1. CS 3630 Database Design and Implementation

2. Second Normal Form (2NF)
A relation R is in 1NF, and
every non-primary-key attribute is fully functionally dependent on the primary key
Then R is in 2NF.
No Partial FDs on PK.

3. Lease (RNo, RName, PNo, PAddress, Start, Finish, Rent, ONo, OName)
Primary Key: PNo, Start
FDs:
PNo, Start  All
PNo  PAddress, ONo, Oname (Partial on PK!)
Lease1 (PNo, PAddress, ONo, OName)
Primary Key: Pno
Alternate Key: None
Foreign Key: None
PNo  All
PAddress  All
ONo  OName
Lease (RNo, RName, PNo, Start, Finish, Rent)
Foreign Key: PNo References Lease1
RNo  RName

4. Better Name for the New Table
Lease (RNo, RName, PNo, PAddress, Start, Finish, Rent, ONo, OName)
Primary Key: PNo, Start
Better Name for the New Table
Lease1 (PNo, PAddress, ONo, OName)
Primary Key: Pno
Property
Lease (RNo, RName, PNo, Start, Finish, Rent)

5. Property (PNo, PAddress, ONo, OName)
Primary Key: PNo
FDs:
PNo  All
PAddress  All
ONo  OName
Table Instance
PNo PAddress ONo OName
P main O Tina
P main O Tony
P first O Tony
P first O Tony
In 2NF?
But still some Redundancy
Reason?

6. Third Normal Form (3NF) No Transitive FDs on PK.
Relation R in 2NF, and
No non-Primary-Key attribute is transitively functionally dependent on the primary key
Then R is in 3NF.
No Transitive FDs on PK.

7. Transitive FDs If A  B and B  C Then A  C is always TRUE
(provided A is not functionally dependent on either B or C)
(NO cycle!)

8. Transitive FDs A B C
If A  B and B  C
Then A  C

9. Transitive FDs A B C If A  B and B  C Then A C
Provided not B  A (A and B are equivalent)
nor C  A (A, B and C are equivalent)
No Cycle!

10. Property (PNo, PAddress, ONo, OName) FDs: PNo  All PAddress  All
Primary Key: PNo
FDs:
PNo  All
PAddress  All
ONo  OName
Table Instance
PNo PAddress ONo OName
P main O Tina
P main O Tony
P first O Tony
P first O Tony
In 2NF, but still some redundancy
Not in 3NF!
Ono  Oname
(Oname is transitively functionally dependent on Pno)

11. Decompose Property into 3NF
Property (PNo, PAddress, ONo, OName)
PNo  All
PAddress  All
ONo  OName
Property1 (ONo, OName)
Primary Key: Ono
Alternate Keys: None
Foreign Keys: ?
FDs:
ONo  Oname
Property (PNo, PAddress, ONo)
Primary Key: PNo
Alternate Keys: PAddress
Foreign Keys: ?
FDs:
PNo  All
PAddress  All

12. Decompose Property into 3NF
Property (PNo, PAddress, ONo, OName)
PNo  All
PAddress  All
ONo  OName
Property1 (ONo, OName)
Primary Key: Ono
Alternate Keys: None
Foreign Keys: None
FDs:
ONo  Oname
Better table name?
Owner
Property (PNo, PAddress, ONo)
Primary Key: PNo
Alternate Keys: PAddress
Foreign Keys: Ono references Property1
FDs:
PNo  All
PAddress  All

13. Decompose Property into 3NF
Property (PNo, PAddress, ONo, OName)
PNo  All
PAddress  All
ONo  OName
Owner (ONo, OName)
Primary Key: Ono
Alternate Keys: None
Foreign Keys: None
FDs:
ONo  Oname
Property (PNo, PAddress, ONo)
Primary Key: PNo
Alternate Keys: PAddress
Foreign Keys: Ono references Owner
FDs:
PNo  All
PAddress  All

14. Table Instances Property PNo PAddress ONo OName
P main O Tina
P main O Tony
P first O Tony
P first O Tony
Property
PNo PAddress Ono
P main O100
P main O109
P first O109
P first O109
Owner
ONo OName
O Tina
O Tony

15. Final Tables
Lease Property Owner

16. Relation R
R (A, B, C, D, E, F)
Primary Key: A, B
Alternate Keys: None
Functional Dependencies:
A, B  All
C  D
E  F
Is it in 2NF?
Is it in 3NF?
How many tables we will have to decompose it into 3NF?

17. R (A, B, C, D, E, F) PK: A, B AK: None FK: None FDs: A, B  All C  D
Table Instance
A B C D E F
x se
y cis
x cis
y ct
Decompose into 3NF
Each table: DBDL
FDs
Table instance

18. Assignment 5-2 Due Monday, March 5 Quiz 2 Wednesday, March 7 Lab 206
Schedule
Assignment 5-2 Due Monday, March 5 Quiz 2 Wednesday, March 7 Lab 206