3
$\begingroup$

You can see the rules for pointing problem at this article. It has been clarified that there is no solution to this question.

But, what if we remove the diagonal rule? is there any solution?

$\endgroup$
5
$\begingroup$
. 1 . | . . . | . . .
. . . | . 2 . | . . .
. . . | . . . | . 7 .
------+-------+------
. . . | . . . | 9 . .
. . . | 3 . . | . . .
4 . . | . . . | . . .
------+-------+------
. . . | . . . | . . 6
. . . | . . 8 | . . .
. . 5 | . . . | . . .

I got 720 solutions having the 1 in the top left bloc.

$\endgroup$
  • 1
    $\begingroup$ And that are all solutions, since the whole trail of numbers is circular, you can assign the name "1" to any Number and edit the rest accordingly. So in the End there are 720 times 9 solutions (just shifting through the numbers) $\endgroup$ – Falco Sep 25 '14 at 15:09
  • $\begingroup$ Exactly. That's why I added this restriction. $\endgroup$ – Florian F Sep 25 '14 at 15:24

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