Midterm Examination CS 265 Spring 2015 Name: I will not use notes, other exams, or any source other than my own brain on this exam: (please sign)

Size: px

Start display at page:

Download "Midterm Examination CS 265 Spring 2015 Name: I will not use notes, other exams, or any source other than my own brain on this exam: (please sign)"

Alan Bishop
6 years ago
Views:

1 Midterm Examination CS 265 Spring 2015 Name: I will not use notes, other exams, or any source other than my own brain on this exam: (please sign 1. (5 pts You have two relations R = [C, F, G] and P = [B, C, D, F, H]. Using only relational algebra operators of selection (σ and cross product (X, express the natural join of R and P. 2. (5 pts Consider the relational schema, R = [C S J D P Q V K ] with FDs J,P à C S,D à P J à S C à S,J,D,P,Q,V Key(s: Give all keys for this relational schema. 3. (5 pts Consider the relation P with 5 attributes, P = [ C D E F G ]. You are given the following functional dependencies: C,D! E and F,G! C,D. Give a dependency-preserving decomposition of P where each relation of the decomposition is in BCNF. Your decomposition should have as few relations as possible, while still satisfying the specifications of the problem. 1

2 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 DELETE CASCADE; Product ProdID... Pid1... Pid2... Pid3... CREATE TABLE Transaction ( TransID Integer, AccntNo Integer, ProdId Integer,... PRIMARY KEY (TransID, FOREIGN KEY (AccntNo REFERENCES Account ON DELETE NO ACTION, FOREIGN KEY (ProdId REFERENCES Product ON DELETE NO ACTION; 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... Cross out all rows of each table that are deleted as a result of performing these delete operations in order (if you are using a pen, go slow: Ssn3 Acct6... DELETE FROM Transaction WHERE ProdID = Pid3; DELETE FROM Customer WHERE SSN = Ssn1; DELETE FROM Customer WHERE SSN = Ssn2; DELETE FROM Customer WHERE SSN = Ssn3; DELETE FROM Product WHERE ProdID = Pid1; DELETE FROM Product WHERE ProdID = Pid2; DELETE FROM Product WHERE ProdID = Pid3; 2

3 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. X 0..* UNIQUE(y implies that y NOT NULL, but not vice versa. PK stands for PRIMARY KEY. FK stands for FOREIGN KEY. x1 PK R r Z z1 PK (A (B (C (D (E CREATE TABLE X ( x1, PK (x1, FK (x1 refs R CREATE TABLE R ( x1, r1, z1 NOT NULL, PK(x1, FK (z1 refs Z, FK (x1 refs X z1, PK (z1 CREATE TABLE X ( x1, PK (x1 CREATE TABLE R ( x1, r1, z1 NOT NULL, PK(x1, FK (z1 refs Z, FK (x1 refs X z1 PK (z1 CREATE TABLE XR ( x1, r1, z1, PK(x1, FK (z1 refs Z z1, PK(z1 CREATE TABLE XR ( x1, r1, z1 NOT NULL, PK(x1, FK (z1 refs Z z1, PK(z1 CREATE TABLE XR ( x1, r1, z1, PK(x1, UNIQUE(z1, FK (z1 refs Z z1, PK(z1 (F None of the above 3

4 6. (5 pts Consider the following table definitions: CREATE TABLE RelA (Aid integer, a1 integer, a2 integer, PRIMARY KEY (Aid CREATE TABLE RelB (Aid integer, Cid integer, b1 integer, PRIMARY KEY (Aid, Cid, b1, FOREIGN KEY (Aid REFERENCES RelA, FOREIGN KEY (Cid REFERENCES RelC CREATE TABLE RelC (Cid integer, c1 integer, c2 integer, c3 integer, PRIMARY KEY (Cid Circle all queries below that are equivalent to: π c1 (( Temp1 Temp2 where Temp1 = π Cid (( σ a2=q RelA Temp2 = π Cid (( σ a2=r RelA RelC RelB and RelB By equivalent, we mean produces the same output given the same input (and we are not referring to efficiency or elegance (a SELECT DISTINCT C.c1, RelB B1, RelA A1, RelB B2, RelA A2 WHERE C.Cid = B1.Cid AND B1.Aid = A1.Aid AND C.Cid = B2.Cid AND B2.Aid = A2.Aid AND A1.a2 = q AND A2.a2 = r (b SELECT DISTINCT C.c1, RelB B, RelA A WHERE C.Cid = B.Cid AND B.Aid = A.Aid AND A.a2 = q INTERSECT SELECT DISTINCT C2.c1 2, RelB B2, RelA A2 WHERE C2.Cid = B2.Cid AND B2.Aid = A2.Aid AND A2.a2 = r (c SELECT DISTINCT C.c1 FROM RelA A, RelB B, RelC C WHERE C.Cid = B.Cid AND B.Aid = A.Aid AND A.a2 = q AND C.Cid IN (SELECT C2.Cid 2, RelA A2, RelB B2 WHERE C2.Cid = B2.Cid AND B2.Aid = A2.Aid AND A2.a2 = r (d SELECT DISTINCT C.c1 WHERE C.Cid IN (( SELECT B.Cid FROM RelA A, RelB B WHERE B.Aid = A.Aid AND A.a2 = q INTERSECT (( SELECT B2.Cid FROM RelA A2, RelB B2 WHERE B2.Aid = A2.Aid AND A2.a2 = r (e π c1 (Temp1 RelB RelC where Temp1 = (σ a2=q RelA (σ a2=r RelA (f None of the above 4

5 7. (5 points Consider the two relational schema, SV and VMo, with specific instances, SV i and Vmo i. SV i S V Và S S1 V4 S1 V5 S2 V6 S V Mo Và S V Mo Và Mo V1 Mo1 V2 Mo2 V3 Mo1 V4 Mo3 V5 Mo4 V6 Mo2 V7 Mo3 VMo i S V Mo NULL V1 Mo1 NULL V2 Mo2 NULL V3 Mo1 S1 V4 Mo3 S1 V5 Mo4 S2 V6 Mo2 NULL V7 Mo3 SVMo i For the example above, circle all variations on the outer join in the FROM clause of an SQL query that would produce SVMo i instance of SVMo from the given instances of SV and VMo. (a SV NATURAL RIGHT OUTER JOIN Vmo (b VMo NATURAL LEFT OUTER JOIN SV (c VMo NATURAL RIGHT OUTER JOIN SV (d SV NATURAL LEFT OUTER JOIN Vmo (e SV NATURAL FULL OUTER JOIN Vmo i.e., SELECT S, V, Mo FROM SV NATURAL RIGHT OUTER JOIN VMo (f None of the above 5

6 8. (5 pts Consider the following table definition: CREATE TABLE RelC (Cid integer, c1 integer, c2 integer, c3 integer, PRIMARY KEY (Cid Circle all queries below that are equivalent to the query: SELECT C.c2, AVG (C.c3 AS avc3 WHERE C.c3 > 5 GROUP BY C.c2 HAVING COUNT (* > 1 By equivalent, we mean would return the same result, without concern for efficiency or elogance. (a SELECT C.c2, AVG (C.c3 AS avc3 WHERE C.c3 > 5 AND COUNT(* > 1 GROUP BY C.c2 (b SELECT C.c2, AVG (C.c3 AS avc3 WHERE C.c3 > 5 GROUP BY C.c2 HAVING 1 < (SELECT COUNT (* 2 WHERE C.c2 = C2.c2 AND C2.c3 > 5 (c SELECT C.c2, AVG (C.c3 AS avc3 GROUP BY C.c2 HAVING COUNT(* > 1 AND C.c3 > 5 (d SELECT Temp.c2, Temp.avc3 FROM (SELECT C.c2, AVG (C.c3 AS avc3, COUNT (* AS c2cnt WHERE C.c3 > 5 GROUP BY C.c2 AS Temp WHERE Temp.c2cnt > 1 (e None of the above 6

7 9. (5 pts Circle all options that would correctly enforce a Complete Coverage constraint of Tab (in Tab1 and Tab2 in an SQL translation of the following UML fragment. Tab Tkey PK Tattr Tab1 Tab2 a CREATE ASSERTION CompleteCoverageOfTab1AndTab2 CHECK (SELECT COUNT (DISTINCT Tab.Tkey FROM Tab = (SELECT COUNT (DISTINCT Tab1.Tkey FROM Tab1 + (SELECT COUNT(DISTINCT Tab2.Tkey FROM Tab2 b CREATE ASSERTION CompleteCoverageOfTab1AndTab2 CHECK (NOT EXISTS (SELECT Tab.Tkey FROM Tab EXCEPT SELECT Tab1.Tkey FROM Tab1 EXCEPT SELECT Tab2.Tkey FROM Tab2 c CREATE ASSERTION CompleteCoverageOfTab1AndTab2 CHECK (NOT EXISTS (SELECT Tab.Tkey FROM Tab EXCEPT (SELECT Tab1.Tkey FROM Tab1 UNION SELECT Tab2.Tkey FROM Tab2 d CREATE ASSERTION CompleteCoverageOfTab1AndTab2 CHECK (NOT EXISTS (SELECT Tab.Tkey FROM Tab WHERE Tab.Tkey NOT IN (SELECT Tab1.Tkey FROM Tab1 AND Tab.Tkey NOT IN (SELECT Tab2.Tkey FROM Tab2 e None of the above the primary keys for Tab1 and Tab2 are not specified. 7

8 10. (5 pts Given tables corresponding to these relations of a wholesale business that sells products and collects payments (bills for those products: Account (SSN: integer, AccntNo: integer, balance: real Transaction (AccntNo: integer, ProductId: integer, date: string, bill: real Write a row-level trigger that increments the appropriate balance, as defined by matching AccntNo, in the Account table by the billed value of a newly inserted transaction to the Transaction table. Use as close to SQLite syntax as you can. CREATE TRIGGER UpdateAccountBalance AFTER INSERT ON Transaction FOR EACH ROW /* Complete the Trigger definition */ 8

Midterm Examination CS 265 Spring 2015 Name: KEY Total on any question cannot exceed 5 and cannot be less than 0 JPK, CK, JDK

Midterm Examination CS 265 Spring 2015 Name: KEY Total on any question cannot exceed 5 and cannot be less than 0 I will not use notes, other exams, or any source other than my own brain on this exam: (please