---
title: "Compute canonical cover"
subject: "Database Management System"
chapter: 4
key: "semester_6/database_management_system:canonical-cover-abc"
source: https://www.bctengineer.com/q/semester_6/database_management_system/canonical-cover-abc
site: BCT Engineering
---

# Compute canonical cover

Database Management System · Chapter 4: Database Constraints & Normalization

- Asked in: year not given
- Video: [Lec-34: Minimal Cover in DBMS With Example | Canonical cover](https://www.youtube.com/watch?v=sS-LJMTVVj8) — Gate Smashers
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/database_management_system/canonical-cover-abc
- Whole chapter: https://www.bctengineer.com/md/semester_6/database_management_system/chapter-4

## Question

Compute canonical cover: R = (A, B, C), F = {A → BC, B → C, A → B, AB → C}.

## Short answer

Fc = {A → B, B → C}. This minimal set is equivalent to F; A is the candidate key.

## Answer

A canonical cover Fc is a minimal set of FDs equivalent to F, with no extraneous attributes on any side, no redundant FDs, and a unique left side for each FD.

F = {A → BC, B → C, A → B, AB → C}

**Step 1: split the right sides (decomposition)**

F = {A → B, A → C, B → C, A → B, AB → C}. Remove the duplicate A → B:

F = {A → B, A → C, B → C, AB → C}

**Step 2: remove extraneous attributes from left sides**

Only AB → C has a multi-attribute left side. Test whether A is extraneous: compute B⁺ using F. B⁺ = {B, C} already contains C, so B → C holds and A is extraneous. Replace AB → C by B → C, which is already present:

F = {A → B, A → C, B → C}

**Step 3: remove redundant FDs**

Test A → C: remove it and compute A⁺ using {A → B, B → C}. A⁺ = {A, B, C}, which contains C. So A → C is implied by A → B and B → C (transitivity) and is redundant.

Test A → B: without it, A⁺ = {A, C}, no B, so it is needed. Test B → C: without it, B⁺ = {B}, so it is needed.

**Step 4: combine right sides with equal left sides**: nothing to combine.

**Canonical cover: Fc = {A → B, B → C}**

Check: A⁺ under Fc = {A, B, C}, so A → BC and A → C still follow. The original F and Fc are equivalent. The candidate key of R is A.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which describes a canonical cover Fc?
   A. Equivalent to F with no extraneous attributes, no redundant FDs and unique left sides
   B. A set that contains every FD in F+
   C. A set of only trivial FDs
   D. The same as F with the right sides split
   Answer: A. Fc is a minimal set equivalent to F with no extraneous attributes, no redundant FDs and each left side unique. (Key point: Canonical cover properties)

2. In AB -> C, why is A extraneous?
   A. Because A is a key
   B. Because B+ = {B, C} already contains C, so B -> C holds
   C. Because A -> B is in F
   D. Because C is on the right
   Answer: B. If B alone determines C (using B -> C), then A adds nothing, so AB -> C becomes B -> C. (Key point: Extraneous attribute)

3. Why is A -> C removed from {A->B, A->C, B->C}?
   A. Because A -> C is trivial
   B. Because A -> C follows from A -> B and B -> C by transitivity
   C. Because C is on the right side of B -> C
   D. Because A -> B is a duplicate
   Answer: B. A+ under {A->B, B->C} is {A,B,C}, so A -> C is implied and redundant. (Key point: Redundant FD)

4. What is the canonical cover of F = {A->BC, B->C, A->B, AB->C}?
   A. {A->BC, B->C}
   B. {A->B, B->C}
   C. {A->B, A->C}
   D. {A->C, B->C}
   Answer: B. After splitting, dropping extraneous A in AB->C and removing redundant A->C, Fc = {A->B, B->C}. (Key point: Final canonical cover)

## Related in Database Management System

- [Lossless and dependency preserving check 2](https://www.bctengineer.com/question.md?key=semester_6%2Fdatabase_management_system%3Adecomp-abcdeg-four-relations) — Database Management System, chapter 4

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