• Please give details of your calculation. A direct answer without explanation is not counted.
• Your answers must be in English.
• Please carefully read problem statements.
• During the exam you are not allowed to borrow others’ class notes.
• Try to work on easier questions first.
• This is the final exam of Prof. Chih-Jen Lin’s Introduction to the Theory of Computation. If you take Prof. Jieh Hsiang’s class, please go to R104.
Problem 1 (30 pts)
Given an example of f and g such that (a) f , g are not polynomials
(b) f = O(g) (c) f = o(g)
Answer
If
f (x) = ex g(x) = e2x, which are not polynomials, then
x→∞lim f (x)
g(x) = lim
x→∞
ex
e2x = lim
x→∞e−x = 0.
As a result, f (x) = o (g(x)). Furthermore, we have ex < e2x for all x > 0, so we have f (x) = O(g(x)) with c = n0 = 1.
Common Mistakes
You cannot just give f (x) and g(x) without any explanation.
1
Problem 2 (30 pts)
Let
CRE = {hR, ki | R is a regular expression. L(R) contains exactly k strings where k ≥ 0 or k = ∞.}.
Is this language decidable?
Answer
Yes.
1. Given input hR, ki, then we can convert R to a DFA D, and remove all state which can not reach qaccept or can not be reached from q0.
2. If there is a loop in D:
• If k = ∞, then accept.
• If k 6= ∞, then reject.
3. Generate all (only finite many) accepted strings of D.
4. If # generated strings = k, then accept. Otherwise, reject.
Another Solution
Check “∗” in regular expression instead of converting the regular expression to a DFA, but also need to check it is neither ∗ nor any other equivalent expression.
Problem 3 (30 pts)
Define
f (n) = 2o(g(n)) if
n→∞lim
log2f (n) g(n) = 0.
Prove or disprove the following statements:
(a)
f (n) = 2o(g(n)) is equivalent to
f (n) = o(2g(n)).
2
(b)
f (n) = 2O(g(n)) is equivalent to
f (n) = O(2g(n)).
Answer
(a) No. If
f (x) = 2x g(x) = 2x, then
x→∞lim f (x)
2g(x) = lim
x→∞
2x
22x = lim
x→∞2−x= 0, which means f (x) = o 2g(x).
However,
x→∞lim
log2f (x)
g(x) = lim
x→∞
x 2x = 1
2, so f (x) 6= 2o(g(x)).
(b) No. If
f (x) = 22x g(x) = x, then
2x = O(x), which means f (x) = 22x = 2O(x) = 2O(g(x)).
However, for all positive integers n0 and c, there exists a n = max (n0 + 1, log2(c)) > n0, such that f (x) = 22x ≥ 2x× 2log2(c) ≥ c2x = c2g(x) (because 2x is strictly increasing). So f (x) 6= O 2g(x).
Common Mistakes
1. To show that two things are not equivalent, you need to give “examples.”
2. We of course assume f (n) and g(n) are positive functions.
3
Problem 4 (10 or 40 pts)
(a) (10 pts) Calculate 2, 011 × 530, 528
(b) (Bonus 30 pts) Find 55, 983, 881 = p × q, where p, q > 1 and p, q ∈ N.
Answer
(a) 2, 011 × 530, 528 = 1, 066, 891, 808.
(b) 55, 983, 881 = 5, 281 × 10, 601.
4
