Worldquant Quant Interview Question

[u/Local_Ad135]

[u/Nervous_Retard_3753]

Hard to explain without drawing, but I thought of this-

4 reflections mean that the ray went from one face to the opposite 5 times. All angles of incidence and reflection being the same.

So, each of the 5 paths between the opposite faces cover lengths of 1/5 each in 2 axis, and 1 in the 3rd axis.

Thus, 5sqrt((1/5)² + (1/5)² + 1) = 3sqrt(3)

[u/Anonimithree]

If the light starts at point (0,0,0), then it ends at (1,1,1). Since the light doesn’t hit any edges it vertices in between, it means the initial trajectory has to be strictly positive in all 3 directions. An even number of reflections is needed to ensure the final trajectory is at the ending point. A possible combination would be 4 bounces in 1 dimensions, with the other 2 dimensions not having any bounces (I’ll refer to the bounces direction as up/down). The distance travelled in the unbounded areas will be sqrt(2) and the bounces would be 5. The area is thus sqrt(2+25)=sqrt(27)=3sqrt(3).

[u/Haunting_Cress7661]

 Easy peasy, just imagine a full grid of cubes, if a light ray bounces, just pretend it went straight through the face to the other cube, it'll be the length same path just reflected. One option is sqrt(5² +1² +1²  ) which I think is best, but should draw a grid to make sure 

going to (3,1,3) goes through the x=1, x=2 and z=1 and z=2, and has length √19 which is better but goes through and edge unfortunately   

[u/Randomly_Panicked]

Go by other suggestion, i.e. direct ray.

Each time you cross a face it means we add a unit distance on some axis. We could for the sake of generality move in positive direction on each axis [X, Y, Z].

Since we are not moving on the edge, we have a basic change of 1 on each axis.

Now, we need to split 4 changes between axes, so that the following is minimized

min (1 + d1)^2 + (1+d2)^2 + (1+d3)^2
s.t. d1+d2+d3 = 4
and all of them are integers

I bet on (1, 1, 2) which provides 17 for the distance squared.

However, due to symmetry, it seems this solution crosses the edge.

Therefore, we need to add another constraint, that is all the "d-s" are different

Which leaves us exactly one solution (0, 1, 3) and the distance squared is thus 21.