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Cubic graphs are graphs where every vertex has degree 3. They have been extensively studied and I'm aware that several NP-hard problems remain NP-hard even restricted to subclasses of cubic graphs, but some others get easier. A superclass of cubic graphs is the class of graphs with maximum degree $\Delta \leq 3$.

Is there any problem that can be solve in polynomial time for cubic graphs but that is NP-hard for graphs with maximum degree $\Delta \leq 3$?

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Degenrate answer that shows there can be different complexities (though neither is NP-Hard): Finding $\delta$ is constant time on cubic graphs but linear on graphs with $\Delta \le 3$. :-) –  William Macrae Nov 24 '12 at 18:09
    
Good point. :-) –  Vinicius dos Santos Nov 24 '12 at 19:56
    
For bad choices of encodings it can even be $NP$-hard when $\Delta \le 3$, but it will be much more valuable to find a problem that doesn't rely on a poor encoding, and even better if that problem is a well-studied one. –  William Macrae Nov 24 '12 at 20:24
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To expand on William's comment, here is an artificial problem. Given a graph $G$, does the degree sequence of $G$, interpreted as the encoding of an instance of 3-SAT, represent a satisfiable instance? (Assuming the encoding is such that the all-3 degree sequence represents a satisfying assignment for every $n$.) :-) –  Neal Young Nov 24 '12 at 20:30
    
See also cstheory.stackexchange.com/questions/1215/… for more inspiration (e.g., problems that are hard on trees of max degree 3, but trivial if there are no leaf nodes). –  Jukka Suomela Nov 25 '12 at 2:24

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up vote 20 down vote accepted

Here's a reasonably natural one: on an input $(G,k)$, determine whether $G$ has a connected regular subgraph with at least $k$ edges. For 3-regular graphs this is trivial, but if max degree is 3 and the input is connected, not a tree, and not regular, then the largest such subgraph is the longest cycle, so the problem is NP-complete.

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"...then the solution is either the longest cycle or a maximum matching...". How does your claim depend on k? It is not true for all k. –  Tyson Williams Nov 24 '12 at 21:25
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@Tyson, it only needs to be hard for one $k$ to be hard, right? E.g. take $k=n$. David, do you need to stipulate that the subgraph should be connected? (Otherwise, any cycle cover (not just a Hamiltonian cycle) will have $n$ edges, and determining the existence of a cycle cover is in $P$.) –  Neal Young Nov 24 '12 at 22:14
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David, a maximum matching (of size greater than 1) in G is not a connected subgraph of G. Do you mean to say "...either the longest cycle or a single edge, ..."? –  Tyson Williams Nov 25 '12 at 3:04
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Ok, ok. Today doesn't seem to be a good day for me to be rigorous — too much turkey probably. I added some language to rule out this special case. –  David Eppstein Nov 25 '12 at 5:41
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@YininCao Since the graph is connected but not regular, there is no way to pick a 3-regular subgraph. Suppose it were. Then there exist a vertex that was not selected since the graph is not regular. Since the graph is connected, this vertex is connected to some 3-regular vertex that was selected. But that means there exists a vertex of degree 4, a contradiction. –  Tyson Williams Nov 25 '12 at 13:33

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