Is there a sequence of undirected graphs $\{C_n\}_{n\in \mathbb N}$, where each $C_n$ has exactly $n$ vertices and the problem

Given $n$ and a graph $G$, is $C_n$ an induced subgraph of $G$?

is known to be in class $\mathsf{P}$? (For example, when $C_n=K_n$, this is the NP-complete clique problem.)

  • $\begingroup$ Crosspost from cs.stackexchange.com/questions/10576 $\endgroup$
    – sdcvvc
    Commented Mar 24, 2013 at 12:40
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    $\begingroup$ So $\{ C_n \}$ is part of the problem definition, $n$ is part of the input, and $G$ is part of the input? $\endgroup$ Commented Mar 24, 2013 at 21:22
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    $\begingroup$ @Andrew D. King: Yes. $\endgroup$
    – sdcvvc
    Commented Mar 24, 2013 at 22:06
  • $\begingroup$ What about if $C_n$ is a star (one central node connected to $n-1$ nodes that form an independent set)? to check, merely enumerate all nodes of degree $n-1$ in $G$, and check if the neighbors form an independent set. $\endgroup$ Commented Mar 25, 2013 at 13:42
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    $\begingroup$ @Suresh: There may be a vertex of degree larger than $n-1$, whose some $n-1$ neighbors form an independent set. Finding them is NP-complete. $\endgroup$
    – sdcvvc
    Commented Mar 25, 2013 at 13:59

1 Answer 1


If I'm not mistaken your question was answered by Chen-Thurley-Weyer-2008 modulo parameterized complexity assumptions.

I didn't read the paper carefully yet, but as far as I understood, there is a dichotomy in the sense that if $C$ is finite then the problem is in $P$, but if $C$ has an infinite number of graphs then the induced subgraph isomorphism is $W[1]$ complete (Corollary 4, page 6).

Thus it seems that unless the first level $W[1]$ of the $W$ hierarchy collapses to $FPT$, there is no such an infinite class of graphs whose induced subgraph isomorphism is in $P$.

There is another interesting result stating that if $P\neq NP$ then there are classes for which the induced isomorphism is neither in $P$ nor $NP$ complete.


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