Exam 1 CS x265 Spring 2018 Name: KEY. I will not use notes, other exams, or any source other than my own brain on this exam:

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1 Exam 1 CS x265 Spring 2018 Name: KEY I will not use notes, other exams, or any source other than my own brain on this exam: To students: To request a regrade or to otherwise appeal your score, submit a one paragraph explanation of why you think the grade was in error and submit your explanation/request together with your exam to Doug after class. To graders: No question answer should receive less than 0, regardless of what rubric otherwise suggests. In general, be forgiving of minor syntax errors, but ask if there is any question 1

2 Exam 1 CS 265 Spring 2018 Name: KEY I will not use notes, other exams, or any source other than my own brain on this exam: 1. (2 pts) In the MoneyThink Mobile app spend or save challenge, choose one of the following groups of relations as most reasonable for inclusion in the database, with potentially still other relations, to support this app. 2 pts for this (a) Decision(PersonId, ItemId, ItemImage, DateOfDecision, Buy?), Friend(Person1Id, Person2Id) (b) Decision(PersonId, StockId, DateOfDecision, StockPrice, Buy?), Broker(BrokerId, ClientId) 1 pt for this instead (i.e., par5al credit) (c) Decision(PersonId, ItemId, ItemImage, DateOfDecision, Buy?), Friend(Person1Id, Person2Id) i.e., no social information recorded (d) Decision(PersonId, StockId, Buy?), Broker(BrokerId, ClientId) (e) None of the above are reasonable choices. Explain: 0 pts for anything else (unless there is a good explana5on for e, then lets discuss) 2 minutes 2

3 2. Consider the following relational schema (underlined variables in each schema make up its 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) (2 pts) Give a relational algebra expression that implements the query specified as Find the pids and pnames of any green part. Project(pid, pname) (select(color = green ) Part) π (σ Part)) Pid, pname color= green 1 pt for a proper selec5on (sigma), including the base table (Part) and condi5on 1 point for a proper projec5on (pi) b) (3 pts) Give a relational algebra expression that implements the query specified as Find the snames of Suppliers that supply any part that costs less than $5 (or just 5 ), and list the pid along with the supplier s name. 1 pt for a proper selec5on (sigma), including condi5on Project(sname, pid) ( select(cost < 5) (Supplier njoin Catalog)) 1 point for a proper join, including both tables njoin means natural join 1 point for proper projec5on π Sname, pid (σ cost<5 (Supplier Catalog)) Joining with Part is unnecessary, but no points off for that extra join π Sname, pid ((σ cost<5 Catalog) Supplier )) Pushing select is correct too Answers can also write a natural join as a theta join (e.g., with the explicit condition of equality between all same named attributes (but to be precisely correct, this would require renaming 4 minutes operators, but be lenient if such an answer forgets renaming) 3

4 3. (3 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) Give a left-deep expression tree that represents 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) ((Supplier njoin Catalog) njoin Part)) This is one possible relational algebra expression on which the tree can be based, and if they gave a Correct RA expression, ensure 1 point partial credit regardless of how incorrect the tree is. (But its not required they separately give the a text RA expression at all). Other candidate expressions might have pushed a selection (or both selections still partial credit, but something we will discuss in class) Project(sid) (select(color=green) (select(city=nashville) (Supplier)) njoin Catalog njoin Part)) Project(sid) ( select(city=nashville) (Supplier njoin Catalog njoin (select(color=green) Part))) Any correct expression tree in which the right child of each join is a base table receives 3 points; and correct tree that is not left deep receives 1 point 5 minutes 4

5 Give a left-deep expression that represents Find the sids of Suppliers that supply any green part and are in the city of Nashville. Possible answers Project(sid) ( select(color=green and city=nashville) (Supplier njoin Catalog njoin Part)) Project(sid) (select(color=green) (select(city=nashville) (Supplier)) njoin Catalog njoin Part)) π sid π sid σ Color= Green and city= Nashville This could be split into two separate selects σ Color= Green Part Part Supplier Catalog σ Catalog city= Nashville Supplier 5

6 Give a left-deep expression that represents Find the sids of Suppliers that supply any green part and are in the city of Nashville. WRONG answer π sid σ Catalog city= Nashville σ color= Green Part This is a valid expression tree that correctly implements the query specifica5on, but it is NOT let deep (because there exists a child of a join that is NOT a base table) Supplier For other examples of let deep trees, though not for the same query specifica5on, see the key for Quiz 2 under week 3 of hyps://my.vanderbilt.edu/cs265/schedule/ 6

7 4. Assume the following relational schemas (underlined variables in each each schema make up its key). Customer ( SSN: integer, name: string, address: string, city: string) Account (SSN: integer, AccntNo: integer) Transaction (AccntNo: integer, ProductId: integer, date: string, quantity: integer) /* quantity is the number of ProductId purchased on given transaction */ /* You can think of transaction as synonymous with a purchase */ Product ( ProductId: integer, ProductName: string, cost: real) Write SQL queries for the following. You may NOT use nested queries a) (1 pt) Write an SQL query that lists the names of all customers living in Nashville (i.e., city = Nashville ). Do NOT use JOIN keywords. SELECT C.Name FROM Customer C WHERE C.city = Nashville Short for Customer AS C need not use column label C, etc b) (2 pt) Write an SQL query that lists the names and account numbers of all customers (regardless of city). Do NOT use JOIN keywords. SELECT C.Name, A.AcctNo FROM Customer C, Account A WHERE C.SSN = A.SSN Or other correct query (e.g., need not use column labels C and A, etc) c) (2 pt) Write an SQL query that lists the names and account numbers of all customers (regardless of city). You MUST use a JOIN keyword. need not use column label C, etc SELECT C.Name, A.AcctNo SELECT C.Name, A.AcctNo FROM Customer C JOIN Account A FROM Customer C NATURAL JOIN Account A ON C.SSN = A.SSN SELECT C.Name, A.AcctNo FROM Customer C INNER JOIN Account A ON C.SSN = A.SSN SELECT C.Name, A.AcctNo FROM Customer C JOIN Account A USING (SSN) 5 minutes 7

8 5. Using the same relational schemas, write SQL queries for the following Customer ( SSN: integer, name: string, address: string, city: string) Account (SSN: integer, AccntNo: integer) Transaction (AccntNo: integer, ProductId: integer, date: string, quantity: integer) Product ( ProductId: integer, ProductName: string, cost: real) a) (1 pts) List the total number (sum) of all quantities for the product identified by ProductId = X, regardless of AccntNo, that were purchased on date = Y. Do NOT use JOIN keywords. SELECT SUM(T.quantity) FROM Transactions T WHERE T.ProductId = X AND T.date = Y Not much variance possible b) (2 pts) For EACH transaction date and ProductId, list the date, ProductId, and the total number (sum) of all a) the product s quantities purchased on that date. Do NOT use JOIN keywords. SELECT T.ProductId, T.date, SUM(T.quantity) FROM Transactions T GROUP BY T.ProductId, T.date -1 if didn t Group By both ProductID and date -2 if no Group By at all -1 if did nt use aggregate operator -0 this 5me only if used COUNT or TOTAL instead of SUM -1 for other errors There were some who were confused between SUM and COUNT on this ques5on, and ques5on 6 5 minutes 8

9 6. Using the same relational schemas, write SQL queries for the following Customer ( SSN: integer, name: string, address: string, city: string) Account (SSN: integer, AccntNo: integer) Transaction (AccntNo: integer, ProductId: integer, date: string, quantity: integer) Product ( ProductId: integer, ProductName: string, cost: real) a) (3 pts) For each product, list the item s ProductId, ProductName, and total (sum) of all item quantities purchased by Nashville customers. Do NOT use JOIN keywords. Giving a name, like Total, to SUM(T.quantity) would be fine SELECT P.ProductId, P.ProductName, SUM(T.quantity) FROM Customers C, Accounts A, Transactions T, Products P WHERE C.city = Nashville AND C.SSN = A.SSN AND A.AcctNo = T.AcctNo AND T.ProductId = P.ProductId GROUP BY P.ProductId, P.ProductName Don t take off this time for missing ProductName in Group By, but -1 for each omission from this answer -0 this time only for COUNT, etc instead of SUM b) (2 pts) For EACH item, list the item s ProductId, ProductName, and total (Sum) of all item quantities purchased by Nashville customers, BUT ONLY for those product s with an individual cost > 50 and having a total purchased quantity of greater than 100. You may answer by stating the addition(s) to your query in (a) if you wish). Do NOT use JOIN keywords. SELECT P.ProductId, P.ProductName, SUM(T.quantity) FROM Customers C, Accounts A, Transactions T, Products P WHERE C.city = Nashville AND C.SSN = A.SSN AND A.AcctNo = T.AcctNo AND T.ProductId = P.ProductId AND P.cost > 50 GROUP BY P.ProductId, P.ProductName HAVING SUM(T.quantity) > point for each addition in red (to whatever query they had from part (a), even if part (a) was incorrect Could also use the answer to (a) as a nested query in the FROM clause, creating a temporary Table, with SUM(T.quantity) AS Total, for B, and checking whether Temp.Total > 100 in the WHERE clause of the outer query of B. 6 minutes 9

10 7. (3 pts) (Inspired by a Widom practice exercise) Consider a database of researchers and their works (like ResearchGate.com), with relational schema Researcher (ID, Name, Institution ) Collaborator (ID1, ID2 ) /* ID1 is a collaborator with ID2. Collaboration is symmetric, so if (abc, wxy) is in the Collaborator table, so is (wxy, abc) */ Follows (ID1, ID2 ) /* ID1 follows the posts of ID2, where Follows is not symmetric, so if (abc, wxy) is in Follows table, there is no guarantee that (wxy, abc) is also present. */ Consider finding all those researchers for whom all of those they Follow are at different institutions than themselves. Return the names and institutions of all such researchers. One way to write this query is SELECT R.Name, R.Institution -2 points if literal replacement of NOT EXISTS by NOT IN FROM Researcher R WHERE NOT EXISTS (SELECT * FROM Follows F, Researcher R2 WHERE R.ID = F.ID1 AND F.ID2 = R2.ID AND R2.Institution = R.Institution) Write the query that satisfies the same English specification using the NOT IN phrase rather than NOT EXISTS : All researchers who don t follow anyone from the same institution The set of all researchers following someone who is at the same institution of R SELECT R.Name, R.Institution FROM Researcher R WHERE R.ID NOT IN (SELECT F.ID1 FROM Follows F, Researcher R2 F.ID2 = R2.ID AND R.Institution = R2.Institution) -1 point if Name used instead of ID; 2 points only if they have * in inner SELECT This is a correlated query (because R was defined in outer query and also referenced in inner query) 4 minutes 10

11 7. (3 pts) (Inspired by a Widom practice exercise) Consider a database of researchers and their works (like ResearchGate.com), with relational schema Researcher (ID, Name, Institution ) Collaborator (ID1, ID2 ) /* ID1 is a collaborator with ID2. Collaboration is symmetric, so if (abc, wxy) is in the Collaborator table, so is (wxy, abc) */ Follows (ID1, ID2 ) /* ID1 follows the posts of ID2, where Follows is not symmetric, so if (abc, wxy) is in Follows table, there is no guarantee that (wxy, abc) is also present. */ Consider finding all those researchers for whom all of those they Follow are at different institutions than themselves. Return the names and institutions of all such researchers. One way to write this query is SELECT R.Name, R.Institution FROM Researcher R WHERE NOT EXISTS (SELECT * FROM Follows F, Researcher R2 WHERE R.ID = F.ID1 AND F.ID2 = R2.ID AND R2.Institution = R.Institution) Write the query that satisfies the same English specification using the NOT IN phrase rather than NOT EXISTS : Basically same as previous, but for each researcher being considered in turn; SELECT R.Name, R.Institution will return empty set if R.ID isn t following anyone at same institution FROM Researcher R WHERE R.ID NOT IN (SELECT F.ID1 /* could be R.ID here */ FROM Follows F, Researcher R2 WHERE R.ID = F.ID1 AND F.ID2 = R2.ID AND R.Institution = R2.Institution) This addition to previouspage query is what causes nested query to evaluate to NULL for Rs who arent following anyone at same institution. This isn t needed, but full credit. 11

12 7. (3 pts) (Inspired by a Widom practice exercise) Consider a database of researchers and their works (like ResearchGate.com), with relational schema Researcher (ID, Name, Institution ) Collaborator (ID1, ID2 ) /* ID1 is a collaborator with ID2. Collaboration is symmetric, so if (abc, wxy) is in the Collaborator table, so is (wxy, abc) */ Follows (ID1, ID2 ) /* ID1 follows the posts of ID2, where Follows is not symmetric, so if (abc, wxy) is in Follows table, there is no guarantee that (wxy, abc) is also present. */ Consider finding all those researchers for whom all of those they Follow are at different institutions than themselves. Return the names and institutions of all such researchers. One way to write this query is SELECT R.Name, R.Institution FROM Researcher R WHERE NOT EXISTS (SELECT * FROM Follows F, Researcher R2 WHERE R.ID = F.ID1 AND F.ID2 = R2.ID AND R2.Institution = R.Institution) Write the query that satisfies the same English specification using the NOT IN phrase rather than NOT EXISTS : IDs of all researchers who follow anyone from the same institution SELECT R.Name, R.Institution FROM Researcher R WHERE R.ID NOT IN (SELECT F.ID1 /* could be R1.ID here */ FROM Follows F, Researcher R1, Researcher R2 WHERE R1.ID = F.ID1 AND F.ID2 = R2.ID AND R1.Institution = R2.Institution) A version that doesn t use a correlated query but still full credit 12

13 7. (3 pts) (Inspired by a Widom practice exercise) Consider a database of researchers and their works (like ResearchGate.com), with relational schema Researcher (ID, Name, Institution ) Collaborator (ID1, ID2 ) /* ID1 is a collaborator with ID2. Collaboration is symmetric, so if (abc, wxy) is in the Collaborator table, so is (wxy, abc) */ Follows (ID1, ID2 ) /* ID1 follows the posts of ID2, where Follows is not symmetric, so if (abc, wxy) is in Follows table, there is no guarantee that (wxy, abc) is also present. */ Consider finding all those researchers for whom all of those they Follow are at different institutions than themselves. Return the names and institutions of all such researchers. One way to write this query is SELECT R.Name, R.Institution FROM Researcher R WHERE NOT EXISTS (SELECT * FROM Follows F, Researcher R2 WHERE R.ID = F.ID1 AND F.ID2 = R2.ID AND R2.Institution = R.Institution) Write the query that satisfies the same English specification using the NOT IN phrase rather than NOT EXISTS : Institutions of all researchers who are followed by a given R 3 pts SELECT R.Name, R.Institution FROM Researcher R WHERE R.Institution NOT IN (SELECT R2.Institution FROM Follows F, Researcher R2 WHERE R.ID = F.ID1 AND R2.ID = F.ID2) Correct answers also stem from variants on a different perspective, exemplified at left All researchers who don t follow anyone from the same institution 13

14 8. (4 pts) Consider the following table definitions (attribute types omitted). CREATE TABLE Reading ( Kilowatts, Date, Time, LocationId, PRIMARY KEY (Date, Time, LocationId), FOREIGN KEY (LocationId) REFERENCES Location ) CREATE TABLE Location ( LocationId, Occupancy, PRIMARY KEY (LocationId) ) We want a query that returns the average Kilowatts over all readings on a given Date (Date = D, where D is a parameter) for each Location. In addition to the averages, the result should list the LocationId and the locations occupancy for each reported average. Moreover, the query should only return the averages for LocationId groups for which the minimum individual Kilowatt reading is greater than 0.5 on date D. Complete the following query so that properly implements the specification. SELECT L.LocationId, L.Occupancy, AVG(R.Kilowatts) FROM Readings R, Location L WHERE L.LocationId = R.LocationId AND R.Date = D GROUP BY L.LocationId, L.Occupancy HAVING MIN(R.Kilowatts) > minutes 2 points for correct GROUP BY (adding LocId) 2 points for correct HAVING 14

15 9. (5 pts) This question uses the same relational schema as question 7. Researcher (ID, Name, Institution ) Collaborator (ID1, ID2 ) Follows (ID1, ID2 ) What is the average number of Collaborators per Researcher? Importantly, compute this average over only those Researchers who have more than 1 Collaborator. (Your result should be just one number.). SELECT AVG(Temp.Cnt) Most will have something very much like this (5 pts) FROM (SELECT COUNT(*) AS Cnt -2 for each main missing element FROM Collaborator C GROUP BY C.ID1 /* or C.ID2 because Collaborator is symmetric */ HAVING COUNT(*) > 1) AS Temp Some will have something more complicated, like this, but still (5 pts) SELECT AVG(Temp.Cnt) FROM (SELECT COUNT(*) AS Cnt FROM Researcher R, Collaborator C WHERE R.ID = C.ID1 /* or C.ID2 because Collaborator is symmetric */ GROUP BY C.ID1 /* or R.ID */ HAVING COUNT(*) > 1) AS Temp SELECT AVG(Temp.Cnt) FROM (SELECT COUNT(*) AS Cnt FROM Researcher R, Collaborator C WHERE R.ID = C.ID1 GROUP BY C.ID1) AS Temp WHERE Temp.Cnt > 1 Don t need rename as Temp in any of these, but could simply reference Cnt in AVG (using AVE instead is fine) 5 minutes 15

16 10. (4 pts) Consider the same relational schemas as problem 2 (without quantity for Transaction). Customer ( SSN: integer, name: string, address: string, city: string) Account (SSN: integer, AccntNo: integer, balance: real) Transaction (AccntNo: integer, ProdId: integer, date: string) Product ( ProdId: integer, ProdName: string, cost: integer) CREATE TABLE Customers ( SSN Integer, Name CHAR[25] NOT NULL, Address CHAR[25] NOT NULL, City CHAR[25] NOT NULL, PRIMARY KEY (SSN) CREATE TABLE Accounts ( SSN Integer NOT NULL, AccntNo Integer, Balance Float NOT NULL, PRIMARY KEY (AccntNo) Consider the CREATE TABLE statements that implement the relations to the left. SSN in Account refers to an SSN in Customer. AcctNo in Transaction refers to an AccntNo in Account. ProdId in Transaction refers to a ProdId in Product. Complete these definitions so that if a Customer is deleted, all the Customer s Accounts are deleted (i.e., all Accounts with an SSN that matches the deleted Customer s SSN), unless one or more of the Customer s Accounts participates in any Transaction (as identified by AccntNo), in which case the attempt to delete the Customer is blocked. Also, the definitions should insure that a product can only be deleted if it does not participate in any transaction (as identified by ProdId in Transactions). While certain default settings might have otherwise applied in answering this problem, I want you to ignore defaults and make explicit all actions in the appropriate definitions. CREATE TABLE Transactions ( AccntNo Integer, ProdId Integer, Date CHAR[6], PRIMARY KEY (AccntNo, ProdId) ) CREATE TABLE Products ( ProdID Integer, ProdName CHAR[15], Cost Integer, PRIMARY KEY (ProdId) ) ) FOREIGN KEY (SSN) REFERENCES Customer ON DELETE CASCADE FOREIGN KEY (AccntNo) REFERENCES Account ON DELETE NO ACTION, FOREIGN KEY (ProductId) REFERENCES Product ON DELETE NO ACTION) or RESTRICT 2 pt for one correct FK (but forgive VERY minor syntax errors this time) 3 pts for two correct FKs 4 pts for three correct FKs 5 minutes -1 for each extra FKs 16

17 11. (4 pts) Using the definitions of two pages ago, and repeated here Accounts (SSN: integer, AccntNo: integer, balance: real) Transactions (AccntNo: integer, ProductId: integer, date: string, bill: real) Write a row-level trigger that increments the appropriate (defined by matching AccntNo) balance in the Accounts table by the bill value of a newly inserted transaction to the Transactions table. You may use SQLite syntax, but not required. CREATE TRIGGER UpdateAccountBalance AFTER INSERT ON Transactions // Complete the Trigger definition CREATE TRIGGER AFTER INSERT ON Transactions FOR EACH ROW FOR EACH ROW is optional BEGIN If this is -, that s fine UPDATE Accounts SET balance = balance + NEW.bill WHERE AccntNo = NEW.AccntNo; END; SQLite version -1 for each missing or mangled component, to include clauses of UPDATE Can preface by Accounts 5 minutes 17

18 11. (4 pts) Using the definitions of two pages ago, and repeated here Accounts (SSN: integer, AccntNo: integer, balance: real) Transactions (AccntNo: integer, ProductId: integer, date: string, bill: real) Write a row-level trigger that increments the appropriate (defined by matching AccntNo) balance in the Accounts table by the bill value of a newly inserted transaction to the Transactions table. You may use SQLite syntax, but not required. CREATE TRIGGER UpdateAccountBalance AFTER INSERT ON Transactions // Complete the Trigger definition CREATE TRIGGER AFTER INSERT ON Transactions REFERENCING NEW ROW AS NewRow FOR EACH ROW UPDATE Accounts SET balance = balance + NewRow.bill WHERE Accounts.AccntNo = NewRow.AccntNo; CREATE TRIGGER AFTER INSERT ON Transactions REFERENCING NEW ROW AS NewRow NEW TABLE AS NewTable FOR EACH ROW UPDATE Accounts A SET balance = (SELECT SUM(bill) FROM Transactions T WHERE T.AccntNo = A.AccntNo) WHERE A.AccntNo = NewRow.AccntNo; These are non-sqlite versions, but still correct (by the SQL standard). Be accepting of close (incorrect) syntactic variants 5 minutes 18

19 12. (3 pts) Consider two tables, Tab1 and Tab2. Tkey is the PRIMARY KEY of both tables. Circle all options that would correctly enforce a No Overlap (aka Disjoint) constraint between Tab1 and Tab2 (i.e., that if a tuple with a particular Tkey value appears in Tab1, then it does not appear in Tab2, and if a tuple with a particular Tkey value appears in Tab2, then it does not appear in Tab1. 1 point for each If you thought EXISTS had higher precedence than set operators, then see me a) CREATE ASSERTION NoOverlapBetweenTab1AndTab2 CHECK ((SELECT COUNT (Tab1.Tkey) FROM Tab1) = (SELECT COUNT (Tab2.Tkey) FROM Tab2)) b) CREATE ASSERTION NoOverlapBetweenTab1AndTab2 CHECK (NOT EXISTS (SELECT Tab1.Tkey FROM Tab1) INTERSECT (SELECT Tab2.Tkey FROM Tab2) ) c) CREATE ASSERTION NoOverlapBetweenTab1AndTab2 CHECK (EXISTS (SELECT * FROM Tab1 T1, Tab2 T2 WHERE T1.Tkey = T2.Tkey)) d) CREATE ASSERTION NoOverlapBetweenTab1AndTab2 CHECK (NOT EXISTS (SELECT * FROM Tab1 T1, Tab2 T2 WHERE T1.Tkey = T2.Tkey)) e) CREATE ASSERTION NoOverlapBetweenTab1AndTab2 CHECK (NOT EXISTS (SELECT Tab1.Tkey FROM Tab1 WHERE Tab1.Tkey IN (SELECT Tab2.Tkey FROM Tab2) UNION SELECT Tab2.Tkey FROM Tab2 WHERE Tab2.Tkey IN (SELECT Tab1.Tkey FROM Tab1))) -1 for each option (a) or (c) (again, a negative score will simply be counted zero/0) ; f) None of the above 0 points if this is circled, with or without other choices 6 minutes 19