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Bookmark or Share- Show that a decidable language is not decided by a decider in a given setI believe I can see both that M here is a decider and that its language is different from that of any machine in A. However, I am struggling to see where the lower bound of 2^i comes from. The algorithm M runs Mi only to check input strings of length i. So, for all such strings the results of running M and Mi on it will be different.
- Flaw in my NP = CoNP Proof? - Computer Science Stack ExchangeMy 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.
- Turing Machines: What is the difference between recognizing, deciding ...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.
- nondeterminism - Non-deterministic Turing machine that halts on at ...Then M clearly cannot be a decider, and is thus only a recognizer. M recognizes the language of all strings, for which any branch of M ends in an accepting state. Since M is a recognizer, it is only guaranteed to accept a string if the string is in L(M). Given a string, that is not in L(M), it may reject the string, or loop forever.
- computability - Show that a language is decidable iff some enumerator ...$\begingroup$ 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."
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