Find the decopmosition of R into R1(A, B, C), R2(B, C,D ) and R3(C, D, E)
The given relation is R (A, B, C, D, E).
The given set of functional dependency F is :
AB -> CD
A -> E
C -> D
D -> E
The closure of the given set is :
AB -> C [ Decomposition rule ]
AB -> D [ Decomposition rule ]
A -> E
C -> D
D -> E
C -> E [ Transitive rule ]
The decomposition of R is done into 2 tables.
For table R1(A, B, C),
The set of functional dependencies F1 is :
AB -> C
For table R2(B, C, D),
The set of functional dependencies F2 is :
C -> D
For table R3 ( C, D, E ),
The set of functional dependencies F3 is :
C -> D
D -> E
C -> E
For checking lossless property :
The common attirbute between R2(B, C, D) and R3 ( C, D, E ) is CD and CD is a superkey in relation R3. When the attributes of R2 and R3 are combined to form a new table R4 ( B, C, D, E), then the common attribute between R1(A, B, C) and R4 ( B, C, D, E), is BC and BC is a superkey in the combined relation R4.
So, the decomposition is a lossless decomposition.
For checking dependency preserving,
Closure of ( F1 U F2 U F3 ) is :
AB -> C
C -> D
D -> E
C -> E
AB -> D [ Transitive rule ]
The closure of the original set of functional dependency F is :
AB -> C [ Decomposition rule ]
AB -> D [ Decomposition rule ]
A -> E
C -> D
D -> E
C -> E [ Transitive rule ]
So, the functional dependency A -> E is not present in the closure of ( F1 U F2 U F3 ) and so this closure is not equal to the closure of F.
The decomposition is not dependency preserving.
Hence, the answer is option b that is lossless and not dependency preserving.
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