EECS 203: DISCRETE MATHEMATICS
Homework 6 Solutions
1. (4 points) A rooted ternary tree is either a single vertex (which is the root) or consists of three ternary
trees
T
1
,T
2
,T
3
with roots
r
1
,r
2
,r
3
, a new root vertex
r
, and edges
{
r,r
1
}
,
{
r,r
2
}
,
{
r,r
3
}
. What is the
maximum number of vertices in a ternary tree with height
h
? Your proof should be by induction.
Solution: Let
V
(
n
) be the maximum number of vertices in a ternary tree with height
n
. Using the
recursive deﬁnition of ternary tree we have:
V
(0) = 1
V
(
n
) = 3
V
(
n

1) + 1 for
n >
0
In order to make an educated guess at what
T
(
n
) might be, note that the maximum number of
vertices at height 0 is 1, at height 1 is 3, and in general, at height
h
the maximum number of vertices
is 3
h
. A good guess is that
V
(
n
) = 3
0
+ 3
1
+
···
+ 3
n
=
3
n
+1

1
2
. We prove this by induction.
Base case:
V
(0) = 1 =
3
1

1
2
.
General case:
n >
0
Assume inductively that
V
(
n
0
) =
3
n
0
+1

1
2
for all
n
0
< n
. Then:
V
(
n
)
=
3
V
(
n

1) + 1
{
def of
V
}
=
3(
3
n

1
2
) + 1
{
inductive assumption
}
=
3
n
+1

3
2
+ 1 =
3
n
+1

1
2
{
algebra
}
2. (4 points) The sequence (
a
0
,a
1
,...
) is deﬁned recursively as follows:
a
0
= 0
and, for
n >
0,
a
n
=
±
a
n/
2
for
n
even
1 +
a
(
n

1)
/
2
for
n
odd
Prove by induction that
a
n
is the number of 1s in the binary representation of
n
.
Solution:
Let (
b
m
b
m

1
...b
1
b
0
)
2
be the binary representation of
n
(each
b
i
is a bit). Note that (i) if
n
is odd
then
b
0
= 1 and
n

1 = (
b
m
...b
1
0)
2
and (ii) if
n
is even then
b
0
= 0 and
n/
2 = (
b
m
...b
1
)
2
.
Proof by induction: The base case is
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 Fall '07
 YaoyunShi
 Graph Theory, Planar graph, CN, 3 g, 4 g

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