Consider the grammar with non-terminals N = {S,C,S1 },terminals T={a,b,i,t,e},…

Correct reason: there is a FIRST/FOLLOW conflict for the nonterminal S1, so a single-token lookahead cannot always choose the right production.

• Productions:

  • S → i C t S S1

  • S → a

  • S1 → e S

  • S1 → ε

  • C → b

Compute FIRST sets:

• FIRST(S) = {i, a}

• FIRST(S1) = {e, ε}

• FIRST(C) = {b}

Compute FOLLOW sets (key points):

• S is the start symbol, so $ ∈ FOLLOW(S).

• In the production S → i C t S S1, the S that appears before S1 has FIRST(S1) − {ε} = {e} in its FOLLOW. Because S1 can be ε, FOLLOW(S) (which contains $) is also in FOLLOW of that S. Hence FOLLOW(S) = {e, $}.

• S1 appears at the end of the S production, so FOLLOW(S1) = FOLLOW(S) = {e, $}.

Why this causes a conflict:

• For nonterminal S1 we have two alternatives: one starts with terminal e (FIRST = {e}) and the other is ε (FIRST contains ε).

• Because ε is an alternative, the parser must consult FOLLOW(S1) to decide when to use the ε-production. But FOLLOW(S1) contains e, and e is also in FIRST of the other alternative.

• Therefore FIRST(e S) ∩ FOLLOW(S1) contains e, creating a predictive parsing table conflict. This makes the grammar not LL(1).

Additional clarifications:

• This is a FIRST/FOLLOW conflict, not necessarily ambiguity. Ambiguity is a different property (multiple parse trees for the same string).

• The grammar is context-free and not left-recursive, so the listed reasons claiming non-context-freeness or left recursion are incorrect.