Question

Consider the following grammar where \(S\) is the start symbol, and \(a\) and \(b\) are terminal symbols.
\[ S \rightarrow aSbS \mid bS \mid \epsilon \]
Which of the following statements is/are true?

• A. The grammar is ambiguous
• B. The string \(abb\) has two distinct derivations in this grammar
• C. The string \(abab\) has only one rightmost derivation
• D. The language generated by the grammar is undecidable

Hint:

Every application of \(aSbS\) fixes one \(a\) together with one \(b\); track how many ways the leftover \(b\)'s can be split between the two sub-copies of \(S\) to see when more than one parse tree exists.

Answer 1

The Correct Option is A, B, C

Step 1: Understand how symbols enter a derivation.
The rule \(S \rightarrow aSbS\) is the only rule that introduces the terminal \(a\), and every time it fires it introduces exactly one \(a\) together with one \(b\). The rule \(S \rightarrow bS\) introduces an extra unmatched \(b\). So in any string generated by \(S\), the number of \(b\)'s is always at least the number of \(a\)'s, and a bare symbol \(a\) by itself can never be generated.

Step 2: Check option (B) using the string \(abb\).
Since \(abb\) begins with \(a\), the first rule applied must be \(S \rightarrow aSbS\), giving the shape \(a\,S_1\,b\,S_2\). The remaining text after the leading \(a\) is \(bb\), so \(\text{str}(S_1) + b + \text{str}(S_2)\) must equal \(bb\). Two assignments work: (i) \(S_1 \Rightarrow \epsilon\), \(S_2 \Rightarrow b\), giving \(S \Rightarrow aSbS \Rightarrow abS \Rightarrow ab(bS) \Rightarrow abbS \Rightarrow abb\); and (ii) \(S_1 \Rightarrow b\), \(S_2 \Rightarrow \epsilon\), giving \(S \Rightarrow aSbS \Rightarrow a(bS)bS \Rightarrow abbS \Rightarrow abb\). Both are valid and give two structurally different parse trees for the same string, so option (B) is true.

Step 3: Check option (A).
A grammar is ambiguous exactly when some string it generates has more than one parse tree. Since \(abb\) has two distinct parse trees as shown in Step 2, the grammar is ambiguous, so option (A) is true.

Step 4: Check option (C) using the string \(abab\).
Again the derivation must start with \(S \rightarrow aSbS\), giving \(a\,S_1\,b\,S_2\), so \(\text{str}(S_1) + b + \text{str}(S_2)\) must equal \(bab\) (length 3). Trying \(|S_1|=0\): the equation becomes \(b+\text{str}(S_2)=bab\), so \(\text{str}(S_2)=ab\); this is achieved only as \(S \Rightarrow aSbS \Rightarrow a\epsilon b\epsilon = ab\), a unique derivation. Trying \(|S_1|=1,|S_2|=1\): the constructed string is \(S_1[0], b, S_2[0]\), so its middle character is forced to be \(b\), but the target's middle character is \(a\); this case is impossible. Trying \(|S_1|=2,|S_2|=0\): then \(\text{str}(S_1)=ba\), but \(S\) can never derive \(ba\) because that would require the inner copy of \(S\) in \(S \rightarrow bS\) to derive the bare symbol \(a\), which Step 1 showed is impossible. So exactly one valid parse exists for \(abab\): \(S \Rightarrow aSbS \Rightarrow abS \Rightarrow ab(aSbS) \Rightarrow ab(a\epsilon b\epsilon) \Rightarrow abab\). Since each distinct parse tree corresponds to exactly one rightmost derivation, and \(abab\) has only one parse tree, option (C) is true.

Step 5: Check option (D).
The language generated by any context-free grammar is a context-free language, and membership in a context-free language is always decidable (for example by the CYK or Earley algorithms, which terminate in finite time and correctly decide whether a given string belongs to the language). Nothing about this grammar changes that fact, so the claim that the generated language is undecidable is false. Option (D) is incorrect.

Step 6: Conclusion.
Options (A), (B) and (C) are true, and option (D) is false.
\[ \boxed{\text{Correct options: (A), (B), (C)}} \]