As an example, if I have 2 vertices V1, V2.
All the possible directed graphs ( there can be indegree 0 vertices as well) are 3 {{V1,V2}, {V1->V2}, {V1<-V2}}.
But, when there are 3 vertices all the possible graphs are found to be very large.
{{V1,V2,V3}, {V1->V2,V3}, {V1<-V2,V3}…..{V1->V2->V3} …. } . Like
- graphs with edges 0
- graphs with 1 edges and their directionality changed
- graphs with 2 edges and their directionality changed
- graphs with 3 edges and there directionality changed
My attempt so far: As mentioned above I simply analyze the questions, but it is found to be growing larger. I was able to found that for any N no. of vertices, all the possible edges ( undirected ) are N*(N-1) /2. It is like I have to analyze all the number of possible edges and their directionality.
But, which is not clear is the possible number of graphs is growing with a possible number of edges.
I would appreciate any suggestions on how to obtain all the possible directed graphs from N number of vertices.
Best Answer
There are $\binom{V}{2}=V(V-1)/2$ distincy pairs of vertices. For any pair of vertices $\{u,v\}$, any one of the 3 options can hold:
Hence total number of directed graphs on $V$ vertices is $3^{V(V-1)/2}$.