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I have a following CFG rule :

  • S -> BSA | epsilon
  • A -> abC | a | c
  • B -> baC | b | epsilon
  • C -> aCc | AB | epsilon

I am on the epsilon elimination stage of the algorithm, I have eliminated following eplsions C -> epsiolon, B -> epsilon and here is what i got so far :

  • S_0 -> S
  • S -> BSA | SA | epsilon
  • A -> abC | a | c| ab
  • B -> baC | b | ba
  • C -> aCc | AB | ac| A Should I also eliminate S-> epsilon(shown in bold) since S is the original start variable?

Should I also copy the epsilon to S_0 at the unit rule stage of the algorithm?

Eric Leschinski
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Solix
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1 Answers1

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I your first grammar, You can derive epsilon from S. So the empty word belong to the described language. Therefore you must have a epsilon in the second equivalent grammar. Now in a normal form grammar, when there is a derivation S -> epsilon, then S can't appear on the right of a derivation. So the rule 

S -> BSA | SA | epsilon

is not allowed is a Chomsky normal form. So you probably want something as

S_0 -> S | epsilon      // initial
S -> BA | A | BSA | SA
[...]
hivert
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