3 votes 3 votes Which of the following is/are undecidable? $L=\left\{\langle M\rangle \mid M\right.$ is a TM, $\mathrm{L}(M) \neq \emptyset$, and $\left.\mathrm{L}(M) \neq \Sigma^*\right\}$. $\{\langle M\rangle \mid M$ is a TM and $\mathrm{L}(M)=\varnothing\}$ $\mathrm{L}=\{\langle\mathrm{M}>| \mathrm{M}$ is a TM and $\mathrm{L}(\mathrm{M})$ is uncountable $\}$ $L=\{\langle M\rangle \mid M$ is a DFA and $L(M)$ is uncountable $\}$ Theory of Computation goclasses2024-mockgate-11 goclasses theory-of-computation decidability multiple-selects 2-marks + – GO Classes asked Jan 13 • retagged Jan 13 by Lakshman Bhaiya GO Classes 596 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply GauravRajpurohit commented Jan 16 reply Follow Share https://www.youtube.com/watch?v=atrZdsQU5RE Time stamp 2:26:30<!-- notionvc: 24e238be-d770-4297-be88-c0a5f2f8f6a3 --> 0 votes 0 votes GO Classes Support commented Jan 16 reply Follow Share $ \large{\colorbox{yellow}{Detailed video solution of this question with direct time stamp}}$ All India Mock Test 2 - Solutions Part 3 0 votes 0 votes Please log in or register to add a comment.
3 votes 3 votes EVERY language is countable. So, Option C $\&$ D are trivial properties (trivially false), hence countable. GO Classes answered Jan 13 • edited Jan 13 by Lakshman Bhaiya 1 flag: ✌ Edit necessary (rajan31 “"hence decidable"”) GO Classes comment Share Follow See all 0 reply Please log in or register to add a comment.