If $G$ is a grammar with productions $$S \rightarrow SaS \mid aSb \mid bSa \mid SS \mid \epsilon$$ where $S$ is the start variable, then which one of the following strings is not generated by $G$?

Let $n_a(w)$ and $n_b(w)$ denote the number of occurrences of 'a' and 'b' in string $w$, respectively.

Analyzing each production of the grammar:

• $S \rightarrow SaS$: introduces one 'a' and zero 'b' ($n_a - n_b$ increases by 1)

• $S \rightarrow aSb$: introduces one 'a' and one 'b' ($n_a - n_b$ unchanged)

• $S \rightarrow bSa$: introduces one 'a' and one 'b' ($n_a - n_b$ unchanged)

• $S \rightarrow SS$: concatenation of two valid strings

• $S \rightarrow \epsilon$: terminal step with zero terminals ($n_a - n_b = 0$)

Therefore, every string $w$ generated by this grammar must satisfy the property:

Checking the options:

• A. $abab$: $n_a = 2, n_b = 2 \implies n_a \ge n_b$ (can be generated)

• B. $aaab$: $n_a = 3, n_b = 1 \implies n_a \ge n_b$ (can be generated)

• C. $abbaa$: $n_a = 3, n_b = 2 \implies n_a \ge n_b$ (can be generated)

• D. $babba$: $n_a = 2, n_b = 3 \implies n_a < n_b$ (cannot be generated)

Hence, the string babba cannot be generated by $G$, making D the correct answer.