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## Find the functional dependencies that are violating BCNF, Find the FDs that are not violating the BCNF rules, Find FD for BCNF decomposition

Question:
Consider a relation schema R with attributes ABCDEFGH with functional dependencies S:
S = {B CD; BF H; C AG; CEH F; CH B}
Which of these functional dependencies violate BCNF (Boyce-Codd Normal Form)?

Solution:

BCNF requires that the LHS of an FD be a super key. Hence, we would find the closure for all the Left Hand Side attributes/sets of attributes to check whether the LHS forms the key or not. [Note: B+ means the closure of the attribute B]
 LHS Result Decision B+ = BCD from B → CD = BCDAG from C → A Result does not include attributes E, F and H. Hence, B is not a super key and B → CD violates BCNF. BF+ = BFH from BF → H = BFHCD from B → CD = BFHCDAG from C → AG Result does not include E. Hence, (BF) is not a super key and BF → H violates BCNF. C+ = CAG from C → AG Result does not include B, D, E, F, and H. Hence C is not a super key and C → AG violates BCNF. CEH+ = CEHF from CEH → F = CEHFB from CH → B = CEHFBAG from C → AG = CEHFBAGD from B → CD Result includes all the attributes of relation R. Hence, (CEH) is the super key of R and CEH → F does not violate BCNF. CH+ = CHB from CH → B = CHBD from B → CD = CHBDAG from C → AG Result does not include E and F. Hence, (CH) is not a super key and CH → B violates BCNF.

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