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Minimal Covers - Free MCQ Practice Test with solutions, GATE CSE (CSE)


MCQ Practice Test & Solutions: Test: Minimal Covers (10 Questions)

You can prepare effectively for Computer Science Engineering (CSE) Database Management System (DBMS) with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Minimal Covers". These 10 questions have been designed by the experts with the latest curriculum of Computer Science Engineering (CSE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 10

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Test: Minimal Covers - Question 1

Consider the following set of functional dependencies on the relation (ABC), {A -> B, AB -> C, A -> BC, B-> C}. Find the canonical cover for the above FD set.

Detailed Solution: Question 1

As, A -> BC =>  A -> B and A -> C so, the FD A -> B is redundant.
Similarly, B -> C can be implied by AB -> C.
So the canonical cover will be {A -> BC, AB ->C}

Test: Minimal Covers - Question 2

What is the canonical cover of the following set F of functional dependencies on the schema(A, B,C)?

A → BC 
B → C 
A → B
AB → C

Detailed Solution: Question 2

There are two functional dependencies with the same set of attributes on the left side of the arrow:

A → BC
A → B

We can combine these functional dependencies into A → BC.
A is extraneous in AB → C. This is true because B → C is already there in the given set of functional dependencies. 
C is extraneous in A → BC because it is logically implied by A → B and B → C.

Thus, the canonical cover is:
A → B
B → C

Test: Minimal Covers - Question 3

What is the canonical cover for F?

F = A → BC, CD → E, B → D, E → A

Detailed Solution: Question 3

There is no extraneous attribute in left hand side of FDs and the FDs cannot be reduced further.

Test: Minimal Covers - Question 4

Consider the following set of functional dependency on the schema (A, B,C)
A→BC, B→C, A→B, AB→C

The canonical cover for this set is:

Detailed Solution: Question 4

All of the FDs are implied by the FDs {A->BC, B->C}
From option b: A+=BC (hence A->BC, A->b and AB->Care implied by FDs set given in option b)

   B+=BC (hence B->C is implied by FDs set given in option b)

Test: Minimal Covers - Question 5

 A _________ Fc for F is a set of dependencies such that F logically implies all dependencies in Fc, and Fc logically implies all dependencies in F.

Detailed Solution: Question 5

A Canonical cover Fc for F is a set of dependencies such that F logically implies all dependencies in Fc, and Fc logically implies all dependencies in F. In Fc, no functional dependency should contain an extraneous attribute and each left side of functional dependency should be unique.

Test: Minimal Covers - Question 6

Consider the following set of functional dependencies F on the schema S(P, Q, R)
F = {P → QR, PQ → R, Q → R, P → Q}
The canonical cover of the above given set is 

Detailed Solution: Question 6

Option 3: P → Q and Q → R
P+ = {P, Q, R}
It covers, P → QR, PQ → R and P → Q
Also, Q → R
It is also minimal
Therefore, P → Q and Q → R is the canonical cover of the above given set.

Test: Minimal Covers - Question 7

Which of the following is true given 2 schemes F & G
F : {A → BC, B → C, AC → B}
G : {AB → C, A → B, A → C}

Detailed Solution: Question 7

i) Check for if F covers G :
checking FDs of F :-
{AB+} = {ABC}
{A+} = {ABC}
∴ F covers G

ii) check for it G covers F :
Cheeking FDs of G:-
{A+} = {ABC}
{B+} = {B} ⇒ B → C not covered
{AC+} = {ABC}
∴ G doesn’t covers F

Hence, a is the correct answer.

Test: Minimal Covers - Question 8

Find the minimal set of FDs.
R(A, B, C, D)
F : { A -> B, C -> B, D -> ABC, AC -> D}

Detailed Solution: Question 8

C → B
A → B
AC → D
D → ABC
Given that, D → ABC

Case I:
B functionally dependent on  A so we replace
B to A
D → AAC
D → AC

Case II:
B functionally dependent on C, so we replace
B to C
D → ACC
D → AC

Test: Minimal Covers - Question 9

Which of the following is true for the below given functional dependencies of the relation R(X, Y,Z,W) and S(X, Y, Z, W)?
R: {X → Y, XY → Z, W → XZ, Z → W}
S: {X → YZ, W → XY}

I. R ⊇ S
II. S ⊇ R

Detailed Solution: Question 9

Check: R ⊇ S
X+ = {Y, Z, E}
∴ X → YZ
W+ = {W, X, Z, Y}
W → XY
Hence R is covering S, that is, R ⊇ S

Check: S ⊇ R
X+ = {X, Y, Z}
∴ X → Y
(XY)+ = {X, Y, Z}
∴ XY → Z
W+ = {W, X, Y, Z}
∴ W → XZ
Z+ = {Z}
Z → W, it cannot be determined by S
and hence, S cannot cover R, that is, S ⊇ R is false
Only I is correct.

Test: Minimal Covers - Question 10

Let F be a set of functional dependencies given as
F={A->BC, B->C, A->B, AB->C}

Find minimum cover for F

Detailed Solution: Question 10

Initialization: {A->B, A->C,B->C,AB->C}
Consider A->B
G={A->C, B->C, AB->C}=G’ SINCE A->B ∉ G’, A->B STAYS
CONSIDER A->C
G={A->B, B->C, AB->C}=G’ SINCE A->B ∈ G’, A->C  REMOVED

CONSIDER B->C
G={A->B, AB->C}=G’ SINCE B->C ∉ G’, B->C STAYS

CONSIDER AB->C
G={A->B, B->C}=G’ SINCE AB->C ∈ G’, AB->C  REMOVED
THUS
G={A->B, B->C}

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