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I have pretty simple question, but still can't find an answer just googling it.

I'm trying to understand Chomsky Normal Form (CNF). There are three production rules:

  1. $A \to BC$
  2. $A \to \alpha$
  3. $S \to \epsilon$

First two I understand. But last one $\epsilon$ doesn't makes sense for me. Why do we need this rule? What is use of having this?

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    $\begingroup$ Should be migrated to cs.stackexchange. $\endgroup$ – Yuval Filmus Apr 21 '14 at 23:32
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    $\begingroup$ To answer the question, without this rule you cannot represent languages containing $\epsilon$. This is the only reason for allowing this rule. $\endgroup$ – Yuval Filmus Apr 21 '14 at 23:33
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The last rule is necessary for languages containing $\epsilon$. A context-free grammar in which $S$ does not generate $\epsilon$ can be converted to Chomsky normal form without the rule $S \to \epsilon$. Conversely, simple induction shows that all other rules do not generate $\epsilon$, so this rule is needed for languages generating $\epsilon$.

The rule arises during the $\epsilon$-elimination step. For each non-terminal generating $\epsilon$, we "optionally" delete it from any production containing in in the right-hand side, and continue the process inductively (eliminating one non-terminal overtly generating $\epsilon$ may reveal that under generates $\epsilon$). Finally, it could happen that $\epsilon$ is pushed all the way to the starting symbol $S$; there is no way to eliminate it there.

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  • $\begingroup$ Thanks. Now I understand it. Just checked epsilon-elimination algorithm, and it became clear. $\endgroup$ – Igor Konoplyanko Apr 22 '14 at 9:04

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