Relational Algebra - University of Waterloo › ~cs348 › F14 › 3_algebra.pdf · 2014-09-02 ·...

Relational Algebra

David R. Cheriton S hool of Computer S ien e

University of Waterloo

CS 348

Introdu tion to Database Management

Fall 2014

CS 348 (Intro to DB Mgmt) Relational Algebra FALL 2014 1 / 1

Database S hema Used in Examples

RespEmpDeptNoProjNo

EmPTime

Project

EmEnDate

Emp_Act

EmpNoMajProj

MidInitLastName

Employee

WorkDeptHireDateSalary

FirstName

EmpNo

DeptNameMgrNoAdmrDept

DeptNo

Department

ProjNoActNoEmStDate


Relational Algebra

�

the relational algebra onsists of a set of operators

�

relational algebra is losed

�

ea h operator takes as input one or more relations

�

ea h operator de�nes a single output relation in terms of its input

relation(s)

�

relational operators an be omposed to form expressions that

de�ne new relations in terms of existing relations.

�

Notation:

R is a relation name; E is a relational algebra expression


Primary Relational Operators

�

Relation Name: R



�

Relation Name: R

�

Sele tion: �

ondition

(E)

�

result s hema is the same as E 's

�

result instan e in ludes the subset of the tuples of E that ea h

satis�es the ondition



�

Relation Name: R

�

Sele tion: �

ondition

(E)

�

result s hema is the same as E 's

�

result instan e in ludes the subset of the tuples of E that ea h

satis�es the ondition

�

Proje tion: �

attributes

(E)

�

result s hema in ludes only the spe i�ed attributes

�

result instan e ould have as many tuples as E , ex ept that

dupli ates are eliminated


Primary Relational Operators ( ont'd)

�

Rename: �(R(F );E)

�

F is a list of terms of the form oldname ! newname

�

returns the result of E with olumns renamed a ording to F .

�

remembers the result as R for future expressions


Primary Relational Operators ( ont'd)

�

Rename: �(R(F );E)

�

F is a list of terms of the form oldname ! newname

�

returns the result of E with olumns renamed a ording to F .

�

remembers the result as R for future expressions

�

Produ t: E

1

� E

2

�

result s hema has all of the attributes of E

1

and all of the attributes

of E

2

�

result instan e in ludes one tuple for every pair of tuples (one from

ea h expression result) in E

1

and E

2

�

sometimes alled ross-produ t or Cartesian produ t

�

renaming is needed when E

1

and E

2

have ommon attributes


Cross Produ t Example

R

AAA BBB

a

1

b

1

a

2

b

2

a

3

b

3

S

CCC DDD

1

d

1

2

d

2

R � S

AAA BBB CCC DDD

a

1

b

1

1

d

1

a

2

b

2

1

d

1

a

3

b

3

1

d

1

a

1

b

1

2

d

2

a

2

b

2

2

d

2

a

3

b

3

2

d

2


Sele t,Proje t,Produ t Examples

�

Note: Use Emp to mean the Employee relation, Proj the proje t

relation

�

Find the last names and hire dates of employees who make more

than $100000.



�


relation

�


than $100000.

�

LastName ;HireDate

(�

Salary>100000

(Emp))



�


relation

�


than $100000.

�

LastName;HireDate

(�

Salary>100000

(Emp))

�

For ea h proje t for whi h department E21 is responsible, �nd the

name of the employee in harge of that proje t.



�


relation

�


than $100000.

�

LastName;HireDate

(�

Salary>100000

(Emp))

�

For ea h proje t for whi h department E21 is responsible, �nd the

name of the employee in harge of that proje t.

�

ProjNo;LastName

(�

DeptNo=E21

(�

RespEmp=EmpNo

(Emp � Proj )))


Joins

�

Conditional join: E

1

⋊⋉ ondition

E

2

�

equivalent to �

ondition

(E

1

� E

2

)

�

spe ial ase: equijoin

Proj ⋊⋉(RespEmp=EmpNo)

Emp


Joins

�

Conditional join: E

1

⋊⋉ ondition

E

2

�

equivalent to �

ondition

(E

1

� E

2

)

�

spe ial ase: equijoin

Proj ⋊⋉(RespEmp=EmpNo)

Emp

�

Natural join (E

1

⋊⋉ E

2

)

�

The result of E

1

⋊⋉ E

2

an be formed by the following steps

1 form the ross-produ t of E

1

and E

2

(renaming dupli ate attributes)

2 eliminate from the ross produ t any tuples that do not have

mat hing values for all pairs of attributes ommon to s hemas E

1

and E

2

3 proje t out dupli ate attributes

�

if no attributes in ommon, this is just a produ t


Example: Natural Join

�

Consider the natural join of the Proje t and Department tables,

whi h have attribute DeptNo in ommon

�

the s hema of the result will in lude attributes ProjName, DeptNo,

RespEmp, MajProj, DeptName, MgrNo, and AdmrDept

�

the resulting relation will in lude one tuple for ea h tuple in the

Proje t relation (why?)


Set-Based Relational Operators

�

Union (R [ S):

�

s hemas of R and S must be �union ompatible�

�

result in ludes all tuples that appear either in R or in S or in both

�

Union Compatible:

�

Same number of �elds.

�

'Corresponding' �elds have the same type



�

Union (R [ S):

�


�


�

Di�eren e (R � S):

�


�

result in ludes all tuples that appear in R and that do not appear

in S

�

Union Compatible:

�


�




�

Union (R [ S):

�


�


�

Di�eren e (R � S):

�


�

result in ludes all tuples that appear in R and that do not appear

in S

�

Interse tion (R \ S):

�


�

result in ludes all tuples that appear in both R and S

�

Union Compatible:

�


�



Relational Division

X

A B C

a

1

b

1

1

a

1

b

1

2

a

1

b

2

2

a

2

b

1

1

a

2

b

1

2

a

2

b

2

2

a

2

b

3

3

a

3

b

1

1

S

B C

b

1

1

b

1

2

b

2

2

X =S

A

a

1

a

2


Division is the Inverse of Produ t

R

A

a

1

a

2

S

B C

b

1

1

b

1

2

b

2

2

R � S

A B C

a

1

b

1

1

a

1

b

1

2

a

1

b

2

2

a

2

b

1

1

a

2

b

1

2

a

2

b

2

2

(R � S)=S

A

a

1

a

2


Summary of Relational Operators

E ::= R

j �

ondition

(E)

j �

attributes

(E)

j �(R(F );E)

j E

1

� E

2

j E

1

⋊⋉ ondition

E

2

j E

1

⋊⋉ E

2

j E

1

[ E

2

j E

1

\ E

2

j E

1

� E

2

j E

1

= E

2


Algebrai Equivalen es

�

This:

�

ProjNo;LastName

(�

DeptNo=E21

(�

RespEmp=EmpNo

(E � P)))



�

This:

�

ProjNo;LastName

(�

DeptNo=E21

(�

RespEmp=EmpNo

(E � P)))

�

is equivalent to this:

�

ProjNo;LastName

(�

DeptNo=E21

(E ⋊⋉RespEmp=EmpNo

P))



�

This:

�

ProjNo;LastName

(�

DeptNo=E21

(�

RespEmp=EmpNo

(E � P)))

�


�

ProjNo;LastName

(�

DeptNo=E21


P))

�


�

ProjNo;LastName


�

DeptNo=E21

(P))



�

This:

�

ProjNo;LastName

(�

DeptNo=E21

(�

RespEmp=EmpNo

(E � P)))

�


�

ProjNo;LastName

(�

DeptNo=E21


P))

�


�

ProjNo;LastName


�

DeptNo=E21

(P))

�


�

ProjNo;LastName

( ( �

LastName;EmpNo

(E)) ⋊⋉RespEmp=EmpNo

( �

ProjNo;RespEmp

(�

DeptNo=E21

(P))))



�

This:

�

ProjNo;LastName

(�

DeptNo=E21

(�

RespEmp=EmpNo

(E � P)))

�


�

ProjNo;LastName

(�

DeptNo=E21


P))

�


�

ProjNo;LastName


�

DeptNo=E21

(P))

�


�

ProjNo;LastName

( ( �

LastName;EmpNo

(E)) ⋊⋉RespEmp=EmpNo

( �

ProjNo;RespEmp

(�

DeptNo=E21

(P))))

�

More on this topi later when we dis uss database tuning. . .


Relational Completeness

De�nition (Relationally Complete)

A query language that is at least as expressive as relational algebra is

said to be relationally omplete.






�

The following languages are all relationally omplete:

�

safe relational al ulus

�

relational algebra

�

SQL






�

The following languages are all relationally omplete:

�

safe relational al ulus

�

relational algebra

�

SQL

�

SQL has additional expressive power be ause it aptures dupli ate

tuples, unknown values, aggregation, ordering, . . .


Relational Algebra - University of Waterloo › ~cs348 › F14 › 3_algebra.pdf · 2014-09-02 ·...

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Transcript of Relational Algebra - University of Waterloo › ~cs348 › F14 › 3_algebra.pdf · 2014-09-02 ·...