Logic Masters Deutschland e.V.

a walk in the park

(Eingestellt am 29. September 2025, 05:30 Uhr von aqjhs)

So Gene asked me if a projected parity party dög puzzle was possible, and it was!

...another sneaky Top Right Boi here
  • Normal sudoku rules apply.
  • Doppel-Dög loop:
    • Draw an orthogonal non-branching closed loop that doesn't cross or touch itself, not even diagonally.
    • The loop "starts" at the cell marked with the dog and is considered to be oriented in the direction indicated by the arrow (i.e. r1c9 and r1c8 are on the loop).
    • A cell on the loop with a digit D will have a value equal to the digit D steps away along the loop in the orientation given by the arrow.
    • Cells not on the loop will have a value equal to the digit in them.
  • Adjacent digits (not values) along the loop must have a difference of at least 5.
  • Projected Parity Party: a number outside the grid shows the value sum of the loop cells in the direction of the clue until the first parity change, including the first cell of the opposite parity.
  • Cells marked with a circle are not part of the loop and have a digit equal to the count of the, up to 8, neighboring loop cells.
  • Cells joined by a ball have the same value.

Online in Sudokupad

More Doppel-Dögs:

Lösungscode: Values of row 4.

Zuletzt geändert Gestern, 00:29 Uhr

Gelöst von Piff, oskode, tuturitu, Azumagao
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Gestern, 00:29 Uhr von aqjhs
updated difficulty and one extra clue for good measure

Zuletzt geändert am 29. September 2025, 22:16 Uhr

am 29. September 2025, 22:10 Uhr von clock
Perhaps the answer to this question can be obtained from the puzzle itself, but I would like to confirm: does "parity change" refer to a change in the parity of the value or a change in the parity of the digit?

***

It is all in values.

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