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Here is the following problem:

What two edges should one remove from the complete graph $K_n$ so that the number of the spanning trees of the new graph is as small as possible?

One can solve this problem of course by computing the determinant of the Laplacian matrix associated to this graph, however I was wondering if there is some more elegant solution involving perhaps some intelligent counting, like say for the problem of counting the number of the spanning trees when we remove just one edge instead.

user223794
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1 Answers1

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Let $A(e_1,e_2)$ be the number of spanning subtrees of $K_n$ containing both $e_1$ and $e_2$.

We want to minimize $A(e_1,e_2)$.

If $e_1$ and $e_2$ are adjacent then we can count exactly how many trees have $e_1$ and $e_2$. To do this contract those edges into a single vertex $v$, for each spanning tree in the new graph with $n-2$ vertices there are $3^{d(v)}$ spanning trees on all $n$ vertices.

So using the prufer code we can see there are $\sum_{i=0}^{n-4}3^{i+1}\binom{n-4}{i}(n-3)^{n-4-i}$ trees in total.

If $e_1$ and $e_2$ are not adjacent then we can also count exactly how many trees have $e_1$ and $e_2$ exactly, we do the same trick (contract each edge into vertices $v_1,v_2$), but now we notice that for each spanning tree on the new graph with $n-2$ vertices there are $2^{d(v_1)+d(v_2)}$ spanning trees on all $n$ vertices

So using the prüfer code we can see there are $\sum_{i=0}^{n-4}\binom{n-4}{i}2^{i+2}2^i(n-4)^{n-4-i}$

Since there are $\binom{n}{i}$ ways to pick the $i$ vertices that are $v_1$ or $v_2$ and $2^i$ ways to split them (between $v_1$ and $v_2$).

So all that remains is to compare:

$\sum_{i=0}^{n-4}\binom{n-4}{i}3^{i+1}(n-3)^{n-4-i}$ and $\sum_{i=0}^{n-4}\binom{n-4}{i}4^{i+1}(n-4)^{n-4-i}$

But using the binomial theorem, the sum on the left is:

$3n^{n-4}$ and the sum on the right is $4n^{n-4}$, so the left is smaller.

Asinomás
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  • This is nice (+1), but the simple results suggest that there should be a more direct argument without the summation. – joriki Jul 24 '16 at 21:10
  • Yeah, there's probably something super cool, but I always suck at finding those sort of solutions (especially when the base of an exponent is increased). Like for example with problem $3$ of the second day of the IMC 2015 – Asinomás Jul 24 '16 at 21:15
  • Actually, I think you want to minimise $A(e_1,e_2)$? The complement of the set of spanning trees that use neither $e_1$ nor $e_2$ is the set of spanning trees that use $e_1$ or $e_2$. If you use inclusion-exclusion to express this as $A(e_1)+A(e_2)-A(e_1,e_2)$ in terms of $A(e_1,e_2)$ and the individual counts $A(e_1)=A(e_2)$, the latter are the same in either case, and minimising $A(e_1,e_2)$ maximises the count, which is the complement of the count the question seeks to minimise. – joriki Jul 25 '16 at 11:40
  • And about increasing the base of the exponent: Actually it should be $(n-3)^{n-4-i}$ and $(n-4)^{n-4-i}$, respectively, since the remaining $n-4-i$ Prüfer entries mustn't be the label of the contracted vertex; so the result is in fact $3n^{n-4}$ and $4n^{n-4}$ (which suggests even more that there should be a more direct argument). – joriki Jul 25 '16 at 12:05
  • Oh yeah, you are right about the raising the base of the exponent thing. But on the inclusion/exclusion thing, I think we want to maximize $n^{n-2}-A(e_1)-A(e_2)+A(e_1,e_2)$ – Asinomás Jul 25 '16 at 14:41
  • nevermins, you are correct in both matters. – Asinomás Jul 25 '16 at 14:42