I want to be Wild and Free!
(Eingestellt am 29. Mai 2021, 21:11 Uhr von panthchesh)
Es gilt die Anti-Ritter-Regel.
Lösungscode: Zeile 1 Spalte 9
Kommentare
am 6. Oktober 2021, 16:27 Uhr von uvo_mod
Labels angepasst.
am 10. August 2021, 04:09 Uhr von panthchesh
Adjustment to German version.
am 10. August 2021, 01:45 Uhr von Krokant
I stumbled upon some deadly patterns that I couldn't resolve no matter what. I got a lot easier after I realised that the german version did not mention the anti-knight rule. You might want to add this information.
Otherwise a very nice puzzle, as usual from you. :)
Panthera: I had someone translate it for me. Thanks for the heads up . I'll use a translator to add in that line!
am 28. Juli 2021, 17:04 Uhr von Vebby
Very nice variation!
A few statements about the rules that may allay any doubts for future solvers:
1. Each cell has 3 options: blue or red or unshaded.
2. Clues outside a row/column indicate the totals of all blue and red groups in that row/column.
3. A row/column with red clues cannot have blue cells in it, and vice versa.
4. For each row/column with black clues, at least one clue must be for a blue group and at least one clue must be for a red group.
am 31. Mai 2021, 03:02 Uhr von jwmpuz
You say "... must be separated by an unshaded cell when the groups are the same color". Should that say AT LEAST one unshaded cell? Or is at always just one?
At least 1. Thank you for asking!
am 29. Mai 2021, 21:12 Uhr von panthchesh
You can watch the live solve on Rangsk's livestream #20: https://youtu.be/q7XT970lndk
Also to clarify, each black number must have a red and a blue clue in it, but it can be something like blue, red, unshaded, red
