Question

Consider the relation R(P, Q, S, T, X, Y, Z, W) with the following functional dependencies.

PQ → X; P → YX; Q → Y; Y → ZW

Consider the decomposition of the relation R into the constituent relations according to the following two decomposition schemes.

D₁ : R = [(P, Q, S, T); (P, T, X); (Q, Y); (Y, Z, W)]

D₂ : R = [(P, Q, S); (T, X); (Q, Y); (Y, Z, W)]

Which one of the following options is correct?

1.

D₁ is a lossy decomposition, but D₂ is a lossless decomposition.

2. Both D₁ and D₂ are lossless decompositions.
3. Both D₁ and D₂ are lossy decompositions.
4. D₁ is a lossless decomposition, but D₂ is a lossy decomposition.

Answer 1

Correct Answer - Option 4 : D₁ is a lossless decomposition, but D₂ is a lossy decomposition.

Answer: Option 4

Concept:

Lossless Decomposition:

for a Decomposition of two Relation, R₁ and R₂ to be lossless 2 condition needs to be satisfied that is

1. R1 ∩ R2 → R1 or R₂ i.e. common attributes must be key to either of the relation.

2. attributes of R1 ∪ attributes of R2 ≡ attributes of R

Explanation:

D1 : R = [(P, Q, S, T); (P, T, X); (Q, Y); (Y, Z, W)]

lets first take 2 relations R₁(P, Q, S, T )  R₂(P, T, X) 

common attributes are PT and PT → TX ( according to augmentation property )

so relation becomes R₁(P, Q, S, T, X) R₂(Q, Y) 

The common attribute is Q and Q→ Y is key to R₂ Hence (P, Q, S, T, X, Y)

 So now relation becomes R1(P, Q, S, T, X, Y) R2(Y, Z, W)

 The common attribute is Y and Y is key to R₂.

Hence all attributes get combined into one relation and hence this Decomposition is lossless.

D2 : R = [(P, Q, S); (T, X); (Q, Y); (Y, Z, W)]

If you observe relation (T, X); Its attributes not common to any other relations.

even if we combined all other attributes R₁(P, Q, S, Y, Z, W)  R₂(T, X) 

still no common attributes Hence this decomposition is lossy.