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Consider the following grammar

$S\rightarrow Aa\mid B $

$B\rightarrow a\mid BC$

$C \rightarrow a\mid \in$

the no of productions in simplified cfg is
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4 votes
4 votes
1.Removal of null production

S->Aa|B
B->a|B|BC
C->a

2.Removal of Unit production
S->Aa|a|BC
B->a|BC
C->a

3.Removal of useless production

S->a|BC
B->a|BC
C->a
so total 5 production in simplified CFG.
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4 votes
4 votes

We can remove Aa as A is not defined.
S->B
B-->a|bC
C-->a|eps


now, S->B is unit production
S-->a|bC
C-->a|eps


removing epsilon production,
S-->a|b|bC
C-->a


So, simplified CFG contains 4 productions.

4 votes
4 votes
It's 4.. firstly epsilon is removed from C.. making C->a and B->a|b|bC and den unit production S->B is replaced by production of B making S->a|b|bC and also Aa is removed from S coz of useless symbol.. so total production will be

S->a|b|bC

C->a

That's 4...

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