Fundamentals of

Database SystemsFourth Edition

El Masri & Navathe

Instructor: Mr. Ahmed Al Astal

Chapter 10

Functional Dependencies and Normalization for Relational

Databases

University Of Palestine

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Normalization

Dependency Preserving:-Suppose that the relation R is decomposed into

R1,R2, and the set of FDs F that holds on R ar distributed on R1,R2 then:

We can say that the decomposition is dependency preserving iff (F1UF2) ≡ F

Otherwise the decomposition is not Dependency Preserving.

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Normalization (Cont.)

Lossless-join decomposition:-Suppose that the relation R is decomposed into

R1,R2, and the set of FDs F that holds on R ar distributed on R1,R2, If the common attribute between R1 and R2 as a key in either R1 or R2 then:

We can say that the decomposition is Lossless-JoinOtherwise the decomposition is Lossy Join

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Normalization (Cont.)Example

Suppose that we have the following relation std_info:

and the following set of FDs that hold on sd_infoF={Sname Major,Phone,

Cname Instructor,

Sname, Cname Score}Then The Decomposition :

F1={ Sname Major, Department} F2={Cname instructor Sname, Cname Score}

This Decomposition is Dependency Preserving and Lossless Join

SnameCnameMajorPhoneInstructorScore

SnamMajorPhone SnameCnameInstructorScore

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Normalization (Cont.)

Normalization:The process of decomposing unsatisfactory "bad"

relations by breaking up their attributes into smaller relations

Normal form:Condition using keys and FDs of a relation to certify

whether a relation schema is in a particular normal form

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Practical Use of Normal Forms

Normalization is carried out in practice so that the resulting designs are of high quality and meet the desirable properties The practical utility of these normal forms becomes questionable when the constraints on which they are based are hard to understand or to detectThe database designers need not normalize to the highest possible normal form

• (usually up to 3NF, BCNF or 4NF)

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Definitions of Keys and Attributes Participating in Keys (1)

A superkey of a relation schema R = {A1, A2, ...., An} is a set of attributes S subset-of R with the property that no two tuples t1 and t2 in any legal relation state r of R will have t1[S] = t2[S]

A key K is a superkey with the additional property that removal of any attribute from K will cause K not to be a superkey any more.

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Definitions of Keys and Attributes Participating in Keys (1)

If a relation schema has more than one key, each is called a candidate key.One of the candidate keys is arbitrarily designated to be the primary key, and the others are called secondary keys.A Prime attribute must be a member of some candidate keyA Nonprime attribute is not a prime attribute—that is, it is not a member of any candidate key.

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First Normal Form •If a relation schema has more than one key, each is

called a candidate key.One of the candidate keys is arbitrarily designated to be the primary key, and the others are called secondary keys.A Prime attribute must be a member of some candidate keyA Nonprime attribute is not a prime attribute—that is, it is not a member of any candidate key.

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First Normal FormDisallows

• composite attributes• multivalued attributes• nested relations; attributes whose values for an

individual tuple are non-atomic

Considered to be part of the definition of relation

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BCNF (Boyce-Codd Normal Form) The strongest Normal Form for the relationGiven arelation R and aset of FDs F hold on R, R is in BCNF (i.e. R is a good relation) iff for each FD in the form α -> β in F : α must be a super key.Each normal form is strictly stronger than the previous one:

• Every 2NF relation is in 1NF• Every 3NF relation is in 2NF• Every BCNF relation is in 3NF

There exist relations that are in 3NF but not in BCNFThe goal is to have each relation in BCNF (or 3NF)

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BCNF (Boyce-Codd Normal Form)

SnameCnamemajorphoneinstructorscore

stdinfo

F={ sname major,phone cname instructor sname,cname score }

Each FD on R must satisfy the condition to be in BCNF, otherwise we must divide the relation R into tow relation R1,R2 as follows:R1(α, β) , R2(R- β) that satisfying the tow conditions

The decomposition is dependency preservingThe decomposition is loosless-join

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BCNF (Boyce-Codd Normal Form) 1.Stdinfo relation is not in BCNF so we must devide the relation as follows:

R1(sname, major, phone), R2(cname, sname,instructor,score)

2.The second step is ditrbuting the FDs to R1,R2F1={snamemajor,phone}, F2={cnameinstructor cname,snamescore}

3.The third step we check the decomposition for dependency preserving and lossless-join conditions.

Snamemajorphone SnameCnameinstructorscore

R1 R2

F1 F2

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BCNF (Boyce-Codd Normal Form) This decomposition is dependency preserving because (F1UF2)=FD and loosless-join because the common attribute which is Sname is a key for R1.

The forth step is repeating checking for BCNF for the resulting relation R1,R2.

• For F1 On R1, α which is Sname is akey for R1, So R1 Is in BCNF

Snamemajorphone

R1

F1

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BCNF (Boyce-Codd Normal Form) • For R2:

Sname, cname score

α is akey for R2

Cname instructor

α is not akey for R2 so,,

We must devide R2 into tow relations R2.1,R2.2

SnameCnameinstructorscore

R2

F2

F2={ Sname,Cname->score,

Cname->instructor}

Cnameinstructor

F2.1

R2.1SnameCnamescore

R2.2

F2.1

R2.1,R2.2 are both in BCNF Form so finally we have 3 relations which are : R1, R2.1, R2.2 which ar all in BCNF

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Summary of BCNF Checking

1.Check Each FD for the condition α is a key.2.If any FD fails then we divide the relation into tow relation R1(α , β) , R2(R- β(.3.Distribute the original FD on R1, R2.4.Check the decomposition for dependency preserving and loosless-join.5.Repeat the checking for every resulting relation till having all the relations in BCNF Form.

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BCNF: another exampleBnameBankerNameCnameOffice#R

FD={ Bname,Sname BnakerName , BankerName Bname,Office# }

FD

Now we must check each FD for the BCNF condition:

Bname,Cname BankerName : √ ( Bname,Sname is a key)

BankerName Bname,Office# : X ( BankerName is not A key)

So we must divide R into tow relations R1,R2

BnameBankerNameOffice#R1 BankerNameCnameR2

F1={ BankerName->Bname,Office# } F2={ Ø }

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BCNF: another example (Cont.)

•Hence BankerName is akey For R1 so the last decomposition is loosless-join, But it is Not dependency Preserving because F1UF2 ≠ FD.•So here we stop and don’t continue with BCNF, we try by 3NF OR 2NF.

•Notes: BCNF has avery restricted condition so not

always we have a good decomposition . BCNF sometimes looses FDs.

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Third Normal Form 3NFGiven arelation R and aset of FDs F hold on R, R is in 3NF iff for each FD in the form α -> β in F at least one of the following conditions satisfies:

1. α must be a key for R OR2. Each attribute in β is Prime attribute.

FD={ Bname,Sname BnakerName,

BankerName Bname,Office# }

BnameBankerNameCnameOffice#R

FD

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3NF (Cont.)

The last relation is not in 3NF Because for the second FD: α is not a key and Office# is not a prime attribute.

So we must divide R into two relations as follows:R1(α , all non prime attributes in β )R2(R- non prime attributes in β )

BankerNameOffice# BnameBankerNameCnameR1 R2

F1 F2

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3NF (Cont.)F1UF2 = FD So the decomposition is dependency preserving.BankerName is a key for R1 so The decomposition is loosless join

R1,R2 are in 3NF so finally we have two relations R1,R2

Important Notes:• 3NF Never ever loose FD.• If R is in BCNF then surely it will be in 3NF• Butt R may be in 3NF but Not in BCNF.

BankerNameOffice# BnameBankerNameCnameR1 R2

F1 F2

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Second Normal Form 2NF

•Given arelation R and aset of FDs F hold on R, R is in 3NF iff for each FD in the form α -> β in F at least one of the following conditions satisfies:

1. α must be a key for R OR2. Each attribute in β is Prime attribute.3. α is not a subset of any key

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2NF: Example

FD={ BC->ADE E->G } For F1 BC is a keyFor F2 E is not a subset of any key

So R is in 2NF

ABCDEFR

FD

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2NF: Other Example

ABCDEFR

FDFD={ BC->ADE

D->B

E->G }F1: BC is akey

F2: E is not a subset of any key

F3: B is prime attribute

So R is in 2NF

The decomposition when the test fails is similar to 3NF