Honors Theory, Spring 2002, Yap Homework 7

This is due on Mon April 29.

1. (10 Points) Let w = (01)^{n} be a string. Prove that K(⟨w⟩) ≤ ℓ(n)+C for some C, and K(⟨w⟩|n) = O(1).

SOLUTION: Given input n, we can compute (01)^{n}. Thus K(⟨(01)^{n}⟩) ≤ ℓ(n) + C. This alsow shows that K(⟨w⟩|n) is O(1).

2. (20 Points) Let A ⊆ ℕ be any set. Let χ_{A} = b_{0}b_{1}b_{2} ⋯ be the ω-string such that b_{i} = 1 iff i ∈ A. Write χ_{A}[i : j] for b_{i}b_{i+1} ⋯ b_{j}, and χ_{A}[j] for χ_{A}[0 : j].

(i) For all recursive A, there exists a constant c = c(A) such that K′(χ_{A}[n]) ≤ c. NOTE: K′(x) here is K(x|ℓ(x)), the length-conditioned Kolmogorov complexity function. (ii) For all r.e. A, there is a constant c = c(A) such that K′(χ_{A}[n]) ≤ ℓ(n) + c.

SOLUTION:

(i) Since this is length-conditioned, we must show how to compute χ_{A}[n] when we are only given n. This is easy. Construct a STM M that on input n, runs some decisive Turing machine for A on the inputs i = 0, 1, …, n in order. Then M output χ_{A}[n].

(ii) In case A is r.e., the previous Turing machine M is now modified to N as follows: N checks that its input has the form ⟨n, m⟩ with m > n + 1. If not, it rejects. Otherwise, it dovetails the computations of an acceptor for A on the inputs i = 0, 1, …, n. When m of these computations accepts, then N outputs a string b_{0}b_{1} ⋯ b_{n} where b_{i} = 1 iff the acceptor for A has accepted i. If less than m of these computations accept, N will loop. Clearly, if χ_{A}[n] has m 1’s, then our machine N on ⟨n, m⟩ will output χ_{A}[n]. Thus, K_{N}(χ_{A}[n]|n) = ℓ(m) ≤ ℓ(n) + C.

REMARKS: The proof of (ii) is very revealing about the non-constructiveness of K(x|y). To show K(x) = ℓ(z) where z is minimal program for x, we need not know how to construct z. We only have to show its existence! The m in the above proof is also purely existential.

3. (15 Points) Show that there is some C_{0} such that every minimal programs is C_{0}-incompressible. More precisely, this says: for any x ∈ ℕ, if z is a minimal program for x (i.e., Φ(z) = x and K(x) = ℓ(z)) then K(z) ≥ ℓ(z) − C. HINT: Consider the function f(z) = Φ(Φ(z)). If there is no such C_{0}, then for every C, we can find x such that K(x) − K_{f}(x) > C.

SOLUTION: Suppose no such C_{0} exists. Using the HINT, we know that for each C, there is a x = x(C) such that K(x) − K_{f}(x) > C. This contradicts the universality of K.

4. (25 Points) Recall the lower bound proof that the time-space product is Ω(n^{2}) for any deterministic Turing machine that accept palindromes. Prove the same result when Turing machine is nondeterministic.

SOLUTION: Recall the proof in the lecture notes. Where does the proof break down if M is now nondeterministic? Only in the definition of the “compatibility” of (w, S), where we simulated M on input w “modulo S”. Now, compatibility is defined to mean that “some computation path of M will produce the crossing sequence S”. We may verify that Lemma 11 still holds with this new definition.