ICOM 5016 Introduction to Database Systems

1
ICOM 5016 Introduction to Database Systems

• Lecture 6
• Dr. Manuel Rodriguez
• Department of Electrical and Computer Engineering
• University of Puerto Rico, Mayagüez

2
Chapter 3 Relational Model

• Structure of Relational Databases
• Relational Algebra
• Tuple Relational Calculus
• Domain Relational Calculus
• Extended Relational-Algebra-Operations
• Modification of the Database
• Views

3
Projection Operation

• Given a relation R, the projection operation is
  used to create a new relation S, such that each
  tuple ts is formed by taking a tuple tR and
  removing one or more columns.
• Formally, the projection of R over columns A1,
  A2, ,An is defined as

4
Project Operation Example
A
B
C

• Relation r

? ? ? ?
10 20 30 40
1 1 1 2
A
C
A
C

• ?A,C (r)

? ? ? ?
1 1 1 2
? ? ?
1 1 2

5
Project Operation

• Notation ?A1, A2, , Ak (r)
• where A1, A2 are attribute names and r is a
  relation name.
• The result is defined as the relation of k
  columns obtained by erasing the columns that are
  not listed
• Duplicate rows removed from result, since
  relations are sets
• E.g. To eliminate the branch-name attribute of
  account ?account-number, balance
  (account)

6
Union Operation Example

• Relations r, s

A
B
A
B
? ? ?
1 2 1
? ?
2 3
s
r
r ? s
A
B
? ? ? ?
1 2 1 3
7
Union Operation

• Notation r ? s
• Defined as
• r ? s t t ? r or t ? s
• For r ? s to be valid.
• 1. r, s must have the same arity (same number
  of attributes)
• 2. The attribute domains must be compatible
  (e.g., 2nd column of r deals with the same
  type of values as does the 2nd column of s)
• E.g. to find all customers with either an account
  or a loan ?customer-name (depositor) ?
  ?customer-name (borrower)

8
Set Difference Operation Example

• Relations r, s

A
B
A
B
? ? ?
1 2 1
? ?
2 3
s
r
r s
A
B
? ?
1 1
9
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

10
Cartesian-Product Operation-Example
A
B
C
D
E
Relations r, s
? ?
1 2
? ? ? ?
10 10 20 10
a a b b
r
s
r x s
A
B
C
D
E
? ? ? ? ? ? ? ?
1 1 1 1 2 2 2 2
? ? ? ? ? ? ? ?
10 10 20 10 10 10 20 10
a a b b a a b b
11
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.
• A tuple is r x s is made by concatenating the
  columns from the first tuple, with the those of
  the second tuple.

12
Composition of Operations

• Can build expressions using multiple operations
• Example ?AC(r x s)
• r x s
• ?AC(r x s)

A
B
C
D
E
? ? ? ? ? ? ? ?
1 1 1 1 2 2 2 2
? ? ? ? ? ? ? ?
10 10 20 10 10 10 20 10
a a b b a a b b
A
B
C
D
E
? ? ?
? ? ?
10 20 20
a a b
1 2 2
13
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.

14
Banking Example

• branch (branch-name, branch-city, assets)
• customer (customer-name, customer-street,
  customer-only)
• account (account-number, branch-name, balance)
• loan (loan-number, branch-name, amount)
• depositor (customer-name, account-number)
• borrower (customer-name, loan-number)

15
Example Queries

• Find all loans of over 1200

• ?amount gt 1200 (loan)

• Find the loan number for each loan of an amount
  greater than
• 1200

• ?loan-number (?amount gt 1200 (loan))

16
Example Queries

• Find the names of all customers who have a loan,
  an account, or both, from the bank

• ?customer-name (borrower) ? ?customer-name
  (depositor)

• Find the names of all customers who have a loan
  and an
• account at bank.

• ?customer-name (borrower) ? ?customer-name
  (depositor)

17
Example Queries

• Find the names of all customers who have a loan
  at the Perryridge branch.

?customer-name (?branch-namePerryridge
(?borrower.loan-number loan.loan-number(borrower
x loan)))

• Find the names of all customers who have a loan
  at the Perryridge branch but do not have an
  account at any branch of the bank.

?customer-name (?branch-name Perryridge
(?borrower.loan-number loan.loan-number(borrower
x loan))) ?customer-name(depos
itor)
18
Example Queries

• Find the names of all customers who have a loan
  at the Perryridge branch. Two possible solutions
  follow

• Query 1 ?customer-name(?branch-name
  Perryridge ( ?borrower.loan-number
  loan.loan-number(borrower x loan)))

• ? Query 2
• ?customer-name(?loan.loan-number
  borrower.loan-number( (?branch-name
  Perryridge(loan)) x borrower))

19
Example Queries

• Find the largest account balance
• Rename account relation as d
• The query is

?balance(account) - ?account.balance
(?account.balance lt d.balance (account x rd
(account)))
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Formal Definition

• A basic expression in the relational algebra
  consists of either one of the following
• A relation in the database
• A constant relation
• Let E1 and E2 be relational-algebra expressions
  the following are all relational-algebra
  expressions
• E1 ? E2
• E1 - E2
• E1 x E2
• ?p (E1), P is a predicate on attributes in E1
• ?s(E1), S is a list consisting of some of the
  attributes in E1
• ? x (E1), x is the new name for the result of E1

21
Additional Operations

• We define additional operations that do not add
  any power to the
• relational algebra, but that simplify common
  queries.
• Set intersection
• Natural join
• Division
• Assignment

22
Set-Intersection Operation

• Notation r ? s
• Defined as
• r ? s t t ? r and t ? s
• Assume
• r, s have the same arity
• attributes of r and s are compatible
• Note r ? s r - (r - s)

23
Set-Intersection Operation - Example

• Relation r, s
• r ? s

A B
A B
? ? ?
1 2 1
? ?
2 3
r
s
A B
? 2
24
Natural-Join Operation

• Notation r s

• Let r and s be relations on schemas R and S
  respectively. Then, r s is a relation on
  schema R ? S obtained as follows
• Consider each pair of tuples tr from r and ts
  from s.
• If tr and ts have the same value on each of the
  attributes in R ? S, add a tuple t to the
  result, where
• t has the same value as tr on r
• t has the same value as ts on s
• Example
• R (A, B, C, D)
• S (E, B, D)
• Result schema (A, B, C, D, E), and R ?S
  (B,D)
• r s is defined as ?r.A, r.B, r.C, r.D,
  s.E (?r.B s.B ? r.D s.D (r x s))

25
Natural Join Operation Example

• Relations r, s

B
D
E
A
B
C
D
1 3 1 2 3
a a a b b
? ? ? ? ?
? ? ? ? ?
1 2 4 1 2
? ? ? ? ?
a a b a b
r
s
A
B
C
D
E
? ? ? ? ?
1 1 1 1 2
? ? ? ? ?
a a a a b
? ? ? ? ?