Attribute Closure and FD-Set Equivalence MCQs: 12 Solved Questions

An FD can become usable only after several closure rounds; FD-set equivalence repeats that fixed-point test in both directions. Write each closure as a sequence of growing sets, then check key minimality or directional coverage separately. The Attribute Closure and Equivalence of Sets question set adds more practice on the same subtopic.

Use one closure loop for every question

For an attribute set X, start with X+ = X. Scan every FD Y -> Z. If Y is contained in the current closure, add Z. Repeat complete scans until one full pass adds nothing. Then accept the result. To test X -> Y, check that every attribute of Y lies in X+. A candidate key needs a full closure and minimality, so removing any attribute must destroy the full closure.

For Questions 3 and 4, use R(A,B,C,D,E,F,G) and F = {A->B, C->D, AB->E, BE->C, EF->G}. The rounds are A+ = {A} -> {A,B} -> {A,B,E} -> {A,B,C,E} -> {A,B,C,D,E}. Stop because F is absent, so EF->G cannot fire. Starting with AF+ = {A,F} runs the same chain, then EF->G adds G, giving {A,B,C,D,E,F,G}.

For equivalence, F covers G only if F implies every FD in G. The sets are equivalent only when coverage also holds from G to F. Read the wider DBMS map first if these relational-model ideas need revision.

Questions 1-3: compute closures and know when to stop

Question 1

Consider the relation R(A, B, C, D, E, F) with the following functional dependencies:

- A -> B

- C -> DE

- AC -> F

What is the closure (AC)+?

• A. ABCDEF

• B. ACFDE

• C. ABCDE

• D. ACFD

Answer: A. ABCDEF. Start at {A,C}. A->B gives {A,B,C}, C->DE gives {A,B,C,D,E}, and AC->F completes {A,B,C,D,E,F}. B omits B, C omits F, and D omits B and E.

Question 2

Let R = ABCDE is a relational scheme with functional dependency set F = {A → B, B → C, AC → D}. The attribute closures of A and E are

• A. ABCD, φ

• B. ABCD, E

• C. Φ, φ

• D. ABC, E

Answer: B. ABCD, E. A+ grows as {A} -> {A,B} -> {A,B,C} -> {A,B,C,D}. The last FD fires after C arrives. E+ = {E} because no FD starts with E; a closure always retains its starting attributes. See the full worked solution.

Question 3

For given FD set of R(ABCDEFG)

{A->B, C->D, AB->E, BE->C, EF->G}, which of the following does NOT belong to A+?

• A. A

• B. C

• C. G

• D. E

Answer: C. G. From the Section 1 trace, A+ = {A,B,C,D,E}. Only EF->G can introduce G, but F never enters the closure.

Questions 4-6: turn closure into candidate-key and superkey decisions

Question 4

For given FD set of R(ABCDEFG) {A->B, C->D, AB->E, BE->C, EF->G}, which of the following does NOT belong to AF+?

• A. C

• B. F

• C. B

• D. None of these

Answer: D. None of these. From {A,F}, derive B, E, C, D and finally G, so AF+ = {A,B,C,D,E,F,G}. AF is minimal because A+ lacks F,G, while F+ = {F}.

Question 5

Let R = (A, B, C, D, E, F) be a relation scheme with the following functional dependencies:

C -> F, E -> A, EC -> D, A -> B

Which of the following is a key of R?

• A. CD

• B. EC

• C. AE

• D. AC

Answer: B. EC. Neither E nor C appears on a RHS, so both must be in every key. EC+ grows through E->A, A->B, C->F and EC->D to {A,B,C,D,E,F}. Removing either starting attribute prevents full closure. See the full worked solution.

Question 6

Consider a relation R(ABCDEF) with the FD set F = {A → B, B → C, C → D, D → E}. The number of super keys of R is:

• A. 8

• B. 6

• C. 16

• D. None of these

Answer: C. 16. A+ = {A,B,C,D,E}, so it lacks F, while AF is the minimal key. Every superkey contains A,F; each of B,C,D,E is optional, giving 2^4 = 16. Examples are {A,F} and {A,B,D,F}. {A,B,C,D,E} is not a superkey because it lacks F.

Questions 7-9: test whether an FD is implied

Question 7

Suppose the following functional dependencies hold on a relation U with attributes P,Q,R,S, and T:

P → QR

RS → T

Which of the following functional dependencies can be inferred from the above functional dependencies?

• A. PS → T

• B. R → T

• C. P → R

• D. PS → Q

Answer: A, C and D. P+ = {P,Q,R}, proving C. PS+ gains Q,R, then RS->T adds T, proving A and D. R+ = {R}, so B fails without S. See the full worked solution.

Question 8

Given the relational schema 𝑅 = (𝑈,𝑉, 𝑊, 𝑋, 𝑌, 𝑍) and the set of functional dependencies:

{𝑈 → 𝑉, 𝑈 → 𝑊, 𝑊𝑋 → 𝑌, 𝑊𝑋 → 𝑍, 𝑉 → 𝑋}

Which of the following functional dependencies can be derived from the above set?

• A. 𝑉𝑊 → 𝑌𝑍

• B. 𝑊𝑋 → 𝑌𝑍

• C. 𝑉𝑊 → 𝑈

• D. 𝑉𝑊 → 𝑌

Answer: A, B and D. Union gives WX->YZ, proving B. From {V,W}, V->X makes both WX dependencies usable, so VW+ = {V,W,X,Y,Z} and A and D hold. U is absent, so C fails. See the full worked solution.

Question 9

In a schema with attributes A, B, C, D and E following set of functional dependencies are given

A → B

A → C

CD → E

B → D

E → A

Which of the following functional dependencies is NOT implied by the above set?

• A. CD → AC

• B. BD → CD

• C. BC → CD

• D. AC → BC

Answer: B. BD → CD. CD+ reaches E,A,B, so it contains AC. BD+ = {B,D} and lacks C. BC+ gains D, while AC+ gains B. Thus only B is not implied. See the full worked solution.

Questions 10-12: detect redundancy and compare two FD sets

Question 10

An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionality dependency set $𝐹$ is given as: F:{A→BC,D→E,BC→D,A→D\}. Consider a universal relation (R(A, B, C, D, E) with functional dependency set 𝐹. Also all attributes are simple and take atomic values only.

Identify the redundant functional dependency in F

• A. BC→D

• B. D→E

• C. A→D

• D. A→BC

Answer: C. A→D. Remove A->D. The remaining FDs take A+ through {A,B,C}, then add D through BC->D and E through D->E, so A->D is still implied. By contrast, without BC->D, BC+ = {B,C}. See the full worked solution.

Question 11

Consider the following sets of functional dependencies over a relation R={p,q,r,s,t}

S1 = {t→q, pr→q, pq→st, r→p}

S2 = {rs→p, q→t, p→s}

Which of the following is true:

• A. S1 covers S2

• B. S2 covers S1

• C. S1 and S2 are equivalent

• D. none of the above

Answer: D. none of the above. Under S2, q+ = {q,t}, but under S1, q+ = {q}, so S1 does not cover S2. Under S1, pr+ = {p,r,q,s,t}; under S2, it reaches only {p,r,s}. Therefore S2 does not cover S1 either.

Question 12

F = {A->B, B->C, C->A}

G = {A->BC, B->AC, BC->A, AB->C}

Then which of the following is true? Here, F covers G when F logically implies every FD in G.

• A. F covers G

• B. G covers F

• C. Both of them cover each other

• D. None of them cover each other

Answer: A. F covers G. Under F, the chains imply A->BC, B->AC, BC->A and AB->C, covering all of G. Under G, C+ = {C}, so G cannot imply C->A from F.

Diagnose the traps, score the set and choose the next step

Use these as study thresholds, not official cutoffs. At 10-12 correct, redo Questions 8 and 11 without the explanations. At 7-9, write closures by rounds and repeat Questions 3, 6 and 9. At 0-6, reproduce A+ and AF+ from the first section, then retry Questions 1-6 before equivalence.

For a semester-oriented CS route, use ZERO TO HERO. For placement-focused DBMS revision, use Computer Science Fundamentals for Placements by Sanchit Sir. Browse the CS Fundamentals collection.

The short version is simple: grow a closure to a fixed point, test the RHS against it, check key minimality separately, and prove FD-set equivalence in both directions. Now hide every answer, write the closure sequence for all 12 items, and circle the first FD that adds a new attribute in each round.