1. 1 IT 244 Database Management System Topic 7 The Relational Data Model Functional Dependencies Ref : -A First Course in Database System (Jeffrey D Ullman, Jennifer Widom) + online.

2. 2 Functional Dependencies Science is the knowledge of consequences, and dependence of one fact upon another. Thomas Hobbes (1588-1679)

3. 3 Functional Dependencies Functional Dependence often abbreviated FD is a unique-value constrain and the most important type of constraint we use for relational schema design. The requirement that a data item be dependent on the entire primary key and not just a part of the primary key.

4. 4 How? It is also possible for database designers to produce relational schemas directly from application requirements although doing so can be very difficult. Regardless of how relational designer are produced, we shall see that frequently it is possible to improve designs systematically based on certain types of constraints.

5. 5 It is vital? Knowledge of this type of constraint is vital for the redesigning of database schemas What it does? - it eliminate redundancy. There are also some other kinds of constraints that help us design good databases schemas such as multivalued dependencies and referential integrity constraints and others.

6. 6 The Evils of Redundancy Redundancy is at the root of several problems associated with relational schemas: –redundant storage, insert/delete/update anomalies Integrity constraints, in particular functional dependencies, can be used to identify schemas with such problems and to suggest refinements. Main refinement technique: decomposition –replacing ABCD with, say, AB and BCD, or ACD and ABD. Decomposition should be used judiciously: –Is there reason to decompose a relation? –What problems (if any) does the decomposition cause?

7. 7 Functional Dependency A functional dependency (FD) on a relation R is a statement of the form “If two tuples of R agree on attributes A1, A2, …. An ( ie the tuples have the same values in their respective components for each of these attributes) then they must agree on another attribute B”. (Book definition)

8. 8 In other words: X  Y means Given any two tuples in r, if the X values are the same, then the Y values must also be the same. (but not vice versa) Can read “  ” as “determines

9. 9 In other words cont’ An FD is a statement about all allowable relations. –Must be identified based on semantics of application. –Given some instance r1 of R, we can check if r1 violates some FD f, but we cannot determine if f holds over R. Question: How related to keys? if “K  all attributes of R” then K is a superkey for R (does not require K to be minimal.) FDs are a generalization of keys.

10. 10 Expression R represent Relation A 1, A 2, A 3 … A n represent attributes X, Y, Z represent sets of attributes Letters at the beginning of the alphabet represent single attributes, and letters at the end of the alphabet represent a set of attributes.

11. 11 Formally we write FD as A 1, A 2, … A n -> B and say that A 1, A 2, …, A n functionally determine B. A set of attributes A 1, A 2, …, A n functionally determines more than one attributes A 1, A 2, …, A n -> B 1 A 1, A 2, …, A n -> B 2 …….. A 1, A 2, …, A n -> B m For short we write this set of FD’s as A 1, A 2, …, A n -> B 1 B 2 … B m

12. 12 In other words 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 }.

13. 13 What this FD’s suggest

14. 14 Example 1 Movie (title, year, length, filmType, studioName, starName) Instance of what we reproduce is TitleYearLengthfilm Typestudio Namestar Name Star wars1977124ColorFoxCarrie Fisher Star wars1977124ColorFoxMark Hamili Star Wars1977124ColorFoxHarrison Ford Mighty Ducks1991104ColorDisneyEmilio Estevez Wayne’s World199295ColorParamountDana Carvey Wayne’s World199295ColorParamountMike Mayers Several FD’s that we can reasonably assert about this Movie Relations title year -> length title year -> filmType title year -> studioName

15. 15 What does this FD’s Says? Informally it says that if two tuples have the same value in their title components, and they also have the same value in their year components then these two tuples must have the same values in their length component, the same value in their filmtype component and the same values in their studioName component. This assertion makes sense if we remember the original design from which this relation schema was developed. Attributes title and year form a key for the Movie entity set. Thus we expect that given a title and year, there is a unique movie, therefore there is a unique length and a unique film type for the movie. There is a many to one relationship from Movies to Studios. Consequently, we expect that given a movie there is only one owing studio.

16. 16 What does this FD’s Says? On the other hand, we observe that the statement title year -> starsName is false, it is not functionally dependency. Given a movie, it is entirely possible that there is more than one star for the movie listed in our database.

17. 17 Example 2 Drinkers(name, addr, beersLiked, manf, favBeer). Think of a: Reasonable FD’s to assert:

18. 18 Drinkers(name, addr, beersLiked, manf, favBeer). Reasonable FD’s to assert: 1.name -> addr 2.name -> favBeer 3.beersLiked -> manf

19. 19 Example Data nameaddr beersLiked manffavBeer ViliKalau VB A.BIkale ViliKalau Ikale R.B Ikale ToluManima VB A.BVB Because name -> addr Because name -> favBeer Because beersLiked -> manf

20. 20 Exercise Look at your own database, see if you can find examples of FD’s from it.

21. 21 Functional Dependencies Tell us about the Schema Recall that FD like any constraint is an assertion about the schema of a relation, not about a particular instance. If we look at an instance we cannot tell for certain that a FD holds we might suppose that a FD like title -> filmtype holds because for every tuple in this particular instance of the relation Movie it happens that any two tuples agreeing on title also agree on filmtype - two version of King Kong –one is color and one is black and white

22. 22 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 -> favBeer become name -> addr favBeer > 1 attribute on left may be essential. –Example: bar beer -> price

23. 23 Another Example Attributes {title, year, starName} form a key for the relation Movies. But first we must show that they functionally determine all the other attributes. That is suppose two tuples agree on these three attributes: title, year, and starName. Because they agree on title and year, they must agree on the other attributes – length, filmType and studioName. Thus, two different tuples can not agree on all of title, year and starName they would in fact be the same tuple,

24. 24 Now we must agree that no proper subset of {title, year, starName} functionally determines all other attributes. Why?- begin by observing that title and year do not determine starName, because many movies have more than one star. Thus {title and year} is not a key – {year, starName} is not a key because we could have a star in two moives in the same year: therefor year starName -> title is not a FD. Also we claim that {title, starName} is not a key because two movies with the same title made in different years, occasionally have a star in common.

25. 25 Therefore… Several FD’s that we can reasonably assert about the Movie Relations title year -> length title year -> filmType title year -> studioName Since each have the same left side title and year we summarize them like Title year -> length filmType studioName.

26. 26 Keys of Relations We say a set of one or more attributes {A1, A2, … An } is a key for a relation R if: 1.Those attributes functionally determine all other attributes of the relation. That is because relations are sets, it is impossible for two distinct tuples of R to agree on all of A1, A2, … An

27. 27 2.No proper subset of {A1, A2, … An } functionally determines all other attributes of R; ie. A key must be minimal *When a key consists of a single attribute A, we often say that A (rather than {A}) is a key.

28. 28 Short form… K is a key for relation R if: –Set K functionally determines all attributes of R –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.

29. 29 Superkeys A set of attributes that contain a key is called a superkey, short for superset of a key. Thus every key is a superkey. Note that every superkey satisfies the first condition of a key; it functionally determines all other attributes of the relation. However, a superkey need not satisfy the second condition: minimality.

30. 30 Example Consider relation Drinkers(name, addr, beersLiked, manf, favBeer). {name, beersLiked} is a superkey because together these attributes determine all the other attributes. –name -> addr favBeer –beersLiked -> manf

31. 31 Example, Cont. {name, beersLiked} is a key because neither {name} nor {beersLiked} is a superkey. –name doesn’t -> manf; beersLiked doesn’t -> addr. In this example, there are no other keys, but lots of superkeys. –Any superset of {name, beersLiked}.

32. 32 Discovering Keys for Relations When a relation schema was developed by converting an E/R design to relations, we can often predict the key of the relation. Rule 1 If the relation comes from an entity set then the key for the relation is the key attributes of this entity set. Eg. Entity set Movies and Stars could be converted to relations. The keys for these entity sets were {title, year} and {name} Movies (title, year, length, filmtype) Stars (name, address)

33. 33 Second Rule concerns binary relationship – if a relation R is constructed form a relationship, then the multiplicity of the relationship affects the key for R. There are three cases - If the relationship is many-many, then the keys of both connected entity sets are the key attributes for R -If the relationship is a many-one from entity set E1 to entity set E2, then the key attributes of E1 are key attributes of R, but those of E2 are not. -If the relationship is one-one, then the key attributes for either of the connected entity sets are key attributes of R, Thus there is not a unique key for R.

34. 34 E/R and Relational Keys Keys in E/R are properties of entities Keys in relations are properties of 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.

35. 35 Example Data 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 Vili entity and 2 tuples for VB entity. nameaddr beersLiked manffavBeer ViliKalau VB A.BIkale ViliKalau Ikale R.B Ikale ToluManima VB A.BVB

36. 36 A continuous Example!!! Constraints on Entity Set Consider relation obtained from Hourly_Emps: Hourly_Emps (ssn, name, lot, rating, wage_per_hr, hrs_per_wk) We sometimes denote a relation schema by listing the attributes: e.g., SNLRWH This is really the set of attributes {S,N,L,R,W,H}. Sometimes, we refer to the set of all attributes of a relation by using the relation name. e.g., “Hourly_Emps” for SNLRWH

37. 37 A continuous Example!!! count’ What are some FDs on Hourly_Emps? –ssn is the key: S  SNLRWH –rating determines wage_per_hr: R  W –lot determines lot: L  L (“trivial” dependnency) Know Trivial Dependency? Find!!!

38. 38 A continuous Example!!! Update anomaly: Can we modify W in only the 1st tuple of SNLRWH? Insertion anomaly: What if we want to insert an employee and don’t know the hourly wage for his or her rating? (or we get it wrong?) Deletion anomaly: If we delete all employees with rating 5, we lose the information about the wage for rating 5! Hourly_Emps Problems Due to R  W

39. 39 A continuous Example!!! Hourly_Emps Q: Why was R  W problematic, but S  W not? Detecting Redundancy

40. 40 A continuous Example!!! Redundancy can be removed by “chopping” the relation into pieces. FD’s are used to drive this process. R  W is causing the problems, so decompose SNLRWH into what relations? Hourly_Emps2 Wages Decomposing a Relation

41. 41 Refining an ER Diagram 1st diagram becomes: Workers(S,N,L,D,Si) Departments(D,M,B) –Lots associated with workers. Suppose all workers in a dept are assigned the same lot: D  L Redundancy; fixed by: Workers2(S,N,D,Si) Dept_Lots(D,L) Departments(D,M,B) Can fine-tune this: Workers2(S,N,D,Si) Departments(D,M,B,L) lot dname budgetdid since name Works_In Departments Employees ssn lot dname budget did since name Works_In Departments Employees ssn Before: After:

42. 42 Where Do Keys Come From? 1.We could simply assert a key K. Then the only FD’s are K -> A for all attributes 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. uE/R gives us FD’s from entity-set keys and many-one relationships.

43. 43 FD’s From “Physics” While most FD’s come from E/R Keynes and many-one relationships, some are really physical laws. Example: “no two courses can meet in the same room at the same time” tells us: hour room -> course.

44. 44 Inferring FD’s: Motivation In order to design relation schemas well, we often need to tell what FD’s hold in a relation. We 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. –Example: If A -> B and B -> C hold, surely A - > C holds, even if we don’t say so.

45. 45 Reasoning About FDs Given some FDs, we can usually infer additional FDs: title  studio, star implies title  studio and title  star title  studio and title  star implies title  studio, star title  studio, studio  star implies title  star But, title, star  studio does NOT necessarily imply that title  studio or that star  studio An FD f is implied by a set of FDs F if f holds whenever all FDs in F hold. F + = closure of F is the set of all FDs that are implied by F. (includes “trivial dependencies”)

46. 46 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 !

47. 47 Basic Idea 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. Then, restrict to those FD’s that involve only attributes of the projected schema.

48. 48 Rules of Inference Armstrong’s Axioms (X, Y, Z are sets of attributes): –Reflexivity: If X  Y, then X  Y –Augmentation: If X  Y, then XZ  YZ for any Z –Transitivity: If X  Y and Y  Z, then X  Z These are sound and complete inference rules for FDs! –i.e., using AA you can compute all the FDs in F+ and only these FDs. Some additional rules (that follow from AA): –Union: If X  Y and X  Z, then X  YZ –Decomposition: If X  YZ, then X  Y and X  Z

49. 49 Inference Test To test if Y -> B, start assuming two tuples agree in all attributes of Y. Use the given FD’s to infer that these tuples must also agree in certain other attributes. 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 two- tuple relation that proves Y -> B does not follow from the given FD’s.

50. 50 Example Contracts(cid,sid,jid,did,pid,qty,value), and: –C is the key: C  CSJDPQV –Proj purchases each part using single contract: JP  C –Dept purchases at most 1 part from a supplier: SD  P Problem: Prove that SDJ is a key for Contracts JP  C, C  CSJDPQV imply JP  CSJDPQV (by transitivity) (shows that JP is a key) SD  P implies SDJ  JP (by augmentation) SDJ  JP, JP  CSJDPQV imply SDJ  CSJDPQV (by transitivity) thus SDJ is a key. Q: can you now infer that SD  CSDPQV (i.e., drop J on both sides)? No! FD inference is not like arithmetic multiplication.

51. 51 Attribute Closure Computing the closure of a set of FDs can be expensive. (Size of closure is exponential in # attrs!) Typically, we just want to check if a given FD X  Y is in the closure of a set of FDs F. An efficient check: –Compute attribute closure of X (denoted X + ) wrt F. X + = Set of all attributes A such that X  A is in F + X + := X Repeat until no change: if there is an fd U  V in F such that U is in X +, then add V to X + –Check if Y is in X + –Approach can also be used to find the keys of a relation. If all attributes of R are in the closure of X then X is a superkey for R. Q: How to check if X is a “candidate key”?

52. 52 Closure Test An easier way to test is to compute the closure of Y, denoted Y +. Basis: Y + = Y. 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 +.

53. 53 Attribute Closure (example) R = {A, B, C, D, E} F = { B  CD, D  E, B  A, E  C, AD  B } Is B  E in F+ ? B+ = B –B+ = BCD –B+ = BCDA –B+ = BCDAE … Yes! and B is a key for R too! Is D a key for R? –D+ = D –D+ = DE –D+ = DEC – … Nope! Is AD a key for R? AD + = AD AD + = ABD and B is a key, so Yes! Is AD a candidate key for R? A + = A, D+ = DEC … A,D not keys, so Yes! Is ADE a candidate key for R? … No! AD is a key, so ADE is a superkey, but not a cand. key

54. 54 Y+Y+ new Y + XA

55. 55 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. 4.Finally, use only FD’s involving projected attributes.

56. 56 A Few Tricks Never need to compute the closure of the empty set or of the set of all attributes. If we find X + = all attributes, don’t bother computing the closure of any supersets of X.

57. 57 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.

58. 58 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 }.

59. 59 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.

60. 60 Example: R(A,B) {(1,2), (3,4)} {} {(1,2), (3,4), (1,3)} {(5,1)}

61. 61 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.

62. 62 Example: A -> B for R(A,B) {(1,2), (3,4)} {} {(1,2), (3,4), (1,3)} {(5,1)} A -> B

63. 63 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.

64. 64 Example A->B B->C CD->A Instances satisfying A->B, B->C, and CD->A

65. 65 Implication of FD’s If 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. –That is, every instance satisfying all the FD’s X i -> A i surely satisfies Y -> B. –But an instance could satisfy Y -> B, yet not be in this intersection.

66. 66 Example A->B B->C A->C

67. 67 Summary Functional Dependencies –definition, -expressions, -applications FD’s with multiple attributes Keys of Relations Superkeys Discovering Keys of a Relations Inferring FD’s Geometric View of FD’s