Let's say I have a bunch of squares of side x and the grid of square sectors, each of side y. I am placing the squares randomly in this grid - the sides of squares are parallel to sides of the grid, no rotation or anything.
y is always greater or equal than x so if I place a square on this grid, it will either end up fully in within one sector, fully in two sectors (if it was placed on the border between them) or in 4 sectors (if it was placed on the corner between 4 sectors).
Let's define A as a sum of areas of sectors that the square lies in - so it will be either y^2, 2y^2 or 4y^2 depending on the case.
I want A to be the least possible which would suggest that y should be the smallest possible, and therefore y = x.
There is an issue though: because the squares are placed randomly within the grid, if y = x, then the probability of a square landing in one or two sectors only is 0 (the square would need to land ideally in place) while the probability of square landing in 4 sectors is 100 %. Which results always in the worst possible case for me (as I want smallest possible A).
On the other hand if I increase y, then the probability of square landing in one or two sectors will be higher but because y is getting bigger, the A will also be slowly getting bigger. That suggests that there must be some y/x ratio that would result with relatively smallest A for the highest probability that this A will actually be achieved..
I would like to ask you guys for help in finding that ratio as I'm not that good at maths.. Thank you in advance for help <3
I'm not sure if this answers the question but I did some calculations and this is what I got:
There are three cases: square lands in one sector, two sectors or four sectors. I calculate probabilities of each of these cases by dividing the possible area where the square's center can be to the area of entire sector.
Case 1: It lands in one sector, then the probability of this happening is: $$ P_{1} = \frac{(y-2*\frac{x}{2})^2}{y^2}=\frac{(y-x)^2}{y^2} $$ Case 2: It lands in two sectors, then the probability of this happening is: $$ P_{2} = \frac{4*\frac{x}{2}*(y-x)}{y^2}=\frac{2x(y-x)}{y^2} $$ Case 3: It lands in four sectors, then the probability of this happening is: $$ P_{4} = \frac{4*(\frac{x}{2})^2}{y^2}=\frac{x^2}{y^2} $$ Now I am not sure if this is good approach but I tried to use the law of expected value to calculate the expected area A: $$ E(A) = P_{1}*A_{1}+P_{2}*A_{2}+P_{4}*A_{4}=\frac{(y-x)^2}{y^2}*y^2+\frac{2x(y-x)}{y^2}*2y^2+\frac{x^2}{y^2}*4y^2 $$ After simplification I get: $$ E(A)=(x+y)^2 $$ Now assuming that the ratio I am looking for (y/x) is a I get: $$ E(A)=(1+a)^2x^2 $$ And because a can't be lower than 1, it would suggest that the solution a = 1 (so y=x) is still correct - which would mean that the possibilities of achieving less sectors than 4 are never worth increasing y more than x.. But I am still not fully persuaded.. what you think guys?