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My question:

Let G(V,E) be a fully connected graph, where V is set of nodes and E is set of links. What is the upper bound (worst case) of the minimum number of spanning trees needed to cover all the links in the graph, if the spanning trees are sorted in lexicographic order?

As an example, for |V|=4, and thus |E|=6, G(V,E) contains the following 16 spanning trees (in lexicograhic order); note that labelling the links differently may produce different order of spanning trees.

1 2 3

1 2 4

1 2 6

1 3 4

1 3 5

1 3 6

1 4 5

1 5 6

2 3 4

2 3 5

2 4 5

2 4 6

2 5 6

3 4 6

3 5 6

4 5 6

In this case, the minimum number of spanning trees needed to cover all the links in the graph will be 5 spanning trees ({1 2 3},{1 2 4 },{1 2 6}, {1 3 4}, {1 3 5}). So all the links are included in these 5 spanning trees.

It is easy to count the number of spanning trees for small graph, but I have problem with larger sized graph, e.g., |V|>4.

Is there any formula to compute the upper bound number for the spanning trees to cover all links in the graph?

Thanks alot

Radi Elshqerat
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1 Answers1

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There are V-1 edges in any MST, and (V)(V-1)/2 total edges. So the lower bound is ceiling(V/2).

I think this is also an exact bound.

You should be able to find "a" combination of MSTs which do not reuse other edges till the the last steps. Think in terms of finding an MST, removing those edges, and still leaving the reduced graph connected, so that new MSTs can be embedded, without destroying the connectivity.

VSOverFlow
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  • I also thought so; but I do not like downvoting, especially folks who make an effort to answer questions which no else is answering, and give a thought process behind their reasoning (even if it has some holes) :D – VSOverFlow Jul 26 '12 at 03:31
  • thanks. i appreciate that. When I tried to prove by induction, i made a silly mistake. For 4 nodes, we just need 2 MSTs instead of 3. – arunmoezhi Jul 26 '12 at 03:37
  • Thank you to VSOverFlow and arunmoezhi for your reply, but what I try to find is the upper bound (worst case). – Radi Elshqerat Jul 26 '12 at 05:49
  • as @VSOverFlow said, it is an exact bound. – arunmoezhi Jul 26 '12 at 09:10