Consider the augmented grammar with $\{+, *, (, ), id\}$ as the set of terminals. $$S^{\prime} \rightarrow S$$ $$S \rightarrow S + R \mid R$$ $$R \rightarrow R * P \mid P$$ $$P \rightarrow (S) \mid id$$ If $I_0$ is the set of two $LR(0)$ items $\{[S^{\prime} \rightarrow S.], [S \rightarrow S. + R]\}$, then $goto(closure(I_0), +)$ contains exactly ________ items.

Given the item set $I_0 = \{[S^{\prime} \rightarrow S.], [S \rightarrow S. + R]\}$.

Since no item in $I_0$ has the dot immediately preceding a non-terminal, $closure(I_0) = I_0$.

To compute $goto(closure(I_0), +)$:

1. Shift the dot over $+$ in the item $[S \rightarrow S. + R]$:

   $$\{[S \rightarrow S + . R]\}$$

2. Find the closure of this item:

   • Since the dot is before $R$, add all productions of $R$ with the dot at the beginning:

     $$[R \rightarrow . R * P]$$

     $$[R \rightarrow . P]$$

   • Since the dot in $[R \rightarrow . P]$ is before non-terminal $P$, add all productions of $P$ with the dot at the beginning:

     $$[P \rightarrow . (S)]$$

     $$[P \rightarrow . id]$$

Collecting all items, $goto(closure(I_0), +)$ consists of the following $5$ items:

1. $[S \rightarrow S + . R]$

2. $[R \rightarrow . R * P]$

3. $[R \rightarrow . P]$

4. $[P \rightarrow . (S)]$

5. $[P \rightarrow . id]$

Hence, the set contains exactly 5 items.