Relational Algebra brief lecture notes for SQL.ppt

CSCD343- Introduction to databases- A. Vaisman
Relational Algebra

Relational Query Languages
 Query languages: Allow manipulation and retrieval
of data from a database.
 Relational model supports simple, powerful QLs:
 Strong formal foundation based on logic.
 Allows for much optimization.
 Query Languages != programming languages!
 QLs not expected to be “Turing complete”.
 QLs not intended to be used for complex calculations.
 QLs support easy, efficient access to large data sets.

Formal Relational Query Languages
 Two mathematical Query Languages form
the basis for “real” languages (e.g. SQL), and
for implementation:
 Relational Algebra: More operational(procedural),
very useful for representing execution plans.
 Relational Calculus: Lets users describe what they
want, rather than how to compute it. (Non-
operational, declarative.)

Preliminaries
 A query is applied to relation instances, and the
result of a query is also a relation instance.
 Schemas of input relations for a query are fixed (but
query will run regardless of instance!)
 The schema for the result of a given query is also
fixed! Determined by definition of query language
constructs.
 Positional vs. named-field notation:
 Positional notation easier for formal definitions,
named-field notation more readable.
 Both used in SQL

Example Instances
sid sname rating age
22 dustin 7 45.0
31 lubber 8 55.5
58 rusty 10 35.0
28 yuppy 9 35.0
31 lubber 8 55.5
44 guppy 5 35.0
58 rusty 10 35.0
sid bid day
22 101 10/10/96
58 103 11/12/96
R1
S1
S2
 “Sailors” and “Reserves”
relations for our examples.
“bid”= boats. “sid”: sailors
 We’ll use positional or
named field notation,
assume that names of fields
in query results are
`inherited’ from names of
fields in query input
relations.

Relational Algebra
 Basic operations:
 Selection ( ) Selects a subset of rows from relation.
 Projection ( ) Deletes unwanted columns from relation.
 Cross-product ( ) Allows us to combine two relations.
 Set-difference ( ) Tuples in reln. 1, but not in reln. 2.
 Union ( ) Tuples in reln. 1 and in reln. 2.
 Additional operations:
 Intersection, join, division, renaming: Not essential, but (very!) useful.
 Since each operation returns a relation, operations can be
composed! (Algebra is “closed”.)






Projection
sname rating
yuppy 9
lubber 8
guppy 5
rusty 10
sname rating
S
,
( )
2
age
35.0
55.5
age S
( )
2
 Deletes attributes that are not in
projection list.
 Schema of result contains exactly
the fields in the projection list, with
the same names that they had in
the (only) input relation.
 Projection operator has to
eliminate duplicates! (Why??, what
are the consequences?)
 Note: real systems typically
don’t do duplicate elimination
unless the user explicitly asks
for it. (Why not?)

Selection
rating
S
8
2
( )
28 yuppy 9 35.0
58 rusty 10 35.0
sname rating
yuppy 9
rusty 10
 
sname rating rating
S
,
( ( ))
8
2
 Selects rows that satisfy
selection condition.
 Schema of result
identical to schema of
(only) input relation.
 Result relation can be
the input for another
relational algebra
operation! (Operator
composition.)

CSCD343- Introduction to databases- A. Vaisman 1
Union, Intersection, 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.
 What is the schema of result?
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
31 lubber 8 55.5
58 rusty 10 35.0
S S
1 2

S S
1 2

22 dustin 7 45.0
S S
1 2


Cross-Product
 Each row of S1 is paired with each row of R1.
 Result schema has one field per field of S1 and R1,
with field names `inherited’ if possible.
 Conflict: Both S1 and R1 have a field called sid.
 ( ( , ), )
C sid sid S R
1 1 5 2 1 1
  
(sid) sname rating age (sid) bid day
22 dustin 7 45.0 22 101 10/10/96
22 dustin 7 45.0 58 103 11/12/96
31 lubber 8 55.5 22 101 10/10/96
31 lubber 8 55.5 58 103 11/12/96
58 rusty 10 35.0 22 101 10/10/96
58 rusty 10 35.0 58 103 11/12/96
 Renaming operator:

Joins
 Condition Join:
 Result schema same as that of cross-product.
 Fewer tuples than cross-product. Filters tuples not
satisfying the join condition.
 Sometimes called a theta-join.
R c S c R S
  
 ( )
(sid) sname rating age (sid) bid day
22 dustin 7 45.0 58 103 11/12/96
31 lubber 8 55.5 58 103 11/12/96
S R
S sid R sid
1 1
1 1

. .


Joins
 Equi-Join: A special case of condition join where
the condition c contains only equalities.
 Result schema similar to cross-product, but only
one copy of fields for which equality is specified.
 Natural Join: Equijoin on all common fields.
sid sname rating age bid day
22 dustin 7 45.0 101 10/10/96
58 rusty 10 35.0 103 11/12/96
)
1
1
(
,..
,
,..,
R
S
sid
bid
age
sid




Division
 Not supported as a primitive operator, but useful for
expressing queries like:
Find sailors who have reserved all
boats.
 Precondition: in A/B, the attributes in B must be included
in the schema for A. Also, the result has attributes A-B.
 SALES(supId, prodId);
 PRODUCTS(prodId);
 Relations SALES and PRODUCTS must be built using
projections.
 SALES/PRODUCTS: the ids of the suppliers supplying ALL
products.

Examples of Division A/B
sno pno
s1 p1
s1 p2
s1 p3
s1 p4
s2 p1
s2 p2
s3 p2
s4 p2
s4 p4
pno
p2
pno
p2
p4
pno
p1
p2
p4
sno
s1
s2
s3
s4
sno
s1
s4
sno
s1
A
B1
B2
B3
A/B1 A/B2 A/B3

Expressing A/B Using Basic Operators
 Division is not essential op; just a useful shorthand.
 (Also true of joins, but joins are so common that systems
implement joins specially. Division is NOT implemented
in SQL).
 Idea: For SALES/PRODUCTS, compute all products
such that there exists at least one supplier not
supplying it.
 x value is disqualified if by attaching y value from B, we
obtain an xy tuple that is not in A.
)
)
Pr
)
(
(( Sales
oducts
Sales
sid
sid
A 

 

The answer is sid(Sales) - A

Find names of sailors who’ve reserved boat #103
 Solution 1:  
sname bid
serves Sailors
(( Re ) )

103

 Solution 2:  
( , Re )
Temp serves
bid
1
103

 ( , )
Temp Temp Sailors
2 1 
 sname Temp
( )
2
 Solution 3:  
sname bid
serves Sailors
( (Re ))

103


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 red
Boats serves Sailors
((
' '
) Re )

 
 A more efficient solution:
   
sname sid bid color red
Boats 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 green
Boats
  
 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 red
Boats serves


 sname Tempred Tempgreen Sailors
(( ) )
 
  
( , ((
' '
) Re ))
Tempgreen
sid 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 ) / ( ))
Tempsids
sid bid
serves
bid
Boats
 sname Tempsids Sailors
( )

 To find sailors who’ve reserved all ‘Interlake’ boats:
/ (
' '
)
 
bid bname Interlake
Boats

.....

Summary
 The relational model has rigorously defined
query languages that are simple and
powerful.
 Relational algebra is more operational; useful
as internal representation for query
evaluation plans.
 Several ways of expressing a given query; a
query optimizer should choose the most
efficient version.

Relational Algebra brief lecture notes for SQL.ppt

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