Made Easy Test Series

Question

Consider a relation R(A,B,C,D,E) and functional dependencies:

F= (AC->B, C->D, A->E, C->B)

Relation R is decomposed into R1(A,B,C) and R2(C,D), then which of the following is correct about the decomposition?

(a) Lossless and dependency preserving

(b) Lossy and dependency preserving

© Lossless and not dependency preserving

(d) Not Lossless and not dependency preserving

Comments

C should be the answer.

As in both the relations C is common and is also key in one of the relations R2.

But after decomposing we lose two functional dependencies A-->E and C-->B

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I too marked C..but they gave D as the answer..

the solution was: R1 U R2 $\neq$ R since attribute E is not preserved. So it is not lossless.

But i guess this is wrong :\

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yeah C should be correct..let others verify

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Yes D should be the answer explanation by  Somoshree Datta 5

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Why so akshat sharma ??It isnt violating the fact of lossless decomposition..decomposition is still lossless since C is the key of relation R2..Why will it be lossy?

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akshat sharma i guess u r saying that it is lossy because when we will combine R1 and R2 to get the relation R, then some tuples would be lost since attribute E wont be present in the result obtained after joining R1 and R2..so the decomposition is lossy, is this the reason?

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Oh how I missed that, attribute E is not preserved after decomposing. This decomposition is lossy.

Sorry for the mistake.

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ya even I made the same mistake during exam..this reason just clicked me now :P

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even I did the same mistake - _-