Solution

The correct answer is: option 2) gives lossless join, but is not dependency preserving

Key Points

Given:

• Relation: R(A, B, C, D)
• Functional Dependencies:
  • A → B
  • B → C
  • C → D
  • D → B
• Decomposition: R1(A, B), R2(B, C), R3(B, D)

Step 1: Candidate Key Identification

• From A → B, and B → C, we get A → C
• From A → C, and C → D, we get A → D
• Therefore, A → B, C, D implies A is a candidate key

Step 2: Check for Lossless Join

• The common attribute B is present in all three relations.
• And we know that A → B, and A is a candidate key.
• Therefore, the decomposition gives a lossless join as per the lossless join property using functional dependency and shared attributes.

Step 3: Check for Dependency Preservation

Functional Dependency	Preserved?	Reason
A → B	✅ Yes	Present in R1 (A, B)
B → C	✅ Yes	Present in R2 (B, C)
C → D	❌ No	C is in R2, D is in R3 – not in a single relation
D → B	✅ Yes	Present in R3 (B, D)

Since C → D is not preserved in any of the decomposed relations, the decomposition is not dependency preserving.

Additional Information

• Lossless Join: Ensures that joining the decomposed relations yields the original relation without any spurious tuples.
• Dependency Preservation: Ensures that all functional dependencies are maintained in the decomposed schema without needing to perform joins to validate them.
• To preserve all dependencies, especially C → D, the decomposition should include a relation like R3(C, D) instead of R3(B, D).

Hence, the correct answer is: option 2) gives lossless join, but is not dependency preserving