This question is motivated by a MathOverflow question by Peng Zhang. Valiant showed that counting maximal cliques in a general graph is #P-complete, but what if we restrict to incomparability graphs (i.e., we want to count maximal antichains in a finite poset)? This question seems natural enough that I suspect that it has been considered before, but I have not been able to locate it in the literature.


According to this abstract for "The Complexity of Counting Cuts and of Computing the Probability that a Graph is Connected" (SIAM J. Comput. 12 (1983), pp. 777-788), counting anti-chains in a partial order is #P-complete. I don't have access to this paper so I can't tell if this result covers maximal anti-chains or not.

  • $\begingroup$ @András: I think that their result is about counting antichains (which are not necessarily maximal). It might be easy to see that counting maximal antichains is also #P-complete, but I cannot see it. $\endgroup$ Aug 6 '11 at 16:14
  • $\begingroup$ @András: The question is about maximal antichains, not maximum-cardinality antichains. I have not studied the reduction in the paper, so maybe their reduction also proves the #P-completeness of counting maximal antichains at the same time, but at least they are different problems. $\endgroup$ Aug 6 '11 at 16:24
  • $\begingroup$ @Tsuyoshi: you are right, the Provan/Ball paper only shows that counting maximum-cardinality antichains is #P-hard. Back to the drawing board... $\endgroup$ Aug 6 '11 at 16:56
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    $\begingroup$ Actually, if you look at the proof, you'll see that #P-completeness is proved for a class of posets in which all maximal antichains have the same cardinality. Namely, start with any bipartite graph $G=(V,E)$ with $n$ vertices and construct a bipartite graph $G'$ with $2n$ vertices by adding $n$ new vertices $\{v' : v\in V\}$ and $n$ new edges $\{(v,v'): v\in V\}$. Then, if $V_1$ and $V_2$ is a bipartition of the vertex set of $G'$, define a poset on $V_1\cup V_2$ by setting $x<y$ if $x\in V_1$ and $y\in V_2$ and $x$ and $y$ are adjacent in $G'$. So this does answer my question. $\endgroup$ Aug 7 '11 at 2:46

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