Test: Minimal Covers - Computer Science Engineering (CSE) MCQ

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.

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}

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.

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.

F⁺ will contain all of the above dependencies but ABC→D.

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

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

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)

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}

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