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CS 319: Theory of Databases: C4. Dr. Alexandra I. Cristea http://www.dcs.warwick.ac.uk/~acristea/. (provisionary) Content. Generalities DB Temporal Data Integrity constraints (FD revisited) Relational Algebra (revisited) Query optimisation Tuple calculus Domain calculus

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Cs 319 theory of databases c4

CS 319: Theory of Databases: C4

Dr. Alexandra I. Cristea

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


Provisionary content
(provisionary) Content

  • Generalities DB

  • Temporal Data

  • Integrity constraints (FD revisited)

  • Relational Algebra (revisited)

  • Query optimisation

  • Tuple calculus

  • Domain calculus

  • Query equivalence

  • LLJ, DP and applications

  • The Askew Wall

  • Datalog



Relational model
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 model1
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



Basic structure
Basic Structure

  • Formally, given sets D1, D2, …. Dn a relation r is a subset of D1 x D2 x … x DnThus, a relation is a set of n-tuples (a1, a2, …, an) where each ai Di


Example
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
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
Relation Schema

  • A1, A2, …, Anare attributes

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

    Example:

    Customer_schema = (customer_name, customer_street, customer_city)

  • r(R) denotes a relationr on the relation schema R

    Example:

    customer (Customer_schema)


Relation instance
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

attributes

(or columns)

customer_name

customer_street

customer_city

Jones

Smith

Curry

Lindsay

Main

North

North

Park

Harrison

Rye

Rye

Pittsfield

tuples

(or rows)

customer


Relations are unordered
Relations are Unordered

  • Order of tuples is irrelevant (tuples may be stored in an arbitrary order)

  • Example: account relation with unordered tuples


Database
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 customer Relation


The depositor relation
The depositor Relation


Keys

  • Let K  R

  • K is a superkeyof 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
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
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
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
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
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
Select Operation

  • Notation: p(r)

  • p is called the selection predicate

  • Defined as:p(r) = {t | t  rand p(t)}

    Where p is a formula in propositional calculus consisting of termsconnected 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
Project Operation – Example

A

B

C

  • Relation r:

10

20

30

40

1

1

1

2

A

C

A

C

A,C (r)

1

1

1

2

1

1

2

=


Project operation
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)


Union operation example
Union Operation – Example

A

B

A

B

  • Relations r, s:

1

2

1

2

3

s

r

A

B

1

2

1

3

  • r  s:


Union operation
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 loan

    customer_name (depositor)  customer_name (borrower)


Set difference operation example
Set Difference Operation – Example

  • Relations r, s:

A

B

A

B

1

2

1

2

3

s

r

  • r – s:

A

B

1

1


Set difference operation
Set Difference Operation

  • Notation r – s

  • Defined as:

    r – s = {t | t rand 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
Cartesian-Product Operation – Example

A

B

C

D

E

  • Relations r, s:

1

2

10

10

20

10

a

a

b

b

r

s

A

B

C

D

E

  • rxs:

1

1

1

1

2

2

2

2

10

10

20

10

10

10

20

10

a

a

b

b

a

a

b

b


Cartesian product operation
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
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

C

D

E

1

1

1

1

2

2

2

2

 

10

10

20

10

10

10

20

10

a

a

b

b

a

a

b

b

A

B

C

D

E

10

10

20

a

a

b

1

2

2


Rename operation
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 .



Relational algebra operators thus
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
Banking Example

branch (branch_name, branch_city, assets)

customer (customer_name, customer_street, customer_city)

account (account_number, branch_name, balance)

loan (loan_number, branch_name, amount)

depositor (customer_name, account_number)

borrower(customer_name, loan_number)


Example queries
Example Queries

Find all loans of over $1200

amount> 1200 (loan)

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

loan_number (amount> 1200 (loan))

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

customer_name (borrower)  customer_name (depositor)


Example queries1
Example Queries

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

customer_name (branch_name=“Perryridge”

(borrower.loan_number = loan.loan_number(borrower x loan)))

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

customer_name (branch_name = “Perryridge”

(borrower.loan_number = loan.loan_number(borrower x loan))) – customer_name(depositor)


Example queries2
Example Queries

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

Query 1

customer_name (branch_name = “Perryridge”

(borrower.loan_number = loan.loan_number (borrower x loan)))

Query 2

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


Example queries3
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(accountxrd(account)))


Formal definition
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 operations
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


Set intersection operation
Set-Intersection Operation

  • Notation: rs

  • Defined as:

  • rs = { t | trandt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
Set-Intersection Operation – Example

  • Relation r, s:

  • r s

A B

A B

1

2

1

2

3

s

r

A B

 2


Natural join operation
Natural-Join Operation

  • Notation: r s

  • Let r and s be relations on schemas R and S respectively. Then, r s is a relation on schema R S obtained as follows:

    • Consider each pair of tuples tr from r and ts from s.

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

      • t has the same value as tr on r

      • t has the same value as ts on s


Natural joint example
Natural joint example

  • Example:

    R = (A, B, C, D)

    S = (E, B, D)

    • Result schema = (A, B, C, D, E)

    • rs 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

Natural Join Operation – Example

  • Relations r, s:

B

D

E

A

B

C

D

1

3

1

2

3

a

a

a

b

b

1

2

4

1

2

a

a

b

a

b

r

s

A

B

C

D

E

1

1

1

1

2

a

a

a

a

b


Division operation
Division Operation

r s

  • 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


Division operation example
Division Operation – Example

A

B

B

1

2

3

1

1

1

3

4

6

1

2

1

2

  • Relations r, s:

s

A

  • r s:

r


Another division example
Another Division Example

  • Relations r, s:

A

B

C

D

E

D

E

a

a

a

a

a

a

a

a

a

a

b

a

b

a

b

b

1

1

1

1

3

1

1

1

a

b

1

1

s

r

  • r s:

A

B

C

a

a


Division operation cont
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
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
Bank Example Queries

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

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

customer_name (borrower)  customer_name (depositor)

customer_name, loan_number, amount (borrower loan)


Bank example queries1
Bank Example Queries

  • Find all customers who have an account from at least the “Downtown” and the “Uptown” branches.

  • Query 1

    customer_name (branch_name = “Downtown” (depositoraccount )) 

    customer_name (branch_name = “Uptown” (depositoraccount))

  • Query 2

    customer_name, branch_name(depositoraccount)  temp(branch_name)({(“Downtown” ), (“Uptown” )})

    Note that Query 2 uses a constant relation.


Bank example queries2
Bank Example Queries

  • Find all customers who have an account at all branches located in Brooklyn city.

customer_name, branch_name(depositoraccount)  branch_name (branch_city = “Brooklyn” (branch))


Library case
Library Case

reservation

book

author

name

department

Initials

title

name

publisher

date

year

cancelled

copy

borrow

name

department

department

from

cpYear

to

present


Library database
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
Library Questions (RA)

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

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

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


Simple employee database
(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
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
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
Exercise 2.5.e

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


Case 1 1 city per company
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
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
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
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 division1
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)



Recognizing types of queries
Recognizing Types of Queries

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

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

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

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

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


Reading queries
Reading Queries

5. customer-name(balance>amount(

account  depositor  borrower  loan))

6. branch-name(branch) 

branch-name(branch-city  customer-city(

branch  account  depositor  customer))

7. X.customer-name (X.account-number = Y.account-number  X.customer-name  Y.customer-name (X(depositor) Y(depositor)))


Beer database
Beer Database

  • visits(drinker, bar)

  • serves(bar, beer)

  • likes(drinker, beer).


Beer questions with a difference
Beer questions with a difference

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

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

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

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

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


Summary
Summary

  • We have learned RA

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


… to follow

Query optimisation


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