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The Relational Data Model

Lecture 6

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Outline

• Relational Data Model• Functional Dependencies• Logical Schema Design

ReadingChapter 8

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The Relational Data Model

DataModeling

Relational Schema

Physicalstorage

E/R diagrams Tables: column names: attributes rows: tuples

Complexfile organizationand index structures.

Have seenthis in SQL

Have seenthis too

Discuss next

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Terminology

Name Price Category Manufacturer

gizmo $19.99 gadgets GizmoWorks

Power gizmo $29.99 gadgets GizmoWorks

SingleTouch $149.99 photography Canon

MultiTouch $203.99 household Hitachi

Tuples or rows or records

Attribute namesTable name or relation name

Products:

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SchemasRelational Schema:

– Relation name plus attribute names– E.g. Product(Name, Price, Category, Manufacturer)– In practice we add the domain for each attribute

Database Schema– Set of relational schemas– E.g. Product(Name, Price, Category, Manufacturer),

Company(Name, Address, Phone), . . . . . . .

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Instances• Relational schema = R(A1,…,Ak)

Instance = relation with k attributes (of “type” R)– values of corresponding domains

• Database schema = R1(…), R2(…), …, Rn(…)Instance = n relations, of types R1, R2, ..., Rn

This is all mathematics, not to be confused with SQL tables !(What's a difference?)

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Example

Name Price Category Manufacturer

gizmo $19.99 gadgets GizmoWorks

Power gizmo $29.99 gadgets GizmoWorks

SingleTouch $149.99 photography Canon

MultiTouch $203.99 household Hitachi

Relational schema:Product(Name, Price, Category, Manufacturer)Instance:

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First Normal Form (1NF)• A database schema is in First Normal

Form if all tables are flat

3.9Carol

3.7Bob

3.8Alice

CoursesGPAName

OS

DB

Math

OS

DB

OS

Math

Student

3.9Carol3.7Bob3.8Alice

GPAName

Student

Course

OSDBMath

OSCarolOSAliceDBBob

AliceCarolAliceStudent Course

DBMathMath

Takes Course

May needto add keys

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Functional Dependencies

• A form of constraint– hence, part of the schema

• Finding them is part of the database design• Also used in normalizing the relations

• Warning: this is the most abstract, and“hardest” part of the course.

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Functional DependenciesDefinition:

If two tuples agree on the attributes

then they must also agree on the attributes

Formally:

A1, A2, …, An B1, B2, …, Bm

A1, A2, …, An

B1, B2, …, Bm

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Examples

• EmpID Name, Phone, Position• Position Phone• but Phone Position

EmpID Name Phone PositionE0045 Smith 1234 ClerkE1847 John 9876 SalesrepE1111 Smith 9876 SalesrepE9999 Mary 1234 Lawyer

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In General• To check A B, erase all other columns

• check if the remaining relation is many-one(called functional in mathematics)

… A … B

X1 Y1

X2 Y2

… …

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Example

EmpID Name Phone PositionE0045 Smith 1234 ClerkE1847 John 9876 SalesrepE1111 Smith 9876 SalesrepE9999 Mary 1234 Lawyer

Position Phone

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Typical Examples of FDs

Product: name price, manufacturer

Person: ssn name, age

Company: name stockprice, president

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Example

Product(name, category, color, department, price)

name colorcategory departmentcolor, category price

Consider these FDs:

What do they say ?

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ExampleFD’s are constraints on relations:• On some instances they hold• On others they don’t

99ToysGreenGadgetTweaker

49ToysGreenGadgetGizmo

pricedepartmentcolorcategoryname

Does this instance satisfy all the FDs ?

name colorcategory departmentcolor, category price

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Example

59Office-supp.GreenStationaryGizmo

99ToysBlackGadgetTweaker

49ToysGreenGadgetGizmo

pricedepartmentcolorcategoryname

What about this one ?

name colorcategory departmentcolor, category price

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Example

If some FDs are satisfied, thenothers are satisfied too

If all these FDs are true:name colorcategory departmentcolor, category price

Then this FD also holds: name, category price

Why ??

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Inference Rules for FD’s

Is equivalent to

Splitting rule and Combining rule

Bm...B1Am...A1

A1, A2, …, An B1, B2, …, Bm

A1, A2, …, An B1A1, A2, …, An B2 . . . . .A1, A2, …, An Bm

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Inference Rules for FD’s(continued)

Trivial Rule

Why ?Am…A1

where i = 1, 2, ..., n

A1, A2, …, An Ai

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Inference Rules for FD’s(continued)

Transitive Closure Rule

If

and

then

Why ?

A1, A2, …, An B1, B2, …, Bm

B1, B2, …, Bm C1, C2, …, Cp

A1, A2, …, An C1, C2, …, Cp

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...C1 CpBm…B1Am…A1

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Example (continued)

Start from the following FDs:

Infer the following FDs:

1. name color2. category department3. color, category price

8. name, category price7. name, category color, category6. name, category category5. name, category color4. name, category name

Which Ruledid we apply ?Inferred FD

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Example (continued)Answers:

Transitivity on 3, 78. name, category priceSplit/combine on 5, 67. name, category color, categoryTrivial rule6. name, category categoryTransitivity on 4, 15. name, category colorTrivial rule4. name, category name

Which Ruledid we apply ?Inferred FD

1. name color2. category department3. color, category price

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Another Example• Enrollment(student, major, course, room, time)

student majormajor, course roomcourse time

What else can we infer ?

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Another Rule

If

then

Augmentation follows from trivial rules and transitivityHow ?

A1, A2, …, An B

A1, A2, …, An , C1, C2, …, Cp B

Augmentation

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Problem: infer ALL FDsGiven a set of FDs, infer all possible FDs

How to proceed ?• Try all possible FDs, apply all 3 rules

– E.g. R(A, B, C, D): how many FDs are possible ?• Drop trivial FDs, drop augmented FDs

– Still way too many• Better: use the Closure Algorithm (next)

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Closure of a set of AttributesGiven a set of attributes A1, …, An

The closure, {A1, …, An}+ , is the set of attributes Bs.t. A1, …, An B

name colorcategory departmentcolor, category price

Example:

Closures: name+ = {name, color} {name, category}+ = {name, category, color, department, price} color+ = {color}

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Closure AlgorithmStart with X={A1, …, An}.

Repeat until X doesn’t change do:

if B1, …, Bn C is a FD and B1, …, Bn are all in X then add C to X. {name, category}+ =

{name, category, color, department, price}

name colorcategory departmentcolor, category price

Example:

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Example

Compute {A,B}+ X = {A, B, }

Compute {A, F}+ X = {A, F, }

R(A,B,C,D,E,F) A, B CA, D EB DA, F B

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Using Closure to Infer ALL FDsA, B CA, D BB D

Example:

Step 1: Compute X+, for every X:

A+ = A, B+ = BD, C+ = C, D+ = DAB+ = ABCD, AC+ = AC, AD+ = ABCDABC+ = ABD+ = ACD+ = ABCD (no need to compute– why ?)BCD+ = BCD, ABCD+ = ABCD

Step 2: Enumerate all FD’s X Y, s.t. Y ⊆ X+ and X∩Y = ∅:

AB CD, ADBC, ABC D, ABD C, ACD B

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Problem: Finding FDs• Approach 1: During Database Design

– Designer derives them from real-world knowledge ofusers

– Problem: knowledge might not be available• Approach 2: From a Database Instance

– Analyze given database instance and find all FD’ssatisfied by that instance

– Useful if designers don’t get enough information fromusers

– Problem: FDs might be artifical for the given instance

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Find All FDs

020CircuitsEEFrank045DBCSEElsa050JavaCSEDan045DBCSECarol040HWEEAlice020C++CSEBob020C++CSEAlice

RoomCourseDeptStudent

Do all FDsmake sensein practice ?

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Answer

Course Dept, RoomDept, Room CourseStudent, Dept Course, RoomStudent, Course Dept, RoomStudent, Room Dept, Course

Do all FDsmake sensein practice ?

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Keys

• A key is a set of attributes A1, ..., An s.t. forany other attribute B, we have A1, ..., An B

• A minimal key is a set of attributes which isa key and for which no subset is a key

• Note: book calls them superkey and key

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Computing Keys

• Compute X+ for all sets X• If X+ = all attributes, then X is a key• List only the minimal keys

Note: there can be many minimal keys !• Example: R(A,B,C), ABC, BCA

Minimal keys: AB and BC

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Examples of Keys• Product(name, price, category, color)

name, category pricecategory color

Keys are: {name, category} and all supersets

• Enrollment(student, address, course, room, time)student addressroom, time coursestudent, course room, time

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Relational Schema Design(or Logical Schema Design)

Main idea:• Start with some relational schema• Find out its FD’s• Use them to design a better relational

schema

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Data AnomaliesWhen a database is poorly designed we get

anomalies:

Redundancy: data is repeated

Update anomalies: need to change in several places

Delete anomalies: may lose data when we don’t want

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Relational Schema Design

Anomalies:• Redundancy = repeat data• Update anomalies = Fred moves to “Bellevue”• Deletion anomalies = Joe deletes his phone number:

what is his city ?

Example: Persons with several phones

SSN Name, City

Westfield908-555-2121987-65-4321JoeSeattle206-555-6543123-45-6789FredSeattle206-555-1234123-45-6789FredCityPhoneNumberSSNName

but not SSN PhoneNumber

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Relation DecompositionBreak the relation into two:

Westfield987-65-4321JoeSeattle123-45-6789FredCitySSNName

908-555-2121987-65-4321206-555-6543123-45-6789206-555-1234123-45-6789PhoneNumberSSN

Anomalies have gone:• No more repeated data• Easy to move Fred to “Bellevue” (how ?)• Easy to delete all Joe’s phone number (how ?)

Westfield908-555-2121987-65-4321Joe

Seattle206-555-6543123-45-6789Fred

Seattle206-555-1234123-45-6789Fred

CityPhoneNumberSSNName

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Relational Schema DesignPersonbuysProduct

name

price name ssn

Conceptual Model:

Relational Model:plus FD’s

Normalization:Eliminates anomalies

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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)

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

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Decomposition

• Sometimes it is correct:

Camera19.99GizmoCamera24.99OneClickGadget19.99Gizmo

CategoryPriceName

19.99Gizmo24.99OneClick19.99Gizmo

PriceName

CameraGizmoCameraOneClickGadgetGizmo

CategoryName

Lossless decomposition

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Incorrect Decomposition

• Sometimes it is not:

Camera19.99GizmoCamera24.99OneClickGadget19.99Gizmo

CategoryPriceName

CameraGizmoCameraOneClickGadgetGizmo

CategoryName

Camera19.99Camera24.99Gadget19.99

CategoryPrice

What’sincorrect ??

Lossy decomposition

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Decompositions in GeneralR(A1, ..., An, B1, ..., Bm, C1, ..., Cp)

If A1, ..., An B1, ..., Bm Then the decomposition is lossless

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

Example: name price, hence the first decomposition is lossless

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

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Normal FormsFirst Normal Form = all attributes are atomic

Second Normal Form (2NF) = old and obsolete

Third Normal Form (3NF) = this lecture

Boyce Codd Normal Form (BCNF) = this lecture

Others...

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Boyce-Codd Normal FormA simple condition for removing anomalies from relations:

In English (though a bit vague):

Whenever a set of attributes of R is determining another attribute, it 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 key for R

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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 R2Until no more violations

R2

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Example

What are the dependencies?SSN Name, City

What are the keys?{SSN, PhoneNumber}

Is it in BCNF?

Westfield908-555-1234987-65-4321JoeWestfield908-555-2121987-65-4321JoeSeattle206-555-6543123-45-6789FredSeattle206-555-1234123-45-6789FredCityPhoneNumberSSNName

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Decompose it into BCNF

Westfield987-65-4321JoeSeattle123-45-6789FredCitySSNName

908-555-1234987-65-4321908-555-2121987-65-4321206-555-6543123-45-6789206-555-1234123-45-6789PhoneNumberSSN

SSN Name, City

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

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Summary of BCNFDecomposition

Find a dependency that violates the BCNF condition:

A’sOthers B’s

R1 R2

Heuristics: choose B , B , … B “as large as possible”1 2 m

Decompose:

2-attribute relations are BCNF

Continue untilthere are noBCNF violationsleft.

A1, A2, …, An B1, B2, …, Bm

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

SSN name, ageage hairColor

Decompose in BCNF (in class):

Step 1: find all keys (How ? Compute S+, for various sets S)

Step 2: now decompose

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Other Example

• R(A,B,C,D) A B, B C

• Key: AD• Violations of BCNF: A B, A C, ABC• Pick A BC: split into R1(A,BC) R2(A,D)• What happens if we pick A B first ?

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Lossless Decompositions A decomposition is lossless if we can recover: R(A,B,C)

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

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

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

Decompose

Recover

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Lossless Decompositions

• Given R(A,B,C) s.t. AB, thedecomposition into R1(A,B), R2(A,C) islossless

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3NF: A Problem with BCNFUnit Company Product

Unit Company

Unit Product

FD’s: Unit → Company; Company, Product → UnitSo, there is a BCNF violation, and we decompose.

Unit → Company

No FDs

Notice: we loose the FD: Company, Product Unit

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So What’s the Problem?

Unit Company Product

Unit Company Unit ProductGalaga99 UW Galaga99 databasesBingo UW Bingo databases

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

Galaga99 UW databasesBingo UW databases

Violates the dependency: company, product -> unit!

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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 } is a 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