Number of leaves in a tree (all types of trees)

$\begingroup$

I have to prove the following claim, given the tree $T=(V,E)$, $|V|=n\geq2$: $$|V_1| = 2 + \sum (j-2)|V_j|$$

where the sum is from $j=3$ up to the highest degree, and $$V_i = \{ x \in V \mid \deg(x) = i\}.$$

This was a bonus question given by my professor. We were sitting on this question for hours and have no idea how to prove it.

Can someone help out with a hint?

Thanks!

$\endgroup$ 1

3 Answers

$\begingroup$

Count the number of edges in two ways.

By the Handshake Lemma: $|E|=\frac{1}{2}\sum\limits_{i=1}^\infty i|V_i|$.

By the characterization of trees: $|E|= \sum\limits_{i=1}^\infty |V_i|-1$.

Equate these two quantities, and remember that $|V_1|$ is the number of leaves.

$\frac{1}{2}\sum\limits_{i=1}^\infty i|V_i|= \sum\limits_{i=1}^\infty |V_i|-1,$

$\sum\limits_{i=1}^\infty (i-2)|V_i|+2=0$

$-|V_1|+0+\sum\limits_{i=3}^\infty (i-2)|V_i|+2=0$

$|V_1|=\sum\limits_{i=3}^\infty (i-2)|V_i|+2$

which is what you are after.

$\endgroup$ 1 $\begingroup$

Prove it by induction on $n$:

It's trivial for the case $n=2$. (Why?)
The inductive step is to add a vertex and, because it's a tree, you've changed at most three of the $V_i$. (Which ones? How have they changed? Understand this before proceeding.) In the sum, then, you're adding $1$ to the left-hand side and $-(k-2)+(k+1-2)$ for some $k$ to the right, which maintains the equality.

$\endgroup$ 2 $\begingroup$

There is a close relationship between your problem and the degree sum formula for graphs. For any finite graph, tree or otherwise, the equation

$$2|E| = \sum_{v\in V} \deg(v)$$

always holds. If you haven't proved this before, it's worth stopping to think about why the equation is correct.

It is possible to transform the equation

$$|V_1| = 2 + \sum (j-2)|V_j|$$

into an explicit statement of the degree sum theorem for trees. Hint: $j|V_j|$ is just another way to write $\sum_{v\in V_j} \deg(v)$.

$\endgroup$