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 . .
Improve this question
$\endgroup$
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 ♦
    Commented Dec 10, 2019 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
    Commented Dec 10, 2019 at 22:38
  • 2
    $\begingroup$ The same with 5 at the top right. $\endgroup$
    – Moti
    Commented Dec 11, 2019 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? $\endgroup$
    – Matthew Jensen
    Commented Dec 11, 2019 at 4:01
  • 1
    $\begingroup$ The puzzle should be solvable now. I updated the puzzle itself and the instructions. Thanks for your feedback. $\endgroup$
    – Joris Schellekens
    Commented Dec 11, 2019 at 10:01

2 Answers 2

Reset to default
5
$\begingroup$

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

enter image description here

Improve this answer
$\endgroup$
2
  • $\begingroup$ Congratulations! $\endgroup$
    – Joris Schellekens
    Commented Dec 14, 2019 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
    Commented Jul 24, 2020 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

Improve this answer
$\endgroup$
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
    Commented Jul 24, 2020 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.