3._Relational_Algebra.pptx:Basics of relation algebra

RELATIONAL ALGEBRA
 There are two types of operations in RDBMS
 Retrieval
 Update
 The set of operations for specifying retrieval
requests (or queries) in relational model is called
Relational Algebra.
 A sequence of relational algebra operations forms a
relational algebra expression.

COMPANY DATABASE
CONSIDERED IN EXAMPLES

SELECT OPERATION(UNARY OPERATION)
 This operation selects a subset of tuples from a relation
that satisfy a selection condition.
 Select is denoted by :  <selection condition>(R)

EXAMPLES : SELECT OPERATION
 Select the employees whose department number is 4:
 DNO = 4 (EMPLOYEE)
 Select all the projects in department 5
 Select the employees whose salary is greater than
$35,000

SELECT OPERATION
 Selection condition is a Boolean expression specified on the
attributes of relation R
 It can include boolean operators AND, OR, NOT applied on relational
operators <, > <=,>=, !=, =
 Select  is commutative:
 <condition1>( < condition2> (R)) =  <condition2> ( < condition1> (R))
 Cascade of Select operations
<cond1>(< cond2> (<cond3>(R)) =  <cond1> AND < cond2> AND < cond3>(R)))

PROJECT OPERATION (UNARY OPERATION)
 This operation selects a subset of columns from the existing
relation.
 Project operation is denoted by <attribute list>R
 It removes duplicate tuples, the result of project is set of tuples
 Example:
 RESULT  LNAME, FNAME, SALARY (EMPLOYEE)
 DN  DNAME, DNUMBER (DEPARTMENT)

PROJECT OPERATION
 Project operation is not commutative
  <list1> ( <list2> (R) ) =  <list1> (R) as long as <list2> contains the
attributes in <list1>
 No of Tuples in the result of projection π <list>(R)
 less or equal to the number of tuples in R
 If the list of attributes includes a key of R, then the no of is equal
to the no of tuples in R

RELATIONAL ALGEBRA EXPRESSIONS
 We may want to apply several relational algebra
operations one after the other
1. We can write the operations as a single
relational algebra expression by nesting the
operations, or
2. We can apply one operation at a time and
create intermediate result relations.

EXAMPLE: SEQUENCE OF OPERATIONS
 To retrieve the first name, last name, and salary of all
employees who work in Department 5
 Result of sequence of operations:
 FNAME, LNAME, SALARY( DNO=5(EMPLOYEE))
 Using intermediate relation:
 D5   DNO=5(EMPLOYEE)
 RESULT   FNAME, LNAME, SALARY (D5)
 Renaming of attributes
 D5   DNO=5(EMPLOYEE)
 R (FirstName,LastName,Salary)   FNAME, LNAME, SALARY (D5)

EXAMPLE OF APPLYING MULTIPLE OPERATIONS
AND RENAME
FNAME, LNAME, SALARY( DNO=5(EMPLOYEE))
D5   DNO=5(EMPLOYEE)
R (First_name,Last_name,Salary)   Fname, Lname, Salary (D5)

RENAME OPEARATION
 Rename operator is denoted by  (rho)
 Rename operation  can be expressed as:
 S(R) rename the relation to S
 (B1, B2, …, Bn )(R) rename the attributes to B1, B2, …..Bn
 S (B1, B2, …, Bn )(R) rename both relation to S, and attributes to
B1, B1, …..Bn
 Example:
  RESULT (First_Name,Last_Name, DNO)(D5)

UNION (BINARY OPERATION)
 The result of 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
 The two relations R and S must be “type compatible” (or
Union compatible)
 R and S must have same number of attributes
 Each pair of corresponding attributes must have same or
compatible domains

UNION 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
DEP5_EMPS  DNO=5 (EMPLOYEE)
RESULT1   SSN(DEP5_EMPS)
RESULT2(SSN)  SUPERSSN(DEP5_EMPS)
RESULT  RESULT1  RESULT2

16
INTERSECTION AND SET DIFFERENCE (BINARY
OPERATIONS)
 INTERSECTION operation: the result of R  S, is a
relation that includes all tuples that are in both R and S
 SET DIFFERENCE operation: the result of R – S, is a
relation that includes all tuples that are in R but not in S
 Two relations R and S must be “type compatible”

RELATIONAL ALGEBRA OPERATIONS FROM SET
THEORY
 Both  and  are commutative operations
 R  S = S  R, and R  S = S  R
 Both  and  can be treated as n-ary operations
 R  (S  T) = (R  S)  T
 (R  S)  T = R  (S  T)
 Minus operation is not commutative
 R – S ≠ S – R

EXAMPLE TO ILLUSTRATE THE RESULT OF
UNION, INTERSECT, AND DIFFERENCE

CARTESIAN PRODUCT
 The result of Cartesian product of two relations
R(A1, A2, . . ., An) x S(B1, B2, . . ., Bm) is given as:
Result(A1, A2, . . ., An, B1, B2, . . ., Bm)
 Let |R| = nR and |S| = nS , then |R x S|= nR * nS
 R and S may NOT be "type compatible”
 Cross Product is a meaningful operation only if it is
followed by other operations

Example (not meaningful):
F   SEX=’F’(EMPLOYEE)
EN   FNAME, LNAME, SSN (F)
E_DP  EN x DEPENDENT
Example (meaningful):
A_DP   SSN=ESSN(E_DP)
R   FNAME, LNAME, DEPENDENT_NAME(A_DP)
Problem:
Retrieve a list of each female
employee’s dependents

JOIN(BINARY OPERATION)
 JOIN denoted by combine related tuples from various
relations
 JOIN combines CARTESIAN PRODECT and SELECT
into a single operation
 General form of a join operation on two relations R(A1,
A2, . . ., An) and S(B1, B2, . . ., Bm) is:
R <join condition>S
21

EXAMPLE OF JOIN OPERATION
 Retrieve the name of the manager of each department.
DEPT_MGR  DEPARTMENT MGRSSN=SSN EMPLOYEE
 The join condition can also be specified as
DEPARTMENT.MGRSSN= EMPLOYEE.SSN
22

COMPLETE SET OF RELATIONAL
OPERATIONS
 The set of operations including
 SELECT ,
 PROJECT  ,
 UNION ,
 DIFFERENCE - ,
 RENAME , and
 CARTESIAN PRODUCT X
is called a complete set because any relational
algebra expression can be expressed using these.
 For example:
 R  S = (R  S ) – ((R - S)  (S - R))
 R <join condition>S =  <join condition> (R X S) 23

SOME PROPERTIES OF JOIN
 Consider the following JOIN operation:
 R(A1, A2, . . ., An) S(B1, B2, . . ., Bm)
R.Ai=S.Bj
 Result is a relation Q with degree n + m attributes:
 Q(A1, A2, . . ., An, B1, B2, . . ., Bm), in that order.
 Relation Q has one tuple for each combination of
tuples—r from R and s from S, but only if they satisfy
the join condition r[Ai]=s[Bj]
 If R has nR tuples, and S has nS tuples, then no of
tuples in join result < nR * nS . 24

THETA-JOIN
 The general case of JOIN operation is called a
Theta-join: R S
theta
 Theta is a boolean expression on the attributes of R
and S; for example:
 R.Ai<S.Bj AND (R.Ak=S.Bl OR R.Ap<S.Bq)
 Theta can have any comparison operators
{=,≠,<,≤,>,≥,}
25

EQUI-JOIN
 EQUIJOIN is a join condition that involves only equality
operator = .
 Example:
 DEPT_MGR  DEPARTMENT MGRSSN=SSN EMPLOYEE
 Retrieve a list of each female employee’s dependents
F   SEX=’F’(EMPLOYEE)
EN   FNAME, LNAME, SSN (F)
E_DP  EN DEPENDENT
SSN=ESSN
26

ISSUE WITH EQUIJOIN OPERATION
 Superfluous column
 Result of EQUIJOIN always have one or more pairs of
attributes that have identical values in every tuple.

NATURAL JOIN OPERATION
 NATURAL JOIN operation (denoted by *) is created to
get rid of the superfluous attribute in an EQUIJOIN
condition.
 The two join attributes, or each pair of corresponding join
attributes must have the same name in both relations
 If this is not the case, a renaming operation is applied first.
28

NATURAL JOIN OPERATION
 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
 Only attribute with the same name is DNUMBER
 An implicit join condition is created based on this attribute:
DEPARTMENT.DNUMBER=DEPT_LOCATIONS.DNUMBER
29

EXAMPLE: NATURAL JOIN
 Another example: Q  R(A,B,C,D) * S(C,D,E)
 The implicit join condition includes each pair of attributes with the
same name, “AND” together:
 R.C=S.C AND R.D=.S.D
 Result keeps only one attribute of each such pair:
 Q(A,B,C,D,E)
30

EXAMPLE OF NATURAL JOIN OPERATION
31

DIVISION (BINARY OPERATION)
 The division operation is applied to two relations
R(Z)  S(X), where X  Z.
 Let Y = Z – X
 We have Z = X  Y and Y is a set of attributes of R that are
not the attributes of S.
32
 The result of DIVISION is a relation T(Y)
 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.

EXAMPLE OF DIVISION
33
 Smith   fname=‘John’ and lname=‘Smith’ (Employee)
 Smith_Pnos   Pno (Works_on essn=ssn Smith)
 Ssn_Pnos   Essn,Pno (Works_on)
 SSNS(ssn) Ssn_Pnos  Smith_Pnos
Retrieve all employees who work on all the project
that John Smith works on

RECAP OF RELATIONAL ALGEBRA
OPERATIONS
34

AGGREGATE FUNCTIONS
 Now we specify mathematical aggregate functions on
collections of values from the database.
 Examples:
 Retrieve the average or total salary of all employees
 Retrieve total number of employee tuples
 Functions applied to collections of numeric values
include
 SUM, AVERAGE, MAXIMUM, and MINIMUM.
 COUNT function is used for counting tuples or values.
35

AGGREGATE FUNCTION OPERATION
 Use of the Aggregate Functional operation ℱ
 ℱMAX Salary (EMPLOYEE)
 ℱMIN Salary (EMPLOYEE)
 ℱSUM Salary (EMPLOYEE)
 ℱCOUNT SSN, AVERAGE Salary (EMPLOYEE)
 computes no of employees and their average salary
 Note: count just counts the number of rows, without
removing duplicates
36

USING GROUPING WITH AGGREGATION
 Grouping can be combined with Aggregate Functions
 Example:
 For each department, retrieve the DNO, COUNT SSN, and
AVERAGE SALARY
 DNO ℱCOUNT SSN, AVERAGE Salary (EMPLOYEE)
37

EXAMPLE: AGGREGATE FUNCTIONS AND GROUPING
38

EXAMPLES OF QUERIES IN RELATIONAL
ALGEBRA
39
 Q1: Retrieve the name and address of all employees who work for the
‘Research’ department.
RESEARCH_DEPT   DNAME=’Research’ (DEPARTMENT)
RESEARCH_EMPS  (RESEARCH_DEPT DNUMBER= DNO EMPLOYEE)
RESULT   FNAME, LNAME, ADDRESS (RESEARCH_EMPS)

ALGEBRA
40
 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)

ALGEBRA
41
 Q5: Retrieve the names of all employees with two or more dependents.
T1(Ssn, No_of_dependents)  Essn ℱ COUNT Dependent_name (DEPENDENT)
T2   No_of_dependents >1(T1)
RESULT   LNAME, FNAME (T2 * EMPLOYEE)

ASSIGNMENT 2
 Section B: Due date 13-Feb 2013
 Section C: Due date 16-Feb 2013

OUTER JOIN OPERATION
 In INNER JOIN, tuples without a matching are
eliminated from the join result
 Tuples with null are also eliminated
 This amounts to loss of information.
 OUTER joins operations are used when we want
to keep
 all the tuples in R in the join result , or
 all tuples in S in the join result, or
 all tuples in both relations R and S in the join result
43

LEFT OUTER JOIN
 List the employees name and the department name that they
manage. If they don’t manage one, then indicate this with a
null value.
 Temp  (Employee Ssn=Mgr_Ssn Department)
 Result   Fname, Minit, Lname, Dname(Temp)
44

OUTER JOIN OPERATION
 Left outer join: keeps every tuple in R, denoted as
R S
 if no matching tuple is found in S, then the attributes of
S in the join result are filled with null values.
 Right outer join: keeps every tuple in S in the
result of R S.
 Full outer join: keeps all tuples in both the left and
the right relations. It is denoted by
45

FULL OUTER JOIN VS CARTESIAN PRODUCT
Employee Ssn=Mgr_Ssn Department
??

OUTER UNION OPERATIONS
 The outer union operation take the union of tuples
in two relations R(X, Y) and S(X, Z) that are
partially compatible,
 Only some of their attributes, say X, are type
compatible.
 The attributes that are type compatible are represented
only once in the result
 The attributes that are not type compatible from either
relation are also kept in the result relation T(X, Y, Z).
47

OUTER JOIN EXAMPLE
 An outer union can be applied to two relations STUDENT(Name,
SSN, Department, Advisor) and INSTRUCTOR(Name, SSN,
Department, Rank).
 Tuples 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.
 Result relation:
STUDENT_OR_INSTRUCTOR (Name, SSN, Department,
Advisor, Rank)
48

RECURSIVE CLOSURE OPERATION
 This can’t be specified in general using Relational Algebra
 Example: Retrieve all SUPERVISEES of an EMPLOYEE e at all
levels — that is,
 all employees 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.
 We can retrieve employees at each level and then take their union,
however, we cannot 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.
49

RECURSIVE CLOSURE OPERATION
50

51
Example of Query Tree
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.

QUERY TREE
 An internal data structure to represent a query
 Standard technique to estimate the work done in executing
the query, and the optimization of execution
 Nodes stand for operations like selection, projection, join,
renaming, division, ….
 Leaf nodes represent base relations
 A tree gives a good visual feel of the complexity of the query
and the operations involved
 Algebraic Query Optimization consists of rewriting the query
or modifying the query tree into an equivalent tree.
52

RELATIONAL ALGEBRA OPERATORS
 Relational Algebra consists of several groups of operations
 Unary Relational Operations
 SELECT (symbol:  (sigma))
 PROJECT (symbol:  (pi))
 RENAME (symbol:  (rho))
 Relational Algebra Operations From Set Theory
 UNION (  ), INTERSECTION (  ), DIFFERENCE (–)
 CARTESIAN PRODUCT ( x )
 Binary Relational Operations
 JOIN (several variations of JOIN exist)
 DIVISION
 Additional Relational Operations
 OUTER JOINS, OUTER UNION
 AGGREGATE FUNCTIONS (These compute summary of
information: for example, SUM, COUNT, AVG, MIN, MAX)

CHAPTER SUMMARY
 Relational Algebra
 Unary Relational Operations
 Relational Algebra Operations From Set Theory
 Binary Relational Operations
 Additional Relational Operations
 Examples of Queries in Relational Algebra
54

3._Relational_Algebra.pptx:Basics of relation algebra

More Related Content

Similar to 3._Relational_Algebra.pptx:Basics of relation algebra (20)

More from ZakriyaMalik2 (8)

Recently uploaded (20)

3._Relational_Algebra.pptx:Basics of relation algebra