无损分解

Question

Consider a schema R(A, B, C, D) and functional dependencies A ⟶ B and C ⟶ D. Then why isn't the decomposition of R into R1(A, B) and R2(C, D) a lossless decomposition? Can you please explain with real life example that what info is lost here?

Answer 1

You certainly need the two relations R1(A,B) and R2(C, D) that you outline in the lossless decomposition, but you've lost the crucial information about which A values are associated with which C values that was present in the original R(A, B, C, D). So you also need R3(A, C) to keep all the original information.

Relation R

A    B    C    D
1    2    13   14
2    2    13   14
3    1    12   15

Relation R1

A    B
1    2
2    2
3    1

Relation R2

C    D
13   14
12   15

Join R1 and R2 (Cartesian product); bogus rows marked ☜

A    B    C    D
1    2    13   14
1    2    12   15   ☜
2    2    13   14
2    2    12   15   ☜
1    3    13   14   ☜
3    1    12   15

Since this join is not the same as R, the proposed decomposition is not lossless.

Relation R3

A   C
1   13
2   13
3   12

Join R1, R2, R3

A    B    C    D
1    2    13   14
2    2    13   14
3    1    12   15

Since this result relation is the same as the original R, the decomposition into R1, R2, and R3 is lossless.

Answer 2

Then why isn't the decomposition of R into R1(A, B) and R2(C, D) a lossless decomposition?

Because now (A,B) and (C,D) are unrelated, which they weren't. You need also a relation between A and C.