DBMS Module 2.2.pdf......................

Module 2.2
The Relational Algebra and Calculus

Chapter Outline
 Example Database Application (COMPANY)
 Relational Algebra
– Unary Relational Operations
– Relational Algebra Operations From Set Theory
– Binary Relational Operations
– Additional Relational Operations
– Examples of Queries in Relational Algebra

Database State for COMPANY
All examples discussed below refer to the COMPANY database shown here.

Relational Algebra
 The basic set of operations for the relational model is known
as the relational algebra. These operations enable a user to
specify basic retrieval requests.
 The result of a retrieval is a new relation, which may have
been formed from one or more relations. The algebra
operations thus produce new relations, which can be further
manipulated using operations of the same algebra.
 A sequence of relational algebra operations forms a
relational algebra expression, whose result will also be a
relation that represents the result of a database query (or
retrieval request).

Unary Relational Operations
 SELECT Operation
SELECT operation is used to select a subset of the tuples from a relation that
satisfy a selection condition. It is a filter that keeps only those tuples that
satisfy a qualifying condition – those satisfying the condition are selected
while others are discarded.
Example: To select the EMPLOYEE tuples whose department number is
four or those whose salary is greater than $30,000 the following notation is
used:
DNO = 4 (EMPLOYEE)
SALARY > 30,000 (EMPLOYEE)
In general, the select operation is denoted by  <selection condition>(R) where the
symbol  (sigma) is used to denote the select operator, and the selection
condition is a Boolean expression specified on the attributes of relation R

Unary Relational Operations
SELECT Operation Properties
– The SELECT operation  <selection condition>(R) produces a relation S that
has the same schema as R
– The SELECT operation  is commutative; i.e.,
 <condition1>( < condition2> ( R)) =  <condition2> ( < condition1> ( R))
– A cascaded SELECT operation may be applied in any order; i.e.,
 <condition1>( < condition2> ( <condition3> ( R))
=  <condition2> ( < condition3> ( < condition1> ( R)))
– A cascaded SELECT operation may be replaced by a single selection
with a conjunction of all the conditions; i.e.,
 <condition1>( < condition2> ( <condition3> ( R))
=  <condition1> AND < condition2> AND < condition3> ( R)))

Unary Relational Operations (cont.)

 PROJECT Operation
This operation selects certain columns from the table and discards the other
columns. The PROJECT creates a vertical partitioning – one with the needed
columns (attributes) containing results of the operation and other containing
the discarded Columns.
Example: To list each employee’s first and last name and salary, the
following is used:
LNAME, FNAME,SALARY(EMPLOYEE)
The general form of the project operation is <attribute list>(R) where 
(pi) is the symbol used to represent the project operation and <attribute list>
is the desired list of attributes from the attributes of relation R.
The project operation removes any duplicate tuples, so the result of the
project operation is a set of tuples and hence a valid relation.

PROJECT Operation Properties
– The number of tuples in the result of projection  <list> (R)is always
less or equal to the number of tuples in R.
– If the list of attributes includes a key of R, then the number of tuples is
equal to the number of tuples in R.
–  <list1> ( <list2> (R) ) =  <list1> (R) as long as <list2> contains
the attributes in <list2>

 Rename Operation
We may want to apply several relational algebra operations one after the other.
Either we can write the operations as a single relational algebra expression
by nesting the operations, or we can apply one operation at a time and create
intermediate result relations. In the latter case, we must give names to the
relations that hold the intermediate results.
Example: To retrieve the first name, last name, and salary of all employees
who work in department number 5, we must apply a select and a project
operation. We can write a single relational algebra expression as follows:
FNAME, LNAME, SALARY( DNO=5(EMPLOYEE))
OR We can explicitly show the sequence of operations, giving a name to each
intermediate relation:
DEP5_EMPS   DNO=5(EMPLOYEE)
RESULT   FNAME, LNAME, SALARY (DEP5_EMPS)

 Rename Operation (cont.)
The rename operator is 
The general Rename operation can be expressed by any of the
following forms:
  S (B1, B2, …, Bn ) ( R) is a renamed relation S based on R with column names B1, B1,
…..Bn.
  S ( R) is a renamed relation S based on R (which does not specify column names).
  (B1, B2, …, Bn ) ( R) is a renamed relation with column names B1, B1, …..Bn which
does not specify a new relation name.

Relational Algebra Operations From
Set Theory
 UNION Operation
The result of this operation, denoted by R  S, is a relation that includes all
tuples that are either in R or in S or in both R and S. Duplicate tuples are
eliminated.
Example: To retrieve the social security numbers of all employees who either
work in department 5 or directly supervise an employee who works in
department 5, we can use the union operation as follows:
DEP5_EMPS  DNO=5 (EMPLOYEE)
RESULT1   SSN(DEP5_EMPS)
RESULT2(SSN)   SUPERSSN(DEP5_EMPS)
RESULT  RESULT1  RESULT2
The union operation produces the tuples that are in either RESULT1 or
RESULT2 or both. The two operands must be “type compatible”.

Set Theory
 Type Compatibility
– The operand relations R1(A1, A2, ..., An) and R2(B1, B2, ..., Bn)
must have the same number of attributes, and the domains of
corresponding attributes must be compatible; that is,
dom(Ai)=dom(Bi) for i=1, 2, ..., n.
– The resulting relation for R1R2,R1  R2, or R1-R2 has the
same attribute names as the first operand relation R1 (by
convention).

Set Theory
 UNION Example
STUDENTINSTRUCTOR

Relational Algebra Operations From Set
Theory (cont.) – use Fig. 6.4

Theory (cont.)
 INTERSECTION OPERATION
The result of this operation, denoted by R  S, is a relation that includes all
tuples that are in both R and S. The two operands must be "type compatible"
Example: The result of the intersection operation (figure below) includes only
those who are both students and instructors.
STUDENT  INSTRUCTOR

Theory (cont.)
 Set Difference (or MINUS) Operation
The result of this operation, denoted by R - S, is a relation that includes all
tuples that are in R but not in S. The two operands must be "type compatible”.
Example: The figure shows the names of students who are not instructors, and
the names of instructors who are not students.
STUDENT-INSTRUCTOR
INSTRUCTOR-STUDENT

Theory (cont.)
 Notice that both union and intersection are commutative
operations; that is
R  S = S  R, and R  S = S  R
 Both union and intersection can be treated as n-ary operations
applicable to any number of relations as both are associative
operations; that is
R  (S  T) = (R  S)  T, and (R  S)  T = R  (S  T)
 The minus operation is not commutative; that is, in general
R - S ≠ S – R

Theory (cont.)
 CARTESIAN (or cross product) Operation
– This operation is used to combine tuples from two relations in a
combinatorial fashion. In general, the result of R(A1, A2, . . ., An) x S(B1,
B2, . . ., Bm) is a relation Q with degree n + m attributes Q(A1, A2, . . ., An,
B1, B2, . . ., Bm), in that order. The resulting relation Q has one tuple for
each combination of tuples—one from R and one from S.
– Hence, if R has nR tuples (denoted as |R| = nR ), and S has nS tuples, then
| R x S | will have nR * nS tuples.
– The two operands do NOT have to be "type compatible”
Example:
FEMALE_EMPS   SEX=’F’(EMPLOYEE)
EMPNAMES   FNAME, LNAME, SSN (FEMALE_EMPS)
EMP_DEPENDENTS  EMPNAMES x DEPENDENT

Theory (cont.)

Binary Relational Operations
 JOIN Operation
– The sequence of cartesian product followed by select is used
quite commonly to identify and select related tuples from two
relations, a special operation, called JOIN. It is denoted by a
– This operation is very important for any relational database
with more than a single relation, because it allows us to process
relationships among relations.
– The general form of a join operation on two relations R(A1, A2,
. . ., An) and S(B1, B2, . . ., Bm) is:
R <join condition>S
where R and S can be any relations that result from general
relational algebra expressions.

Binary Relational Operations (cont.)
Example: Suppose that we want to retrieve the name of
the manager of each department. To get the manager’s
name, we need to combine each DEPARTMENT tuple
with the EMPLOYEE tuple whose SSN value matches
the MGRSSN value in the department tuple. We do this
by using the join operation.
DEPT_MGR  DEPARTMENT MGRSSN=SSN
EMPLOYEE

 EQUIJOIN Operation
The most common use of join involves join conditions with equality comparisons only.
Such a join, where the only comparison operator used is =, is called an EQUIJOIN. In
the result of an EQUIJOIN we always have one or more pairs of attributes (whose
names need not be identical) that have identical values in every tuple.
The JOIN seen in the previous example was EQUIJOIN.
 NATURAL JOIN Operation
Because one of each pair of attributes with identical values is superfluous, a new
operation called natural join—denoted by *—was created to get rid of the second
(superfluous) attribute in an EQUIJOIN condition.
The standard definition of natural join requires that the two join attributes, or each pair
of corresponding join attributes, have the same name in both relations. If this is not the
case, a renaming operation is applied first.

Example: To apply a natural join on the DNUMBER attributes of
DEPARTMENT and DEPT_LOCATIONS, it is sufficient to write:
DEPT_LOCS  DEPARTMENT * DEPT_LOCATIONS

Complete Set of Relational Operations
The set of operations including select ,
project  , union , set difference - , and
cartesian product X is called a complete set
because any other relational algebra expression
can be expressed by a combination of these five
operations.
For example:
R  S = (R  S ) – ((R  S)  (S  R))
R <join condition>S =  <join condition> (R X S)

 DIVISION Operation
– The division operation is applied to two relations
R(Z)  S(X), where X subset Z.
– Let Y = Z - X (and hence Z = X  Y); that is, let Y be the
set of attributes of R that are not attributes of S.
– The result of DIVISION is a relation T(Y) that includes a
tuple t if tuples tR appear in R with tR [Y] = t, and with
tR [X] = ts for every tuple ts in S.
– For a tuple t to appear in the result T of the DIVISION, the
values in t must appear in R in combination with every tuple
in S.

Notation for Query Trees
 The notation is called a query tree or sometimes it is known as a query
evaluation tree or query execution tree.
 It includes the relational algebra operations being executed and is used
as a possible data structure for the internal representation of the query in
an RDBMS.
 A query tree is a tree data structure that corresponds to a relational
algebra expression.
 It represents the input relations of the query as leaf nodes of the tree,
and represents the relational algebra operations as internal nodes.
 An execution of the query tree consists of executing an internal node
operation whenever its operands (represented by its child nodes) are
available, and then replacing that internal node by the relation that
results from executing the operation.
 The execution terminates when the root node is executed and produces
the result relation for the query.

 Figure 6.9 shows a query tree for a Query.
 “ For every project located in ‘Stafford’, list the project
number, the controlling department number, and the
department manager’s last name, address, and birth
date.”
 This query corresponds to the following relational
algebra expression:
Chapter 6-33
Notation for Query Trees
(Continued..)

Recap of Relational Algebra Operations

Additional Relational Operations
 Aggregate Functions and Grouping
– A type of request that cannot be expressed in the basic relational algebra
is to specify mathematical aggregate functions on collections of values
from the database.
– Examples of such functions include retrieving the average or total salary
of all employees or the total number of employee tuples. These functions
are used in simple statistical queries that summarize information from
the database tuples.
– Common functions applied to collections of numeric values include
SUM, AVERAGE, MAXIMUM, and MINIMUM. The COUNT
function is used for counting tuples or values.

Additional Relational Operations (cont.)

Functional operator ℱ
ℱMAX Salary (Employee) retrieves the maximum salary value from the
Employee relation
ℱMIN Salary (Employee) retrieves the minimum Salary value from the
Employee relation
ℱSUM Salary (Employee) retrieves the sum of the Salary from the Employee
relation
DNO ℱCOUNT SSN, AVERAGE Salary (Employee) groups employees by DNO
(department number) and computes the count of employees and average
salary per department.
[ Note: count just counts the number of rows, without removing duplicates]

 Recursive Closure Operations
– Another type of operation that, in general, cannot be specified in the
basic original relational algebra is recursive closure. This operation is
applied to a recursive relationship.
– An example of a recursive operation is to retrieve all SUPERVISEES of
an EMPLOYEE e at all levels—that is, all EMPLOYEE e’ directly
supervised by e; all employees e’’ directly supervised by each employee
e’; all employees e’’’ directly supervised by each employee e’’; and so
on .
– Although it is possible to retrieve employees at each level and then take
their union, we cannot, in general, specify a query such as “retrieve the
supervisees of ‘James Borg’ at all levels” without utilizing a looping
mechanism.
– The SQL3 standard includes syntax for recursive closure.

 The OUTER JOIN Operation
– In NATURAL JOIN tuples without a matching (or related) tuple are eliminated
from the join result. Tuples with null in the join attributes are also eliminated.
This amounts to loss of information.
– A set of operations, called outer joins, can be used when we want to keep all the
tuples in R, or all those in S, or all those in both relations in the result of the
join, regardless of whether or not they have matching tuples in the other relation.
– The left outer join operation keeps every tuple in the first or left relation R in
R S; if no matching tuple is found in S, then the attributes of S in the join
result are filled or “padded” with null values.
– A similar operation, right outer join, keeps every tuple in the second or right
relation S in the result of R S.
– A third operation, full outer join, denoted by keeps all tuples in both the
left and the right relations when no matching tuples are found, padding them
with null values as needed.

Left Outer Join
Retrieve the list of all employee names as well as the name of
the departments they manage if they happen to manage a
department; if they do not manage one, indicate it with NULL
value.

 OUTER UNION Operations
– The outer union operation was developed to take the union of tuples from two
relations if the relations are not union compatible.
– This operation will take the union of tuples in two relations R(X, Y) and S(X, Z)
that are partially compatible, meaning that only some of their attributes, say X,
are union compatible.
– The attributes that are union compatible are represented only once in the result,
and those attributes that are not union compatible from either relation are also
kept in the result relation T(X, Y, Z).
– Example: An outer union can be applied to two relations whose schemas are
STUDENT(Name, SSN, Department, Advisor) and INSTRUCTOR(Name, SSN,
Department, Rank).
– Tuples from the two relations are matched based on having the same combination
of values of the shared attributes—Name, SSN, Department. If a student is also an
instructor, both Advisor and Rank will have a value; otherwise, one of these two
attributes will be null.
The result relation STUDENT_OR_INSTRUCTOR will have the following
attributes:
STUDENT_OR_INSTRUCTOR (Name, SSN, Department, Advisor, Rank)

OUTER UNION Operations
 STUDENT (Name, SSN, Department, Advisor)
 INSTRUCTOR (Name, SSN, Department, Rank)
After applying OUTER UNION operation on STUDENT table and
INSTRUCTOR table,
STUDENT_OR_INSTRUCTOR (Name, SSN, Department, Advisor,
Rank)

Query 2. For every project located in ‘Stafford’, list the project
number, the controlling department number, and the department
manager’s last name, address, and birth date.
Chapter 6-54
• In this example, we first select the projects located in Stafford, then
join them with their controlling departments, and then join the result
with the department managers.
• Finally, we apply a project operation on the desired attributes.

Query 3. Find the names of employees who work on all
the projects controlled by department number 5.
• In this query, we first create a table DEPT5_PROJS that contains the project
numbers of all projects controlled by department 5.
• Then we create a table EMP_PROJ that holds (Ssn, Pno) tuples, and apply the
division operation.
• Notice that we renamed the attributes so that they will be correctly used in the
division operation.
• Finally, we join the result of the division, which holds only Ssn values, with the
EMPLOYEE table to retrieve the Fname, Lname attributes from EMPLOYEE.

Query 4. Make a list of project numbers for projects that involve an
employee whose last name is ‘Smith’, either as a worker or as a
manager of the department that controls the project.
• In this query, we retrieved the project numbers for projects that involve an employee
named Smith as a worker in SMITH_WORKER_PROJS.
• Then we retrieved the project numbers for projects that involve an employee named Smith
as manager of the department that controls the project in SMITH_MGR_PROJS.
• Finally, we applied the UNION operation on SMITH_WORKER_PROJS and
SMITH_MGR_PROJS.
• As a single in-line expression, this query becomes: (Method 2)

Query 5. List the names of all employees with
two or more dependents.
Chapter 6-57
• This query cannot be done in the basic (original) relational algebra. We have
to use the AGGREGATE FUNCTION operation with the COUNT aggregate
function.
• We assume that dependents of the same employee have distinct
Dependent_name values.

Examples of Queries in Relational Algebra
 Q6: Retrieve the names of employees who have no
dependents.
ALL_EMPS   SSN(EMPLOYEE)
EMPS_WITH_DEPS(SSN)   ESSN(DEPENDENT)
EMPS_WITHOUT_DEPS  (ALL_EMPS - EMPS_WITH_DEPS)
RESULT   LNAME, FNAME (EMPS_WITHOUT_DEPS * EMPLOYEE)

DBMS Module 2.2.pdf......................

More Related Content

Similar to DBMS Module 2.2.pdf......................

Recently uploaded

DBMS Module 2.2.pdf......................