Chapter 3

Topics to be discusses

• Functional Dependency• Key• Closure of Function Dependency

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Chapter 3

Functional Dependency• A functional dependency (FD) is a constraint

between two sets of attributes in a relation from a database.

• Require that the value for a certain set of attributes determines uniquely the value for another set of attributes.

• A functional dependency is a generalization of the notion of a key.– Notation:– α→β (α determines β)– (α→β may take the form AB→C, A→BC, etc.)

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Chapter 3

• Let R be a relation schema R and R

• The functional dependency

holds on R if and only if for any legal relations r(R), whenever any two tuples t1 and t2 of r agree on the attributes , they also agree on the attributes . That is,

t1[] = t2 [] t1[ ] = t2 [ ] • Example: Consider r(A,B ) with the following instance of r.

• On this instance, A B does NOT hold, but B A does hold.

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1 41 53 7

Chapter 3 4

A B C

A1 B1 C1

A1 B1 C2

A2 B1 C1

A2 B1 C3

A->BC->BAC->B

Find the functional dependency

Chapter 3

A B C D

A1 B1 C1 D1

A1 B2 C1 D2

A2 B2 C2 D2

A2 B3 C2 D3

A3 B3 C2 D4

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• A->C• AB->C• AB->D• AD->B• AD->C

Find the functional dependency

Chapter 3

Example

Drinkers(name, addr, beersLiked, manf, favBeer)• Reasonable FD’s to assert:

1. name -> addr2. name -> favBeer3. beersLiked -> manf

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Chapter 3

Example Data

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name addr beersLiked manf favBeerJaneway Voyager Bud A.B. WickedAleJaneway Voyager WickedAle Pete’s WickedAleSpock Enterprise Bud A.B. Bud

Because name -> addr Because name -> favBeer

Because beersLiked -> manf

Chapter 3

FD’s With Multiple Attributes

• No need for FD’s with > 1 attribute on right.– But sometimes convenient to combine FD’s as

a shorthand.– Example: name -> addr and name ->

favBeer become name -> addr favBeer

• > 1 attribute on left may be essential.– Example: bar beer -> price

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Chapter 3

Keys of Relations

• K is a superkey for relation R if K functionally determines all of R.

• K is a Candidate key for R if K is a superkey, but no proper subset of K is a superkey.

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Chapter 3

Example

Drinkers(name, addr, beersLiked, manf,favBeer)

• {name, beersLiked} is a superkey because together these attributes determine all the other attributes.– name -> addr favBeer– beersLiked -> manf

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Chapter 3

Example, Cont.

• {name, beersLiked} is a key because neither {name} nor {beersLiked} is a superkey.– name doesn’t -> manf; beersLiked doesn’t -> addr.

• There are no other keys, but lots of superkeys.– Any superset of {name, beersLiked}.

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Chapter 3 12

A B C

A1 B1 C1

A1 B1 C2

A2 B1 C1

A2 B1 C3

A->BC->BAC->BCANDIDATE KEY AC->BPRIMARY KEY CAN BE ANY ONE

Find the super key ,candidate key and primary key

Chapter 3

• Find the super key ,candidate key and primary key• R(A,B,C,D)• A->C• AB->C• AB->D• AD->B• AD->C• Super key AB,AD,ABC,ABD,ABCD,• ABD,ACB,ABCD• Candidate key AB,AD• Primary key can be AB or AD

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Chapter 3

Example Data

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name addr beersLiked manf favBeerJaneway Voyager Bud A.B. WickedAleJaneway Voyager WickedAle Pete’s WickedAleSpock Enterprise Bud A.B. Bud

Relational key = {name beersLiked}But in E/R, name is a key for Drinkers, and beersLiked is a key

for Beers.Note: 2 tuples for Janeway entity and 2 tuples for Bud entity.

Chapter 3

CLOUSRE OF A SET OF FUNCTIONAL DEPENDENCY

• Armstrong’s Axioms:– Reflexivity rule:• if α is a set of attributes and β is contained in α then

α→β• i.e. given AB→C, then A→B

– Augmentation rule:• given α→β and another set of attributes γ, then

γα→γβ

– Transitivity rule:• if α→β and β→γ , then α→γ

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Chapter 3

• Other Rules:– Union rule:• if α→β and α→γ , then α→βγ

– Decomposition rule:• if α→βγ , then α→β and α→γ

– Pseudotransitivity rule:• if α→β and γβ→δ , then γα→δ

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Chapter 3

Examp

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ExampleExampleR=(A,B,C,G,H,I)R=(A,B,C,G,H,I)A->BA->BA->CA->CCG->HCG->HCG->ICG->IB->HB->H

Chapter 3

• Closure (F+)

• A->H Since A->B and B->H holds, we apply the transitivity rule.

• CG->HI. since cg->h and CG->I the union rule • AG->I. since A->C and CG->I pseudo transitivity

rule

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Chapter 3

• To compute the closure of a set of functional dependencies F:

F + = Frepeat

for each functional dependency f in F+

apply reflexivity and augmentation rules on f add the resulting functional dependencies to F +

for each pair of functional dependencies f1and f2 in F +

if f1 and f2 can be combined using transitivity then add the resulting functional dependency to F +

until F + does not change any further

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Chapter 3

• We can further simplify manual computation of F+ by using the following additional rules.– If holds and holds, then

holds (union)– If holds, then holds and

holds (decomposition)– If holds and holds, then

holds (pseudotransitivity)

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Chapter 3

Closure of Attribute Sets• To test a set is super key or not• Given a set of attributes define the closure of under F (denoted by +)

as the set of attributes that are functionally determined by under F

• Algorithm to compute +, the closure of under F

result := ;while (changes to result) do

for each in F dobegin

if result then result := result end

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Chapter 3

Example of Attribute Set Closure• R = (A, B, C, G, H, I)• F = {A B

A C CG HCG IB H}

• (AG)+

1. result = AG2. result = ABCG (A C and A B)3. result = ABCGH (CG H and CG AGBC)4. result = ABCGHI (CG I and CG AGBCH)

• Is AG a candidate key? 1. Is AG a super key?

1. Does AG R? == Is (AG)+ R2. Is any subset of AG a superkey?

1. Does A R? == Is (A)+ R2. Does G R? == Is (G)+ R

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Chapter 3

Uses of Attribute ClosureThere are several uses of the attribute closure algorithm:• Testing for superkey:

– To test if is a superkey, we compute +, and check if + contains all attributes of R.• Testing functional dependencies

– To check if a functional dependency holds (or, in other words, is in F+), just check if +.

– That is, we compute + by using attribute closure, and then check if it contains . – Is a simple and cheap test, and very useful

• Computing closure of F– For each R, we find the closure +, and for each S +, we output a functional

dependency S.

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Chapter 3 24

Excercises

Chapter 3

StudentID Semester Lecture TA

1234 6 Numerical Methods John

1200 4 Numerical Methods Peter1234 6 Visual Computing Thomas

1201 4 Numerical Methods Peter

1201 4 Physics II Simone

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Excercise

Chapter 3

Solution

• StudentID → Semester. • {StudentID, Lecture} → TA • {StudentID, Lecture} → {TA, Semester}

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Chapter 3

Find the super key of given function dependencyR(A,B,C,D)

• A->D• D->B• B->C• E->B

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AE is super key

Chapter 3

Find the super key of given functional dependencyR(A,B,C,D)AB->CC->BC->D

A+ =AAB+ = ABCDAC+=ABCD

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L M R

A B,C D