What is the maximum number of reduce moves that can be taken by a bottom-up parser for a grammar with no epsilon- and unit-production (i.e., of type $\text{A} \rightarrow \epsilon$ and $\text{A} \rightarrow \text{a}$) to parse a string with $n$ tokens?

In a bottom-up parser, each reduce move replaces a handle (the right-hand side of a production rule) with a non-terminal.

• The input initially consists of $n$ terminal tokens.

• Since the grammar contains no epsilon-productions ($A \rightarrow \epsilon$) and no single-symbol productions ($A \rightarrow a$), every production rule must have at least two symbols on its right-hand side (i.e., length $\ge 2$).

• A reduction using a rule of length $k \ge 2$ reduces the total count of symbols on the stack by $k - 1 \ge 1$.

• To reduce $n$ tokens to a single start symbol, the net reduction in symbol count must be $n - 1$.

• The maximum number of reductions is achieved when every reduction consumes the minimum possible number of symbols, which is $2$ symbols (reducing $2$ symbols to $1$, yielding a reduction of $1$ symbol per move):

  $$\text{Maximum reduce moves} = n - 1$$

Thus, option (B) is correct.