Problems - Normalization

 Problems

Given:

Relation:
R(A, B, C, D, E)

Functional Dependencies:

• A → B
• A → C
• BC → D
• D → E

i. Identify the candidate key(s).

ii.Determine whether R is in 1NF, 2NF, and 3NF. Give reasons for each.

iii. If R is not in 3NF, decompose it into 3NF relations.

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(i) Find Candidate Key(s)

Step 1: Find closure of A

Compute $A^+$:

• A → B, C ⇒ {A, B, C}
• BC → D ⇒ {A, B, C, D}
• D → E ⇒ {A, B, C, D, E}

👉 $A^+ = \{A, B, C, D, E\}$

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✅ Conclusion:

• A determines all attributes
• So, A is a candidate key

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Check others (brief):

• No smaller subset exists
• So only one candidate key

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✅ Answer:

Candidate Key = {A}

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(ii) Normal Forms

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✅ 1NF (First Normal Form)

• Assumes atomic attributes (given in question)

✔ R is in 1NF

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✅ 2NF (Second Normal Form)

Rule:

• No partial dependency on candidate key

👉 Candidate key = A (single attribute)

• Partial dependency happens only with composite keys
• Here key is single → no partial dependency possible

✔ R is in 2NF

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❌ 3NF (Third Normal Form)

Rule:

For every FD $X → Y$, either:

1. X is a super key, OR
2. Y is a prime attribute

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Check each FD:

FD	Status
A → B	OK (A is key)
A → C	OK (A is key)
BC → D	❌ BC is not key
D → E	❌ D is not key

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❌ Violations:

• BC → D
• D → E

👉 Non-key attributes determining other attributes

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❌ Conclusion:

R is NOT in 3NF

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(iii) Decomposition into 3NF

We decompose based on violating FDs.

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Step 1: Create relations for each FD

From A → B, C:

R1(A, B, C)

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From BC → D:

R2(B, C, D)

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From D → E:

R3(D, E)

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Step 2: Ensure key is preserved

• A is present in R1 → OK

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✅ Final 3NF Relations:

R1(A, B, C)
R2(B, C, D)
R3(D, E)