GATE CSE 2014 | Set 1 | Question: 30

Question

Given the following two statements: 

S1: Every table with two single-valued attributes is in $\text{1NF, 2NF, 3NF}$ and $\text{BCNF}.$ 

S2: $AB \to C, D \to E, E \to C$ is a minimal cover for the set of functional dependencies $AB \to C, D \to E, AB \to E, E \to C$.

Which one of the following is CORRECT?

• A. S1 is TRUE and S2 is FALSE.
• B. Both S1 and S2 are TRUE.
• C. S1 is FALSE and S2 is TRUE.
• D. Both S1 and S2 are FALSE.

Comments

for statement 2 if we replace AB ->C with AB->E in minimal cover by removing AB->C then it will be a minimal cover

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yes.   
AB ->E
E ->C
D ->E

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S1 is true and S2 is false

Option A is true

Answer 1

(A) S1 is TRUE and S2 is FALSE.

A relation with $2$ attributes is always in BCNF

The two sets of functional dependencies are not the same. We can not derive $AB \to E$ from the $1^{\text{st}}$ set.

Comments

Minimal Cover = { AB->E, D->E, E->C }
Thanks @rajoramanoj

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Original Set of Dependencies: {AB → C, D → E, AB → E, E → C}

Proposed Minimal Cover (S2): {AB → C, D → E, E → C}

In the original set, the dependency AB → E is redundant. Here's why:

• We have AB → C and E → C. These tell us about what determines C, but not E.

• We also have D → E. This establishes a way to determine E.

However, if we consider the possibility of transitivity, we'd need a chain like X → Y and Y → Z to get X → Z. In the original set, there's no such chain that produces AB → E from the other dependencies.

But, if we look at what AB can determine using the other dependencies in the proposed minimal cover {AB → C, D → E, E → C}, we can see:

• AB determines C.

• E determines C.

• D determines E.

There is no way to get from AB to E using {AB → C, D → E, E → C}. Therefore, AB → E from the original set is not redundant and should be part of the minimal cover. This means that the proposed minimal cover S2 in the image is incorrect because it has removed a non-redundant dependency. A correct minimal cover would need to re-evaluate all dependencies for redundancy and extraneous attributes. A likely minimal cover for the original set is {D → E, AB → E, E → C} after removing the redundant AB → C (since AB → E and E → C implies AB → C).

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got this this time :-)

Answer 2

S1 is true bcoz if there are only 2 attributes then relation is always in bcnf.

let R(A B) possible cases are:

(1) {A->B } here A is cand key so BCNF

(2) {B->A}  here B is cand key so BCNF

(3) { A->B  B->A} here A and B both are cand key so BCNF

​​​​​​​(4) no non-trivial FD's here AB is cand key so BCNF

S2 is false bcoz minimal cover should be { D->E , AB->E , E->C }

so ans is A

Answer 3

Table with 2 single valued attributes will be in 1NF,

2NF because there can not be a partial key dependencies. Key can be 1 attribute or 2 attributes. If one is key then other can be dependent, if 2 attributes make key then it is trivial.

3NF, there is no question of Transitive dependencies.

BCNF, because if at all a dependency exists, the determinant will be a Key.

Answer 4

s1 is true and s2 is false.In s2 AB->E is not covered.

Answer 5

COMMENT BELOW IF YOU HAVE ANY DOUBT