Anza Power

Divide the board into diagonal lines like so:

Now let's define c{i} to be the number of amoebas at diagonal i, let's give weights to the diagonals w{i}=0.5^i-1, so the weight of the first diagonal is 1 and the weight of every diagonal after that is half the weight of diagonal before it...

Now define board_sum to be the sum of the number of amoebas on each diagonal times the weight of that diagonal, boardsum_ = sum(c{i}w{i})...

At the beginning we have board_sum=1, whenever we make a move we split an amoaba at diagonal i into two amoebas at diagonal i+1, since the weight of diagonal i+1 is half that of diagonal i we can see that the board_sum has not changed...

So the board_sum is always constantly 1.

Now imagine the case where there is an infinite number of amoebas one in every cell, the board_sum in that case is:

(the convergence was calculated using the formula for the mean of a geometric distribution with parameter 0.5)


Now take away the amoebas at the cells we want to clear, their weight is:
1 + 0.5 + 0.5 + 0.25 + 0.25 + 0.25 + 0.125 + 0.125 = 3.
So the board_sum of the board when there is an infinite number of amoebas filling every cell except the ones that we need to clear is 1, which means that if the number of amoebas was finite then the board_sum would be less than 1 which is impossible...

If the ant was to stand still, let's say exactly in the middle of the band, it will always be in the middle of the band, if it stood at 1 quarter of the length of the band, it will always be at one quarter of the length of the band and so on...

You can imagine splitting the band to N points and marking them, now if the ant stood between any two points when the band stretches it will always be between those two points

get a marker and mark N points on the band at equal distances, the band grows linearly, it's length at time t would be L(t) = L₀ + v₀t, you can see that the distance between every two points would be exactly 1/N * L(t), take the dirivative and you'll see that the growth rate of that piece is v₀/N.

So if we take an N that is big enough so that v₀/N is less than the ant's walking speed then the ant could get from mark k to mark k+1 in finite time, and since there is a finite number of points it will get to the last point in finite time as well,,,