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There are $n$ green axis-parallel squares on the plane. You may scale and translate each square arbitrarily, as long as no two of them intersect. Now you have to put paiwise-disjoint red squares on the plane, such that each green square intersects 4 red squares. What is the minimum number of red squares required (as a function of $n$)?

The most compact arrangement of green squares that I could come up with is in a grid, like this:

enter image description here

The number of red squares required (assuming $n$ is a square of an integer) is $(\sqrt{n}+1)^2 = n+O(\sqrt{n})$.

Is there an arrangement that requires a smaller number of red squares?

(NOTE: squares can be of different sizes).

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up vote 1 down vote accepted

For four green squares, your formula suggests nine red squares are needed.

You can certainly do it with eight: pair the green squares with a narrow corridor in each pair, and put four small red squares in each corridor.

enter image description here

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Thanks! But when $n$ is large, this method will be asymptotically worse than the $n+O(\sqrt{n})$, don't you think? – Erel Segal-Halevi Jan 23 '14 at 15:54
Yes - but it shows that other patterns are possible at least for small $n$. Looking and the middle of my picture, $7$ red squares is possible – Henry Jan 23 '14 at 17:26

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