Solving 5V 4E Graph Theory Problem

[dobry_den]

Hi! I was given the following task:

(a) Exhibit three nonisomorphic, connected graphs with five vertices and four edges.
(b) Argue that every connected graph with five vertices and four edges is isomomorphic to one of the three in part (a).

It's not very difficult to draw the graphs:
http://i83.photobucket.com/albums/j315/dobry_den/graphs.jpg

but i can't think of how i would argue... Do you have any idea? Thanks in advance!

 

[dobry_den]

Maybe I've come up with something.. The sum of deg(v) has to be 8 (I have 4 edges, each is shared by two vertices). Moreover, there are 5 vertices, so the only combinations of degrees are: (1,1,1,1,4), (1,1,2,2,2) and (1,1,1,2,3).

(i) There's no other way how to connect (1,1,1,1,4) vertices to yield a nonisomorphic graph with the (1,1,1,1,4) graph from point (a). This is quite clear.

(ii) The same for a (1,1,2,2,2) graph.. This is a path and there's no other nonisomorphic connected graph that could be constructed from these vertices (eg. a graph consisting of a triangle and a path of length 1 has also vertices of degrees (1,1,2,2,2)..but it isn't connected).

(iii) To ensure that a (1,1,1,2,3) graph is connected, the vertices with degrees 3 and 2 must share an edge (otherwise, there would be no path from the former to the latter.. since the rest of the vertices are of deg=1.. and that would yield a graph that's not connected).. thus all graphs with the degree set (1,1,1,2,3) are isomorphic with the one on the picture.

Do you think this is sufficient?