Answer – A
In Relation R (A, B, C, D) –
- Closure of A = { A, B, C, D }
- Closure of B = { B, C, D }
- Closure of C = { B, C, D }
- Closure of D = { B, C, D }
Now, R is decomposed into –
- R1 (A, B) having FD { A → B }; A is the super key.
- R2 (B, C) having FD’s { B → C, C → B }; B and C both are super keys.
- R3 (B, D) having FD’s { B → D, D → B }; B and D both are super keys.
If the common attribute of 2 relation is a super key of any relation, then their join is lossless.
Here, B is common in all 3 relations and B is a super key of 2 relations among them. Thus, join is lossless.
Original relation R has 4 non-trivial functional dependencies.
- A → B. This can again be obtained from FD of R1.
- B → C. This can again be obtained from FD’s of R2.
- C → D. This can again be obtained from FD’s of R2 and R3 ie C → B and B → D thus C → D.
- D → B. This can again be obtained from FD’s of R3.
Thus, join is dependency preserving also.