Functional Dependencies

1. Functional Dependencies Why FD's Meaning of FD’s Keys and Superkeys Inferring FD’s Source: slides by Jeffrey Ullman

2. Improving Relation Schemas • Set of relation schemas obtained by translating from an E/R diagram might need improvement • modeling with E/R diagrams is an art, not a science; relies on experience and intuition • multiple alternative designs are possible, how to choose? • Problems caused by redundant storage of info • wasted space • anomalies when updating, inserting or deleting tuples • Basic idea: replace a relation schema with a collection of "smaller" schemas. Called decomposition

3. Improving Relation Schemas • There is a theory for systematically guiding the improvement of relational designs, called normalization • Normalization uses the notion of "constraints" on the information • functional dependencies • multi-valued dependencies • referential integrity constraints

4. Functional Dependencies • 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.” • 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 }.

5. Example Consumers(name, addr, candiesLiked, manf, favCandy) • Reasonable FD’s to assert: • name -> addr • name -> favCandy • candiesLiked -> manf

6. Because name -> favCandy Because name -> addr Because candiesLiked -> manf Example Data name addr candiesLiked manf favCandy Janeway Voyager Twizzlers Hershey Smarties Janeway Voyager Smarties Nestle Smarties Spock Enterprise Twizzlers Hershey Twizzlers

7. FD’s With Multiple Attributes • No need for FD’s with > 1 attribute on right. • But sometimes convenient to combine FD’s as a shorthand. • Example: name -> addr and name -> favCandy become name -> addr favCandy • > 1 attribute on left may be essential. • Example: store candy -> price

8. Keys of Relations • K is a superkey for relation R if K functionally determines all of R. • K is a key for R if K is a superkey, but no proper subset of K is a superkey.

9. Example Consumers(name, addr, candiesLiked, manf, favCandy) • {name, candiesLiked} is a superkey because together these attributes determine all the other attributes. • name -> addr favCandy • candiesLiked -> manf

10. Example, Cont. • {name, candiesLiked} is a key because neither {name} nor {candiesLiked} is a superkey. • name doesn’t -> manf; candiesLiked doesn’t -> addr. • There are no other keys, but lots of superkeys. • Any superset of {name, candiesLiked}.

11. E/R and Relational Keys • Keys in E/R concern entities. • Keys in relations concern tuples. • Usually, one tuple corresponds to one entity, so the ideas are the same. • But --- in poor relational designs, one entity can become several tuples, so E/R keys and Relational keys are different.

12. Example Data name addr candiesLiked manf favCandy Janeway Voyager Twizzlers Hershey Smarties Janeway Voyager Smarties Nestle Smarties Spock Enterprise Twizzlers Hershey Twizzlers Relational key = {name candiesLiked} But in E/R, name is a key for Consumers, and candiesLiked is a key for Candies. Note: 2 tuples for Janeway entity and 2 tuples for Twizzlers entity.

13. Discovering Keys • Suppose schema was obtained from an E/R diagram. • If relation R came from an entity set E, then key for R is the keys of E • If R came from a binary relationship from E1 to E2: • many-many: key for R is the keys of E1 and E2 • many-one: key for R is the keys for E1 (but not the keys for E2) • one-one: key for R is the keys for E1; another key for R is the keys for E2

14. Key Example name name addr manf Con- sumers Likes Candies husband 2 1 Likes(consumer, candy) Favorite Buddies key: consumer candy Favorite(consumer, candy) wife Married key: consumer Married(husband, wife) Buddies(name1, name2) keys: husband or wife key: name1 name2

15. More FD’s From Application • Example: “no two courses can meet in the same room at the same time” tells us: hour room -> course. • Ultimately, FD's and keys come from the semantics of the application. • FD's and keys apply to the schema, not to specific instantiations of the relations

16. Manipulating FD's • Need to be able to reason about FD's in order to support the normalization process (improving relational schemas) • Some preliminaries: • can split FD X -> A B into X -> A, X -> B • can combine FD's X -> A, X -> B into FD X -> A B • Since A -> A is trivially true, no point in having any attribute on the RHS that is also on the LHS

17. Closure of a Set of FD's • If we have FD's A -> B and B -> C, then it is also true that A -> C. • Ex: If name -> address and address -> phone, then name -> phone. • What about a chain of such deductions? • Called closure

18. FD Closure Algorithm Input: a set of attributes {A1,…,An} and a set of FD's S • Z := {A1,…,An} • find an FD in S of the form X -> C such that all the attributes in X are in Z but C is not in Z. Add C to Z • repeat step 2 until there is nothing more that can be put in Z • return Z as the closure of {A1,…,An}

19. Example of Closure Algorithm • Given relation with attributes A, B, C, D, E, F and FD's • AB -> C • BC -> A, BC -> D • D -> E • CF -> B • Compute closure of {A,B}. • Answer: • Z := {A,B} • add C • add A and D • add E • final answer is Z = {A,B,C,D,E}

20. Use of Closure Algorithm • Now we can check if a particular FD A1 … An -> B follows from a set of FD's S: • compute {A1,…,An}+ using S • if B is in the closure, then the FD follows • otherwise it does not

21. Example • Given same set of attributes and FD's as in previous example: • Does AB -> D follow? • compute {A,B}+ = {A,B,C,D,E}. Since D is in {A,B}+, YES. • Does D -> A follow? • compute {D}+ = {D,E}. Since A is not in {D}+, NO.

22. Finding All Implied FD’s • Motivation: “normalization,” the process where we break a relation schema into two or more schemas. • 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 !

23. d1=d2 because C -> D a1b1cd1 a2b2cd2 comes from a1=a2 because D -> A Why? ABCD ABC a1b1c a2b2c Thus, tuples in the projection with equal C’s have equal A’s; C -> A.

24. Basic Idea • Start with given FD’s and find all nontrivial FD’s that follow from the given FD’s. • Nontrivial = left and right sides disjoint. • Restrict to those FD’s that involve only attributes of the projected schema.

25. Simple, Exponential Algorithm • For each set of attributes X, compute X+. • Add X ->A for all A in X + - X. • However, drop XY ->A whenever we discover X ->A. • Because XY ->A follows from X ->Ain any projection. • Finally, use only FD’s involving projected attributes.

26. A Few Tricks • No need to compute the closure of the empty set or of the set of all attributes. • If we find X + = all attributes, so is the closure of any superset of X.

27. Example • ABC with FD’s A ->B and B ->C. Project onto AC. • A +=ABC ; yields A ->B, A ->C. • We do not need to compute AB + or AC +. • B +=BC ; yields B ->C. • C +=C ; yields nothing. • BC +=BC ; yields nothing.

28. Example --- Continued • Resulting FD’s: A ->B, A ->C, and B ->C. • Projection onto AC : A ->C. • Only FD that involves a subset of {A,C }.

29. A Geometric View of FD’s • Imagine the set of all instances of a particular relation. • That is, all finite sets of tuples that have the proper number of components. • Each instance is a point in this space.

30. Example: R(A,B) {(1,2), (3,4)} {(5,1)} {} {(1,2), (3,4), (1,3)}

31. An FD is a Subset of Instances • For each FD X ->A there is a subset of all instances that satisfy the FD. • We can represent an FD by a region in the space. • Trivial FD = an FD that is represented by the entire space. • Example: A -> A.

32. Example: A -> B for R(A,B) A -> B {(1,2), (3,4)} {(5,1)} {} {(1,2), (3,4), (1,3)}

33. Representing Sets of FD’s • 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.

34. Instances satisfying A->B, B->C, and CD->A Example A->B B->C CD->A

35. Implication of FD’s • 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-> Aisurely satisfies Y -> B. • But an instance could satisfy Y -> B, yet not be in this intersection.

36. Example B->C A->C A->B