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Relational Algebra

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After this lecture, you should be able to:

Understand the differences between SQL (Structured Query Language) and Relational Algebra expressions.

Build queries in Relational Algebra.

Understand how DBMS’s process the SQL queries.

Make simple query optimization.

Be ready for Quiz 4 and Complete Assignment 7 (Part I).

- Relational algebra deals with a set of relationsclosed under operators. Operators operate on one or more relations to yield a relation.
- Algebraic Expressions
(a + b) * c

(A B) C

- Relational algebra is a basis for database query languages.

- Operational
- The order of operations can be specified.

- Set-at-a-time
- Multiple rows are processed by one operation.

- Not too detailed
- How to access data is not specified.
- Good for query optimization.

- Who supplied P2?
- SQL
Select SName from S, SP

where S.S# = SP.S# and P# = 'P2' ;

- Relational Algebra
SName (S ⋈ (p#=‘P2‘ SP ))

- SQL

- Union: A B
- Intersection: A B
- Difference: A ̶ B
- Cartesian Product: S × SP
- Division: A / B
- Selection: city=‘Rome‘ S
- Projection: SName,StatusS
- Join: S ⋈ SP
- Renamingρ(citypcity) P

Relation ~ Table

Attribute ~ Column

Tuple ~ Row (Record)

Relation schema ~ Structure of table

Relation Instance ~ Data of table

P

Q

P Q

P Q

P

Q

P Q = ?

P Q = ?

Q

P

P ̶ Q

Q

P

P ̶ Q = ?

Q

P

P × Q = {(p, q)|p P and q Q}

Q

P

P × Q = ?

R

B = 2 R

Table S

sno | sname | status | city

-----------------------------

s1 | Smith | 20 | London

s2 | Jones | 10 | Paris

s3 | Blake | 30 | Paris

s4 | Clark | 20 | London

s5 | Adams | 30 | Athens

select *

from s

where city = ‘Paris’;

city = ‘Paris‘ S

R

C, AR

Table S

sno | sname | status | city

-----------------------------

s1 | Smith | 20 | London

s2 | Jones | 10 | Paris

s3 | Blake | 30 | Paris

sno | status

------------

s1 | 20

s2 | 10

s3 | 30

select sno, status

from s;

sno, statusS

Table S

sno | sname | status | city

-----------------------------

s1 | Smith | 20 | London

s2 | Jones | 10 | Paris

s3 | Blake | 30 | Paris

s4 | Clark | 20 | London

s5 | Adams | 30 | Athens

sno | status

------------

s2 | 10

s3 | 30

select sno, status

from s

where city = 'Paris;

sno, status(city = ‘Paris‘ S)

Get the names and weights of the green parts.

pno pname color weight city

---- ------ ------- ------ ------

p1 nut red 12 London

p2 bolt green 15 Paris

p3 screw green 17 Rome

pname weight

----- ------

bolt 15

screw 17

select pname, weight

from p

where color = ‘green’;

pname, weight(color = ‘green‘ P)

P

Q

B1 ≥ B2(P × Q)

Q

P

B1 = B2(P × Q)

P

Q

P ⋈ Q

S

SP

S ⋈ SP = ?

What are the part numbers of the parts supplied by the

suppliers located in Taipei?

SQL:

select SP.pno

from S, SP

where S.city = ‘Taipei’

and S.sno = SP.sno;

Algebra:

pno((S.City=‘Taipei’S)⋈ SP)

What are the names of the suppliers who supplied green

parts?

SQL:

select sname

from P, SP, S

where P.color = ‘green’

and P.pno = SP.pno and SP.sno = S.sno;

Algebra:

sname((pno(color=‘green‘P))⋈ SP ⋈ S)

Get the supplier numbers of the suppliers who did not

supply red parts.

- The supplier numbers of all the suppliers:
AS = s# (S)

- The supplier numbers of the suppliers who supplied some red parts:
RPS = s# (SP ⋈ (Color = ‘Red‘ (P))

- The answer is:
AS - RPS

Get the names of the suppliers who supplied only red

parts.

- The supplier numbers of the suppliers who supplied some parts:
SS = s# (SP)

- The supplier numbers of the suppliers who supplied at least one non-red parts:
NRS = s# (SP ⋈ (color <> ‘Red‘ (P))

- The answer is:
SName ((SS – NRS) ⋈ S)

Ρ(B->B2, 3C2)

Get the pairs of the supplier numbers of the suppliers

who supply at least one identical part.

SQL:

select SPX.sno, SPY.sno

from SP SPX, SP SPY

where SPX.pno = SPY.pno

and SPX.sno < SPY.sno

Algebra:

snox,snoy(snox < snoy

((ρ(snosnox)sno,pnoSP)⋈

(ρ(snosnoy)sno,pnoSP)

)

)

What are the colors of the parts supplied by the suppliers

located in Taipei?

Table S Table P

sno | sname | sts | city pno | pname | color | wgt | city

-------------------------- ----------------------------------

s1 | Smith | 20 | London p1 | nut | red | 12 | London

s2 | Jones | 10 | Taipei p2 | bolt | green | 17 | Taipei

p3 | screw | blue | 17 | Rome

Table SP

sno | pno | qty

---------------

s1 | p1 | 300

s2 | p3 | 200

What are the colors of the parts supplied by the suppliers

located in Taipei?

SQL:

select color

From S, SP, P

where S.city = ‘Taipei’

and S.sno = SP.sno and SP.pno = P.pno;

Algebra:

color(S.City=‘Taipei‘(S ⋈ SP ⋈ P))

Joined also by S.city = P.city !

Not Correct:

color(S.City=‘Taipei‘(S ⋈ SP ⋈ P))

Renamed:

color(City=‘Taipei‘

(S ⋈ SP ⋈ρ(citypcity)P))

Optimized:

color((pno((sno(City=‘Taipei‘S))

⋈ SP))⋈ P)

color

⋈

pno

⋈

sno

city=‘Taipei‘S

P

S

SP

- An SQL query is transformed into an algebraic form for query optimization.
- Query optimization is the major task of an SQL query proccessor.
- select A1, A2, … , Am
from T1, T2, … , Tn

where C

is converted to

A1, A2, …, AmC (T1 x T2 x … x Tn)

and then optimized.

- JOIN operations are associative and commutative
- (R ⋈ S) ⋈ T = R ⋈ ( S ⋈ T)
- R ⋈ S = S ⋈ R

- JOIN and SELECT operations are associative and commutative if the SELECT operations are still applicable.
- PROJECTIONs can be performed to remove column values not used.
- These properties can be used for query optimization.
- The order of operations can be changed to produce smaller intermediate results.

What are the part numbers of the parts supplied by the

suppliers located in Taipei?

Not Optimized:

pno (S.City=‘Taipei‘(S ⋈ SP))

Improved:

pno ((S.City=‘Taipei‘S)⋈ SP)

Optimized:

pno((sno(S.City=‘Taipei‘S))⋈(sno,pnoSP))