BCNF and Normalization - Computer Science atcourses.cs.vt.edu/~cs4604/Fall08/lectures/lecture14.pdf · BCNF and Normalization Zaki Malik October 21, 2008. Relational Schema Design

BCNF and NormalizationBCNF and Normalization

Zaki Malik

October 21, 2008October 21, 2008

Relational Schema DesignRelational Schema Design

• Goal of relational schema design is to avoidGoal of relational schema design is to avoid redundancy and anomalies.

Bad DesignBad Design

name addr beersLiked manf favBeername addr beersLiked manf favBeer

Janeway Voyager Export Molson G.I. LagerJaneway Voyager G.I. Lager Gr. Is. G.I. LagerSpock Enterprise E port Molson E portSpock Enterprise Export Molson Export

•Redundancyy

• Update anomaly• if Janeway is transferred to Intrepid, will we remember to change each of her tuples?of her tuples?

• Deletion anomaly•If nobody likes Export, we lose track of the fact that Molson manufactures Export.

Another ExampleAnother Example

Relational DecompositionRelational Decomposition

Example of DecompositionExample of Decomposition

Triviality of FDs

7

Boyce‐Codd Normal FormBoyce Codd Normal Form

Closures of FDs vs. Closures of AttributesClosures of FDs vs. Closures of Attributes

Checking for BCNF Violations

Decomposition into BCNF

Decomposing CoursesDecomposing Courses

Decomposing Courses

Another Example of DecompositionAnother Example of Decomposition

Another Example of Decomposition (2)

BCNFs and Two‐Attribute Relationships

Decomposition into BCNFDecomposition into BCNF

Candidate Normalization AlgorithmCandidate Normalization Algorithm

Joining RelationsJoining Relations

Recovering Information from a Decomposition

Correct Decompositions

A decomposition is lossless if we can recover:p

R(A,B,C)

Decompose

R1(A,B) R2(A,C)

Decompose

R1(A,B)      R2(A,C)

Recover

R’(A,B,C)   should be the same asR(A,B,C)R(A,B,C)

R’ is in general larger than R. Must ensure R’ = R

Example of Lossy‐Join Decomposition 

• Example: Decomposition of R = (A, B)R1 = (A) R2 = (B)

A B A B

ααβ

121

αβ

12

∏ ( ) ∏β

r∏A(r) ∏B(r)

∏A (r) ∏B (r) A B

ααββ

1212β 2

Example: BCNF DecompositionExample: BCNF Decomposition

Drinkers(name addr beersLiked manf favBeer)Drinkers(name, addr, beersLiked, manf, favBeer)FDs = name‐>addr,   name ‐> favBeer,   beersLiked‐>manf

• Pick BCNF violation name >addr• Pick BCNF violation name‐>addr.• Close the left side: {name}+ = {name, addr, 

favBeer}favBeer}.• Decomposed relations:

1 i k 1( dd f )1. Drinkers1(name, addr, favBeer)2. Drinkers2(name, beersLiked, manf)

Example ‐‐ ContinuedExample  Continued

• We are not done; we need to check Drinkers1 and Drinkers2 for BCNF.

• Is Drinkers1 in BCNF ?– For Drinkers1(name, addr, favBeer), relevant FD’s are name‐>addr and   name‐>favBeer.

– Thus, {name} is the only key and Drinkers1 is in BCNFBCNF.

Example ‐‐ ContinuedExample  Continued

• For Drinkers2(name beersLiked manf) the• For Drinkers2(name, beersLiked, manf), the only FD is   beersLiked‐>manf, and the only key iskey is 

{name, beersLiked}.Violation of BCNF ?– Violation of BCNF ? 

• beersLiked+ = {beersLiked, manf}, so we decompose Drinkers2 into:decompose Drinkers2 into:1. Drinkers3(beersLiked, manf)2 Drinkers4(name beersLiked)2. Drinkers4(name, beersLiked)

Example ‐‐ ConcludedExample  Concluded

• The resulting decomposition of Drinkers :The resulting decomposition of Drinkers :1. Drinkers1(name, addr, favBeer)

2 Drinkers3(beersLiked manf)2. Drinkers3(beersLiked, manf)

3. Drinkers4(name, beersLiked)

• Note: D i k 1 t ll b t d i k– Drinkers1 tells us about drinkers,

– Drinkers3 tells us about beers, and 

– Drinkers4 tells us the relationship between drinkers andDrinkers4 tells us the relationship between drinkers and the beers they like.

Summary of BCNF DecompositionSummary of BCNF Decomposition

Find a dependency that violates the BCNF condition:

A , A , … A 1 2 n B , B , … B

1 2 m

Decompose:

A’sOthers B’sIs there a 2 attribute

Continue untilthere are noBCNF violationsleft

R1 R2

2-attribute relation that isnot in BCNF ?

left.