Chapter 6 Normalization

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An example Consider the HasAccount(AccountNumber, ClientId, OfficeId) and the FD: { ClientId, OfficeId → AccountNumber, AccountNumber → OfficeId.} It is not in BCNF since AccountNumber in the second FD is not a super key of the relation. We can thus use this FD as a guidance to split the relation. Letting X be AccountNumber, and Y be OfficeId., the While loop will come up with two subschemas: S 1 =({AccountNumber, OfficeId) }; { AccountNumber → OfficeId }) and S 2 =({ClientId, AccountNumber)}; ∅). Note that Client → AccountNumber is not implied by the original FDs. Question: Is the last one in BCNF 46
Via 3NF If we settle for 3NF, it is always possible to come up with a decomposition that preserves dependency, although as we already saw, 3NF could contain redundancy. A detailed process for generating a 3NF table is given in §6.8, which leads to decomposition both lossless and dependency preserving. If the result of this procedure is already in BCNF, we are done. Otherwise, we can further apply the BCNF procedure to it until all of the components are in BCNF. The point is that if there is a lossless and dependency preserving decomposition, the 3NF procedure is likely to find it. On the other hand, if some components are not in BCNF after the 3NF process, loss of some FD will be inevitable. 47
Page 1 and 2: Chapter 6 Normalization Conceptual
Page 3 and 4: Remember this one Consider the foll
Page 5 and 6: Another abnormally Assume that we w
Page 7 and 8: Decomposition The solution to the a
Page 9 and 10: What have we learned 1. Decompositi
Page 11 and 12: Technically, ... ..., a functional
Page 13 and 14: FD examples Given the following val
Page 15 and 16: What do we really want We notice th
Page 17 and 18: Key is a special FD Note that the k
Page 19 and 20: Identify the FDs... Consider the ta
Page 21 and 22: The bad guy Consider the following
Page 23 and 24: What is happening As a result, we h
Page 25 and 26: What does that mean Every RDB is in
Page 27 and 28: The BCNF A relation schema S = (R,
Page 29 and 30: BCNF states that if this is the cas
Page 31 and 32: 4. A relation instance that is in a
Page 33 and 34: 5. BCNF relations with more than on
Page 35 and 36: 3NF could be redundant We saw earli
Page 37 and 38: A plan Since there is no redundancy
Page 39 and 40: An example Consider the following s
Page 41 and 42: An example The decomposition of Has
Page 43 and 44: A bit too late ..., which we will o
Page 45: BCNF Decomposition Let S 0 = (R, F)

Chapter 6 Normalization

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