0 votes 0 votes L=0^i 1^j 2^k | i=j or j=k is the grammar DCFG? Theory of Computation theory-of-computation context-free-language + – aditi19 asked Oct 23, 2018 aditi19 443 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply Hemanth_13 commented Oct 23, 2018 reply Follow Share No, I don't think so. Push 0 when you encounter it Pop 0 when you encounter 1. Here we lost the track of 'j' so we will not able to compare with 'k' if the first condition(i==j) failed. But we can construct a NPDA for this. 0 votes 0 votes Verma Ashish commented Oct 23, 2018 reply Follow Share Yes i also getting same. 0 votes 0 votes Shivani gaikawad commented Oct 23, 2018 reply Follow Share it is non deterministic CFL 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes it is the union of 0^i 1^i 2^k union 0^i 1^j 2^j so for j is not clear that we r to consider it equal to i or j NPDA can do that not DPDA Devvrat Tyagi answered Nov 18, 2018 Devvrat Tyagi comment Share Follow See all 0 reply Please log in or register to add a comment.
