Molina Exercise-3.3 BCNF decomposition

Question

Decompose into BCNF

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

FD: AB->C, C->D, D>B, D->E

Comments

F={BCD→A,BC→E,A→F,F→G,C→D,A→G}
BC+=BCAD
yes you're correct, D is extraneous attribute. So BCD->A can be reduces to BC->A
then the relation R1(B, C, A, E) will be in BCNF

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yes, hectic to think such question
How possible to think within 2 minutes in GATE Exam :(

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I've the same question😂

Answer 1

yes, there will be $3$ tables

$\left ( A,B,C \right )$,$\left ( B,D,E \right )$,$\left ( C,D \right )$

In $\left ( A,B,C \right )$ here $\left ( {\color{Red} {A,B}},C \right )$ $AB$ is the key of the table

In $\left ( B,D,E \right )$ here $\left ( B,{\color{Red} {D}},E \right )$ $D$ is the key attribute

In $\left ( C,D \right )$ here $\left ( {\color{Red} {C}},D \right )$ $C$ is key attribute

In BCNF decomposition, if there are no key present in a relation, break it in a table and make a key attribute for that relation.

So, here $3$ tables will be enough

Ref:http://www.cs.colostate.edu/~cs430dl/yr2019sp/more_examples/Ch8/Decomposition%20into%20BCNF.pdf

Comments

R1(CD),C->D

R2(BDE),D->B,D->E

R3(AC) in BCNF DEPENDENCY PRESERVING ALWAYS NOT POSSIBLE.IS THIS CORRECT ????

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if you will find closure of C = {B,C,D,E} you will see that there is derived dependency C — >B  

so your (A.B.C) is not in BCNF you have to further devide it into (A,B) (C,B) making it lossy decomposition

Answer 2

We get 3 BCNF Relations(DBE, CD, AC) which is lossless but not fd preserving due to AB→ C

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From given R(ABCDE) and FD’s : { AB->C, C->D, D>B, D->E} => CK={ AB,AC,AD} and Prime attributes = A,B,C,D .

So, D→ E is not in 2nf and C→ D, D→ B are not in BCNF.

find closure of all these  C+ = CDBE and D+ = DBE

1^st decompose into CDBE and AC.

in R1(CDBE) only C->D, D->B,D->E holds. we get CK ={C} and now again D->B,D->E are not in 3nf. so decompose into DBE(D->BE) and CD(C->D)

Therefore we get total 3 Relations(DBE, CD, AC) which are in BCNF.