Relational Algebra

transcript

Relational Algebra

Operations and Set operations

Relational Algebra

By relieving the brain of all unnecessary work, a good notation sets it free to

concentrate on more advanced problems, and, in effect,

increases the mental power of the race.

-- Alfred North Whitehead (1861 - 1947)

What is Relational Algebra

• It defines basic set of operations for manipulating relational data.

• It enables users for basic retrieval operations

• Result of any retrieval always gives another relation to manipulate

• It specifies relations to be performed on relations to derive result relations.

Relational algebra : Features• It is a procedural query language.

• It consists of a collection of operators, such as join, union, and intersect,that take relations as their operands and return relations as their result.

• It is the formal description of how a relational database operates.

• Relational algebra is the mathematics which underpin SQL operations.

• Operators in relational algebra are not necessarily the same as SQL operators, even if they have the same name.

The BIG picture

Its place in the big picture:

Declarativequery

language

Declarativequery

languageAlgebraAlgebra ImplementationImplementation

SQL,relational calculus

Relational algebra

RELATIONAL ALGEBRA OPERATORS

Specific Operators

Set Operators

Set Operators: Introduction

Relational algebra received little attention until the publication of E.F. Codd’s relational model of data in 1970. Codd proposed such an algebra as a basis for database query languages.

E.F. Codd’s Operators

Set operators from mathematical set theory

Intersection

Difference

Cartesian Product

Set Operators

Relation is a set of tuples, so set operations should apply: , , (set difference)

Result of combining two relations with a set operator is a relation

All its elements must be tuples having same structure

Hence, scope of set operations limited to union union compatible relations.compatible relations.

Union Compatible Relations

Two relations are union compatibleunion compatible ifBoth have same degree (number of columns)Names of attributes are the same in bothAttributes with the same name in both relations have the same domain (i.e.the data type of the attributes should be same like character,integer etc. )

Union compatible relations can be combined using unionunion, intersectionintersection, and setset difference.difference.

• Returns a relation consisting of all tuples appearing in either or both of two specified relations.

• Duplicates are eliminated• Both relations must be union compatible• It is denoted by symbol .

Example

S# SNAME STATUS CITY

S1 Smith 20 London

S4 Clark 20 London

S1 Smith 20 London

S4 Clark 20 London

S2 Jones 20 Paris

S1 Smith 20 London

S2 Jones 20 Paris

Intersection

• Returns a relation consisting of all tuples appearing in both of two specified relations.

• It is denoted by symbol .• Both relations must be union compatible.

Duplicates are eliminated

Example

S # SNAME STATUS CITY

S1 Smith 20 London

S4 Clark 20 London

S1 Smith 20 London

S4 Jones 20 Paris

A INTERSECTION B (A B)S # SNAME STATUS CITY

S1 Smith 20 London

Difference

Returns a relation consisting of all tuples appearing in the first and not the second of two specified relations.It is denoted by symbol .

Example

S1 Smith 20 London

S4 Jones 20 Paris

S1 Smith 20 London

S4 Clark 20 London

A DIFFERENCE B (A - B)S # SNAME STATUS CITY

S4 Clark 20 London

Cartesian ProductCross product or cross join

Result relation will have degree= sum of degrees of both source relations

Result relation will have tuples= product of tuples of both source relations

Returns a relation consisting of all possible tuples that are a combination of two tuples, one from each of two specified relations.It is denoted by the symbol X.

RR and SS need not be union compatible

Example

a x a x

b * y = a y

(A) (B) b y

(A X B)

Cross-Product : detailed example• Each row of R is paired with each row of S.• Result schema has one field per field of S and

R, with field names `inherited’ if possible.

E 5F 4D 3B 2A 1

A 1C 2D 3E 4

A 1 A 1C 2D 3E 4

B 2 A 1C 2D 3E 4A 1C 2D 3E 4

F 4 A 1C 2D 3E 4

E 5 A 1C 2D 3E 4

A 1A 1A 1

B 2B 2B 2

D 3D 3D 3

F 4F 4F 4

E 5E 5E 5

R CROSS S

• Union and Intersection are cumulative and associative operations therefore A U B = B U A

A B = B A

A U ( B U C ) = ( A U B ) U C

A ( B C) = ( A B ) C

– But difference operation is not thereforeA – B ≠ B - A

Operators Developed Specifically For Relational Databases

Project

Select

Project

Retains only wanted columns from relation.

Projection operator has to eliminate duplicates

Denoted by symbol π.

Produces table containing subset of columns of argument table

attribute list(relation)

Example

sid sname rating age 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0

sname rating

SelectSelects rows that satisfy specific conditions.No duplicates in result Result relation can be the input for another relational algebra operation (Operator composition.)Denoted by symbol σ Produce table containing subset of rows of argument table satisfying condition

condition (relation)Operators: <, , , >, =,

For multiple conditions use : and (^), or (V)

Example

sid sname rating age 28 yuppy 9 35.0 58 rusty 10 35.0

sid sname rating age 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0

rating

S82( )

Rename

• Used as alias name for result relation• Often used to assign simple name to the result

relation or attributes when it is to be used again in some other expression.

• Symbol used is ←• example:

(B400)←(bookid,bname)(price>400)(books)

• Now B400 can be used again• Example : (bookid)(B400)

Joins are compound operators involving cross product, selection, and (sometimes) projection.

Most common type of join is a “natural join” R S conceptually is:

Compute R X S

Project all unique attributes and one copy of each of the common ones.

Example

sid sname rating age

22 dustin 7 45.0

31 lubber 8 55.5 58 rusty 10 35.0

sid bid day

22 101 10/10/96 58 103 11/12/96

R1 S1 =sid sname rating age bid day

22 dustin 7 45.0 101 10/ 10/ 96 58 rusty 10 35.0 103 11/ 12/ 96

Division

• Not supported as a primitive operator, but useful for certain

• Let A have 2 fields, x and y; B have only field y:– A/B = – i.e., A/B contains all x tuples such that for every y

tuple in B, there is an xy tuple in A.

Examples of Division A/B

sno pnos1 p1s1 p2s1 p3s1 p4s2 p1s2 p2s3 p2s4 p2s4 p4

pnop2p4

pnop1p2p4

snos1s2s3s4

snos1s4

A/B1 A/B2 A/B3

Practical Examples

Example Instances

• “Sailors” and “Reserves” relations for our examples.

sid sname rating age

22 dustin 7 45.0

31 lubber 8 55.558 rusty 10 35.0

sid sname rating age28 yuppy 9 35.031 lubber 8 55.544 guppy 5 35.058 rusty 10 35.0

sid bid day

22 101 10/10/9658 103 11/12/96

Find names of sailors who’ve reserved boat #103

• Solution 1: sname bidserves Sailors(( Re ) )

Solution 2: ( , Re )Temp servesbid

( , )Temp Temp Sailors2 1

sname Temp( )2

Solution 3: sname bidserves Sailors( (Re ))

Find names of sailors who’ve reserved a red boat

• Information about boat color only available in Boats; so need an extra join:

sname color redBoats serves Sailors((

' ') Re )

A more efficient solution:

sname sid bid color redBoats s Sailors( ((

' ') Re ) )

A query optimizer can find this given the first solution!

Find sailors who’ve reserved a red or a green boat

• Can identify all red or green boats, then find sailors who’ve reserved one of these boats: ( , (

' ' ' '))Tempboats

color red color greenBoats

sname Tempboats serves Sailors( Re )

Can also define Tempboats using union! (How?)

What happens if is replaced by in this query?

Find sailors who’ve reserved a red and a green boat

• Previous approach won’t work! Must identify sailors who’ve reserved red boats, sailors who’ve reserved green boats, then find the intersection (note that sid is a key for Sailors): ( , ((

' ') Re ))Tempred

sid color redBoats serves

sname Tempred Tempgreen Sailors(( ) )

( , ((' '

) Re ))Tempgreensid color green

Boats serves

Find the names of sailors who’ve reserved all boats

• Uses division; schemas of the input relations to / must be carefully chosen:

( , (,

Re ) / ( ))Tempsidssid bid

servesbid

sname Tempsids Sailors( )

To find sailors who’ve reserved all ‘Interlake’ boats:

/ (' '

) bid bname Interlake

Boats.....

Relational Algebra

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