---
title: "Lossless and dependency preserving check 4"
subject: "Database Management System"
chapter: 4
key: "semester_6/database_management_system:decomp-abcde-abc-ade"
source: https://www.bctengineer.com/q/semester_6/database_management_system/decomp-abcde-abc-ade
site: BCT Engineering
---

# Lossless and dependency preserving check 4

Database Management System · Chapter 4: Database Constraints & Normalization

- Asked in: year not given
- Video: [Illustration on Lossless Join Decomposition and Dependency Preserving Decomposition. | DBMS](https://www.youtube.com/watch?v=iAarosX8ees) — Edredo for Learners
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/database_management_system/decomp-abcde-abc-ade
- Whole chapter: https://www.bctengineer.com/md/semester_6/database_management_system/chapter-4

## Question

R = (A, B, C, D, E), R1 = (A, B, C), R2 = (A, D, E). F = {A → BC, CD → E, B → D, E → A}. Check whether the decomposition is lossless and dependency preserving.

## Short answer

Lossless join (A is a key). Not dependency preserving: CD→E and B→D span both relations and are lost.

## Answer

R = (A, B, C, D, E), R1 = (A, B, C), R2 = (A, D, E), F = {A → BC, CD → E, B → D, E → A}.

#### Lossless join test
R1 ∩ R2 = {A}. A⁺ = {A, B, C, D, E} (A → BC, B → D, CD → E), so A is a key of R1 and R2 and A → R1 holds.

Chase table:

| | A | B | C | D | E |
|---|---|---|---|---|---|
| R1 | a | a | a | b | b |
| R2 | a | b | b | a | a |

A → BC: both rows have A = a, so B and C in R2 become a. B → D: both rows now have B = a, so D in R1 becomes a. CD → E: both rows have C = a and D = a, so E in R1 becomes a. Both rows are all "a".

**The decomposition is lossless.**

#### Dependency preservation test
Projections:
- F1 on (A, B, C): A → BC (A⁺ ∩ R1 = ABC). B⁺ = BD and C⁺ = C add nothing inside R1. So F1 = {A → BC}.
- F2 on (A, D, E): E → A (E⁺ = ABCDE, so E → AD), A → DE (A⁺ ∩ R2 = ADE), AD → E.

Check:
- A → BC: in F1. Preserved.
- E → A: in F2. Preserved.
- CD → E: C is in R1 only and D in R2 only; (C)⁺ ∩ R1 = C, (D)⁺ ∩ R2 = D. E is not reached through A either (neither C nor D gives A). **Lost.**
- B → D: B in R1 only; B⁺ ∩ R1 = {B}. D not reached. **Lost.**

**The decomposition is NOT dependency preserving** (CD → E and B → D are lost).

**Answer: lossless, but not dependency preserving.**

## Check yourself

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

1. Why is R1 = (A,B,C), R2 = (A,D,E) lossless?
   A. Common attribute A has A+ = ABCDE, so it is a key
   B. E -> A holds
   C. R1 and R2 have 3 attributes each
   D. CD -> E holds
   Answer: A. A -> BC, B -> D and CD -> E give A+ = R, so the common attribute is a key. (Key point: A as key)

2. In the chase, how does R1's D become 'a'?
   A. Via CD -> E
   B. Via B -> D after A -> BC makes B an 'a' in R2
   C. Via E -> A
   D. It stays b
   Answer: B. A -> BC makes B and C equal to a in R2, then B -> D (both rows have B = a) sets D to a in R1. (Key point: Chase order)

3. Which FDs are lost?
   A. A -> BC and E -> A
   B. CD -> E and B -> D
   C. Only E -> A
   D. None
   Answer: B. C and D are in different relations, and B is only in R1 without D, so CD -> E and B -> D cannot be checked within one relation. (Key point: Lost FDs)

4. What is the conclusion?
   A. Lossless and dependency preserving
   B. Lossless but not dependency preserving
   C. Lossy but dependency preserving
   D. Lossy and not dependency preserving
   Answer: B. A is a key of the join (lossless), but CD -> E and B -> D are lost. (Key point: Final verdict)

## Close variants set in other exams

- [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
- [Lossless and dependency preserving check 3](https://www.bctengineer.com/question.md?key=semester_6%2Fdatabase_management_system%3Adecomp-abc-ab-ac) — Database Management System, chapter 4
- [Lossless and dependency preserving check 5](https://www.bctengineer.com/question.md?key=semester_6%2Fdatabase_management_system%3Adecomp-abcd-abc-cd) — Database Management System, chapter 4

## Related in Database Management System

- [Lossless and dependency preserving check 1](https://www.bctengineer.com/question.md?key=semester_6%2Fdatabase_management_system%3Adecomp-abcde-r1-r2-cde) — Database Management System, chapter 4

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