Chapter 6 Normalization

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Which one do you like better We have so far discussed two important features of decomposition: losslessness and dependency preservation. We have also seen a BCNF decomposition that is lossless but not dependency preserving. Unfortunately, there is no BCNF for this example that honors both properties. Question: Which property is more important Answer: losslessness is mandatory and dependency preserving is optional. Now we are ready to get into the decomposition business. 44
BCNF Decomposition Let S 0 = (R, F) be a relation schema, the following algorithm constructs a decomposition of S by repeatedly splitting R into smaller subschemas. At each step the new database schema has strictly fewer FDs that violate BCNF than the one in the previous iteration. When the algorithm terminates, all schemas in the result are in BCNF. 1. D = {S 0 } 2. While S = (S; F ′ ) ∈ D is not in BCNF do 3. Let X → Y ∈ F ′ + such that XY ⊆ S and it violates BCNF in S 4. Replace S with S 1 = (XY ; F ′ 1) and S 2 = ((S − Y ) ∪ X; F ′ 2), where F ′ 1 = π XY (F ′ ), F ′ 2 = π (S−Y )∪X (F ′ ). Both F ′ 1 and F′ 2 must be implied by F′ + . It can be shown this decomposition is always lossless. 45
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: A bit too late ..., which we will o
Page 47 and 48: Via 3NF If we settle for 3NF, it is

Chapter 6 Normalization

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