Let's take CNF as our benchmark. Productions in CNF are of the form:-
$A\rightarrow BC$ (allowed as per question)
$A\rightarrow b$ (not allowed as per question)
Since $A\rightarrow b$ isn't allowed, take two samples
- $A\rightarrow bb$
- $A\rightarrow bbb$
We know, CNF takes $2|w|-1$ steps to derive the string. => It takes $2|w|-1$ reductions.
Take
$A\rightarrow bb$ in modified CNF
Instead of deriving $w$ symbols in $2|w|-1$ steps, we actually derived $2w$ symbols in $2|w|-1$ steps. So for w symbols, we took $|w|-1$ steps.
Take
$A\rightarrow bbb$ in modified CNF
Instead of deriving $w$ symbols in $2|w|-1$ steps, we actually derived $3w$ symbols in $2|w|-1$ steps. So for $w$ symbols, we took $\frac{2|w|}{3}-1$ steps.
Clearly if we keep on adding terminals on the RHS, we'll keep getting lesser steps.
To maximise the number, $A\rightarrow bb$ is most suitable (as lower than that is illegal as per question)
And $A\rightarrow bb$ fused with CNF takes $|w|-1$ steps for w tokens, as already seen.
Option B