KORTH Standard book question - DBMS -Normalization doubt- chapter 7

Question

hi sir,

Below is the question

Give an example of a relation schema R and set F of functional dependencies such that there are at least three distinct lossless-join decompositions of R into BCNF.

My Solution : Sir, Using BCNF decompostion I am able to get only one set of decomposition and not 3 decompostions , how to get rest 2 decompositions for below example , I am getting only  R1 = {(A, B), (C, D), (B, C)}  but not R2 and R3 as show below 

Solution in KORTH:

Given the relation R =(A, B, C, D) the set of functional dependen- cies F = A → B, C → D, B → C allows three distinct BCNF decomposi- tions. R1 = {(A, B), (C, D), (B, C)} is in BCNF as is R2 = {(A, B), (C, D), (A, C)} R2 = {(A, B), (C, D), (A, C)} R3 = {(B, C), (A, D), (A, B)}

Comments

plz answer above question

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You followed a procedure which is guaranteed to give ONE of the possible decomposition. You did not follow a procedure which gives all such decompositions (such a procedure is not there I suppose other than brute force).

Answer 1

I think relation R with following set of dependencies should work:

F = {A->B,  B->C, C->A}

first decomposition:  R1(AB)  R2(BC) R3(AC)

second decomposition: R1(AC) R2(BC)

third decomposition: R1(AB) R2(BC)