In such question we try to find candidate key first and hence prime and non prime attributes..But for knowing about BCNF , knowing candidate key is sufficient as left hand side of the FD has to be superkey for BCNF to hold.
Given FD set : F:{B ->D,
CD ->E,
E->A,
A->BC}
Hence we have candidate keys : { CD , BC , A , E }
So the FD B --> D is creating problem..So the way to remove is to separate it..We do this as :
B+ = BD
So let we have relation BD which contains the FD : B --> D .Other possible FD is D --> B..But the attribute closure of D w.r.t. original FD set gives D only, so we cannot add it in implied FD of the subrelation.
Hence let R1( B , D ) be subrelation with B as the superkey..
Now to have lossless property we place A , C , E in other subrelation as well as attribute B as for maintaining lossless property we require that the intersection of attributes should be the superkey for at least one relation which is B here.
So let we have R2( A , B , C , E)..Now we define the FDs that can be included in this subrelation with the help of original FD set.
We had A , E , BC as keys earlier , so we can write : A --> BCE and
E --> ABC and
BC --> AE
Now we see that all the FDs that we have found in R1 and R2 will give us back all the original FDs
except one FD which is CD --> E as CD+ = CD only w.r.t the FDs which we have found in the subrelations.
Hence there does not exist a decomposition which satisfies all of the properties i.e. BCNF , lossless decomposition and dependency preserving in this case as CD --> E is not satisfied ..
In fact this is a well known fact that :
BCNF decomposition does not guarantee dependency preservation.
