Dependency Preservation

Question

Doubts : -

1. R(A,B,C,D) and The Functional Dependencies are =  AB->C, C->AD . So when I'll decompose it then it will be R1(A,B,C) and R2(C,D).We cannot do R1(A,B,C) and R2(C,A,D) because the Production C->A will become redundant because this production is already available in R1 isn't it?

2. Say R(A,B,C,D) and FD are = A->B, B->C, C->D . We decompose it as R1 = (A,B), R2 = (B,C),R3 = (C,D).

Now to check whether its lossless we will find common attribute between relation which should be the candidate key in any one of the relations So :-

R1 intersection  R2 = B  --->which is candidate key in R2

R2 intersection  R3 = C  --->which is candidate key in R3

Do we have to check R1 intersection R3 as well ? because if we do then it wont be lossless as there isn't any common attribute.

Please tell the right way to check.

Comments

your first point is true.

for 2) there is no need to have common key(to any one table) attribute  b/w R1 and R3, why do we need common attribute ? so that we keep track of attribute so we do not get the extra data after merging table.

So if any two table have common key(to any one table) attribute our job is done.

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Sir let me be clear you say :-

If R is decomposed to R1,R2 and R3 then there should be a common attribute between any one of the relation (or may be more ) R1 and R2,R2 and R3,R1 and R3 right Sir? And that common attribute is a super key.

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yes..

Answer 1

if a relation R is decomposed into two relation R1 and R2 then it must be lossless and dependency preserving.

1) both decomposition is right bcz both are lossless and dependency preserving.

2)if a relation R is decomposed into R1 , R2 and R3 . FOR LOSSLESS we intersect R1 and R2 . if this is key then at first join it and it become R12. now we can again intersect R12 and R3 . if we get candidate key then we can join it. 

look at this solution.

Comments

Sir i got little doubt R was decomposed to R1(A,B,C) and R2(C,D).

For functional dependecny preservation FDs should be preserved here in R1 we have A->BC,in R2 we have C->D If we take union       AB-> C  and C->D but C->A is lost..

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We can derive fd C----->A From R1.

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Oh nooo. Sorry i didn't saw that coming thanku :)