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2 Answers

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i) R(A, B, C, D, E)

FD's ABC -> DEF, BC -> E, C ->B.

Now through the split rule we can write the functional dependencies given above is :-

ABC -> D, ABC -> E, ABC -> F, BC -> E, C -> B.

Now take the decomposition given above

R1(A, B, C, D, F)

FD's {ABC -> D, ABC -> F, ABC -> BC (Trivial functional dependency )}

R2(B, C, E)

FD's {BC -> E, C -> B}

Now let's check dependencies preserving:-

ABC -> D, ABC -> F, BC -> E, C -> B (these are present in the decomposition)

ABC -> BC and BC -> E (transitive dependency) hence ABC -> E

HENCE IT IS DEPENDENCY PRESERVING.

ii.)

R(A, B, C, D, E)

FD's ABC -> DEF, BC -> EF, EF ->B.

Now take the decomposition given above in the question.

R1(A, B, C, D, F)

FD's {ABC -> D, ABC -> F, ABC -> BC (Trivial functional dependency ), BC -> F}

R2(B, C, D, E)

now the functional dependency EF -> B, this is not satisfying because E and F are in different relations hence this is not dependency preserving decomposition.

HENCE ONLY i) IS SATISFYING DEPENDENCY PRESERVING DECOMPOSITION CONDITION.

ANS IS a) only (i)
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Option A is the answer.

  1. after removing redundancies from the FD’s. we get  ABC->DF, BC->E, C->B . So, ABC->DF goes into R1 , BC->E goes into R2 and  C->B goes into both R1 & R2. All FD’s are preserved. True.
  2. after removing redundancies from the FD’s. we get  ABC->D, BC->EF, EF->B . So, ABC->D, BC->F goes into R1 , BC->E goes into R2 but EF→ B can’t be preserved through R1 or R2. False.

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