CS 173: Discrete Structures, Fall 2010
Homework 8 Solutions
This homework contains 3 problems worth a total of 37 regular points and one bonus point.
1.
More on recurrences [12 points]
(a) (6 points) Derive closed form solutions for the following two recurrences.
i.
T
(0) = 2 and
T
(
n
) = (
n
+ 1)
T
(
n

1)
.
[Solution:]
We will solve this problem using unrolling.
T
(
n
) = (
n
+ 1)
T
(
n

1)
= (
n
+ 1)
·
n
·
T
(
n

2)
= (
n
+ 1)
·
n
·
(
n

1)
·
T
(
n

3)
=
.....
= (
n
+ 1)
·
n
·
(
n

1)
·
(
n

2)
...
(
n

k
)
·
T
(
n

(
k
+ 2))
In order to reach the base case we need
n

(
k
+ 2) = 0. So
k
=
n

2 and we
get follwing.
T
(
n
) = (
n
+ 1)
·
n
·
(
n

1)
...
·
n

(
n

2)
·
T
(0)
= (
n
+ 1)
·
n
·
(
n

1)
...
·
2
·
2
= 2
·
(
n
+ 1)!
ii.
T
(0) = 1 and
T
(
n
) =
T
(
n

1) + 3(
n

1)
2
+ 3(
n

1) + 1
.
[Solution:]
T
(
n
) =
T
(
n

1) + 3(
n

1)
2
+ 3(
n

1) + 1
=
T
(
n

2) + 3(
n

2)
2
+ 3(
n

1)
2
+ 3(
n

2) + 3(
n

1) + 2
=
T
(
n

3) + 3(
n

3)
2
+ 3(
n

2)
2
+ 3(
n

1)
2
+3(
n

3) + 3(
n

2) + 3(
n

1) + 3
=
...
=
T
(
n

k
)
·
3(
n

k
)
2
+
...
+ 3(
n

1)
2
+ 3(
n

k
) +
...
+ 3(
n

1) +
k
In order to reach the base case
n

k
= 0
.
So
n
=
k
and we get following.
1
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View Full DocumentT
(
n
) =
T
(0) + 3(0
2
+ 1
2
+
...
+ (
n

1)
2
) + 3(0 + 1 + 2 +
...
+ (
n

1)) +
n
= 1 + 3(
n

1
X
i
=1
i
2
) + 3(
n

1
X
i
=1
i
) + 3
n

2
n
= 1 + 3(
n

1
X
i
=1
i
2
) + 3(
n
X
i
=1
i
)

2
n
= 1 + 3(
(
n

1)(
n
)(2
n

1)
6
) + 3(
n
(
n
+ 1)
2
)

2
n
= 1 + 3
·
n
(
(
n

1)(2
n

1)
6
+
n
+ 1
2
)

2
n
= 1 + 3
·
n
(
(
n

1)(2
n

1) + 3(
n
+ 1)
6
)

2
n
= 1 + 3
·
n
(
2
n
2
+ 4
6
)

2
n
= 1 +
n
(
n
2
+ 2)

2
n
= 1 +
n
3
+ 2
n

2
n
= 1 +
n
3
(b) (6 points) Give closed form of the following recurrence using recursion tree.
i.
T
(1) = 1 and
T
(
n
) = 4
T
(
n/
2) +
n
3
Assume
n
is power of 2.
[Solution:]
The recursion tree is shown in Figure 1. The height of the tree is
log
2
n
. At each level
i
there are 4
i
nodes and total work done at each level is 4
i
(
n
2
i
)
3
.
T
(
n
) =
log
2
n
X
i
=0
4
i
(
n
2
i
)
3
=
n
3
·
log
2
n
X
i
=0
(
4
i
2
3
i
) =
n
3
·
log
2
n
X
i
=0
(
4
8
)
i
=
n
3
·
log
2
n
X
i
=0
(
1
2
)
i
=
n
3
(
(1
/
2)
log
2
n
+1

1
(1
/
2)

1
) = 2
n
3
(1

(1
/
2)
log
2
n
+1
)
=
n
3
(2

(1
/
2)
log
2
n
) = 2
n
3

n
3
(1
/n
)
= 2
n
3

n
2
2
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 Control flow, Big O notation, Recurrence relation, procedure foo

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