1 votes 1 votes Design PDA for i) L={a^n b^m c^n|m,n>=1} ii) L={a^m b^n c^p|m+n=p} iii) L={a^i b^i c^j|i,j=1} iv) L={a^i b^j c^j|i,j>=1} I have made an attempt to draw pda Please can someone crosscheck the answer? Theory of Computation theory-of-computation pushdown-automata context-free-language dcfl + – pricool84 asked Aug 6, 2017 pricool84 2.0k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Your Logic is correct. PDA’s are all wrong. your are ignoring just one b correct except move on Q0 to Q1 as i=1 only one a should be accepted as i>=1 one a must be present but you are ignoring all a’s, aaaakash001 answered Oct 6, 2022 aaaakash001 comment Share Follow See all 0 reply Please log in or register to add a comment.