Solutions
Homework 1
1. Solution 1. Notice that for any value of j and k, i iterates k − j + 1 times. So given a
specific value of j, since k goes from j to n, the number of times i iterates is
1 + 2 + · · · + (n − j + 1) =
(n − j + 1)(n − j + 2)
(n − j + 1)(n − j)
≥
2
2
So
total number of iterations ≥
n
X
(n − j + 1)(n − j + 2)
2
j=1
n
1X 2
=
(n + n) − (2n + 1)j + j 2
2 j=1
n(n + 1) n(n + 1)(2n + 1)
1
2
+
n(n + n) − (2n + 1)
=
2
2
6
2
n(n − 1)
=
6
Now set
n(n2 − 1)
≥ cn3
6
and solve this inequality for n. We get
1
n≥ √
.
1 − 6c
Therefore, if we let c = 1/12 and n0 = 2, we have that
total number of iterations ≥ cn3
whenever n ≥ n0 .
So the runtime is Ω(n3 ).
Solution 2. We have
total number of iterations =
n X
n X
k
X
1
j=1 k=j i=j
≥
n/3
2n/3
n
X
X
X
1
j=1 k=2n/3 i=n/3
≥
n3
.
27
Therefore, if we let c = 1/27 and n0 = 1, we have that
total number of iterations ≥ cn3
whenever n ≥ n0 .
So the runtime is Ω(n3 ).
1
Solutions
Homework 1
Solution 3. Notice that the pseudocode iterates through all possible triplets (j, i, k) such
that j ≤ i ≤ k and such i, j, k each has a value between 1 and n. If we have ignore the
duplicate values such as (2, 2, 3), then the total number of such triplets is
n
n(n − 1)(n − 2)
=
.
3
6
Set
n(n − 1)(n − 2)
≥ cn3 .
6
Let c = 1/12. This simplifies to
n2 − 6n + 4 ≥ 0
Now it’s easy to see that this inequality holds for all n ≥ 6. Therefore, if we let c = 1/12
and n0 = 6, we have that
total number of iterations ≥ cn3
whenever n ≥ n0 .
So the runtime is Ω(n3 ).
2. If n0 = 59, then 10n log n < n2 whenever n ≥ n0 .
3. All answers below are approximate.
1 Second
log n
10300000
√
n
1012
n
106
n log n 6 × 104
n2
103
n3
102
n
2
20
n!
9
1 Hour
9
1010
2 × 1019
4 × 109
108
6 × 104
2 × 103
32
12
1 Month
11
109×10
9 × 1024
3 × 1012
8 × 1010
2 × 106
2 × 104
41
15
1 Century
14
109×10
9 × 1030
3 × 1015
7 × 1013
5 × 107
2 × 105
51
17
4. reverse(A):
s = stack()
for i from 1 to len(A):
s.push(A[i])
for i from 1 to len(A):
A[i] = s.pop()
The running time is 2n or O(n).
2