1. 1 Functional Dependencies Meaning of FD’s Keys and Superkeys Inferring FD’s

2. 2 Functional Dependencies uX -> 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. wSay “X -> A holds in R.” wConvention: …, X, Y, Z represent sets of attributes; A, B, C,… represent single attributes. wConvention: no set formers in sets of attributes, just ABC, rather than {A,B,C }.

3. 3 Example Drinkers(name, addr, beersLiked, manf, favBeer) uReasonable FD’s to assert: 1.name -> addr 2.name -> favBeer 3.beersLiked -> manf

4. 4 Example Data nameaddr beersLiked manffavBeer JanewayVoyager Bud A.B.WickedAle JanewayVoyager WickedAle Pete’sWickedAle SpockEnterprise Bud A.B.Bud Because name -> addr Because name -> favBeer Because beersLiked -> manf

5. 5 FD’s With Multiple Attributes uNo need for FD’s with > 1 attribute on right. wBut sometimes convenient to combine FD’s as a shorthand. wExample: name -> addr and name -> favBeer become name -> addr favBeer u > 1 attribute on left may be essential. wExample: bar beer -> price

6. 6 Keys of Relations uK is a superkey for relation R if K functionally determines all of R. uK is a key for R if K is a superkey, but no proper subset of K is a superkey.

7. 7 Example Drinkers(name, addr, beersLiked, manf, favBeer) u {name, beersLiked} is a superkey because together these attributes determine all the other attributes. wname -> addr favBeer wbeersLiked -> manf

8. 8 Example, Cont. u{name, beersLiked} is a key because neither {name} nor {beersLiked} is a superkey. wname doesn’t -> manf; beersLiked doesn’t -> addr. uThere are no other keys, but lots of superkeys. wAny superset of {name, beersLiked}.

9. 9 E/R and Relational Keys uKeys in E/R concern entities. uKeys in relations concern tuples. uUsually, one tuple corresponds to one entity, so the ideas are the same. uBut --- in poor relational designs, one entity can become several tuples, so E/R keys and Relational keys are different.

10. 10 Example Data nameaddr beersLiked manffavBeer JanewayVoyager Bud A.B.WickedAle JanewayVoyager WickedAle Pete’sWickedAle SpockEnterprise Bud A.B.Bud 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.

11. 11 Where Do Keys Come From? 1.Just assert a key K. wThe only FD’s are K -> A for all attributes A. 2.Assert FD’s and deduce the keys by systematic exploration. wE/R model gives us FD’s from entity-set keys and from many-one relationships.

12. 12 More FD’s From “Physics” uExample: “no two courses can meet in the same room at the same time” tells us: hour room -> course.

13. 13 Inferring FD’s uWe are given FD’s X 1 -> A 1, X 2 -> A 2,…, X n -> A n, and we want to know whether an FD Y -> B must hold in any relation that satisfies the given FD’s. wExample: If A -> B and B -> C hold, surely A -> C holds, even if we don’t say so. uImportant for design of good relation schemas.

14. 14 Inference Test uTo test if Y -> B, start by assuming two tuples agree in all attributes of Y. Y 0000000... 0 00000??... ?

15. 15 Inference Test – (2) uUse the given FD’s to infer that these tuples must also agree in certain other attributes. wIf B is one of these attributes, then Y -> B is true. wOtherwise, the two tuples, with any forced equalities, form a two-tuple relation that proves Y -> B does not follow from the given FD’s.

16. 16 Closure Test uAn easier way to test is to compute the closure of Y, denoted Y +. uBasis: Y + = Y. uInduction: 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 +.

17. 17 Y+Y+ new Y + XA

18. 18 Finding All Implied FD’s uMotivation: “normalization,” the process where we break a relation schema into two or more schemas. uExample: ABCD with FD’s AB ->C, C ->D, and D ->A. wDecompose into ABC, AD. What FD’s hold in ABC ? wNot only AB ->C, but also C ->A !

19. 19 Why? a1b1ca1b1c ABC ABCD a2b2ca2b2c Thus, tuples in the projection with equal C’s have equal A’s; C -> A. a 1 b 1 cd 1 a 2 b 2 cd 2 comes from d 1 =d 2 because C -> D a 1 =a 2 because D -> A

20. 20 Basic Idea 1.Start with given FD’s and find all nontrivial FD’s that follow from the given FD’s. wNontrivial = left and right sides disjoint. 2.Restrict to those FD’s that involve only attributes of the projected schema.

21. 21 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 for the minimal basis uBecause XY ->A follows from X ->A in any projection. uNeed to include it for a complete set 4.Finally, use only FD’s involving projected attributes.

22. 22 A Few Tricks uNo need to compute the closure of the empty set or of the set of all attributes. uIf we find X + = all attributes, so is the closure of any superset of X.

23. 23 Example uABC with FD’s A ->B and B ->C. Find non-trivial FDs when projected onto AC. wA + =ABC ; yields A ->B, A ->C. We do not need to compute AB + or AC +. wB + =BC ; yields B ->C. wC + =C ; yields nothing. wBC + =BC ; yields nothing.

24. 24 Example --- Continued uResulting the minimal basis of non- trivial FD’s: A ->B, A ->C, and B ->C. uProjection onto AC : A ->C. wOnly FD that involves a subset of {A,C }. uKey for such AC: A uFor complete set of non-trivial FDs in ABC, add AB->C, AC->B

25. 25 More Examples u ABCDE with FD’s AB ->DE, C->E, D->C, and E ->A, project onto ABC, please give the non- trivial FDs uInference the disclosures wA+ = A, B+ = B, C+ = ACE wAB+ = ABCDE wAC+ = ACE wBC+ = ABCDE uIgnore D and E, for wA minimal basis: C->A, AB->C wA complete set: also include BC->A

26. 26 A Geometric View of FD’s uImagine the set of all instances of a particular relation. uThat is, all finite sets of tuples that have the proper number of components. uEach instance is a point in this space.

27. 27 Example: R(A,B) {(1,2), (3,4)} {} {(1,2), (3,4), (1,3)} {(5,1)}

28. 28 An FD is a Subset of Instances uFor each FD X -> A there is a subset of all instances that satisfy the FD. uWe can represent an FD by a region in the space. uTrivial FD = an FD that is represented by the entire space. wExample: A -> A.

29. 29 Example: A -> B for R(A,B) {(1,2), (3,4)} {} {(1,2), (3,4), (1,3)} {(5,1)} A -> B

30. 30 Representing Sets of FD’s uIf each FD is a set of relation instances, then a collection of FD’s corresponds to the intersection of those sets. wIntersection = all instances that satisfy all of the FD’s.

31. 31 Example A->B B->C CD->A Instances satisfying A->B, B->C, and CD->A

32. 32 Implication of FD’s uIf an FD Y -> B follows from FD’s X 1 -> A 1,…,X n -> A n, then the region in the space of instances for Y -> B must include the intersection of the regions for the FD’s X i -> A i. wThat is, every instance satisfying all the FD’s X i -> A i surely satisfies Y -> B. wBut an instance could satisfy Y -> B, yet not be in this intersection.

33. 33 Example A->B B->C A->C

34. 34 Backup Slides

35. 35 More Examples u ABCD with FD’s AB ->C, C->D, and D ->A, uWhat are the non-trivial FDs in ABCD ? wA+ = A, B+ = B, C+ = ACD, D+ = AD, gives C->A wAB+ = ABCD, gives AB->D, no need for ABC+, or ABD+ wAC+ = ACD, gives AC->D wAD+ = AD wBC+ = ABCD, gives BC->A, BC->D wBD+ = ABCD, gives BD->A, BD->C wCD+ = ACD, gives CD->A