LL(1) Check for the Grammar (S \rightarrow iEtSA \mid a,; A \rightarrow eS \mid \epsilon,; E \rightarrow b)

To check whether a context-free grammar is LL(1), we verify that for every nonterminal (X), the parser has at most one production to choose based on the next input token. Concretely, we use the LL(1) condition based on FIRST and FOLLOW sets and the behavior of (\epsilon)-productions:

• For any nonterminal (X), productions (X \rightarrow \alpha) and (X \rightarrow \beta) must satisfy
  (\text{FIRST}(\alpha) \cap \text{FIRST}(\beta) = \varnothing).
• If (\epsilon \in \text{FIRST}(\alpha)), then (\text{FIRST}(\beta)) must not intersect (\text{FOLLOW}(X)), equivalently the LL(1) parsing-table row for (X) must have no conflicts.

Key terms used below:

• LL(1) grammar
• FIRST set
• FOLLOW set
• Parsing table
• Epsilon production

We are given the grammar: [ S \rightarrow iEtSA \mid a ] [ A \rightarrow eS \mid \epsilon ] [ E \rightarrow b ] Terminals are ({i,e,b,a}). The start symbol is (S).

We will compute (\text{FIRST}) and (\text{FOLLOW}) sets, then show whether the LL(1) parsing table for (S) and (A) has conflicts.

Important: The grammar is LL(1) iff, for each nonterminal, the productions chosen by a given lookahead token are unique (no multiple entries per table cell).

1) FIRST sets

We compute:

FIRST((E))

Since (E \rightarrow b), [ \text{FIRST}(E)={b}. ]

FIRST((S))

(S) has two productions: (S \rightarrow iEtSA) and (S \rightarrow a).

• From (S \rightarrow iEtSA), the first terminal is (i), so (i \in \text{FIRST}(S)).
• From (S \rightarrow a), (a \in \text{FIRST}(S)).

Also, neither right-hand side can derive (\epsilon). Therefore: [ \text{FIRST}(S)={i,a}. ]

FIRST((A))

(A \rightarrow eS \mid \epsilon).

• From (A \rightarrow eS): (e \in \text{FIRST}(A)).
• From (A \rightarrow \epsilon): (\epsilon \in \text{FIRST}(A)).

So: [ \text{FIRST}(A)={e,\epsilon}. ]

2) FOLLOW sets

Let ($) denote end-of-input marker. Since (S) is the start symbol: [ $ \in \text{FOLLOW}(S). ]

Now scan productions:

From (S \rightarrow iEtSA)

Right-hand side: (i; E; t; S; A)

• (E) is followed by terminal (t), but note: there is no terminal (t) listed in the grammar statement. In the given production it appears as (t). For LL(1) analysis we must treat every symbol on the RHS that is not a nonterminal as a terminal. So (t) is a terminal symbol here. Hence: [ t \in \text{FOLLOW}(E). ]

• (S) is followed by (A) (since the RHS has (S A)). Therefore: [ \text{FIRST}(A)\setminus{\epsilon} \subseteq \text{FOLLOW}(S). ] Since (\text{FIRST}(A)={e,\epsilon}), we get: [ e \in \text{FOLLOW}(S). ] Also, because (A\Rightarrow^* \epsilon), we must add: [ \text{FOLLOW}(S) \subseteq \text{FOLLOW}(S) ] (which is tautological for this self-recursive position). Concretely, after (S) comes a symbol that may vanish, so whatever can follow the left-hand side (S) can also follow that inner (S). That is already captured by ($ \in \text{FOLLOW}(S)) and propagated consistently; no contradiction arises.

From (A \rightarrow eS)

RHS: (e; S). Since (S) is at the end of the production, we add: [ \text{FOLLOW}(A)\subseteq \text{FOLLOW}(S)\quad\text{does not directly apply here;} ] rather, it means: [ \text{FOLLOW}(S) \text{ gets } \text{FOLLOW}(A) \text{ when } A \text{ is the LHS. } ] So we should determine (\text{FOLLOW}(A)).

Determine FOLLOW((A))

(A) occurs in (S \rightarrow iEtSA) at the end. Therefore: [ \text{FOLLOW}(A) \subseteq \text{FOLLOW}(S). ] So: [ $\in \text{FOLLOW}(A) \quad \text{(since } $\in \text{FOLLOW}(S)\text{)}. ]

Summary of FOLLOW sets

• (\text{FOLLOW}(S)) contains at least ({$, e}).
• (\text{FOLLOW}(A)) contains at least ({$}).
• (\text{FOLLOW}(E)) contains ({t}) (where (t) is the terminal appearing after (E) in the production (S \rightarrow iEtSA)).

So we can take: [ \text{FOLLOW}(S)={$, e},\quad \text{FOLLOW}(A)={$},\quad \text{FOLLOW}(E)={t}. ]

3) LL(1) conflict checks

Nonterminal (S)

Productions:

• (S \rightarrow iEtSA)
• (S \rightarrow a)

We check FIRST/FIRST disjointness: [ \text{FIRST}(iEtSA)={i},\quad \text{FIRST}(a)={a}. ] Since ({i}\cap{a}=\varnothing), there is no LL(1) conflict in the (S) row.

Nonterminal (A)

Productions:

• (A \rightarrow eS)
• (A \rightarrow \epsilon)

We check FIRST/FIRST: [ \text{FIRST}(eS)={e},\quad \text{FIRST}(\epsilon)={\epsilon}. ] Their intersection is empty (terminals only), so FIRST/FIRST is fine.

Now apply epsilon rule: because (A \rightarrow \epsilon), we must ensure: [ \text{FIRST}(eS)\cap \text{FOLLOW}(A) = \varnothing. ] We have: [ \text{FIRST}(eS)={e},\quad \text{FOLLOW}(A)={$}. ] Thus: [ {e}\cap{$}=\varnothing. ] So the (A) row is also conflict-free.

Conclusion

All LL(1) conditions for the grammar are satisfied:

• For (S), the FIRST sets of alternatives are disjoint: ({i}) vs ({a}).
• For (A), the ε-production does not cause a conflict: (\text{FIRST}(eS)={e}) is disjoint from (\text{FOLLOW}(A)={$}).

Therefore, the given grammar is LL(1).