Databases. Exercises on Relational Algebra

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1 Databases Exercises on Relational Algebra

2 The Lab Sessions Giacomo Bergami 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

3 Keys def. A superkey K in r(s) (K S) univocally identifies tuples in r. t 1 t 2 r. t 1 [K]=t 2 [K] t 1 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.

4 and Superkeys def. 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]) t 1 t 2 r. t 1 t 2 r. t 1 [K]=t 2 [K]

5 Recap: operators Set Operations: /,,, σ θ (R): Filter R by θ π L (R): Reduce R s schema to the attributes in L R S, R θ S: combining two relations using a predicate θ ρ B A (R): Renames A as B in R s schema

6 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 )?

7 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

8 Exercise 1 Solution A ρ C1 Code,S Start,Sw End,M1 miles (Train) Train(C1, S, Sw, M1) S= Boston AND E= Chicago ρ C2 Code,Sw Start,E End,M2 miles (Train) Train(C2, Sw, E, M2) Solution B ρ C1 Code,S Start,Sw End,M1 miles (σ Start= Boston (Train)) ρ C2 Code,Sw Start,E End,M2 miles (σ End= Chicago (Train)) Which is the most efficient solution?

9 Exercise 2(a) Given the following relation: Employee(Code, Name Surname) BelongsTo(Employee, Office) Associate to each employee its office, only if it exists. R θ S combines the tuples from both relations iff. both of them satisfy a predicate θ

10 Exercise 2(a) Employee Code Name Surname BelongsTo Employee Office 0123 Johann Sebastian Bach 4567 Wolfgang Amadeus Mozart 0571 Edvard Grieg 0123 Baroque 4567 Classical 3573 Impressionism 3573 Claude Debussy π Code,Name,Surname,Office (Employee Code=Employee BelongsTo) Code Name Surname Office 0123 Johann Sebastian Bach Baroque 4567 Wolfgang Amadeus Mozart Classical 3573 Claude Debussy Impressionism

11 Exercise 2(b) Given the following relation: 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

12 Exercise 2(b) Employee Code Name Surname BelongsTo Employee Office 0123 Johann Sebastian Bach 4567 Wolfgang Amadeus Mozart 0571 Edvard Grieg 0123 Baroque 4567 Classical 3573 Impressionism 3573 Claude Debussy π Code,Name,Surname,Office (Employee Code=Employee BelongsTo) Code Name Surname Office 0123 Johann Sebastian Bach Baroque 4567 Wolfgang Amadeus Mozart Classical 0571 Edvard Grieg NULL 3573 Claude Debussy Impressionism

13 Exercise 2(b) π Code,Name,Surname,Office (Employee Code=Employee BelongsTo) Code Name Surname Office 0123 Johann Sebastian Bach Baroque 4567 Wolfgang Amadeus Mozart Classical 0571 Edvard Grieg NULL 3573 Claude Debussy Impressionism π Code,Name,Surname (σ Office is NULL (Employee Code=Employee BelongsTo)) Code Name Surname 0571 Edvard Grieg

14 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 inhabitants. How to solve this query?

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

16 Exercise 3(b) Given the following relations: State(Name, Area) City(Code, Name, Inhabitants) FormedOf(State, City) Return the U.S.A States names having cities with more than inhabitants.

17 Exercise 3(b) Given the following relations State(Name, Area) City(Code, Name, Inhabitants) FormedOf(State, City) State FormedOf City Name Area State City Code Name Inhabitants Rhode Island 1,545 New Jersey 8,723 California 163,696 Alaska 665,384 Rhode Island 401 New Jersey 862 California 213 Alaska 907 Alaska Providence 178, Newark 277, Los Angeles 4,030, Anchorage 291, North Pole 2,117

18 Exercise 3(b) State Name Return the U.S.A States names having cities with more than inhabitants (θ).? Area FormedOf State City City Code Name Inhabitants Rhode Island 1,545 Rhode Island Providence 178,042 New Jersey 8,723 Florida 65,758 New Jersey 862 California Newark 277, Los Angeles 4,030,904 Alaska 665,384 Alaska Anchorage 291,826 Alaska North Pole 2,117? FormedOf already has the State s name information. City and FormedOf are enough σ Inhabitants> (City) Code Name Inhabitants 213 Los Angeles 4,030,904

19 Exercise 3(b) FormedOf State City Rhode Island 401 New Jersey 862 California 213 City=Code σ Inhabitants> (City) Code Name Inhabitants 213 Los Angeles 4,030,904 Alaska 907 Alaska 908 π State (FormedOf City=Code σ Inhabitants> (City)) ρ Code City (FormedOf) State Code Rhode Island 401 New Jersey 862 California 213 City=Code σ Inhabitants> (City) Code Name Inhabitants 213 Los Angeles 4,030,904 Alaska 907 Alaska 908 π State (ρ Code City (FormedOf) σ Inhabitants> (City))

20 Exercise 4 Given the following relations: 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.

21 Exercise 4 Given the following relations: ρ Person ID (Person) Registration ρ Lecture Name (Lecture) Person(Person, Name,, Age) Registration(Person,Lecture) Lecture(Lecture,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.

22 Exercise 4 π Name (σ Cost>10*Age ( ρ Person ID (Person) Registration ρ Lecture Name (Lecture) )) Return the people s names which are registred for a lecture that costs more than 10 times their age.

23 Exercise 5 Given the following relations: State(Name, Area) City(Code, Name, Inhabitants) FormedOf(State, City) return the City names belonging to states larger than squared miles

24 ) Exercise 5 ρ State Name (σ Area>1000 (State)) ρ City Code (City) State(State, Area) City(City, Name, Inhabitants) FormedOf FormedOf(State, City) return the City names belonging to states larger than squared miles π Name ( ρ City Code (City) FormedOf ρ State Name (σ Area>1000 (State))

25 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

26 Exercise 6(a) Student(Number, Surname, Name, Dept) Exam(Student, Subject, Grade, Day) π Surname,Name ( Student Number=Student π Student (σ Grade= A (Exam)) ) Using natural joins π Surname,Name ( Student π Number (ρ Number Student (σ Grade= A (Exam))) )

27 Exercise 6(b) Given the following relations: Student(Number, Surname, Name, Dept) Exam(Student, Subject, Grade, Day) Provide the students surnames and names that passed no exam with grade A.

28 Example 6(b) Remember that S =U\ S, where U is the universe relation U: Student (id) S: Students (id) that gave exams with grade A π Surname,Name (Student ( π Number (Student) \ π Number (ρ Number Student (σ Grade= A (Exam) ) ) U \ S

29 Exercise 6(c) Given the following relations: Student(Number, Surname, Name, Dept) Exam(Student, Subject, Grade, Day) 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?

30 Exercise 6(c) Student Number Surname Name Dept. M1 Rossi Ugo Computer Science M2 Bianchi Mario Computer Science Exam Student Subject Grade Day M1 DB A 08/05/2012 M1 Compl. DB A 10/05/2012 M1 Lambda Calc. A 06/06/2012 M1 ALGEBRA B 07/01/2011 M2 OS B 07/02/2012

31 Defining the min. operator 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. min C (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 (π C (S)\π C (π C (S) C>C π C (ρ C C (S))))

32 Defining the local min. operator 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 locmin A,C (S)= =S (π A,C (S)\π A,C (π A,C (S) C>C π A,C (ρ C C (S))))

33 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!

34 Exercise 6(c): Final Result Provide the students surnames, names and first exam (by date) passed with grade A. S = σ Grade= A (Exam) T = π Student,Day (S)\π Student,Day (π Student,Day (S) Day>Day π Student,Day (ρ Day Day (S))) π Surname,Name,Day (Student Number=Student T)

35 Local minimum for student/day 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). S = σ Grade= A (Exam) π Student,Day (S)\π Student,Day (π Student,Day (S) Day>Day π Student,Day (ρ Day Day (S)))

36 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?

37 Exercise 7(a) 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 never booked a covered field Use the negated semantics.

38 Exercise 7(a) return the Tax code of the users that have never booked a covered field Use the negated semantics. π Tax (User)\π Tax ( User Bookings σ IsCovered=True (Field))

39 Exercise 7(b) 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 Is there a way to evaluate such function using the previoulsy defined functions?

40 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). ex C (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) ex2 C (S)=ex C (S\ex C (S))=min C (S\min C (S))

41 At least two locmin (1) locmin A,C (S\locmin A,C (S))=? First of all, exclude the minimum from S: S\locmin A,C (S)= = S\(S (π A,C (S)\π A,C (π A,C (S) C>C π A,C (ρ C C (S)) ) )) = S π A,C (π A,C (S) C>C π A,C (ρ C C (S))) = S π A,C ( S)

42 At least two minloc (2) locmin A,C (S\locmin A,C (S))=? T = S π A,C ( S) = S\locmin A,C (S) locmin A,C (S)= =S (π A,C (S)\π A,C (π A,C (S) C>C π A,C (ρ C C (S)))) locmin A,C (S\locmin A,C (S))= = locmin A,C (S π A,C ( S)) = (S π A,C ( S)) (π A,C (T)\π A,C (π A,C (T) C>C π A,C (ρ C C (T)))) = S (π A,C (T)\π A,C (π A,C (T) C>C π A,C (ρ C C (T))))

43 At least two minloc (3) locmin A,C (S\locmin A,C (S))= = S (π A,C (T)\π A,C (π A,C (T) C>C π A,C (ρ C C (T)))) Recall: π A,C (ρ C C (T))=ρ C C (π A,C (T)) = S (π A,C (T)\π A,C (π A,C (T) C>C ρ C C (π A,C (T))))

44 At least two minloc (4) locmin A,C (S\locmin A,C (S))= = S (π A,C (T)\π A,C (π A,C (T) C>C ρ C C (π A,C (T)))) Recall: π A,C (S π A,C ( S))=π A,C ( S) Recall: S = π A,C (S) C>C π A,C (ρ C C (S)) = S (π A,C ( S)\π A,C (π A,C ( S) C>C ρ C C (π A,C ( S)))).

45 Exercise 7(b) [ ] Return the Tax code of the users that have booked at least two times a non covered field S=π FCode,,Tax (Bookings σ IsCovered=True (Field)) π Tax (ex2 Tax,(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 )

46 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 = π Tax (User)\π Tax (User Bookings σ IsCovered=True (Field)) S = π FCode,,Tax (Bookings σ IsCovered=True (Field)) A π Tax (ex2 Tax,(FCode,Day,timeStart) (S))

47 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.

Database Applications (15-415)

Database Applications (15-415) Relational Calculus Lecture 5, January 27, 2014 Mohammad Hammoud Today Last Session: Relational Algebra Today s Session: Relational algebra The division operator and summary