1. Functional Dependencies and Normalization for Relational Databases
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Fundamentals of Database Systems Fourth Edition El Masri & Navathe Instructor: Mr. Ahmed Al Astal
Chapter 10
Functional Dependencies and Normalization for Relational Databases

2. Normalization Dependency Preserving:-
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Database Management CH 10
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.

3. Normalization (Cont.) Lossless-join decomposition:-
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Database Management CH 10
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-Join
Otherwise the decomposition is Lossy Join

4. Normalization (Cont.) Example
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Database Management CH 10
Normalization (Cont.)
Example
Suppose that we have the following relation std_info:
and the following set of FDs that hold on sd_info
F={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
Score
Instructor
Phone
Major
Cname
Sname
Phone
Major
Snam
Score
Instructor
Cname
Sname

5. Normalization (Cont.) Normalization:
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Database Management CH 10
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

6. Practical Use of Normal Forms
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Database Management CH 10
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 detect
The database designers need not normalize to the highest possible normal form
(usually up to 3NF, BCNF or 4NF)

7. Definitions of Keys and Attributes Participating in Keys (1)
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Database Management CH 10
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.

8. Definitions of Keys and Attributes Participating in Keys (1)
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Database Management CH 10
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 key
A Nonprime attribute is not a prime attribute—that is, it is not a member of any candidate key.

9. University Of Palestine
Database Management CH 10
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 key
A Nonprime attribute is not a prime attribute—that is, it is not a member of any candidate key.

10. First Normal Form Disallows composite attributes
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First Normal Form
Disallows
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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Database Management CH 10

12. BCNF (Boyce-Codd Normal Form)
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Database Management CH 10
BCNF (Boyce-Codd Normal Form)
The strongest Normal Form for the relation
Given 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 BCNF
The goal is to have each relation in BCNF (or 3NF)

13. BCNF (Boyce-Codd Normal Form)
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Database Management CH 10
BCNF (Boyce-Codd Normal Form)
stdinfo
score
instructor
phone
major
Cname
Sname
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 preserving
The decomposition is loosless-join

14. BCNF (Boyce-Codd Normal Form)
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BCNF (Boyce-Codd Normal Form)
Stdinfo relation is not in BCNF so we must devide the relation as follows:
R1(sname, major, phone), R2(cname, sname,instructor,score)
The second step is ditrbuting the FDs to R1,R2
F1={snamemajor,phone}, F2={cnameinstructor
cname,snamescore}
The third step we check the decomposition for dependency preserving and lossless-join conditions. R1 R2
phone
major
Sname
score
instructor
Cname
Sname F1 F2

15. BCNF (Boyce-Codd Normal Form)
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Database Management CH 10
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 R1
phone
major
Sname F1

16. BCNF (Boyce-Codd Normal Form)
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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 R2
score
instructor
Cname
Sname F2
F2={ Sname,Cname->score,
Cname->instructor}
R2.1
R2.2
instructor
Cname
score
Cname
Sname
F2.1
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

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

18. BCNF: another example R FD R1 R2 F2={ Ø }
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BCNF: another example R
Office#
Cname
BankerName
Bname FD
FD={ Bname,Sname  BnakerName , BankerName  Bname,Office# }
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 R1 R2
Office#
BankerName
Bname
Cname
BankerName
F2={ Ø }
F1={ BankerName->Bname,Office# }

19. BCNF: another example (Cont.)
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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 3NF
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:
α must be a key for R OR
Each attribute in β is Prime attribute.
FD={ Bname,Sname  BnakerName,
BankerName  Bname,Office# } R
Office#
Cname
BankerName
Bname FD

21. University Of Palestine
Database Management CH 10
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 ) R1
Office#
BankerName R2
Cname
BankerName
Bname F1 F2

22. 3NF (Cont.) F1UF2 = FD So the decomposition is dependency preserving.
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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. R1
Office#
BankerName R2
Cname
BankerName
Bname F1 F2

23. University Of Palestine
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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:
α must be a key for R OR
Each attribute in β is Prime attribute.
α is not a subset of any key

24. 2NF: Example FD={ BC->ADE E->G } For F1 BC is a key
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2NF: Example
FD={ BC->ADE
E->G }
For F1 BC is a key
For F2 E is not a subset of any key
So R is in 2NF R F E D C B A FD

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2NF: Other Example R F E D C B A
FD={ BC->ADE
D->B
E->G } FD
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