## Is the given decomposition a lossless join decomposition?

####
**Question:**

10. Assume a relation R(A, B, C, D, E)
with set F of functional dependencies as follows;

F = {A → BC, CD → E, B → D, E → A}

If R is decomposed in to R1(A, B, C)
and R2(A, D, E), which of the following holds?

(a) Decomposition is Loss-less Join Decomposition

(b) Decomposition is Dependency Preserving Decomposition

(c) Decomposition is both (a) and
(b)

(d) Decomposition is Lossy-Join
Decomposition

####
**Answer:**

(a) Decomposition is Loss-less Join
Decomposition

__Discussion:__**Is the given decomposition is loss-less?**

**YES**

If either R1 ⋂ R2 → R1 is true or R1 ⋂ R2 → R2 is true for the given decomposition, then we would say that the decomposition is loss-less join decomposition.

What is R1 ⋂ R2? – The common
attribute(s) between R1 and R2.

In our case, R1 ⋂ R2 is A. We have
to check whether A can determine all attributes of R1 or all attributes of R2
uniquely using the given FDs.

From F we can determine A+ as
follows using the closure finding algorithm;

Result = A

Result = ABC from A → BC

Result = ABCD from B → D

Result = ABCDE from CD → E

From this, we would know that A is a
candidate key for R.

A determines A, B, and C which is
R1. So, R1 ⋂ R2 → R1 is true.

Hence, the given decomposition is
loss-less join decomposition.

**Is the given decomposition is dependency preserving?**

**NO**

The above said decomposition of R into R1 and R2 is a dependency preserving decomposition if (F1 U F2)

^{+}= F

^{+}, where F1 is set of FDs hold by R1, F2 is set of FDs hold by R2, and F is the set of FDs hold by R. (F1 U F2)

^{+}is the closure of (F1 U F2), and F

^{+}is the closure of F.

Step 1: for R1, the derivable
non-trivial functional dependency is A → BC. Hence, F1 = {A →
BC}

Step 1: for R1, the derivable
non-trivial functional dependency is E → A. Hence, F2 = {E → A}

(F1 U F2)

^{+}= {A → BC, E → A}
F

^{+}= {A → BC, CD → E, B → D, E → A, E → BC, A → E}
A → BC is there in F

^{+}, hence preserved.
E → A is there in F

^{+}, hence preserved.
CD → E, and B → D are not there in
F

^{+}, hence not preserved.
Hence, the decomposition is not
dependency preserving.