8
$\begingroup$ 
 .  .  . | .  .  . | .  .  . 
 6  .  . | .  .  1 | .  2  . 
 .  .  . | .  .  . | .  .  . 
---------+---------+--------
 .  .  7 | .  .  . | .  .  . 
 .  .  . | .  .  . | .  .  . 
 .  .  . | .  .  . | .  .  5 
---------+---------+--------
 1  .  . | 9  .  . | .  6  . 
 .  .  . | .  .  . | .  .  4 
 .  .  . | .  1  3 | .  .  .
  • Normal Sudoku rules apply
  • All box centers appear in ascending order when read anti-clockwise
  • The diagonals contain the numbers [1..9] exactly once

clarification

This is an example of the box center constraint. All box centers appear in ascending order when read anti-clockwise. The obvious exception being the 9 - 1 centers (where the chain resets). I did not mention the middle square, as it does not participate in this constraint.

 .  .  . | .  .  . | .  .  . 
 .  2  . | .  1  . | .  9  . 
 .  .  . | .  .  . | .  .  . 
---------+---------+--------
 .  .  . | .  .  . | .  .  . 
 .  3  . | .  4  . | .  8  . 
 .  .  . | .  .  . | .  .  . 
---------+---------+--------
 .  .  . | .  .  . | .  .  . 
 .  5  . | .  6  . | .  7  . 
 .  .  . | .  .  . | .  .  .
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$\endgroup$
3
  • 1
    $\begingroup$ There are several ways to read box centres clockwise, but none of them includes all the 9 centres. Does that mean that it's just the outer 8 centres that are ordered? (Probably not, since that seems impossible.) $\endgroup$
    – Bass
    Dec 16, 2019 at 19:51
  • $\begingroup$ You haven't used no-computers tag, but I want to ask to make it sure - are you OK with answers found by computer programs? $\endgroup$
    – Annosz
    Dec 17, 2019 at 8:29
  • 1
    $\begingroup$ My only goal in creating these puzzles is providing people with some entertainment. If you find it enjoyable to create a computer-program that looks for a solution, I am more than pleased to accept your answer. $\endgroup$
    – Joris Schellekens
    Dec 17, 2019 at 9:00

1 Answer 1

Reset to default
2
$\begingroup$

5 7 2 | 4 6 9 | 8 3 1
6 4 8 | 5 3 1 | 9 2 7
9 1 3 | 8 2 7 | 4 5 6
------------------------
2 3 7 | 1 5 8 | 6 4 9
4 5 6 | 7 9 2 | 3 1 8
8 9 1 | 3 4 6 | 2 7 5
------------------------
1 2 5 | 9 8 4 | 7 6 3
3 6 9 | 2 7 5 | 1 8 4
7 8 4 | 6 1 3 | 5 9 2

I arrived here by querying a prolog db for all constraints and then a few tries with the box center constraints - only one worked for me in which for these blocks.

1 | 2 | 3
4 | 5 | 6
7 | 8 | 9

The one which worked for me was in which center for $B6 < B9$

The script can be found here.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ What I tried was to fill in the centres first. 1 cannot be the centre of the middle box, so it has to go below where the '2' is. 9, 8 and 7 can be in any order, but 6 must go in the bottom left box, and the other numbers follow. $\endgroup$
    – Toby Mak
    Dec 17, 2019 at 13:37

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