---
title: "Lossless and dependency preserving check 2"
subject: "Database Management System"
chapter: 4
key: "semester_6/database_management_system:decomp-abcdeg-four-relations"
source: https://www.bctengineer.com/q/semester_6/database_management_system/decomp-abcdeg-four-relations
site: BCT Engineering
---

# Lossless and dependency preserving check 2

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-abcdeg-four-relations
- Whole chapter: https://www.bctengineer.com/md/semester_6/database_management_system/chapter-4

## Question

R = (A, B, C, D, E, G), R1 (A, B), R2 (B, C), R3 (A, B, D, E), R4 (E, G). F = {AB → C, AC → B, AD → E, B → D, BC → A, E → G}. Check whether the decomposition is lossless and dependency preserving.

## Short answer

The decompositon is not lossless; no chase row becomes all 'a' (missing C in R1,R3). It is not dependency preserving because AB → C, AC → B, and BC → A are lost.

## Answer

R = (A, B, C, D, E, G); R1 = (A, B), R2 = (B, C), R3 = (A, B, D, E), R4 = (E, G).
F = {AB → C, AC → B, AD → E, B → D, BC → A, E → G}.

#### Lossless join test (chase)
Rows are the relations, columns the attributes. a = distinguished value, b = unknown value.

Initial table:

| | A | B | C | D | E | G |
|---|---|---|---|---|---|---|
| R1 | a | a | b | b | b | b |
| R2 | b | a | a | b | b | b |
| R3 | a | a | b | a | a | b |
| R4 | b | b | b | b | a | a |

1. AB → C: R1 and R3 agree on A, B. Make their C equal (b13). R1 and R3 still have no "a" in C.
2. B → D: R1, R2, R3 all have B = a. R3 has D = a, so D becomes a in R1 and R2.
3. AD → E: R1 and R3 agree on A and D (both a). R3 has E = a, so R1's E becomes a.
4. E → G: R1, R3, R4 have E = a. R4 has G = a, so G becomes a in R1 and R3.
5. AC → B, BC → A: no new rows agree on their left sides (only R1 and R3 share A and C, and their B is already a). No further change.

Final table:

| | A | B | C | D | E | G |
|---|---|---|---|---|---|---|
| R1 | a | a | b | a | a | a |
| R2 | b | a | a | a | b | b |
| R3 | a | a | b | a | a | a |
| R4 | b | b | b | b | a | a |

No row is all "a" (R1 and R3 are missing C).

**The decomposition is NOT lossless.**

#### Dependency preservation test
Projections:
- R1 (A, B): no non-trivial FD (A⁺ = A, B⁺ = BD).
- R2 (B, C): no non-trivial FD.
- R3 (A, B, D, E): B → D, AD → E (and AB → DE, which follows).
- R4 (E, G): E → G.

FD by FD:
- AD → E: in R3. Preserved.
- B → D: in R3. Preserved.
- E → G: in R4. Preserved.
- AB → C: start Z = {A, B}. R3: Z ∩ R3 = AB, (AB)⁺ ∩ R3 = ABDE, so Z = ABDE. R4: E⁺ ∩ R4 = EG, so Z = ABDEG. R1 and R2: Z ∩ R2 = {B}, B⁺ ∩ R2 = {B}; nothing new. C is never reached. **Lost.**
- AC → B: start Z = {A, C}. R1 gives A⁺ ∩ R1 = {A}, R2 gives C⁺ ∩ R2 = {C}, R3 gives {A}, R4 nothing. A and C are never together in one relation, so B is never reached. **Lost.**
- BC → A: start Z = {B, C}. R2: (BC)⁺ ∩ R2 = {B, C}; R3: B⁺ ∩ R3 = {B, D}, so Z = BCD; R1 and R4 add nothing. A is never reached. **Lost.**

**The decomposition is NOT dependency preserving** (AB → C, AC → B, BC → 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. In the chase test for lossless join, what shows the decomposition is lossless?
   A. Some row ends up with all distinguished values (a)
   B. Every column has at least one a
   C. All rows become identical
   D. No FD can be applied
   Answer: A. After applying the FDs, a row of all a's means the join recovers an original tuple, so it is lossless. (Key point: Chase method)

2. In the final table no row is all 'a'. Which attribute is missing in R1 and R3?
   A. D
   B. E
   C. C
   D. G
   Answer: C. C appears as a only in R2; no FD application made C distinguished in R1 or R3, so the decomposition is lossy. (Key point: Why the chase fails)

3. Which FDs are preserved inside single relations?
   A. AB -> C, AC -> B, BC -> A
   B. AD -> E, B -> D, E -> G
   C. Only E -> G
   D. All six FDs
   Answer: B. AD -> E and B -> D sit in R3, and E -> G sits in R4. (Key point: Preserved dependencies)

4. Why is AC -> B lost?
   A. C is only in R2 and A is not in R2, so no relation holds both on its left side
   B. B is a key
   C. A and C are in R4
   D. It is trivial
   Answer: A. No relation contains both A and C with B, and closures inside relations never reach B, so AC -> B, AB -> C and BC -> A are lost. (Key point: Lost dependencies)

## Close variants set in other exams

- [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 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 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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