10
$\begingroup$
 8  .  . | .  .  . | .  .  6 
 .  .  4 | .  .  . | .  .  . 
 .  .  . | .  .  . | 9  1  . 
---------+---------+--------
 .  .  . | .  2  . | .  .  . 
 .  .  9 | 3  .  . | .  .  . 
 .  .  . | .  .  . | .  .  . 
---------+---------+--------
 .  .  . | 9  .  7 | .  .  . 
 9  .  . | 6  3  . | 4  .  . 
 .  .  . | .  .  . | 1  .  . 
  • Normal sudoku rules apply.
  • Long diagonals contain only unique values.
  • If A is a box-center, and A and B are a knight's move apart, then A and B ought to be different.

A knight's move constitutes going up/down by two followed by going left/right by one, or going up/down by one followed by going left/right by two:

. . B . B . .
. B . . . B .
. . . A . . .
. B . . . B .
. . B . B . .
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8
  • 1
    $\begingroup$ Does "diagonals" mean just the two long diagonals, or every line at 45 degrees? [EDITED to add:] Duh, must be the former since there are two diagonally-adjacent 1s in the grid already. $\endgroup$
    – Gareth McCaughan
    Dec 10 '19 at 22:35
  • 14
    $\begingroup$ It looks like the 9s in R2C4 and R4C5 contradict the pseudo-knight move restriction. Is that intended? $\endgroup$
    – HTM
    Dec 10 '19 at 22:38
  • 2
    $\begingroup$ The same with 5 at the top right. $\endgroup$
    – Moti
    Dec 11 '19 at 1:25
  • 2
    $\begingroup$ I believe this is unsolvable. I've tried twice to solve it deterministically, both times resulting in contradictions. Are you able to check that a solution exists satisfying the conditions? $\endgroup$ Dec 11 '19 at 4:01
  • 1
    $\begingroup$ The puzzle should be solvable now. I updated the puzzle itself and the instructions. Thanks for your feedback. $\endgroup$ Dec 11 '19 at 10:01
5
$\begingroup$

A very tough Sudoku, but I think I got it:

enter image description here

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2
  • $\begingroup$ Congratulations! $\endgroup$ Dec 14 '19 at 21:06
  • $\begingroup$ Well done. Can you show your working to prove this solution is unique? (I've seen some dud puzzles on PSE but I have to give OP the benefit of the doubt until proven otherwise) $\endgroup$
    – happystar
    Jul 24 '20 at 23:48
0
$\begingroup$

(1) the 9s in columns 3 and 4 contradict the knight's-move restriction.

(2) a 9 cannot be placed in the middle 3x3.

  • column 4 has already a 9 in row 7.

  • the two spaces in column 5 are both restricted by the 9 in column 3 (same row and knight's move).

  • r4c6 is in a main diagonal that already contains a 9 in column 7.

  • r5c6 is within a row with a 9 in column 3.

  • r6c6 is within knight's move from 9 in c4

there must be an error in the initial values.enter image description here

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1
  • 1
    $\begingroup$ But the puzzle's criterion states "If A is a box-center, and A and B are a knight's move apart, then A and B ought to be different." The knight's move element is only relevant for box-centre-positioned numbers. Numbers in other spaces are not constrained by this. With this in mind, I don't think your objection holds... Please correct me if I'm wrong :) $\endgroup$
    – Stiv
    Jul 24 '20 at 22:28

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