LECTURE 4 RELATIONAL ALGEBRA. Introduction Relational algebra defines the theoretical way of manipulating table contents through a number of relational

LECTURE 4

RELATIONAL ALGEBRA

Introduction•Relational algebra defines the theoretical

way of manipulating table contents through a number of relational operators

• These relational operators include;▫SELECT (or RESTRICT)▫PROJECT▫ JOIN▫PRODUCT▫ INTERSECT▫UNION▫DIFFERENCE▫DIVIDE

Introduction (cont’d)•The relational operators have the property

of closure, i.e., relational algebra operators are used on existing tables to produce new tables

•The relational operators are classed as being unary or binary

•Unary operators such as SELECT and PROJECT, can be applied to one relation

•Binary operators such as JOIN are applied on two relations

SELECTion•SELECT or RESTRICT, can be used to list all

of the row values, or return only the row values that match a specified criterion.

•The SELECT operator is denoted by and is formally defined as:▫(R) or (RELATION) where (R) is the set of specified tuples of the relation R and is the predicate (or criterion) to extract the required tuples.

• It is possible to create more complex criteria by using the logical operators AND, OR, and NOT

SELECT (cont’d)

OrderNum PartNum21608 AT9421610 DR9321610 DW1121613 KL6221614 KT0321617 BV0621617 CD5221619 DR9321623 KV29

OrderLine (OrderLine)

(OrderLine)


OrderNum PartNum

21608 AT94

PROJECTion

•This operator returns all values for

selected attributes.

•The PROJECT operator is denoted by and

formally defined as:

▫ or

where the projection of the relation R,

denoted by is the set of specified attributes

a1…..an of the relation R

PROJECT (cont’d)


OrderLine OrderNum(OrderLine)OrderNum216082161021610216132161421617216172161921623

UNION

• The UNION set operator combines all tuples from

two relations, excluding duplicate tuples.

• The relations must have the same attribute

characteristics (the columns and domains must

be identical) to be used in the UNION.

• When two or more tables share the same

number of columns, i.e., have the same degree,

and when the share the same (or compatible)

domains, they are said to be union-

compatible

UNION(cont’d)

• The UNION operator is denoted by and formally

defined as:

▫ The union of relations and denoted by with

degree n, is the relation where for each i (i = 1,

2, …n), ai and bi must have compatible domains.

▫ The degree of R3 is the same as that of R1 and

R2. However, the cardinality of R3 is a+b, only if

a and b are the cardinalities of R1 and R2

respectively

UNION (cont’d)

OrderNum PartNum21608 AT9421610 DR9321610 DW1121613 KL62

OrderLine1OrderNum PartNum21614 KT0321617 BV0621617 CD5221619 DR9321623 KV29

OrderLine2


OrderLine1 OrderLine2

INTERSECT

• The INTERSECT operator denoted as , returns only

the tuples that appear in both relations

• The tables must be union-compatible to give valid

results.

• The INTERSECT operator is formally defined as:

▫ The intersect of relations and denoted by with

degree n, is the relation that includes only those

tuples of R1 that also appear in R2 where for each i (i

= 1, 2, …n), ai and bi must have compatible domains.

INTERSECT(cont’d)

PartNum

DR93

DW11

BV06

CD52

PartNumAT94DR93DW11KL62KT03KV29

Part1 Part2Part1 Part2

PartNumDR93DW11

DIFFERENCE

• The DIFFERENCE operator returns all tuples in one

relation that are not found in the other relation.

• The DIFFERENCE operator requires that the two

relations be union-compatible.

• The DIFFERENCE operator is formally defined as:

▫ The difference of relations and denoted by with

degree m, is the relation that includes all the tuples

that are in R1 but not in R2 where for each i (i = 1,

2, …m), ai and bi must have compatible domains.

DIFFERENCE (cont’d)

PartNum

DR93

DW11

BV06

CD52

PartNumAT94DR93DW11KL62KT03DR93KV29

Part1 Part2Part1 Part2

PartNumAT94KL62KT03KV29

CARTESIAN PRODUCT

• The CARTESIAN PRODUCT is usually written as with the

new resulting relation containing all the attributes which

are present in and along with all possible combinations of

tuples from both and

• It can be formally defined as:

▫ The CARTESIAN PRODUCT of two relations with cardinality i and

with cardinality j is a relation with degree , cardinality and

attributes . This can be denoted as

• The CARTESIAN PRODUCT is not a very useful operation by

itself, as it combines many tuples that have no association

with each other. However, when used in conjunction with

the RESTRICT (SELECT) operator, it becomes a very

important operator known as a JOIN

CARTESIAN PRODUCT (cont’d)

OrderNum

OrderDate

21608 10/20/2010

21610 10/20/2010

21613 10/21/2010

PartNum Description

DR93 Gas Range

DW11 Washer

OrderNum OrderDate PartNum Description21608 10/20/2010 DR93 Gas Range21610 10/20/2010 DR93 Gas Range21613 10/21/2010 DR93 Gas Range21608 10/20/2010 DW11 Washer21610 10/20/2010 DW11 Washer21613 10/21/2010 DW11 Washer

Part

Orders

Product of Orders and Part

DIVISION

• The division operation produces a new relation by selecting

the tuples in one relation, , that match every row in another

relation, . It is the inverse of the CARTESIAN PRODUCT.

• DIVISION, denoted by , can be formally defined as:

▫ The DIVISION of two relations with cardinality i and with

cardinality j is a relation with degree and cardinality .

Division (cont’d)

OrderNum21610

PartNum

DR93

DW11

OrderNum PartNum

21608 AT9421610 DR9321610 DW1121613 KL6221614 KT0321617 BV0621617 CD5221619 DR9321623 KV29

OrderLine Part

OrderLine Part

JOIN•The JOIN operation is one of the essential

operations of relational algebra. • It is said to be the real power behind the

relational database, allowing the use of independent tables linked by common attributes

•The JOIN of two relations R1 and R2 is a restriction on their Cartesian product R1XR2 to meet a specified criterion.

•The join itself is defined on an attribute a of R1 and b of R2 where the attributes a and b share the same domain.

JOIN(cont’d)

• The JOIN operator is formally defined as:

▫ The join of two relations R1(a1, a2,…., an) and R2(b1, b2,

…, bm) is a relation R3 with degree and attributes (a1,

a2,…, an, b1, b2,…, bm)

• Types of join operations

▫ THETA JOIN

▫ EQUIJOIN

▫ NATURAL JOIN

▫ LEFT OUTER JOIN

▫ RIGHT OUTER JOIN

THETA JOIN AND EQUIJOIN

• EQUIJOIN is on of the most commonly used joins

which links tables on the basis of an equality

condition that compares specified columns of

each table

• The outcome of the equijoin does not eliminate

duplicate columns , and the condition or

criterion used must be explicitly defined.

• The equijoin takes its name from the equality

comparison operator (=) used in the condition

THETA JOIN AND EQUIJOIN(cont’d)

• If any other comparison operator is used the join

is called a THETA JOIN, denoted by θ (θ-join)

▫ Therefore, theta represents a predicate which

consists of the following comparison operators (<,

<=, >=, <>)

• EQUIJOIN is a special type of THETA JOIN

• Let R1(a1, a2,…., an) and R2(b1, b2,…, bm) be

relations which may have different schemas.

▫ Then, θ-join of R1 and R2 is denoted as R1 θR2

▫ The equijoin is denoted as R1 R1.a = R2.bR2

THETA JOIN AND EQUIJOIN(cont’d)

• It is possible to express the θ-join and the equijoin in

terms of restriction and Cartesian product operations

▫ Eg; equijoin R1 R1.a = R2.bR2 may also be written as

R1.a = R2.b(R1 XR2 )

• Looking at the θ-join and the equijoin in this way allows

for some rules to be created which helps in the

computation of such joins on two relations:

▫ Compute R1 XR2 . This first performs a Cartesian product to

form all possible combinations of the rows of R1 and R2

▫ Restrict the Cartesian product to only those rows where

the values in certain columns match

Equijoin Example

DEPT_CODE DEPT_NAMEACCT AccountingBIOL BiologyCIS Computer Info. SystemsENGL English

STU_NUM

STU_LNAME

STU_FNAME

STU_DOB DEPT_CODE

321452 Bowser William 12 February 1972

BIOL

324257 Smithson Anne 15 November 1977

CIS

324258 Brewer Juliette 23 August 1966 ACCT

324269 Oblonski Walter 16 September 1973

CIS

324273 smith John 30 December 1955

ENGL

STUDENT

DEPARTMENT

Equijoin Example (cont’d)

STU_NUM STU_LNAME

STU_FNAME

STU_DOB S.DEPT_CODE

D.DEPT_CODE DEPT_NAME


BIOL ACCT Accounting


BIOL BIOL Biology


BIOL CIS Computer Info. Systems


BIOL ENGL English


CIS ACCT Accounting


CIS BIOL Biology


CIS CIS Computer Info. Systems


CIS ENGL English

324258 Brewer Juliette 23 August 1966 ACCT ACCT Accounting

324258 Brewer Juliette 23 August 1966 ACCT BIOL Biology

324258 Brewer Juliette 23 August 1966 ACCT CIS Computer Info. Systems

324258 Brewer Juliette 23 August 1966 ACCT ENGL English


CIS ACCT Accounting


CIS BIOL Biology




CIS ENGL English


ENGL ACCT Accounting


ENGL BIOL Biology


ENGL CIS Computer Info. Systems


ENGL ENGL English

Cartesian product (STUDENT X DEPARTMENT)

Equijoin Example (cont’d)

STU_NUM

STU_LNAME STU_FNAME

STU_DOB S.DEPT_CODE

D.DEPT_CODE

DEPT_NAME


BIOL BIOL Biology



324258 Brewer Juliette 23 August 1966 ACCT ACCT Accounting




ENGL ENGL English

STUDENT_IN_DEPT = STUDENT.DEPT_CODE = DEPARTMENT.DEPT_CODE(STUDENT X DEPARTMENT )

NATURAL JOIN

• The natural join operation is the most common

variant of the joins and requires that the two

operant relations must have at least one common

attribute, i.e., attributes that share the same

domain. The common column(s) is (are) referred to

as the join column(s)

• The natural join is in fact an equijoin, however, in

addition, the duplicate attributes are dropped with

the resulting relation containing one less column

than that of the equijoin

NATURAL JOIN(cont’d)

• Let R1 be a relation having attributes (a1, a2,…,

an, y), R2 be another relation having attributes

(b1, b2,…, bmy) where y is a set of common

attributes (join column(s)) which share the

same domain

• The natural join operator is defined as:

▫ The natural join of R1 and R2, denoted R1 X R2,

consists of combining the tuples of R1 and R2 to

build a new relation R3, such that if , ,, then

NATURAL JOIN(cont’d)

• The common set of attributes y appears only once in R3; the

notation correspond to the attribute value of a tuple of R1

• The steps required to perform the natural join of two

relations are:

1. Compute R1 X R2.

2. Select those tuples where . Only the rows are selected

where the attribute values in the join column(s) are equal

3. Perform a PROJECT operation on either to the result in step

2, and call it y in the final relation. This is to ensure that the

final relation results in a single copy of each attribute in the

joining column, thereby eliminating duplicates

Natural Join ExampleCUSTOMER

AGENT

CUS_CODE CUS_LNAME

CUS_POSTCODE

AGENT_CODE

1132445 Walker M1 5RT 2311217782 Adares NW6 4RT 1251312243 Rakowski 678954 1671321242 Rodriguez NW6 2WS 1251542311 Smithson N4 3YP 4211657399 Vanloo 67543W 231

AGENT_CODE AGENT_PHONE

125 01812439887167 01813426778231 01812431124333 01131234445

Natural Join Example (cont’d)

Cartesian product (CUSTOMER X AGENT)C.CUS_CODE C.CUS_LNAME C.CUS_POSTCODE C.AGENT_CODE A.AGENT_CODE A.AGENT_PHONE

1132445 Walker M1 5RT 231 125 018124398871132445 Walker M1 5RT 231 167 018134267781132445 Walker M1 5RT 231 231 018124311241132445 Walker M1 5RT 231 333 011312344451217782 Adares NW6 4RT 125 125 018124398871217782 Adares NW6 4RT 125 167 018134267781217782 Adares NW6 4RT 125 231 018124311241217782 Adares NW6 4RT 125 333 011312344451312243 Rakowski 678954 167 125 018124398871312243 Rakowski 678954 167 167 018134267781312243 Rakowski 678954 167 231 018124311241312243 Rakowski 678954 167 333 011312344451321242 Rodriguez NW6 2WS 125 125 018124398871321242 Rodriguez NW6 2WS 125 167 018134267781321242 Rodriguez NW6 2WS 125 231 018124311241321242 Rodriguez NW6 2WS 125 333 011312344451542311 Smithson N4 3YP 421 125 018124398871542311 Smithson N4 3YP 421 167 018134267781542311 Smithson N4 3YP 421 231 018124311241542311 Smithson N4 3YP 421 333 011312344451657399 Vanloo 67543W 231 125 018124398871657399 Vanloo 67543W 231 167 018134267781657399 Vanloo 67543W 231 231 018124311241657399 Vanloo 67543W 231 333 01131234445

Natural Join Example (cont’d)CUSTOMER X AGENT

CUS_CODE CUS_LNAME CUS_POSTCODE

AGENT_CODE

AGENT_PHONE

1132445 Walker M1 5RT 231 01812431124

1217782 Adares NW6 4RT 125 01812439887

1312243 Rakowski 678954 167 01813426778

1321242 Rodriguez NW6 2WS 125 01812439887

1657399 Vanloo 67543W 231 01812431124

OUTER JOIN

• When using the theta and natural join, it is possible

that some tuples in the joined relations do not have

identical values for the common attributes. As a

result these tuples will be ‘lost’.

• If it is required that all tuples from the original

tables be shown in the resulting relation, then it is

necessary to have a join which keeps all the tuples

in R1 which have no corresponding values in R2. The

tuples in the R2 will have null values. This type of

join is known as the outer join

OUTER JOIN (cont’d)

• There are 3 common types of the outer join

▫ Left outer join – keeps data from the left-hand relation

▫ Right outer join – keeps data from the right-hand

relation

▫ Full outer join – keeps data from both relations

• The steps for determining an outer join are very

similar to those for computing a natural join, except

that data from the left or right side of the relation,

depending on whether one is performing a left or

right outer join is included.

OUTER JOIN (cont’d)

• The stages in performing a left outer join are:

1. Compute R1 X R2.

2. Select those tuples where . Only the rows are selected where

the attribute values in the join column(s) are equal

3. Select those tuples in R1 that do not have matching values in

R2 , so

4. Perform a PROJECT operation on either to the result in step

2, and call it y in the final relation. This is to ensure that the

final relation results in a single copy of each attribute in the

joining column, thereby eliminating duplicates. Finally,

project the rest of attributes in R1 and R2 , except y, and drop

the R1 and R2 in the final relation

Left Outer Join Example (cont’d)

•A left outer join of CUSTOMER and

AGENT, will return all the tuples in the

CUSTOMER relation, including those that

do not have a matching value in the

AGENT relation.CUS_CODE CUS_LNAME CUS_POSTCODE

AGENT_CODE AGENT_PHONE

1132445 Walker M1 5RT 231 01812431124

1217782 Adares NW6 4RT 125 01812439887

1312243 Rakowski 678954 167 01813426778

1321242 Rodriguez NW6 2WS 125 01812439887

1657399 Vanloo 67543W 231 01812431124

1542311 Smithson N4 3YP 421 NULL

Right Outer Join Example (cont’d)

•A right outer join of CUSTOMER and

AGENT, will return all the tuples in the

AGENT relation, including those that do

not have a matching value in the

CUSTOMER relation.CUS_CODE CUS_LNAME CUS_POSTCOD

EAGENT_CODE AGENT_PHON

E1132445 Walker M1 5RT 231 01812431124

1217782 Adares NW6 4RT 125 01812439887

1312243 Rakowski 678954 167 01813426778

1321242 Rodriguez NW6 2WS 125 01812439887

1657399 Vanloo 67543W 231 01812431124

NULL NULL NULL 333 01131234445

CONSTRUCTING QUERIES USING RELATIONAL ALGEBRAIC EXPRESSIONS

• The main purpose of relational algebra is to

provide a way to create and manipulate relations

(tables) in a database.

• During the lifetime of a database, users will ask

many different kinds of queries. The task of

building a query involves breaking the query

down into a number of smaller steps, where

each step generates a set of intermediate results

which are then used in the steps of the query.

CONSTRUCTING QUERIES USING RELATIONAL ALGEBRAIC EXPRESSIONS

• The following steps should be followed when

building a query using relational algebraic

expressions

1. List all the attributes needed to give the

answer

2. Select all the relations needed based on the

list of attributes.

3. Specify the relational operators and the

intermediate results that are needed

Building Queries Example - Car Maintenance Database

• In a small database that stores information about the

maintenance of cars, each car is required to undergo an

inspection each year to test to see if it is roadworthy.

After each inspection a maintenance record is created

and any repairs that are needed are recorded. A repair

can require new parts to be purchased and fitted. If a car

needs a repair then the EVALUATION is set to FAIL until

all the repairs are completed and then it is set to PASS.

The Entity Relationship Diagram (ERD) and tables are

shown below.

Car Maintenance Example (cont’d)Car inspection ERD

Car Maintenance Example (cont’d)

REGISTRATION

CAR_MAKE

CAR_MODEL

CAR_COLOUR MODEL_YEAR

LINCENCE_NO

3679MR82 Toyota Corolla Blue 2006 1967fr89768

E-TS865 Nissan Micro Red 2004 1973Smith121

PE57UVP Peugeot 407 Blue 2007 1990byt3212

PISE567 Volkswagen

Eos Lime 2006 DF-678-WV

ROMA482 Volkswagen

Golf GT Black 2007 AQ-123-AV

Z-BA975 Peugeot 207 Black 2007 1980vrt7312PART_NO PART_NAME PART_COST12390 Paint

sealantsGhȻ14.95

12391 Wiper GhȻ19.9512392 Brake pads GhȻ24.9912393 Brake Discs GhȻ49.5412395 Spark Plugs GhȻ0.9912396 Airbag GhȻ24.9512397 Tyres GhȻ25.00

CAR

PART


INSPECTION_CODE REGISTRATION INSPECTION_DATE EVALUATION

100036 PE57UVP 10/05/2008 FAIL100390 ROMA482 01/09/2008 106750 E-TS865 01/03/2006 PASS122456 Z-BA975 03/10/2008 FAIL145678 PISE567 30/09/2007 PASS200450 E-TS865 21/02/2005 PASS200456 E-TS865 01/04/2007 FAIL

MAINTENANCE_RECORD

INSPECTION_CODE PART_NO106750 12396106750 12397100036 12393200450 12391100036 12397200450 12392200456 12397

REPAIR


•Consider the following query asked by a user:▫‘List all information about cars where the

model year is after 2006’•To answer this query, one must first

interpret that ‘List all information about cars’ means list all attributes in the relation CAR. The user only wants to see information on cars where the attribute MODEL_YEAR>2006. using the relational operator SELECT we can write the query as a relational algebraic expression as:


REGISTRATION

CAR_MAKE CAR_MODEL

CAR_COLOUR MODEL_YEAR

LINCENCE_NO

PE57UVP Peugeot 407 Blue 2007 1990byt3212

ROMA482 Volkswagen

Golf GT Black 2007 AQ-123-AV

Z-BA975 Peugeot 207 Black 2007 1980vrt7312

Car Maintenance Assignment

1. Display all the part names and their

prices where the cost of the part is

greater than GhȻ20.00

2. List the car registration and model

details and part numbers for all cars

where the model year is 2007, where an

inspection was carried out after

01/03/2008, which resulted in a part

being required for the repair.

Documents

LECTURE 4 RELATIONAL ALGEBRA. Introduction Relational algebra defines the theoretical way of manipulating table contents through a number of relational