This question has been posed to us by our Algorithms teacher, anyone who solved it would've gotten a bonus directly on the final grade and they did not state that only 1 person may solve it, we were given more than 3 months yet still no one in the entire course was able to solve it, can you?
The puzzle is a game, on the XY plane you have points and each point has a weight, we start off with two points at (1,0) and (0,1) each with weight of 0.5.
The object of the game is to use a set of allowed operations to move/split/join/rotate the points in order to end up with exactly 1 point at coordinates (x,y) where x,y<=2/3 (in other words your point has to be in the square defined by (0,0) (2/3,2/3) )
The x y coordinates of the points must remain non-negative, and weights must remain positive.
The allowed moves are as follows (note the writing p(x,y) or [p](x,y) means a point at coordinates (x,y) with weight p), for moves that are done on two or more points all the points must be on the same horizontal or vertical line. Move: Join: join two points at their center of mass. Split: split a point into 2, notice this is not the inverse move to 2. Rotate, some more explanation can be found below.
Explanation for Rotate: for k=1 the operation is useless, for k=2 we can calculate that:
START = [1/12](r-2d) , [1/6](r+d)
END = [1/12](r+2d) , [1/6](r-d)
Since we can multiply the weights by epsilon the exact weights themselves don't matter what matters is the relationship between the weights, now we can see that if you have point P1 with weight w and point P2 with weight 2w and they are 2d units apart (P1 is closer to 0 than P2), then we can rotate them around their center of mass, moving P1 4d units ahead but point P2 2d units back.
For k=3 you can calculate that the relations are:
START = [1](r-3d) , [5](r-d) , [4](r+2d)
END= [1](r+3d) , [5](r+d) , [4](r-2d)
And so on.
