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

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}

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)

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.

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.

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.

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

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}