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Canonical Cover in DBMS – Top 10 MCQs with Answers and Explanations
Canonical Cover in DBMS is the minimal set of functional dependencies obtained by removing redundancy and extraneous attributes while preserving equivalence. It is widely used in normalization and database design.
Understanding Canonical Cover (Minimal Cover) is essential in Database Management Systems (DBMS), especially for normalization and functional dependency optimization. It helps simplify a set of functional dependencies by removing redundancy while preserving equivalence.
In this page, you will find top 10 carefully selected MCQs on Canonical Cover designed for university exams, GATE, and technical interviews. Each question is explained step-by-step to help you master key concepts like splitting dependencies, removing extraneous attributes, and eliminating redundant functional dependencies.
These practice questions will strengthen your problem-solving skills and help you quickly identify minimal covers in exam scenarios.
What is Canonical Cover in DBMS?
A Canonical Cover is a simplified set of functional dependencies that is equivalent to the original set but has:
- No redundant dependencies
- No extraneous attributes
- Single attribute on the right-hand side
It plays a crucial role in database normalization and helps in identifying candidate keys efficiently.
If you are new to canonical cover (minimal cover), I strongly suggest you to learn the required properties of every functional dependency to be in a canonical or minimal cover here.
Explanation:
Canonical cover requires single attribute on RHS. So A → BC is split into A → B and A → C.
Explanation:
A → C is redundant since it can be derived using A → B and B → C.
Explanation:
AB → C is redundant because A alone determines C.
Explanation:
A → D is redundant since it can be derived through A → B → C → D.
Explanation:
Since A → B, AB reduces to A. Hence AB → C becomes A → C.
Explanation:
Split A → BC into A → B and A → C. A → C is redundant since A → B and B → C imply it.
Explanation:
AB → C is redundant because A alone determines C.
Explanation:
Both dependencies are necessary; neither is redundant.
Explanation:
Since A → B, AB reduces to A. Thus AB → D becomes A → D.
Explanation:
A → BC splits into A → B and A → C. A → D is redundant since A → B and B → D imply it.
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