Suppose we have a graph whose edges are coloured. It's not necessarily a proper colouring: a given node may have 0, 1, or several incident edges of a given colour.

Is the following problem NP-complete? Determine whether there are two edge-disjoint spanning trees, such that in each individual tree, no colour appears twice.

I am curious because the variant "determine whether there are two edge-disjoint spanning trees, such that in the union of the trees, no colour appears twice" is solvable in polynomial time, for example using matroid theory.

  • $\begingroup$ I can see how to find one rainbow spanning tree using matroid intersection. Can you say a bit more about how to find two disjoint spanning trees whose union is rainbow? $\endgroup$ – Tony Huynh May 24 '10 at 23:05
  • $\begingroup$ What I had in mind is to use matroid union and intersection. First, take the matroid union of the graphic matroid with itself; this yields another matroid M1 whose bases are those edge sets which are partitionable into two trees. Second, consider the partition matroid M2 whose parts are the colour classes. Now use matroid intersection on M1 and M2: what we want to know is whether the largest common independent set has size 2(|V|-1). $\endgroup$ – Dave Pritchard May 25 '10 at 6:58
  • $\begingroup$ Btw, I might be wrong, but the rank function of M1 seems to be NP-hard to compute (see problem 10 in my answer), so I do not see why you could solve the problem in P. Where am I mistaken? $\endgroup$ – domotorp May 25 '10 at 14:28
  • $\begingroup$ Sorry I should mis-spoke; M1 is the matroid whose bases are those edge sets which are partitionable into two spanning trees. Schrijver's book sections 51.4, 42.3 talks about polynomial-time algorithms for this (corrected) version. $\endgroup$ – Dave Pritchard May 26 '10 at 12:17

Here are some nice lemmas that you can use: http://www.cs.elte.hu/egres/qp/egresqp-10-04.pdf

Dave pointed out my mistake, the subgraphs of spanning trees do not have to be trees. So I have no clue about the answer.

Espacially Problem 4 (or 10) seems promising. Take the graph from their construction such that all of its edges have a different color, suppose it has e edges. Then if we allow multigraphs, adding every edge with multiplicity 2n-2-e, all of a different color but same for each edge (thus in total we have 2n-2 colors) shows that your question solvable in P is NP-hard for multigraphs. Am I right? I would guess that with some further tricks you can make a simple graph from this for the variant you asked.

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  • $\begingroup$ Will check out this paper when I pause travelling long enough. Due to forgetting "spanning" in the comment, I need to correct that actually the "rainbow union-of-2-spanning-trees" problem is in P even for multigraphs. I would be equally happy to resolve the "2 individually-rainbow spanning-trees" problem whether for simple or multigraphs. To reiterate, in all cases, the only trees I care about are spanning. $\endgroup$ – Dave Pritchard May 26 '10 at 22:16
  • $\begingroup$ Okay. I am still confused about the matroid M1, I do not think that it has a poly time computable rank function, since deciding the bases does not give an algorithm for deciding independent sets. Plus there is my above mentioned result that shows that it is NP-hard. $\endgroup$ – domotorp May 27 '10 at 2:24

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