Database Systems: Chapter 6: Relational Algebra

Dr. Taysir Hassan Abdel Hamid

Lecturer, IS Department

Faculty of Computer and Information

Assiut University

Contact: taysir_soliman@hotmail.com

March 30, 2008

Operations in Relational Algebra

Group one: includes set operations from mathematical set theory:

UNION, INTERSECTION, SET DIFFERENCE, AND CARTESIAN PRODUCT.

Group two: includes operations specifically for relational databases:

SELECT, PROJECT, AND JOIN Common database requests cannot be

performed with the original relational operations.

Unary Relational OperationsSELECT SELECT operation is used to select a subset

of the tuples from a relation that satisfies a selection condition.

One can consider the SELECT operation to be a filter that keeps only those tuples that satisfy a qualifying condition.

SELECT can also be visualized as a horizontal partition of the relation into two sets of tuples.

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Select Operator Produce table containing subset of rows of

argument table satisfying condition

<condition>(<relation>)

• Example: H o b b y = ’ s t a m p s ’ ( P e r s o n )

1123 John 123 Main stamps

1123 John 123 Main coins

5556 Mary 7 Lake Dr hiking

9876 Bart 5 Pine St stamps

1123 John 123 Main stamps

9876 Bart 5 Pine St stamps

Person

Id Name Address Hobby Id Name Address Hobby

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Selection Condition

Operators: <, , , >, =, Simple selection condition:

<attribute> operator <constant> <attribute> operator <attribute>

<condition> AND <condition> <condition> OR <condition> NOT <condition> NOTE: Each attribute in condition must be one

of the attributes of the relation name

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Id>3000 OR Hobby=‘hiking’ (Person)

Id>3000 AND Id <3999 (Person)

NOT(Hobby=‘hiking’) (Person)

Hobby‘hiking’ (Person)

(Id > 3000 AND Id < 3999) OR Hobby‘hiking’ (Person)

Selection Condition - Examples

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Selection Example

EmployeeSSN Name DepartmentID Salary999999999 John 1 30,000777777777 Tony 1 32,000888888888 Alice 2 45,000

SSN Name DepartmentID Salary888888888 Alice 2 45,000

Find all employees with salary more than $40,000.Salary > 40000 (Employee)

Projection Extracts some of the columns from a relation.

SexSalary := (EMPLOYEE, (SEX, SALARY))

No attribute may occur more than once. Duplicate will be removed.

Projection and selection combined.

((EMPLOYEE, CITY = “LONDON”), (NAME,SSN))

This does a selection followed by a projection. The DBMS may re-organise this for faster retrieval.

Union

Produce a relation which combines two relations by containing all of the tuples from each - removing duplicates.

The two relations must be "union compatible" i.e. have the same number of attributes drawn from the same domain (but maybe having different names).

LondonOrRich: = EMPLOYEE, CITY = “LONDON”) (EMPLOYEE, SALARY > 60K)

If attribute names differ, the names from the first one are taken.

Same can be done using disjuncts!

Intersection

Similar to union but returns tuples that are in both relations.

FemalesInLondon := EMPLOYEE, CITY = “LONDON”) (EMPLOYEE, SEX = “F”)

Same can be done with conjuncts!

Difference

Similar to union but returns tuples that are in the first relation but not the second.

NonLocals := EMPLOYEE - LOCALS

Intersection and difference both require union compatibility.

Intersection and difference use column names from the first relation.

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Union Compatible Relations

Two relations are union compatible if Both have same number of columns Names of attributes are the same in both Attributes with the same name in both relations

have the same domain Union compatible relations can be combined

using union, intersection, and set difference

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Example

Tables:

Person (SSN, Name, Address, Hobby)

Professor (Id, Name, Office, Phone)

are not union compatible. However Name (Person) and Name (Professor)are union compatible and Name (Person) - Name (Professor)makes sense.

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Cartesian Product If R and S are two relations, R S is the set

of all concatenated tuples <x,y>, where x is a tuple in R and y is a tuple in S (R and S need not be union compatible)

R S is expensive to compute: Factor of two in the size of each row Quadratic in the number of rows

a b c d a b c d

x1 x2 y1 y2 x1 x2 y1 y2

x3 x4 y3 y4 x1 x2 y3 y4

x3 x4 y1 y2

R S x3 x4 y3 y4

R S

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Renaming

Result of expression evaluation is a relation Attributes of relation must have distinct names.

This is not guaranteed with Cartesian product e.g., suppose in previous example a = c

Renaming operator tidies this up. To assign the names A1, A2,… An to the attributes of the n column relation produced by expression use

expression [A1, A2, … An]

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Example

Transcript (StudId, CrsCode, Semester, Grade)

Teaching (ProfId, CrsCode, Semester)

StudId, CrsCode (Transcript)[StudId, SCrsCode] ProfId, CrsCode(Teaching) [ProfId, PCrscode]

This is a relation with 4 attributes:

StudId, SCrsCode, ProfId, PCrsCode

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Derived Operation: JoinThe expression :

join-condition´ (R S)

where join-condition´ is a conjunction of terms:

Ai oper Bi

in which Ai is an attribute of R, Bi is an attribute of S, and oper is one of =, <, >, , , is referred to as the (theta) join of R and S and denoted:

R join-condition S

Where join-condition and join-condition´ are (roughly) the same …

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Join and Renaming

Problem: R and S might have attributes with the same name – in which case the Cartesian product is not defined

Solution: Rename attributes prior to forming the product

and use new names in join-condition´. Common attribute names are qualified with

relation names in the result of the join

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Theta Join – Example

Output the names of all employees that earnmore than their managers.Employee.Name (Employee MngrId=Id AND Salary>Salary Manager)The join yields a table with attributes: Employee.Name, Employee.Id, Employee.Salary, MngrId Manager.Name, Manager.Id, Manager.Salary

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Equijoin Join - Example

Name,CrsCode(Student Id=StudId Grade=‘A’(Transcript))

Id Name Addr Status111 John ….. …..222 Mary ….. …..333 Bill ….. …..444 Joe ….. …..

StudId CrsCode Sem Grade 111 CSE305 S00 B 222 CSE306 S99 A 333 CSE304 F99 A

Mary CSE306Bill CSE304

The equijoin is commonlyused since it combines related data in different relations.

Student Transcript

Equijoin: Join condition is a conjunction of equalities.

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Natural Join Special case of equijoin:

join condition equates all and only those attributes with the same name (condition doesn’t have to be explicitly stated)

duplicate columns eliminated from the resultTranscript (StudId, CrsCode, Sem, Grade)Teaching (ProfId, CrsCode, Sem)

Transcript Teaching = StudId, Transcript.CrsCode, Transcript.Sem, Grade, ProfId ( Transcript

CrsCode=CrsCode AND Sem=SemTeaching)

[StudId, CrsCode, Sem, Grade, ProfId ]

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Natural Join (con’t) More generally:

R S = attr-list (join-cond (R × S) )

where

attr-list = attributes (R) attributes (S)

(duplicates are eliminated) and join-cond has

the form:

A1 = A1 AND … AND An = An

where

{A1 … An} = attributes(R) attributes(S)

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Natural Join Example

List all Id’s of students who took at least two different courses:

StudId ( CrsCode CrsCode2 (

Transcript Transcript [StudId, CrsCode2, Sem2, Grade2])) (don’t join on CrsCode, Sem,

and Grade attributes)

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Division

Goal: Produce the tuples in one relation, r, that match all tuples in another relation, s r (A1, …An, B1, …Bm)

s (B1 …Bm)

r/s, with attributes A1, …An, is the set of all tuples <a> such that for every tuple <b> in s, <a,b> is in r

Can be expressed in terms of projection, set difference, and cross-product

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Division (con’t)

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Division - Example List the Ids of students who have passed

all courses that were taught in spring 2000 Numerator: StudId and CrsCode for every

course passed by every student StudId, CrsCode (Grade ‘F’ (Transcript) )

Denominator: CrsCode of all courses taught in spring 2000 CrsCode (Semester=‘S2000’ (Teaching) )

Result is numerator/denominator