Thread: Table of Connected Graphs

Does anyone know where I can find a table that displays all connected graphs up to some order? (Of course the higher the better). For example there is this nice table of prime knots and links http://www.math.unl.edu/~mbrittenham2/ldt/table9.gif and any thing like that would be awesome.

I'm doing starting some research on graph theory and this would be good to have a round as reference. Thanks in advance guys

While I understand your question, I am still puzzled by it.
Do you understand that given any simple graph, , then (complement) is connected?
I cannot see how any list could help.

Originally Posted by Plato

While I understand your question, I am still puzzled by it.
Do you understand that given any simple graph, , then (complement) is connected?
I cannot see how any list could help.

This is not true as both of which are clearly disconnected. It may be the case that this is the only example, but since I am not sure I won't jump to conclusions.

The list will help me because for what I am doing it is quite pointless to study disconnected graphs as their properties (for this situation) can quite easily be derived from their components. Having a list of connected graphs will save me the trouble of finding those graphs for studying.

If it helps, I just want a large variety of low order graphs so that I can play with them myself and gain intuition into the problem I have.

We must be using different definitions of complement and/or union of graphs.
That statemenent is a well-known property of simple graphs.

Originally Posted by Plato

We must be using different definitions of complement and/or union of graphs.
That statemenent is a well-known property of simple graphs.

Oh wow... my bad. This is definitely true now that I think about it. For some reason I missed the middle cross when computing the complement. My apologies.

Anyhow, It would still be useful to have a table of connected graphs such that each element is not isomorphic to another. I just don't want to create the list myself if someone already has...