Multiple choice

Eliminate left recursion S $\rightarrow$ Aa|b A $\rightarrow$Ac|Sd|f

  1. $S \rightarrow Aa | b \\\\ A \rightarrow bdA'| fA' \\\\A | \rightarrow cA'| adA'| \epsilon$
  2. $S \rightarrow Aa | b \\\\ A \rightarrow cdA'| dA' \\\\ A' \rightarrow bA'| adA' |\epsilon $
  3. $S\rightarrow Aa|b \\\\ A\rightarrow bdA'|fA'$
  4. $S \rightarrow Aa| b$ $A \rightarrow bdA'|fA'$ $A'\rightarrow cA'|adA'$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$A\rightarrow Ac|Sd|f \rightarrow Ac|Aad|bd|f $ Eliminating the immediate left recursion in A $A \rightarrow bdA'| fA'$ $A| \rightarrow |adA' | \epsilon$ So the resulting grammar is option A.   

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