My understanding of decider is that a machine is a decider if it always halts with output $0$ or $1$ (or whatever you prefer meaning "YES" and "NO"). However, I am confused by the definition that decider is a Turing machine that always halts since there are machines that do always halt with different than $0$ and $1$ outputs.

Further, I also know that a Decider is a TM that halts on all possible inputs. Now, my question is how can I visualize a Non determistic Decider? Does a non-determistic decider mean a TM where. All the copies must halt, (OR) At-least one copy halts. Kindly explain in …

2019年6月30日 · A total Turing machine or a decider is a machine that always halts regardless of the input. If a TM decides a language, then it is decider by definition or a total Turing Machine. Edit: To answer some of the questions in the OP's comments: A language does not define a Turing Machine.

2021年7月4日 · Fallowing are some definitions from book "introduction to theory of computation" by sipser. a nondeterministic turing machine is a decider if all its computation branches halt on all inp...

2018年12月27日 · An oracle doesn’t have an implementation, it is just a black box giving answer to any particular question (most importantly, the ones we cannot answer), whereas a decider has to be well defined Turing machine. In other words, we must know how a decider gets the answer, meaning it cannot answer any question.

Yes, it is possible to do so. Although a given PDA may have $\varepsilon$ loops that can induce infinite computation, we can sidestep this by converting the PDA to a CFG, then back to a PDA (using the standard methods).

$\begingroup$ Could you clarify further how one can make sure that a language has a decider, given only its lexicographical enumerator, and no other information on whether the language is finite or infinite? I imagine receiving a printer that prints strings with a label on top saying that "Fact: It prints in lexicographical order."

You want to prove that the language of deciders—the language of Turing machines whose language is decidable—is not recognizable.

2014年10月5日 · My definition of a decider is that I accept if input is in L and reject if input is not in L. This decider can be deterministic or non-deterministic. However, I say that L is in NP, and therefore if I am using a non-deterministic TM then I will take polynomial time. Also, may I know why flipping the bit does not necessarily complement the language.

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