BCNF form

Question

consider the relation R(ABCDE) and functional dependency F={AB->C , C->D , D->E ,E->A}

if we convert the given relation in BCNF then

given answer =>R(EA) R(ED) R(CD) R(ABC)

my answer =>R(EA) R(ED) R(CD) R(BC)

if mine is wrong please explain the right one

Comments

how?

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based on what you kept BC together? what is key for this Relation?

most important question when DP can preserved, then why to lose it?

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Data redundancy will be eliminated after converting into BCNF.

DP is not a major concern but yes loss less property should be followed...

Answer 1

Possible CK for this are : BC , AB, EB , BD . now only AB->C is satisfying BCNF property so possible decomposition is (ABC) (CD) (DE) (AE). So given answer is correct.

Comments

F = { AB -> C , C->D , D->E , E->A }

candidate keys = { AB ,CB , DB , EB }

AB -> C satisfies

C->D doesnt , take C+ = { C , D ,E , A}

ABCDE divided as ACDE and CB

now take D+ and divide ACDE as ADE and CD

now take E+ and divide ADE as EA and ED

so the ans R(EA)  R(ED) R(CD) R(CB)

the queestion was if its converted to bcnf is it dependency preserving or not

so i decomposed this way and it should be " not dependency preserving "

but the answer given , they have decomposed the other way and then the answer is "dependency preserving "

 

can someone help with where i went wrong

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F = { AB -> C , C->D , D->E , E->A }

candidate keys = { AB ,CB , DB , EB }

AB -> C satisfies

C->D doesnt , take C+ = { C , D ,E , A}

ABCDE divided as ACDE and CB

now take D+ and divide ACDE as ADE and CD

now take E+ and divide ADE as EA and ED

so the ans R(EA)  R(ED) R(CD) R(CB)

the question was if its converted to bcnf is it dependency preserving or not

so i decomposed this way and it should be " not dependency preserving "

but the answer given , they have decomposed the other way and then the answer is "dependency preserving "

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If they ask you that decomposition is dependency preserving or not then your answer is wrong.

If there exist atleast 1 decomposition which is Dependency preserving then we can say relation is Dependency preserving in your case you took the one which is not, but there exist an decomposition which is DP and lossless.

If no decomposition exist which is dependency preserving then it is we can say it's not

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thanks :)