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2003 Ungraded Homework P1

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Create a table Together(sid1,sid2,day) that contains all pairs ... Additionally, for players their weight, height, position and birth dates are of importance. ... – PowerPoint PPT presentation

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Title: 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(?day10/13/2003 (Reserves)) ?
    ?sid(?day10/17/2003 (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(?colorred
    (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), ?day11/11/2003
    Reserves)
  • ?(2R, (?sid(?day11/11/2003(Reserves)
    Xsidsid2 and bid?bid2 Reserves2))
  • ?sid,sname((?sid(Reserves) ? 2R) 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 bids 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
empl.
name
ssn
Sal
(0,4)
name
Manager
isa
(1,3)
Person
Team
isa
city
contr
(24,99)
(0,1)
Player
birthd
Home
Visit
NFL E/R Problem
(0,)
(0,)
weight
pos
height
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.bidR2.bid AND R1.Sid ltgt 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.bidB.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.bidB.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 FDs!!
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)
R21(A,B,C)
E ? A good
A ? BC good C ? A good
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.
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