Homework 8 Solutions

1. Homework 8 Solutions Problem 1. Draw a diagram showing the various classes of languages that we have discussed and alluded to in terms of which class contains which other classes. Indicate (by words if necessary) which containments are not known to be strict. Note: this is an open-ended question, but at a minimum include the Kleene hierarchy, the polynomial-time hierarchy, PSPACE, NP-complete languages, and regular languages

2. P NP co-NP logspace NL Regular Languages PSPACE

3. NP = 1p 2p co-NP = 1p P = 0p = 0p 2p • SAT • Clique • co-SAT • co-VC • #SAT #P Non-Collapse has not been proved • TQBF PSPACE complete

4. Decidable=0 = 0 1 1 PSPACE 2 2 EXPTIME

5. Problem 2: State the time hierarchy theorem (consult Sipser). Show the error in the following argument that NP = P:Proof by contradiction. Assume NP = P. Then SAT is in P. Therefore there exists a value k such that SAT is in TIME(nk). Because every language in NP is polynomial-time reducible to SAT, we know that P is a subset of TIME(nk). Yet the time hierarchy theorem tells us thatTIME(nk+1) contains a language that is not in TIME(nk). Therefore our assumption that NP = P must be false.

6. Because every language in NP is polynomial-time reducible to SAT, we know that P is a subset of TIME(nk). Time hierarchy Theorem: If f is a time-constructable function and f(n)>n, then TIME(f(n)) properly contains TIME(f(n)/log(f(n))) This is incorrect because a reduction from L to SAT may transform inputs of size n to a larger size (e.g. nb). So the running time of an algorithm that employs the reduction will be O(nbk) not O(nk)

7. Problem 3: Define “uniform family of circuits” and NC (consult Sipser). A uniform family of circuits {Ci}i=1 is a set of circuits that are can be output by a logspace machine NC = Ui=1 NC(i) where NC(i) is all languages L decidable by some uniform family of polynomial size depth logi(n) circuits. Problem 4: Define “trapdoor function” (consult Sipser). A trapdoor function, primarily used in cryptography, is a function f such that it is efficient to compute f(x) for any x but, given random y, it is not possible to find any value x such that f(x) = y in polynomial time without knowledge of a key k. With knowledge of key k, it is possible to find x efficiently. 8 8

8. Problem 5: Prove that NPSPACE = PSPACE Savitch’s theorem implies that  k, NSpace(nk)  DSpace(n2k) Therefore, NPSpace =  NSpace(nk)   DSpace(n2k)   DSpace(nj) = PSpace Because we already know that PSpace  NPSpace, this implies PSpace = NPSpace    k k j

9. Problem 6. Prove that TQBF restricted to formulas where the part following the quantifiers is in conjunctive normal form is still PSPACE-complete (Sipser 8.13) The idea is simple: in the proof that TQBF is PSPACE-complete, the boolean expression following the quantifiers can always be put in conjunctive normal form. This expression says that “machine M can more from configuration X to configuration Y in one step.”

10. Problem 7. Let EQREX = {(R,S): R an S are equivalent regular expressions}. Show that EQREX PSPACE (Sipser 8.16) A PSPACE algorithm that decides can be written in 2 steps: 1. Convert (R,S) to (Rnfa, Snfa) where (Rnfa, Snfa) are NFAs that accept R and S respectively. This done by using the 3 proof tricks that show regular languages are closed under concatenation, union, and Kleene star, building the NFAs directly from the regular expressions. This is a polynomial time (and hence obviously PSPACE) operation 2. Decide whether Rnfa and Snfa accept the same languages. The idea here is that it is sufficient to determine whether there are any strings w of length less than n -- the max of the 2 NFAs’ pumping lengths -- such that w is accepted by one NFA but not the other. To do this, one can run through all strings of length less than n. This requires exponential time, but polynomial space.