Odd-Even Parity Cages Sudoku
(Eingestellt am 11. März 2021, 09:00 Uhr von stephane.bura)
Normal Sudoku rules apply.
If a cell is in a cage, its digit is the number of digits of the same parity (odd or even) in this cell and all its neighboring cells, orthogonally and diagonally.
Digits in the same cage have the same parity.
Orthogonally adjacent caged digits in different cages have different parities.
Some possible cages may be omitted.
Enjoy!
Lösungscode: Column 7, column 8
Kommentare
am 17. März 2021, 07:43 Uhr von Arcadiotl
very nice puzzle
am 14. März 2021, 20:47 Uhr von Rollo
Spannend bis zum Ende :-).
am 13. März 2021, 13:25 Uhr von filuta
It was a real pleasure to solve this one, thanks!
am 13. März 2021, 05:01 Uhr von mackerel
A really delightful puzzle. I found the rules clear (so your revisions must have been good) and I had a great time working out what that meant for the grid. Thanks!
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I'm glad :) Thanks!
am 12. März 2021, 23:02 Uhr von askel083
Thanks! I really enjoyed that! I’m exhausted now, but in a good way!
am 12. März 2021, 21:13 Uhr von Grumpy
I'll admit that the rules were a little confusing at first (although far from the worst offender on this site hehe), but when I finally got it it was fun :)
am 12. März 2021, 07:48 Uhr von stephane.bura
Final final rewording.
am 12. März 2021, 05:10 Uhr von Ryx
@ Dandelo - thank you! Now I understand the rule. But still no idea where to break in ;-) I will stare at it some more ...
am 11. März 2021, 21:14 Uhr von Dandelo
Maybe it would be much easier to read, esp. for non-mathematicians:
Cells in the same cage have the same parity and orthogonal neighbours in different cages have different parities.
But I don't want to keep the discussion alive.
am 11. März 2021, 19:55 Uhr von stephane.bura
Rewording. Last one! ;)
Thanks, Dandelo.
am 11. März 2021, 19:45 Uhr von Dandelo
This is wrong. It must read: Two caged cells which are orthogonal neighbours are in the same cage, if and only if they have the same parity.
The actual formulation would imply, that R2C1 and R3C3 have a different parity.
am 11. März 2021, 18:25 Uhr von stephane.bura
"If and only if..."
am 11. März 2021, 18:18 Uhr von henrypijames
Now, about the puzzle itself: I like the ruleset, and I can't help but wonder if this one is another gateway drug paving the way for much harder stuff.
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I'm not sure :)
Giving the parity of the digits makes it too easy and not indicating the neighbors of same parity makes it too hard and unfun.
Maybe combined with another ruleset...
am 11. März 2021, 17:15 Uhr von henrypijames
Well, a strict interpretation of the "if ... then ..." clause wouldn't even provide what it is supposed to say:
If two digits of same parity are neighbors, they belong to the same cage. Okay. A strict interpretation of that would be non-exclusive (i. e. the reverse is not automatically true), therefore digits with other properties may also belong to the same cage. In particular, digits of *different* parities may be in the same cage - nothing in the strict interpretation is precluding that.
am 11. März 2021, 13:44 Uhr von stephane.bura
Added change in the rules too.
am 11. März 2021, 13:42 Uhr von Dandelo
I think parity in mathematics and computer sciences means odd/even. And the "if ... then ..."-relation clearly states if there is a negative constraint or not. IMHO the rules were ok.
Nevertheless you can add it to clarify the rules, esp. for solvers who translate it automatically to other languages. There such subtleties could be lost.
am 11. März 2021, 13:38 Uhr von stephane.bura
Following henrypijames' advice: renamed and precised that no use of the negative constraint was intended.
am 11. März 2021, 13:20 Uhr von henrypijames
I suggest specifying "odd-even parity", because there are other kinds of parities (e. g. high-low).
Also, a clarification on negative constraints would always be helpful.
am 11. März 2021, 13:19 Uhr von marcmees
An easy stephane bura puzzle ? nice.
am 11. März 2021, 10:31 Uhr von Dandelo
Very nice, the idea and the puzzle.
