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How to remove left recursion in the following Grammar:

S→Bb/a
B→Bc/Sd/e

Im new to this, below is the way I'm doing it:

S-> Sd/a
B-> Bc/Bb/e

I replaced the values of S and B with Bb and Sd to make the grammar easier to deal with. Then:

S-> aS'
S'-> dS' / ε

Is this the correct approach to do this ?

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  • $\begingroup$ Now do the same for $B$. $\endgroup$
    – vonbrand
    May 6, 2023 at 18:09
  • $\begingroup$ @vonbrand B-> eB' B'-> cB' / dB' / ε Is this the correct way of doing this? Someone told me that I cant replace the values of S and B with Bb and Sd. $\endgroup$
    – whoAsked
    May 7, 2023 at 14:59
  • $\begingroup$ The custom is to use | where you're writing /. $\endgroup$ Jan 10 at 9:51

1 Answer 1

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S->Bb|a
B->SdB'|eB'
B'->cB'|ε
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  • $\begingroup$ We're not looking for an answer that just contains a grammar. We would like to see explanation, justification, etc. $\endgroup$
    – D.W.
    Jan 10 at 8:29

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