slides 7 relational algebra

8/3/2019 Slides 7 Relational Algebra

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Relational Algebra

Lecture 7

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Outline

Relational Algebra (Section 6.1)


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Relational Algebra

Formalism for creating new relationsfrom existing ones

Its place in the big picture:

Declarative

querylanguage

Algebra Implementation

SQL,

relational calculus

Relational algebra

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Relational Algebra Five operators:

Union: !

Difference: -

Selection:s Projection: P Cartesian Product: "

Derived or auxiliary operators: Intersection, complement

Joins (natural,equi-join, theta join, semi-join)

Renaming:r


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1. Union and 2. Difference

R1 ! R2Example:

ActiveEmployees ! RetiredEmployees

R1 R2

Example: AllEmployees ! RetiredEmployees

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What about Intersection?

It is a derived operatorR1 # R2 = R1 (R1 R2)

Also expressed as a join (will see later)

Example

UnionizedEmployees# RetiredEmployees


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3. Selection

Returns all tuples which satisfy a condition

Notation: sc(R) (sometimes also !)

Examples

s Salary > 40000(Employee)

s name = Smith(Employee)

The condition c can be =, , %,

[in SQL: SELECT * FROM Employee

WHERE Salary > 40000]

condition

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Selection Example

Employee

SSN Name DepartmentID Salary

999999999 John 1 30,000

777777777 Tony 1 32,000888888888 Alice 2 45,000


888888888 Alice 2 45,000

Find all employees with salary more than $40,000.s Salary > 40000(Employee)


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4. Projection

Eliminates columns, then removes duplicates

Notation: PA1,,An(R) (sometimes !) Example:

project to social-security number and names:

PSSN, Name (Employee)

Output schema: Answer(SSN, Name)

[In SQL: SELECT DISTINCT SSN, Name FROM Employee]

attributes

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Projection Example

Employee


999999999 John 1 30,000

777777777 Tony 1 32,000

888888888 Alice 2 45,000

SSN Name

999999999 John

777777777 Tony888888888 Alice

PSSN, Name (Employee)


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5. Cartesian Product

Combine each tuple in R1 with each tuple inR2

Notation: R1 " R2

Example: Employee " Dependents

Very rare in practice; mainly used to expressjoins

[In SQL: SELECT * FROM R1, R2]

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Cartesian Product Example

Employee

Name SSN

John 999999999

Tony 777777777

Dependents

EmployeeSSN Dname999999999 Emily

777777777 Joe

Employee x Dependents

Name SSN EmployeeSSN Dname

John 999999999 999999999 Emily

John 999999999 777777777 Joe

Tony 777777777 999999999 Emily

Tony 777777777 777777777 Joe

Note the output


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Cartesian Product and SQLselect * fromEmployee, Dependents;

+------+-----------+-------------+-------+| Name | SSN | EmployeeSSN | Dname |

+------+-----------+-------------+-------+

| John | 999999999 | 999999999 | Emily |

| Tony | 777777777 | 999999999 | Emily |

| John | 999999999 | 777777777 | Joe |

| Tony | 777777777 | 777777777 | Joe |

+------+-----------+-------------+-------+

select * from Employee, Dependentswhere SSN=EmpoyeeSSN;

+------+-----------+-------------+-------+| Name | SSN | EmployeeSSN | Dname |

+------+-----------+--------------+-------+

| John | 999999999 | 999999999 | Emily |

| Tony | 777777777 | 777777777 | Joe |

+------+-----------+-------------+-------+

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Relational Algebra Five operators:

Union: !

Difference: -

Selection:s Projection: P Cartesian Product: "

Derived or auxiliary operators: Intersection, complement

Joins (natural,equi-join, theta join, semi-join)

Renaming:r


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Renaming

Changes the schema, not the instance

Schema: R(A1, , An )

Notation: r B1,,Bn (R)

Example:

r LastName, SocSocNo (Employee)

Output schema: Answer(LastName, SocSocNo)

[in SQL: SELECT Name AS LastName, SSN AS SocSocNo

FROM Employee]

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Renaming Example

Employee

Name SSNJohn 999999999

Tony 777777777

LastName SocSocNoJohn 999999999Tony 777777777

rLastName, SocSocNo (Employee)


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Natural Join

Notation: R1 ! R2

Meaning: R1 ! R2 = PA(sC(R1 " R2)) Equality on common attributes names

Where: The selection sC checks equality of all common

attributes

The projection eliminates the duplicate commonattributes

[in SQL: SELECT DISTINCT R1.A, R1. B, R2.C FROM R1, R2WHERE R1.B = R2.BSchema: R1(A,B), R2(B,C)]

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Natural Join Example

Employee

Name SSNJohn 999999999Tony 777777777

Dependents

SSN Dname999999999 Emily777777777 Joe

Name SSN DnameJohn 999999999 EmilyTony 777777777 Joe

Employee Dependents =

PName, SSN, Dname(s SSN=SSN2(Employee x r SSN2, Dname(Dependents))


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Natural Join

R= S=

R ! S=

VZ

ZY

ZX

YX

BA

VZ

WV

UZ

CB

WVZ

VZY

UZY

VZX

UZX

CBA

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Natural Join

Given the schemas R(A, B, C, D), S(A, C,

E), what is the schema of R ! S ?

Given R(A, B, C), S(D, E), what is R ! S ?

Given R(A, B), S(A, B), what is R ! S ?


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Theta Join

A join that involves a predicate

R1 !q R2 = sq (R1 " R2)

Here qcan be any condition

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Eq-join

A theta join where qis an equalityR1 !A=B R2 = s A=B (R1 " R2)

Equality on two different attributes

Example:

Employee !SSN=SSN Dependents

Most useful join in practice(difference to natural join?)


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Semijoin

R " S = PA1,,An (R ! S)

Where A1, , An are the attributes in R

Example:

Employee " Dependents

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Employee

Name EmpId DeptNameHarry 3415 Finance

Sally 2241 Sales

George 3401 FinanceHarriet 2202 Sales

Dept

DeptName Manager

Sales HarrietProduction Charles

Semijoin - Example

Employee!DeptName EmpId DeptNameSally 2241 Sales

Harriet 2202 Sales

Only attributes of Employee are selected


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Semijoins in DistributedDatabases

Semijoins are used in distributed databases

. . .. . .

NameSSN

Dname

. . .. . .

AgeSSNEmployee

Dependents

network

Employee !ssn=ssn (s age>71 (Dependents))T = PSSNs age>71 (Dependents)

R = Employee "!TAnswer = R! Dependents

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Complex RA Expressions

Person Purchase Person Product

sname=fred sname=gizmoPpidPssn

seller-ssn=ssn

pid=pid

buyer-ssn=ssn

Pname


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Application:Query Rewriting for Optimization

Reserves Sailors

sid=sid

bid=100 rating > 5

sname

Reserves Sailors

sid=sid

bid=100

sname

rating > 5(Scan;write totemp T1)

(Scan;write totemp T2)

The earlier we process selections, the fewer tuples we need tomanipulate higher up in the tree (predicate pushdown)Disadvantages?

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Algebraic Laws (Examples)

Commutative and Associative Laws R "S = S "R, R "(S "T) = (R "S) "T

R S = S R, R (S T) = (R S) T

Laws involving selection s C AND C"(R) = sC(s C"(R)) = sC(R) "sC"(R)

s C (R S) = sC (R) S When C involves only attributes of R

Laws involving projections PM(PN(R)) = PM,N(R)

! ! ! ! ! !

! !


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Operations on Bags

A bag = a set with repeated elementsAll operations need to be defined carefully on bags

{a,b,b,c}!{a,b,b,b,e,f,f}={a,a,b,b,b,b,b,c,e,f,f}

{a,b,b,b,c,c} {b,c,c,c,d} = {a,b,b}

sC(R): preserve the number of occurrences

PA(R): no duplicate elimination

Cartesian product, join: no duplicate elimination

Important: Relational Engines work on bags, not sets!

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Finally: RA has Limitations !

Cannot compute transitive closure

Find all direct and indirect relatives of Fred

Cannot express in RA !!! Need to write Cprogram

SisterLouNancy

SpouseBillMary

CousinJoeMary

FatherMaryFred

RelationshipName2Name1


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Conclusion

Relational algebra is important for SQL

We mainly need projects, selections,

joins

Joins are typically time consuming for

large databases

slides 7 relational algebra

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