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Reducible language ( Decidability)
Na462
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Theory of Computation
Jan 21, 2019
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Na462
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Jan 21, 2019
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Na462
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Shobhit Joshi
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Jan 21, 2019
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$M$ is reducible to $L$, so $M$ can be recursive enumerable. So, $M\bigcap L$ is not recursive always.
Correct me if i'm wrong.
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Na462
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Jan 21, 2019
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See we get that L is R.E.L and may or may not be Recursive though
Now if M <= L (Say L i assumed as Recursive)
then M will also be recursive then : M intersection L = Recursive intersection Recursive = Recursive
If I assume L as Recursively enumerable but not recusive and M is reducible to L them M can be recursive
So M intersection L will be Recusively enumerable not recursive
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Magma
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Jan 21, 2019
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thanks
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amitarp818
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Dec 28, 2023
237
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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
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Dec 28, 2023
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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 $L\left ( M_{2} \right )$
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Ayush Upadhyaya
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Theory of Computation
Jan 13, 2019
426
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Decidability
From http://www.cs.rice.edu/~nakhleh/COMP481/final_review_sp06_sol.pdf $L_{26}=\{<M>|$ M is a TM such that both L(M) and $\lnot L(M)$ are infinite $\}$ I was unable to get proof given in pdf above.Can anyone explain, if someone got it.
Ayush Upadhyaya
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Theory of Computation
Jan 13, 2019
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