## Find the correct BCNF decomposition for a given database table

####
**Question:**

11. Consider the relation R(A, B, C, D,
E) with set of functional dependencies F = {A → B, C → D}. Let us assume that R
is in 1NF. Which of the following is the correct BCNF decomposition of R?

(a) R1 (A, C, D, E) and R2 (A, B)

(b) R1 (A, B, C, E) and R2 (C, D)

(c) R1 (A, B), R3 (C, D), and R3 (A,
C, E)

(d) All of the above.

####
**Answer:**

(c)
R1 (A, B), R3 (C, D), and R3 (A, C, E)

__Discussion:__
It is given that R is in 1NF. To
proceed further, we need to find the key of R.

*Where to start?*
From the given set F of functional
dependencies, we would understand that the attributes A, C and E are not
present in the RHS of functional dependencies. Hence, we shall start with
finding the closure of (ACE). Refer here to know how to find closure of anattribute.

__Find (ACE)__^{+}
Result = ACE

Result = ABCE (from the functional dependency A → B)

Result = ABCDE
(from the FD C → D)

At this stage, result = ABCDE which
is equal to R. Hence, ACE is a super key.

*Is (ACE) a candidate key?*
To check this, we need to find the
closure of proper subset of (ACE). If the proper subset does not contain
another super key, then ACE is a candidate key.

Proper subset of (ACE) consists of
(A), (C), (E), (AC), (CE), and (AE). Find the closure for each component. None
of them forms the super key. Hence, (ACE) is the candidate key.

*Is R in 2NF?*
No. The reason is the partial
dependency A → B. So, we need to decompose R using the FD A → B. We shall get relations
as follows;

**R1 (A, B)**– it is in 2NF.

R2 (A, C, D, E) – it is not in 2NF
because of the violating FD C → D.

We shall break R2 as follows;

**R2 (C, D)**– it is in 2NF.

**R3 (A, C, E)**– it is in 2NF.

*Are R1, R2 and R3 are in 3NF?*
3NF – Transitive dependencies must
not present in a relation.

Neither of R1, R2, and R3 have
transitive dependencies. Hence, all these relations are in 3NF.

*Are R1, R2 and R3 are in 3NF?*
BCNF – the LHS of all FDs should be
candidate key.

For R1, A → B is the only FD and A
is the candidate key.

For R2, C → D is the only FD and C
is the candidate key.

For R3, no FDs. Hence, all
attributes (ACE) form the key.

So, R1, R2, and R3 are in BCNF.

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