Question

Consider the context-free grammar \(G\) below: \[ S \;\to\; aSb \;|\; X \] \[ X \;\to\; aX \;|\; Xb \;|\; a \;|\; b \] where \(S\) and \(X\) are non-terminals, and \(a\) and \(b\) are terminal symbols. The starting non-terminal is \(S\). Which one of the following statements is CORRECT?

• A. The language generated by \(G\) is \((a+b)^*\)
• B. The language generated by \(G\) is \(a^*(a+b)b^*\)
• C. The language generated by \(G\) is \(a^*b^*(a+b)\)
• D. The language generated by \(G\) is not a regular language

Hint:

When analyzing CFGs, first identify the “core string” generated by one non-terminal, then check how other rules wrap or extend it. Here, \(X\) gives \(a^*(a+b)b^*\), and \(S\) just reinforces this same structure.

Answer 1

The Correct Option is B

Step 1: Analyze production \(S \to aSb \;|\; X\)
This rule allows wrapping equal numbers of \(a\)'s at the left and \(b\)'s at the right around whatever is produced by \(X\). Hence, any string generated by \(S\) has the form: \[ a^n \; w \; b^n \text{where } w \in L(X),\; n \geq 0 \] Step 2: Analyze production of \(X\)
The productions of \(X\) are: \[ X \to aX \;|\; Xb \;|\; a \;|\; b \] This means \(X\) can produce:
- Start with a base case: \(a\) or \(b\).
- Then prepend any number of \(a\)'s or append any number of \(b\)'s.
Thus, \(X\) generates all strings of the form: \[ X = a^i \; \alpha \; b^j \text{where } \alpha \in \{a, b\}, \; i,j \geq 0 \] In other words, \(X\) produces a string with some leading \(a\)'s, then a single terminal \(a\) or \(b\), followed by some trailing \(b\)'s.
Step 3: Combine with \(S\)
Since \(S \to a^n \; w \; b^n\), with \(w \in L(X)\), the final language is: \[ L(G) = \{\, a^n \; (a^i \alpha b^j) \; b^n \mid n,i,j \geq 0,\; \alpha \in \{a,b\}\,\} \] This simplifies to: \[ L(G) = a^*(a+b)b^* \] because the total left \(a\)'s combine, and the total right \(b\)'s combine. Step 4: Conclusion
The language generated is regular and corresponds exactly to option (B): \[ L(G) = a^*(a+b)b^* \] \[ \boxed{\text{Correct Option: (B)}} \]