1. Lecture 09: Functional Dependencies, Database Design
Friday, October 17, 2003

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

3. 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’

4. Inference If some FDs are satisfied, then others are satisfied too
name  color
category  department
color, category  price
If all these FDs are true:
Then this FD also holds:
name, category  price
Why ??

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

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

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

8. Example (continued) Which Rule did we apply ? Inferred FD
1. name  color
2. category  department
3. color, category  price
Start from the following FDs:
Infer the following FDs:
Inferred FD
Which Rule did we apply ?
4. name, category  name
5. name, category  color
6. name, category  category
7. name, category  color, category
8. name, category  price

9. Example (continued) 1. name  color 2. category  department
3. color, category  price
Answers:
Inferred FD
Which Rule did we apply ?
4. name, category  name
Trivial rule
5. name, category  color
Transitivity on 4, 1
6. name, category  category
7. name, category  color, category
Split/combine on 5, 6
8. name, category  price
Transitivity on 3, 7

10. Another Example Enrollment(student, major, course, room, time)
course  time
What else can we infer ? [in class, or at home]

11. Another Rule Augmentation A1, A2, …, An  B If then
A1, A2, …, An , C1, C2, …, Cp  B
Augmentation follows from trivial rules and transitivity How ?

12. Problem: infer ALL FDs Given a set of FDs, infer all possible FDs
How to proceed ?
Try all possible FDs, apply all 3 rules
E.g. R(A, B, C, D): how many FDs are possible ?
Drop trivial FDs, drop augmented FDs
Still way too many
Better: use the Closure Algorithm (next)

13. Closure of a set of Attributes
Given a set of attributes A1, …, An
The closure, {A1, …, An}+ , is the set of attributes B s.t. A1, …, An  B
Example:
name  color
category  department
color, category  price
Closures: name+ = {name, color} {name, category}+ = {name, category, color, department, price} color+ = {color}

14. Closure Algorithm Start with X={A1, …, An}. Example:
Repeat until X doesn’t change do:
if B1, …, Bn  C is a FD and
B1, …, Bn are all in X
then add C to X.
Example:
name  color
category  department
color, category  price
{name, category}+ = {name, category, color, department, price}

15. Example In class: A, B  C R(A,B,C,D,E,F) A, D  E B  D A, F  B
Compute {A,B}+ X = {A, B, }
Compute {A, F}+ X = {A, F, }

16. Using Closure to Infer ALL FDs
Example:
A, B  C A, D  B
B  D
Step 1: Compute X+, for every X:
A+ = A, B+ = BD, C+ = C, D+ = D
AB+ = ABCD, AC+ = AC, AD+ = ABCD
ABC+ = ABD+ = ACD+ = ABCD (no need to compute– why ?)
BCD+ = BCD, ABCD+ = ABCD
Step 2: Enumerate all FD’s X  Y, s.t. Y  X+ and XY = :
AB  CD, ADBC, ABC  D, ABD  C, ACD  B

17. Problem: find FD from the data
Given a database instance
Find all FD’s satisfied by that instance
Useful if we don’t get enough information from our users: need to reverse engineer a data instance

18. In Class: Find All FDs Student Dept Course Room Alice CSE C++ 020 BobEE HW
040
Carol DB
045
Dan
Java
050
Elsa
Frank
Circuits
Do all FDs make sense in practice ?
Course  Dept, Room
Dept, Room  Course
Student, Dept  Course, Room
Student, Course  Dept, Room
Student, Room  Dept, Course

19. Answer Course  Dept, Room Do all FDs make sense in practice ?
Dept, Room  Course
Student, Dept  Course, Room
Student, Course  Dept, Room
Student, Room  Dept, Course
Do all FDs make sense in practice ?

20. Keys
A key is a set of attributes A1, ..., An s.t. for any other attribute B, we have 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

21. Computing Keys Compute X+ for all sets X
If X+ = all attributes, then X is a key
List only the minimal keys
Note: there can be many minimal keys !
Example: R(A,B,C), ABC, BCA Minimal keys: AB and BC

22. 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

23. FD’s for E/R Diagrams
Given a relation constructed from an E/R diagram, what is its key?
Rule 1: If the relation comes from an entity set,
the key of the relation is the set of attributes which is the
key of the entity set.
address
name
ssn
Person
Person(address, name, ssn)

24. FD’s for E/R Diagrams buys(name, ssn, date)
Rule 2: If the relation comes from a many-many relationship,
the key of the relation is the set of all attribute keys in the
relations corresponding to the entity sets
name
buys
Person
Product
price
name
ssn
date
buys(name, ssn, date)

25. FD’s for E/R Diagrams Purchase(name , sname, ssn, card-no)
Except: if there is an arrow from the relationship to E, then
we don’t need the key of E as part of the relation key.
Purchase
Product
Person
Store
CreditCard
sname
name
card-no
ssn
Purchase(name , sname, ssn, card-no)

26. FD’s for E/R Diagrams More rules:
Many-one, one-many, one-one relationships
Multi-way relationships
Weak entity sets
(Try to find them yourself, or check book)

27. Incomplete (what does it say ?)
FD’s for E/R Diagrams
Say: “the CreditCard determines the Person”
Product
sname
Purchase
name
Store
Incomplete (what does it say ?)
card-no
CreditCard
Person
ssn
Purchase(name , sname, ssn, card-no)
card-no  ssn

28. 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

29. Data Anomalies When a database is poorly designed we get anomalies:
Redundancy: data is repeated
Updated anomalies: need to change in several places
Delete anomalies: may lose data when we don’t want

30. 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 “Bellevue”
Deletion anomalies = Joe deletes his phone number: what is his city ?

31. Relation Decomposition
Break the relation into two:
Name
SSN
PhoneNumber
City
Fred
Seattle
Joe
Westfield
Name
SSN
City
Fred
Seattle
Joe
Westfield
SSN
PhoneNumber
Anomalies have gone:
No more repeated data
Easy to move Fred to “Bellevue” (how ?)
Easy to delete all Joe’s phone number (how ?)

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

33. Decompositions in General
R(A1, ..., An, B1, ..., Bm, C1, ..., Cp)
R1(A1, ..., An, B1, ..., Bm)
R2(A1, ..., An, C1, ..., Cp)
R1 = projection of R on A1, ..., An, B1, ..., Bm
R2 = projection of R on A1, ..., An, C1, ..., Cp

34. Decomposition Sometimes it is correct: Lossless decomposition Name
Price
Category
Gizmo
19.99
Gadget
OneClick
24.99
Camera
Name
Price
Gizmo
19.99
OneClick
24.99
Name
Category
Gizmo
Gadget
OneClick
Camera
Lossless decomposition

35. Incorrect Decomposition
Sometimes it is not:
Name
Price
Category
Gizmo
19.99
Gadget
OneClick
24.99
Camera
What’s incorrect ??
Name
Category
Gizmo
Gadget
OneClick
Camera
Price
Category
19.99
Gadget
24.99
Camera
Lossy decomposition

36. Decompositions in General
R(A1, ..., An, B1, ..., Bm, C1, ..., Cp)
R1(A1, ..., An, B1, ..., Bm)
R2(A1, ..., An, C1, ..., Cp)
If A1, ..., An  B1, ..., Bm
Then the decomposition is lossless
Note: don’t need necessarily A1, ..., An  C1, ..., Cp
Example: name  price, hence the first decomposition is lossless