CS 405G: Introduction to Database Systems

CS 405G: Introduction to Database Systems 18. Normal Forms and Normalization

Last class • Functional Dependency. • Normalization • Decomposition Chen Qian @ University of Kentucky

Review • Functional dependencies • X->Y: X “determines” Y • If two rows agree on X, they must agree on Y • Attribute on the LHS is known as the determinant • X is a determinant of Y

Normalization • A normalization is the process of organizing the fields and tables of a relational database to minimize redundancy and dependency. • A normal form is acertification that tells whether a relation schema is in a particular state Chen Qian @ University of Kentucky

First Normal Form ( 1NF ) • A relation is in first normal form if the domain of each attribute contains only atomic values, and the value of each attribute contains only a single value from that domain. Chen Qian @ University of Kentucky

2nd Normal Form • An attribute A of a relation R is a nonprimary attributeif it is not part of any key in R, otherwise, A is a primary attribute. • R is in (general) 2nd normal form if every nonprimary attribute A in R is not partially functionally dependent on any key of R Chen Qian @ University of Kentucky

Redundancy Example • If a key will result a partial dependency of a nonprimary attribute. • e.g. EID, PID->Ename • In this case, the attribute (Ename) should be separated with its full dependency key (EID) to be a new table. • So, to check whether a table includes redundancy. Try every nonprimary attribute and check whether it fully depends on any key. Chen Qian @ University of Kentucky

Decomposition • Decomposition eliminates redundancy • To get back to the original relation, use natural join. Decomposition Foreign key Chen Qian @ University of Kentucky

Decomposition • Decomposition may be applied recursively Chen Qian @ University of Kentucky

Questions about decomposition • When to decompose • How to come up with a correct decomposition (i.e., lossless join decomposition) Chen Qian @ University of Kentucky

Third normal form • 3NF requires that there are no non-trivial functional dependencies of non-key attributes on something other than a superset of a candidate key. • Recall: non-trivial FD means LHS has no intersection with RHS. • In summary, all non-key attributes are mutually independent. Chen Qian @ University of Kentucky

Ename and email has no FD Chen Qian @ University of Kentucky

Boyce-Codd normal form (BCNF) • BCNF requires that there are no non-trivial functional dependencies of attributes on something other than a superset of a candidate key (called a superkey). • All attributes are dependent on a key, a whole key and nothing but a key (excluding trivial dependencies, like A->A). Chen Qian @ University of Kentucky

A table is said to be in the BCNF if and only if it is in the 3NF and every non-trivial, left-irreducible functional dependency has a candidate key as its determinant. • In more informal terms, a table is in BCNF if it is in 3NF and the only determinants are the candidate keys. Chen Qian @ University of Kentucky

Non-key FD’s • Consider a non-trivial FD X->Y where X is not a super key • Since X is not a super key, there are some attributes (say Z) that are not functionally determined by X That b is always associated with a is recorded by multiple rows:redundancy, update anomaly, deletion anomaly Chen Qian @ University of Kentucky

Dealing with Nonkey Dependency: BCNF • A relation R is in Boyce-Codd Normal Form if • For every non-trivial FD X->Y in R, X is a super key • That is, all FDs follow from “key -> other attributes” • When to decompose • As long as some relation is not in BCNF • How to come up with a correct decomposition • Always decompose on a BCNF violation (details next) • Then it is guaranteed to be a lossless join decomposition Chen Qian @ University of Kentucky

BCNF decomposition algorithm • Find a BCNF violation • That is, a non-trivial FD X->Y in R where X is not a super key of R • Decompose R into R1 and R2, where • R1 has attributes XY • R2 has attributes XZ, where Z contains all attributes of R that are in neither X nor Y (i.e. Z = attr(R) – X – Y) • Repeat until all relations are in BCNF Chen Qian @ University of Kentucky

Grade (EID, PID, hours) Student (EID, Ename, email) BCNF BCNF BCNF decomposition example WorkOn (EID, Ename, email, PID, hours) BCNF violation: EID->Ename, email Chen Qian @ University of Kentucky

StudentID (email, EID) StudentGrade’ (email, Ename, PID, hours) StudentName (email, Ename) Grade (email, PID, hours) BCNF BCNF Another example WorkOn (EID, Ename, email, PID, hours) BCNF violation: email->EID BCNF BCNF violation: email->Ename Chen Qian @ University of Kentucky

Exercise • Property(Property_id#, County_name, Lot#, Area, Price, Tax_rate) • Property_id#-> County_name, Lot#, Area, Price, Tax_rate • County_name, Lot# -> Property_id#, Area, Price, Tax_rate • County_name -> Tax_rate • area -> Price Chen Qian @ University of Kentucky

LOTS1 (County_name, Tax_rate ) LOTS2A (Area, Price) LOTS2 (Property_id#, County_name, Lot#, Area, Price) LOTS2B (Property_id#, County_name, Lot#, Area) Exercise Property(Property_id#, County_name, Lot#, Area, Price, Tax_rate) BCNF violation: County_name-> Tax_rate BCNF BCNF violation: Area-> Price BCNF BCNF Chen Qian @ University of Kentucky

Why is BCNF decomposition lossless Given non-trivial X->Y in R where X is not a super key of R, need to prove: • Anything we project always comes back in the join:R πXY ( R ) πXZ ( R ) • Sure; and it doesn’t depend on the FD • Anything that comes back in the join must be in the original relation:RπXY ( R ) πXZ ( R ) • Proof makes use of the fact that X->Y Chen Qian @ University of Kentucky

Recap • Functional dependencies: a generalization of the key concept • Partial dependencies: a source of redundancy • Use 2nd Normal form to remove partial dependency • Non-key functional dependencies: a source of redundancy • BCNF decomposition: a method for removing ALL functional dependency related redundancies • Plus, BNCF decomposition is a lossless join decomposition Chen Qian @ University of Kentucky

Normalization • 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 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 2NF 3NF BCNF

First Normal Form The following is not in 1NF EmpNum EmpPhone EmpDegrees 123 233-9876 333 233-1231 BA, BSc, PhD 679 233-1231 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 EmployeeDegree Employee EmpNum EmpDegree EmpNum EmpPhone 333 BA 123 233-9876 333 BSc 333 233-1231 333 PhD 679 233-1231 679 BSc 679 MSc An outer join between Employee and EmployeeDegree will produce the information we saw before

Second Normal Form Consider this InvLine table (in 1NF): InvNum LineNum ProdNum Qty InvDate InvNum, LineNum ProdNum There are two candidate keys. Qty InvNum, ProdNum LineNum Qty is the only non-key attribute, and it is dependent on InvNum InvNum InvDate 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 InvLine is only in 1NF

Second Normal Form InvLine InvNum LineNum ProdNum Qty InvDate 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: InvNum LineNum ProdNum Qty InvNum InvDate

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

EmpNum EmpName DeptNum DeptName We correct the situation by decomposing the original relation into two 3NF relations. Note the decomposition is lossless. DeptNum DeptName EmpNum EmpName DeptNum Third Normal Form Verify these two relations are in 3NF. Are they in BCNF?

Boyce-Codd Normal Form • 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.

In 3NF, but not in BCNF: Instructor teaches one course only. student_no course_no instr_no Student takes a course and has one instructor. {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.

student_no course_no instr_no BCNF student_no instr_no course_no instr_no {student_no, instr_no}  student_no {student_no, instr_no}  instr_no instr_no  course_no

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

inv_no line_no prod_no prod_desc qty

inv_no line_no prod_no prod_desc qty 2NF, but not in 3NF, nor in BCNF: since prod_no is not a candidate key and we have: prod_no  prod_desc.

EmployeeDept ename ssn bdate address dnumber dname

BCNF = no redundancy? • Student (SID, CID, club) • Suppose your classes have nothing to do with the clubs you join • FDs? • None • BCNF? • Yes • Redundancies? • Tons!

Multivalued dependencies • A multivalued dependency (MVD) has the formX->>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 Must be in R too

4NF violation: SID ->>CID Enroll (SID, CID) Join (SID, club) 4NF 4NF 4NF decomposition example Student (SID, CID, club)

3NF, BCNF, 4NF, and beyond • Of historical interests • 1NF: All column values must be atomic • 2NF: Slightly more relaxed than 3NF

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!

Next Class • Quiz again Chen Qian @ University of Kentucky

CS 405G: Introduction to Database Systems