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Unformatted text preview: The Relational Data Model
Functional Dependencies 1 Functional Dependencies
x X > A is an assertion about a relation R that whenever two tuples of R agree on all the attributes of X, then they must also agree on the attribute A. Say "X > A holds in R." Notice convention: ...,X, Y, Z represent sets of attributes; A, B, C,... represent single attributes. Convention: no set formers in sets of attributes, just ABC, rather than {A,B,C }. 2 Example
x Drinkers(name, addr, beersLiked, manf, favBeer). x Reasonable FD's to assert:
1. name > addr 2. name > favBeer 3. beersLiked > manf 3 Example Data
addr beersLiked manf favBeer Voyager Bud A.B. WickedAle Voyager WickedAle Pete's WickedAle Enterprise Bud A.B. way way k Because name > addr Because name > favBeer Because beersLiked > manf 4 FD's With Multiple Attributes
x No need for FD's with > 1 attribute on right. x > 1 attribute on left may be essential. Example: bar beer > price But sometimes convenient to combine FD's as a shorthand. Example: name > addr and name > favBeer become name > addr favBeer 5 Keys of Relations
x K is a key for relation R if:
1. Set K functionally determines all attributes of R 2. For no proper subset of K is (1) true. If K satisfies (1), but perhaps not (2), then K is a superkey. Note E/R keys have no requirement for minimality, as in (2) for relational keys. 6 Example
x Consider relation Drinkers(name, addr, beersLiked, manf, favBeer). x {name, beersLiked} is a superkey because together these attributes determine all the other attributes. name > addr favBeer beersLiked > manf 7 Example, Cont.
x {name, beersLiked} is a key because neither {name} nor {beersLiked} is a superkey. x In this example, there are no other keys, but lots of superkeys. Any superset of {name, beersLiked}.
8 name doesn't > manf; beersLiked doesn't > addr. E/R and Relational Keys
x Keys in E/R are properties of entities x Keys in relations are properties of tuples. x Usually, one tuple corresponds to one entity, so the ideas are the same. x But in poor relational designs, one entity can become several tuples, so E/R keys and Relational keys are different. 9 Example Data
addr beersLiked manf favBeer Voyager Bud A.B. WickedAle Voyager WickedAle Pete's WickedAle Enterprise Bud A.B. Relational key = name beersLiked But in E/R, name is a key for Drinkers, and beersLiked is a key for Beers. Note: 2 tuples for Janeway entity and 2 tuples for Bud entity. way way k 10 Where Do Keys Come From?
1. We could simply assert a key K. Then the only FD's are K > A for all atributes A, and K turns out to be the only key obtainable from the FD's. 2. We could assert FD's and deduce the keys by systematic exploration.
x E/R gives us FD's from entityset keys and manyone relationships. 11 FD's From "Physics"
x While most FD's come from E/R keyness and manyone relationships, some are really physical laws. x Example: "no two courses can meet in the same room at the same time" tells us: hour room > course. 12 Inferring FD's: Motivation
x In order to design relation schemas well, we often need to tell what FD's hold in a relation. x We are given FD's X1 > A1, X2 > A2,..., Xn > An , and we want to know whether an FD Y > B must hold in any relation that satisfies the given FD's. Example: If A > B and B > C hold, surely A > C holds, even if we don't say so.
13 Inference Test
x To test if Y > B, start assuming two tuples agree in all attributes of Y. x Use the given FD's to infer that these tuples must also agree in certain other attributes. x If B is eventually found to be one of these attributes, then Y > B is true; otherwise, the two tuples, with any forced equalities form a twotuple relation that proves Y > B does not follow from the given FD's.
14 Closure Test
x An easier way to test is to compute the closure of Y, denoted Y +. x Basis: Y + = Y. x Induction: Look for an FD's left side X that is a subset of the current Y +. If the FD is X > A, add A to Y +. 15 X Y+ A new Y+ 16 Finding All Implied FD's
x Motivation: "normalization," the process where we break a relation schema into two or more schemas. x Example: ABCD with FD's AB >C, C >D, and D >A. Decompose into ABC, AD. What FD's hold in ABC ? Not only AB >C, but also C >A !
17 Basic Idea
x To know what FD's hold in a projection, we start with given FD's and find all FD's that follow from given ones. x Then, restrict to those FD's that involve only attributes of the projected schema. 18 Simple, Exponential Algorithm
1. For each set of attributes X, compute X +. 2. Add X >A for all A in X + X. 3. However, drop XY >A whenever we discover X >A. 4. Finally, use only FD's involving projected attributes.
x Because XY >A follows from X >A. 19 A Few Tricks
x Never need to compute the closure of the empty set or of the set of all attributes. x If we find X + = all attributes, don't bother computing the closure of any supersets of X. 20 Example
x ABC with FD's A >B and B >C. Project onto AC. A +=ABC ; yields A >B, A >C. B +=BC ; yields B >C. C +=C ; yields nothing. BC +=BC ; yields nothing. We do not need to compute AB + or AC +. 21 Example, Continued
x Resulting FD's: A >B, A >C, and B >C. x Projection onto AC : A >C. Only FD that involves a subset of {A,C }. 22 A Geometric View of FD's
x Imagine the set of all instances of a particular relation. x That is, all finite sets of tuples that have the proper number of components. x Each instance is a point in this space. 23 Example: R(A,B) {(1,2), (3,4)} {(5,1)} {} {(1,2), (3,4), (1,3)}
24 An FD is a Subset of Instances
x For each FD X > A there is a subset of all instances that satisfy the FD. x We can represent an FD by a region in the space. x Trivial FD : an FD that is represented by the entire space. Example: A > A. 25 Example: A > B for R(A,B) {(1,2), (3,4)} {} A > B {(5,1)} {(1,2), (3,4), (1,3)}
26 Representing Sets of FD's
x If each FD is a set of relation instances, then a collection of FD's corresponds to the intersection of those sets. Intersection = all instances that satisfy all of the FD's. 27 Example
Instances satisfying A>B, B>C, and CD>A A>B B>C CD>A 28 Implication of FD's
x If an FD Y > B follows from FD's X1 > A1,..., Xn > An , then the region in the space of instances for Y > B must include the intersection of the regions for the FD's Xi > Ai . That is, every instance satisfying all the FD's Xi > Ai surely satisfies Y > B. But an instance could satisfy Y > B, yet not be in this intersection. 29 Example A>B A>C B>C 30 ...
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 Spring '08
 CHIRKOVA

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