Exam. Question: Total Points: Score:

Transcription

1 FS 2016 Data Modelling and Databases Date: August 17, 2016 ETH Zurich Systems Group Prof. Gustavo Alonso Exam Name: Question: Total Points: Score: Rules (please read carefully) You have 120 minutes for the exam. Please write your name and Legi number on this cover page. Please write your Legi number on all other pages, including the additional pages you possibly use. Please write your answers on the exam sheets. Use blue or black ink, DO NOT USE red ink. DO NOT USE pencils. Write as clearly as possible and cross out everything that you do not consider to be part of your solution. Answers can be given in either English or German. Remarks Most questions are designed such that the fastest way to solve them is to solve the problem without looking at the answers and only then to find the given answer that corresponds to your solution. We expect it to be slower to try out all combinations of questions and solutions in order to find the ones that match.

2 1 Entity-Relationship Model (11 points) (a) (5 points) Consider the following Entity-Relationship (ER) model. passport number person 1 1 has address name is_a man 1 father mother 1 woman father 1 N son daughter N 1 mother isson isdaughter For each of the following statements say whether it is true or false in the model. Each correct answer gives points, each wrong answer gives 0.25 points. i. Each person has at least one address. True False ii. Each person has at most one address. True False iii. Each address belongs to at most one person. True False iv. Each address belongs to at least one person. True False v. Each person has at least one passport number. True False vi. Each person has at most one passport number. True False vii. Each passport number belongs to at most one person. True False viii. There are no two persons with the same name. True False ix. Each person who is not a man is a woman. True False x. Each person who is a man is not a woman. True False xi. Each woman is a person. True False xii. Each man has at least one father. True False xiii. Each man has at most one father. True False xiv. A man can be his own father. True False xv. If a woman has a father, she also has a mother. True False xvi. Each man is a father. True False xvii. Each father is a man. True False xviii. Each son is a man. True False xix. Each woman has at least one daughter. True False xx. Each woman has at most one daughter. True False

3 (b) (6 points) Consider the following Entity-Relationship (ER) model, where the cardinalities of relationship R are given as variables M A, M B, and M C. For example, if M A = M B = M C = 1, then R is a 1:1:1 relationship. A M A M B B R M C C The table below shows entities of type A, B, and C respectively that are connected through relationship R. For example, the first tuple m, o, x indicates that entity m of type A, entity o of type B, and entity x of type C are connected through R. We are now interested in the question which are valid sets of tuples for R for different cardinalities M A, M B, and M C. For each assignment of cardinalities in the table, select a subset of tuples that form a valid set R. Select a tuple by adding a checkmark to the corresponding cell. If a tuple cannot be in R, write down the number of the tuple with which it conflicts. tuple number tuple A B C A, B, C R for given multiplicity? M A = N M B = N M C = N M A = 1 M B = N M C = N M A = 1 M B = 1 M C = N M A = 1 M B = 1 M C = 1 1 m o x 2 n o z 3 m q y 4 n p z 5 n o x 6 m q z

4 2 Relational Algebra (11 points) (a) (9 points) Consider two relations: R = A B 1 x 2 y 2 z 3 x 9 a S = B C D x 0 3 y 2 1 y 3 3 w 3 0 y 2 0 Fill out for the following relational algebra expressions how many tuples each of them returns, based on the data given above. Expression 1) R x S 2) R S 3) R S 4) R S 5) R A=D S 6) ρ C A (R) S 7) Π B (R) - Π B (σ C<3 (S)) 8) Π A (R) ρ A D (Π D (S)) 9) Π D (S) S Cardinality of result

5 (b) (2 points) Consider two relations: P = A B Q = B C D 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]). 1. P Q : a) [ 1, 1, 7, 2 ] b) [ 1, 2, 5, 2 ] c) [ 3, 2, 5, 0 ] d) [ 3, 1, 4, 0 ] e) [ 2, 4, 2, 2 ] 2. P Q : a) [ 3, 1, 7, 2 ] b) [ 4, 1, 4, 0 ] c) [ 2, 4, -, - ] d) [ 3, 1, 4, 0 ] e) [ 1, 3, 2, 2 ] f) [ -, 3, 2, 2 ]

6 3 SQL I (10 points) Given is the following schema (based on the ZVV timetable) with two tables trips and stop_times where trips contain a list of trips trams make within one day, and stop_times contains all the stops made for every trip: CREATE TABLE trips( trip_id INTEGER NOT NULL PRIMARY KEY, tram_number INTEGER NOT NULL ); CREATE TABLE stop_times( trip_id INTEGER NOT NULL REFERENCES trips(trip_id), stop_sequence INTEGER NOT NULL, stop_name VARCHAR(50) NOT NULL, arrival_time TIMESTAMP NOT NULL, departure_time TIMESTAMP NOT NULL, PRIMARY KEY (trip_id, stop_sequence) ); (a) (5 points) The following queries can be executed on the schema. All queries are functional and return some result. Q1. SELECT COUNT(*) FROM trips; Q2. SELECT COUNT(*) FROM trips t, stop_times st WHERE t.trip_id = st.trip_id; Q3. SELECT COUNT(*) FROM stop_times GROUP BY trip_id; Q4. SELECT COUNT(*) FROM trips JOIN stop_times USING(trip_id) WHERE stop_name LIKE % ; Q5. SELECT SUM(valA) FROM (SELECT stop_name, COUNT(*) as vala FROM stop_times GROUP BY stop_name) taba; Q6. SELECT COUNT(DISTINCT trip_id) FROM stop_times; Identify all equivalence classes for the given queries and write them down in the table below. Two queries are equivalent if they return the same set of results for any data that the database may contain. Note that there may be fewer than 6 equivalence classes. Equivalence Queries

7 (b) (5 points) A tram stop (stop_name) is terminal if it is the last stop for any trip (as identified through the trip_id). The last stop of a trip can be identified by its stop_sequence. For example, if a tram makes a trip along six tram stops, stop_sequence=6 identifies the last stop of the trip. Fill in the blanks below to obtain a SQL query that finds all terminal tram stops (stop_name). Also make sure the result does not contain duplicate entries. SELECT FROM stop_times st1, (SELECT FROM stop_times GROUP BY ) st2 WHERE AND

8 4 SQL Part 2 (12 points) Consider three tables: Actor: id name 1 John Doe 2 John Smith 3 Johanna Doe 4 Daniela Johannsen 5 Jack Jackson Director: id name 1001 Adam West 1002 Carla East 1003 Charlie Smith 1004 Lucas Doe Movie: id dir_id lead_id rating title We have run seven queries on this data (see next page) and received a number as a result for each of them. These numbers are shown in the table below. List in the box by the result which queries give which results (some of them will have multiple queries associated). Result value: List queries returning this result: Queries, A) through F) (see also next page): A) SELECT max(rating) from movie where lead_id!= 1 B) SELECT min(rating) from movie where dir_id in (select id from director where name not like %S% )

9 C) select min(rating) from movie, actor where lead_id=actor.id and actor.name like Jo% D) select (2+avg(rating)) from movie m1, director where m1.dir_id=director.id and director.name like %l%t and (select count(*) from movie m2 where m2.lead_id=m1.lead_id)>1 E) select (min(rating)) from movie m1 where (select count(*) from movie m2 where m2.lead_id=m1.lead_id)=1 F) select avg(m2.rating) from movie m1, movie m2, actor a1, actor a2 where m1.lead_id=a1.id and m2.lead_id=a2.id and m1.dir_id=m2.dir_id and a1.id=a2.id and a1.name like %a%do%

10 5 Integrity constraints (9 points) (a) (2 points) Given a database with multiple tables, which of the following constraints can be used in a way to ensure, or will by definition not allow NULL values to be inserted? Mark all correct options below: UNIQUE NOT NULL FOREIGN KEY PRIMARY KEY CHECK (b) (3 points) Are the following statements true or false? Place a checkmark in the correct column. Statement True False a) In principle a check constraint could be added to a table that makes it impossible to insert any data. b) A primary key column can have up to one NULL value. c) The primary key constraint can be augmented with an on delete set null d) The on delete cascade constraint specified for a column in a table leads to record deletions in other tables. e) The check constraint can not alter values in a table.

11 (c) (4 points) Consider the following schema creation statement for a university: CREATE TABLE student ( id INT PRIMARY KEY, age INT NOT NULL ); CREATE TABLE lecture ( id INT PRIMARY KEY, name VARCHAR(32) NOT NULL ); CREATE TABLE attends ( sid INT, lid INT, constraint fk1 FOREIGN KEY (sid) REFERENCES student(id) ON DELETE CASCADE ON UPDATE CASCADE, constraint fk2 FOREIGN KEY (lid) REFERENCES lecture(id) ON DELETE NO ACTION ON UPDATE SET NULL ) Initially, the three tables contain the following entries: student: id age lecture: id name 2 Databases 3 Literature 4 Biology attends: sid lid Fill in the table, showing how many tuples the tables contain after executing each set of SQL statements. Assume that each set of SQL statements is executed on the initial state of the database above and that individual SQL statements are executed in a separate transaction. SQL student lecture attends INSERT INTO lecture VALUES (1, Math I ); INSERT INTO lecture VALUES (2, Math II ); INSERT INTO student VALUES (9, 30); INSERT INTO attends VALUES (9, 2); DELETE FROM lecture WHERE id>2; UPDATE lecture SET id=99 WHERE id<=2; DELETE FROM student WHERE age > (SELECT average(age) FROM student);

12 6 Normal Forms & Decomposition (10 points) 1. Given the following WorksFor relation: WorksFor: (employee_id, department_id, start_date, end_date, department_name, department_manager) with the following functional dependencies: employee_id, department_id start_date, end_date department_id department_name, department_manager (a) (1 point) Identify the candidate keys of the relation WorksFor. (b) (1 point) Which is the highest normal form the relation WorksFor is in? 1NF 2NF 3NF BCNF 4NF (c) (1 point) Which are the problematic functional dependencies which prevent the relation WorksFor from being in the next higher normal form? employee_id, department_id start_date, end_date department_id department_name, department_manager (d) (2 points) Propose a lossless decomposition of WorksFor into R 1 and R 2 such that R 1 and R 2 are in 3NF. Prove that your decomposition is lossless.

13 2. Given the relation R(A, B, C, D, E), and two sets of functional dependencies: F 1: AB C F 2: AB CE ABC DE ABC D AE B E B For each of them, there are no further non-trivial functional dependencies. (a) (2 points) Determine the highest normal form for each set of functional dependencies. Indicate the functional dependency that prevents the relation from being in the next higher normal form. Set of FDs 1NF 2NF 3NF BCNF Problematic FD F 1 F 2 (b) (3 points) If the relations are not in BCNF, apply the decomposition algorithm to bring the relations into BCNF. Indicate if your decomposition preserves all the functional dependencies. Set of FDs Is in BCNF? (Yes/No) F 1 F 2 Result of Decomposition Algorithm Dependency preserving? (Yes/No)

14 7 Functional Dependencies & 3NF (11 points) Given the relation R(A, B, C, D, E), there exist two sets of functional dependencies: F 1: A B F 2: A CE D E D AB AE D AB DE BC AD AC D For each of them, there are no further non-trivial functional dependencies. (a) (4 points) For each set of attributes in the table below, decide whether it is a candidate key of R given the respective functional dependencies of F 1 and F 2. F 1 F 2 Possible Is Cand. Is not Is Cand. Is not Candidate Key Key Cand. Key Key Cand. Key A D AB AC BC BCE (b) (3 points) Determine whether the functional dependencies above are minimal or whether you can do a right or left reduction. If the dependencies are not minimal, write down one violating functional dependency and what kind of reduction (left or right) could be applied to this dependency. Is minimal. Is not minimal. Violating FD and Explanation F 1 F 2 (c) Apply the synthesis algorithm to R given F 1 and F 2 respectively. i. (2 points) Select the correct decomposition of R given F 1 according to the synthesis algorithm: R 1(A, B), R 2(A, D, E), R 3(A, B, C, D), R 4(D, E) R 1(A, D, E), R 2(A, B, C, D) R 1(A, D, E), R 2(B, C, D) ii. (2 points) Select the correct decomposition of R given F 2 according to the synthesis algorithm: R 1(A, B, D), R 2(A, C, D, E) R 1(A, B, D, E), R 2(A, C) R 1(A, B, D), R 2(A, C, E), R 3(A, C, D)

15 8 Query Processing (10 points) Consider the following three relations: Relation Director (did, dname, birth_date) Category (cid, ctitle, description) Movie (mid, mtitle, did, cid, release_date, rating) Size 1k tuples 100 tuples 10k tuples Table Movie references table Director (did is a foreign key to Director.did) and table Category (cid is a foreign key to Category.cid). The following query uses these tables and determines the average rating of directors per movie category: SELECT name, ctitle, AVG(rating) FROM (SELECT ctitle, did, rating FROM movie AS M JOIN category AS C USING (cid)) MPC JOIN director AS D USING(did) GROUP BY name, cat_title; (a) (2 points) In order to execute this query, the database system can use query plans with the following two join orderings. For both orderings, write the maximum number of tuples in the intermediate results into the boxes. name,ctitle AVG(rating) name,ctitle AVG(rating) 100 tuples category tuples cid Π name,ctitle,rating movie did director 10k tuples Join Ordering 1 tuples 1k tuples 100 tuples Π name,ctitle,rating category cid 10k tuples movie tuples did Join Ordering 2 tuples director 1k tuples

16 (b) (3 points) In order to compute a join of two tables R and S, the database system may choose between several implementations. Complete the following table by filling in the maximum costs in terms of operations on tuples in O-notation of the following join algorithms as discussed in the lecture. Use the number of tuples R and S in R and S respectively as variables. Join algorithm Cost Sort-Merge-Join (SMJ) Grace-Hash-Join (GHJ) Nested-Loop-Join (NLJ) (c) (5 points) We now compare query plans using different join orderings and different join implementations. We refer to a plan using Join Ordering 1 from above as (C impl M) impl D and a plan using Join Ordering 2 as C impl (M impl D), where impl is one of the join implementations SMJ, GHJ, or NLJ. Complete the table below by giving an estimate of the order of magnitude of the costs of each plan. Assume the costs of a plan is the sum of the maximum costs of its joins. Assume the maximum number of tuples for intermediate results from Part a. To estimate the costs of a join, you may use the algorithm complexities from Part b using 1 as constant and logarithms with base 10. For example, sorting a list of n = 1000 tuples would be estimated to have a cost of n log n = 1000 log = = 3000 operations. Query plan Cost N ops N ops < N ops < N ops < N ops (C NLJ M) SMJ D C NLJ (M SMJ D) C SMJ (M GHJ D) C SMJ (M NLJ D) C GHJ (M SMJ D)

17 9 Transactions (15 points) Notation used in this exercise: b i - Begin of transaction i r i(a) - Transaction i reads data object A. w i(a) - Transaction i writes to data object A. c i - Transaction i commits successfully. a i - Transaction i aborts or is aborted. (a) (4 points) Determine if the following histories are possible under snapshot isolation (SI), strict two-phase locking (S2PL), or both. Write either YES or NO in each of the boxes. History SI S2PL b 1 r 1(A) w 1(B) b 2 w 2(A) r 2(B) b 3 c 2 r 3(A) c 1 c 3 b 1 r 1(A) b 2 r 2(A) w 2(B) c 2 w 1(A) c 1 b 1 w 1(B) r 1(A) b 2 w 2(B) c 1 a 2 b 1 w 1(A) b 2 r 1(B) r 2(B) c 1 w 2(A) c 2 (b) (1 point) Can phantoms occur in snapshot isolation? (Explain the answer in one or two sentences) Yes No (c) (2 points) In Two Phase Locking (2PL), one can release all locks at the point when the transaction has finished but has not issued a commit. Will the result be serializable? Yes No Are histories generated by this approach always recoverable? Yes No

18 (d) (2 points) Provide a serialization order for the transactions of the following histories. If the history is not serializable, write down "not serializable". History Serialized History r 1(A) w 2(A) c 1 r 2(B) w 3(B) c 2 c 3 w 2(A) r 1(A) w 3(A) a 3 c 2 c 1 w 2(A) r 3(B) r 1(A) w 1(B) c 3 w 2(B) c 2 c 1 r 2(C) w 1(A) r 2(B) w 1(B) c 1 w 2(C) c 2 (e) (2 points) Using two conflicting transactions: T 1: r 1(A) w 1(B) c 1 T 2: r 2(B) w 2(A) c 2 Find a history with c 1 < c 2 (i.e. T 1 commits before T 2) and which corresponds to the serial execution T 2, T 1. Is this history possible under Two Phase Locking (2PL) and Strict Two Phase Locking (S2PL)? 2PL S2PL Yes No (f) (2 points) Find the most strict recoverability class (not recoverable, recoverable, avoid-cascadingaborts, strict) for the following histories. History Recoverability class w 2(B) w 1(A) r 2(A) w 2(A) c 2 c 1 w 3(A) w 2(A) c 3 w 1(A) r 1(B) w 2(A) c 2 w 1(B) c 1 w 1(A) r 2(B) w 1(B) c 1 w 2(A) c 2 w 1(A) w 2(A) r 2(A) w 3(C) c 3 c 1 c 2

19 (g) (2 points) Below you find the Venn diagram of the recoverability classes for transactions with read and write operations. Now, assume that the only operation possible in a transaction is an atomic read-write (represented as rw i(x)), where the read and the write happen in one single atomic step. Starting from the Venn diagram on the left, modify the diagram for such transactions. Draw your answer in the box on the right. Diagram for transactions with reads r i(a) and writes w i(a): Diagram for transactions with atomic read-writes rw i(a): Recoverable ACA Strict

20 10 Commit protocols (10 points) (a) (6 points) A coordinator C and two participants P 1, P 2 run the three-phase-commit (3PC) protocol. The coordinator also acts as participant. We model the execution of the protocol as a series of events. An event can be one of the following: A message event of the form (X, Y, M) means that node X sends the message M to node Y, where X, Y {A, B 1, B 2} and the message M {request, yes, no, pre-commit, ack, abort, commit}, meaning request to vote, voting yes, voting no, pre-commit, acknowledge last message, request to abort, and request to commit respectively. A group communication event of the form (X, ask around abort) or (X, ask around commit) means that node X initiates a round of group communication where all reachable nodes exchange all relevant information and then decide to abort or to commit accordingly. We assume that no failures occur during group communication. A failure event of the form (X, fail) means that node X fails. The following sequence of events shows an execution of the 3PC protocol where no failures occur: time step event 1 (C, P 1, request) 2 (C, P 2, request) 3 (P 1, C, yes) 4 (P 2, C, yes) 5 (C, P 1, pre-commit) 6 (C, P 2, pre-commit) 7 (P 1, C, ack) 8 (P 2, C, ack) 9 (C, P 1, commit) 10 (C, P 2, commit) We now modify this sequence of events starting from some time step. Complete each new sequence with one possible next event such that it models a valid execution of the 3PC protocol. Sequence (i): time step event 4 (P 2, C, no) 5 Sequence (iv): time step event 6 (C, fail) 7 Sequence (ii): time step event 2 (C, fail) 3 (P 1, C, yes) 4 Sequence (iii): time step event 5 (C, fail) 6 Sequence (v): time step event 4 (P 2, fail) 5 Sequence (vi): time step event 6 (P 2, fail) 7 (C, P 2, pre-commit) 8 (P 1, C, ack) 9 (b) (2 points) Let us assume that the three nodes C, P 1, and P 2 are connected through a network with the links L 0, L 1, and L 3 respectively:

21 C L 0 L 1 L 2 P 1 P 2 If a link fails, messages sent over the links are lost. To model link failures, we extend our definition of failure events to links, i.e., (X, fail) means the failure of X, where X may be a node or a link. Complete the following sequence of events by filling in the blanks such that it models a valid execution of the 3PC protocol: time step event 6 (L 2, fail) 7 (C, P 2, pre-commit) 8 (P 2, ) 9 (P 1, C, ack) 10 (C, ) (c) (2 points) Give a sequence of events modeling an execution of the two-phase commit (2PC) protocol with the nodes C, P 1, P 2 where no failures occur and all nodes want to commit using the notation above. You may not need all rows of the table. time step event

22 11 Replication (8 points) (a) (4 points) Rank the replication strategies by response times starting from the one that provides shortest response times. Replication strategy Synchronous primary copy Asynchronous primary copy Synchronous update everywhere Asynchronous update everywhere Rank (b) (4 points) For each statement below, mark the replication strategies for which the statement is valid. Statement Replicas may have inconsistent data Distributed coordination of updates is not necessary Load is evenly distributed Updates can be lost (needs reconciliation) Primary copy Synchronous Update everywhere Primary copy Asynchronous Update everywhere

23 Empty Pages You can use the following pages at your convenience. If you use them for a solution, mark as clearly as possible to which question your solution belongs and what is part of that solution and what is not.

24

25