Relational Algebra-Relational Calculus-SQL

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Database Systems

Session 5 – Main Theme

Relational Algebra, Relational Calculus, and SQL

Dr. Jean-Claude Franchitti

New York University

Computer Science Department

Courant Institute of Mathematical Sciences

Presentation material partially based on textbook slides

Fundamentals of Database Systems (7th Edition)

by Ramez Elmasri and Shamkant Navathe

Slides copyright © 2016 and on slides produced by Zvi

Kedem copyight © 2014

http://www.cs.nyu.edu/kedem/DataBaseClass/

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Agenda

1 Session Overview

5 Summary and Conclusion

2 Relational Algebra and Relational Calculus

3 Relational Algebra Using SQL Syntax

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Session Agenda

Session Overview

Relational Algebra and Relational Calculus

Relational Algebra Using SQL Syntax

Summary & Conclusion

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What is the class about?

Course description and syllabus:

» http://www.nyu.edu/classes/jcf/CSCI-GA.2433-001

» http://cs.nyu.edu/courses/spring16/CSCI-GA.2433-001/

Textbooks: » Fundamentals of Database Systems (7th Edition)

Ramez Elmasri and Shamkant Navathe

Addition Wesley

ISBN-10: 0133970779, ISBN-13: 978-0133970777 6th Edition (06/08/15)

http://www.nyu.edu/classes/jcf/CSCI-GA.2433-001













http://cs.nyu.edu/courses/spring16/CSCI-GA.2433-001/

















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Icons / Metaphors

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Common Realization

Information

Knowledge/Competency Pattern

Governance

Alignment

Solution Approach

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Agenda

1 Session Overview




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Agenda

Relational Algebra » Unary Relational Operations

» Relational Algebra Operations From Set Theory

» Binary Relational Operations

» Additional Relational Operations

» Examples of Queries in Relational Algebra

Relational Calculus » Tuple Relational Calculus

» Domain Relational Calculus

Example Database Application (COMPANY)

Overview of the QBE language (appendix D)

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The Relational Algebra and Relational Calculus

Relational algebra

Basic set of operations for the relational model

Relational algebra expression

Sequence of relational algebra operations

Relational calculus

Higher-level declarative language for specifying

relational queries

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Relational Algebra Overview

Relational algebra is the basic set of

operations for the relational model

These operations enable a user to specify basic

retrieval requests (or queries)

The result of an operation is a new relation,

which may have been formed from one or

more input relations

» This property makes the algebra “closed” (all

objects in relational algebra are relations)

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Relational Algebra Overview (continued)

The algebra operations thus produce new

relations

» These can be further manipulated using

operations of the same algebra

A sequence of relational algebra operations

forms a relational algebra expression » The result of a relational algebra expression is also a

relation that represents the result of a database

query (or retrieval request)

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Brief History of Origins of Algebra

Muhammad ibn Musa al-Khwarizmi (800-847 CE) – from Morocco wrote a book titled al-jabr about arithmetic of variables » Book was translated into Latin.

» Its title (al-jabr) gave Algebra its name.

Al-Khwarizmi called variables “shay” » “Shay” is Arabic for “thing”.

» Spanish transliterated “shay” as “xay” (“x” was “sh” in Spain).

» In time this word was abbreviated as x.

Where does the word Algorithm come from? » Algorithm originates from “al-Khwarizmi"

» Reference: PBS (http://www.pbs.org/empires/islam/innoalgebra.html)

http://www.pbs.org/empires/islam/innoalgebra.html

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Relational Algebra Overview

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 (or

MINUS, – )

• 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)

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Database State for Company

All examples discussed below refer to the COMPANY

database shown here.

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Unary Relational Operations: SELECT (1/3)

The SELECT operation (denoted by (sigma)) is used to select a

subset of the tuples from a relation based on a selection condition.

» The selection condition acts as a filter

» Keeps only those tuples that satisfy the qualifying condition

» Tuples satisfying the condition are selected whereas the other tuples are discarded (filtered out)

Examples:

» Select the EMPLOYEE tuples whose department number is 4:

DNO = 4 (EMPLOYEE)

» Select the employee tuples whose salary is greater than $30,000:

SALARY > 30,000 (EMPLOYEE)

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» In general, the select operation is denoted by

<selection condition>(R) where

• the symbol (sigma) is used to denote the select

operator

• the selection condition is a Boolean (conditional)

expression specified on the attributes of relation R

• tuples that make the condition true are selected

– appear in the result of the operation

• tuples that make the condition false are filtered out

– discarded from the result of the operation

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SELECT Operation Properties

» The SELECT operation <selection condition>(R) produces a relation S

that has the same schema (same attributes) as R

» SELECT is commutative:

<condition1>( < condition2> (R)) = <condition2> ( < condition1> (R))

» Because of commutativity property, a cascade (sequence) of SELECT operations may be applied in any order:

<cond1>(<cond2> (<cond3> (R)) = <cond2> (<cond3> (<cond1> ( R)))

» A cascade of SELECT operations may be replaced by a single selection with a conjunction of all the conditions:

<cond1>(< cond2> (<cond3>(R)) = <cond1> AND < cond2> AND <

cond3>(R)))

» The number of tuples in the result of a SELECT is less than (or equal to) the number of tuples in the input relation R

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Unary Relational Operations: SELECT and PROJECT (2/3)

Example:

<selection condition> applied independently

to each individual tuple t in R

If condition evaluates to TRUE, tuple selected

Boolean conditions AND, OR, and NOT

Unary

Applied to a single relation

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Unary Relational Operations: SELECT and PROJECT (3/3)

Selectivity

Fraction of tuples selected by a selection

condition

SELECT operation commutative

Cascade SELECT operations into a single

operation with AND condition

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The Following Query Results Refer to this Database State

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Unary Relational Operations: PROJECT (1/3)

PROJECT Operation is denoted by (pi)

This operation keeps certain columns (attributes) from a relation and discards the other columns.

» PROJECT creates a vertical partitioning • The list of specified columns (attributes) is kept in

each tuple

• The other attributes in each tuple are discarded

Example: To list each employee’s first and last name and salary, the following is used:

LNAME, FNAME,SALARY(EMPLOYEE)

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The general form of the project operation is:

<attribute list>(R) » (pi) is the symbol used to represent the

project operation

» <attribute list> is the desired list of attributes from relation R.

The project operation removes any duplicate tuples » This is because the result of the project

operation must be a set of tuples • Mathematical sets do not allow duplicate elements.

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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 in the result of PROJECT is equal

to the number of tuples in R

» PROJECT is not commutative

<list1> ( <list2> (R) ) = <list1> (R) as long as <list2>

contains the attributes in <list1>

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Examples of Applying SELECT and PROJECT Operations

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Relational Algebra Expressions

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.

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Single Expression vs. Sequence of Relational Operations (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)

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Unary Relational Operations: RENAME (1/3)

The RENAME operator is denoted by

(rho)

In some cases, we may want to rename the

attributes of a relation or the relation name

or both

» Useful when a query requires multiple

operations

» Necessary in some cases (see JOIN operation

later)

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The general RENAME operation can be

expressed by any of the following forms:

» S (B1, B2, …, Bn )(R) changes both:

• the relation name to S, and

• the column (attribute) names to B1, B1, …..Bn

» S(R) changes:

• the relation name only to S

» (B1, B2, …, Bn )(R) changes:

• the column (attribute) names only to B1, B1, …..Bn

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For convenience, we also use a shorthand

for renaming attributes in an intermediate

relation:

» If we write: • RESULT FNAME, LNAME, SALARY (DEP5_EMPS)

• RESULT will have the same attribute names as

DEP5_EMPS (same attributes as EMPLOYEE)

• If we write: • RESULT (F, M, L, S, B, A, SX, SAL, SU, DNO)

RESULT (F.M.L.S.B,A,SX,SAL,SU, DNO)(DEP5_EMPS)

• The 10 attributes of DEP5_EMPS are renamed to

F, M, L, S, B, A, SX, SAL, SU, DNO, respectively

Note: the symbol is an assignment operator

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Example of Applying Multiple Operations and RENAME

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Relational Algebra Operations from Set Theory: UNION (1/2)

UNION Operation

» Binary operation, denoted by

» 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 operand 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 be type compatible (have same or compatible domains)

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Relational Algebra Operations from Set Theory: UNION (2/2)

Example: » To retrieve the social security numbers of all employees who

either work in department 5 (RESULT1 below) or directly supervise an employee who works in department 5 (RESULT2 below)

» 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

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Result of the UNION Operation RESULT ← RESULT1 ∪ RESULT2

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Relational Algebra Operations from Set Theory

Type Compatibility of operands is required for the binary

set operation UNION , (also for INTERSECTION , and

SET DIFFERENCE –, see next slides)

R1(A1, A2, ..., An) and R2(B1, B2, ..., Bn) are type

compatible if:

» they have the same number of attributes, and

» the domains of corresponding attributes are type compatible

(i.e. dom(Ai)=dom(Bi) for i=1, 2, ..., n).

The resulting relation for R1R2 (also for R1R2, or R1–

R2, see next slides) has the same attribute names as the

first operand relation R1 (by convention)

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Relational Algebra Operations from Set Theory: INTERSECTION

INTERSECTION is denoted by

The result of the operation R S, is a

relation that includes all tuples that are in

both R and S

»The attribute names in the result will be the

same as the attribute names in R

The two operand relations R and S must be

“type compatible”

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Relational Algebra Operations from Set Theory: SET DIFFERENCE

SET DIFFERENCE (also called MINUS or

EXCEPT) is denoted by –

The result of R – S, is a relation that

includes all tuples that are in R but not in S

»The attribute names in the result will be the

same as the attribute names in R

The two operand relations R and S must be

“type compatible”

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Example to Illustrate the Result of UNION/INTERSECTION/DIFFERENCE

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Some Properties of UNION/INTERSECT/DIFFERENCE

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

» (R S) T = R (S T)

The minus operation is not commutative; that is, in

general

» R – S ≠ S – R

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Relational Algebra Operations from Set Theory: CARTESIAN PRODUCT

CARTESIAN (or CROSS) PRODUCT Operation

» This operation is used to combine tuples from two relations

in a combinatorial fashion.

» Denoted by R(A1, A2, . . ., An) x S(B1, B2, . . ., Bm)

» Result is a relation Q with degree n + m attributes:

• Q(A1, A2, . . ., An, B1, B2, . . ., Bm), in that order.

» The resulting relation state 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”

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Generally, CROSS PRODUCT is not a meaningful operation

» Can become meaningful when followed by other operations

Example (not meaningful): » FEMALE_EMPS SEX=’F’(EMPLOYEE)

» EMPNAMES FNAME, LNAME, SSN (FEMALE_EMPS)

» EMP_DEPENDENTS EMPNAMES x DEPENDENT

EMP_DEPENDENTS will contain every combination of EMPNAMES and DEPENDENT

» whether or not they are actually related

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To keep only combinations where the DEPENDENT is related to the EMPLOYEE, we add a SELECT operation as follows

Example (meaningful): » FEMALE_EMPS SEX=’F’(EMPLOYEE)

» EMPNAMES FNAME, LNAME, SSN (FEMALE_EMPS)

» EMP_DEPENDENTS EMPNAMES x DEPENDENT

» ACTUAL_DEPS SSN=ESSN(EMP_DEPENDENTS)

» RESULT FNAME, LNAME, DEPENDENT_NAME (ACTUAL_DEPS)

RESULT will now contain the name of female employees and their dependents

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The CARTESIAN PRODUCT (CROSS PRODUCT) Operation (1/3)

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Binary Relational Operations: JOIN (1/2)

JOIN Operation (denoted by ) » The sequence of CARTESIAN PRODECT followed by

SELECT is used quite commonly to identify and select related tuples from two relations

» A special operation, called JOIN combines this sequence into a single operation

» This operation is very important for any relational database with more than a single relation, because it allows us combine related tuples from various 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.

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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

MGRSSN=SSN is the join condition » Combines each department record with the employee who

manages the department

» The join condition can also be specified as DEPARTMENT.MGRSSN= EMPLOYEE.SSN

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Result of the JOIN Operation DEPT_MGR ← DEPARTMENT|X| Mgr_ssn=SsnEMPLOYEE

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Some Properties of JOIN (1/2)

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.

» The resulting relation state 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]

» Hence, if R has nR tuples, and S has nS tuples, then the join

result will generally have less than nR * nS tuples.

» Only related tuples (based on the join condition) will appear

in the result

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The general case of JOIN operation is called a Theta-join: R S

theta

The join condition is called theta

Theta can be any general 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)

Most join conditions involve one or more equality conditions “AND”ed together; for example:

» R.Ai=S.Bj AND R.Ak=S.Bl AND R.Ap=S.Bq

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Binary Relational Operations: EQUIJOIN

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 an EQUIJOIN.

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Binary Relational Operations: NATURAL JOIN (1/2)

NATURAL JOIN Operation

» Another variation of JOIN called NATURAL JOIN —

denoted by * — was created to get rid of the second

(superfluous) attribute in an EQUIJOIN condition.

• because one of each pair of attributes with identical values is

superfluous

» 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.

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Binary Relational Operations: NATURAL JOIN (2/2)

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

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”ed 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)

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Example of NATURAL JOIN Operation

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A Complete Set of Relational Algebra Operations

The set of operations including SELECT ,

PROJECT , UNION , DIFFERENCE - ,

RENAME , 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)

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Binary Relational Operations: DIVISION

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.

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Example of DIVISION

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Operations of Relational Algebra (1/2)

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Operations of Relational Algebra (2/2)

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Query Tree Notation

Query Tree

» An internal data structure to represent a query

» Standard technique for estimating the work involved in executing the query, the generation of intermediate results, 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

» (See Textbook Chapter 15)

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Example of Query Tree

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Additional Relational Operations: Aggregate Functions & 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.

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Aggregate Function Operation

Use of the Aggregate Functional operation ℱ

» ℱMAX Salary (EMPLOYEE) retrieves the maximum salary value

from the EMPLOYEE relation

» ℱMIN Salary (EMPLOYEE) retrieves the minimum Salary value


» ℱSUM Salary (EMPLOYEE) retrieves the sum of the Salary


» ℱCOUNT SSN, AVERAGE Salary (EMPLOYEE) computes the count

(number) of employees and their average salary

• Note: count just counts the number of rows, without removing

duplicates

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Using Grouping with Aggregation

The previous examples all summarized one or more attributes for a set of tuples » Maximum Salary or Count (number of) Ssn

Grouping can be combined with Aggregate Functions

Example: For each department, retrieve the DNO, COUNT SSN, and AVERAGE SALARY

A variation of aggregate operation ℱ allows this: » Grouping attribute placed to left of symbol

» Aggregate functions to right of symbol

» DNO ℱCOUNT SSN, AVERAGE Salary (EMPLOYEE)

Above operation groups employees by DNO (department number) and computes the count of employees and average salary per department

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The Aggregate Function Operation

a.ρR(Dno, No_of_employees, Average_sal)(Dno ℑ COUNT Ssn,

AVERAGE Salary (EMPLOYEE)). b. Dno ℑ alary(EMPLOYEE). c. ℑ COUNT Ssn, AVERAGE Salary(EMPLOYEE).

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Results of GROUP BY and HAVING (in SQL) – Figure 7.1a (Q24)

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Additional Relational Operations: Recursive Closure (1/2)

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.

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Additional Relational Operations: Recursive Closure (2/2)

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.

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A Two-Level Recursive Query (Figure 8.11)

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Additional Relational Operations: OUTER JOIN (1/2)

The OUTER JOIN Operation

» In NATURAL JOIN and EQUIJOIN, 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.

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Additional Relational Operations: OUTER JOIN (2/2)

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.

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The Result of a LEFT OUTER JOIN Operation

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Additional Relational Operations: OUTER UNION (1/2)

OUTER UNION Operations » The outer union operation was developed to

take the union of tuples from two relations if the relations are not type 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 type compatible.

» The attributes that are type compatible are represented only once in the result, and those attributes that are not type compatible from either relation are also kept in the result relation T(X, Y, Z).

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Additional Relational Operations: OUTER UNION (2/2)

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)

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Examples of Queries in Relational Algebra – Procedural Form

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= DNOEMPLOYEEEMPLOYEE)

RESULT FNAME, LNAME, ADDRESS (RESEARCH_EMPS)

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)

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Examples of Queries in Relational Algebra – Single Expressions

As a single expression, these queries become:

Q1: Retrieve the name and address of all employees who work for the

‘Research’ department.

Fname, Lname, Address (σ Dname= ‘Research’

(DEPARTMENT Dnumber=Dno(EMPLOYEE))

Q6: Retrieve the names of employees who have no dependents.

Lname, Fname(( Ssn (EMPLOYEE) − ρ Ssn ( Essn

(DEPENDENT))) ∗ EMPLOYEE)

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Additional Examples of Queries in Relational Algebra

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Relational Calculus (1/2)

A relational calculus expression creates a new relation, which is specified in terms of variables that range over rows of the stored database relations (in tuple calculus) or over columns of the stored relations (in domain calculus).

In a calculus expression, there is no order of operations to specify how to retrieve the query result—a calculus expression specifies only what information the result should contain.

» This is the main distinguishing feature between relational algebra and relational calculus.

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Relational Calculus (2/2)

Relational calculus is considered to be a

nonprocedural or declarative language.

This differs from relational algebra, where

we must write a sequence of operations to

specify a retrieval request; hence relational

algebra can be considered as a procedural

way of stating a query.

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Tuple Relational Calculus (1/2)

The tuple relational calculus is based on specifying a

number of tuple variables.

Each tuple variable usually ranges over a particular

database relation, meaning that the variable may take as

its value any individual tuple from that relation.

A simple tuple relational calculus query is of the form

{t | COND(t)}

» where t is a tuple variable and COND (t) is a conditional

expression involving t.

» The result of such a query is the set of all tuples t that

satisfy COND (t).

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Tuple Relational Calculus (2/2)

Example: To find the first and last names of all employees whose salary is above $50,000, we can write the following tuple calculus expression:

{t.FNAME, t.LNAME | EMPLOYEE(t) AND t.SALARY>50000}

The condition EMPLOYEE(t) specifies that the range relation of tuple variable t is EMPLOYEE.

The first and last name (PROJECTION FNAME, LNAME) of each EMPLOYEE tuple t that satisfies the condition t.SALARY>50000 (SELECTION SALARY >50000) will be retrieved.

80

Tuple Variables and Range Relations

Tuple variables

Ranges over a particular database relation

Satisfy COND(t):

Specify:

Range relation R of t

Select particular combinations of tuples

Set of attributes to be retrieved (requested

attributes)

81

Expressions and Formulas in Tuple Relational Calculus

General expression of tuple relational

calculus is of the form:

Truth value of an atom

Evaluates to either TRUE or FALSE for a

specific combination of tuples

Formula (Boolean condition)

Made up of one or more atoms connected via

logical operators AND, OR, and NOT

82

The Existential and Universal Quantifiers (1/2)

Two special symbols called quantifiers can appear in formulas; these are the universal quantifier () and the existential quantifier ().

Informally, a tuple variable t is bound if it is quantified, meaning that it appears in an ( t) or ( t) clause; otherwise, it is free.

If F is a formula, then so are ( t)(F) and ( t)(F), where t is a tuple variable. » The formula ( t)(F) is true if the formula F evaluates to true

for some (at least one) tuple assigned to free occurrences of t in F; otherwise ( t)(F) is false.

» The formula ( t)(F) is true if the formula F evaluates to true for every tuple (in the universe) assigned to free occurrences of t in F; otherwise ( t)(F) is false.

83

The Existential and Universal Quantifiers (2/2)

is called the universal or “for all” quantifier because

every tuple in “the universe of” tuples must make F true to

make the quantified formula true.

is called the existential or “there exists” quantifier

because any tuple that exists in “the universe of” tuples

may make F true to make the quantified formula true.

84

Example Query Using Existential Quantifier

Retrieve the name and address of all employees who work for the ‘Research’ department. The query can be expressed as :

{t.FNAME, t.LNAME, t.ADDRESS | EMPLOYEE(t) and ( d) (DEPARTMENT(d) and d.DNAME=‘Research’ and d.DNUMBER=t.DNO) }

The only free tuple variables in a relational calculus expression should be those that appear to the left of the bar ( | ).

» In above query, t is the only free variable; it is then bound successively to each tuple.

If a tuple satisfies the conditions specified in the query, the attributes FNAME, LNAME, and ADDRESS are retrieved for each such tuple.

» The conditions EMPLOYEE (t) and DEPARTMENT(d) specify the range relations for t and d.

» The condition d.DNAME = ‘Research’ is a selection condition and corresponds to a SELECT operation in the relational algebra, whereas the condition d.DNUMBER = t.DNO is a JOIN condition.

85

Example Query Using Universal Quantifier

Find the names of employees who work on all the projects controlled by department number 5. The query can be:

{e.LNAME, e.FNAME | EMPLOYEE(e) and ( ( x)(not(PROJECT(x)) or not(x.DNUM=5)

OR ( ( w)(WORKS_ON(w) and w.ESSN=e.SSN and x.PNUMBER=w.PNO))))}

Exclude from the universal quantification all tuples that we are not interested in by making the condition true for all such tuples. » The first tuples to exclude (by making them evaluate automatically to true) are

those that are not in the relation R of interest.

In query above, using the expression not(PROJECT(x)) inside the universally quantified formula evaluates to true all tuples x that are not in the PROJECT relation. » Then we exclude the tuples we are not interested in from R itself. The

expression not(x.DNUM=5) evaluates to true all tuples x that are in the project relation but are not controlled by department 5.

Finally, we specify a condition that must hold on all the remaining tuples in R.

( ( w)(WORKS_ON(w) and w.ESSN=e.SSN and x.PNUMBER=w.PNO)

86

Sample Queries in Tuple Relational Calculus

87

Notation for Query Graphs

88

Transforming the Universal and Existential Quantifiers

Transform one type of quantifier into other

with negation (preceded by NOT)

AND and OR replace one another

Negated formula becomes un-negated

Un-negated formula becomes negated

89

Using the Universal Quantifier in Queries

90

Safe Expressions

Guaranteed to yield a finite number of

tuples as its result

Otherwise expression is called unsafe

Expression is safe

If all values in its result are from the domain of

the expression

91

Languages Based on Tuple Relational Calculus (1/2)

The language SQL is based on tuple calculus. It uses the basic block structure to express the queries in tuple calculus:

» SELECT <list of attributes>

» FROM <list of relations>

» WHERE <conditions>

SELECT clause mentions the attributes being projected, the FROM clause mentions the relations needed in the query, and the WHERE clause mentions the selection as well as the join conditions.

» SQL syntax is expanded further to accommodate other operations. (See Textbook Chapter 8).

92

Languages Based on Tuple Relational Calculus (2/2)

Another language which is based on tuple calculus is

QUEL which actually uses the range variables as in tuple

calculus. Its syntax includes:

» RANGE OF <variable name> IS <relation name>

Then it uses

» RETRIEVE <list of attributes from range variables>

» WHERE <conditions>

This language was proposed in the relational DBMS

INGRES. (system is currently still supported by Computer

Associates – but the QUEL language is no longer there).

93

The Domain Relational Calculus (1/2)

Another variation of relational calculus called the domain relational calculus, or simply, domain calculus is equivalent to tuple calculus and to relational algebra.

The language called QBE (Query-By-Example) that is related to domain calculus was developed almost concurrently to SQL at IBM Research, Yorktown Heights, New York.

» Domain calculus was thought of as a way to explain what QBE does.

Domain calculus differs from tuple calculus in the type of variables used in formulas:

» Rather than having variables range over tuples, the variables range over single values from domains of attributes.

To form a relation of degree n for a query result, we must have n of these domain variables— one for each attribute.

94

The Domain Relational Calculus (2/2)

An expression of the domain calculus is of

the form

{ x1, x2, . . ., xn |

COND(x1, x2, . . ., xn, xn+1, xn+2, . . ., xn+m)}

» where x1, x2, . . ., xn, xn+1, xn+2, . . ., xn+m are

domain variables that range over domains (of

attributes)

» and COND is a condition or formula of the

domain relational calculus.

95

Example Query Using Domain Calculus

Retrieve the birthdate and address of the employee whose name is ‘John B. Smith’.

Query :

{uv | ( q) ( r) ( s) ( t) ( w) ( x) ( y) ( z)

(EMPLOYEE(qrstuvwxyz) and q=’John’ and r=’B’ and s=’Smith’)}

Abbreviated notation EMPLOYEE(qrstuvwxyz) uses the

variables without the separating commas: EMPLOYEE(q,r,s,t,u,v,w,x,y,z)

Ten variables for the employee relation are needed, one to range over the domain of each attribute in order.

» Of the ten variables q, r, s, . . ., z, only u and v are free.

Specify the requested attributes, BDATE and ADDRESS, by the free domain variables u for BDATE and v for ADDRESS.

Specify the condition for selecting a tuple following the bar ( | )—

» namely, that the sequence of values assigned to the variables qrstuvwxyz be a tuple of the employee relation and that the values for q (FNAME), r (MINIT), and s (LNAME) be ‘John’, ‘B’, and ‘Smith’, respectively.

96

QBE: A Query Language Based on Domain Calculus (1/2)

This language is based on the idea of giving an example

of a query using “example elements” which are nothing

but domain variables.

Notation: An example element stands for a domain

variable and is specified as an example value preceded

by the underscore character.

P. (called P dot) operator (for “print”) is placed in those

columns which are requested for the result of the query.

A user may initially start giving actual values as examples,

but later can get used to providing a minimum number of

variables as example elements.

See Textbook Appendix C

97

QBE: A Query Language Based on Domain Calculus (2/2)

The language is very user-friendly, because it uses minimal syntax.

QBE was fully developed further with facilities for grouping, aggregation, updating etc. and is shown to be equivalent to SQL.

The language is available under QMF (Query Management Facility) of DB2 of IBM and has been used in various ways by other products like ACCESS of Microsoft, and PARADOX.

For details, see Appendix C in the textbook.

98

QBE Examples

QBE initially presents a relational schema

as a “blank schema” in which the user fills

in the query as an example:

99

Example Schema as a QBE Query Interface

100

QBE Examples

The following domain calculus query can be

successively minimized by the user as

shown:

Query :

{uv | ( q) ( r) ( s) ( t) ( w) ( x) ( y) (

z)

(EMPLOYEE(qrstuvwxyz) and q=‘John’

and r=‘B’ and s=‘Smith’)}

101

Four Successive Ways to Specify a QBE Query

102

QBE Examples

Specifying complex conditions in QBE:

A technique called the “condition box” is

used in QBE to state more involved

Boolean expressions as conditions.

The C.4(a) gives employees who work on

either project 1 or 2, whereas the query in

C.4(b) gives those who work on both the

projects.

103

Complex Conditions with and without a Condition Box in QBE Query

104

Handling AND conditions in a QBE Query

105

JOIN in QBE: Examples

The join is simply accomplished by using

the same example element (variable with

underscore) in the columns being joined

from different (or same as in C.5 (b))

relation.

Note that the Result is set us as an

independent table to show variables from

multiple relations placed in the result.

106

Performing Join with Common Ex. Elements & Use of RESULT Relation

107

Aggregation in QBE

Aggregation is accomplished by using .CNT

for count,.MAX, .MIN, .AVG for the

corresponding aggregation functions

Grouping is accomplished by .G operator.

Condition Box may use conditions on

groups (similar to HAVING clause in SQL –

see Textbook Section 8.5.8)

108

Aggregation in QBE: Examples

109

Negation in QBE: Examples

110

UPDATING in QBE: Examples

111

Summary

Relational Algebra

» Unary Relational Operations

» Relational Algebra Operations From Set Theory

» Binary Relational Operations

» Additional Relational Operations

» Examples of Queries in Relational Algebra

Relational Calculus

» Tuple Relational Calculus

» Domain Relational Calculus

Overview of the QBE language (see Textbook Appendix C)

112

Agenda

1 Session Overview




113

Agenda

Relational Algebra and SQL

Basic Syntax Comparison

Sets and Operations on Relations

Relations in Relational Algebra

Empty Relations

Relational Algebra vs. Full SQL

Operations on Relations

» Projection

» Selection

» Cartesian Product

» Union

» Difference

» Intersection

From Relational Algebra to Queries (with Examples)

Microsoft Access Case Study

Pure Relational Algebra

114

Relational Algebra And SQL

SQL is based on relational algebra with many extensions

» Some necessary

» Some unnecessary

“Pure” relational algebra, use mathematical notation with Greek

letters

It is covered here using SQL syntax; that is this unit covers

relational algebra, but it looks like SQL

And will be really valid SQL

Pure relational algebra is used in research, scientific papers, and

some textbooks

So it is good to know it, and material is provided at the end of this

unit material from which one can learn it

But in anything practical, including commercial systems, you will be

using SQL

115

Key Differences Between SQL And “Pure” Relational Algebra

SQL data model is a multiset not a set; still rows in tables (we

sometimes continue calling relations)

» Still no order among rows: no such thing as 1st row

» We can (if we want to) count how many times a particular row appears

in the table

» We can remove/not remove duplicates as we specify (most of the time)

» There are some operators that specifically pay attention to duplicates

» We must know whether duplicates are removed (and how) for each

SQL operation; luckily, easy

Many redundant operators (relational algebra had only one:

intersection)

SQL provides statistical operators, such as AVG (average)

» Can be performed on subsets of rows; e.g. average salary per

company branch

116

Key Differences Between Relational Algebra And SQL

Every domain is “enhanced” with a special

element: NULL

» Very strange semantics for handling these

elements

“Pretty printing” of output: sorting, and

similar

Operations for

» Inserting

» Deleting

» Changing/updating (sometimes not easily

reducible to deleting and inserting)

117

Basic Syntax Comparison (1/2)

Relational Algebra SQL

a, b SELECT a, b

(d > e) (f =g ) WHERE d > e AND f = g

p q FROM p, q

a, b (d > e) (f =g ) (p q) SELECT a, b

FROM p, q

WHERE d > e AND f = g;

{must always have SELECT even if all

attributes are kept, can be written as:

SELECT *}

renaming AS {or blank space}

p := result INSERT INTO p

result {assuming p was empty}

a, b (p) (assume a, b are the only

attributes)

SELECT *

FROM p;

118

Basic Syntax Comparison (2/2)

Relational Algebra SQL

p q SELECT *

FROM p

UNION

SELECT *

FROM q

p − q SELECT *

FROM p

EXCEPT

SELECT *

FROM q

Sometimes, instead, we have DELETE

FROM

p q SELECT *

FROM p

INTERSECT

SELECT *

FROM q

119

Sets And Operations On Them

If A, B, and C are sets, then we have the operations

Union, A B = { x x A x B }

Intersection, A B = { x x A x B }

- Difference, A - B = { x x A x B }

Cartesian product, A B = { (x,y) x A y B }, A

B C = { (x,y,z) x A y B z B }, etc.

The above operations form an algebra, that is you can

perform operations on results of operations, such as (A

B) (C A)

So you can write expressions and not just programs!

120

Relations in Relational Algebra

Relations are sets of tuples, which we will also call

rows, drawn from some domains

Relational algebra deals with relations (which look like

tables with fixed number of columns and varying

number of rows)

We assume that each domain is linearly ordered, so for

each x and y from the domain, one of the following

holds

» x < y

» x = y

» x < y

Frequently, such comparisons will be meaningful even if

x and y are drawn from different columns

» For example, one column deals with income and another with

expenditure: we may want to compare them

121

Reminder: Relations in Relational Algebra

The order of rows and whether a row appears once or

many times does not matter

The order of columns matters, but as our columns will

always be labeled, we will be able to reconstruct the

order even if the columns are permuted.

The following two relations are equal:

R A B

1 10

2 20

R B A

20 2

10 1

20 2

20 2

122

Many Empty Relations

In set theory, there is only one empty set

For us, it is more convenient to think that for each

relation schema, that for specific choice of column

names and domains, there is a different empty relation

And of, course, two empty relations with different

number of columns must be different

So for instance the two relations below are different

The above needs to be stated more precisely to be

“completely correct,” but as this will be intuitively clear,

we do not need to worry about this too much

123

Relational Algebra Versus Full SQL

Relational algebra is restricted to querying the database

Does not have support for

» Primary keys

» Foreign keys

» Inserting data

» Deleting data

» Updating data

» Indexing

» Recovery

» Concurrency

» Security

» …

Does not care about efficiency, only about specifications

of what is needed

124

Operations on relations

There are several fundamental operations on

relations

We will describe them in turn:

» Projection

» Selection

» Cartesian product

» Union

» Difference

» Intersection (technically not fundamental)

The very important property: Any operation on

relations produces a relation

This is why we call this an algebra

125

Projection: Choice Of Columns

SQL statement

SELECT B, A, D

FROM R

We could have removed the duplicate

row, but did not have to

R A B C D

1 10 100 1000

1 20 100 1000

1 20 200 1000

B A D

10 1 1000

20 1 1000

20 1 1000

126

Selection: Choice Of Rows

SQL statement:

SELECT * (this means all columns)

FROM R

WHERE A <= C AND D = 4; (this is a predicate, i.e., condition)

R A B C D

5 5 7 4

5 6 5 7

4 5 4 4

5 5 5 5

4 6 5 3

4 4 3 4

4 4 4 5

4 6 4 6

A B C D

5 5 7 4

4 5 4 4

127

Selection

In general, the condition (predicate) can be

specified by a Boolean formula with

NOT, AND, OR on atomic conditions, where a

condition is:

» a comparison between two column names,

» a comparison between a column name and a

constant

» Technically, a constant should be put in quotes

» Even a number, such as 4, perhaps should be put in

quotes, as ‘4’, so that it is distinguished from a

column name, but as we will never use numbers for

column names, this not necessary

128

Cartesian Product

SQL statement

SELECT A, R.B, C, S.B, D

FROM R, S; (comma stands for Cartesian product)

A R.B C S.B D

1 10 40 10 10

1 10 50 20 10

2 10 40 10 10

2 10 50 20 10

2 20 40 10 10

2 20 50 20 10

R A B

1 10

2 10

2 20

S C B D

40 10 10

50 20 10

129

A Typical Use Of Cartesian Product

SQL statement: SELECT ID#, R.Room#, Size

FROM R, S

WHERE R.Room# = S.Room#;

R Size Room#

140 1010

150 1020

140 1030

S ID# Room# YOB

40 1010 1982

50 1020 1985

ID# R.Room# Size

40 1010 140

50 1020 150

130


After the Cartesian product, we got

This allowed us to correlate the information from the two original tables by examining each tuple in turn

Size R.Room# ID# S.Room

#

YOB

140 1010 40 1010 1982

140 1010 50 1020 1985

150 1020 40 1010 1982

150 1020 50 1020 1985

140 1030 40 1010 1982

140 1030 50 1020 1985

131


This example showed how to correlate information from

two tables

» The first table had information about rooms and their sizes

» The second table had information about employees including

the rooms they sit in

» The resulting table allows us to find out what are the sizes of

the rooms the employees sit in

We had to specify R.Room# or S.Room#, even though

they happen to be equal due to the specific equality

condition

We could, as we will see later, rename a column, to get

Room# ID# Room# Size

40 1010 140

50 1020 150

132

Union

SQL statement

(SELECT *

FROM R)

UNION

(SELECT *

FROM S);

Note: We happened to choose to remove duplicate

rows

Note: we could not just write R UNION S (syntax quirk)

R A B

1 10

2 20

S A B

1 10

3 20

A B

1 10

2 20

3 20

133

Union Compatibility

We require same -arity (number of columns),

otherwise the result is not a relation

Also, the operation “probably” should make

sense, that is the values in corresponding

columns should be drawn from the same

domains

Actually, best to assume that the column names

are the same and that is what we will do from

now on

We refer to these as union compatibility of

relations

Sometimes, just the term compatibility is used

134

Difference

SQL statement

(SELECT *

FROM R)

MINUS

(SELECT *

FROM S);

Union compatibility required

EXCEPT is a synonym for MINUS

R A B

1 10

2 20

S A B

1 10

3 20

A B

2 20

135

Intersection

SQL statement

(SELECT *

FROM R)

INTERSECT

(SELECT *

FROM S);


Can be computed using differences only: R – (R – S)

R A B

1 10

2 20

S A B

1 10

3 20

A B

1 10

136

From Relational Algebra to Queries

These operations allow us to define a large

number of interesting queries for relational

databases.

In order to be able to formulate our examples,

we will assume standard programming

language type of operations:

» Assignment of an expression to a new variable;

In our case assignment of a relational expression to

a relational variable.

» Renaming of a relations, to use another name to

denote it

» Renaming of a column, to use another name to

denote it

137

A Small Example

The example consists of 3 relations:

Person(Name,Sex,Age)

This relation, whose primary key is Name, gives information about the

human’s sex and age

Birth(Parent,Child)

This relation, whose primary key is the pair Parent,Child, with both being

foreign keys referring to Person gives information about who is a parent of

whom. (Both mother and father would be generally listed)

Marriage(Husband,Wife, Age) or

Marriage(Husband,Wife, Age)

This relation listing current marriages only, requires choosing which spouse

will serve as primary key. For our exercise, it does not matter what the

choice is. Both Husband and Wife are foreign keys referring to Person. Age

specifies how long the marriage has lasted.

For each attribute above, we will frequently use its first letter to refer to

it, to save space in the slides, unless it creates an ambiguity

Some ages do not make sense, but this is fine for our example

138

Relational Implementation

Two options for selecting the primary key of Marriage

The design is not necessarily good, but nice and simple

for learning relational algebra

Because we want to focus on relational algebra, which

does not understand keys, we will not specify keys in

this unit

Marriage

PK,FK2 W

FK1 H

Person

PK N

S

A

Birth

PK,FK1 P

PK,FK2 C

Marriage

PK,FK1 H

FK2 W

Person

PK N

S

A

Birth

PK,FK1 P

PK,FK2 C

139

Microsoft Access Database

Microsoft Access Database with this example

has been posted

» The example suggests that you download and install

Microsoft Access 2007

» The examples are in the Access 2000 format so that

if you have an older version, you can work with it

Access is a very good tool for quickly learning

basic constructs of SQL DML, although it is not

suitable for anything other than personal

databases

140

Microsoft Access Database

The database and our queries (other than the one with

MINUS at the end) are the appropriate “extras” directory

on the class web in “slides”

» MINUS is frequently specified in commercial databases in a

roundabout way

» We will cover how it is done when we discuss commercial

databases

Our sample Access database: People.mdb

The queries in Microsoft Access are copied and pasted

in these notes, after reformatting them

Included copied and pasted screen shots of the results

of the queries so that you can correlate the queries with

the names of the resulting tables

141

Our Database With Sample Queries - Open In Microsoft Access

142

Our Database

Person N S A

Albert M 20

Dennis M 40

Evelyn F 20

John M 60

Mary F 40

Robert M 60

Susan F 40

Birth P C

Dennis Albert

John Mary

Mary Albert

Robert Evelyn

Susan Evelyn

Susan Richard

Marriage H W A

Dennis Mary 20

Robert Susan 30

143

Our Instance In Microsoft Access

144

A Query

Produce the relation Answer(A) consisting of all

ages of people

Note that all the information required can be

obtained from looking at a single relation,

Person

Answer:=

SELECT A

FROM Person;

Recall that whether duplicates are removed is

not important (at least for the time being in our

course)

A

20

40

20

60

40

60

40

145

The Query In Microsoft Access

The actual query was copied and pasted

from Microsoft Access and reformatted for

readability

The result is below

146

A Query

Produce the relation Answer(N) consisting of all

women who are less or equal than 32 years old.

Note that all the information required can be

obtained from looking at a single relation,

Person

Answer:=

SELECT N

FROM Person

WHERE A <= 32 AND S =‘F’;

N

Evelyn

147




readability

The result is below

148

A Query

Produce a relation Answer(P, Daughter) with

the obvious meaning

Here, even though the answer comes only from

the single relation Birth, we still have to check in

the relation Person what the S of the C is

To do that, we create the Cartesian product of

the two relations: Person and Birth. This gives

us “long tuples,” consisting of a tuple in Person

and a tuple in Birth

For our purpose, the two tuples matched if N in

Person is C in Birth and the S of the N is F

149

A Query

Answer:=

SELECT P, C AS Daughter

FROM Person, Birth

WHERE C = N AND S = ‘F’;

Note that AS was the attribute renaming

operator

P Daughter

John Mary

Robert Evelyn

Susan Evelyn

150

Cartesian Product With Condition: Matching Tuples Indicated

Person N S A

Albert M 20

Dennis M 40

Evelyn F 20

John M 60

Mary F 40

Robert M 60

Susan F 40

Birth P C

Dennis Albert

John Mary

Mary Albert

Robert Evelyn

Susan Evelyn

Susan Richard

151




readability

The result is below

152

A Query

Produce a relation Answer(Father, Daughter) with the

obvious meaning.

Here we have to simultaneously look at two copies of

the relation Person, as we have to determine both the S

of the Parent and the S of the C

We need to have two distinct copies of Person in our

SQL query

But, they have to have different names so we can

specify to which we refer

Again, we use AS as a renaming operator, these time

for relations

Note: We could have used what we have already

computer: Parent,Daughter

153

A Query

Answer :=

SELECT P AS Father, C AS Daughter

FROM Person, Birth, Person AS Person1

WHERE P = Person.N AND C = Person1.N

AND Person.S = ‘M’ AND Person1.S = ‘F’;

Father Daughter

John Mary

Robert Evelyn

154


Person N S A

Albert M 20

Dennis M 40

Evelyn F 20

John M 60

Mary F 40

Robert M 60

Susan F 40

Birth P C

Dennis Albert

John Mary

Mary Albert

Robert Evelyn

Susan Evelyn

Susan Richard

Person N S A

Albert M 20

Dennis M 40

Evelyn F 20

John M 60

Mary F 40

Robert M 60

Susan F 40

155




readability

The result is below

156

A Query

Produce a relation: Answer(Father_in_law,

Son_in_law).

A classroom exercise, but you can see the

solution in the posted database.

Hint: you need to compute the Cartesian

product of several relations if you start from

scratch, or of two relations if you use the

previously computed (Father, Daughter) relation

F_I_L S_I_L

John Dennis

157




readability

The result is below

158

A Query

Produce a relation:

Answer(Grandparent,Grandchild)

A classroom exercise, but you can see

the solution in the posted database

G_P G_C

John Albert

159


Birth P C

Dennis Albert

John Mary

Mary Albert

Robert Evelyn

Susan Evelyn

Susan Richard

Birth P C

Dennis Albert

John Mary

Mary Albert

Robert Evelyn

Susan Evelyn

Susan Richard

160




readability

The result is below

161

Further Distance

How to compute (Great-grandparent,Great-grandchild)?

Easy, just take the Cartesian product of the

(Grandparent, Grandchild) table with (Parent,Child)

table and specify equality on the “intermediate” person

How to compute (Great-great-grandparent,Great-great-

grandchild)?

Easy, just take the Cartesian product of the

(Grandparent, Grandchild) table with itself and specify

equality on the “intermediate” person

Similarly, can compute (Greatx-grandparent,Greatx-

grandchild), for any x

Ultimately, may want (Ancestor,Descendant)

162

Relational Algebra Is Not Universal:Cannot Compute (Ancestor,Descendant)

Standard programming languages are universal

This roughly means that they are as powerful as Turing machines, if unbounded amount of storage is permitted (you will never run out of memory)

This roughly means that they can compute anything that can be computed by any computational machine we can (at least currently) imagine

Relational algebra is weaker than a standard programming language

It is impossible in relational algebra (or standard SQL) to compute the relation Answer(Ancestor, Descendant)

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Relational Algebra Is Not Universal: Cannot Compute (Ancestor,Descendant)

It is impossible in relational algebra (or standard SQL) to compute the relation Answer(Ancestor, Descendant)

Why?

The proof is a reasonably simple, but uses cumbersome induction.

The general idea is: » Any relational algebra query is limited in how many relations or

copies of relations it can refer to

» Computing arbitrary (ancestor, descendant) pairs cannot be done, if the query is limited in advance as to the number of relations and copies of relations (including intermediate results) it can specify

This is not a contrived example because it shows that we cannot compute the transitive closure of a directed graph: the set of all paths in the graph

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A Sample Query

Produce a relation Answer(A) consisting

of all ages of visitors that are not ages of

marriages

SELECT

A FROM Person

MINUS

SELECT

A FROM MARRIAGE;

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We do not show this here, as it is done in

a roundabout way and we will do it later

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It Does Not Matter If We Remove Duplicates

Removing duplicates

- =

Not removing duplicates

- =

A

20

40

20

60

40

60

40

A

20

30

A

40

60

40

60

40

A

20

40

60

A

20

30

A

40

60

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It Does Not Matter If We Remove Duplicates

The resulting set contains precisely ages: 40, 60

So we do not have to be concerned with whether the

implementation removes duplicates from the result or not

In both cases we can answer correctly

» Is 50 a number that is an age of a marriage but not of a person

» Is 40 a number that is an age of a marriage but not of a person

Just like we do not have to be concerned with whether it

sorts (orders) the result

This is the consequence of us not insisting that an

element in a set appears only once, as we discussed

earlier

Note, if we had said that an element in a set appears

once, we would have to spend effort removing

duplicates!

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Now To “Pure” Relational Algebra

This was described in several slides

But it is really the same as before, just the

notation is more mathematical

Looks like mathematical expressions, not

snippets of programs

It is useful to know this because many

articles use this instead of SQL

This notation came first, before SQL was

invented, and when relational databases

were just a theoretical construct

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: Projection: Choice Of Columns

SQL statement Relational

Algebra

SELECT B, A, D B,A,D (R)

FROM R

We could have removed the duplicate row,

but did not have to

R A B C D

1 10 100 1000

1 20 100 1000

1 20 200 1000

B A D

10 1 1000

20 1 1000

20 1 1000

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: Selection: Choice Of Rows

SQL statement: Relational Algebra

SELECT * A C D=4 (R) Note: no need for

FROM R

WHERE A <= C AND D = 4;

R A B C D

5 5 7 4

5 6 5 7

4 5 4 4

5 5 5 5

4 6 5 3

4 4 3 4

4 4 4 5

4 6 4 6

A B C D

5 5 7 4

4 5 4 4

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Selection

In general, the condition (predicate) can be

specified by a Boolean formula with

and on atomic conditions, where a

condition is:

» a comparison between two column names,

» a comparison between a column name and a

constant

» Technically, a constant should be put in quotes

» Even a number, such as 4, perhaps should be put in

quotes, as ‘4’ so that it is distinguished from a

column name, but as we will never use numbers for

column names, this not necessary

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: Cartesian Product

SQL statement Relational Algebra

SELECT A, R.B, C, S.B, D R S

FROM R, S

A R.B C S.B D

1 10 40 10 10

1 10 50 20 10

2 10 40 10 10

2 10 50 20 10

2 20 40 10 10

2 20 50 20 10

R A B

1 10

2 10

2 20

S C B D

40 10 10

50 20 10

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SQL statement: Relational Algebra SELECT ID#, R.Room#, Size ID#, R.Room#, Size (R.B = S.B (R S))

FROM R, S

WHERE R.Room# = S.Room#

R Size Room#

140 1010

150 1020

140 1030

S ID# Room# YOB

40 1010 1982

50 1020 1985

ID# R.Room# Size

40 1010 140

50 1020 150

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: Union

SQL statement Relational Algebra

(SELECT * R S

FROM R)

UNION

(SELECT *

FROM S)

Note: We happened to choose to remove duplicate

rows


R A B

1 10

2 20

S A B

1 10

3 20

A B

1 10

2 20

3 20

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-: Difference


Algebra

(SELECT * R - S

FROM R)

MINUS

(SELECT *

FROM S)


R A B

1 10

2 20

S A B

1 10

3 20

A B

2 20

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: Intersection


Algebra

(SELECT * R S

FROM R)

INTERSECT

(SELECT *

FROM S)


R A B

1 10

2 20

S A B

1 10

3 20

A B

1 10

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Summary (1/2)

A relation is a set of rows in a table with labeled columns

Relational algebra as the basis for SQL

Basic operations:

» Union (requires union compatibility)

» Difference (requires union compatibility)

» Intersection (requires union compatibility); technically not a basic

operation

» Selection of rows

» Selection of columns

» Cartesian product

These operations define an algebra: given an expression

on relations, the result is a relation (this is a “closed”

system)

Combining this operations allows production of

sophisticated queries

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Summary (2/2)

Relational algebra is not universal: We can write

programs producing queries that cannot be

specified using relational algebra

We focused on relational algebra specified

using SQL syntax, as this is more important in

practice

The other, “more mathematical” notation came

first and is used in research and other venues,

but not commercially

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Agenda

1 Session Overview




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Summary

Relational algebra and relational calculus are formal

languages for the relational model of data

A relation is a set of rows in a table with labeled columns

Relational algebra and associated operations are the

basis for SQL

These relational operations define an algebra

Relational algebra is not universal as it is possible to write

programs producing queries that cannot be specified

using relational algebra

Relational algebra can be specified using SQL syntax

The other, “more mathematical” notation came first and is

used in research and other venues, but not commercially

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Assignments & Readings

Readings

» Slides and Handouts posted on the course web site

» Textbook: Chapters 6, 8

Assignment #5

» Textbook exercises: Textbook exercises: 8.15, 8.18, 8.22, 8.30, 6.8, 6.12

Project Framework Setup (ongoing)

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Next Session: Standard Query Language (SQL) Features

SQL as implemented in commercial

databases

183

Any Questions?

Documents

Relational Algebra-Relational Calculus-SQL