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I've been watching a youtube channel hosted by puzzlers and attempting the sudoku variation they post each day. Today's I found particularly interesting and while sharing it around it proved quite the challenge.

The puzzle was shared with the hosts of the youtube channel, and is set, by Mitchell Lee.

The rules:

  1. The classic sudoku rules apply: Place numbers on the grid below such that each row, column and 3×3 box contain the numbers 1 to 9.
  2. Any cells separated by a chess knight's move (1 across, two up/down or 2 across 1 up/down) can not contain the same number.
  3. Any cells separated by a chess king's move (any of the 8 cells directly adjacent to/surrounding a cell) can not contain the same number.
  4. Any two orthogonally adjacent cells (above, below, to the left or to the right) can not contain consecutive numbers.

Chess sudoku variant grid

Source. Note: The youtuber does solve the puzzle on the channel.

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  • $\begingroup$ Yo Ed, any chance of an accept? :-) (Lest it be said that I'm begging for rep, in all fairness: Xnero's answer was first, although mine is more complete with full step-by-step explanation.) $\endgroup$
    – Rand al'Thor
    Commented Jan 12, 2021 at 21:11

2 Answers 2

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Final solution

enter image description here

Step-by-step explanation

We start off by placing most of the $1$ and $2$ cells across the whole board, using all the given conditions to narrow down the possibilities. They can both be placed in the left-middle, centre, and right-middle boxes right away, and then we can make some more deductions in the upper and lower boxes:

enter image description here

Here the grey shaded cells are the ones which can be $3$, using the given condition 4. Notice that the $3$ in the left-middle box can only be in one of two possible places.

Now a slightly sneaky observation in the top-right box:

The $2$ cannot be in the top-middle cell, because then the $1$ couldn't be anywhere. So the $2$ must be one of the lower two possibilities, which means the $1$ can't be in the central cell, and must therefore be on the top row. That means we know where the $1$ is in the top-middle box, namely in the middle-left cell.

A similarly sneaky observation in the top-middle box:

The $2$ must be in either the top-middle cell, the top-right cell, or the middle-right cell. But putting it in the middle-right cell would leave no possibilities for $2$ in the top-right box, so it must be in the top row. Now we can place the $2$ in the top-left and top-right boxes.

Looking again at $3$ in the middle-left box:

if it's on the bottom, we get this with no possible position for $3$ in the top-right box. So $3$ is in the top-middle cell of the middle-left box, and we can immediately reduce the reminaing possibilities for placing $3$ by a LOT:

enter image description here

Now there's only one place for $3$ in the seventh row. Also, in the bottom-middle box,

if the $3$ is on the right-hand side, then it can't be anywhere in the bottom-right box, contradiction. Now we can place ALL the remaining $1$, $2$, and $3$ cells, and shade in the possibilities for $4$:

enter image description here

The fifth-row $4$ must be in the centre box, and can't be on the left as that would leave no possibilities for the middle-left box. Then we can fill in all the $4$ cells and shade possibilities for $5$:

enter image description here

Starting from the middle-right box, it's easy to fill in all the $5$ cells. Shading possibilities for $6$:

enter image description here

Starting from the centre and top-right boxes, it's easy to fill in all the $6$ cells. Then just keep going similarly with the $7$, $8$, and $9$ cells to get the final solution.

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  • $\begingroup$ From where you said "looking again at the 3 in the middle left box", you can place a 1 in R7C5 because it is the only place for it in row 7. After that you can place a 1 in R8C2. After that, if you check all the places to put 3's, you can eliminate places using the strategy "if a 3 must be one of these 2-3 places in a sector, it eliminates this place in another sector because that place is connected to all 2-3 places". Eventually, you can actually place a 3 on the board (without using guess and check at any point). Disclaimer: I did not watch the video, this is just what I did. $\endgroup$
    – JS1
    Commented May 20, 2020 at 23:58
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Solved the sudoku by logical deduction starting with the 1's and 2's:

solved sudoku

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  • $\begingroup$ Wow that was very quick! $\endgroup$
    – Edlothiad
    Commented May 11, 2020 at 14:22

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