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Database Systems and

ApplicationsLecture 16

Normalization

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Types of Normal forms

First normal form Second normal form

Third normal form

Boyce-Codd normal form Fourth normal form

Fifth normal form

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Steps in normalizationTable with multi-valued attributes

1st normal form

2nd normal form

3rd normal form

Boyce-Coddnormal form

4th normal form

5th normal form

Remove multi-valued attributes

Remove partialdependencies

Removeremaining anomalies

resulting from functionaldependencies

Remove transitive

dependencies

Remove multi-valued dependencies

Remove remaining

anomalies

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Third Normal Form (3NF)

To fulfill 3NF a table should fulfill 2NFand in addition no non-key attributeshould be FFD of any other non-keyattribute.

That means ..

A relation is in 3NF if it is in 2NF andno transitive dependencies exist.

A transitive dependency in a relationis a functional dependency betweentwo or more non key attributes.

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3NF (cont.)

Due to the transitivedependency, there will beupdate anomalies, like

Insertion anomaly:

Deletion anomaly:Modification anomaly:

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Consider the following relation:(st_id, st_name, course_no,

course_name, sec_no,classroom,inst_id, inst_name,no_of_sections)

with the functional dependencies st_id st_name course_no course_name

,no_of_sections st_id , course_no, sec_no course_no, sec_no classroom, inst_id inst_id inst_name

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R1( st_id ,st_name)

R2(course_no,course_name,no_of_sections,)

R3(st_id,course_no,sec_no,classroom,inst_id ,inst_name)

Now it is in 2 NF

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st_id , course_no, course_no ,sec_no

course_no, sec_no classroom,inst_id

Therefore

st_id , course_no, classroom,inst_id

In the same way

course_no, sec_no classroom,inst_id

inst_id inst_name

Therefore

course no sec no inst name

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(st_id , course_no, sec_no)

(course_no, sec_no ,classroom,

inst_id )

(inst_id ,inst_name )

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cno cname , cno ins_ic

ins_ic chamber no

Decompose into

(cno,cname,ins_ic) (ins_ic,chamber no)

Example II for convertinginto 3NF

cno cname

ins_ic chamber

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Boyce-Codd Normal Form(BCNF)

Boyce-Codd normal form (BCNF)

A relation is in BCNF, if and only if, every determinant is a

candidate key.

The difference between 3NF and BCNF is that for a functionaldependency A B, 3NF allows this dependency in a relationif B is a primary-key attribute and A is not a candidate key,

whereas BCNF insists that for this dependency to remain in arelation, A must be a candidate key.

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FD1 clientNo, interviewDate interviewTime, staffNo, roomNo(Primary Key)

FD2 staffNo, interviewDate, interviewTime clientNo , roomNo(Candidate key)

FD3 roomNo, interviewDate, interviewTime clientNo, staffNo(Candidate key)

FD4 staffNo, interviewDate roomNo (not a candidate key)

As a consequece the ClientInterview relation may suffer from updateanmalies.

For example, two tuples have to be updated if the roomNo need be changedfor staffNo SG5 on the 13-May-02.

ClientInterview

ClientNo interviewDate interviewTime staffNo roomNo

CR76 13-May-02 10.30 SG5 G101

CR75 13-May-02 12.00 SG5 G101

CR74 13-May-02 12.00 SG37 G102

CR56 1-Jul-02 10.30 SG5 G102

CR55 13-May-02 10:30 SG6 G103

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Example of BCNFTo transform the ClientInterview relation to BCNF, we must remove

the violating functional dependency by creating two new relations

called Interview and StaffRoom as shown below,

Interview (clientNo, interviewDate, interviewTime, staffNo)

StaffRoom(staffNo, interviewDate, roomNo)

ClientNoClientNo interviewDateinterviewDate interviewTimeinterviewTime staffNostaffNoCR76CR76 13-May-0213-May-02 10.3010.30 SG5SG5

CR75CR75 13-May-0213-May-02 12.0012.00 SG5SG5

CR74CR74 13-May-0213-May-02 12.0012.00 SG37SG37

CR56CR56 1-Jul-021-Jul-02 10.3010.30 SG5SG5

CR55 13-May-02 10:30 SG6

staffNostaffNo interviewDateinterviewDate roomNoroomNoSG5SG5 13-May-0213-May-02 G101G101

SG37SG37 13-May-0213-May-02 G102G102

SG5SG5 1-Jul-021-Jul-02 G102G102SG6 13-May-02 G103

Interview

StaffRoom

BCNF Interview and StaffRoom relations

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Another BCNF Example

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Fourth Normal Form

Multivalued Dependency is a rule that twoattributes or set of attributes areindependent of one another. i.e., when

there are atleast three attributes(e.g., A,B, C) in a relation, with well defined set ofB and C values for each A value, but thoseB and C values are independent of each

other. A relation is in 4NF if it is in BCNF and

contains no multivalued dependencies.

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Multi ValuedDependency

Let R be a relation and A and Barbitrary sets of attributes of R.

There is a multi-valued dependency

from A to BA Bif and only each A value exactly

determines a set of B values,independently of the other attributes

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4NF (cont.)COURSE_OFFERING

c_no teacher_id textbook_code

CSC352 245 T-1

CSC352 245 T-2

ISC332 300 T-1

ISC332 300 T-2CSC374 250 T-4

CSC374 250 T-5

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4NF (cont.) Ex: (c_no, teacher_id, textbook_code)

COURSE_OFFERING

In this example all the 3 attributes are candidate key,but teacher_id and textbook_code dont have anyrelationship. To remove the multivalued dependency,decompose the relation in two new relations.

course_no techer_id textbook_code

course_no techer_id course_no textbook_code