Divide the grid into 9 orthogonally connected regions of size 9, and place the digits 1-9 in the grid such that each row, column, and region has one copy of each digit.
Then, divide each region into some polyominoes such that no two polyominoes of the same size share an edge.
Digits in a circle indicate the size of the polyomino it is in, and digits in a square indicate the number of polyominoes in its region.
Penpa solvers only: Answer check requires all digits to be filled out, green edges for region borders, and double edges for pentomino borders that are NOT region borders. (All region borders are also pentomino borders because of the rules.)
Lösungscode: Column 1, but also include an X for every fillomino barrier. It might look something like 1X234X5678X9.
am 3. Juli 2023, 07:01 Uhr von OldCaptainWZJ
Very intricate and rewarding puzzle. It took me a long time in a few tough spots, but it was very satisfying to be able to make it through. Incredible setting!
am 25. Juni 2023, 23:55 Uhr von Christounet
Incredible puzzle ! The kind of one that makes you feel that you've broken something so many times. But each time, there's just that tiny possibility that doesn't break it after all. Thrilling until the very end ! Thanks :)
am 8. Juni 2023, 15:30 Uhr von Silverstep
The colouring on the CTC version interacts weirdly with the shapes, so I used lines instead. Very smooth and relaxing solve, overall not difficult. Perhaps a low 3-starer.
am 1. Juni 2023, 09:46 Uhr von twobear
Great one!
Broke it soooooo many times…
am 29. Mai 2023, 22:03 Uhr von zhall12570
I messed this up about 8 times before I could figure out the solution. Very rewarding solve, and the logic was always switching up. So fun! Thanks. Please make more
> Thanks for solving and glad you were able to figure it out in the end. Another FCC is coming up soon --- just need to get it tested :)
am 29. Mai 2023, 02:22 Uhr von dennischen
Glad to see that so many more people are enjoying this puzzle :)
Some part of me wishes I made this an instructionless puzzle, given the year-long drought LMD has had and how pliant the ruleset was to instructionless. Oh well, maybe next time.
am 26. Mai 2023, 22:14 Uhr von prodigis
Very beautiful and very difficult, but enjoyable to the end. Thanks!
am 26. Mai 2023, 13:09 Uhr von madhupt
What an intricate and beautiful puzzle! It was hard for me but once you see the logic, it’s very smooth. In my case, the logic only dawned on breaking the puzzle. It refused to yield till the very end.
Thanks a lot for sharing.
am 26. Mai 2023, 08:47 Uhr von Xendari
Beautiful puzzle, thanks!
am 26. Mai 2023, 05:13 Uhr von bodemeister
Awesome puzzle! This belongs in my top 5 most favorite chaos constructions. Thanks!
am 26. Mai 2023, 04:09 Uhr von Agent
I can't count the number of times I broke it, very cool puzzle though.
am 25. Mai 2023, 13:19 Uhr von marcmees
Fantastic CC. Got tricked several times. thanks
am 25. Mai 2023, 03:48 Uhr von bodemeister
Just to confirm, does each cell in the grid need to be part of a polyomino?
> Yup.
am 25. Mai 2023, 02:35 Uhr von dennischen
Edit links to my new website puzz.dennisc.net
(This has virtually 0 effect for all solvers)
am 24. Mai 2023, 22:43 Uhr von peacherwu2
Genius! My new favorite!
> Thanks for the comment! I'm glad you enjoyed this puzzle so much.
am 24. Mai 2023, 22:01 Uhr von dennischen
Thanks for all the comments, glad you guys all enjoyed! :D
am 24. Mai 2023, 20:24 Uhr von GrizzledStoat
Must the fillomino constraint (no two polyominoes share an edge) hold everywhere or just within the regions?
> Everywhere.
am 24. Mai 2023, 18:40 Uhr von KNT
enjoyable and deceptively tricky chaos construction!
am 24. Mai 2023, 09:55 Uhr von RJBlarmo
Great puzzle, had a couple of pretty tricky steps!
am 24. Mai 2023, 09:28 Uhr von Jesper
Nice concept, enjoyed it, thanks!
am 23. Mai 2023, 08:49 Uhr von Chefofdeath
This is a really cool concept with a really cool break in :) Definitely tough in a few spots, with some really cool deductions!
Schwierigkeit: | |
Bewertung: | 100 % |
Gelöst: | 60 mal |
Beobachtet: | 12 mal |
ID: | 000DY4 |