Functional Dependency Consider the following instance: Does the functional depen

Question

Functional Dependency

Consider the following instance: Does the functional dependency C rightarrow D hold in this instance? Does the functional dependency D rightarrow C hold in this instance? Does the functional dependency A rightarrow BCD hold in this instance? Given a relation schema R = (A, B, C, D, G, H) and a set of functional dependencies F = {D rightarrow G, CD rightarrow G, D rightarrow C, H rightarrow G} Find (BD)^+ in F. Find a candidate key for relations on schema R, and prove that your answer is a candidate key. Consider a relation schema R = (A, B, C, D, E) with functional dependencies: F = {A rightarrow E, C rightarrow D, D rightarrow BE}. Is the decomposition (A, C, E), (B, C, D) a lossless-join decomposition? Why or why not? Is the decomposition (A, B, D), (C, D, E) a lossless-join decomposition? Why or why not?

Explanation / Answer

Answer:

1.

(1)

The functional dependency does not hold in C->D because,

5->2 in first row matches with the 5->2 in fourth row of given instances.

4->4 in second row does not match with the 4->3 in third row of given instances.

So, the answer is No.

(2)

Similarly for the second bit,

The functional dependency does not hold in D->Cbecause,

2->5in first row matches with the 2->5 in fourth row of given instances.

4->4 in second row does not match with the 3->4 in third row of given instances.

So, the answer is No.

(3)

The functional dependency holds for the A->BCD because, it is consists of unique instances.

3.

A decomposition is a lossless join if, for all relations r on schema R that are legal under the given set of functional dependency constraints,

r = R1 (r) ? R2 (r) ? R3 (r) ? R4 (r)

In-order to prove or disprove that the (A, C, E) and (B, C, D) are lossless-join decomposition, it has to be prove either of the following statements.

(1)

Let

So,

Given,

So, C is in set R2.

Thus, this decomposition is in lossless-join decomposition.

(2)

Let

So,

Given,

So, B is does not depend on any other key sets.

So, the decomposition is not possible in this case.

Thus, this decomposition is not lossless-join decomposition.

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