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The 8th and 9th installments in this series are going to be particularly large and difficult, so for both I am going to post examples by themselves prior to their larger counterpart. This ruleset is quite complex and I am happy to clarify any points. The solution will be posted alongside the main puzzle next week.

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__Rules:__

Clues outside the grid are Japanese sums, i.e. they give the sum of the digits between empty cells in the respective row or column in the correct order. The filled cells of the grid form a cave, they are all orthogonally connected and all empty cells have an orthogonal connection through other empty cells to the perimeter of the grid. In any row or column, the digits 1-6 may be placed at most once.

Clues are ciphered, same letters stand for same digits, different letters for different digits. However, the cipher of the clues within the grid (ABC) is different from the cipher of the clues outside the grid (EXAMPLE). More explicitly, B may not equal C, and E may not equal X, but B may equal E. Only the letter A occurs both inside and outside the grid, and it appears 6 times in the puzzle, each time as a different digit from 1 to 6. Because this is an example puzzle, all letters from either cipher may only be values from 1-6. As standard, a question mark represents any value from 0 to 9, and for any multi-digit clue, the tens digit may not be zero.

Clues inside the grid are twilight cave clues. If an internal clue is filled, it gives the correct cave clue at that cell, which is the total number of filled cells seen in all directions where empty cells block vision. If an internal clue is empty, then it indicates the size of the orthogonally connected block of empty cells it belongs to. A block of empty cells may have multiple clues as long as they are all identical in value.

In the case for a filled clue cell, the number that must be placed in the cell must be one different in value from the clue itself, which may be outside the 1-6 range, or even a double digit number. If a double digit number is placed, it contributes to any relevant Japanese sums, and one must not worry about the digits comprising the double digit value repeating in the row or column. That is, if in a cell is the number 12, this has no effect on the 1s and 2s in the relevant rows or columns.

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__Clarifications:__

-) The cave clues are correct, but the number placed on them is not. That is, the digit placed on the cave clue must be one different from the value of the cave clue itself. For example, if B=3, then if any B was inside the cave, 3 cells would be seen, but the digit placed on the cell must be 2 or 4.

-) Empty clue cells, just like other empty cells, do not contain digits.

-) It is possible for more than 6 cells in a row or column to be filled, as long as the additional cells to the digits 1-6 come from the "one-different" rule pushing numbers on cave clues outside the 1-6 range. For example, if B=6 and is inside the cave, you may place a 7 on that cell. If AC=23 and inside the cave, you may place a 22 on that cell.

Enjoy :)

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Lösungscode: Row 4, Column 2, S for empty cells

Zuletzt geändert am 4. Dezember 2022, 03:10 Uhr

Gelöst von Vebby, Jesper, MagnusJosefsson, ONeill, ascension, Koalagator2, polar, Alex, ibag, marcmees, cmb, jkuo7, harrison, dunder, Niverio, Counterfeitly, RJBlarmo, Mark Sweep, ffricke, lerroyy, Gliperal, Jaych, akamchinjir

**am 9. Dezember 2022, 05:21 Uhr von cmb**

Beautiful.

I'm looking forward to the long one.

**am 8. Dezember 2022, 13:24 Uhr von marcmees**

Once the rules understood completely (thanks to polar) it solved smooth without being easy. thanks

Zuletzt geändert am 8. Dezember 2022, 12:08 Uhr

**am 8. Dezember 2022, 12:08 Uhr von polar**

@marcmees:

Wall segments are always the correct value, whether a regular digit, or as an A in this example. As long as A is assigned each digit exactly once. So for example, you could have A = 1,2,3 in the outside clues, A = 4 as a cave clue (value in grid would be 3 or 5, but the cave sees 4) and A = 5,6 as correct wall segments. For the wall segments, you may still have a digit 5/6 in the respective rows / columns as part of the sums, which is what I meant about walls not interacting with the other numbers. Hope that makes sense. As to your last question - yes that's right.

Zuletzt geändert am 8. Dezember 2022, 09:41 Uhr

**am 8. Dezember 2022, 09:39 Uhr von polar**

@marcmees: In the case of a cave cell, it will see the original number of cells (3), but the digit that is placed (2/4) contributes to the sum.

Also for clarity (since I messed this part up in testing), the value of any wall clues do not interact with the numbers/sums within the cave.

Zuletzt geändert am 7. Dezember 2022, 20:00 Uhr

**am 7. Dezember 2022, 17:36 Uhr von ibag**

Very nice little puzzle. I hope you don't want to really ruin it next week by posting the solution???

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After your comment (and comments beforehand from my tester) I've decided to include an example diagram for this and the next one. I agree, giving away the solution may ruin some of the fun for those that want to do this smaller one in the future.

Also, regarding your hidden comment about the rules, if, say B in R7C4 is a 5 replaced by 4, it would not be possible to repeat a 4 in the row or column.

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