Pro-Diagonal
(Eingestellt am 7. Juli 2025, 13:03 Uhr von vitaminz)
Please leave a comment if you enjoy it!
Latin Square rules apply (place the digits 1 to 9 once each in every row and column).
Parity Parity: The grid is covered in a checkerboard pattern of yellow and blue cells. Each digit appears in one of the colors exactly as many times as its value. For example, 7 will appear either exactly seven times in blue or exactly seven times in yellow.
Anti-Diagonal (Negative): Digits may not repeat along any negative diagonal (those going from the top left to the bottom right).
Anti-Knight: Cells a knight’s move (in chess) apart cannot contain the same digit.
Odd Circles: Digits in grey circles must be odd.
Lösungscode: The columns in which the digit 5 appears, from top to bottom
Kommentare
Gestern, 07:06 Uhr von kagiso
Man I thought this was very difficult initially. Took me a while to get the trick I left it and when I got back it was so comical to see how straight forward it is once you see it.
am 10. Juli 2025, 05:41 Uhr von Exigus
Wow that was almost miraculous. Super smooth having done a puzzle with similar global constraint recently. Thanks!
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Thank you for your incredibly kind comments here and on other puzzles! -Justin
am 7. Juli 2025, 20:46 Uhr von Mattisahuman
I don’t understand this constraint. If each digit appears in each color exactly as many times as its value, then there would be two 1s in the completed grid; one in blue and one in yellow. But then how am I supposed to place 1 in every row and column if I can only place two?
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For each digit, that constraint holds for one color but not necessarily (and as you have shown, not possibly) for both colors. In the example, it’s true for 7 EITHER in yellow OR in blue.
-Justin
am 7. Juli 2025, 13:22 Uhr von vitaminz
Added anti-knight to conflict checker
