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Think about it like this. If you go 1/2 of the way before noticing that the other guy moved then you will travel 1/2 unit. Now look. They are all still in a square (but it's a rotated square) and the side of the new square is sqrt(2)/2 (pythagorean theorem) since the diagonal of the square is 1/2.

So the solution for square of size 1 is S(1)

S(1) = 1/2 + S(sqrt(2)/2)

This is 1/2(1 + r^2 + r^3 + r^4 ...) where r is sqrt(2)/2.

Now if r > 0 and r < 1 then 1 + r^2 + r^3 + ... is a geometric series and equals 1/(1-r)

So in this case the answer is 1 + sqrt(2)/2

But what if we go only 1/3 of the way?

Using the pythagorean theorem again it is 1/3(1 + r^2 + r^3 ...) where r is now sqrt(5)/9

So the geometric series gives 1/3(1/(1-sqrt(5)/9)) = 3/(9 - sqrt(5))

So what if we go 1/n of the way and n tends toward infinity?

1/n(1 + r^2 + r^3 + r^4....) where r is sqrt(n^2 - 2n + 2)/n

Again geometric series makes it 1/n(1/(1-r)) or 1/(n - sqrt(n^2 - 2n + 2))

And (if you know limits from calculus) as n goes to infinity that goes to 1