RELATIONAL ALGEBRA SECTION 5 An Introduction. Introduction E.F. Codd in 1970 A revolutionary advance...

RELATIONAL ALGEBRA

SECTION 5

An Introduction

Introduction

• E.F. Codd in 1970

• A revolutionary advance in data manipulation

• Very significant. Why?

• Manipulation of data solely on their logical characteristics

• Codd proposed two data manipulation languages

• Change in terminology emphasized logical meaning

– procedural

• Relational calculus– nonprocedural

• Relational algebra

Relational Algebra

• Manipulation of relations

• Creation of new relations

• Consists of nine operations:– union, intersection, difference, product, select, project, join, divide, and assignment

Union ()

• Consider the two following relations:

– Representations of sales person

– SALES_SUBORD subordinate salespeople– SALES_MGR sales managers

• Combine data from two relations

SALES_SUBORD

SALPERS ID

SALPERS NAME MANAGER ID OFFICE COMM %

10 Rodney Jones 27 Chicago 10

14 Masaji Matsu 44 Tokyo 11

23 Francois Moire 35 Brussels 9

37 Elena Hermana 12 B. A. 13

39 Goro Azuma 44 Tokyo 10

44 Albert Ige 27 Tokyo 12

35 Brigit Bovary 27 Brussels 11

12 Buster Sanchez 27 B. A. 10

SALES_MGR

SALPERS ID


27 Terry Cardon Chicago 15




If A = SALES_SUBORD and B = SALES_MGR

What is A B?

A B

SALESPERSON

SALPERS ID











• Union compatible

• SALESPERSON:= SALES_SUBORD SALES_MGR

Intersection ()

• Using the original two union-compatible tables

• Allows the identification of rows that are common to two relations


What is A B?

A B

SALES_SUBORD_MGR

SALPERS ID





• If C is the intersection of A and B, C := A B

• What does C consist of?

Difference (–)

• Again using the original two-union compatible tables

• SALES_MGR_MGR := SALES_ MGR – SALES_ SUBORD

• Creates the set difference of two union-compatible relations


What is A - B?

A B

SALES_MGR_MGR

SALPERS ID



• If C is the difference between A and B, C := A – B

• Then?

• Consider SALES_SUBORD – SALES_ MGR

• What does this give?

• Note:

A – B B – A

A B

Product ()

• Creates the Cartesian product of two relations

C := A B

X Y W Z

10 22 33 54

10 22 37 98

10 22 42 100

11 25 33 54

11 25 37 98

11 25 42 100

A

X Y

10 22

11 25

B

W Z

33 54

37 98

42 100

• Consider the following example of a PRODUCT table and a SALE table

• Want PS := PRODUCT SALE

• Note:– Adjoins the attributes of the two relations

– Attaches to each row in A, each of the rows in B

PRODUCT

PROD ID PROD DESC MANUFACTR ID

COST PRICE

1035 Sweater 210 12.50 20.00

2241 Table Lamp 317 22.50 32.50

2249 Table Lamp 317 35.50 48.00

2518 Brass Sculpture 253 11.00. 22.00

SALE

DATE CUST ID SALPERS ID PROD ID QTY

02/28 100 10 2241 200

02/12 101 23 2518 300

02/15 101 23 1035 150

02/19 100 39 2518 200

02/02 101 23 1035 200

02/05 105 10 2241 100

02/22 110 37 2518 150

02/14 105 10 2249 50

02/01 101 23 2249 75

02/04 101 23 2241 250

• Any problems?

• Any obvious application for a product operation?

• How many columns and rows would PS have?

PROD ID

PROD DESC MANUFACTR ID

COST PRICE

1035 Sweater 210 12.50 20.00

1035 Sweater 210 12.50 20.00

1035 Sweater 210 12.50 20.00

2241 Table Lamp 317 22.50 32.50

2241 Table Lamp 317 22.50 32.50

2241 Table Lamp 317 22.50 32.50

DATE CUST ID SALPERS ID

PROD ID

QTY

02/28 100 10 2241 200

02/12 101 23 2518 300

02/15 101 23 1035 150

02/28 100 10 2241 200

02/12 101 23 2518 300

02/15 101 23 1035 150

Select

• Query:– Give all information for salespeople in the Tokyo office.

• SALES_TOKYO := SELECT (SALESPERSON: OFFICE = ‘TOKYO’)

• Creates a relation from another relation by selecting only those rows that satisfy a given condition

SALES_TOKYO

SALPERS ID





• Essentially an ‘IF’ statement

• Other examples of a SELECT– SALPERS_ID = 23– SALPERS_NAME = ‘Brigit Bovary’– MANAGER_ID >= 20– OFFICE not = ‘B.A.’– COMM_% < 11

• Query:– Which salesperson has ID 23?

SALPERS ID



Solution: SELECT (SALESPERSON: SALPERS_ID = 23)

• Query:– Give all information about salesperson Brigit Bovary

SALPERS ID



Solution: SELECT (SALESPERSON: SALPERS_NAME = ‘Brigit Bovary’)

• Query:– Who are the sales people working for managers having an ID greater

than 20?

SALPERS ID









Solution: SELECT (SALESPERSON:MANAGER_ID > 20)

• Query:– Give information on all salespeople except those in the Buenos Aries

office.

SALPERS ID









Solution: SELECT (SALESPERSON:OFFICE not = ‘B.A.’)

• Query:– Who are the salespeople in Tokyo getting more thn 10% commission?

SALPERS ID




Solution: SELECT (SALESPERSON:OFFICE = ‘Tokyo’ and COMM_% > 10)

• Query:– Who is reporting to manager 27 or getting over 10% commission?

SALPERS ID









Solution: SELECT (SALESPERSON:MANAGER_ID = 27 or COMM_% > 10)

Project

• A projection

• Used to eliminate unwanted columns

SALES_TOKYO

SALPERS ID





• Who are the salespeople in the Tokyo office?

Solution: SALES_TOKYO [SALPERS_NAME]

SALPERS NAME

Masaji Matsu

Goro Azuma

Albert Ige

• You can project any number of columns

E.g. SALES_TOKYO [SALESPERS_ID, SALPERS_NAME, MANAGER_ID]

SALPERS ID

SALPERS NAME MANAGER ID

14 Masaji Matsu 44

39 Goro Azuma 44

44 Albert Ige 27

Another Example: SALESPERSON [COMM_%]

COMM %

10

11

9

13

15

12

• What is noticeable in this new relation?

• Query:– Which salespeople are getting less than 11% commission?

SALPERS NAME

Rodney Jones

Francois Moire

Goro Azuma

Buster Sanchez

Solution: SELECT (SALESPERSON:COMM% < 11) [SALPERS_NAME]

Join

• The most important function in any database language

• Natural join– Connects relations when common columns have equal values

• Used to connect data across a number of relations

CUSTOMER

CUST ID

CUST NAME ADDRESS COUNTRY BEGINNING

BALANCE

CURRENT

BALANCE

100 Watabe Bros Box 241, Tokyo Japan 45,551 52,113

101 Maltzl Salzburg Austria 75,314 77,200

105 Jefferson Box 918, Chicago USA 49,333 57,811

110 Gomez Santiago Chile 27,400 35,414

An example:

Logical connections between SALESPERSON, PRODUCT, and CUSTOMER

•Query:–What are the names of the customers who have made purchases from salesperson 10?

SALE


02/28 100 10 2241 200

02/12 101 23 2518 300

02/15 101 23 1035 150

02/19 100 39 2518 200

02/02 101 23 1035 200

02/05 105 10 2241 100

02/22 110 37 2518 150

02/14 105 10 2249 50

02/01 101 23 2249 75

02/04 101 23 2241 250

•The SALE relation

Solution:1. Select from SALES those sales belonging to salesperson 10

SALE _SP10


02/28 100 10 2241 200

02/05 105 10 2241 100

02/14 105 10 2249 50

2. Join SALE_SP10 and CUSTOMER

Syntax JOIN (SALE_SP10, CUSTOMER)

• The operation proceeds as follows

1. The product of SALE_SP10 and CUSTOMER is created

SALE _SP10


02/28 100 10 2241 200

02/05 105 10 2241 100

02/14 105 10 2249 50

CUSTOMER

CUST ID

CUST NAME ADDRESS COUNTRY BEGINNING

BALANCE

CURRENT

BALANCE





• Want SALE_SP10 CUSTOMER

DATE CUST ID

SALPERS ID

PROD ID

QTY

02/28 100 10 2241 200

02/28 100 10 2241 200

02/28 100 10 2241 200

02/28 100 10 2241 200

02/05 105 10 2241 100

02/05 105 10 2241 100

02/05 105 10 2241 100

02/05 105 10 2241 100

02/14 105 10 2249 50

02/14 105 10 2249 50

02/14 105 10 2249 50

02/14 105 10 2249 50

CUST ID

CUST NAME

ADDRESS COUNTRY BEGINNING

BALANCE

CURRENT

BALANCE













The product of SALE_SP10 and CUSTOMER

DATE CUST ID

SALPERS ID

PROD ID

QTY

02/28 100 10 2241 200

02/28 100 10 2241 200

02/28 100 10 2241 200

02/28 100 10 2241 200

02/05 105 10 2241 100

02/05 105 10 2241 100

02/05 105 10 2241 100

02/05 105 10 2241 100

02/14 105 10 2249 50

02/14 105 10 2249 50

02/14 105 10 2249 50

02/14 105 10 2249 50

CUST ID

CUST NAME


BALANCE

CURRENT

BALANCE













Notice that the customer id’s do not match. Thus these are invalid rows and must be deleted. This leads to step 2.

2. Eliminate all rows except those in which CUST_ID from SALE_SP10 is equal to CUST_ID from CUSTOMER

DATE CUST ID

SALPERS ID

PROD ID

QTY

02/28 100 10 2241 200

02/05 105 10 2241 100

02/14 105 10 2249 50

CUST ID

CUST NAME


BALANCE

CURRENT

BALANCE




3. Eliminate one of the CUST_ID columns from this new relation

DATE CUST ID

SALPERS ID

PROD ID

QTY

02/28 100 10 2241 200

02/05 105 10 2241 100

02/14 105 10 2249 50

CUST NAME


BALANCE

CURRENT

BALANCE

Watabe Bros Box 241, Tokyo Japan 45,551 52,113

Jefferson Box 918, Chicago USA 49,333 57,811

Jefferson Box 918, Chicago USA 49,333 57,811

1. Take the product of A and B. The resulting relation will have two columns for each of C1 ….. Cn

2. Eliminate all rows from the product except those on which the values of the columns C1 ….. Cn in A are equal, respectively, to the values of those in columns in B.

3. Eliminate duplicate columns.

4. Project out one copy of the columns C1 ….. Cn

• General definition of a natural join of two relations A and B which have columns C1 ….. Cn in common.

•Query:–What is the name of the customer involved in each sale?

Solution: A := CUSTOMER [CUST_ID, CUST_NAME]

B := JOIN (SALE, A)

B

DATE SALPERS ID PROD ID QTY CUST ID CUST NAME

02/28 10 2241 200 100 Watabe Bros

02/12 23 2518 300 101 Maltzl

02/15 23 1035 150 101 Maltzl

02/19 39 2518 200 100 Watabe Bros

02/02 23 1035 200 101 Maltzl

02/05 10 2241 100 105 Jefferson

02/22 37 2518 150 110 Gomez

02/14 10 2249 50 105 Jefferson

02/01 23 2249 75 101 Maltzl

02/04 23 2241 250 101 Maltzl

•Query:

Gives the names of customers who have purchased product 2518.

Solution: A := SELECT (SALE: PROD_ID = 2518)

B := JOIN (A, CUSTOMER) [CUST_NAME]

B

CUST NAME

Maltzl

Watabe Bros

Gomez

•Query:

Who has bought table lamps?

Solution:

A := SELECT (PRODUCT: PROD_DESC = ‘Table Lamp’)

B := JOIN (A, SALE)

C := JOIN (B, CUSTOMER) [CUST_NAME]

C

CUST NAME

Watabe Bros

Jefferson

Maltzl

MANUFACTURER

MANUFACTR ID MANUFACTR NAME

ADDRESS COUNTRY

210 Kiwi Klothes Auckland New Zealand

253 Brass Works Lagos Nigeria

317 Llama Llamps Lima Peru

• Add another table: The manufacturer of the products

•Query:

Which salespeople have sold products manufactured in Peru?

Solution:

A := SELECT (MANUFACTURER: COUNTRY = ‘Peru’)

B := JOIN (A, PRODUCT)

C := JOIN (B, SALE)

D := JOIN (C, SALESPERSON) [SALPERS_NAME]

D

SALPERS NAME

Rodney Jones

Francois Moire

Theta Join

– Query: Identify salespeople whose manager gets a commission rate exceeding 11%.

• Consider the following:

SALESPERSON

SALPERS ID











• Cannot be solved by a simple select operation

• Have to join a relation to itself. How?

• Have to match SALPERS_ID to MANAGER_ID

• Create two copies of SALESPERSON– SP1 := SALESPERSON– SP2 := SALESPERSON

• Then:A := JOIN (SP1, SP2: SP1.MANAGER_ID = SP2.SALESPERS_ID)

• Have a new version of a JOIN– The Theta Join

• The result

SP1.

SALPERS ID

SP1.SALPERS NAME

SP1.

MANAGER ID

SP1.

OFFICE

SP1.

COMM %









SP2.

SALPERS ID

SP2.SALPERS NAME

SP2.

MANAGER ID

SP2.

OFFICE

SP2.

COMM %









A

• Whose manager gets a commission rate exceeding 11%?B := SELECT (A: SP2.COMM_% > 11) [SP1.SALPERS_NAME]

• Not yet finished

B

SP1.SALPERS NAME

Rodney Jones

Masaji Matsu

Goro Azuma

Albert Ige

Brigit Bovary

Buster Sanchez

• The comparison operators:=, not =, <, >, <=, >=

• General form:– JOIN (A, B: X δ Y)

• The Theta Join is a join with a specified condition involving a column from each relation

Divide

• Query:– List salespeople who have sold every product

• Thus for each product there must be at least one row in SALE containing the SALE_ID of that salesperson

• Selects rows in one relation that match every row in another relation

• Different requirement here

• Have to look at the whole relation

• Key word is every

• Using the example relations

• Obtain a relation consisting of all product ids– PI := PRODUCT [PROD_ID]

• Need instances of salesperson and product– PISI := SALE [PROD_ID, SALPERS_ID]

• Result: Two projections

PI

PROD ID

1035

2241

2249

2518

PISI

SALPERS

ID

PROD ID

10 2241

23 2518

23 1035

39 2518

37 2518

10 2249

23 2249

23 2241

• A := PISI / PI

• What do we want from these relations now?

A

SALPERS

ID

23

• General Description of a Divide

• Assume A, B, and C are relations where B/C = A

1. Columns of C must be a subset of B

2. The columns of A are all and only those columns of B that are not columns of C

Assignment

• Have been using it all the time

A := SELECT (SALESPERSON: COMM_% > 11)

• An operation that gives a name to a relation

RELATIONAL ALGEBRA SECTION 5 An Introduction. Introduction E.F. Codd in 1970 A revolutionary advance...

Documents

Transcript of RELATIONAL ALGEBRA SECTION 5 An Introduction. Introduction E.F. Codd in 1970 A revolutionary advance...