Computer Science 124: Problem Set 1

Collaborators
Computer Science 124: Problem Set 1
Christian Anderson
February 11, 2013
I collaborated with Boom the Thunder Dragon on problem 4.
1
No solution released. I include this section so that the example L ATEX is realistic.
2
No solution released. I include this section so that the example L ATEX is realistic.
3
No solution released. I include this section so that the example L ATEX is realistic.
4
For this problem, I will use the following two definitions:
4.1
• Definition of O: f is O(g) when there exist a positive real number c and a positive
integer N such that for all n > N, f(n) ≤ cg(n).
• Definition of o: f is o(g) when
f(n)
lim
n→∞ g(n)
Let f1(n) = n. We claim that f1(2n) is O(f1(n)). Let c = 2 and let N be any positive
integer. For all n > N, f1(2n) = 2n ≤ 2n = cf1(n).
1
= 0.

4.2
Let f2(n) = 2 n . We claim that f2(2n) is not O(f2(n)). We give a proof by contradiction.
Suppose that f2(2n) is O(f2(n)). Then there exist c and N satisfying the above definition.
This gives that for all n > N, 2 2n ≤ c·2 n . Dividing both sides by 2 n gives that for all n > N,
2 n ≤ c. This is a contradiction.
4.3
We know that there exist (c1, N1) such that for k > N1, f(k) ≤ c1g(k). Similarly, there
exist (c2, N2) such that for ℓ > N2, g(ℓ) ≤ c2h(ℓ). Let N = max(N1, N2). Then for n > N,
f(n) ≤ c1g(n) and g(n) ≤ c2h(n). Composing these equations gives that for n > N,
f(n) ≤ c1c2h(n). Thus f(n) is O(h(n)).
4.4
This statement is false. We give a counterexample. Let f(n) = n. Let g(n) be defined
differently on the even and odd positive integers. For even n, g(n) = 1. For odd n, g(n) = n2 .
An alternative way to state this is
1 for even n
g(n) =
n2 for odd n.
We see that f(n) is not O(g(n)) because n is not O(1). Likewise, g(n) is not O(f(n)) because
n 2 is not O(n).
4.5
This statement is true. We give a proof. Suppose that f is o(g). Then we know that
f(n)
lim
n→∞ g(n)
The definition of the limit (available on Wikipedia and elsewhere) lets us re-write the above
equation as the following: for any ɛ > 0, there exists C such that for all x > C, |f(x)/g(x)| <
ɛ. Choose some c > 0. Then the previous statement implies that there exists a positive
integer N such that for all x > N, f(x) < cg(x). Thus f is O(g).
2
= 0.