Question

The square of a directed graph G = (V, E) is the graph G2 = (V, E2) such that u→w is in E2 if and only if u ≠ w and there is a vertex v such that both u→v and v→w are in E2. The input file simply lists the edges in arbitrary order as ordered pairs of vertices, with each edge on a separate line. The vertices are numbered in order from 1 to the total number of vertices.

*self-loops and duplicate/parallel edges are not allowed

If we look at the an example of input data:

1 6
1 4
1 3
2 4
2 8
2 6
2 5
3 5
3 2
3 6
4 7
4 5
4 6
4 8
5 1
5 8
5 7
6 3
6 4
7 5
7 4
7 6
8 1

Then the output would be:

1: 3 4 7 8 5 2 6
2: 5 6 3 4 1 8 7
3: 1 7 8 6 5 4
4: 5 6 8 7 3 1
5: 3 1 4 6
6: 2 7 5 8
7: 1 5 6 8 3 4
8: 6 4 3

I’m writing the code in C.

My thoughts are to run through the file, see how many vertices they are and then allocate an array of pointers. Proceed to go through the list again searching for just where the line has a 1 in it, then look at where those corresponding numbers lead. If its not a duplicate or the same number(1) then I’ll add it to a linked list, from the array of pointers. I will do this for every number vertex number in the file.

However, I feel this is terribly inefficient, and not the best way to go about doing this. If anyone has any other suggestions I would be extremely grateful.

Answer 1

if I get it right, you want to build a result set for each node where all nodes with a distance of one and two for each node are stated.

therefore, one can hold the edges in an adjacency matrix of bit arrays, where a bit is one when an edge exists and zero if not.

now one can multiply this matrix with itself. in this case multiply means you can make an AND on row and column.

A small example (sorry, don’t know how to insert a matrix properly):

0 1 0   0 1 0   0 0 1
0 0 1 x 0 0 1 = 1 1 0
1 1 0   1 1 0   0 1 1

This matrix contains a one for all nodes reachable in two steps. simply it’s the adjacency matrix for two instead of one steps. If you now OR this matrix with your initial matrix you have a matrix which holds all paths of length one and two.

this approach has multiple advantages. at first bit operations are very fast. the cpu parallyzes your calculations and you can stop for the result matrix cell if one pair is found where the results gives one.

furthermore it is well documented how to calculate matrix multiplication in parallel.

you can easily calculate all other length of pathes. for a length k one has to calculate:

A^k = A^(k-1) * A

hope that helped