There exists a Turing machine that accepts $\epsilon$. You are probably confusing it with a decider.

Awesome q2a theme

+1 vote

As per the definition of a Turing Machine, it does not accept Epsilon. But, A → ϵ can be generated by a Unrestricted Grammar. So, how can we say that Turing Machines are the acceptors for Recursively Enumerable languages generated by Unrestricted Grammar?

0

There exists a Turing machine that accepts $\epsilon$. You are probably confusing it with a decider.

0

it does not accept Epsilon

**Citation needed.**

There very much exist TM's that recognize $\epsilon$. Trivial example would be one that has a transition $q_{initial}, B, B, R/L, q_{final}$

If you cannot digest that because it is too trivial - you can also build a machine that upon reading $\epsilon$, performs a prime factorization of some unique large number while going through a hundred different states and then accepts. Net result would still recognizing $\epsilon$.

8,998 questions

3,130 answers

14,378 comments

95,816 users