Lecture 09: Functional Dependencies

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: name buys Product Person Relational Model: plus FD’s price name ssn date Normalization: Eliminates anomalies

4. Functional Dependencies Definition: A1, ..., Am →B1, ..., Bn holds in R if: t, t’  R, (t.A1=t’.A1  ...  t.Am=t’.Am  t.B1=t’.B1  ...  t.Bm=t’.Bm ) R t if t, t’ agree here then t, t’ agree here t’

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

7. Formal definition of a key • A key is a set of attributes A1, ..., An s.t. for any other attribute B, A1, ..., An → 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 A1 , A2 , … An→B1, B2, … Bm Splitting rule and Combing rule Is equivalent to A1 , A2 , … An →B1 A1 , A2 , … An →B2 AND … A1 , A2 , … An →Bm

10. Inference Rules for FD’s(continued) A1 , A2 , … An → Ai Trivial Rule where i = 1, 2, ..., n Why ?

11. Inference Rules for FD’s(continued) Transitive Closure Rule If A1 , A2 , … An → B1, B2, … Bm B1, B2, … Bm → C1 , C2 , … Ck and A1 , A2 , … An → C1 , C2 , … Ck then Why?

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 {A1, …, An} and a set of dependencies S. Problem: find all attributes B such that: any relation which satisfies S also satisfies: A1 , A2 , … An → B The closure of {A1, …, An}, denoted {A1, …, An}+, is the set of all such attributes B

15. Closure Algorithm Start with X={A1, …, An}. UntilX doesn’t change do: if B1, B2, … Bm → C is in S, and B1, B2, … Bmare 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: • A1 , A2 , … An → B • Initially X = {A1 , A2 , … An } holds • Induction step: B1, …, Bm in X • Implies A1 , A2 , … An → B1, B2, … Bm • We also have B1, B2, … Bm → C • By transitivity we have A1 , A2 , … An → 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 Recall set attributes (persons with several phones): SSN→Name, City, but not SSN→PhoneNumber • Anomalies: • Redundancy = repeated data • Update anomalies = Fred moves to “Bellevue” • Deletion anomalies = Fred drops all phone numbers: what is his city ?

20. Relation Decomposition Break the relation into two:

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

22. Decompositions in General R(A1, ..., An) Create two relations R1(B1, ..., Bm) and R2(C1, ..., Ck) such that: B1, ..., Bm C1, ..., Ck= A1, ..., An and: R1 = projection of R on B1, ..., Bm R2 = projection of R on C1, ..., Ck

23. Incorrect Decomposition • Sometimes it is incorrect: Decompose on : Name, Category and Price, Category

24. Incorrect Decomposition 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: A relation R is in BCNF if: Whenever there is a nontrivial dependency A1, ..., An→B in R , {A1, ..., An} is a key for R Including superkeys In English: Whenever a set of attributes of Rdetermines one moreattribute, it should determine all the attributes of R.

27. Example What are the dependencies? SSN → Name, City What are the keys? {SSN, PhoneNumber} Is it in BCNF?

28. Decompose it into BCNF SSN → Name, City

29. Summary of BCNF Decomposition Find a dependency that violates the BCNF condition: A1 , A2 , … An → B1, B2, … Bm Heuristics: choose B1, B2, … Bm “as large as possible” Decompose: Continue until there are no BCNF violations left. Others A’s B’s Exercise: Is there a 2-attribute relation that is not in BCNF ? R1 R2

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 R1(A,B,C), R2(A,D) • What happens if we pick A → Bfirst ?

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) Decompose Recover R’ is in general larger than R. Must ensure R’ = R

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

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: What happened here? Violates the dependency: Film, Actor Character!

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 A1, A2, ..., An Bfor R , then {A1, A2, ..., An } 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)