1. Database Design Theory CS405G: Introduction to Database Systems Jinze Liu 3/15/20161

2. Normalization We discuss four normal forms: first, second, third, and Boyce-Codd normal forms 1NF, 2NF, 3NF, and BCNF Normalization is a process that “improves” a database design by generating relations that are of higher normal forms. The objective of normalization: “to create relations where every dependency is on the key, the whole key, and nothing but the key”. Normalization 3/15/20162

3. Normalization 3/15/20163 There is a sequence to normal forms: 1NF is considered the weakest, 2NF is stronger than 1NF, 3NF is stronger than 2NF, and BCNF is considered the strongest Also, any relation that is in BCNF, is in 3NF; any relation in 3NF is in 2NF; and any relation in 2NF is in 1NF.

4. Normalization 3/15/20164 BCNF 3NF 2NF 1NF a relation in BCNF, is also in 3NF a relation in 3NF is also in 2NF a relation in 2NF is also in 1NF

5. Normalization 3/15/20165 The benefit of higher normal forms is that update semantics for the affected data are simplified. This means that applications required to maintain the database are simpler. A design that has a lower normal form than another design has more redundancy. Uncontrolled redundancy can lead to data integrity problems. First we introduce the concept of functional dependency

6. Review Functional dependencies – X -> Y: X “determines” Y If two rows agree on X, they must agree on Y  A generalization of the key concepts 3/15/20166 XYA xy1a1 x?a2 y1

7. Functional Dependencies 3/15/20167 If EmpNum is the PK then the FDs: EmpNum  EmpEmail EmpNum  EmpFname EmpNum  EmpLname must exist.

8. Functional Dependencies 3/15/20168 EmpNum  EmpEmail EmpNum  EmpFname EmpNum  EmpLname EmpNum EmpEmail EmpNum EmpEmail EmpFname EmpLname 3 different ways you might see FDs depicted

9. Determinant 3/15/20169 Functional Dependency EmpNum  EmpEmail Attribute on the LHS is known as the determinant EmpNum is a determinant of EmpEmail

10. EmployeeProject ssnpnumberhoursenameplocation Essn pno hours WORKS ON What functional dependencies? 3/15/201610

11. Transitive dependency 3/15/201611 Transitive dependency Consider attributes A, B, and C, and where A  B and B  C. Functional dependencies are transitive, which means that we also have the functional dependency A  C We say that C is transitively dependent on A through B.

12. Transitive dependency 3/15/201612 EmpNum EmpEmail DeptNum DeptNname DeptName is transitively dependent on EmpNum via DeptNum EmpNum  DeptName EmpNum  DeptNum DeptNum  DeptName

13. Partial dependency 3/15/201613 A partial dependency exists when an attribute B is functionally dependent on an attribute A, and A is a component of a multipart candidate key. InvNumLineNumQtyInvDate Candidate keys: {InvNum, LineNum} InvDate is partially dependent on {InvNum, LineNum} as InvNum is a determinant of InvDate and InvNum is part of a candidate key

14. First Normal Form 3/15/201614 First Normal Form We say a relation is in 1NF if all values stored in the relation are single-valued and atomic. 1NF places restrictions on the structure of relations. Values must be simple.

15. First Normal Form 3/15/201615 The following is not in 1NF EmpNumEmpPhoneEmpDegrees 123233-9876 333233-1231BA, BSc, PhD 679233-1231BSc, MSc EmpDegrees is a multi-valued field: employee 679 has two degrees: BSc and MSc employee 333 has three degrees: BA, BSc, PhD

16. First Normal Form 3/15/201616 To obtain 1NF relations we must, without loss of information, replace the above with two relations - see next slide What would the ERD be for the above situation with EmpNum, EmpPhone, EmpDegrees. Would we have generated the above table using our “mapping algorithm”?

17. First Normal Form 3/15/201617 EmpNumEmpDegree 333BA 333BSc 333PhD 679BSc MSc679 EmpNumEmpPhone 123233-9876 333233-1231 679233-1231 An outer join between Employee and EmployeeDegree will produce the information we saw before Employee EmployeeDegree

18. Boyce-Codd Normal Form 3/15/201618 LineNumProdNumQtyInvNum InvNum, LineNumProdNum InvNum, ProdNumLineNum There are two candidate keys. Since every determinant is a candidate key, the relation is in BCNF This relation is about Invoice lines only. Qty {InvNum, LineNum} and {InvNum, ProdNum} are the two candidate keys.

19. Second Normal Form 3/15/201619 Second Normal Form A relation is in 2NF if it is in 1NF, and every non-key attribute is fully dependent on each candidate key. (That is, we don’t have any partial functional dependency.) 2NF (and 3NF) both involve the concepts of key and non-key attributes. A key attribute is any attribute that is part of a key; any attribute that is not a key attribute, is a non-key attribute. Relations that are not in BCNF have data redundancies A relation in 2NF will not have any partial dependencies

20. Second Normal Form 3/15/201620 LineNumProdNumQtyInvNum InvNum, LineNumProdNum InvNum, ProdNumLineNum Since there is a determinant that is not a candidate key, InvLine is not BCNF InvLine is not 2NF since there is a partial dependency of InvDate on InvNum Qty InvDate InvNum There are two candidate keys. Qty is the only non- key attribute, and it is dependent on InvNum InvLine is only in 1NF Consider this InvLine table (in 1NF):

21. Second Normal Form 3/15/201621 LineNumProdNumQtyInvNumInvDate InvLine The above relation has redundancies: the invoice date is repeated on each invoice line. We can improve the database by decomposing the relation into two relations: LineNumProdNumQtyInvNum InvDateInvNum Question: What is the highest normal form for these relations? 2NF? 3NF? BCNF?

22. 2NF, but not in 3NF, nor in BCNF: inv_noline_noprod_noprod_descqty since prod_no is not a candidate key and we have: prod_no  prod_desc. 3/15/201622

23. 2NF, but not in 3NF, nor in BCNF: since dnumber is not a candidate key and we have: dnumber  dname. EmployeeDept enamessnbdateaddressdnumberdname 3/15/201623

24. Third Normal Form 3/15/201624 Third Normal Form a relation is in 3NF if the relation is in 1NF and all determinants of non-key attributes are candidate keys That is, for any functional dependency: X  Y, where Y is a non-key attribute (or a set of non-key attributes), X is a candidate key. this definition of 3NF differs from BCNF only in the specification of non-key attributes - 3NF is weaker than BCNF. (BCNF requires all determinants to be candidate keys.) A relation in 3NF will not have any transitive dependencies

25. Boyce-Codd Normal Form 3/15/201625 Boyce-Codd Normal Form BCNF is defined very simply: a relation is in BCNF if it is in 1NF and if every determinant is a candidate key. If our database will be used for OLTP (on line transaction processing), then BCNF is our target. Usually, we meet this objective. However, we might denormalize (3NF, 2NF, or 1NF) for performance reasons.

26. Third Normal Form 3/15/201626 EmpNumEmpNameDeptNumDeptName EmpName, DeptNum, and DeptName are non-key attributes. DeptNum determines DeptName, a non-key attribute, and DeptNum is not a candidate key. Consider this Employee relation Is the relation in 3NF? … no Is the relation in 2NF? … yes Is the relation in BCNF? … no Candidate keys are? …

27. Third Normal Form 3/15/201627 EmpNumEmpNameDeptNumDeptName We correct the situation by decomposing the original relation into two 3NF relations. Note the decomposition is lossless. EmpNumEmpNameDeptNumDeptNameDeptNum Verify these two relations are in 3NF. Are they in BCNF?

28. student_nocourse_noinstr_no Instructor teaches one course only. Student takes a course and has one instructor. In 3NF, but not in BCNF: {student_no, course_no}  instr_no instr_no  course_no since we have instr_no  course-no, but instr_no is not a Candidate key. 3/15/201628

29. course_noinstr_no student_noinstr_no student_nocourse_noinstr_no BCNF {student_no, instr_no}  student_no {student_no, instr_no}  instr_no instr_no  course_no 3/15/201629

30. Another example: 3NF Address (street_address, city, state, zip) – street_address, city, state -> zip – zip -> city, state Keys – {street_address, city, state} – {street_address, zip} BCNF? – Violation: zip -> city, state 3/15/201630

31. To decompose or not to decompose Address 1 (zip, city, state) Address 2 (street_address, zip) FD’s in Address 1 – zip -> city, state FD’s in Address 2 – None! Hey, where is street_address, city, state -> zip? – Cannot check without joining Address 1 and Address 2 back together Dilemma: Should we get rid of redundancy at the expense of making constraints harder to enforce? 3/15/201631

32. BCNF = no redundancy? Student (SID, CID, club) – Suppose your classes have nothing to do with the clubs you join – FD’s? None – BCNF? Yes – Redundancies? Tons! 3/15/201632 SIDCIDclub 142CPS116ballet 142CPS116sumo 142CPS114ballet 142CPS114sumo 123CPS116chess 123CPS116golf...

33. Multivalued dependencies A multivalued dependency (MVD) has the form X ->>Y, where X and Y are sets of attributes in a relation R X ->>Y means that whenever two rows in R agree on all the attributes of X, then we can swap their Y components and get two new rows that are also in R 3/15/201633 Must be in R too XYZ ab1c1 ab2c2 ab1c2 ab2c1

34. 4NF decomposition example 3/15/201634 Student (SID, CID, club) 4NF violation: SID ->>CID Enroll (SID, CID)Join (SID, club) 4NF SIDCIDclub 142 CPS11 6ballet 142 CPS11 6sumo 142 CPS11 4ballet 142 CPS11 4sumo 123 CPS11 6chess 123 CPS11 6golf... SIDCID 142CPS116 142CPS114 123CPS116... SIDclub 142ballet 142sumo 123chess 123golf...

35. 3NF, BCNF, 4NF, and beyond Of historical interests – 1NF: All column values must be atomic – 2NF: Slightly more relaxed than 3NF 3/15/201635 Anomaly/normal form3NFBCNF4NF Lose FD’s?NoPossible Redundancy due to FD’sPossibleNo Redundancy due to MVD’sPossible No

36. Summary Philosophy behind BCNF, 4NF: Data should depend on the key, the whole key, and nothing but the key! Philosophy behind 3NF: … But not at the expense of more expensive constraint enforcement! 3/15/201636