I have the following girth-counting algorithm on a large 0-1 matrix and I need to optimize it for speed:

function CountG8()
M = 512
N = 1024
H = rand(0:1,M,N)
girth_8 = 0

for i=1:M-3
    for j=1:N-1
        if H[i,j]==1
            for j1=j+1:N
                if H[i,j1]==1
                    for i1=i+1:M
                        if H[i1,j1]==1
                            for j2=1:N
                                if H[i1,j2]==1&&j2!=j1
                                    for i2=i+1:M
                                        if H[i2,j2]==1&&i2!=i1
                                            for j3=1:N
                                                if H[i2,j3]==1&&j3!=j2
                                                    for i3=i+1:M
                                                        if H[i3,j3]==1&&i3!=i2
                                                            if H[i3,j]==1
                                                                girth_8 += 1

The algorithm runs without errors, but it is too slow.

The code is supposed to count every cycle of length 8 in a bipartite graph.

A bipartite graph is a graph of two groups of nodes. Left group, Check Nodes (CNs) and right group, Variable Nodes (VNs). Every CN from left group is allowed to be connected to a VN from right group. This is known as Directed Graph (DG). A DG has only cycles of even length 4,6,8,10,.. The Girth of the graph is the length of shortest cycle in it.

The algorithm walks through the graph in a BFS fashion searching for all possible nodes combination forming a cycle of the specified length (here 8). Matrix H in the code represents the graph. Each row is a CN and each column is a VN in the graph.

Here is a Bipartite Graph with its corresponding matrix as well as 2 cycles of length 4 and 6 shown:

enter image description here

EDIT: A comment on matrix power approach given in the answer.

In my case, the adjacency matrix is (M+N)x(M+N) = 1536 x 1536. Watch out for the complexity of (1536x1536)^8.

  • \$\begingroup\$ Can you mentally run through this code and keep your variables straight? your naming is not so good. \$\endgroup\$
    – Malachi
    Commented Sep 29, 2014 at 20:28
  • \$\begingroup\$ @Malachi Every i is a row index and every j is a column index. \$\endgroup\$
    – AboAmmar
    Commented Sep 29, 2014 at 21:04

1 Answer 1


What you should be using is a neighboring matrix.

The key is graph theory.

You represent the graph with a matrix.

For example:
graph \$ \begin{bmatrix} 0 & 1 & 1 & 0 & 2 \\ 1 & 0 & 2 & 0 & 0 \\ 1 & 2 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 2 & 0 & 1 & 1 & 0 \\ \end{bmatrix} \$
Here is a graph and its corresponding neighbouring matrix. If you look at the first row you see that A has no one-step connection to itself. But it has a one-step connection to every other node. Node B has a one-step connection to A and 2 different one-step connections to node C. As you can see it all sums up if you keep going like this.

Now then, here's the mathemagic. If you multiply the matrix with itself the result will be a matrix which tells you all the connections of two-stop length. Continue multiplying with the original matrix until you've done it 8 times and you will have a matrix which tells you how many connections of eight-step length you have.

How is matrix multiplication done, you might ask?
Another site (with examples)
Basically, you multiply rows with columns.(It would take another whole answer to explain it)

When you've gotten the neighbouring matrix of the eight power, you just sum up all the elements of the diagonal. That will tell you the number of cycles of length 8. (because the diagonal holds the information of a n-step way back to itself, where n is the power of which the matrix has been multiplied)

This method works for all graphs. All you need to do is make sure the graph is bipartite to begin with.

  • \$\begingroup\$ Am I missing something, or is it a 2-step from A to D, so A/D should be 2, not 1, in the matrix? \$\endgroup\$
    – rolfl
    Commented Sep 29, 2014 at 21:35
  • 1
    \$\begingroup\$ You're correct. I copied the graph from another site and assumed the matrix was corresponding. I'll see if i can find a better matrix. Though it should be a 0 not a 2 because there are no one-step ways from A to D \$\endgroup\$
    – Tejpbit
    Commented Sep 29, 2014 at 21:42
  • \$\begingroup\$ @AndréSamuelsson Here you use the adjacency matrix which is a square matrix of size n x n, where n is the total number of nodes in the graph. I use the so-called Tanner Graph matrix, which divides the nodes into 2 groups (rows & columns) as I demonstrated in my post. \$\endgroup\$
    – AboAmmar
    Commented Sep 29, 2014 at 22:07

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