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? A) D1 is a lossless decomposition, but D2 is a lossy decomposition. B) D1 is a lossy decomposition, but D2 is a lossless decomposition. C) Both D1 and D2 are lossless decompositions. D) Both D1 and D2 are lossy decompositions.

Official answer: A) D1 is a lossless decomposition, but D2 is a lossy decomposition.. Explanation: A decomposition of relation R into R1, R2, , Rk is lossless if we can successively join pairs of subrelations such that the common attributes form a superkey of at least one of the subrelations. ### Analysis of Decomposition D1: Constituent subrelations: R1(P, Q, S, T), R2(P, T, X), R3(Q, Y), R4(Y, Z, W). 1. Join R3(Q, Y) and R4(Y, Z, W): - Common attribute: R3 R4 = \Y\. - Given Y ZW, Y is a candidate key for R4. - Thus, R3 R4 = R34(Q, Y, Z, W) is a lossless join. 2. Join R1(P, Q, S, T) and R2(P, T, X): - Common attributes: R1 R2 = \P, T\. - Since P YX P X, it follows that PT X. Hence, \P, T\ is a superkey of R2. - Thus, R1 R2 = R12(P, Q, S, T, X) is a lossless join. 3. Join R12(P, Q, S, T, X) and R34(Q, Y, Z, W): - Common attribute: R12 R34 = \Q\. - Attribute closure of Q under the given FDs: Q+ = \Q, Y, Z, W, X\. - Therefore, Q \Q, Y, Z, W\, meaning Q is a superkey of R34. - Thus, R12 R34 = R(P, Q, S, T, X, Y, Z, W) is lossless. Hence, D1 is a lossless decomposition. --- ### Analysis of Decomposition D2: Constituent subrelations: R1(P, Q, S), R2(T, X), R3(Q, Y), R4(Y, Z, W). - Notice attribute T appears only in R2(T, X). - The other relations contain attributes \P, Q, S, Y, Z, W\. - The intersection of R2(T, X) with the union of all other relations is \X\. - For a lossless join with R2, either X T or X (P, Q, S, Y, Z, W) must hold. However, X+ = \X\, so X cannot determine T or the other attributes. Hence, D2 is a lossy decomposition. Therefore, A is the correct option.

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

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

C) Both D1 and D2 are lossless decompositions.

D) Both D1 and D2 are lossy decompositions.

Databases · Integrity Constraints and Normal Forms

Official IIT answer key · IIT Bombay · Audited Aug 2026

Take topic test

GATE CSE 2021 Set 1 Q33 · MCQ · 2 marks

Consider the relation $R(P, Q, S, T, X, Y, Z, W)$ with the following functional dependencies. $PQ \rightarrow X;\ P \rightarrow YX;\ Q \rightarrow Y;\ Y \rightarrow ZW$ Consider the decomposition of the relation $R$ into the constituent relations according to the following two decomposition schemes. $D_1 :\quad R = [(P, Q, S, T);\ (P, T, X);\ (Q, Y);\ (Y, Z, W)]$ $D_2 :\quad R = [(P, Q, S);\ (T, X);\ (Q, Y);\ (Y, Z, W)]$ Which one of the following options is correct?

Select an option to check

Key concept

Integrity Constraints and Normal Forms

Step-by-step verified solution

Examiner trap & misconception

Topic notes

How GATE tests Integrity Constraints and Normal Forms

In GATE, Normal Forms questions test a candidate's ability to identify candidate keys, classify relational schemas into appropriate normal forms (such as 3NF and BCNF), and evaluate decomposition properties (lossless join and dependency preservation). Questions combine normal form verification with decomposition analysis, frequently formatted as multi-select questions (MSQs) requiring rigorous evaluation of all properties across both the original and decomposed relations.

Full Integrity Constraints and Normal Forms guide →

More on Integrity Constraints and Normal Forms

All 30 questions on Integrity Constraints and Normal Forms →

Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.

Databases · Integrity Constraints and Normal Forms

Official IIT answer key · IIT Bombay · Audited Aug 2026

Take topic test

GATE CSE 2021 Set 1 Q33 · MCQ · 2 marks

Consider the relation $R(P, Q, S, T, X, Y, Z, W)$ with the following functional dependencies. $PQ \rightarrow X;\ P \rightarrow YX;\ Q \rightarrow Y;\ Y \rightarrow ZW$ Consider the decomposition of the relation $R$ into the constituent relations according to the following two decomposition schemes. $D_1 :\quad R = [(P, Q, S, T);\ (P, T, X);\ (Q, Y);\ (Y, Z, W)]$ $D_2 :\quad R = [(P, Q, S);\ (T, X);\ (Q, Y);\ (Y, Z, W)]$ Which one of the following options is correct?

Select an option to check

Key concept

Integrity Constraints and Normal Forms

Step-by-step verified solution

Examiner trap & misconception

Topic notes

How GATE tests Integrity Constraints and Normal Forms

Full Integrity Constraints and Normal Forms guide →

More on Integrity Constraints and Normal Forms

All 30 questions on Integrity Constraints and Normal Forms →

Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.

Step-by-step verified solution

Check you can still solve Integrity Constraints and Normal Forms under a clock

More from GATE CSE 2021 Set 1

More on Integrity Constraints and Normal Forms

Step-by-step verified solution

Check you can still solve Integrity Constraints and Normal Forms under a clock

More from GATE CSE 2021 Set 1

More on Integrity Constraints and Normal Forms

Step-by-step verified solution

Check you can still solve Integrity Constraints and Normal Forms under a clock

More from GATE CSE 2021 Set 1

More on Integrity Constraints and Normal Forms

Share this solution

Step-by-step verified solution

Check you can still solve Integrity Constraints and Normal Forms under a clock

More from GATE CSE 2021 Set 1

More on Integrity Constraints and Normal Forms

Share this solution