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## Which of the FDs are correct for R to be in BCNF?, Boyce-Codd Normal Form

#### Question:

13. Let us consider a relation R with the schema R (A, B, C, D). If we expect R to be in BCNF, then which of the following sets of functional dependencies must hold in R?

(a) A → C, A → D, AD → C, B → A
(b) C → B, C → D, A → C, D → A
(c) C → D, CD → A, BD → A, AB → C
(d) A → D, C → A, AC → B, D → B

(b) C → B, C → D, A → C, D → A

Discussion/Reason:

For a relation to be in BCNF, then all the FDs must have candidate keys on their left hand side.
Let us find the closure (refer here for attribute closure) for LHS of each FD of option (b).
 Closure of C (C+) FD causes the result result = CB C → B result = CBD C → D result = CBDA D → A
From the above table, the result of closure of C = ABCD = R. Hence, C is one of the candidate keys.
 Closure of A (A+) FD causes the result result = AC A → C result = ACB C → B result = ACBD C → D
From the above table, the result of closure of A = ABCD = R. Hence, A is one of the candidate keys.
 Closure of D (D+) FD causes the result result = DA D → A result = DAC A → C result = DACB C → B
From the above table, the result of closure of D = ABCD = R. Hence, D is one of the candidate keys.
For option (b), all the FDs left hand sides are candidate keys. So, R should hold these FDs to be in Boyce Codd Normal Form.

You try for other FDs given in other options. The LHS will not form a key.

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