Cs141 Home/Hwk1Soln

1. Ordered from slowest growing to fastest growing:

1/n < 2¹⁰⁰ < log log n < log² n < n^0.01 < ⌈ sqrt(n) ⌉ < 2^{log₂ n} < 6n log n < 4n^3/2 < 2²ⁿ

2.

	1 Hour	2 Hours
log n	236*108	272*108
sqrt(n)	1296*1019	5184*1019
n	36*108	72*108
n log n	1.33*108	2.58*108
n2	60000	8.49*104
n3	1532.62	1930.98
2n	31	32
n!	12	13

3.

We want the following inequality to hold for a constant c when n ≥ n₀:

n⁵ + 5n⁴ + 10n³ + 10n² + 5n + 1 ≤ cn⁵.

Dividing both sides by n⁵, this is the same as

(n⁵ + 5n⁴ + 10n³ + 10n² + 5n + 1)/n⁵ ≤ (cn⁵)/n⁵.

Simplifying, this is the same as

1 + 5/n + 10/(n²) + 10/(n³) + 5/(n⁴) + 1/(n⁵) ≤ c.

The inequality holds when n=1 and c=32 (check).
All terms except 1 on the left side are decreasing functions of n.
Therefore when n ≥ n₀ = 1, c = 32, the inequality holds. This completes the proof.

4.

Algorithm (Input: X = (a₁, a₂, a₃,.. a_n), Output: A = (A₁, A₂, ..., A_n)
1. A[1] = X[1]
2. print A[1]
3. for i=2 to n
4. ___ A[i] = A[i-1] + X[i]
5. ___ print A[i]
6. end for

Complexity: The algorithm clearly runs in O(n) time.

Correctness: Clearly, A[i] = a[1] + a[2] + ... + a[i]. (prove by induction on i if you want to be formal).