Multiple Fortnights (Blank Fog of War)
(Eingestellt am 29. Juni 2023, 04:58 Uhr von Farkov)
Rules:
Normal sudoku rules apply.
Fog of War rules apply. The grid is completely covered in fog. Placing correct digits will clear the fog from surrounding cells.
Orthogonally adjacent cells may not sum to 14.
No two-digit number (read left to right and top to bottom) may be a multiple of its row, column or box number if that row, column or box number is greater than 2.
Disambiguation:
If we call all of the legal two-digit combinations 'dominoes' then each row has 8 dominoes, each column has 8 dominoes and each box has 12 dominoes.
Example:
If we place a 1 on R7C8 then the following combinations are illegal: 8 on R7C7 makes 81 in box 9. 8 on R7C9 makes 18 in box 9. 8 on R8C8 makes 18 in box 9. 2 on R7C7 makes 21 in row 7. 9 on R7C7 makes 91 in row 7. 4 on R7C9 makes 14 in row 7. 6 on R8C8 makes 16 in column 8.
All feedback welcome and appreciated.
Lösungscode: Row 8, column 9.
Kommentare
am 3. Juli 2023, 19:04 Uhr von thoughtbyte
That was really tough but I enjoyed it very much, thanks!
@thoughtbyte Thanks for your feedback. It was a bit of a divisive puzzle this one. I can see how it appealed to some brains and not others. Welcome to the warped brain club :)
am 30. Juni 2023, 23:28 Uhr von Gammon88
Isn't this broken. Doesn't Column 5, Rows 2 and 3, make a number divisible by 3?
@Gammon88 No, it is not broken. I think you might want to take a look at the rules again. In column 5 you have 8 dominoes and none of them are allowed to be a multiple of 5. The lowest two of those dominoes are in box 8, so they are not be allowed to be multiples of 8 either.
am 30. Juni 2023, 15:25 Uhr von Ragna
Wow! What a puzzle! Unbelievable tricky.
Thanks @ sanabas, your help brought light in the fog of my brain ;-)
@ Farkov: Thank you for surely much much time. :-))
@Ragna Thanks man. Glad you stuck with it :)
am 30. Juni 2023, 01:42 Uhr von sanabas
@ragna - think about box3 and row3, and use 3 different colours for the 3 different remainders when divided by 3. Which colours are allowed to border which? When you also colour in the known digits in box 1 & 2, you should get somewhere.
am 30. Juni 2023, 00:07 Uhr von Ragna
This puzzle makes me a lot of problems. Cells are not available. Surely not so very important. 155 minutes, and I got only the break-in. Bifurcations are not very helpful. Cheating is not my way.
What is the secret here?
Can you or other clever solver show a logical way?
am 29. Juni 2023, 22:34 Uhr von SenatorGronk
Not sure why this was a fog of war. None of the hidden digits make the break-in any easier and all are revealed very early in the solve.
@SenatorGronk I bet you don't like your Christmas presents wrapped either :)
am 29. Juni 2023, 15:03 Uhr von mihel111
I got the same problem. Impossible to fill R1C6, R2C25, R4C9 and R6C2. Not with central or corner marking or with plein numbers. It doesn't also work with letters. Only colouring is possible.
@mihel111 Without wanting to give away too much of a hint...<stop reading now if you're worried>...but after the initial break-ins there is literally only one digit given by revealing the fog. This is why this issue never came up in testing. Apparently the 'fix' is to swap given digits for a single cage that sums to that total...which is not very pretty. I don't think this bug should affect the solve as far as I can see.
am 29. Juni 2023, 12:21 Uhr von sanabas
Just made a weird discovery, I can't pencil-mark certain squares, presumably because they have a given digit under the fog.
@sanabas That is a good pickup. I believe that is the reason, yes. It just never came up during testing because of the nature of those hidden numbers. If there is a work-around I will post a new link. Thanks.
