# Complexity of finding disjoint 2-factors with equal cardinality in cubic graphs?

Finding a connected 2-factor that contains every node in cubic graphs is $NP$-complete since it is equivalent to the Hamiltonian cycle problem. I'm interested in the complexity of finding vertex disjoint 2-factors with equal cardinality in cubic graphs.

I suspect that it is $NP$-complete but I'm not able to find a reference. Also, What is the complexity of finding vertex disjoint 2-factors with equal cardinality in planar cubic bipartite graphs? Is it $NP$-complete?

Providing references is highly appreciated.

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$NP$-completeness refers to the decision version of the problem. – Mohammad Al-Turkistany Apr 13 '11 at 10:49