Thread: Show that every finite automaton is a one-state pushdown automaton

I need help with Showing that every finite automaton is a one-state pushdown automaton

Thanks

The best place for such questions is the Discrete Math subforum.

Strictly speaking, a finite automaton is not a pushdown automaton because it does not have a stack. You probably need to emulate a finite automaton with a one-state pushdown automaton. What do you think about having a single symbol in the stack that represents the state of the finite automaton?

Thanks for the reply
But what if the finite automaton has more than one state? what about accepting state?

Originally Posted by user30669

But what if the finite automaton has more than one state? what about accepting state?

My suggestion addresses both of these issues. I suggested storing the state of the finite automaton in the stack.