Galloping Queens #3

Define a Galloping Queen as a chess piece whose legal move is that of a Knight , and that of a Queen .

What is the minimum value of integer n > 1 n > 1 such that you can place n n non-attacking Galloping Queens on an n × n n \times n chessboard?

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2 solutions

It is possible with 10 Galloping Queens: This is actually the only arrangement, if you don't count rotations and reflections as new instances (otherwise there are 4).

Using a brute-force search it is also possible to exclude n 9 n \leq 9 as possible options. Note that all solutions to the "n Galloping Queen problem" must also be solutions to the "n Queen problem", which has been very well studied. This means that a brute-force search in this case can be done by studying all known solutions and checking whether there are no knight moves between queens. There are e.g. only 46 fundamental solutions for the 9-Queen problem, so the required effort is small enough that you could check them by hand. And by the list of multiple choice answers for this problem, you can easily exclude n < 9 n<9 as these are not even given as options.

Michael Mendrin
Aug 4, 2015

And of course, why is that a minimum?

Calvin Lin Staff - 5 years, 10 months ago

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The picture isn't showing up, just a request from Photobucket requesting some update the account to allow 3rd-party hosting

Bert Seegmiller - 3 years, 3 months ago

Yes, Kudos, it is 10. Could you prove why? Also, can you determine exactly how many arrangements are possible? You've listed one. Is that all or are there more?

Vishnu Bhagyanath - 5 years, 10 months ago

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I have no idea how many other solutions are possible! I couldn't find any for a 9x9, but I found this one for a 10x10, so I went with it. My chances of being right was then at least 50% It helped for me to look at some well-known 8-queens solutions to see that I could just "expand" some of them to avoid knight-like attacks.

Proving this to be the minimum is an interesting problem. Hmm...

Meanwhile, here's my starting point to develop the 10x10 solution

Michael Mendrin - 5 years, 10 months ago

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Interesting! You're close. There aren't too many 10x10 solutions. I'm still wondering if it's possible to prove this without the help of computers or brute force!

Vishnu Bhagyanath - 5 years, 10 months ago

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@Vishnu Bhagyanath Well, then, how do you know that 10 is the minimum? I guess it's an open question? We know that a 10x10 solution exists.

Michael Mendrin - 5 years, 10 months ago

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@Michael Mendrin It's been computationally proven that none of the permutations upto 9x9 will hold true.

Except 1x1 of course.

Vishnu Bhagyanath - 5 years, 10 months ago

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