dbms- a walk through 2

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NORMALIZATION

(3NF AND BCNF)



EX AMPL

E



Insertion anomaly: if we want to insert the informationof a new salesperson, we cannot insert this information

until a customer is assigned to that personDeletion anomaly: if we delete the only customerassigned to a salesperson, we drop the informationabout the salesperson

Update anomaly: if the region of a salesperson changes, we have to search and find each customer assigned tohim.

ANOMALIES



If an attribute is indirectly dependent on primary key,

then we say, there exists a transitive dependency.

For example:

If AB, BC then by transitivity theorem A C.

i.e. If A is the primary key, C is transitively dependent on

A and functionally dependent on B which is a nonprimeattribute.

TRANS

ITIV E

DE

PE

NDE

NCY



A relation r (R) is in third normal form (3NF) if and

only if the following conditions are satisfied

simultaneously.1. R elation R is already in 2NF

2. No non-prime attribute is transitively dependent on

the key or

No non-prime attribute functionally determines any

other non-prime attribute

THIRD NORMAL FORM



THIRD NORMAL FORM (3NF)

1) Meet all the requirements of the 1NF

2) Meet all the requirements of the 2NF

3) Remove columns that are dependent upon thenon primary key.



DECOMPOSING THE SALES RELA



CustID Name

CustID

Salesperson



EX AMPL

E

Beer

Beer Brewery Strength City Region

Choice Websters XX York North West

Old Bob Websters XXX York North West

Landlord Taylors XXX Leeds North West

Directors Fremlins X London South East

Wobbly Joy Sam

Smith

XXXX York North West



functional dependencies(beer, brewery, strength, city, region)

brewery p city

cityp region

beer p brewery, strength, city, region

EX AMPL

E..



Sales order

(SalesOrderNo, Date, CustomerNo, CustomerName,

CustomerAdd, ClerkNo, ClerkName)

EXERCI

SE



CERTIFICATION-PROGRAM table(area, course, section, time, location)



This relation is in 3NF because neither of the

nonprime attributes functionally determines the

other attributes.However, if there is only one location per city then

Location Area. If we delete the last tuple of the

relation we lose information about the location of

the educational center.

ANOMALY



Student number Course Teacher

101 History Singh

101 Art Asha102 Geography Aanchal

103 Sociology Smith

104 Economics Pradeep

105 History Singh

106 History Rekha

EX AMPL

E



A relation r (R) is in Boyce-Codd form (BCNF) if

and only if the following conditions are met

simultaneously1.The relation is in 1NF

2.For every functional dependency X A, we

have that either A X or X is a super key of r.

In other words, every functional dependency is

a trivial dependency or if it is not trivial

dependency then X must be a super key.

BOYCE-CODD NORMAL FORM



A relation is in BCNF if every determinant is a

candidate key.

A determinant is a column on which some of the

columns are fully functional dependent.

ALTE

RNATE

DE

FINITION



Student

number(pk)

Teacher(pk)

101 Singh

101 Asha

102 Aanchal

103 Smith

104 Pradeep

105 Singh

106 Rekha

Teacher (pk) Course

Singh History

Asha Art Aanchal Geography

Smith Sociology

Pradeep Economics

Rekha History



EX

AMPLE



Set 1:

Manufacturer (id, name)

M-part (id, item-no, qty)

Set 2:

Manufacturer (id, name)

M-part (name, item-no, qty)

S

OLUTION



ClientInterview

ClientClient

NoNo

interviewinterview

DateDate

interviewinterview

TimeTimestaffNostaffNo roomNoroomNo

CR76 13-May-02 10.30 SG5 G101

CR76 13-May-02 12.00 SG5 G101

CR74 13-May-02 12.00 SG37 G102

CR56 1-Jul-02 10.30 SG5 G102



FD1 clientNo, interviewDate interviewTime, staffNo,

roomNo (Primary Key)

FD2 staffNo, interviewDate, interviewTime

clientNo(Candidate key)

FD3 roomNo, interviewDate, interviewTime clientNo,

staffNo (Candidate key)

FD4 staffNo, interviewDate roomNo (not a candidatekey) As a consequence the ClientInterview relation may

suffer from update anomalies. For example, two tuples have to be

updated if the roomNo need be changed for staffNo SG5 on the

13-May-02.



To transform the ClientInterview relation to BCNF, wemust remove the violating functional dependency bycreating two new relations called Interview and StaffRoomas shown below

Interview (clientNo, interviewDate, interviewTime,staffNo)

StaffRoom(staffNo, interviewDate, roomNo)

BCNF Interview and StaffRoom relations

EX

AMPLE



ClientNoClientNo interviewDateinterviewDate interviewTimeinterviewTime staffNostaffNo

CR76 13-May-02 10.30 SG5

CR76 13-May-02 12.00 SG5

CR74 13-May-02 12.00 SG37

CR56 1-Jul-02 10.30 SG5

staffNostaffNo interviewDateinterviewDate roomNoroomNo

SG5 13-May-02 G101

SG37 13-May-02 G102

SG5 1-Jul-02 G102

Interview

StaffRoom



CONCURR ENCY MANAGEMENT

Serializability Test



Read lock (shared locks): a transaction T wishing

only to read an item A executes RLOCK A which preventsany other transaction from writing a new value of A while

T is reading A.

Write Lock (exclusive locks): a transaction wishing to

change the value of item A first obtains a write lock byexecuting WLOCK A. when some transaction holds a write

lock on an item no other transaction can obtain a readlock or write lock on that item.

Both locks can be removed using UNLOCK statement.

LOCK S



Input: a schedule S for a set of transactions T1, Tk

Output: a determination whether S is serializable and if

so an equivalent serial schedule

Method: we construct a precedence graph G as follows

the nodes correspond to the transactions as before. The

arcs are determined by the following rules

1)Suppose in S transaction Ti, read locks A and Tj is the

next transaction to write lock A then place an arc from Ti

to Tj.

SERIALIZABILITY TEST FOR SERIALIZABILITY TEST FOR

SCHEDULES WITH READ/WRITE SCHEDULES WITH READ/WRITE

LOCKSLOCKS



2. Suppose in S, transaction Ti write locks A and Tj is thenext transaction to write lock A. then draw an arc Ti to

Tj. Further Tm be any transaction that reads locks Aafter Ti unlocks its write lock but before Tj write locks A. if there is no Tj, then Tm is any transaction to readlock A after Ti unlocks A. then draw an arc Ti to Tm.

If G has a cycle, then S is not serializable. If G is acyclicthen any topological sort of G is a serial order for thetransaction

CONT..



1 WLOCK A

2 RLOCK B

3 UNLOCK A

4 RLOCK A5 UNLOCK B

6 WLOCK B

7 RLOCK A

8 UNLOCK B9 WLOCK B

10 UNLOCK A

11 UNLOCK A

12 WLOCK A

13 UNLOCK B

14 RLOCK B

15 UNLOCK A

16 UNLOCK B

T1 T2 T3 T4



T17

T19

T18



If G is acyclic, find a serial schedule as follows;

a.Initialize the serial schedule empty

b.Find a transaction Ti, such that there are no arcs enteringTi. Ti is the next transaction in the serial schedule

c.Remove Ti and all edges emerging from it. If theremaining set is empty, serial schedule is complete,otherwise return to b

SOLUTION

dbms- a walk through 2

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