While searching for an answer to this question I found out that there is an unambiguous grammar for every regular language. But is there a regular language for every unambiguous grammar? How can I prove that this is/isn't true?

  • $\begingroup$ Have you tried finding an unambiguous grammar for $a^nb^n$? $\endgroup$
    – greybeard
    Sep 16 '20 at 6:58

The following grammar is unambiguous yet generates a non-regular language: $$ S \to aSb \mid \epsilon $$

  • 3
    $\begingroup$ As a side note, this works for every $a,b \in \Sigma$ where $a \neq b$. A very notable example would be the language of matched parenthesis $(^n)^n$. $\endgroup$
    – Polygnome
    Sep 15 '20 at 22:53
  • $\begingroup$ It's been some time since my theory courses. Is this the trivial PDA? $\endgroup$ Sep 16 '20 at 6:10
  • 1
    $\begingroup$ This is a context-free grammar, not a push down automaton. $\endgroup$ Sep 16 '20 at 6:16

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