Relational Algebra, R. Ramakrishnan and J. Gehrke 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))

Relational Algebra, R. Ramakrishnan and J. Gehrke 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=11/11/2003 Reserves)  (2+R, (  sid (  day=11/11/2003 (Reserves) |X| sid=sid2 and bid  bid2 Reserves2))  sid,sname ((  sid (Reserves)  2+R) |X| Sailors))

Relational Algebra, R. Ramakrishnan and J. Gehrke3 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.

Relational Algebra, R. Ramakrishnan and J. Gehrke4 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!

Team empl. contr name city Manager Player Person name ssn weightpos birthd height Game play played-in. Day min score HomeVisit Sal (1,3) (0,4) (1,1) (24,99) (0,1) (22,*) (0,*) NFL E/R Problem (0,*) isa Scoring: 1.Play relationship a Set: 3 2.Person/Player/Manager: 3 3.Weak Game Entity: 3 4.Played-in: 2 5.Can Only Play once on a day: 1 6.Contract: 3 7.Salary, score, min attribute: 3

Relational Algebra, R. Ramakrishnan and J. Gehrke6 Using the Default Mapping to Map the E/R Diagram to the Relational Data Model Game(home, visit, day, score) Played_in(home,visit, day, ssn, min) Player(ssn, birthd, pos,…) Team(name, city) Person(ssn, name) Contract(Team,Player,Salary)

Relational Algebra, R. Ramakrishnan and J. Gehrke7 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

Relational Algebra, R. Ramakrishnan and J. Gehrke8 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’)

Relational Algebra, R. Ramakrishnan and J. Gehrke9 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’