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On the following chessboard every white square meets at least two red squares.

enter image description here

Let's paint a new chessboard so that every white square meets exactly two red square. Every square should be painted either red or white. It is clearly possible from the following chessboard.

enter image description here

But, too many squares were painted red. What is the minimal number of squares should be painted red?

P.-S. Park
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3 Answers3

15

I can do

29

like so:

enter image description here

The pattern might be even prettier than that of @hexomino's answer. :-)

EDIT: sadly, no circle pattern anymore, but scores none better: (RE-EDIT: bold letter added because of Jaap's keen eyes; h3 must be red)

28:
enter image description here

And another one with the same amount of redness (RE-EDIT: my only decent solution), this time with more boring symmetry (and also optimal if we are to trust @2012rcampion's comment below):

enter image description here

Bass
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  • Oooh, very nice! +1 – hexomino Sep 24 '20 at 15:10
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    I think your second solution is the best possible (based on my computer results). – 2012rcampion Sep 24 '20 at 15:22
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    Your first solution of 28 is incorrect. The white square in row 5 column 8 has only one red neighbour. When you fix that, it is essentially your first 29 solution rotated 180 degrees. Your symmetric solution looks correct. – Jaap Scherphuis Sep 24 '20 at 15:46
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    I confirmed via integer linear programming that 28 is optimal. – RobPratt Sep 24 '20 at 16:06
  • Thank you very much @JaapScherphuis! (and also the computer solvers!) – Bass Sep 24 '20 at 17:14
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    For $n\in{1,\dots,17}$, the optimal values are ${1,2,5,8,11,17,21,28,35,42,51,60,69,80,91,102,115}$, from which I conjecture general formula $\lceil (n+2)^2/3 \rceil - 6$ for $n \ge 7$. – RobPratt Sep 24 '20 at 20:13
  • @Bass Right. 28 is the answer. There are two more patterns other than yours when symmetries are taken into accounts. – P.-S. Park Sep 25 '20 at 00:34
  • @RobPratt Oh, thank you for your search. Is it possible to check the number of solutions? I just know that 8x8 board has exactly 3 solutions and 9x9 board has a unique solution. – P.-S. Park Sep 25 '20 at 00:38
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    @P.-S.Park Here are the counts for $n\in{1,\dots,10}$, ignoring symmetry: $1, 2, 5, 4, 2, 16, 2, 8, 2, 8$ – RobPratt Sep 25 '20 at 02:08
  • @RobPratt Thank you! – P.-S. Park Sep 25 '20 at 04:03
10

Not sure if this is minimal but there is a nice symmetric solution for

$32$ red squares

as follows

enter image description here

hexomino
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2

OP's comment.

When I propose this problem, many people used to hand in the following patterns with 32 red squares.

enter image description here

That's a beautiful symmetric pattern. But, they are not a minimal answer. This problem was proposed to deceive people who were satisfied with the symmetric solutions. :-)

The correct answers with 28 red squares are as follows:

enter image description here

Note that Quadrants II and IV are symmetric and Quadrants I and III on each pattern are mixed with 2 types. So, ignoring symmetry, the total number of solutions of 28 red squares is 8 as commented by RobPratt.

The minimal answer for $9 \times 9$ board is a pattern with 35 red squares.

enter image description here

This pattern is symmetric, so the total number of solutions is just 2 as also commented.

P.-S. Park
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