To solve this question, we need to determine the language generated by the context-free grammar \(G\) provided in the question. Let's examine the productions step-by-step.
The grammar is given as follows:
- \(S \to aSb \mid X\)
- \(X \to aX \mid Xb \mid a \mid b\)
We must understand what strings the grammar can generate.
Step-by-Step Analysis of the Grammar
- Production \(S \to aSb\): This production rule implies that for any string derived from \(S\), it can be surrounded by an equal number of \(a\)s and \(b\)s. This means the string will be of the form \(a^m w b^m\), where \(w\) is a string generated by \(X\).
- Production \(S \to X\): This allows the variable \(S\) to derive any string that can be generated by \(X\).
- Productions for \(X\):
- \(X \to aX\): Generates any string starting with \(a\)
- \(X \to Xb\): Generates any string ending with \(b\)
- \(X \to a\) and \(X \to b\): Generates the strings \(a\) and \(b\)
Combining these observations, any string generated by \(X\) can be any configuration of \(a\)s and \(b\)s, but crucially:
- The initial or starting string can be anything, as \(X\) allows any combination of \(a\) and \(b\).
- Using the rule \(S \to aSb\), strings can be symmetrically extended with \(a\) and \(b\) on the respective sides.
Conclusion
This allows us to conclude that the language generated is of the form \(a^*(a+b)b^*\). This means the strings can have any number of \(a\)s initially, followed by any mix of \(a\)s and \(b\)s in the middle, and ending with any number of \(b\)s.
Thus, the correct statement about the language generated by the grammar is:
The language generated by \(G\) is \(a^*(a+b)b^*\).