0 votes 0 votes Which of the following languages is regular? $L = \{ bba (ba)^* a^{n-1} \mid n> 0 \}$ $L = \{a^nb^n \mid n < 1000 \}$ $L = \{a^nb^k \mid \text{ n is odd or k is even} \}$ $L = \{ wxw^R \mid w,x \in (0+1)^* \}$ $1$, $3$ and $4$ $2, 3, 4$ $2, 3$ $1, 2, 3, 4$ Theory of Computation tbb-toc-2 theory-of-computation regular-language + – Bikram asked Aug 12, 2017 • retagged Sep 17, 2020 by ajaysoni1924 Bikram 450 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply vupadhayayx86 commented Nov 13, 2019 reply Follow Share How a is regular? Someone please explain. 0 votes 0 votes Shaik Masthan commented Nov 29, 2019 reply Follow Share it's a typo in options, now those are corrected :) 0 votes 0 votes Please log in or register to add a comment.
Best answer 1 votes 1 votes Sir Plz explain . The number of strings in L2 are limited so L2 should be regular.Option D should be correct. Shiv Gaur answered Jul 2, 2018 • selected Aug 18, 2019 by Bikram Shiv Gaur comment Share Follow See all 4 Comments See all 4 4 Comments reply Mr.OOPs commented Jul 2, 2018 reply Follow Share yes you are correct. 1 votes 1 votes logan1x commented Dec 4, 2019 reply Follow Share Can anyone explain how option 1 & 4th are correct. 0 votes 0 votes KUSHAGRA गुप्ता commented Jan 3, 2020 reply Follow Share @logan1x $1)\ bba(ba)^*a^*$ $4)\ wxw^r=\epsilon(0+1)^*\epsilon=Complete\ language: Regular$ 1 votes 1 votes logan1x commented Jan 4, 2020 reply Follow Share Thank you @Kushagra गुप्ता I was confused with L(4) because I was did mugged up wX$w^r$ . Small doubt - won't $1^*(0+1)^*1^*$ or any other form which makes it $a^nb^n$ will make it irregular or cfl? becuase if it fails one case than it can't be regular? 0 votes 0 votes Please log in or register to add a comment.