# What is the name for this type of graph? [closed]

(old image at bayimg.com/image/iaeidaacn.jpg just there for the post to still make sense)

Both graphs have a braching factor of 3. Graph A is a "tree". Is there a name for the type of graphs as B? (you can ignore the colours)

edit ok, my bad. Clearly I have not as much knowledge of graphs. I am indeed referring to graph searching, as that is something I have more knowledge of. So I suppose I should have drawn arrows, as the graphs are directed from left to right.

• Is it correct that the first graph is called a tree?
• Then, the second graph was supposed to be analogous to a tree. Only with the childnodes of sibling nodes merged, so that the number of nodes grows at a constant rate (3 in my example, but $n$ in the general case) rather than exponential as is the case in the tree.
• I've drawn a new graph image to illustrate more clearly what I mean. Also left out the left-most single node as I think this is a more general graph.

I hope this clears up my drawings a bit.

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## closed as no longer relevant by Scott Morrison♦Nov 13 '09 at 20:06

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The branching factor that you refer to is called the degree of a vertex of a graph. Graphs whose vertices all have the same degree are called regular. 3-regular graphs are often called "cubic" graphs. You probably want to read this en.wikipedia.org/wiki/Glossary_of_graph_theory – j.c. Nov 13 '09 at 18:46
You really need to say what you mean, rather than saying "the type of graphs as B". There's no way we can guess from B what class of graphs you have in mind. – Reid Barton Nov 13 '09 at 18:57
Actually, these are not 3-regular graphs. In graph B, the middle three vertices have degree 4, not 3. I am guessing that branching factor means something like the number of times a breadth first search branches, but I don't know. – David Speyer Nov 13 '09 at 19:10
Seems to be a terminology question which has now been cleared up, so I've closed. Feel free to disagree (preferably at meta, or by flagging for moderator attention). – Scott Morrison Nov 13 '09 at 20:07