Let's solve the problem step by step: --- ### Step 1: Understanding Arc Consistency **Arc consistency** for A -> B means: For every value in the domain of A, there must be **some** value in the domain of B such that the pair is consistent with the constraint between A and B. --- ### Step 2: Initial Domains After enforcing unary constraints, the domains are: | A | B | |-------|-------| | (1,1) | | | (1,3) | (1,3) | | (2,2) | | | (2,3) | (2,3) | So, for both A and B: A: {(1,1), (1,3), (2,2), (2,3)} B: {(1,3), (2,3)} --- ### Step 3: Enforcing Arc Consistency on A -> B For each value in A's domain, check if there is a value in B's domain that is consistent. #### Let's check each value in A: - **A = (1,1):** Is there a (1,3) or (2,3) in B that satisfies the constraint with (1,1)? *We don't know the constraint, but typically, it's equality on the second element or something similar.* Let's check if (1,1) matches the second value of (1,3) or (2,3): - (1,1) vs (1,3) → second values: 1 vs 3 (not equal) - (1,1) vs (2,3) → second values: 1 vs 3 (not equal) So, **(1,1) has no support in B**. - **A = (1,3):** (1,3) vs (1,3): 3 vs 3 (match) So, **(1,3) stays**. - **A = (2,2):** (2,2) vs (1,3): 2 vs 3 (not equal) (2,2) vs (2,3): 2 vs 3 (not equal) So, **(2,2) has no support in B**. - **A = (2,3):** (2,3) vs (2,3): 3 vs 3 (match) So, **(2,3) stays**. --- **After enforcing A -> B**, the domain of A is reduced to only those values that have a corresponding value in B: A: {(1,3), (2,3)} B: {(1,3), (2,3)} (no change in B) --- ### Step 4: Compare with the Options Let's see which option matches: #### i A: (1,1), (1,3), (2,2), (2,3) B: (1,3), (2,3) *-- A is not reduced, so this is not correct.* #### ii A: (1,3), (2,3) B: (1,3), (2,3) *-- This matches our result!* #### iii A: (1,1), (1,3), (2,2), (2,3) B: (1,3), (2,3) *-- Same as i, so not correct.* #### iv A: (1,1), (1,3), (2,2), (2,3) B: (1,3) *-- B is reduced incorrectly, not correct.* --- ## **Final Answer** **ii** is the correct option. After enforcing arc consistency on A -> B, the domains are: | A | B | |-------|-------| | (1,3) | (1,3) | | (2,3) | (2,3) |

Let's analyze the problem carefully, step by step, to understand how enforcing arc consistency on A → B affects the domains and to identify the correct option among the given choices. --- ### **1. Understanding the Initial Domains after Unary Constraints** Given the table: | A | B | |--------|--------| | (1,1) | | | (1,3) | (1,3) | | (2,2) | | | (2,3) | (2,3) | The initial domains are: - **Domain of A:** {(1,1), (1,3), (2,2), (2,3)} - **Domain of B:** {(1,3), (2,3)} *(Note: The empty cells imply that the unary constraints have limited the possible values. The domains are as above.)* --- ### **2. Enforcing Arc Consistency on A → B** **Goal:** For each value in A’s domain, ensure there exists **at least one** value in B’s domain satisfying the constraint between A and B. *Assumption:* The constraint between A and B is that their second values must be equal (a common scenario in such problems). --- ### **3. Check each value in A:** #### **a. A = (1,1):** - **B values:** {1, 3} - **Constraint:** second value of A = second value of B - Does (1,1) support any B value? - (1,3): second value 3 ≠ 1 → **No support** - (2,3): second value 3 ≠ 1 → **No support** **Result:** (1,1) has **no support** in B. **Thus, (1,1) is eliminated from A’s domain.** --- #### **b. A = (1,3):** - B: { (1,3), (2,3) } - Check support: - (1,3): second value 3 = 3 → **Support** - (2,3): second value 3 = 3 → **Support** **Result:** (1,3) **remains** in A’s domain. --- #### **c. A = (2,2):** - B: { (1,3), (2,3) } - Check support: - (1,3): second value 3 ≠ 2 → **No support** - (2,3): second value 3 ≠ 2 → **No support** **Result:** (2,2) has **no support** and is eliminated. --- #### **d. A = (2,3):** - B: { (1,3), (2,3) } - Check support: - (1,3): second value 3 = 3 → **Support** - (2,3): second value 3 = 3 → **Support** **Result:** (2,3) **remains**. --- ### **4. Updated Domains after enforcing A → B** - **A:** {(1,3), (2,3)} - **B:** remains {(1,3), (2,3)} (no change) --- ### **5. Comparing with the options** Let's interpret the options: | Option | A Domain after enforcement | B Domain after enforcement | Comments | |---------|------------------------------|------------------------------|----------| | **i** | (1,1), (1,3), (2,2), (2,3) | (1,3), (2,3) | No reduction in A, so **incorrect** | | **ii** | (1,3), (2,3) | (1,3), (2,3) | Matches our result — **correct** | | **iii** | Same as i | (1,3), (2,3) | No reduction in A, so **incorrect** | | **iv** | (1,1), (1,3), (2,2), (2,3) | (1,3) | B domain incorrectly reduced, **incorrect** | ### **6. Final answer:** **Option ii** correctly reflects the updated domains after enforcing arc consistency on A → B. --- ### **Summary:** - The key step was checking each value in A's domain against B's domain for support. - Values (1,1) and (2,2) are eliminated because they lack supporting values in B. - The final domains are **A: {(1,3), (2,3)}** and **B: {(1,3), (2,3)}**. **Therefore, the correct choice is:** # **Option ii**

Certainly! Here's a detailed explanation of each option with subheadings, clarifying why each is correct or incorrect, followed by the final answer. --- ## **Initial Setup and Assumptions** - The initial domains after unary constraints are: | Variable | Domain | |------------|----------------------------| | A | {(1,1), (1,3), (2,2), (2,3)} | | B | {(1,3), (2,3)} | - The arc we're enforcing is **A → B**, meaning for each value in A's domain, there must be **supporting** values in B's domain satisfying the constraint. - **Assumption:** The constraint between A and B is that their second elements are equal, i.e., `A.second = B.second`. --- ## **Step-by-step enforcement of A → B** ### Check each value in A: - **(1,1):** Support in B? - B = (1,3): second value 3 ≠ 1 → **No** - B = (2,3): second value 3 ≠ 1 → **No** **Result:** No support; eliminate (1,1). - **(1,3):** Support? - B = (1,3): 3 = 3 → **Yes** - B = (2,3): 3 = 3 → **Yes** **Result:** Keep (1,3). - **(2,2):** Support? - B = (1,3): 3 ≠ 2 → **No** - B = (2,3): 3 ≠ 2 → **No** **Result:** No support; eliminate (2,2). - **(2,3):** Support? - B = (1,3): 3 = 3 → **Yes** - B = (2,3): 3 = 3 → **Yes** **Result:** Keep (2,3). --- ## **Final Domains after enforcing A → B** | Variable | Domain | |------------|----------------------------| | A | {(1,3), (2,3)} | | B | {(1,3), (2,3)} | --- ## **Analysis of Options** ### **Option i** **Domains:** - A: (1,1), (1,3), (2,2), (2,3) - B: (1,3), (2,3) **Explanation:** This option **does not** reflect the elimination of unsupported values in A's domain. Since (1,1) and (2,2) have no support in B, they should have been removed. Therefore, **this option is incorrect**. --- ### **Option ii** **Domains:** - A: (1,3), (2,3) - B: (1,3), (2,3) **Explanation:** This matches exactly the domains after applying arc consistency: unsupported values (1,1) and (2,2) are removed. **This is the correct option.** --- ### **Option iii** **Domains:** - A: (1,1), (1,3), (2,2), (2,3) - B: (1,3), (2,3) **Explanation:** This is identical to Option i; it fails to eliminate unsupported values in A. **Incorrect**. --- ### **Option iv** **Domains:** - A: (1,1), (1,3), (2,2), (2,3) - B: (1,3) **Explanation:** Here, B's domain is reduced to only (1,3), which is inconsistent with the support check; (2,3) is supported in B and should remain. **Incorrect**. --- ## **Final Conclusion** The only option that correctly reflects the updated domains after enforcing arc consistency on A → B is: ### **Answer:** **Option ii** --- If you need further clarifications or additional steps, feel free to ask!