3 votes 3 votes Consider the following languages : $L_{1}=\left\{a^{k} b^{m} c^{n} \mid(k=m\right.$ or $m=n)$ and $\left.k+m+n \geq 2\right\}$ $L_{2}=\left\{a^{k} b^{m} c^{n} \mid(k=m\right.$ or $m=n)$ and $\left.k+m+n \leq 2\right\}$ $L_{3}=\left\{a^{k} b^{m} c^{n} \mid k+m+n \geq 2\right\}$ Which of the above languages are Regular? $\mathrm{L}_{2}$ Only $\mathrm{L}_{3}$ Only $L_{2}$ and $L_{3}$ only All Theory of Computation goclasses2024-toc-2-weekly-quiz goclasses theory-of-computation regular-language 1-mark + – GO Classes asked Jun 22, 2022 • retagged Jun 17, 2023 by Lakshman Bhaiya GO Classes 350 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
6 votes 6 votes The point to be considered here is : if comparison → L is NON-REGULAR, → Except L is finite In L1: comparison & nonfinite → Non-regular In L2 : comparison & finite → Regular In L3: No comparison : >= k : regular Udhay_Brahmi answered Jun 26, 2022 Udhay_Brahmi comment Share Follow See all 0 reply Please log in or register to add a comment.
5 votes 5 votes In $\mathrm{L}_{1}$, there is a matching between $\mathrm{k}, \mathrm{m}$ Or $\mathrm{m}, \mathrm{n}$; so, it is Non-regular but CFL. $\mathrm{L}_{2}$ is finite, so regular. $\mathrm{L}_{3}$ is regular because No matching required. GO Classes answered Jun 22, 2022 GO Classes comment Share Follow See all 0 reply Please log in or register to add a comment.
