Question

19. (1 point) Suppose that L is undecidable and L is recognizable. Which of the following could be false? A. I is co-Turing recognizable. B. I is not recognizable. C. I is undecidable. D. L* is not recognizable. E. None of the above. 20. (2 points) Let ETM {(M)|L(M) = 0} and EQTM = {(M1, M2)|L(Mi) = L(M2)}. We want to show that EQTM is undecidable by reducing Etm to EQTM and we do this by assuming R is a decider for EQTM and construct a decider, S for Etm which contradicts that ETM is undecidable and so EQTM must also be undecidable. Which of the following is a correct construction for S? A. 1. On input (M) where M is a TM, construct M1 where Mi is a TM which rejects all strings. 2. Run R on (M, Mı). If R rejects, reject. If R accepts, accept. B. 1. On input (M) where M is a TM, construct M1 where M is a TM which accepts all strings. 2. Run R on (M, Mı). If R rejects, accept. If R accepts, reject. C. 1. On input (M) where M is a TM, construct M1 where M is a TM which accepts a single string w # €. 2. Run R on (M, M1). If R rejects, accept. If R accepts, reject. D. Both A and B. E. All of A, B and C.

Answer 1

19.

Option B

B is false, as it goes against the rules of Recognizable.

20.

Option D

Both A and B are valid if you pay attention to the way TM rejects or accepts all the strings. It makes it undecidable.

comment for any further clarifications!