Exercises on Relational Algebramontesi/BD/ES2016-17/01. Relational Algebra.pdf · that is missing...

Databases

Exercises on Relational Algebra

The Lab Sessions

Giacomo Bergami ([email protected])bergami.co.nr

▪ 2016/10/07▪ Keys and Superkeys▪ Relational Algebra (I)

▪ Negation▪ Minimum

▪ 2016/10/14▪ Relational Algebra (II)

▪ At least 2…▪ More exercises + Questions

mailto:[email protected]

Keys…

def. A superkey K in r(S) (K⊆S) univocally identifies tuples in r.

¬∃t1≠t

2 ∈ r. t

1[K]=t

2[K]

∀t1≠t

2 ∈ r. t

1[K]≠t

2[K]

▪ Recap: within the relational model, each tuple is unique. ∀t

1∈r. ∀t

2∈ r\{t

1}. t

1≠t

2 (HP)

▪ Hence, having the aforementioned condition is “sufficient” for unicity.

…and Superkeysdef. K⊆S is minimal for a property P(.) if ¬∃T⊂K.P(T)

▪ This means that any other key smaller than K does not satisfy a predicate.

def. A key K in r(S) (K⊆S) is a minimal superkey.▪ This means that a key is already minimal, since

there is no other subset of K satisfying the property “being a superkey”. (A key is also a “minimal key”)¬∃T⊂K.(¬∃t

1≠t

2 ∈ r. t

1[K]=t

2[K])

∀t1≠t

2 ∈ r.∃t

1≠t

2 ∈ r. t

1[K]=t

2[K]

Recap: operators

▪ Set Operations: /, ∪, ∩, ×▪ σθ(R): Filter R by θ▪ πL(R): Reduce R’s schema to the attributes in

L▪ R⋈S, R⋈θS: combining two relations using a

predicate θ▪ ρB←A(R): Renames A as B in R’s schema

Why?

▪ Query expressions’ equivalences: query optimizations in current (relational) databases use algebras to rewrite rules into an equivalent expression that takes less time to compute.▪ In this course “correctness” and “readable” results

are preferred to “quick” and “efficient” ones (see above).

▪ A way to express queries in a logical framework.▪ How to express “not forall”?▪ How to express “minimum”?▪ How to “count” (there exists at least two…)?

Exercise 1

Given the following relation:

Train(Code, Start, End, miles)

Provide all the routes between Boston and Chicago with one switch.

▪ In order to switch train, you have to reach an end and then start again the journey.

▪ We have to join the Train relation.▪ We have to rename the fields before joining them

Exercise 1

ρC1←Code,S←Start,Sw←End,M1←miles

(Train) Train(C1, S, Sw, M1) ⋈

S=”Boston” AND E=”Chicago”

ρC2←Code,Sw←Start,E←End,M2←miles

(Train) Train(C2, Sw, E, M2)

Which is the most efficient solution?

▪ Solution BρC1←Code,S←Start,Sw←End,M1←miles

(σStart=”Boston”

(Train)) ⋈ρC2←Code,Sw←Start,E←End,M2←miles

(σEnd=”Chicago”

(Train))

▪ Solution A

Exercise 2(a)


Employee(Code, Name Surname)

BelongsTo(Employee, Office)

Associate to each employee its office, only if it exists.

▪ R⋈θS combines the tuples from both relations iff. both of them satisfy a predicate θ

Exercise 2(a)

Code Name Surname

0123 Johann Sebastian Bach

4567 Wolfgang Amadeus Mozart

0571 Edvard Grieg

3573 Claude Debussy

Employee

Employee Office

0123 Baroque

4567 Classical

3573 Impressionism

BelongsTo

Code Name Surname Office

0123 Johann Sebastian Bach Baroque

4567 Wolfgang Amadeus Mozart Classical

3573 Claude Debussy Impressionism

πCode,Name,Surname,Office

(Employee ⋈ Code=Employee BelongsTo)

Exercise 2(b)


Employee(Code, Name Surname)

BelongsTo(Employee, Office)

Return employees’ codes with with no office.

▪ In this case we have to use a left join in order to get the complete list of employees

Exercise 2(b)

Code Name Surname

0123 Johann Sebastian Bach

4567 Wolfgang Amadeus Mozart

0571 Edvard Grieg

3573 Claude Debussy

Employee

Employee Office

0123 Baroque

4567 Classical

3573 Impressionism

BelongsTo




0571 Edvard Grieg NULL



(Employee Code=Employee BelongsTo)

Exercise 2(b)




0571 Edvard Grieg NULL



(Employee Code=Employee BelongsTo)

πCode,Name,Surname

(σOffice is NULL

(Employee Code=Employee BelongsTo))

Code Name Surname

0571 Edvard Grieg

Exercise 3(a)

Given the following relations:

State(Name, Area)

City(Code, Name, Inhabitants)

FormedOf(State, City)

Return the U.S.A. States’ names having more than 1.000.000 inhabitants.

How to solve this query?

Exercise 3(a)


State(Name, Area)



Return the U.S.A. States’ names having more than 1.000.000 inhabitants.

■ This query requires the group by operator (Γ,γ), that is missing in the proposed relational algebra. Let’s change the query.

Exercise 3(b)


State(Name, Area)



Return the U.S.A States’ names having cities with more than 1.000.000 inhabitants.

Exercise 3(b)

Name Area

Rhode Island 1,545

New Jersey 8,723

California 163,696

Alaska 665,384

State

State City

Rhode Island 401

New Jersey 862

California 213

Alaska 907

Alaska 908

Code Name Inhabitants

401 Providence 178,042

862 Newark 277,140

213 Los Angeles 4,030,904

907 Anchorage 291,826

908 North Pole 2,117

FormedOf City

Given the following relations…

State(Name, Area)



Exercise 3(b)

Name Area

Rhode Island 1,545

New Jersey 8,723

Florida 65,758

Alaska 665,384

State

State City

Rhode Island 401

New Jersey 862

California 213

Alaska 907

Alaska 908


401 Providence 178,042

862 Newark 277,140


907 Anchorage 291,826

908 North Pole 2,117

FormedOf City

…Return the U.S.A States’ names having cities with more than 1.000.000 inhabitants (θ).



σInhabitants>1000000

(City)? FormedOf already has the State’s name information. City and FormedOf are

enough

?

⋈

Exercise 3(b)State City

Rhode Island 401

New Jersey 862

California 213

Alaska 907

Alaska 908

FormedOf




(City)

⋈City=Code

State Code

Rhode Island 401

New Jersey 862

California 213

Alaska 907

Alaska 908

ρCode←City(FormedOf)




(City)

⋈City=Code

πState

(FormedOf ⋈City=Code


(City))

πState

(ρCode←City(FormedOf)⋈ σInhabitants>1000000(City))

Exercise 4


Person(ID, Name, Surname, Age)

Registration(Person,Lecture)

Lecture(Name,Price)

Return the people’s names which are registred for a lecture that costs more than 10 times their age.

■ Let’s just use natural joins, projection, selections and renamings.

Exercise 4


ρPerson←ID

(Person) Person(Person, Name,…, Age)

Registration Registration(Person,Lecture)

ρLecture←Name

(Lecture) Lecture(Lecture,Price)


■ Let’s just use natural joins, projection, selections and renamings.

Exercise 4

πName

(σCost>10*Age

(

ρPerson←ID

(Person) ⋈

Registration ⋈

ρLecture←Name

(Lecture)

))


Exercise 5


State(Name, Area)



return the City names belonging to states larger than 10000 squared miles

Exercise 5ρState←Name

(σArea>1000

(State)) State(State, Area)

ρCity←Code

(City) City(City, Name, Inhabitants)

FormedOf FormedOf(State, City)

return the City names belonging to states larger than 10000 squared miles

ρCity←Code

(City) ⋈

FormedOf ⋈

ρState←Name

(σArea>1000

(State))

πName

(

)

Exercise 6(a)Given the following relations:

Student(Number, Surname, Name, Dept)

Exam(Student, Subject, Grade, Day)

Provide the students’ surnames and names that passed at least one exam with grade A.

▪ Please note that, in this case, we could select all the exams with grade A, then grasp the student

▪ This time we’re going to use theta-joins

Exercise 6(a)

πSurname,Name

(

Student ⋈ π

Number(ρ

Number←Student(σ

Grade=“A”(Exam)))

)



πSurname,Name

(

Student ⋈Number=Student

πStudent

(σGrade=“A”

(Exam))

)

▪ Using natural joins

Exercise 6(b)Given the following relations:



Provide the students’ surnames and names that passed no exam with grade A.

Example 6(b)

πSurname,Name

(Student⋈(

πNumber

(Student)

\ πNumber

(ρNumber←Student

(σGrade=“A”

(Exam))

)

U

\

⟦S⟧

▪ Remember that ⟦¬S⟧=U\⟦S⟧, where U is the universe relation▪ U: Student (id)▪ S: Students (id) that gave exams with

grade A

Exercise 6(c)Given the following relations:



Provide the students’ surnames, names and first exam (by date) passed with grade A.

▪ Is it possible to define a minimum operator in relational algebra?

Number Surname Name Dept.M1 Rossi Ugo Computer

ScienceM2 Bianchi Mario Computer

Science

Student

ExamStudent Subject Grade Day

M1 DB A 08/05/2012M1 Compl. DB A 10/05/2012M1 Lambda Calc. A 06/06/2012M1 ALGEBRA B 07/01/2011M2 OS B 07/02/2012

Exercise 6(c)

Defining the min. operator

minC(S)={t|∃t,t’∈S. t[C]≤t’[C]}

={t|∃t∈S. ¬∀t’∈S. t[C]>t’[C]}

=S\{t |∃t∈S. ∀t’∈S. t[C]>t’[C]}

=S⋈(πC(S)\π

C(π

C(S)⋈

C>C’ πC’(ρ

C’←C(S))))

■ Return the tuple in S having the minimum value of C. ■ In this case we return the tuple t from S having

the minimum C value.

Defining the local min. operator

locminA,C

(S)=

=S⋈(πA,C

(S)\πA,C

(πA,C

(S)⋈C>C’

πA,C’

(ρC’←C

(S))))

■ Return the tuple in S having the minimum value of C within the entries with the same A value. ■ By doing so, we perform the comparison

C>C’among all the tuples with the same A values

Warning!

▪ “min” and “local min.” operators are “non standard” operators

▪ When you do the exam, you have to provide the algebra expression associated with it.▪ Problem: it is too difficult to “keep in mind”▪ Any kind of “cards”, “texts” and “notes” are

forbidden.▪ Logical language is a way to obtain a “safe“

expressions without risks. Use them!

Exercise 6(c): Final Result

S = σGrade=“A”

(Exam)

T = πStudent,Day

(S)\πStudent,Day

(πStudent,Day

(S)

⋈Day>Day’

πStudent,Day’

(ρDay’←Day

(S)))

πSurname,Name,Day

(Student ⋈Number=Student

T)

Provide the students’ surnames, names and first exam (by date) passed with grade A.

Local minimum for student/day

S = σGrade=“A”

(Exam)

πStudent,Day

(S)\πStudent,Day

(πStudent,Day

(S)

⋈Day>Day’

πStudent,Day’

(ρDay’←Day

(S)))

Provide the students’ surnames, names and first exam (by date) passed with grade A.■ Return the student (id) and exam day having

the minimum value of exam date with grade A within the entries for the same student (id).

Exercise 7

Given the following relations:User(Tax,Surname,Birth)

Field(FCode,IsCovered)

Bookings(FCode,Day,TimeStart,TimeEnd,Tax)

return the Tax code of the users that have booked at least two times a “non covered field” and that have never booked a covered field.▪ and = intersection▪ never booked a covered field = negation▪ at least two times = how to “count”?

Exercise 7(a)




return the Tax code of the users that have never booked a covered field

▪ Use the negated semantics.

Exercise 7(a)

return the Tax code of the users that have never booked a covered field

▪ Use the negated semantics.

πTax

(User)\πTax

( User⋈

Bookings ⋈

σIsCovered=True

(Field))

Exercise 7(b)




return the Tax code of the users that have booked at least two times a “non covered field”

▪ Is there a way to evaluate such function using the previoulsy defined functions?

At least one, at least two…

▪ First: if there exists a minimum, then it means that there exists at least one element (that is the minimum).

exC(S)=min

C(S)

▪ Second: if there exsits another element immediately following the minimum, then there exists two elements (the minimum and the element immediately following it)

ex2C(S)=ex

C(S\ex

C(S))=min

C(S\min

C(S))

At least two – locmin (1)

locminA,C

(S\locminA,C

(S))=?

▪ First of all, exclude the minimum from S:S\locmin

A,C(S)=

= S\(S⋈(πA,C

(S)\πA,C

(πA,C

(S) ⋈C>C’

πA,C’

(ρC’←C

(S))

)

))

= S⋈ πA,C

(πA,C

(S)⋈C>C’

πA,C’

(ρC’←C

(S)))

= S⋈ πA,C

(∂S)

At least two – minloc (2)

locminA,C

(S\locminA,C

(S))=?

▪ T = S ⋈ πA,C

(∂S) = S\locminA,C

(S)▪ locmin

A,C(S)=

=S⋈(πA,C

(S)\πA,C

(πA,C

(S)⋈C>C’

πA,C’

(ρC’←C

(S))))

locminA,C

(S\locminA,C

(S))=

= locminA,C

(S⋈ πA,C

(∂S))= (S⋈π

A,C(∂S))⋈

(πA,C

(T)\πA,C

(πA,C

(T)⋈C>C’

πA,C’

(ρC’←C

(T))))

= S⋈(πA,C

(T)\πA,C

(πA,C

(T)⋈C>C’

πA,C’

(ρC’←C

(T))))


locminA,C

(S\locminA,C

(S))=

= S⋈(πA,C

(T)\πA,C

(πA,C

(T)⋈C>C’

πA,C’

(ρC’←C

(T))))

▪ Recall: πA,C’

(ρC’←C

(T))=ρC’←C

(πA,C’

(T))

= S⋈(πA,C

(T)\πA,C

(πA,C

(T)⋈C>C’

ρC’←C

(πA,C’

(T))))


locminA,C

(S\locminA,C

(S))=

= S⋈(πA,C

(T)\πA,C

(πA,C

(T)⋈C>C’

ρC’←C

(πA,C’

(T))))

▪ Recall: πA,C

(S⋈πA,C

(∂S))=πA,C

(∂S) ▪ Recall: ∂S = π

A,C(S)⋈

C>C’πA,C’

(ρC’←C

(S))

= S⋈(πA,C

(∂S)\πA,C

(πA,C

(∂S)⋈C>C’

ρC’←C

(πA,C

(∂S)))).

Exercise 7(b)[…] Return the Tax code of the users that have booked at least two times a “non covered field”

S=πFCode,…,Tax

(Bookings⋈σIsCovered=True

(Field))

πTax

(ex2Tax,(FCode,Day,timeStart)

(S))

▪ For ordering multiple attributes, use the lexicographical order:

(FCode<FCode’) OR

(FCode=FCode’ AND Day<Day’) OR

(FCode=FCode’ AND Day=Day’ AND

timeStart<timeStart’)

Exercise 7 (finally!)

[…] Return the Tax code of the users that have booked at least two times a “non covered field” and that have never booked a covered field.

▪ Use the intersection.

A = πTax

(User)\πTax

(User⋈Bookings⋈σIsCovered=True

(Field))

S = πFCode,…,Tax

(Bookings⋈σIsCovered=True

(Field))

A ∩ πTax

(ex2Tax,(FCode,Day,timeStart)

(S))

Warning!

▪ If logic is used as a “formal specification”, you could derive a “correct” relational algebra expression from it.▪ Abstract reasoning (see functional

programming) allows to reason on the result and not on “how” to evaluate the solution.

▪ Intermediate variables and operators allow to have a more readable code.

▪ Any non standard operator (min, locmin, ex, ex2,…) shall be defined.

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