Shaftless Arrow Sudoku
(Published on 27. August 2021, 23:03 by SenatorGronk)
UPDATE: I've added an additional killer cage box that should make things easier after the break-in. Despite that, based on feedback, I've bumped the difficulty to 4 stars.
UPDATE 2: Apologies for the multiple edits. I've walked through the solution path several times now, and am confident that while difficult, it doesn't require any steps that feel like bifurcating. It does however require you to lean quite heavily on the negative constraint (which was the whole idea of the puzzle to begin with). I've swapped the image back to the original version, but have kept the link to the one with the extra killer cage, for people who'd like some extra help.
I've also removed references to an additional constraint that I thought was necessary for one part late in the solve, but I found it could be solved without it. I think this is a change for the better because lots of people were hung up on the ramifications of this constraint and without it, solvers will be forced to focus on the primary (negative) constraint in the puzzle.
With the extra starting cage, I gave the puzzle 4 stars, so I'll bump the final (original) version to 5 stars.
Rules
- Standard sudoku and killer sudoku rules apply.
- Standard arrow sudoku rules also apply, with an asterisk: Each circled cell is the base of a horizontal or vertical arrow whose shaft is for you to determine (that is, the circled digit is the sum of the next 2 or 3 digits in its row and/or column). One circle can have multiple shafts extending off of it in different directions: Example 4-5-(9)-7-2.
- NEGATIVE CONSTRAINT: Every possible circled cell has been provided. In other words, the given circles contain the _only_ digits in the completed grid that could form a horizontal or vertical arrow. For example, (8)-1-4-3 is a valid way to fulfill a circled 8, but (8)-1-3-4 isn't because the 1 and 3 sum to 4 and the 4 isn't in a circle.
F-puzzles (original version)
F-puzzles (version with an extra cage)
Solution code: Row 8 + Column 6 (18 digits no spaces)
Last changed on on 2. February 2022, 19:14
Solved by OutOfMyMindBRB, Steven R, Jesper, henrypijames, Dentones, bernhard, polar, soroush, Gnosis66
Comments
on 31. August 2021, 17:37 by soroush
I did the "easy" version, but wow. So intricate! 4 star seems right to me for the version with the extra cage
on 29. August 2021, 18:47 by SenatorGronk
Removing an extra constraint and making the version with the extra cage an alternate link.
Last changed on 28. August 2021, 19:15on 28. August 2021, 15:06 by henrypijames
@SKORP17: Your question answers itself when you get a bit into the puzzle.
I have a different question on the rules: Is a sequence like (8)431 or 4(3)12 allowed (circled digit in brackets)? On one hand, in both cases, 4=3+1 forms an "unsanctioned" arrow; but on the other hand, the rules say arrows cannot overlap, so maybe that prevents the unsanctioned arrow from being formed? In other words, while the rules forbid both uncircled arrow and overlapping arrow, when a sequence create both those possibilities at the same time, do they cancel each other out (like a double negative)?
So far, the puzzle is at least 4 stars for me, probably getting up to 5 stars in the end - if I manage to solve it at all.
RESPONSE: @henrypijames, Neither of your examples would be valid because as you mention, the 4 isn't circled. The negative constraint is just about the circles -- they are the _only_ digits in the completed grid that will sum to a string of adjacent numbers in their row and/or column.
Last changed on 28. August 2021, 12:34on 28. August 2021, 04:23 by bigger
What do you mean by all possible circle are given? As in before all the given circles' shaft were drawn or after?
Edit: I mean every digit that would be the base of a straight arrow in a row or column is marked as such. Ex: 9-2-7 could only appear adjacently in that order in a row or column if the 9 was circled.
I mean does the negative constraint came in later or before. There could be two understanding, one is all possible base digit is given, the given base don't cross shaft. Second is after all circled shafts are drawn, all other digits have no shaft. So in the case of 9-4-5, If 5 is already on a shaft, 94 is not, is it valid?