Detecting cycles in an adjacency matrix

Question

Let A be the adjacency matrix for the graph G = (V,E). A(i,j) = 1 if the nodes i and j are connected with an

Answer 1

This approach uses DFS, but is very efficient, because we don't repeat nodes in subsequent DFS's.

High-level approach:

Initialize the values of all the nodes to -1.

Do a DFS from each unexplored node, setting that node's value to that of an auto-incremented value starting from 0.

For these DFS's, update each node's value with previous node's value + i/n^k where that node is the ith child of the previous node and k is the depth explored, skipping already explored nodes (except for checking for a bigger value).

So, an example for n = 10:

   0.1   0.11   0.111
   j   - k    - p
0 /    \ 0.12
i \ 0.2  l
    m

1   1.1
q - o
...

You can also use i/branching factor+1 for each node to reduce the significant digits of the numbers, but that requires additional calculation to determine.

So above we did a DFS from i, which had 2 children j and m. m had no children, j had 2 children, .... Then we finished with i and started another DFS from the next unexplored node q.

Whenever you encounter a bigger value, you know that a cycle occurred.

Complexity:

You check every node at most once, and at every node you do n checks, so complexity is O(n^2), which is the same as looking at every entry in the matrix once (which you can't do much better than).

Note:

I'll also just note that an adjacency list will probably be faster than an adjacency matrix unless it's a very dense graph.