
graph + tree
If a graph has
8 verticies:
2 with 1 degree
1 with 2 degrees
5 with 3 degrees.
If it is not a tree I must prove why.
The only way I can think of to do this is to draw the graph (which I can't replicate here).
When I draw the graph I find I have 2 nontrivial circuits, therefore the graph is not a tree.
Is this the only way to prove the graph is not a tree?

For your graph to be a tree then $\displaystyle \epsilon = v1$ where $\displaystyle \epsilon$ = number of edges and v = number of vertices which is given to be 8.
To check you need to find the number of edges. $\displaystyle \epsilon = \frac{1}{2}\sum d(v)$ where $\displaystyle d(v)$ is the degree of each vertex. Does this equal $\displaystyle 81 = 7 $ ?

ok,
I understand that the edges = vertices 1 for a tree.
So for 8 vertices there should be 7 edges.
(I can draw this, bu t I get nontrivial circuits. Therefore not a tree)
But to prove it, I use
edge = vertices  1.
edge = $\displaystyle \frac{1}{2} *$ v1 degrees + v2 degrees... vn degrees
edge = 7
edge = 0.5 * (1+1+2+3+3+3+3+3)
= 0.5 * 19
= 9.5
7 != 9.5, therefore not a tree.
correct?
btw, thanks for taking the time to help pickslides.
