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

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

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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)

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