Description
1. Show that for any π β₯ 3, if a tree π has fewer than π leaves, then the
maximum degree Ξ(π) among the vertices of π must satisfy Ξ(π) < π. It
can help to consider the summations
π = βππ
β
π=1
; 2(π β 1) = total degree = βπππ
β
π=1
.
The phrase βπ has fewer than π leavesβ means π1 < π.
The two sums can be combined into the single sum
β(2 β π)ππ = 2
π
π=1
It suffices to show that ππ = 0 for all π β₯ π.
2. Let (π, π) be a rooted tree. Recall that the level of a vertex π₯ is πΏ(π₯) =
π·(π, π₯). Also, the height of a rooted tree π» is the maximum of the levels of
its vertices.
a. Show that if π is on the unique π’, π£-path, then π·(π’, π£) = πΏ(π’) +
πΏ(π£).
b. Show that if πΏ(π’) + πΏ(π£) = π·(π’, π£), then π must be on the unique
π’, π£-path.
c. Show that for any two vertices π’ and π£, π·(π’, π£) β€ 2π».
d. Show that if π·(π’, π£) = 2π», then π’ and π£ must be non-parents.
Equivalently, you can show that if either π’ or π£ is a parent, then
π·(π’, π£) < 2π».
3. Suppose (π, π) is a rooted π-ary tree where every parent has exactly π
children; such a tree is said to be saturated.
a. Show that π has ππ edges for some integer π.
b. Find a formula for the number of vertices of π in terms of π, π.
c. Find a formula for the number of non-parents in terms of π, π.
4. Suppose (π, π) is a rooted tree with exactly 1012 edges. Recall that a lower
bound or an upper bound on π» is tight if there exists an example π where
that bound is attained.
a. Find tight lower and upper bounds for π», the height of π.
b. Find tight lower and upper bounds for π» if π is a saturated rooted
binary tree. Recall that saturated means every parent has the
maximum allowed number of children; here, that number is 2.