Infinite Language Decidable. This involves using the pumping lemma for 1 The problem of dete

This involves using the pumping lemma for 1 The problem of determining whether a recursively enumerable language is empty or infinite cannot be solved. f. Does this work? The above was an example of a language, didn't think it would be recognizable. The Church-Turing Thesis formalized the notion of an algorithm, as a procedure The complement of a decidable language is decidable because a decidable language has a TM that answers "yes" or "no" on every string. All semi . If L3 was not decidable, no possible Turing machine could decide L3. So my idea how to solve this problem is the following: k = Every decidable theory or logical system is semidecidable, but in general the converse is not true; a theory is decidable if and only if both it and its complement are semidecidable. To be undecidable, there must be individual strings in the language that cause the TM to fail. Therefore, some languages are not recognized by any Turing machine. By taking that TM and reversing its answer, you If there is an algorithm that can decide membership in a language, that language is in R. An example of this is Hilbert's 10th problem, which is recognizable and undecidable, but we If so, L (CFG) is infinite, else finite. Indeed, there must be infinitely many such strings (otherwise, the poorly behaved An algorithm or recipe for recognizing languages is anything that can be fed a word over the given alphabet, and that depending on its input either "accepts" the input after some time, or That is impossible because the set of languages is uncountable, and the set of Turing machines is countable. By the Church-Turing thesis, any effective model of computation is equivalent in power to a Turing Yet, with no upper limit, we can have $1^\infty$ - meaning we will never halt. In this homework we'll explore decidability: the property of a language that guarantees that it is both Turing recognizable and that it will never throw the Turing machine into an infinite loop. However, we only know of one TM (M) that does not decide L3. These languages are called decidable languages, and TMs that • End of the course By far most important and interesting material in CS 360. Languages decided by a TM are called decidable. A For a language L if there is some Turing Machine that accepts every string in L and rejects every string not in L, then L is a decidable language if there is some Turing machine that accepts Infinite union is a powerful concept because every language is an infinite union of singletons, as noted in the answer. For example, The set INFINITEPDA is decidable because there exists an algorithm to determine if a given PDA recognizes an infinite language. Let’s examine the class of decidable languages in a bit more detail to see if we believe that it does correspond to the intuitive notion of computability that we want it to capture. 1 shows each problem’s decidability property. Any language outside Dec is undecidable. The proof goes by reduction to the problem of decidability, which is known Show that every infinite Turing-recognizable language has an infinite decidable subset. The existence of undecidable languages follows by a counting argument: The set of all languages is uncountable whereas the set of decidable There are some languages for which a Turing machine can be written that will halt on all input, either to accept or reject. So above problem (2) is decidable . I am currently struggling with figuring out the following problem: Given decidable languages L1, L2, L3, L4, Is the infinite union of Languages L1, decidable? We will identify which of the 24 problems are decidable and, if they do not appear decidable, prove that they are undecidable. I'm not getting above explanation of this problem : How we decide for a given context free grammar The set $\ {L' : L' \subseteq L\}$ is an uncountable set of languages, and since there is only countably many decidable languages, it has to contain an undecidable language. Table 7. Why is the set of decidable languages, R, a subset of Every TM for a semi-decidable+ language halts in the accept state for strings in the language but loops for some strings not in the language. This question has been asked on each "language level": -- I have following problem: INFPDA={ A |A is PDA and L(A)=infinite language} Prove that this is decidable problem. • L is I read this in a book on computability: (Kleene's Theorem) A language is regular if and only if it can be obtained from finite languages by applying the three operations union, Decidable Problems Concerning Context-Free Languages Topics Problem 1: describe algorithms to test whether a CFG generates a particular string Problem 2 describe algorithms to test So it is my understanding that Recursively Decidable languages are the languages that we can build a Turing Machine such that given an input w from that language, Turing Languages recognized by a TM are called recognizable. This question has been asked on each "language level": -- Infinite union is a powerful concept because every language is an infinite union of singletons, as noted in the answer. • L is decidable or recursive if there exists a Turing machine M with L = L(M) where M halts on all inputs.

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