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I am reading graph theroy. Here author mentions that the number of possible digraphs is truly huge. Each of the $V^2$ possible directed edges (including self-loops) could be present or not, so the total number of graphs is $2^{V^2}$. For undirected graphs the number is given by $2^{V(V+1)/2}$.

For $V =2$ we have $8$ undirected graphs and $16$ digraphs.

My question , can any write all combinations for $V =2$ asuming vertices are $"a" $and $"b"$. I tried but not able to find all. For example in directed $\emptyset$, ${a}, {(a,a)}, {(b,b)}, {b}, {(a,b)}, {(b,a)}, {(a,b), (b,a)} $I came up with 8, but according to formula above we should get 16

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Here are them:

  • without loops:

$\emptyset$, $\{(a,b)\}$, $\{(b,a)\}$, $\{(a,b),(b,a)\}$

  • with $(a,a)$ loop but without $(b, b)$ loop:

$\{(a,a)\}$, $\{(a,a), (a,b)\}$, $\{(a,a), (b,a)\}$, $\{(a,a), (a,b), (b,a)\}$

  • with $(b,b)$ loop but without $(a,a)$ loop

this case is symmetric with the second case

  • with both $(a,a)$ and $(b,b)$ loops

$\{(a,a),(b,b)\}$, $\{(a,a),(b,b),(a,b)\}$, $\{(a,a),(b,b),(b,a)\}$, $\{(a,a),(b,b),(a,b),(b,a)\}$

Can you observe how other three cases relate to the first case?

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  • $\begingroup$ Yes. We are combining loop to first case. By the way for your time and answer. $\endgroup$ – venkysmarty Jun 23 '15 at 14:08

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