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

Here in question they have mentioned that the table contains two single valued attributes. But the table to be in 1NF, the attributes should be atomic and single valued. So, can we conclude that S1 is TRUE ???

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how about a Relation R(A,B) with FD's like

A->B

B->A

? which normal form the table is in?

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it is in BCNF form because both A and B are candidate keys ,uniquely defines table.Also there is no partial dependency and transitive dependencey

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A	B
1	2
1	3
2	2

So in above schema neither

A$\rightarrow$ B hold nor 

B $\rightarrow$ A holds

Then how we can say this is in 1NF ?

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https://www.iitg.ac.in/awekar/teaching/cs344fall11/lecturenotes/august%2029.pdf

@Dharmendra Lodhi

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@shivam kasat .....From above PDF

If a relation has only two attributes then it is in BCNF.

Proof: Let a, b be the two attributes in a relation R.

The possible functional dependencies are:

case 1: LHS contains both attributes

a b -> ---

clearly this is a trivial functional dependency because RHS attributes form subset of LHS attributes .

Case 2 : LHS contains only one attribute.

a->---

b->---

clearly in this case, the LHS attribute will be candidate key

how can you decide it is candidate key.......plz explain

 

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if single-valued not given then also statement 1 correct?As any relation with two attributes will always be in BCNF so it will also be in 1NF,2NF and 3NF

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@Abhishek Rauthan If single-valued not given then is it even in 1NF?

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Yes it would not be in 1 nf also so statement 1 would be false

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$S1$ Video Explanation with Proof: https://youtu.be/SffSx6Kyv_s?feature=shared

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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