DBMS Class 2

Attribute Types
• Each attribute of a relation has a name
• The set of allowed values for each attribute is called the
domain of the attribute
• Attribute values are (normally) required to be atomic;
that is, indivisible
– E.g. the value of an attribute can be an account number,
but cannot be a set of account numbers
• Domain is said to be atomic if all its members are atomic
• The special value null is a member of every domain
• The null value causes complications in the definition of
many operations
– We shall ignore the effect of null values in our main
presentation and consider their effect later

Relation Schema
• Formally, given domains D1, D2, …. Dn a relation
r is a subset of
D1 x D2 x … x Dn
Thus, a relation is a set of n-tuples
(a1, a2, …, an) where each ai Di
• Schema of a relation consists of
– attribute definitions
• name
• type/domain
– integrity constraints

Relation Instance
• The current values (relation instance) of a relation are
specified by a table
• An element t of r is a tuple, represented by a row in a table
• Order of tuples is irrelevant (tuples may be stored in an
arbitrary order)
attributes
(or columns)
customer_name customer_street customer_city

Jones Main Harrison
Smith North Rye tuples
Curry North Rye (or rows)
Lindsay Park Pittsfield

customer

Database
• A database consists of multiple relations
• Information about an enterprise is broken up
into parts, with each relation storing one part
of the information
• E.g.
account : information about accounts
depositor : which customer owns which
account
customer : information about customers

Why Split Information Across
Relations?
• Storing all information as a single relation such as
bank(account_number, balance, customer_name, ..)
results in
– repetition of information
• e.g.,if two customers own an account (What gets repeated?)

– the need for null values
• e.g., to represent a customer without an account

• Normalization theory (Chapter 7) deals with how to
design relational schemas

Keys
• Let K R
• K is a superkey of R if values for K are sufficient to
identify a unique tuple of each possible relation r(R)
– by “possible r ” we mean a relation r that could exist in the
enterprise we are modeling.
– Example: {customer_name, customer_street} and
{customer_name}
are both superkeys of Customer, if no two customers can
possibly have the same name
• In real life, an attribute such as customer_id would be used instead
of customer_name to uniquely identify customers, but we omit it to
keep our examples small, and instead assume customer names are
unique.

Keys (Cont.)
• K is a candidate key if K is minimal
Example: {customer_name} is a candidate key for
Customer, since it is a superkey and no subset of it is a
superkey.
• Primary key: a candidate key chosen as the principal
means of identifying tuples within a relation
– Should choose an attribute whose value never, or very
rarely, changes.
– E.g. email address is unique, but may change

Foreign Keys
• A relation schema may have an attribute that
corresponds to the primary key of another relation.
The attribute is called a foreign key.
– E.g. customer_name and account_number attributes of
depositor are foreign keys to customer and account
respectively.
– Only values occurring in the primary key attribute of the
referenced relation may occur in the foreign key attribute
of the referencing relation.

Query Languages
• Language in which user requests information
from the database.
• Categories of languages
– Procedural
– Non-procedural, or declarative
• “Pure” languages:
– Relational algebra
– Tuple relational calculus
– Domain relational calculus
• Pure languages form underlying basis of query
languages that people use.

Relational Algebra
• Procedural language
• Six basic operators
– select:
– project:
– union:
– set difference: –
– Cartesian product: x
– rename:
• The operators take one or two relations as
inputs and produce a new relation as a result.

Select Operation – Example
 Relation r
A B C D

1 7
5 7
12 3
23 10

 A=B ^ D > 5 (r)
A B C D

1 7
23 10

Project Operation – Example
• Relation r:
A B C

10 1
20 1
30 1
40 2

A,C (r) A C A C

1 1
1 = 1
1 2
2

•
Union Operation – Example
Relations r, s:
A B A B

1 2
2 3
1 s
r
A B
 r s:
1
2
1
3

Set Difference Operation – Example
• Relations r, s:
A B A B

1 2
2 3
1 s
r

 r – s:
A B

1
1

Cartesian-Product Operation –
Example
 Relations r, s:
A B C D E

1 10 a
10 a
2
20 b
r 10 b
s
 r x s:
A B C D E
1 10 a
1 10 a
1 20 b
1 10 b
2 10 a
2 10 a
2 20 b
2 10 b

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 A , A2 , …., An .
1

Composition of Operations
• Can build expressions using multiple operations
• Example: A A=C(r x s)D E
B C
• rxs
1 10 a
1 10 a
1 20 b
1 10 b
2 10 a
2 10 a
2 20 b
2 10 b

A B C D E

1 10 a
• A=C(r x s) 2 10 a
2 20 b

Banking Example
branch (branch_name, branch_city, assets)
customer
(customer_name, customer_street, customer
_city)
account
(account_number, branch_name, balance)
loan (loan_number, branch_name, amount)
depositor (customer_name, account_number)
borrower (customer_name, loan_number)

Example Queries

• Find all loans of over $1200

amount > 1200 (loan)

 Find the loan number for each loan of an amount greater than
$1200
loan_number ( amount > 1200 (loan))

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

customer_name (borrower) customer_name (depositor)

Example Queries
• Find the names of all customers who have a loan at
the Perryridge branch.
customer_name ( branch_name=“Perryridge”
( 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(depositor)

Example Queries
• Find the names of all customers who have a loan at the Perryridge
branch.
 customer_name ( branch_name = “Perryridge” (
borrower.loan_number = loan.loan_number (borrower x loan)))

 customer_name( loan.loan_number =
borrower.loan_number (
( branch_name = “Perryridge” (loan)) x borrower))

DBMS Class 2

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