BCNF Decomposition
-------------------

Given a relation R and a minimal set F of fds. R is not in BCNF,
we compute the following decomposition:

      - Find any X->Y in F that violates BCNF., compute X+ in F.

       Create two new relation instead of R:
       R1 contains attributes in X+, and R2 contains all attributes in R except for
       attributes in (X+-X).

     - Compute projection of fds for R1 and R2, and if either one is not in
     BCNF, then apply BCNF decomposition recursively until all resulting relations
     are in BCNF.


BCNF Decompositions are:

     - Guaranteed to be lossless!
     - Not guaranteed to be dependency preserving

Ex:

S(A,B,C,D)
AB->C
A->D    x violates BCNF
C->A    X violates BCNF
Key:AB


A->D
A+ = {A,D}

S1(A,D)    F1={A->D}   in BCNF

S2(A,B,C)       (everything in S except for A+-{A} = {D})
          F2 = {AB->C, C->A}, not in BCNF

Decompose further using C->A:

S21(A,C)   F21 ={C->A} in BCNF
S22(B,C)   F22 ={}  in BCNF

Done! Resulting functional dependencies:

S1(A,D)    F1={A->D}   in BCNF
S21(A,C)   F21 ={C->A} in BCNF
S22(B,C)   F22 ={}  in BCNF


--->

F1 union F21 union F22 = {A->D, C->A}
clearly: AB->C in F is lost.


-------------------------------
Additional Exercise for BCNF Decomposition!

Ex:

R(A,B,C,D,E,F)
AB->C
CD->EF
F->A

Keys: ABD, BDF, BCD

Decompose using: CD->EF

CD+ = {C,D,E,F,A}

R1(A,C,D,E,F)   F1 = {CD->EF, F->A},  Keys: CD,  F->A violates BCNF

R2(B,C,D)       F2 = {}  Key: BCD
         in BCNF
everything in R, except CD+-{CD} = {E,F,A}

Decompose using F->A

R11(A,F)        F11 = {A->F}    Key: A , in BCNF
R12(C,D,E,F)    F12 = {CD->EF}   Key: CD, in BCNF

Resulting relations:

AF
CDEF
BCD

Check if using different sequence of fds, do I get different results?