1. Lecture 09: Functional Dependencies

2. Outline Functional dependencies (3.4) Rules about FDs (3.5) Design of a Relational schema (3.6)

3. Relational Schema Design Conceptual Model: Relational Model: plus FD’s Normalization: Eliminates anomalies Person buys Product name price namessn date

4. Functional Dependencies Definition: A 1,..., A m → B 1,..., B n holds in R if:  t, t’  R, (t.A 1 =t’.A 1 ...  t.A m =t’.A m  t.B 1 =t’.B 1 ...  t.B m =t’.B m ) A1...AmB1...Bm if t, t’ agree here then t, t’ agree here t t’ R

5. Important Point! Functional dependencies are part of the schema! They constrain the possible legal data instances. At any point in time, the actual database may satisfy additional FD’s.

6. Examples EmpID → Name, Phone, Position Position → Phone but Phone → Position EmpIDNamePhonePosition E0045John Smith2376HR E1847Zheng Li2376QA E3123Lucy Larsen1264Developer E9923Zheng Li1264Developer

7. Formal definition of a key A key is a set of attributes A 1,..., A n s.t. for any other attribute B, A 1,..., A n → B A minimal key is a set of attributes which is a key and for which no subset is a key Note: book calls them superkey and key

8. Examples of Keys Product(name, price, category, color) name, category → price category → color Keys are: {name, category} and all supersets Enrollment(student, address, course, room, time) student → address room, time → course student, course → room, time Keys are: [in class]

9. Inference Rules for FD’s A 1, A 2, … A n →B 1, B 2, … B m Is equivalent to A 1, A 2, … A n →B 1 A 1, A 2, … A n →B 2 A 1, A 2, … A n →B m … Splitting rule and Combing rule A1...AmB1...Bm AND

10. Inference Rules for FD’s (continued) Trivial Rule Why ? A1...Am where i = 1, 2,..., n A 1, A 2, … A n → A i

11. Inference Rules for FD’s (continued) Transitive Closure Rule If and then Why? A 1, A 2, … A n → B 1, B 2, … B m B 1, B 2, … B m → C 1, C 2, … C k A 1, A 2, … A n → C 1, C 2, … C k

12. A1...AmB1...BmC1C1...CkCk

13. Enrollment(student, major, course, room, time) student → major major, course → room course → time What else can we infer ? [in class]

14. Closure of a set of Attributes Given a set of attributes {A 1, …, A n } and a set of dependencies S. Problem: find all attributes B such that: any relation which satisfies S also satisfies: A 1, A 2, … A n → B The closure of {A 1, …, A n }, denoted {A 1, …, A n } +, is the set of all such attributes B

15. Closure Algorithm Start with X={A 1, …, A n }. Until X doesn’t change do: if B 1, B 2, … B m → C is in S, and B 1, B 2, … B m are all in X and C is not in X then add C to X

16. Example The schema - R(A,B,C,D,E,F) A, B→C A,D→ E B→ D A,F→B Closure of {A,B}:X = {A, B,} Closure of {A, F}:X = {A, F,}

17. Why Is the Algorithm Correct ? Show the following by induction: –For every B in X: A 1, A 2, … A n → B Initially X = {A 1, A 2, … A n } holds Induction step: B 1, …, B m in X –Implies A 1, A 2, … A n → B 1, B 2, … B m –We also have B 1, B 2, … B m → C –By transitivity we have A 1, A 2, … A n → C This shows that the algorithm is sound; need to show it is complete

18. Relational Schema Design (or Logical Design) Main idea: Start with some relational schema Find out its FD’s Use them to design a better relational schema

19. Relational Schema Design Anomalies: Redundancy= repeated data Update anomalies= Fred moves to “Bellevue” Deletion anomalies= Fred drops all phone numbers: what is his city ? Recall set attributes (persons with several phones): SSN → Name, City, but not SSN → PhoneNumber NameSSNPhoneNumberCity Fred123-45-6789206-555-1234Seattle Fred123-45-6789206-555-6543Seattle Joe987-65-4321908-555-2121Westfield Joe987-65-4321908-555-1234Westfield

20. Relation Decomposition Break the relation into two: NameSSNCity Fred123-45-6789Seattle Joe987-65-4321Westfield SSNPhoneNumber 123-45-6789206-555-1234 123-45-6789206-555-6543 987-65-4321908-555-2121 987-65-4321908-555-1234

21. Relational Schema Design Conceptual Model: Relational Model: plus FD’s Normalization: Eliminates anomalies Person buys Product name price namessn date

22. Decompositions in General R(A 1,..., A n ) Create two relations R 1 (B 1,..., B m ) and R 2 (C 1,..., C k ) such that: B 1,..., B m  C 1,..., C k = A 1,..., A n and: R 1 = projection of R on B 1,..., B m R 2 = projection of R on C 1,..., C k

23. Incorrect Decomposition Sometimes it is incorrect: NamePriceCategory Gizmo19.99Gadget OneClick24.99Camera DoubleClick29.99Camera Decompose on : Name, Category and Price, Category

24. Incorrect Decomposition NameCategory GizmoGadget OneClickCamera DoubleClickCamera PriceCategory 19.99Gadget 24.99Camera 29.99Camera NamePriceCategory Gizmo19.99Gadget OneClick24.99Camera OneClick29.99Camera DoubleClick24.99Camera DoubleClick29.99Camera When we put it back: Cannot recover information

25. Normal Forms First Normal Form = all attributes are atomic Second Normal Form (2NF) = old and obsolete Third Normal Form (3NF) = this lecture Boyce Codd Normal Form (BCNF) = this lecture Others...

26. Boyce-Codd Normal Form A simple condition for removing anomalies from relations: In English: Whenever a set of attributes of R determines one more attribute, it should determine all the attributes of R. A relation R is in BCNF if: Whenever there is a nontrivial dependency A 1,..., A n → B in R, {A 1,..., A n } is a key for R A relation R is in BCNF if: Whenever there is a nontrivial dependency A 1,..., A n → B in R, {A 1,..., A n } is a key for R Including superkeys

27. Example What are the dependencies? SSN → Name, City What are the keys? {SSN, PhoneNumber} Is it in BCNF? NameSSNPhoneNumberCity Fred123-45-6789206-555-1234Seattle Fred123-45-6789206-555-6543Seattle Joe987-65-4321908-555-2121Westfield Joe987-65-4321908-555-1234Westfield

28. Decompose it into BCNF NameSSNCity Fred123-45-6789Seattle Joe987-65-4321Westfield SSNPhoneNumber 123-45-6789206-555-1234 123-45-6789206-555-6543 987-65-4321908-555-2121 987-65-4321908-555-1234 SSN → Name, City

29. Summary of BCNF Decomposition Find a dependency that violates the BCNF condition: A 1, A 2, … A n → B 1, B 2, … B m A’s Others B’s R1R2 Heuristics: choose B 1, B 2, … B m “as large as possible” Decompose: Exercise: Is there a 2- attribute relation that is not in BCNF ? Continue until there are no BCNF violations left.

30. Example Decomposition Person(name, SSN, age, hairColor, phoneNumber) SSN → name, age age → hairColor Decompose in BCNF (in class): Step 1: find all keys Step 2: now decompose

31. Other Example R(A,B,C,D) A → B, B → C Key: A, D Violations of BCNF: A → B, B → C, A → C, A → B,C Pick A → B,C: split R into R 1 (A,B,C), R 2 (A,D) What happens if we pick A → B first ?

32. Correct Decompositions A decomposition is lossless if we can recover: R(A,B,C) R1(A,B) R2(A,C) R’(A,B,C) should be the same as R(A,B,C) R’ is in general larger than R. Must ensure R’ = R Decompose Recover

33. Correct Decompositions Given R(A,B,C) s.t. A→B, the decomposition into R1(A,B), R2(A,C) is lossless

34. 3NF: A Problem with BCNF FD’s: Character  Film;Film, Actor  Character So, there is a BCNF violation, and we decompose. Character  Film No FDs FilmActorCharacter ActorCharacter FilmCharacter

35. So What’s the Problem? No problem so far. All local FD’s are satisfied. Let’s put all the data back into a single table again: Violates the dependency: Film, Actor  Character! FilmCharacter Austin Powers Dr. Evil ActorCharacter Mike MyersAustin Powers Mike MyersDr. Evil FilmActorCharacter Austin PowersMike MyersAustin Powers Mike MyersDr. Evil What happened here?

36. Solution: 3rd Normal Form (3NF) A simple condition for removing anomalies from relations: 3NF has a dependency-preserving decomposition A relation R is in 3rd normal form if : Whenever there is a nontrivial dependency A 1, A 2,..., A n  B for R, then {A 1, A 2,..., A n } a key for R, or B is part of a key. A relation R is in 3rd normal form if : Whenever there is a nontrivial dependency A 1, A 2,..., A n  B for R, then {A 1, A 2,..., A n } a key for R, or B is part of a key. Including superkeys

37. Decomposition into 3NF Easy when we know BCNF decomposition (how?) Not-so-easy: if we want a dependency- preserving decomposition…  In the book (3.5)