Chapter 8: Relational Database Design
■ Features of Good Relational Design
■ Atomic Domains and First Normal Form
■ Decomposition Using Functional Dependencies ■ Functional Dependency Theory
■ Algorithms for Functional Dependencies
■ Decomposition Using Multivalued Dependencies ■ More Normal Forms
■ Database-Design Process
■ Some Examples
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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.
■ Design Goals:
● Avoid redundant data
● Ensure that relationships among attributes are properly represented
● Facilitate the checking of updates for violation of database integrity constraints
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Combine Schemas?
■ Suppose we combine instructor and department into inst_dept
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Combine Schemas?
■ Suppose we combine instructor and department into inst_dept
■ Result is possible repetition of information
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A Combined Schema Without Repetition
■ Consider combining relations
● section(course_id, sec_id, semester, year)
and
● sec_classroom(sec_id, building, room_number)
into one relation
● section(course_id, sec_id, semester, year,
building, room_number)
■ No repetition in this case
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What About Smaller Schemas?
■ Suppose we had started with inst_dept. How would we know to split up (decompose) it into instructor and department?
■ Write a rule “if there were a table (dept_name, building, budget), then dept_name would be a candidate key”
■ Denote as a functional dependency: dept_name → building, budget
■ In inst_dept, because dept_name is not a candidate key, the building and budget of a department may have to be repeated.
● This indicates the need to decompose inst_dept
■ Not all decompositions are good. Suppose we decompose
employee(ID, name, street, city, salary) into employee1 (ID, name)
employee2 (name, street, city, salary)
■ This loses information — we cannot reconstruct the original employee relation — and so, this is a lossy decomposition.
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A Lossy Decomposition
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Example of Lossless-Join Decomposition
■ Lossless join decomposition ■ Decomposition of R = (A, B, C)
R1 =(A,B)
R2 =(B,C)
A
B
C
A
B
B
C
α β
1 2
A B
α β
1 2
1 2
A B
r ∏A,B(r) ∏A (r) ∏B (r)
∏B,C(r)
A
B
C
α β
1 2
A B
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First Normal Form
■ Domain is atomic if its elements are considered to be indivisible units
● Examples of non-atomic domains:
! Set of names, composite attributes
! Identification numbers like CS101 that can be broken up into parts
■ A relational schema R is in first normal form if the domains of all attributes of R are atomic
■ Non-atomic values complicate storage and encourage redundant (repeated) storage of data
● Example: Set of accounts stored with each customer, and set of owners stored with each account
● We assume all relations are in first normal form
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First Normal Form (Cont’d)
■ Atomicityisactuallyapropertyofhowtheelementsofthe
domain are used.
● Example: Strings would normally be considered indivisible
● Suppose that students are given roll numbers which are strings of the form CS0012 or EE1127
● If the first two characters are extracted to find the department, the domain of roll numbers is not atomic.
● Doing so is a bad idea: leads to encoding of information in application program rather than in the database
● but nonetheless, sometimes it makes sense to violate atomicity
● typical issues: dates, IDs, addresses, text clobs (contains)
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Goal — Devise a Theory for the Following
■ Decide whether a particular relation R is in “good” form. ■ If 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
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Functional Dependencies
■ Constraints on the set of legal relations.
■ Require that the value for a certain set of attributes determines uniquely the value for another set of attributes.
■ A functional dependency is a generalization of the notion of a key.
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Functional Dependencies (Cont.)
■ Let R be a relation schema α⊆R and β⊆R
■ The functional dependency
α → β
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 α, they also agree on the attributes β. That is,
t1[α]=t2[α] ⇒ t1[β] =t2[β]
■ Example: Consider r(A,B ) with the following instance of r.
■ On this instance, A → B does NOT hold, but B → A does hold
■ But does it hold on all instances (well, depends ….)
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Functional Dependencies (Cont.)
■ K is a superkey for relation schema R if and only if K → R
■ KisacandidatekeyforRifandonlyif
● K→R,and
● fornoα⊂K,α→R
■ Functional dependencies allow us to express constraints that
cannot be expressed using superkeys. Consider the schema: inst_dept (ID, name, salary, dept_name, building, budget ).
We expect these functional dependencies to hold: dept_name→ building
and ID à building
but would not expect the following to hold:
dept_name → salary
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Use of Functional Dependencies
■ We use functional dependencies to:
● test relations to see if they are legal under a given set of
functional dependencies.
! If a relation r is legal under a set F of functional dependencies,
we say that r satisfies F.
● specify constraints on the set of legal relations
! We say that F holds on R if all legal relations on R satisfy the set of functional dependencies F.
■ Note: A specific instance of a relation schema may satisfy a functional dependency even if the functional dependency does not hold on all legal instances.
● For example, a specific instance of instructor may, by chance, satisfy
name → ID.
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Functional Dependencies (Cont.)
■ A functional dependency is trivial if it is satisfied by all instances of a relation
● Example:
! ID, name → ID ! name → name
● In general, α → β is trivial if β ⊆ α
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Closure of a Set of Functional Dependencies
■ Given a set F of functional dependencies, there are certain other functional dependencies that are logically implied by F.
● Forexample: If A→Band B→C, thenwecan infer that A → C
■ The set of all functional dependencies logically implied by F is the closure of F.
■ We denote the closure of F by F+.
■ F+ is a superset of F.
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Boyce-Codd Normal Form
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
α→β
where α ⊆ R and β ⊆ R, at least one of the following holds: ■ α→β istrivial(i.e.,β⊆α)
■ αisasuperkeyforR
Example schema not in BCNF:
instr_dept (ID, name, salary, dept_name, building, budget )
because dept_name→ building, budget
holds on instr_dept, but dept_name is not a superkey
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Decomposing a Schema into BCNF
■ Suppose we have a schema R and a non-trivial dependency α →β causes a violation of BCNF.
We decompose R into: • (α U β )
• ( R – ( β – α ) )
■ In our example,
● α = dept_name
● β = building, budget
and inst_dept is replaced by
● (α U β ) = ( dept_name, building, budget )
● ( R – ( β – α ) ) = ( ID, name, salary, dept_name )
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BCNF and Dependency Preservation
■ Constraints, including functional dependencies, are costly to check in practice unless they pertain to only one relation
■ If it is sufficient to test only those dependencies on each individual relation of a decomposition in order to ensure that all functional dependencies hold, then that decomposition is dependency preserving.
■ Because it is not always possible to achieve both BCNF and dependency preservation, we consider a weaker normal form, known as third normal form.
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Third Normal Form
■ A relation schema R is in 3rd normal form (3NF) if for all: α → β in F+
at least one of the following holds:
● α → β is trivial (i.e., β ∈ α)
● α is a superkey for R
● Each attribute A in β – α is contained in a candidate key for R.
(NOTE: each attribute may be in a different cand. key)
■ If a relation is in BCNF it is in 3NF (since in BCNF one of the first two conditions above must hold).
■ Third condition is a minimal relaxation of BCNF to ensure dependency preservation (will see why later).
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Goals of Normalization
■ Let R be a relation scheme with a set F of functional dependencies.
■ Decide whether a relation scheme R is in “good” form.
■ In the case that a relation scheme R is not in “good” form, decompose it into a set of relation scheme {R1, R2, …, Rn} such that
● each relation scheme is in good form
● the decomposition is a lossless-join decomposition
● Preferably, the decomposition should be dependency preserving.
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How good is BCNF?
■ There are database schemas in BCNF that do not seem to be sufficiently normalized
■ Consider a relation
inst_info (ID, child_name, phone)
● where an instructor may have more than one phone and can have multiple children
ID
child_name
phone
99999
99999
99999
99999
David David William William
512-555-1234 512-555-4321 512-555-1234 512-555-4321
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How good is BCNF? (Cont.)
■ There are no non-trivial functional dependencies and therefore the relation is in BCNF
■ Insertion anomalies – i.e., if we add a phone 981-992-3443 to 99999, we need to add two tuples
(99999, David, 981-992-3443)
(99999, William, 981-992-3443)
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How good is BCNF? (Cont.)
■ Therefore, it is better to decompose inst_info into:
inst_child
ID
child_name
99999
99999
99999
99999
David David William William
ID
phone
99999
99999
99999
99999
512-555-1234 512-555-4321 512-555-1234 512-555-4321
inst_phone
This suggests the need for higher normal forms, such as Fourth
Normal Form (4NF), which we shall see later.
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Functional-Dependency Theory
■ We now consider the formal theory that tells us which functional dependencies are implied logically by a given set of functional dependencies.
■ We then develop algorithms to generate lossless decompositions into BCNF and 3NF
■ We then develop algorithms to test if a decomposition is dependency-preserving
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Closure of a Set of Functional Dependencies
■ Given a set F set of functional dependencies, there are certain other functional dependencies that are logically implied by F.
● Fore.g.: If A→Band B→C, thenwecaninferthatA→C
■ The set of all func. dependencies logically implied by F is the closure of F.
■ We denote the closure of F by F+.
■ We can find F+, the closure of F, by repeatedly applying
Armstrong’s Axioms:
● if β ⊆ α, then α → β (reflexivity)
● ifα→β,thenγα→ γβ (augmentation) ● ifα→β,andβ→γ,thenα→ γ (transitivity)
■ These rules are
● sound (generate only functional dependencies that actually hold), and ● complete(generateallfunctionaldependenciesthathold).
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■ R=(A,B,C,G,H,I)
F={ A→B
A → C
CG → H
CG → I
B → H}
■ some members of F+
Example
● A → H
! by transitivity from A → B and B → H
● AG → I
! by augmenting A → C with G, to get AG → CG
and then transitivity with CG → I
● CG→HI
! by augmenting CG → I to infer CG → CGI,
and augmenting of CG → H to infer CGI → HI, and then transitivity
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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
NOTE: We shall see an alternative procedure for this task later
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Closure of Functional Dependencies (Cont.)
■ Additional rules:
● Ifα→βholdsandα→γholds, thenα→βγholds
(union)
● Ifα→βγholds,thenα→β holdsandα→γholds
(decomposition)
● Ifα→β holdsandγβ→δholds,thenαγ→δholds
(pseudotransitivity)
These rules can be inferred from Armstrong’s axioms.
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Closure of Attribute Sets
■ Given a set of attributes α, define the closure of α under F (denoted by α+) as the set of attributes that are functionally determined by α under F
■ Algorithmtocomputeα+,theclosureofαunderF
result := α;
while (changes to result) do
for each β → γ in F do
begin
if β ⊆ result then result := result ∪ γ
end
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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
3. result = ABCGH (CG → H and CG ⊆ AGBC) 4. result = ABCGHI (CG → I and CG ⊆ AGBCH)
■ Is AG a candidate key? 1. IsAGasuperkey?
1. DoesAG→R?==Is(AG)+⊆R 2. IsanysubsetofAGasuperkey?
1. DoesA→R?==Is(A)+⊆R
2. DoesG→R?==Is(G)+⊆R
(A→CandA→B)
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Uses of Attribute Closure
There are several uses of the attribute closure algorithm: ■ Testing for superkey:
● To test if α is a superkey, we compute α+, and check if α+ contains all attributes of R.
■ Testing functional dependencies
● To check if a functional dependency α → β holds (or, in other
words, is in F+), just check if β ⊆ α+.
● That is, we compute α+ by using attribute closure, and then
check if it contains β.
● Is a simple and cheap test, and very useful
■ Computing closure of F
● Foreachγ⊆R,wefindtheclosureγ+,andforeachS⊆γ+,we
output a functional dependency γ → S.
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Canonical Cover
■ Sets of functional dependencies may have redundant dependencies that can be inferred from the others
● Forexample: A→Cisredundantin: {A→B, B→C,AàC} ● Parts of a functional dependency may be redundant
! E.g.:onRHS: {A→B, B→C, A→CD} canbesimplifiedto
{A→B, B→C, A→D}
! E.g.:onLHS: {A→B, B→C, AC→D} canbesimplifiedto
{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
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Extraneous Attributes
■ Consider a set F of functional dependencies and the functional dependency α → β in F.
● AttributeAisextraneousinαifA∈α
andFlogicallyimplies(F–{α→β})∪{(α –A)→β}.
● AttributeAisextraneousinβifA∈β
and the set of functional dependencies
(F – {α → β}) ∪ {α →(β – A)} logically implies F.
■ Note: implication in the opposite direction is trivial in each of the cases above, since a “stronger” functional dependency always implies a weaker one
■ Example:GivenF={A→C,AB→C}
● BisextraneousinAB→Cbecause{A→C,AB→C}logicallyimplies
A → C (I.e. the result of dropping B from AB → C).
■ Example: GivenF={A→C,AB→CD}
● C is extraneous in AB → CD since AB → C can be inferred even after deleting C
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Testing if an Attribute is Extraneous
■ Consider a set F of functional dependencies and the functional dependency α → β in F.
■ To test if attribute A ∈ α is extraneous in α
1. compute ({α} – A)+ using the dependencies in F
2. check that ({α} – A)+ contains β; if it does, A is extraneous in α
■ To test if attribute A ∈ β is extraneous in β
1. compute α+ using only the dependencies in
F’=(F –{α→β})∪{α→(β–A)},
2. check that α+ contains A; if it does, A is extraneous in β
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Canonical Cover
■ A canonical cover for F is a set of dependencies Fc such that ● F logically implies all dependencies in Fc, and
● FclogicallyimpliesalldependenciesinF,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
α1 →β1 andα1 →β2 withα1 →β1 β2
Find a functional dependency α → β with an
extraneous attribute either in α or in β
/* Note: test for extraneous attributes done using Fc, not F*/
If an extraneous attribute is found, delete it from α → β
until F does not change
■ Note: Union rule may become applicable after some extraneous attributes have been deleted, so it has to be re-applied
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Computing a Canonical Cover
■ R=(A,B,C)
F={A→BC, B→C, A→B, AB→C}
■ CombineA→BCandA→BintoA→BC ● Setisnow{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!
● Setisnow{A→BC,B→C}
■ C is extraneous in A → BC
● CheckifA→CislogicallyimpliedbyA→Bandtheotherdependencies ! Yes: using transitivity on A → B and B → C.
– Can use attribute closure of A in more complex cases
■ Thecanonicalcoveris: A→B, B→C
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Lossless-join Decomposition
■ For the case of R = (R1, R2), we require that 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 at least
one of the following dependencies is in F+: ● R1 ∩R2 →R1
● R1 ∩R2 →R2
■ The above functional dependencies are a sufficient condition for lossless join decomposition; the dependencies are a necessary condition only if all constraints are functional dependencies
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■ R=(A,B,C)
F = {A → B, B → C)
Example
● Can be decomposed in two different ways ■ R1 =(A,B), R2 =(B,C)
● Lossless-join decomposition:
R1 ∩R2 ={B}andB→BC
● Dependency preserving ■ R1=(A,B), R2 =(A,C)
● Lossless-join decomposition:
R1 ∩R2 ={A}andA→AB
● Not dependency preserving
(cannot check B → C without computing R1 R2)
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Dependency Preservation
■ Let Fi be the set of dependencies F + that include only attributes in Ri.
! A decomposition is dependency preserving if (F1 ∪F2∪…∪Fn)+ =F+
! that is, if all dependency can still be inferred if we project all dependencies down into the Ri
! If it is not, then checking updates for violation of functional dependencies may require computing joins, which is expensive.
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Testing for Dependency Preservation
■ To check if a dependency α → β is preserved in a decomposition of R into R1, R2, …, Rn we apply the following test (with attribute closure done with respect to F)
● result = α
while (changes to result) do
for each Ri in the decomposition
t = (result ∩ Ri)+ ∩ Ri
result = result ∪ t
● If result contains all attributes in β, then the functional dependency
α → β 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)+
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■ 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
Example
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Testing for BCNF
■ To check if a non-trivial dependency α →β causes a violation of BCNF 1. compute α+ (the attribute closure of α), 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+.
● IfnoneofthedependenciesinFcausesaviolationofBCNF,then none of the dependencies in F+ will cause a violation of BCNF either.
■ However, simplified test using only F is incorrect when testing a relation in a decomposition of R
● ConsiderR=(A,B,C,D,E),withF={A→B,BC→D}
! Decompose R into R1 = (A,B) and R2 = (A,C,D, E)
! Neither of the dependencies in F contain only attributes from
(A,C,D,E) so we might be mislead into thinking R2 satisfies BCNF.
! In fact, dependency AC → D in F+ shows R2 is not in BCNF.
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Testing Decomposition for BCNF
■ To check if a relation Ri in a decomposition of R is in BCNF,
● Either test Ri for BCNF with respect to the restriction of F to Ri
(that is, all FDs in F+ that contain only attributes from Ri)
● or use the original set of dependencies F that hold on R, but with
the following test:
– for every set of attributes α ⊆ Ri, check that α+ (the attribute closure of α) either includes no attribute of Ri- α, or includes all attributes of Ri.
! If the condition is violated by some α → β in F, the dependency
α → (α+ – α) ∩ Ri
can be shown to hold on Ri, and Ri violates BCNF.
! We use above dependency to decompose Ri
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8.45 ©Silberschatz, Korth and Sudarshan
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 α → β be a nontrivial functional dependency that
holdsonRi suchthatα→Ri isnotinF+,
andα∩β =∅;
result:=(result–Ri)∪(Ri –β)∪(α,β);
end
else done := true;
Note: each Ri is in BCNF, and decomposition is lossless-join. Also note: this can lead to many different compositions
Database System Concepts – 6th Edition
Modified by T. Suel for CS6083, NYU Tandon, Spring 2020
8.46 ©Silberschatz, Korth and Sudarshan
Example of BCNF Decomposition
■ R=(A,B,C)
F = {A → B
B → C }
Key = {A }
■ RisnotinBCNF(B→CbutBisnot superkey) ■ Decomposition
● R1 =(B,C) ● R2 = (A,B)
Database System Concepts – 6th Edition
Modified by T. Suel for CS6083, NYU Tandon, Spring 2020
8.47 ©Silberschatz, Korth and Sudarshan
Example of BCNF Decomposition
■ class (course_id, title, dept_name, credits, sec_id, semester, year, building, room_number, capacity, time_slot_id)
■ Functional dependencies:
● course_id→ title, dept_name, credits
● building, room_number→capacity
● course_id, sec_id, semester, year→building, room_number, time_slot_id
■ Acandidatekey{course_id,sec_id,semester,year}. ■ BCNF Decomposition:
● course_id→ title, dept_name, credits holds ! but course_id is not a superkey.
● We replace class by:
! course(course_id, title, dept_name, credits)
! class-1 (course_id, sec_id, semester, year, building, room_number, capacity, time_slot_id)
Database System Concepts – 6th Edition
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8.48 ©Silberschatz, Korth and Sudarshan
BCNF Decomposition (Cont.)
■ course is in BCNF
● How do we know this?
■ building, room_number→capacity holds on class-1 ● but {building, room_number} is not a superkey
for class-1.
● We replace class-1 by:
! classroom (building, room_number, capacity) ! section (course_id, sec_id, semester, year,
building, room_number, time_slot_id) ■ classroom and section are in BCNF.
Database System Concepts – 6th Edition
Modified by T. Suel for CS6083, NYU Tandon, Spring 2020
8.49 ©Silberschatz, Korth and Sudarshan
BCNF and Dependency Preservation
It is not always possible to get a BCNF decomposition that is dependency preserving
■ R=(J,K,L)
F = {JK → L
L → K }
Two candidate keys = JK and JL
■ RisnotinBCNF
■ AnydecompositionofRwillfailtopreserve
JK → L
This implies that testing for JK → L requires a join
Database System Concepts – 6th Edition
Modified by T. Suel for CS6083, NYU Tandon, Spring 2020
8.50 ©Silberschatz, Korth and Sudarshan
Third Normal Form: Motivation
■ There are some situations where
● BCNF is not dependency preserving, and
● efficient checking for FD violation on updates is important
■ Solution: define a weaker normal form, called Third Normal Form (3NF)
● Allows some redundancy (with resultant problems; we will see examples later)
● But functional dependencies can be checked on individual relations without computing a join.
● There is always a lossless-join, dependency-preserving decomposition into 3NF.
Database System Concepts – 6th Edition
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8.51 ©Silberschatz, Korth and Sudarshan
Third Normal Form
■ A relation schema R is in 3rd normal form (3NF) if for all: α → β in F+
at least one of the following holds:
● α → β is trivial (i.e., β ∈ α)
● α is a superkey for R
● Each attribute A in β – α is contained in a candidate key for R.
(NOTE: each attribute may be in a different cand. key)
■ If a relation is in BCNF it is in 3NF (since in BCNF one of the first two conditions above must hold).
Database System Concepts – 6th Edition
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8.52 ©Silberschatz, Korth and Sudarshan
3NF Example
■ Relation dept_advisor:
● dept_advisor (s_ID, i_ID, dept_name)
F = {s_ID, dept_name → i_ID, i_ID → dept_name }
● Note: student can have one advisor in each department
● Two candidate keys: s_ID, dept_name, and i_ID, s_ID
● Risin3NF
! s_ID, dept_name → i_ID
– s_ID, dept_name is a superkey
! i_ID → dept_name
– dept_name is contained in a candidate key
Database System Concepts – 6th Edition
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8.53 ©Silberschatz, Korth and Sudarshan
Redundancy in 3NF
■ There is some redundancy in this schema
■ Example of problems due to redundancy in 3NF
● R=(J,K,L)
F = {JK → L, L → K }
J
L
K
j1
j2 j3
null
l1 l1 l1
l2
k1 k1 k1
k2
■ repetition of information (e.g., the relationship l1, k1) l (i_ID,dept_name)
■ need to use null values (e.g., to represent the relationship
l2, k2 where there is no corresponding value for J).
l (i_ID,dept_nameI)ifthereisnoseparaterelationmapping instructors to departments
Database System Concepts – 6th Edition
Modified by T. Suel for CS6083, NYU Tandon, Spring 2020
8.54 ©Silberschatz, Korth and Sudarshan
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 α → β, if α is a superkey.
■ If α is not a superkey, we have to verify if each attribute in β 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
Database System Concepts – 6th Edition
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8.55 ©Silberschatz, Korth and Sudarshan
3NF Decomposition Algorithm
Let Fc be a canonical cover for F;
i := 0;
for each functional dependency α → β in Fc do
ifnoneoftheschemasRj,1≤j ≤icontains αβ
then begin
i:=i +1;
Ri :=αβ
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
/* Optionally, remove redundant relations */
repeat
if any schema Rj is contained in another schema Rk
then /* delete Rj */
Rj = R;;
i=i-1;
return (R1, R2, …, Ri)
■ Above algorithm ensures:
● each relation schema Ri is in 3NF
● decompositionisdependencypreservingandlossless-join
Database System Concepts – 6th Edition
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8.56 ©Silberschatz, Korth and Sudarshan
3NF Decomposition: An Example
■ Relation schema:
cust_banker_branch = (customer_id, employee_id, branch_name, type )
■ The functional dependencies for this relation schema are: 1. customer_id,employee_id→branch_name,type
2. employee_id→branch_name
3. customer_id,branch_name→employee_id
■ We first compute a canonical cover
● branch_name is extraneous in the r.h.s. of the 1st dependency
● No other attribute is extraneous, so we get FC =
customer_id, employee_id → type
employee_id → branch_name
customer_id, branch_name → employee_id
Database System Concepts – 6th Edition
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8.57 ©Silberschatz, Korth and Sudarshan
3NF Decomposition Example (Cont.)
■ The for loop generates following 3NF schema: (customer_id, employee_id, type )
(employee_id, branch_name) (customer_id, branch_name, employee_id)
● Observethat(customer_id,employee_id,type)containsa candidate key of the original schema, so no further relation schema needs be added
■ At end of for loop, detect and delete schemas, such as (employee_id, branch_name), which are subsets of other schemas
● resultwillnotdependontheorderinwhichFDsareconsidered
■ The resultant simplified 3NF schema is:
(customer_id, employee_id, type) (customer_id, branch_name, employee_id)
Database System Concepts – 6th Edition
Modified by T. Suel for CS6083, NYU Tandon, Spring 2020
8.58 ©Silberschatz, Korth and Sudarshan
Comparison of BCNF and 3NF
■ It is always possible to decompose a relation into a set of relations that are in 3NF such that:
● the decomposition is lossless
● the dependencies are preserved
■ It is always possible to decompose a relation into a set of relations that are in BCNF such that:
● the decomposition is lossless
● it may not be possible to preserve dependencies.
■ But sometimes, 3NF algorithm may give you BCNF (with dep. pres.) even when BDNF algorithm does not!
Database System Concepts – 6th Edition
Modified by T. Suel for CS6083, NYU Tandon, Spring 2020
8.59 ©Silberschatz, Korth and Sudarshan
Design Goals
■ Goal for a relational database design is: ● BCNF.
● Losslessjoin.
● Dependencypreservation.
■ If we cannot achieve this, we accept one of
● Lackofdependencypreservation
● Redundancyduetouseof3NF
■ Interestingly, SQL does not provide a direct way of specifying functional
dependencies other than superkeys.
Can specify FDs using assertions, but they are expensive to test, (and currently not supported by any of the widely used databases!)
■ Even if we had a dependency preserving decomposition, using SQL we would not be able to efficiently test a functional dependency whose left hand side is not a key.
Database System Concepts – 6th Edition
Modified by T. Suel for CS6083, NYU Tandon, Spring 2020
8.60 ©Silberschatz, Korth and Sudarshan
Multivalued Dependencies
■ Suppose we record names of children, and phone numbers for instructors:
● inst_child(ID, child_name)
● inst_phone(ID, phone_number)
■ If we were to combine these schemas to get
● inst_info(ID, child_name, phone_number)
● Example data:
(99999, David, 512-555-1234)
(99999, David, 512-555-4321)
(99999, William, 512-555-1234)
(99999, William, 512-555-4321)
■ This relation is in BCNF ● Why?
Database System Concepts – 6th Edition
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8.61 ©Silberschatz, Korth and Sudarshan
Multivalued Dependencies (MVDs)
■ Let R be a relation schema and let α ⊆ R and β ⊆ R. The multivalued dependency
α →→ β
holds on R if in any legal relation r(R), for all pairs for tuples t1 and t2 in r such that t1[α] = t2 [α], there exist tuples t3 and t4 in r such that:
t1[α]=t2[α]=t3 [α]=t4[α]
t3[β] = t1 [β]
t3[R –β]= t2[R –β]
t4 [β] = t2[β]
t4[R –β]= t1[R –β]
Database System Concepts – 6th Edition
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8.62 ©Silberschatz, Korth and Sudarshan
Fourth Normal Form
■ ArelationschemaRisin4NFwithrespecttoasetDoffunctional and multivalued dependencies if for all multivalued dependencies inD+ oftheformα→→β,whereα⊆Randβ⊆R,atleastoneofthe following hold:
● α→→βistrivial(i.e.,β⊆αorα∪β=R)
● α is a superkey for schema R
■ Ifarelationisin4NFitisinBCNF
Restriction of 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
α →→ (β ∩ Ri)
whereα⊆Ri and α→→βisinD+
Database System Concepts – 6th Edition
Modified by T. Suel for CS6083, NYU Tandon, Spring 2020
8.63 ©Silberschatz, Korth and Sudarshan
Further Normal Forms
■ Join dependencies generalize multivalued dependencies ● lead to project-join normal form (PJNF) (also called
fifth normal form)
■ A class of even more general constraints, leads to a
normal form called domain-key normal form.
■ Problem with these generalized constraints: are hard to reason with, and no set of sound and complete set of inference rules exists.
■ Hence rarely used
Database System Concepts – 6th Edition
Modified by T. Suel for CS6083, NYU Tandon, Spring 2020
8.64 ©Silberschatz, Korth and Sudarshan
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.
Database System Concepts – 6th Edition
Modified by T. Suel for CS6083, NYU Tandon, Spring 2020
8.65 ©Silberschatz, Korth and Sudarshan
ER Model and Normalization
■ When an E-R diagram is carefully designed, identifying all entities correctly, the tables generated from the E-R diagram should not need further normalization.
■ However, in a real (imperfect) design, there can be functional dependencies from non-key attributes of an entity to other attributes of the entity
● Example: an employee entity with attributes
department_name and building,
and a functional dependency
department_name→ building
● Good design would have made department an entity
■ Functional dependencies from non-key attributes of a relationship
set possible, but rare — most relationships are binary
Database System Concepts – 6th Edition
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8.66 ©Silberschatz, Korth and Sudarshan
Denormalization for Performance
■ May want to use non-normalized schema for performance
■ For example, displaying prereqs along with course_id, and title
requires join of course with prereq
■ Alternative 1: Use denormalized relation containing attributes of course as well as prereq with all above attributes
● faster lookup
● extra space and extra execution time for updates
● extra coding work for programmer and possibility of error in extra code
■ Alternative2:useamaterializedviewdefinedas
course prereq
● Benefits and drawbacks same as above, except no extra coding work for programmer and avoids possible errors
Database System Concepts – 6th Edition
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8.67 ©Silberschatz, Korth and Sudarshan
Other Design Issues
■ Some aspects of database design are not caught by normalization ■ Examples of bad database design, to be avoided:
Instead of earnings (company_id, year, amount ), use
● earnings_2004, earnings_2005, earnings_2006, etc., all on the schema (company_id, earnings).
! Above are in BCNF, but make querying across years difficult and needs new table each year
● company_year (company_id, earnings_2004, earnings_2005,
earnings_2006)
! Also in BCNF, but also makes querying across years difficult and requires new attribute each year.
! Is an example of a crosstab, where values for one attribute become column names
! Used in spreadsheets, and in data analysis tools
Database System Concepts – 6th Edition
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Example Problem
■ R(A,B,C,D,E) ! AàBC
! CDàE ! BàD ! EàA
● Candidate keys? ● Canonical cover?
● Lossless decomposition into BCNF?
● Is this decomposition dependency preserving? ● If not, decompose into 3NF
Database System Concepts – 6th Edition
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8.69 ©Silberschatz, Korth and Sudarshan
Example Problem
■ ActorMovie (aid, aname, mid, mtitle, role, hours, payph) ! Unique aid per actor, and unique mid per movie ! Actor can play several roles in one movie
! But these roles must have different names
! And different actors playing same role in same movie must be paid the same per hour
● Functional dependencies?
● Lossless decomposition into BCNF?
● What if we also require actors with same name must get same pay per hour in same movie?
Database System Concepts – 6th Edition
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8.70 ©Silberschatz, Korth and Sudarshan
Another Example
■ Customer data in online store:
(cid, cname, street, stnum, zip, city, state, country)
● state == state or province
● Does zip à city hold? How about city à state?
● Assume: {zip,city}àstate and stateàcountry ■ How should we normalize this relation?
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Example continued:
■ Normalizing: {zip, city} à state
● Should we have extra table {zip, city, state} ?
● Or is this too cumbersome? (state is small)
■ Normalized solution:
● (cid, name, street, stnum, zip, city) ● (zip, city, state)
● (state, country)
■ What could go wrong here?
Database System Concepts – 6th Edition
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8.72 ©Silberschatz, Korth and Sudarshan
Example continued:
■ Where does the data come from? ● Probably the customer!
● When determining functional dependencies, need to consider the data supplied by the customers
● It is not about “the official zip code rules” ■ Remember our “normalized” solution:
● (cid, name, street, stnum, zip, city) ● (zip, city, state)
● (state, country)
■ 1st customer inputs :11201, Brooklyn, Ohio, Mexico
Database System Concepts – 6th Edition
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8.73 ©Silberschatz, Korth and Sudarshan
Example continued:
■ Functional dependencies may not really hold ● On customer data
● Note: databases have high rates of incorrect data
■ 1st solution: keep things non-normalized
■ 2nd solution: preload small tables with data and check customer inputs:
● Use a “standard” database of {zip, city à state} and stateàcountry mappings
● Reject customer input violating these mappings ■ Lesson: thing about the data and who makes it
Database System Concepts – 6th Edition
Modified by T. Suel for CS6083, NYU Tandon, Spring 2020
8.74 ©Silberschatz, Korth and Sudarshan