Database System Concepts, 6th Ed.©Silberschatz, Korth and .docx

Database System Concepts, 6th Ed.
©Silberschatz, Korth and Sudarshan
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Chapter 6: Formal Relational Query Languages
6.*
Database System Concepts - 6th Edition
Chapter 6: Formal Relational Query LanguagesRelational
AlgebraTuple Relational CalculusDomain Relational Calculus
6.*
Relational AlgebraProcedural languageSix basic
operators – Cartesian
inputs and produce a new relation as a result.

6.*
Select Operation –
6.*
predicateDefined as:
Where p is a formula in propositional calculus consisting
Each term is one of:
<attribute> op <attribute> or <constant>
Example of selection:
structor)

6.*
Project Operation – ExampleRelation r:
6.*
Project OperationNotation:
where A1, A2 are attribute names and r is a relation
name.The result is defined as the relation of k columns obtained
by erasing the columns that are not listedDuplicate rows
removed from result, since relations are setsExample: To
eliminate the dept_name attribute of instructor

6.*
Union Operation –
6.*
1. r, s must have the same arity (same number of
attributes)
2. The attribute domains must be compatible (example:
2nd column
of r deals with the same type of values as does the 2nd
column of s)Example: to find all courses taught in the Fall
2009 semester, or in the Spring 2010 semester, or in both

6.*
Set difference of two relationsRelations r, s:r – s:
6.*
Set Difference OperationNotation r – sDefined as:
r –
Set differences must be taken between compatible relations.r
and s must have the same arityattribute domains of r and s must
be compatibleExample: to find all courses taught in the Fall
2009 semester, but not in the Spring 2010 semester
6.*

Cartesian-Product Operation – ExampleRelations r, s:r x s:
6.*
Cartesian-Product OperationNotation r x sDefined as:
r x s = {t
Assume that attributes of r(R) and s(S) are disjoint. (That is, R
renaming must be used.
6.*
Composition of OperationsCan build expressions using multiple
r x s

6.*
Rename OperationAllows us to name, and therefore to refer to,
the results of relational-algebra expressions.Allows us to refer
to a relation by more than one name.Example:
returns the expression E under the name XIf a relational-
algebra expression E has arity n, then
returns the result of expression E under the name X, and
with the
attributes renamed to A1 , A2 , …., An .
6.*
Example QueryFind the largest salary in the universityStep 1:
find instructor salaries that are less than some other instructor
salary (i.e. not maximum)

Find t –
6.*
Example QueriesFind the names of all instructors in the Physics
department, along with the course_id of all courses they have
taughtQuery 1
teaches)))Query 2
6.*
Formal DefinitionA basic expression in the relational algebra

consists of either one of the following:A relation in the
databaseA constant relationLet E1 and E2 be relational-algebra
expressions; the following are all relational-algebra
– ate
6.*
Additional Operations
We define additional operations that do not add any power to
the
relational algebra, but that simplify common queries.Set
intersectionNatural joinAssignmentOuter join
6.*
Set-Intersection Ope
– (r – s)

6.*
Set-Intersection Operation – ExampleRelation r, s:
6.*
Natural-Join OperationLet r and s be relations on schemas R and
S respectively.
Then, r s is a relation on schema R
follows:Consider each pair of tuples tr from r and ts from s. If
add a tuple t to the result, wheret has the same value as tr on rt
has the same value as ts on sExample:
R = (A, B, C, D)
S = (E, B, D)Result schema = (A, B, C, D, E)r s is defined
as:
Notation: r s

6.*
Natural Join ExampleRelations r, s:r s
6.*
Natural Join and Theta JoinFind the names of all instructors in
the Comp. Sci. department together with the course titles of all
dept_name=“Comp. Sci.” (instructor teaches
course))Natural join is associative(instructor teaches)
course is equivalent to
instructor (teaches course)Natural join is
commutativeinstruct teaches is equivalent to
6.*

Assignment Operati
convenient way to express complex queries. Write query as a
sequential program consisting ofa series of assignments
followed by an expression whose value is displayed as a result
of the query.Assignment must always be made to a temporary
relation variable.
6.*
Outer JoinAn extension of the join operation that avoids loss of
information.Computes the join and then adds tuples form one
relation that does not match tuples in the other relation to the
result of the join. Uses null values:null signifies that the value
is unknown or does not exist All comparisons involving null are
(roughly speaking) false by definition.We shall study precise
meaning of comparisons with nulls later
6.*
Outer Join – ExampleRelation instructor1Relation teaches1
ID
course_id
10101

12121
76766
CS-101
FIN-201
BIO-101
Comp. Sci.
Finance
Music
ID
dept_name
10101
12121
15151
name
Srinivasan
Wu
Mozart
6.*
Outer Join – ExampleJoin
instructor teaches Left Outer Join
instructor teaches
ID
dept_name
10101
12121
Comp. Sci.

Finance
course_id
CS-101
FIN-201
name
Srinivasan
Wu
ID
dept_name
10101
12121
15151
Comp. Sci.
Finance
Music
course_id
CS-101
FIN-201
null
name
Srinivasan
Wu
Mozart
6.*
Outer Join – Example

Full Outer Join
instructor teaches Right Outer Join
instructor teaches
ID
dept_name
10101
12121
76766
Comp. Sci.
Finance
null
course_id
CS-101
FIN-201
BIO-101
name
Srinivasan
Wu
null
ID
dept_name
10101
12121
15151
76766
Comp. Sci.
Finance
Music
null
course_id
CS-101
FIN-201
null
BIO-101
name
Srinivasan

Wu
Mozart
null
6.*
Outer Join using JoinsOuter join can be expressed using basic
operationse.g. r s can be written as
(r s) U (r – ∏R(r s) x {(null, …, null)}
6.*
Null ValuesIt is possible for tuples to have a null value, denoted
by null, for some of their attributesnull signifies an unknown
value or that a value does not exist.The result of any arithmetic
expression involving null is null.Aggregate functions simply

ignore null values (as in SQL)For duplicate elimination and
grouping, null is treated like any other value, and two nulls are
assumed to be the same (as in SQL)
6.*
Null ValuesComparisons with null values return the special
truth value: unknownIf false was used instead of unknown, then
not (A < 5)
would not be equivalent to A >= 5Three-
valued logic using the truth value unknown:OR: (unknown or
true) = true,
(unknown or false) = unknown
(unknown or unknown) = unknownAND: (true and
unknown) = unknown,
(false and unknown) = false,
(unknown and unknown) = unknownNOT: (not
unknown) = unknownIn SQL “P is unknown” evaluates to true if
predicate P evaluates to unknownResult of select predicate is
treated as false if it evaluates to unknown

6.*
Division OperatorGiven
-S) such that
(takes ) and
then r
-S (r )
-S ((temp1 x s ) – -S,S (r ))
result = temp1 –
is assigned to the relation variable
variable in subsequent expressions.
6.*
Extended Relational-Algebra-OperationsGeneralized
ProjectionAggregate Functions

6.*
Generalized ProjectionExtends the projection operation by
allowing arithmetic functions to be used in the projection list.
E is any relational-algebra expressionEach of F1, F2, …, Fn are
are arithmetic expressions involving constants and attributes in
the schema of E.Given relation instructor(ID, name, dept_name,
salary) where salary is annual salary, get the same information
but with monthly salary
6.*
Aggregate Functions and OperationsAggregation function takes
a collection of values and returns a single value as a result.
avg: average value
min: minimum value
max: maximum value
sum: sum of values

count: number of valuesAggregate operation in relational
algebra
E is any relational-algebra expressionG1, G2 …, Gn is a list of
attributes on which to group (can be empty)Each Fi is an
aggregate functionEach Ai is an attribute nameNote: Some
6.*
Aggregate Operation – ExampleRelation r:
A
B
C
7
7
3
10 sum(c) (r)
sum(c )

27
6.*
Aggregate Operation – ExampleFind the average salary in each
department
dept_name avg(salary) (instructor)
avg_salary
6.*
Aggregate Functions (Cont.)Result of aggregation does not have
a nameCan use rename operation to give it a nameFor
convenience, we permit renaming as part of aggregate operation
dept_name avg(salary) as avg_sal (instructor)
6.*

Modification of the DatabaseThe content of the database may be
modified using the following
operations:DeletionInsertionUpdatingAll these operations can
be expressed using the assignment operator
6.*
Multiset Relational AlgebraPure relational algebra removes all
duplicates e.g. after projectionMultiset relational algebra retains
duplicates, to match SQL semanticsSQL duplicate retention was
initially for efficiency, but is now a featureMultiset relational
algebra defined as followsselection: has as many duplicates of a
tuple as in the input, if the tuple satisfies the
selectionprojection: one tuple per input tuple, even if it is a
duplicatecross product: If there are m copies of t1 in r, and n
copies of t2 in s, there are m x n copies of t1.t2 in r x sOther
operators similarly defined E.g. union: m + n copies,
intersection: min(m, n) copies
difference: min(0, m – n) copies
6.*
SQL and Relational Algebraselect A1, A2, .. An

from r1, r2, …, rm
where P
is equivalent to the following expression in multiset
relational algebra
sum(A3)
where P
group by A1, A2
relational algebra
6.*
SQL and Relational AlgebraMore generally, the non-aggregated
attributes in the select clause may be a subset of the group by
attributes, in which case the equivalence is as follows:
select A1, sum(A3)
where P
group by A1, A2
relational algebra

(r1 x r2 x
.. x rm)))
6.*
Tuple Relational Calculus
6.*
Tuple Relational CalculusA nonprocedural query language,
where each query is of the form
{t | P (t ) }It is the set of all tuples t such that
predicate P is true for tt is a tuple variable, t [A ] denotes the
relation rP is a formula similar to that of the predicate calculus
6.*
Predicate Calculus Formula
1. Set of attributes and constants

2.
3. Se
4.
5. Set of quantifiers:
such that predicate Q (t ) is true
6.*
Example QueriesFind the ID, name, dept_name, salary for
instructors whose salary is greater than $80,000 As in the
previous query, but output only the ID attribute value
80000)}
Notice that a relation on schema (ID) is implicitly defined
by
the query

6.*
Example QueriesFind the names of all instructors whose
department is in the Watson building
Find the set of all courses taught in the Fall 2009 semester, or
in
the Spring 2010 semester, or both
6.*
Example Queries

Find the set of all courses taught in the Fall 2009 semester, and
in
the Spring 2010 semester
s [semester] = “Fal
Find the set of all courses taught in the Fall 2009 semester, but
not in
6.*
Safety of ExpressionsIt is possible to write tuple calculus
domain of any
attribute of relation r is infiniteTo guard against the problem,
we restrict the set of allowable expressions to safe
expressions.An expression {t | P (t )} in the tuple relational
calculus is safe if every component of t appears in one of the
relations, tuples, or constants that appear in PNOTE: this is

not safe --- it defines an infinite set with attribute values that do
not appear in any relation or tuples or constants in P.
6.*
Universal QuantificationFind all students who have taken all
s [course_id] = u [course_id]))}Note that
without the existential quantification on student, the above
query would be unsafe if the Biology department has not offered
any courses.
6.*
Domain Relational Calculus

6.*
Domain Relational CalculusA nonprocedural query language
equivalent in power to the tuple relational calculusEach query is
an expression of the form:
x1, x2, …, xn represent domain variablesP represents a formula
similar to that of the predicate calculus
6.*
Example QueriesFind the ID, name, dept_name, salary for
instructors whose salary is greater than $80,000{< i, n, d, s> | <
department is in the Watson building
instructor
))}

6.*
Example Queries
courses taught in the Fall 2009 semester, or in
the Spring 2010 semester, or both
This case can also be written as
2010))} Find the set of all courses taught in the Fall 2009
semester, and in

6.*
Safety of Expressions
The expression:
is safe if all of the following hold:
All values that appear in tuples of the expression are values
from dom (P ) (that is, the values appear either in P or in a
tuple of a relation mentioned in P ).
the subformula is true if and only if there is a value of x in
dom (P1) such that P1(x ) is true.
subformula is true if and only if P1(x ) is true for all values x
from dom (P1).
6.*
Universal QuantificationFind all students who have taken all

=“Biology”
))}Note that without the existential quantification on student,
the above query would be unsafe if the Biology department has
not offered any courses.
* Above query fixes bug in page 246, last query
End of Chapter 6
6.*
Figure 6.01
6.*
Figure 6.02

6.*
Figure 6.03
6.*
Figure 6.04
6.*
Figure 6.05

6.*
Figure 6.06
6.*
Figure 6.07
6.*
Figure 6.08

6.*
Figure 6.09
6.*
Figure 6.10
6.*
Figure 6.11
6.*

Figure 6.12
6.*
Figure 6.13
6.*
Figure 6.14
6.*
Figure 6.15

6.*
Figure 6.16
6.*
Figure 6.17
6.*
Figure 6.18

6.*
Figure 6.19
6.*
Figure 6.20
6.*
Figure 6.21

6.*
DeletionA delete request is expressed similarly to a query,
except instead of displaying tuples to the user, the selected
tuples are removed from the database.Can delete only whole
tuples; cannot delete values on only particular attributesA
deletion is expressed in relational algebra by:
– E
where r is a relation and E is a relational algebra query.
6.*
Deletion ExamplesDelete all account records in the Perryridge
branch. Delete all accounts at branches located in Needham.
Delete all loan records with amount in the range of 0 to 50
–
–
branch )
– r2
– r3

6.*
InsertionTo insert data into a relation, we either:specify a tuple
to be insertedwrite a query whose result is a set of tuples to be
insertedin relational algebra, an insertion is expressed by:
where r is a relation and E is a relational algebra
expression.The insertion of a single tuple is expressed by letting
E be a constant relation containing one tuple.
6.*
Insertion ExamplesInsert information in the database specifying
that Smith has $1200 in account A-973 at the Perryridge branch.
Provide as a gift for all loan customers in the Perryridge
branch, a $200 savings account. Let the loan number serve
as the account number for the new savings account.
acc -973”, “Perryridge”, 1200)}
-973”)}
number (r1)

6.*
UpdatingA mechanism to change a value in a tuple without
charging all values in the tupleUse the generalized projection
operator to do this task
Each Fi is either the I th attribute of r, if the I th attribute is not
updated, or,if the attribute is to be updated Fi is an expression,
involving only constants and the attributes of r, which gives the
new value for the attribute
6.*
Update ExamplesMake interest payments by increasing all
balances by 5 percent. Pay all accounts with balances over
$10,000 6 percent interest
and pay all others 5 percent
r, branch_name, balance * 1.06
(account)

6.*
Example QueriesFind the name of all customers who have a
loan at the bank and the loan amountFind the names of all
customers who have a loan and an account at bank.
(depositor)
6.*
Example QueriesFind all customers who have an account from
at least the “Downtown” and the Uptown” branches.Query 1
account))
Query 2
wntown” ),
(“Uptown” )})
Note that Query 2 uses a constant relation.

6.*
Bank Example QueriesFind all customers who have an account
at all branches located in Brooklyn city.
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Chapter 1: Introduction
*
1.*
Database Management System (DBMS)DBMS contains
information about a particular enterpriseCollection of
interrelated dataSet of programs to access the data An
environment that is both convenient and efficient to
useDatabase Applications:Banking: transactionsAirlines:
reservations, schedulesUniversities: registration, gradesSales:
customers, products, purchasesOnline retailers: order tracking,
customized recommendationsManufacturing: production,
inventory, orders, supply chainHuman resources: employee
records, salaries, tax deductionsDatabases can be very
large.Databases touch all aspects of our lives
*

1.*
University Database ExampleApplication program examplesAdd
new students, instructors, and coursesRegister students for
courses, and generate class rostersAssign grades to students,
compute grade point averages (GPA) and generate transcriptsIn
the early days, database applications were built directly on top
of file systems
*
1.*
Drawbacks of using file systems to store data
Data redundancy and inconsistencyMultiple file formats,
duplication of information in different filesDifficulty in
accessing data Need to write a new program to carry out each
new taskData isolation — multiple files and formatsIntegrity
problemsIntegrity constraints (e.g., account balance > 0)
become “buried” in program code rather than being stated
explicitlyHard to add new constraints or change existing ones
*

1.*
Drawbacks of using file systems to store data (Cont.)
Atomicity of updatesFailures may leave database in an
inconsistent state with partial updates carried outExample:
Transfer of funds from one account to another should either
complete or not happen at allConcurrent access by multiple
usersConcurrent access needed for performanceUncontrolled
concurrent accesses can lead to inconsistencies
Example: Two people reading a balance (say 100) and updating
it by withdrawing money (say 50 each) at the same timeSecurity
problemsHard to provide user access to some, but not all, data
Database systems offer solutions to all the above problems
*
1.*
Levels of AbstractionPhysical level: describes how a record
(e.g., customer) is stored.Logical level: describes data stored in
database, and the relationships among the data.
type instructor = record
ID : string;
name : string;
dept_name : string;

salary : integer;
end;View level: application programs hide details of data types.
Views can also hide information (such as an employee’s salary)
for security purposes.
*
1.*
View of Data
An architecture for a database system
*
1.*
Instances and SchemasSimilar to types and variables in
programming languagesSchema – the logical structure of the
database Example: The database consists of information about a
set of customers and accounts and the relationship between
themAnalogous to type information of a variable in a
programPhysical schema: database design at the physical
levelLogical schema: database design at the logical
levelInstance – the actual content of the database at a particular
point in time Analogous to the value of a variablePhysical Data

Independence – the ability to modify the physical schema
without changing the logical schemaApplications depend on the
logical schemaIn general, the interfaces between the various
levels and components should be well defined so that changes in
some parts do not seriously influence others.
*
1.*
Data ModelsA collection of tools for describing Data Data
relationshipsData semanticsData constraintsRelational
modelEntity-Relationship data model (mainly for database
design) Object-based data models (Object-oriented and Object-
relational)Semistructured data model (XML)Other older
models:Network model Hierarchical model
*
1.*
Relational ModelRelational model (Chapter 2)Example of
tabular data in the relational model
Columns
Rows

*
1.*
A Sample Relational Database
*
1.*
Data Manipulation Language (DML)Language for accessing and
manipulating the data organized by the appropriate data
modelDML also known as query languageTwo classes of
languages Procedural – user specifies what data is required and
how to get those data Declarative (nonprocedural) – user
specifies what data is required without specifying how to get
those dataSQL is the most widely used query language
*
1.*

Data Definition Language (DDL)Specification notation for
defining the database schema
Example: create table instructor (
ID char(5),
name varchar(20),
dept_name varchar(20),
salary numeric(8,2))DDL compiler
generates a set of table templates stored in a data
dictionaryData dictionary contains metadata (i.e., data about
data)Database schema Integrity constraintsPrimary key (ID
uniquely identifies instructors)Referential integrity (references
constraint in SQL)
e.g. dept_name value in any instructor tuple must appear in
department relationAuthorization
*
1.*
SQLSQL: widely used non-procedural languageExample: Find
the name of the instructor with ID 22222
select name
from instructor

where instructor.ID = ‘22222’Example: Find the ID
and building of instructors in the Physics dept.
select instructor.ID, department.building
from instructor, department
where instructor.dept_name = department.dept_name and
department.dept_name = ‘Physics’
Application programs generally access databases through one
ofLanguage extensions to allow embedded SQLApplication
program interface (e.g., ODBC/JDBC) which allow SQL queries
to be sent to a databaseChapters 3, 4 and 5
*
1.*
Database Design
The process of designing the general structure of the database:
Logical Design – Deciding on the database schema. Database
design requires that we find a “good” collection of relation
schemas.Business decision – What attributes should we record
in the database?Computer Science decision – What relation
schemas should we have and how should the attributes be
distributed among the various relation schemas?
Physical Design – Deciding on the physical layout of the
database

*
1.*
Database Design?Is there any problem with this design?
*
1.*
Design ApproachesNormalization Theory (Chapter 8)Formalize
what designs are bad, and test for themEntity Relationship
Model (Chapter 7)Models an enterprise as a collection of
entities and relationshipsEntity: a “thing” or “object” in the
enterprise that is distinguishable from other objects
Described by a set of attributesRelationship: an association
among several entitiesRepresented diagrammatically by an
entity-relationship diagram:
*

1.*
The Entity-Relationship ModelModels an enterprise as a
collection of entities and relationshipsEntity: a “thing” or
“object” in the enterprise that is distinguishable from other
objectsDescribed by a set of attributesRelationship: an
association among several entitiesRepresented diagrammatically
by an entity-relationship diagram:
What happened to dept_name of instructor and student?
*
1.*
Object-Relational Data ModelsRelational model: flat, “atomic”
valuesObject Relational Data ModelsExtend the relational data
model by including object orientation and constructs to deal
with added data types.Allow attributes of tuples to have
complex types, including non-atomic values such as nested
relations.Preserve relational foundations, in particular the
declarative access to data, while extending modeling
power.Provide upward compatibility with existing relational
languages.
*

1.*
XML: Extensible Markup LanguageDefined by the WWW
Consortium (W3C)Originally intended as a document markup
language not a database languageThe ability to specify new
tags, and to create nested tag structures made XML a great way
to exchange data, not just documentsXML has become the basis
for all new generation data interchange formats.A wide variety
of tools is available for parsing, browsing and querying XML
documents/data
*
1.*
Storage ManagementStorage manager is a program module that
provides the interface between the low-level data stored in the
database and the application programs and queries submitted to
the system.The storage manager is responsible to the following
tasks: Interaction with the file manager Efficient storing,
retrieving and updating of dataIssues:Storage accessFile
organizationIndexing and hashing
*
1.*

Query Processing
1. Parsing and translation
2. Optimization
3. Evaluation
*
1.*
Query Processing (Cont.)Alternative ways of evaluating a given
queryEquivalent expressionsDifferent algorithms for each
operationCost difference between a good and a bad way of
evaluating a query can be enormousNeed to estimate the cost of
operationsDepends critically on statistical information about
relations which the database must maintainNeed to estimate
statistics for intermediate results to compute cost of complex
expressions
*
1.*
Transaction Management What if the system fails?What if more
than one user is concurrently updating the same data?A

transaction is a collection of operations that performs a single
logical function in a database applicationTransaction-
management component ensures that the database remains in a
consistent (correct) state despite system failures (e.g., power
failures and operating system crashes) and transaction
failures.Concurrency-control manager controls the interaction
among the concurrent transactions, to ensure the consistency of
the database.
*
1.*
Database Users and Administrators
Database
*
1.*
Database System Internals

*
1.*
Database Architecture
The architecture of a database systems is greatly influenced by
the underlying computer system on which the database is
running:CentralizedClient-serverParallel (multi-
processor)Distributed
*
1.*
History of Database Systems1950s and early 1960s:Data
processing using magnetic tapes for storageTapes provided only
sequential accessPunched cards for inputLate 1960s and
1970s:Hard disks allowed direct access to dataNetwork and
hierarchical data models in widespread useTed Codd defines the
relational data modelWould win the ACM Turing Award for this
workIBM Research begins System R prototypeUC Berkeley
begins Ingres prototypeHigh-performance (for the era)
transaction processing
*

1.*
History (cont.)1980s:Research relational prototypes evolve into
commercial systemsSQL becomes industrial standardParallel
and distributed database systemsObject-oriented database
systems1990s:Large decision support and data-mining
applicationsLarge multi-terabyte data warehousesEmergence of
Web commerceEarly 2000s:XML and XQuery
standardsAutomated database administrationLater 2000s:Giant
data storage systemsGoogle BigTable, Yahoo PNuts, Amazon, ..
*
1.*
End of Chapter 1
*
1.*
Figure 1.02

*
1.*
Figure 1.04
*
1.*
Figure 1.06
*
Chapter 2: Intro to Relational Model

*
2.*
Example of a Relation
attributes
(or columns)
tuples
(or rows)
*
2.*
Attribute TypesThe set of allowed values for each attribute is
called the domain of the attributeAttribute values are (normally)
required to be atomic; that is, indivisibleThe special value null
is a member of every domainThe null value causes
complications in the definition of many operations
*

2.*
Relation Schema and InstanceA1, A2, …, An are attributes
R = (A1, A2, …, An ) is a relation schema
Example:
instructor = (ID, name, dept_name, salary)Formally,
given sets D1, D2, …. Dn a relation r is a subset of
D1 x D2 x … x Dn
Thus, a relation is a set of n-tuples (a1, a2, …, an) where each
The current values (relation instance) of a relation are specified
by a tableAn element t of r is a tuple, represented by a row in a
table
*
2.*
Relations are Unordered Order of tuples is irrelevant (tuples
may be stored in an arbitrary order) Example: instructor relation
with unordered tuples
*

2.*
DatabaseA database consists of multiple relationsInformation
about an enterprise is broken up into parts
instructor
student
advisorBad design:
univ (instructor -ID, name, dept_name, salary, student_Id,
..)
results inrepetition of information (e.g., two students have the
same instructor)the need for null values (e.g., represent an
student with no advisor)Normalization theory (Chapter 7) deals
with how to design “good” relational schemas
*
2.*
sufficient to identify a unique tuple of each possible relation
r(R) Example: {ID} and {ID,name} are both superkeys of
instructor.Superkey K is a candidate key if K is minimal
Example: {ID} is a candidate key for InstructorOne of the
candidate keys is selected to be the primary key.which

one?Foreign key constraint: Value in one relation must appear
in anotherReferencing relationReferenced relation
*
2.*
Schema Diagram for University Database
2.*
Relational Query LanguagesProcedural vs.non-procedural, or
declarative“Pure” languages:Relational algebraTuple relational
calculusDomain relational calculusRelational operators
*
2.*
Selection of tuplesRelation rSelect tuples with A=B and D > 5σ
A=B and D > 5 (r)

*
2.*
Selection of Columns (Attributes)Relation r:
Select A and CProjectionΠ A, C (r)
*
2.*
Joining two relations – Cartesian ProductRelations r, s:r x s:
*
2.*
Uni

*
2.*
Set difference of two relationsRelations r, s:r – s:
*
2.*
Set Intersection of two relationsRelation r, s:
*
2.*

Joining two relations – Natural JoinLet r and s be relations on
schemas R and S respectively.
Then, the “natural join” of relations R and S is a relation on
tr from r and ts from s. If tr and ts have the same value on each
as
the same value as tr on rt has the same value as ts on s
*
2.*
Natural Join ExampleRelations r, s:
Natural Joinr s
*
2.*
Figure in-2.1
*

End of Chapter 2
*
2.*
Figure 2.01
*
2.*
Figure 2.02
*

2.*
Figure 2.03
*
2.*
Figure 2.04
*
2.*
Figure 2.05
*
2.*

Figure 2.06
*
2.*
Figure 2.07
*
2.*
Figure 2.10
*
2.*
Figure 2.11

*
2.*
Figure 2.12
*
2.*
Figure 2.13
*
Pg. 03
Question Two
Assignment #1
Deadline: Day 04/10/2019 @ 23:59

[Total Mark for this Assignment is 5]
Fundamentals of Databases
IT403
College of Computing and Informatics
Question One
1 Mark
Learning Outcome(s): LO1
Describe fundamental data and database concepts
1. Give an example of 3 relations illustrating primary keys and
foreign keys.
Question Two
4 Marks
Learning Outcome(s): LO7
Use Structured Query Language to perform queries and to
perform relational operations.

Consider the following relations:
Student(sID, surName, firstName, campus, email)
Course(dept, cNum, name)
Offering(oID, dept, cNum, term, instructor)
Took(sID, oID, grade)
Such as:
Offering[dept, cNum] ⊆ Course[dept, cNum]
Took[sID] ⊆ Student[sID]
Took[oID] ⊆ Offering[oID]
Answer the following query using relational algebra:
1. Give the Student number of all students who have taken the
course number “343” by the department “CS”.
2. Find sID of all students who have earned some grade over 80
and some grade below 50.
3. Find the Terms when the course number “369” by the
department “CS” was not offered.
4. Find the Department and course number of courses that have
never been offered.

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