dr. alexandra i. cristea acristea/ cs 319: theory of databases: c4

Dr. Alexandra I. Cristea

http://www.dcs.warwick.ac.uk/~acristea/

CS 319: Theory of Databases: C4

(provisionary) Content1. Generalities DB2. Temporal Data3. Integrity constraints (FD revisited)4. Relational Algebra (revisited)5. Query optimisation6. Tuple calculus7. Domain calculus8. Query equivalence9. LLJ, DP and applications10. The Askew Wall11. Datalog

… previous

FD

Relational model

• E.F. Codd: check wikipedia!• Papers:• http://www.cl.cam.ac.uk/Teaching/2003/

Databases/codd.pdf – Reprint at:

A relational Model of Data for Large Shared Data Banks

• Relational Completeness of Data Base Sublanguages

Relational Model*

• Structure of Relational Databases

• Fundamental Relational-Algebra-Operations

• Additional Relational-Algebra-Operations

*based on Silberschatz, Korth, Sudarshan Database System Concepts 5th Edition, Chapter 2

Example of a Relation Instance

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, …}

/* Set of all customer names */

– customer_street = {Main, North, Park, …} /* set of all street names*/

– customer_city = {Harrison, Rye, Pittsfield, …} /* set of all city names */

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

Attribute Types• Each attribute of a relation has a name• The set of allowed values for each attribute is called the

domain of the attribute• Attribute values are (normally) required to be atomic; that is,

indivisible– E.g. the value of an attribute can be an account number,

but cannot be a set of account numbers

• Domain is said to be atomic if all its members are atomic• The special value null is a member of every domain• The null value causes complications in the definition of many

operations– We shall ignore the effect of null values in our main presentation and

consider their effect later

Relation Schema• A1, A2, …, An are attributes

• R = (A1, A2, …, An ) is a relation schema

Example:

Customer_schema = (customer_name,

customer_street, customer_city)• r(R) denotes a relation r on the relation schema R

Example:

customer (Customer_schema)

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

JonesSmithCurryLindsay

customer_name

MainNorthNorthPark

customer_street

HarrisonRyeRyePittsfield

customer_city

customer

attributes(or columns)

tuples(or rows)

Relations are Unordered Order of tuples is irrelevant (tuples may be stored in an arbitrary order)

Example: account relation with unordered tuples

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

Storing all information as a single relation such as bank(account_number, balance, customer_name, ..)results in

– repetition of information

• e.g.,if two customers own an account (What gets repeated?)

– the need for null values

• e.g., to represent a customer without an account

• Normalization theory deals with how to design relational schemas

The customer Relation

The depositor Relation

Keys• Let K R• K is a superkey of R if values for K are

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

Example: {customer_name, customer_street} and {customer_name} are both superkeys of Customer, if no two customers can possibly have the same name

Keys (Cont.)

• K is a candidate key if K is minimal

• Example: {customer_name} : superkey + no subset of it is a superkey.

• Primary key: a candidate key chosen as the principal means of identifying tuples within a relation

Foreign Keys• A relation schema may have an attribute that

corresponds to the primary key of another relation. The attribute is called a foreign key.

Query Languages• Language in which user requests information

from the database.• Categories of languages

– Procedural– Non-procedural, or declarative

• “Pure” languages:– Relational algebra– Tuple relational calculus– Domain relational calculus

• Pure languages form underlying basis of query languages that people use.

Relational Algebra• Procedural language

• Six basic operators

– select: – project: – union: – set difference: – – Cartesian product: x

– rename: • The operators take one or two relations as

inputs and produce a new relation as a result.

Select Operation – Example Relation r A B C D

1

5

12

23

7

7

3

10

A=B ^ D > 5 (r) A B C D

1

23

7

10

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)

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)

Project Operation• Notation:

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: To eliminate the branch_name attribute of

account account_number, balance (account)

)( ,,, 21r

kAAA

Union Operation – Example• Relations r, s:

r s:

A B

1

2

1

A B

2

3

rs

A B

1

2

1

3

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 (example: 2nd column of r deals with the same type of values as does the 2nd

column of s)• Example: to find all customers with either an account

or a loancustomer_name (depositor) customer_name (borrower)

Set Difference Operation – Example

• Relations r, s:

r – s:

A B

1

2

1

A B

2

3

r

s

A B

1

1

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

Cartesian-Product Operation – Example

Relations r, s:

r x s:

A B

1

2

A B

11112222

C D

1010201010102010

E

aabbaabb

C D

10102010

E

aabbr

s

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.

Composition of Operations• Can build expressions using multiple

operations

• Example: A=C(r x s)

• r x sA=C(r x s)

A B

11112222

C D

1010201010102010

E

aabbaabbA B C D E

122

101020

aab

Rename Operation• Allows us to refer to results of RA expressions & to

refer to a relation by more than one name.

• Example: x (E)returns the expression E under the name X

• expression E under name X, with attributes renamed to A1 , A2 , …., An .

)(),...,,( 21E

nAAAx

• Revise basic operators!

Relational Algebra Operators thus …

• Procedural language with six basic operators

– select: – project: – union: – set difference: – – Cartesian product: x

– rename: • The operators take one or two relations as

inputs and produce a new relation as a result.

Banking Example

branch (branch_name, branch_city, assets)

customer (customer_name, customer_street, customer_city)

account (account_number, branch_name, balance)

loan (loan_number, branch_name, amount)

depositor (customer_name, account_number)

borrower (customer_name, loan_number)

Example QueriesFind 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))

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

customer_name (borrower) customer_name (depositor)

Example QueriesFind 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)))

Example QueriesFind 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)))

Example Queries• Find the largest account balance

– Strategy:• Find those balances that are not the largest

– Rename account relation as d so that we can compare each account balance with all others

• Use set difference to find those account balances that were not found in the earlier step.

– The query is:

balance(account) - account.balance

(account.balance < d.balance (account x d (account)))

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

Additional OperationsWe define additional operations that do not add any power

to the

relational algebra, but that simplify common queries.

• Set intersection• Natural join• Division• Assignment

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)

Set-Intersection Operation – Example

• Relation r, s:

• r s

A B

121

A B

23

r s

A B

2

Notation: r sNatural-Join Operation

• Let r and s be relations on schemas R and S respectively. Then, r s is a relation on schema R S obtained as follows:– Consider each pair of tuples tr from r and ts from s.

– If tr and ts have the same value on each of the attributes in R S, add a tuple t to the result, where• t has the same value as tr on r

• t has the same value as ts on s

Natural joint example

• 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))

Natural Join Operation – Example

• Relations r, s:

A B

12412

C D

aabab

B

13123

D

aaabb

E

r

A B

11112

C D

aaaab

E

s

r s

Division Operation• Notation: • 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 ) } Where tu means the concatenation of tuples t and u to

produce a single tuple

r s

Division Operation – Example

Relations r, s:

r s: A

B

1

2

A B

12311134612

r

s

Another Division ExampleA B

aaaaaaaa

C D

aabababb

E

11113111

Relations r, s:

r s:

D

ab

E

11

A B

aa

C

r

s

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 operationLet 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.

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.

Bank Example Queries Find the names of all customers who have a loan

and an account at the bank.

customer_name (borrower) customer_name (depositor)

• Find the name of all customers who have a loan at the

bank and the loan amount

customer_name, loan_number, amount (borrower loan)

Query 1

customer_name (branch_name = “Downtown” (depositor account ))

customer_name (branch_name = “Uptown” (depositor account))

Query 2

customer_name, branch_name (depositor account)

temp(branch_name) ({(“Downtown” ), (“Uptown” )})

Note that Query 2 uses a constant relation.

Bank Example Queries• Find all customers who have an account from at least the

“Downtown” and the “Uptown” branches.

• Find all customers who have an account at all

branches located in Brooklyn city.

Bank Example Queries

customer_name, branch_name (depositor account)

branch_name (branch_city = “Brooklyn” (branch))

Library Case

author bookreservation

copy borrow

title

publisheryear

date nameInitials

name

department

cancelled

departmentdepartment

name

from

topresent cpYear

Library database

book ( ISBN, title, publisher, year )

author (ISBN, initials, name )

copy (barcode, ISBN, department, cpYear, present )

reservation (name, department, ISBN, date, cancelled )

borrow (name, department, barcode, from, to )

Library Questions (RA)

1 List all the authors of books of which the library has a copy that has never been borrowed.

2 List all the authors of books that have never been borrowed.

3 List all the authors of which no book has ever been borrowed.

(simple) Employee database

• employee(person_name, street, city)

• works(person_name, company_name, salary)

• company(company_name, city)

• manages(person_name, manager_name)

Exercise 2.1.c• Find the names of all employees who earn

more than every employee of “SBC”.– The “more than every...” clause cannot be

expressed directly in the relational algebra.– For all other employees there is an employee of

“SBC” earning more. This can be described using a selection on a cartesian product (of works with works).

– We then use the difference to find the correct result.

Solution 2.1.c

person-name (W) –

- person-name( (w.salary<=w2.salary ^ w2.company-name = ‘SBC’ (W x W2(W) ))

Exercise 2.5.e

• Find all companies located in every city in which “SBC” is located.

Case 1: 1 city per company

C.company-name (

(C.city=SBC.city

(C (company ) x (SBC.company-name=‘SBC’ (SBC (company ) ) )

Trivial!

Case 2: multiple cities per company• Find all companies located in every city in which

“SBC” is located.– The “every city...” clause cannot be expressed directly

in the relational algebra (except for division).– Let’s see …

Case 2a: multiple cities: division• All cities where SBC is located:

• SBCCity = city (company-name=‘SBC’ (company ) )

• Is SBCCity constant?• Yes!• So we can use it on the right hand side of the

division.• companies located in every city in which “SBC”

is located:• company SBCCity• Still quite neat!!

Case 2b: multiple cities: no division• Find all companies located in every city in which

“SBC” is located.– The “every city...” clause cannot be expressed directly

in the relational algebra (except for division).– For the other companies there is a city in which “SBC”

is located and the other company is not.– The cities where a company is not located are all the

cities except the ones where the company is located.– We thus need 2 times a difference in this query.– Can also be formulated using a difference operator

Case 2b: multiple cities: no division

• For the other companies there is a city in which “SBC” is located and the other company is not = IntRes.

• E = company-name (company ) x SBCCity

• IntRes = E – company

• We want not (IntRes):company-name (company ) - company-name (IntRes)

What is the meaning of:

SBCCity

)(_

companynamecompany

))()((,_

companySBCCitycompanycnamecitycnamenamecompany

))((_

companySBCnamecompanycity

Recognizing Types of Queries

• Identify the type of the following queries, and afterwards also translate them to the algebra (PSJ, union, intersection, difference, division):

1. Give the name of customers that have a loan with a branch where they also have an account.

2. Give the name of customers who have a loan at a branch where they do not have an account.

3. Give the name of customers who have a loan at every branch where they have an account.

4. Give the name of customers who have loans only at branches where they have an account.

Reading Queries

customer-name(

balance>amount(

account depositor borrower loan))

branch-name

(branch)

branch-name

(branch-city customer-city

(

branch account depositor customer))

X.customer-name

(X.account-number = Y.account-number X.customer-name Y.customer-name

(X(depositor)

Y(depositor)))

Beer Database

• visits(drinker, bar)

• serves(bar, beer)

• likes(drinker, beer).

Beer questions with a difference

1. Give all drinkers that visit bars that don’t serve any beer they like

2. Give all drinkers that only visit bars that serve a beer they like

3. Give all drinkers that only visit bars that serve no beer they like

4. Give all drinkers that only visit bars that serve all beers they like (and maybe other beers as well)

5. Give all drinkers that only visit bars that only serve beers they like (and thus serve nothing else)

Summary

• We have learned RA

• We have learned to perform simple and more complex queries in RA

… to follow

Query optimisation