Test your understanding of normalization concepts.
A table where every attribute is atomic is in:
- A) 1NF
- B) 2NF
- C) 3NF
- D) BCNF
Correct Answer: A) 1NF
Explanation: 1NF requires that every column contains atomic (indivisible) values. No repeating groups allowed. Higher NFs require 1NF as a prerequisite plus additional conditions.
What is removed to transition from 1NF to 2NF?
- A) Transitive Dependencies
- B) Partial Functional Dependencies
- C) Multi-valued attributes
- D) All of the above
Correct Answer: B) Partial Functional Dependencies
Explanation: 2NF removes partial dependencies — where a non-prime attribute depends on only part of a composite candidate key. Transitive dependencies are removed in 3NF.
A table is in 3NF if it is in 2NF and has no:
- A) Primary Keys
- B) Partial Dependencies
- C) Transitive Dependencies
- D) Foreign Keys
Correct Answer: C) Transitive Dependencies
Explanation: 3NF removes transitive dependencies — where a non-key attribute depends on another non-key attribute. Example: Course_ID → Instructor → Instructor_Office (Instructor_Office depends transitively on Course_ID via Instructor).
Given R(A,B,C) and FDs {A → B, B → C}. What is the highest normal form?
- A) 1NF
- B) 2NF
- C) 3NF
- D) BCNF
Correct Answer: B) 2NF
Explanation: A is the Candidate Key (A⁺ = {A,B,C}). B → C is a transitive dependency (non-prime B determines non-prime C). This violates 3NF but not 2NF (no partial dependency since PK is single attribute).
Which normal form requires every determinant to be a candidate key?
- A) 1NF
- B) 2NF
- C) 3NF
- D) BCNF
Correct Answer: D) BCNF
Explanation: BCNF (Boyce-Codd Normal Form) requires that for every non-trivial FD X → Y, X must be a Super Key. This is stricter than 3NF.
A Lossless Join decomposition requires:
- A) All tables have the same number of rows
- B) The common attribute is a super key in at least one table
- C) All FDs are preserved
- D) The tables have no common attributes
Correct Answer: B) The common attribute is a super key in at least one table
Explanation: For a decomposition to be lossless, the common attribute(s) between the two resulting tables must be a super key in at least one of them. Otherwise, joining them back produces spurious rows.
What is the main reason to denormalize a database?
- A) To improve write performance
- B) To improve read performance by avoiding JOINs
- C) To reduce storage space
- D) To satisfy 3NF requirements
Correct Answer: B) To improve read performance by avoiding JOINs
Explanation: Denormalization intentionally adds redundancy to reduce the number of JOINs needed for queries. This sacrifices write performance and integrity for read speed, common in OLAP/data warehouse systems.
A table that is in 3NF but violates BCNF has:
- A) A non-key attribute determining another non-key attribute
- B) A determinant that is not a candidate key
- C) A composite primary key
- D) Multi-valued attributes
Correct Answer: B) A determinant that is not a candidate key
Explanation: BCNF violation occurs when an attribute (like Advisor) determines another attribute (Major) but Advisor is not a candidate key. The table may still be in 3NF if there are no transitive dependencies on non-prime attributes.
Quick Revision
- 1NF: Single atomic values
- 2NF: No partial PK dependency
- 3NF: No transitive dependencies
- BCNF: Every determinant is a candidate key
- Decomposition should be lossless and dependency preserving
- Denormalize only when read performance is critical
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