Chapter 8: Relational Database Design Chapter 8: Relational Database Design

(1)

Database System Concepts, 6^th Ed.

©Silberschatz, Korth and Sudarshan See www.db-book.com for conditions on re-use

Chapter 8: Relational Database Design

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Chapter 8: Relational Database Design Chapter 8: Relational Database Design



Features of Good Relational Design



Atomic Domains and First Normal Form



Decomposition Using Functional Dependencies



Functional Dependency Theory



Algorithms for Functional Dependencies



Decomposition Using Multivalued Dependencies



More Normal Form



Database-Design Process



Modeling Temporal Data

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©Silberschatz, Korth and Sudarshan 8.3

Database System Concepts - 6^th Edition

Combine Schemas?



Suppose we combine instructor and department into inst_dept



(No connection to relationship set inst_dept)



Result is possible repetition of information

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A Combined Schema Without Repetition A Combined Schema Without Repetition



Consider combining relations



sec_class(sec_id, building, room_number) and



section(course_id, sec_id, semester, year) into one relation



section(course_id, sec_id, semester, year, building, room_number)



No repetition in this case

(5)

What About Smaller Schemas?



Suppose we had started with inst_dept. How would we know to split up (decompose) it into instructor and department?



Write a rule “if there were a schema (dept_name, building, budget), then dept_name would be a candidate key”



Denote as a functional dependency:

dept_name  building, budget



In inst_dept, because dept_name is not a candidate key, the building and budget of a department may have to be repeated.



This indicates the need to decompose inst_dept



Not all decompositions are good. Suppose we decompose employee(ID, name, street, city, salary) into

employee1 (ID, name)

employee2 (name, street, city, salary)



The next slide shows how we lose information -- we cannot reconstruct

the original employee relation -- and so, this is a lossy decomposition.

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A Lossy Decomposition

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Example of Lossless-Join Decomposition Example of Lossless-Join Decomposition



Lossless join decomposition



Decomposition of R = (A, B, C)

R

1

= (A, B) R

2

= (B, C)

A B

  1 2

A

 

B 1 2

r 

_B,C

(r)



A

(r) 

B

(r) ^A ^B

  1 2

C A B B

1 2

C A B C

A B



_A,B

(r)

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First Normal Form 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 all attributes of R are atomic



Non-atomic values complicate storage and encourage redundant (repeated) storage of data



Example: Set of accounts stored with each customer, and set of owners stored with each account



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

Chapter 22: Object Based Databases)

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First Normal Form (Cont’d) First Normal Form (Cont’d)



Atomicity is actually a property of how the elements of the domain are used.



Example: 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.

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Goal — Devise a Theory for the Following 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 {R

1

, R

2

, ..., R

n

} such that



each relation is in good form



the decomposition is a lossless-join decomposition



Our theory is based on:



functional dependencies



multivalued dependencies

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Functional Dependencies 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.

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Functional Dependencies (Cont.) 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 t

1

and t

2

of r agree on the attributes , they also agree on the attributes . That is,

t

1

[] = t

2

[]  t

1

[ ] = t

2

[ ]



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



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

1 4

1 5

3 7

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Functional Dependencies (Cont.) 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 allow us to express constraints that cannot be expressed using superkeys. Consider the schema:

inst_dept (ID, name, salary, dept_name, building, budget ).

We expect these functional dependencies to hold:

dept_name building and ID  building

but would not expect the following to hold:

dept_name  salary

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Use of Functional Dependencies 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 all legal instances.



For example, a specific instance of instructor may, by chance, satisfy

name  ID.

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Functional Dependencies (Cont.) Functional Dependencies (Cont.)



A functional dependency is trivial if it is satisfied by all instances of a relation



Example:



ID, name  ID



name  name



In general,    is trivial if   

(16)

Closure of a Set of Functional Closure of a Set of Functional

Dependencies Dependencies



Given a set F of functional dependencies, there are certain other functional dependencies that are logically implied by F.



For example: If A  B and B  C, then we can infer that A  C



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



We denote the closure of F by F

⁺

.



F

⁺

is a superset of F.

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Boyce-Codd Normal Form Boyce-Codd Normal Form



   is trivial (i.e.,   )



 is a superkey for R

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:

Example schema not in BCNF:

instr_dept (ID, name, salary, dept_name, building, budget ) because dept_name building, budget

holds on instr_dept, but dept_name is not a superkey

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Decomposing a Schema into BCNF Decomposing a Schema into BCNF



Suppose we have a schema R and a non-trivial dependency 

causes a violation of BCNF.

We decompose R into:

• ^{(U  )}

• ( R - (  -  ) )



In our example,



 = dept_name



 = building, budget

and inst_dept is replaced by



(U  ) = ( dept_name, building, budget )



( R - (  -  ) ) = ( ID, name, salary, dept_name )

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BCNF and Dependency Preservation BCNF and Dependency Preservation



Constraints, including functional dependencies, are costly to check in practice unless they pertain to only one relation



If it is sufficient to test only those dependencies on each individual relation of a decomposition in order to ensure that all functional dependencies hold, then that decomposition is dependency preserving.



Because it is not always possible to achieve both BCNF and

dependency preservation, we consider a weaker normal form, known

as third normal form.

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Third Normal Form Third Normal Form



A 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.

(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 (will see why later).

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Goals of Normalization Goals of Normalization



Let R be a relation scheme with a set F of functional dependencies.



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



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

decompose it into a set of relation scheme {R

1

, R

2

, ..., R

n

} such that



each relation scheme is in good form



the decomposition is a lossless-join decomposition



Preferably, the decomposition should be dependency preserving.

(22)

How good is BCNF?



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



Consider a relation

inst_info (ID, child_name, phone)



where an instructor may have more than one phone and can have multiple children

ID child_name phone

99999 99999 99999 99999

David David William Willian

512-555-1234 512-555-4321 512-555-1234 512-555-4321

inst_info

(23)



There are no non-trivial functional dependencies and therefore the relation is in BCNF



Insertion anomalies – i.e., if we add a phone 981-992-3443 to 99999, we need to add two tuples

(99999, David, 981-992-3443) (99999, William, 981-992-3443)

How good is BCNF? (Cont.)

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

Therefore, it is better to decompose inst_info into:

This suggests the need for higher normal forms, such as Fourth Normal Form (4NF), which we shall see later.

How good is BCNF? (Cont.) How good is BCNF? (Cont.)

ID child_name

99999 99999 99999 99999

David David William Willian

inst_child

ID phone

99999 99999 99999 99999

512-555-1234 512-555-4321 512-555-1234 512-555-4321

inst_phone

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Functional-Dependency Theory Functional-Dependency Theory



We now consider the formal theory that tells us which functional dependencies are implied logically by a given set of functional dependencies.



We then develop algorithms to generate lossless decompositions into BCNF and 3NF



We then develop algorithms to test if a decomposition is dependency-

preserving

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Closure of a Set of Functional Closure of a Set of Functional

Dependencies Dependencies



Given a set F set of functional dependencies, there are certain other functional dependencies that are logically implied by F.



For e.g.: If A  B and B  C, then we can infer that A  C



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



We denote the closure of F by F

⁺

.

(27)

Closure of a Set of Functional Closure of a Set of Functional

Dependencies Dependencies



We can find F

^+,

the closure of F, by repeatedly 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).

(28)

Example 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



by augmenting CG  I to infer CG  CGI,

and augmenting of CG  H to infer CGI  HI,

and then transitivity

(29)

Procedure for Computing F 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 f

1

and f

2

in F

⁺

if f

1

and f

2

can be combined using transitivity

then add the resulting functional dependency to F

⁺

until F

⁺

does not change any further

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

(30)

Closure of Functional Dependencies Closure of Functional Dependencies

(Cont.) (Cont.)



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.

(31)

Closure of Attribute Sets Closure of Attribute Sets



Given a set of attributes  define the closure of  under F (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 do

begin

if   result then result := result  

end

(32)

Example of Attribute Set Closure 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

(33)

Uses of Attribute Closure 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.

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Canonical Cover Canonical 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, 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

(35)

Extraneous Attributes 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

(36)

Testing if an Attribute is Extraneous Testing 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 

(37)

Canonical Cover Canonical Cover



A canonical cover for F is a set of dependencies F

_c

such that



F logically implies all dependencies in F

c,

and



F

c

logically implies all dependencies in F, and



No functional dependency in F

_c

contains an extraneous attribute, and



Each left side of functional dependency in F

_c

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 

/ Note: test for extraneous attributes done using F*

_c,

not F*/

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

(38)

Computing a Canonical Cover 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

(39)

Lossless-join Decomposition Lossless-join Decomposition



For the case of R = (R

1

, R

2

), we require that for all possible relations r on schema R

r = 

R1

(r ) 

R2

(r )



A decomposition of R into R

1

and R

2

is lossless join if at least one of the following dependencies is in F

⁺

:



R

₁

 R

2

 R

1



R

1

 R

2

 R

2



The above functional dependencies are a sufficient condition for

lossless join decomposition; the dependencies are a necessary

condition only if all constraints are functional dependencies

(40)

Example Example



R = (A, B, C)

F = {A  B, B  C)



Can be decomposed in two different ways



R

1

= (A, B), R

2

= (B, C)



Lossless-join decomposition:

R

1

 R

2

= {B} and B  BC



Dependency preserving



R

1

= (A, B), R

2

= (A, C)



Lossless-join decomposition:

R

1

 R

2

= {A} and A  AB



Not dependency preserving

(cannot check B  C without computing R

1

R

2

)

(41)

Dependency Preservation Dependency Preservation



Let F

i

be the set of dependencies F

⁺

that include only attributes in R

i

.



A decomposition is dependency preserving, if (F

1

 F

2

 …  F

n

)

⁺

= F

⁺



If it is not, then checking updates for violation of functional dependencies may require computing joins, which is

expensive.

(42)

Testing for Dependency Preservation Testing for Dependency Preservation



To check if a dependency    is preserved in a decomposition of R into R

₁

, R

₂

, …, R

_n

we apply the following test (with attribute closure done with respect to F)



result = 

while (changes to result) do

for each R

i

in the decomposition t = (result  R

i

)

⁺

 R

i

result = result  t



If result contains all attributes in , then the functional dependency

   is preserved.



We apply the test on all 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 (F

₁

 F

₂

 …  F

_n

)

⁺

(43)

Example Example



R = (A, B, C ) F = {A  B

B  C}

Key = {A}



R is not in BCNF



Decomposition R

1

= (A, B), R

2

= (B, C)



R

1

and R

2

in BCNF



Lossless-join decomposition



Dependency preserving

(44)

Testing for BCNF Testing 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 all 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, simplified test using only F is incorrect when testing a relation in a decomposition of R



Consider R = (A, B, C, D, E), with F = { A  B, BC  D}



Decompose R into R

1

= (A,B) and R

2

= (A,C,D, E)



Neither of the dependencies in F contain only attributes from (A,C,D,E) so we might be mislead into thinking R

₂

satisfies BCNF.



In fact, dependency AC  D in F

⁺

shows R

2

is not in BCNF.

(45)

Testing Decomposition for BCNF Testing Decomposition for BCNF



To check if a relation R

_i

in a decomposition of R is in BCNF,



Either test R

i

for BCNF with respect to the restriction of F to R

i

(that is, all FDs in F

⁺

that contain only attributes from R

i

)



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

– for every set of attributes   R

i

, check that 

⁺

(the

attribute closure of ) either includes no attribute of R

i

- , or includes all attributes of R

i

.



If the condition is violated by some  in Rj, the dependency  (

⁺

- )  R

i

can be shown to hold on R

i

, and R

i

violates BCNF.



We use above dependency to decompose R

i

(46)

BCNF Decomposition Algorithm BCNF Decomposition Algorithm

result := {R };

done := false;

compute F

⁺

;

while (not done) do

if (there is a schema R

i

in result that is not in BCNF) then begin

let    be a nontrivial functional dependency that holds on R

i

such that   R

i

is not in F

⁺

,

and    = ;

result := (result – R

i

)  (R

i

– )  (,  );

end

else done := true;

Note: each R

_i

is in BCNF, and decomposition is lossless-join .

(47)

Example of BCNF Decomposition Example of BCNF Decomposition



R = (A, B, C ) F = {A  B

B  C}

Key = {A}



R is not in BCNF (B  C but B is not superkey)



Decomposition



R

1

= (B, C)



R

2

= (A,B)

(48)

Example of BCNF Decomposition Example of BCNF Decomposition



class (course_id, title, dept_name, credits, sec_id, semester, year, building, room_number, capacity, time_slot_id)



Functional dependencies:



course_id→ title, dept_name, credits



building, room_number→capacity



course_id, sec_id, semester, year→building, room_number, time_slot_id



A candidate key {course_id, sec_id, semester, year}.



BCNF Decomposition:



course_id→ title, dept_name, credits holds



but course_id is not a superkey.



We replace class by:



course(course_id, title, dept_name, credits)



class-1 (course_id, sec_id, semester, year, building,

room_number, capacity, time_slot_id)

(49)

BCNF Decomposition (Cont.) BCNF Decomposition (Cont.)



course is in BCNF



How do we know this?



building, room_number→capacity holds on class-1



but {building, room_number} is not a superkey for class-1.



We replace class-1 by:



classroom (building, room_number, capacity)



section (course_id, sec_id, semester, year, building, room_number, time_slot_id)



classroom and section are in BCNF.

(50)

BCNF and Dependency Preservation BCNF and Dependency Preservation



R = (J, K, L ) F = {JK  L

L  K }

Two candidate keys = JK and JL



R is not in BCNF



Any decomposition of R will fail to preserve JK  L

This implies that testing for JK  L requires a join

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

dependency preserving

(51)

Third Normal Form: Motivation 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, called Third Normal Form (3NF)



Allows some redundancy (with resultant problems; we will see examples later)



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



There is always a lossless-join, dependency-preserving

decomposition into 3NF.

(52)

3NF Example 3NF Example



Relation dept_advisor:



dept_advisor (s_ID, i_ID, dept_name)

F = {s_ID, dept_name  i_ID, i_ID  dept_name}



Two candidate keys: s_ID, dept_name, and i_ID, s_ID



R is in 3NF



s_ID, dept_name  i_ID s_ID – dept_name is a superkey



i_ID  dept_name

– dept_name is contained in a candidate key

(53)

Redundancy in 3NF Redundancy in 3NF

J j

₁

j

₂

j

₃

null

L l

₁

l

₁

l

₁

l

₂

k K

₁

k

₁

k

₁

k

₂

 repetition of information (e.g., the relationship l1, k1)

 (i_ID, dept_name)

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

 (i_ID, dept_nameI) if there is no separate relation mapping instructors to departments



There is some redundancy in this schema



Example of problems due to redundancy in 3NF



R = (J, K, L)

F = {JK  L, L  K }

(54)

Testing for 3NF Testing for 3NF



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

(55)

3NF Decomposition Algorithm 3NF Decomposition Algorithm

Let F

_c

be a canonical cover for F;

i := 0;

for each functional dependency    in F

_c

do if none of the schemas R

_j

, 1  j  i contains   then begin

i := i + 1;

R

_i

:=  

end if none of the schemas R

_j

, 1  j  i contains a candidate key for R then begin

i := i + 1;

R

_i

:= any candidate key for R;

end

/* Optionally, remove redundant relations */

repeat

if any schema R

_j

is contained in another schema R

_k

**then /* delete R**

j

*/

R

_j

= R;;

i=i-1;

return (R

₁

, R

₂

, ..., R

_i

)

(56)

3NF Decomposition Algorithm (Cont.) 3NF Decomposition Algorithm (Cont.)



Above algorithm ensures:



each relation schema R

i

is in 3NF



decomposition is dependency preserving and lossless-join



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

(57)

3NF Decomposition: An Example 3NF Decomposition: An Example



Relation schema:

cust_banker_branch = (customer_id, employee_id, branch_name, type )



The functional dependencies for this relation schema are:

1.

customer_id, employee_id  branch_name, type

2.

employee_id  branch_name

3.

customer_id, branch_name  employee_id



We first compute a canonical cover



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

^st

dependency



No other attribute is extraneous, so we get F

C

= customer_id, employee_id  type

employee_id  branch_name

customer_id, branch_name  employee_id

(58)

3NF Decompsition Example (Cont.) 3NF Decompsition Example (Cont.)



The for loop generates following 3NF schema:

(customer_id, employee_id, type )

(employee_id, branch_name)

(customer_id, branch_name, employee_id)



Observe that (customer_id, employee_id, type ) contains a

candidate key of the original schema, so no further relation schema needs be added



At end of for loop, detect and delete schemas, such as (employee_id, branch_name), which are subsets of other schemas



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



The resultant simplified 3NF schema is:

(customer_id, employee_id, type)

(customer_id, branch_name, employee_id)

(59)

Comparison of BCNF and 3NF Comparison 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.

(60)

Design Goals 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, (and currently not supported by any of the widely used databases!)



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.

(61)

Multivalued Dependencies Multivalued Dependencies



Suppose we record names of children, and phone numbers for instructors:



inst_child(ID, child_name)



inst_phone(ID, phone_number)



If we were to combine these schemas to get



inst_info(ID, child_name, phone_number)



Example data:

(99999, David, 512-555-1234) (99999, David, 512-555-4321) (99999, William, 512-555-1234) (99999, William, 512-555-4321)



This relation is in BCNF



Why?

(62)

Multivalued Dependencies (MVDs) 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 all pairs for tuples t

1

and t

2

in r such that t

1

[] = t

2

[], there exist tuples t

3

and t

4

in r such that:

t

1

[] = t

2

[] = t

3

[] = t

4

[]

t

3

[] = t

1

[]

t

3

[R – ] = t

2

[R – ]

t

4

[] = t

2

[]

t

4

[R – ] = t

1

[R – ]

(63)

MVD (Cont.) MVD (Cont.)



Tabular representation of   

(64)

Example 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 all possible relations r (R )

< y

1

, z

1

, w

1

>  r and < y

1

, z

2

, w

2

>  r then

< y

1

, z

1

, w

2

>  r and < y

1

, z

2

, w

1

>  r



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

Y  Z if Y  W

(65)

Example (Cont.) Example (Cont.)



In our example:

ID  child_name ID  phone_number



The above formal definition is supposed to formalize the notion that given a particular value of Y (ID) it has associated with it a set of values of Z (child_name) and a set of values of W (phone_number), 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) Z

1

= Z

2

The claim follows.

(66)

Use of Multivalued Dependencies 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.

(67)

Theory of MVDs Theory of MVDs



From the definition of multivalued dependency, we can derive the following rule:



If   , then   

That is, every functional dependency is also a multivalued dependency



The closure D

⁺

of D is the set of all functional and multivalued dependencies logically 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).

(68)

Fourth Normal Form Fourth Normal Form



A relation schema R is in 4NF with respect to a set D of functional and multivalued dependencies if for all multivalued dependencies in D

⁺

of the form   , where   R and   R, at least one of the following hold:



   is trivial (i.e.,    or    = R)



 is a superkey for schema R



If a relation is in 4NF it is in BCNF

(69)

Restriction of Multivalued Dependencies Restriction of Multivalued Dependencies



The restriction of D to R

i

is the set D

i

consisting of



All functional dependencies in D

⁺

that include only attributes of R

i



All multivalued dependencies of the form   (  R

i

)

where   R

i

and    is in D

⁺

(70)

4NF Decomposition Algorithm 4NF Decomposition Algorithm

result: = {R};

done := false;

compute D

⁺

;

Let D

i

denote the restriction of D

⁺

to R

i

while (not done)

if (there is a schema R

_i

in result that is not in 4NF) then begin

let    be a nontrivial multivalued dependency that holds on R

i

such that   R

i

is not in D

i

, and ;

result := (result - R

i

)  (R

i

- )  (, );

end

else done:= true;

Note: each R

i

is in 4NF, and decomposition is lossless-join

(71)

Example 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) R

₁

= (A, B) (R

₁

is in 4NF)

b) R

₂

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

₂

is not in 4NF, decompose into R

₃

and R

₄

) c) R

3

= (C, G, H) (R

3

is in 4NF)

d) R

₄

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

₄

is not in 4NF, decompose into R

₅

and R

₆

)



A  B and B  HI  A  HI, (MVD transitivity), and



and hence A  I (MVD restriction to R

4

) e) R

₅

= (A, I) (R

₅

is in 4NF)

f)R

6

= (A, C, G) (R

6

is in 4NF)

(72)

Further Normal Forms 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 called 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

(73)

Overall Database Design Process 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.

(74)

ER Model and Normalization ER 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.



However, in a real (imperfect) design, there can be functional

dependencies from non-key attributes of an entity to other attributes of the entity



Example: an employee entity with attributes department_name and building,

and a functional dependency department_name building



Good design would have made department an entity



Functional dependencies from non-key attributes of a relationship set

possible, but rare --- most relationships are binary

(75)

Denormalization for Performance Denormalization for Performance



May want to use non-normalized schema for performance



For example, displaying prereqs along with course_id, and title requires join of course with prereq



Alternative 1: Use denormalized relation containing attributes of course as well as prereq with all 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 course prereq



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

for programmer and avoids possible errors

(76)

Other Design Issues 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_2004, earnings_2005, earnings_2006, 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_2004, earnings_2005, earnings_2006)



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

(77)

Modeling Temporal Data Modeling Temporal Data



Temporal data have an association time interval during which the data are valid.



A snapshot is the value of the data at a particular point in time



Several proposals to extend ER model by adding valid time to



attributes, e.g., address of an instructor at different points in time



entities, e.g., time duration when a student entity exists



relationships, e.g., time during which an instructor was associated with a student as an advisor.



But no accepted standard



Adding a temporal component results in functional dependencies like ID  street, city

not to hold, because the address varies over time



Chapter 8: Relational Database Design Chapter 8: Relational Database Design

Chapter 8: Relational Database Design

Chapter 8: Relational Database Design

Chapter 8: Relational Database Design Chapter 8: Relational Database Design

Features of Good Relational Design

Atomic Domains and First Normal Form

Decomposition Using Functional Dependencies

Functional Dependency Theory

Algorithms for Functional Dependencies

Decomposition Using Multivalued Dependencies

More Normal Form

Database-Design Process

Modeling Temporal Data

Combine Schemas?

Combine Schemas?

Suppose we combine instructor and department into inst_dept

(No connection to relationship set inst_dept)

Result is possible repetition of information

A Combined Schema Without Repetition A Combined Schema Without Repetition

Consider combining relations

sec_class(sec_id, building, room_number) and

section(course_id, sec_id, semester, year) into one relation

section(course_id, sec_id, semester, year, building, room_number)

No repetition in this case

What About Smaller Schemas?

What About Smaller Schemas?

Suppose we had started with inst_dept. How would we know to split up (decompose) it into instructor and department?

Write a rule “if there were a schema (dept_name, building, budget), then dept_name would be a candidate key”

Denote as a functional dependency:

dept_name  building, budget

In inst_dept, because dept_name is not a candidate key, the building and budget of a department may have to be repeated.

This indicates the need to decompose inst_dept

Not all decompositions are good. Suppose we decompose employee(ID, name, street, city, salary) into

employee1 (ID, name)

employee2 (name, street, city, salary)

The next slide shows how we lose information -- we cannot reconstruct

the original employee relation -- and so, this is a lossy decomposition.

A Lossy Decomposition

A Lossy Decomposition

Example of Lossless-Join Decomposition Example of Lossless-Join Decomposition

Lossless join decomposition

Decomposition of R = (A, B, C)

R

= (A, B) R

= (B, C)

A B

  1 2

A

 

B 1 2

r 

(r)



(r) 

(r) A B

  1 2

C A B B

1 2

C A B C

A B



(r)

First Normal Form 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 all attributes of R are atomic

Non-atomic values complicate storage and encourage redundant (repeated) storage of data

Example: Set of accounts stored with each customer, and set of owners stored with each account

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

Chapter 22: Object Based Databases)

First Normal Form (Cont’d) First Normal Form (Cont’d)

Atomicity is actually a property of how the elements of the domain are used.

Example: 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.

Goal — Devise a Theory for the Following Goal — Devise a Theory for the Following

(r) ^A ^B