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CS 590 Final Name: This is a closed b o ok, closed notes examination, with the following ex- ception: you may refer to a single 8.5x11 inch (or A4) sheet of notes|with anything on it which you think might b e useful. You may also use as much blank scratch pap er as you need. This exam is worth 100 p oints, and each problem is worth 10 p oints. It is a two hour exam. Note: answers with pro ofs, or at least plausible arguments, will receive more credit than those without.
CS 590 Theory of Computation. Test T
May 1993
Problem T-1: Let F in = fe : We is niteg. Is F in recursive?
Answer: F in is a non-trivial index set, so it is not recursive by Rice's theorem.
Problem T-2: Construct a set which is undecidable relative to K |that is, it is unde- cidable even when K is used as an oracle. (Hint: diagonalize).
Answer: Let We (K ) b e the set you can recognize with Turing machine Me and oracle K. Let K (K ) = fe : e 62 We (K )g. K (K ) cannot b e recognized by any machine Me even with the oracle, since it di ers from every such set on the value e.
Problem T-3: Prove that there is a total, recursive function m such that
Em(x) = f 0 ; 1 ; 2 x^ ; 3 x^ ; : : :g
Answer: Let m b e such that m(x) (y ) = y x^. There is an f as required.
Problem T-4: Prove that if LO G = N LO G then D S P AC E (n) = N S P AC E (n).
Answer: A simple padding argument.
Problem T-5: Prove that there is an oracle A such that P (A) = P H (A). (Hint: P H (A) P S P AC E (A) for all A).
Answer: Let A = QB F. Then P (A) P H (A) P S P AC E (A) = P S P AC E P (A) The rst relation follows from the fact that 0 p (A) = P (A) pk (A) P H (A) for any oracle A|which follows from the fact that the pro of of Wrathall's theorem relativizes. Similarly, P H (A) P S P AC E (A) for any oracle, since the pro of that P H P S P AC E relativizes. We know that P S P AC E (QB F ) = P S P AC E as we saw in class. The last relation follows from the fact that every set in PSPACE is decidable in p olynomial time with access to QBF, since QBF is PSPACE-complete.
Problem T-6: Supp ose the function f is computable using p olynomial time (i.e., f 2 P F ). How complex is the set Gf = fy : f (x) = y for some xg?
Answer: It is in N P by Wrathall's theorem, since
Gf = fy : 9 p^ x(f (x) = y )g
and the predicate in parentheses is p olynomial time decidable. Note that f (x) is at most p olynomial longer than x, since only p olynomial time is available to compute f. That is, for some p olynomial p, jxj p(jf (x)j).
Problem T-7: Prove that pm is re exive and transitive (i.e., for any A, B , and C , A pm A and if A pm B and B pm C then A pm C ).
Answer: See the b o ok.
Reject Else reject Accept
This algorithm is clearly p olynomial time, since each simulation is p olynomial time. It is correct b ecause it can guess an accepting path only if one exists, in which case x 62 A. On the other hand, if no accepting path exists, in which case x 2 A, then this algorithm accepts. So, this is an N P algorithm for A. So, A 2 N P P and N P 6 = P.
Answer: First, the enumeration of DTMs may not b e e ective in p oly time, so verifying the guess may not b e p oly time since just getting a description of Mn from n may take to o long. Second, even if a go o d enumeration were chosen, the length of the compu- tation path is the p olynomial b ounding Mn , and this will di er from machine to machine. So, we cannot conclude that there is a single p olynomial which b ounds the runtime of this algorithm.
