1. Relational Algebra
Friday, 11/14/2003

2. Outline Relational Algebra: 5.2, 5.3, 5.4 Reading assignment:
Extensible hash tables: ,
Linear hash tables: ,

3. Some History The Relational Data Model
Invented by Ted Codd, circa 1970
Main idea:
Data = mathematical relations
Queries = mathematical formulas (“abstract SQL”)
Elegant, clean
But practical ?

4. History Continued State of the art in databases in 1970s:
The COBOL/CODASYL approach
Data = a file of records, with pointers
No queries, only pointer navigation
Very efficient
But hard to write queries/programs

5. More History
For the relational model to prove itself it needed to show how to implement mathematical “queries” efficiently. Three ideas:
Intermediate language = relational algebra
Efficient implementation of each operator
Optimization
First proof of concept: System R (1979)

6. Relational Algebra Operates on relations, i.e. sets Five operators:
Later: we discuss how to extend this to bags
Five operators:
Union: 
Difference: -
Selection: s
Projection: P
Cartesian Product: 
Derived or auxiliary operators:
Intersection, complement
Joins (natural,equi-join, theta join, semi-join)
Renaming: r

7. 1. Union and 2. Difference R1  R2 Example: R1 – R2
ActiveEmployees  RetiredEmployees
R1 – R2
AllEmployees – RetiredEmployees

8. What about Intersection ?
It is a derived operator
R1  R2 = R1 – (R1 – R2)
Also expressed as a join (will see later)
Example
UnionizedEmployees  RetiredEmployees

9. 3. Selection Returns all tuples which satisfy a condition
Notation: sc(R)
Examples
sSalary > (Employee)
sname = “Smithh” (Employee)
The condition c can be =, <, , >, , <>

10. Find all employees with salary more than $40,000.
s Salary > (Employee)

11. 4. Projection Eliminates columns, then removes duplicates
Notation: P A1,…,An (R)
Example: project social-security number and names:
P SSN, Name (Employee)
Output schema: Answer(SSN, Name)

12. P SSN, Name (Employee)

13. 5. Cartesian Product Each tuple in R1 with each tuple in R2
Notation: R1  R2
Example:
Employee  Dependents
Very rare in practice; mainly used to express joins

15. Renaming Changes the schema, not the instance Notation: r B1,…,Bn (R)
Example:
rLastName, SocSocNo (Employee)
Output schema: Answer(LastName, SocSocNo)

16. LastName, SocSocNo (Employee)
Renaming Example
Employee
Name
SSN
John
Tony
LastName, SocSocNo (Employee)
LastName
SocSocNo
John
Tony

17. Natural Join Notation: R1 ⋈ R2 Meaning: R1 ⋈ R2 = PA(sC(R1  R2))
Where:
The selection sC checks equality of all common attributes
The projection eliminates the duplicate common attributes

18. Natural Join Example
Employee
Name
SSN
John
Tony
Dependents
SSN
Dname
Emily
Joe
Employee Dependents =
PName, SSN, Dname(s SSN=SSN2(Employee x rSSN2, Dname(Dependents))
Name
SSN
Dname
John
Emily
Tony
Joe

19. Natural Join
R= S=
R ⋈ S= A B X Y Z V B C Z U V W A B C X Z U V Y W

20. Natural Join
Given the schemas R(A, B, C, D), S(A, C, E), what is the schema of R ⋈ S ?
Given R(A, B, C), S(D, E), what is R ⋈ S ?
Given R(A, B), S(A, B), what is R ⋈ S ?

21. Theta Join A join that involves a predicate R1 ⋈ q R2 = s q (R1  R2)
Here q can be any condition: =, <, , , > 

22. Eq-join A theta join where q is an equality
R1 ⋈A=B R2 = s A=B (R1  R2)
Example:
Employee ⋈SSN=SSN Dependents
Most useful join in practice

23. Semijoin R ⋉ S = P A1,…,An (R ⋈ S)
Where A1, …, An are the attributes in R
Example:
Employee ⋉ Dependents

24. Semijoins in Distributed Databases
Semijoins are used in distributed databases
Dependents
Employee
SSN
Dname
Age
. . .
SSN
Name
. . .
network
Employee ⋈ssn=ssn (s age>71 (Dependents))
T = P SSN s age>71 (Dependents)
R = Employee ⋉ T
Answer = R ⋈ Dependents

25. Complex RA Expressions
P name
buyer-ssn=ssn
pid=pid
seller-ssn=ssn
P ssn
P pid
sname=fred
sname=gizmo
Person Purchase Person Product

26. Operations on Bags A bag = a set with repeated elements
Relational Engines work on bags, not sets !
All operations need to be defined carefully on bags
{a,b,b,c}{a,b,b,b,e,f,f}={a,a,b,b,b,b,b,c,e,f,f}
{a,b,b,b,c,c} – {b,c,c,c,d} = {a,b,b,d}
sC(R): preserve the number of occurrences
PA(R): no duplicate elimination
Cartesian product, join: no duplicate elimination

27. Logical Operators in the Bag Algebra
Union, intersection, difference
Selection s
Projection P
Join ⋈
Duplicate elimination d
Grouping g
Sorting t
Relational
Algebra (on bags)

28. Example SELECT city, count(*) FROM sales P city, c GROUP BY city
HAVING sum(price) > 100
P city, c
s p > 100
T(city,p,c)
g city, sum(price)→p, count(*) → c
sales

29. Physical Operators Query Plan: logical tree
SELECT S.buyer
FROM Purchase P, Person Q
WHERE P.buyer=Q.name AND
Q.city=‘seattle’ AND
Q.phone > ‘ ’
Purchase
Person
Buyer=name
City=‘seattle’
phone>’ ’
buyer
(Simple Nested Loops) 
(Table scan)
(Index scan)
Query Plan:
logical tree
implementation choice at every node
scheduling of operations.
Some operators are from relational
algebra, and others (e.g., scan)
are not.

30. Architecture of a Database Engine
SQL query
Parse Query
Logical plan
Select Logical Plan
Query optimization
Select Physical Plan
Physical plan
Query Execution

31. Question in Class Logical operator:
Product(pname, cname) ⋈ Company(cname, city)
Propose three physical operators for the join, assuming the tables are in main memory:

32. Question in Class Product(pname, cname) ⋈ Company(cname, city)
products
1000 companies
How much time do the following physical operators take if the data is in main memory ?
Nested loop join time =
Sort and merge = merge-join time =
Hash join time =

33. Cost Parameters
In database systems the data is on disks, not in main memory
The cost of an operation = total number of I/Os
Cost parameters:
B(R) = number of blocks for relation R
T(R) = number of tuples in relation R
V(R, a) = number of distinct values of attribute a

34. Cost Parameters Clustered table R: Unclustered table R:
Blocks consists only of records from this table
B(R)  T(R) / blockSize
Unclustered table R:
Its records are placed on blocks with other tables
When R is unclustered: B(R)  T(R)
When a is a key, V(R,a) = T(R)
When a is not a key, V(R,a)

35. Cost Cost of an operation = number of disk I/Os needed to:
read the operands
compute the result
Cost of writing the result to disk is not included on the following slides
Question: the cost of sorting a table with B blocks ?
Answer:

36. Scanning Tables The table is clustered: The table is unclustered
Table-scan: if we know where the blocks are
Index scan: if we have a sparse index to find the blocks
The table is unclustered
May need one read for each record

37. Sorting While Scanning
Sometimes it is useful to have the output sorted
Three ways to scan it sorted:
If there is a primary or secondary index on it, use it during scan
If it fits in memory, sort there
If not, use multi-way merge sort

38. Cost of the Scan Operator
Clustered relation:
Table scan:
Unsorted: B(R)
Sorted: 3B(R)
Index scan
Unsorted: B(R)
Sorted: B(R) or 3B(R)
Unclustered relation
Unsorted: T(R)
Sorted: T(R) + 2B(R)