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I have a new kind of grid puzzle (you could call it a variant of sudoku) I will be calling a Parity Sudoku.

Parity Sudoku is played on a 6x6 grid of integer cells with numbers ranging from 1 to 6. It has two main rules:

  • For every cell with number k, every cell exactly k steps from it, either right, left, up or down has to have a different parity than it. (Oh, and since cells with number k=6 won't do anything with the rule above, their rule is changed so that they see cells exactly 5 steps from it right, left, up or down.)

  • This grid also has one additional rule: Every number from 1 through 6 can occur only once in each 2x3 sextet of the grid - the location of the sextets is shown below.

***|*** ***|*** ------- ***|*** ***|*** ------- ***|*** ***|***
  • The classic Latin square restriction is not necessary, but you could certainly include it. While it would make the rules more tight, it would be easier to find a contradiction that way.

The main question is:

With the rules above, is the Parity Sudoku solvable?

Bonus question: If we remove the sextet-rule, will it then be solvable?

Hint:

Make a parity grid.

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  • $\begingroup$ The same rules apply as a normal sudoku that every number should occur once in every column and row as well? $\endgroup$ Commented Jul 24, 2023 at 17:44
  • $\begingroup$ @Lezzup well, no. Safe to say, the rules are already very strict; there's no need to make them harsher ;) $\endgroup$ Commented Jul 24, 2023 at 17:46
  • $\begingroup$ In that case, glad I asked :) $\endgroup$ Commented Jul 24, 2023 at 17:47
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    $\begingroup$ I think that without the basic Latin square restriction, it's not really a sudoku variant. That doesn't mean it's not an interesting puzzle in its own right, though. $\endgroup$ Commented Jul 24, 2023 at 19:10
  • $\begingroup$ @fljx True... I was going to include the Latin square-restriction, but then it turned out that would be way too strict, so I eventually threw that away. What other names do you think I could change the name to? $\endgroup$ Commented Jul 24, 2023 at 19:22

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This is rather cumbersome and I fully expect someone will make me look silly by posting a much simpler answer.

First, we observe that a 1 placed on one of the two middle squares of a sextet will

bind all even numbers available on that sextet including the 2. If there is a neighbour of 1 opposite the 2 it cannot satisfy both parity constraints, therefore such a neighbour cannot exist. This rules out a 1 on these squares:

 . . . . . . . x . . x . . x . . x . . x . . x . . x . . x . . . . . . .

Now let's also rule out the remaining sextet middle positions:

Focusing on a stack of two sextets there is up to symmetry only one case

 . x . x O x E O x . x . . x . 1 E . ==> ==> . 2 . . 2 . E 2 . E 1 E E 1 E E 1 E

Here E means (must be) even, O odd and x can't be 1. We see that a 1 placed in the bottom middle position ultimately forces four even numbers on that sextet, contradiction.

Ruled out so far:

 . x . . x . . x . . x . . x . . x . . x . . x . . x . . x . . x . . x .

Next, we rule out the following configuration of two 1s.

 . x . E x . 1 x . 1 E E ==> . x 1 E E 1 . x . . x E

As not both 1s can be hugging the boundary at most one of the Es can be a 2, contradiction.

From this we can conclude that given a 1 in the corner of a sextet, the vertically adjacent sextet (if it exists) must have even numbers in two specific positions:

 . . . E . . E . . 1 . . 1 E . 1 E . ==> ==>
 x x x E x x E . . . x . . E . 1 E .

Furthermore, none of the four indicated Es can be a 2.

We also observe that of the two candidate positions for the second 1 only the left one is viable, because the top neighbour of the right one cannot be a 2 but no other evens are left.

And, finally, we remark that a three-storey version of this configuration is not possible because some of the Es have to be 4s.

Next we zoom in on the two middle sextets:

This configuration is impossible

 . x . . x . 1 x . 1 x . || \/ E x . E x . 1 E E 1 E .

because none of the three indicated Es in the left sextet can be a 2. Please note that this argument also holds for the bottom row of sextets if the middle left sextet has its 1 in its lower half.

 . x . . x . 1 x . . x 1 || \/ 4 x . . x 4 1 6 . . 6 1

In this configuration of 1s the indicated 4s are 4s because they cannot be 2s or 6s, leaving for the other non-2s only 6s. But now none of the remaining squares can be a 2, contradiction.

What we know so far:

Both the left and the right halves must have 1s in the top and bottom rows; the respective third 1s cannot be on the same row. Also they must be in the same column as the closer one of the bottom or top row 1s.

Furthermore, the ones farther away can only be at the very corner.

This rules out that the closer ones are in the corner because

 1 E . . . 1 E . . . . E 1 . . . . . . . . . . 1 . . . . . . 1 E . . . 1

we remember that one of the two top left Es has to be a 6.

But they cannot be in the centre either

 . . 1 . E 1 . . . . . E . E 1 E . . . . E 1 . . . . . . . . 1 . . 1 . .

because there cannot be a 2 in the middle right sextet.

So the answer is

No, it is not solvable.

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  • $\begingroup$ Doesn't your conclusion follow immediately from your fourth spoiler-block and the fact that every column must contain the digit you mention? $\endgroup$ Commented Jul 25, 2023 at 7:07
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    $\begingroup$ @RosieF there is no Latin square constraint. (See comments on the question.) $\endgroup$ Commented Jul 25, 2023 at 7:35
  • $\begingroup$ Great work! I was hoping for something more elegant, but the proof is definitely correct. $\endgroup$ Commented Jul 25, 2023 at 11:12

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