1. CS 3630 Database Design and Implementation

2. Unnormalized Form (UNF) student courses John CS363 CS334 CS323 Multi-Value attribute Common in reports 2

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

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. 4

5. 5 Partial FDs X, Y  W is partial (W partially depends on X or Y) if X  W or Y  W is true Example: Sno, Bno  staffAddress, staffPhone staffAddress and staffPhone are Partially depend on Sno, since the following is true Sno  staffAddress, staffPhone

6. 6 Full FDs A FD is a full FD if it’s not partial. A, B  X, Y is a full FD if both of the following are false: A  X, Y B  X, Y Sno  staffAddress, staffPhone Cannot be partial!

7. 7 Full FDs A FD is a full FD if it’s not partial. Pno, Start  Rno, Rent, Finish Both of the following are false: Pno  Rno, Rent, Finish Start  Rno, Rent, Finish

8. Second Normal Form (2NF) No Partial FDs on the PK. PK ---> All other attributes Assume a table with a simple PK of Att1 Att1 ---> All other attributes Cannot have partial FD on simple PK Assume a table with a composite PK of Att1 and Att2 Att1, Att2 ---> Att3, Att4, Att5 (All other attributes) It’s possible Att2 ---> Att5 Att1, Att2 ---> Att3, Att4 8

9. Example Lease (RNo, RName, PNo, PAddress, Start, Finish, Rent, ONo, OName) PK: PNo, Start AK: PNo, Finish PAddress, Start PAddress, Finish FDs: PNo, Start ---> All PNo, Finish ---> All PAddress, Start ---> All PAddress, Finish ---> All RNo ---> RName ONo ---> Oname PAddress ---> PNo, ONo, OName PNo ---> PAddress, ONo, OName (Partial on Primary Key) 9

10. Table Instance RNo RName PNo PAddress Start Finish Rent Ono OName R101 John P1001 1001 main 1-1-04 12-31-04 350 O100 Tina R102 Mike P1001 1001 main 1-1-05 2-28-05 380 O100 Tina R103 Kay P1001 1001 main 3-1-05 12-31-05 380 O100 Tina R101 John P1002 2001 main 1-1-05 12-31-05 550 O110 Tony R105 Scot P1009 1009 first 1-1-04 5-31-04 350 O109 Tony PNo, Start ---> All other attributes (Partial on Primary Key) PNo ---> PAddress, ONo, OName Not in 2NF! Poor Performance Redundancy Inconsistency What to Do? New Table! 10

11. Decompose Lease into 2NF Lease (RNo, RName, PNo, PAddress, Start, Finish, Rent, ONo, OName) PK: PNo, Start Partial FD on Primary Key PNo ---> PAddress, ONo, OName Remove partial dependent on Primary Key Create a new relation Which attributes are in the new relation? All attributes in the partial FD (both side of) Lease1 (PNo, PAddress, ONo, OName) Which attributes are left in the original relation? PK remains the same Lease2 (RNo, RName, PNo, Start, Finish, Rent) PK and FK? 11

12. Lease1 (PNo, PAddress, ONo, OName) PK: PNo AK: PAddress FK: None (from this conversion Functional Dependencies: PNo ---> All other attributes PAddress ---> PNo, ONo, OName ONo ---> OName In 2NF? Lease2 (RNo, RName, PNo, Start, Finish, Rent) PK: PNo, Start AK: PNo, Finish FK: PNo references Lease1 Functional Dependencies: PNo, Start ---> All other attributes PNo, Finish ---> All other attributes RNo ---> Rname In 2NF? 12

13. Relation/Table Instances Lease Rno RName PNo PAddress Start Finish Rent Ono OName R101 John P1001 1001 main 1-1-04 12-31-04 350 O100 Tina R102 Mike P1001 1001 main 1-1-05 2-28-05 380 O100 Tina R103 Kay P1001 1001 main 3-1-05 12-31-05 380 O100 Tina R101 John P1002 2001 main 1-1-05 12-31-05 550 O110 Tony R105 Scot P1009 1009 first 1-1-04 5-31-04 350 O109 Tony Lease1 PNo PAddress ONo OName P1001 1001 main O100 Tina P1002 2001 main O110 Tony P1009 1009 first O109 Tony Lease1 =  PNo, PAddress, Ono, OName (Lease) 13

14. Removing Duplicates Lease1 PNo PAddress ONo OName P1001 1001 main O100 Tina P1002 2001 main O110 Tony P1009 1009 first O109 Tony Lease1 P1001 1001 main O100 Tina P1002 2001 main O110 Tony P1009 1009 first O109 Tony 14

15. Lease RNo RName PNo PAddress Start Finish Rent Ono OName R101 John P1001 1001 main 1-1-04 12-31-04 350 O100 Tina R102 Mike P1001 1001 main 1-1-05 2-28-05 380 O100 Tina R103 Kay P1001 1001 main 3-1-05 12-31-05 380 O100 Tina R101 John P1002 2001 main 1-1-05 12-31-05 550 O110 Tony R105 Scot P1009 1009 first 1-1-04 5-31-04 350 O109 Tony Lease2 RNo RName PNo Start Finish Rent R101 John P1001 1-1-04 12-31-04 350 R102 Mike P1001 1-1-05 2-28-05 380 R103 Kay P1001 3-1-05 12-31-05 380 R101 John P1002 1-1-05 12-31-05 550 R105 Scot P1009 1-1-04 5-31-04 350 Lease2 =  RNo, RName, RNo, Start, Finish, Rent (Lease) 15

16. Some FDs are lost Lease (RNo, RName, PNo, PAddress, Start, Finish, Rent, ONo, OName) PAddress, Start ---> All other attributes PAddress, Finish ---> All other attributes Lease1 (PNo, PAddress, ONo, OName) Lease2 (RNo, RName, PNo, Start, Finish, Rent) 16

17. Example Relation R (A, B, C, D, E, F) PK: A, B AK: None FK: None Functional Dependencies: A, B ---> All A ---> C B ---> E, F In 2NF? NO! 17

18. Table Instance A B C D E F 1 x 10 100 cs se 1 y 10 200 cs ee 2 x 20 300 cs se 2 y 20 400 cs ee Functional Dependencies: A, B ---> All A ---> C B ---> E, F 18

19. R1 (A, C) Primary Key: A Alternate Key: None Foreign Key: None Functional Dependencies: A ---> C R2 (B, E, F) Primary Key: B Alternate Key: None Foreign Key: None Functional Dependencies: B ---> E, F R3 (A, B, D) Primary Key: A, B Alternate Key: None Foreign Key: A References R1 B References R2 Functional Dependencies: A, B ---> D 19 R (A, B, C, D, E, F) Functional Dependencies: A, B ---> All A ---> C B ---> E, F Decompose table R into 2NF

20. Table Instance R A B C D E F 1 x 10 100 cs se 1 y 10 200 cs ee 2 x 20 300 cs se 2 y 20 400 cs ee R1 A C 1 10 2 20 R2 B E F x cs se y cs ee R3 A B D 1 x 100 1 y 200 2 x 300 2 y 400 20

21. Schedule Assignment 5-1 Due Wednesday Assignment 6-1 Due Monday, March 2 Quiz 2 Friday, March 6 21

22. Assignment 4 DBDL Style Will Lose Points on Quiz/Test 22