Chapter 6 Normalization

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There are some natural relationships between the attributes. For example, it seems that every S# lives in exactly one city. Thus, {S# } → {CITY}. Moreover, this instance of SCP also satisfies several other FDs: 1. {S#,P# } → {QTY} 2. {S#,P# } → {CITY} 3. {S#,P# } → {CITY, QTY} 4. {S#,P# } → {S#} 5. {S#,P# } → {S#,P#,CITY,QTY} 6. {S# } → {QTY} 7. {QTY } → {S#} The left- and right-hand sides of an FD are sometimes called the determinant and the dependent, respectively. When such a set consists of just one attribute, we often drop the set braces. E.g., S# → CITY. 14
What do we really want We notice that QTY → CITY holds for the above instance, but it does not have to for another one. What we are really interested is the general case: A set of FDs not only hold for some instance of a table, but for all the possible instances that table could have. For example, S# → CITY holds for all possible values of SCP, since at any given time, a given supplier has to stay in just one place. Thus, any two tuples appearing in SCP at the same time with the same supplier number must necessarily have the same city as well. From now on, by FD, we mean this more demanding, time-independent sense. Below are some examples: {S#,P# } → {QTY} {S#} → {CITY} 15
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: FD examples Given the following val
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 and 46: BCNF Decomposition Let S 0 = (R, F)
Page 47 and 48: Via 3NF If we settle for 3NF, it is

Chapter 6 Normalization

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