Lossless Decomposition
Let r(R) be a relation schema.
Let R be decomposed into R1 and R2. The corresponding instances of R1 and R2 are r1 and r2 respectively.
Here, the decomposition of R into R1 and R2 is said to be ‘Lossless decomposition’, if
πR1 (r ) |X| πR2 (r) = r .
Note :
Let R1 and R2 be the decomposition of R, F be a set of FDs on R.
Here, this decomposition of R into R1 and R2 holds good, if at least one of the following satisfies.
R1 ∩ R2 → R1 (or) R1 ∩ R2 → R2
Example-1:
Let R = (A, B, C, D, E), R1 = (A, B, C), R2 = (C, D, E)
r :
A | B | C | D | E |
5 | 8 | 11 | 13 | 15 |
7 | 9 | 13 | 14 | 16 |
8 | 10 | 14 | 15 | 17 |
Prove that the decomposition of R in R1 and R2 is lossless.
Output :
R1 | R2 | |||||
A | B | C | C | D | E | |
5 | 8 | 11 | 11 | 13 | 15 | |
7 | 9 | 13 | 13 | 14 | 16 | |
8 | 10 | 14 | 14 | 15 | 17 |
Here, R = R1 U R2.
Here, C is a common attribute. And
Let r ’ = r1 |X| r2 is as follows :
A | B | C | D | E |
5 | 8 | 11 | 13 | 15 |
7 | 9 | 13 | 14 | 16 |
8 | 10 | 14 | 15 | 17 |
Hence proved, r = r ‘
Therefore, this is lossless decomposition.
Reference