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Topic: One of each (Read 441 times) |
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Noke Lieu
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One of each
« on: Sep 5th, 2004, 11:45pm » |
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another bubbling away in my head puzzle. Need to make room.
Can't find it, so here we go...
n x n grid, n colours/numbers etc and n of each of them- ie if n=3, 3 red, 3 blue and 3 green counters.
You can place these counters in the grid so that there is only one of each in every row and every column. That's not so hard. (at least, for small values of n)
It gets trickier when you have to take into account the main diagonals. Then, clearly, n=2 is impossible. As is n=3. n=4 works.
I can not get n=6 to work. I have come up with little explanations, which I like, but question their validity.
So... is it possible? When is the next value of n not possible?
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| « Last Edit: Sep 5th, 2004, 11:54pm by Noke Lieu » |
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asterix
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Is it only the 2 main diagonals that are not allowed a repeat? In that case a solution is
134625
261534
325461
452316
516243
643152
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Noke Lieu
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Re: One of each
« Reply #2 on: Sep 6th, 2004, 5:50pm » |
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My admiration for you grows Asterix, whomever you are.
Presumably you did that with the help of silicon, rather than just nutting it out? please
so, what is the next value of n? Is there one?
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towr
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Re: One of each
« Reply #3 on: Sep 7th, 2004, 3:50am » |
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I conjecture it's possible for all n>3
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| « Last Edit: Sep 7th, 2004, 5:10am by towr » |
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towr
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Re: One of each
« Reply #4 on: Sep 7th, 2004, 5:44am » |
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I think I have a general solution for odd n > 3, moreso, the same number never occurs twice on _any_ diagonal.
It's just a matter of overlaying N different solutions of the N-queens problems. (Which is easy to do for odd numbers, but harder for even ones)
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| « Last Edit: Sep 7th, 2004, 5:45am by towr » |
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Aryabhatta
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Re: One of each
« Reply #5 on: Sep 8th, 2004, 4:28pm » |
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For odd n, this seems to work:
Let the first row, R1 be 1,2,3,...,n
We form the second row R2 by cyclic right shift of R1 by 2.
R3 = cyclic right shift of R2 by 2
and so on.
Even n is tougher as towr noted.
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asterix
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That method doesn't work for all odd n. If n is divisible by 3, the Southwest-Northeast diagonal will repeat multiples of 3.
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Aryabhatta
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Re: One of each
« Reply #7 on: Sep 9th, 2004, 7:58am » |
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on Sep 8th, 2004, 8:02pm, asterix wrote:| That method doesn't work for all odd n. If n is divisible by 3, the Southwest-Northeast diagonal will repeat multiples of 3. |
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You are right.
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