4 votes 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 asked Jun 22, 2018 srestha 1.5k views answer comment Share Follow See all 18 Comments See all 18 18 Comments reply Show 15 previous comments Abhishek Rai 2 commented Jun 25, 2018 reply Follow Share 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 votes 0 votes srestha commented Jun 25, 2018 reply Follow Share here 2nd one? 0 votes 0 votes ankitgupta.1729 commented Jun 25, 2018 reply Follow Share @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 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes $\boldsymbol{\textbf{1.CSL 2.DCFL 3.CSL}}$ Prince Sindhiya answered Jun 23, 2018 Prince Sindhiya comment Share Follow See 1 comment See all 1 1 comment reply srestha commented Jun 23, 2018 reply Follow Share no see the comment section 0 votes 0 votes Please log in or register to add a comment.
0 votes 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 See all 0 reply Please log in or register to add a comment.