decomposition made easy mock

Question

Which of the following relation can decompose into BCNF with dependency preserving and lossless join decomposition.
(i) R(ABCDE) {AB → C, C → AB, C → D, D → E}
(ii) R(ABCDE) {AB → C, C → A, C → D, D → E}

A) only i

B) Only II

C) Both

D) Neither

Answer 1

----for first relation,candidate key is AB and C.

so,all the FDs except D->E are in BCNF

now we will make a separate relation for D->E

so,now we got two relation R1(A,B,C,D) R2(D,E)

so,it is both lossless as well as dependency preserving.

---for second relation,candidate key is AB aand CB

so,here,except AB -> C,all are not in BCNF

so,we need to make separate relation for C->A ,C->D , D->E

now,we get 4 relations,R1(B,C) ,R2(C,A) ,R3(C,D) ,R4(D,E)

now we wont be able to derive id AB->C

and hence it is not a dependency preserving BCNF decomposition.

am i corect??

Comments

I think 2nd can be also dependency preserving as well lossless

we have AB and BC as the key

then we can decompose ABCDE into ABC and ACDE ryt?

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then ABC will contain dependency AB-->C C-->A which is not in BCNF

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is this wrong?

if i decompose 2nd as follows:-

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

FOR D->E : R1(D,E) ; R(A,B,C,D)

FOR C->D : R2(C,D); R(A,B,C)

FOR AB->C : R3(A,B); R3(A,B,C) 

FOR C->A : R4(A,B,C) R5(B,C)

total 6 relations are formed:-

R1(D,E)

R2(C,D)

R3(A,B)

R4(B,C)

R5(A,B,C)

HENCE DEPENDENCY PRESERVING!!