Consider the relation R(V, with functional dependencies {Z→Y, Y→Z, X→Y,
Consider the relation R(V, W, X, Y, Z) with functional dependencies {Z→Y, Y→Z, X→Y, X→V, VW→X}.
Suppose that relation R is decomposed into two relations, R1(V, W, X) and R2(X, Y, Z). Is this decomposition a lossless decomposition?
- A.
Lossless due to X → Y,Z.
- B.
Lossless due to VW → X.
- C.
Lossy because X is insufficient.
- D.
Lossy due to dependency break.
Attempted by 117 students.
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Correct answer: A
Step 1: Identify the common attributes.
R1 ∩ R2 = {X}.
Step 2: Compute the closure of the common attribute X using the given FDs.
Start: X+ = {X}.
From X → Y, add Y: X+ = {X, Y}.
From Y → Z, add Z: X+ = {X, Y, Z}.
No FD adds V or W to X+, so closure stops at {X, Y, Z}.
Conclusion: X+ = {X, Y, Z}, which equals all attributes of R2. Therefore the common attribute X functionally determines R2, so the decomposition R1(V, W, X) and R2(X, Y, Z) is lossless.
Note: Although VW → X is a given FD, it does not help the lossless-join test here because it shows VW determines X, not that the intersection X determines the other attributes.