decidable languages are closed under permute and half swap
Turing decidable languages are closed under intersection Proof
Theorem: Turing decidable languages are closed under complement Proof: Let M be a TM which decides L It is easy to construct the machine schema |
Further discussion of computability
The decidable and semidecidable languages are closed under many (but not all) of the operations on languages that we have considered thus far in the course |
Why are decidable languages closed under complement?
The decidable languages are also closed under complementation, as the next proposition states.
Perhaps this one is simple enough that we can safely skip the proof; it is clear that if a DTM M decides A, and we simply swap the accept and reject states of M, we obtain a DTM deciding A.Turing recognizable languages are closed under union and intersection.
Explanation: A recognizer of a language is a machine that recognizes that language.
A decider of a language is a machine that decides that language.
Are decidable languages closed under concatenation?
Decidable languages are closed under concatenation and Kleene Closure.
Are recursive languages not closed under homomorphism?
We know that Lu is not recursive.
So in three words, L is recursive, h(L) = Lu is RE but not recursive.
So recursive languages are not closes under homomorphism.
Durée : 11:34
Postée : 5 mai 2020Termes manquants : permute | Afficher les résultats avec :permuteAutres questions
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