1. CS 3630 Database Design and Implementation

2. Assignment 6-1 2. staffNo  dentistName patNo  patName
staffNo, Date, Time  All
patNo, Date, Time  All
surgeryNo, Date, Time  All
staffNo, Date  surgeryNo 3.
staffNo, Date, Time
(any one of the three ALL FDs) 4.
patNo, Date, Time
surgeryNo, Date, Time

3. First Normal Form (1NF) No multi-value attributes
Done when mapping E-R model to relational schema
DBDL

4. 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 the PK.

5. 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.

6. Boyce-Codd Normal Form (BCNF)
Definition
R in 1NF and
The determinant of each FD is a candidate key.
Review: 1NF
determinant
candidate key

7. BCNF is stronger than 3NF
BCNF and 3NF
BCNF is stronger than 3NF
If R in BCNF, then
R in 3NF.
If R not in 3NF, then
R not in BCNF.

8. Proof If R not in 3NF, then PK  B, and B  C, (PK  C)
NO cycle for transitive FD
B  PK : False
B is not a candidate key
but a determinant (B  C )
So, R is not in BCNF.

9. Example
Lease (RNo, RName, PNo, PAddress, Start, Finish, Rent, ONo, OName)
Primary Key: PNo, Start
Alternate Key: PNo, Finish
PAddress, Start
PAddress, Finish
FDs:
PNo, Start  All other attributes
PNo, Finish  All other attributes
PAddress, Start  All other attributes
PAddress, Finish  All other attributes
PNo  PAddress, ONo, OName (Pno not a candidate key)
PAddress  PNo, ONo, Oname (Paddress not a candidate key)
RNo  Rname (Rno not a candidate key)
ONo  OName (Ono not a candidate key)
Not in BCNF.
How many tables in order to make it BCNF?

10. Decompose Lease into BCNF
Lease (RNo, RName, PNo, PAddress, Start, Finish, Rent, ONo, OName)
PNo  PAddress, ONo, OName (Pno not a candidate key)
PAddress  PNo, ONo, Oname (Paddress not a candidate key)
RNo  Rname (Rno not a candidate key)
ONo  OName (Ono not a candidate key)
Property (PNo, PAddress, Ono, Oname)
PNo  PAddress, Ono, Oname
PAddress  PNo, Ono, Oname
The Same Table!

11. Decompose Lease into BCNF
Lease (RNo, RName, PNo, PAddress, Start, Finish, Rent, ONo, OName)
PNo  PAddress, ONo, OName (Pno not a candidate key)
(PAddress  PNo, ONo, Oname (Paddress not a candidate key) )
RNo  Rname (Rno not a candidate key)
ONo  OName (Ono not a candidate key)
Property (PNo, PAddress, Ono, Oname)
PNo  PAddress, Ono, Oname
PAddress  PNo, ONo, Oname
(Ono  Oname)
Renter (Rno, Rname)
Rno  Rname
Lease (RNo, PNo, Start, Finish, Rent)
PNo, Start  All
PNo, Finish  All

12. Decompose Lease into BCNF
Lease (RNo, RName, PNo, PAddress, Start, Finish, Rent, ONo, OName)
Property (PNo, PAddress, Ono, Oname)
PNo  PAddress, Ono, Oname
Ono  Oname
BCNF?
Renter (Rno, Rname)
Rno  Rname
Lease (RNo, PNo, Start, Finish, Rent)
PNo, Start  All
PNo, Finish  All

13. Decompose Lease into BCNF
Lease (RNo, RName, PNo, PAddress, Start, Finish, Rent, ONo, OName)
Property (PNo, PAddress, Ono, Oname)
PNo  PAddress, Ono, Oname
PAddress  PNo, Ono, Oname
Ono  Oname (determinant is not a candidate key)
Property (PNo, PAddress, Ono)
PNo  PAddress, Ono
PAddress  PNo, Ono
Owner (Onp, Oname)
Ono  Oname

14. Decompose Lease into BCNF
Lease (RNo, RName, PNo, PAddress, Start, Finish, Rent, ONo, OName)
PNo  PAddress, ONo, OName (Pno not a candidate key)
PAddress  PNo, ONo, Oname (Paddress not a candidate key)
RNo  Rname (Rno not a candidate key)
ONo  OName (Ono not a candidate key)
Property (PNo, PAddress, ONo)
PNo  PAddress, ONo
PAddress  PNo, Ono
Renter (RNo, RName)
RNo  RName
Owner (ONo, OName)
ONo  Oname
Only 4 tables, not 5.
Ono will not be in Lease.
Pno  Ono
Lease (RNo, PNo, Start, Finish, Rent)
PNo, Start  All
PNo, Finish  All

15. Example
R (A, B, C, D, E, F) PK: A, B, C AK: B, C, D FK: None FDs: A, B, C  All B, C, D  All B, D  A
A: staffNo
B, C: Date, Time
D: surgeryNo

16. Table Instance
A B C D E F 2 10 x u ct y v cis z u se x v cs 4 FDs: A, B, C  All B, C, D  All B, D  A
A: staffNo
B, C: Date, Time
D: surgeryNo

17. Decomposing to BCNF
R (A, B, C, D, E, F) PK: A, B, C AK: B, C, D FK: None FDs: A, B, C  All B, C, D  All B, D  A B, D and A should be in a new table with (B, D) as PK B and D should remain in the original table as FK A should not remain in the original table PK of the original table must be changed to B, C, D.
A: staffNo
B, C: Date, Time
D: surgeryNo

18. Decomposing to BCNF
R (A, B, C, D, E, F) PK: A, B, C AK: B, C, D FK: None FDs: A, B, C  All B, C, D  All B, D  A
R1 (A, B, D)
PK: B, D
AK: NONE
FK: None
FDs: B, D  A
Does R1 have a FK?
R (B, C, D, E, F)
PK: B, C, D
AK: NONE
FK: B, D References R1
FDs: B, C, D  All
Does R2 have a FK?
A: staffNo
B, C: Date, Time
D: surgeryNo

19. Table Instance
A B C D E F 2 10 x u ct y v cis z u se x v cs 4 FDs: A, B, C  All B, C, D  All B, D  A
A B D u v
B C D E F
10 x u ct 1
20 y v cis 2
10 z u se 3
20 x v cs 4
A: staffNo
B, C: Date, Time
D: surgeryNo

20. Selecting B, C, D as PK at the Beginning
R (A, B, C, D, E, F) PK: A, B, C AK: B, C, D FK: None FDs: A, B, C  All B, C, D  All B, D  A
R (A, B, C, D, E, F)
PK: B, C, D
AK: A, B, C
FK: None
FDs: A, B, C  All
B, C, D  All
B, D  A
Partial FD on PK!
A: staffNo
B, C: Date, Time
D: surgeryNo

21. Review: Normalization
1NF
Remove multi-value attributes
Why: each element can not be a set (first order logic)
2NF
Remove partial FDs on PK
Why: remove redundant data
3NF
Remove transitive FDs on PK
BCNF
Stronger than 3NF
Any candidate keys
Why: better PK
remove redundant data
In most cases, BCNF is enough.

22. Lossless Decomposition
After a relation is normalized into two or more relations, the original relations could be obtained by joining new relations
Primary Key and Foreign Key

23. Decompose Lease into BCNF
Lease (RNo, RName, PNo, PAddress, Start, Finish, Rent, ONo, OName)
Owner (ONo, OName)
ONo ---> OName
Renter (RNo, RName)
RNo ---> RName
Property (PNo, PAddress, ONo)
PNo ---> PAddress, ONo
PAddress ---> PNo, Ono
Lease (RNo, PNo, Start, Finish, Rent)
PNo, Start ---> All other attributes
PNo, Finish ---> All other attributes
How to get Property data for a lease?
Lease Property
How to get Renter data for a lease?
Lease Renter
How to get Owner data for a lease?
Lease Property Owner

24. De-Normalization Normalized relations Minimal redundancy
Need join operation to get results
How far should we go?
Where to stop?

25. Review: Database Design
A structured approach that uses procedures, techniques, tools, and documentation aids to support and facilitate the process of design.
Three main phases
Conceptual database design
Understanding client data
E-R (EER) Model
Contract between clients and designers
Logical database design
Mapping E-R Model to (relational) database schema
(Derive relational schema from E-R Model)
DBDL
Normalization
Physical database design

26. Schedule Assignment_52 Assignment_62 Quiz 2 Test 1
Monday, March 5, by 5 PM
Assignment_62
Friday, March 9, by 5 PM
Quiz 2
Wednesday, March 7
Test 1
Wednesday, March 14

27. Quiz 2 Wednesday, March 7 Lab 206 D2L
Must go to Lab 206 during your section to take the quiz
Derive table schemas from E-R Model
(Mapping E-R Model to Database Schema)
DBDL
Functional Dependency
Note11 – Note14
Assignment 4, 5-1 and 6-1