So what is the next logical step that I fail to see? Of course I could guess...
Posted 11th Oct 2020 at 20:30
winterer Daily subscriber Rated puzzle: ModerateBest completion time: 5:10Time on first attempt: 6:53Used 'show wrong moves' Used 'check puzzle' when incorrect
JoergWausW,
you need to guess, in order to get a contradiction. Sometimes, it's necessary.
If you set the 4 in the open cell on the left of row 6, you'll have only one cell left in the fourth row to place a 4. This will force the 3 in column 1, which will lead to a contradiction in row 2. Hence, the 4 must be placed in the right-hand corner.
Posted 11th Oct 2020 at 21:33
gareth Administrator Daily subscriber Completion time: 31:54Used 'auto remove' Used 'check puzzle' when incorrect
The puzzle can be solved by some combination of the following: * Usual sudoku/Latin-square solving techniques up to but not exceeding x-wing * Evaluating what combination of digits can fit in each square covered by each individual skyscraper clue, given the existing pencil marks, on a separate clue by clue basis (and not simultaneously considering the opposite clue, if present, either)
I know this because the code I use to check the puzzles only has those operations. :)
Posted 11th Oct 2020 at 22:07
gareth Administrator Daily subscriber Completion time: 31:54Used 'auto remove' Used 'check puzzle' when incorrect
I've run this through my skyscraper solver, and the easiest next step is to eliminate the '4' option from the fourth cell down in the second column. If this was a '4', then it would force the column to have 4 visible skyscrapers instead of 3. This is because placing this 4 would force every other square in the column.
So you can progress by considering only a single skyscraper clue at a time, and while this is not an obvious deduction it is definitely something you can see in your head if you try thinking 'what if this was a 4?' about that cell. :)
After that single deduction, the rest of the puzzle is completely trivial
Posted 11th Oct 2020 at 22:24
JoergWausW Daily subscriber Completion time: 0:56
Thanks to all.
I trusted that you don't have to guess (I can do that, but I don't really want to). That's why I asked. I checked "what, if?" most cells in clue-rows and -colums. Starting at the border and with low numbers. But that square in the second column was "hidden" behind the 5 below.
I did this for the 3-clue in the 2nd column: -> 3 or 4 in bottom row doesn't matter because of the 5. -> Which one of those two cells has the five? Doesn't matter, because the bottom cell can have the 4. I even thought this: -> If the 4th row has a 4, the bottom one is 3 -> the 5 goes directly above(in between) -> no decision possible. I did NOT see, that this 4 would have more impact in this column.
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Last edited by JoergWausW 11th Oct 2020 at 14:44
stuck
So what is the next logical step that I fail to see? Of course I could guess...
you need to guess, in order to get a contradiction. Sometimes, it's necessary.
If you set the 4 in the open cell on the left of row 6, you'll have only one cell left in the fourth row to place a 4.
This will force the 3 in column 1, which will lead to a contradiction in row 2.
Hence, the 4 must be placed in the right-hand corner.
* Usual sudoku/Latin-square solving techniques up to but not exceeding x-wing
* Evaluating what combination of digits can fit in each square covered by each individual skyscraper clue, given the existing pencil marks, on a separate clue by clue basis (and not simultaneously considering the opposite clue, if present, either)
I know this because the code I use to check the puzzles only has those operations. :)
So you can progress by considering only a single skyscraper clue at a time, and while this is not an obvious deduction it is definitely something you can see in your head if you try thinking 'what if this was a 4?' about that cell. :)
After that single deduction, the rest of the puzzle is completely trivial
I trusted that you don't have to guess (I can do that, but I don't really want to).
That's why I asked.
I checked "what, if?" most cells in clue-rows and -colums. Starting at the border and with low numbers. But that square in the second column was "hidden" behind the 5 below.
I did this for the 3-clue in the 2nd column:
-> 3 or 4 in bottom row doesn't matter because of the 5.
-> Which one of those two cells has the five? Doesn't matter, because the bottom cell can have the 4.
I even thought this:
-> If the 4th row has a 4, the bottom one is 3 -> the 5 goes directly above(in between) -> no decision possible.
I did NOT see, that this 4 would have more impact in this column.
Thank you!
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