# Why Deterministic PDA accepts $\epsilon$ input but DFA not

I was going through a deterministic PDA that accepts $wcw^R$ (described in Ullman's textbook), in which the last transition is given as $(q_1,\epsilon, Z_0)\to(q_2,Z_0)$, where $q_2$ is the final state.

In DFAs we don't consider $\epsilon$ transitions, while in PDAs we do include them. Why?

• It's a question of definition; $\epsilon$ transitions can be eliminated from PDAs in the same way they are eliminated from $\epsilon$-DFAs. Mar 10 '14 at 2:37
• @YuvalFilmus Oh We have $\epsilon$-DFA! Mar 10 '14 at 4:59
• @YuvalFilmus: As far as I know, there is no standard construction for elimination $\varepsilon$-transitions from PDA other than going via grammars. Do you know one?
– Raphael
Mar 10 '14 at 9:07
• @Raphael You're right, they are more powerful in PDAs because of the stack. Mar 10 '14 at 12:04

There are definitely definitions of DFA that allow $\varepsilon$-transitions. Arguably, they are not very useful for three reasons.
2. If you have them, you have to require (for DFA) that every $\varepsilon$-transition is the only transition leaving its source state.
In PDAs, $\varepsilon$-transitions can be used to deal with stack content without consuming any input. Even in DPDA there can be multiple choices in any given state, depending on stack content. So they are useful here for constructing PDA and are thus usually included in the definition.