You are given two graphs G and H , and want to know if H is a subgraph of G.

You know that H has a max vertex degree K (constant integer).

What can you say about the complexity of this?

I know that Isomorphism of Graphs of Bounded Valence Can Be tested in Polynomial Time - Luks 1982, that's the best I currently have...

  • $\begingroup$ (1) What is the constraint and what is the objective function? If you want to maximize the number of vertices of the chosen subgraph while bounding the maximum degree of the chosen subgraph (as is suggested in the question), then obviously the optimal way is to choose all vertices and no edges. (2) What is “unbounded graph”? $\endgroup$ Apr 12, 2012 at 11:32
  • $\begingroup$ Hi @Tsuyoshi Ito, thanks for the help, I want to know if the bounded graph is a subgraph of the unbounded graph, both graphs are given. I'll edit my question... $\endgroup$ Apr 12, 2012 at 11:38
  • $\begingroup$ I see, but I do not know what you mean by “max vertex” then. Can you edit the question so that people do not have to read comments to understand the question? “Both graphs are given” is a very important piece of information which was missing from the question. $\endgroup$ Apr 12, 2012 at 11:39
  • $\begingroup$ Edited, @Tsuyoshi Ito , thank you very much for the feedback, would be happy to know if further editing is needed. $\endgroup$ Apr 12, 2012 at 11:51
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2 Answers 2


If H is the cycle on |G| vertices (and is of maximum degree 2), then this is the Hamiltonian cycle problem. Seems like your problem is NP-hard.


To add to the answer from Marcin Kaminski - H could even be the complement of a complete graph (i.e., regular graph with degree = 0). So your problem is at least as hard as the independent set problem, and is thus NP-hard. The max degree of G is immaterial.

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    $\begingroup$ It is a subgraph not an induced subgraph so the independent set argument does not seem to work. $\endgroup$ Apr 13, 2012 at 6:00
  • $\begingroup$ Although if you set a degree limit of 1 for H then the problem reduces to the independent set problem on a graph G' whose vertices are the edges of G and are connected iff they share a vertex in G, so the only tractable case is a degree limit of 0 (in which case it's just a case of checking $|H| \le |G|$). $\endgroup$ Apr 13, 2012 at 7:53
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    $\begingroup$ It is the independent set problem in line graphs, and this is computationally easy. I think that if deg(H) <=1, then the subgraph problem can be solved in polynomial time. $\endgroup$ Apr 13, 2012 at 8:48
  • $\begingroup$ @MarcinKamiński, I stand corrected. $\endgroup$ Apr 13, 2012 at 14:04
  • $\begingroup$ Good point Marcin Kaminski. An argument, such as yours, where H is a Hamiltonian circuit of G, or one where H is a clique, is a correct one. $\endgroup$
    – Ankur
    Apr 16, 2012 at 18:41

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