Sudoku variants #1: Schrödinger's Grave (6x6 grid)

Question

_{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)}

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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.

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

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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.

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The puzzle (only 6x6)

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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.

Answer 1

(partial answer, I remember nothing about this puzzle and have lost the solution paper)

TL;DR:

No, it's not solvable, but this is only because of the orthogonal line requirement. The sudoku might still be solvable otherwise.

Some initial thoughts:

• It appears that a number cannot be replaced by any other number other than 0.
  • I think I imposed this restriction on myself when creating the puzzle because I didn't exactly understand the concept of a number other than 0 replacing a number 1-6
• I do also remember that there is only one 0 in each column, row, and box.

About Box 4

We need the numbers in the boxes (not including the 0 replacement) to add to 21. However, the (valid according to the rules) addition combinations yield:

• 11+8=19
• 11+23=34
• 11+28=39
• 27+8=35
• 27+23=50
• 27+28=55
• 28+8=36
• 28+27=55
• 28+28=56

and admittedly, this caused some confusion, because none of those results equal 21.

However, this is where the concept of 0 replacing a number comes in, because we have that 21-19=2 (19 being the result closest to 21 that does not go over), and therefore we have that 2 is being replaced by 0.

So we have that 0/2 could either go in R3C4, R3C5, R4C4, or R4C5. This also implies that we have a 3 and a 5 going in either R3C6 or R4C6.

Addressing the orthogonal line from the green cell to the red cell

I know, making an edit almost 6 months later to a post can be frowned upon at times, but I'm going to have to remove this rule.

If I am understanding the rules correctly, it is literally impossible to satisfy this condition.

Proof:

Explanation:

• First off, the rule (which I find extremely stupid in the context of this puzzle) says that we cannot cross the highest or the lowest digit in a grave.
• This means that for any grave we are drawing the line through, we can only cross through $n-2$ cells ($n$ is the number of cells in that grave). Also, we can only go straight through 3-cell graves and have to avoid 2-cell graves at all costs.
• However, there's a problem. Drawing a line starting at the red cell and going through the 5/11/12 grave, we run into the problem where either we head up and hit the 2-cell 8/23/28 grave, or we head left and hit the 2-cell 8/11/27 grave.
• Also going from the green cell, we see that no matter what, we end up with the problem of running out of cells to go through the 12/24/87 grave, or the same thing happens with the 11/27/28 grave.

So yes, the puzzle's impossible. However I would still like to know whether or not the Sudoku clues can be deduced as normal.

Final notes - About Box 5

We first deduce that since we cannot have three distinct natural numbers $n_1,n_2,n_3$ where $1\le n_1,n_2,n_3\le6$ such that $n_1+n_2+n_3$ is equal to any of $1,16,27$, we have that the numbers in the 1/7/27 grave add up to 7 (and therefore is a 1/2/4 pair), we have that $6+5+3_{=0}$ goes in the 11/13/16 grave.

My final progress on the Sudoku portion so far: