Exam 2 CS 265 Spring 2018 Name: KEY. 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 Exam CS 65 Spring 018 Name: KEY 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 = [C, S, J, D, P, Q, V, K ] with FDs K is not on RHS of any FD; J,P à C So K must be part of any key S,D à P J à S C à S,J,D,P,Q,V Give all keys for this relational schema. JPK, CK, JDK 3 pt for one, 4 for two, 5 for all three -1 for each of any others Key(s:. (5 pts Consider the relation P with 5 attributes, P = [ C, D, E, F, G ]. You are given the following functional dependencies C,D à E and F,G à 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. C D E F G CD à E F, G not on RHS of any FD. So F,G must be part of any key. Attribute closure of F,G is all attributes, so F,G is the only key C D E F G FG à CD This splits on an FD That does not violate BCNF. In any case, its not dependency preserving ( pt only C D E F G C D F G C D F G E All that is needed in ellipse (5 pts 8 minutes

2 3. (5 pts Consider the relation R with 7 attributes, R = [ A, B, C, D, E, F]. You are given the following functional dependencies: Q = {A à B A, B, C, D, à E C, F à A, B}. pts for ACDàE pts for CFàA 1 pt for AàB - for any extras Give a minimal set of FDs equivalent to Q. If Q is already a minimal set, then say so. BE CLEAR! Can LHS of any FD be simplified? ABCD can be simplified to ACD, because B can be inferred from A CF can t be simplified; C cannot be inferred from F, or vice versa Can B be inferred from A without AàB? No keep AàB Can E be inferred from CD without ACDàE? No keep ACDàE Can A be inferred from CF without CFàA? No keep CFàA Can B be inferred from CF without CFàB? Yes CFàAàB, so CFàB redundant Minimal set in green: AàB, ACD à E, CFàA 4. (5 pts Consider the relation P with 5 attributes, R = [ A, B, C, D, E ]. You are given the two functional dependencies: A à B B à C Circle all true statements. (g [A, B], [A, C, D, E] is a dependency preserving decomposition (a [A, B] is in BCNF (b [B, C] is in BCNF (h [A, B, D, E], [B, C] is a dependency preserving decomposition (c [A, D, E] is in BCNF (i [A, B], [A, D, E], [B, C] is a dependency preserving decomposition (d [A, C, D, E] is in BCNF (j {A à B, B à C} is a minimal set of FDs (e (f [A, B, D, E] is in BCNF [A, B, C, D, E] is in BCNF (k 5 points for all 6 and nothing else -1 for each missing -1 for each extra [A, B], [C, D, E] is a lossless decomposition 8 minutes

3 4. Consider the relation P with 5 attributes, R = [ A, B, C, D, E ]. You are given the functional dependencies: A à B B à C. Circle all true statements. (a [A, B] is in BCNF AàB and A a key ADE is the only key (b [B, C] is in BCNF BàC and B a key Explana,ons for Ques,on 4 A B C D E A à B (c [A, D, E] is in BCNF No FDs assignable, thus no violation of BCNF A B A C D E (d [A, C, D, E] is in BCNF AàC (transitivity but A not key of [A C D E] (e (f [A, B, D, E] is in BCNF AàB and A NOT a key [A, B, C, D, E] is in BCNF AàB and BàC,and neither A or B a key A B C D E B à C (g [A, B], [A, C, D, E] is a dependency preserving decomposition BàC not covered B C A B D E A à B (h [A, B, D, E], [B, C] is a dependency preserving decomposition AàB and BàC both covered (i [A, B], [A, D, E], [B, C] is a dependency preserving decomposition AàB and BàC both covered (j {A à B, B à C} is a minimal set of FDs (k [A, B], [C, D, E] is a lossless decomposition No basis for info preserving join

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 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 CREATE TABLE Z ( z1, PK (z1 - points CREATE TABLE X ( x1, PK (x1 CREATE TABLE R ( x1, r1, z1 NOT NULL, PK(x1, FK (z1 refs Z, FK (x1 refs X CREATE TABLE Z ( z1 PK (z1 CREATE TABLE XR ( x1, r1, z1, PK(x1, FK (z1 refs Z CREATE TABLE Z ( z1, PK(z1 CREATE TABLE XR ( x1, r1, z1 NOT NULL, PK(x1, FK (z1 refs Z CREATE TABLE Z ( z1, PK(z1-1 points 3 points for one right choice and 5 pts for both CREATE TABLE XR ( x1, r1, z1, PK(x1, UNIQUE(z1, FK (z1 refs Z CREATE TABLE Z ( z1, PK(z1 - points (F None of the above 0 points if circled, with or without other choices 5 minutes

5 6. (5 pts Circle all options that would correctly enforce a Complete Coverage constraint of Tab (in Tab1 and Tab in an SQL translation of the following UML fragment. Tab Tkey PK Tattr Tab1 Tab a CREATE ASSERTION CompleteCoverageOfTab1AndTab CHECK (SELECT COUNT (DISTINCT Tab.Tkey FROM Tab = (SELECT COUNT (DISTINCT Tab1.Tkey FROM Tab1 + (SELECT COUNT(DISTINCT Tab.Tkey FROM Tab b CREATE ASSERTION CompleteCoverageOfTab1AndTab CHECK (NOT EXISTS (SELECT Tab.Tkey FROM Tab EXCEPT SELECT Tab1.Tkey FROM Tab1 EXCEPT SELECT Tab.Tkey FROM Tab 3 points for one; 4 points for two; 5 points for three; - for (a c CREATE ASSERTION CompleteCoverageOfTab1AndTab CHECK (NOT EXISTS (SELECT Tab.Tkey FROM Tab EXCEPT (SELECT Tab1.Tkey FROM Tab1 UNION SELECT Tab.Tkey FROM Tab d CREATE ASSERTION CompleteCoverageOfTab1AndTab CHECK (NOT EXISTS (SELECT Tab.Tkey FROM Tab WHERE Tab.Tkey NOT IN (SELECT Tab1.Tkey FROM Tab1 AND Tab.Tkey NOT IN (SELECT Tab.Tkey FROM Tab e None of the above the primary keys for Tab1 and Tab are not specified. 0 points for this 5 minutes

6 7. (5 pts Vanderbilt s academic record system (VARS of the future shares much in common with the system of today, but there will be notable differences. In the database to support VARS, the following are specified. Each academic Program (e.g., CS, History includes at least one Course. Each Course is included in 0 or more Programs. Each Course is composed of at least one Module (or mini-course; each Module is included in 0 or more Courses. Each Module is created by exactly one Individual; each Individual creates 0 or more Modules. Each Module has 0 or more other Prerequisite Modules; each Module may be a Prerequisite to 0 or more other Modules. Each Module is Equivalent to 0 or more other Modules. Each Individual designs 0 or more Courses; each Course is designed by at least one Individual. Each Individual endorses 0 or more Courses; each Course is endorsed by 0 or more Individuals. Complete the following UML snippet by adding cardinalities, associations, and/or classes, so that it satisfies all constraints in the specification above. ModulePreReq PreReq PostReq Program Course Module 1..* Pname PK Cname PK mid PK Includes ComposedOf Mname -0.5 for each error in cardinality, and for similar errors Design Endorse 1..* 1..* Individual iid PK iname Creates Module or 1..1 ModuleEquiv 8 minutes

7 8. (5 pts Consider the two tables below. Write a CREATE VIEW statement that lists the average water readings for each building of each day, but only for daily averages computed over more than values. The view, call it Maintenance, should list ReadingDate, BuildingName, and the average reading for that date/building, listed as AverageValue. CREATE TABLE WaterSensor ( BuildingName VARCHAR(35 NOT NULL, WaterSensorID INTEGER, WaterSensorOnLineDate DATE, PRIMARY KEY (WaterSensorID; CREATE TABLE WReading ( WaterSensorID INTEGER, WReadingDate DATE, WReadingTime TIME, WValue INTEGER NOT NULL, PRIMARY KEY (WaterSensorID, WReadingDate, WReadingTime, FOREIGN KEY (WaterSensorID REFERENCES WaterSensor; op,onal CREATE VIEW Maintenance (ReadingDate, BuildingName, AverageValue AS SELECT WR.WReadingDate, WS.BuildingName, AVERAGE(WR.HRWReadingValue AS AverageValue FROM WaterSensor WS, WReading WR WHERE WS.WaterSensorID = WR.WaterSensorID GROUP BY WR.WReadingDate, WS.BuildingName HAVING COUNT(* > - for first missing or incorrect clause, and -1 for each a^er that (can use JOIN keyword in the FROM clause 5 minutes

8 9. (5 pts Consider the relational schema Student (StuID, Name Instructor (InsID, Name Assessment (StuID, InsID, Rating, Term and a view called AssessmentSummary CREATE VIEW AssessmentSummary AS SELECT DISTINCT A.StuID, S.Name, A.Rating FROM Assessment A, Student S WHERE A.StuID = S.StuID AND Term = Fall09 Write an INSTEAD OF TRIGGER that implements UPDATEs to the StuID attribute of AssessmentSummary by updating Student.StuID and Assessment.StuID for the corresponding student. Update all Assessment tuples with the old StuID, not just the ones contributing to the view. Don't worry about updates to other attributes. CREATE TRIGGER updateassessmentsummary INSTEAD OF UPDATE OF StuID ON AssessmentSummary FOR EACH ROW /* implied by SQLite */ BEGIN UPDATE Student SET StuID = New.StuID WHERE StuID = Old.StuID; UPDATE Assessment SET StuID = New.StuID WHERE StuID = Old.StuID; END; -1 for each error Make sure that Old and New references are correct Finish the trigger 5 minutes

9 10. (5 pts Consider the B+ tree below * 9* 37* 41* 45* 53* 61* 65* 73* 81* 89* 94* Note that this tree does not show data nodes, and you do not need to see the data nodes to answer this question. At each leaf, N* is an index of the form <N, <page id, slot #>>, where N is the value of the search key. Show the tree that results from inserting a record with search key 84 (do not use redistribution. If you can do so clearly and unambiguously, then you can circle and label subtrees in this diagram that do not change and use those labels in your answer on the next page. Answer on the next page 8 minutes

10 64 Only correct tree * 9* 37* 41* 45* 53* 61* 65* 73* 81* 84* 89* 94* - if 84 or 81 does not appear at leaf -3 at least for any tree that isn t 4 levels. Use discretion on partial credit. Make sure that there are no syntax errors, or constraint errors (e.g., make sure that a value V, such as 67, does not appear to the right subtree of a value larger than V, such as 81.

11 Intermediate steps * 9* 84* 37* 41* 45* 53* 61* 65* 73* 81* 89* 94* 73* 81* 84* (will be passed up, and split the root by middle value of * 89* 94* 73* 81* 84* * 89* 94* 73* 81* 84* 11

12 11. (5 pts Consider the extendible hash table to the left. Assume Hash(x = x. Show the result of inserting the following keys in order: 41, for each misplaced key (order within a bin not important Answer here 8 minutes

13 Insert Insert Intermediate answer 13