dr. t. y. lin | sjsu | cs 157a | fall 2015 chapter 3 database normalization 1

Dr. T. Y. Lin | SJSU | CS 157A | Fall 2015

Chapter 3

Database Normalization

1


Database Design Theory

Different Levels of Anomaly Problems

Normalization

2


Anomaly Problems

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S #

Salary

STATUS CITY P # QTY

S1 40000

20 LONDON P1 300

S1 40000

20 LONDON P2 200

S1 40000

20 LONDON P3 400

S1 40000

20 LONDON P4 200

S1 40000

20 LONDON P5 100

S1 40000

20 LONDON P6 100

S2 30000

10 PARIS P1 300

S2 30000

10 PARIS P2 400

S3 30000

10 PARIS P2 200

S4 40000

20 LONDON P2 200

S4 40000

20 LONDON P4 300

S4 40000

20 LONDON P5 400

Initial

Dr. T. Y. Lin | SJSU | CS 157A | Fall 2015 4

Deletion/insertion anomalyS # Salary STATUS CITY P # QTY

S1 40000 20 LONDON P1 300

S1 40000 20 LONDON P2 200

S1 40000 20 LONDON P3 400

S1 40000 20 LONDON P4 200

S1 40000 20 LONDON P5 100

S1 40000 20 LONDON P6 100

S2 30000 10 PARIS P1 300

S2 30000 10 PARIS P2 400

S3 30000 10 PARIS P2 200

S4 40000 20 LONDON P2 200

S4 40000 20 LONDON P4 300

S4 40000 20 LONDON P5 400

S5 60000 30 ATHENS -


Insertion/update anomaly

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S # Salary STATUS CITY P # QTY

S1 40000 20 LONDON P1 300

S1 40000 20 LONDON P2 200

S1 40000 20 LONDON P3 400

S1 40000 20 LONDON P4 200

S1 40000 20 LONDON P5 100

S1 40000 20 LONDON P6 100

S2 30000 10 PARIS P1 300

S2 30000 10 PARIS P2 400

S3 30000 10 PARIS P2 200

S4 40000 20 LONDON P2 200

S4 40000 20 LONDON P4 300

S4 40000 20 LONDON P5 400


Further Normalization

The problem of database design involves the decision of a suitable logical structure for that data. In other words, the decision is what relations are needed and what attributes they should use.

Codd defined three Normal Forms ( 1NF, 2NF, 3NF ) to remove some undesirable properties from relations. Later, both Boyce and Codd defined an even stronger Normal Form called Boyce - Codd (BCNF ). Later, Fagin introduced 4NF and finally 5NF ( PJ/NF ).

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Dr. T. Y. Lin | SJSU | CS 157A | Fall 2015 7

UNIVERSE OF RELATIONS (NORMALIZED AND UNNORMALIZED)

1NF RELATIONS (NORMALIZED RELATIONS)

2NF RELATIONS

3NF RELATIONS

BCNF RELATIONS

4NF RELATIONS

5NF RELATIONS

PJ/NF

NORMAL FORMS


Functional Dependencies (FD)

Given a relation R, attribute Y of R is functionally dependent on attribute X of R if each X - value in R has associated with it precisely one Y - value in R (at any one time).

(no X-values are mapped to two Y-values)

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Functional Dependencies (FD)

A functional dependency is a special form of integrity constraint. In other words, every legal extension ( tabulation ) of that relation satisfies that constraint.

An attribute Y is said to be fully functionally dependent on X if Y functionally depends on X but not any proper subset of X. From now on, by FD, we mean full FD.

9


First Normal Form Relations (1NF)

A relation is said to be 1NF if all underlying domains contain atomic values only. so any normalized relation is in 1NF.

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G # SNAME STATUS CITYG1 SMITH,

ADAMS20,30

LONDON,ATHENS

G2 JONES,BLAKE

10,30

PARIS

G3 BLAKE 30 PARISG4 CLARK 20 LONDONG5 ADAMS 30 ATHENS


First Normal Form Relations (1NF) Normalized (1NF)

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G # SNAME STATUS CITYG1 SMITH,

20,

LONDON,

G1 SMITH,

20 ATHENS

G1 SMITH,

30 LONDON

G1 SMITH,

30 ATHENS

G1 SMITH,ADAMS

20,30

LONDON,ATHENS

G2 JONES,BLAKE

10,30

PARIS

G3 BLAKE 30 PARISG4 CLARK 20 LONDONG5 ADAMS 30 ATHENS



First Normal Form Relations(1 NF)

All relations will be in 1NF

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First

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S # STATUS CITY P # QTYS1 20 LONDON P1 300

S1 20 LONDON P2 200

S1 20 LONDON P3 400

S1 20 LONDON P4 200

S1 20 LONDON P5 100

S1 20 LONDON P6 100

S2 10 PARIS P1 300

S2 10 PARIS P2 400

S3 10 PARIS P2 200

S4 20 LONDON P2 200

S4 20 LONDON P4 300

S4 20 LONDON P5 400


Functional Dependencies In The Relation First

We can verify the FD by SQL; but this is merely a NECESSARY condition (SEE “group by” in Ch6)

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P#

S#

CITY

STATUS

QTY


Second Normal Form (2NF)

A relation is in 2NF if it is in 1NF and every nonkey (not part of CK) attribute is fully functionally dependent (ffd) on the primary key.

W=a * Sin X + b * Cos Y

(a and b are two parameters)

W is ffd on X and Y, if both a and b are on-zero

W is not ffd on X and Y, if one of a and b

are zero; W=0 * Sin X + b * Cos Y 15


BCNF (Boyce-Codd Normal Form)

For Relations with Equal or More Than One Candidate Key,

A relation R is said to be in BCNF if and only if every determinant is a candidate key.

A determinant is an attribute, possibly composite, on which some other attribute is fully functionally dependent.

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2NF And SP

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S#

STATUS

CITY

S#

P#

QTY


2NF and SP

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S # STATUS CITYS1 20 LONDONS2 10 PARISS3 10 PARISS4 20 LONDONS5 30 ATHENS

S # P # QTYS1 P1 300S1 P2 200S1 P3 400S1 P4 200S1 P5 100S1 P6 100S2 P1 300S2 P2 400S3 P2 200S4 P2 200S4 P4 300S4 P5 400


2NF and SP

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S #

STATUS CITY AMSTERDAM

S1 20 LONDON S2 10 PARIS S3 10 PARIS S4 20 LONDON S5 30 ATHENS S

#STATUS CITY

S1 20 LONDONS2 10 PARISS3 10 PARISS4 20 LONDONS5 30 ATHENS

Insertion anomaly is fixed

Update anomaly is fixed


2NF and SP

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Deletion anomaly is fixed


“Degree Two” Problems

Second (Update, deletion and insertion anomaly)

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S # STATUS CITYS1 20 LONDONS2 10 PARISS3 10 PARISS4 20 LONDON 60 ROME



Functional Dependencies In The Third Normal Form (3NF)

Definition 1 A relation is in 3NF if it is in 2NF and every

non-key attribute is non transitively dependent on the candidate key.

Definition 2 A relation is in 3NF if for every non-trivial FD,

it either starts from super-key or end at part of the CK.

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Functional Dependencies In The Third Normal Form (3NF)

Definition 3 A relation is in 3NF iff the non-key attributes of

R are

a) mutually independent

b) fully dependent on the primary key of R.

Definition 3 (In other words) A relation R is in 3NF if, for all time, each tuple

consists of a primary key value that identifies some entity, together with a set of zero or more mutually independent attribute values that describe that entity in some way.

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Sample Tabulations OfSC and CS

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S # CITYS1 LONDONS2 PARISS3 PARISS4 LONDONS5 ATHENS

CITY STATUSATHENS 30LONDON 20PARIS 10

SC

CS


Functional DependenciesIn The Relations SC and CS

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S# CITY

CITY STATUS


Another set of examples (Skip 2012)

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LOTS

PROPERTY_ID # COUNTY_NAME LOT # AREA PRICE TAX_RATE

fd1

fd2

fd3

fd4

PROPERTY_ID # COUNTY_NAME LOT # AREA PRICE

fd1

fd2

fd4

fd3

LOTS1

LOTS2

COUNTY_NAME TAX_RATE



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LOTS1

PROPERTY_ID # COUNTY_NAME LOT # AREA PRICE

fd1

fd2

fd4LOTS2

COUNTY_NAME TAX_RATE

fd3

LOTS1A

PROPERTY_ID # COUNTY_NAME LOT # AREA

LOTS1B

AREA PRICE

fd1

fd2fd4

LOTS

LOTS1 LOTS2

LOTS1A LOTS1BLOTS2

1NF

2NF

3NF



Figure 13.11 Example to illustrate normalization to 2NF and 3NF. (a) The LOTS relation schema and its

functional dependencies fd1 through fd4.

(b) Decomposing LOTS into the 2NF relations LOTS1 and LOTS2.

(c) Decomposing LOTS1 into the 3NF relations LOTS1A and LOTS1B.

(d) Summary of normalization of LOTS.

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Boyce-Codd Normal Form (BCNF)

Codd did not deal satisfactorily, in 3NF, with the case of a relation that (a) had multiple CKs

(b) CKs were composite

(c) CKs overlapped

The 3NF was subsequently replaced by BCNF.

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Relations with Equal or More Than One Candidate Key

A relation R is said to be in BCNF iff every determinant is a candidate key.

A determinant is an attribute, possibly composite, on which some other attribute is fully functionally dependent.

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Consider a relation SJT with attributes S(student), J(subject), and T(teacher). The meaning of the tuple (s,j,t) is that student s is taught subject j by teacher t. Suppose, in addition, that the following constraints apply. For each subject, each student of that

subject is taught by only one teacher.

Each teacher teaches only one subject.

Each subject is taught by several teachers.

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Problem If we delete the student 'Jones' and the subject

'Physics', we will lose the information that 'Brown' teaches 'Physics' (Professor get fired?).

Solution Split SJT into

ST (S,T) and TJ (T, J)

This decomposition avoids the above problem but introduces different problems, what are they? What are the candidate keys?

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Sample Tabulation Of The Relation SJT

S J TSMITH MATH Prof. WHITESMITH PHYSICS Prof. GREENJONES MATH Prof. WHITEJONES PHYSICS Prof.

BROWN

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S

J

T


Sample Tabulations

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J TMATH Prof. WHITEPHYSICS Prof. GREENPHYSICS Prof.

BROWN

S TSMITH Prof. WHITESMITH Prof. GREENJONES Prof. WHITEJONES Prof.

BROWN

JTST



Consider the relation EXAM with overlapping candidate keys (S, J) and (J, P), and with attributes S (student), J (subject), and P (position). The meaning of an EXAM tuple (s, j, p) is that student s was examined in subject j and achieved position P in the class list. Let us assume that the following constraint holds.

There are no ties; that is, no two students obtained the same position in the same subject.

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Note that update anomalies such as those associated with relation SJT do not apply to relation EXAM, Why?

Overlapping candidate keys do not necessarily lead to problems. In what normal form is relation EXAM?

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Sample Tabulation Of SJP

S was examined in subject J and achieved position P

There are no ties; no students obtained The same position in the same subject

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S J PSMITH MATH FIRST (M)SMITH PHYSICS FIRST (P)JONES MATH SECOND (M)JONES PHYSICS SECOND (P)



Illustrating BCNF: (a) BCNF normalization

with the dependency of fd2 being "lost" in the decomposition.

(b) A relation R in 3NF but not in BCNF.

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LOTS1A

PROPERTY_ID # COUNTY_NAME LOT# AREA

fd1

fd2

fd5

BCNF Normalization

LOTS1AX

R

A B C

fd1

fd2

LOTS1AY

AREA COUNTY_NAMEPROPERTY_ID # AREA LOT #

(b)

(a)


Good and Bad Decomposition

In decomposition (A), the two projections are independent of one another, in the following sense :

Updates can be made to either one without regard for the other, provided that it does not violate the primary key uniqueness constraint for that projection. Actually, if attribute CITY of relation SC is

regarded as a foreign key matching the primary key CITY of relation CS, then a certain amount of cross - checking between the two projections will be required on updates after all 39


Independent Components In decomposition (B), by contrast, update

to either of the two projections must be monitored to ensure that the FD CITY STATUS is not violated (if two suppliers have the same city, they must have the same status). Consider, for example : What is involved in decomposition (B) in

moving supplier S1 from London to Paris ?

In decomposition (B), the FD CITY STATUS has become an inter-relational constraint.

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Independent Components

Rissanen shows that projections R1 and R2 of a relation R are independent if and only if 1. Every FD in R can be logically deduced from

those in R1 and R2, and

2. The common attributes of R1 and R2 form a candidate key for at least one of the pair.

Recall the relation SJT with its two projections ST (S, T) and TJ (T,J)

These two projections are not independent. By Rissanen's Theorem, the FD (S, J) T cannot be deduced from the FD TJ

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Independent Components

Relations which cannot be decomposed into independent components are said to be atomic. Thus, SJT is atomic, even though it is not in BCNF.

Unfortunately, we are forced to the unpleasant conclusion that the two objections of decomposing a relation into BCNF components and decomposing it into independent components may occasionally be in conflict.

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dr. t. y. lin | sjsu | cs 157a | fall 2015 chapter 3 database normalization 1

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