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Relational Algebra
Chapter 4, Part A
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
1
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.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
2
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. (Non-operational,
declarative.)
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
3
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.
Schema for the result
SELECT S.name, E.grade
FROM Students S, Enrolled E
WHERE S.sid = E.sid and S.age<20
Fixed
Schema of input: Students(SSN: CHAR(9), Name: CHAR(40), Age: INTEGER)
Enrolled(stuid CHAR(20), cid CHAR(20), grade CHAR(10))
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
4
Notation
SQL uses both positional and named-field
notations:

Positional notation – referring to fields by
their positions. Easier for formal definitions,

Named-field notation – using field names to
refer to fields. Make query more readable.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
5
Example Instances
R1 sid
“Sailors” and
“Reserves” relations for S1 sid
22
our examples.
31
 We’ll use positional or
58
named field notation,
assume that names of S2 sid
fields in query results
28
are `inherited’ from
31
names of fields in query
44
input relations.
58

Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
22
58
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
6
Relational Algebra

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”.)
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
7
Projection
sid
28
31
44
58
sname
rating
yuppy
lubber
guppy
rusty
9
8
5
10
sname rating age
yuppy
9
35.0
lubber
8
55.5
guppy
5
35.0
rusty
10 35.0
 sname,rating(S2)
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
8
Projection



sid
Deletes attributes that are not in
28
projection list.
31
44
Schema of result contains exactly
58
the fields in the projection list, with
the same names that they had in the
(only) input relation.
Projection operator has to eliminate
duplicates.
 Note: real systems typically don’t
do duplicate elimination unless
the user explicitly asks for it.
Duplicate
sname
yuppy
lubber
guppy
rusty
elimination
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
rating
9
8
5
10
age
35.0
55.5
35.0
35.0
 age(S2)
age
35.0
55.5
9
Selection
sid
28
31
44
58

sname rating age
yuppy
9
35.0
lubber
8
55.5
guppy
5
35.0
rusty
10 35.0
Selects rows that satisfy
selection condition.

Schema of result identical
to schema of (only) input
relation

No duplicates in result!
(Why?)
 rating 8(S2)
sid sname rating age
28 yuppy 9
35.0
58 rusty
10
35.0
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
10
Operator
Composition
Result relation can
be the input for
another relational
algebra operation !
sname rating
yuppy 9
rusty
10
S2 sid
28
31
44
58
sname rating age
yuppy
9
35.0
lubber
8
55.5
guppy
5
35.0
rusty
10 35.0
 rating 8(S2)
sid
28
58
sname
yuppy
rusty
rating
9
10
age
35.0
35.0
 sname,rating( rating 8(S2))
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
11
Union
S1
sid
22
31
58
sname rating age
dustin
7
45.0
lubber
8
55.5
rusty
10 35.0
S2
sid
28
31
44
58
sname rating age
yuppy
9
35.0
lubber
8
55.5
guppy
5
35.0
rusty
10 35.0
sid sname rating age
22
31
58
44
28
dustin
lubber
rusty
guppy
yuppy
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
7
8
10
5
9
45.0
55.5
35.0
35.0
35.0
S1 S2
12
Intersection
S1
sid
22
31
58
sname rating age
dustin
7
45.0
lubber
8
55.5
rusty
10 35.0
S2
sid
28
31
44
58
sname rating age
yuppy
9
35.0
lubber
8
55.5
guppy
5
35.0
rusty
10 35.0
Does not go to
the output
sid sname rating age
31 lubber 8
55.5
58 rusty
10
35.0
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
S1 S2
13
Set-Difference
S1
sid
22
31
58
sname rating age
dustin
7
45.0
lubber
8
55.5
rusty
10 35.0
sid sname
22 dustin
S1 S2
S2
sid
28
31
44
58
rating age
7
45.0
sname rating age
yuppy
9
35.0
lubber
8
55.5
guppy
5
35.0
rusty
10 35.0
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
No “22”
14
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.
22
31
58
44
28
sid sname
22 dustin
sid sname rating age
31 lubber 8
55.5
58 rusty
10
35.0
rating age
7
45.0
dustin
lubber
rusty
guppy
yuppy
7
8
10
5
9
45.0
55.5
35.0
35.0
35.0
S1 S2
S1 S2
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
S1 S2
15
Cross-Product
S1
sid
22
31
58
R1
sname rating age
dustin
7
45.0
lubber
8
55.5
rusty
10 35.0
R1 × S1
Each row of R1
is paired with
each row of S1.
sid bid
day
22 101 10/10/96
58 103 11/12/96
(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
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
16
Cross-Product
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
17
Joins
All
combinations
Condition Join: R  c S   c ( R  S)
Selected
combinations
Example: S1 
S1.sid  R1.sid
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
R1
18
Joins
S1
(sid)
22
58
sname
dustin
rusty
R1 × S1
Not in
join
results
S1.sid  R1.sid
rating
7
10
age
45.0
35.0
R1
(sid)
58
58
(sid) sname rating age
bid
103
103
day
11/12/96
11/12/96
(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
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
19
Joins
 Result schema
same as that of cross-
product.
 Fewer
tuples than cross-product,
might be able to compute more
efficiently
 Sometimes called a
theta-join.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
20
Equi-Join
Foreign key
S1
sid
22
31
58
sname rating age
dustin
7
45.0
lubber
8
55.5
rusty
10 35.0
R1
sid bid
day
22 101 10/10/96
58 103 11/12/96
Equi-Join: A special case of condition join
where the condition c contains only
equalities.
sid
22
58
sname
dustin
rusty
rating age
7
45.0
10
35.0
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
S1 
bid
101
103
sid
R1
day
10/10/96
11/12/96
21
Equi-Joins & Natural Join
 Equi-Join:
A special case of condition join where
the condition c contains only equalities.
sid
22
58
sname
dustin
rusty
rating age
7
45.0
10
35.0
S1 
sid
bid
101
103
day
10/10/96
11/12/96
R1
 Result
schema similar to cross-product, but only
one copy of fields for which equality is specified.
 Natural Join: Equijoin on all common fields.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
22
Joins
Condition Join: R  S   ( R  S)
c
c
S1 sid sname rating age
22
31
58
dustin
lubber
rusty
7
8
10
S1 
(sid)
22
31
sname
dustin
lubber
45.0
55.5
35.0
R1 sid bid
day
22 101 10/10/96
58 103 11/12/96
S1. sid  R1. sid
rating
7
8
age
45.0
55.5
R1
(sid)
58
58
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
bid
103
103
day
11/12/96
11/12/96
23
Division
Not supported as a primitive operator, but useful for
expressing queries like:
Division
Find sailors who have reserved all boats
A/B X
A X
Y
B Y
No
yellow
nor green
No red
nor
green
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Division
 Let



A have 2 fields, x and y; B have only field y:
|  x, y  A  y  B
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.
A/B =
x
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.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
25
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
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
26
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.)
Idea: For A/B, compute all x values in A that are not
`disqualified’ by some y value in B.

x value is disqualified if by attaching y value from B, we obtain
xy
an xy tuple that is not in A.
All possible xy Combinations
combinations
in A
Disqualified x values:  x (( x ( A)  B)  A)
A/B:
 x ( A) 
Possible xy combinations
not in A
all disqualified x values
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
27
Find names of sailors who’ve reserved boat #103
 sname((
(3) Find out the
names of those
sailors
bid 103
Reserves)  Sailors)
(1) Find all
reservations for
boat 103
(2) Follow the foreign-key
pointer to identify sailors who
made these reservations
Sailors
sid
22
31
58
sname rating age
dustin
7
45.0
lubber
8
55.5
rusty
10 35.0
Foreign key
Reserves
sid bid
day
22 101 10/10/96
58 103 11/12/96
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
28
Find names of sailors who’ve reserved boat #103
Solution 2:
 (Temp1, 
bid  103
Re serves)
 ( Temp2, Temp1  Sailors)
(3) Find out the
 sname (Temp2)
names of those
(1) Find all
reservations for
boat 103
(2) Follow the foreignkey pointer to identify
sailors who made
these reservations
sailors
Sailors
sid
22
31
58
sname rating age
dustin
7
45.0
lubber
8
55.5
rusty
10 35.0
Foreign key
Reserves
sid bid
day
22 101 10/10/96
58 103 11/12/96
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
29
Find names of sailors who’ve reserved boat #103
Solution 3:
 sname (
bid 103
(3) Find out the
names of sailors who
made those sailors
(Re serves  Sailors))
(2) Retain only
reservations for
boat 103
Sailors
sid
22
31
58
sname rating age
dustin
7
45.0
lubber
8
55.5
rusty
10 35.0
(1) Find reservations
for every sailor
Foreign key
Reserves
sid bid
day
22 101 10/10/96
58 103 11/12/96
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
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Which one is better ?
 sname((
 sname (
bid 103
bid 103
Reserves)  Sailors)
(Re serves  Sailors))
Need optimization before query execution
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
31
Find names of sailors who’ve reserved a red boat
Start with “boat”
because we have a
predicate about boat
 sname ((
Boats)  Re serves  Sailors)
color ' red '
1
2
Sailors(sid, sname, rating, age)
Foreign key
2nd join
Reserves(sid, bid, day)
Foreign key
1st join
Boats(bid, bname, color)
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
32
Find names of sailors who’ve reserved a red boat
 sname ((
Boats)  Re serves  Sailors)
color ' red '

A more efficient solution:
 sname ( ((

Boats)  Re s)  Sailors)
sid bid color ' red '
Make Join operations
less expensive
A query optimizer can find this, given the first solution!
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
33
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
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
34
Find sailors who’ve reserved a red and a green boat
This won’t work!
Empty Relation
ρ(Tempboats, (Ϭ color=‘red’˄ color=‘green’ Boats))
 sname (Tempboats  Reserves  Sailors)
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
35
Find sailors who’ve reserved a red and a green boat

Identify sailors who’ve reserved red boats
 (Tempred, 

sid
color ' red '
Boats)  Re serves))
Identify sailors who’ve reserved green boats
 (Tempgreen, 

((
sid
((
color ' green'
Boats)  Re serves))
Find the intersection
 sname((Tempred  Tempgreen)  Sailors)
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
36
Find the names of sailors who’ve reserved all boats

Uses division; schemas of the input relations to division
must be carefully chosen:
sid of sailors who have reserved all boats
 (Tempsids,(
sid ,bid
Re serves)/(
All boats
bid
Boats))
 sname (Tempsids  Sailors)

To find sailors who’ve reserved all ‘Interlake’ boats:
.....
/
bid
(
bname ' Interlake'
Boats)
‘Interlake’ boats
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
37
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.

SQL
DBMS
Algebra
Several ways of expressing a given
query; a query optimizer should
choose the most efficient version.
Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke
38