Chapter 7: Relational Database Design

1.1

Chapter 7: Relational Database DesignChapter 7: Relational Database Design

Where are we?

We can build an ER model for information in an enterprise

convert it to relation schema (tables)

Are we done? (no)

Relational database design requires that we find a “good” collection of relation schemas

Design goals: avoid redundant data

ensure that relationships among entities are represented

facilitate checking updates for violation of integrity constraints

1.2

ExampleExampleLending_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 wasted space

complicated updating with possibility of inconsistency of assets value

Using nulls to store information about a branch if no loans exist

Normalization: decomposing into more relations, e.g.

(branch_name, branch_city, assets) and

(branch_namecustomer_name, loan_number, amount)

1.3

7.1.1 Design Alternative: Larger Schemas7.1.1 Design Alternative: Larger Schemas(not usually a good idea)(not usually a good idea)

Suppose we combine borrower and loan to get

bor_loan = (customer_id, loan_number, amount )

Result is possible repetition of information (L-100 in example below)

Normally db design goes from the one table to two

1.4

More Important: Decomposing SchemasMore Important: Decomposing Schemas

Start with bor_loan = (customer_id, loan_number, amount )

If more than one customer has the same loan, the amount is repeated

Option: split (decompose) it into borrower (customer_id, loan_number) and loan= (loan_number, amount )

How would we know to do this?

The value of amount depends on loan_number

called a functional dependency:

loan_number amount

This suggests decomposing bor_loan into borrower and loan

Nice property: borrower loan = bor_loan

1.5

Not all Decompositions are GoodNot all Decompositions are Good

employee = (employee_id, employee_name,telephone_number,

start_date) decomposes into

employee1

= (employee_id, employee_name)

employee2

= (employee_name, telephone_number,

start_date)

We lose information:

employee1 employee2 ≠ employee

Can’t reconstruct the original employee

relation

This is a lossy decomposition

1.6

Normal FormsNormal Forms

Requirements on relations that reflect good design

Most basic: first normal form

A domain is atomic if its elements are considered indivisible units non-atomic: a set of names, composite attributes

Relational schema R is in first normal form if the domains of all attributes are atomic

Non-atomic values encourage redundant data and complicate storage e.g. multivalued attribute children might be stored with each parent

example from text: employeeID concatenates dept# and unique ID

These are two different pieces of information that might need to be split apart

Assume all relations are in first normal form e.g. person(name, birthday) should be defined as

person(name, b_month, b_year, b_day)

1.7

Decide what good form means

If R is not in good form, decompose it into a set of relations

{R1, R2, ..., Rn}

such that

each Ri is in good form

the decomposition is a lossless-join decomposition

preferably, the decomposition is dependency preserving

Theory is based on functional dependencies generalization of the notion of a key

allow us to decompose a relation meaningfully

Goal:Goal: A Theory A Theory

1.8

Functional DependenciesFunctional Dependencies

Constraints on the set of legal relations

When the value for a set of attributes uniquely determines the value for another set of attributes e.g. city and state imply the value for sales_tax in

(city_name, state_name, sales_tax)

expressed as

city_name, state_name sales_tax

Does every relation have a functional dependency? primary key any set of attributes

trivial dependencies (satisfied by all instances of a relation):

customer_name, loan_number customer_name

customer_name customer_name

is trivial if

1.9

Functional Dependencies (cont.)Functional Dependencies (cont.)

Let R be a relation schema and R , R

The functional dependency

holds on R if and only if for any legal relations r(R), when any two tuples t1 and t2 of r agree on the attributes , they also agree on the attributes :

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

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

On this instance, A B does not hold, but B A does

Does this make B A a functional dependency?

branch_name branch_city holds on slide 2. Will it always?

1 41 53 7

1.10


Revisiting candidate and superkeys:

K is a superkey for relation schema R iff K R

K is a candidate key for R iff

K R, and

for no K, R

Some constraints cannot be expressed using superkeys.

Loan_info_schema =

(branch_name, loan_number, customer_name, amount)

might have functional dependency

customer_name branch_name

but customer_name is not a superkey

Functional dependencies can express constraints such as this

Updates to the db can be tested to see if they satisfy the given set of functional dependencies

1.11


A specific instance of a schema may satisfy a functional dependency, even if the functional dependency does not hold on all legal instances. e.g., some, but not all, instances of loan_schema may satisfy

loan_number customer_name.

Database design goal: identify and enforce FDs that should hold

1.12

Boyce-Codd Normal FormBoyce-Codd Normal Form

A relation schema R is in BCNF with respect to a set F of functional dependencies if for all functional dependencies in F+ of the form

, where R and R,

at least one of the following holds: is trivial (i.e., )

is a superkey for R

Example: R = (A, B, C), key = {A}

F = {A B B C}

R is not in BCNF (why?)

Decompose R into R1 = (A, B), R2 = (B, C)

R1 and R2 are in BCNF (why?)

This decomposition is dependency-preserving

1.13

Decomposing a Schema into BCNFDecomposing a Schema into BCNF

Suppose we have a schema R and a non-trivial dependency causes a violation of BCNF. Decompose R into: (U )

( R - ( - ) )

Back to the example

R = (A, B, C), key = {A}

F = {A B, B C}

B C causes the violation, so

= B

= C

and R = (A, B, C) is replaced by

(U ) = ( B,C )

( R - ( - ) ) = ( A, B )

1.14

BCNF and Dependency PreservationBCNF and Dependency Preservation

Consistency constraints need to be checked with database update

E.g. attribute types

uniqueness of the primary key

foreign key/primary key consistency

others expressed in SQL: assertions, triggers, check constraints

functional dependencies (dependency preserving)

Functional dependencies that span more than one relation are costly to check

Example: R = (A, B, C), key = {A} F = {A B, B C}:

R1(A,B), R2(B,C) – can check dependencies easily

R1(A,B), R2(A,C) – need to join R1 and R2 to check B C

1.15

BCNF ExampleBCNF Example

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}

1. R is not BCNF, use branch_name assets, branch_city to obtain

R1 = (branch_name, branch_city, assets)

R2 = (branch_name, customer_name, loan_number, amount)

2. R2 is not BCNF, use loan_number amount, branch_name

R3 = (branch_name, loan_number, amount)

R4 = (customer_name, loan_number)

3. Final decomposition: R1, R3, R4

1.16

Third Normal Form: MotivationThird Normal Form: Motivation

In some situations BCNF is not dependency preserving, and

efficient checking for FD violation on updates is important

Example

R = (J, K, L) F = {JK L, L K}candidate keys are JK and JL R is not in BCNF (L is not a superkey)

possible decomposition: R1 = (L, K), R2 = (L, J)

does not preserve JK L

Solution: a weaker normal form, third normal form (3NF) allows some redundancy (with resultant problems)

but functional dependencies can be checked on individual relations without computing a join

There is always a 3NF lossless-join, dependency-preserving decomposition

1.17

Third Normal FormThird Normal Form

Relation schema R is in third normal form (3NF) if for all:

in F+

at least one of the following holds: is trivial (i.e., )

is a superkey for R

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

(each attribute may be in a different candidate key)

If a relation is in BCNF it is in 3NF one of the first two conditions above always holds in BCNF

Third condition is a minimal relaxation of BCNF to ensure dependency preservation (will see why later)

1.18

3NF Example3NF Example

R = (J, K, L )F = {JK L, L K }

Candidate keys JK, JL

R is in 3NF JK L JK is a superkey

L K K is contained in a candidate key

Possible problems in this schema: repetition of information

(e.g., the relationship l1, k1)

need to use null values (e.g., to represent the relationship l2, k2 where there is no corresponding value for J)

Jj1

j2

j3

null

Ll1

l1

l1

l2

Kk1

k1

k1

k2

1.19

Design GoalsDesign 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

Next step: functional-dependency theory a formal theory that tells us which functional dependencies are

implied logically by a given set of functional dependencies

algorithms to generate lossless decompositions into BCNF and 3NF

algorithms to test if a decomposition is dependency-preserving

1.20

Closure of a Set of FDsClosure of a Set of FDs

Given a set F set of FDs, other FDs are logically implied by F example: If A B and B C, we can infer that A C

The set of all functional dependencies logically implied by F is the closure of F, denoted F+.

We can find all 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)

1.21

ExampleExample

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

by augmenting CG I to infer CG CGI,

and augmenting of CG H to infer CGI HI,

and then transitivity

1.22

Procedure for Computing FProcedure for Computing F++

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

F + = Frepeat

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

Alternative procedure for this task later

CS 325 spring ‘02

1.23

Example AgainExample Again

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

A CCG HCG I B H}

reflexivity and augmentation give AA, ... ,AG BG, AG CG,

CGI HI, CG CGI,

transitivity gives A H from A B and B H

CG HI from CG CGI, and CGI HI

... (takes forever)

1.24

Functional Dependencies Functional Dependencies Simplify manual computation of F+ by additional rules

If and hold, then holds (union)

If holds, then holds and holds (decomposition)

If and hold, then holds (pseudotransitivity)

These rules can be inferred from Armstrong’s axioms

The closure under F of a set of attributes, (denoted by +) as the set of attributes that are functionally determined by under F

Algorithm to compute +, the closure of under F

result := ;while (changes to result) do

for each in F dobeginif result then result := result end

1.25

Example of Attribute Set ClosureExample of Attribute Set Closure

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

F = {A BA C CG HCG IB 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?

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

Is any subset of AG a superkey?1. Does A R? == Is (A)+ R

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

1.26

Uses of Attribute ClosureUses of Attribute Closure

Testing for superkey: to test if is a superkey, 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, compute + by using attribute closure, and then check if it contains .

a simple, useful, and cheap testl

Computing F+

for each R, find the closure +, and for each S +, output a functional dependency S.

CS 325 spring ‘02

1.27

Finding F+Finding F+

R = (A, B, C, G, H, I)F = { A B, A C, CG H, CG I, B H}

Find the closure of attribute combinations A+ = ABCH

B+ = BH

C+ = C

G+ = G

H+ = H

I+ = I

CG+ = CGHI ...

Subset of F+

A ABCH

B BH

CG CGHI

1.28

Canonical CoverCanonical Cover

Sets of functional dependencies may have redundant dependencies that can be inferred from the others For example: A C is redundant in: {A B, B 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

Why do we want this? To have a minimal set to check when performing an update.

1.29

Extraneous AttributesExtraneous 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: 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: F = {A C, AB CD} C is extraneous in AB CD since AB C can be inferred even

after deleting C

1.30

Testing if an Attribute is ExtraneousTesting 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 ; if it does, A is extraneous in

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 in

1.31

Canonical CoverCanonical 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 2 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

(Union rule may apply again after deleting extraneous attributes)

1.32

Computing a Canonical CoverComputing a Canonical Cover

R = (A, B, C)F = {A BC

B C A BAB 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 BB C

1.33

Lossless-join DecompositionLossless-join Decomposition

For the decomposition R = (R1, R2), we require that for all

possible relations r on schema R

r = R1 (r ) R2 (r )

(lossless join)

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

1.34

ExampleExample

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)

1.35

Dependency PreservationDependency Preservation

Let Fi be the set of dependencies F + that include only attributes in Ri.

A decomposition is dependency preserving, if

(F1 F2 … Fn )+ = F +

Why is this nice? each dependency can be checked using only one relation

Using F is not sufficient:

R = (A, B, C )

F = {A BC}

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

(F1 F2 )+ = (φ) +

But: F + includes A B and A C

(F1 F2 )+ =( (A B) (A C) ) + = F +

1.36

Dependency PreservationDependency Preservation

Computing F + is expensive.

Alternate: check each dependency in F

To check if is preserved in the decomposition of R into R1, R2,

…, Rn :

result = while (changes to result) do

for each Ri in the decomposition

t = (result Ri)+ Ri //whatever in Ri is dependent on

result = result t

If result contains all attributes in , then the functional dependency is preserved

1.37

ExampleExample

R = (A, B, C )

F = {A BC B C}

R is not in BCNF because of B C

R1 = (A, B), R2 = (B, C) is BCNF.

Check dependency preserving for A BC:

result = A

Use R1: t = (result Ri)+ Ri = (A (A,B))+ (A, B)

= (A, B, C) (A, B) = (A, B)

result = (A,B)

Use R2: t = (result R2)+ R2 = ((A,B) (B,C))+ (B, C)

= (B)+ (B, C) = (B, C) (B, C) = (B, C)

result = (A,B) (B, C) = (A,B,C) which contains BC

result = while (changes to result) do

for each Ri in the decomposition (attribute closure is wrt F) t = (result Ri)+ Ri

result = result t

If result contains all attributes in , then the functional dependency

is preserved

1.38

Testing for BCNFTesting for BCNF

To check if a non-trivial dependency causes a violation of BCNF compute + (the attribute closure of ), and verify that it includes all attributes of R (it is a superkey of R).

Simplified test: check only the dependencies in the given set F for violation of BCNF, rather than checking all dependencies in F+.

Even simpler: use Fc

Problem: using only Fc is incorrect for testing relations in a decomposition of R example: R = (A, B, C, D, E), with F = { A B, BC D}

decompose R into R1 = (A,B) and R2 = (A,C,D, E)

neither of the dependencies in F contain only attributes from (A,C,D,E) so we might think R2 satisfies BCNF

dependency AC D in F+ shows R2 is not in BCNF

1.39

Testing A Decomposition for BCNFTesting A Decomposition for BCNF

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

test Ri for BCNF with respect to the restriction of F+ to Ri (all FDs in F+ that contain only attributes

from Ri)

or use the original set F with the following test:

for every set of attributes Ri, check that + (the attribute closure of ) either includes no attribute of Ri- , or includes all attributes of Ri.

if the condition is violated by some in F,

the dependency (+ - ) Ri holds on Ri and

Ri violates BCNF.

use to decompose Ri

Apply to R = (A, B, C, D, E), with F = { A B, BC D} and

R1 = (A,B) and R2 = (A,C, D, E): a possible problem is AC in R2

(AC)+ = (ABCD), thus AC D is a dependency.

Decompose R2 to R3 = (ACD) and R4 =(ACB)

1.40

BCNF Decomposition AlgorithmBCNF 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 beginlet be a nontrivial functional dependency that holds

on Ri such that Ri is not in F +,

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

endelse done := true;

Each Ri is now in BCNF and the decomposition is lossless-join

1.41

Example of BCNF DecompositionExample of BCNF Decomposition

Original relation R and functional dependency F

R = (A, B, C, D, E, F)

F = {B A, ACD, BC E, BC AF}

Key = {BC}

Canonical cover Fc:

{B A, ACD, BC EF}

Use B A to decompose R:

R1 = (A,B )

R2 = (B,C,D,E,F)

Use BC EF to decompose R2

R3 = (BCEF)

R4 = (BCD)

Use attribute closure tests to see if done, e.g. (BC) on R4

Final decomposition: R1, R3, R4

1.42

BCNF SummaryBCNF Summary

Algorithm says decompose using F +

This takes forever

Easier: use Fc

But this may miss some FDs

So, use Fc but check attribute closure:

for every set of attributes Ri, check that + (the attribute closure of ) either includes no attribute of Ri- , or includes all attributes of Ri.

1.43

3NF Decomposition Algorithm3NF Decomposition AlgorithmLet 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)

Each relation schema Ri is in 3NF (Ri := : → is in the canonical cover, the last Ri contains a candidate key and nothing else so can't be decomposed

Decomposition is dependency preserving (each → in the canonical cover is in one of the Ri

Decomposition is lossless-join (one of the Ri s contains a candidate key)

1.44

3NF Decomposition3NF Decomposition

Original relation R and functional dependency F

R = (A, B, C, D, E, F)

F = {B A, ACD, BC E, BC AF}

Key = {BC}

Canonical cover Fc: {B A, ACD, BC EF}

Decomposition

R1 = (A,B )

R2 = (A,C,D)

R3 = (BCEF)

R3 contains a candidate key so we are done

Compare with BCNF

R1 = (AB)

R3 =(BCEF)

R3 =(BCD) which is not dependency preserving

1.45

3NF Decomposition: Example3NF Decomposition: Example

Relation schema:

cust_banker_branch = (customer_id, employee_id, branch_name, type )

The functional dependencies for this relation schema are:

customer_id, employee_id branch_name, type

employee_id branch_name

customer_id, branch_name employee_id

Compute a canonical cover:

branch_name is extraneous in the r.h.s. of the 1st dependency

no other attribute is extraneous, so we get FC =

customer_id, employee_id type employee_id branch_name

customer_id, branch_name employee_id

1.46

3NF Decomposition Example (Cont.)3NF Decomposition Example (Cont.)

The for loop generates following 3NF schema:

(customer_id, employee_id, type )

(employee_id, branch_name)

(customer_id, branch_name, employee_id)

(customer_id, employee_id, type ) contains a candidate key of the original schema, so no further relation schema needs be added

The second schema is redundant

Minor extension of the 3NF decomposition algorithm: at end of for loop, detect and delete schemas which are subsets of other schemas

If the FDs were considered 1st, 3rd, 2nd (employee_id, branch_name) would not be included in the decomposition because it is a subset of (customer_id, branch_name, employee_id)

Final result will not depend on the order in which FDs are considered

1.47

Comparison of BCNF and 3NFComparison of BCNF and 3NF

It is always possible to decompose a relation into a set of relations that are in 3NF such that: the decomposition is lossless

the dependencies are preserved

It is always possible to decompose a relation into a set of relations that are in BCNF such that: the decomposition is lossless

it may not be possible to preserve dependencies

1.48

ER Model and NormalizationER Model and Normalization

When an E-R diagram is carefully designed, identifying all entities correctly, the tables generated from the E-R diagram should not need further normalization

In a real (imperfect) design, there can be functional dependencies from non-key attributes of an entity example: an employee entity with attributes department_number

and department_address, and a functional dependency department_number department_address

good design would have made department an entity

1.49

Denormalization for PerformanceDenormalization for Performance

May want to use non-normalized schema for performance when queries span two relations (avoid a join)

Alternative 1: Use denormalized relation containing attributes that are usually combined 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 with the specific attributes benefits and drawbacks same as above, except no extra coding work for

programmer and avoids possible errors

view: a virtual table representing the result of a query. A query or update of

the view's table is against the underlying base tables.

materialized view: stored as a concrete table that is infrequently updated

from the original base tables

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Chapter 7: Relational Database Design