Question

PDA for {a^i b^j i != j}

PDA for {a^i b^j c^k, i = j or j = k}

PDA for # of a's = # of b's

PDA for # b's = twice # of a's

Answer 1

Note: A in first and z in rest represent start stack symbol and / operation shows the popping function

PDA for {a^i b^j i != j}

PDA for {a^i b^j c^k, i = j or j = k}

PDA for # of a's = # of b's

PDA for # b's = twice # of a's

(Feel free to ask any doubts in the comments and throw an upvote)