1. Lecture 09: Functional Dependencies

2. Outline Functional dependencies (3.4) Rules about FDs (3.5)
Design of a Relational schema (3.6)

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

4. Functional Dependencies
Definition: A1, ..., Am → B1, ..., Bn holds in R if:
t, t’  R, (t.A1=t’.A1  ...  t.Am=t’.Am  t.B1=t’.B1  ...  t.Bm=t’.Bm ) R A1
... Am B1 Bm t
if t, t’ agree here
then t, t’ agree here t’

5. Important Point! Functional dependencies are part of the schema!
They constrain the possible legal data instances.
At any point in time, the actual database may satisfy additional FD’s.

6. Examples EmpID → Name, Phone, Position Position → Phone
John Smith
2376 HR
E1847
Zheng Li QA
E3123
Lucy Larsen
1264
Developer
E9923
EmpID → Name, Phone, Position
Position → Phone
but Phone → Position

7. Formal definition of a key
A key is a set of attributes A1, ..., An s.t. for any other attribute B, A1, ..., An → B
A minimal key is a set of attributes which is a key and for which no subset is a key
Note: book calls them superkey and key

8. Examples of Keys Product(name, price, category, color)
name, category → price
category → color
Keys are: {name, category} and all supersets
Enrollment(student, address, course, room, time)
student → address
room, time → course
student, course → room, time
Keys are: [in class]
Keys: {student, room, time}, {student, course} and all supersets

9. Inference Rules for FD’s
A1 , A2 , … An→B1, B2, … Bm
Splitting rule
and
Combing rule
Is equivalent to
A1 , A2 , … An →B1
A1 , A2 , … An →B2
A1 , A2 , … An →Bm … A1
... Am B1 Bm
AND

10. Inference Rules for FD’s (continued)
A1 , A2 , … An → Ai
Trivial Rule
where i = 1, 2, ..., n A1
... Am
Why ?

11. Inference Rules for FD’s (continued)
Transitive Closure Rule If
A1 , A2 , … An → B1, B2, … Bm
and
B1, B2, … Bm → C1 , C2 , … Ck
then
A1 , A2 , … An → C1 , C2 , … Ck
Why?

12. A1
... Am B1 Bm C1 Ck

13. Enrollment(student, major, course, room, time)
course → time
What else can we infer ? [in class]

14. Closure of a set of Attributes
Given a set of attributes {A1, …, An} and a set of dependencies S.
Problem: find all attributes B such that:
any relation which satisfies S also satisfies:
A1 , A2 , … An → B
The closure of {A1, …, An}, denoted {A1, …, An}+,
is the set of all such attributes B

15. Closure Algorithm Start with X={A1, …, An}. Until X doesn’t change do:
if B1, B2, … Bm → C is in S, and
B1, B2, … Bm are all in X and
C is not in X
then
add C to X

16. Example The schema - R(A,B,C,D,E,F) A, B → C A,D → E B → D A,F → B
Closure of {A,B}: X = {A, B, }
Closure of {A, F}: X = {A, F, }

17. Why Is the Algorithm Correct ?
Show the following by induction:
For every B in X:
A1 , A2 , … An → B
Initially X = {A1 , A2 , … An } holds
Induction step: B1, …, Bm in X
Implies A1 , A2 , … An → B1, B2, … Bm
We also have B1, B2, … Bm → C
By transitivity we have A1 , A2 , … An → C
This shows that the algorithm is sound; need to show it is complete

18. Relational Schema Design (or Logical Design)
Main idea:
Start with some relational schema
Find out its FD’s
Use them to design a better relational schema

19. 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 = repeated data
Update anomalies = Fred moves to “Bellevue”
Deletion anomalies = Fred drops all phone numbers: what is his city ?

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

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

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

23. 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

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

25. 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...

26. 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
Including superkeys
In English:
Whenever a set of attributes of R determines one more attribute,
it should determine all the attributes of R.

27. 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?

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

29. Summary of BCNF Decomposition
Find a dependency that violates the BCNF condition:
A1 , A2 , … An → B1, B2, … Bm
Heuristics: choose B1, B2, … Bm “as large as possible”
Decompose:
Continue until
there are no
BCNF violations
left.
Others
A’s
B’s
Exercise: Is there a 2-attribute relation that is not in BCNF ? R1 R2

30. 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

31. Other Example R(A,B,C,D) A → B, B → C Key: A, D
Violations of BCNF: A → B, B → C, A → C, A → B,C
Pick A → B,C: split R into R1(A,B,C), R2(A,D)
What happens if we pick A → B first ?

32. 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

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

34. 3NF: A Problem with BCNF Film Actor Character
FD’s: Character Film; Film, Actor Character
So, there is a BCNF violation, and we decompose.
Film
Character
Character Film
Actor
Character
No FDs

35. So What’s the Problem? Film Character Austin Powers Dr. Evil Actor
Mike Myers
Austin Powers
Dr. Evil
No problem so far. All local FD’s are satisfied.
Let’s put all the data back into a single table again:
What happened here?
Film
Actor
Character
Austin Powers
Mike Myers
Dr. Evil
Violates the dependency: Film, Actor Character!

36. Solution: 3rd Normal Form (3NF)
A simple condition for removing anomalies from relations:
3NF has a dependency-preserving decomposition
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 key for R,
or B is part of a key.
Including superkeys

37. Decomposition into 3NF Easy when we know BCNF decomposition (how?)
Not-so-easy: if we want a dependency-preserving decomposition…
In the book (3.5)