1. RELATIONAL ALGEBRA (Chapter 2)
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2. Overall Organization Query Parser Query Optimizer Query Interpretor
Relational Algebra operators: , , , , , , , , 
Index structures
Abstraction of records
Buffer Pool Manager
File System
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3. RELATIONAL ALGEBRA INTRODUCTION
Assume the following two relations:
Emp (SS#, name, age, salary, dno)
Dept (dno, dname, floor, mgrSS#)
Relational algebra is a procedural query language, i.e., user must define both “how” and “what” to retrieve.
Relational algebra consists of a set of operators that consume either one or two relations as input. An operator produces one relation as its output.
Unary operators include: select, project, and rename
Binary operators include: cartesian product, equality join, natural join, join, semi-join, division, union, and set difference.
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4. < < SELECT OPERATOR бSalary>30,000 age<25(Employee) <
Select (б): selects tuples that satisfy a predicate; e.g., retrieve the employees whose salary is 30,000
бSalary=30,000(Employee)
Conjunctive ( ) and disjunctive ( ) selection predicates are allowed;
e.g., retrieve employees whose salary is higher than 30,000 and are younger than 25 years old:
бSalary>30, age<25(Employee)
Note that only selection predicates are allowed. A selection predicate is either (1) a comparison (=, ≠, ≤, ≥, <, >) between an attribute and a constant (e.g., salary = 30,000) or (2) a comparison between two different attributes of the same relation (e.g., salary = age × 100).
Note: This operator is different than the SELECT command of SQL.
<
<
<
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5. EXAMPLE Emp table: SS# Name Age Salary dno 1 Joe 24 20000 2 Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Emp table:
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6. EXAMPLE бSalary=30,000(Employee) Emp table: SS# Name Age Salary dno 1
Joe 24
20000 2
Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Emp table:
бSalary=30,000(Employee)
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7. EXAMPLE бSalary=30,000(Employee) Emp table: SS# Name Age Salary dno 1
Joe 24
20000 2
Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Emp table:
бSalary=30,000(Employee)
SS#
Name
Age
Salary
dno 4
Kathy 30
30000 5
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8. EXAMPLE бAge>22(Employee) Emp table: SS# Name Age Salary dno 1 Joe24
20000 2
Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Emp table:
бAge>22(Employee)
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9. EXAMPLE бAge>22(Employee) Emp table: SS# Name Age Salary dno 1 Joe24
20000 2
Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Emp table:
бAge>22(Employee)
SS#
Name
Age
Salary
dno 1
Joe 24
20000 2 4
Kathy 30
30000 5
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10. PROJECT OPERATOR
Project (∏) retrieves a column. It is a unary operator that eliminate duplicates.
e.g., name of employees:
∏ name(Employee)
e.g., name of employees earning more than 30,000:
∏ name(бSalary>30,000(Employee))
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11. EXAMPLE Emp table: SS# Name Age Salary dno 1 Joe 24 20000 2 Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Emp table:
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12. EXAMPLE Emp table: ∏ age(Emp) SS# Name Age Salary dno 1 Joe 24 20000 2
Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Emp table:
∏ age(Emp)
Age 24 20 22 30 4
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13. EXAMPLE Emp table: ∏ name,age(бSalary=4000 (Emp) ) SS# Name Age Salary
dno 1
Joe 24
20000 2
Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Emp table:
∏ name,age(бSalary=4000 (Emp) )
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14. EXAMPLE Emp table: ∏ name,age(бSalary=4000 (Emp) ) SS# Name Age Salary
dno 1
Joe 24
20000 2
Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Emp table:
∏ name,age(бSalary=4000 (Emp) )
SS#
Name
Age
Salary
dno 5
Shideh 4
4000 1
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15. EXAMPLE Emp table: ∏ name,age(бSalary=4000 (Emp) ) SS# Name Age Salary
dno 1
Joe 24
20000 2
Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Emp table:
∏ name,age(бSalary=4000 (Emp) )
Name
Age
Shideh 4
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16. CARTESIAN PRODUCT
Cartesian Product (R1 × R2) combines two relations by concatenating their tuples together, evaluating all possible combinations. If the name of a column is identical for two relations, this ambiguity is resolved by attaching the name of each relation to a column. e.g., Emp × Dept
(SS#, name, age, salary, Emp.dno, Dept.dno, dname, floor, mgrSS#)
If t(Emp) and t(Dept) is the cardinality of the Employee and Dept relations respectively, then the cardinality of Emp × Dept is: t(Emp) × t(Dept)
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17. CARTESIAN PRODUCT (Cont…)
Example:
Emp table:
Dept table:
SS#
Name
age
salary
dno
345
John Doe 23
25,000 1
943
Jane Java 25
28,000 2
876
Joe SQL 22
32,000
dno
dname
floor
mgrSS# 1
Toy
345 2
Shoe
943
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18. CARTESIAN PRODUCT (Cont…)
Cartesian product of Emp and Dept: Emp × Dept:
SS#
Name
age
salary
Emp.dno
Dept.dno
dname
floor
mgrSS#
345
John Doe 23
25,000 1
Toy
943
Jane Java 25
28,000 2
876
Joe SQL 22
32,000
Shoe
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19. CARTESIAN PRODUCT
Example: retrieve the name of employees that work in the toy department:
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20. CARTESIAN PRODUCT
Example: retrieve the name of employees that work in the toy department:
∏name(бEmp.dno=Dept.dno(Emp × бdname=‘toy’(Dept)))
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21. CARTESIAN PRODUCT (Cont…)
∏name(бdname=‘toy’ (б Emp.dno=Dept.dno(Emp × Dept)))
SS#
Name
age
salary
Emp.dno
Dept.dno
dname
floor
mgrSS#
345
John Doe 23
25,000 1
Toy
943
Jane Java 25
28,000 2
876
Joe SQL 22
32,000
Shoe
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22. CARTESIAN PRODUCT (Cont…)
∏name(бdname=‘toy’ (б Emp.dno=Dept.dno(Emp × Dept)))
SS#
Name
age
salary
Emp.dno
Dept.dno
dname
floor
mgrSS#
345
John Doe 23
25,000 1
Toy
876
Joe SQL 22
32,000
943
Jane Java 25
28,000 2
Shoe
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23. CARTESIAN PRODUCT (Cont…)
∏name(бdname=‘toy’ (б Emp.dno=Dept.dno(Emp × Dept)))
SS#
Name
age
salary
Emp.dno
Dept.dno
dname
floor
mgrSS#
345
John Doe 23
25,000 1
Toy
876
Joe SQL 22
32,000
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24. CARTESIAN PRODUCT (Cont…)
∏name(бdname=‘toy’ (б Emp.dno=Dept.dno(Emp × Dept)))
Name
John Doe
Joe SQL
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25. RENAME OPERATOR
Rename operator changes the name of its input table to its subscript,
ρe2(Emp)
Changes the name of Emp table to e2
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26. EQUALITY JOIN, NATURAL JOIN, JOIN, SEMI-JOIN
Equality join connects tuples from two relations that match on certain attributes. The specified joining columns are kept in the resulting relation.
∏name(бdname=‘toy’(Emp Dept)))
Natural join connects tuples from two relations that match on the specified common attributes
How is an equality join between Emp and Dept using dno different than a natural join between Emp and Dept using dno?
Equality join: SS#, name, age, salary, Emp.dno, Dept.dno, …
Natural join: SS#, name, age, salary, dno, dname, …
Join is similar to equality join using different comparison operators
A S op = {=, ≠, ≤, ≥, <, >}
att op att
(dno)
(dno)
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27. EXAMPLE JOIN Equality Join, (Emp Dept))) EMP Dept SS# Name Age Salary
dno 1
Joe 24
20000 2
Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Equality Join, (Emp Dept)))
dno
dname
floor
mgrss# 1
Toy 5 2
Shoe
EMP
Dept
(dno)
SS#
Name
Age
Salary
EMP.dno
Dept.dno
dname
floor
mgrss# 1
Joe 24
20000 2
Shoe
Mary 20
25000
Toy 5 3
Bob 22
27000 4
Kathy 30
30000
Shideh
4000
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28. EXAMPLE JOIN Natural Join, (Emp Dept))) EMP Dept SS# Name Age Salary
dno 1
Joe 24
20000 2
Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Natural Join, (Emp Dept)))
dno
dname
floor
mgrss# 1
Toy 5 2
Shoe
EMP
Dept
(dno)
SS#
Name
Age
Salary
dno
dname
floor
mgrss# 1
Joe 24
20000 2
Shoe
Mary 20
25000
Toy 5 3
Bob 22
27000 4
Kathy 30
30000
Shideh
4000
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29. EXAMPLE JOIN Join, (Emp ρx(Emp)))) EMP Dept SS# Name Age Salary dno 1
Joe 24
20000 2
Mary 20
25000 3
Bob 22
27000 4
Kathy 30
30000 5
Shideh
4000
Join, (Emp ρx(Emp))))
dno
dname
floor
mgrss# 1
Toy 5 2
Shoe
EMP
Dept
Salary > 5 * salary
SS#
Name
Age
Salary
dno
x.SS#
x.Name
x.Age
x.Salary
x.dno 2
Mary 20
25000 1
Shideh 4
4000 3
Bob 22
27000
Kathy 30
30000
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30. EQUALITY JOIN, NATURAL JOIN, JOIN, SEMI-JOIN (Cont…)
Example: retrieve the name of employees who earn more than Joe:
∏name(Emp (sal>x.sal)бname=‘Joe’(ρ x(Emp)))
Semi-Join selects the columns of one relation that joins with another. It is equivalent to a join followed by a projection:
Emp (dno)Dept ≡∏SS#, name, age, salary, dno(Emp Dept)
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31. UNION, SET DIFFERENCE & SET INTERSECT
Union puts all tuples of two relations in one relation. To use this operator, two conditions must hold:
The two relations must be of the same arity.
The domain of ith attribute of the two participating relation must be the same.
Set difference operator computes tuples that are in one relation, but not in another.
Set intersect operator computes tuples that are common in two relations:
The five fundamental operations of the relational algebra are: select, project, cartesian product, Union, and set difference
All other operators can be constructed using these operators
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32. EXAMPLE Assume a database with the following three relations:
Sailors (sid, sname, rating)
Boats (bid, bname, color)
Reserve (sid, bid, date)
Query 1: Find the bid of red colored boats:
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33. EXAMPLE Assume a database with the following three relations:
Sailors (sid, sname, rating)
Boats (bid, bname, color)
Reserve (sid, bid, date)
Query 1: Find the bid of red colored boats:
∏bid(бcolor=red(Boats))
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34. EXAMPLE Assume a database with the following three relations:
Sailors (sid, sname, rating)
Boats (bid, bname, color)
Reserve (sid, bid, date)
Query 1: Find the name of sailors who have reserved Boat number 2.
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35. EXAMPLE Assume a database with the following three relations:
Sailors (sid, sname, rating)
Boats (bid, bname, color)
Reserve (sid, bid, date)
Query 1: Find the name of sailors who have reserved Boat number 2.
∏sname(бbid=2(Sailors (sid)Reserve))
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36. EXAMPLE Assume a database with the following three relations:
Sailors (sid, sname, rating)
Boats (bid, bname, color)
Reserve (sid, bid, date)
Query 1: Find the name of sailors who have reserved both a red and a green boat.
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37. EXAMPLE Assume a database with the following three relations:
Sailors (sid, sname, rating)
Boats (bid, bname, color)
Reserve (sid, bid, date)
Query 1: Find the name of sailors who have reserved both a red and a green boat.
∏sid( бcolor=red(Boats) (Sailors Reserve) ) 
∏sid( бcolor=green(Boats) (Sailors Reserve) )
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38. EXAMPLE Assume a database with the following three relations:
Emp(SS#, name, salary, age, mgrss#)
Query 3: Find employees whose salary is higher than their manager’s salary.
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39. EXAMPLE Assume a database with the following relations:
Emp(SS#, name, salary, age, mgrss#)
Query 3: Find name of those employees whose salary is higher than their manager’s salary.
∏Emp.name( бemp.salary>mgr.salary(Emp ρmgr(Emp)) )
mgrSS#=SS#
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40. EXAMPLE SS# Name Age Salary mgrSS# 1 Joe 30 30000 Null 2 Bob 23 25000
Shideh 4
40000
Kathy 15
32000
∏Emp.name( бemp.salary>mgr.salary(Emp ρmgr(Emp)) )
mgrSS#=SS#
SS#
Name
Age
Salary
mgrSS#
Mgr.SS#
Mgr.Name
Mgr.Age
Mgr.Salary
Mgr.mgrSS# 2
Bob 23
25000 1
Joe 30
30000
Null 3
Shideh 4
40000
Kathy 15
32000
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41. EXAMPLE SS# Name Age Salary mgrSS# 1 Joe 30 30000 Null 2 Bob 23 25000
Shideh 4
40000
Kathy 15
32000
∏Emp.name( бemp.salary>mgr.salary(Emp ρmgr(Emp)) )
mgrSS#=SS#
SS#
Name
Age
Salary
mgrSS#
Mgr.SS#
Mgr.Name
Mgr.Age
Mgr.Salary
Mgr.mgrSS# 3
Shideh 4
40000 1
Joe 30
30000
Null
Kathy 15
32000
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42. EXAMPLE SS# Name Age Salary mgrSS# 1 Joe 30 30000 Null 2 Bob 23 25000
Shideh 4
40000
Kathy 15
32000
∏Emp.name( бemp.salary>mgr.salary(Emp ρmgr(Emp)) )
mgrSS#=SS#
Name
Shideh
Kathy
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43. OTHER ALGEBRAIC OPERATORS
Assignment: One may use the assignment operator to assign the relation produced by an algebraic expression to a new relation, e.g.,
ToyEmp  Emp бname=‘toy’ (Dept)
Deletion is performed using the assignment operator:
R  R – E
Example: Fire all employees whose salary is higher than 100,000
Emp  Emp – бsalary>100,000 (Emp)
Insertion is also performed using the assignment operator:
R  R  E
Example: Hire Joe
Emp  Emp  (5, Joe, 15, 30000, 1)
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44. OTHER ALGEBRAIC OPERATORS
Update operator changes the value of records based on its input expression
AE(R)
Example, Give all employees a 10% raise:
salary1.1*salary(Emp)
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45. A PROGRAM Recall Give the toy employees a 10% raise:
Emp(SS#,name,age,salary, dno)
Dept(dno,dname,floor,mgrss#)
Give the toy employees a 10% raise:
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