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Question

ADE, ABC, BD, ACB, BCA, EF
the decomposition of R into R1(ABDE), R2(ABC) and R3(EF) is

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Solution

Option (a)
  • R(A, B, C, D, E, F)
R1(ABCE) R2(ABC)
BD
ADE
ABC
BCA
ACB

So, it is Dependency Preserving.
  • It is lossless join because in every relation there is

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