Lossless Join Decomposition

Question

I am studying for a test, and this is on the study guide sheet. This is not homework, and will not be graded.

Relation Schema R = (A,B,C,D,E)

Functional Dependencies = (AB->E, C->AD, D->B, E->C)

Is r1 = (A,C,D) r2 = (B,C,E) OR

x1 = (A,C,D) x2 = (A,B,E) a lossless join decomposition? and why?

Answer 1

My relational algebra is horribly rusty, but here is how I remember it to go

If r1 ∩ r2 -> r1 - r2 or r1 ∩ r2 -> r2 - r1 in FDs then you have lossless decomposition.

r1 ∩ r2 = C
r1 - r2 = AD

C->AD is in functional dependencies => lossless

for x1 and x2

x1 ∩ x2 = A
x1 - x2 = CD

A->CD is not in FDs now check x2 - x1

x2 - x1 = BE

A->BE is not in FDs either, therefore lossy

references here, please check for horrible mistakes that I might have committed