Assignment 3: Relational Algebra Solution

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1 Data Modelling and Databases Exercise dates: March 15/March 16, 2018 Ce Zhang, Gustavo Alonso Last update: April 13, 2018 Spring Semester 2018 Head TA: Ingo Müller Assignment 3: Relational Algebra Solution This assignment will be discussed during the exercise slots indicated above. If you want feedback for your copy, hand it in during the lecture on the Wednesday before (preferably stapled and with your address). You can also annotate your copy with questions you think should be discussed during the exercise session. If you have questions that are not answered by the solution we provide, send them to Claude 1 Library Database Consider the following relational schema: Reader: (ReaderNr, Surname, Firstname, City, Birthdate) Book: (ISBN, Title, Author, NoPages, PubYear, PubName) Publisher: (PubName, PubCity) Category: (CategoryName, BelongsTo) Copy: (ISBN, CopyNumber, Shelf, Position) Loan: (ReaderNr, ISBN, Copy, ReturnDate) BookCategory: (ISBN, CategoryName) Formulate the following queries in relational algebra: 1. Which are the last names of the readers in Zurich? Surname (σ City= Zurich (Reader)) 2. Which books (Author, Title) are from publishers in Zurich, Bern, or New York?

2 Author,Title (Book (σ PubCity= Zurich PubCity= Bern PubCity= NewYork (Publisher))) Alternative (join first, then project): Author,Title (σ PubCity= Zurich PubCity= Bern PubCity= NewYork (Book Publisher)) 3. Which books (Author, Title) has the reader Lemmi Schmöker borrowed? Author,Title (Book Loan (σ Surname= Schmoker Firstname= Lemmi (Reader))) 4. Which books in the category Alps do not belong to the category Switzerland? Do not take into account subcategories! ( ISBN (σ CategoryName= Alps (Book BookCategory))) ( ISBN (σ CategoryName= Switzerland (Book BookCategory))) The following is not correct: ISBN (σ CategoryName= Alps CategoryName Switzerland (Book BookCategory)) The reason is that the selection operator is defined on tuples. It will thus select book, category pairs that are where the category is both equal to Alps and not equal to Switzerland. The input, which is the result of the join, however, may contain several book, category pairs of the same book. In particular, a book that is in both categories, Alps and Switzerland, has two such tuples, out of which the one in the Alps category matches the selection and the other one does not. Such a book would thus be in the result although it should not. 5. Which readers (Surname, Firstname) have borrowed books that were published in their home town? Firstname,Surname (σ City=PubCity(Publisher Book Loan Reader)) 6. Which readers (Surname, Firstname) have borrowed at least a book that has been borrowed also by the reader Lemmi Schmöker (the reader Lemmi Schmöker should not be included in the results)?

3 R1.Firstname,R1.Surname (ρ R1(Reader) ρ L1 (Loan)) R1.ReaderNr<>R2.ReaderNr,L1.ISBN=L2.ISBN (ρ R2 (σ Surname= Schmoker Surname= Lemmi Reader) ρ L2 (Loan)))

4 2 Olympic Games Consider the following relational schema: Runner: (Name, Birthday, Country) Run: (Name, Distance, Time) A runner can run in several runs over different race distances. Thanks to high-speed cameras, two runners cannot have the exact same time in the same run. 1. For every description find a matching relational algebra query. Some descriptions may not have a matching query. Descriptions: 1. All 100m race distance runs in which only runners from Switzerland (CH ) participated. 2. All runs with a distance greater than 100m in which only runners from Switzerland participated. 3. All runs in which only runners from Switzerland participated. 4. All 100m race distance runs in which the runners were not from Switzerland. 5. All runs in which the runners were not from Switzerland. Relational Algebra Queries: 1. Name,Distance,Time ((Runner σ Country<> CH (Runner)) (Run σ Distance<100 (Run))) 2. Name,Distance,Time ((σ Country= CH (Runner) σ Country<> CH (Runner)) σ Distance<>100 (Run σ Distance<100 (Run))) 3. Name,Distance,Time ((σ Country<> CH (Runner) σ Country= CH (Runner)) σ Distance=100 (Run σ Distance>100 (Run))) 4. Name,Distance,Time ((σ Country= CH (Runner) σ Country<> CH (Runner)) σ Distance=100 (Run σ Distance<100 (Run))) 5. Name,Distance,Time (σ Country<> CH (Runner)) σ Distance>100 (Run σ Distance<100 (Run))) Fill in the table below by writing to every description on the left the right query letter on the right. If there is no matching query for a description, put a cross.

5 Description Query Explanation 1 4 Right side of the join: all 100m runs; left side of the join: all swiss runners (the second/inner term of each side has no effect); the joins gives all 100m runs of swiss runners. 2 2 Left side of the join: all swiss runners (subtracting a disjoint set has no effect); right side of the join: all runs strictly longer than 100m; the join gives all runs of swiss runners with distance strictly greater than 100m. 3 1 Left side of the join: all swiss runners; right side of the join: all runs (the union with a subset has no effect); the join gives all runs of swiss runners. 4 3 Left side of the join: all non-swiss runners (subtracting a disjoint subset has no effect); right side of the join: all 100m runs (the inner subtraction has no effect since it does not remove elements selected outside); the join gives all 100m runs of non-swiss runners. 5-5 Right side of the join: all runs strictly longer than 100m (this inner subtraction has no effect since it does not remove elements selected outside); left side of the join: all non-swiss runners; the join gives all runs of non-swiss runners with a distance strictly longer than 100m. This does not correspond to any description. 2. Which of the following relational algebra expressions finds all runners which only participated in 100m race distance runs? Name (σ Distance=100Run) Name (Runner) Name (σ Distance=100Run) Name (Run) Name (σ Distance=100Run) Name (Run) Name (σ Distance<>100Run) Name (Runner) Name (σ Distance<>100Run) Note: The last query differs in that it contains runners that only participated in 100m races and runners that never participated in any race. Be aware that an entry in the Runner table does not need to appear in the Run table, i.e., some people registered as runners but never ran a race. 3. We want to find winners for every distance. A winner has the shortest time for a given distance. Mark all the queries that find the winners. Name,Country,Distance,Time (Runner (Run Run1.Name,Run1.Distance,Run1.Time ( σ Run1.time>Run2.time (ρ Run1 (Run) Name ρ Run2 (Run))))) Name,Country,Distance,Time (Runner (Run Run1.Name,Run1.Distance,Run1.Time ( σ Run1.time>Run2.time (ρ Run1 (Run) Distance ρ Run2 (Run)))))

6 Name,Country,Distance,Time (Runner (Run Run1.Name,Run1.Distance,Run1.Time ( σ Run1.time<Run2.time (ρ Run1 (Run) Name ρ Run2 Run)))) Name,Country,Distance,Time (Runner (Run Run1.Name,Run1.Distance,Run1.Time ( σ Run1.time<Run2.time (ρ Run1 (Run) Distance ρ Run2 (Run)))))

7 3 Result Cardinality Consider the following two relations: A B B C D 1 x x 0 3 R = 2 y 2 z S = y 2 1 y x w a y 2 0 For the following relational algebra expressions, fill out the number of tuples each of them produces using the data given above. Expression R S 25 R S 5 R S 7 R S 6 R A=D S 3 ρ C A (R) S 2 B (R) - B (σ C<3(S)) 2 A (R) ρ A D( D (S)) 2 D (S) S 5 Size of result (number of tuples)

8 4 Join and Division Operators 1. Consider the following two relations: A B 1 1 B C D P = 1 3 Q = For each of the following expressions circle all the tuples that are not in its result set (the tuples contain all four columns: [A, B, C, D]). A. P Q : B. P Q : (a) [ 1, 1, 7, 2 ] (a) [ 3, 1, 7, 2 ] (b) [ 1, 2, 5, 2 ] (c) [ 3, 2, 5, 0 ] not contained (d) [ 3, 1, 4, 0 ] (e) [ 2, 4, 2, 2 ] not contained (b) [ 4, 1, 4, 0 ] not contained (c) [ 2, 4, -, - ] not contained (d) [ 3, 1, 4, 0 ] (e) [ 1, 3, 2, 2 ] 2. Let {(,..., ) A } be the relation with one tuple containing null-values with the schema of A. Which of the following relational algebra expressions represents 1 a left outer join (R S), 2 a right outer join (R S), and 3 a full outer join (R S). 2 (R S) ((S S (R S)) {(,..., ) R S}) 1 (R S) ((R R (R S)) {(,..., ) S R}) 3 (R S) ((R R (R S)) {(,..., ) S R}) ((S S (R S)) {(,..., ) R S}) 3. Which tuples are included in the output of the following relational division? Student Lecture Fred Databases Tom Databases Lecture Lisa Software Engineering Databases Lisa Compiler Design Networking Tom Compiler Design Tom Networking Fred Fred, Lisa Lisa, Fred Tom Lisa Tom, Lisa

9 5 Train Connections Consider the following relational schema: Cities: (Name, State) Stations: (Name, NoPlatforms, CityName, State) Itinerary: (ItNr, Length, StartStation, DestinationStation) Connections: (FromStation, ToStation, ItNr, Departure, Arrival) Suppose that the relation Connections already contains the transitive closure for each given train, e.g., if there is a direct train from Zurich to Geneva with a stop in Bern, then there exists a relation tuple for Zurich Bern, Bern Geneva, and Zurich Geneva. Formulate the following queries in relational algebra: 1. Find all the direct connections from Zurich (any station) to Geneva (any station) (ρ FromName Name ( Name (σ CityName=Zurich(Stations)))) FromName=FromStation Connections ToName=ToStation (ρ ToName Name ( Name (σ CityName=Geneva(Stations)))) 2. Find all the single-transfer connections from Zurich to Locarno. The transfer station can be any of the stations but the connecting trains should run on the same day. (You can use a function DAY() on the attributes Departure and Arrival in order to determine the day.) (ρ FromName Name ( Name (σ CityName=Zurich(Stations)))) FromName=c1.FromStation ρ c1 (Connections) c1.tostation=c2.fromstation c1.arrival<c2.departure DAY(c1.Arrival)=DAY(c2.Departure) c1.itnr<>c2.itnrρ c2 (Connections) ToName=c2.ToStation (ρ ToName Name ( Name (σ CityName=Locarno(Stations)))) 3. Is it possible to find all possible connections between two stations independent on the number of transfers? Relational algebra does not offer infinite recursion. Therefore, we can only find connections up to a specific maximum search depth.

Assignment 2: Relational Model

Data Modelling and Databases Exercise dates: March 8/March 9, 208 Ce Zhang, Gustavo Alonso Last update: March 6, 208 Spring Semester 208 Head TA: Ingo Müller Assignment 2: Relational Model This assignment