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Given a Deterministic Finite Automata (DFA) M_1, does there always exist a Pushdown Automata (PDA) M_2 that accepts the same language as M_1? I.e. can any DFA be simulated by a PDA? Intuitively, it makes sense to me that a PDA is more powerful since it has an arbitrary amount of memory and can therefore accept more languages than a DFA, but how could this be formally proven?

Paradox
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    Rather, an NFA **is** a PDA that just doesn't use its stack. NFAs and DFAs are computationally equivalent. A PDA **is** a nondeterministic Turing Machine that only looks at the most recently written item on its tape. The way to formalize the proof is to map an NFA's definition into a PDA's definition. I.e., given an NFA ` -> qk, sigma, F>`, create PDA ` -> qk, q0, null, F>`. This PDA accepts the same language as the NFA. Hand-wavely, this PDA can only follow transitions to accepting states the way the original NFA would. – Welbog Jan 31 '18 at 20:07

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A finite automaton is nothing but a Turing machine that does not use its memory. A PDA is nothing but a Turing machine that restricts its access to its memory to LIFO.

So, a finite automaton is a PDA.

akim
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