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Consider the relation R(A,B,C,D,E) with the set of F=(A->C,B->C,C->D,DC->C,CE->A) Suppose the relation has been Decomposed by the relations R1(A,D),R2(A,B),R3(B,E),R4(C,D,E),R5(A,E)

Is this decomposition lossy or lossless?

i tried solving this question using the matrix method and i am getting the answer as lossless because i managed to get a row in the 5*5 matrix filled with one variable however the book from which i am solving gives the answer as lossy. which one is the correct answer??

batman007
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1 Answers1

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It is a lossless decomposition for sure. The row corresponding to R3 gets filled with one variable.

As an aside, if you have the above decomposition obtained using Bernstein Synthesis then just checking whether any of the decomposed relations consists of all the attributes of the key of the original relation R will ensure that it's a lossless decomposition. For example, BE is the key for the relation R in the example above. The decomposed relation R3 consists of both the primary attributes B and E and hence this ensures a lossless decomposition.

Karup
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  • thanks mate.i appreciate your help.i was getting my r3 filled with one variable as you said eariler so now i can be sure that that answer is lossless. its never good to trust the books blindly. also i didnt know about Bernstein synthesis.after your post i had a look at the topic and it has finally cleared all my doubts about this question. – batman007 Apr 30 '16 at 15:55
  • @batman007 Happy to help, and welcome to Stack Overflow. If this answer or any other one solved your issue, please mark it as accepted. :) – Karup Apr 30 '16 at 15:58