Lecture 09: Functional Dependencies

Outline

• Functional dependencies (3.4)• Rules about FDs (3.5)• Design of a Relational schema (3.6)

Relational Schema Design

Conceptual Model:

Relational Model:plus FD’s

Normalization:Eliminates anomalies

PersonbuysProduct

name

pricename ssndate

Functional Dependencies

Definition: A1, ..., Am → B1, ..., Bn holds in R if:

t, t’ R, (t.A1=t’.A1 ... t.Am=t’.Am t.B1=t’.B1 ... t.Bm=t’.Bm )

A1 ... Am B1 ... Bm

if t, t’ agree here then t, t’ agree here

t

t’

R

Important Point!

• Functional dependencies are part of the schema!

• They constrain the possible legal data instances.

• At any point in time, the actual database may satisfy additional FD’s.

Examples

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

EmpID Name Phone PositionE0045 John Smith 2376 HRE1847 Zheng Li 2376 QAE3123 Lucy Larsen 1264 DeveloperE9923 Zheng Li 1264 Developer

Formal definition of a key

• A key is a set of attributes A1, ..., An s.t. for any other attribute B, A1, ..., An → B

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

• Note: book calls them superkey and key

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

Keys are: [in class]

Inference Rules for FD’s

A1 , A2 , … An→B1, B2, … Bm

Is equivalent to

A1 , A2 , … An →B1

A1 , A2 , … An →B2

A1 , A2 , … An →Bm

…

Splitting rule and Combing rule

A1 ... Am B1 ... Bm

AND

Inference Rules for FD’s(continued)

Trivial Rule

Why ?A1 ... Am

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

A1 , A2 , … An → Ai

Inference Rules for FD’s(continued)

Transitive Closure Rule

If

and

then

Why?

A1 , A2 , … An → B1, B2, … Bm

B1, B2, … Bm → C1 , C2 , … Ck

A1 , A2 , … An → C1 , C2 , … Ck

A1 ... Am B1 ... Bm C1 ... Ck

• Enrollment(student, major, course, room, time)student → majormajor, course → roomcourse → time

What else can we infer ? [in class]

Closure of a set of Attributes

Given a set of attributes {A1, …, An} and a set of dependencies S.Problem: find all attributes B such that:

any relation which satisfies S also satisfies: A1 , A2 , … An → B

The closure of {A1, …, An}, denoted {A1, …, An}+,is the set of all such attributes B

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

Until X doesn’t change do:

if B1, B2, … Bm → C is in S, and

B1, B2, … Bm are all in X and

C is not in X

thenadd C to X

ExampleThe schema - R(A,B,C,D,E,F)

A, B → CA,D → EB → DA,F → B

Closure of {A,B}: X = {A, B, }

Closure of {A, F}: X = {A, F, }

Why Is the Algorithm Correct ?• Show the following by induction:

– For every B in X:• A1 , A2 , … An → B

• Initially X = {A1 , A2 , … An } holds• Induction step: B1, …, Bm in X

– Implies A1 , A2 , … An → B1, B2, … Bm

– We also have B1, B2, … Bm → C– By transitivity we have A1 , A2 , … An → C

• This shows that the algorithm is sound; need to show it is complete

Relational Schema Design(or Logical Design)

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

schema


Anomalies:• Redundancy = repeated data• Update anomalies = Fred moves to “Bellevue”• Deletion anomalies = Fred drops all phone numbers:

what is his city ?

Recall set attributes (persons with several phones):

SSN → Name, City, but not SSN → PhoneNumber

Name SSN PhoneNumber CityFred 123-45-6789 206-555-1234 SeattleFred 123-45-6789 206-555-6543 Seattle

Joe 987-65-4321 908-555-2121 WestfieldJoe 987-65-4321 908-555-1234 Westfield

Relation DecompositionBreak the relation into two:

Name SSN CityFred 123-45-6789 SeattleJoe 987-65-4321 Westfield

SSN PhoneNumber123-45-6789 206-555-1234123-45-6789 206-555-6543

987-65-4321 908-555-2121987-65-4321 908-555-1234


Conceptual Model:

Relational Model:plus FD’s

Normalization:Eliminates anomalies

PersonbuysProduct

name

pricename ssndate

Decompositions in GeneralR(A1, ..., An)

Create two relations R1(B1, ..., Bm) and R2(C1, ..., Ck)

such that: B1, ..., Bm C1, ..., Ck = A1, ..., An

and:R1 = projection of R on B1, ..., Bm

R2 = projection of R on C1, ..., Ck

Incorrect Decomposition

• Sometimes it is incorrect:

Name Price Category

Gizmo 19.99 Gadget

OneClick 24.99 Camera

DoubleClick 29.99 Camera

Decompose on : Name, Category and Price, Category

Incorrect Decomposition

Name Category

Gizmo Gadget

OneClick Camera

DoubleClick Camera

Price Category

19.99 Gadget

24.99 Camera

29.99 Camera

Name Price Category

Gizmo 19.99 Gadget





When we put it back:

Cannot recover information

Normal Forms

First 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...

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

In English:

Whenever a set of attributes of R determines one more attribute, it should determine all the attributes of R.

A relation R is in BCNF if:

Whenever there is a nontrivial dependency A1, ..., An → B

in R , {A1, ..., An} is a key for R

Including superkeys

Example

What are the dependencies?SSN → Name, City

What are the keys?{SSN, PhoneNumber}

Is it in BCNF?

Name SSN PhoneNumber CityFred 123-45-6789 206-555-1234 SeattleFred 123-45-6789 206-555-6543 Seattle

Joe 987-65-4321 908-555-2121 WestfieldJoe 987-65-4321 908-555-1234 Westfield

Decompose it into BCNF

Name SSN CityFred 123-45-6789 SeattleJoe 987-65-4321 Westfield

SSN PhoneNumber123-45-6789 206-555-1234123-45-6789 206-555-6543987-65-4321 908-555-2121987-65-4321 908-555-1234

SSN → Name, City

Summary of BCNF Decomposition

Find a dependency that violates the BCNF condition:

A1 , A2 , … An → B1, B2, … Bm

A’sOthers B’s

R1 R2

Heuristics: choose B1, B2, … Bm “as large as possible”

Decompose:

Exercise: Is there a 2-attribute relation that is not in BCNF ?

Continue untilthere are noBCNF violationsleft.

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

SSN → name, ageage → hairColor

Decompose in BCNF (in class):

Step 1: find all keys

Step 2: now decompose

Other Example

• R(A,B,C,D) A → B, B → C• Key: A, D• Violations of BCNF: A → B, B → C, A → C,

A → B,C• Pick A → B,C:

split R into R1(A,B,C), R2(A,D)• What happens if we pick A → B first ?

Correct 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

Correct Decompositions

• Given R(A,B,C) s.t. A→B, the decomposition into R1(A,B), R2(A,C) is lossless

3NF: A Problem with BCNF

FD’s: Character Film; Film, Actor CharacterSo, there is a BCNF violation, and we decompose.

Character Film

No FDs

Film Actor Character

Actor Character

Film Character

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:

Violates the dependency: Film, Actor Character!

Film CharacterAustin Powers Austin PowersAustin Powers Dr. Evil

Actor CharacterMike Myers Austin PowersMike Myers Dr. Evil

Film Actor CharacterAustin Powers Mike Myers Austin PowersAustin Powers Mike Myers Dr. Evil

What happened

here?

Solution: 3rd Normal Form (3NF)

A simple condition for removing anomalies from relations:

3NF has a dependency-preserving decomposition

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 key for R, or B is part of a key.

Including superkeys

Decomposition into 3NF

• Easy when we know BCNF decomposition (how?)

• Not-so-easy: if we want a dependency-preserving decomposition… In the book (3.5)

Lecture 09: Functional Dependencies

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