EPTTo play the game you must submit a rectangle in the unit square. That is four numbers between 0 and 1. The first two numbers are the x and y coordinates respectively for the bottom left corner of your rectangle and the next two numbers are x and y coordinates respectively for the upper right corner of your rectangle.
If someone plays the exact same numbers as you then you are automatically on the same “crew.”
The mark you receive on this assignment will be proportional to the area successfully claimed by your crew divided by 1.2^(n-1) where n is the number of people in your crew.
If your crew’s rectangle overlaps the rectangle of another crew then that overlapped area is disputed. To resolve how much of the disputed area your crew receives the following rule is used. The length of the perimeter of your undisputed territory is calculated. Say it is L. Then the number of people in your crew is counted up. Say it is N. Then N/L is the density of crew folks along the perimeter of your undisputed area. There is a border between your crew’s undisputed territory and the disputed territory. Say that border has length B. Then the force which your crew puts into the disputed area is
force into disputed area = B * N/L
The other crews that are disputing this area with you do a similar calculation and which ever crew applies the greater force to the disputed area counts the disputed area as part of their total area and everyone else does not get to count it. In the event of a tie, the area is split up evenly between the tied crews. Note that this resolution method is not sensitive to the order in which the conflicts are resolved.
Three weird cases, (and hopefully no more but if you find one let me know, and include it in your analysis, if you do one, for bonus marks)
The first weird case is when a crew has no undisputed area. (The simplest way for this to happen is for one crews rectangle to be completely enclosed within another’s). In this case the enclosed crew’s original perimeter is treated as their undisputed perimeter and the calculations are then made as above.
The second weird case is when there is a disputed area, and all the other area’s that border on it are also disputed. In this case the conflict is temporarily unresolvable, so it is left along with any other unresolvable situations, until all the resolvable disputes have been resolved by the above rules. The formerly disputed areas are now treated as undisputed areas making some of the formerly unresolvable conflicts resolvable. This process is iterated through until all conflicts are resolved.
The third weird case is when two crews share some but not all of their borders. Say Crew 1 has chosen the rectangle with lower left corner (0.1,0.1) and upper right corner (0.3,0.3), and Crew 2 has chosen the rectangle with lower left corner (0.1,0.1), and upper right corner (0.2,0.2). The total perimeter of Crew 2 is the same as the perimeter applied to their disputed area which is 0.4 (four sides of length 0.1 each). Crew 1 on the other hand has a total perimeter of 1 (four sides of length 0.2 each plus two sides of “interior” perimeter each of length 0.1). Crew 1 is considered to have 0.4 units of perimeter bordering the disputed area.
One thing to note is that there will inevitably be some people who try to band the whole class together to get the "socially optimal" solution.
So the question is what kind of square would you choose (and how many people should you group together with) to get the best mark?
