I can't figure out how to find a graphs with this properties:
I have to find 2 non-isomorphic plane graphs which (each of them) have 7 faces, two of which are of size 3 and the rest of size 4.
This is my partial solution:
EDIT: I've figured out that this Euler calculation is wrong. Do anybody knows how to get number of vertices from that? Is it possible?
The first thing I've thought about is that I could count number of vertices in this graph using Euler which says: |V(G)-7 (number of faces)- E(G)| = 2. So |V(G) = 5 + E(G) and now I'm lost. What to do next?
Another solution (which is obviously incorrect) is this: Since there are 7 faces, 2 of them with size 3 and 5 of them with size 4, the outer face should be of size 4. But if the outer face was of size 4, the "outer edge" of graph should be like square (C4 Graph). But it is not possible to find 2 faces of size 3 and 4 faces of size 4 in the C4 because we can draw only 2 edges there.
Do you know where is the problem? Thank you very much
It is is easy to show that $\sum\limits_{\text{f is a face}}e(f)=2E$. (here $e(f)$ is the number of edges in the face $f$)
So there are $(3+3+4+4+4+4+4)/2=13$ edges
From here we apply Euler's formula, so $V=13-7+2=8$ vertices.
I found one way so far: