Answer Problem1(15pts)

• Please give details of your answer. 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.

Problem 1 (15 pts)

Is (log n)³ = o(n)?

You can either use the formal definition of limit or calculate

n→∞lim

(log n)³

n (1)

However, you must show your steps of calculating (1) rather than just give your answer. We assume natural log here.

Answer

Yes.

n→∞lim

(log n)³

n = lim

n→∞

3(log n)²

n = lim

n→∞

6(log n)

n = lim

n→∞

6 n = 0

You should always be careful in writing mathematical derivations. Otherwise, we should have proved P = PN or not. Here is an example.

n→∞lim log n

n = 0

n→∞lim

(log n)³

n = 0 · lim

n→∞(log n)² = 0 We don’t have

0 · ∞ = 0

Problem 2 (30 pts)

1. Assume

f₁(n) = O(g₁(n)) and f₂(n) = O(g₂(n)).

Is

f₁(n) × f₂(n) = O(g₁(n) × g₂(n))?

2. Assume

f₁(n) = o(g₁(n)) and f₂(n) = o(g₂(n)).

Is

f1(n) + f2(n) = o(g1(n) + g2(n))?

You need to formally prove your answer.

Answer

1. Yes. By f₁(n) = O(g₁(n)) and f₂(n) = O(g₂(n)), there exist positive c₁, c₂, n₁, n₂ such that

f₁(n) ≤ c₁g₁(n), ∀n ≥ n₁, and f₂(n) ≤ c₂g₂(n), ∀n ≥ n₂.

Take c = c₁c₂ and n₀ = max(n₁, n₂), then for every integer n ≥ n₀, f₁(n) × f₂(n) ≤ c₁c₂g₁(n)g₂(n) ≤ c(g₁(n) × g₂(n)).

Therefore, f₁(n) × f₂(n) = O(g₁(n) × g₂(n)).

2. Yes. By f₁(n) = o(g₁(n)) and f₂(n) = o(g₂(n)),

n→∞lim f₁(n)

g1(n) = 0, and

n→∞lim f₂(n) g₂(n) = 0.

Then

n→∞lim

f₁(n) + f₂(n) g₁(n) + g₂(n)

= lim

n→∞

f₁(n)

g₁(n) + g₂(n) + lim

n→∞

f₂(n) g₁(n) + g₂(n)

≤ lim

n→∞

f₁(n)

g₁(n) + lim

n→∞

f₂(n) g₂(n)

=0 + 0 = 0

Therefore, f1(n) + f2(n) = o(g1(n) + g2(n)).

Another Solution

∀c > 0, ∃n₁ such that f₁(n) ≤ cg₁(n), ∀n ≥ n₁

∀c > 0, ∃n2 such that f2(n) ≤ cg2(n), ∀n ≥ n2

Therefore,

∀c > 0, ∃n₃ = max(n₁, n₂) such that f₁(n) + f₂(n) ≤ c(g₁(n) + g₂(n)), ∀n ≥ n₃

Common Mistakes

1.

∀c₁ > 0, · · ·

∀c₂ > 0, · · · Choose

c = max(c₁, c₂) · · · 2. You need to explain why

n→∞lim

f₁(n)

f₁(n) + g₁(n) = 0

Problem 3 (15 pts)

If

f (n) = 2^O(1), then is

f (n) = O(1)?

Give a formal proof or a counter example.

Answer

Yes. By f (n) = 2^O(1), there exist positive c and n₀ such that for every integer n ≥ n₀ f (n) ≤ 2^c1.

Take c⁰ = 2^c, then for every integer n ≥ n₀,

f (n) ≤ 2^c= c⁰1 Therefore, f (n) = O(1).

Common Mistakes

A wrong solution

O(1) = c, where c is a constant f (n) = 2^O(1) = 2^c

2^c= O(1)

f (n) = 2^O(1) = O(1) However

O(1) 6= c

when we say

f (n) = O(1) we know

f (n) ≤ a constant.

So we cannot write

O(1) = c, where c is a constant.

For example,

f (n) =

(1, if n is even 2, if n is odd then

f (n) = O(1).

But

f (n) 6= any constant.

Problem 4 (10 pts)

Give an example which is not Turing-recognizable. You need to explain why.

Answer

Denote the halting problem as H. ¯H is an example that is not Turing-recognizable. By Theorem 4.22, a language is not decidable iff either it or its complement must be not Turing-recognizable.

H is Turing-recognizable, but not decidable. Therefore, ¯H is not Turing-recognizable.

Common Mistakes

L = {w|w is an infinite string satisfying . . . } Because w is infinite, the language is not Turing-recognizable.

Note that any string in a language must be finite.

Problem 5 (15 pts)

If A and B are Turing-recognizable, is A ◦ B Turing-recognizable? Please prove it or give a counter example.

Answer

Let M_A, M_B be Turing machines that recognize A, B, respectively. We prove that A ◦ B is Turing-recognizable by constructing a Turing machine M_AB that recognizes A ◦ B as follows.

i Assume the input string is s.

ii Non-deterministically split s into sub-strings s_a and s_b in all possible ways.

iii Run M_A on s_a. If it rejects, reject.

iv Run M_B on s_b. If it accepts, accept. If it rejects, reject.

Common Mistakes

1. Enumerate all possible splits of the input string. The problem is that TM may get to an

∞ loop in one split, but a later one is accepted. What you can do is to run all splits up to only i steps, where i = 0, 1, 2, · · · .

2. Run M_A on w until accept, and then run B on w until remaining string. If M_B accept, then accept.

This is incorrect. For example,

A = {0^k1^k|k ∈ N } B = {11}

w = 001111 MA will not be in accept state after it read 0011.

Problem 6 (15 pts)

If A is Turing-decidable, is A^∗ Turing-decidable? Please prove it or give a counter example.

Answer

Let MA be a Turing machine that decides A. We prove that A^∗ is Turing-decidable by con- structing a Turing machine MA^∗ that decides A^∗ as follows.

i. Assume the input string is s.

ii. Non-deterministically split s into sub-strings s1, s2, · · · in all possible ways.

iii. Run M_A on s₁, s₂, · · · , respectively.

iv. If M_A accepts each substring, then accept. Otherwise, reject.

For this problem you can instead enumerating all possibilities without worrying ∞ loops.

Problem 7 (5 or -10 pts)

In Chapter Seven (or in the lecture) we mentioned one of the greatest unsolved computer science problem. What is it?

If you correctly answer this question, you get 5 points. Otherwise, you get −10 points.

Answer

NP = P or NP 6= P

Problem 8 (5 or 25 pts)

1. (5 pts) Calculate 38, 239 × 7, 283.

2. (Bonus 20 pts) Find 33, 742, 691 = p × q, where p, q > 1 and p, q ∈ N . If you think your solution is correct, you need to let TAs know during the exam. They will give you another similar problem. If both are correct, you get the bonus 20 points.

Answer

1. 38, 239 × 7, 283 = 278, 494, 637.

2. 33, 742, 691 = 3, 779 × 8, 929