CS 70
Discrete Mathematics for CS
Spring 2005
Clancy/Wagner
Final Solns
Problem 1. (8 points)
Let
a
(
n
)
denote the number of ternary strings of length
n
with no pair of adjacent 2’s. Example: “2011” is a
ternary string with no pair of adjacent 2’s.
(a) What are
a
(
1
)
and
a
(
2
)
?
Solution:
a
(
1
) =
3,
a
(
2
) =
8.
Grading:
1 pt for correct answer.
Solution #1:
a
(
n
) =
2
a
(
n

1
) +
2
a
(
n

2
)
for
n
≥
3.
JustiFcation: If the Frst digit is 0 or 1 (2 choices), then the remaining
n

1 digits can be any ternary
string with no 22s (
a
(
n

1
)
choices). If the Frst digit is 2 (1 choice), then the second digit can be 0 or 1
(2 choices), and the remaining
n

2 digits can be any ternary string with no 22s (
a
(
n

2
)
choices).
Solution #2:
a
(
n
) =
3
a
(
n

1
)

2
a
(
n

3
)
for
n
≥
4.
JustiFcation: Take any ternary string of length
n

1 with no 22s, and append 0, 1, or 2 (3
a
(
n

1
)
choices). We need to exclude strings of the form
···
022 and
···
122 (2
a
(
n

3
)
choices); once we’ve
eliminated these, what remains must be exactly the set of ternary strings of length
n
with no 22s.
Common mistakes:
Overcounting. ±or example, some said
a
(
n
) =
3
n

3
n

2
(
n

1
)
, reasoning that
there are 3
n

2
(
n

1
)
ternary strings with a 22, since there are
n

1 places where the 22 could start and
3
n

2
options for the other
n

2 digits in the string. The problem with this reasoning is that it overcounts
strings with multiple 22s; for instance, the string 22022 is counted twice, and the string 22222 is counted
four times.
Grading:
3 pts for a correct recurrence; 1 pt for something with a partial explanation.
(b) Prove that
a
(
n
)
≥
2
n
+
1
for all
n
≥
2.
Solution #1:
Proof by strong induction.
Base cases:
a
(
2
) =
8
≥
2
2
+
1
.
a
(
3
) =
22
≥
2
3
+
1
.
Induction step: Assume
a
(
k
)
≥
2
k
+
1
for
k
=
2
,
3
,...,
n
. Then for
n
≥
3,
a
(
n
+
1
) =
2
a
(
n
) +
2
a
(
n

1
)
(by part (b))
≥
2
·
2
n
+
1
+
2
·
2
n

1
+
1
(by the inductive hypothesis)
≥
2
·
2
n
+
1
=
2
(
n
+
1
)+
1
.
CS 70, Spring 2005, Final Solns
1
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View Full DocumentSolution #2:
For each ternary string of length
n
with no 22s, we can get two unique ternary strings of length
n
+
1 with no 22s by appending either a 0 or a 1. This shows that
a
(
n
+
1
)
≥
2
a
(
n
)
for all
n
≥
1. Now use
induction to show the desired result.
Solution #3:
There are 2
n
ternary strings (of length
n
) with no 2.
There are
n
×
2
n

1
more that have
exactly one 2. Thus there are at least 2
n
+
n
×
2
n

1
ternary strings of length
n
with no 22. Moreover,
2
n
+
n
×
2
n

1
≥
2
n
+
2
×
2
n

1
≥
2
n
+
1
for
n
≥
2.
Grading:
4 pts for an impeccable proof. For induction proofs: 1 pt for all needed base cases and a correctly
stated induction predicate; 3 pts for a correct and fully justi±ed inductive step.

1 pt for backwards reason
ing, for steps with no justi±cation, for small algebraic errors, for confusing what you can assume vs. what
you must prove, or for confusion between integers and booleans (e.g.,
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 Correct Answer, common mistakes, Final Solns

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