The $q$-Dimensional $p$-Matching is defined as follows:

Given disjoint universes $U_1,\ldots,U_q$, think of an element in $U_1\times\ldots\times U_q$ as a set that contains exactly one element from each universe $U_i$.

Now, given a family ${\cal S}\subseteq U_1\times \ldots\times U_q$ and a parameter $k\in\mathbb{N}$, determine if there exists a subfamily of $k$ pairwise-disjoint sets in ${\cal S}$.

We abbreviate by $\#q$-Dimensional $p$-Matching (or $\#(q,p)$-matching in short) the counting version of the problem, asking how many matchings exist in $\cal S$.

In their famous paper from 2004, Flum and Grohe have developed the $\#W$ hierarchy, the counting equivalent of the $W$ hierarchy for parameterized problem.

Among their results in the paper, they have shown that counting the number of $k$-paths is $\#W[1]$-complete, and conjectured that counting the number of matchings of size $k$ in a bipartite graph is hard as well.

I'm unaware of any results addressing this conjecture, but was wondering about even a weaker result:

Is #$q$-Dimensional $p$-Matching (counting the number of instances in a $(q,p)$-matching instace) $\#W[1]$-complete, where the parameter is $p+q$?

Such result would be weaker than the FG's conjuncture, as counting bipartite matchings is the same as $\#(2,p)$-matching.

Related result:

Recent result by Bläser and Curticapean shows that weighted counting of $k$-matching in bipartite graphs is $\#W[1]$-complete.


2 Answers 2


Our recent paper shows that counting k-matchings is #W[1]-hard even in bipartite graphs. This answers your question.

Radu Curticapean, Dániel Marx: Complexity of counting subgraphs: only the boundedness of the vertex-cover number counts. CoRR abs/1407.2929 (2014)


Radu Curticapean showed that counting matchings of size k is #W[1]-Hard. See http://link.springer.com/chapter/10.1007/978-3-642-39206-1_30

  • 1
    $\begingroup$ True, but this doesn't answer the question.. This result is for general graphs and it's not translatable to #$q$-dimensional $p$-matching, but rather to #$(2,p)$-Set Packing (the problem is that the sets (edges by the graph formulation) may be arbitrary and not two items from two disjoint universes). $\endgroup$
    – R B
    Commented Aug 12, 2014 at 16:27
  • 1
    $\begingroup$ You are right. Sorry I missed the "bipartite" constraint... $\endgroup$
    – Heng Guo
    Commented Aug 12, 2014 at 17:37

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