Question

1. Let n be a positive integer. Classify the languages (i) R = {(M)IM is a TM and L(M) contains exactly n strings) (ii) S- (M)|M is a TM and L(M) contains more than n strings as (a) decidable, (b) Turing-recognizable but not co-Turing-recognizable, (c) co-Turing-recognizable but not Turing-recognizable, (d) neither Turing-recognizable nor co-Turing-recognizable. Justify your answers.

Answer 1

Solution Answer: R -> Neither Turing-recognizable nor co-Turing-recognizable S -> Turing-recognizable Explanation S is Turing-recognizable because we can introduce implementations of M on every feasible nput tape We identify that R is not Turing-recognizable since, if it is, the language S-f(M)M is a TM and L(M)contains no more than n strings) This will be Turing-recognizable which means that both S and S' were decidable given that S ought to be co-Turing-recognizable If R' is co-Turing-recognizable, then we may possibly identify the machines that allow a number of strings dissimilar than n. In specific, assumen - 1. Now we are able to execute the recognizer for machines which languages have apart from n strings on a Turing Machine built to recognize L(M) \{k for all feasible string k.

At every phase, we execute single step of every active machines, and then create a new machine, and do again, thus interleaving implementations and verifying all TMs finally get to execute randomly various steps. Considering-1, accuratdly one of given 's will at-accept and the remaining vwil any halt-reject or execute forever. As a result, a recognizer for L(Mis able to build a recognizer for L(M1. This simplify to L(M)!, I and L(M)-p by deducting every feasible sets of p input strings. Therefore, if R is co-Turing-recognizable, then it will be Turing-recognizaЫе, hence decidable But for given question, it is incorrect. So, it is concluding that R is neither co-Turing- recognizable nor luring-recognizable