Chapter 7,Relational Database Design
?Silberschatz,Korth and Sudarshan7.2Database System Concepts
Chapter 7,Relational Database Design
? First Normal Form
? Pitfalls in Relational Database Design
? Functional Dependencies
? Decomposition
? Boyce-Codd Normal Form
? Third Normal Form
? Multivalued Dependencies and Fourth Normal
Form
? Overall Database Design Process
?Silberschatz,Korth and Sudarshan7.3Database System Concepts
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
?Suppose that students are given roll numbers which
are strings of the form CS0012 or EE1127
? A relational schema R is in first normal form（第一范式，
1NF） if the domains of all attributes of R are atomic
?Silberschatz,Korth and Sudarshan7.4Database System Concepts
Pitfalls in Relational Database Design
? Relational database design requires that we find a
“good” collection of relation schemas,A bad design may
lead to
? Repetition of Information.
? Inability to represent certain information.
? Consider the relation schema:
Lending-schema = (branch-name,branch-city,
assets,customer-name,loan-number,amount)
?Silberschatz,Korth and Sudarshan7.5Database System Concepts
Example
? Redundancy（冗余）,
? Data for branch-name,branch-city,assets are repeated
for each loan that a branch makes
? Wastes space
? Complicates updating,introducing possibility of
inconsistency of assets value
? Null values
? Cannot store information about a branch if no loans exist
? Can use null values,but they are difficult to handle.
?Silberschatz,Korth and Sudarshan7.6Database System Concepts
Decomposition
? Decompose the relation schema Lending-schema into:
Branch-schema = (branch-name,branch-city,assets)
Loan-info-schema = (customer-name,loan-number,
branch-name,amount)
? All attributes of an original schema (R) must appear in the
decomposition (R1,R2):
R = R1 ? R2
? Lossless-join decomposition（无损连接分解）,
For all possible relations r on schema R
r = ?R1 (r) ?R2 (r)
?Silberschatz,Korth and Sudarshan7.7Database System Concepts
Example of Non Lossless-Join Decomposition
? Decomposition of R = (A,B)
R2 = (A) R2 = (B)
A B
?
?
?
1
2
1
A
?
?
B
1
2
r
?A(r) ?B(r)
?A (r) ?B (r) A B
?
?
?
?
1
2
1
2
?Silberschatz,Korth and Sudarshan7.8Database System Concepts
Goal — Devise a Theory for the Following
? Decide whether a particular relation R is in,good”
form.
? In the case that a relation R is not in,good” form,
decompose it into a set of relations {R1,R2,...,Rn}
such that
? each relation is in good form
? the decomposition is a lossless-join decomposition
? Our theory is based on:
? functional dependencies
? multivalued dependencies
?Silberschatz,Korth and Sudarshan7.9Database System Concepts
Functional Dependencies
（函数依赖）
? 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.
?Silberschatz,Korth and Sudarshan7.10Database System Concepts
Functional Dependencies (Cont.)
? Let R be a relation schema
?? R and ? ? R
? The functional dependency
? ? ?
holds on R if and only if for any legal relations r(R),
whenever any two tuples t1 and t2 of r agree on the
attributes ?,they also agree on the attributes ?,That is,
t1[?] = t2 [?] ? t1[? ] = t2 [? ]
? 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
?Silberschatz,Korth and Sudarshan7.11Database System Concepts
Functional Dependencies (Cont.)
? Functional dependencies allow us to express constraints that
cannot be expressed using superkeys,（函数依赖使我们可
以表示不能用超玛表示的约束） Consider the schema:
Loan-info-schema = (customer-name,loan-number,
branch-name,amount).
We expect this set of functional dependencies to hold:
loan-number ? amount
loan-number ? branch-name
but would not expect the following to hold,
loan-number ? customer-name
?Silberschatz,Korth and Sudarshan7.12Database System Concepts
Use of Functional Dependencies
? We use functional dependencies to:
? test relations to see if they are legal under a given set
of functional dependencies,（用于判定关系是否在给定
函数依赖集上合法）
? 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（用于指
明合法关系集上的约束）
?We say that F holds on R if all legal relations on R
satisfy the set of functional dependencies F.
?A?C,C?A
A B C D
a1 b1 c1 d1
a1 b2 c1 d2
a2 b2 c2 d2
a2 b2 c2 d3
a3 b3 c2 d4
?Silberschatz,Korth and Sudarshan7.13Database System Concepts
Functional Dependencies (Cont.)
? A functional dependency is trivial(平凡的） if it is
satisfied by all instances of a relation
? E.g.
? customer-name,loan-number ? customer-
name
? customer-name ? customer-name
? In general,? ? ? is trivial if ? ??
?Silberschatz,Korth and Sudarshan7.14Database System Concepts
The customer Relation
Customer-name?customer-city?
Customer-street?customer-city?
?Silberschatz,Korth and Sudarshan7.15Database System Concepts
The loan Relation
Loan-number?amount?
Loan-number?branch-name?
?Silberschatz,Korth and Sudarshan7.16Database System Concepts
The branch Relation
Branch-name?assets?
assets?Branch-name?
?Silberschatz,Korth and Sudarshan7.17Database System Concepts
Closure of a Set of Functional
Dependencies（函数依赖集的闭包）
? Given a set F set of functional dependencies,there are certain other
functional dependencies that are logically implied（逻辑蕴含） by F.
? 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+.
? We can find all of F+ by applying Armstrong’s Axioms（公理）,
? if ? ??,then ? ?? (reflexivity)自反律
? if ? ??,then ?? ? ?? (augmentation)增补率
? if ? ??,and ? ??,then ? ? ? (transitivity)传递率
? These rules are
? Sound（保真的） (不会产生错误的函数依赖 )
? complete （完备的） (对一个给定函数依赖集 F，它们能产生整个 F+).
?Silberschatz,Korth and Sudarshan7.18Database System Concepts
Example
? R = (A,B,C,G,H,I)
F = { A ? B
A ? C
CG ? H
CG ? I
B ? H}
? some members of F+
? A ? H
?by transitivity from A ? B and B ? H
? AG ? I
?by augmenting A ? C with G,to get AG ? CG
and then transitivity with CG ? I
? CG ? HI
?from CG ? H and CG ? I,,union rule” can be inferred from
– definition of functional dependencies,or
– Augmentation of CG ? I to infer CG ? CGI,augmentation of
CG ? H to infer CGI ? HI,and then transitivity
?Silberschatz,Korth and Sudarshan7.19Database System Concepts
Procedure for Computing F+
? To compute the closure of a set of functional dependencies
F:
F+ = F
repeat
for each functional dependency f in F+
apply reflexivity and augmentation rules on f
add the resulting functional dependencies to F+
for each pair of functional dependencies f1and f2 in F+
if f1 and f2 can be combined using transitivity
then add the resulting functional dependency to
F+
until F+ does not change any further
2× 2n＝ 2n+1
NOTE,We will see an alternative procedure for this task later
?Silberschatz,Korth and Sudarshan7.20Database System Concepts
Closure of Functional Dependencies
(Cont.)
? We can further simplify manual computation of F+ by using
the following additional rules.
? If ? ? ? holds and ? ? ? holds,then ? ? ? ? holds
(union)合并率
? If ? ? ? ? holds,then ? ? ? holds and ? ? ? holds
(decomposition)分解率
? If ? ? ? holds and ? ? ? ? holds,then ? ? ? ? holds
(pseudotransitivity)伪传递率
The above rules can be inferred from Armstrong’s axioms.
?Silberschatz,Korth and Sudarshan7.21Database System Concepts
Closure of Attribute Sets（属性集的闭包）
? 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:
? ? ? is in F+ ? ? ? ?+
? Algorithm to compute ?+,the closure of ? under F
result,= ?;
while (changes to result) do
for each ? ? ? in F do
begin
if ? ? result then result,= result ? ?
end
?Silberschatz,Korth and Sudarshan7.22Database System Concepts
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
?Silberschatz,Korth and Sudarshan7.23Database System Concepts
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.
?Silberschatz,Korth and Sudarshan7.24Database System Concepts
Canonical Cover（正则覆盖）
? Sets of functional dependencies may have redundant
dependencies that can be inferred from the others
? Eg,A ? C is redundant in,{A ? B,B ? C,A ? C}
? Parts of a functional dependency may be redundant
?E.g,on RHS,{A ? B,B ? C,A ? CD} can be
simplified to
{A ? B,B ? C,A ? D}
?E.g,on LHS,{A ? B,B ? C,AC ? D} can be
simplified to
{A ? B,B ? C,A ? D}
? Intuitively,a canonical cover of F is a,minimal” set of
functional dependencies equivalent to F,having no
redundant dependencies or redundant parts of
dependencies
?Silberschatz,Korth and Sudarshan7.25Database System Concepts
Extraneous Attributes（无关属性）
? 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.
? 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
?Silberschatz,Korth and Sudarshan7.26Database System Concepts
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 A; if it does,A is
extraneous
? To test if attribute A ? ? is extraneous in ?
1,compute ?+ using only the dependencies in
F’ = (F – {? ? ?}) ? {? ?(?– A)},
2,check that ?+ contains A; if it does,A is
extraneous
?Silberschatz,Korth and Sudarshan7.27Database System Concepts
Canonical Cover
? A canonical cover for F is a set of dependencies Fc such that
? F logically implies all dependencies in Fc,and
? Fc logically implies all dependencies in F,and
? No functional dependency in Fc contains an extraneous
attribute,and
? Each left side of functional dependency in Fc is unique.
? To compute a canonical cover for F:
repeat
Use the union rule to replace any dependencies in F
?1 ? ?1 and ?1 ? ?1 with ?1 ? ?1 ?2
Find a functional dependency ? ? ? with an
extraneous attribute either in ? or in ?
If an extraneous attribute is found,delete it from ? ? ?
until F does not change
?Silberschatz,Korth and Sudarshan7.28Database System Concepts
Example of Computing a Canonical Cover
? R = (A,B,C)
F = {A ? BC
B ? C
A ? B
AB ? C}
? Combine A ? BC and A ? B into A ? BC
? Set is now {A ? BC,B ? C,AB ? C}
? A is extraneous in AB ? C
? Check if the result of deleting A from AB ? C is implied by the other
dependencies
?Yes,in fact,B ? C is already present!
? Set is now {A ? BC,B ? C}
? C is extraneous in A ? BC
? Check if A ? C is logically implied by A ? B and the other
dependencies
?Yes,using transitivity on A ? B and B ? C,
– Can use attribute closure of A in more complex cases
? The canonical cover is,A ? B,B ? C
?Silberschatz,Korth and Sudarshan7.29Database System Concepts
Decomposition
? Decompose the relation schema Lending-schema into:
Branch-schema = (branch-name,branch-city,assets)
Loan-info-schema = (customer-name,loan-number,
branch-name,amount)
? All attributes of an original schema (R) must appear in the
decomposition (R1,R2):
R = R1 ? R2
? Lossless-join decomposition.
For all possible relations r on schema R
r = ?R1 (r) ?R2 (r)
? A decomposition of R into R1 and R2 is lossless join if and only if at
least one of the following dependencies is in F+:
? R1 ? R2 ? R1
? R1 ? R2 ? R2
?Silberschatz,Korth and Sudarshan7.30Database System Concepts
The Relation branch-customer
?Silberschatz,Korth and Sudarshan7.31Database System Concepts
The Relation customer-loan
?Silberschatz,Korth and Sudarshan7.32Database System Concepts
The Relation branch-customer
customer-loan
?Silberschatz,Korth and Sudarshan7.33Database System Concepts
Example of Lossy-Join Decomposition
? Lossy-join decompositions result in information loss.
? Example,Decomposition of R = (A,B)
R2 = (A) R2 = (B)
A B
?
?
?
1
2
1
A
?
?
B
1
2
r
?A(r) ?B(r)
?A (r) ?B (r) A B
?
?
?
?
1
2
1
2
?Silberschatz,Korth and Sudarshan7.34Database System Concepts
Normalization Using Functional Dependencies
? When we decompose a relation schema R with a set of
functional dependencies F into R1,R2,..,Rn we want
? Lossless-join decomposition,Otherwise decomposition
would result in information loss.
? No redundancy,The relations Ri preferably should be in
either Boyce-Codd Normal Form or Third Normal Form.
? Dependency preservation（保持依赖）, Let Fi be the set
of dependencies F+ that include only attributes in Ri,
? Preferably the decomposition should be dependency
preserving,that is,(F1 ? F2 ? … ? Fn)+ = F+
?Otherwise,checking updates for violation of functional
dependencies may require computing joins,which is
expensive.
?Silberschatz,Korth and Sudarshan7.35Database System Concepts
Example
? R = (A,B,C)
F = {A ? B,B ? C)
? Can be decomposed in two different ways
? R1 = (A,B),R2 = (B,C)
? Lossless-join decomposition:
R1 ? R2 = {B} and B ? BC
? Dependency preserving
? R1 = (A,B),R2 = (A,C)
? Lossless-join decomposition:
R1 ? R2 = {A} and A ? AB
? Not dependency preserving
(cannot check B ? C without computing R1 R2)
?Silberschatz,Korth and Sudarshan7.36Database System Concepts
Testing for Dependency Preservation
? To check if a dependency ??? is preserved in a decomposition of R
into R1,R2,…,R n we apply the following simplified test (with attribute
closure done w.r.t,F)
? result = ?
while (changes to result) do
for each Ri in the decomposition
t = (result ? Ri)+ ? Ri
result = result ? t
? If result contains all attributes in ?,then the functional dependency
?? ? is preserved.
? We apply the test on 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 (F1 ? F2 ? … ? Fn)+
?Silberschatz,Korth and Sudarshan7.37Database System Concepts
Boyce-Codd Normal Form（ BC范式 ）
? ?? ? ? 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:
?Silberschatz,Korth and Sudarshan7.38Database System Concepts
Example
? R = (A,B,C)
F = {A ? B
B ? C}
Key = {A}
? R is not in BCNF
? Decomposition R1 = (A,B),R2 = (B,C)
? R1 and R2 in BCNF
? Lossless-join decomposition
? Dependency preserving
?Silberschatz,Korth and Sudarshan7.39Database System Concepts
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,using only F is incorrect when testing a relation in a
decomposition of R
? E.g,Consider R (A,B,C,D),with F = { A ?B,B ?C}
?Decompose R into R1(A,B) and R2(A,C,D)
?Neither of the dependencies in F contain only attributes from
(A,C,D) so we might be mislead into thinking R2 satisfies BCNF,
?In fact,dependency A ? C in F+ shows R2 is not in BCNF,
?Silberschatz,Korth and Sudarshan7.40Database System Concepts
BCNF Decomposition Algorithm
result,= {R};
done,= false;
compute F+;
while (not done) do
if (there is a schema Ri in result that is not in BCNF)
then begin
let ?? ? ? be a nontrivial functional
dependency that holds on Ri
such that ?? ? Ri is not in F+,
and ? ? ? = ?;
result,= (result – Ri ) ? (Ri – ?) ? (?,? );
end
else done,= true;
Note,each Ri is in BCNF,and decomposition is lossless-
join.
?Silberschatz,Korth and Sudarshan7.41Database System Concepts
Example of BCNF Decomposition
? R = (branch-name,branch-city,assets,
customer-name,loan-number,amount)
F = {branch-name ? assets branch-city
loan-number ? amount branch-name}
Key = {loan-number,customer-name}
? Decomposition
? R1 = (branch-name,branch-city,assets)
? R2 = (branch-name,customer-name,loan-number,amount)
? R3 = (branch-name,loan-number,amount)
? R4 = (customer-name,loan-number)
? Final decomposition
R1,R3,R4
?Silberschatz,Korth and Sudarshan7.42Database System Concepts
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
It is not always possible to get a BCNF decomposition that is
dependency preserving
?Silberschatz,Korth and Sudarshan7.43Database System Concepts
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.
? Allows some redundancy (with resultant problems;
we will see examples later)
? But FDs can be checked on individual relations
without computing a join.
? There is always a lossless-join,dependency-
preserving decomposition into 3NF.
?Silberschatz,Korth and Sudarshan7.44Database System Concepts
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).
?Silberschatz,Korth and Sudarshan7.45Database System Concepts
3NF (Cont.)
? Example
? R = (J,K,L)
F = {JK ? L,L ? K}
? Two candidate keys,JK and JL
? R is in 3NF
JK ? L JK is a superkey
L ? K K is contained in a candidate key
? BCNF decomposition has (JL) and (LK)
?Testing for JK ? L requires a join
? There is some redundancy in this schema
? Equivalent to example in book:
Banker-schema = (branch-name,customer-name,banker-name)
banker-name ? branch-name
branch-name customer-name ? banker-name
?Silberschatz,Korth and Sudarshan7.46Database System Concepts
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
?Silberschatz,Korth and Sudarshan7.47Database System Concepts
3NF Decomposition Algorithm
Let Fc be a canonical cover for F;
i,= 0;
for each functional dependency ? ? ? in Fc do
if none of the schemas Rj,1 ? j ? i contains ? ?
then begin
i,= i + 1;
Ri,= ? ?
end
if none of the schemas Rj,1 ? j ? i contains a candidate
key for R
then begin
i,= i + 1;
Ri,= any candidate key for R;
end
return (R1,R2,...,Ri)
?Silberschatz,Korth and Sudarshan7.48Database System Concepts
3NF Decomposition Algorithm (Cont.)
? Above algorithm ensures:
? each relation schema Ri is in 3NF
? decomposition is dependency preserving and
lossless-join
? Proof of correctness is at end of this file
?Silberschatz,Korth and Sudarshan7.49Database System Concepts
Example
? Relation schema:
Banker-info-schema = (branch-name,customer-
name,banker-name,office-number)
? The functional dependencies for this relation schema
are:
banker-name ? branch-name office-number
customer-name branch-name ? banker-name
? The key is:
{customer-name,branch-name}
?Silberschatz,Korth and Sudarshan7.50Database System Concepts
Applying 3NF to Banker-info-schema
? The for loop in the algorithm causes us to include the
following schemas in our decomposition:
Banker-office-schema = (banker-name,branch-
name,office-number)
Banker-schema = (customer-name,branch-name,
banker-name)
? Since Banker-schema contains a candidate key for
Banker-info-schema,we are done with the
decomposition process.
?Silberschatz,Korth and Sudarshan7.51Database System Concepts
Comparison of BCNF and 3NF
? It is always possible to decompose a relation into
relations in 3NF and
? the decomposition is lossless
? the dependencies are preserved
? It is always possible to decompose a relation into
relations in BCNF and
? the decomposition is lossless
? it may not be possible to preserve dependencies.
?Silberschatz,Korth and Sudarshan7.52Database System Concepts
Comparison of BCNF and 3NF (Cont.)
J
j1
j2
j3
null
L
l1
l1
l1
l2
K
k1
k1
k1
k2
A schema that is in 3NF but not in BCNF has the problems of
? repetition of information (e.g.,the relationship l1,k1)
? need to use null values (e.g.,to represent the relationship
l2,k2 where there is no corresponding value for J).
? Example of problems due to redundancy in 3NF
? R = (J,K,L)
F = {JK ? L,L ? K}
?Silberschatz,Korth and Sudarshan7.53Database System Concepts
Design Goals
? Goal for a relational database design is:
? BCNF.
? Lossless join.
? Dependency preservation.
? If we cannot achieve this,we accept one of
? Lack of dependency preservation
? Redundancy due to use of 3NF
? Interestingly,SQL does not provide a direct way of specifying
functional dependencies other than superkeys.
Can specify FDs using assertions,but they are expensive to test
? Even if we had a dependency preserving decomposition,using SQL
we would not be able to efficiently test a functional dependency whose
left hand side is not a key.
?Silberschatz,Korth and Sudarshan7.54Database System Concepts
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.
?Silberschatz,Korth and Sudarshan7.55Database System Concepts
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 FDs
from non-key attributes of an entity to other attributes of
the entity
? E.g,employee entity with attributes department-number
and department-address,and an FD department-number
? department-address
? Good design would have made department an entity
? FDs from non-key attributes of a relationship set possible,
but rare --- most relationships are binary
?Silberschatz,Korth and Sudarshan7.56Database System Concepts
Universal Relation Approach
? Dangling tuples –Tuples that,disappear” in computing a
join.
? Let r1 (R1),r2 (R2),….,rn (Rn) be a set of relations
? A tuple r of the relation ri is a dangling tuple if r is not in
the relation:
?Ri (r1 r2 … rn)
? The relation r1 r2 … rn is called a universal relation
since it involves all the attributes in the,universe” defined
by
R1 ? R2 ?… ? Rn
? If dangling tuples are allowed in the database,instead of
decomposing a universal relation,we may prefer to
synthesize a collection of normal form schemas from a
given set of attributes.
?Silberschatz,Korth and Sudarshan7.57Database System Concepts
Universal Relation Approach
? Dangling tuples may occur in practical database
applications.
? They represent incomplete information
? E.g,may want to break up information about loans into:
(branch-name,loan-number)
(loan-number,amount)
(loan-number,customer-name)
? Universal relation would require null values,and have
dangling tuples
?Silberschatz,Korth and Sudarshan7.58Database System Concepts
Universal Relation Approach (Contd.)
? A particular decomposition defines a restricted form of
incomplete information that is acceptable in our database.
? Above decomposition requires at least one of
customer-name,branch-name or amount in order
to enter a loan number without using null values
? Rules out storing of customer-name,amount without an
appropriate loan-number (since it is a key,it can't be
null either!)
? Universal relation requires unique attribute names unique
role assumption
? e.g,customer-name,branch-name
? Reuse of attribute names is natural in SQL since relation
names can be prefixed to disambiguate names
?Silberschatz,Korth and Sudarshan7.59Database System Concepts
Denormalization for Performance
? May want to use non-normalized schema for performance
? E.g,displaying customer-name along with account-number and
balance requires join of account with depositor
? Alternative 1,Use denormalized relation containing attributes of
account as well as depositor 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
account depositor
? Benefits and drawbacks same as above,except no extra coding
work for programmer and avoids possible errors
?Silberschatz,Korth and Sudarshan7.60Database System Concepts
Other Design Issues
? Some aspects of database design are not caught by normalization
? Examples of bad database design,to be avoided,
Instead of earnings(company-id,year,amount),use
? earnings-2000,earnings-2001,earnings-2002,etc.,all on the
schema (company-id,earnings).
?Above are in BCNF,but make querying across years difficult
and needs new table each year
? company-year(company-id,earnings-2000,earnings-2001,
earnings-2002)
?Also in BCNF,but also makes querying across years difficult
and requires new attribute each year.
?Is an example of a crosstab,where values for one attribute
become column names
?Used in spreadsheets,and in data analysis tools
Proof of Correctness of 3NF
Decomposition Algorithm
?Silberschatz,Korth and Sudarshan7.62Database System Concepts
Correctness of 3NF Decomposition
Algorithm
? 3NF decomposition algorithm is dependency preserving (since
there is a relation for every FD in Fc)
? Decomposition is lossless join
? A candidate key (C) is in one of the relations Ri in decomposition
? Closure of candidate key under Fc must contain all attributes in R,
? Follow the steps of attribute closure algorithm to show there is only
one tuple in the join result for each tuple in Ri
?Silberschatz,Korth and Sudarshan7.63Database System Concepts
Correctness of 3NF Decomposition
Algorithm (Contd.)
Claim,if a relation Ri is in the decomposition generated by the
above algorithm,then Ri satisfies 3NF.
? Let Ri be generated from the dependency ? ??
? Let ? ? B be any non-trivial functional dependency on Ri,(We
need only consider FDs whose right-hand side is a single
attribute.)
? Now,B can be in either ? or ? but not in both,Consider each
case separately.
?Silberschatz,Korth and Sudarshan7.64Database System Concepts
Correctness of 3NF Decomposition
(Contd.)
? Case 1,If B in ?:
? If ? is a superkey,the 2nd condition of 3NF is satisfied
? Otherwise ? must contain some attribute not in ?
? Since ?? B is in F+ it must be derivable from Fc,by using attribute
closure on ?.
? Attribute closure not have used ? ?? - if it had been used,? must
be contained in the attribute closure of ?,which is not possible,since
we assumed ? is not a superkey.
? Now,using ?? (?- {B}) and ?? B,we can derive ? ?B
(since ?? ? ?,and B ?? since ?? B is non-trivial)
? Then,B is extraneous in the right-hand side of ? ??; which is not
possible since ? ?? is in Fc.
? Thus,if B is in ? then ? must be a superkey,and the second
condition of 3NF must be satisfied.
?Silberschatz,Korth and Sudarshan7.65Database System Concepts
Correctness of 3NF Decomposition
(Contd.)
? Case 2,B is in ?.
? Since ? is a candidate key,the third alternative in the definition of
3NF is trivially satisfied.
? In fact,we cannot show that ? is a superkey.
? This shows exactly why the third alternative is present in the
definition of 3NF.
Q.E.D.
End of Chapter
?Silberschatz,Korth and Sudarshan7.67Database System Concepts
An Instance of Banker-schema
?Silberschatz,Korth and Sudarshan7.68Database System Concepts
Tabular Representation of ??????
?Silberschatz,Korth and Sudarshan7.69Database System Concepts
Relation bc,An Example of Reduncy in a BCNF Relation
?Silberschatz,Korth and Sudarshan7.70Database System Concepts
An Illegal bc Relation
?Silberschatz,Korth and Sudarshan7.71Database System Concepts
Decomposition of loan-info
?Silberschatz,Korth and Sudarshan7.72Database System Concepts
Relation of Exercise 7.4
?Silberschatz,Korth and Sudarshan7.2Database System Concepts
Chapter 7,Relational Database Design
? First Normal Form
? Pitfalls in Relational Database Design
? Functional Dependencies
? Decomposition
? Boyce-Codd Normal Form
? Third Normal Form
? Multivalued Dependencies and Fourth Normal
Form
? Overall Database Design Process
?Silberschatz,Korth and Sudarshan7.3Database System Concepts
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
?Suppose that students are given roll numbers which
are strings of the form CS0012 or EE1127
? A relational schema R is in first normal form（第一范式，
1NF） if the domains of all attributes of R are atomic
?Silberschatz,Korth and Sudarshan7.4Database System Concepts
Pitfalls in Relational Database Design
? Relational database design requires that we find a
“good” collection of relation schemas,A bad design may
lead to
? Repetition of Information.
? Inability to represent certain information.
? Consider the relation schema:
Lending-schema = (branch-name,branch-city,
assets,customer-name,loan-number,amount)
?Silberschatz,Korth and Sudarshan7.5Database System Concepts
Example
? Redundancy（冗余）,
? Data for branch-name,branch-city,assets are repeated
for each loan that a branch makes
? Wastes space
? Complicates updating,introducing possibility of
inconsistency of assets value
? Null values
? Cannot store information about a branch if no loans exist
? Can use null values,but they are difficult to handle.
?Silberschatz,Korth and Sudarshan7.6Database System Concepts
Decomposition
? Decompose the relation schema Lending-schema into:
Branch-schema = (branch-name,branch-city,assets)
Loan-info-schema = (customer-name,loan-number,
branch-name,amount)
? All attributes of an original schema (R) must appear in the
decomposition (R1,R2):
R = R1 ? R2
? Lossless-join decomposition（无损连接分解）,
For all possible relations r on schema R
r = ?R1 (r) ?R2 (r)
?Silberschatz,Korth and Sudarshan7.7Database System Concepts
Example of Non Lossless-Join Decomposition
? Decomposition of R = (A,B)
R2 = (A) R2 = (B)
A B
?
?
?
1
2
1
A
?
?
B
1
2
r
?A(r) ?B(r)
?A (r) ?B (r) A B
?
?
?
?
1
2
1
2
?Silberschatz,Korth and Sudarshan7.8Database System Concepts
Goal — Devise a Theory for the Following
? Decide whether a particular relation R is in,good”
form.
? In the case that a relation R is not in,good” form,
decompose it into a set of relations {R1,R2,...,Rn}
such that
? each relation is in good form
? the decomposition is a lossless-join decomposition
? Our theory is based on:
? functional dependencies
? multivalued dependencies
?Silberschatz,Korth and Sudarshan7.9Database System Concepts
Functional Dependencies
（函数依赖）
? 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.
?Silberschatz,Korth and Sudarshan7.10Database System Concepts
Functional Dependencies (Cont.)
? Let R be a relation schema
?? R and ? ? R
? The functional dependency
? ? ?
holds on R if and only if for any legal relations r(R),
whenever any two tuples t1 and t2 of r agree on the
attributes ?,they also agree on the attributes ?,That is,
t1[?] = t2 [?] ? t1[? ] = t2 [? ]
? 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
?Silberschatz,Korth and Sudarshan7.11Database System Concepts
Functional Dependencies (Cont.)
? Functional dependencies allow us to express constraints that
cannot be expressed using superkeys,（函数依赖使我们可
以表示不能用超玛表示的约束） Consider the schema:
Loan-info-schema = (customer-name,loan-number,
branch-name,amount).
We expect this set of functional dependencies to hold:
loan-number ? amount
loan-number ? branch-name
but would not expect the following to hold,
loan-number ? customer-name
?Silberschatz,Korth and Sudarshan7.12Database System Concepts
Use of Functional Dependencies
? We use functional dependencies to:
? test relations to see if they are legal under a given set
of functional dependencies,（用于判定关系是否在给定
函数依赖集上合法）
? 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（用于指
明合法关系集上的约束）
?We say that F holds on R if all legal relations on R
satisfy the set of functional dependencies F.
?A?C,C?A
A B C D
a1 b1 c1 d1
a1 b2 c1 d2
a2 b2 c2 d2
a2 b2 c2 d3
a3 b3 c2 d4
?Silberschatz,Korth and Sudarshan7.13Database System Concepts
Functional Dependencies (Cont.)
? A functional dependency is trivial(平凡的） if it is
satisfied by all instances of a relation
? E.g.
? customer-name,loan-number ? customer-
name
? customer-name ? customer-name
? In general,? ? ? is trivial if ? ??
?Silberschatz,Korth and Sudarshan7.14Database System Concepts
The customer Relation
Customer-name?customer-city?
Customer-street?customer-city?
?Silberschatz,Korth and Sudarshan7.15Database System Concepts
The loan Relation
Loan-number?amount?
Loan-number?branch-name?
?Silberschatz,Korth and Sudarshan7.16Database System Concepts
The branch Relation
Branch-name?assets?
assets?Branch-name?
?Silberschatz,Korth and Sudarshan7.17Database System Concepts
Closure of a Set of Functional
Dependencies（函数依赖集的闭包）
? Given a set F set of functional dependencies,there are certain other
functional dependencies that are logically implied（逻辑蕴含） by F.
? 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+.
? We can find all of F+ by applying Armstrong’s Axioms（公理）,
? if ? ??,then ? ?? (reflexivity)自反律
? if ? ??,then ?? ? ?? (augmentation)增补率
? if ? ??,and ? ??,then ? ? ? (transitivity)传递率
? These rules are
? Sound（保真的） (不会产生错误的函数依赖 )
? complete （完备的） (对一个给定函数依赖集 F，它们能产生整个 F+).
?Silberschatz,Korth and Sudarshan7.18Database System Concepts
Example
? R = (A,B,C,G,H,I)
F = { A ? B
A ? C
CG ? H
CG ? I
B ? H}
? some members of F+
? A ? H
?by transitivity from A ? B and B ? H
? AG ? I
?by augmenting A ? C with G,to get AG ? CG
and then transitivity with CG ? I
? CG ? HI
?from CG ? H and CG ? I,,union rule” can be inferred from
– definition of functional dependencies,or
– Augmentation of CG ? I to infer CG ? CGI,augmentation of
CG ? H to infer CGI ? HI,and then transitivity
?Silberschatz,Korth and Sudarshan7.19Database System Concepts
Procedure for Computing F+
? To compute the closure of a set of functional dependencies
F:
F+ = F
repeat
for each functional dependency f in F+
apply reflexivity and augmentation rules on f
add the resulting functional dependencies to F+
for each pair of functional dependencies f1and f2 in F+
if f1 and f2 can be combined using transitivity
then add the resulting functional dependency to
F+
until F+ does not change any further
2× 2n＝ 2n+1
NOTE,We will see an alternative procedure for this task later
?Silberschatz,Korth and Sudarshan7.20Database System Concepts
Closure of Functional Dependencies
(Cont.)
? We can further simplify manual computation of F+ by using
the following additional rules.
? If ? ? ? holds and ? ? ? holds,then ? ? ? ? holds
(union)合并率
? If ? ? ? ? holds,then ? ? ? holds and ? ? ? holds
(decomposition)分解率
? If ? ? ? holds and ? ? ? ? holds,then ? ? ? ? holds
(pseudotransitivity)伪传递率
The above rules can be inferred from Armstrong’s axioms.
?Silberschatz,Korth and Sudarshan7.21Database System Concepts
Closure of Attribute Sets（属性集的闭包）
? 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:
? ? ? is in F+ ? ? ? ?+
? Algorithm to compute ?+,the closure of ? under F
result,= ?;
while (changes to result) do
for each ? ? ? in F do
begin
if ? ? result then result,= result ? ?
end
?Silberschatz,Korth and Sudarshan7.22Database System Concepts
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
?Silberschatz,Korth and Sudarshan7.23Database System Concepts
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.
?Silberschatz,Korth and Sudarshan7.24Database System Concepts
Canonical Cover（正则覆盖）
? Sets of functional dependencies may have redundant
dependencies that can be inferred from the others
? Eg,A ? C is redundant in,{A ? B,B ? C,A ? C}
? Parts of a functional dependency may be redundant
?E.g,on RHS,{A ? B,B ? C,A ? CD} can be
simplified to
{A ? B,B ? C,A ? D}
?E.g,on LHS,{A ? B,B ? C,AC ? D} can be
simplified to
{A ? B,B ? C,A ? D}
? Intuitively,a canonical cover of F is a,minimal” set of
functional dependencies equivalent to F,having no
redundant dependencies or redundant parts of
dependencies
?Silberschatz,Korth and Sudarshan7.25Database System Concepts
Extraneous Attributes（无关属性）
? 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.
? 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
?Silberschatz,Korth and Sudarshan7.26Database System Concepts
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 A; if it does,A is
extraneous
? To test if attribute A ? ? is extraneous in ?
1,compute ?+ using only the dependencies in
F’ = (F – {? ? ?}) ? {? ?(?– A)},
2,check that ?+ contains A; if it does,A is
extraneous
?Silberschatz,Korth and Sudarshan7.27Database System Concepts
Canonical Cover
? A canonical cover for F is a set of dependencies Fc such that
? F logically implies all dependencies in Fc,and
? Fc logically implies all dependencies in F,and
? No functional dependency in Fc contains an extraneous
attribute,and
? Each left side of functional dependency in Fc is unique.
? To compute a canonical cover for F:
repeat
Use the union rule to replace any dependencies in F
?1 ? ?1 and ?1 ? ?1 with ?1 ? ?1 ?2
Find a functional dependency ? ? ? with an
extraneous attribute either in ? or in ?
If an extraneous attribute is found,delete it from ? ? ?
until F does not change
?Silberschatz,Korth and Sudarshan7.28Database System Concepts
Example of Computing a Canonical Cover
? R = (A,B,C)
F = {A ? BC
B ? C
A ? B
AB ? C}
? Combine A ? BC and A ? B into A ? BC
? Set is now {A ? BC,B ? C,AB ? C}
? A is extraneous in AB ? C
? Check if the result of deleting A from AB ? C is implied by the other
dependencies
?Yes,in fact,B ? C is already present!
? Set is now {A ? BC,B ? C}
? C is extraneous in A ? BC
? Check if A ? C is logically implied by A ? B and the other
dependencies
?Yes,using transitivity on A ? B and B ? C,
– Can use attribute closure of A in more complex cases
? The canonical cover is,A ? B,B ? C
?Silberschatz,Korth and Sudarshan7.29Database System Concepts
Decomposition
? Decompose the relation schema Lending-schema into:
Branch-schema = (branch-name,branch-city,assets)
Loan-info-schema = (customer-name,loan-number,
branch-name,amount)
? All attributes of an original schema (R) must appear in the
decomposition (R1,R2):
R = R1 ? R2
? Lossless-join decomposition.
For all possible relations r on schema R
r = ?R1 (r) ?R2 (r)
? A decomposition of R into R1 and R2 is lossless join if and only if at
least one of the following dependencies is in F+:
? R1 ? R2 ? R1
? R1 ? R2 ? R2
?Silberschatz,Korth and Sudarshan7.30Database System Concepts
The Relation branch-customer
?Silberschatz,Korth and Sudarshan7.31Database System Concepts
The Relation customer-loan
?Silberschatz,Korth and Sudarshan7.32Database System Concepts
The Relation branch-customer
customer-loan
?Silberschatz,Korth and Sudarshan7.33Database System Concepts
Example of Lossy-Join Decomposition
? Lossy-join decompositions result in information loss.
? Example,Decomposition of R = (A,B)
R2 = (A) R2 = (B)
A B
?
?
?
1
2
1
A
?
?
B
1
2
r
?A(r) ?B(r)
?A (r) ?B (r) A B
?
?
?
?
1
2
1
2
?Silberschatz,Korth and Sudarshan7.34Database System Concepts
Normalization Using Functional Dependencies
? When we decompose a relation schema R with a set of
functional dependencies F into R1,R2,..,Rn we want
? Lossless-join decomposition,Otherwise decomposition
would result in information loss.
? No redundancy,The relations Ri preferably should be in
either Boyce-Codd Normal Form or Third Normal Form.
? Dependency preservation（保持依赖）, Let Fi be the set
of dependencies F+ that include only attributes in Ri,
? Preferably the decomposition should be dependency
preserving,that is,(F1 ? F2 ? … ? Fn)+ = F+
?Otherwise,checking updates for violation of functional
dependencies may require computing joins,which is
expensive.
?Silberschatz,Korth and Sudarshan7.35Database System Concepts
Example
? R = (A,B,C)
F = {A ? B,B ? C)
? Can be decomposed in two different ways
? R1 = (A,B),R2 = (B,C)
? Lossless-join decomposition:
R1 ? R2 = {B} and B ? BC
? Dependency preserving
? R1 = (A,B),R2 = (A,C)
? Lossless-join decomposition:
R1 ? R2 = {A} and A ? AB
? Not dependency preserving
(cannot check B ? C without computing R1 R2)
?Silberschatz,Korth and Sudarshan7.36Database System Concepts
Testing for Dependency Preservation
? To check if a dependency ??? is preserved in a decomposition of R
into R1,R2,…,R n we apply the following simplified test (with attribute
closure done w.r.t,F)
? result = ?
while (changes to result) do
for each Ri in the decomposition
t = (result ? Ri)+ ? Ri
result = result ? t
? If result contains all attributes in ?,then the functional dependency
?? ? is preserved.
? We apply the test on 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 (F1 ? F2 ? … ? Fn)+
?Silberschatz,Korth and Sudarshan7.37Database System Concepts
Boyce-Codd Normal Form（ BC范式 ）
? ?? ? ? 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:
?Silberschatz,Korth and Sudarshan7.38Database System Concepts
Example
? R = (A,B,C)
F = {A ? B
B ? C}
Key = {A}
? R is not in BCNF
? Decomposition R1 = (A,B),R2 = (B,C)
? R1 and R2 in BCNF
? Lossless-join decomposition
? Dependency preserving
?Silberschatz,Korth and Sudarshan7.39Database System Concepts
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,using only F is incorrect when testing a relation in a
decomposition of R
? E.g,Consider R (A,B,C,D),with F = { A ?B,B ?C}
?Decompose R into R1(A,B) and R2(A,C,D)
?Neither of the dependencies in F contain only attributes from
(A,C,D) so we might be mislead into thinking R2 satisfies BCNF,
?In fact,dependency A ? C in F+ shows R2 is not in BCNF,
?Silberschatz,Korth and Sudarshan7.40Database System Concepts
BCNF Decomposition Algorithm
result,= {R};
done,= false;
compute F+;
while (not done) do
if (there is a schema Ri in result that is not in BCNF)
then begin
let ?? ? ? be a nontrivial functional
dependency that holds on Ri
such that ?? ? Ri is not in F+,
and ? ? ? = ?;
result,= (result – Ri ) ? (Ri – ?) ? (?,? );
end
else done,= true;
Note,each Ri is in BCNF,and decomposition is lossless-
join.
?Silberschatz,Korth and Sudarshan7.41Database System Concepts
Example of BCNF Decomposition
? R = (branch-name,branch-city,assets,
customer-name,loan-number,amount)
F = {branch-name ? assets branch-city
loan-number ? amount branch-name}
Key = {loan-number,customer-name}
? Decomposition
? R1 = (branch-name,branch-city,assets)
? R2 = (branch-name,customer-name,loan-number,amount)
? R3 = (branch-name,loan-number,amount)
? R4 = (customer-name,loan-number)
? Final decomposition
R1,R3,R4
?Silberschatz,Korth and Sudarshan7.42Database System Concepts
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
It is not always possible to get a BCNF decomposition that is
dependency preserving
?Silberschatz,Korth and Sudarshan7.43Database System Concepts
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.
? Allows some redundancy (with resultant problems;
we will see examples later)
? But FDs can be checked on individual relations
without computing a join.
? There is always a lossless-join,dependency-
preserving decomposition into 3NF.
?Silberschatz,Korth and Sudarshan7.44Database System Concepts
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).
?Silberschatz,Korth and Sudarshan7.45Database System Concepts
3NF (Cont.)
? Example
? R = (J,K,L)
F = {JK ? L,L ? K}
? Two candidate keys,JK and JL
? R is in 3NF
JK ? L JK is a superkey
L ? K K is contained in a candidate key
? BCNF decomposition has (JL) and (LK)
?Testing for JK ? L requires a join
? There is some redundancy in this schema
? Equivalent to example in book:
Banker-schema = (branch-name,customer-name,banker-name)
banker-name ? branch-name
branch-name customer-name ? banker-name
?Silberschatz,Korth and Sudarshan7.46Database System Concepts
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
?Silberschatz,Korth and Sudarshan7.47Database System Concepts
3NF Decomposition Algorithm
Let Fc be a canonical cover for F;
i,= 0;
for each functional dependency ? ? ? in Fc do
if none of the schemas Rj,1 ? j ? i contains ? ?
then begin
i,= i + 1;
Ri,= ? ?
end
if none of the schemas Rj,1 ? j ? i contains a candidate
key for R
then begin
i,= i + 1;
Ri,= any candidate key for R;
end
return (R1,R2,...,Ri)
?Silberschatz,Korth and Sudarshan7.48Database System Concepts
3NF Decomposition Algorithm (Cont.)
? Above algorithm ensures:
? each relation schema Ri is in 3NF
? decomposition is dependency preserving and
lossless-join
? Proof of correctness is at end of this file
?Silberschatz,Korth and Sudarshan7.49Database System Concepts
Example
? Relation schema:
Banker-info-schema = (branch-name,customer-
name,banker-name,office-number)
? The functional dependencies for this relation schema
are:
banker-name ? branch-name office-number
customer-name branch-name ? banker-name
? The key is:
{customer-name,branch-name}
?Silberschatz,Korth and Sudarshan7.50Database System Concepts
Applying 3NF to Banker-info-schema
? The for loop in the algorithm causes us to include the
following schemas in our decomposition:
Banker-office-schema = (banker-name,branch-
name,office-number)
Banker-schema = (customer-name,branch-name,
banker-name)
? Since Banker-schema contains a candidate key for
Banker-info-schema,we are done with the
decomposition process.
?Silberschatz,Korth and Sudarshan7.51Database System Concepts
Comparison of BCNF and 3NF
? It is always possible to decompose a relation into
relations in 3NF and
? the decomposition is lossless
? the dependencies are preserved
? It is always possible to decompose a relation into
relations in BCNF and
? the decomposition is lossless
? it may not be possible to preserve dependencies.
?Silberschatz,Korth and Sudarshan7.52Database System Concepts
Comparison of BCNF and 3NF (Cont.)
J
j1
j2
j3
null
L
l1
l1
l1
l2
K
k1
k1
k1
k2
A schema that is in 3NF but not in BCNF has the problems of
? repetition of information (e.g.,the relationship l1,k1)
? need to use null values (e.g.,to represent the relationship
l2,k2 where there is no corresponding value for J).
? Example of problems due to redundancy in 3NF
? R = (J,K,L)
F = {JK ? L,L ? K}
?Silberschatz,Korth and Sudarshan7.53Database System Concepts
Design Goals
? Goal for a relational database design is:
? BCNF.
? Lossless join.
? Dependency preservation.
? If we cannot achieve this,we accept one of
? Lack of dependency preservation
? Redundancy due to use of 3NF
? Interestingly,SQL does not provide a direct way of specifying
functional dependencies other than superkeys.
Can specify FDs using assertions,but they are expensive to test
? Even if we had a dependency preserving decomposition,using SQL
we would not be able to efficiently test a functional dependency whose
left hand side is not a key.
?Silberschatz,Korth and Sudarshan7.54Database System Concepts
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.
?Silberschatz,Korth and Sudarshan7.55Database System Concepts
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 FDs
from non-key attributes of an entity to other attributes of
the entity
? E.g,employee entity with attributes department-number
and department-address,and an FD department-number
? department-address
? Good design would have made department an entity
? FDs from non-key attributes of a relationship set possible,
but rare --- most relationships are binary
?Silberschatz,Korth and Sudarshan7.56Database System Concepts
Universal Relation Approach
? Dangling tuples –Tuples that,disappear” in computing a
join.
? Let r1 (R1),r2 (R2),….,rn (Rn) be a set of relations
? A tuple r of the relation ri is a dangling tuple if r is not in
the relation:
?Ri (r1 r2 … rn)
? The relation r1 r2 … rn is called a universal relation
since it involves all the attributes in the,universe” defined
by
R1 ? R2 ?… ? Rn
? If dangling tuples are allowed in the database,instead of
decomposing a universal relation,we may prefer to
synthesize a collection of normal form schemas from a
given set of attributes.
?Silberschatz,Korth and Sudarshan7.57Database System Concepts
Universal Relation Approach
? Dangling tuples may occur in practical database
applications.
? They represent incomplete information
? E.g,may want to break up information about loans into:
(branch-name,loan-number)
(loan-number,amount)
(loan-number,customer-name)
? Universal relation would require null values,and have
dangling tuples
?Silberschatz,Korth and Sudarshan7.58Database System Concepts
Universal Relation Approach (Contd.)
? A particular decomposition defines a restricted form of
incomplete information that is acceptable in our database.
? Above decomposition requires at least one of
customer-name,branch-name or amount in order
to enter a loan number without using null values
? Rules out storing of customer-name,amount without an
appropriate loan-number (since it is a key,it can't be
null either!)
? Universal relation requires unique attribute names unique
role assumption
? e.g,customer-name,branch-name
? Reuse of attribute names is natural in SQL since relation
names can be prefixed to disambiguate names
?Silberschatz,Korth and Sudarshan7.59Database System Concepts
Denormalization for Performance
? May want to use non-normalized schema for performance
? E.g,displaying customer-name along with account-number and
balance requires join of account with depositor
? Alternative 1,Use denormalized relation containing attributes of
account as well as depositor 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
account depositor
? Benefits and drawbacks same as above,except no extra coding
work for programmer and avoids possible errors
?Silberschatz,Korth and Sudarshan7.60Database System Concepts
Other Design Issues
? Some aspects of database design are not caught by normalization
? Examples of bad database design,to be avoided,
Instead of earnings(company-id,year,amount),use
? earnings-2000,earnings-2001,earnings-2002,etc.,all on the
schema (company-id,earnings).
?Above are in BCNF,but make querying across years difficult
and needs new table each year
? company-year(company-id,earnings-2000,earnings-2001,
earnings-2002)
?Also in BCNF,but also makes querying across years difficult
and requires new attribute each year.
?Is an example of a crosstab,where values for one attribute
become column names
?Used in spreadsheets,and in data analysis tools
Proof of Correctness of 3NF
Decomposition Algorithm
?Silberschatz,Korth and Sudarshan7.62Database System Concepts
Correctness of 3NF Decomposition
Algorithm
? 3NF decomposition algorithm is dependency preserving (since
there is a relation for every FD in Fc)
? Decomposition is lossless join
? A candidate key (C) is in one of the relations Ri in decomposition
? Closure of candidate key under Fc must contain all attributes in R,
? Follow the steps of attribute closure algorithm to show there is only
one tuple in the join result for each tuple in Ri
?Silberschatz,Korth and Sudarshan7.63Database System Concepts
Correctness of 3NF Decomposition
Algorithm (Contd.)
Claim,if a relation Ri is in the decomposition generated by the
above algorithm,then Ri satisfies 3NF.
? Let Ri be generated from the dependency ? ??
? Let ? ? B be any non-trivial functional dependency on Ri,(We
need only consider FDs whose right-hand side is a single
attribute.)
? Now,B can be in either ? or ? but not in both,Consider each
case separately.
?Silberschatz,Korth and Sudarshan7.64Database System Concepts
Correctness of 3NF Decomposition
(Contd.)
? Case 1,If B in ?:
? If ? is a superkey,the 2nd condition of 3NF is satisfied
? Otherwise ? must contain some attribute not in ?
? Since ?? B is in F+ it must be derivable from Fc,by using attribute
closure on ?.
? Attribute closure not have used ? ?? - if it had been used,? must
be contained in the attribute closure of ?,which is not possible,since
we assumed ? is not a superkey.
? Now,using ?? (?- {B}) and ?? B,we can derive ? ?B
(since ?? ? ?,and B ?? since ?? B is non-trivial)
? Then,B is extraneous in the right-hand side of ? ??; which is not
possible since ? ?? is in Fc.
? Thus,if B is in ? then ? must be a superkey,and the second
condition of 3NF must be satisfied.
?Silberschatz,Korth and Sudarshan7.65Database System Concepts
Correctness of 3NF Decomposition
(Contd.)
? Case 2,B is in ?.
? Since ? is a candidate key,the third alternative in the definition of
3NF is trivially satisfied.
? In fact,we cannot show that ? is a superkey.
? This shows exactly why the third alternative is present in the
definition of 3NF.
Q.E.D.
End of Chapter
?Silberschatz,Korth and Sudarshan7.67Database System Concepts
An Instance of Banker-schema
?Silberschatz,Korth and Sudarshan7.68Database System Concepts
Tabular Representation of ??????
?Silberschatz,Korth and Sudarshan7.69Database System Concepts
Relation bc,An Example of Reduncy in a BCNF Relation
?Silberschatz,Korth and Sudarshan7.70Database System Concepts
An Illegal bc Relation
?Silberschatz,Korth and Sudarshan7.71Database System Concepts
Decomposition of loan-info
?Silberschatz,Korth and Sudarshan7.72Database System Concepts
Relation of Exercise 7.4
