Answer:
Given, FDs are:
S -> D
I -> B
IS -> Q
B -> O
a)
"I" and "S" must be there in any candidate key because they do not appear on the right side of any functional dependency.
The only candidate key is: IS
IS -> ISBDQO
b)
Decomposition of R into 3NF: (I, B), (S, D), (B, O), (I, S, Q)
c)
Decomposition of R into BCNF:
Decompose R by I → B into R1 = (I, B) and R2 = (I, O, S, Q, D).
R1 is in BCNF
Decompose R2 by S → D into R21 = (S, D) and R22 = (O, I, S, Q).
R21is in BCNF
Decompose R22 by I → O into R221 = (I, O) and R222 = (I, S, Q).
R221 is in BCNF.
R222 is in BCNF.
The decomposition is: (I, B), (S, D), (I, O), (I, S, Q)
We can also write it as: (I, B), (S, D), (B, O), (I, S, Q)
Explanation:
The answer above is rendered in a very explanatory way.
