database systems

Database SystemsDatabase Systems

DataBase System

Haichang Gao , Software School , Xidian University 2

Major Content & GradeMajor Content & Grade Introduction * The Relational Model

*** SQL **** Transaction Management *** Database Design (E-R) *** Database Design (Normalization) ***

DataBase System


Unit 2 Unit 2 The Relational ModelThe Relational ModelRelational ModelRelational Algebra(关系代数 )Relational Calculus(关系演算 )

DataBase System


Relational ModelRelational ModelLaid down in 1969-1970 by E.F.CODD

“ A Relational Model of Data for Large Shared Data Bank” C ACM 1970

Mathematical Bases ( Relational Theory) Developed in 1980s Most commercial DBMS are Relational

DataBase System


Mathematical RelationsMathematical Relations Formally, given sets D1, D2, …. Dn, a relation r is a subset of

Cartesian product(笛卡尔积 ) D1 × D2 × … × Dn

Thus a relation( 关系 ) is a set of n-tuples (a1, a2, …, an) where 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 × customer-street × customer-city.

DataBase System


Mathematical RelationsMathematical Relations Each attribute of a relation has a name

The set of allowed values for each attribute is called the domain （域） of the attribute

Attribute values are (normally) required to be atomic; that is, indivisible Note: multivalued attribute( 多值属性 ) values are not

atomic Note: composite attribute ( 组合属性 ) 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

DataBase System


Mathematical RelationsMathematical Relations A1, A2, …, An are attributes

R = (A1, A2, …, An ) is a relation schema(关系模式 )

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)

DataBase System


Mathematical RelationsMathematical Relations The current values (relation instance, 关系实例 ) of a

relation are specified by a table An element t of r is a tuple(元组 ), represented by a row in a

table Order of tuples is irrelevant ( 元组是无序的 tuples may be

stored in an arbitrary order)

JonesSmithCurryLindsay

customer_name

MainNorthNorthPark

customer_street

HarrisonRyeRyePittsfield

customer_city

customer

attributes(or columns)

tuples(or rows)

DataBase System


Relational DatabaseRelational Database A Relational database consists of multiple relations Information about an enterprise is broken up into parts, with

each relation storing one part of the information Example:

The customer table stores information about customers

DataBase System


Relational DatabaseRelational Database

(b) The account table(c) The depositor table

The account table stores information about accounts.

The depositor table containing information about which customer owns which account

Storing all information as a single relation is NOT a good

idea!

DataBase System


KeyKey Let K R K is a superkey(超码 ) of R if values for K are sufficient to

identify a unique tuple of each possible relation r(R).

Example: {customer_id, customer_name} and {customer_id} are both superkeys of Customer.

by “possible r” we mean a relation r that could exist in the enterprise we are modeling.

DataBase System


KeyKey Let K R K is a superkey(超码 ) of R if values for K are sufficient to

identify a unique tuple of each possible relation r(R). K is a candidate key(候选码 ) if K is minimal.

Example: {customer_id} is a candidate key for Customer, since it is a superkey, and no subset of it is a superkey.

{customer_name} is a candidate key also, assuming no two customers can possibly have the same name.

DataBase System


KeyKey Primary key(主码 ): The candidate key that is selected to

identify tuples uniquely within the relation. And the others candidate key is called Alternate Keys(替换码 ). Example: for relation

customer (customer_id, customer_name, customer_street, customer_city) Candidate key : {customer_id}, {customer_name} Primary key: {customer_id}

DataBase System


KeyKey For the banking database, we have several relations to store data:

Foreign Key(外码 ): An attribute, or set of attributes,within one relation R1 that matches the candidate key of some(possibly the same) relation R2. Example: Attribute customer_name in table depositor is a

Foreign Key references to customer_name in customer.

DataBase System


KeyKey Usually, Primary Key and Foreign Key was indicated using

undirected line and dashed line. Example:

Find out PK and FK of the following schema:

Student(sno, sname, ssex, sbirth, sdept)

Course(cno, cname, cpno, ccredit)

SC(sno, cno, grade)

DataBase System


Integrity of Relational ModelIntegrity of Relational Model Entity integrity(实体完整性约束 ): In a base relation, NO

attribute of a primary key can be null.

Referential integrity(参照完整性约束 ): If a foreign key exists in a relation, either the foreign key value must match a candidate key value of some tuple in its home relation(被参照关系 ) or the foreign key value must be wholly null. Example:

customer (customer_name, customer_street, customer_city)

depositor (customer_name, account_number)

account (account_number, branch_name, balance)

DataBase System


Integrity of Relational ModelIntegrity of Relational Model Null(空值 ): Represents a value for an attribute that is

currently unknown or is not applicable for this tuple.

It is possible for tuples to have a null value, denoted by null,

for some of their attributes

The result of any arithmetic expression involving null is null.

Aggregate functions simply ignore null values (as in SQL)

For duplicate elimination and grouping, null is treated like

any other value, and two nulls are assumed to be the same

(as in SQL)

Null value is NOT 0 of integer, NOT empty string , it is

JUST NULL.

DataBase System


Integrity of Relational ModelIntegrity of Relational Model Comparisons with null values return the special truth value:

unknown If false was used instead of unknown, then not (A < 5)

would not be equivalent to A >= 5 Three-valued logic using the truth value unknown:

OR: (unknown or true) = true, (unknown or false) = unknown (unknown or unknown) = unknown

AND: (true and unknown) = unknown, (false and unknown) = false, (unknown and unknown) = unknown

NOT: (not unknown) = unknown Result of select predicate is treated as false if it evaluates to

unknown

DataBase System



DataBase System


Manipulation of Relational ModelManipulation of Relational Model Query Language is a Language in which user requests

information from the database. Categories of languages

Procedural( 过程化的 ) Non-procedural, or declarative( 非过程化的，声明性的 )

Mathematical languages: Relational algebra( 关系代数 ) Tuple relational calculus( 元组演算 ) Domain relational calculus ( 域演算 )

Characteristics of Relational languages all set-at-a-time ( 一次一集合 )

operands & results are whole tables Closure property （封闭性）

the output from one operation can become input to another.

DataBase System


Relational AlgebraRelational Algebra Relational algebra is a Procedural language

Basic operators select( 选择 ): project( 投影 ): π union( 并 ): ∪ set difference( 集合差 ): – Cartesian product( 笛卡尔积 ): × rename:

The operators take two or more relations as inputs and give a new relation as a result.

DataBase System


Basic operatorsBasic operators Select Operation(选择操作 )

Notation: p(r)

p is called the selection predicate( 条件谓词 ) Defined as:

p(r) = {t | t r p(t)}

Where p is a formula in propositional calculus (命题演算 ) consisting of terms connected by : (and), (or), (not)Each term is one of:

<attribute> op <attribute> (or <constant>)

where op is one of: =, , >, , <, (comparison operator)

Example of selection: branch_name=“Perryridge”(account)

DataBase System


Basic operatorsBasic operators

• Relation r A B C D

15

1223

773

10

A=B D > ‘5’ (r)A B C D

123

710

Select Operation(选择操作 )

DataBase System


Basic operatorsBasic operators Project Operation(投影操作 )

Notation: π 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 Example: For relation

account (account_number, branch_name, balance)

To eliminate the branch_name attribute of account

π account_number, balance (account)

DataBase System


Basic operatorsBasic operators Project Operation(投影操作 )

Relation r: A B C

10203040

1112

A C

1112

=

A C

112

π A,C (r)

DataBase System


Basic operatorsBasic operators Union Operation(并操作 )

Notation: r s r union s Defined as:

r s = {t | t r t s} For r s to be valid:

1. r, s must have the same degree( 列数相同 )

2. The attribute domains must be compatible ( 域相容 ) Example: find all customers with either an account or a loan

π customer_name (depositor) π customer_name (borrower)

DataBase System


Basic operatorsBasic operators Union Operation(并操作 )

Relations r, s:

r s:

A B

121

r

A B

23

s

A B

1213

The 1st column of r deals with the same type of values(domain) as does the 1st column of s ,and 2nd column also, and also for all attributes.

DataBase System


Basic operatorsBasic operators Difference Operation(差操作 )

Notation: r – s r minus s Defined as:

r – s = {t | t r t s} Set differences must be taken between compatible relations.

r and s must have the same degree attribute domains of r and s must be compatible

DataBase System


Basic operatorsBasic operators Difference Operation (差操作 )

Relations r, s: A B

121

r

A B

23

s

r – s : A B

11

DataBase System


Basic operatorsBasic operators Cartesian-Product Operation(笛卡尔积操作 )

Notation: r × s r times s Defined as:

r × s = {tq | t r q s} Assume that attributes of r(R) and s(S) are disjoint. (That is,

R S = , NO same attribute in R and S). If attributes of r(R) and s(S) are not disjoint, then renaming

must be used.

DataBase System


Basic operatorsBasic operators Cartesian-Product Operation(笛卡尔积操作 )

Relations r, s:

r × s:A B

11112222

C D

1019201010102010

E

aabbaabb

A B

12

r

C D

10102010

E

aabb

s

DataBase System


Basic operatorsBasic operators Rename Operation(重命名操作 )

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 degree n, then

x (A1, A2, …, An) (E)

A1, A2, …., An returns the result of expression E under the name x, and with the attributes renamed to A1, A2, …., An.

DataBase System

Haichang Gao , Software School , Xidian University

33

Example QueriesExample Queries Find the largest account balance.

account_number

Branch_name

balance

A-101 Downtown 500

A-215 Mianus 700

A-102 Perryridge 400

… …… …

account

A1.balance >A2.balance (A1(account) × A2(account) ) ?

πA1.balance (A1.balance < A2.balance (A1(account) × A2(account) ) )πbalance(account) –

最小值不被选中 !

DataBase System



DataBase System


Relational Algebra expressionRelational Algebra expression 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 × E2

p(E1), P is a predicate on attributes in E1

πs(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

DataBase System


Example QueriesExample Queries Banking DataBase:

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

DataBase System


Example QueriesExample Queries Find the loan number for each loan of an amount greater than

$1200. loan_number

branch_name amount

L-11 Round Hill 900L-14 Downtown 1500L-15 Perryridge 1500L-16 Perryridge 1300L-17 Downtown 1000L-23 Redwood 2000L-93 Mianus 500

loanamount > ‘1200’ (loan)

loan_number

L-14L-15L-16L-23

注：除特别要求外，写查询表达式时不需要写出结果集！

πloan_number ( )

DataBase System


Example QueriesExample Queries

Find the customers name who have at least one deposit of a balance greater than $700.

account_number

branch_name

balance

A-101 Downtown 500A-215 Mianus 700A-102 Perryridge 400A-305 Round Hill 350A-201 Brighton 900A-222 Redwood 700A-217 Brighton 750

account

customer_name

account_number

Hayes A-102Johnson A-101Johnson A-201Jones A-217Lindsay A-222Smith A-215Turner A-305

depositor

balance>’700’(account)π account_number( )×depositor

πCustomer_name(account.account_number = depositor.account_number (

) )πCustomer_name(account.account_number = depositor.account_number balance>’700’ (

account ×depositor ) )

A1:

A2:

DataBase System



Find all customers who have at least two deposits.

customer_name

account_number


depositor

D1.cname D1.a# D2.cname D2.a#Hayes A-102 Hayes A-102Hayes A-102 Johnson A-101Hayes A-102 Johnson A-201

…… …… …… ……Johnson A-101 Johnson A-201

…… …… …… ……Johnson A-201 Johnson A-101

…… …… …… ……

D1(cname,a#)(depositor) × D2(cname,a#)( depositor)

πD1.cname(D1.cname=D2.cname D1.a#<>D2.a#∧ (

) )

DataBase System



account_number

Branch_name

balance

A-101 Downtown 500

A-215 Mianus 700


… …… …

accountA.ano A.bn A.b B.ano B.bn B.bA-101 Downtown 500 A-101 Downtown 500A-101 Downtown 500 A-215 Mianus 700A-101 Downtown 500 A-102 Perryridge 400A-215 Mianus 700 A-101 Downtown 500A-215 Mianus 700 A-215 Mianus 700A-215 Mianus 700 A-102 Perryridge 400A-102 Perryridge 400 A-101 Downtown 500A-102 Perryridge 400 A-215 Mianus 700A-102 Perryridge 400 A-102 Perryridge 400

… ……. … … …… …

A.b > B.b (A × B)?

πA1.balance (A1.balance < A2.balance (A1(account) × A2(account) ) )πbalance(account) –

A.b < B.b (A × B)?

DataBase System


Additional OperationsAdditional Operations We define additional operations that do not add any

power to the relational algebra, but that simplify common queries.

Set intersection( 集合交 ) Natural join( 自然连接 ) Division( 除法 )

DataBase System


Additional OperationsAdditional Operations Intersection Operation(交操作 )

Notation: r s Assume:

r, s have the same degree attributes of r and s are compatible

Defined as:

r s ={ t | t r t s } Note: r s = r - (r - s)

πcustomer_name (borrower ) πcustomer_name (depositor )

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

DataBase System


Additional OperationsAdditional Operations -Join Operation(-连接 )

-Join is a general join which don’t need there are same attribute for both operators, and the join condition can be any compare operations. That is:

R times S WHERE F

R S = { | tr R ts S tr[A] ts[B] } AθB

tr ts

Equi-join is another join. Its join condition is equal operator.

R S = { | tr R ts S tr[A]= ts[B] } A=B

tr ts

DataBase System


Additional OperationsAdditional Operations -Join Operation(-连接 )

Relations r, s:RA B C

a1 b15

a1 b26

a2 b38

a2 b412

SB E

b13

b27

b310

b32

b52

R SB = B

A R.B C S.B E

a1 b1 5 b1 3

a1 b2 6 b2 7

a2 b3 8 b3 10

a2 b3 8 b3 2

R SC < E

ES.BCR.BA

10b35b1a1

7b25b1a1

10b38b3a2

10b36b2a1

7b26b2a1

DataBase System


Additional OperationsAdditional Operations Natural join Operation(自然连接 )

Notation: r s Let r and s be relations on schemas R and S respectively.The

result is a relation on schema R S which is obtained by considering 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 RS, a tuple t is added 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) r s is defined as:π r.A, r.B, r.C, r.D, s.E (r.B = s.B r.D = s.D (r × s))

DataBase System


Additional OperationsAdditional Operations Natural join Operation(自然连接 )

Relations r, s:

A B

12412

C D

aabab

r

A B C D E

B

13123

D

aaabb

E

s

r s

1 a 1 a 1 a 1 a 2 b

DataBase System


Example QueriesExample QueriesAdditional Operations simplify the relational

algebra expressions !

balance>’700’(account)π account_number( )×depositor

πCustomer_name(account.account_number = depositor.account_number (

) )

A1:

πCustomer_name(account.account_number = depositor.account_number balance>’700’ (

account ×depositor ) )

A2:


account depositorA3: πCustomer_name(balance>’700’ ( ) )

DataBase System



(exercise)Find all customers who have at least two deposits in different branch.

customer_name

account_number


depositor

account_number

branch_name

balance


account

πD1.Customer_name( D1.Customer_name =D2. Customer_name

D1. Account_number<>D2. Account_number D1.branch_name<>D2. branch_name(

)

× D2(depositor account)D1(depositor account)

DataBase System




customer_name

account_number


depositor

account_number

branch_name

balance


account

depositor.account_number=account.account_number(depositor × account))D1(cname,a#,bname)(πdepositor.customer_name,account_number,branch_name

D2(cname,a#,bname)(πdepositor.customer_name,account_number,branch_name

depositor.account_number=account.account_number(depositor × account)) ×

πD1.cname(D1.cname=D2.cname D1.a#<>D2.a# D1.bname<>D2.bname(

)

DataBase System


Additional OperationsAdditional Operations Division Operation(除 )

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)

Defined as:

r s = { t | t π R-S(r) u s → tu r ) }

DataBase System



Relations r, s:

r s:

B

A

12

A B

12311134612

r

s

DataBase System



Another Defined as:

)}(|][{ SYRtXtSR Yxrr

Yx={y | tr R ∈ y=tr[Y] x=tr[X]}

A B

aaaaaaaa

C D

aabababb

E

11113111

D

ab

E

11

r sr s:

A B C

Relations r, s:

a a

DataBase System



Property Let q is the result of r s Then q is the largest relation satisfying q × s r

Definition in terms of the basic algebra operation

Let r(R) and s(S) be relations, and let S R

r s = π R-S (r) –πR-S ( (π R-S (r) × s) – πR-S,S(r))

DataBase System


Example QueriesExample Queries Find all customers who have an account at all branches

located in Brooklyn city.

branch_name branch_city assets

Brighton Brooklyn 7100000Downtown Brooklyn 9000000Mianus Horseneck 400000North Town Rye 3700000Perryridge Horseneck 1700000Pownal Bennington 300000Redwood PaloAlto 2100000Round Hill Horseneck 8000000

branch

πbranch_name (branch_city = “Brooklyn” (branch))

account_number

branch_name

balance


account

customer_name

JohnsonSmithHayesTurnerJohnsonLindsayJones

π customer_name, branch_name (depositor account)

DataBase System


Extended OperationsExtended Operations Outer join Operation(外连接 )

The (left) Outer join is a join on which tuples from R that do not have matching values in the common attributes of S are alse included in the result relation. Missing values in the second relation are set to null.

A B

12

s

A C

10102010

D

aabb

r

A C D B

10 a 1 10 a 2 20 b 2 10 b

r s

Notation: r s (left outer join )

DataBase System


Extended OperationsExtended Operationsloan_numbe

rbranch_name amount

L-170 Downtown 3000L-230 Redwood 4000L-260 Perryridge 1700

loan

customer_name loan_number

Jones L-170Smith L-230Hayes L-155

borrower

loan_number branch_Name amount customer_nameL-170 Downtown 3000 JonesL-230 Redwood 4000 Smith

Inner Join: loan borrower

Left Outer Join : loan borrower

loan_number branch_Name amount customer_name

L-170 Downtown 3000 Jones

L-230 Redwood 4000 Smith

L-260 Perryridge 1700 NULL

DataBase System


Extended OperationsExtended Operationsloan_numbe

rbranch_name amount

L-170 Downtown 3000L-230 Redwood 4000L-260 Perryridge 1700

loan

customer_name Loan_number

Jones L-170Smith L-230Hayes L-155

borrower

loan_number nranch_name amount customer_nameL-170 Downtown 3000 JonesL-230 Redwood 4000 SmithL-155 Hayes

loan_number branch_name amount customer_nameL-170 Downtown 3000 JonesL-230 Redwood 4000 SmithL-260 Perryridge 1700L-155 Hayes

Right Outer Join : loan borrower

Full Outer Join : loan borrower

DataBase System


Homework 1Homework 1 2.1

2.5

DataBase System


Unit 2 Unit 2 The Relational ModelThe Relational ModelOverviewRelational ModelRelational Algebra(关系代数 )Relational Calculus(关系演算 )

(教材 5.1 节 )

DataBase System


Relational calculusRelational calculus Relational Calculus Based on a branch of mathematical

logical called the predicate calculus(谓词演算 ).

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, called component of t on attribute A R(t) denotes that tuple t is in relation R P is a formula of the predicate calculus

DataBase System


Relational calculusRelational calculus Tuple-relational-Calculus Formula

R(t) is an atom formula(原子公式 ) <t[A] θ t[B]> (or <t[A] θ c> or <c θ t[A] > ) is an atom

formula

θ is one of the comparison operators: (, , , , , ) if P1, P2 are formulae, then

P1, P1 P2, P1 P2, P1 P2 are formulae x(P(x)), x (P(x)) are formulae

Sets of equivalent expressions: P1 P2 (P1 P2 )

P1 P2 P1 P2

t ( P(t) ) t ( P(t) )

DataBase System


Relational calculusRelational calculus Example

987

654

321

CBA

R

965

643

321

CBA

S R1={ t | S(t) t[1]> ‘2’ }

CBA

965

643

R2={ t | R(t) S(t)}

A B C

4 5 6

7 8 9

R3={ t | S(t) u(R(u) t[3]<u[2]) }

A B C

1 2 3

3 4 6

DataBase System



987

654

321

CBA

R

965

643

321

CBA

S R4={ t | R(t) u(S(u)t[3]>u[1]) }

A B C

4 5 6

7 8 9

DataBase System



987

654

321

CBA

R

965

643

321

CBA

S R4={ t | R(t) u(S(u)t[3]>u[1]) }

A B C

4 5 6

7 8 9

R5={ t | u v( R(u) S(v) u[1]>v[2] t[1]=u[2]

t[2]=v[3] t[3]=u[1] )}

Process：1. Ensure the schema of result；2. Obtain tuples according the

predicate calculus .

R.AS.CR.B

435

798

768

738

DataBase System


Example QueriesExample Queries Banking DataBase:

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

DataBase System


Example QueriesExample Queries Find the loan_number, branch_name, and amount for loans of

over $1200. loan_number

branch_name amount

L-11 Round Hill 900L-14 Downtown 1500L-15 Perryridge 1500L-16 Perryridge 1300L-17 Downtown 1000L-23 Redwood 2000L-93 Mianus 500

loan{t | loan(t) t[amount] 1200}

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

{t | s( loan(s) s[amount ] 1200 t [loan_number] = s [loan_number ] )}

DataBase System




account_number

branch_name

balance


account

customer_name

account_number


depositor

{t | u( depositor(u) v( account(v) u[account_number ]

= v[account_number] v[balance ] > ‘700’ )

t[customer_name] = u[customer_name ] ) }

DataBase System



Find all customers who have at least two deposits.

customer_name

account_number


depositor

customer_name

account_number


depositor

{t | u v( depositor(u) depositor(v) u[customer_name ] =

v[customer_name] u[account_number] v [account_number]

t[customer_name] = u [customer_name ] ) }

u v

DataBase System



account_number

branch_name

balance

A-101 Downtown 500

A-215 Mianus 700


… …… …

account

account_number

branch_name

balance

A-101 Downtown 500

A-215 Mianus 700


… …… …

account

u v

{t | u ( account(u) v( account(v) u[balance] ≥

v[balance]) t[balance] = u [balance] ) }

A1:

A2: {t | u ( account(u) v( account(v) u[balance]

v[balance]) t[balance] = u [balance] ) }

x( P(x) Q(x) )

x(P(x) Q(x) )

DataBase System




customer_name

account_number


depositor

account_number

branch_name

balance


account

{t | u x ( depositor(u) account(x) u[account_number] =

x[account_number] v y ( depositor(v) account(y)

v[account_number] = y[account_number] u[customer_name] =

v[customer_name] x[branch_name] y[branch_name]

u[account_number] v[account_number] ) t[customer_name] =

u [customer_name] ) }

u

v

x

y

DataBase System



located in Brooklyn city.branch_name branch_city …

Brighton Brooklyn …Downtown Brooklyn …Mianus Horseneck …North Town Rye …Perryridge Horseneck …Pownal Bennington …Redwood PaloAlto …Round Hill Horseneck …

branch

account_number

branch_name

…

A-101 Downtown …A-215 Mianus …A-102 Perryridge …A-305 Round Hill …A-201 Brighton …A-222 Redwood …A-217 Brighton …

account{t | u ( depositor(u) v (branch(v) v[branch_city] = ‘Brooklyn’

w( account(w) w[account_number] = u[account_number]

w[branch_name] = v[branch_name]) ) t[customer_name] =

u[customer_name] ) }

customer_name

account_number


depositor

DataBase System



located in Brooklyn city.

{t | u ( depositor(u) v (branch(v) v[branch_city] = ‘Brooklyn’


w[branch_name] = v[branch_name]) ) t[customer_name] = u

[customer_name] ) }

等价转换

x( P(x) Q(x) )

x(P(x) Q(x) )

{t | u ( depositor(u) v (branch(v) v[branch_city] = ‘Brooklyn’


w[branch_name] = v[branch_name]) ) t[customer_name] =

u[customer_name] ) }

DataBase System


Example QueriesExample Queries Find the names of all customers having a loan from the

Perryridge branch, and the cities in which they live.

Banking DataBase:branch (branch_name, branch_city, assets)customer (customer_name, customer_street,

customer_city)depositor (customer_name, account_number)account (account_number, branch_name, balance)borrower (customer_name, loan_number)loan (loan_number, branch_name, amount)

{t | u v w (customer(u) borrower(v) loan(w) u[customer_name]

= v[customer_name] v[loan_number] = w[loan_number]

w[branch_name] = ‘Perryridge’ t[customer_name] =

u[customer_name] t[customer_city] = u[customer_city] ) }

DataBase System


Safety of ExpressionsSafety of ExpressionsIt is possible to write tuple calculus expressions

that generate infinite(无穷的 ) relations. For example, {t | R(t)} 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

IN Relational Algebra Operations we defined is safety.

DataBase System


Homework 2Homework 2 5.1 5.6

database systems

Documents

software school

xidian university

customerstreet customercity

single relation

relation schemae

possible relation rr

relational modellaid

database design er