# Can a deterministic language be accepted by a deterministic Push Down Automaton?

I have a question that asks me to show that the PDA of the language L is not deterministic, but that the language is nevertheless deterministic. I was under the assumption that any deterministic language contains a PDA that is deterministic.

The language in question is: $L = \{w \in \{a,b\}^* : n_a(w) = n_b(w)\}$

• Is $n_a(w)$ the amount of a's in $w$? – Renato Sanhueza Nov 23 '15 at 2:32
• "the PDA of the language L is not X" -- there are many. Is one given? "language contains an automaton" -- that's misleading terminology. Languages like the one you give contain words. Automata accept languages. Also, regarding your tag choice: the given language is not regular. – Raphael Nov 23 '15 at 10:21

\begin{align} \delta(q, a, A) = \{(p, \alpha)\} \end{align}
you can add the following, without changing the language accepted (the modified PDA isn't deterministic, it is enabled to do completely pointless $\epsilon$ moves):
\begin{align} \delta(q, \epsilon, A) = \{(q, A)\} \end{align}