1 relational query languages relational algebra (procedural) relational calculus (non-procedural)

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1 Relational Query Languages Relational Query Languages Relational Algebra (procedural) Relational Calculus (non-procedural)

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Relational Query LanguagesRelational Query Languages

Relational Algebra (procedural)

Relational Calculus (non-procedural)

Relational LanguagesRelational Languages

• Relational Algebra (procedural)– defines operations on tables

• Relational Calculus (declarative)– based on first-order predicate calculus

• Every relational algebra query can be translated to relational calculus

• Every safe relational calculus query can be translated to relational algebra.

• Any language that is atleast as expressive as relational algebra is said to be relationally complete.

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Relational Algebra OperatorsRelational Algebra Operators

• Select: given a relation R and a predicate P, select tuples from R that satisfy P.

• Project: given a relation R and a subset of its attributes X, return a relation which is the same as R except that all columns not in X are left out.

• Rename: given a relation R and a name N, return a relation that is exactly the same as R except that it has a name N.

• Cartesian Product: Given 2 relations R1and R2,.return a relation R3 whose tuples are the concatenation of tuples in R1 and R2

• Union: Given relations R1 and R2, return a relation R3 which contains all tuples in R1 and R2

• Set Difference: Given relations R1 and R2, return a relation R3 containing all tuples in R1 that are not in R2

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• selection cond(R) or select[selection cond]R

• Example: Employee(name, dept, sal)

Employee select [sal > 20,000] Employee

name dept saljane pharmacy 30,000

jack hardware 30,000

jill pharmacy 75,000

select [(dept = toy) or (sal < 20,000)] Employee

name dept sal

joe toy 20,000

bill toy 12,000

Selection OperationSelection Operation

name dept sal

jane pharmacy 30,000

jack hardware 30,000

jill pharmacy 75,000

joe toy 20,000

bill toy 12,000

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ProjectionProjection

• Proj [list of attr of R] (R ) or list of attr of R (R)

R A B C S A C

Jane Toy 10,000 Jane Toy

Jim Toy 20,000 John Complaint

June Complaint 20,000

Proj[A]R Proj[CB]R

A C B

Jane 10,000 Toy

Jim 20,000 Complaint

June 20,000 Toy

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Cartesian ProductCartesian Product

• Denoted by R x S

R: A B C S: A D

joe toy 10K joe jill

jack com 20K jack jill

RxS: R.A B C S.A D

joe toy 10K joe jill

joe toy 10K jack jill

jack com 20K joe jill

jack com 20K jack jill

• Notice attribute naming strategy to disambiguate attribute names

attributes get the name, R.A, where A is attrib name, and R is the relation name from which attrib originates. If there is no possible ambiguity, relation name is dropped!

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Set DifferenceSet Difference

• Denoted by R - S( Illegal if R & S have different numbers of attributes or if respective domains

mismatch!)

R A B S A C

Jane Toy Jane Toy

Jim Toy John Complaint

June Complaint

R - S = A B Jim Toy

June Complaint

Note attributes in resulting relation take name from the first relation

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Rename OperatorRename Operator

• Strategy used to disambiguate attribute names:– For union and set difference operators, the resulting relation takes

the attribute names of the first relation

– For cartesian product, attributes are named as Relation-name.attribute-name, where Relation name refers to the relation from which the attribute originally came.

– Strategy will not disambiguate when the same relation appears multiple times in the relational query.

• Let R(A,B) be a relation. Consider R x R ---- what to name the attributes of the resulting relation?

• Define a rename operator:• denoted by rename[N]R or by N(R)

– returns a relation which is exactly same as R except it has the name N

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Rename OperatorRename Operator

• Consider relation Employee(name, dept, sal)

• List all the employees who work in the same department as Jill

• We first find the department(s) for which Jill works

– Proj[dept](select[name = Jill] Employee) ---list of departments for which Jill works

• To find out about all Employees working for this department, we need to reference the Employee table again:

– select[P] ( Employee x Proj[dept](select[name = Jill] Employee) )

– where P is a selection predicate which requires dept values to be equal.

• If we use Employee.dept in P it is ambiguous which instance of Employee relation in the query the attribute refers to.

• To disambiguate, use rename operator:

• Proj[Employee.name](select[Employee.dept = Employee2.dept]

Employee x (Proj[dept] (select[name = Jill](

rename[Employee2](Employee))))

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Formal Definition of Relational AlgebraFormal Definition of Relational Algebra

• Basic Expressions:– Relation in a database

– Constant Relations

• General Expressions: constructed from basic ones. Let E1 and E2 be relational algebra expressons. Then the following are also expressions:– E1 U E2, if arity of E1 = E2 and corresponding attributes are of the

same type

– E1 - E2, if arity of E1 = E2 and corresponding attributes are of the same type

– E1 x E2, if attributes in E1 and E2 have different names

– Select[P](E1), where P is a predicate on attributes in E1

– Proj[S](E1), where S is a list of some attributes in E1

– rename[X](E1), where X is a new name for relation E1

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AdditionalAdditional OperatorsOperators

Basic Relational Algebra operators are like assembly language.

Define more powerful operators that make the task of writing relational algebra queries easier

Each of these operators can be expressed in relational algebra and do not increase the expressibility of the language

Example: Intersection

R S = R - (R - S)

= { t | t R & t S }

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R join condition S = select[ join condition] (R x S)

join condition is of the form:

<condition> AND <condition> AND <condition>

where each condiition is of the form Ai Bj, where

– Ai is attribute of R

– Bj is attribute of S

– is a comparison operator {=, <, >, <=, >=, <>}

JoinsJoins

Example:

E(emp, dept) M(dept, mgr)

List all employees and their managers.

Proj[emp, mgr](select[E.dept = M.dept] (ExM))

can be represented as:

Proj[emp,mgr] ( E E.dept = M.dept M )

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Types of JoinsTypes of Joins

• Theta-Join: if a join condition uses some comparison operator other than equality.

– E.g., list names of all managers who manage departments other than Jill’s– Proj[mgr]( select[emp = Jill](E ) (E.dept M.dept) M)

• Equi-Join: if join conditions use only equality operator.– E.g., list the manager’s name of Jill’s department– Proj[mgr]( select[emp = Jill](E ) (E.dept M.dept) M)

• Natural Join: special type of equi-join..– Let R and S be relations. Let attributes of R and S be denoted by R and S

respectively. – Denote by R U S the union of the list of attributes of R and S– Let list of attributes common to both R and S be {A1, A2, …, An}– Natural join of R and S (denoted R S) is:– Proj[R U S ] (R R.A1 = S.A1 and R.A2 = S.A2 and … and R.An = S.An S)– E.g., Proj[mgr]( select[emp = Jill](E ) M)

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Assignment OperatorAssignment Operator

• Lots of time convenient to write relational algebra expressions in parts using assignment to temporary relational variables.

• For this purpose use assignment operator, denoted by :=

• E.g., Who makes more than their manager?

E(emp, dept, sal) M(mgr, dept)

ESM(emp, sal, mgr) := Proj[emp, sal, mgr] (E M)

(Proj[ESM.emp](ESM [mgr = E.emp & ESM.sal >E.sal] E) )

• With the assignment operator, a query can be written as a sequential program consisting of a series of assignments followed by an expression whose value is the result of the query.

ExamplesExamples

• A query is a composition of basic relational algebra operators

• Consider relations:– customer(ssno, name, street, city)

– account(acctno, custid, balance)

• list account balance of Sharad

• list names of all customers who have a higher balance than all customers in Champaign

balance ssnocustid

customeraccount ))((

)))))))((

)(((()((

(

, .1,.1 1..1

customercustomeraccount

acctaccount

Champaigncityssnocustid

custidacctno custidacctacctnoacct acctbalanceacctbalanceacct

name ssnocustid

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Diag Pat Dis

Winslett Strep

Liu Mono

Harandi Meningitis

Harandi Hepatitis

Liu Hepatitis

Outcome Pat Test Outcome

Winslett a T

Winslett b F

Liu b T

Harandi f T

Winslett e F

Harandi e F

Harandi g F

Winslett e T

ToDiag Dis Test

Strep A

Mono B

Meningitis C

Hepatitis D

Encehhalitis E

Meningitis F

Meningitis G

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1. Who has what disease?Diag

2. Who has a disease they have been tested for?– Proj[pat](Diag | ToDiag | Outcome)

3. Who has a disease they tested positively for?Proj[pat](Diag | ToDiag | (select[outcome = ‘T’])Outcome))

4. Who has a disease that they tested both positively & negatively for?

Temp1(pat, dis) := Proj[pat,dis](Diag | ToDiag select[outcome = ‘T’])Outcome)

Temp2(pat, dis) : = Proj[pat,dis](Diag | ToDiag select[outcome = ‘F’])Outcome)

Proj[pat](Temp1 Temp2)

Use better names!!

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Example of Queries in Relational AlgebraExample of Queries in Relational Algebra

5. Who tested both positively and negatively for a disease, whether or not they have it?

Testpos(pat, dis) = Proj[pat,dis](ToDiag | select[outcome = ‘T’]) Outcome)

Testneg(pat, dis) = Proj[pat,dis](ToDiag | select[outcome = ‘F’]) Outcome)

(Testpos Testneg)[pat]

6. Who tested both positively & negatively for the same test? (Winslett)

Proj[pat](Outcome | condition (rename[Outcome2](Outcome))

where condition is:

[Outcome.pat = Outcome2.pat & Outcome.test = Outcome2.test & Outcome.outcome = Outcome2. outcome]

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7. What testable disease does no one have?

(encephalitis)

Proj[dis]ToDiag Proj[dis]Diag

Note technique: compute opposite of what you want, take a difference. Use in hard queries with negation (‘no one’)

8. What disease does more than one person have?

Proj[dis](Diag condition rename[Diag2](Diag))

where, condition is [Diag.pat Diag2.pat & Diag.dis = Diag2dis]

9. What disease does everyone have?

clue that query is very hard

Disease(dis) := diag[dis]

Patients(pat) := diag[pat]

DiseasesNotEveryoneHas(dis) := Proj[dis]((Patients x Disease) - Diag)

Disease - Diseases Not Everyone Has

Note technique used! A very hard query might require taking the difference several times.

Outer JoinsOuter Joins

E emp dept sal M mgr dept

Jones Missiles 10K Mendez Tanks

Chu Tanks 20K Frank Explosive

Swami Tanks 50K Jones Missiles

Barth Revolver 100K

Right outer join of E with M

emp dept sal mgr Jones Missiles 10K JonesChu Tanks 20K Mendeznull Explosives null Frank

left outer join of E with M

emp dept sal mgr Jones Missiles 10K JonesChu Tanks 20K MendezSwami Tanks 50K MendezBarth Revolver 100K null

Full outer join of E with M

emp dept sal mgr Jones Missiles 10K JonesChu Tanks 20K MendezSwami Tanks 50K MendezBarth Revolver 100K nullnull Explosives null Frank

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Aggretate FunctionsAggretate Functions

• Mathematical aggregate functions (e.g., sum, count, average, min, max)

on a collection of values from the database.• [grouping attributes] F [function list] (<relation expression>)

where

grouping attributes specify how to group the tuples

function list specify the aggregate functions to be applied to the individual groups

relation expression is the expression that results in a relation.• Example:

Consider a relation Employee(dno, ssno, sal).

An expression

[dno] F [COUNT ssno, AVERAGE salary] (Employee)

results in a relation:

dno #employee average_salary

1 10 10,000

2 23 8000

Queries involving aggregate functions cannot be expressed in pure relational algebra!

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Recursive Closure QueriesRecursive Closure Queries

Consider parent relation

parent Child Mom Dad

sam Anda Chuck

Chuck Donna Harvey

Harvey Betty Reggie

Reggie Cristie John

It may be interesting to query the relation for the following:

retrieve all the female anscestors of sam.

retrieve all male ansestors of chuck

retrieve harvey’s family tree

retrieve all the descendants of cristie

Such queries (in general) require an unbounded application of

joins of the parent relation to itself and CANNOT be represented

in relational algebra.