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- Dictionaryrecognizable/ˌrɛkəɡˈnʌɪzəbl/
adjective
- 1. able to be recognized or identified from previous encounters or knowledge: "there was no recognizable photograph of him"
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Mar 8, 2011 · Def : A Language is called Turing Recognizable if some Turing Machine recognizes it. Now, consider a Turing Machine M M and a language L L (over input alphabet Σ Σ) that is recognized by M M. Thus, L L is a Turing Recognizable Language (since the TM M M recognizes it). Consider the set of strings that are not in L L (we call it L¯¯¯¯ L ¯).
Dec 18, 2015 · $\begingroup$ Don't know about hangs, that is an unnecessary concept. Turing decidable means it halts in an accepting state if the input word is in the language, and halts in a rejecting state if the word is not in the language, Turing recognizable means it halts in an accepting state if the word is in the language, and in a rejecting state or fails to halt if the word is not in the language.
Nov 9, 2021 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
The basic idea is: The set K = { M : M( M ) halt } is recognizable but not decidable. Hence its complement ˉK = { M : M( M ) does not halt} is not recognizable. Let Mx denote the Turing machine with code x. You want to construct a Turing machine U that reduces ˉK to L. Given M define TM to the be Turing machine such that:
Aug 21, 2021 · According to my understanding: Turing-recognizable languages are languages whice are accepted by a Turing machine; decidable languages are languages for which a Turing machines halts, i.e. either accepts or rejects, but never loops. This would make me think that decidable languages include Turing-recognizable languages, and not viceversa.
Is the union of undecidable languages not Turing-recognizable? 0 I understand Turing Machine things about languages but I don't understand same things about problems and their inputs
Feb 25, 2019 · 1. We know that if a language L L is decidable, then the complement L¯¯¯¯ L ¯ is also decidable, since we can simply reverse the accept and reject conditions in the Turing machine deciding L L. Furthermore, if L1 L 1 and L2 L 2 are decidable languages, then their intersection L1 ∩L2 L 1 ∩ L 2 is decidable, since we can accept if both ...
Apr 27, 2022 · I was reviewing for an exam and I found this question: Let A and B be two disjoint languages (that is, A ∩ B = ∅). Say that a language C separates A and B iff A ⊆ C and B ⊆ (not C) . Define two disjoint languages by. A = { M,w : M is a TM and M accepts w} B = { M,w : M isaTMandM rejects w}
Dec 6, 2022 · If y ∈ Σ∗ ∖ L y ∈ Σ ∗ ∖ L, then if L ⊆Σ∗ L ⊆ Σ ∗ is decidable, then M M rejects y y, therefore y ∉ L(M) y ∉ L (M). However dom(f) = Σ∗ d o m (f) = Σ ∗ so it holds that L(M) ≠ dom(f) L (M) ≠ d o m (f) and therefore IL I L is not computable. And i get he same problem for the implication "from right to left ...
Apr 12, 2015 · In the functional models, a machine can directly read the input "all at once", but in the Turing machine model it takes time just to read the whole input. This is why Turing machines are used to define computational complexity, because the time that they take to read the input is important in that context. $\endgroup$ –
