What this topic covers
21 subtopics- Anomalies5
- Finding the normal form5
- Lossless join4
- 2NF3
- BCNF3
- Minimal cover3
- 1NF2
- 3NF2
- Armstrong's axioms2
- Decomposition2
- Denormalization2
- Finding candidate keys2
Show all 21 subtopics
- Functional dependency2
- 3NF synthesis1
- 4NF1
- 5NF1
- Dependency preservation1
- Design1
- Normal forms1
- Trade-offs1
- Why normalise1
- Beginner 15
- Intermediate 15
- Advanced 15
About this topic
Normalization organises tables so that each fact is stored once, removing the update, insertion and deletion anomalies that redundancy causes. Each normal form forbids a specific kind of dependency: 1NF repeating groups, 2NF partial dependencies on part of a key, 3NF transitive dependencies, and BCNF any dependency whose left side is not a super key. GATE tests these with functional-dependency sets; interviews test whether you can spot the problem in a real table design.
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Sample questions
Try each one first, then open the answer for the explanation and why the other options are wrong.
What is the main goal of normalization?
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Explanation
Normalization stores each fact in one place, so it cannot be updated inconsistently, cannot be lost by deleting an unrelated row and does not require dummy data to insert.
Why the other options are wrong
- BIt usually increases the number of tables.
- CNormalised schemas often need more joins; reads can get slower.
- DSecurity is unrelated.
In employees(emp_id, dept_id, dept_name), emp_id → dept_id and dept_id → dept_name. Which normal form is violated?
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Explanation
The key is a single attribute, so 2NF holds. But a non-prime attribute (dept_name) depends on another non-prime attribute (dept_id): a transitive dependency. Move department data to its own table.
Why the other options are wrong
- AAll values are atomic.
- CThe transitive dependency breaks 3NF.
- DWith a single-attribute key there are no partial dependencies.
R(A, B, C) has the FDs AB → C and C → B. What is the highest normal form of R?
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Explanation
Candidate keys: AB and AC, so every attribute is prime. C → B has a non-super-key left side, which breaks BCNF, but its right side B is prime, which 3NF allows.
Why the other options are wrong
- AIt is in 3NF.
- BC is not a super key.
- DIt is also in 3NF.