1 Design Theory for Relational Databases Functional Dependencies Decompositions Normal Forms

2 Functional Dependencies uX ->Y 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 all attributes in set Y. wSay “X ->Y 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 Splitting Right Sides of FD’s uX->A 1 A 2 …A n holds for R exactly when each of X->A 1, X->A 2,…, X->A n hold for R. uExample: A->BC is equivalent to A->B and A->C. u There is no splitting rule for left sides. uWe’ll generally express FD’s with singleton right sides.

4 Example: FD’s Drinkers(name, addr, beersLiked, manf, favBeer) uReasonable FD’s to assert: 1.name -> addr favBeer wNote this FD is the same as name -> addr and name -> favBeer. 2.beersLiked -> manf

5 Example: Possible 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

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 Example: Superkey 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 Example: Key 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 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.

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

11 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.

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

13 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.

14 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 +.

15 Y+Y+ new Y + XA

16 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 !

17 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

18 Basic Idea 1.Start with given FD’s and find all nontrivial FD’s that follow from the given FD’s. wNontrivial = right side not contained in the left. 2.Restrict to those FD’s that involve only attributes of the projected schema.

19 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. uBecause XY ->A follows from X ->A in any projection. 4.Finally, use only FD’s involving projected attributes.

20 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.

21 Example: Projecting FD’s uABC with FD’s A ->B and B ->C. Project 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.

22 Example -- Continued uResulting FD’s: A ->B, A ->C, and B ->C. uProjection onto AC : A ->C. wOnly FD that involves a subset of {A,C }.

23 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.

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

25 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.

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

27 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.

28 Example A->B B->C CD->A Instances satisfying A->B, B->C, and CD->A

29 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.

30 Example A->B B->C A->C

31 Relational Schema Design uGoal of relational schema design is to avoid anomalies and redundancy. wUpdate anomaly : one occurrence of a fact is changed, but not all occurrences. wDeletion anomaly : valid fact is lost when a tuple is deleted.

32 Example of Bad Design Drinkers(name, addr, beersLiked, manf, favBeer) nameaddrbeersLikedmanffavBeer JanewayVoyagerBudA.B.WickedAle Janeway???WickedAlePete’s??? SpockEnterpriseBud???Bud Data is redundant, because each of the ???’s can be figured out by using the FD’s name -> addr favBeer and beersLiked -> manf.

33 This Bad Design Also Exhibits Anomalies nameaddrbeersLikedmanffavBeer JanewayVoyagerBudA.B.WickedAle JanewayVoyagerWickedAlePete’sWickedAle SpockEnterpriseBudA.B.Bud Update anomaly: if Janeway is transferred to Intrepid, will we remember to change each of her tuples? Deletion anomaly: If nobody likes Bud, we lose track of the fact that Anheuser-Busch manufactures Bud.

34 Boyce-Codd Normal Form uWe say a relation R is in BCNF if whenever X ->Y is a nontrivial FD that holds in R, X is a superkey. wRemember: nontrivial means Y is not contained in X. wRemember, a superkey is any superset of a key (not necessarily a proper superset).

35 Example Drinkers(name, addr, beersLiked, manf, favBeer) FD’s: name->addr favBeer, beersLiked->manf uOnly key is {name, beersLiked}. uIn each FD, the left side is not a superkey. uAny one of these FD’s shows Drinkers is not in BCNF

36 Another Example Beers(name, manf, manfAddr) FD’s: name->manf, manf->manfAddr uOnly key is {name}. uname->manf does not violate BCNF, but manf->manfAddr does.

37 Decomposition into BCNF uGiven: relation R with FD’s F. uLook among the given FD’s for a BCNF violation X ->Y. wIf any FD following from F violates BCNF, then there will surely be an FD in F itself that violates BCNF. uCompute X +. wNot all attributes, or else X is a superkey.

38 Decompose R Using X -> Y uReplace R by relations with schemas: 1. R 1 = X R 2 = R – (X + – X ). uProject given FD’s F onto the two new relations.

39 Decomposition Picture R-X +R-X + XX +-XX +-X R2R2 R1R1 R

40 Example: BCNF Decomposition Drinkers(name, addr, beersLiked, manf, favBeer) F = name->addr, name -> favBeer, beersLiked->manf uPick BCNF violation name->addr. uClose the left side: {name} + = {name, addr, favBeer}. uDecomposed relations: 1.Drinkers1(name, addr, favBeer) 2.Drinkers2(name, beersLiked, manf)

41 Example -- Continued uWe are not done; we need to check Drinkers1 and Drinkers2 for BCNF. uProjecting FD’s is easy here. uFor Drinkers1(name, addr, favBeer), relevant FD’s are name->addr and name->favBeer. wThus, {name} is the only key and Drinkers1 is in BCNF.

42 Example -- Continued uFor Drinkers2(name, beersLiked, manf), the only FD is beersLiked->manf, and the only key is {name, beersLiked}. wViolation of BCNF. ubeersLiked + = {beersLiked, manf}, so we decompose Drinkers2 into: 1.Drinkers3(beersLiked, manf) 2.Drinkers4(name, beersLiked)

43 Example -- Concluded uThe resulting decomposition of Drinkers : 1.Drinkers1(name, addr, favBeer) 2.Drinkers3(beersLiked, manf) 3.Drinkers4(name, beersLiked) uNotice: Drinkers1 tells us about drinkers, Drinkers3 tells us about beers, and Drinkers4 tells us the relationship between drinkers and the beers they like.

44 Third Normal Form -- Motivation uThere is one structure of FD’s that causes trouble when we decompose. uAB ->C and C ->B. wExample: A = street address, B = city, C = zip code. uThere are two keys, {A,B } and {A,C }. uC ->B is a BCNF violation, so we must decompose into AC, BC.

45 We Cannot Enforce FD’s uThe problem is that if we use AC and BC as our database schema, we cannot enforce the FD AB ->C by checking FD’s in these decomposed relations. uExample with A = street, B = city, and C = zip on the next slide.

46 An Unenforceable FD street (A) zip (C) 545 Tech Sq Tech Sq city(B) zip (C) Cambridge02138 Cambridge02139 Join tuples with equal zip codes. street (A) city(B) zip (C) 545 Tech Sq.Cambridge Tech Sq.Cambridge02139 Although no FD’s were violated in the decomposed relations, FD street city -> zip is violated by the database as a whole. AB ->C and C ->B.

47 3NF Let’s Us Avoid This Problem u3 rd Normal Form (3NF) modifies the BCNF condition so we do not have to decompose in this problem situation. uAn attribute is prime if it is a member of any key. uX ->A violates 3NF if and only if X is not a superkey, and also A is not prime.

48 Example: 3NF uIn our problem situation with FD’s AB ->C and C ->B, we have keys AB and AC. uThus A, B, and C are each prime. uAlthough C ->B violates BCNF, it does not violate 3NF.

49 What 3NF and BCNF Give You uThere are two important properties of a decomposition: 1.Lossless Join : it should be possible to project the original relations onto the decomposed schema, and then reconstruct the original. 2.Dependency Preservation : it should be possible to check in the projected relations whether all the given FD’s are satisfied.

50 3NF and BCNF -- Continued uWe can get (1) with a BCNF decomposition. uWe can get both (1) and (2) with a 3NF decomposition. uBut we can’t always get (1) and (2) with a BCNF decomposition. wstreet-city-zip is an example.

51 Testing for a Lossless Join uIf we project R onto R 1, R 2,…, R k, can we recover R by rejoining? uAny tuple in R can be recovered from its projected fragments. uSo the only question is: when we rejoin, do we ever get back something we didn’t have originally?

52 The Chase Test uSuppose tuple t comes back in the join. uThen t is the join of projections of some tuples of R, one for each R i of the decomposition. uCan we use the given FD’s to show that one of these tuples must be t ?

53 The Chase – (2) uStart by assuming t = abc…. uFor each i, there is a tuple s i of R that has a, b, c,… in the attributes of R i. us i can have any values in other attributes. uWe’ll use the same letter as in t, but with a subscript, for these components.

54 Example: The Chase uLet R = ABCD, and the decomposition be AB, BC, and CD. uLet the given FD’s be C->D and B ->A. uSuppose the tuple t = abcd is the join of tuples projected onto AB, BC, CD.

55 The Tableau ABCDabc1d1a2bcd2a3b3cdABCDabc1d1a2bcd2a3b3cd d Use C->D a Use B ->A We’ve proved the second tuple must be t. The tuples of R pro- jected onto AB, BC, CD.

56 Summary of the Chase 1.If two rows agree in the left side of a FD, make their right sides agree too. 2.Always replace a subscripted symbol by the corresponding unsubscripted one, if possible. 3.If we ever get an unsubscripted row, we know any tuple in the project-join is in the original (the join is lossless). 4.Otherwise, the final tableau is a counterexample.

57 Example: Lossy Join uSame relation R = ABCD and same decomposition. uBut with only the FD C->D.

58 The Tableau ABCDabc1d1a2bcd2a3b3cdABCDabc1d1a2bcd2a3b3cd d Use C->D These three tuples are an example R that shows the join lossy. abcd is not in R, but we can project and rejoin to get abcd. These projections rejoin to form abcd.

59 3NF Synthesis Algorithm uWe can always construct a decomposition into 3NF relations with a lossless join and dependency preservation. uNeed minimal basis for the FD’s: 1.Right sides are single attributes. 2.No FD can be removed. 3.No attribute can be removed from a left side.

60 Constructing a Minimal Basis 1.Split right sides. 2.Repeatedly try to remove an FD and see if the remaining FD’s are equivalent to the original. 3.Repeatedly try to remove an attribute from a left side and see if the resulting FD’s are equivalent to the original.

61 3NF Synthesis – (2) uOne relation for each FD in the minimal basis. wSchema is the union of the left and right sides. uIf no key is contained in an FD, then add one relation whose schema is some key.

62 Example: 3NF Synthesis uRelation R = ABCD. uFD’s A->B and A->C. uDecomposition: AB and AC from the FD’s, plus AD for a key.

63 GIVENGENERATED FD SELIMINATION AB -> ACDAB -> A x (trivial) B -> DAB -> CAB -> C x (redundant) A -> ACAB -> DAB -> D x (redundant) C -> ACB -> DB -> D A - > Ax (trivial)A -> CC -> A C -> Cx (trivial) MINIMAL BASIS B -> D A -> C C -> A KEYS AB and BC DECOMPOSITION BD, AC comes from minimal basis. (we do not add a relation if it is a subset of another one) None of the schemas of basis vectors is a key. Therefore, we need to add one of the keys, say, AB. So the final decomposition is AB, AC, BD. 3NF Decomposition Example (2)

64 Why It Works uPreserves dependencies: each FD from a minimal basis is contained in a relation, thus preserved. uLossless Join: use the chase to show that the row for the relation that contains a key can be made all- unsubscripted variables. u3NF: hard part – a property of minimal bases.

References in your textbook uChapter 3 w3.1 w3.2 w3.3 w3.4 w3.5 65

Homework from your textbook u3.1.1 u3.1.2 (2.1.1 of second edition, solution given in course homepage) u3.2.1 (3.5.1 of second edition, solution given in course homepage) u3.2.2 u3.3.1 a,b (3.6.1 of second edition, solution given in course homepage) u3.3.1 c,d,e,f u3.4.1 a u