BCNF Decomposition

(recall BCNF: For every functional dependency X → Y, X must be a superkey)

Step 1

Check which functional dependencies of R are not in BCNF, if they're all in BCNF we do not need to decompose R.

AB -> CD

D -> E

A -> C

B -> D

{AB}⁺ = {A,B,C,D,E} -> CK: AB

AB -> CD

D -> E (violates BCNF)

A -> C

B -> D

Step 2

Create two sub-relations:

{X}⁺
R - {X}⁺ + X

AB -> CD

D -> E -> {D}⁺ = {D, E} -> R₁(DE), R₂(ABCD)

A -> C

B -> D

Step 3

For each sub-relation we check if all of its functional dependencies are now in BCNF

if they are we keep the sub-relation as it is
if not, do Step 2 again for the sub-relation's dependency that does not satisfy BCNF

R₁(DE)

CK = D

D -> E (BCNF)

R₂(ABCD)

CK = AB

AB -> CD (BCNF)

A -> C (not in BCNF)

B -> D

R₃(AC)

CK = A

A -> C (BCNF)

R₄(ABD)

CK = AB

AB -> D (BCNF)

B -> D (not in BCNF)

R₅(BD)

CK = B

B -> D (BCNF)

R₆(AB)

CK = AB

AB -> AB (BCNF)

Step 4

Final decomposition is composed of sub-relations that all their functional dependencies hold BCNF

R₁(DE)

D -> E

R₃(AC)

A -> C

R₅(BD)

B -> D

R₆(AB)

AB -> AB