---
title: "Lossless and dependency preserving check 1"
subject: "Database Management System"
chapter: 4
key: "semester_6/database_management_system:decomp-abcde-r1-r2-cde"
source: https://www.bctengineer.com/q/semester_6/database_management_system/decomp-abcde-r1-r2-cde
site: BCT Engineering
---

# Lossless and dependency preserving check 1

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-r1-r2-cde
- Whole chapter: https://www.bctengineer.com/md/semester_6/database_management_system/chapter-4

## Question

Suppose R = (A, B, C, D, E) is decomposed into R1 (A, B, C) and R2 (C, D, E). Is it lossless? Is it dependency preserving? Consider F = {A → BC, CD → E, B → D, E → A}.

## Short answer

The decomposition is not lossless because the common attribute C is not a key for either relation, and the chase test fails. It is not dependency preserving because B → D and E → A are lost.

## Answer

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

#### Lossless join test
R1 ∩ R2 = {C}. The decomposition is lossless if C → R1 or C → R2 holds, that is, if C is a key of R1 or R2.

C⁺ = {C}, since no FD has C alone on the left side. So C → R1 and C → R2 both fail.

Chase table (a = known value, b = unknown):

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

The two rows never agree on the left side of any FD (A: a vs b, CD: R1 has b in D, B: a vs b, E: b vs a), so nothing can be changed and neither row becomes all "a".

**The decomposition is NOT lossless.**

#### Dependency preservation test
Projections of F:
- F1 on (A, B, C): A → BC (and the FDs derived from it: A → B, A → C, AB → C, ...)
- F2 on (C, D, E): CD → E, E → CD (E⁺ = ABCDE, so E → C, E → D)

Check each FD of F:
- A → BC: in F1. Preserved.
- CD → E: in F2. Preserved.
- B → D: B is in R1 but D is not, and the closure of B inside R1 is B⁺ ∩ R1 = {B, D} ∩ {A, B, C} = {B}. Never reaches D. **Lost.**
- E → A: E⁺ ∩ R2 = {C, D, E}, then Z = {E, C, D}; Z ∩ R1 = {C}, C⁺ = {C}. A is never reached. **Lost.**

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

**Answer: not lossless and not dependency preserving.**

## Check yourself

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

1. R1 = (A,B,C) and R2 = (C,D,E) share only C. When is the decomposition lossless?
   A. Whenever they share any attribute
   B. When C -> R1 or C -> R2 holds
   C. When R1 and R2 have equal size
   D. When every FD fits inside R1
   Answer: B. A binary decomposition is lossless if the common attributes are a key of at least one of the two relations. (Key point: Lossless join condition)

2. Why does the lossless test fail here?
   A. C+ = {C}, since no FD has C alone on its left side
   B. C+ = {C, D}
   C. C+ = {A, B, C}
   D. Because R1 and R2 overlap
   Answer: A. No FD starts with C alone, so C determines neither R1 nor R2 and the join is lossy. (Key point: Closure of C)

3. Why is B -> D lost in the decomposition?
   A. B and D are in the same relation
   B. B is only in R1, D is only in R2, and B+ restricted to R1 never reaches D
   C. D is a key
   D. B -> D is trivial
   Answer: B. B and D lie in different relations and the restricted closure of B inside R1 is just {B}. (Key point: Lost dependency B -> D)

4. What is the conclusion for this decomposition?
   A. Lossless and dependency preserving
   B. Lossless but not dependency preserving
   C. Not lossless and not dependency preserving
   D. Not lossless but dependency preserving
   Answer: C. The common attribute C is not a key (lossy), and B -> D and E -> A are lost. (Key point: Final verdict)

## Related in Database Management System

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

- [3rd order Butterworth LPF, R1 = 1 Ω, R2 = 4 Ω](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abutterworth-lp-third-order-unequal-termination) — Filter Design, chapter 5, asked 4×
- [Significance of reflection coefficient; R1 = 1 Ω, R2 = 4 Ω](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Areflection-significance-unequal-termination) — Filter Design, chapter 5
- [IDMT grading: R1, R2 with 1000/5 CTs](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aidmt-r1-r2-1000-5) — Switchgear and Protection, chapter 7, asked 2×
- [Show i1 = i2 = -V/(R1+R2+R3) at t=0+](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aswitch-a-to-b-initial-values) — Electrical Circuit Theory, chapter 2