related to BCNF Decomposition

Question

quest: consider the Rn R(C,S,Z) with FDs CS->Z and Z->C

then their BCNF decomposition will be lossless join or Dependency preserving or both of them?

Answer 1

CS and ZS are  candidate keys

Fd Z->C violates Bcnf

So bcnf decomposition is

R1(S,Z) and R2(Z,C) is lossless  as  Z is common attr in both reln but not dependency preserving as CS->Z not preserved

Comments

How CS->Z is getting preserved ?? I think it is lossless but dependency is not preserved.

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Pooja i think it's only lossless not dependency preserving , as here CS->Z is attaining a constant value corresponding to both the decomposed realtions. correct me if m wrong!

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yeah here  CS->Z is attaining a constant value corresponding to both the decomposed relations.