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Question 3. Given the following relation, functional dependencies and decomposit

ID: 3749749 • Letter: Q

Question

Question 3. Given the following relation, functional dependencies and decomposition, answer the following questions: Relation R(A, B, C, D, E, F) with F-AB- F, BD C,CE F,F+D Decomposition: R1(A, B, D), R2(A, B,C, E), R3(B, D,E,F) (a) Is this decomposition lossless? Show yes or no using Chase algorithm. (b) Is this decomposition dependency preserving? Show your work. Note: two show that two sets of functional dependencies, F and F2 are equivalent, it is sufficient to show that (1) all functional dependencies in F1 are implied by F2, and (2) all all functional dependencies in F2 are implied by F1

Explanation / Answer

DEPENDENCY PRESERVATION CHECK-

given relation R(A,B,C,D,E,F) and FD's are- {AB->F, BD->C, CE->F ,F->D}

1.For lossless decomposition-

we are creating a table ,if any of the three rows has all alpha then the decomposition is lossless otherwise it is lossy

rule to enter alpha-

dependencies-

AB->F, BD->C, CE->F ,F->D

here ,above for dependency- AB->F we don't see any row satisfying it .Similarly for the other dependencies too.

so the given relation decomposition is lossy as no row is having all alpha's full .

here,we use alpha to check whether we get the the original table by joining the two tables or not.

2.Decomposition- R1(A,B,D), R2(A,B,C,E), R3(B,D,E,F)

dependency preservation is when the dependencies after decomposition are same the original dependencies are not.So,we can check it as-

i)for,AB->F, A,B and F together does not belong to any of the decomposition so this dependency can't be preserved.

ii)for,BD->C,B,D and C also does not belong to any single decomposition so so this dependency also can't be preserved.

iii)CE->F , C ,E and F also does not belong to anydecomposition and ,

iv) for ,F->D F and D belong to relation R3

so, above i,ii and iii are not preserved and hence dependency preservation is not there.

A B C D E F r1 alpha alpha alpha r2 alpha alpha alpha alpha r3 alpha alpha alpha alpha
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