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

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

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.

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

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

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

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

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

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

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

Transitive dependency 3/15/ 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.

Transitive dependency 3/15/ EmpNum Emp DeptNum DeptNname DeptName is transitively dependent on EmpNum via DeptNum EmpNum  DeptName EmpNum  DeptNum DeptNum  DeptName

Partial dependency 3/15/ 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

First Normal Form 3/15/ 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.

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

First Normal Form 3/15/ 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”?

First Normal Form 3/15/ EmpNumEmpDegree 333BA 333BSc 333PhD 679BSc MSc679 EmpNumEmpPhone An outer join between Employee and EmployeeDegree will produce the information we saw before Employee EmployeeDegree

Boyce-Codd Normal Form 3/15/ 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.

Second Normal Form 3/15/ 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

Second Normal Form 3/15/ 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):

Second Normal Form 3/15/ 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?

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

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

Third Normal Form 3/15/ 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

Boyce-Codd Normal Form 3/15/ 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.

Third Normal Form 3/15/ 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? …

Third Normal Form 3/15/ 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?

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

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

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

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

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/ SIDCIDclub 142CPS116ballet 142CPS116sumo 142CPS114ballet 142CPS114sumo 123CPS116chess 123CPS116golf...

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/ Must be in R too XYZ ab1c1 ab2c2 ab1c2 ab2c1

4NF decomposition example 3/15/ 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 142CPS CPS CPS SIDclub 142ballet 142sumo 123chess 123golf...

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

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