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.

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Third Normal Form (3NF)

Relation R in 2NF, and No non-Primary-Key attribute is transitively

functionally dependent on the primary keyThen R is in 3NF.

No Transitive FDs on PK.

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Assignment 5-2

Due Wednesday2NF, then 3NF

(No BCNF)

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Boyce-Codd Normal Form (BCNF)

Definition R in 1NF andThe determinant of each FD is a candidate key.

Review: 1NF determinant candidate key

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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.

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

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ExampleLease (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?

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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)

Owner (ONo, OName) ONo ---> Oname

Renter (RNo, RName) RNo ---> RName

Lease (RNo, PNo, Start, Finish, Rent) PNo, Start ---> All PNo, Finish ---> All

Property (PNo, PAddress, ONo) PNo ---> PAddress, ONo PAddress ---> PNo, Ono

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Only 4 tables, not 5.Pno PaddressPaddress Pno

Ono will not be in Lease.

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

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Table Instance

A B C D E F 2 10 x u ct 1 1 20 y v cis 2 2 10 z u se 3 1 20 x v cs 4

FDs: A, B, C All B, C, D All B, D A

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Decomposing to BCNFR (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 PKB and D should remain in the original table as FKA should not remain in the original tablePK of the original table must be changed to B, C, D.

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Decomposing to BCNFR (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

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R2 (B, C, D, E, F) PK: B, C, D AK: NONE FK: B, D References R1 FDs: B, C, D AllDoes R2 have a FK?

R1 (A, B, D) PK: B, D AK: NONE FK: None FDs: B, D ADoes R1 have a FK?

Table Instance A B C D E F 2 10 x u ct 1 1 20 y v cis 2 2 10 z u se 3 1 20 x v cs 4

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A B D 2 10 u 1 20 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

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

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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 AA is Partial on PK!

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 Why: remove redundant data • BCNF Stronger than 3NF Any candidate keys Why: PK is a good choice

In most cases, BCNF is enough.

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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

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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

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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

De-Normalization

• Normalized relations Minimal redundancy Need join operation to get results • How far should we go? • Where to stop?

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Review: Database DesignA structured approach that uses procedures, techniques, tools, and

documentation aids to support and facilitate the process of design.

Three main phases 1. Conceptual database design Understanding client data E-R (EER) Model Contract between clients and designers

2. Logical database design Mapping E-R Model to (relational) database schema (Derive relational schema from E-R Model) DBDL Normalization

3. Physical database design

Schedule

• WednesdayAssignment5-2

• FridayNo Class

• Monday, March 11Quiz2

• Wednesday, March 20Test1

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Quiz 2 (25 points)

• Monday, March 11• Lab 206• Using Computers and Drop Files• Derive table schemas from E-R Model (Mapping E-R Model to Database Schema) • DBDL• Functional Dependency• Note11 – Note15• Assignment 4, 5-1 and 6-1

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