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In the picture below the square is surrounded by four black squares and three white squares are surrounded by seven black squares:

enter image description here

What is the maximum number of white squares that can be surrounded by $ n $ black squares?

What I tried: I analyzed white formats that don't change the amount of blacks to maximize whites

And I looked at the maximum that can increase from maximum amount from $ n $ to $ n + 1 $

Then I created a figure that increased as much as possible from $n$ to $n +1$, and saw that from a certain n the maximum amount of whites is standardized.

Lambert macuse
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    "I analyzed white formats that don't change the amount of blacks to maximize whites": what does this mean? – xxxxxxxxx Nov 14 '19 at 16:33
  • @MorganRodgers To maximize white balance – Lambert macuse Nov 14 '19 at 16:36
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    Please describe your thoughts in a clearer fashion. "White balance", "White format" are not automatically defined mathematical terms and has no meaning outside your mind until you define them for us. "The maximum that can increase from maximum amount from n to n+1", "amount of whites is standardized" these are quite vague phrases too. Elaborate in clearer terms or this will get closed for being unclear. – Jack's wasted life Nov 15 '19 at 03:41
  • I have posted a very specific special case of your question which should get you the answer you require. As part of this I have sketched proofs for even $n$. –  Nov 15 '19 at 16:33
  • See https://math.stackexchange.com/questions/3437002/how-many-white-squares-can-be-surrounded-by-n-black-squares –  Nov 15 '19 at 16:35

1 Answers1

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A good guess would be that for square numbers $n^2$ of white squares with $n \geq 2$, the minimal number of black squares is $4 \cdot n$. All of this is postulate; you'd have to prove that this is the optimal method.

Between squares, replace the black border squares with white squares and complete the border in this order:

Sample replacement for <span class=$4x4$ square">

  1. Start with replacing the green squares with white squares; this doesn't change the amount of black squares you need to complete the boundary.
  2. Replace the orange squares; the first one adds one black square to the boundary (indicated in red), the second one doesn't.
  3. Replace the blue squares; both add one black square to the boundary.
  4. Replace the red square; this adds one black square to the boundary.

So by 'extending' the boundary from a $n \times n$ square to a $n + 1 \times n + 1$ square, you're up net $4$ black squares, which fits with the formula I postulated.

You can derive a formula from the four steps.

Alexander Geldhof
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  • So for n black squares the maximum of white is $ 4n $? – Lambert macuse Nov 14 '19 at 17:12
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    It is more efficient to use diagonal arrangement of black squares. This shows $25$ white squares enclosed by only $16$ black squares, where an orthogonal arrangement would require $20$ black squares. – Mike Earnest Nov 14 '19 at 17:33
  • @MikeEarnest Great find. One could similarly extend your 'stars' using my method, having to add a black square sometimes while keeping the amount of black stars constant at other times. – Alexander Geldhof Nov 14 '19 at 17:36