Midterm Examination CS 265 Spring 2016 Name: I will not use notes, other exams, or any source other than my own brain on this exam: (please sign)

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1 Midterm Examination CS 265 Spring 2016 Name: I will not use notes, other exams, or any source other than my own brain on this exam: (please sign 1. (5 pts Consider the relational schema, R = [A B C D E F G] with applicable FDs A, B à C, D A, E à G F à A A à F Key(s: Give all keys for this relational schema. Be clear. 2. (5 pts Consider the relational schema, R = [A B C ] with applicable FDs A à B B à C C à A B à A C à B A à C Give TWO minimal set of FDs consistent with those given. Do not take any short cuts in presenting your answers. Be clear. First Minimal Set Second Minimal Set 1

2 3. (5 pts Consider the relational schema, R = [A B C D E F G H I J ] with applicable FDs A, F à C, I B, E à G, H B à J B, C, E à D C à A, F Reflect on the task of decomposing relation R into a set of BCNF relations. For example, I might use B à J as the first step to decompose R: A B C D E F G H I J B à J B J A B C D E F G H I Circle the FD(s in the list above that would NOT be used as the first step to decompose R using the decomposition procedure described in class. Beside each circled FD, give a brief explanation of why you would not use that FD to decompose the relation R. 2

3 4. (5 points Consider the following four table definitions, together with all entries in each of the four tables. CREATE TABLE Customer ( SSN Integer,... PRIMARY KEY (SSN; CREATE TABLE Product ( ProdID Integer,... PRIMARY KEY (ProdId; Customer SSN... Ssn1... Ssn2... Ssn3... CREATE TABLE Account ( SSN Integer NOT NULL, AccntNo Integer,... PRIMARY KEY (AccntNo, FOREIGN KEY (SSN REFERENCES Customer ON UPDATE CASCADE; Product ProdID... Pid1... Pid2... Pid3... CREATE TABLE Transaction ( TransID Integer, AccntNo Integer, ProdId Integer,... PRIMARY KEY (TransID, FOREIGN KEY (AccntNo REFERENCES Account ON UPDATE NO ACTION, /* aka RESTRICT */ FOREIGN KEY (ProdId REFERENCES Product ON UPDATE CASCADE; 3 Account SSN AccntNo... Ssn1 Acct1... Ssn2 Acct4... Ssn1 Acct2... Transaction TransID AcctNo ProdID... Tid1 Acct6 Pid3... Tid2 Acct3 Pid2... Tid3 Acct3 Pid3... Ssn2 Acct3... Ssn2 Acct5... Ssn3 Acct6... Change all attribute values as a result of performing these UPDATE operations in order (BE NEAT!!!. If an operation fails, and has no effect as a result, then move to the next operation. UPDATE Transaction SET AccntNo = Acct4 WHERE TransID = Tid2; UPDATE Account SET AccntNo = Acct1 WHERE AccntNo = Acct5; UPDATE Account SET SSN = Ssn1 WHERE AccntNo = Acct3; UPDATE Product SET ProdID = Pid4 WHERE ProdID = Pid3; UPDATE Customer SET SSN = Ssn5 WHERE SSN = Ssn2; UPDATE Account SET AccntNo = Acct7 WHERE SSN = Ssn3;

4 5. (5 pts Consider the UML fragment to the right and identify (circle all equivalent table translations (i.e., those translations that faithfully enforce the constraints implied by the UML without regard to elegance from those given below. You might receive partial credit for a brief explanation of your choices. UNIQUE(y implies that y NOT NULL, but not vice versa. PK stands for PRIMARY KEY. FK stands for FOREIGN KEY. X x1 PK role R is an associabon class role1 0..* R r1 (A (B (C (D (E CREATE TABLE XR ( x1, r1, role2 NOT NULL, PK(x1, FK (role2 refs XR CREATE TABLE X ( x1, PK (x1 CREATE TABLE R ( r1, role1, role2 NOT NULL, PK(role1, FK (role2 refs X, FK (role1 refs X CREATE TABLE XR ( x1, r1, role2, PK(x1, UNIQUE(role2, FK (role2 refs XR CREATE TABLE XR ( x1, r1, role2, PK(x1, FK (role2 refs XR CREATE TABLE X ( x1, PK (x1, FK (x1 refs R CREATE TABLE R ( x1, r1, role1 NOT NULL, PK(r1, FK (role1 refs X, FK (x1 refs X (F None of the above 4

5 6. (5 points A health facility wants to measure the AVERAGE (AVG spread in the weight of clients between the ages of 45 and 60, inclusive, who have at least 100 weight entries in the facility s DB. The spread of a client s weight is the maximum weight on record in the facility s DB minus the minimum weight on record in the DB (regardless of age at the time the weight entries were made. The DB contains two tables representing these two relations (among others. Client (cid, name, age, address, phone,... DailyRecord(cid, date, weight,... Primary key attributes of each table (relation are bold-face and underlined. In the actual table definitions, the attribute of age is declared as NOT NULL, and weight can be NULL. Complete the following skeletal query to compute the AVERAGE (AVG spread in the weight of clients between the ages of 45 and 60, inclusive, with at least 100 weight entries, by filling in the blanks. SELECT AVG(Temp.Flux FROM (SELECT MAX(D.weight MIN(D.weight As Flux FROM Client C, DailyRecord D WHERE C.id = D.id AND D.weight IS NOT NULL AND C.age >= 45 AND C.age <= 60 GROUP BY C.id HAVING COUNT(* >= 100 AS Temp As an aside, it is likely that instead of represenbng age as an explicit aaribute, we would alternabvely (or perhaps addibonally, store BirthDate and compute age on demand or somehat proacbvely with a Trigger on a regular basis. 5

6 7. (5 pts Consider the following table definition: CREATE TABLE HRel (ID integer, name integer, PRIMARY KEY (ID Circle all queries below that return the tuples of HRel with the top 5 values of ID (a SELECT H.ID, H.name WHERE NOT EXISTS (SELECT * 1, HRel H2 WHERE H.ID = H1.ID AND H1.ID < H2.ID GROUP BY H1.ID HAVING COUNT(* >= 5 ORDER BY ID DESC (b SELECT H.ID, H.name WHERE (SELECT COUNT (* 2 WHERE H.ID < H2.ID < 5 ORDER BY H.ID DESC (c SELECT H.ID, H.name EXCEPT SELECT H.ID, H.name WHERE (SELECT COUNT (* 2 WHERE H.ID < H2.ID >= 5 ORDER BY H.ID DESC (d SELECT H.ID, H.name EXCEPT SELECT H1.ID, H1.name 1, HRel H2 WHERE H1.ID < H2.ID GROUP BY H1.ID, H1.name HAVING COUNT(*+1 > 5 ORDER BY ID DESC (e None of the above 6

7 8. (5 pts Consider the following relational schema (underlined variables in each schema make up its primary key Supplier ( sid: integer, sname: string, address: string, city: string Catalog (sid: integer, pid: integer, cost: real Part (pid: integer, pname: string, color: string a (3 pts Give a relational algebra expression that implements the query specified as Find the sids of Suppliers that supply any green part and are in the city of Nashville. Project(sid ( select(color=green and city=nashville (Suppliers njoin Catalog njoin Parts b (2 pts Give a left-deep expression tree for the relational algebra expression that you give in (a 7

8 9. (5 pts Assume that you have a DB with a table, Likes(a, b (read a likes b,with Primary Key (a,b. Write a trigger so that when a row is inserted into Likes of the form Likes(X, 'Friendly, where Friendly is a constant and X can match any value, a tuple that indicates that Friendly Likes X is inserted into Likes (unless it is already in the table. So if Likes( Abe, Friendly is inserted, then Likes( Friendly, Abe is inserted. MOREOVER, for each person, Y, who X Likes, the trigger also inserts Likes ( Friendly, Y (unless it is already there. So if Likes( Abe, Mary and Likes( Abe, Hua are in Likes, and Likes( Abe, Friendly is inserted into Likes, then Likes( Friendly, Abe is inserted, and so is Likes( Friendly, Mary and Likes( Friendly, Hua. Use as close to SQLite syntax as you can. CREATE TRIGGER RinsFriendly AFTER INSERT ON Likes FOR EACH ROW /* Complete the Trigger definition */ WHEN new.name = 'Friendly' BEGIN INSERT INTO Likes(ID1, ID2 SELECT new.id, S.ID FROM Highschooler S WHERE new.grade = S.grade AND new.id <> S.ID; END; 8

9 10. (5 points Consider the following definitions, and give a UML diagram (on the right of the page that is consistent with the definitions. CREATE TABLE M ( m1, PRIMARY KEY (m1 CREATE TABLE Q ( q1, PRIMARY KEY (q1 CREATE TABLE R ( r1, m1, q1, PRIMARY KEY (m1, q1, FOREIGN KEY (m1 REFERENCES M, FOREIGN KEY (q1 REFERENCES Q CREATE ASSERTION MparticipatesQ CHECK (NOT EXISTS (SELECT * FROM M WHERE M.m1 NOT IN (SELECT m1 FROM R 9