Session 3
Welcome: To session 3-the third learning sequence
“ Relational algebra “
Recap : In the previous learning sequences, we
discussed the four operators of relational algebra.
Present learning: We shall explore the following topic:
- Another four operators.
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Relational Algebra
2
Relational Algebra
Four basic operators




set difference
Cartesian product
rename
assignment
The operators take one or more
relations as inputs and give a new
relation as a result.
Set Difference Operation
Notation r – s
Defined as:
r – s = {t | t  r and t  s}
Set differences must be taken
between compatible relations.
 r and s must have the same arity.
 attribute domains of r and s must be compatible.
Set Difference Operation –
Example
For example, find all customers who have an account
and haven't a loan.
 customer-name (depositor) - customer-name (borrower)
Accou
nt-no.
A-101
A-201
A-217
A-222
A-305
borrower
depositor
Customer
-name
Ali
Mahmood
Ahmid
Linda
Rana
Customer- Loanname
no.
Ali
L-11
Kasim
L-11
Ahmid
L-25
Linda
L-11
5
Rana
L-34
Set Difference Operation –
Example
 customer-name (depositor)
Customername
Ali
Mahmood
Ahmid
Linda
Rana
….
customer-name (borrower)
Customer
-name
Ali
Kasim
Ahmid
Linda
Rana
…..
Set Difference Operation –
Example
 The result relation is:
Customername
Mahmood
Set Difference Operation –
Example
For example, find all customers who have
a loan, but they haven't an account.
 customer-name (borrower) - customer-name (depositor)
depositor
borrower
Customer- Loanname
no.
Ali
L-11
Kasim
L-11
Ahmid
L-25
Linda
L-11
Rana
L-34
Customer- Accountname
no.
Ali
A-101
Mahmood A-201
Ahmid
A-217
Linda
A-222
Rana
A-305
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Set Difference Operation –
Example
 customer-name (borrower)
Customer
-name
Ali
Kasim
Ahmid
Linda
Rana
…..
customer-name (depositor)
Customername
Ali
Mahmood
Ahmid
Linda
Rana
….
Set Difference Operation –
Example
 The result relation is:
Customername
Kasim
Cartesian-Product Operation
Notation r x s
Defined as:
r x s = {t q | t  r and q  s}
Assume that attributes of r(R) and
s(S) are disjoint. (That is,
R  S = ).
If attributes of r(R) and s(S) are not
disjoint, then renaming must be
used.
Cartesian-Product Operation
Relations r and s :
r
A
B
a1 1
a2 2
s
C
D
E
c1
d1
1
c2
d2
2
c3
d3
3
c4
d4
2
Cartesian-Product Operation
The relation r x s is:
A
a1
a1
a1
a1
a2
a2
a2
a2
B
1
1
1
1
2
2
2
2
C
c1
c2
c3
c4
c1
c2
c3
c4
D
d1
d2
d3
d4
d1
d2
d3
d4
E
1
2
3
2
1
2
3
2
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Composition of Operations
Can build expressions using multiple
operations
Example: B=E(r x s)
A
B
C
D
E
a1
a2
a2
1
2
2
c1
c2
c4
d1
d2
d4
1
2
2
Rename Operation
 Allows us to name, and therefore to refer to, the results
of relational-algebra expressions.
 Allows us to refer to a relation by more than one name.
Example:
 x (E)
returns the expression E under the name X
If a relational-algebra expression E has arity n, then
x (A1, A2, …, An) (E)
returns the result of expression E under the name X, and
with the attributes renamed to A1, A2, …., An.
Rename Operation
R d (account)
d
Assignment Operation
The assignment operation ()
provides a convenient way to express
complex queries.
 Write query as a sequential program consisting of
• a series of assignments.
• followed by an expression whose value is displayed as a
result of the query.
 Assignment must always be made to a temporary relation
variable.
Assignment Operation
 Example: Write
 customer-name (borrower)-customer-name (depositor)
temp1  customer-name (borrower)
temp2  customer-name (depositor)
result = temp1 – temp2
 The result to the right of the  is assigned to the
relation variable on the left of the .
 May use variable in subsequent expressions.
Relational Algebra
Summary: In this learning sequence, we discussed
another four basic operators of the topic relational
algebra.
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END
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