Can you place 22 red, 22 white and 22 black bishops on a 10x10 grid, such that no two bishops of different colours attack each other?
Here is a similar question for 8x8 grid: Peaceable Bishops on an 8x8 grid
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asked Nov 24, 2019 at 9:17
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2 Answers
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Answer
Yes, it is possible.
Attempt 1
At first I thought
the best we can do is $20,20,24$ instead of $22,22,22$.
This can be achieved as follows, using the same technique as Daniel Mathias previously:
Note that we're filling each corner in a checkerboard way: two opposite corners A flanked by B and the other two C flanked by B. This means the B colour plays the same role in all corners, and we end up with $24$ B bishops and only $20$ each of A and C.
Adjusting this a little, we can get
$20,22,22$, losing a bit of symmetry in the overall configuration, removing four B bishops and adding two each of A and C:
After a lot more fiddling around with this, I realised it seems that
it's impossible to get $22,22,22$ this way. But A flanked by B with C flanked by B is not the only way to play it!
Attempt 2
I then tried
A flanked by B in two opposite corners and B flanked by C in the other two, changing the role of B. It's then easily possible to get $22$ B bishops ($4+7+4+7$ from the four corners), but the A and C bishops then have to share all the squares of one colour, and I couldn't manage to fit in $22$ of each.
Attempt 3
Finally, the third option is
A flanked by B in two opposite corners and A flanked by C in the other two. Like in the previous puzzle, there's one colour of bishop playing the same role in all corners, but this time it's the one being flanked instead of the one flanking like before. Again we can easily get $22$ A bishops by doing $4+7+4+7$ in the four corners:
Then we fill in B and C bishops around these, to get the following symmetric configuration:
But now we can only fill in more B bishops (blue circles) and no more C bishops.
The mistake was
to make the A configurations in opposite corners the same and in adjacent corners different, because that's always going to advantage one of B, C over the other. Instead we lay out the A bishops like this:
Again we fill in B and C bishops around these, to get the following symmetric configuration:
And finally we break symmetry by putting B bishops in two of the blue circles and C bishops in the other two, to get the desired $22,22,22$.
answered Nov 24, 2019 at 13:39
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1$\begingroup$ I enjoyed reading the ups and downs in your solution. You got the correct answer in the end. $\endgroup$ Commented Nov 24, 2019 at 21:16
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3$\begingroup$ Wow, what a snatch, not only did you pass me in votes, you got the tick too! :-) Guess the "here's a lot of stuff that doesn't work" section is interesting to surprisingly many. $\endgroup$– BassCommented Nov 24, 2019 at 21:48
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$\begingroup$ @Bass I confess that I gave you the tick first. However after some thought I realised that this answer provides more explanation and shows how to arrive at a solution. If I could I would give you both the ticket as you have both answered the question correctly. So please don't feel bad. $\endgroup$ Commented Nov 25, 2019 at 10:22
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1$\begingroup$ @Bass Sorry :-( I wasn't expecting that to happen, and I feel like you deserve the tick for being first to get the correct answer. At least you've now overtaken me in votes (and you have my upvote, of course). The only interesting extra thing I could say my solution contains is sort-of-showing (not rigorously of course) that this is the only way to do it, by eliminating all the other possible ways of starting from the four corners. $\endgroup$ Commented Nov 25, 2019 at 12:36
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$\begingroup$ I think the best solution is for you guys to solve another bishop puzzle ;) puzzling.stackexchange.com/questions/91522/… $\endgroup$ Commented Nov 26, 2019 at 1:20
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Here's my go (click to enlarge):
The pattern of reds is
left-right symmetrical
and the white and black patterns are
right-to-left mirror images of each other
so it's enough to count half the pieces only. :-)
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3$\begingroup$ Welp, you beat me to it by almost half an hour :-) Your solution seems equivalent to mine, and I'm pretty sure there's just one possible solution, up to rotations/reflections. $\endgroup$ Commented Nov 24, 2019 at 13:44
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1$\begingroup$ As expected, this is similar to my solution to the previous question. With white and black on opposite colors and red split evenly between, it is sufficient to find 22+11 on one color. $\endgroup$ Commented Nov 24, 2019 at 14:32
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