A tree is a graph with no vertex contained in a cycle.
A non-tree is a graph with some vertex contained in a cyle.
What's the name of graphs with each vertex contained in a cycle?
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asked Jan 11, 2010 at 0:29
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1$\begingroup$ I'm wondering why no one had voted up this question (until I just did), even though people were answering and voting up answers. $\endgroup$– Jonas MeyerCommented Jan 11, 2010 at 1:34
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5 Answers
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Undirected graphs in which every edge is contained in a cycle are called bridgeless or 2-edge-connected. But I don't know of a word for the analogous concept for vertices.
answered Jan 11, 2010 at 0:38
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$\begingroup$ Could this mean that this is an unfruitful concept (since if it were fruitful it would have a name X, e.g. to be able to formulate statements like "all graphs that are X are Y" or vice versa?) The other way round: Do you know an interesting theorem which implies graphs with the above mentioned property? $\endgroup$ Commented Jan 11, 2010 at 0:48
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$\begingroup$ If this graph property was fruitless: How could this be explained, compared to the "unreasonable" fruitfulness of trees? $\endgroup$ Commented Jan 11, 2010 at 0:52
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7$\begingroup$ Nice pun! Was it intentional? $\endgroup$ Commented Jan 11, 2010 at 1:56
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$\begingroup$ I guess it was not (if you refer to the "fruits" of "trees"). But can you - who has used the term "unreasonable" too (in another context) - appreciate the question? $\endgroup$ Commented Jan 11, 2010 at 2:21
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$\begingroup$ @David: If every edge is contained in a cyle, doesn't every vertex have to be contained in a cycle, too? And vice versa? That is, isn't the concept "self-dual"? $\endgroup$ Commented Oct 29, 2012 at 18:17
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These are the graphs that admit "vertex cycle covers". http://en.wikipedia.org/wiki/Vertex_cycle_cover
answered Jan 25, 2010 at 4:40
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$\begingroup$ +1, even if this is not really a name but just another characterization. (At least it's something to google for.) $\endgroup$ Commented Jan 25, 2010 at 7:27
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I don't know a name, but I'll give you a different characterization. Biconnectivity is sufficient but too strong, while "having minimum degree at least 2" is necessary but too weak. I'm almost certain this is a necessary and sufficient condition:
$G$ has minimum degree at least 2, and if v is a cutvertex of $G$, then there is some new connected component of $G - v$ with at least two vertices adjacent to v.
Here's a proof of sufficiency: If v is not a cutvertex of $G$, then pick any two vertices adjacent to v. There's a path between them not going through v (since $G - v$ is connected), so v is contained in a cycle.
If v is a cutvertex of $G$, then pick the two vertices adjacent to v that are in the same connected component of $G - v$. There's a path between them that extends to a cycle containing v.
Now, a proof of necessity. Suppose that $G$ has a cutvertex $v$ whose removal does create deg(v)-1 new connected components. Then $v$ can't lie in a cycle. (This is easy to check.)
This characterization is equivalent to: Removing any vertex of degree d increases the total number of connected components by at most $d-2$. Some generalization of this property may have a name.
answered Jan 11, 2010 at 1:04
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$\begingroup$ Thanks. Please see my comments to David's answer. They apply also to yours. $\endgroup$ Commented Jan 11, 2010 at 1:07
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$\begingroup$ How can vertices of G-v be adjacent to v? $\endgroup$ Commented Jan 11, 2010 at 1:10
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$\begingroup$ Sorry, I meant adjacent in the original graph. $\endgroup$ Commented Jan 11, 2010 at 1:16
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$\begingroup$ Sorry for my part, I should have understood it this way. (You treat the one counter-example of David below - the 8-graph - and we are done?) $\endgroup$ Commented Jan 11, 2010 at 1:24
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$\begingroup$ I am so silly, but why a minimum degree of 2 is required? $\endgroup$– feireCommented Sep 15 at 16:21
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Assuming that $G$ is connected, we can get a characterization by looking at the block decomposition $B(G)$ of $G$ into 2-connected pieces. That is, we simply look at which vertices of $B(G)$ are $K_2$'s. The required characterization is:
Every vertex of $G$ is contained in a block of $G$ which is not a $K_2$.
Of course, this is the same answer as Harrison's, but it gives a more global view. Also, necessity and sufficiency are trivial.
answered Feb 17, 2010 at 23:48
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2-connected or biconnected
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$\begingroup$ of course, this works only for connected graphs. $\endgroup$ Commented Jan 11, 2010 at 0:32
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$\begingroup$ A figure-eight graph has every vertex in a cycle but is not biconnected. $\endgroup$ Commented Jan 11, 2010 at 0:36
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$\begingroup$ Even if those graphs were biconnected, their biconnectivity would have to be proved, but would not be the defining property, for which I am looking for a name. $\endgroup$ Commented Jan 11, 2010 at 0:55