Consider following relation S (S1, S2, S3, E1, E2) and functional dependencies…
2024
Consider following relation S (S1, S2, S3, E1, E2) and functional dependencies hold on the schema are as – FD1: S1 → {S2, S3} FD2: {E1, S3, S2} → S1 FD3: E2 → {S3, S2, E1} What is the highest normal form of the relation S?
Answer: B. 2NF — Explanation: Find Candidate Key(s) --> Given: FD1: S1 → {S2, S3} FD2: (E1, S3, S2) → S1 FD3: E2 → {S3, S2, E1} So: S1⁺ = {S1, S2, S3} E2⁺ = {E2, S1, S2, S3,…
- A.
1NF
- B.
2NF
- C.
BCNF
- D.
3NF
- E.
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Correct answer: B
Explanation:
Find Candidate Key(s) -->
Given:
FD1: S1 → {S2, S3}
FD2: (E1, S3, S2) → S1
FD3: E2 → {S3, S2, E1}
So:
S1⁺ = {S1, S2, S3}
E2⁺ = {E2, S1, S2, S3, E1}
{E1, S2, S3}⁺ = {E1, S2, S3, S1}
E2 is a candidate key
Prime vs Non-prime Attributes
Prime attribute: E2
Non-prime attributes: S1, S2, S3, E1
1.CHECK BCNF
Condition: For every FD X → Y:
X is a superkey, Y can be anything
Check FD1:
S1 → S2, S3
S1 is not a superkey
So, it violates the condition of BCNF.
2.CHECK 3NF
Condition: For every FD X → Y:
X is a superkey OR
Y is prime
Check FD1:
S1 → S2, S3
S1 is not a superkey
S2, S3 are non-prime
So, it violates the condition of 3NF.
3.CHECK 2NF
Candidate key is single attribute (E2)
No partial dependency possible
Relation is in 2NF
Final Answer: 2NF