COMP7640 Assignment 2

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Dwayne Newman
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1 COMP7640 Assignment 2 Due Date: 23:59, 14 November 2014 (Fri) Description Question 1 (20 marks) Consider the following relational schema. An employee can work in more than one department; the pct time attribute of the Works relation shows the percentage of time that a given employee works in a given department. pct time values are integers between 0 to 100. managerid is a foreign key pointing to the primary key of Emp. Write the following queries in SQL. Emp (eid, ename, age, salary) Works (eid, did, pct time) Dept (did, dname, budget, managerid) (a) (5 marks) Find the names of employees who work in either the Development department or the Testing department. (b) (5 marks) For every department with more than ten FTE (full-time-equivalent) employees (computed as the sum the percentage of each employee working in that department), return the did and the number of employees that work in that department. (c) (5 marks) Find the name of each employee whose salary exceeds 5% of the sum of budget of all the department(s) that he or she works in. (d) (5 marks) Find the managerids of managers who manage all departments with budgets greater than five hundred thousand. For example, if there are five such departments, the returned person, if exists, must manage all of the five departments; it is entirely possible that he or she may also manage deparments other than the five above though. Solution: 1

2 (a) SELECT DISTINCT E.ename FROM Emp E, Works W, Dept D WHERE E.eid = W.eid AND W.did = D.did AND (D.dname = Development OR D.dname = Testing ) (b) SELECT D.did, COUNT(W.eid) FROM Works W, Dept D WHERE W.did = D.did GROUP BY D.did HAVING SUM(pct_time) > 10*100 (c) Note the we need to use ALL or ANY before the comparison with value(s) returned from the subquery (in this query, either is okay). SELECT E.ename FROM Emp E WHERE E.salary > ALL ( SELECT 0.05 * SUM(D.budget) FROM Works W2, Dept D WHERE W2.eid = E.eid AND W2.did = D.did) Or you can use a join SELECT E.ename FROM Emp E, (SELECT W2.eid, 0.05 * SUM(D.budget) AS b FROM Works W2, Dept D WHERE W2.eid = E.eid GROUP BY W2.did) AS tmp WHERE E.eid = tmp.eid AND E.salary > tmp.b (d) (Note the all constraint) SELECT DISTINCT D. managerid FROM Dept D WHERE NOT EXISTS ( SELECT * FROM Dept D2 WHERE D2.budget > AND D2. managerid <> D. managerid) or SELECT DISTINCT D. managerid FROM Dept D WHERE NOT EXISTS ( SELECT D2.did FROM Dept D2 WHERE D2.budget > EXCEPT SELECT D3.did FROM Dept D3 WHERE D3. managerid = D. managerid ) or use this one if EXCEPT is not supported 2

3 SELECT DISTINCT D. managerid FROM Dept D WHERE ( SELECT COUNT(DISTINCT D2.did) FROM Dept D2 WHERE D2.budget > ) = ( SELECT COUNT(DISTINCT D3.did) FROM Dept D3 WHERE D3. managerid = D. managerid); Question 2 (20 marks) Prove or disprove the following inference rules for functional dependencies. A proof must be made by using the three Armstrong inference rules only (i.e., you are not allowed to use the rules of union, decomposition or pseudo-transitivity). A disproof should be done by demonstrating a relation instance which serves as an counterexample of the inference rule. 1. If W Y, X Z, then WX Y. 2. If XY Z and Y W, then XW Z. 3. If X Z, Y Z, then X Y. 4. If X Y, X W and WY Z, then X Z. Solution: 1 1. WX YX, from W Y 2. YX Y, from Y YX 3. WX Y, from (1) and (2) 2 Disproof: X Y Z W x 1 y 1 z 1 w 1 x 1 y 2 z 2 w 1 The above two tuples satisfy XY Z and Y W but do not satisfy XW Z. 3

4 3 Disproof: X Y Z x 1 y 1 z 1 x 1 y 2 z 1 The above two tuples satisfy X Y and Y Z but do not satisfy X Y X XY, from X Y 2. XY WY, from X W 3. X WY, from (1) and (2) 4. X Z, from WY Z and (3) Question 3 (20 marks) Consider the attribute set R = ABCDEG and the FD set F = {AB C, AC B, E BD, B D, BC A, C G} 1. Find all the candidate keys for R. 2. Consider R 1 = ABCD resulting from a decomposition of R. (1) Write down a minimal cover for F projected on R 1. (2) Is R 1 in BCNF? Justify your answer. 3. Which of the following decompositions of R = ABCDEG, with the same set of dependencies F, is (a) dependency-preserving? (b) lossless-join? (a) {ACDE, BEG} (b) {ABCE, BD, CG} Solution: 1 Notices that all candidate keys must include E, as it does not appear on the right hand side of any FD. We test all attribute subsets containing E from size one upwards. Size 1: 4

5 Size 2: E+ = BDE, not a candidate key. Since E must appear in any candidate key and E+ = BDE, in the following, we only need to consider subsets involving E and a subset of {A, C, G}. AE + = ABCDEG, so it is a candidate key. CE + = ABCDEG, so it is a candidate key. EG + = BDEG, not a candidate key. Size 3: there is no valid combination of 3 attributes satisfying the constraints: containing E and a subset of {A, C, G} not a superset of any candidate key discovered so far. Therefore, we stop and all the candidate keys are: {AE, CE}. 2 (1) It is easy to see that the projected FDs on R 1 are: It is already a minimal cover. {AB C, AC B, B D, BC A} (2) It is easy to verify that all the candidate keys are {AB, AC, BC}. Hence, R 1 is not in BCNF because B D (i.e., B is not a super key and D is not a subset of B) (a) Lossless-join: No. Since E is not a key in either decomposed relation, it is easy to come up with an example where spurious tuples will occur if joined back. (b) FD-preserving: No. C G is lost. 2. Lossless-join: Yes. ABCDEGH {ABCDE, CG} is lossless join, due to C G; ABCDE {ABCE, BD} is lossless join too, due to B D. FD-preserving: Yes. Easy to check that the all except the third FD (E BD) is trivially preserved in the decomposed relations. Notice that E B is preserved in ABCE and B D is preserved in BD, hence E BD is preserved. 5

6 Question 4 (10 marks) What will be the output of SELECT query in the following script if run on a MySQL (ver 5.x) database? DROP TABLE t; CREATE TABLE t (a INT, b INT, c INT); INSERT INTO t VALUES (1, 1, NULL), (1, 1, 1), (1, NULL, 2), (NULL, 1, 3), (a, 1, 4), ("a", 1, 5); SELECT a, COUNT(b) FROM t GROUP BY a ORDER BY a; Solution: Output for MySQL 5.0, as it accepts the eccentric input a COUNT(b) NULL For higher versions of MySQL, the INSERT statement will cause an error. Late penalty Maximum marks reduced by 10% for the first day, and 30% every day onwards. 6

University of Massachusetts Amherst Department of Computer Science Prof. Yanlei Diao

University of Massachusetts Amherst Department of Computer Science Prof. Yanlei Diao CMPSCI 445 Midterm Practice Questions NAME: LOGIN: Write all of your answers directly on this paper. Be sure to clearly