1. Lecture 11: Functional Dependencies
January 31st, 2003

2. Outline
Relational decomposition
Normal forms
Begin relational algebra

3. Relational Schema Design
Recall set attributes (persons with several phones):
Name
SSN
PhoneNumber
City
Fred
Seattle
Joe
Westfield
SSN  Name, City, but not SSN  PhoneNumber
Anomalies:
Redundancy = repeat data
Update anomalies = Fred moves to “Bellvue”
Deletion anomalies = Fred drops all phone numbers: what is his city ?

4. Relation Decomposition
Break the relation into two:
Name
SSN
City
Fred
Seattle
Joe
Westfield
SSN
PhoneNumber

5. Relational Schema Design
Person
buys
Product
name
price
ssn
Conceptual Model:
Relational Model:
plus FD’s
Normalization:
Eliminates anomalies

6. Decompositions in General
R(A1, ..., An)
Create two relations R1(B1, ..., Bm) and R2(C1, ..., Cp)
such that: B1, ..., Bm  C1, ..., Cp = A1, ..., An
and:
R1 = projection of R on B1, ..., Bm
R2 = projection of R on C1, ..., Cp

7. Incorrect Decomposition
Sometimes it is incorrect:
Name
Price
Category
Gizmo
19.99
Gadget
OneClick
24.99
Camera
DoubleClick
29.99
Decompose on : Name, Category and Price, Category

8. Incorrect Decomposition
Name
Category
Gizmo
Gadget
OneClick
Camera
DoubleClick
Price
Category
19.99
Gadget
24.99
Camera
29.99
Name
Price
Category
Gizmo
19.99
Gadget
OneClick
24.99
Camera
29.99
DoubleClick
When we put it back:
Cannot recover information

9. Normal Forms First Normal Form = all attributes are atomic
Second Normal Form (2NF) = old and obsolete
Third Normal Form (3NF) = this lecture
Boyce Codd Normal Form (BCNF) = this lecture
Others...

10. Boyce-Codd Normal Form
A simple condition for removing anomalies from relations:
A relation R is in BCNF if:
Whenever there is a nontrivial dependency A1, ..., An  B
in R , {A1, ..., An} is a key for R
In English (though a bit vague):
Whenever a set of attributes of R is determining another attribute,
should determine all the attributes of R.

11. Example What are the dependencies? SSN  Name, City What are the keys?
PhoneNumber
City
Fred
Seattle
Joe
Westfield
What are the dependencies?
SSN  Name, City
What are the keys?
{SSN, PhoneNumber}
Is it in BCNF?

12. Decompose it into BCNF SSN  Name, City Name SSN City Fred 123-45-6789
Seattle
Joe
Westfield
SSN  Name, City
SSN
PhoneNumber

13. Summary of BCNF Decomposition
Find a dependency that violates the BCNF condition:
A , A , … A
B , B , … B 1 2 n 1 2 m
Heuristics: choose B , B , … B “as large as possible” 1 2 m
Decompose:
Continue until
there are no
BCNF violations
left.
Others
A’s
B’s
Is there a
2-attribute
relation that is
not in BCNF ? R1 R2

14. Example Decomposition
Person(name, SSN, age, hairColor, phoneNumber)
SSN  name, age
age  hairColor
Decompose in BCNF (in class):
Step 1: find all keys
Step 2: now decompose

15. Other Example
R(A,B,C,D) A B, B C
Keys:
Violations of BCNF:

16. Correct Decompositions
A decomposition is lossless if we can recover:
R(A,B,C)
R1(A,B) R2(A,C)
R’(A,B,C) should be the same as R(A,B,C)
Decompose
Recover
R’ is in general larger than R. Must ensure R’ = R

17. Correct Decompositions
Given R(A,B,C) s.t. AB, the decomposition into R1(A,B), R2(A,C) is lossless

18. 3NF: A Problem with BCNF Unit Company Product
FD’s: Unit  Company; Company, Product  Unit
So, there is a BCNF violation, and we decompose.
Unit Company
Unit  Company
Unit Product
No FDs

19. So What’s the Problem? Unit Company Unit Product
Galaga UW Galaga databases
Bingo UW Bingo databases
No problem so far. All local FD’s are satisfied.
Let’s put all the data back into a single table again:
Unit Company Product
Galaga UW databases
Bingo UW databases
Violates the dependency: company, product -> unit!

20. Solution: 3rd Normal Form (3NF)
A simple condition for removing anomalies from relations:
A relation R is in 3rd normal form if :
Whenever there is a nontrivial dependency A1, A2, ..., An  B for R , then {A1, A2, ..., An } a super-key for R,
or B is part of a key.

21. Declartive query language
Relational Algebra
Formalism for creating new relations from existing ones
Its place in the big picture:
Declartive query language
Algebra
Implementation
Relational algebra Relational bag algebra
SQL, relational calculus

22. Relational Algebra Five operators: Derived or auxiliary operators:
Union: 
Difference: -
Selection: s
Projection: P
Cartesian Product: 
Derived or auxiliary operators:
Intersection, complement
Joins (natural,equi-join, theta join, semi-join)
Renaming: r

23. 1. Union and 2. Difference R1  R2 Example: R1 – R2
ActiveEmployees  RetiredEmployees
R1 – R2
AllEmployees -- RetiredEmployees

24. What about Intersection ?
It is a derived operator
R1  R2 = R1 – (R1 – R2)
Also expressed as a join (will see later)
Example
UnionizedEmployees  RetiredEmployees

25. 3. Selection Returns all tuples which satisfy a condition
Notation: sc(R)
Examples
sSalary > (Employee)
sname = “Smithh” (Employee)
The condition c can be =, <, , >, , <>

26. Find all employees with salary more than $40,000.
s Salary > (Employee)

27. 4. Projection Eliminates columns, then removes duplicates
Notation: P A1,…,An (R)
Example: project social-security number and names:
P SSN, Name (Employee)
Output schema: Answer(SSN, Name)

28. P SSN, Name (Employee)

29. 5. Cartesian Product Each tuple in R1 with each tuple in R2
Notation: R1  R2
Example:
Employee  Dependents
Very rare in practice; mainly used to express joins

31. Relational Algebra Five operators: Derived or auxiliary operators:
Union: 
Difference: -
Selection: s
Projection: P
Cartesian Product: 
Derived or auxiliary operators:
Intersection, complement
Joins (natural,equi-join, theta join, semi-join)
Renaming: r

32. Renaming Changes the schema, not the instance Notation: r B1,…,Bn (R)
Example:
rLastName, SocSocNo (Employee)
Output schema: Answer(LastName, SocSocNo)

33. LastName, SocSocNo (Employee)
Renaming Example
Employee
Name
SSN
John
Tony
LastName, SocSocNo (Employee)
LastName
SocSocNo
John
Tony

34. Natural Join Notation: R1 ⋈ R2 Meaning: R1 ⋈ R2 = PA(sC(R1  R2))
Where:
The selection sC checks equality of all common attributes
The projection eliminates the duplicate common attributes

35. Natural Join Example
Employee
Name
SSN
John
Tony
Dependents
SSN
Dname
Emily
Joe
Employee Dependents =
PName, SSN, Dname(s SSN=SSN2(Employee x rSSN2, Dname(Dependents))
Name
SSN
Dname
John
Emily
Tony
Joe

36. Natural Join
R= S=
R ⋈ S= A B X Y Z V B C Z U V W A B C X Z U V Y W

37. Natural Join
Given the schemas R(A, B, C, D), S(A, C, E), what is the schema of R ⋈ S ?
Given R(A, B, C), S(D, E), what is R ⋈ S ?
Given R(A, B), S(A, B), what is R ⋈ S ?

38. Theta Join A join that involves a predicate R1 ⋈ q R2 = s q (R1  R2)
Here q can be any condition

39. Eq-join A theta join where q is an equality
R1 ⋈A=B R2 = s A=B (R1  R2)
Example:
Employee ⋈SSN=SSN Dependents
Most useful join in practice

40. Semijoin R ⋉ S = P A1,…,An (R ⋈ S)
Where A1, …, An are the attributes in R
Example:
Employee ⋉ Dependents

41. Semijoins in Distributed Databases
Semijoins are used in distributed databases
Dependents
Employee
SSN
Dname
Age
. . .
SSN
Name
. . .
network
Employee ⋈ssn=ssn (s age>71 (Dependents))
T = P SSN s age>71 (Dependents)
R = Employee ⋉ T
Answer = R ⋈ Dependents

42. Complex RA Expressions
P name
buyer-ssn=ssn
pid=pid
seller-ssn=ssn
P ssn
P pid
sname=fred
sname=gizmo
Person Purchase Person Product

43. Operations on Bags A bag = a set with repeated elements
All operations need to be defined carefully on bags
{a,b,b,c}{a,b,b,b,e,f,f}={a,a,b,b,b,b,b,c,e,f,f}
{a,b,b,b,c,c} – {b,c,c,c,d} = {a,b,b,d}
sC(R): preserve the number of occurrences
PA(R): no duplicate elimination
Cartesian product, join: no duplicate elimination
Important ! Relational Engines work on bags, not sets !
Reading assignment: 5.3 – 5.4

44. Finally: RA has Limitations !
Cannot compute “transitive closure”
Find all direct and indirect relatives of Fred
Cannot express in RA !!! Need to write C program
Name1
Name2
Relationship
Fred
Mary
Father
Joe
Cousin
Bill
Spouse
Nancy
Lou
Sister