minimum no of tables in the decomposed form.
R={A->D,B->D,D->BC}
KEY IS {A}
decomposed into two relations R1(AD) and R2(DBC)
R1(AD) =A->D A ISTHE KEY.
R2(DBC) =D->BC,B->D. (B,D ARE KEYS.)
SO THE ABOVE RELATION SATISFIES BCNF DECOMPOSITION
MINIMUM NO OF RELATIONS ARE 2