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Here Candidate keys are (AB, BC, BD, BE) and As you can see Relation {A, B, C, D, E} is already in 3NF but NOT in BCNF because of  Functional dependencies { C -> D, D->E, E->A} So We need to take the closure 

                                                            (c)+ ={CDEA} 

Now

We need to take the decomposition of the relation R(ABCDE) into R1(CDEA) and R2(BC)

                                                                                R(ABCDE)

                                                            R1(CDEA)                      R2(BC)

                                                             C -> D              {As you can see here A attribute not present So we are not taking AB->C}

                                                             D -> E

                                                             E -> A

                                                    Here C.K.= {C}

So further decomposition in (CD) and (DEA), but

Again you find (DEA) have problem So gain decomposition

relation (DEA) into (DE) and (EA)

Finally What we get 

                                                                     

                                                                              R(ABCDE)

                                     R1(CDEA)                                                                                     R2(BC)

               R11(CD)                         R12(DEA)

                                       R121(DE)                     R122(EA)

          {C=>D}                {D=>E}                            {E=>A}       

But Here you can see that {AB=>C} is not present that mean Dependency not preserving, But we can somehow manage to achieve Dependency preserving by Adding attribute  "A" ( this would not be always possible for every case, but here somehow possible) here Finally, we achieve {BCNF, LL, DP } 

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