Revision for Final Exam

Prof. Sin-Min Lee

Department of Computer Science

Relational Query Languages

• Languages for describing queries on a relational database

• Structured Query LanguageStructured Query Language (SQL)– Predominant application-level query language– Declarative

• Relational AlgebraRelational Algebra– Intermediate language used within DBMS– Procedural

What is an Algebra?

• A language based on operators and a domain of values

• Operators map values taken from the domain into other domain values

• Hence, an expression involving operators and arguments produces a value in the domain

• When the domain is a set of all relations (and the operators are as described later), we get the relational algebrarelational algebra

• We refer to the expression as a queryquery and the value produced as the queryquery resultresult

Relational Algebra

• Domain: set of relations• Basic operators: selectselect, projectproject, unionunion, setset

differencedifference, CartesianCartesian productproduct• Derived operators: set intersectionset intersection, divisiondivision, joinjoin• Procedural: Relational expression specifies query

by describing an algorithm (the sequence in which operators are applied) for determining the result of an expression

The Role of Relational Algebra in a DBMS

Select Operator

• Produce table containing subset of rows of argument table satisfying condition

condition (relation)

• Example:

Person Person Hobby=‘stamps’(PersonPerson)

1123 John 123 Main stamps1123 John 123 Main coins5556 Mary 7 Lake Dr hiking9876 Bart 5 Pine St stamps

1123 John 123 Main stamps9876 Bart 5 Pine St stamps

Id Name Address HobbyId Name Address Hobby

Selection Condition

• Operators: <, , , >, =, • Simple selection condition:

– <attribute> operator <constant>– <attribute> operator <attribute>

• <condition> AND <condition>

• <condition> OR <condition>• NOT <condition>

Selection Condition - Examples

Id>3000 OR Hobby=‘hiking’ (PersonPerson)

Id>3000 AND Id <3999 (PersonPerson)

NOT(Hobby=‘hiking’) (PersonPerson)

Hobby‘hiking’ (PersonPerson)

Project Operator• Produces table containing subset of columns of the

table taken as argument attribute list(relation)

• Example: PersonPerson Name,Hobby(PersonPerson)

1123 John 123 Main stamps1123 John 123 Main coins5556 Mary 7 Lake Dr hiking9876 Bart 5 Pine St stamps

John stampsJohn coinsMary hikingBart stamps

Id Name Address Hobby Name Hobby

Project Operator

1123 John 123 Main stamps1123 John 123 Main coins5556 Mary 7 Lake Dr hiking9876 Bart 5 Pine St stamps

John 123 MainMary 7 Lake DrBart 5 Pine St

Result is a table (no duplicates); can have fewer tuplesthan the original

Id Name Address Hobby Name Address

• Example: PersonPerson Name,Address(PersonPerson)

Expressions

1123 John 123 Main stamps1123 John 123 Main coins5556 Mary 7 Lake Dr hiking9876 Bart 5 Pine St stamps

1123 John9876 Bart

Id Name Address Hobby Id Name

PersonPerson

ResultResult

Id, Name ( Hobby=’stamps’ OR Hobby=’coins’ (PersonPerson) )

Set Operators

• Relation is a set of tuples, so set operations should apply: , , (set difference)

• Result of combining two relations with a set operator is a relation; hence all its elements must be tuples having the same structure

• Hence, scope of set operations limited to union compatible relationsunion compatible relations

Union Compatible Relations

• Two relations are union compatibleunion compatible if– Both have same number of columns– Names of attributes are the same in both– Attributes with the same name in both relations

have the same domain

• Union compatible relations can be combined using unionunion, intersectionintersection, and setset differencedifference

ExampleTables: PersonPerson (SSN, Name, Address, Hobby) ProfessorProfessor (Id, Name, Office, Phone)are not union compatible. But Name (PersonPerson) and Name (ProfessorProfessor)

are union compatible so

Name (PersonPerson) - Name (ProfessorProfessor)

makes sense.

Cartesian Product• If RR and SS are two relations, RR SS is the set of all

concatenated tuples <x,y>, where x is a tuple in RR and y is a tuple in S S (but see naming problem next)(but see naming problem next)

• RR SS is expensive to compute:– Factor of two in the size of each row– Quadratic in the number of rows

A B C D A B C D x1 x2 y1 y2 x1 x2 y1 y2 x3 x4 y3 y4 x1 x2 y3 y4 x3 x4 y1 y2 RR SS x3 x4 y3 y4 RR SS

Renaming in Cartesian ProductResult of expression evaluation is a relation.

Attributes of relation must have distinct names.

So what do we do if they don’t?E.g., suppose R(A,B) and S(A,C)

and we wish to compute RR SS .

One solution is to rename the attributes of the answer:

RR S( R.A, R.B, S.A, S.C)S( R.A, R.B, S.A, S.C)

Although onlyAlthough only A A needs to be renamed, it is“cleaner” to needs to be renamed, it is“cleaner” to rename them all.rename them all.

Renaming Operator• Previous solution is used whenever possible but it

won’t work when R is the same as S.

• Renaming operator resolves this. It allows to assign any desired names, say A1, A2,… An , to the attributes of the n column relation produced by expression expr with the syntax

expr [A1, A2, … An]

Example

This is a relation with 4 attributes: StudId, CrsCode1, ProfId, CrsCode2

TranscriptTranscript (StudId, CrsCode, Semester, Grade)

TeachingTeaching (ProfId, CrsCode, Semester)

  StudId, CrsCode (TranscriptTranscript)[StudId, CrsCode1]

ProfId, CrsCode(TeachingTeaching) [ProfId, CrsCode2]

Derived Operation: Join

A (generalgeneral or thetatheta) join join of R and S is the expression R join-condition S

where join-condition is a conjunction of terms: Ai oper Bi

in which Ai is an attribute of R; Bi is an attribute of S; and oper is one of =, <, >, , .

The meaning is:

join-condition´ (R S)

where join-condition and join-condition´ are the same, except for possible renamings of attributes caused by the Cartesian product.

Theta Join – Example

Employee(Employee(Name,Id,MngrId,SalaryName,Id,MngrId,Salary) Manager(Manager(Name,Id,SalaryName,Id,Salary)

Output the names of all employees that earnmore than their managers.EmployeeEmployee.Name (EmployeeEmployee MngrId=Id AND Salary>Salary ManagerManager)

The join yields a table with attributes: EmployeeEmployee.Name, EmployeeEmployee.Id, EmployeeEmployee.Salary, Employee.MngrId

ManagerManager.Name, ManagerManager.Id, ManagerManager.Salary

Relational Algebra

• Relational algebra operations operate on relations and produce relations (“closure”)f: Relation -> Relation f: Relation x Relation

-> Relation• Six basic operations:

– Projection (R)– Selection (R)– Union R1 [ R2

– Difference R1 – R2

– Product R1 £ R2

– (Rename) (R)

Equijoin Join - Example

Name,CrsCode(StudentStudent Id=StudId Grade=‘A’ (TranscriptTranscript))

Id Name Addr Status111 John ….. …..222 Mary ….. …..333 Bill ….. …..444 Joe ….. …..

StudId CrsCode Sem Grade 111 CSE305 S00 B 222 CSE306 S99 A 333 CSE304 F99 A

Mary CSE306Bill CSE304

The equijoin is used veryfrequently since it combinesrelated data in different relations.

StudentStudent TranscriptTranscript

EquijoinEquijoin: Join condition is a conjunction of equalities.

Natural Join• Special case of equijoin + a special projection

– join condition equates all and only those attributes with the same name (condition doesn’t have to be explicitly stated)

– duplicate columns eliminated (projected out) from the result

TranscriptTranscript (StudId, CrsCode, Sem, Grade)Teaching (Teaching (ProfId, CrsCode, Sem)

TranscriptTranscript TeachingTeaching = StudId, Transcript.CrsCode, Transcript.Sem, Grade, ProfId

( TranscriptTranscript CrsCode=CrsCode AND Sem=Sem Sem Teaching Teaching ) [StudId, CrsCode, Sem, Grade, ProfId ]

Natural Join (cont’d)

• More generally:RR SS = attr-list (join-cond (RR × SS) )

where attr-list = attributes (RR) attributes (SS)(duplicates are eliminated) and join-cond has the form: A1 = A1 AND … AND An = An

where {A1 … An} = attributes(RR) attributes(SS)

Natural Join Example

• List all Ids of students who took at least two different courses:

StudId ( CrsCode CrsCode2 ( TranscriptTranscript

TranscriptTranscript [StudId, CrsCode2, Sem2, Grade2] )) We don’t want to join on CrsCode, Sem, and Grade attributes, hence renaming!

Example Data Instance

sid name

1 Jill

2 Qun

3 Nitin

4 Marty

fid name

1 Ives

2 Saul

8 Roth

sid exp-grade cid

1 A 550-0103

1 A 700-1003

3 A 700-1003

3 C 500-0103

4 C 500-0103

cid subj sem

550-0103 DB F03

700-1003 AI S03

501-0103 Arch F03

fid cid

1 550-0103

2 700-1003

8 501-0103

STUDENT Takes COURSE

PROFESSOR Teaches

Natural Join and IntersectionNatural join: special case of join where is implicit – attributes with same name must be equal:

STUDENT Takes ⋈ ´ STUDENT ⋈STUDENT.sid = Takes.sid Takes

Intersection: as with set operations, derivable from difference

A-B B-A

A B

A B

Division

• A somewhat messy operation that can be expressed in terms of the operations we have already defined

• Used to express queries such as “The fid's of faculty who have taught all subjects”

• Paraphrased: “The fid’s of professors for which there does not exist a subject that they haven’t taught”

Division Using Our Existing Operators

• All possible teaching assignments: Allpairs:

• NotTaught, all (fid,subj) pairs for which professor fid has not taught subj:

• Answer is all faculty not in NotTaught:

fid,subj (PROFESSOR £ subj(COURSE))

Allpairs - fid,subj(Teaches COURSE)⋈fid(PROFESSOR) - fid(NotTaught)

´ fid(PROFESSOR) - fid(fid,subj (PROFESSOR £ subj(COURSE)) -

fid,subj(Teaches COURSE))⋈

Division: R1 R2

• Requirement: schema(R1) ¾ schema(R2)• Result schema: schema(R1) – schema(R2)• “Professors who have taught all courses”:

• What about “Courses that have been taught by all faculty”?

fid (fid,subj(Teaches ⋈ COURSE) subj(COURSE))

Division

• Goal: Produce the tuples in one relation, r, that match all tuples in another relation, s– rr (A1, …An, B1, …Bm)– ss (B1 …Bm)– rr/ss, with attributes A1, …An, is the set of all

tuples <a> such that for every tuple <b> in ss, <a,b> is in rr

• Can be expressed in terms of projection, set difference, and cross-product

Division (cont’d)

Division - Example• List the Ids of students who have passed all

courses that were taught in spring 2000• Numerator:

– StudId and CrsCode for every course passed by every student:

StudId, CrsCode (Grade ‘F’ (TranscriptTranscript) )

• Denominator:– CrsCode of all courses taught in spring 2000

CrsCode (Semester=‘S2000’ (TeachingTeaching) )

• Result is numerator/denominator

Relational Calculus

• Important features:– Declarative formal query languages for relational mode

l

– Based on the branch mathematical logic known as predicate calculus

– Two types of RC:• 1) tuple relational calculus

• 2) domain relational calculus

– A single statement can be used to perform a query

Tuple Relational Calculus

• based on specifying a number of tuple variables

• a tuple variable refers to any tuple

Generic Form

• {t | COND (t)} – where– t is a tuple variable and – COND(t) is Boolean expression involving t

Simple example 1

• To find all employees whose salary is greater than $50,000– {t| EMPLOYEE(t) and t.Salary>5000}

• where

• EMPLOYEE(t) specifies the range of tuple variable t

– The above operation selects all the attributes

Simple example 2

• To find only the names of employees whose salary is greater than $50,000– {t.FNAME, t.NAME| EMPLOYEE(t) and

t.Salary>5000}

• The above is equivalent to• SELECT T.FNAME, T.LNAME• FROM EMPLOYEE T• WHERE T.SALARY > 5000

Elements of a tuple calculus

• In general, we need to specify the following in a tuple calculus expression:– Range Relation (I.e, R(t)) = FROM– Selected combination= WHERE– Requested attributes= SELECT

More Example:Q0

• Retrieve the birthrate and address of the employee(s) whose name is ‘John B. Smith’

• {t.BDATE, t.ADDRESS| EMPLOYEE(t) AND t.FNAME=‘John’ AND t.MINIT=‘B” AND t.LNAME=‘Smith}

Formal Specification of tuple Relational Calculus

• A general format:• {t1.A1, t2.A2,…,tn.An |COND ( t1 ,t2 ,…, tn, tn+1,

tn+2,…,tn+m)}– where– t1,…,tn+m are tuple var

– Ai : attributeR(ti)– COND (formula)

• Where COND corresponds to statement about the world, which can be True or False

Elements of formula

• A formula is made of Predicate Calculus atoms:– an atom of the from R(ti)– ti.A op tj.B op{=, <,>,..}– F1 And F2 where F1 and F2 are formulas– F1 OR F2– Not (F1)– F’=(t) (F) or F’= (t) (F)

Y friends (Y, John) X likes(X, ICE_CREAM)

•

Example Queries Using the Existential Quantifier

• Retrieve the name and address of all employees who work for the ‘ Research ’ department

• {t.FNAME, t.LNAME, t.ADDRESS| EMPLOYEE(t) AND ( d) (DEPARTMENT (d) AND d.DNAME=‘Research’ AND d.DNUMBER=t.DNO)}

More Example

• For every project located in ‘Stafford’, retrieve the project number, the controlling department number, and the last name, birthrate, and address of the manger of that department.

Cont.

• {p.PNUMBER,p.DNUM,m.LNAME,m.BDATE, m.ADDRESS|PROJECT(p) and EMPLOYEE(M) and P.PLOCATION=‘Stafford’ and ( d) (DEPARTMENT(D) AND P.DNUM=d.DNUMBER and d.MGRSSN=m.SSN))}

Safe Expressions

• A safe expression R.C:– An expression that is guaranteed to generate a fi

nite number of rows (tuples)

• Example:– {t | not EMPLOYESS(t))} results values not bei

ng in its domain (I.e., EMPLOYEE)

Domain Relational Calculus (DRC)

• Another type of formal predicate calculus-based language

• QBE is based on DRC

• The language shares a lot of similarities with the tuple calculus

DRC

• The only difference is the type of variables:– variables range over singles values from domai

ns of attributes

• An expression of DRC is:– {x1, x2,…,xn|COND(x1,x2,…,xn, xn+2,…,xn+m)}

• where x1,x2,…,xn+m are domain var range over attrib

uters

• COND is a condition (or formula)

Examples

• Retrieve the birthdates and address of the employee whose name is ‘John B. Smith’

• {uv| (q)(r)(s) (EMPLOYEE(qrstuvwxyz) and q=‘John’ and r=‘B’ and s=‘Smith’

Alternative notation

• Ssign the constants ‘John’, ‘B’, and ‘Smith’ directly

• {uv|EMPLOYEE (‘John’, ’B’, ’Smith’ ,t ,u ,v ,x ,y ,z)}

More example

• Retrieve the name and address of all employees who work for the ‘Reseach’ department

• {qsv | ( z) EMPLOYEE(qrstuvwxyz) and ( l) ( m) (DEPARTMENT (lmno) and l=‘Research’ and m=z))}

More example

• List the names of managers who have at least on e dependent

• {sq| ( t) EMPLOYEE(qrstuvwxyz) and (( j)( DEPARTMENT (hijk) and (( l) | (DEPENTENT (lmnop) and t=j and t=l))))}

QBE

• Query-By-Example– Supports graphical query language based on

DRC– Implemented in commercial db such as

Access/Paradox– Query can be specified by filling in templates

of relations– Fig 9.5

Summary

• It can be shown that any query that can be expressed in the relational algebra, it can also be expressed in domain and tuple relational calculus

Quiz

• In what sense doe R.C differ from R.A, and in what sense are they similar?

The Big Picture: SQL to Algebra toQuery Plan to Web Page

SELECT * FROM STUDENT, Takes, COURSE

WHERE STUDENT.sid = Takes.sID AND Takes.cID = cid

STUDENT

Takes COURSE

Merge

Hash

by cid by cidOptimizer

ExecutionEngine

StorageSubsystem

Web Server / UI / etc

Query Plan – anoperator tree

Relational Calculus: A Logical Way of

Expressing Query Operations• First-order logic (FOL) can also be thought of as a

query language, and can be used in two ways:– Tuple relational calculus– Domain relational calculus– Difference is the level at which variables are used: for at

tributes (domains) or for tuples

• The calculus is non-procedural (declarative) as compared to the algebra– More like what we’ll see in SQL– More convenient to express certain things

Domain Relational Calculus

Queries have form:

{<x1,x2, …, xn>| p}

Predicate: boolean expression over x1,x2, …, xn

– Precise operations depend on the domain and query language – may include special functions, etc.

– Assume the following at minimum:<xi,xj,…> R X op Y X op const const op X

where op is , , , , ,

xi,xj,… are domain variables

domain variables

predicate

More Complex Predicates

Starting with these atomic predicates, build up new predicates by the following rules:– Logical connectives: If p and q are predicates, then so are

pq, pq, p, and pq• (x>2) (x<4)• (x>2) (x>0)

– Existential quantification: If p is a predicate, then so is x.p

x. (x>2) (x<4)

– Universal quantification: If p is a predicate, then so is x.p

x.x>2 x. y.y>x

Some Examples

• Faculty ids

• Course names for courses with students expecting a “C”

• Courses taken by Jill

Logical Equivalences

• There are two logical equivalences that will be heavily used:– pq p q

(Whenever p is true, q must also be true.) x. p(x) x. p(x)

(p is true for all x)

• The second can be a lot easier to check!

Free and Bound Variables

• A variable v is bound in a predicate p when p is of the form v… or v…

• A variable occurs free in p if it occurs in a position where it is not bound by an enclosing or

• Examples: – x is free in x>2– x is bound in x.x>y

Can Rename Bound Variables Only

• When a variable is bound one can replace it with some other variable without altering the meaning of the expression, providing there are no name clashes

• Example: x.x>2 is equivalent to y.y>2

• Otherwise, the variable is defined outside our “scope”…

Safety

• Pitfall in what we have done so far – how do we interpret: {<sid,name>| <sid,name> STUDENT}

– Set of all binary tuples that are not students: an infinite set (and unsafe query)

• A query is safe if no matter how we instantiate the relations, it always produces a finite answer– Domain independent: answer is the same regardless of the domain

in which it is evaluated– Unfortunately, both this definition of safety and domain independ

ence are semantic conditions, and are undecidable

Safety and Termination Guarantees

• There are syntactic conditions that are used to guarantee “safe” formulas– The definition is complicated, and we won’t discuss it;

you can find it in Ullman’s Principles of Database and Knowledge-Base Systems

– The formulas that are expressible in real query languages based on relational calculus are all “safe”

• Many DB languages include additional features, like recursion, that must be restricted in certain ways to guarantee termination and consistent answers

Mini-Quiz

How do you write:– Which students have taken more than one

course from the same professor?

– What is the highest course number offered?

Translating from RA to DRC

• Core of relational algebra: , , , x, -• We need to work our way through the structure of

an RA expression, translating each possible form.– Let TR[e] be the translation of RA expression e into D

RC.

• Relation names: For the RA expression R, the DRC expression is {<x1,x2, …, xn>| <x1,x2, …, xn> R}

Selection: TR[ R]

• Suppose we have (e’), where e’ is another RA expression that translates as:

TR[e’]= {<x1,x2, …, xn>| p}• Then the translation of c(e’) is

{<x1,x2, …, xn>| p’}where ’ is obtained from by replacing each attribute with the corresponding variable

• Example: TR[#1=#2 #4>2.5R] (if R has arity 4) is

{<x1,x2, x3, x4>|< x1,x2, x3, x4> R x1=x2 x4>2.5}

Projection: TR[i1,…,im(e)]

• If TR[e]= {<x1,x2, …, xn>| p} then TR[i1,i2,…,im

(e)]=

{<x i1,x i2

, …, x im >| xj1,xj2

, …, xjk.p},

where xj1,xj2

, …, xjk are variables in x1,x2, …, xn

that are not in x i1,x i2

, …, x im

• Example: With R as before,#1,#3 (R)={<x1,x3>| x2,x4. <x1,x2, x3,x4> R}

Union: TR[R1 R2]

• R1 and R2 must have the same arity• For e1 e2, where e1, e2 are algebra expressions

TR[e1]={<x1,…,xn>|p} and TR[e2]={<y1,…yn>|q}

• Relabel the variables in the second:TR[e2]={< x1,…,xn>|q’}

• This may involve relabeling bound variables in q to avoid clashesTR[e1e2]={<x1,…,xn>|pq’}.

• Example: TR[R1 R2] = {< x1,x2, x3,x4>| <x1,x2, x3,x4>R1 <x1,x2, x3,x4>R2

Other Binary Operators

• Difference: The same conditions hold as for unionIf TR[e1]={<x1,…,xn>|p} and TR[e2]={< x1,…,xn>|q}

Then TR[e1- e2]= {<x1,…,xn>|pq}

• Product: If TR[e1]={<x1,…,xn>|p} and TR[e2]={< y1,…,ym>|q}

Then TR[e1 e2]= {<x1,…,xn, y1,…,ym >| pq}

• Example: TR[RS]= {<x1,…,xn, y1,…,ym >|

<x1,…,xn> R <y1,…,ym > S }

Summary

• Can translate relational algebra into (domain) relational calculus.

• Given syntactic restrictions that guarantee safety of DRC query, can translate back to relational algebra

• These are the principles behind initial development of relational databases– SQL is close to calculus; query plan is close to algebra– Great example of theory leading to practice!

Limitations of the Relational Algebra / Calculus

Can’t do:– Aggregate operations– Recursive queries– Complex (non-tabular) structures

• Most of these are expressible in SQL, OQL,