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Consider the following context free grammar:

$S \rightarrow ASA | aB$

$A \rightarrow B | S$

$B \rightarrow b |  \epsilon$

How many productions will be there in the modified grammar if we remove null-productions and unit-productions from this grammar? 

My Solution:
Step 1: Remove epsilon transitions
Nullable variables = $ \big\{A,B\big\} $

Modified grammar after removal of null productions:
$S \rightarrow ASA | aB | a | AS | SA | S$
$A \rightarrow B | S $
$B \rightarrow b$

Step 2: Remove Unit Productions
$S \rightarrow ASA |aB |a | AS | SA | \mathbf{S}$
$A \rightarrow \mathbf {B} | \mathbf{S}$ 
$B  \rightarrow b$

Modified grammar after the removal of the unit productions:
$S \rightarrow ASA |aB |a | AS | SA$
$A \rightarrow b | ASA |aB |a | AS | SA$   
$B  \rightarrow b$

I am getting 12 productions. can someone please confirm if it's correct?
 

in Theory of Computation by Boss
edited by | 534 views
0
getting same.
0
12 correct
0

 mam

If the qsn was about simplify the grammer, then Can I write A->ab instead of A->aB & remove the production B->b ???

In that case we'll have 11 productions.

please confirm this.

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