Question

Consider the following CFG :

(Where S, A, B, C are non -terminals and a, b, c, e, d are terminal)
$S\to SABC$
$A\to aAd|\in$
$B\to bB|\in$
$C\to Cc|Be|\in$


Number of elements in the FOLLOW(S) are _____

• A. 5

Answer 1

The correct option is A 5
$S\to SABC$
$A\to aAd|\in$
$B\to bB|\in$
$C\to Cc|Be|\in$

FOLLOWS(S) = FIRST(A)
$\Rightarrow \left[a\right]\cup \left[First\right(B\left)\right]=\left[a\right]\cup \left[b\right]\cup \left[FIRST\right(C\left)\right]$
$=\left[a\right]\cup \left[b\right]\cup b,c,e,e$ ($\in$ can not be part of FOLLOW)
= {a, b, c, e} $\cup$ (S) (S is start symbol so its FOLLOW contain S)
= {a, b, c, e, S}

It contain 5 elements.