I have a relation R(A,B,C,G,H,I)
with the functional dependencies (FDs for brevity) F = {A→B; A→C; C,G→H; C,G→I; B→H}
. I want to decompose it into BCNF.
So, we have AG
as candidate key. I consider the first FD A→B
. I see that A
is not a super key, so I determine its closure as ABCH
and decompose it as R1(A,B,C,H)
and R2(A,G,I)
.
Now, R1
satisfies the FD A→B, A→C
, but violates B→H
, so I decompose it further as R3(B,H)
and R4(A,B,C)
.
Now, my question is: How do I fit satisfy the other FD C,G→I
and C,G→H
in R2
, as R2
does not contain C
or H
?