Not in conjunction with my Minesweeper puzzles - that would be concerning!

Inspired by Cracking The Cryptic's video "The Graveyard Of Schrödinger" (original gimmick by fjam)

How this gimmick works:

Normal sudoku rules almost apply. Place the digits 0-6 into cells such that each digit appears once in every row, column and region. To accommodate this, one cell in each row, column and region is a Schrödinger cell, which contains two digits. In graves (cages), digits must sum to the day, month or year of the date on the grave. Digits cannot repeat within graves. Escape the graveyard by carefully plotting an orthogonally connected path between the green and red cells. The path may not cross the highest or lowest digits in a grave.

Some things to clarify (as best as I can):

  1. What happens when there is only a grave with two cells?

Here's my take on this. Say we have this table (ripped directly from the video) (the dates mean that the numbers in the cage add up to either the day, month, or year listed on it, for example R1C2 and R1C3 add up to either 2 (the month), 13 (the day), or 20 (the year)):

Green cell 13/02/20 13/02/20
14/01/23 31/08/30 31/08/30
14/01/23 31/08/30 31/08/30

Which Simon from Cracking the Cryptic solved this box as such:

3 7 4/9
8 1 2
6 5 0

With a line going through R1C1, R1C2, R2C2, R2C3 before going into Box 2.

This is valid because we have:

  1. All digits 0-9 ✓
  2. Orthogonal line not going through the highest and lowest digits in each "grave" ✓
  3. The digits in each "grave" adding up to either the year, month, or day for the grave listed ✓

Which brings us to our second question: How is this valid? Because

sure, we have$$7+4\overset{✓}=13$$but what we also have is$$7+9\ne13,2,20$$

I honestly think this is because of our "Schrödinger" cell in box 1, so I think that's an exception to the rule "digits must sum to the day, month or year of the date on the grave" however if someone could give me a more in depth explanation on this that would be good.

The puzzle (only 6x6)

How the 6 boxes are going to be split up:

  1. Box 1: R1C1, R1C2, R1C3, R2C1, R2C2, R2C3
  2. Box 2: R1C4, R1C5, R1C6, R2C4, R2C5, R2C6
  3. Box 3: R3C1, R3C2, R3C3, R4C1, R4C2, R4C3
  4. Box 4: R3C4, R3C5, R3C6, R4C4, R4C5, R4C6
  5. Box 5: R5C1, R5C2, R5C3, R6C1, R6C2, R6C3
  6. Box 6: R5C3, R5C5, R5C6, R6C4, R6C5, R6C6
$\color{green}✓$ 9/12/14 10/15/16 10/15/16 10/15/16 10/15/16
8/13/25 9/12/14 9/12/14 10/15/16 12/06/67 12/06/67
8/13/25 12/24/87 12/24/87 11/27/28 11/27/28 8/23/28
8/13/25 12/24/87 12/24/87 11/27/28 11/27/28 8/23/28
1/7/27 11/13/16 11/13/16 8/11/27 5/11/12 5/11/12
1/7/27 1/7/27 11/13/16 8/11/27 5/11/12 $\color{red}✓$

Note: Please give me any feedback you have about this puzzle, I have never done something like this before and I think that this is honestly a really fun variant of sudoku.

  • 1
    $\begingroup$ I assume 8/23/25 is a typo and should be 8/13/25. Also, unless I'm missing something, boxes 4 and 5 are broken - the numbers in them need to sum to 21, but the gravestones that cover them precisely do not have combinations of numbers that can sum to 21. $\endgroup$
    – bobajob
    Nov 6 at 16:43
  • $\begingroup$ @bobajob 8/23/25 is a typo, also I don't think boxes 4/5 are broken (since 0 exists in this puzzle, read the bit about the Schrödinger cells to understand what is up with that) $\endgroup$
    – CrSb0001
    Nov 6 at 16:47


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