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Consider R(A,B,C,D,E)with the FD Set F(A->B, A->C, DE->C, DE->B, C->D)

Consider this decomposition : R1(A,B,C), R2(B,C,D,E) and R3(A,E)

Then, the decompositions is

  1. Lossless Decomposition
  2. Dependency Preserving

Which of the following is correct?

A) 1 is TRUE but 2 is FALSE

B) 1 is FALSE but 2 is TRUE

C) Both 1 and 2 are TRUE

D) Both 1 and 2 are FALSE 

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2 Answers

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Best answer
(C)  both 1 and 2 are true.

Decomposition of R(ABCDE) into R1(ABC), R2(BCDE) and R3(AE) is lossless since the common attribute between R1 and R3(i.e. A) is the primary key for the table R1 and the common attributes between R(ABCE) and R3(BCDE) form the primary key for R3. So it is lossless.

Dependencies applicable on R1 are

A->B

A->C

dependencies applicable on R2 are

DE->C

DE->B

C->D

there are no dependencies applicable on R3.

Since all the FDs that are given are applicable on some table after decomposition, so it is decomposition preserving too.
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edited by

@Somoshree Datta 5 @Balaji Jegan I think BC->D, CE->D and BCE->D are also there in R2 Functional Dependencies. Although, it won't affect the answer. 

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Option C :1 and 2 both True

1.In R1 and R2 give all attributes and have a common attribute so it is lossless

and FD preserving

2.R1,R2 and R3 give all attributes and have a common attribute so it is lossless

and FD preserving

So R1,R2 and R3 are Lossless And FD preserving

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