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Lossless and dependency preserving check 2
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.
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.
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 |
- AB → C: R1 and R3 agree on A, B. Make their C equal (b13). R1 and R3 still have no "a" in C.
- B → D: R1, R2, R3 all have B = a. R3 has D = a, so D becomes a in R1 and R2.
- AD → E: R1 and R3 agree on A and D (both a). R3 has E = a, so R1's E becomes a.
- E → G: R1, R3, R4 have E = a. R4 has G = a, so G becomes a in R1 and R3.
- 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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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