chapter 6 relational database design

Chapter 6 Relational Database Design

6.1 Pitfalls in Relational-Database Design

6.2 Decomposition

6.3 Normalization Using Functional Dependence

6.4 BCNF decomposition

6.5 3NF decomposition

6.6 Fourth Normal Form

6.7 Exercise

Chapter 6 Relational Database Design

In general, the goal of a relational-database design is to generate a set of relation schemas that allows us to store information without unnecessary redundancy, yet also allows us to retrieve information easily. One approach is to design schemas that are in an appropriate normal form.

Among the undesirable properties that a bad design may have are.

① Repetition of information② Inability to represent certain information

Example:

The information concerning loans is now kept in one single relation, lending, which is defined over the relation schema. Lending-schema=(branch-name, branch-city, assets, customer-name, loan-number, amount)

branch-name

branch-city assets customer-name

loan-number

amount

Downtown Brooklyn 9000000 Jones L-17 1000

Redwood Palo Alto 2100000 Smith L-23 2000

Perryridge Horseneck 1700000 Hayes L-15 1500

Downtown Brooklyn 9000000 Jackson L-14 1500

Mianus Horseneck 400000 Jones L-93 500

Round Hill Horseneck 8000000 Turner L-11 900

Lending Figure A

Problems:① add a new loan

The loan is made by the Perryridge branch to Adams in the amount of $1500. Let the loan-number be L-31.

We add the tuple

(Perryridge, Horseneck, 1700000,Adams, L-31, 1500)

ⅰRepeating information wastes space.

ⅱRepeating information complicates updating the database.

② we cannot represent directly the information concerning a branch.(branch-name, branch-city,assets) unless there exists at least one loan at the branch. The problem is that tuples in the lending relation require values for loan-number, amount and customer-name.

Solution: introduce null values to handle updates through views.(difficult)

6.2 Decomposition

A bad design suggests that we should decompose a relation schema that has many attributes into several schemas with fever attributes. Careless decomposition, however, many lead to another form of bad design.

6.2 Decomposition

Example:

Consider an alternative design in which Lending-schema is decomposed into the following two schemas:

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

Customer-loan-schema=(customer-name,loan-number, amount)

6.2 Decomposition

branch-name

Downtown Brooklyn 9000000 Jones

Redwood Palo Alto 2100000 Smith

Perryridge Horseneck 1700000 Hayes

Downtown Brooklyn 9000000 Jackson

Mianus Horseneck 400000 Jones

Round Hill Horseneck 8000000 Turner

customer-name

loan-number

amount

Jones L-17 1000

Smith L-23 2000

Hayes L-15 1500

Jackson L-14 1500

Jones L-93 500

Turner L-11 900

Branch-customer B Customer-loan C

6.2 Decomposition

Suppose that we wish to find all branches that have loans with amount less than $1000. We should write.

branch-name

loan-number

amount

Redwood Palo Alto 2100000 Smith L-23 2000

Perryridge Horseneck 1700000 Hayes L-15 1500

Downtown Brooklyn 9000000 Jackson L-14 1500

Round Hill Horseneck 8000000 Turner L-11 900

Figure D

6.2 Decomposition

Problems:

Additional tuples:(Downtown, Brooklyn, 9000000,Jones, L-93, 500) (Mianus, Horsereck, 400000,Jones, L-17, 1000)

Consider the query, “Find all branches that have made a loan in an amount less than $1000”.

Figure A Mianus Round Hill

Figure D Mianus Round Hill Downtown

∏branch-name(amount<1000(branch-customer customer-loan))

6.2 Decomposition

Loss information: If a customer happens to have several loans from different branches, we cannot tell which loan belongs to which branch. Although we have more tuples in branch-customer customer-loan, we actually have less information. We are no longer able, in general, to represent in the database information about which customers are borrowers form which branch.

6.2 Decomposition

Lossy decomposition: Because of this loss of information, we call the decomposition of Lending-schema into Branch-customer-schema and customer-loan-schema a lossy decomposition, or a lossy-join decomposition.

6.2 Decomposition

Lossless-join decomposition: A decomposition that is not a lossy-join decomposition is a lossless-join decomposition.

Let C represent a set of constrains on the database. A decomposition {R1,R2……Rn} of a relation schema R is a lossless-join decomposition for R if for all relation r on schema R that are legal under C.

r=∏R1(r) ∏R2(r) …… ∏Rn(r)

Lossless-join Decomposition:

We must first present a criterion for determining whether a decomposition is lossy.

Let R be a relation schema, and let F be a set of functional dependencies on R. This decomposition is a lossless-join decomposition of R if at least one of the following functional dependencies are in F+.

① R1∩R2→R1

② R1∩R2→R2

Example:

We now show that our decomposition of Lending-schema is a lossless-join decomposition by showing a sequence of steps that generate the decomposition

① Branch-schema=(branch-name,branch-city, assets)

Loan-info-schema=(branch-name,customer-name,loan-number, amount)

branch-name→{branch-name,branch-city,assets}

Branch-schema∩Loan-info-schema={branch-name}

② Loan-info-schema

Loan-schema=(branch-name, loan-number, amount)

Borrower-schema=(customer-name, loan-number)

This step results is a lossless-join decomposition.

loan-number→{loan-number,branch-name,amount}

Loan-schema∩Borrower-schema={loan-number}

Example:

Suppose that we decompose the schema R=(A,B,C,D,E,F) into R1

={CF,BE,ECD,AB}; R2={ABE,CDEF}. Determine which decomposition is a lossless-join decomposition if the following set F of functional dependencies holds.

{ABE }∩{ CDEF}= {E}

E→A A→B E→B

F={A→B,C→F,E→A,CE→D}

E→A E→B E→AB E→ABE R2 is a lossless-join decomposition

A B C D E F

CF b11 b12 a3 b14 b15 a6

BE b21 a2 b23 b24 a5 b26

ECD b31 b32 a3 a4 a5 b36

AB a1 a2 b43 b44 b45 b46

b21a6a2

Another Algorithm:

Lossy-join decomposition

A B C D E F

ABE a1 a2 b13 b14 a5 b16

CDEF b21 b22 a3 a4 a5 a6a1 a2

Lossless-join decomposition

Dependency Preservation:

There is another goal in relational-database design: dependency preservation.

Let F be a set of functional dependencies on a schema R, and let R1,R2……Rn be a decomposition of R. The restriction of F to Ri is the set Fi of all functional dependencies in F+ that include only attributes of Ri.

Let F’=F1 F∪ 2 …… F∪ ∪ n. F’ is a set of functional dependencies on schema R. if F’+= F+ is true, then every dependency in F is logically implied by F’, and, if we verify that F’ is satisfied, we have verified that F is satisfied. We say that a decomposition having the property F’+= F+ is a dependency-preserving decomposition.

Normal Form:

The process of further normalization—here in after abbreviated to just normalization—is built around the concept of normal forms. A relvar is said to be in a particular normal form if it satisfies a certain prescribed set of conditions.

① First normal form: A relvar is in 1NF if and only if, in every legal value of that relvar, every tuple contains exactly one value for each attribute.

Customer-name

Customer-city

Customer-street

Jones Brooklyn Ray

Hayes Palo Alto Heroes

Customer-name Customer-

cityCustomer-street

Brooklyn

Palo Alto

Heroes

② Second normal form (definition assuming only one candidate key, which we assume is the primary key): A relvar is in 2NF if and only if it is in 1NF and every nonkey attribute is irreducibly dependent on the primary key.

Example:

Relation R=(BNO,Title, Author, Price,Lo-No, Card, Name, Dept,Date)

F={BNO→Title, Author, Price, Lo-NoCard →Name, DeptBNO,Card →Date}

Candidate key {BNO,Card}

{BNO,Card} →Title

BNO →Title F∈

So, it not in 2NF

BNO Title Card

Redundancy

③ Third normal form(definition assuming only one candidate key, which we further assume is the primary key): A relvar is in 3NF if and only if it is in 2NF and every nonkey attribute is nontransitively dependent on the primary key. Note: “No transitive dependences” implies no mutual dependencies.

Example:

Relvar student=(SNO, SNAME, SAGE,MASTER,CLASS, SEX)

F={SNO→SNAME, SNO →SAGE, SNO →SEX,SNO →CLASS,CLASS →MASTER}

Candidate key {SNO}

SNO →CLASSCLASS →MASTER

MASTER transitively dependent on SNO.

④ Boyce/codd normal form(BCNF): A relvar is in BCNF if and only if every nontrivial, left-irreducible FD has a candidate key as its determinant.

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:

ⅰ → is a trivial functional dependency (that is ) ⅱ is a superkey for schema R

Examples:

① customer-schema=(customer-name,customer- street, customer-city)

② branch-schema=(branch-name,assets,branch-city)

customer-name→{customer-street,customer-city}

branch-name→ {assets,branch-city}

customer-name is a candidate key

customer-schema is in BCNF

branch-schema is a candidate key

branch-schema is in BCNF

③ loan-info-schema=(branch-name, customer-name,loan-number, amount)

loan-number→{amount,branch-name}

loan-number is not a candidate key

loan-number → {amount,branch-name} is not a trivial dependency.

loan-info-schema is not in BCNF

(Downtown,John Bell,L-44,1000)

(Downtown,Jane Bell,L-44,1000)

loan-info-schema=(branch-name, customer-name,loan-number, amount)

loan-info-schema is not in BCNF. Redundancy exist, so we should decompose it.

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

borrower-schema=(customer-name,loan-number)

loan-number is a candidate key of loan-schema

loan-schema and borrower-schema are both in BCNF

If R is not in BCNF, we can decompose R into a collection of BCNF schemas R1,R2……Rn. Which not only a BCNF decomposition, but also a lossless-join 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 thatholds on Ri such that Ri is not in F+ ,and ∩ = ;result:=(result-Ri) ∪(Ri- ) ∪( ,);endelse done:=true;

Example:Lending-schema={branch-name,branch-city, assets,customer-name,loan-number,amount}

A candidate key for this schema is:

branch-name is not a superkey. Thus ,Lending-schema is not in BCNF, so decompose it.

BCNF decomposition :

{loan-number,customer-name}

branch-name→{branch-city,assets}

loan-number→{amount,branch-name}F =

R1=branch-name {∪ branch-city assets}

branch-name→{branch-city,assets}

{branch-name,branch-city, assets,customer-name,loan-number,amount}{branch-name}+

① branch-name→ {branch-name,branch-city, assets,customer-name,loan-number,amount} is not in F+

② branch-name∩{branch-city,assets}=

R2={branch-name,branch-city, assets,customer-name,loan-number,amount}-{branch-city,assets}

Branch-schema={branch-name,branch-city, assets}

Loan-info-schema={branch-name,customer-name,loan-number,amount}

branch-name is a key for Branch-schema, Thus ,Branch-schema is in BCNF.

loan-number is not a key for Loan-info-schema. Thus,Loan-info-schema is not in BCNF

loan-number→{amount,branch-name}F’ =

Decompose Loan-info-schema

R2= loan-number {∪ amount,branch-name}

{branch-name,customer-name,loan-number,amount}{loan-number}+

① branch-name→{branch-name,customer-name,loan-number,amount} is not in F’+

② loan-number∩{amount,branch-name}=

R3= {branch-name,customer-name,loan-number,amount}-{amount,branch-name}

Loan-schema={branch-name,loan-number,amount}

Borrower-schema={customer-name,loan-number}

Branch-schema and Borrower-schema are in BCNF.

Branch-schema={branch-name,branch-city, assets}

Loan-schema={branch-name,loan-number,amount}

Lending-schema=(branch-name,branch-city, assets,customer-name,loan-number,amount)

Algorithm:

Input: a schema R and R satisfied the FDs F.

Output: R is a lossless-join decomposition which satisfies the FDs F and for each subschemas Ri is a BCNF de

composition which satisfies Fi=∏Ri(F)

① initialize the result closure to {R}, ={R}

② if all subschemas in are BCNF then turn to ④. Let us find a subschemas S in that is not in BCNF. According to the definition of BCNF, there must has a FD X→A and A-X=, X do not include any candidate key of S, S-X-A≠ .

③ Let S={S1,S2}, S1 is XA, S2 is S-A (A∩X)∪ , replace the S with {S1,S2} turn to ②

④ Stop decomposing and output

Compute candidate key:

If x+=u and y+ is a proper subset of x+ ( y+ ≠u). then the attributes in R can be classified into four groups:

L: the attributes appear only on the left-hand side of FD

R: the attributes appear only on the right-hand side of FD

LR: the attributes appear both on the left-hand side and left-hand side of FD

N: the attributes do not appear on the FD

③ L+N unknown:

② the classification of (L+N) must appear in candidate key

① the classification of R are not candidate key

ⅰ L+N =x1 x1+=u then have only one candidate key

ⅱ L+N =x1 x1+≠u find an attribute in R and add it to the x1.repeat!

Example:

Relvar R=(O,I,S,Q,B,D) satisfies the following FDs.

F={S→D, I →B, IS →Q, B →O}

Give a lossless-join decomposition into BCNF of the schema R

L: S, I R: D, Q, O LR: B

(SI)+=SIDBQO=U have only one candidate key.

S→D ISS not in BCNF

S→D D-S≠ S do not include IS

Decompose: R1=SD R2=OISQB

FR2={I→B, IS →Q, B →O}

L: S, I R: Q, O LR: B

(SI)+=SIBOQ=U have only one candidate key.

I→B ISI B-I≠ not in BCNF

Decompose: R21=IB R22=OISQ

F R21={I→B}F R22 ={IS →Q}

=(R1, R21, R22)

Example:

Relvar R=(A,B,C,D) satisfies the following FDs.F={A→D, C→D, D →B}

Give a lossless-join decomposition into BCNF of the schema R

L: A,C R: B LR: D

(AC)+=ACDB=U have only one candidate key.

Decompose: R1=AD R2=ACB

FR2={C→B, A →B}

……

Not every BCNF decomposition is dependency preserving.

Example:Banker-schema={branch-name,customer-name,banker-name}

banker-name→{branch-name}

{branch-name,customer-name}→banker-nameF =

banker-name is not a key for Banker-schema.

Thus, Banker-schema is not in BCNF. Decompose it.

Banker-branch-schema={banker-name,branch-name}

Customer-banker-schema={customer-name,banker-name}

F1={banker-name→branch-name}

{branch-name,customer-name}→banker-name in F+

{branch-name,customer-name}→banker-name not in {F1 F∪ 2}+

{F1 F∪ 2}+<>F+ and the decomposition is not dependency preserving

This example demonstrates that not every BCNF decomposition is dependency preserving. Moreover, it demonstrates that we cannot always satisfy all three design goals:

③ Dependency preservation

② Lossless join

① BCNF

A relation schema R is in 3NF 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:

① → is a trivial functional dependency (that is )② is a superkey for schema R

③ each attribute A in - is contained in a candidate key for R.

Banker-schema={branch-name,customer-name,banker-name}

banker-name→ {branch-name}

{branch-name,customer-name}→banker-nameF =

banker-name is not a key for Banker-schema.

Thus, Banker-schema is not in BCNF.

Example:

{branch-name,customer-name} is a candidate key for Banker-schema. Banker-schema is in 3NF.

If R is not in 3NF, we can decompose R into a collection of 3NF schemas R1,R2……Rn, which is not only a lossless-join decomposition, but also a dependency-preserving 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, j=1,2……,i contains

then begini:=i+1;Ri:= ;

endIf none of the schemas Rj, j=1,2……,i contains a candidate key for R

then begini:=i+1;Ri:= any candidate key for R ;

endReturn(R1,R2,……,Ri)

Example:

Relvar R=(O,I,S,Q,B,D) satisfies the following FDs.

F={S→D, I→B, IS →Q, B →O}

Give a lossless-join, dependency-preserving decomposition into 3NF of the schema R.

={SD,IB,ISQ,BO}

L: S,I R: D,Q,O LR: B(IS)+=ISDBQO=U have only one candidate key.

Example:Banker-info-schema={branch-name,customer-name,banker-name,office-number}

Banker-name→{branch-name,office-number}

{customer-name,branch-name}→banker-name

R1= banker-name∪{branch-name,office-number}

R2= {customer-name,branch-name}∪banker-name

3NF decomposition:

Banker-office-schema={banker-name,branch-name,office-number}

Banker-schema={customer-name,branch-name,banker-name}

banker-schema contains a candidate key for banker-info-schema

Banker-info-schema={branch-name,customer-name,banker-name,office-number}

Banker-office-schema={banker-name,branch-name,office-number}

Banker-schema={customer-name,branch-name,banker-name}

6.5 comparison of BCNF and 3NF

Advantage of 3NF:

It is always possible to obtain a 3NF design without sacrificing a lossless join or dependency preservation.

If we do not eliminate all transitive dependencies, we may have to use null values to represent some of the possible meaningful relationships among data items, and there is the problem of repetition of information.

Disadvantage of 3NF:

If a schema is in 3NF but not in BCNF,then redundancy will occur.

Consider banking example:

BC-schema={loan-number,customer-name,customer-street,customer-city}

customer-name→{customer-street,customer-city}

customer-name is not a key for BC-schema, so BC-schema is not a BCNF schema.

Remove this functional dependency, BC-schema is a BCNF schema. But repetition of information are still exist.

Solution:

multivalued dependencies

As we did for functional dependencies, we shall use multivalued dependencies to define a normal form for relation schemas. This normal form, called fourth normal form(4NF), is more restrictive than BCNF.

4NF schema is in BCNF, BCNF schema is not in 4NF

6.6.1 Multivalued Dependencies

Functional dependencies rule out certain tuples from being in a relation. If A→B, then we cannot have two tuples with the same A value but different B values. Multivalued dependencies do not rule out the existence of certain tuples. Instead, they require that other tuples of a certain form be present in the relation. For this , functional dependencies sometimes are referred to as equality-generating dependencies, and multivalued dependencies are referred to as tuple-generating dependencies.

Definition: Let R be a relation schema and let R and R. then

multivalued dependency →→ holds on R if , in any legal relation r( R ), for all pairs of

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--]

Intuitively, the multivalued dependency →→ says that the relationship between and is independent of the relationship between and R-. If the multivalued dependency →→ is satisfied by all relations on schema R, then →→ is a trivial multivalued dependency on schema R. thus, →→ is trivial if or ∪=R.

R--t1 1… I i+1… j j+1… n

t2 1… I bi+1… bj bj+1… bn

t3 1… I i+1… j bj+1… bn

t4 1… I bi+1… bj j+1… n

Example:Relvar R-schema={C,T,H,R,S,G}

C: course name T: teacher H: time R: classroom S: student G: grade

time classroom

C→→HRT→→HR

C T H R S G

C1 T1 H1 R1 S1 G1

C1 T1 H2 R2 S1 G1

C1 T1 H1 R1 S2 G2

C1 T1 H2 R2 S2 G2

6.6.2 Theory of Multivalued Dependencies

Let D denote a set of functional and multivalued dependencies. The closure D+ of D is the set of all functional and multivalued dependencies logically implied by D.

1) Reflexivity rule: if is a set of attributes and , then holds.

2) Augmentation rule: if holds and is a set of attributes, then holds.

The following list of inference rules for functional and multivalued dependencies is sound and complete.

3) Transitivity rule: if holds and holds, then holds.

4) Complementation: if holds, then R-- holds.

5) Multivalued augmentation rule: if holds and R and then holds.

6) Multivalued Transitivity rule: if holds and then - holds.

7) Replication rule: if holds then holds.

8) Coalescence rule: if holds and , and there is a such that R, and ∩= , and then holds.

Multivalued union rule: if holds and then holds.

Intersection rule: if holds and then ∩ holds.

Difference rule: if holds and then - holds and - holds.

Example:Let R={A,B,C,G,H,I} with the following set of dependencies D given:

AB BHI CGH

AHI : AB BHI AHI-B (HI) AHI (multivalued transitivity rule)

ACGHI : AB AR-B-A (CGHI) ACGHI (complementation rule)

BH : BHI HHI and CG∩HI= BH BH (coalescence rule)

ACG : ABCGHI AHI ACGHI-HI (CG) ACG (difference rule)

6.6.3 Fourth Normal Form

Definition: 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 holds

is a trivial multivalued dependency

is a superkey for schema R

a database design is in 4NF if each member of the set of relation schemas that constitutes the design is in 4NF

6.6.3 4NF decomposition

Algorithm:result:={R};done:=false;compute D+;while (not done) do

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

let be a nontrivial functional dependency that

result:=(result-Ri) ∪(Ri- ) ∪( ,);end

else done:=true;

holds on Ri such that Ri is not in D+ and ∩= ;

Example:

customer-nameloan-number

BC-schema={loan-number,customer-name,customer-street,customer-city}

customer-name is not a superkey for BC-schema

customer-name{loan-number,customer-name,customer-street,customer-city} not exist in D+

customer-name∩loan-number=

R2={loan-number,customer-name,customer-street,customer-city}- loan-number

Customer-schema={customer-name,customer-street,customer-city}

R1=customer-name∪loan-number

This algorithm generates only lossless-join decompositions:

let R be a relation schema, and let D be a set of functional and multivalued dependencies of R. Let R1 and R2 form a decomposition of R. This decomposition is a lossless-join decomposition of R if and only if at least one of the following multivalued dependencies is in D+:

R1∩R2 R1

R1∩R2 R2

consider a set of D of both functional and 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

Where Ri and is in D+

If we are given a set of multivalued and functional dependencies, it is advantageous to find a database design that meets the three criteria of

③ Dependency preservation

② Lossless join

① 4NF (all we have are functional dependencies, BCNF)

The 4NF decomposition can not ensure dependency preservation.

When we can not achieve our three goals, we compromise on 4NF,and accept BCNF or even 3NF, if necessary, to ensure dependency preservation.

Exercise:

① Suppose that we decompose the schema R=(U,V,W,X,Y,Z) into 1=

{WZ,VY,WXY,UVY}; 2={UVY,WXYZ} determine which decomposition is a lossless-join decomposition if the following set F of functional dependencies holds

F={U→V, W →Z, Y →U, WY →X}

② Suppose that we decompose the schema ={R1(A1A4), R2(A1A2), R3(A2A3), R4(A3A4A5),R5(A1A5)} determine whether this decomposition is a lossless-join decomposition. If the following set F of functional dependencies holds

F={A1→A3, A3→A4, A2→A3, A4A5→A3, A3A5→A1}

Exercise:

③ Relvar R=(A,B,C,D,E) satisfies the following FDs:

F={A→C,C →D,B →C,DE →C,CE →A}

ⅰFind an irreducible equivalent for this set of FDs

ⅱFind all candidate keys of the schema R.

ⅲDetermine whether the decomposition ={AD,AB,BC,CDE,AE} is a lossless-join decomposition or not.

ⅳGive a lossless-join decomposition into BCNF of the schema R.

ⅴGive a lossless-join, dependency-preserving decomposition into 3NF of the schema R.

chapter 6 relational database design

relational database

relationaldatabase design6

perryridge branch

new loan

alternative design

form of bad design

lending relation

loannumb amount6

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