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How to define a non ambiguous grammar for regular expressions on the $\Sigma = \{a,b\}$ alphabet?

Given that:

If $\Theta = \{+, ^*, (,),\cdot, \emptyset\}$ is a set of symbols

A regular expression on $\Sigma$ is a string such that one of the following condition applies

  1. $r = \emptyset$
  2. $r \in \Sigma $
  3. $r = (s+t)$ or $ r = (s\cdot t)$ or $r = s^*$
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  • $\begingroup$ Ok fade2black so now the grammar $G$ should recognize regular expressions. I wonder if i can use $\epsilon$ instead of $\emptyset$ in the grammar $\endgroup$ – Jack Nov 3 '17 at 20:03
  • $\begingroup$ $\epsilon$ cannot be used instead of $\emptyset$ since they have different meanings in regular expressions. The first one corresponds to a regular set with a single element $\epsilon$, the empty string, while the other is just an empty set containing no strings. $\endgroup$ – fade2black Nov 3 '17 at 20:35
  • $\begingroup$ I think your question already answers itself. You're on the right track. $\endgroup$ – reinierpost Nov 3 '17 at 20:55
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I need to define a context free grammar to recognize regular expression. I know also that CF grammars are used to recognize mathematical expressions. so the grammar $G$ should recognize regular expressions.

$G:$

$ E \rightarrow E + T \mid T $

$ T \rightarrow T \cdot F \mid F $

$ F \rightarrow(E) \mid E^* \mid a\mid b\mid\emptyset$

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  • $\begingroup$ $F^*$, not $E^*$ $\endgroup$ – rici Nov 3 '17 at 22:53
  • $\begingroup$ @rici $F^*$ or $E^*$, the grammar will generate the same set, won't it? $\endgroup$ – fade2black Nov 3 '17 at 23:01
  • $\begingroup$ @fade yes, but only one of them is unambiguous. $\endgroup$ – rici Nov 3 '17 at 23:09
  • $\begingroup$ @rici did you find a string having two different leftmost derivations? Could you share it? $\endgroup$ – fade2black Nov 3 '17 at 23:22
  • $\begingroup$ @fade: I didn't look very hard :-). Try $a+b^*$ $\endgroup$ – rici Nov 3 '17 at 23:50

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