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decidability
sushmita
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Theory of Computation
Feb 10, 2017
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Jul 4, 2017
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Arjun
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is intersection of two context sensitive decidable??
I think since they are closed under intersection, it must be decidable.
please answer
decidability
theory-of-computation
turing-machine
recursive-and-recursively-enumerable-languages
sushmita
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Theory of Computation
Feb 10, 2017
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Arjun
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srestha
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Feb 10, 2017
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yes here only closer property is enough to say decidable
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sushmita
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thanx sretha :)
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bhargav9873
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Feb 12, 2017
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Can any one please explain me why is it closed under intersection
I mean i know it is
Buy why??
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Yes They are ...
Anup patel
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Feb 10, 2017
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sushmita
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thanx a lot
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Anup patel
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Feb 11, 2017
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One question of 2 Marks from this Table ..Session 6
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When we say that some Language 'X' is close under some function 'f' then here we basically mean that there exists an Algorithm to prove it. Eventually when Algo exists then it would become REC(Decidable).
Rupendra Choudhary
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May 22, 2017
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amitarp818
asked
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Dec 28, 2023
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Decidability
L(M)={0} We can have Tyes for {0} and Tno for Σ∗ ({0}⊂Σ∗{0}⊂Σ∗). Hence, L={M ∣ L(M)={0}} is not Turing recognizable (not recursively enumerable) I don’t understand why this is not decidable. We can easily create a turing that accepts this language
amitarp818
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Dec 28, 2023
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decidability
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ajaysoni1924
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Theory of Computation
Jul 15, 2019
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Self Doubt: Decidability
$L=\left \{\langle M_{1},M_{2}\rangle \text{ such that L}(M_{1})\prec L(M_{2}) \right \}$ is it recursive enumerable? here $L\left ( M_{1} \right )\prec L\left ( M_{2} \right )$ signifies language $L\left ( M_{1} \right )$ is reducible to $L\left ( M_{2} \right )$
ajaysoni1924
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Jul 15, 2019
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theory-of-computation
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Rahul_Rathod_
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Dec 24, 2018
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decidability
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Dec 24, 2018
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decidability
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Mk Utkarsh
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Theory of Computation
Nov 23, 2018
416
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Decidability Doubt
$L_1 = \{ \text{<M>} | \ \text{M is a TM, } \text{M}_0 \ \text{is a TM that halts on all inputs and, } \text{M}_0 \in L(M) \}$ ... because $\exists$ infinite number of TM's which accept $\Sigma^*$. Now $L(M)$ is not even RE. If that then how $L_1$ is RE. Please explain both.
Mk Utkarsh
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Nov 23, 2018
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