Consider the canonical $LR(0)$ parsing of the grammar below using terminals $\{a, b, c\}$ and non-terminals $\{A, B, C, S\}$ with $S$ as the start symbol. $$S \rightarrow ACB$$ $$A \rightarrow aA \mid \epsilon$$ $$C \rightarrow cC \mid \epsilon$$ $$B \rightarrow bB \mid b$$ Which one of the following options gives the number of shift-reduce conflicts that will occur in the $LR(0)$ ACTION table?

To find the number of shift-reduce (S-R) conflicts in the canonical $LR(0)$ ACTION table, we first construct the augmented grammar and its canonical collection of $LR(0)$ items:

Augmented Grammar:

1. $S^{\prime} \rightarrow S$

2. $S \rightarrow ACB$

3. $A \rightarrow aA$

4. $A \rightarrow \epsilon$

5. $C \rightarrow cC$

6. $C \rightarrow \epsilon$

7. $B \rightarrow bB$

8. $B \rightarrow b$

Canonical Collection of $LR(0)$ Sets of Items:

• State $I_0$:

  $$\text{CLOSURE}(\{S^{\prime} \rightarrow .S\}) = \{S^{\prime} \rightarrow .S,\ S \rightarrow .ACB,\ A \rightarrow .aA,\ A \rightarrow .\}$$

  • Shift on terminal $a$ (from $A \rightarrow .aA$).

  • Reduce on all terminals including $a$ (from $A \rightarrow .$).

  • S-R conflict #1:$(I_0, a)$

• State $I_1 = \text{GOTO}(I_0, S)$: $$\{S^{\prime} \rightarrow S.\}$$ (Accept state, no conflict)

• State $I_2 = \text{GOTO}(I_0, A)$:

  $$\{S \rightarrow A.CB,\ C \rightarrow .cC,\ C \rightarrow .\}$$

  • Shift on terminal $c$ (from $C \rightarrow .cC$).

  • Reduce on all terminals including $c$ (from $C \rightarrow .$).

  • S-R conflict #2:$(I_2, c)$

• State $I_3 = \text{GOTO}(I_2, C)$:

  $$\{S \rightarrow AC.B,\ B \rightarrow .bB,\ B \rightarrow .b\}$$

  (No reduce items, so no S-R conflict)

• State $I_4 = \text{GOTO}(I_2, c)$:

  $$\{C \rightarrow c.C,\ C \rightarrow .cC,\ C \rightarrow .\}$$

  • Shift on terminal $c$ (from $C \rightarrow .cC$).

  • Reduce on all terminals including $c$ (from $C \rightarrow .$).

  • S-R conflict #3:$(I_4, c)$

• State $I_5 = \text{GOTO}(I_0, a)$:

  $$\{A \rightarrow a.A,\ A \rightarrow .aA,\ A \rightarrow .\}$$

  • Shift on terminal $a$ (from $A \rightarrow .aA$).

  • Reduce on all terminals including $a$ (from $A \rightarrow .$).

  • S-R conflict #4:$(I_5, a)$

• State $I_6 = \text{GOTO}(I_3, B)$: $$\{S \rightarrow ACB.\}$$ (Reduce item only, no conflict)

• State $I_7 = \text{GOTO}(I_3, b)$:

  $$\{B \rightarrow b.B,\ B \rightarrow b.,\ B \rightarrow .bB,\ B \rightarrow .b\}$$

  • Shift on terminal $b$ (from $B \rightarrow .bB,\ B \rightarrow .b$).

  • Reduce on all terminals including $b$ (from $B \rightarrow b.$).

  • S-R conflict #5:$(I_7, b)$

• State $I_8 = \text{GOTO}(I_4, C)$:$\{C \rightarrow cC.\}$

• State $I_9 = \text{GOTO}(I_5, A)$:$\{A \rightarrow aA.\}$

• State $I_{10} = \text{GOTO}(I_7, B)$:$\{B \rightarrow bB.\}$

Thus, there are a total of 5 shift-reduce conflicts in the $LR(0)$ ACTION table:

1. In state $I_0$ on terminal $a$

2. In state $I_2$ on terminal $c$

3. In state $I_4$ on terminal $c$

4. In state $I_5$ on terminal $a$

5. In state $I_7$ on terminal $b$

Hence, the correct option is (D).