1. Chapter 7 Normalization Chapter 14 & 15 in Textbook

2. 2 Database Design Steps in building a database for an application: Real-world domain Conceptual model DBMS data model Create Schema (DDL) Modify data (DML) Normalization

3. 3 How to produce a good relation schema? 1. Start with a set of relation. 2. Define the functional dependencies for the relation to specify the PK. 3. Transform relations to normal form. Normalization

4. 4 Data Redundancy SL21 SG37 SG14 SA9 SG5 StaffNo John Ann David Mary Susan FName White Beech Ford Howe Brand LNameposition Manager Assistant Supervisor Assistant Manager Salary 30000 12000 18000 9000 24000 BrnNo B005 B003 B007 B003 City London Glasgow Aberdeen Glasgow SL41 JulieLeeAssistant 9000B005 London Address 22 Deer Rd 163 Main St 16 Arglly St 22 Deer Rd 163 Main St Relations that have redundant data may have update anomalies (insert, modify, delete) STAFFBRANCH B003 Glasgow 163 Main St B003 Glasgow 163 Main St B003 Glasgow 163 Main St Normalization

5. 5 Data Redundancy SL21 SG37 SG14 SA9 SG5 StaffNo John Ann David Mary Susan FName White Beech Ford Howe Brand LNameposition Manager Assistant Supervisor Assistant Manager Salary 30000 12000 18000 9000 24000 SL41 JulieLeeAssistant 9000 BrnNo B005 B003 B007 City London Glasgow Aberdeen Address 22 Deer Rd 163 Main St 16 Arglly St STAFF BRANCH BrnNo B005 B003 B007 Normalization

6. 6 Relation Decomposition Normalization process involve decomposing a relation. Decomposition require to be reversible. Functional dependencies guarantee decomposition to be reversible. While normalization, two important properties associated with decomposition: 1. Lossless-join 2. Dependency preservation Normalization

7. 7 Data Redundancy SL21 SG37 SG14 SA9 SG5 StaffNo John Ann David Mary Susan FName White Beech Ford Howe Brand LNameposition Manager Assistant Supervisor Assistant Manager Salary 30000 12000 18000 9000 24000 SL41 JulieLeeAssistant 9000 BrnNo B005 B003 B007 City London Glasgow London Address 22 Deer Rd 163 Main St 16 Arglly St STAFF BRANCH City London Glasgow London Glasgow Normalization

8. 8 Data Redundancy SL21 SG37 SG14 SA9 SG5 StaffNo John Ann David Mary Susan FName White Beech Ford Howe Brand LNameposition Manager Assistant Supervisor Assistant Manager Salary 30000 12000 18000 9000 24000 BrnNo B005 City London SL41 JulieLeeAssistant 9000B005 London Address 22 Deer Rd STAFFBRANCH B003 Glasgow 163 Main St B003 Glasgow 163 Main St B003 Glasgow 163 Main St SL21 JohnWhiteManager 30000 London B007 16 Arglly St SA9 Mary HoweAssistant 9000B007 London 16 Arglly St SL41 JulieLeeAssistant 9000B007 London 16 Arglly St Normalization

9. 9 Functional Dependencies Describes the relationship between attributes in a relation. If A and B are attributes of relation R, B is functionally dependent on A, denoted by A B, if each value of A is associated with exactly one value of B. B may have several values of A. Determinant Dependent Functional dependency is identifies between attributes in a relation at different times (all time functional dependency). AB B is functionally dependent on A Normalization

10. 10 A B t u If t & u agree hereThen they must agree here Functional Dependencies A B whenever two tuples t & u agree on all attributes of A, then they must agree on attribute B. Normalization

11. 11 Functional Dependencies Example StaffNoposition Position is functionally dependent on Staffno positionStaffNo StaffNo is NOT functionally dependent on position SL21 Manager Manager SL21 SG5 1:1 or M:1 relationship between attributes in a relation 1:M relationship between attributes in a relation Normalization

12. 12 Trivial Functional Dependencies A B is trivial if B  A StaffNo, Sname SName StaffNo, SName StaffNo We are not interested in trivial functional dependencies as it provides no genuine integrity constraints on the value held by these attributes. Normalization

13. 13 StaffBranch Example Functional dependencies on StaffBranch relation: StaffNo FName, Lname, position, salary, brnNo, Address, city BranchNo Address, city Address, city BranchNo BranchNo, position salary Address, city, position salary Determinants: StaffNo, BranchNo, (Address, city), (branchNo, position), and (address, city, position) Normalization

14. 14 Identifying the PK Purpose of functional dependency, specify the set of integrity constraints that must hold on a relation. The determinant attribute(s) are candidate of the relation, if: 1:1 relationship between determinant & dependent. No subset of determinant attribute(s) is a determinant. (nontrivial) If (A, B) C, then NOT A B, and NOT B A All attributes that are not part of the CK should be functionally dependent on the key: CK all attributes of R Hold for all time. PK is the candidate attribute(s) with the minimal set of functional dependency. Normalization

15. 15 Closure Closure (inferred from) X + : The set of functional dependencies that are implied by a given set of functional dependencies X. A B t u If t & u agree hereThen they must agree here C So surely they will agree here C  B X A B X + A C Normalization

16. 16 Closure Example SBranchNo (Address, city) S + BranchNo Address BranchNo city Implied by Normalization

17. 17 Inference Rules for Functional Dependencies Armstrong’s aximos (inference rules): The set of inference rules specifies how functional dependencies can be inferred from given one. Inference rules: ReflexivityIf B  A, then A B Augmentation If A B, then A,C B,C Transitivity If A B and B C, then A C Self-DeterminationA A DecompositionIf A B,C, then A B and A C UnionIf A B and A C, then A B,C Normalization

18. 18 Minimal Sets of Functional Dependencies Complete set of functional dependencies for a relation can be very large. We need to reduce the set to a manageable size, by applying the inference rules repeatedly until they stop producing new FDs. Assume S1 & S2 are set of dependencies: S1  S2, then (S2 is a cover for S1) OR (S1 is covered by S2) if S2 is a cover for S1 & S1 is a cover for S2 S1 equivalent to S2 Normalization

19. 19 Minimal Sets of Functional Dependencies A set of functional dependencies X is minimal if it satisfies the following: 1.Every dependency in X has a single attribute for its right-hand side. 2.Can’t replace any dependency A B in X with C B, where C  A, & still have a set of dependencies equivalent to X. 3.Can’t remove any dependency from X and still have a set of dependencies that is equivalent to X. Normalization

20. 20 Minimal Sets of Functional Dependencies 1. For each X {A1, A2,.. An}, create X A1, X A2, …., X An. 2. A, B C is equivalent to B C, then replace A, B C with B C. 3. X - {A B} equivalent to X, then remove A B. Normalization

21. Question Find the minimal set of the following FDs: Fd1: B  A Fd2: D  A Fd3: A,B  D 21Normalization

22. Question Find FDs of the relation shown below that lists dentist/patient appointment data; known that: A patient is given an appointment at a specific time and date with a dentist located at a particular surgery. On each day of patient appointments, a dentist is allocated to a specific surgery for that day. Dentist-patient (staffNo, dentistName, aDate, aTime, patNo, patName, surgeryNo) 22Normalization

23. 23 The Purpose of Normalization Normalization is a bottom-up approach to database design that begins by examining the relationships between attributes. It is performed as a series of tests on a relation to determine whether it satisfies or violates the requirements of a given normal form. Purpose: Guarantees no redundancy due to FDs Guarantees no update anomalies Normal Forms: First Normal Form (1NF) Second Normal Form (2NF) Third Normal Form (3NF) Boyce-Codd Normal Form (BCNF) Fourth Normal Form (4NF) Fifth Normal Form (5NF)

24. 24 The Process of Normalization Normalization is a technique for analyzing relations based on their CK & FD. 5NF 4NF BCNF 3NF 2NF 1NF Higher Normal Form Stronger in format Less vulnerable to update anomalies Normalization

25. 25 First Normal Form (1NF) Unnormalized form (UNF): A relation that contains one or more repeating groups. First normal form (1NF): A relation in which the intersection of each row and column contains one & only one value. Unnormalized relation ClientNo CR76 PropertyNo PG4 Name John Key CLIENT_PROPERTY PG16 PG4 PG36 PG16 CR56 Aline Stewart Normalization

26. 26 UNF 1NF Approach 1 Expand the key so that there will be a separate tuple in the original relation for each repeated attribute(s). Primary key becomes the combination of primary key and redundant value. 1NF relation Disadvantage: introduce redundancy in the relation. ClientNo CR76 PropertyNo PG4 Name John Key CLIENT_PROPERTY PG16 PG4 PG36 PG16 CR56 Aline Stewart CR76 John Key CR56 Aline Stewart CR56 Aline Stewart Normalization

27. 27 If the maximum number of values is known for the attribute, replace repeated attribute (PropertyNo) with a number of atomic attributes (PropertyNo1, PropertyNo2, PropertyNo3). 1NF relation Disadvantage: introduce NULL values in the relation. UNF 1NF Approach 2 ClientNo CR76 PropertyNo1 PG4 Name John Key CLIENT_PROPERTY PG16 PG4 PG36 CR56 Aline Stewart PropertyNo2PropertyNo3 NULL PG16 Normalization

28. 28 UNF 1NF Approach 3 Remove the attribute that violates the 1NF and place it in a separate relation along with a copy of the primary key. ClientNo CR76 Name John Key CLIENT CR56 Aline Stewart ClientNo CR76 PropertyNo PG4 PROPERTY PG16 PG4 PG36 PG16 CR56 CR76 CR56 1NF relation Normalization

29. 29 Full Functional Dependency If A and B are attributes of a relation. B is fully functionally dependent on A if B is functionally dependent on A, but not on any proper subset of A. B is partial functional dependent on A if some attributes can be removed from A & the dependency still holds. StaffNo, Sname BranchNo Partial dependency ClientNo, PropertyNo RentDate Full dependency Normalization

30. 30 Second Normal Form (2NF) Second normal form (2NF): A 1NF relation in which every attribute is fully nontrivial functionally dependent on the PK. (non-prime attributes fully dependent on PK.) Applies to relations with composite primary keys & partial dependencies. 1NF relation ClientNo cName PropertyNo CLIENT_RENTAL pAddressRentStartRentFinishRentOwnerNoOName Normalization

31. 31 1NF 2NF 1. Start with 1NF relation. 2. Find the FDs of a relation. 3. Test the FDs whose determinant attribute is part of the PK. Normalization

32. ClientNo cName PropertyNo CLIENT_RENTAL pAddressRentStartRentFinishRentOwnerNoOName (ClientNo, PropertyNo) PK ClientNo, PropertyNo RentStart, RentFinish Full Dependency ClientNo CNamePartial Dependency PropertyNo Paddress, Rent, OwnerNo, Oname Partial Dependency OwnerNo OName ClientNo, RentStart PropertyNo, pAddress, RentFinish, Rent, OwnerNo, Oname PropertyNo, RentStart ClientNo, cName, RentFinish 1NF 2NF 32Normalization

33. 33 1NF 2NF 4. Remove partial dependencies by placing the functionally dependent attributes in a new relation along with a copy of their determinants. 2NF relation 2NF relation 2NF relation ClientNo cName CLIENT ClientNoPropertyNoRentStartRentFinish RENTAL PropertyNo PROPERTY_OWNER pAddressRentOwnerNoOName Normalization

34. 34 Transitive Dependency A, B, C are attributes of a relation, such that: If A B and B C, then C is transitively dependent on A via B. Provided A is NOT functionally dependent on B or C (nontrivial FD). Example: StaffNo BranchNo, BranchNo Address StaffNo Address Normalization

35. 35 Third Normal Form (3NF) Third normal form (3NF): A 2NF relation in which NO non-prime attribute is transitively dependent on the PK. 3NF relation 3NF relation 2NF relation ClientNo cName CLIENT ClientNoPropertyNoRentStartRentFinish RENTAL PropertyNo PROPERTY_OWNER pAddressRentOwnerNoOName Normalization

36. 36 2NF 3NF 1. Identify the PK in the 2NF relation. 2. Identify FDs in this relation. 3. If transitive dependencies exist, place transitively dependent attributes in a new relation along with a copy of their determinants. 3NF relation 3NF relation OwnerNo OName OWNER PropertyNopAddressrentOwnerNo PROPERTY_FOR_RENT Normalization

37. 37 Review of Decompositions CLIENT_RENTAL CLIENTRENTAL OWNER PROPERTY_FOR_RENT PROPERTY_OWNER 1NF 2NF 3NF RENTALCLIENT Normalization

38. 38 General Definition of 2NF & 3NF Second normal form (2NF): A 1NF relation in which every non-primary-key attribute is fully functionally dependent on the CK. Third normal form (3NF): A 2NF relation in which NO non-primary-key attribute in a nontrivial FD is transitively dependent on the CK. Normalization

39. 39 Boyce-Codd Normal Form (BCNF) Boyce-Codd normal form (BCNF): A 3NF relation in which every determinant in a nontrivial FD is a CK. Difference between 3NF & BCNF: A B 3NF allows A NOT CK. BCNF insists on A is a CK. Potential to violate BCNF may occur in a relation that: Contain two (or more) composite CKs. CKs overlap. (at least one attribute in common). Normalization

40. 40 Boyce-Codd Normal Form (BCNF) A BCD 3NF but not BCNF Normalization

41. 41 ClientNo CLIENT_INTERVIEW Int_DateInt_TimeStaffNo RoomNo 3NF BCNF ClientNo, Int_Date Int_Time, StaffNo, RoomNo StaffNo, Int_Date, Int_Time ClientNo RoomNo, Int_Date, Int_Time StaffNo, ClientNo StaffNo, Int_Date RoomNo 1. Examine FDs for a relation. 2. If determinant is NOT a CK, decompose relation into 2 relations. Normalization

42. 42 3NF BCNF 3. Remove non-CK dependencies by placing the functionally dependent attributes in a new relation along with a copy of their determinants. BCNF relation BCNF relation Int_Date RoomNo STAFF_ROOM ClientNoInt_dateInt_timeStaffNo INTERVIEW StaffNo Normalization

43. 43 Review Example PG4 PG16 Pno pAddress 18-Oct-00 22-Apr-01 1-Oct-01 22-Apr-01 24-Oct-01 iDateiTime 10:00 09:00 12:00 13:00 14:00 comments Replace crockery Good order Damp rot Replace carpet Good condition StaffNo SG37 SG14 SG37 CarReg M23JGR M53HDR N72HFR M53HDR N72HFR Lawrence St, Glasgow 5 Novar Dr., Glasgow sName Ann David Ann STAFF_PROPERTY_INSPECTION Unnormalized relation Normalization

44. 44 UNF 1NF PG4 PG16 Pno pAddress 18-Oct-00 22-Apr-01 1-Oct-01 22-Apr-01 24-Oct-01 iDateiTime 10:00 09:00 12:00 13:00 14:00 comments Replace crockery Good order Damp rot Replace carpet Good condition StaffNo SG37 SG14 SG37 CarReg M23JGR M53HDR N72HFR M53HDR N72HFR Lawrence St, Glasgow 5 Novar Dr., Glasgow sName Ann David Ann STAFF_PROPERTY_INSPECTION 1NF Normalization

45. 45 1NF 2NF Pno pAddressiDateiTime commentsStaffNo CarReg sName STAFF_PROPERTY_INSPECTION Pno, iDate iTime, comments, StaffNo, sName, carReg Pno pAddressPartial Dependency StaffNo Sname iDate, StaffNo CarReg iDate, iTime, CarReg Pno, pAddress, comments, StaffNo, Sname iDate, iTime, StaffNo Pno, pAddress, Comments

46. 46 1NF 2NF Pno iDateiTime commentsStaffNo CarReg sName PROPERTY_INSPECTION Pno, iDate iTime, comments, StaffNo, Sname, CarReg StaffNo Sname Transitive Dependency iDate, StaffNo CarReg iDate, iTime, CarReg Pno, comments, StaffNo, Sname iDate, iTime, StaffNo Pno, comments 2NF Normalization Pno pAddress PROPERTY 2NF Pno pAddress

47. 47 2NF 3NF Pno iDateiTime commentsStaffNo CarReg PROPERTY_INSPECTION PROPERTY(Pno, pAddres) STAFF(StaffNo, sName) PROPERTY_INSPECT(Pno, iDate, iTime, comments, staffNo, CarReg) 3NF Normalization Pno pAddress PROPERTY 3NF StaffNo sName STAFF 3NF

48. 48 3NF BCNF Pno iDateiTime commentsStaffNo CarReg PROPERTY_INSPECTION Pno, iDate iTime, comments, staffNo, CarReg StaffNo, iDate carReg CarReg, iDate, iTime pno, comments, staffNo StaffNo, iDate, iTime pno, comments PROPERTY(Pno, pAddres) STAFF(StaffNo, sName) STAFF_CAR(StaffNo, iDate, CarReg) PROPERTY_INSPECT(pno, iDate, iTime, comments, StaffNo) 3NF Normalization

49. 49 Multi-Valued Dependency (MVD) Represents a dependency between attributes A, B, C in a relation, such that for each value of A, there is a set of values for B and a set of values for C. However, the set of values for B & C are independent of each others. Denoted by: A B, A C Example: BranchNo SName, BranchNo OName SNameOName BRANCH_STAFF_OWNER BranchNo B003 Ann David Ann David Carol Tina Normalization

50. 50 Trivial MVD A B trivial MVD if: B  A OR A  B = R Normalization

51. 51 Fourth Normal Form (4NF) Fourth normal form (4NF): A BCNF relation with NO nontrivial MVD. BCNF relation SNameOName BRANCH_STAFF_OWNER BranchNo B003 Ann David Ann David Carol Tina Normalization

52. 52 BCNF 4NF 1.Start with a BCNF relation. 2.Examine FDs for a relation. 3.If nontrivial MVD exists, remove the MVD by placing the attributes in a new relation along with a copy of their determinant. 4NF 4NF SName BRANCH_STAFF BranchNo B003 Ann David OName BRANCH_OWNER BranchNo B003 Carol Tina Normalization

53. 53 Lossless-Join Dependency A property of decompostion, which ensures that no spurious tuples are generated when relations are reunited through a natural join operation. Objectives: Preserve all the data in the original relation Does not result in the creation of additional spurious tuples Normalization

54. 54 Join Dependency A, B,.., Z attributes in relation R satisfies join dependency if Every legal value of R is equal to the join of its projections on A, B,.., Z Normalization

55. 55 Fifth Normal Form (5NF) Fifth normal form (5NF): A 4NF relation with NO join dependency. I_DescriptionSupplierNo PROPERTY_ITEM_SUPPLIER PropertyNo PG4 PG16 Bed Chair Bed S1 S2 Normalization Illegal State

56. 56 4NF 5NF I_Description PROPERTY_ITEM PropertyNo PG4 PG16 Bed Chair Bed SupplierNo ITEM_SUPPLIER I_Description Bed Chair Bed S1 S2 SupplierNo PROPERTY_ITEM PropertyNo PG4 PG16 S1 S2 I_Description SupplierNo PROPERTY_ITEM_SUPPLIER PropertyNo PG4 PG16 PG4 Bed Chair Bed S1 S2 Legal State

57. Given the following Dentist-patient database schema: Dentist-patient (staffNo, dentistName, aDate, aTime, patNo, patName, surgeryNo) Normalize the above relation, showing appropriate dependency diagrams to justify decomposition.Question 57Normalization