Relational Algebra - Chapter 6.1-6.5(7th ed. 8.1-8.5)

Relational Algebra

• Theoretical basis for SQL (E.F. Codd) • Relational algebra (algebraic notation) and

relational calculus (logical notation)• Created to demonstrate the potential for a

query language of the relational model• Algebra and calculus are equivalent in

expressive power• Can represent complex queries compactly,

but too mathematical for the average person

What is an “Algebra”?

• Mathematical system consisting of:– Operands --- variables or values from which new

values can be constructed.– Operators --- symbols denoting procedures that

construct new values from given values.

What does it do?

• Provide DML and DDL • In relational algebra, a series of operations are

combined to form a relational algebra expression (query)

Set Theoretic Operations

Set theoretic operations • Union, Intersection, Difference - Binary

– Applied to 2 sets (relations) - no duplicates in result - mathematical set– Must be same type of tuples - Compatibility

• same degree n • dom(Ai) = dom(Bi)

– Fig. 6.4– Resulting relation - same attribute names as first relation – Which operations are:

• Commutative ? – R U S = S U R, R ∩ S = S ∩ R– R - S = S - R ?

• Associative ? – R U (S U T) = (R U S) U T, R ∩ (S ∩ T) = (R ∩ S) ∩ T – (R - S) – T ?

Cartesian Product X - Binary

• Also binary, but does not require union compatibility – R(A1, A2, … An) X S(B1, B2, …, Bm)

• Creates a tuple with the combined attributes of 2 tables – Q(A1, A2, …, An, B1, B2, …, Bm) Fig. 6.5

• Degree of resulting relation? – n+m

Relational Algebra Operations

Select Operation s

• s - unary operation (Where in SQL) • A subset of tuples satisfying a selection

condition • Selects rows • Equivalent to select condition in WHERE clause s <selection condition> (<Relation name>)

s dno=4(Employee)

s salary>30000(Employee)

Select Operation

• Select condition is a Boolean expression <attr_name> <comparison op> <constant value> <attr_name> <comparison op> <attr_name> comparison op - =, <, <=, etc. • You can use boolean conditions to connect clauses • Can combine cascade of selects into single select with ANDs s(dno=4 and salary>25000) or (dno=5 and salary>30000)(Employee) Fig. 6.1

Select Operation

• Degree of resulting relation?

• Selectivity - the fraction of tuples selected– number of tuples total tuples

• Is Select is commutative?

σc1 (σc2(R)) = σc2 (σc1(R))

Project Operation p or Õ• - p unary operation • Equivalent to the SELECT clause in SQL(Select) • Keeps only certain attributes (columns) from a relation • Selects columns • Form of operation: p <attr_list> (<relation name>)

p fname, lname, salary Employee Fig 6.1

• Resulting relation has only those attributes specified • Degree of relation ?

– attributes in attr_list

Project Operation

• The project operation eliminates duplicate tuples in the resulting relation so that it remains a mathematical set p sex, salary Employee Fig 6.1

• If several male employees have salary $30,000 only single tuple <M, 30000> is kept in the resulting relation.

• Is the project operation commutative?– No, why?

Sequences of operations

• Several relation algebra operations can be combined to form a relational algebra expression (query).

• Retrieve the names and salaries of employees who work in department 5.

Q ← p fname, lname, salary (s dno=5 Employee) • Alternately, explicit intermediate relations can

be specified for each step: Dept5 ← s dno=5 Employee

R ← p fname,lname,salary Dept5

Write the following in Relational Algebra

• Select * from Employee• Select bdate from Employee• Select * from Employee where sex='F'

Write the following in Relational Algebra

List SSN of employees who do not have dependents

(Select ssn from employee) minus (Select essn from dependent)

Renaming

• Attributes can optionally be renamed in the resulting relation:

Dept5 ¬ s dno=5 Employee

T(firstname,lastname,salary) ¬ p fname,lname,salary Dept5 Fig. 6.2

• Alternative notation in textbook, can rename attributes and/or table

rR(firstname, lastname, salary) p fname,lname,salary Dept5

Operations

• DML Operations:– set theory operations

• Union• Intersection• Difference

– relational DB operations• Select • Project • Anything else?

– Now write SSN, lname of employees who do not have dependents

• Join

The Join operation |X|

• Similar to a Cartesian Product followed by a select • Form of operation: R |X|<join condition> S

• Result is: Q (A1, A2, …, An, B1, B2, …, Bm) A1, A2 … are the attributes of R B1, B2, .. are the attributes of S

• For all tuples that satisfy the join condition • join condition: <cond> and <cond> and … Fig. 6.6• Resulting number of tuples? • Different types of joins - theta join, natural join, equijoin

Equijoin

• R |X|Ai=Bi S• requires identical values in every tuple for each pair

of join attributes • (one or more equality comparisons) • Join conditions are all of the form Ai = Bi and Aj = Bj …• Retrieve each department’s name and manager’s

name. T¬ Department |X|mgrssn=ssnEmployee

Result ¬ p dname,fname,lname (T)

Order of precedence

• Unary: – S elect, project (highest precedence)

• Binary: – Joins, Cartesian product– Intersection– Union, minus

• Use lots of parenthesis!

Write the following in Relational Algebra

• Select * from Employee, Department where dno=dnumber • List employee SSNs who are female and work for the research

department• Select * From Employee, dept_locations Where dno =

dnumber and dlocation = 'Houston'

• List lname, essn of employees who work on pno=10• List lname, essn of employees who work on project

‘Computerization’

Theta Join

• R |X|Ai q Bi S• where the join condition is of the form: Ai q Bi q is =, < , £ , etc.

• Example:

Scholarship(SName GPA_Req Desc) Student (Name CWID GPA Major)

Select Name, SName From Student, Scholarship Where GPA >= GPA_Req

Natural join

• We will use the * notation (some others use |X| without subscript)

• Like an equijoin, except attributes for the equijoin in the second relation are deleted from result

• (Why have 2 columns with the same value?)

Q ¬ R * (<list1>),(<list2>) S Fig 6.7

• Equivalent to equijoin but keep only list1

• If attributes have the same name in both relations, list1 and list2 are not needed.

• In the original definition of natural join, the join attributes required to have the same names in both relations.

Write the following in Relational Algebra

• List locations of ‘Research’ department (use natural join)• List SSN, lname of employees who do not have dependents

When renaming is needed• A relation can have a set of join attributes with itself• List all employee names and their supervisor names S(soc, first, last)¬ p ssn,fname,lname Employee

Temp ¬ Employee |X|superssn=soc S

Result ¬ p fname,lname,first,last (Temp)

//alternativelyResult ¬ p fname,lname,first,last((rS(soc, first, last) p ssn,fname,lname Employee) |X|

superssn=soc Employee)

• Usually, don't see qualification of attributes in relational algebra

Complete Set of Relational Algebra Operations

{ s , p , È , - , ×} • All other relational algebra operations can be

expressed as a sequence of operations from this set. • Other operations are for convenience.

R |X| S =

s <cond> (R X S)

R Ç S = (RÈ S) - ((R - S) È (S - R))

• Do we need anymore relational algebra operations to satisfy queries?

• How about?Select COUNT(*) From Project Select pname, COUNT(ssn) From Project, Works_on Where pnumber=pnoGroup by pname

Additional relation algebra Operations

• Aggregate function - SUM, COUNT, AVG, MIN, MAX

[<grouping attribute>] Á <function list> (<relation name>)

R ¬ Á count ssn, avg salary(Employee)

The following uses the optional grouping attribute R ¬ dno Á count ssn, avg salary(Employee) Fig. 6.9

• The attributes returned from an aggregate function are the attributes in the function list and any grouping attributes listed

• Find out how many projects there are

Write in Relational Algebra

• Retrieve the number of projects• For each project, retrieve the project number,

project name, and the number of employees who work on that project.

Select pnumber, pname, COUNT(*) From Project, Works_on Where pnumber =pno Group By pnumber, pname

Outer Join

• Extension of join and union• In a regular equijoin or natural join, tuples in

R1 or R2 that do not have matching tuples in the other relation do not appear in the result.

• Some queries require all tuples in R1 (or R2 or both) to appear in the result

• When no matching tuples are found, nulls are placed for the missing attributes.

Outer Join

• Left outer join: R1 ]X| R2 keeps every tuple in R1 in result. • List all employees and if they are a manager, list dname

Temp <- (Employee ]X| ssn=mgrssn Department) R <- p fname, minit, lname, dname (Temp) Fig. 6.11

• Right outer join: R1 |X[ R2 keeps every tuple in R2 in result.

• Full outer join: R1 ]X[ R2 keeps every tuple in R1 and R2 in result.

• Think about how this is different from R1 X R2.

Division operation • Part of original relational algebra• T(Y) = R(Z) ¸ S(X) • tuple t is in result if t is in R for every tuple in S • More generally, result is a relation T(Y) that includes t

if t appears in R with the value of X for every tuple in S. Fig. 6.8

• The attributes Y in table T = attributes of R in Z - attributes S in X, where Y is the set of attributes in R not in S.

Result <- R ¸ S

Division operation

• For example, Retrieve ssn of employees who work on all projects John smith works on.

smith_pnos <- p pno Works_on |X| ssn=essn s fname='John' and lname='Smith'Employee ssn_pnos <- p essn, pno Works_on ssns <- ssn_pnos ¸ smith_pnos

//if wanted names would have to do another join results <- p fname, lname (ssns |X| ssn=essn Employee)

// Using minusSelect ssnfrom Employee where lname <> ‘Smith’ and

not exists ((Select pno from Works_on, Employeewhere ssn=essn and lname = ‘Smith’) minus

(select pno from Works_on where ssn=essn));

/ / if result of minus is empty, work on same projects

Write the following in Relational Algebra

• For every project located in 'Stafford' list the project number, the controlling department number and department manager's last name, address and birthdate.

Select pnumber, dnum, lname, bdate, address From Project, Department, Employee Where dnum = dnumber and mgrssn = ssn and plocation = 'Stafford'

Write in Relational Algebra

• For each project on which more than two employees work, retrieve the project number, project name, and the number of employees who work on that project.

Select pnumber, pname, COUNT(*) From Project, Works_on Where pnumber =pno Group By pnumber, pname Having COUNT(*) > 2

Write the following in Relational Algebra

• Compute the average number of dependents over employees with dependents

• Think about how you would write this:

Select * From Employee Where salary > all (Select salary From Employee Where sex = 'F')

DDL - Also provided

• Declare Schema for database• Declare Relation for Schema• Insert <values> into Relation• Delete Relation tuple with specified condition• Modify col. of Relation tuple with specified

condition