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).

5.

(tasters) Let n be the number of bottles. Let k = ⌈ log₂ n ⌉ and choose k testers, call them 1,2,..,k. There are 2^k ≥ n subsets of {1,2,...,k}. For i = 1,2,...,n, pair the i'th bottle with the i'th subset S_i of {1,2,...,k}, and have the tasters in that subset taste the i'th bottle.

If the i'th bottle is poisoned, then at the end of the month, exactly those tasters in the set S_i will die. This will identify which bottle is poisoned.

(missing number) Here are two algorithms. Let the array be A[1..n-1].

Algorithm 1:

1. Compute T = 1+2+3+...+n and S = A[1] + A[2] + ... + A[n-1]. Then T-S is the missing number.

Algorithm 2:

1. Use the partition procedure from quicksort to reorder the array so that the numbers less than n/2 are in the first half of the array, while the other numbers are in the second half of the array.
2. By counting the numbers in the two halves of the array, determine whether the missing number is less than n/2 or ≥ n/2.
3. Recurse on whichever half of the array the missing number is missing from.

Total time: n + n/2 + n/4 + n/8 + ... = O(n)