Consider the following functional dependencies over the attribute set BCGHMVWY:

W ? V
WY ? BV
WC ? V
X ? B
BG ? M
BV ? Y
BYH ? V
M ? W
Y ? H
CY ? W

Find a minimal cover, then decompose into lossless 3NF. After that, check if all the resulting relations are in BCNF. If you find a schema that is not, decompose it into a lossless BCNF. Explain all steps.

Minimal cover: First split WY ? BV into WY ? B and WY ? V. The next step is to reduce the left-hand sides of the FDs. Since W ? B and W ? V are implied by the given set of FDs, we can replace WY ? B, WY ? V, and WC ? V in the original set with W ? B (W ? V already exists in the original set, so we discard the duplicate). Likewise V ? Y and BY ? V can be derived from the original set so we can replace BV ? Y and BYH ? V with V ? Y and BY ? V.

In the next step, we remove redundant FDs, of which we find only W B. The final result is therefore

W ? V
V ? B
BG ? M
V ? Y
BY ? V
M ? W
Y ? H
CY ? W

Lossless 3NF: (WV; {W ? V}), (VBY; {BY ? V, V ? Y, V ? B}),
(BGM; {BG ? M}), (MW; {M ? W}), (YH; {Y ? H}), (CYW; {CY ? W).
Since BGM is a superkey of the original schema, we don’t need to add anything to this decom- position.

BCNF: The schema (BGM; {BG ? M}) is not in BCNF because M ? B is entailed by the original set of FDs and (CYW; {CY ? W) is not in BCNF because W ? Y is entailed.
We decompose BGM with respect to M ? B into (BM; {M ? B}) and (GM; {}), thereby loosing the FD BG ? M.
Similarly CYW is decomposed using W ? Y into (WY; {W ? Y}) and (WC; {}), loosing CY ? W.