2003 Ungraded Homework P1

1. 2003 Ungraded Homework P1 Reminder: Reserves(sid,bid,day) a.Create a table Together(sid1,sid2,day) that contains all pairs of sailors that have a reservation for the same boat on the same day. (Reserves1 (1sid1),Reserves) (Reserves2 (1sid2),Reserves) (Together, sid1,sid2,day(Reserves1 |X| Reserves2)) b. Give the sids of all sailors that have a reservation for 10/13/2003 but not for 10/17/2003. sid(day=10/13/2003 (Reserves)) sid(day=10/17/2003 (Reserves))

2. 2003 Ungraded Homework P1 (cont) c.Give the name and sid of all sailors that have reservations for all red boats (Tempsids, sid,bid(Reserves) /bid(color=red (Boats)) ) sid,sname(Tempsids |X| Sailors) d.Give the name and sid of all sailors that have exactly one reservation for 11/11/2003 (Reserves2 (1sid2,2bid2), day=11/11/2003Reserves) (2+R, (sid(day=11/11/2003(Reserves) |X|sid=sid2 and bidbid2 Reserves2)) sid,sname((sid(Reserves) 2+R) |X|Sailors))

3. P2 All but one boat problem SELECT R.sid FROM Reserves R GROUPED BY R.sid HAVING COUNT(DISTINCT R.bid)) + 1= (SELECT count(*) FROM Boats Bo) Remarks: still needs to be tested if the query runs on ORACLE9i; counting, instead of checking the real bid’s is okay because of referential integrity.

4. P3 NFL E/R Design Problem Design an Entity-Relationship Diagram that models the following objects and relationships in the world of football (NFL): teams, players, games, managers and contracts. Each (NFL-) team has a unique team name, and a city it plays in. Each person being part of the NFL-world has a unique ssn and a name. Additionally, for players their weight, height, position and birth dates are of importance. Players have a contract with at most one team and receive a salary for their services, and teams have at least 24 and at most 99 players under contract. Each team has one to three managers; managers can work for at most 4 teams and receive a salary for each of their employments. Players cannot be managers. A game involves a home-team and visiting-team; additionally, the day of the game, and the score of the game are of importance; teams play each other several times in a season (not on the same day!). Moreover, for each game played we like to know which players participated in the game and how many minutes they played. Indicate the cardinalities for each relationship type; assign roles (role names) to each relationship if there are ambiguities! Use sub-types, if helpful to express constraints!

5. Sal empl. ssn name (0,4) name Manager isa (1,3) Person Team isa city (24,99) contr (0,1) Player birthd Home Visit NFL E/R Problem (0,*) (0,*) weight height pos Sal (0,*) play played-in. score min (1,1) (22,*) • Scoring: • Play relationship a Set: 3 • Person/Player/Manager: 3 • Weak Game Entity: 3 • Played-in: 2 • Can Only Play once on a day: 1 • Contract: 3 • Salary, score, min attribute: 3 Game Day

6. Using the Default Mappingto Map the E/R Diagram to theRelational Data Model Contract(Team,Player,Salary) Team(name, city) Player(ssn, birthd, pos,…) Person(ssn, name) Game(home, visit, day, score) Played_in(home,visit, day, ssn, min)

7. SQL-Queries B1) “Give the dates of all reservations for red boats” [2] SELECT R.day FROM Reserve R, Boat B WHERE R.bid = B. bid AND B.color = ‘red’ B2) “Give the boats (return bid ) that have at least 2 reservations for 5/5/2003” [4] SELECT DISTINCT R1.bid FROM Reserve R1, Reserve R2 WHERE R1.day= ‘5/5/03’ AND R2.day = ‘5/5/03’ AND R1.bid=R2.bid AND R1.Sid <> R2.Sid

8. SQL-Queries B2) “Give the name and sid of all sailors that do not have any reservations for green boat(“There is no green boat that is reserved by this sailor”)[4] Wrong: SELECT S.sname, S.sid FROM Sailor S, Reserve R, Boat B WHERE S.sid = R.sid AND R.bid=B.bid AND not(B.color = ‘green’)

9. SQL-Queries B2) “Give the name and sid of all sailors that do not have any reservations for green boat(“There is no green boat that is reserved by this sailor”)[4] Correct: SELECT S.sname, S.sid FROM Sailor S EXECPT SELECT S.sname S.sid FROM Sailor S, Reserve R, Boat B WHERE S.sid = R.sid AND R.bid=B.bid AND B.color = ‘green’

10. Homework2 --- Problem1 R(A, B, C, D, E) with the following functional dependencies is given: (1)   A  BC bad (2)   C  A bad (3)   E  A good (4)   B D bad What is (are) the candidate key(s) for R? [2] E The schema R1, R21, R22 is in BCNF and has no lost FD’s!! R(A,B,C,D,E) BD A  BC bad E  A good C  A bad R2(A,B,C,E) R1(B,D) B  D good ABC R22(A, E) A  BC good C  A good E  A good R21(A,B,C)

11. Homework2 --- Problem1 R(A, B, C, D, E) with the following functional dependencies is given: (1)   A  BC bad (2)   C  A bad (3)   E  A good (4)   B D bad What is (are) the candidate key(s) for R? [2] E The schema R3, R41, R42 is in BCNF and but BD is lost! R(A,B,C,D,E) ABC BD lost!! A  D bad E  A good R4(A,D,E) R3(A,B,C) A  BC good C  A good R42(A, E) R41(A,D)

12. Problem2a Homework2 • 2)    Properties of Decompositions [19] R(A,B,C,D,E) with (1) A  C (2) BC  E • a)      Assume , we decompose R into R1(A,E) and R2(A,B,C,D). Does this decomposition have the lossless join property --- is it possible to reconstruct R from R1 and R2 using a natural join? Give reasons if your answer is yes; give a counter example if your answer is no (hint: tuples of your counter example cannot violate (1) and (2))! [9] • Counter example: Assume R(A,B,C,D,E) contains (a,b1,c,d1,e1) and (a,b2,c,d2, e2); obviously, R satisfies the functional dependencies (1) and (2). • However, A,,E(R)|X| A,,B,C,D(R)(R) additionally contains (a,b2,c,d2,e1) and (a,b1,c,d1,e2) which are not in R. Therefore, the decomposition of R into R1 and R2 does not have the lossless join property. • Remark: (a,b1,c1,d1,e1) and (a,b2,c2,d2,e2) is not a correct counter example, because it violates AC (c1 and c2 have to be the same!!); also be aware of the fact that AE not necessary holds for R.