CSE 544 Principles of Database Management Systems

Transcription

1 CSE 544 Principles of Database Management Systems Lecture 3 Schema Normalization CSE Winter

2 Announcements Project groups due on Friday First review due on Tuesday (makeup lecture) Run git pull to get the review subdirectory Place your review there Commit, push (see instructions in hw1.md); no need to tag Homework 1 due next Friday Run turninhw.sh hw1 Or manually: git commit..., git tag..., git push... CSE Winter

3 Database Design The relational model is great, but how do I design my database schema? CSE Winter

4 Outline Conceptual db design: entity-relationship model Problematic database designs Functional dependencies Normal forms and schema normalization CSE Winter

5 Conceptual Schema Design Conceptual Model: name Patient patien_of Doctor zip name dno Relational Model: plus FD s (FD = functional dependency) Normalization: Eliminates anomalies CSE Winter

6 Entity-Relationship Diagram Attributes Entity sets Relationship sets name Patient patient_of 6

7 Entity-Relationship Diagram Patient Doctor Attributes Entity sets Relationship sets name Patient patient_of 7

8 Entity-Relationship Diagram name pno dno Patient Doctor zip specialty name Attributes Entity sets Relationship sets name Patient patient_of 8

9 Entity-Relationship Diagram name pno dno Patient patient_of Doctor zip specialty name Attributes Entity sets Relationship sets name Patient patient_of 9

10 Entity-Relationship Diagram pno name since dno Patient patient_of Doctor zip specialty name Attributes Entity sets Relationship sets name Patient patient_of 10

11 Entity-Relationship Model Typically, each entity has a key ER relationships can include multiplicity One-to-one, one-to-many, etc. Indicated with arrows Many One Can model multi-way relationships Can model subclasses And more... CSE Winter

12 Subclasses to Relations price name category Product isa isa Software Product Educational Product platforms Age Group 12

13 Subclasses to Relations price name category Product Name Price Category Gizmo 99 gadget Product Camera 49 photo Toy 39 gadget isa isa Software Product Educational Product platforms Age Group 13

14 Subclasses to Relations price name category Product Name Price Category Gizmo 99 gadget Product Camera 49 photo Toy 39 gadget isa isa Sw.Product Name platforms Software Product Educational Product Gizmo unix platforms Age Group 14

15 Subclasses to Relations price name category Product Name Price Category Gizmo 99 gadget Product Camera 49 photo Toy 39 gadget isa isa Sw.Product Name platforms Software Product platforms Educational Product Age Group Gizmo Name unix Age Group Gizmo toddler Ed.Product Toy senior

16 General approach to Translating Diagram into Relations Each entity set becomes a relation with a key Each relationship set becomes a relation with foreign keys except many-one relationships: just add a fk Each isa relationship becomes another relation, with both a key and foreign key CSE Winter

17 Outline Conceptual db design: entity-relationship model Problematic database designs Functional dependencies Normal forms and schema normalization CSE Winter

18 Relational Schema Design Name SSN PhoneNumber City Fred Seattle Fred Seattle Joe Westfield One person may have multiple phones, but lives in only one city Primary key is thus (SSN, PhoneNumber) What is the problem with this schema? CSE Winter

19 Relational Schema Design Name SSN PhoneNumber City Fred Seattle Fred Seattle Joe Westfield Anomalies: Redundancy = repeat data for Fred Update anomalies = what if Fred moves to Bellevue? Deletion anomalies = what if Joe deletes his phone number? CSE Winter

20 Relation Decomposition Break the relation into two: Name SSN PhoneNumber City Fred Seattle Fred Seattle Joe Westfield Name SSN City Fred Joe Seattle Westfield SSN PhoneNumber Anomalies have gone: No more repeated data Easy to move Fred to Bellevue (how?) Easy to delete all Joe s phone numbers (how?) 20

21 Relational Schema Design (or Logical Design) How do we do this systematically? Start with some relational schema Find out its functional dependencies (FDs) Use FDs to normalize the relational schema CSE Winter

22 Functional Dependencies (FDs) Definition If two tuples agree on the attributes A 1, A 2,, A n then they must also agree on the attributes Formally: B 1, B 2,, B m A1 An determines B1..Bm A 1, A 2,, A n à B 1, B 2,, B m CSE Winter

23 Functional Dependencies (FDs) Definition A 1,..., A m à B 1,..., B n holds in R if: t, t R, (t.a 1 = t.a 1... t.a m = t.a m à t.b 1 = t.b 1... t.b n = t.b n ) R A 1... A m B 1... B n t t if t, t agree here then t, t agree here 23

24 Example An FD holds, or does not hold on an instance: EmpID Name Phone Position E0045 Smith 1234 Clerk E3542 Mike 9876 Salesrep E1111 Smith 9876 Salesrep E9999 Mary 1234 Lawyer EmpID à Name, Phone, Position Position à Phone but not Phone à Position CSE Winter

25 Example EmpID Name Phone Position E0045 Smith 1234 Clerk E3542 Mike 9876 ß Salesrep E1111 Smith 9876 ß Salesrep E9999 Mary 1234 Lawyer Position à Phone CSE Winter

26 Example EmpID Name Phone Position E0045 Smith 1234 à Clerk E3542 Mike 9876 Salesrep E1111 Smith 9876 Salesrep E9999 Mary 1234 à Lawyer But not Phone à Position CSE Winter

27 Example name à color category à department color, category à price name category color department price Gizmo Gadget Green Toys 49 Tweaker Gadget Green Toys 99 Which FD s hold? CSE Winter

28 Buzzwords FD holds or does not hold on an instance If we can be sure that every instance of R will be one in which a given FD is true, then we say that R satisfies the FD If we say that R satisfies an FD, we are stating a constraint on R CSE Winter

29 An Interesting Observation If all these FDs are true: Then this FD also holds: name à color category à department color, category à price name, category à price Find out from application domain some FDs, Compute all FD s implied by them CSE Winter

30 Closure of a set of Attributes Given a set of attributes A 1,, A n The closure is the set of attributes B, denoted {A 1,, A n } +, s.t. A 1,, A n à B CSE Winter

31 Closure of a set of Attributes Given a set of attributes A 1,, A n The closure is the set of attributes B, denoted {A 1,, A n } +, s.t. A 1,, A n à B Example: 1. name à color 2. category à department 3. color, category à price CSE Winter

32 Closure of a set of Attributes Given a set of attributes A 1,, A n The closure is the set of attributes B, denoted {A 1,, A n } +, s.t. A 1,, A n à B Example: 1. name à color 2. category à department 3. color, category à price Closures: name + = {name, color} CSE Winter

33 Closure of a set of Attributes Given a set of attributes A 1,, A n The closure is the set of attributes B, denoted {A 1,, A n } +, s.t. A 1,, A n à B Example: 1. name à color 2. category à department 3. color, category à price Closures: name + = {name, color} {name, category} + = {name, category, color, department, price} CSE Winter

34 Closure of a set of Attributes Given a set of attributes A 1,, A n The closure is the set of attributes B, denoted {A 1,, A n } +, s.t. A 1,, A n à B Example: 1. name à color 2. category à department 3. color, category à price Closures: name + = {name, color} {name, category} + = {name, category, color, department, price} color + = {color} CSE Winter

35 Keys A superkey is a set of attributes A 1,..., A n s.t. for any other attribute B, we have A 1,..., A n à B A key is a minimal superkey (no subset is a superkey) CSE Winter

36 Computing (Super)Keys For all sets X, compute X + If X + = [all attributes], then X is a superkey Try reducing to the minimal X s to get the key CSE Winter

37 Example Product(name, price, category, color) name, category à price category à color What is the key? CSE Winter

38 Example Product(name, price, category, color) name, category à price category à color What is the key? (name, category) + = { name, category, price, color } CSE Winter

39 Example Product(name, price, category, color) name, category à price category à color What is the key? (name, category) + = { name, category, price, color } Hence (name, category) is a key CSE Winter

40 Key or Keys? Can we have more than one key? CSE Winter

41 Key or Keys? Can we have more than one key? A à B B à C C à A what are the keys here? CSE Winter

42 Key or Keys? Can we have more than one key? A à B B à C C à A ABàC BCàA what are the keys here? AàBC BàAC CSE Winter

43 Eliminating Anomalies Main idea: X à A is OK if X is a (super)key X à A is not OK otherwise Need to decompose the table, but how? CSE Winter

44 Boyce-Codd Normal Form There are no bad FDs: Definition. A relation R is in BCNF if: Whenever Xà B is a non-trivial dependency, then X is a superkey. Equivalently: Definition. A relation R is in BCNF if: X, either X + = X or X + = [all attributes] CSE Winter

45 BCNF Decomposition Algorithm Normalize(R) find X s.t.: X X + and X + [all attributes] if (not found) then R is in BCNF let Y = X + - X; Z = [all attributes] - X + decompose R into R1(X Y) and R2(X Z) Normalize(R1); Normalize(R2); Y X Z X + CSE Winter

46 Example Name SSN PhoneNumber City Fred Seattle Fred Seattle Joe Westfield Joe Westfield SSN à Name, City The only key is: {SSN, PhoneNumber} Hence SSN à Name, City is a bad dependency Name, City SSN + SSN Phone- Number In other words: SSN+ = SSN, Name, City and is neither SSN nor All Attributes

47 Find X s.t.: X X + and X + [all attributes] Example BCNF Decomposition Person(name, SSN, age, haircolor, phonenumber) SSN à name, age age à haircolor CSE Winter

48 Find X s.t.: X X + and X + [all attributes] Example BCNF Decomposition Person(name, SSN, age, haircolor, phonenumber) SSN à name, age age à haircolor Iteration 1: Person: SSN+ = SSN, name, age, haircolor Decompose into: P(SSN, name, age, haircolor) Phone(SSN, phonenumber) name, age, haircolor SSN phonenumber CSE Winter

49 Find X s.t.: X X + and X + [all attributes] Example BCNF Decomposition Person(name, SSN, age, haircolor, phonenumber) SSN à name, age What are age à haircolor the keys? Iteration 1: Person: SSN+ = SSN, name, age, haircolor Decompose into: P(SSN, name, age, haircolor) Phone(SSN, phonenumber) Iteration 2: P: age+ = age, haircolor Decompose: People(SSN, name, age) Hair(age, haircolor) Phone(SSN, phonenumber) CSE Winter

50 Find X s.t.: X X + and X + [all attributes] Example BCNF Decomposition Person(name, SSN, age, haircolor, phonenumber) SSN à name, age Note the keys! age à haircolor Iteration 1: Person: SSN+ = SSN, name, age, haircolor Decompose into: P(SSN, name, age, haircolor) Phone(SSN, phonenumber) Iteration 2: P: age+ = age, haircolor Decompose: People(SSN, name, age) Hair(age, haircolor) Phone(SSN, phonenumber) CSE Winter

51 R(A,B,C,D) Example: BCNF A à B B à C R(A,B,C,D) CSE Winter

52 R(A,B,C,D) Example: BCNF A à B B à C Recall: find X s.t. X X + [all-attrs] R(A,B,C,D) CSE Winter

53 R(A,B,C,D) Example: BCNF A à B B à C R(A,B,C,D) A + = ABC ABCD CSE Winter

54 R(A,B,C,D) Example: BCNF A à B B à C R(A,B,C,D) A + = ABC ABCD R 1 (A,B,C) R 2 (A,D) CSE Winter

55 R(A,B,C,D) Example: BCNF A à B B à C R(A,B,C,D) A + = ABC ABCD R 1 (A,B,C) B + = BC ABC R 2 (A,D) CSE Winter

56 R(A,B,C,D) Example: BCNF A à B B à C R(A,B,C,D) A + = ABC ABCD R 11 (B,C) R 1 (A,B,C) B + = BC ABC R 12 (A,B) R 2 (A,D) What are the keys? What happens if in R we first pick B +? Or AB +? 56

57 Decompositions in General R(A 1,..., A n, B 1,..., B m, C 1,..., C p ) S 1 (A 1,..., A n, B 1,..., B m ) S 2 (A 1,..., A n, C 1,..., C p ) S 1 = projection of R on A 1,..., A n, B 1,..., B m S 2 = projection of R on A 1,..., A n, C 1,..., C p CSE Winter

58 Lossless Decomposition Name Price Category Gizmo Gadget OneClick Camera Gizmo Camera Name Price Name Category Gizmo OneClick Gizmo Gizmo OneClick Gizmo Gadget Camera Camera CSE Winter

59 Lossy Decomposition What is lossy here? Name Price Category Gizmo Gadget OneClick Camera Gizmo Camera Name Gizmo OneClick Gizmo Category Gadget Camera Camera Price Category Gadget Camera Camera CSE Winter

60 Lossy Decomposition Name Price Category Gizmo Gadget OneClick Camera Gizmo Camera Name Gizmo OneClick Gizmo Category Gadget Camera Camera Price Category Gadget Camera Camera CSE Winter

61 Decomposition in General R(A 1,..., A n, B 1,..., B m, C 1,..., C p ) S 1 (A 1,..., A n, B 1,..., B m ) S 2 (A 1,..., A n, C 1,..., C p ) Let: S 1 = projection of R on A 1,..., A n, B 1,..., B m S 2 = projection of R on A 1,..., A n, C 1,..., C p The decomposition is called lossless if R = S 1 S 2 Fact: If A 1,..., A n à B 1,..., B m then the decomposition is lossless It follows that every BCNF decomposition is lossless

62 Testing for Lossless Join If we decompose R into Π S1 (R), Π S2 (R), Π S3 (R), Is it true that S1 S2 S3 = R? That is true if we can show that: R S1 S2 S3 always holds (why?) R S1 S2 S3 neet to check

63 Example from textbook Ch The Chase Test for Lossless Join R(A,B,C,D) = S1(A,D) S2(A,C) S3(B,C,D) R satisfies: AàB, BàC, CDàA S1 = Π AD (R), S2 = Π AC (R), S3 = Π BCD (R), hence R S1 S2 S3 Need to check: R S1 S2 S3

64 Example from textbook Ch The Chase Test for Lossless Join R(A,B,C,D) = S1(A,D) S2(A,C) S3(B,C,D) R satisfies: AàB, BàC, CDàA S1 = Π AD (R), S2 = Π AC (R), S3 = Π BCD (R), hence R S1 S2 S3 Need to check: R S1 S2 S3 Suppose (a,b,c,d) S1 S2 S3 Is it also in R?

65 Example from textbook Ch The Chase Test for Lossless Join R(A,B,C,D) = S1(A,D) S2(A,C) S3(B,C,D) R satisfies: AàB, BàC, CDàA S1 = Π AD (R), S2 = Π AC (R), S3 = Π BCD (R), hence R S1 S2 S3 Need to check: R S1 S2 S3 Suppose (a,b,c,d) S1 S2 S3 Is it also in R? R must contain the following tuples: A B C D Why? a b1 c1 d (a,d) S1 = Π AD (R)

66 Example from textbook Ch The Chase Test for Lossless Join R(A,B,C,D) = S1(A,D) S2(A,C) S3(B,C,D) R satisfies: AàB, BàC, CDàA S1 = Π AD (R), S2 = Π AC (R), S3 = Π BCD (R), hence R S1 S2 S3 Need to check: R S1 S2 S3 Suppose (a,b,c,d) S1 S2 S3 Is it also in R? R must contain the following tuples: A B C D Why? a b1 c1 d (a,d) S1 = Π AD (R) a b2 c d2 (a,c) S2 = Π BD (R)

67 Example from textbook Ch The Chase Test for Lossless Join R(A,B,C,D) = S1(A,D) S2(A,C) S3(B,C,D) R satisfies: AàB, BàC, CDàA S1 = Π AD (R), S2 = Π AC (R), S3 = Π BCD (R), hence R S1 S2 S3 Need to check: R S1 S2 S3 Suppose (a,b,c,d) S1 S2 S3 Is it also in R? R must contain the following tuples: A B C D Why? a b1 c1 d (a,d) S1 = Π AD (R) a b2 c d2 (a,c) S2 = Π BD (R) a3 b c d (b,c,d) S3 = Π BCD (R)

68 Example from textbook Ch The Chase Test for Lossless Join R(A,B,C,D) = S1(A,D) S2(A,C) S3(B,C,D) R satisfies: AàB, BàC, CDàA S1 = Π AD (R), S2 = Π AC (R), S3 = Π BCD (R), hence R S1 S2 S3 Need to check: R S1 S2 S3 Suppose (a,b,c,d) S1 S2 S3 Is it also in R? R must contain the following tuples: Chase them (apply FDs): AàB A B C D a b1 c1 d a b1 c d2 a3 b c d A B C D Why? a b1 c1 d (a,d) S1 = Π AD (R) a b2 c d2 (a,c) S2 = Π BD (R) a3 b c d (b,c,d) S3 = Π BCD (R)

69 Example from textbook Ch The Chase Test for Lossless Join R(A,B,C,D) = S1(A,D) S2(A,C) S3(B,C,D) R satisfies: AàB, BàC, CDàA S1 = Π AD (R), S2 = Π AC (R), S3 = Π BCD (R), hence R S1 S2 S3 Need to check: R S1 S2 S3 Suppose (a,b,c,d) S1 S2 S3 Is it also in R? R must contain the following tuples: Chase them (apply FDs): AàB BàC A B C D Why? a b1 c1 d (a,d) S1 = Π AD (R) a b2 c d2 (a,c) S2 = Π BD (R) a3 b c d (b,c,d) S3 = Π BCD (R) A B C D a b1 c1 d a b1 c d2 a3 b c d A B C D a b1 c d a b1 c d2 a3 b c d

70 Example from textbook Ch The Chase Test for Lossless Join R(A,B,C,D) = S1(A,D) S2(A,C) S3(B,C,D) R satisfies: AàB, BàC, CDàA S1 = Π AD (R), S2 = Π AC (R), S3 = Π BCD (R), hence R S1 S2 S3 Need to check: R S1 S2 S3 Suppose (a,b,c,d) S1 S2 S3 Is it also in R? R must contain the following tuples: Chase them (apply FDs): A B C D Why? a b1 c1 d (a,d) S1 = Π AD (R) a b2 c d2 (a,c) S2 = Π BD (R) a3 b c d (b,c,d) S3 = Π BCD (R) AàB BàC CDàA A B C D a b1 c1 d a b1 c d2 A B C D a b1 c d a b1 c d2 A B C D a b1 c d a b1 c d2 Hence R contains (a,b,c,d) a3 b c d a3 b c d a b c d

71 Schema Refinements = Normal Forms 1st Normal Form = all tables are flat 2nd Normal Form = obsolete Boyce Codd Normal Form = no bad FDs 3rd Normal Form = see book BCNF is lossless but can cause loss of ability to check some FDs 3NF fixes that (is lossless and dependency-preserving), but some tables might not be in BCNF i.e., they may have redundancy anomalies CSE Winter