Is Eqcfg Turing recognizable?

Otherwise, exactly one of the CFGs generates the string and the other CFG does not, so the CFGs are not equivalent, and the TM accepts. Thus, D is Turing- recognizable. We showed in a previous homework that the class of Turing-recognizable languages is closed under union, so EQCFG is Turing-recognizable.

Similarly one may ask, is ETM Turing recognizable?

Since ETM is Turing-recognizable, this means that ATM is also Turing-recognizable (using Theorem 5.2 in the text), a contradiction (to Corollary 4.17). Alternatively, we could use Corollary 5.23 to derive a contradiction to (a). (b) L2 = {(M) | M is a Turing machine that halts on an empty input}.

Subsequently, question is, is ATM recognizable? Because we know that ATM is recognizable, our theorem implies that ATM and ATM are both decidable. But we know that ATM is not decidable. This is a contradiction, hence ATM cannot be recognizable. The language ATM and its undecidability (including proof).

Additionally, what is co Turing recognizable?

Intuitively, if a language is co-Turing-recognizable, it means that there is a computer program that, given a string not in the language, will eventually confirm that the string is not in the language. It might loop infinitely if the string is indeed within the language, though.

Is EQTM Decidable?

Theorem: EQTM is undecidable. We are getting tired of reducing A TM to everything. Let's try instead a reduction from EMPTY TM to EQTM.

Can a Turing machine accept the empty string?

If a TM(Turing Machine) accepts NO input string(even the blank), then its language is empty. If a TM ONLY accepts the blank string(meaning that there is nothing on the tape except for the default blank characters), then its language has only one item and it is the blank string.

How do you prove a language is decidable?

By definition, a language is decidable if there exists a Turing machine that accepts it, that is, halts on all inputs, and answers "Yes" on words in the language, "No" on words not in the language. Therefore one way of showing that a language is decidable is by describing a Turing machine that accepts it.

How do you prove Turing recognizable?

Prove that the language it recognizes is equal to the given language and that the algorithm halts on all inputs. To prove that a given language is Turing-recognizable: Construct an algorithm that accepts exactly those strings that are in the language. It must either reject or loop on any string not in the language.

Are all languages Turing recognizable?

3 Answers. A language is Recognizable iff there is a Turing Machine which will halt and accept only the strings in that language and for strings not in the language, the TM either rejects, or does not halt at all. Note: there is no requirement that the Turing Machine should halt for strings not in the language.

How do you prove Unrecognizability?

Proof: “Only-If” direction: if L is decidable then it is automatically Turing-recognizable. Also, if L is decidable, then L is also decidable, and so L is also Turing-recognizable.

How do you calculate Decidability?

A language is decidable if and only if it and its complement are recognizable. Proof. If a language is decidable, then its complement is decidable (by closure under complementation). Either w ∈ L, or w ∈ L .

What is meant by halting problem?

In computability theory, the halting problem is the problem of determining, from a description of an arbitrary computer program and an input, whether the program will finish running, or continue to run forever.

What is undecidable language?

undecidable language. (definition) Definition: A language for which the membership cannot be decided by an algorithm --- equivalently, cannot be recognized by a Turing machine that halts for all inputs. See also decidable language, undecidable problem, decidable problem.

What does it mean to be Undecidable?

Definition: A decision problem is a problem that requires a yes or no answer. Definition: A decision problem that admits no algorithmic solution is said to be undecidable. No undecidable problem can ever be solved by a computer or computer program of any kind. It means we can never find an algorithm for the problem.

Is halt recognizable?

The halting problem is not in co-recognizable. In other words, no Turing machine can recognize all Turing machines that never halt. Proof. The halting problem is recognizable but not decidable.

What is the difference between a decidable language and a Turing recognizable language?

Recognizable Language A Turing machine M recognizes language L if L = L(M). We say L is Turing-recognizable (or simply recognizable) if there is a TM M such that L = L(M). Decidable Language A Turing machine M decides language L if L = L(M) and M halts on all inputs.

What is mapping reducible?

Mapping Reducibility is the use of a computable function to convert instances of problem A to instances of problem B. A function f : Σ*→Σ* is a computable function if some Turing machine M, on every input w, halts with just f (w) on its tape.