1. CSE255 Midterm Exam Statistics (Fa07)
Notes:
Final Exam 120 points
MT - range from 40-50%;
FE - range from 60-50%
Track Performance from MT to Final
Exam Differential from MT to Final
If Increase - Weighted Average
If Stay Same - that is your Performance
If Decrease - that is your Performance
Consider P1 points if Close to next range
Worry
Worry a lot!

2. Problem 1a - Relational Algebra (Fa07)
List the name and address of all vendors that provide both hardware and software.
BOTHV = HVendorID,SVendorID (
 HWFlag=T and SWFlag=T (Vendor))
ANS = HVName, HVAddr (HardwareVendor * BOTHV)
HVendorID
 SVName, SVAddr (SardwareVendor * BOTHV)
SVendorID

3. Problem 1b - Relational Algebra (Fa07)
List the versions and vendor names of all C++ software installed on computers made by the vendor DEC.
COMPV = VendorID,CInventNum (Computer * Inventory)
CInventNum=InventNum
HWVEND = HVendorID,CInventNum (COMPV * Vendor)
VendorID
DEC = CInventNum ( HVName=DEC
(HWVEND * HardwareVendors))
HVendorID
DECSW = SInventNum (DEC * InstalledSW)
CInventNum
DEC++SW = SVendorID,SWVersion ( SWName=C++
DECSW * Software)
SInventNum
ANS = SWName,SWVersion (DEC++SW * SoftwareVendor)
SVendorID

4. Problem 1c - Relational Algebra (Fa07)
List all purchase orders from the vendor Dell that have been ordered but not delivered.
DELL = VendorID ( HVName=DELL
(Vendor * HardwareVendors))
HVendorID
ANS = PONumber, PODate, POCost ( DeliveredDate=Null
DELL * Inventory)
VendorID

5. Problem 2 - ER Specialization (Fa07)
Each Specialization was worth 5 pts – deductions for
Not indicating Total
Not supplying attributes
Most Popular Answers:
Disjoint: Inventory with Computer, Accessory, and Software as Children
Overlap Vendor with HWVendor and SWVendor as Children
Union:
Accessory m
Software
Computer 1 m u 1
DeployedPC

6. Problem 3- Functional Dependencies (Fa07)
Computer( CInventNum, ComputerName, ComputerType, AccID);
CInventNum  ComputerName, ComputerType
CInventNum, AccID  ComputerName, ComputerType
Inventory( InvenNum, SerialNum, PONum, PODate, DeliveredDate, POCost, VendorID);
PONum  InventNum, SerialNum, PODate, DeliveredDate, etc.
InventoryNum  SerialNum (or some other one not involving PO)
SoftwareVendor (SVendorID, SVName, SVAddr, SWName, SWVersion, SWDesc);
SVendorID  SVName, SVAddr
SVendorID, SVName, SWName, SWVersion SWDesc, SWAddr
SVendorID   SWName (first two basically equivalent)
SVName   SWName
SWName   SWVersion

7. Prob 4 – Relational Schema Analysis (Fa07)
Computer( CInventNum, ComputerName, ComputerType, AccID);
Insert - For a new computer, you must always insert an accessory (since it is part of the key). If there are N accessories, there are N rows for each computer.
Update – if you change the Name or Type, you must change all N tuples.
Delete - No obvious delete anomalies.
Conclusion: Computer represents two different entities (Guideline 1) – the computer and its accessories, and as a result, violates Guideline 2 in regards to insert anomalies. A better design would separate accessories in a similar manner to the Installed Software table.
Accessory( AccID, AInventNum, HVendorID, AccName, AccType, AccSize);
From an inventory control perspective, there is no way to track the total number of each accessory that has been purchased. You may have 10 USB 120 Gig external hard drives, and each one would have its own AccID and AIventNum. The other problem is related to Guideline 3 due to null values for AccSize (limited problem).
Insert, Delete, and Update: No obvious anomalies.
Conclusion: The table is OK – but it could be improved by separating out the different types of accessories (that have been purchased). It may also make sense not to track this at all in their gory detail – many companies (UConn included) don’t track equipment that is less than $1000, and many of these fit into that category.

8. Prob 4 – Relational Schema Analysis (Fa07)
Software( SInventNum, SVendorID, SWName, SWVersion);
The only real problem in this table is that SVendorID, SWName, and SWVersion are foreign keys into the SoftwareVentor table, and as a result, this information is replicated in both tables.
Conclusion: There may be a better way to design the Software, Installed Software, and SoftwareVendor tables, particularly in regards to reducing the key size (and hence the foreign key linkages).
Inventory(InvenNum, SerialNum,
PONum, PODate, DeliveredDate, POCost, VendorID);
This table suffers violates two guidelines: Guideline 1 in regards to representing two different entities (inventory and purchase orders), and Guideline 3 in regard to an excessive amount of null values.
Insert, Delete, and Update: No obvious anomalies.
Conclusion: Split into two different tables: Inventory (InvenNum, SerialNum, PONum) and PurchaseOrder (PONum, PODate, DeliveredDate, POCost, VendorID) which will address Guideline 1 and will not result in a Inventory tuple until the item is actually received. DeliveredDate will still be null for all outstanding orders.

9. Prob 4 – Relational Schema Analysis (Fa07)
InstalledSoftware( CInventNum, SWInventNum);
Vendor( VendorID, HWFlag, HWVendorID, SWFlag, SWVendorID);
InstalledSoftware is dealing with two foreign key references to the Computer and Software tables, respectively. Vendor is allowing us to unify the different ID tracking systems for software and hardware vendors. The only problem with Vendor is that there are potentially null values for companies that sell either hardware or software but not both. You could argue that the flags are not needed in Vendor as well, since the null values (or not-null) has this information.
Conclusion: In the case of Vendor (and VendorID, SVendorID, HVendorID), this may be a poor design and if the database has not been deployed, it may make sense
to totally redesign this identifier to have a single identifier. This would allow the Vendor table to be eliminated. This would separate vendor common information
into a single Vendor (VendorID, VName, VAddr). This would eliminate the null value (Guideline 3) problem of Vendor.
Thus, changes to Vendor would impact both the HardwareVendor and SoftwareVendor Tables.

10. Prob 4 – Relational Schema Analysis (Fa07)
HardwareVendor ( HVendorID, HVName, HVAddr, ModelNum, ModelName, ModelDescr);
SoftwareVendor (SVendorID, SVName, SVAddr, SWName, SWVersion, SWDesc);
Both tables have insert anomalies (can’t insert HWV or SWV without inserting a product), delete anomalies (if you delete the last item for a vendor you delete the vendor), and update anomalies (if you change an address, or a name – buyout, you need to change multiple tuples) – Guideline 2 is a real issue in this regard.
Both tables are representing two different entities: the contact information for a vendor (id, name, and address) and the vendors products – violating Guideline 1 for having a relation only represent a single entity. As a result, the keys are convoluted – you need to have a ModelNum for HardwareVendors and a compound key for SoftwareVendors; this makes the foreign key references more complicated.
Conclusion: As mentioned, redesign the tables Vendor, HWVendor, and SWVendor to pull out their commonalities and unify the identifier. Use Vendor as defined on the previous slide, use VendorID in the HWVendor and SWVendor tables while dropping Name and Addr from those tables.

11. Problem 5- Normalization (Fa07)
INVOICE( OrderID, OrderDate, CustID, CustName, CustAddr,
ProdID, ProdDesc, UnitPrice, OrderedQuantity)
FULL A. {OrderID, ProductID}  OrderedQuantity
PART B. OrderID  {OrderDate, CustID, CustName, CustAddr}
TRANS C. CustID  {CustName, CustAddr}
PART D. ProdID  {ProdDesc, UnitPrice}
Remove Partial Dependencies
ORDER_LINE( OrderID, ProdID, OrderedQuantity)
PRODUCT(ProdID, ProdDesc, UnitPrice)
CUST_ORDER( OrderID, OrderDate, CustID, CustName, CustAddr)
Remove Transitive Dependency in CUST_ORDER
ORDER_LINE( OrderID, ProdID, OrderedQuantity)
PRODUCT(ProdID, ProdDesc, UnitPrice)
ORDER( OrderID, OrderDate, CustID)
CUSTOMER( CustID, CustName, CustAddr)

12. CSE255 Midterm Exam Statistics (Fa04)
Notes:
Final Exam 120 points
MT - range from 40-50%; FE - range from 60-50%
Track Performance from MT to Final
Exam Exam Differential from MT to Final
If Increase - Weighted Average
If Stay Same - that is your Performance
If Decrease - that is your Performance
Worry
Worry
Worry
a lot!
Worry
a lot!

13. Problem 1- Relational to EER (Fa04)
Person
PersonID
Lname
Fname
StartYear o  
Player
NumYears
UniformNum
Coach
EndYear
Record
Wins
Losses  d
RSRecord
PORecord n 1 m
PPG
ROSTERS
STATISTICS
RPG k
APG m
Team
TeamID
Year
Squad m
TITLES
TitleType 1

14. Problem 2 - EER to Relational (Fa04)
Step 1 - Strong Entities
Item(UPC, Name, Wcost, Rcost, …)
DailySales(Date, Total)
CustomerOrder(AcctNum, Total)
Step 8A - Option A
DeliItem(UPC, Weight, Costlb, Increment)
Step 7 - m-n-k
Sales(ItemUPC, DeliItemUPC, Date)
Step 7 - m-n-k - collapse DeliOrder and Order
Order(AcctNum, ItemUPC, DeliItemUPC)

15. Problem 3a- Functional Dependencies (Fa04)
BOOKAUTHORS(BookId, SSN, LastName, FirstName, );
BookID, SSN  Lname, Fname,
SSN  Lname, Fname,
 Lname, FName
BORROWER(CardNo, SSN, LastName, FirstName, Address, Phone);
Card#  Lname, Fname, etc.
SSN  Lname, Fname, etc.
Card# SSN  Lname, Fname, etc.
LIBRARYBRANCH(BranchId, BranchName, Address);
BranchID  Bname, Addr
Other Single FDs

16. Problem 3a - Functional Dependencies(Fa04)
BOOKS(BookId, Title, PubName, PubAddress, PubPhone);
Title   BookID
PubName   Title BookID
Others Possible
BOOKLOANS(BookId, BranchId, CardNo, DateOut, DueDate);
BookID   BranchID
BranchID   BookID
CardNo   BookID
CardNo   Dateout, DueDate
BookID   CardNo
Others Possible
BOOKCOPIES(BookId, BranchId, NoOfCopies);
BookID   NoCopies
BranchID   BookID
BookID   BranchID

17. Problem 4a - Relational Algebra (Fa04)
Find the names of all TV directors (first and last) who were (are) also movie actors, who directed a TVShow after their first movie role.
TVD = PersonID,ShowID, StartYear (TVDirectors * TVShows)
ShowID
MVA = PersonID,ShowID, Year (MovieRoles * Movies)
ShowID
TVDMVA=PersonID ( StartYear>Year (TVD * MVA))
PersonID
ANS4a = LName, FName (Person * ( TVDMVA))
PersonID

18. Problem 4b - Relational Algebra (Fa04)
Find the Movie names and gross revenues, and names of all Movie directors (first and last) who had roles in at least 3 TV episodes.
X = M.PersonID,TVR.PersonID, EpisodeID (MovieDirs*TVRoles)
PersonID
MVD=PersonID(MD.PID=TVD.PID and Count(EpisodeID(*))>2(X)
DIRS=ShowID, Lname, FName(Person*(MVD*MovieDirectors)
PersonID
PersonID
ANS4b = MovieName, Gross, LName, FName (DIRS*Movies)
ShowID

19. Problem 5a - SQL Queries (Fa04)
Find the names (first and last) and role names (first and last) for all actors who had their first movie role before their first TV role.
LOTS OF WAYS TO SOLVE THIS PROBLEM.
I LOOKED FOR:
1. Use of Person, Roles, TVShows, TVRoles, Movies, MovieRoles Tables.
2. Join Conditions to Link all of the Tables
3. Checking FirstRole = True
4. Comparing StartYear to Year
I WAS PRETTY GENEROUS IN GRADING THIS PROBLEM.

20. Problem 5b - SQL Queries (Fa04)
Find the show names and names of all TV directors (first and last) who have directed at least 4 different TV episodes for which they have won emmys, sorted and grouped by TV show name.
SELECT ShowName, Lname, FName
FROM Person AS P, TVShows AS TVS, TVDirectors AS TVD
WHERE TVD.EmmyFlag = True AND TVS.SHowID = TVD.ShowID AND
TVD.PersonID = P.PersonID
GROUP BY ShowName, EpisodeID
HAVING Count (*) > 4
ORDER BY ShowName;

21. CSE255 Midterm Exam Statistics (Fa03)
Note:
I have Included but Have Hidden Project Grades.
Find Yourself Based on HW and Exam Grades.
Worry

22. Problems 1-3 - Relational Tables (Fa03)

23. Problem 1 - ER Specialization (Fa03)
Each Specialization was worth 5 pts
3 pts for the entities
1 pt for having attributes
1 pt for having assumptions
Basically, the answers were good for disjoint and overlapping, and were somewhat lacking (absent) for unions/categories.
Some Union Were
Directors (TV and Movie)
Productions (TVShows and Movies)

24. Problem 2a - Relational Algebra (Fa03)
List the full name of an actor that has been a Star in a Movie and a Guest-Star in a TVShow.
STAR = PersonID (MovieRoles * ( RoleType=Star Roles))
RoleID,
ShowID
GSTAR=PersonID (TVRoles * ( RoleType=Guest-Star Roles))
RoleID,
ShowID
ANS2a = LName, FName (Person * ( Star * GStar))
PersonID

25. Problem 2b - Relational Algebra (Fa03)
List the full name and the movie name for all directors who have won an Oscar for a movie that grossed less money than it cost to make.
MVS = MovieName, ShowID ( ( Gross < Cost Movies))
DIRS= (MVS * ( OscarFlag=True MovieDirectors))
ShowID
ANS2b = MovieName, LName, FName (DIRS * Person)
PersonID

26. Problem 3a - SQL Queries (Fa03)
List the full name of all actors who have never been directors.
First SELECT - All Actors
NOT IN SELECT - All Directors
SELECT LName, FName
FROM Person AS P, TVRoles AS TR, MoiveRoles AS MR
WHERE ( (TR.PersonID = P.PersonID) OR (MR.PersonID = P.PersonID)) AND P.PersonID NOT IN
(SELECT PersonID
FROM TVDirectors, MovieDirectors)

27. Problem 3b - SQL Queries (Fa03)
List the TV show name and StartYear for all TV shows that have been on TV for at least 5 years and that have the episode name ``The Pilot''.
SELECT ShowName, StartYear
FROM Episodes AS E, TVShows AS T
WHERE E.Ename = ‘The Pilot’ AND T.NuMSeasons >= 5 AND
E.ShowID = T.ShowID)

28. Problem 3c - SQL Queries (Fa03)
List the full name of Actors that have won at least one Oscar and at least one Emmy.
SELECT LName, FName
FROM PERSON
WHERE PersonID IN
( SELECT PersonID
FROM TVRoles AS T, MovieRoles AS M
WHERE T.PersonID = M.PersonID AND T.EmmyFlag =True and M.OscarFlag= True)
FROM Person AS P, TVRoles AS T, MovieRoles AS M
WHERE T.PersonID = M.PersonID AND P.PersonID = M.PersonID AND T.EmmyFlag =True and M.OscarFlag= True)

29. Problem 3d - SQL Queries (Fa03)
List the role name (first and last) of all Movies and TV Shows that ``Steve Martin'' has had a role in.
SELECT RLName, RFName
FROM Roles
WHERE RoleID IN
( SELECT RoleID
FROM TVRoles AS T, MovieRoles AS R, Person AS P
WHERE P.LName = ‘Martin’ AND P.FName = ‘Steve’ AND (P.PersonID = T.PersonID OR
AND P.PersonID = M.PersonID))
FROM Roles AS R, TVRoles AS T, MovieRoles AS R, Person WHERE P.LName = ‘Martin’ AND P.FName = ‘Steve’ AND ((P.PersonID = T.PersonID AND R.RoleID = T.RoleID )
OR (P.PersonID = M.PersonID AND R.RoleID = M.RoleID )))

30. Problem 4 - Update Anomalies (Fa03)
No anomalies in either table - they each represent a single concept with some nulls and no repetitive values
Insert - For a new episode, the role type for a given RLName/RFName/ShowID must be consistent with earlier value. Insert new episode, all data must be replicated and exact for all roles.
Update - Change role name (first/last), roletype, PersonId, etc. multiple tuples must be modified. Values of TVFlag and EmmyFlag, and Mflag/OscarFlag must all be consistent with one another whenever there is a change.
Delete - No obvious delete anomalies.

31. Problem 4 - Update Anomalies (Fa03)
Insert - For new episodes, exact values for Showname, Showyear, NumSeasons, ShowID must be consistent with earlier episodes of same show. Can’t insert Show without an Episode, nor Episode without Show.
Update - Change Showname/Startyear, NumSeasons, NumEmmy, ShowID affects all tuples for all Episodes.
Delete - Delete the last episode, you remove information on a show.
Insert - No strongly obvious anomalies. However, since EpisodeID is “faked” for movies (set to E1), this is a potential problem.
Update - Values of TVFlag and EmmyFlag, and Mflag/OscarFlag must all be consistent with one another whenever there is a change. For movies, must prohibit updates that set EpisodeID to any value except E1.
Delete - No obvious anomalies.

32. Problem 5a- Functional Dependencies (Fa03)
RLName, RFName, ShowID, EpisodeID, TVFlag  EmmyFlag
RLName, RFName, ShowID, MFlag  OscarFlag
RoleID, EpisodeID, TVFlag  EmmyFlag;RoleID, EpisodeID OscarFlag
RLName, RFName, ShowID, EpisodeID, RoleID, PersonID  RoleType, TVFlag, MFlag
EpisodeID, RoleID  PersonID, RoleType
RoleID  RLName, RFName, ShowID, EpisodeID, RoleID, PersonID, RoleType, TVFlag, Mflag
RLName, RFName, ShowID  RoleID
ShowName, StartYear  NumSeasons, ShowID, NumEmmy
ShowName, StartYear  ShowID
ShowID  ShowName, StartYear
ShowID, EpisodeID  EName, Edescr
or ShowName, StartYear, EpisodeID  EName, Edescr

33. Problem 5a - Functional Dependencies(Fa03)
PersonID, ShowID, EpisodeId, TVFlag  EmmyFlag
PersonID, ShowID, MFlag  OscarFlag
PersonID, ShowID, EpisodeId  TVFlag, Mflag or
PersonID, ShowID, EpisodeId  TVFlag
PersonID, ShowID  Mflag

34. Problem 5b - Functional Dependencies(Fa03)
Showname, Startyear  EpisodID, ShowId
EName  ShowId
ShowID  EpisodeID
ShowName  Numseasons, StartYear, ShowID, NumEmmy
EpisodeID  Ename, Edescr
ShowName  EpisodeID, Ename, Edesc
PersonID  ShowID
PersondID  EpisodeID
PersonID, ShowID  EpisodeID
PersonID  ShowID, EpisodeID (combo of first two)

35. CSE255 Midterm Exam Statistics (Sp03)
Out of 75
All Problems
are Counted
and
Problem 4’s
Bonus
Worry
Out of 65
Only 1a,b
and 2a,b
with
Problem 4’s
Bonus
Worry
I will take
your
maximum
percentage
of both
versions

36. CSE255 Midterm Exam Solution (Sp03)
Problems 1 and 2 - Relational Tables

37. Problem 1a - Relational Algebra (Sp03)
List the full names of all mens players that have/had the uniform number 5.
PL = PLName, PFName ( UniformNumber=5 (PLAYER))
MT = TeamID ( Squad=‘Mens’ (TEAM))
ANS1a = PLName, PFName (PL * ( MT * ROSTERS))

38. Problem 1b - Relational Algebra (Sp03)
List the full names and points per game of all mens players that averaged more than 20 points per game and at least 5 rebounds per game in one year on teams with winning regular season records.
MT = TeamID ( Squad=‘Mens’ (TEAM))
WR = TeamID ( Wins > Losses (RSRECORD * MT ))
PL = ( PPG > 20 and RPG  5 (STATISTICS * WR ))
ANS1b = PLName, PFName, PPG (PT * PLAYER)

39. Problem 1c - Relational Algebra (Sp03)
List the full names and uniform numbers of all womens players that played for Geno Auriemma in 1996.
WT = TeamID ( Squad=‘Womens’ and Year =1996 (TEAM ))
GAP = ( CLName=‘Auriemma’ (ROSTERS * WT))
ANS1c = PLName, PFName, UniformNumber (PLAYER*GAP)

40. Version 1 - Problem 2a - SQL Queries (Sp03)
Find the coaches (full names) of all teams that won a NCAA title without winning a BigEastRS (Regular Season) title.
SELECT CLName, CFName
FROM COACH AS C, ROSTERS AS R
WHERE C.CLName = R.CLName AND EXISTS
( SELECT *
FROM TITLES AS T
WHERE TitleType = ‘NCAA’ and T.TeamID = R.TeamID)
AND NOT EXISTS
WHERE TitleType = ‘BigEastRS’ and T.TeamID = R.TeamID)

41. Version 2 - Problem 2a - SQL Queries (Sp03)
Find the coaches (full names) of all teams that won a NCAA title without winning a BigEastRS (Regular Season) title.
SELECT CLName, CFName
FROM COACH AS C, ROSTERS AS R, TITLES AS T
WHERE C.CLName = R.CLname AND
T.TeamID = R.TeamID AND
T.TitleType = ‘NCAA’ AND
T.TeamID NOT IN
( SELECT TeamID
FROM TITLES AS S
WHERE S.TitleType = ‘BigEastRS’)

42. Version 3 - Problem 2a - SQL Queries (Sp03)
Find the coaches (full names) of all teams that won a NCAA title without winning a BigEastRS (Regular Season) title.
SELECT CLName, CFName
FROM COACH AS C, ROSTERS AS R, TITLES AS T
WHERE C.CLName = R.CLname AND
T.TeamID = R.TeamID AND
T.TitleType = ‘NCAA’ AND
NOT EXISTS
( SELECT *
FROM TITLES AS S
WHERE S.TitleType = ‘BigEastRS’ AND S.TeamID = T.TeamID)

43. Problem 2b - SQL Queries (Sp03)
Find the average PPG, RPG, and APG for their entire careers, for all mens players (full names) who started their careers during the 1980s.
SELECT PLName, PFName, AVG(PPG), AVG(RPG), AVG(APG)
FROM TEAM AS T, PLAYER AS P, STATISTICS AS S
WHERE P.StartYear > 1979 and P.StartYear < 1990 and
P.PLName = S.PLName and T.Squad = ‘Mens’ and T.TeamID = S.TeamID
GROUP BY PLName

44. Problem 2c - SQL Queries (Sp03)
Find the total wins of all coaches (full names) for their entire careers.
SELECT CLName, CFName, SUM(R.Wins + P.Wins)
FROM COACH AS C, ROSTERS AS R,
RSRECORD AS X, PORECORD AS Y
WHERE C.CLName = R.CLName and
R.TeamID = X.TeamID and
R.TeamID = Y.TeamID

45. Problem 3 - Update Anomalies (Sp03)
No Modify Anomaly - only one entry per player/coach (unique LNames)
Insert Anomaly - Player in past can’t be coach in future - Player or coach leaves for 1 (or more years) and then returns - no way to store his/her return.
No Delete Anomaly-only one entry per player/coach (unique LNames)
Note: Lots of null values due to capturing two types of people.

46. Problem 3 - Update Anomalies (Sp03)
There are numerous problems for this table, since many values are replicated.
For Example, teams that win multiple titles (NCAA and BigEastRS) must have each player and coach listed twice to capture this data. If you insert a player for a past team (that was omitted), you would have to make sure you inserted the player for all Titles of TeamID.
Specifically: Insert: Can’t have a Team without having a player. Can’t have TitleType unless there is a Team with that title Delete: Last Player on a team - loose the team Modify: Change PLName, impact all TeamIDs for the player
Basically - I looked for reasoning and a solid argument for this table.

47. Problem 3 - Update Anomalies (Sp03)
Modify Anomaly - Whenever RSWins or RSLosses is modified (for a win or a loss), TTLWins or TTLLosses must be incremented.
No Insert Anomaly - only one entry per team.
No Delete Anomaly - no information is lost on anything but the team.
No Modify Anomaly - PPG, RPG, and APG are independent of one another and no values in common across different players.
No Insert Anomaly - only one entry per player.
No Delete Anomaly- no information is lost on anything but the player.

48. Problem 4 - Functional Dependencies (Sp03)
LName  FName, StartYear
LName, PFlag  NumYears, UniformNumber
LName, CFlag  EndYear
Year, Squad  CLName
CLName  TitleType
TeamID  Year, Squad
PLName  TitleType
CLName  TeamId, Year, Squad
TeamID, Year, Squad  TitleType
TeamID, Year, Squad  PLName

49. Problem 4 - Functional Dependencies (Sp03)
TeamID  RSWins, RSLosses, TTLWins, TTLLosses
PLName, TeamID  PPG, RPG, APG
PLName   PPG, RPG, APG
PLName   TeamID
TeamID   PLName

50. Problem 5 (Sp03)
WRITES
Author
SSN
FirstName
LastName
Person