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Lecture 10:Database Design

Wednesday, January 26, 2005

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Outline

• Design of a Relational schema (3.6)

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Decompositions in General

R1 = projection of R on A1, ..., An, B1, ..., Bm

R2 = projection of R on A1, ..., An, C1, ..., Cp

R(A1, ..., An, B1, ..., Bm, C1, ..., Cp) R(A1, ..., An, B1, ..., Bm, C1, ..., Cp)

R1(A1, ..., An, B1, ..., Bm)R1(A1, ..., An, B1, ..., Bm) R2(A1, ..., An, C1, ..., Cp)R2(A1, ..., An, C1, ..., Cp)

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Decomposition

• Sometimes it is correct:Name Price Category

Gizmo 19.99 Gadget

OneClick 24.99 Camera

Gizmo 19.99 Camera

Name Price

Gizmo 19.99

OneClick 24.99

Gizmo 19.99

Name Category

Gizmo Gadget

OneClick Camera

Gizmo Camera

Lossless decomposition

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Incorrect Decomposition

• Sometimes it is not:

Name Price Category

Gizmo 19.99 Gadget

OneClick 24.99 Camera

Gizmo 19.99 Camera

Name Category

Gizmo Gadget

OneClick Camera

Gizmo Camera

Price Category

19.99 Gadget

24.99 Camera

19.99 Camera

What’sincorrect ??

Lossy decomposition

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Decompositions in General

R(A1, ..., An, B1, ..., Bm, C1, ..., Cp) R(A1, ..., An, B1, ..., Bm, C1, ..., Cp)

If A1, ..., An B1, ..., Bm

Then the decomposition is lossless

R1(A1, ..., An, B1, ..., Bm)R1(A1, ..., An, B1, ..., Bm) R2(A1, ..., An, C1, ..., Cp)R2(A1, ..., An, C1, ..., Cp)

Example: name price, hence the first decomposition is lossless

Note: don’t need A1, ..., An C1, ..., Cp

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Normal Forms

First Normal Form = all attributes are atomic

Second Normal Form (2NF) = old and obsolete

Third Normal Form (3NF) = will discuss

Boyce Codd Normal Form (BCNF) = will discuss

Others...

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Boyce-Codd Normal Form

A simple condition for removing anomalies from relations:

In English:

Whenever a set of attributes of Ris determining another attribute, should determine all the attributes of R.

A relation R is in BCNF if:

If A1, ..., An B is a non-trivial dependency

in R , then {A1, ..., An} is a superkey for R

A relation R is in BCNF if:

If A1, ..., An B is a non-trivial dependency

in R , then {A1, ..., An} is a superkey for R

Yet another way:

for any set X, either X+ = X, or X+ = all attributes

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BCNF Decomposition Algorithm

A’s OthersB’s

R1

Is there a 2-attribute relation that isnot in BCNF ?

Repeat choose A1, …, Am B1, …, Bn that violates the BNCF condition split R into R1(A1, …, Am, B1, …, Bn) and R2(A1, …, Am, [others]) continue with both R1 and R2

Until no more violations

Repeat choose A1, …, Am B1, …, Bn that violates the BNCF condition split R into R1(A1, …, Am, B1, …, Bn) and R2(A1, …, Am, [others]) continue with both R1 and R2

Until no more violations

R2

In practice, we havea better algorithm (next)

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Example

What are the dependencies?SSN Name, City

What are the keys?{SSN, PhoneNumber}

Is it in BCNF?

Name SSN PhoneNumber City

Fred 123-45-6789 206-555-1234 Seattle

Fred 123-45-6789 206-555-6543 Seattle

Joe 987-65-4321 908-555-2121 Westfield

Joe 987-65-4321 908-555-1234 Westfield

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Decompose it into BCNF

Name SSN City

Fred 123-45-6789 Seattle

Joe 987-65-4321 Westfield

SSN PhoneNumber

123-45-6789 206-555-1234

123-45-6789 206-555-6543

987-65-4321 908-555-2121

987-65-4321 908-555-1234

SSN Name, City

Let’s check anomalies:• Redundancy ?• Update ?• Delete ?

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Example Decomposition Person(name, SSN, age, hairColor, phoneNumber)

SSN name, ageage hairColor

Decompose in BCNF (in class):

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BCNF Decomposition Algorithm

BCNF_Decompose(R)

find X s.t.: X ≠X+ ≠ [all attributes]

if (not found) then “R is in BCNF”

let Y = X+ - X let Z = [all attributes] - X+ decompose R into R1(X Y) and R2(X Z) continue to decompose recursively R1 and R2

BCNF_Decompose(R)

find X s.t.: X ≠X+ ≠ [all attributes]

if (not found) then “R is in BCNF”

let Y = X+ - X let Z = [all attributes] - X+ decompose R into R1(X Y) and R2(X Z) continue to decompose recursively R1 and R2

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Example BCNF DecompositionPerson(name, SSN, age, hairColor, phoneNumber)

SSN name, ageage hairColor

Iteration 1: PersonSSN+ = SSN, name, age, hairColorDecompose into: P(SSN, name, age, hairColor) Phone(SSN, phoneNumber)

Iteration 2: Page+ = age, hairColorDecompose: People(SSN, name, age) Hair(age, hairColor) Phone(SSN, phoneNumber)

Iteration 1: PersonSSN+ = SSN, name, age, hairColorDecompose into: P(SSN, name, age, hairColor) Phone(SSN, phoneNumber)

Iteration 2: Page+ = age, hairColorDecompose: People(SSN, name, age) Hair(age, hairColor) Phone(SSN, phoneNumber)

Find X s.t.: X ≠X+ ≠ [all attributes]

What isthe key ?

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Example

• R(A,B,C,D) A B, B C

• In R: A+ = ABC ≠ ABCD– Split into R1(A,B,C), R2(A,D)

• In R1: B+ = BC ≠ ABC– Split into R11(B,C), R12(A,B)

• Final decomposition: R11(B,C), R12(A,B), R2(A,D)

• What happens if in R we first pick B+ ? Or AB+ ?

What arethe keys ?

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3NF: A Problem with BCNF

Unit CompanyCompany, Product Unit

Unit CompanyCompany, Product Unit

Unit+ = Unit, Company

We loose the FD: Company, Product Unit !!

Unit Company Product

Unit Company Unit Product

Unit CompanyUnit Company

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So What’s the Problem?

No problem so far. All local FD’s are satisfied.Let’s put all the data back into a single table again:

Unit Company

Galaga99 UW

Bingo UW

Unit Product

Galaga99 Databases

Bingo Databases

Unit Company Product

Galaga99 UW Databases

Bingo UW Databases

Unit CompanyUnit Company

Company, Product UnitCompany, Product UnitViolates the FD:

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The Problem

• We started with a table R and FD

• We decomposed R into BCNF tables R1, R2, …with their own FD1, FD2, …

• We can reconstruct R from R1, R2, …

• But we cannot reconstruct FD from FD1, FD2, …

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Solution: 3rd Normal Form (3NF)

A simple condition for removing anomalies from relations:

A relation R is in 3rd normal form if :

Whenever there is a nontrivial dependency A1, A2, ..., An Bfor R , then {A1, A2, ..., An } a super-key for R, or B is part of a key.

A relation R is in 3rd normal form if :

Whenever there is a nontrivial dependency A1, A2, ..., An Bfor R , then {A1, A2, ..., An } a super-key for R, or B is part of a key.

Tradeoff:BCNF = no anomalies, but may lose some FDs3NF = keeps all FDs, but may have some anomalies

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3NF Decomposition Algorithm3NF_Decompose(R) let K = [all attributes that are part of some key]

find X s.t.: X+ - X - K ≠ and X+ ≠ [all attributes] if (not found) then “R is already in 3NF” let Y = X+ - X - K let Z = [all attributes] - (X Y) decompose into R1(X Y) and R2(X Z) decompose, recursively, R1 and R2

3NF_Decompose(R) let K = [all attributes that are part of some key]

find X s.t.: X+ - X - K ≠ and X+ ≠ [all attributes] if (not found) then “R is already in 3NF” let Y = X+ - X - K let Z = [all attributes] - (X Y) decompose into R1(X Y) and R2(X Z) decompose, recursively, R1 and R2

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Example of 3NF decompositionR(A,B,C,D,E):R(A,B,C,D,E):

AB CC DD BD E

AB CC DD BD E

Keys: (need to compute X+, for several Xs) AB, AC, AD

K = {A, B, C, D}

Pick X = CC+ = BCDEC BDE is a BCNF violationFor 3NF: remove B, D (part of K):C E is a 3NF violationDecompose: R1(C, E), R2(A,B,C,D)

R1 is in 3NFR2 is in 3NF (because its keys: AB, AC, AD)

22BCNF

3NF v.s. BCNF DecompositionA B C D E F G H K

A B C D E E F G H K

E F G G H KA B C C D E

A B A B A B A B A B A B A BA B

3NF