Chapter 6 Relational Database Design
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Transcript of 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.
6.1 Pitfalls in Relational-Database Design
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)
6.1 Pitfalls in Relational-Database Design
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
6.1 Pitfalls in Relational-Database Design
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.
6.1 Pitfalls in Relational-Database Design
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
branch-city assets customer-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
branch-city assets customer-name
loan-number
amount
Downtown Brooklyn 9000000 Jones L-17 1000
Downtown Brooklyn 9000000 Jones L-93 500
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-17 1000
Mianus Horseneck 400000 Jones L-93 500
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)
6.3 Normalization Using Functional Dependence
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
6.3 Normalization Using Functional Dependence
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}
6.3 Normalization Using Functional Dependence
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}
6.3 Normalization Using Functional Dependence
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
6.3 Normalization Using Functional Dependence
F={A→B,C→F,E→A,CE→D}
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
6.3 Normalization Using Functional Dependence
A B C D E F
ABE a1 a2 b13 b14 a5 b16
CDEF b21 b22 a3 a4 a5 a6a1 a2
Lossless-join decomposition
F={A→B,C→F,E→A,CE→D}
6.3 Normalization Using Functional Dependence
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.
6.3 Normalization Using Functional Dependence
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.
6.3 Normalization Using Functional Dependence
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.
6.3 Normalization Using Functional Dependence
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
Jones
Hayes
Brooklyn
Palo Alto
Ray
Heroes
City
6.3 Normalization Using Functional Dependence
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}
6.3 Normalization Using Functional Dependence
Candidate key {BNO,Card}
{BNO,Card} →Title
BNO →Title F∈
So, it not in 2NF
BNO Title Card
1 A W
2 B W
3 C W
Redundancy
6.3 Normalization Using Functional Dependence
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.
6.3 Normalization Using Functional Dependence
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.
6.3 Normalization Using Functional Dependence
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
6.3 Normalization Using Functional Dependence
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
6.3 Normalization Using Functional Dependence
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)
6.4 BCNF decomposition
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→{amount,branch-name}
loan-number is a candidate key of loan-schema
loan-schema and borrower-schema are both in BCNF
6.4 BCNF decomposition
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.
6.4 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 thatholds on Ri such that Ri is not in F+ ,and ∩ = ;result:=(result-Ri) ∪(Ri- ) ∪( ,);endelse done:=true;
6.4 BCNF decomposition
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 =
6.4 BCNF decomposition
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}
6.4 BCNF decomposition
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
6.4 BCNF decomposition
R2= loan-number {∪ amount,branch-name}
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}
6.4 BCNF decomposition
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}
Borrower-schema={customer-name,loan-number}
Lending-schema=(branch-name,branch-city, assets,customer-name,loan-number,amount)
6.4 BCNF decomposition
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)
6.4 BCNF decomposition
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
6.4 BCNF decomposition
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
6.4 BCNF decomposition
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!
6.4 BCNF decomposition
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
6.4 BCNF decomposition
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)
6.4 BCNF decomposition
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}
……
6.4 BCNF decomposition
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.
6.4 BCNF decomposition
Banker-branch-schema={banker-name,branch-name}
Customer-banker-schema={customer-name,banker-name}
F1={banker-name→branch-name}
F2=
{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
6.4 BCNF decomposition
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
6.5 3NF decomposition
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.
6.5 3NF decomposition
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.
6.5 3NF decomposition
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.
6.5 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, 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)
6.5 3NF decomposition
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.
F=FC
={SD,IB,ISQ,BO}
L: S,I R: D,Q,O LR: B(IS)+=ISDBQO=U have only one candidate key.
6.5 3NF decomposition
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:
6.5 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.
6.6 Fourth Normal Form
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.
6.6 Fourth Normal Form
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.
6.6.1 Multivalued 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--]
6.6.1 Multivalued Dependencies
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
6.6.1 Multivalued Dependencies
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.
6.6.2 Theory of Multivalued Dependencies
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.
6.6.2 Theory of Multivalued Dependencies
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.
6.6.2 Theory of Multivalued Dependencies
Example:Let R={A,B,C,G,H,I} with the following set of dependencies D given:
AB BHI CGH
R+
AHI : AB BHI AHI-B (HI) AHI (multivalued transitivity rule)
ACGHI : AB AR-B-A (CGHI) ACGHI (complementation rule)
6.6.2 Theory of Multivalued Dependencies
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 ∩= ;
6.6.3 4NF decomposition
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=
6.6.3 4NF decomposition
R2={loan-number,customer-name,customer-street,customer-city}- loan-number
Customer-schema={customer-name,customer-street,customer-city}
Borrower-schema={customer-name,loan-number}
R1=customer-name∪loan-number
6.6.3 4NF decomposition
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
6.6.3 4NF decomposition
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
∩Ri
Where Ri and is in D+
6.6.3 4NF decomposition
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.