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Relational algebra
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Relational algebra Notation
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Unary Operations
Selection course = ‘Computing’ StudentsIn SQL:Select *From StudentsWhere course = ‘Computing’;
Projection stud#, name Students
In SQL:Select stud#, nameFrom Students;
Selection & Projection stud#, name ( course = ‘Computing’ Students)
In SQL:Select stud#, nameFrom studentsWhere course = ‘Computing’;
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Binary Operations/Joins
Cartesian Product: Students X CoursesIn SQL: Select * From Students, Courses;
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Rename
• RENAME operator (): Renames the input relation and attributes with a new relation name & attributes specified.
S(B1, B2, …, BN) (R)
Example,TEMP NAME, MAJOR(STUDENT)
STUD_INFO (FULL_NAME,M_DEPT ) TEMP
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Renaming
TEMP DNO=5(EMPLOYEE)R(FIRSTNAME, LASTNAME, SALARY) FNAME, LNAME, SALARY (TEMP)
Example
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Union and Set-Difference
• All of these operations take two input relations, which must be union-compatible:• Same number of fields.• Corresponding’ fields have the same type.
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Set Operators
•Given two relations R1, R2 that are union-compatible, we have that R1 R2 returns the set of tuples that
are in R1 or R2. [UNION]∆R1 R2 returns the set of tuples that
are both in R1 and R2. [INTERSECTION]∆R1 - R2 returns the set of tuples that are
in R1, but not in R2. [SET DIFFERENCE]8
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Set Operators
Name (FACULY) Name (STUDENT) Address (FACULY) Address (STUDENT) CrsCode (CLASS) - CrsCode (TRANSCRIPT)
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Union
S S1 2
S1
S2
sid sname rating age
22 dustin 7 45.0
31 lubber 8 55.5 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
sid sname rating age
22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0 44 guppy 5 35.0 28 yuppy 9 35.0
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Set Difference
S1
S2
S S1 2
S2 – S1
sid sname rating age
22 dustin 7 45.0
31 lubber 8 55.5 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
sid sname rating age
22 dustin 7 45.0
sid sname rating age 28 yuppy 9 35.0 44 guppy 5 35.0
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Joins
Three new join operators are introduced:• Left Outer Join (denoted as )
• Right Outer Join (denoted as )
• Full Outer Join (denoted as )
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Join…
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Students ⋈ <stud# =200> Courses
In SQL: Select * From Students, CoursesWhere stud# = 200;
Left Outer Join
Left Outer Join : A <join condition> B
• ensures that all tuples in the in the relation A are present in the result set.
• The tuples in A without matching tuples in B are filled with null values for B’s attributes
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Left Outer Join - ExampleStudents Coursesstud# name course course# name100 Fred PH PH Pharmacy200 Dave CM CM Computing 400 Peter EN CH Chemistry
Students <course = course#> Courses
stud# Students.name course course# Courses.name
100 Fred PH PH Pharmacy200 Dave CM CM Computing400 Peter EN NULL NULL
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Right Outer Join
Right Outer Join: A <join condition> B
• Reverse of left outer join. Retrieves all tuples of B and null values for attributes of A in non-matching tuples of B
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Right Outer Join - ExampleStudents Coursesstud# name course course# name100 Fred PH PH Pharmacy200 Dave CM CM Computing 400 Peter EN CH Chemistry
Students <course = course#> Courses
stud# Students.name course course# Courses.name
100 Fred PH PH Pharmacy200 Dave CM CM ComputingNULL NULL NULL CH Chemistry
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Combination of Unary and Join Operations
Students Coursesstud# name address course course# name100 Fred Aberdeen PH PH Pharmacy200 Dave Dundee CM CM Computing 300 Bob Aberdeen CM
Show the names of students (from Aberdeen) and the names Show the names of students (from Aberdeen) and the names of their coursesof their courses
R1= Students ⋈ <course=course#> CoursesR2= <address=“Aberdeen”> R1R3= <Students.name, Course.name> R2
Students.name Courses.nameFred PharmacyBob Computing
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Full Outer Join
Full Outer Join: A <join condition> B
• ensures that all tuples of A and B are present in the result set
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Exercise 1
• Query 1: List customers whose cred_lim is greater than £500.• Query 2: List customers whose cred_lim is greater than £500
and lives in London.
Example: Customer
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Answers• Query 1: List customers whose cred_lim is greater than £500.
(cred_lim > 500)(customer)
• Query 2: List customers whose cred_lim is greater than £500 and lives
in London.(cred_lim>500) AND (city=London) (customer)
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Exercise 2 Reserves
SailorsBoats
sid bid day
22 101 10/10/11 58 103 11/12/11
sid sname rating age
22 Jesly 7 45.0
31 Mishail 8 55.5 58 Raj 10 35.0
bid bname color 101 Interlake Blue 102 Interlake Red 103 Clipper Green 104 Marine Red
1.Find names of sailors who’ve reserved boat #1032.Find names of sailors who’ve reserved a red boat3.Find sailors who’ve reserved a red or a green boat4.Find sailors who’ve reserved a red and a green boat5. Find the names of sailors who’ve reserved all boats
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1.Find names of sailors who’ve reserved boat #103
• Solution 1: sname bidserves Sailors(( Re ) )
103
• Solution 2: sname bidserves Sailors( (Re ))
103
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2.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'red '
Boats))Res)Sailors)
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3.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:
( , (' ' ' '
))Tempboatscolor red color green
Boats
sname Tempboats serves Sailors( Re )
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4.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 ))Tempredsid color red
Boats serves
sname Tempred Tempgreen Sailors(( ) )
( , ((' '
) Re ))Tempgreensid color green
Boats serves
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5. 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
Boats
sname Tempsids Sailors( )
To find sailors who’ve reserved all ‘Interlake’ boats:
/ (' '
) bid bname Interlake
Boats
.....
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Aggregate Functions and Operations• Aggregation function takes a collection of values and returns a
single value as a result.avg: average valuemin: minimum valuemax: maximum valuesum: sum of valuescount: number of values
• Aggregate operation in relational algebra G1, G2, …, Gn g F1( A1), F2( A2),…, Fn( An) (E)
• E is any relational-algebra expression• G1, G2 …, Gn is a list of attributes on which to group (can be
empty)• Each Fi is an aggregate function• Each Ai is an attribute name
Aggregate Operation – Example
• Relation account grouped by branch-name:
branch-name g sum(balance) (account)
branch-name account-number balance
PerryridgePerryridgeBrightonBrightonRedwood
A-102A-201A-217A-215A-222
400900750750700
branch-name balance
PerryridgeBrightonRedwood
13001500700
Aggregate Functions • Result of aggregation does not have a name• Can use rename operation to give it a name• For convenience, we permit renaming as part of
aggregate operation
branch-name g sum(balance) as sum-balance (account)
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