When does graph minor containment imply subgraph containment? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T19:32:30Z http://mathoverflow.net/feeds/question/34060 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34060/when-does-graph-minor-containment-imply-subgraph-containment When does graph minor containment imply subgraph containment? Robin Kothari 2010-07-31T21:45:25Z 2010-08-01T15:00:40Z <p>Consider a path of length 3. Any graph G which contains this graph as a minor must also contain it as a subgraph. For paths of any length this is easy to prove.</p> <p>In general this happens for any graph in which each connected component is either a path or a subdivision of the claw grah. (The claw graph is the star graph on 4 vertices, or the complete bipartite graph $K_{1,3}$.) </p> <p>Does anyone know where I can find a proof of this fact? I know how to prove it, but if it has already appeared in some book/paper, it's easier for me to cite the result instead.</p> <p>I am also interested in the following more general question, whose answer I do not know: How does one characterize sets of graphs H={H<sub>1</sub>,H<sub>2</sub>,...,H<sub>k</sub>}, such that containing any of the H<sub>i</sub> as a minor is equivalent to containing some graph from a finite set G={G<sub>1</sub>,G<sub>2</sub>,...,G<sub>m</sub>} as a subgraph. In other words, for which sets H is H-minor containment equivalent to G-subgraph containment for some finite set G. (Note that this is trivial if we allow G to be infinite.)</p> <p>For example, containing any graph from the set {path of length 3, claw} as a minor is equivalent to containing any graph from that set as a subgraph. As a non-trivial example, containing any from from {path of length 4, cycle of length 3} as a minor is equivalent to containing one of {path of length 4, cycle of length 3, cycle of length 4} as a subgraph.</p> <p><b>Edit</b>: For the single graph problem, I think I have stated a complete characterization of such graphs. I only wish to know if this appears in the literature somewhere. For the second problem I do not know a characterization (other than some special cases), and would welcome any information about the problem.</p> http://mathoverflow.net/questions/34060/when-does-graph-minor-containment-imply-subgraph-containment/34078#34078 Answer by Nathann Cohen for When does graph minor containment imply subgraph containment? Nathann Cohen 2010-08-01T01:28:26Z 2010-08-01T05:49:56Z <p>Hello !</p> <p>Well, first I do not think I read about this in any book I read. Which is not really a surprise, as I did not find many books mentionning graph minors. You will find a chapter about minors at the end of Diestel's book (http://www.math.uni-hamburg.de/home/diestel/), and I think we will otherwise have to wait for Bruce Reed's book on Graph Minors.</p> <p>To focus on your problem, you can for example say that this property does not hold when your graph is not a forest : you can through edge subdivision arbitrarily increase the girth (length of a smallest cycle) of any graph.It means that if you think finding H (with a cycle) as a minor if G is the same as finding it as a subgraph, then you expect to find in H a cycle of length k. You can now subdivide the edges of G k times, to remove any possible cycle of length &lt;= k, which of course changes nothing to minor containment.</p> <p><strong>Topological Minors :</strong></p> <p>The same fact that tells you $K_{1,3}$ has this property can be used to prove that for any $k$, $K_{1,k}$ also has this property.</p> <p>Actually, let $H$ be any graph with two vertices $u,v$ of degree at least 3. Let $d(u,v)$ be their distance in $H$. Now take a graph $G$ having $H$ as a minor, and subdivide all its edges $d(u,v)+1$ times. Well, now you will not find two vertices of $G$ of degree larger than 3 at distance less than $d(u,v)$, so even though $G$ still contains $H$ as a minor, it does not contain it as a subgraph.</p> <p>So in order to have this property, you must have a most one vertex of degree larger than 3, and I think this is an equivalence.</p> <p><strong>Usual minors :</strong></p> <p>In this case, for any graph $H$, there exists a graph $G$ having $H$ as a minor such that $G$ has maximal degree $3$. So there is no need to look at anything different from $K_{1,3}$ or a path to answer your question :-)</p> <p>I can not help with your set version for the moment, though... :-)</p> <p>Nathann</p> http://mathoverflow.net/questions/34060/when-does-graph-minor-containment-imply-subgraph-containment/34083#34083 Answer by David Eppstein for When does graph minor containment imply subgraph containment? David Eppstein 2010-08-01T04:05:10Z 2010-08-01T07:35:22Z <p>The case when H has more than one forbidden minor is messy, but here's a partial analysis.</p> <p>If H contains a linear forest (disjoint union of paths) then some path P_k is forbidden and the H-minor-free graphs are obviously describable by a finite set of forbidden subgraphs (P_k and the expansions of H that do not have a P_k subgraph).</p> <p>If H does not contain a linear forest or disjoint union of subdivided claws, then the H-minor-free graphs include all subdivided claws. In this case the H-minor-free graphs are obviously not describable by a finite set of forbidden subgraphs -- any minimal forbidden minor of H can have its vertices blown up to degree-three trees and then its edges blown up to long paths so that the local neighborhood of any vertex looks the same as a subdivided claw, producing a graph that is not H-minor-free but looks locally like an H-minor-free graph.</p> <p>If H does contain a cycle, and doesn't contain a linear forest, then the H-minor-free graphs are again obviously not describable by a finite set of forbidden subgraphs -- long cycles are not H-minor-free but look locally like paths, which are H-minor-free.</p> <p>If H doesn't contain a linear forest or a cycle, but does contain a linear forest + claw in which only one of the claw edges is subdivided, then the H-minor-free graphs are describable by a finite set of forbidden subgraphs (the subdivided claw and the expansions of H that do not contain it -- because H doesn't contain a path or a cycle, each edge in each forbidden minor of H can only be expanded to a length shorter than the subdivided edge of the forbidden claw).</p> <p>The remaining case is when H contains linear forest + claw subgraphs, but only those in which two or three of the claw edges have been subdivided, or when it contains graphs with more than one claw component.</p>