4 4 votes $\left \{ a^{m+n}b^{m+n}c^{n}|m,n\geq 1 \right \}$ $\left \{ a^{m+n}b^{m+n}c^{k} |m,n,k\geq 1\right \}$ $\left \{ a^{m+n}b^{m+k}c^{n+k} |m,n,k\geq 1\right \}$ Which one DCFL, CFL or CSL? Theory of Computation theory-of-computation dcfl context-free-language pushdown-automata + – srestha 3.5k views answer comment Share Follow Print See all 18 Comments 18 18 Comments reply Show 15 previous comments Abhishek Rai 2 commented Jun 25, 2018 reply Follow flag Yes because when b comes after few pop operation(as human we can see that there is only m no. of b's to pop) but machine don't know either to pop 'b' again or start pushing it but this can be guessed (for every b )on each step using Non-deterministicPDA . 0 0 replyShare srestha commented Jun 25, 2018 reply Follow flag here 2nd one? 0 0 replyShare ankitgupta.1729 commented Jun 25, 2018 reply Follow flag @Shaik , push a's until we see 'b' ..so there are m+n symbols onto the stack. now pop stack by b's until we see 'c'..now on c , do nothing onto the stack and go to final state..it will be accepted by DPDA.. 0 0 replyShare Please log in or register to add a comment.
0 0 votes $\boldsymbol{\textbf{1.CSL 2.DCFL 3.CSL}}$ Prince Sindhiya answered Jun 23, 2018 Prince Sindhiya comment Share Follow See 1 comment 1 1 comment reply srestha commented Jun 23, 2018 reply Follow flag no see the comment section 0 0 replyShare Please log in or register to add a comment.
0 0 votes first one is CSL second one is CFL third one is also CFL as when we are removing “a” from stack then only “m” a’s can be popped so that we are left with “k” a’s in stack. Then we push n b’s in the stack making total n+k b’s which can be further popped for c’s Hence 1 stack is sufficient. rish1602 answered Aug 4, 2021 rish1602 comment Share Follow 0 reply Please log in or register to add a comment.