7
$\begingroup$

So I have a process to cover a circle:

Start with an empty circle. Pick a uniformly random point thats not covered in the cirlce. Cover the largest circle with that center that does not cover any already covered area. Repeat.

The process is demonstrated in the following graphic:

ding

So let $A_n$ be the expected area covered after $n$ circles have been colored. Does

$$\lim_{n \to \infty} A_n = \text{ All the Area}$$

(Here's a nice visualisation)

  • 4
    I think the visualization is not what you intend because it has overlapping circles? – 2017-01-16
  • 0
    It might be useful to first consider the 1-dimensional case, i.e. nested intervals. – 2017-01-16
  • 0
    In the 1-d case, it is possible (but unlikely) to fill a given gap by selecting the exact midpoint. In the case of filling with circles, you pick a point that lies in an unfilled section. That section is concave, and you put a circle into it. It is impossible for the new circle to completely fill the region you put it into (Except in the trival case that your first circle is placed dead centre). If after placement of new circle, there is guaranteed to be unfilled space, there always be unfilled space, so your limit is 0. – 2017-01-16
  • 0
    @ladams: The OP says : 'Repeat'. I assume they mean 'create a countable sequence of centers of circles that satisfy the criteria'. It is certainly possible to create such a countable sequence that does not cover the entire circle (imagine all centers fall on a diameter). So the question is not 'can I always find a new center?' – 2017-01-16
  • 0
    On the other hand, at each step $n$ we can certainly make $A_{n+1}>A_n$. So the question is not $equivalent$ to 'can I always find a new center?' – 2017-01-16
  • 0
    Suppose we allow the randomly selected circle centers to be inside old circles, but if such a center is chosen in a specific step, don't add a circle. This is clearly equivalent (in fact, if we make it add the largest circle contained in the smallest circle it is contained in, we get the visualization he gave). Then the sequence of points is almost surely dense. Is it possible to have a dense sequence of points such that this fails? – 2017-01-19
  • 0
    @alphacapture: That seems like a fruitful approach. Proceed! – 2017-01-19

1 Answers 1

1

New to this, so not sure about the rigor, but here goes.

Let $A_k$ be the $k$th circle. Assume the area of $\bigcup_{k=1}^n A_k$ does not approach the total area of the circle $A_T$ as $n$ tends towards infinity. Then there must be some area $K$ which is not covered yet cannot harbor a new circle. Let $C = \bigcup_{k=1}^\infty A_k$. Consider a point $P$ in such that $d(P,K)=0$ and $d(P,C)>0$. If no such point exists, then $K \subset C$, as $C$ is a clearly a closed set of points. If such a point does exist, then another circle with center $P$ and nonzero area can be made to cover part of $K$, and the same logic applies to all possible $K$. Therefore there is no area $K$ which cannot contain a new circle, and by consequence $$\lim_{n\to\infty}\Bigg[\bigcup_{k=1}^n A_k\Bigg] = \big[A_T\big]$$ Since the size of circles is continuous, there must be a set of circles $\{A_k\}_{k=1}^\infty$ such that $\big[A_k\big]=E(\big[A_k\big])$ for each $k \in \mathbb{N}$, and therefore $$\lim_{n\to\infty} E(\big[A_k\big]) = \big[A_k\big] $$

EDIT: This proof is wrong becuase I'm bad at probability, working on a new one.

  • 0
    You may use [`\limits` in MathJax](http://meta.math.stackexchange.com/a/12850/290189) to show $\bigcup\limits_{k=1}^n A_k$ – 2017-01-17
  • 0
    I know, I prefer the inline version when in paragraph form like that. – 2017-01-17
  • 0
    How is $K$ being defined? Or, what is the difference in definition between $K$ and $C$? – 2017-01-18
  • 0
    C is the union of all colored circles, while K is an arbitrary non-covered area in which a new circle can't be made. – 2017-01-18
  • 0
    The existence of a specific sequence that covers the area doesn't imply that the *expectation* of the sequence does it. I think you jumped the most important step there. – 2017-01-18
  • 0
    Shouldn't this prove it for all sequences though, given the generality of $A_k$? Since if you had any specific sequence of circles, you could substitue that in for $\{A_k\}_{k=1}^\infty$. – 2017-01-18
  • 0
    @Vedvart1 clearly there exist sequences that don't fill the region. The problem is to show that you will tend to avoid these in expectation. Check Chas Brows's comment in OP. – 2017-01-18
  • 0
    Are there sequences of nonzero probability that don't fill the region? Also, the points don't have to be chosen, since the proof simply relies on there not being any possibility for the theorem to be false for ANY points, random OR chosen. – 2017-01-18
  • 0
    @Vedvart1 there are infinitely many of sequences that don't fill the region. Any sequence where each new radius is smaller than $1/(\pi n)$ for example. It will converge to a area smaller than $\pi$. Just because each of them has probability 0 doesn't imply convergence of expectation. – 2017-01-18
  • 0
    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51973/discussion-between-vedvart1-and-slug-pue). – 2017-01-18