010

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);
                }
            }
        }
    }
}