Lossless Join Decomposition MCQs: 12 Solved Questions with Explanations

Lossless-join questions turn on one small set: the attributes shared by the fragments. Write the intersection, compute its closure, and check whether it determines either fragment. Use the Lossless Join Decomposition question hub when a question does not carry its own inline solution link.

Lossless join decomposition criterion: Questions 1 and 2

For R -> R1, R2, R1 ∪ R2 = R proves coverage, not losslessness. With X = R1 ∩ R2, the test is X -> R1 or X -> R2 in F+.

For R(A,B,C,D), F = {A→B, C→D}, R1(A,B,C), R2(C,D), the intersection is {C} and C+ = {C,D}. It determines R2, so the join is lossless. R1(A,B) and R2(C,D) have an empty intersection, producing a possible Cartesian product. Revise Attribute Closure and Candidate Keys for GATE if needed.

Question 1

Which of the following conditions must hold for a decomposition of a relation R into R1​ and R2​ to be lossless?

• A. R1∩R2→R1

• B. R1∩R2→R2

• C. Either (A) or (B)

• D. Both (A) and (B)

Answer: C. Either (A) or (B).

X = R1 ∩ R2 must determine one fragment, not both. Hence either condition works.

Question 2

Consider a relation R(A,B,C,D) with FDs A→B and C→D. Which of the following decompositions is lossless?

• A. R1​(A,B), R2​(C,D)

• B. R1​(A,C), R2​(B,D)

• C. R1​(A,B,C), R2(C,D)

• D. None of the above

Answer: C. R1​(A,B,C), R2(C,D).

In C, C+ = {C,D} determines R2. A and B have empty intersections, so neither guarantees losslessness.

Lossy decomposition and spurious tuples: Question 3

Always r ⊆ πR1(r) ⋈ πR2(r). Lossiness makes this containment strict.

For r = {(1,x,p,10), (1,y,q,20)}, project r1 = {(1,x,p), (1,y,q)} and r2 = {(1,10), (1,20)}. Their join is (1,x,p,10), (1,x,p,20), (1,y,q,10), (1,y,q,20). The middle tuples are spurious, so r ⊂ S. Rejoining retains every original tuple.

Question 3

Let r be a relation instance with schema R = (A, B, C, D).

Define r1 = πA,B,C(r) and r2 = πA,D(r). Let S = r1 ⋈ r2, where ⋈ denotes natural join.

Given that the decomposition of R into r1 and r2 is lossy, which one of the following is true?

• A. S ⊂ r

• B. r ∪ S

• C. r ⊂ S

• D. r ⋈ S = S

Answer: C. r ⊂ S.

The join adds two cross-combinations to the originals, so r ⊂ S. Open the exact solved question.

Multivalued dependencies and lossless joins: Questions 4 and 5

For R(X,Y,Z), X ↠ Y gives lossless fragments R1(X,Y) and R2(X,Z): for fixed X, Y varies independently of Z.

Question 4

Consider a relation R(A,B,C) with the MVD A↠B. Which of the following decompositions is guaranteed to be a lossless join?

• A. R into R1​(A,B) and R2​(A,C)

• B. R into R1​(A,B,C) and R2​(B,C)

• C. R into R1​(A,B) and R2​(B,C)

• D. None

Answer: A. R into R1​(A,B) and R2​(A,C).

Here X=A, Y=B, Z=C. Projections (A,B) and (A,C) preserve the combinations required by A↠B.

Question 5

The relation schemas 𝑅1 and 𝑅2 form a Lossless join decomposition of 𝑅 if and only if

(a) $R_1 \cap R_2 \twoheadrightarrow (R_1-R_2)$

(b) $R_1 \rightarrow R_2$

(c) $R_1 \cap R_2 \twoheadrightarrow (R_2-R_1)$

(d) $R_2 \rightarrow R_1 \cap R_2$

Codes :

• A. (a) and (b) happens

• B. (a) and (d) happens

• C. (a) and (c) happens

• D. (b) and (c) happens

Answer: C. (a) and (c) happens.

For X = R1 ∩ R2, complementation pairs X ↠ (R1-R2) with X ↠ (R2-R1). Hence (a) and (c). Open the exact solved question.

Attribute closure proves the join: Questions 6 to 8

Use the same four checks each time:

Question 6

Consider the table R with attributes A, B and C. The functional dependencies that hold on

R are : A → B, C → AB. Which of the following statements is/are True ?

I. The decomposition of R into R1(C, A) and R2(A, B) is lossless.

II. The decomposition of R into R1(A, B) and R2(B, C) is lossy.

• A. Only I

• B. Only II

• C. Both I and II

• D. Neither I nor II

Answer: C. Both I and II.

For I, A+ = {A,B} determines R2. For II, B+ = {B} determines neither fragment. Both hold. Open the exact solved question.

Question 7

Which of the following statements is TRUE ?

D1 : The decomposition of the schema R(A, B, C) into R1(A, B) and R2 (A, C) is always lossless.

D2 : The decomposition of the schema R(A, B, C, D, E) having AD → B, C → DE, B → AE and AE → C, into R1 (A, B, D) and R2 (A, C, D, E) is lossless.

• A. Both D1 and D2

• B. Neither D1 nor D2

• C. Only D1

• D. Only D2

Answer: D. Only D2.

D1 gives no dependency from A. In D2, intersection {A,D} determines R1(A,B,D) through AD→B. Open the exact solved question.

Question 8

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.

Answer: A. Lossless due to X → Y,Z.

From {X}, apply X→Y, Y→Z, X→V: X+ = {X,Y,Z,V} contains all of R2. Missing W is irrelevant.

BCNF and dependency preservation: Questions 9 to 11

Losslessness asks whether joining the fragments reconstructs exactly the original relation. Dependency preservation asks whether projected FDs can enforce F without a join, while BCNF asks whether every non-trivial determinant is a fragment superkey. For projected-FD closures and preservation proofs, work through Dependency Preserving Decomposition MCQs; Questions 9 to 11 use preservation only as a separate verdict after the lossless-join test.

Question 9

Consider a relation R = (A, B, C, D, E) with the set of functional dependency FDs

{A -> ABCDE, B -> C }. Which of the following statement is true?

• A. R1 = (A, C, D, E) and R2 = (B, C) are both in BCNF and preserve lossless-join.

• B. R1 = (A, B, D, E) and R2 = (B, C) are both in BCNF and preserve lossless-join.

• C. both (a) and (b)

• D. None of the above.

Answer: B. R1 = (A, B, D, E) and R2 = (B, C) are both in BCNF and preserve lossless-join.

In B, the intersection is {B} and B→C determines R2(B,C), so the join is lossless. A is a key of R1 and B is a key of R2, so both fragments are in BCNF. The projected dependencies A→ABDE in R1 and B→C in R2 together preserve A→ABCDE.

Question 10

Consider a schema R(A,B,C,D) and functional dependencies A→B and C→D. Then the decomposition of R into R₁(AB) and R₂(CD) is

• A. dependency preserving and lossless join

• B. lossless join but not dependency preserving

• C. dependency preserving but not lossless join

• D. not dependency preserving and not lossless join

Answer: C. dependency preserving but not lossless join.

Both FDs remain local. The fragments have an empty intersection, so their join is a Cartesian product and the decomposition is lossy. DBMS Normalization MCQs uses this same GATE 2001 item inside the broader 1NF-to-BCNF ladder; here the decisive step is the empty-intersection proof. Open the exact solved question.

Question 11

Consider relation R(A,B,C,D,E,F,G) with functional dependencies F = {AD→BF, CD→EGC, BD→F, E→D, F→C, D→F}. After finding the minimal cover, R is decomposed into R1(A,B,C,D,E) and R2(A,D,F,G). Determine whether this decomposition is lossless and dependency preserving.

• A. Decomposition is lossless and dependency preserving

• B. Decomposition is lossy and dependency preserving

• C. Decomposition is lossy and not dependency preserving

• D. Decomposition is lossless and not dependency preserving

Answer: D. Decomposition is lossless and not dependency preserving.

From {A,D}, apply AD→B, D→F, F→C, then CD→E,G. Thus {A,D}+ = {A,B,C,D,E,F,G}, proving losslessness. Preservation fails because projected dependencies cannot enforce cross-fragment F→C without a join.

Multi-relation lossless decomposition: Question 12

For several fragments, join a pair whose intersection determines one member, replace it by their union, and repeat. Disjointness creates a Cartesian product.

Question 12

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

$PQ\rightarrow X;\quad P\rightarrow YX;\quad Q\rightarrow Y; \quad Y\rightarrow ZW$

 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?

• A. $D_1$ is a lossless decomposition, but $D_2$ is a lossy decomposition

• B. $D_1$ is a lossy decomposition, but $D_2$ is a lossless decomposition

• C. Both $D_1$ and $D_2$ are lossless decompositions

• D. Both $D_1$ and $D_2$ are lossy decompositions

Answer: A. $D_1$ is a lossless decomposition, but $D_2$ is a lossy decomposition.

For D1, join (Q,Y) to (Y,Z,W) by Y→ZW, add (P,Q,S,T) by Q→Y→ZW, then add (P,T,X) by P→X. Each step is lossless. In D2, (T,X) is disjoint and forces Cartesian combinations. Open the exact solved question.

The short version

Lossless-join analysis involves the binary criterion, spurious tuples, MVDs, closure, BCNF, dependency preservation and multi-relation assembly. Miss two together, then revise that method.

Write R1 ∩ R2, compute its closure, then check whether it determines either fragment. Test preservation separately. GATE Guidance by Sanchit Sir provides the full DBMS sequence; for a shorter revision loop, redo only the questions whose intersection closure you wrote incorrectly.