Relational Data Model

1
Relational Data Model

• Lecture 3

2
Relational Model

• Domain a set of atomic values. Example set of
  integers
• Data Type Description of a form that domain
  values can be represented. Each domain has a null
  value
• Cartesian Product D1 x D2 a set of pairs
  ltp1,p2gt where p1 belongs to D1 and p2 belongs to
  D2.
• D1 x D2 x D3 x x Dk cartesian product of
  k domains.
• Relation a subset of the cartesian product of
  one or more domains. Elements of relation are
  called tuples. The number of domains in the
  relation is called relation arity
• Relational Schema a set of domain names along
  with theirs types.
• Database collection of relations
• Database Schema set of all relation schemas in
  the database

3
A Relation is a Table

• Relation
• Relational Scheme Student(SSN, Name, Year)

SSN Name Year
111-222-333 Jim
senior 222-111-444 Jane
junior 333-222-555 Joe
freshman 213-343-565 Kyle
junior
4
Relational Operators

• Projection (R)
• Natural join of R1 and R2 is a table that
  contains all attributes from R1 and from R1\R2
  and tuples from r1 have the same values on
  attributes that are in both R1 and R2

5
Reduction of an E-R Schema to Tables

• Primary keys allow entity sets and relationship
  sets to be expressed uniformly as tables which
  represent the contents of the database.
• A database which conforms to an E-R diagram can
  be represented by a collection of tables.
• For each entity set and relationship set there is
  a unique table which is assigned the name of the
  corresponding entity set or relationship set.
• Each table has a number of columns (generally
  corresponding to attributes), which have unique
  names.

6
Representing Entity Sets as Tables

• A strong entity set reduces to a table with the
  same attributes.

7
Composite and Multivalued Attributes

• Composite attributes are flattened out by
  creating a separate attribute for each component
  attribute
• E.g. given entity set customer with composite
  attribute name with component attributes
  first-name and last-name the table corresponding
  to the entity set has two attributes
  name.first-name and name.last-name
• A multivalued attribute M of an entity E is
  represented by a separate table EM
• Table EM has attributes corresponding to the
  primary key of E and an attribute corresponding
  to multivalued attribute M
• E.g. Multivalued attribute dependent-names of
  employee is represented by a table
  employee-dependent-names( employee-id, dname)
• Each value of the multivalued attribute maps to a
  separate row of the table EM

8
Representing Relationship Sets as Tables

• A many-to-many relationship set is represented as
  a table with columns for the primary keys of the
  two participating entity sets, and any
  descriptive attributes of the relationship set.
• E.g. table for relationship set borrower

9
Additional Rules for Translating Relationship
into Relation
If one entity set participates several times in
the relationship with different roles, its key
attributes must be listed as many times and with
different names for each role. Studies(SSN,
Name) Favorite(SSN, Name) Friends(SSN1, SSN2)
Name
SSN
subject
studies
Student
friends
favorite
10
Redundancy of Tables

• Many-to-one relationship sets that are total on
  the many-side can be represented by adding an
  extra attribute to the many side, containing the
  primary key of the one side
• Example We eliminate relation Favorite and we
  extend relation for Student as follows
• Student(SSN, Name, Subject.name)
• If, however, the relationship is many-to-many we
  cannot do that since it leads to redundancy
• For example relation Studies cannot be
  eliminated since otherwise we may end up with
• 111-222-333 John OS
• 111-222-333 John DBMS

11
Representing Weak Entity Sets

• A weak entity set becomes a table that includes a
  column for the primary key of the identifying
  strong entity set

12
Representing Weak Entity Sets(Additional Rules)

• The relation for any relationship in which the
  weak W entity participates must use as a key for
  W all of its key attributes including those of
  strong entities that contribute to the W key
• Weak entity set W participating in the
  relationship should not be converted into a
  relation.

13
Representing Specialization as Tables

• Form a table for the higher level entity
• Form a table for each lower level entity set,
  include primary key of higher level entity set
  and local attributes table table
  attributesperson name, street, city
  customer name, credit-ratingemployee name,
  salary
• Drawback getting information about, e.g.,
  employee requires accessing two tables

14
Relations Corresponding to Aggregation

• To represent aggregation, create a table
  containing
• primary key of the aggregated relationship,
• the primary key of the associated entity set
• Any descriptive attributes

15
Example
ssn
name
ISA
person
passenger
age
booked
ISA
date
departure
assigned
pilot
fhrs
gate
instantof
canfly
flight
plane
dtime
atime
F
man
model
16
Relational schema for the ER diagram

• Passenger(ssn)
  Passenger(ssn, f, date)
• Departure(f, date, gate)
• 
  departure(f,date,gate,man,model,ssn)
• Booked(f, ssn)
• Flight(f, dtime, atime)
  Flight(f, dtime,atime)
• Assigned(f, man, model, ssn)
• Person(ssn, name, age)
  Person(ssn,name,age)
• Pilot(ssn, hrs) Pilot(ssn,hrs,man,mo
  del,f,date)
• Plane(man, model) Plane(man,model)
• Canfly(man, model, ssn)

17
Functional Dependencies

• Let R(A1, A2, .Ak) be a relational schema X and
  Y are subsets of A1, A2, Ak. We say that X-gtY,
• if any two tuples that agree on X, then
  they agree on Y.
• Example
• Student(SSN,Name,Addr,subjectTaken,favSubject,Prof
  )
• SSN-gtName
• SSN-gtAddr
• subjectTaken-gtProf
• Assign(Pilot,Flight,Date,Departs)
• Pilot,Date,Departs-gtFlight
• Flight,Date-gtPilot

18
Functional Dependencies

• No need for FDs with more than one attribute on
  right side. But it maybe convenient
• SSN-gtName
• SSN-gtAddr combine into SSN-gt Name,Addr
• More than one attribute on left is important and
  we may not be able to eliminate it.
• Flight,Date-gtPilot

19
Functional Dependencies

• A functional dependency X-gtY is trivial if it is
  satisfied by any relation that includes
  attributes from X and Y
• E.g.
• customer-name, loan-number ? customer-name
• customer-name ? customer-name
• In general, ? ? ? is trivial if ? ? ?

20
Keys of Relations

• X is a superkey of R if and only if X-gtR
• X is a candidate key if X is a superkey and there
  is no subset of X that is also a superkey for R
• One of the candidate keys is selected as a
  primary key
• Example SSN is a key for
• Student(SSN,NAME, ADDR)
• How to determine keys of a relation
• One can assert a key K.
• Then the only FD on R is K-gtR
• One can be given a set of FDs and keys can be
  found from these dependencies

21
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.
• 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.

22
Closure of a Set of Functional Dependencies

• An inference axiom is a rule that states if a
  relation satisfies certain FDs, it must also
  satisfy certain other FDs
• Set of inference rules is sound if the rules lead
  only to true conclusions
• Set of inference rules is complete, if it can be
  used to conclude every valid FD on R
• We can find all of F by applying Armstrongs
  Axioms
• if ? ? ?, then ? ? ?
  (reflexivity)
• if ? ? ?, then ? ? ? ? ?
  (augmentation)
• if ? ? ?, and ? ? ?, then ? ? ? (transitivity)
• These rules are
• sound and complete

23
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

24
Procedure for Computing F

• To compute the closure of a set of functional
  dependencies F
• F Frepeat 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
  Funtil F does not change any further

25
Closure of Functional Dependencies

• We can further simplify manual computation of F
  by using the following additional rules.
• If ? ? ? holds and ? ? ? holds, then ? ? ? ?
  holds (union)
• If ? ? ? ? holds, then ? ? ? holds and ? ? ?
  holds (decomposition)
• If ? ? ? holds and ? ? ? ? holds, then ? ? ? ?
  holds (pseudotransitivity)
• The above rules can be inferred from Armstrongs
  axioms.

26
Closure of Attribute Sets

• Given a set of attributes a, define the closure
  of a under F (denoted by a) as the set of
  attributes that are functionally determined by a
  under F a ? ? is in F ? ? ? a
• Algorithm to compute a, the closure of a under
  F result a while (changes to result)
  do for each ? ? ? in F do begin if ? ?
  result then result result ? ? end

27
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.

28
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 key?
• Is AG a super key?
• Does AG ? R? Is (AG) ? R
• Is any subset of AG a superkey?
• Does A ? R? Is (A) ? R
• Does G ? R? Is (G) ? R

29
Extraneous Attributes

• Consider a set F of functional dependencies and
  the functional dependency ? ? ? in F.
• Attribute A is extraneous in ? if A ? ? and
• (F ? ? ?) ? (? A) ? ? logically implies
  F,or
• 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

30
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 ?
• compute (? A) using the dependencies in F
• check that (? A) contains A if it does, A
  is extraneous
• To test if attribute A ? ? is extraneous in ?
• compute ? using only the dependencies in
  F (F ? ? ?) ? ? ?(? A),
• check that ? contains A if it does, A is
  extraneous

31
Canonical Cover

• 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
• A canonical cover of F is a minimal set of
  functional dependencies equivalent to F, having
  no redundant dependencies or redundant parts of
  dependencies

32
Canonical Cover(Formal Definition)

• 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.

33
Canonical CoverComputation

• To compute a canonical cover for Frepeat 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

34
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
• A is extraneous in AB ? C
• 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.
• The canonical cover is A ? B B ? C

35
Foreign Keys

• Let R1 and R2 be two relational schemas. Let K1
  and K2 be primary keys of R1 and R2,
  respectively. If R1 contains all attributes from
  K2, then we say that K2 is a foreign key of R1.
• Integrity Constraints
• Domain Constraints
• Key Constraints
• Interdomain Constraints
• Database Schema S is a set of relational schemas
  and constraints defined on them

36
Constraints

• Insert Constraints
• No tuple should be inserted into a relation r1
  with foreign keys of r2 that are not listed as
  primary key in r2 (referential integrity)
• No tuples should be inserted with duplicate
  primary key(primary key constraint)
• No primary key value can contain nulls (primary
  key constraint)
• Delete Constraint Tuple should not be deleted
  from r2 with foreign key values for r2, if a
  deletion of this tuple will result in referential
  integrity constraint violation
• Update should respect referential and primary key
  constraints

37
Relational Database Design Problem

• Problem Given a set of attributes and a set of
  FDs, generate a set of relational schemas
  describing the enterprise.
• Approach 1 Make one big relational schema that
  contains all attributes (universal relation
  approach)
• Problems
• Repetition of information (name, addr, dep_name)
  address is repeated for each dependent.
• Inability to represent certain information,
  unless nulls are used (name, position,sal,
  comission)
• Loss of information referential integrity
  violations

38
Relational Database Design Problem

• If one big relational schema is not good, then we
  need to decompose it into smaller relational
  schemas so that no loss of information will occur
• Issue how to decompose without loosing the
  information?
• (person_name, loan, balance, branch_name)
• Decompose into
• (person_name,loan) (loan,balance,branch_n
  ame)
• Information gets lost!
• Thus, we need to find a lossless decomposition

39
Decomposition

• 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

40
Example of Lossy-Join Decomposition

• Lossy-join decompositions result in information
  loss.
• Example Decomposition of R (A, B)
• R1 (A) R2 (B)

A
B
A
B
? ? ?
1 2 1
? ?
1 2
?B(r)
?A(r)
r
A
B
?A (r) ?B (r)
? ? ? ?
1 2 1 2
41
Goals of Normalization

• 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 satisfies a referential integrity
  constraints
• the decomposition is a lossless-join
  decomposition
• the decomposition preserves the set of functional
  dependencies
• Our theory initially is based on
• functional dependencies

42
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.
• Dependency preservation Let Fi be the set of
  dependencies F that include only attributes in
  Ri.
• (F1 ? F2 ? ? Fn) F
• .

43
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)

44
Testing for Dependency Preservation

• To check if a dependency ??? is preserved in a
  decomposition of R into R1, R2, , Rn 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)

45
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 ??? ?,
where ? ? R and ? ? R, at least one of the
following holds

• ?? ? ? is trivial (i.e., ? ? ?)
• ? is a superkey for R

46
Example

• R (A, B, C)F A ? B B ? CKey A
• R is not in BCNF
• Decomposition R1 (A, B), R2 (B, C)
• R1 and R2 in BCNF
• Lossless-join decomposition
• Dependency preserving

47
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
• 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.

48
BCNF Decomposition Algorithm

• result Rdone falsecompute Fwhile
  (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
• Each Ri is in BCNF, and decomposition is
  lossless-join.

49
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

50
BCNF and Dependency Preservation
It is not always possible to get a BCNF
decomposition that is dependency preserving

• R (A, B, C)F AB ? C C ? BTwo candidate
  keys AB and AC
• R is not in BCNF
• Any decomposition of R will fail to preserve
• AB ? C

51
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.
• FDs can be checked on individual relations
  without computing a join.
• There is always a lossless-join,
  dependency-preserving decomposition into 3NF.

52
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.
• 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.

53
Third Normal Form

• Example
• R (A,B,C)F AB ? C, C ? B
• Two candidate keys AB and AC
• R is in 3NF
• AB ? C AB is a superkey C ? B B is contained
  in a candidate key
• BCNF decomposition has (AC) and (BC)
• Testing for AB ? C requires a join

54
Testing for 3NF

• 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
• However, decomposition into third normal form can
  be done in polynomial time

55
3NF Decomposition Algorithm

• Let Fc be a canonical cover for Fi 0for
  each functional dependency ? ? ? in Fc do if
  none of the schemas Rj, 1 ? j ? i contains ? ?
  then begin i i 1 Ri ? ?
  endif 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)

56
3NF Decomposition Algorithm

• Decomposition algorithm ensures
• each relation schema Ri is in 3NF
• decomposition is dependency preserving and
  lossless-join

57
Example

• Relation schema R(A, B, C, D)
• The functional dependencies for this relation
  schema are C ? AD AB ? C
• The keys are
• BC, AB

58
Applying 3NF

• The for loop in the algorithm causes us to
  include the following schemas in our
  decomposition R1(ACD), R2(ABC)
• Since R2 contains a candidate key for R1, we are
  done with the decomposition process.

59
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.

60
Comparison of BCNF and 3NF

• Example of problems due to redundancy in 3NF
• R (A, B, C)F AB ? C, C ? B

C
A
B
c1 c1 c1 c2
b1 b1 b1 b2
a1 a2 a3 null

• A schema that is in 3NF but not in BCNF has the
  problems of
• repetition of information (e.g., the relationship
  c1, b1)
• need to use null values (e.g., to represent the
  relationship c2, b2 where there is no
  corresponding value for A).

61
Design Goals(revisited)

• 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

62
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 t of the relation ri is a dangling tuple
  if t 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.

63
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

64
Universal Relation Approach

• 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

65
Multivalued Dependencies

• There are database schemas in BCNF that do not
  seem to be sufficiently normalized
• Consider a database
• classes(course, teacher, book)such that
  (c,t,b) ? classes means that t is qualified to
  teach c, and b is a required textbook for c
• The database is supposed to list for each course
  the set of teachers any one of which can be the
  courses instructor, and the set of books, all of
  which are required for the course (no matter who
  teaches it).

66
Multivalued Dependencies
classes
course
teacher
book
database database database database database datab
ase operating systems operating systems operating
systems operating systems
Avi Avi Hank Hank Sudarshan Sudarshan Avi Avi
Jim Jim
DB Concepts Ullman DB Concepts Ullman DB
Concepts Ullman OS Concepts Shaw OS Concepts Shaw

• There are no non-trivial functional dependencies
  and therefore the relation is in BCNF
• Insertion anomalies i.e., if Sara is a new
  teacher that can teach database, two tuples need
  to be inserted
• (database, Sara, DB Concepts) (database, Sara,
  Ullman)

67
Multivalued Dependencies

• Therefore, it is better to decompose classes into

course
teacher
database database database operating
systems operating systems
Avi Hank Sudarshan Avi Jim
teaches
course
book
database database operating systems operating
systems
DB Concepts Ullman OS Concepts Shaw
text
68
Multivalued Dependencies (MVDs)

• Let R be a relation schema and let ? ? R and ? ?
  R. The multivalued dependency
• ? ?? ?
• holds on R if in any legal relation r(R), for
  all pairs for tuples t1 and t2 in r such that
  t1? t2 ?, there exist tuples t3 and t4 in r
  such that
• t1? t2 ? t3 ? t4 ? t3?
  t1 ? t3R ? t2R ? t4
  ? t2? t4R ? t1R ?

69
MVD (Tabular illustration)

• Tabular representation of ? ?? ?

70
Example

• Let R be a relation schema with a set of
  attributes that are partitioned into 3 nonempty
  subsets.
• A, B, C
• We say that A?? B (A multidetermines B)if and
  only if for all possible relations r(R)
• lt a1, b1, c1 gt ? r and lt a2, b2, c2 gt ? r
• then
• lt a1, b1, c2 gt ? r and lt a2, b2, c1 gt ? r
• Note that since the behavior of B and C are
  identical it follows that A ?? B if A?? C

71
Example

• In our example
• course ?? teacher course ?? book
• The above formal definition is supposed to
  formalize the notion that given a particular
  value of A(course) it has associated with it a
  set of values of B(teacher) and a set of values
  of C (book), and these two sets are in some sense
  independent of each other.
• Note
• If A ? B then A ?? B
• Indeed we have (in above notation) B1 B2The
  claim follows.

72
Another Example
A B C D

• A B
• C
  D
• 
  a1b1c1d2
• 
  a2b2c1d1
• but
• 
  a1b1c1d1
• 
  a2b2c1d2 are not in the relation
• Multivalued dependency is a semantic notion
  
  

a1 b1 c1 d2 a1 b2 c2
d1 a1 b2 c1 d2 a1 b1 c2
d1 a2 b2 c1 d1 a2 b3 c2
d2 a2 b2 c2 d2
73
One more example
SSN EducDeg Age Dept
100 BS 32
CS 100 BS 32
CS 200 BS 26
Physics 200 MS 26
Physics 200 PhD
26 Physics
EducDeg
SSN
Every relation with only two attributes has a
multivalued dependency between these attributes
74
Derivation Rules for Functional and Multivalued
Dependencies

• If Y is a subset of X, then X Y reflexivity
• X Y, then XZ YZ augmentation
• X Y and Y Z, then X Z
  transitivity
• If X Y, then X U-X-Y -
  complementation
• If X Y and V is a subset of W, then XW
  VY
• 
  augmentation
• If X Y and Y Z, then X
  YZ - transitivity
• If X Y, then X Y
• If X Y, Z is a subset of Y and
  intersection of W and Y empty, and W Z, then
  X Z

75
Use of Multivalued Dependencies

• We use multivalued dependencies in two ways
• 1. To test relations to determine whether they
  are legal under a given set of functional and
  multivalued dependencies
• 2. To specify constraints on the set of legal
  relations. We shall thus concern ourselves only
  with relations that satisfy a given set of
  functional and multivalued dependencies.

76
Theory of MVDs

• From the definition of multivalued dependency, we
  can derive the following rule
• If ? ? ?, then ? ?? ?
• That is, every functional dependency is also a
  multivalued dependency
• The closure D of D is the set of all functional
  and multivalued dependencies logically implied by
  D.
• We can compute D from D, using the formal
  definitions of functional dependencies and
  multivalued dependencies.
• We can manage with such reasoning for very simple
  multivalued dependencies, which seem to be most
  common in practice
• For complex dependencies, it is better to reason
  about sets of dependencies using a system of
  inference rules.

77
Fourth Normal Form

• A relation schema R is in 4NF with respect to a
  set D of functional and multivalued dependencies
  if for all multivalued dependencies in D of the
  form ? ?? ?, where ? ? R and ? ? R, at least one
  of the following hold
• ? ?? ? is trivial (i.e., ? ? ? or ? ? ? R)
• ? is a superkey for schema R
• If a relation is in 4NF it is in BCNF

78
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
• ? ?? (? ? Ri)
• where ? ? Ri and ? ?? ? is in D

79
4NF Decomposition Algorithm

• result Rdone falsecompute
  DLet Di denote the restriction of D to Ri
• while (not done) if (there is a schema
  Ri in result that is not in 4NF) then
  begin
• let ? ?? ? be a nontrivial multivalued
  dependency that holds on Ri such that
  ? ? Ri is not in Di, and ?????
  result (result - Ri) ? (Ri - ?) ? (?, ?)
  end else done true
• Note each Ri is in 4NF, and decomposition is
  lossless-join

80
Example

• R (A, B, C, G, H, I)
• F A ?? B
• B ?? HI
• CG ?? H
• R is not in 4NF since A ?? B and A is not a
  superkey for R
• Decomposition
• a) R1 (A, B) (R1 is in 4NF)
• b) R2 (A, C, G, H, I) (R2 is not in 4NF)
• c) R3 (C, G, H) (R3 is in 4NF)
• d) R4 (A, C, G, I) (R4 is not in 4NF)
• Since A ?? B and B ?? HI, A ?? HI, A ?? I
• e) R5 (A, I) (R5 is in 4NF)
• f)R6 (A, C, G) (R6 is in 4NF)

81
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

82
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.
• 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.

83
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

84
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

85
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.