O
Ryan Davis
EECS 310 Problem Set 7
1.a) The maximum number of nodes of height h in a kary tree is k
h
for all h <= H and 0 for all h > H.
We will first show that the number of nodes of height h in a kary tree of arbitrary k is maximized when all interior nodes
in the tree have k children. We show this by contradiction. Assume that the number of nodes of height h <= H is not
maximized when each interior node has k children. Then it must be possible to increase the number of nodes at height h.
But every interior node has k children, so we cannot add any nodes to the tree at any height h <= H, contradicting our
initial assumption. Thus, the number of nodes of height h in the tree is maximized when all interior nodes in the tree have
k children.
Next, we build on the previous argument to prove by induction that the maximum number of nodes of height h in a kary
tree is k
h
.
Base Case: H = 0. The height of the tree is zero, so the tree is simply the root node. Since there is only one node in the
graph, the maximum number of nodes at h = 0 is 1, which equals k
0
for an arbitrary k. For all h > 0, the number of nodes
of degree h is also zero.
Inductive Step: Assume that the maximum number of nodes of height h in a kary tree is k
h
for all h <= H and 0 for all h >
H. We now perform induction on the maximum height of the tree and consider a graph with maximum height H + 1 (call
this G
H+1
and the graph with maximum height H G
H
). By the inductive hypothesis, we know that the maximum number of
nodes at height H in G
H
is k
H
. For G
H+1
, each leaf of the graph with maximum height H becomes an interior node with k
children, because we know that each interior node must have k children and each leaf of G
H
has height H. For each of the
k
H
leaves in G
H
, we add k nodes as children, so the total number of nodes at height H+1 in G
H+1
is k
H
*k = k
H+1
. Since G
H
is a subgraph of G
H+1
, the number of nodes at height h <= H in G
H+1
remains the same as in G
H
, k
h
. For h=H+1, we have
shown that the maximum number of nodes at h=H+1 is k
H+1
, thus completing the proof. QED.
b) A ternary tree is a kary tree with k =3. We know from part a) that the total number of nodes at height h <= H is k
h
. We
can find the total number of nodes in the tree by summing over all heights of the tree the number of nodes at that height:
Σ
h=1,…, H
3
h
The closed form solution of the above summation is (3
H+1
–
1)/2. We prove this by induction on H for the general case,
Σ
h=1,…, H
k
h
= (k
H+1
–
1)/(k
–
1)
Base Case: H = 0
Σ
h=1,…, H
k
h
= k
0
= 1 = (k
0+1
–
1)/(k
–
1) = (k
–
1)/(k
–
1) = 1
Inductive Step: Assume
Σ
h=1,…, H
k
h
= (k
H+1
–
1)/(k
–
1) holds for all k. We now prove for h+1.
Σ
h=1,…, H
k
h+1
= (k
H+1
–
1)/(k
–
1) + k
H+1
(k
H+1
–
1)/(k
–
1) + (k
–
1)k
H+1
/(k
–
1)
(k
H+1
–
1 + (k
–
1)k
H+1
)/(k
–
1)
(k
H+1
–
1 + k
H+2
–
k
H+1
) )/(k
–
1)
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(k
H+2
–
1)/(k
–
1) = (k
H+1
+1
–
1)/(k
–
1). QED
2.a) The number of squares in the grid is
Σ
i = 1,…, n
i
2
. We can use the same argument as in part b, except that both
summation operators sum over the same set. The total number of rectangles in the grid is
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 Fall '11
 N/A
 Economics, Summation, Natural number, GH+1

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