1. Database
Normalization

2. Designing Good Schemas
We know how to create schemas, but ...
how do we create good schemas?
what does good mean?
Schema quality measurements:
semantics of the attributes
minimal redundancy
minimal frequency of null values

3. Functional Dependences
A column Y of relational table R is functionally dependent up on column X of relational table R if and only if:
Each value of X in R associated with each value of Y at any given time

4. Functional dependences
Y is functional dependent up on X same as values of X identify values of Y
If X  Y then XZYZ
IF XY and Y  Z then XZ
X Y means that Y depend on X or
X identify Y

5. Examples S#  Ename {S#, P#}  Hours
If for each value of S#, there are exactly one corresponding value for sname, state, city then: S#
Sname
Sate
City

6. Example
If {S#, p#}  Qty S# P#
QTY

7. Redundancy Example
Where’s the redundancy?

8. Redundancy Example

9. Example FDs
Proper FDs
Transitive FDs
Partial Key FD
Partial Key FDs

10. Normal Forms
Each normal form is a set of conditions on a schema that guarantees certain properties (relating to redundancy and update anomalies)
The two commonly used normal forms are third normal form (3NF) and Boyce-Codd normal form (BCNF)

11. Normalization 0NF 1NF 2NF 3NF BCNF 4NF 5NF remove multi-valued
attributes
1NF
2NF
3NF
partial
dependencies
transitive
BCNF
4NF
5NF
remove
remaining
FD anomal
dependencies
multivalue
anomalies

12. 1 NF First normal form is NO multi-valued attributes
No composite attribute
No nested relation
We create new table or new field (telephone, visiting)

13. 1NF Normalization
Proper translation from ER multi-value attributes will achieve 1NF.
Still not a good solution, since we have redundancy in Dnumber and Dmgr_ssn. (This will be handled by 2NF.)

14. 2 NF form
Second normal form that if primary key is multiple attribute and non-key attribute depend on part of primary key S# P#
Hours
Cname
pname
Loc

15. 2NF Normalization
Move the partial key and dependent attributes to a new relation.

16. Transitive Dependencies
X → Y is a transitive dependency (PD) if there exists Z ⊈ any key such that X → Z → Y
TDs can cause redundancy if there are multiple values of X that determine the same value of Z
the value of Y for that value of Z is stored multiple times
3NF normalization: move (Z,Y) to new relation in which Z is the primary key

17. 3 NF
The relation in 3NF if it is 2 NF and every non-key attribute is non-transitively dependent on primary key

18. 3NF Normalization
Create new relation to hold the attributes in the transitive FD.
LHS of transitive FD becomes PK of new relation.

19. Transitive Dependency Example
DEPT
COURSE
SECTION
ROOM
INSTR
I_OFFICE
I_OFFICE (instructor's office) is determined
by the non-PK attribute INSTR
DEPT
COURSE
SECTION
COMP 51 1 2
163 53
ROOM
WPC122
WPC219
WPC130
INSTR
DOHERTY
CLIBURN
BOWRING
CARMAN
I_OFFICE
CSB109
CSB107
CSB108
CSB104

20. NF Decomposition: Foreign Keys
DEPT
COURSE
SECTION
ROOM
INSTR
I_OFFICE
DEPT
COURSE
SECTION
ROOM
INSTR
Decomposition:
INSTR
I_OFFICE

21. Why? Normalization Goal = BCNF = Boyce-Codd Normal Form =
all FD’s follow from the fact “key  everything.”
Formally, R is in BCNF if for every nontrivial FD for R, say X  A, then X is a superkey.
“Nontrivial” = right-side attribute not in left side.
Why?
1. Guarantees no redundancy due to FD’s.
2. Guarantees no update anomalies = one occurrence of a fact is updated, not all.
3. Guarantees no deletion anomalies = valid fact is lost when tuple is deleted.
Arthur Keller – CS 180

22. Boyce-Codd Normal Form
Sample data for Course Section table
Because Prefix  Department, we know that (Prefix, Num, SecNum) could also be a primary key for this table.
Department
Prefix
Num
SecNum
CourseName
Instructor
Mathematics
Math
101 1
Algebra I
Al Jeebra 2
201
Calculus I
Kal Kuelus
Philosophy
Phil
Greek Thought
Arie Stottle
202
Euro Thought
Mike Angelo
Marketing
Mktg
410
Marketing Strategy
Marc Ekking
SpMkg
401
Advanced Sports Marketing
Hulk Hogan

23. Example Students(name, addr, phones, CarLiked)
A student’s phones are independent of the cars they like.
Thus, each of a student’s phones appears with each of the cars they like in all combinations.
This repetition is unlike redundancy due to FD’s, of which name->addr is the only one.

24. Example Only key is {name, CarsLiked}.
Students(name, addr, CarLiked, manf, favCar)
FD’s: name->addr favCar, carsLiked->manf
Only key is {name, CarsLiked}.
In each FD, the left side is not a superkey.
Any one of these FD’s shows Students is not in BCNF

25. Boyce-Codd Normal Form
We say a relation R is in BCNF if whenever X ->A is a nontrivial FD that holds in R, X is a superkey.
Remember: nontrivial means A is not a member of set X.
Remember, a superkey is any superset of a key (not necessarily a proper superset).

26. Example Students(name, addr, CarsLiked, manf, favCar)
F = name->addr, name -> favCar, CarsLiked->manf
Pick BCNF violation name->addr.
Close the left side: {name}+ = {name, addr, favCar}.
Decomposed relations:
Students1(name, addr, favCar)
Students2(name, CarsLiked, manf)

27. 3NF and BCNF
3rd Normal Form (3NF) modifies the BCNF condition so we do not have to decompose in this problem situation.
X ->A violates 3NF if and only if X is not a superkey, and also A is not prime.

28. Exercises
The following relation schema is not in third normal form (3NF) Is this an example of a transitive dependency or a partial key dependency? Give an equivalent schema that is in 3NF.
SID
FROM_CITY
TO_CITY
DISTANCE
SHIPMENT
WEIGHT

29. Exercises
This relation has been proposed to track Pacific alumni: Alumni( SID, LastName, FirstName, Degree, YearAwarded, Phone). Pacific allows students to receive multiple degrees, possibly in different years. Identify all FDs. Give a new schema that is in third normal form.

30. Exercises
Consider the following relation schema: Movie(title, genre, length, actor, sag_id, studio, studio_addr)  
Every movie has a unique title.
A movie may have multiple actors.
Each actor has a unique sag_id.
An actor may appear in multiple movies.
A movie has exactly one studio, but a studio may produce more than one movie.
Each studio has exactly one address.
Identify all functional dependencies.
Normalize the schema to 3NF.