这是有损联接和依赖性保留吗

Question

I am reading this topic Functional dependency and Normalization in Database Management Subject. I came across this example.

Relation R(A,B,C,D) Which one is Lossy join but Dependency Preserving BCNF Decomposition?

a. A ->B, B -> CD

b. A -> B, B -> C, C->D

c. AB -> C, C -> AD

d. A -> BCD

Now answer given is option C.

How can option C. be a lossy decomposition . if you do ABC union CAD = ABCD This satisfies first condition. if we do ABC intersection CAD = AC which is perfectly fine, since in AC, C is key for (CAD) C -> AD decomposition . which also satisfies the second condition. Am i making any mistake in understanding this concept.

Answer 1

Usually for a Normalisation/decomposition exercise, you are given:

The full relation and its attributes. [yes: R(A, B, C, D)]

The Functional dependencies. [yes? it looks like a., b., c., d. are possible sets of Fun Deps.]

The proposed decomposition. [Often named R1, R2, etc. I don't see those. I can't interpret option d. to be proposing a decomposition.]

Perhaps your post has missed out part of the exercise? Perhaps the exercise wants you to decide which decomp preserves the dependencies in BCNF? (But results in a lossy join.)

[editted in response to Nikhil's comment] Note that the list of FD's alone doesn't amount to a decomposition: the FD C -> AD is short-hand for C -> A, C -> D. Does that mean two decomposing relations? No, because A and C are already in the FD AB -> C. So we have R1= (A, B, C), R2 = (C, D). But I don't know if that is what the exercise is asking. Think about it. What does option d. mean in terms of decompositions?

Perhaps the exercise is asking (for example): given a proposed decomposition into R1 = (A, B) and R2 = (B, C, D), which of the sets of FD's would give a lossy decomposition?

There's a worked example here: http://en.wikipedia.org/wiki/Lossless-Join_Decomposition .

It points to a previous q Lossless Join Property .

And there's further references.

By the way, options a., b., include the same Fun Deps as option d., by the transitivity of dependencies (Armstrong's Axioms http://en.wikipedia.org/wiki/Armstrong%27s_axioms see also http://en.wikipedia.org/wiki/Heath%27s_theorem ). This is a clue.