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You have a standard chessboard $(8\times8)$.
And you have a lot of numbers from the set $[-1, 0, 1]$.
You have to place one number in each square so that the sums on each row, column, and the two diagonals are different.

To clarify, they must all be unique, regardless of direction (not just different among rows or columns).

If you can do that, what's the strategy?
If not, why?

feelinferrety
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Marius
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  • We'have had this question before, but without the diagonal rule... Not sure if I can dig it up. – Tim Couwelier May 09 '16 at 17:47
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    hey...why the downvote? what's wrong with my question? – Marius May 09 '16 at 17:47
  • @TimCouwelier You might be thinking of this one, but that was for an 11x11 board (the result was that it's impossible for odd side lengths). – f'' May 09 '16 at 17:51
  • @f'' That's the one indeed. Good thing I was trying to confirm that before attempting to actually flag it as duplicate. I have to say, the added rule about the diagonals does make it rather trivial.. – Tim Couwelier May 09 '16 at 17:52
  • @TimCouwelier The "primes and composites" version was also trivial, but it seems to have been better received... – f'' May 09 '16 at 17:54
  • @Marius, I think possibly the reason for the down votes is that the answer is 'it s impossible' but its not that hard to work out. Not sure though – Beastly Gerbil May 09 '16 at 18:13
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    Just for the record, I'm new here at posting questions. If you keep downvoting without stating a reason, you will not help me improve the quality of my questions and you, most probably, will discourage me from posting future questions. Please educate me before taking destructive measures. – Marius May 10 '16 at 07:54

1 Answers1

9

answer:

not possible

because:

There are 8 rows + 8 columns + 2 diagonals = 18 unique sums required. The possible sums range from -8 to +8, so there are only 17 unique sums available. If only the rows and columns have to be different then there is a solution, described here: Filling an 11-by-11 square

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astralfenix
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