Transcript MYCH4

Relational Algebra
Chapter 4
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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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 algebra/logic.
Allows for much optimization.
Query Languages != programming languages!



QLs not expected to be “Turing complete”.
QLs not intended to be used for calculations.
QLs support easy, efficient access to large data sets.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Formal Relational Query Languages

Two mathematical Query Languages form
the basis for “real” languages (e.g. SQL), and
for implementation:
 Relational Algebra: More operational, very useful
for representing execution plans.
 Relational Calculus: Lets users describe what they
want, rather than how to compute it. (Nonoperational, declarative.) We’ll skip this for now.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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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
The schema for the result of a given query is also
fixed! Determined by definition of query language
constructs.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Preliminaries

Positional vs. named-attribute notation:

Positional notation
•
•

Named-attribute notation
•
•

Ex: Sailor(1,2,3,4)
easier for formal definitions
Ex: Sailor(sid, sname, rating,age)
more readable
Advantages/disadvantages of one over the
other?
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Example Instances


R1 sid
22
58
“Sailors” and “Reserves”
sid
S1
relations for our examples.
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We’ll use positional or
named field notation,
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assume that names of fields
58
in query results are
`inherited’ from names of
S2 sid
fields in query input
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relations.
31
44
58
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
bid
day
101 10/10/96
103 11/12/96
sname rating age
dustin
7
45.0
lubber
8
55.5
rusty
10 35.0
sname rating age
yuppy
9
35.0
lubber
8
55.5
guppy
5
35.0
rusty
10 35.0
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Algebra
In math, algebraic operations like +, -, x, /.
 Operate on numbers: input are numbers,
output are numbers.
 Can also do Boolean algebra on sets, using
union, intersect, difference.
 Focus on algebraic identities, e.g.

 x (y+z) = xy + xz.

(Relational algebra lies between
propositional and 1st-order logic.)
Database 
Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Relational Algebra
Every operator takes one or two relation
instances
 A relational algebra expression is recursively
defined to be a relation

 A combination of relations is a relation
 Result is also a relation
 Can apply operator to
• Relation from database
• Relation as a result of another operator
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Relational Algebra Operations

Basic operations:








Additional 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.
Intersection, join, division, renaming: Not essential, but
(very!) useful.
Since each operation returns a relation, operations
can be composed! (Algebra is “closed”.)
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Projection



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??)
 Note: real systems typically
don’t do duplicate elimination
unless the user explicitly asks
for it. (Why not?)
sname
rating
yuppy
lubber
guppy
rusty
9
8
5
10
 sname,rating(S2)
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
age
35.0
55.5
 age(S2)
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Selection




Selects rows that satisfy selection
condition.
No duplicates in result! (Why?)
Schema of result identical to
schema of (only) input relation.
Selection conditions:
 simple conditions comparing
attribute values (variables)
and / or constants or
 complex conditions that
combine simple conditions
using logical connectives
AND and OR.
sid sname rating age
28 yuppy 9
35.0
58 rusty
10
35.0
 rating 8(S2)
sname rating
yuppy 9
rusty
10
 sname,rating( rating 8(S2))
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Union, Intersection, Set-Difference
sid sname rating age


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?
sid sname
22 dustin
rating age
7
45.0
22
31
58
44
28
dustin
lubber
rusty
guppy
yuppy
7
8
10
5
9
45.0
55.5
35.0
35.0
35.0
S1 S2
sid sname
31 lubber
58 rusty
S1 S2
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rating age
8
55.5
10
35.0
S1 S2
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Exercise on Union
Num shape
ber
holes
1
round
2
2
3
square
4
rectangle 8
Num shape
ber
4
round
5
6
Blue blocks (BB)
bottom top
Stacked(S)
4
2
4
6
6
2
holes
2
square
4
rectangle 8
Yellow blocks(YB)
1. Which tables are unioncompatible?
2. What is the result of the
possible unions?
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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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.

(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:
 (C(1 sid1, 5  sid 2), S1 R1)
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Exercise on Cross-Product
Num shape
ber
holes
1
round
2
2
3
square
4
rectangle 8
Num shape
ber
4
round
5
6
holes
2
square
4
rectangle 8
Blue blocks (BB)
bottom top
Stacked(S)
4
2
4
6
6
2
1. Write down 2 tuples in
BB x S.
2. What is the cardinality
of BB x S?
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Joins

R  c S   c ( R  S)
Condition Join:
(sid)
22
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sname
dustin
lubber
rating age
7
45.0
8
55.5
S1




(sid)
58
58
S1.sid  R1.sid
bid
103
103
day
11/12/96
11/12/96
R1
Result schema same as that of cross-product.
Fewer tuples than cross-product, might be able to compute
more efficiently. How?
Sometimes called a theta-join.
Π-σ-x = SQL in a nutshell.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Exercise on Join
Num shape
ber
holes
1
round
2
2
3
square
4
rectangle 8
Num shape
ber
4
round
5
6
Blue blocks (BB)
BB
holes
2
square
4
rectangle 8
Yellow blocks(YB)
BB.holes YB.holes
YB
Write down 2 tuples in this join.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Joins

Equi-Join: A special case of condition join where
the condition c contains only equalities.
sid
22
58
S1><
sname
dustin
rusty
rating age
7
45.0
10
35.0
bid
101
103
day
10/10/96
11/12/96
R1
R.sid S.sid
 Result schema similar to cross-product, but only

one copy of fields for which equality is specified.
 Natural Join: Equijoin on all common fields.
Without specified, condition A>< B
means the natural join of A and B.

Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Example for Natural Join
Num shape
ber
1
round
2
3
holes
2
square
4
rectangle 8
Blue blocks (BB)
shape
holes
round
2
square
4
rectangle 8
Yellow blocks(YB)
What is the natural join of BB and YB?
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Find names of sailors who’ve reserved boat #103

Solution 1:

Solution 2:
 sname((
bid 103
 (Temp1, 
Reserves)  Sailors)
bid  103
Re serves)
 ( Temp2, Temp1  Sailors)
 sname (Temp2)

Solution 3:
 sname (
bid 103
(Re serves  Sailors))
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Find names of sailors who’ve reserved a red boat

Information about boat color only available in
Boats; so need an extra join:
 sname ((
Boats)  Re serves  Sailors)
color ' red '

A more efficient solution:
 sname ( (( 
Boats)  Re s)  Sailors)
sid bid color ' red '
A query optimizer can find this, given the first solution!
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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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  Re serves  Sailors)

Can also define Tempboats using union! (How?)

What happens if  is replaced by  in this query?
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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):
 (Tempred, 
sid
 (Tempgreen, 
((
sid
color ' red '
((
Boats)  Re serves))
color ' green'
Boats)  Re serves))
 sname((Tempred  Tempgreen)  Sailors)
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Division
Not supported as a primitive operator, but useful for
expressing queries like:
Find sailors who have reserved all boats.
 Typical set-up: A has 2 fields (x,y) that are foreign key
pointers, B has 1 matching field (y).
 Then A/B returns the set of x’s that match all y values
in B.
 Example: A = Friend(x,y). B = set of 354 students.
Then A/B returns the set of all x’s that are friends
with all 354 students.

Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Examples of Division A/B
sno
s1
s1
s1
s1
s2
s2
s3
s4
s4
pno
p1
p2
p3
p4
p1
p2
p2
p2
p4
A
pno
p2
p4
pno
p2
B1
B2
pno
p1
p2
p4
B3
sno
s1
s2
s3
s4
sno
s1
s4
sno
s1
A/B1
A/B2
A/B3
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Find the names of sailors who’ve reserved all boats

Uses division; schemas of the input relations
to / must be carefully chosen:
 (Tempsids, (
sid, bid
Re serves) / (
bid
Boats))
 sname (Tempsids  Sailors)

To find sailors who’ve reserved all ‘Interlake’ boats:
.....
/
bid
(
bname ' Interlake'
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
Boats)
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Division in General
Let A have 2 fields, x and y; B have only field y:
 A/B = {x: for all y in B. the tuple xy is in A}.
 i.e., A/B contains all x tuples (sailors) such that
for every y tuple (boat) in B, there is an xy tuple
in A.
 Or: If the set of y values (boats) associated with an
x value (sailor) in A contains all y values in B, the x
value is in A/B.
 In general, x and y can be any lists of fields; y is the
list of fields in B, and (x,y) is the list of fields of A.
 Then A/B returns the set of all x-tuples such that for
every y-tuple in B, the tuple (x,y) is in A.

Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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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.
 Book has lots of query examples.

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