このページは http://www.slideshare.net/koolkampus/ch07 の内容を掲載しています。

掲載を希望されないスライド著者の方は、こちらよりご連絡下さい。

- Chapter 7: Relational Database Design
- Chapter 7: Relational Database Design

First Normal Form

Pitfal s in Relational Database Design

Functional Dependencies

Decomposition

Boyce-Codd Normal Form

Third Normal Form

Multivalued Dependencies and Fourth Normal Form

Overal Database Design Process

Database System Concepts

7.2

©Silberschatz, Korth and Sudarshan - First Normal Form

Domain is atomic if its elements are considered to be indivisible

units

Examples of non-atomic domains:

Set of names, composite attributes

Identification numbers like CS101 that can be broken up into

parts

A relational schema R is in first normal form if the domains of al

attributes of R are atomic

Non-atomic values complicate storage and encourage redundant

(repeated) storage of data

E.g. Set of accounts stored with each customer, and set of owners

stored with each account

We assume all relations are in first normal form (revisit this in

Chapter 9 on Object Relational Databases)

Database System Concepts

7.3

©Silberschatz, Korth and Sudarshan - First Normal Form (Contd.)

Atomicity is actual y a property of how the elements of the

domain are used.

E.g. Strings would normally be considered indivisible

Suppose that students are given roll numbers which are strings of

the form CS0012 or EE1127

If the first two characters are extracted to find the department, the

domain of roll numbers is not atomic.

Doing so is a bad idea: leads to encoding of information in

application program rather than in the database.

Database System Concepts

7.4

©Silberschatz, Korth and Sudarshan - Pitfalls in Relational Database Design

Relational database design requires that we find a

“good” col ection of relation schemas. A bad design

may lead to

Repetition of Information.

Inability to represent certain information.

Design Goals:

Avoid redundant data

Ensure that relationships among attributes are

represented

Facilitate the checking of updates for violation of database

integrity constraints.

Database System Concepts

7.5

©Silberschatz, Korth and Sudarshan - Example

Consider the relation schema:

Lending-schema = (branch-name, branch-city, assets,

customer-name, loan-number, amount)

Redundancy:

Data for branch-name, branch-city, assets are repeated for each loan that a

branch makes

Wastes space

Complicates updating, introducing possibility of inconsistency of assets value

Null values

Cannot store information about a branch if no loans exist

Can use null values, but they are difficult to handle.

Database System Concepts

7.6

©Silberschatz, Korth and Sudarshan - Decomposition

Decompose the relation schema Lending-schema into:

Branch-schema = (branch-name, branch-city,assets)

Loan-info-schema = (customer-name, loan-number,

branch-name, amount)

Al attributes of an original schema (R) must appear in

the decomposition (R1, R2):

R = R1 R2

Lossless-join decomposition.

For al possible relations r on schema R

r = R1 (r) R2 (r)

Database System Concepts

7.7

©Silberschatz, Korth and Sudarshan - Example of Non Lossless-Join Decomposition

Decomposition of R = (A, B)

R2 = (A)

R2 = (B)

A B

A

B

1

1

2

2

1

(r)

A

B(r)

r

A B

(r) (r)

A

B

1

2

1

2

Database System Concepts

7.8

©Silberschatz, Korth and Sudarshan - Goal — Devise a Theory for the Following

Decide whether a particular relation R is in “good” form.

In the case that a relation R is not in “good” form, decompose it

into a set of relations {R1, R2, ..., Rn} such that

each relation is in good form

the decomposition is a lossless-join decomposition

Our theory is based on:

functional dependencies

multivalued dependencies

Database System Concepts

7.9

©Silberschatz, Korth and Sudarshan - Functional Dependencies

Constraints on the set of legal relations.

Require that the value for a certain set of attributes determines

uniquely the value for another set of attributes.

A functional dependency is a generalization of the notion of a

key.

Database System Concepts

7.10

©Silberschatz, Korth and Sudarshan - Functional Dependencies (Cont.)

Let R be a relation schema

R and R

The functional dependency

holds on R if and only if for any legal relations r(R), whenever any

two tuples t1 and t2 of r agree on the attributes , they also agree

on the attributes . That is,

t1[ ] = t2 [ ] t1[ ] = t2 [ ]

Example: Consider r(A,B) with the following instance of r.

1

4

1 5

3

7

On this instance, A B does NOT hold, but B A does hold.

Database System Concepts

7.11

©Silberschatz, Korth and Sudarshan - Functional Dependencies (Cont.)

K is a superkey for relation schema R if and only if K R

K is a candidate key for R if and only if

K R, and

for no K, R

Functional dependencies al ow us to express constraints that

cannot be expressed using superkeys. Consider the schema:

Loan-info-schema = (customer-name, loan-number,

branch-name, amount).

We expect this set of functional dependencies to hold:

loan-number amount

loan-number branch-name

but would not expect the fol owing to hold:

loan-number customer-name

Database System Concepts

7.12

©Silberschatz, Korth and Sudarshan - Use of Functional Dependencies

We use functional dependencies to:

test relations to see if they are legal under a given set of functional

dependencies.

If a relation r is legal under a set F of functional dependencies, we

say that r satisfies F.

specify constraints on the set of legal relations

We say that F holds on R if all legal relations on R satisfy the set of

functional dependencies F.

Note: A specific instance of a relation schema may satisfy a

functional dependency even if the functional dependency does not

hold on al legal instances.

For example, a specific instance of Loan-schema may, by chance,

satisfy

loan-number customer-name.

Database System Concepts

7.13

©Silberschatz, Korth and Sudarshan - Functional Dependencies (Cont.)

A functional dependency is trivial if it is satisfied by al instances

of a relation

E.g.

customer-name, loan-number customer-name

customer-name customer-name

In general, is trivial if

Database System Concepts

7.14

©Silberschatz, Korth and Sudarshan - Closure of a Set of Functional

Dependencies

Given a set F set of functional dependencies, there are certain

other functional dependencies that are logical y implied by F.

E.g. If A B and B C, then we can infer that A C

The set of al functional dependencies logical y implied by F is the

closure of F.

We denote the closure of F by F+.

We can find al of F+ by applying Armstrong’s Axioms:

if , then (reflexivity)

if , then (augmentation)

if , and , then (transitivity)

These rules are

sound (generate only functional dependencies that actually hold) and

complete (generate all functional dependencies that hold).

Database System Concepts

7.15

©Silberschatz, Korth and Sudarshan - Example

R = (A, B, C, G, H, I)

F = { A B

A C

CG H

CG I

B H}

some members of F+

A H

by transitivity from A B and B H

AG I

by augmenting A C with G, to get AG CG

and then transitivity with CG I

CG HI

from CG H and CG I : “union rule” can be inferred from

– definition of functional dependencies, or

– Augmentation of CG I to infer CG CGI, augmentation of

CG H to infer CGI HI, and then transitivity

Database System Concepts

7.16

©Silberschatz, Korth and Sudarshan - Procedure for Computing F+

To compute the closure of a set of functional dependencies F:

F+ = F

repeat

for each functional dependency f in F+

apply reflexivity and augmentation rules on f

add the resulting functional dependencies to F+

for each pair of functional dependencies f1and f2 in F+

if f1 and f2 can be combined using transitivity

then add the resulting functional dependency to F+

until F+ does not change any further

NOTE: We wil see an alternative procedure for this task later

Database System Concepts

7.17

©Silberschatz, Korth and Sudarshan - Closure of Functional Dependencies

(Cont.)

We can further simplify manual computation of F+ by using

the fol owing additional rules.

If holds and holds, then holds (union)

If holds, then holds and holds

(decomposition)

If holds and holds, then holds

(pseudotransitivity)

The above rules can be inferred from Armstrong’s axioms.

Database System Concepts

7.18

©Silberschatz, Korth and Sudarshan - Closure of Attribute Sets

Given a set of attributes define the closure of under F

(denoted by +) as the set of attributes that are functional y

determined by under F:

is in F+ +

Algorithm to compute +, the closure of under F

result := ;

while (changes to result) do

for each in F do

begin

if result then result := result

end

Database System Concepts

7.19

©Silberschatz, Korth and Sudarshan - Example of Attribute Set Closure

R = (A, B, C, G, H, I)

F = {A B

A C

CG H

CG I

B H}

(AG)+

1. result = AG

2. result = ABCG

(A C and A B)

3. result = ABCGH

(CG H and CG AGBC)

4. result = ABCGHI

(CG I and CG AGBCH)

Is AG a candidate key?

1. Is AG a super key?

1. Does AG R? == Is (AG)+ R

2. Is any subset of AG a superkey?

1. Does A R? == Is (A)+ R

2. Does G R? == Is (G)+ R

Database System Concepts

7.20

©Silberschatz, Korth and Sudarshan - Uses of Attribute Closure

There are several uses of the attribute closure algorithm:

Testing for superkey:

To test if is a superkey, we compute +, and check if + contains all

attributes of R.

Testing functional dependencies

To check if a functional dependency holds (or, in other words,

is in F+), just check if +.

That is, we compute + by using attribute closure, and then check if it

contains .

Is a simple and cheap test, and very useful

Computing closure of F

For each R, we find the closure +, and for each S +, we

output a functional dependency S.

Database System Concepts

7.21

©Silberschatz, Korth and Sudarshan - Canonical Cover

Sets of functional dependencies may have redundant

dependencies that can be inferred from the others

Eg: A C is redundant in: {A B, B C, A C}

Parts of a functional dependency may be redundant

E.g. on RHS: {A B, B C, A CD} can be simplified to

{A B, B C, A D}

E.g. on LHS: {A B, B C, AC D} can be simplified to

{A B, B C, A D}

Intuitively, a canonical cover of F is a “minimal” set of functional

dependencies equivalent to F, having no redundant

dependencies or redundant parts of dependencies

Database System Concepts

7.22

©Silberschatz, Korth and Sudarshan - Extraneous Attributes

Consider a set F of functional dependencies and the functional

dependency in F.

Attribute A is extraneous in if A

and F logically implies (F – { }) {( – A) }.

Attribute A is extraneous in if A

and the set of functional dependencies

(F – { }) { ( – A)} logically implies F.

Note: implication in the opposite direction is trivial in each of

the cases above, since a “stronger” functional dependency

always implies a weaker one

Example: Given F = {A C, AB C }

B is extraneous in AB C because {A C, AB C} logically

implies A C (I.e. the result of dropping B from AB C).

Example: Given F = {A C, AB CD}

C is extraneous in AB CD since AB C can be inferred even

after deleting C

Database System Concepts

7.23

©Silberschatz, Korth and Sudarshan - Te

T sting if an Attribute is Extraneous

Consider a set F of functional dependencies and the functional

dependency in F.

To test if attribute A is extraneous in

1. compute ({ } – A)+ using the dependencies in F

2. check that ({ } – A)+ contains A; if it does, A is extraneous

To test if attribute A is extraneous in

1. compute + using only the dependencies in

F’ = (F – { }) { ( – A)},

2. check that + contains A; if it does, A is extraneous

Database System Concepts

7.24

©Silberschatz, Korth and Sudarshan - Canonical Cover

A canonical cover for F is a set of dependencies Fc such that

F logically implies all dependencies in Fc, and

Fc logically implies all dependencies in F, and

No functional dependency in Fc contains an extraneous attribute, and

Each left side of functional dependency in Fc is unique.

To compute a canonical cover for F:

repeat

Use the union rule to replace any dependencies in F

1 1 and 1 1 with 1 1 2

Find a functional dependency with an

extraneous attribute either in or in

If an extraneous attribute is found, delete it from

until F does not change

Note: Union rule may become applicable after some extraneous

attributes have been deleted, so it has to be re-applied

Database System Concepts

7.25

©Silberschatz, Korth and Sudarshan - Example of Computing a Canonical Cover

R = (A, B, C)

F = {A BC

B C

A B

AB C}

Combine A BC and A B into A BC

Set is now {A BC, B C, AB C}

A is extraneous in AB C

Check if the result of deleting A from AB C is implied by the other

dependencies

Yes: in fact, B C is already present!

Set is now {A BC, B C}

C is extraneous in A BC

Check if A C is logically implied by A B and the other dependencies

Yes: using transitivity on A B and B C.

– Can use attribute closure of A in more complex cases

The canonical cover is: A B

B C

Database System Concepts

7.26

©Silberschatz, Korth and Sudarshan - Goals of Normalization

Decide whether a particular relation R is in “good” form.

In the case that a relation R is not in “good” form, decompose it

into a set of relations {R1, R2, ..., Rn} such that

each relation is in good form

the decomposition is a lossless-join decomposition

Our theory is based on:

functional dependencies

multivalued dependencies

Database System Concepts

7.27

©Silberschatz, Korth and Sudarshan - Decomposition

Decompose the relation schema Lending-schema into:

Branch-schema = (branch-name, branch-city,assets)

Loan-info-schema = (customer-name, loan-number,

branch-name, amount)

All attributes of an original schema (R) must appear in the

decomposition (R1, R2):

R = R1 R2

Lossless-join decomposition.

For all possible relations r on schema R

r = R1 (r) R2 (r)

A decomposition of R into R1 and R2 is lossless join if and only if

at least one of the following dependencies is in F+:

R1 R2 R1

R1 R2 R2

Database System Concepts

7.28

©Silberschatz, Korth and Sudarshan - Example of Lossy-Join Decomposition

Lossy-join decompositions result in information loss.

Example: Decomposition of R = (A, B)

R2 = (A)

R2 = (B)

A B

A

B

1

1

2

2

1

(r)

A

B(r)

r

A B

(r) (r)

A

B

1

2

1

2

Database System Concepts

7.29

©Silberschatz, Korth and Sudarshan - Normalization Using Functional Dependencies

When we decompose a relation schema R with a set of

functional dependencies F into R1, R2,.., Rn we want

Lossless-join decomposition: Otherwise decomposition would result in

information loss.

No redundancy: The relations Ri preferably should be in either Boyce-

Codd Normal Form or Third Normal Form.

Dependency preservation: Let Fi be the set of dependencies F+ that

include only attributes in Ri.

Preferably the decomposition should be dependency preserving,

that is, (F1 F2 … Fn)+ = F+

Otherwise, checking updates for violation of functional dependencies

may require computing joins, which is expensive.

Database System Concepts

7.30

©Silberschatz, Korth and Sudarshan - Example

R = (A, B, C)

F = {A B, B C)

Can be decomposed in two different ways

R1 = (A, B), R2 = (B, C)

Lossless-join decomposition:

R1 R2 = {B} and B BC

Dependency preserving

R1 = (A, B), R2 = (A, C)

Lossless-join decomposition:

R1 R2 = {A} and A AB

Not dependency preserving

(cannot check B C without computing R1 R2)

Database System Concepts

7.31

©Silberschatz, Korth and Sudarshan - Tes

T ting for Dependency Preservation

To check if a dependency is preserved in a decomposition of

R into R1, R2, …, Rn we apply the fol owing simplified test (with

attribute closure done w.r.t. F)

result =

while (changes to result) do

for each Ri in the decomposition

t = (result Ri)+ Ri

result = result t

If result contains all attributes in , then the functional dependency

is preserved.

We apply the test on al dependencies in F to check if a

decomposition is dependency preserving

This procedure takes polynomial time, instead of the exponential

time required to compute F+ and (F1 F2 … Fn)+

Database System Concepts

7.32

©Silberschatz, Korth and Sudarshan - Boyce-Codd Normal Form

A relation schema R is in BCNF with respect to a set F of functional

dependencies if for al functional dependencies in F+ of the form

, where R and R, at least one of the fol owing holds:

is trivial (i.e., )

is a superkey for R

Database System Concepts

7.33

©Silberschatz, Korth and Sudarshan - Example

R = (A, B, C)

F = {A B

B C}

Key = {A}

R is not in BCNF

Decomposition R1 = (A, B), R2 = (B, C)

R1 and R2 in BCNF

Lossless-join decomposition

Dependency preserving

Database System Concepts

7.34

©Silberschatz, Korth and Sudarshan - Test

T

ing for BCNF

To check if a non-trivial dependency causes a violation of

BCNF

1. compute + (the attribute closure of ), and

2. verify that it includes all attributes of R, that is, it is a superkey of R.

Simplified test: To check if a relation schema R is in BCNF, it

suffices to check only the dependencies in the given set F for

violation of BCNF, rather than checking al dependencies in F+.

If none of the dependencies in F causes a violation of BCNF, then

none of the dependencies in F+ will cause a violation of BCNF either.

However, using only F is incorrect when testing a relation in a

decomposition of R

E.g. Consider R (A, B, C, D), with F = { A B, B C}

Decompose R into R1(A,B) and R2(A,C,D)

Neither of the dependencies in F contain only attributes from

(A,C,D) so we might be mislead into thinking R2 satisfies BCNF.

In fact, dependency A C in F+ shows R2 is not in BCNF.

Database System Concepts

7.35

©Silberschatz, Korth and Sudarshan - BCNF Decomposition Algorithm

result := {R};

done := false;

compute F+;

while (not done) do

if (there is a schema Ri in result that is not in BCNF)

then begin

let be a nontrivial functional

dependency that holds on Ri

such that Ri is not in F+,

and = ;

result := (result – Ri ) (Ri – ) ( , );

end

else done := true;

Note: each Ri is in BCNF, and decomposition is lossless-join.

Database System Concepts

7.36

©Silberschatz, Korth and Sudarshan - Example of BCNF Decomposition

R = (branch-name, branch-city, assets,

customer-name, loan-number, amount)

F = {branch-name assets branch-city

loan-number amount branch-name}

Key = {loan-number, customer-name}

Decomposition

R1 = (branch-name, branch-city, assets)

R2 = (branch-name, customer-name, loan-number, amount)

R3 = (branch-name, loan-number, amount)

R4 = (customer-name, loan-number)

Final decomposition R1, R3, R4

Database System Concepts

7.37

©Silberschatz, Korth and Sudarshan - Te

T sting Decomposition for BCNF

To check if a relation Ri in a decomposition of R is in BCNF,

Either test Ri for BCNF with respect to the restriction of F to Ri (that is,

al FDs in F+ that contain only attributes from Ri)

or use the original set of dependencies F that hold on R, but with the

fol owing test:

– for every set of attributes Ri, check that + (the attribute

closure of ) either includes no attribute of Ri- , or includes al

attributes of Ri.

If the condition is violated by some in F, the dependency

( + - ) Ri

can be shown to hold on Ri, and Ri violates BCNF.

We use above dependency to decompose Ri

Database System Concepts

7.38

©Silberschatz, Korth and Sudarshan - BCNF and Dependency Preservation

It is not always possible to get a BCNF decomposition that is

dependency preserving

R = (J, K, L)

F = {JK L

L K}

Two candidate keys = JK and JL

R is not in BCNF

Any decomposition of R wil fail to preserve

JK L

Database System Concepts

7.39

©Silberschatz, Korth and Sudarshan - Third Normal Form: Motivation

There are some situations where

BCNF is not dependency preserving, and

efficient checking for FD violation on updates is important

Solution: define a weaker normal form, cal ed Third Normal Form.

Allows some redundancy (with resultant problems; we will see

examples later)

But FDs can be checked on individual relations without computing a

join.

There is always a lossless-join, dependency-preserving decomposition

into 3NF.

Database System Concepts

7.40

©Silberschatz, Korth and Sudarshan - Third Normal Form

A relation schema R is in third normal form (3NF) if for al :

in F+

at least one of the fol owing holds:

is trivial (i.e., )

is a superkey for R

Each attribute A in – is contained in a candidate key for R.

(NOTE: each attribute may be in a different candidate key)

If a relation is in BCNF it is in 3NF (since in BCNF one of the first

two conditions above must hold).

Third condition is a minimal relaxation of BCNF to ensure

dependency preservation (wil see why later).

Database System Concepts

7.41

©Silberschatz, Korth and Sudarshan - 3NF (Cont.)

Example

R = (J, K, L)

F = {JK L, L K}

Two candidate keys: JK and JL

R is in 3NF

JK L

JK is a superkey

L K

K is contained in a candidate key

BCNF decomposition has (JL) and (LK)

Testing for JK L requires a join

There is some redundancy in this schema

Equivalent to example in book:

Banker-schema = (branch-name, customer-name, banker-name)

banker-name branch name

branch name customer-name banker-name

Database System Concepts

7.42

©Silberschatz, Korth and Sudarshan - Tes

T ting for 3NF

Optimization: Need to check only FDs in F, need not check al

FDs in F+.

Use attribute closure to check for each dependency , if is

a superkey.

If is not a superkey, we have to verify if each attribute in is

contained in a candidate key of R

this test is rather more expensive, since it involve finding candidate

keys

testing for 3NF has been shown to be NP-hard

Interestingly, decomposition into third normal form (described

shortly) can be done in polynomial time

Database System Concepts

7.43

©Silberschatz, Korth and Sudarshan - 3NF Decomposition Algorithm

Let Fc be a canonical cover for F;

i := 0;

for each functional dependency in Fc do

if none of the schemas Rj, 1 j i contains

then begin

i := i + 1;

Ri :=

end

if none of the schemas Rj, 1 j i contains a candidate key for R

then begin

i := i + 1;

Ri := any candidate key for R;

end

return (R1, R2, ..., Ri)

Database System Concepts

7.44

©Silberschatz, Korth and Sudarshan - 3NF Decomposition Algorithm (Cont.)

Above algorithm ensures:

each relation schema Ri is in 3NF

decomposition is dependency preserving and lossless-join

Proof of correctness is at end of this file (click here)

Database System Concepts

7.45

©Silberschatz, Korth and Sudarshan - Example

Relation schema:

Banker-info-schema = (branch-name, customer-name,

banker-name, office-number)

The functional dependencies for this relation schema are:

banker-name branch-name office-number

customer-name branch-name banker-name

The key is:

{customer-name, branch-name}

Database System Concepts

7.46

©Silberschatz, Korth and Sudarshan - Applying 3NF to Banker-inf

er

o-schema

The for loop in the algorithm causes us to include the

fol owing schemas in our decomposition:

Banker-office-schema = (banker-name, branch-name,

office-number)

Banker-schema = (customer-name, branch-name,

banker-name)

Since Banker-schema contains a candidate key for

Banker-info-schema, we are done with the decomposition

process.

Database System Concepts

7.47

©Silberschatz, Korth and Sudarshan - Comparison of BCNF and 3NF

It is always possible to decompose a relation into relations in

3NF and

the decomposition is lossless

the dependencies are preserved

It is always possible to decompose a relation into relations in

BCNF and

the decomposition is lossless

it may not be possible to preserve dependencies.

Database System Concepts

7.48

©Silberschatz, Korth and Sudarshan - Comparison of BCNF and 3NF (Cont.)

Example of problems due to redundancy in 3NF

R = (J, K, L)

F = {JK L, L K}

J

L

K

j

l

k

1

1

1

j

l

k

2

1

1

j

l

k

3

1

1

null l

k

2

2

A schema that is in 3NF but not in BCNF has the problems of

repetition of information (e.g., the relationship l , k )

1

1

need to use nul values (e.g., to represent the relationship

l , k where there is no corresponding value for J).

2

2

Database System Concepts

7.49

©Silberschatz, Korth and Sudarshan - Design Goals

Goal for a relational database design is:

BCNF.

Lossless join.

Dependency preservation.

If we cannot achieve this, we accept one of

Lack of dependency preservation

Redundancy due to use of 3NF

Interestingly, SQL does not provide a direct way of specifying

functional dependencies other than superkeys.

Can specify FDs using assertions, but they are expensive to test

Even if we had a dependency preserving decomposition, using

SQL we would not be able to efficiently test a functional

dependency whose left hand side is not a key.

Database System Concepts

7.50

©Silberschatz, Korth and Sudarshan - Te

T sting for FDs Across Relations

If decomposition is not dependency preserving, we can have an extra

materialized view for each dependency in Fc that is not preserved

in the decomposition

The materialized view is defined as a projection on of the join of the

relations in the decomposition

Many newer database systems support materialized views and database

system maintains the view when the relations are updated.

No extra coding effort for programmer.

The functional dependency is expressed by declaring as a

candidate key on the materialized view.

Checking for candidate key cheaper than checking

BUT:

Space overhead: for storing the materialized view

Time overhead: Need to keep materialized view up to date when

relations are updated

Database system may not support key declarations on

materialized views

Database System Concepts

7.51

©Silberschatz, Korth and Sudarshan - Multivalued Dependencies

There are database schemas in BCNF that do not seem to be

sufficiently normalized

Consider a database

classes(course, teacher, book)

such that (c,t,b) classes means that t is qualified to teach c,

and b is a required textbook for c

The database is supposed to list for each course the set of

teachers any one of which can be the course’s instructor, and the

set of books, al of which are required for the course (no matter

who teaches it).

Database System Concepts

7.52

©Silberschatz, Korth and Sudarshan - Multivalued Dependencies (Cont.)

course

teacher

book

database

Avi

DB Concepts

database

Avi

Ullman

database

Hank

DB Concepts

database

Hank

Ullman

database

Sudarshan

DB Concepts

database

Sudarshan

Ullman

operating systems Avi

OS Concepts

operating systems Avi

Shaw

operating systems Jim

OS Concepts

operating systems Jim

Shaw

classes

There are no non-trivial functional dependencies and therefore

the relation is in BCNF

Insertion anomalies – i.e., if Sara is a new teacher that can teach

database, two tuples need to be inserted

(database, Sara, DB Concepts)

(database, Sara, Ul man)

Database System Concepts

7.53

©Silberschatz, Korth and Sudarshan - Multivalued Dependencies (Cont.)

Therefore, it is better to decompose classes into:

course

teacher

database

Avi

database

Hank

database

Sudarshan

operating systems

Avi

operating systems

Jim

teaches

course

book

database

DB Concepts

database

Ullman

operating systems

OS Concepts

operating systems

Shaw

text

We shal see that these two relations are in Fourth Normal

Form (4NF)

Database System Concepts

7.54

©Silberschatz, Korth and Sudarshan - Multivalued Dependencies (MVDs)

Let R be a relation schema and let R and R.

The multivalued dependency

holds on R if in any legal relation r(R), for al pairs for

tuples t1 and t2 in r such that t1[ ] = t2 [ ], there exist

tuples t3 and t4 in r such that:

t1[ ] = t2 [ ] = t3 [ ] t4 [ ]

t3[ ] = t1 [ ]

t3[R – ] = t2[R – ]

t4 [ ] = t2[ ]

t4[R – ] = t1[R – ]

Database System Concepts

7.55

©Silberschatz, Korth and Sudarshan - MVD (Cont.)

Tabular representation of

Database System Concepts

7.56

©Silberschatz, Korth and Sudarshan - Example

Let R be a relation schema with a set of attributes that are

partitioned into 3 nonempty subsets.

Y, Z, W

We say that Y Z (Y multidetermines Z)

if and only if for al possible relations r(R)

< y1, z1, w1 > r and < y2, z2, w2 > r

then

< y1, z1, w2 > r and < y2, z2, w1 > r

Note that since the behavior of Z and W are identical it fol ows

that Y Z if Y W

Database System Concepts

7.57

©Silberschatz, Korth and Sudarshan - Example (Cont.)

In our example:

course teacher

course book

The above formal definition is supposed to formalize the

notion that given a particular value of Y (course) it has

associated with it a set of values of Z (teacher) and a set

of values of W (book), and these two sets are in some

sense independent of each other.

Note:

If Y Z then Y Z

Indeed we have (in above notation) Z1 = Z2

The claim follows.

Database System Concepts

7.58

©Silberschatz, Korth and Sudarshan - Use of Multivalued Dependencies

We use multivalued dependencies in two ways:

1. To test relations to determine whether they are legal under a

given set of functional and multivalued dependencies

2. To specify constraints on the set of legal relations. We shall

thus concern ourselves only with relations that satisfy a given

set of functional and multivalued dependencies.

If a relation r fails to satisfy a given multivalued

dependency, we can construct a relations r that does

satisfy the multivalued dependency by adding tuples to r.

Database System Concepts

7.59

©Silberschatz, Korth and Sudarshan - Theory of MVDs

From the definition of multivalued dependency, we can derive the

fol owing rule:

If , then

That is, every functional dependency is also a multivalued

dependency

The closure D+ of D is the set of al functional and multivalued

dependencies logical y implied by D.

We can compute D+ from D, using the formal definitions of functional

dependencies and multivalued dependencies.

We can manage with such reasoning for very simple multivalued

dependencies, which seem to be most common in practice

For complex dependencies, it is better to reason about sets of

dependencies using a system of inference rules (see Appendix C).

Database System Concepts

7.60

©Silberschatz, Korth and Sudarshan - Fourth Normal Form

A relation schema R is in 4NF with respect to a set D of

functional and multivalued dependencies if for al multivalued

dependencies in D+ of the form , where R and R,

at least one of the fol owing hold:

is trivial (i.e., or = R)

is a superkey for schema R

If a relation is in 4NF it is in BCNF

Database System Concepts

7.61

©Silberschatz, Korth and Sudarshan - Restriction of Multivalued Dependencies

The restriction of D to Ri is the set Di consisting of

All functional dependencies in D+ that include only attributes of Ri

All multivalued dependencies of the form

( Ri)

where Ri and is in D+

Database System Concepts

7.62

©Silberschatz, Korth and Sudarshan - 4NF Decomposition Algorithm

result: = {R};

done := false;

compute D+;

Let Di denote the restriction of D+ to Ri

while (not done)

if (there is a schema Ri in result that is not in 4NF) then

begin

let be a nontrivial multivalued dependency that holds

on Ri such that Ri is not in Di, and ;

result := (result - Ri) (Ri - ) ( , );

end

else done:= true;

Note: each Ri is in 4NF, and decomposition is lossless-join

Database System Concepts

7.63

©Silberschatz, Korth and Sudarshan - Example

R =(A, B, C, G, H, I)

F ={ A B

B HI

CG H }

R is not in 4NF since A B and A is not a superkey for R

Decomposition

a) R1 = (A, B) (R1 is in 4NF)

b) R2 = (A, C, G, H, I) (R2 is not in 4NF)

c) R3 = (C, G, H)

(R3 is in 4NF)

d) R4 = (A, C, G, I)

(R4 is not in 4NF)

Since A B and B HI, A HI, A I

e) R5 = (A, I) (R5 is in 4NF)

f)R6 = (A, C, G)

(R6 is in 4NF)

Database System Concepts

7.64

©Silberschatz, Korth and Sudarshan - Further Normal Forms

Join dependencies generalize multivalued dependencies

lead to project-join normal form (PJNF) (also called fifth normal

form)

A class of even more general constraints, leads to a normal form

cal ed domain-key normal form.

Problem with these generalized constraints: are hard to reason

with, and no set of sound and complete set of inference rules

exists.

Hence rarely used

Database System Concepts

7.65

©Silberschatz, Korth and Sudarshan - Overall Database Design Process

We have assumed schema R is given

R could have been generated when converting E-R diagram to a set of

tables.

R could have been a single relation containing all attributes that are of

interest (called universal relation).

Normalization breaks R into smaller relations.

R could have been the result of some ad hoc design of relations, which

we then test/convert to normal form.

Database System Concepts

7.66

©Silberschatz, Korth and Sudarshan - ER Model and Normalization

When an E-R diagram is careful y designed, identifying al entities

correctly, the tables generated from the E-R diagram should not need

further normalization.

However, in a real (imperfect) design there can be FDs from non-key

attributes of an entity to other attributes of the entity

E.g. employee entity with attributes department-number and

department-address, and an FD department-number department-

address

Good design would have made department an entity

FDs from non-key attributes of a relationship set possible, but rare ---

most relationships are binary

Database System Concepts

7.67

©Silberschatz, Korth and Sudarshan - Universal Relation Approach

Dangling tuples – Tuples that “disappear” in computing a join.

Let r1 (R1), r2 (R2), …., rn (Rn) be a set of relations

A tuple r of the relation ri is a dangling tuple if r is not in the relation:

Ri (r1 r2 … rn)

The relation r1 r2 … rn is cal ed a universal relation since it

involves al the attributes in the “universe” defined by

R1 R2 … Rn

If dangling tuples are al owed in the database, instead of

decomposing a universal relation, we may prefer to synthesize a

col ection of normal form schemas from a given set of attributes.

Database System Concepts

7.68

©Silberschatz, Korth and Sudarshan - Universal Relation Approach

Dangling tuples may occur in practical database applications.

They represent incomplete information

E.g. may want to break up information about loans into:

(branch-name, loan-number)

(loan-number, amount)

(loan-number, customer-name)

Universal relation would require nul values, and have dangling

tuples

Database System Concepts

7.69

©Silberschatz, Korth and Sudarshan - Universal Relation Approach (Contd.)

A particular decomposition defines a restricted form of

incomplete information that is acceptable in our database.

Above decomposition requires at least one of customer-name,

branch-name or amount in order to enter a loan number without

using null values

Rules out storing of customer-name, amount without an appropriate

loan-number (since it is a key, it can't be null either!)

Universal relation requires unique attribute names unique role

assumption

e.g. customer-name, branch-name

Reuse of attribute names is natural in SQL since relation names

can be prefixed to disambiguate names

Database System Concepts

7.70

©Silberschatz, Korth and Sudarshan - Denormalization for Performance

May want to use non-normalized schema for performance

E.g. displaying customer-name along with account-number and

balance requires join of account with depositor

Alternative 1: Use denormalized relation containing attributes of

account as wel as depositor with al above attributes

faster lookup

Extra space and extra execution time for updates

extra coding work for programmer and possibility of error in extra code

Alternative 2: use a materialized view defined as

account depositor

Benefits and drawbacks same as above, except no extra coding work

for programmer and avoids possible errors

Database System Concepts

7.71

©Silberschatz, Korth and Sudarshan - Other Design Issues

Some aspects of database design are not caught by

normalization

Examples of bad database design, to be avoided:

Instead of earnings(company-id, year, amount), use

earnings-2000, earnings-2001, earnings-2002, etc., all on the

schema (company-id, earnings).

Above are in BCNF, but make querying across years difficult and

needs new table each year

company-year(company-id, earnings-2000, earnings-2001,

earnings-2002)

Also in BCNF, but also makes querying across years difficult and

requires new attribute each year.

Is an example of a crosstab, where values for one attribute

become column names

Used in spreadsheets, and in data analysis tools

Database System Concepts

7.72

©Silberschatz, Korth and Sudarshan - Proof of Correctness of 3NF

Decomposition Algorithm - Correctness of 3NF Decomposition

Algorithm

3NF decomposition algorithm is dependency preserving (since

there is a relation for every FD in Fc)

Decomposition is lossless join

A candidate key (C) is in one of the relations Ri in decomposition

Closure of candidate key under Fc must contain all attributes in R.

Follow the steps of attribute closure algorithm to show there is only

one tuple in the join result for each tuple in Ri

Database System Concepts

7.74

©Silberschatz, Korth and Sudarshan - Correctness of 3NF Decomposition

Algorithm (Contd.)

Claim: if a relation Ri is in the decomposition generated by the

above algorithm, then Ri satisfies 3NF.

Let Ri be generated from the dependency

Let B be any non-trivial functional dependency on Ri. (We

need only consider FDs whose right-hand side is a single

attribute.)

Now, B can be in either or but not in both. Consider each

case separately.

Database System Concepts

7.75

©Silberschatz, Korth and Sudarshan - Correctness of 3NF Decomposition

(Contd.)

Case 1: If B in :

If is a superkey, the 2nd condition of 3NF is satisfied

Otherwise must contain some attribute not in

Since B is in F+ it must be derivable from Fc, by using attribute

closure on .

Attribute closure not have used - if it had been used, must

be contained in the attribute closure of , which is not possible, since

we assumed is not a superkey.

Now, using ( - {B}) and B, we can derive B

(since , and B since B is non-trivial)

Then, B is extraneous in the right-hand side of ; which is not

possible since is in Fc.

Thus, if B is in then must be a superkey, and the second

condition of 3NF must be satisfied.

Database System Concepts

7.76

©Silberschatz, Korth and Sudarshan - Correctness of 3NF Decomposition

(Contd.)

Case 2: B is in .

Since is a candidate key, the third alternative in the definition of

3NF is trivially satisfied.

In fact, we cannot show that is a superkey.

This shows exactly why the third alternative is present in the

definition of 3NF.

Q.E.D.

Database System Concepts

7.77

©Silberschatz, Korth and Sudarshan - End of Chapter
- Sample lending Relation

Database System Concepts

7.79

©Silberschatz, Korth and Sudarshan - Sample Relation r

Database System Concepts

7.80

©Silberschatz, Korth and Sudarshan - The customer Relation

Database System Concepts

7.81

©Silberschatz, Korth and Sudarshan - The loan Relation

Database System Concepts

7.82

©Silberschatz, Korth and Sudarshan - The branch Relation

Database System Concepts

7.83

©Silberschatz, Korth and Sudarshan - The Relation branch-customer

Database System Concepts

7.84

©Silberschatz, Korth and Sudarshan - The Relation customer-loan

Database System Concepts

7.85

©Silberschatz, Korth and Sudarshan - The Relation branch-customer customer-loan

Database System Concepts

7.86

©Silberschatz, Korth and Sudarshan - An Instance of Banker-schema

Database System Concepts

7.87

©Silberschatz, Korth and Sudarshan - Tab

T ular Representation of

Database System Concepts

7.88

©Silberschatz, Korth and Sudarshan - Relation bc: An Example of Reduncy in a BCNF Relation

Database System Concepts

7.89

©Silberschatz, Korth and Sudarshan - An Illegal bc Relation

Database System Concepts

7.90

©Silberschatz, Korth and Sudarshan - Decomposition of loan-info

Database System Concepts

7.91

©Silberschatz, Korth and Sudarshan - Relation of Exercise 7.4

Database System Concepts

7.92

©Silberschatz, Korth and Sudarshan