?Silberschatz,Korth and Sudarshan3.1Database System Concepts
Chapter 3,Relational Model
? Structure of Relational Databases
? Relational Algebra
? Extended Relational-Algebra-Operations
? Modification of the Database
? Views
? Tuple Relational Calculus
? Domain Relational Calculus
?Silberschatz,Korth and Sudarshan3.2Database System Concepts
Example of a Relation
?Silberschatz,Korth and Sudarshan3.3Database System Concepts
Basic Structure
? Formally,given sets 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
? Example,if
customer-name = {Jones,Smith,Curry,Lindsay}
customer-street = {Main,North,Park}
customer-city = {Harrison,Rye,Pittsfield}
Then r = { (Jones,Main,Harrison),
(Smith,North,Rye),
(Curry,North,Rye),
(Lindsay,Park,Pittsfield)}
is a relation over customer-name x customer-street x
customer-city
?Silberschatz,Korth and Sudarshan3.4Database System Concepts
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,multivalued attribute values are not atomic
? E.g,composite attribute values are not 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
?Silberschatz,Korth and Sudarshan3.5Database System Concepts
Relation Schema
? A1,A2,…,An are attributes
? R = (A1,A2,…,An ) is a relation schema
E.g,Customer-schema =
(customer-name,customer-street,customer-city)
? r(R) is a relation on the relation schema R
E.g,customer (Customer-schema)
?Silberschatz,Korth and Sudarshan3.6Database System Concepts
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
Jones
Smith
Curry
Lindsay
customer-name
Main
North
North
Park
customer-street
Harrison
Rye
Rye
Pittsfield
customer-city
customer
attributes
(or columns)
tuples
(or rows)
?Silberschatz,Korth and Sudarshan3.7Database System Concepts
Relations are Unordered
?Order of tuples is irrelevant (tuples may be stored in an
arbitrary order)
? E.g,account relation with unordered tuples
?Silberschatz,Korth and Sudarshan3.8Database System Concepts
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, stores information about accounts
depositor, stores information about which customer
owns which account
customer, stores information about customers
? Storing all information as a single relation such as
bank(account-number,balance,customer-name,..)
results in
? repetition of information (e.g,two customers own an account)
? the need for null values (e.g,represent a customer without an
account)
? Normalization theory (Chapter 7) deals with how to design
relational schemas
?Silberschatz,Korth and Sudarshan3.9Database System Concepts
The customer Relation
?Silberschatz,Korth and Sudarshan3.10Database System Concepts
The depositor Relation
?Silberschatz,Korth and Sudarshan3.11Database System Concepts
E-R Diagram for the Banking Enterprise
?Silberschatz,Korth and Sudarshan3.12Database System Concepts
Keys
? 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.
? K is a candidate key if K is minimal
Example,{customer-name} is a candidate key for
Customer,since it is a superkey (assuming no two
customers can possibly have the same name),and no
subset of it is a superkey.
?Silberschatz,Korth and Sudarshan3.13Database System Concepts
Determining Keys from E-R Sets
? Strong entity set,The primary key of the entity set
becomes the primary key of the relation.
? Weak entity set,The primary key of the relation
consists of the union of the primary key of the strong
entity set and the discriminator of the weak entity set.
? Relationship set,The union of the primary keys of the
related entity sets becomes a super key of the relation.
? For binary many-to-one relationship sets,the primary
key of the,many” entity set becomes the relation’s
primary key.
? For one-to-one relationship sets,the relation’s
primary key can be that of either entity set.
? For many-to-many relationship sets,the union of the
primary keys becomes the relation’s primary key
?Silberschatz,Korth and Sudarshan3.14Database System Concepts
Schema Diagram for the Banking Enterprise
?Silberschatz,Korth and Sudarshan3.15Database System Concepts
Query Languages
? Language in which user requests information from the
database.
? Categories of languages
? procedural
? non-procedural
?,Pure” languages:
? Relational Algebra
? Tuple Relational Calculus
? Domain Relational Calculus
? Pure languages form underlying basis of query
languages that people use.
?Silberschatz,Korth and Sudarshan3.16Database System Concepts
Relational Algebra
? Procedural language
? Six basic operators
? select
? project
? union
? set difference
? Cartesian product
? rename
? The operators take two or more relations as inputs and
give a new relation as a result.
?Silberschatz,Korth and Sudarshan3.17Database System Concepts
Select Operation – Example
? Relation r A B C D
?
?
?
?
?
?
?
?
1
5
12
23
7
7
3
10
??A=B ^ D > 5 (r)
A B C D
?
?
?
?
1
23
7
10
?Silberschatz,Korth and Sudarshan3.18Database System Concepts
Select Operation
? Notation,? p(r)
? p is called the selection predicate
? Defined as:
?p(r) = {t | t ? r and p(t)}
Where p is a formula in propositional calculus
consisting of terms connected by, ? (and),? (or),?
(not)
Each term is one of:
<attribute> op <attribute> or
<constant>
where op is one of,=,?,>,?,<,?
? Example of selection:
? branch-name=“Perryridge”(account)
?Silberschatz,Korth and Sudarshan3.19Database System Concepts
Result of ? branch-name =,Perryridge” (loan)
?Silberschatz,Korth and Sudarshan3.20Database System Concepts
Project Operation – Example
? Relation r,A B C
?
?
?
?
10
20
30
40
1
1
1
2
A C
?
?
?
?
1
1
1
2
=
A C
?
?
?
1
1
2
? ?A,C (r)
?Silberschatz,Korth and Sudarshan3.21Database System Concepts
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)
?Silberschatz,Korth and Sudarshan3.22Database System Concepts
Loan Number and the Amount of the
Loan
?Silberschatz,Korth and Sudarshan3.23Database System Concepts
Union Operation – Example
? Relations r,s:
r ? s:
A B
?
?
?
1
2
1
A B
?
?
2
3
r
s
A B
?
?
?
?
1
2
1
3
?Silberschatz,Korth and Sudarshan3.24Database System Concepts
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)
?Silberschatz,Korth and Sudarshan3.25Database System Concepts
Names of All Customers Who Have
Either a Loan or an Account
?Silberschatz,Korth and Sudarshan3.26Database System Concepts
Set Difference Operation – Example
? Relations r,s:
r – s:
A B
?
?
?
1
2
1
A B
?
?
2
3
r
s
A B
?
?
1
1
?Silberschatz,Korth and Sudarshan3.27Database System Concepts
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
?Silberschatz,Korth and Sudarshan3.28Database System Concepts
Customers With An Account But No Loan
?Silberschatz,Korth and Sudarshan3.29Database System Concepts
Cartesian-Product Operation-Example
Relations r,s:
r x s:
A B
?
?
1
2
A B
?
?
?
?
?
?
?
?
1
1
1
1
2
2
2
2
C D
?
?
?
?
?
?
?
?
10
10
20
10
10
10
20
10
E
a
a
b
b
a
a
b
b
C D
?
?
?
?
10
10
20
10
E
a
a
b
br
s
?Silberschatz,Korth and Sudarshan3.30Database System Concepts
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.
?Silberschatz,Korth and Sudarshan3.31Database System Concepts
Result of borrower ? loan
?Silberschatz,Korth and Sudarshan3.32Database System Concepts
Result of ? branch-name =,Perryridge”
(borrower ? loan)
?Silberschatz,Korth and Sudarshan3.33Database System Concepts
Composition of Operations
? Can build expressions using multiple operations
? Example,?A=C(r x s)
? r x s
? ?A=C(r x s)
A B
?
?
?
?
?
?
?
?
1
1
1
1
2
2
2
2
C D
?
?
?
?
?
?
?
?
10
10
20
10
10
10
20
10
E
a
a
b
b
a
a
b
b
A B C D E
?
?
?
1
2
2
?
?
?
10
20
20
a
a
b
?Silberschatz,Korth and Sudarshan3.34Database System Concepts
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,
?Silberschatz,Korth and Sudarshan3.35Database System Concepts
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)
?Silberschatz,Korth and Sudarshan3.36Database System Concepts
Example Queries
? Find all loans of over $1200
?Find the loan number for each loan of an amount greater
than $1200
?amount > 1200 (loan)
?loan-number (?amount > 1200 (loan))
?Silberschatz,Korth and Sudarshan3.37Database System Concepts
Example Queries
? Find the names of all customers who have a loan,an
account,or both,from the bank
?Find the names of all customers who have a loan and an
account at bank.
?customer-name (borrower) ??customer-name (depositor)
?customer-name (borrower) ??customer-name (depositor)
?Silberschatz,Korth and Sudarshan3.38Database System Concepts
Example Queries
? Find the names of all customers who have a loan at the
Perryridge branch.
? 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)
?customer-name (?branch-name=“Perryridge”
(?borrower.loan-number = loan.loan-number(borrower x loan)))
?Silberschatz,Korth and Sudarshan3.39Database System Concepts
Example Queries
? Find the names of all customers who have a loan at the
Perryridge branch.
? Query 2
?customer-name(?loan.loan-number = borrower.loan-number(
(?branch-name =,Perryridge”(loan)) x borrower))
?Query 1
?customer-name(?branch-name =,Perryridge” (
?borrower.loan-number = loan.loan-number(borrower x loan)))
?Silberschatz,Korth and Sudarshan3.40Database System Concepts
Example Queries
Find the largest account balance
? Rename account relation as d
? The query is:
?balance(account) - ?account.balance
(?account.balance < d.balance (account x ?d (account)))
?Silberschatz,Korth and Sudarshan3.41Database System Concepts
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
?Silberschatz,Korth and Sudarshan3.42Database System Concepts
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
?Silberschatz,Korth and Sudarshan3.43Database System Concepts
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)
?Silberschatz,Korth and Sudarshan3.44Database System Concepts
Set-Intersection Operation - Example
? Relation r,s:
? r ? s
A B
?
?
?
1
2
1
A B
?
?
2
3
r s
A B
? 2
?Silberschatz,Korth and Sudarshan3.45Database System Concepts
? Notation,r s
Natural-Join Operation
? Let r and s be relations on schemas R and S respectively,
Then,r s is a relation on schema R ? S,
? r s=?R ?S (?r.A1== s.A1 ?r.A2 = s.A2 ?… ?r.An = s.An (r x s))
Where is R ?S={A1,A2,…,An}.
? Example:
R = (A,B,C,D)
S = (E,B,D)
? Result schema = (A,B,C,D,E)
? 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))
?Silberschatz,Korth and Sudarshan3.46Database System Concepts
?customer-name,loan-number,amount
(borrower loan)
?customer-name,loan-number,amount (?borrower.loan-
number = loan.loan-number(borrower x loan))
?Silberschatz,Korth and Sudarshan3.47Database System Concepts
Natural Join Operation – Example
? Relations r,s:
A B
?
?
?
?
?
1
2
4
1
2
C D
?
?
?
?
?
a
a
b
a
b
B
1
3
1
2
3
D
a
a
a
b
b
E
?
?
?
?
?
r
A B
?
?
?
?
?
1
1
1
1
2
C D
?
?
?
?
?
a
a
a
a
b
E
?
?
?
?
?
s
r s
?Silberschatz,Korth and Sudarshan3.48Database System Concepts
Division Operation
? Suited to queries that include the phrase,for all”.
? Let r and s be relations on schemas R and S respectively
where
? R = (A1,…,Am,B1,…,Bn)
? S = (B1,…,Bn)
The result of r ? s is a relation on schema
R – S = (A1,…,Am)
r ? s = { t | t ?? R-S(r) ?? u ? s ( tu ? r ) }
r ? s
?Silberschatz,Korth and Sudarshan3.49Database System Concepts
Division Operation – Example
Relations r,s:
r ? s,A
B
?
?
1
2
A B
?
?
?
?
?
?
?
?
?
?
?
1
2
3
1
1
1
3
4
6
1
2
r
s
?Silberschatz,Korth and Sudarshan3.50Database System Concepts
Another Division Example
A B
?
?
?
?
?
?
?
?
a
a
a
a
a
a
a
a
C D
?
?
?
?
?
?
?
?
a
a
b
a
b
a
b
b
E
1
1
1
1
3
1
1
1
Relations r,s:
r ? s:
D
a
b
E
1
1
A B
?
?
a
a
C
?
?
r
s
?Silberschatz,Korth and Sudarshan3.51Database System Concepts
Division Operation (Cont.)
? Property
? Let q – r ? s
? Then q is the largest relation satisfying q x s ? r
? Definition in terms of the basic algebra operation
Let r(R) and s(S) be relations,and let S ? R
r ? s = ?R-S (r) –?R-S ( (?R-S (r) x s) –?R-S,S(r))
To see why
? ?R-S,S(r) simply reorders attributes of r
? ?R-S(?R-S (r) x s) –?R-S,S(r)) gives those tuples t in
?R-S (r) such that for some tuple u ? s,tu ? r.
?Silberschatz,Korth and Sudarshan3.52Database System Concepts
Division Operation – Example
Relations r,s:
r ? s,A
B
?
?
1
2
A B
?
?
?
?
?
?
?
?
?
?
?
1
2
3
1
1
1
3
4
6
1
2
r
s
(?R-S (r) x s)
A B
?
?
?
?
?
?
?
?
?
?
1
2
1
2
1
2
1
2
1
2
(?R-S (r) x s) –?R-S,S(r)
A B
?
?
?
2
2
2
r ? s = ?R-S (r) –?R-S ( (?R-S (r) x s) –?R-S,S(r))
?Silberschatz,Korth and Sudarshan3.53Database System Concepts
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.
? Example,Write r ? s as
temp1??R-S (r)
temp2 ??R-S ((temp1 x s) – ?R-S,S (r))
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.
?Silberschatz,Korth and Sudarshan3.54Database System Concepts
Example Queries
? Find all customers who have an account from at least
the,Downtown” and the Uptown” branches.
where CN denotes customer-name and BN denotes
branch-name.
Query 1
?CN(?BN=“Downtown”(depositor account)) ?
?CN(?BN=“Uptown”(depositor account))
Query 2
?customer-name,branch-name(depositor account)
??temp(branch-name) ({(“Downtown”),(“Uptown”)})
?Silberschatz,Korth and Sudarshan3.55Database System Concepts
? Find all customers who have an account at all branches
located in Brooklyn city.
Example Queries
?customer-name,branch-name (depositor account)
??branch-name (?branch-city =,Brooklyn” (branch))
?Silberschatz,Korth and Sudarshan3.56Database System Concepts
Extended Relational-Algebra-Operations
? Generalized Projection
? Outer Join
? Aggregate Functions
?Silberschatz,Korth and Sudarshan3.57Database System Concepts
Generalized Projection
? Extends the projection operation by allowing
arithmetic functions to be used in the projection list.
? F1,F2,…,Fn (E)
? E is any relational-algebra expression
? Each of F1,F2,…,Fn are are arithmetic expressions
involving constants and attributes in the schema of E.
? Given relation credit-info(customer-name,limit,credit-
balance),find how much more each person can
spend,
?customer-name,limit – credit-balance (credit-info)
?Silberschatz,Korth and Sudarshan3.58Database System Concepts
Aggregate Functions and Operations
? Aggregation function takes a collection of values and
returns a single value as a result.
avg,average value
min,minimum value
max,maximum value
sum,sum of values
count,number of values
? Aggregate operation in relational algebra
G1,G2,…,Gn g F1( A1),F2( A2),…,Fn( An) (E)
? E is any relational-algebra expression
? G1,G2 …,Gn is a list of attributes on which to group (can
be empty)
? Each Fi is an aggregate function
? Each Ai is an attribute name
?Silberschatz,Korth and Sudarshan3.59Database System Concepts
Aggregate Operation – Example
? Relation r:
A B
?
?
?
?
?
?
?
?
C
7
7
3
10
g sum(c)(r)
sum-C
27
?Silberschatz,Korth and Sudarshan3.60Database System Concepts
Aggregate Operation – Example
? Relation account grouped by branch-name:
branch-name g sum(balance) (account)
branch-name account-number balance
Perryridge
Perryridge
Brighton
Brighton
Redwood
A-102
A-201
A-217
A-215
A-222
400
900
750
750
700
branch-name balance
Perryridge
Brighton
Redwood
1300
1500
700
?Silberschatz,Korth and Sudarshan3.61Database System Concepts
Aggregate Functions (Cont.)
? Result of aggregation does not have a name
? Can use rename operation to give it a name
? For convenience,we permit renaming as part of
aggregate operation
branch-name g sum(balance) as sum-balance (account)
?Silberschatz,Korth and Sudarshan3.62Database System Concepts
Outer Join
? An extension of the join operation that avoids loss of
information.
? Computes the join and then adds tuples form one relation
that does not match tuples in the other relation to the result
of the join,
? Uses null values:
? null signifies that the value is unknown or does not exist
? All comparisons involving null are (roughly speaking)
false by definition.
?Will study precise meaning of comparisons with nulls
later
?Silberschatz,Korth and Sudarshan3.63Database System Concepts
Outer Join – Example
? Relation loan
? Relation borrower
customer-name loan-number
Jones
Smith
Hayes
L-170
L-230
L-155
3000
4000
1700
loan-number amount
L-170
L-230
L-260
branch-name
Downtown
Redwood
Perryridge
?Silberschatz,Korth and Sudarshan3.64Database System Concepts
Outer Join – Example
? Inner Join(内连接 ）
loan Borrower
loan-number amount
L-170
L-230
3000
4000
customer-name
Jones
Smith
branch-name
Downtown
Redwood
Jones
Smith
null
loan-number amount
L-170
L-230
L-260
3000
4000
1700
customer-namebranch-name
Downtown
Redwood
Perryridge
? Left Outer Join（左外连接）
loan Borrower
?Silberschatz,Korth and Sudarshan3.65Database System Concepts
Outer Join – Example
? Right Outer Join（右外连接）
loan borrower
loan borrower
? Full Outer Join（全连接）
loan-number amount
L-170
L-230
L-155
3000
4000
null
customer-name
Jones
Smith
Hayes
branch-name
Downtown
Redwood
null
loan-number amount
L-170
L-230
L-260
L-155
3000
4000
1700
null
customer-name
Jones
Smith
null
Hayes
branch-name
Downtown
Redwood
Perryridge
null
?Silberschatz,Korth and Sudarshan3.66Database System Concepts
Modification of the Database
? The content of the database may be modified using
the following operations:
? Deletion
? Insertion
? Updating
? All these operations are expressed using the
assignment operator.
?Silberschatz,Korth and Sudarshan3.67Database System Concepts
Deletion
? A delete request is expressed similarly to a query,
except instead of displaying tuples to the user,the
selected tuples are removed from the database.
? Can delete only whole tuples; cannot delete values on
only particular attributes
? A deletion is expressed in relational algebra by:
r ? r – E
where r is a relation and E is a relational algebra
query.
?Silberschatz,Korth and Sudarshan3.68Database System Concepts
Deletion Examples
? Delete all account records in the Perryridge branch.
?Delete all accounts at branches located in Needham.
r1 ? ??branch-city =,Needham” (account branch)
r2 ? ?branch-name,account-number,balance (r1)
account ? account – r2
?Delete all loan records with amount in the range of 0 to 50
loan ? loan –??amount ??0?and amount ? 50 (loan)
account ? account – ??branch-name =,Perryridge” (account)
?Silberschatz,Korth and Sudarshan3.69Database System Concepts
Insertion
? To insert data into a relation,we either:
? specify a tuple to be inserted
? write a query whose result is a set of tuples to be
inserted
? in relational algebra,an insertion is expressed by:
r ? r ? E
where r is a relation and E is a relational algebra
expression.
? The insertion of a single tuple is expressed by letting
E be a constant relation containing one tuple,
?Silberschatz,Korth and Sudarshan3.70Database System Concepts
Insertion Examples
? Insert information in the database specifying that Smith
has $1200 in account A-973 at the Perryridge branch.
? Provide as a gift for all loan customers in the Perryridge
branch,a $200 savings account,Let the loan number serve
as the account number for the new savings account.
account ? account ?{(“Perryridge”,A-973,1200)}
depositor ? depositor ? {(“Smith”,A-973)}
r1 ? (?branch-name =,Perryridge” (borrower loan))
account ? account ? ?branch-name,loan-number,200 (r1)
depositor ? depositor ??customer-name,loan-number(r1)
?Silberschatz,Korth and Sudarshan3.71Database System Concepts
Updating
? A mechanism to change a value in a tuple without
charging all values in the tuple
? Use the generalized projection operator to do this task
r ? ? F1,F2,…,FI,(r)
? Each Fi is either
? the ith attribute of r,if the ith attribute is not
updated,or,
? if the attribute is to be updated Fi is an expression,
involving only constants and the attributes of r,
which gives the new value for the attribute
?Silberschatz,Korth and Sudarshan3.72Database System Concepts
Update Examples
? Make interest payments by increasing all balances by 5
percent.
? Pay all accounts with balances over $10,000 6
percent interest
and pay all others 5 percent
account ? ? AN,BN,BAL * 1.06 (? BAL ? 10000 (account))
? ?AN,BN,BAL * 1.05 (?BAL ? 10000 (account))
account ?? AN,BN,BAL * 1.05 (account)
where AN,BN and BAL stand for account-number,
branch-name and balance,respectively.
?Silberschatz,Korth and Sudarshan3.73Database System Concepts
Views
? In some cases,it is not desirable for all users to see
the entire logical model (i.e.,all the actual relations
stored in the database.)
? Consider a person who needs to know a customer’s
loan number but has no need to see the loan amount,
This person should see a relation described,in the
relational algebra,by
?customer-name,loan-number (borrower loan)
? Any relation that is not of the conceptual model but is
made visible to a user as a,virtual relation” is called a
view.
?Silberschatz,Korth and Sudarshan3.74Database System Concepts
View Definition
? A view is defined using the create view statement which has
the form
create view v as <query expression>
where <query expression> is any legal relational algebra
query expression,The view name is represented by v.
? Once a view is defined,the view name can be used to refer
to the virtual relation that the view generates.
? View definition is not the same as creating a new relation by
evaluating the query expression
? Rather,a view definition causes the saving of an
expression; the expression is substituted into queries
using the view.
?Silberschatz,Korth and Sudarshan3.75Database System Concepts
View Examples
? Consider the view (named all-customer) consisting of
branches and their customers.
? We can find all customers of the Perryridge branch by writing:
create view all-customer as
?branch-name,customer-name (depositor account)
??branch-name,customer-name (borrower loan)
?customer-name
(?branch-name =,Perryridge” (all-customer))
?Silberschatz,Korth and Sudarshan3.76Database System Concepts
Updates Through View
? Database modifications expressed as views must be
translated to modifications of the actual relations in the
database.
? Consider the person who needs to see all loan data in the
loan relation except amount,The view given to the person,
branch-loan,is defined as,
create view branch-loan as
?branch-name,loan-number (loan)
? Since we allow a view name to appear wherever a relation
name is allowed,the person may write:
branch-loan ? branch-loan ?{(“Perryridge”,L-37)}
?Silberschatz,Korth and Sudarshan3.77Database System Concepts
Updates Through Views (Cont.)
? The previous insertion must be represented by an insertion into the
actual relation loan from which the view branch-loan is constructed.
? An insertion into loan requires a value for amount,The insertion can
be dealt with by either.
? rejecting the insertion and returning an error message to the
user.
? inserting a tuple (“L-37”,“Perryridge”,null) into the loan relation
? Some updates through views are impossible to translate into
database relation updates
? create view v as ?branch-name =,Perryridge” (account))
v ? v ? (L-99,Downtown,23)
? Others cannot be translated uniquely
? all-customer? all-customer ?{(“Perryridge”,“John”)}
?Have to choose loan or account,and
create a new loan/account number!
?Silberschatz,Korth and Sudarshan3.78Database System Concepts
Views Defined Using Other Views
? One view may be used in the expression defining
another view
? A view relation v1 is said to depend directly(直接依赖 )
on a view relation v2 if v2 is used in the expression
defining v1
? A view relation v1 is said to depend on (依赖 ) view
relation v2 if either v1 depends directly to v2 or there is
a path of dependencies from v1 to v2
? A view relation v is said to be recursive (递归 ) if it
depends on itself.
?Silberschatz,Korth and Sudarshan3.79Database System Concepts
View Expansion（ 视图展开 ）
? A way to define the meaning of views defined in terms of
other views.
? Let view v1 be defined by an expression e1 that may itself
contain uses of view relations.
? View expansion of an expression repeats the following
replacement step:
repeat
Find any view relation vi in e1
Replace the view relation vi by the expression defining vi
until no more view relations are present in e1
? As long as the view definitions are not recursive,this loop will
terminate
?Silberschatz,Korth and Sudarshan3.80Database System Concepts
Tuple Relational Calculus
? A nonprocedural query language,where each query is of
the form
{t | P (t) }
? It is the set of all tuples t such that predicate P is true for t
? t is a tuple variable,t[A] denotes the value of tuple t on
attribute A
? t ? r denotes that tuple t is in relation r
? P is a formula similar to that of the predicate calculus
?Silberschatz,Korth and Sudarshan3.81Database System Concepts
Predicate Calculus Formula
1,Set of attributes and constants
2,Set of comparison operators,(e.g.,?,?,?,?,?,?)
3,Set of connectives,and (?),or (v)? not (?)
4,Implication (?),x ? y,if x if true,then y is true
x ? y ???x v y
5,Set of quantifiers:
???t ??r (Q(t)) ??”there exists” a tuple in t in relation r
such that predicate Q(t) is true
??t ??r (Q(t)) ??Q is true,for all” tuples t in relation r
?Silberschatz,Korth and Sudarshan3.82Database System Concepts
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)
?Silberschatz,Korth and Sudarshan3.83Database System Concepts
Example Queries
? Find the loan-number,branch-name,and amount
for loans of over $1200
?Find the loan number for each loan of an amount greater
than $1200
Notice that a relation on schema [loan-number] is implicitly
defined by the query
{t | ? s ??loan (t[loan-number] = s[loan-number] ? s [amount] ? 1200)}
{t | t ? loan ? t [amount] ? 1200}
?Silberschatz,Korth and Sudarshan3.84Database System Concepts
Example Queries
? Find the names of all customers having a loan,an account,
or both at the bank
{t | ?s ? borrower( t[customer-name] = s[customer-name])
??u ? depositor( t[customer-name] = u[customer-name])
? Find the names of all customers who have a loan and an
account at the bank
{t | ?s ? borrower( t[customer-name] = s[customer-name])
??u ? depositor( t[customer-name] = u[customer-name])
?Silberschatz,Korth and Sudarshan3.85Database System Concepts
Example Queries
? Find the names of all customers having a loan at the
Perryridge branch
{t | ?s ? borrower( t[customer-name] = s[customer-name]
??u ? loan(u[branch-name] =,Perryridge”
? u[loan-number] = s[loan-number]))
? not ?v ? depositor (v[customer-name] =
t[customer-name]) }
? Find the names of all customers who have a loan at the
Perryridge branch,but no account at any branch of the
bank
{t | ?s ? borrower(t[customer-name] = s[customer-name]
??u ? loan(u[branch-name] =,Perryridge”
? u[loan-number] = s[loan-number]))}
?Silberschatz,Korth and Sudarshan3.86Database System Concepts
Example Queries
? Find the names of all customers having a loan from
the Perryridge branch,and the cities they live in
{t | ?s ? loan(s[branch-name] =,Perryridge”
??u ? borrower (u[loan-number] = s[loan-number]
? t [customer-name] = u[customer-name])
?? v ? customer (u[customer-name] = v[customer-name]
? t[customer-city] = v[customer-city])))}
?Silberschatz,Korth and Sudarshan3.87Database System Concepts
Example Queries
? Find the names of all customers who have an account
at all branches located in Brooklyn:
{t | ? c ? customer (t[customer.name] = c[customer-name]) ?
? s ? branch(s[branch-city] =,Brooklyn” ?
? u ? account ( s[branch-name] = u[branch-name]
?? s ? depositor ( t[customer-name] = s[customer-name]
? s[account-number] = u[account-number] )) )}
?Silberschatz,Korth and Sudarshan3.88Database System Concepts
Safety of Expressions
? It is possible to write tuple calculus expressions that
generate infinite relations.
? For example,{t | ? t?? r} results in an infinite relation if the
domain of any attribute of relation r is infinite
? To guard against the problem,we restrict the set of
allowable expressions to safe expressions.
? An expression {t | P(t)} in the tuple relational calculus is safe
if every component of t appears in one of the relations,
tuples,or constants that appear in P
? NOTE,this is more than just a syntax condition,
?E.g,{ t | t[A]=5 ? true } is not safe --- it defines an
infinite set with attribute values that do not appear in
any relation or tuples or constants in P,
?Silberschatz,Korth and Sudarshan3.89Database System Concepts
Domain Relational Calculus
? A nonprocedural query language equivalent in power
to the tuple relational calculus
? Each query is an expression of the form:
{ ? x1,x2,…,x n ? | P(x1,x2,…,x n)}
? x1,x2,…,x n represent domain variables
? P represents a formula similar to that of the
predicate calculus
?Silberschatz,Korth and Sudarshan3.90Database System Concepts
Example Queries
? Find the loan-number,branch-name,and amount for
loans of over $1200
{? c,a ? | ? l (? c,l ?? borrower ??b(? l,b,a ?? loan ?
b =,Perryridge”))}
or {? c,a ? | ? l (? c,l ?? borrower ? ?l,“Perryridge”,a ?? loan)}
? Find the names of all customers who have a loan from
the Perryridge branch and the loan amount:
{? c ? | ? l,b,a (? c,l ?? borrower ?? l,b,a ?? loan ? a > 1200)}
? Find the names of all customers who have a loan of
over $1200
{? l,b,a ? | ? l,b,a ?? loan ? a > 1200}
?Silberschatz,Korth and Sudarshan3.91Database System Concepts
Example Queries
? Find the names of all customers having a loan,an
account,or both at the Perryridge branch:
{? c ? | ? s,n (? c,s,n ?? customer) ?
? x,y,z(? x,y,z ?? branch ? y =,Brooklyn”) ?
? a,b(? x,y,z ?? account ?? c,a ?? depositor)}
? Find the names of all customers who have an account at
all branches located in Brooklyn:
{? c ? | ? l ({? c,l ?? borrower
?? b,a(? l,b,a ?? loan ? b =,Perryridge”))
? ? a(? c,a ?? depositor
?? b,n(? a,b,n ?? account ? b =,Perryridge”))}
?Silberschatz,Korth and Sudarshan3.92Database System Concepts
Safety of Expressions
{ ? x1,x2,…,x n ? | P(x1,x2,…,x n)}
is safe if all of the following hold:
1.All values that appear in tuples of the expression are
values from dom(P) (that is,the values appear either in P or
in a tuple of a relation mentioned in P).
2.For every,there exists” subformula of the form ? x (P1(x)),
the subformula is true if an only if P1(x) is true for all values
x from dom(P1).
3,For every,for all” subformula of the form ?x (P1 (x)),the
subformula is true if and only if P1(x) is true for all values x
from dom (P1).
End of Chapter 3
Chapter 3,Relational Model
? Structure of Relational Databases
? Relational Algebra
? Extended Relational-Algebra-Operations
? Modification of the Database
? Views
? Tuple Relational Calculus
? Domain Relational Calculus
?Silberschatz,Korth and Sudarshan3.2Database System Concepts
Example of a Relation
?Silberschatz,Korth and Sudarshan3.3Database System Concepts
Basic Structure
? Formally,given sets 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
? Example,if
customer-name = {Jones,Smith,Curry,Lindsay}
customer-street = {Main,North,Park}
customer-city = {Harrison,Rye,Pittsfield}
Then r = { (Jones,Main,Harrison),
(Smith,North,Rye),
(Curry,North,Rye),
(Lindsay,Park,Pittsfield)}
is a relation over customer-name x customer-street x
customer-city
?Silberschatz,Korth and Sudarshan3.4Database System Concepts
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,multivalued attribute values are not atomic
? E.g,composite attribute values are not 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
?Silberschatz,Korth and Sudarshan3.5Database System Concepts
Relation Schema
? A1,A2,…,An are attributes
? R = (A1,A2,…,An ) is a relation schema
E.g,Customer-schema =
(customer-name,customer-street,customer-city)
? r(R) is a relation on the relation schema R
E.g,customer (Customer-schema)
?Silberschatz,Korth and Sudarshan3.6Database System Concepts
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
Jones
Smith
Curry
Lindsay
customer-name
Main
North
North
Park
customer-street
Harrison
Rye
Rye
Pittsfield
customer-city
customer
attributes
(or columns)
tuples
(or rows)
?Silberschatz,Korth and Sudarshan3.7Database System Concepts
Relations are Unordered
?Order of tuples is irrelevant (tuples may be stored in an
arbitrary order)
? E.g,account relation with unordered tuples
?Silberschatz,Korth and Sudarshan3.8Database System Concepts
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, stores information about accounts
depositor, stores information about which customer
owns which account
customer, stores information about customers
? Storing all information as a single relation such as
bank(account-number,balance,customer-name,..)
results in
? repetition of information (e.g,two customers own an account)
? the need for null values (e.g,represent a customer without an
account)
? Normalization theory (Chapter 7) deals with how to design
relational schemas
?Silberschatz,Korth and Sudarshan3.9Database System Concepts
The customer Relation
?Silberschatz,Korth and Sudarshan3.10Database System Concepts
The depositor Relation
?Silberschatz,Korth and Sudarshan3.11Database System Concepts
E-R Diagram for the Banking Enterprise
?Silberschatz,Korth and Sudarshan3.12Database System Concepts
Keys
? 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.
? K is a candidate key if K is minimal
Example,{customer-name} is a candidate key for
Customer,since it is a superkey (assuming no two
customers can possibly have the same name),and no
subset of it is a superkey.
?Silberschatz,Korth and Sudarshan3.13Database System Concepts
Determining Keys from E-R Sets
? Strong entity set,The primary key of the entity set
becomes the primary key of the relation.
? Weak entity set,The primary key of the relation
consists of the union of the primary key of the strong
entity set and the discriminator of the weak entity set.
? Relationship set,The union of the primary keys of the
related entity sets becomes a super key of the relation.
? For binary many-to-one relationship sets,the primary
key of the,many” entity set becomes the relation’s
primary key.
? For one-to-one relationship sets,the relation’s
primary key can be that of either entity set.
? For many-to-many relationship sets,the union of the
primary keys becomes the relation’s primary key
?Silberschatz,Korth and Sudarshan3.14Database System Concepts
Schema Diagram for the Banking Enterprise
?Silberschatz,Korth and Sudarshan3.15Database System Concepts
Query Languages
? Language in which user requests information from the
database.
? Categories of languages
? procedural
? non-procedural
?,Pure” languages:
? Relational Algebra
? Tuple Relational Calculus
? Domain Relational Calculus
? Pure languages form underlying basis of query
languages that people use.
?Silberschatz,Korth and Sudarshan3.16Database System Concepts
Relational Algebra
? Procedural language
? Six basic operators
? select
? project
? union
? set difference
? Cartesian product
? rename
? The operators take two or more relations as inputs and
give a new relation as a result.
?Silberschatz,Korth and Sudarshan3.17Database System Concepts
Select Operation – Example
? Relation r A B C D
?
?
?
?
?
?
?
?
1
5
12
23
7
7
3
10
??A=B ^ D > 5 (r)
A B C D
?
?
?
?
1
23
7
10
?Silberschatz,Korth and Sudarshan3.18Database System Concepts
Select Operation
? Notation,? p(r)
? p is called the selection predicate
? Defined as:
?p(r) = {t | t ? r and p(t)}
Where p is a formula in propositional calculus
consisting of terms connected by, ? (and),? (or),?
(not)
Each term is one of:
<attribute> op <attribute> or
<constant>
where op is one of,=,?,>,?,<,?
? Example of selection:
? branch-name=“Perryridge”(account)
?Silberschatz,Korth and Sudarshan3.19Database System Concepts
Result of ? branch-name =,Perryridge” (loan)
?Silberschatz,Korth and Sudarshan3.20Database System Concepts
Project Operation – Example
? Relation r,A B C
?
?
?
?
10
20
30
40
1
1
1
2
A C
?
?
?
?
1
1
1
2
=
A C
?
?
?
1
1
2
? ?A,C (r)
?Silberschatz,Korth and Sudarshan3.21Database System Concepts
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)
?Silberschatz,Korth and Sudarshan3.22Database System Concepts
Loan Number and the Amount of the
Loan
?Silberschatz,Korth and Sudarshan3.23Database System Concepts
Union Operation – Example
? Relations r,s:
r ? s:
A B
?
?
?
1
2
1
A B
?
?
2
3
r
s
A B
?
?
?
?
1
2
1
3
?Silberschatz,Korth and Sudarshan3.24Database System Concepts
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)
?Silberschatz,Korth and Sudarshan3.25Database System Concepts
Names of All Customers Who Have
Either a Loan or an Account
?Silberschatz,Korth and Sudarshan3.26Database System Concepts
Set Difference Operation – Example
? Relations r,s:
r – s:
A B
?
?
?
1
2
1
A B
?
?
2
3
r
s
A B
?
?
1
1
?Silberschatz,Korth and Sudarshan3.27Database System Concepts
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
?Silberschatz,Korth and Sudarshan3.28Database System Concepts
Customers With An Account But No Loan
?Silberschatz,Korth and Sudarshan3.29Database System Concepts
Cartesian-Product Operation-Example
Relations r,s:
r x s:
A B
?
?
1
2
A B
?
?
?
?
?
?
?
?
1
1
1
1
2
2
2
2
C D
?
?
?
?
?
?
?
?
10
10
20
10
10
10
20
10
E
a
a
b
b
a
a
b
b
C D
?
?
?
?
10
10
20
10
E
a
a
b
br
s
?Silberschatz,Korth and Sudarshan3.30Database System Concepts
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.
?Silberschatz,Korth and Sudarshan3.31Database System Concepts
Result of borrower ? loan
?Silberschatz,Korth and Sudarshan3.32Database System Concepts
Result of ? branch-name =,Perryridge”
(borrower ? loan)
?Silberschatz,Korth and Sudarshan3.33Database System Concepts
Composition of Operations
? Can build expressions using multiple operations
? Example,?A=C(r x s)
? r x s
? ?A=C(r x s)
A B
?
?
?
?
?
?
?
?
1
1
1
1
2
2
2
2
C D
?
?
?
?
?
?
?
?
10
10
20
10
10
10
20
10
E
a
a
b
b
a
a
b
b
A B C D E
?
?
?
1
2
2
?
?
?
10
20
20
a
a
b
?Silberschatz,Korth and Sudarshan3.34Database System Concepts
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,
?Silberschatz,Korth and Sudarshan3.35Database System Concepts
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)
?Silberschatz,Korth and Sudarshan3.36Database System Concepts
Example Queries
? Find all loans of over $1200
?Find the loan number for each loan of an amount greater
than $1200
?amount > 1200 (loan)
?loan-number (?amount > 1200 (loan))
?Silberschatz,Korth and Sudarshan3.37Database System Concepts
Example Queries
? Find the names of all customers who have a loan,an
account,or both,from the bank
?Find the names of all customers who have a loan and an
account at bank.
?customer-name (borrower) ??customer-name (depositor)
?customer-name (borrower) ??customer-name (depositor)
?Silberschatz,Korth and Sudarshan3.38Database System Concepts
Example Queries
? Find the names of all customers who have a loan at the
Perryridge branch.
? 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)
?customer-name (?branch-name=“Perryridge”
(?borrower.loan-number = loan.loan-number(borrower x loan)))
?Silberschatz,Korth and Sudarshan3.39Database System Concepts
Example Queries
? Find the names of all customers who have a loan at the
Perryridge branch.
? Query 2
?customer-name(?loan.loan-number = borrower.loan-number(
(?branch-name =,Perryridge”(loan)) x borrower))
?Query 1
?customer-name(?branch-name =,Perryridge” (
?borrower.loan-number = loan.loan-number(borrower x loan)))
?Silberschatz,Korth and Sudarshan3.40Database System Concepts
Example Queries
Find the largest account balance
? Rename account relation as d
? The query is:
?balance(account) - ?account.balance
(?account.balance < d.balance (account x ?d (account)))
?Silberschatz,Korth and Sudarshan3.41Database System Concepts
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
?Silberschatz,Korth and Sudarshan3.42Database System Concepts
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
?Silberschatz,Korth and Sudarshan3.43Database System Concepts
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)
?Silberschatz,Korth and Sudarshan3.44Database System Concepts
Set-Intersection Operation - Example
? Relation r,s:
? r ? s
A B
?
?
?
1
2
1
A B
?
?
2
3
r s
A B
? 2
?Silberschatz,Korth and Sudarshan3.45Database System Concepts
? Notation,r s
Natural-Join Operation
? Let r and s be relations on schemas R and S respectively,
Then,r s is a relation on schema R ? S,
? r s=?R ?S (?r.A1== s.A1 ?r.A2 = s.A2 ?… ?r.An = s.An (r x s))
Where is R ?S={A1,A2,…,An}.
? Example:
R = (A,B,C,D)
S = (E,B,D)
? Result schema = (A,B,C,D,E)
? 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))
?Silberschatz,Korth and Sudarshan3.46Database System Concepts
?customer-name,loan-number,amount
(borrower loan)
?customer-name,loan-number,amount (?borrower.loan-
number = loan.loan-number(borrower x loan))
?Silberschatz,Korth and Sudarshan3.47Database System Concepts
Natural Join Operation – Example
? Relations r,s:
A B
?
?
?
?
?
1
2
4
1
2
C D
?
?
?
?
?
a
a
b
a
b
B
1
3
1
2
3
D
a
a
a
b
b
E
?
?
?
?
?
r
A B
?
?
?
?
?
1
1
1
1
2
C D
?
?
?
?
?
a
a
a
a
b
E
?
?
?
?
?
s
r s
?Silberschatz,Korth and Sudarshan3.48Database System Concepts
Division Operation
? Suited to queries that include the phrase,for all”.
? Let r and s be relations on schemas R and S respectively
where
? R = (A1,…,Am,B1,…,Bn)
? S = (B1,…,Bn)
The result of r ? s is a relation on schema
R – S = (A1,…,Am)
r ? s = { t | t ?? R-S(r) ?? u ? s ( tu ? r ) }
r ? s
?Silberschatz,Korth and Sudarshan3.49Database System Concepts
Division Operation – Example
Relations r,s:
r ? s,A
B
?
?
1
2
A B
?
?
?
?
?
?
?
?
?
?
?
1
2
3
1
1
1
3
4
6
1
2
r
s
?Silberschatz,Korth and Sudarshan3.50Database System Concepts
Another Division Example
A B
?
?
?
?
?
?
?
?
a
a
a
a
a
a
a
a
C D
?
?
?
?
?
?
?
?
a
a
b
a
b
a
b
b
E
1
1
1
1
3
1
1
1
Relations r,s:
r ? s:
D
a
b
E
1
1
A B
?
?
a
a
C
?
?
r
s
?Silberschatz,Korth and Sudarshan3.51Database System Concepts
Division Operation (Cont.)
? Property
? Let q – r ? s
? Then q is the largest relation satisfying q x s ? r
? Definition in terms of the basic algebra operation
Let r(R) and s(S) be relations,and let S ? R
r ? s = ?R-S (r) –?R-S ( (?R-S (r) x s) –?R-S,S(r))
To see why
? ?R-S,S(r) simply reorders attributes of r
? ?R-S(?R-S (r) x s) –?R-S,S(r)) gives those tuples t in
?R-S (r) such that for some tuple u ? s,tu ? r.
?Silberschatz,Korth and Sudarshan3.52Database System Concepts
Division Operation – Example
Relations r,s:
r ? s,A
B
?
?
1
2
A B
?
?
?
?
?
?
?
?
?
?
?
1
2
3
1
1
1
3
4
6
1
2
r
s
(?R-S (r) x s)
A B
?
?
?
?
?
?
?
?
?
?
1
2
1
2
1
2
1
2
1
2
(?R-S (r) x s) –?R-S,S(r)
A B
?
?
?
2
2
2
r ? s = ?R-S (r) –?R-S ( (?R-S (r) x s) –?R-S,S(r))
?Silberschatz,Korth and Sudarshan3.53Database System Concepts
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.
? Example,Write r ? s as
temp1??R-S (r)
temp2 ??R-S ((temp1 x s) – ?R-S,S (r))
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.
?Silberschatz,Korth and Sudarshan3.54Database System Concepts
Example Queries
? Find all customers who have an account from at least
the,Downtown” and the Uptown” branches.
where CN denotes customer-name and BN denotes
branch-name.
Query 1
?CN(?BN=“Downtown”(depositor account)) ?
?CN(?BN=“Uptown”(depositor account))
Query 2
?customer-name,branch-name(depositor account)
??temp(branch-name) ({(“Downtown”),(“Uptown”)})
?Silberschatz,Korth and Sudarshan3.55Database System Concepts
? Find all customers who have an account at all branches
located in Brooklyn city.
Example Queries
?customer-name,branch-name (depositor account)
??branch-name (?branch-city =,Brooklyn” (branch))
?Silberschatz,Korth and Sudarshan3.56Database System Concepts
Extended Relational-Algebra-Operations
? Generalized Projection
? Outer Join
? Aggregate Functions
?Silberschatz,Korth and Sudarshan3.57Database System Concepts
Generalized Projection
? Extends the projection operation by allowing
arithmetic functions to be used in the projection list.
? F1,F2,…,Fn (E)
? E is any relational-algebra expression
? Each of F1,F2,…,Fn are are arithmetic expressions
involving constants and attributes in the schema of E.
? Given relation credit-info(customer-name,limit,credit-
balance),find how much more each person can
spend,
?customer-name,limit – credit-balance (credit-info)
?Silberschatz,Korth and Sudarshan3.58Database System Concepts
Aggregate Functions and Operations
? Aggregation function takes a collection of values and
returns a single value as a result.
avg,average value
min,minimum value
max,maximum value
sum,sum of values
count,number of values
? Aggregate operation in relational algebra
G1,G2,…,Gn g F1( A1),F2( A2),…,Fn( An) (E)
? E is any relational-algebra expression
? G1,G2 …,Gn is a list of attributes on which to group (can
be empty)
? Each Fi is an aggregate function
? Each Ai is an attribute name
?Silberschatz,Korth and Sudarshan3.59Database System Concepts
Aggregate Operation – Example
? Relation r:
A B
?
?
?
?
?
?
?
?
C
7
7
3
10
g sum(c)(r)
sum-C
27
?Silberschatz,Korth and Sudarshan3.60Database System Concepts
Aggregate Operation – Example
? Relation account grouped by branch-name:
branch-name g sum(balance) (account)
branch-name account-number balance
Perryridge
Perryridge
Brighton
Brighton
Redwood
A-102
A-201
A-217
A-215
A-222
400
900
750
750
700
branch-name balance
Perryridge
Brighton
Redwood
1300
1500
700
?Silberschatz,Korth and Sudarshan3.61Database System Concepts
Aggregate Functions (Cont.)
? Result of aggregation does not have a name
? Can use rename operation to give it a name
? For convenience,we permit renaming as part of
aggregate operation
branch-name g sum(balance) as sum-balance (account)
?Silberschatz,Korth and Sudarshan3.62Database System Concepts
Outer Join
? An extension of the join operation that avoids loss of
information.
? Computes the join and then adds tuples form one relation
that does not match tuples in the other relation to the result
of the join,
? Uses null values:
? null signifies that the value is unknown or does not exist
? All comparisons involving null are (roughly speaking)
false by definition.
?Will study precise meaning of comparisons with nulls
later
?Silberschatz,Korth and Sudarshan3.63Database System Concepts
Outer Join – Example
? Relation loan
? Relation borrower
customer-name loan-number
Jones
Smith
Hayes
L-170
L-230
L-155
3000
4000
1700
loan-number amount
L-170
L-230
L-260
branch-name
Downtown
Redwood
Perryridge
?Silberschatz,Korth and Sudarshan3.64Database System Concepts
Outer Join – Example
? Inner Join(内连接 ）
loan Borrower
loan-number amount
L-170
L-230
3000
4000
customer-name
Jones
Smith
branch-name
Downtown
Redwood
Jones
Smith
null
loan-number amount
L-170
L-230
L-260
3000
4000
1700
customer-namebranch-name
Downtown
Redwood
Perryridge
? Left Outer Join（左外连接）
loan Borrower
?Silberschatz,Korth and Sudarshan3.65Database System Concepts
Outer Join – Example
? Right Outer Join（右外连接）
loan borrower
loan borrower
? Full Outer Join（全连接）
loan-number amount
L-170
L-230
L-155
3000
4000
null
customer-name
Jones
Smith
Hayes
branch-name
Downtown
Redwood
null
loan-number amount
L-170
L-230
L-260
L-155
3000
4000
1700
null
customer-name
Jones
Smith
null
Hayes
branch-name
Downtown
Redwood
Perryridge
null
?Silberschatz,Korth and Sudarshan3.66Database System Concepts
Modification of the Database
? The content of the database may be modified using
the following operations:
? Deletion
? Insertion
? Updating
? All these operations are expressed using the
assignment operator.
?Silberschatz,Korth and Sudarshan3.67Database System Concepts
Deletion
? A delete request is expressed similarly to a query,
except instead of displaying tuples to the user,the
selected tuples are removed from the database.
? Can delete only whole tuples; cannot delete values on
only particular attributes
? A deletion is expressed in relational algebra by:
r ? r – E
where r is a relation and E is a relational algebra
query.
?Silberschatz,Korth and Sudarshan3.68Database System Concepts
Deletion Examples
? Delete all account records in the Perryridge branch.
?Delete all accounts at branches located in Needham.
r1 ? ??branch-city =,Needham” (account branch)
r2 ? ?branch-name,account-number,balance (r1)
account ? account – r2
?Delete all loan records with amount in the range of 0 to 50
loan ? loan –??amount ??0?and amount ? 50 (loan)
account ? account – ??branch-name =,Perryridge” (account)
?Silberschatz,Korth and Sudarshan3.69Database System Concepts
Insertion
? To insert data into a relation,we either:
? specify a tuple to be inserted
? write a query whose result is a set of tuples to be
inserted
? in relational algebra,an insertion is expressed by:
r ? r ? E
where r is a relation and E is a relational algebra
expression.
? The insertion of a single tuple is expressed by letting
E be a constant relation containing one tuple,
?Silberschatz,Korth and Sudarshan3.70Database System Concepts
Insertion Examples
? Insert information in the database specifying that Smith
has $1200 in account A-973 at the Perryridge branch.
? Provide as a gift for all loan customers in the Perryridge
branch,a $200 savings account,Let the loan number serve
as the account number for the new savings account.
account ? account ?{(“Perryridge”,A-973,1200)}
depositor ? depositor ? {(“Smith”,A-973)}
r1 ? (?branch-name =,Perryridge” (borrower loan))
account ? account ? ?branch-name,loan-number,200 (r1)
depositor ? depositor ??customer-name,loan-number(r1)
?Silberschatz,Korth and Sudarshan3.71Database System Concepts
Updating
? A mechanism to change a value in a tuple without
charging all values in the tuple
? Use the generalized projection operator to do this task
r ? ? F1,F2,…,FI,(r)
? Each Fi is either
? the ith attribute of r,if the ith attribute is not
updated,or,
? if the attribute is to be updated Fi is an expression,
involving only constants and the attributes of r,
which gives the new value for the attribute
?Silberschatz,Korth and Sudarshan3.72Database System Concepts
Update Examples
? Make interest payments by increasing all balances by 5
percent.
? Pay all accounts with balances over $10,000 6
percent interest
and pay all others 5 percent
account ? ? AN,BN,BAL * 1.06 (? BAL ? 10000 (account))
? ?AN,BN,BAL * 1.05 (?BAL ? 10000 (account))
account ?? AN,BN,BAL * 1.05 (account)
where AN,BN and BAL stand for account-number,
branch-name and balance,respectively.
?Silberschatz,Korth and Sudarshan3.73Database System Concepts
Views
? In some cases,it is not desirable for all users to see
the entire logical model (i.e.,all the actual relations
stored in the database.)
? Consider a person who needs to know a customer’s
loan number but has no need to see the loan amount,
This person should see a relation described,in the
relational algebra,by
?customer-name,loan-number (borrower loan)
? Any relation that is not of the conceptual model but is
made visible to a user as a,virtual relation” is called a
view.
?Silberschatz,Korth and Sudarshan3.74Database System Concepts
View Definition
? A view is defined using the create view statement which has
the form
create view v as <query expression>
where <query expression> is any legal relational algebra
query expression,The view name is represented by v.
? Once a view is defined,the view name can be used to refer
to the virtual relation that the view generates.
? View definition is not the same as creating a new relation by
evaluating the query expression
? Rather,a view definition causes the saving of an
expression; the expression is substituted into queries
using the view.
?Silberschatz,Korth and Sudarshan3.75Database System Concepts
View Examples
? Consider the view (named all-customer) consisting of
branches and their customers.
? We can find all customers of the Perryridge branch by writing:
create view all-customer as
?branch-name,customer-name (depositor account)
??branch-name,customer-name (borrower loan)
?customer-name
(?branch-name =,Perryridge” (all-customer))
?Silberschatz,Korth and Sudarshan3.76Database System Concepts
Updates Through View
? Database modifications expressed as views must be
translated to modifications of the actual relations in the
database.
? Consider the person who needs to see all loan data in the
loan relation except amount,The view given to the person,
branch-loan,is defined as,
create view branch-loan as
?branch-name,loan-number (loan)
? Since we allow a view name to appear wherever a relation
name is allowed,the person may write:
branch-loan ? branch-loan ?{(“Perryridge”,L-37)}
?Silberschatz,Korth and Sudarshan3.77Database System Concepts
Updates Through Views (Cont.)
? The previous insertion must be represented by an insertion into the
actual relation loan from which the view branch-loan is constructed.
? An insertion into loan requires a value for amount,The insertion can
be dealt with by either.
? rejecting the insertion and returning an error message to the
user.
? inserting a tuple (“L-37”,“Perryridge”,null) into the loan relation
? Some updates through views are impossible to translate into
database relation updates
? create view v as ?branch-name =,Perryridge” (account))
v ? v ? (L-99,Downtown,23)
? Others cannot be translated uniquely
? all-customer? all-customer ?{(“Perryridge”,“John”)}
?Have to choose loan or account,and
create a new loan/account number!
?Silberschatz,Korth and Sudarshan3.78Database System Concepts
Views Defined Using Other Views
? One view may be used in the expression defining
another view
? A view relation v1 is said to depend directly(直接依赖 )
on a view relation v2 if v2 is used in the expression
defining v1
? A view relation v1 is said to depend on (依赖 ) view
relation v2 if either v1 depends directly to v2 or there is
a path of dependencies from v1 to v2
? A view relation v is said to be recursive (递归 ) if it
depends on itself.
?Silberschatz,Korth and Sudarshan3.79Database System Concepts
View Expansion（ 视图展开 ）
? A way to define the meaning of views defined in terms of
other views.
? Let view v1 be defined by an expression e1 that may itself
contain uses of view relations.
? View expansion of an expression repeats the following
replacement step:
repeat
Find any view relation vi in e1
Replace the view relation vi by the expression defining vi
until no more view relations are present in e1
? As long as the view definitions are not recursive,this loop will
terminate
?Silberschatz,Korth and Sudarshan3.80Database System Concepts
Tuple Relational Calculus
? A nonprocedural query language,where each query is of
the form
{t | P (t) }
? It is the set of all tuples t such that predicate P is true for t
? t is a tuple variable,t[A] denotes the value of tuple t on
attribute A
? t ? r denotes that tuple t is in relation r
? P is a formula similar to that of the predicate calculus
?Silberschatz,Korth and Sudarshan3.81Database System Concepts
Predicate Calculus Formula
1,Set of attributes and constants
2,Set of comparison operators,(e.g.,?,?,?,?,?,?)
3,Set of connectives,and (?),or (v)? not (?)
4,Implication (?),x ? y,if x if true,then y is true
x ? y ???x v y
5,Set of quantifiers:
???t ??r (Q(t)) ??”there exists” a tuple in t in relation r
such that predicate Q(t) is true
??t ??r (Q(t)) ??Q is true,for all” tuples t in relation r
?Silberschatz,Korth and Sudarshan3.82Database System Concepts
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)
?Silberschatz,Korth and Sudarshan3.83Database System Concepts
Example Queries
? Find the loan-number,branch-name,and amount
for loans of over $1200
?Find the loan number for each loan of an amount greater
than $1200
Notice that a relation on schema [loan-number] is implicitly
defined by the query
{t | ? s ??loan (t[loan-number] = s[loan-number] ? s [amount] ? 1200)}
{t | t ? loan ? t [amount] ? 1200}
?Silberschatz,Korth and Sudarshan3.84Database System Concepts
Example Queries
? Find the names of all customers having a loan,an account,
or both at the bank
{t | ?s ? borrower( t[customer-name] = s[customer-name])
??u ? depositor( t[customer-name] = u[customer-name])
? Find the names of all customers who have a loan and an
account at the bank
{t | ?s ? borrower( t[customer-name] = s[customer-name])
??u ? depositor( t[customer-name] = u[customer-name])
?Silberschatz,Korth and Sudarshan3.85Database System Concepts
Example Queries
? Find the names of all customers having a loan at the
Perryridge branch
{t | ?s ? borrower( t[customer-name] = s[customer-name]
??u ? loan(u[branch-name] =,Perryridge”
? u[loan-number] = s[loan-number]))
? not ?v ? depositor (v[customer-name] =
t[customer-name]) }
? Find the names of all customers who have a loan at the
Perryridge branch,but no account at any branch of the
bank
{t | ?s ? borrower(t[customer-name] = s[customer-name]
??u ? loan(u[branch-name] =,Perryridge”
? u[loan-number] = s[loan-number]))}
?Silberschatz,Korth and Sudarshan3.86Database System Concepts
Example Queries
? Find the names of all customers having a loan from
the Perryridge branch,and the cities they live in
{t | ?s ? loan(s[branch-name] =,Perryridge”
??u ? borrower (u[loan-number] = s[loan-number]
? t [customer-name] = u[customer-name])
?? v ? customer (u[customer-name] = v[customer-name]
? t[customer-city] = v[customer-city])))}
?Silberschatz,Korth and Sudarshan3.87Database System Concepts
Example Queries
? Find the names of all customers who have an account
at all branches located in Brooklyn:
{t | ? c ? customer (t[customer.name] = c[customer-name]) ?
? s ? branch(s[branch-city] =,Brooklyn” ?
? u ? account ( s[branch-name] = u[branch-name]
?? s ? depositor ( t[customer-name] = s[customer-name]
? s[account-number] = u[account-number] )) )}
?Silberschatz,Korth and Sudarshan3.88Database System Concepts
Safety of Expressions
? It is possible to write tuple calculus expressions that
generate infinite relations.
? For example,{t | ? t?? r} results in an infinite relation if the
domain of any attribute of relation r is infinite
? To guard against the problem,we restrict the set of
allowable expressions to safe expressions.
? An expression {t | P(t)} in the tuple relational calculus is safe
if every component of t appears in one of the relations,
tuples,or constants that appear in P
? NOTE,this is more than just a syntax condition,
?E.g,{ t | t[A]=5 ? true } is not safe --- it defines an
infinite set with attribute values that do not appear in
any relation or tuples or constants in P,
?Silberschatz,Korth and Sudarshan3.89Database System Concepts
Domain Relational Calculus
? A nonprocedural query language equivalent in power
to the tuple relational calculus
? Each query is an expression of the form:
{ ? x1,x2,…,x n ? | P(x1,x2,…,x n)}
? x1,x2,…,x n represent domain variables
? P represents a formula similar to that of the
predicate calculus
?Silberschatz,Korth and Sudarshan3.90Database System Concepts
Example Queries
? Find the loan-number,branch-name,and amount for
loans of over $1200
{? c,a ? | ? l (? c,l ?? borrower ??b(? l,b,a ?? loan ?
b =,Perryridge”))}
or {? c,a ? | ? l (? c,l ?? borrower ? ?l,“Perryridge”,a ?? loan)}
? Find the names of all customers who have a loan from
the Perryridge branch and the loan amount:
{? c ? | ? l,b,a (? c,l ?? borrower ?? l,b,a ?? loan ? a > 1200)}
? Find the names of all customers who have a loan of
over $1200
{? l,b,a ? | ? l,b,a ?? loan ? a > 1200}
?Silberschatz,Korth and Sudarshan3.91Database System Concepts
Example Queries
? Find the names of all customers having a loan,an
account,or both at the Perryridge branch:
{? c ? | ? s,n (? c,s,n ?? customer) ?
? x,y,z(? x,y,z ?? branch ? y =,Brooklyn”) ?
? a,b(? x,y,z ?? account ?? c,a ?? depositor)}
? Find the names of all customers who have an account at
all branches located in Brooklyn:
{? c ? | ? l ({? c,l ?? borrower
?? b,a(? l,b,a ?? loan ? b =,Perryridge”))
? ? a(? c,a ?? depositor
?? b,n(? a,b,n ?? account ? b =,Perryridge”))}
?Silberschatz,Korth and Sudarshan3.92Database System Concepts
Safety of Expressions
{ ? x1,x2,…,x n ? | P(x1,x2,…,x n)}
is safe if all of the following hold:
1.All values that appear in tuples of the expression are
values from dom(P) (that is,the values appear either in P or
in a tuple of a relation mentioned in P).
2.For every,there exists” subformula of the form ? x (P1(x)),
the subformula is true if an only if P1(x) is true for all values
x from dom(P1).
3,For every,for all” subformula of the form ?x (P1 (x)),the
subformula is true if and only if P1(x) is true for all values x
from dom (P1).
End of Chapter 3
