Question

2. Suppose a language is decidable.

• Prove the language is recognizable.

• Prove the language’s complement is also recognizable.

Answer 1

Recognizable Language

A electronic computer M acknowledges language L if L = L(M). we are saying L is Turing-recognizable (or merely recognizable) if there's a metal M specified L = L(M).

Decidable Language

A electronic computer M decides language L if L = L(M) and M halts on all inputs. We are saying L is decidable if there's a metal M that decides L. Call issues area unit issues that need a yes/no answer on a given input

They need a precise correspondence to languages: L could be a illustration of downside P if And provided that an input x ∈ L if declare x is affirmative in downside P.

Prove the language’s complement is additionally recognizable.

Definition: A language is named semi-decidable (or recognizable) if there exists AN algorithmic program that accepts a given string if and provided that the string belongs thereto language. just in case the string doesn't belong to the language, the algorithmic program either rejects it or runs forever.

Clearly, any decidable language is recognizable. we tend to still have to be compelled to see whether or not or not there area unit recognizable languages that aren't decidable, and whether or not or not there area unit languages that aren't recognizable.

Proposition: The recognizable languages area unit closed underneath union and intersection.

Proof: Let L and M be languages that area unit recognized by algorithms A and B severally. so as to make a decision their union (or intersection) merely run A and B in parallel on a similar given input string. The input string is accepted once either one (or each, respectively) accepts it.

Proof: sure, a decidable language is recognizable. Moreover, if a language is decidable, and then therefore is its complement, and thus that complement is recognizable.

Now suppose a language L and its complement is recognizable. Let A be a recognizer for L, and B for its complement. a call technique for L is obtained by running A and B in parallel on a given input string. just in case A accepts the string, it's accepted as a member of L, and just in case B accepts it, it's rejected as member of L. one in all these outcomes can occur among a finite quantity of your time.