Logic Masters Deutschland e.V.

Deja vu

(Published on 9. June 2024, 12:32 by Al Fresco)

Abnormal Sudoku rules apply. Each 3x3 box contains a full set of digits 1-9. However each row and column contains 8 different digits.

Each row, column and box contains a repeater cell. A digit in a repeater cell must repeat in it’s row and column. Each digit 1-9 appears in one repeater cell.

Cages may not contain the same digit more than once and the digits sum to the given cage total. Digits separated by an x sum to 10.

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Solution code: Row 2

Last changed on on 9. June 2024, 16:50

Solved by loat, OutOfMyMindBRB, Briks
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Comments

on 9. June 2024, 20:47 by Briks
Finally... liked it very much

Last changed on 9. June 2024, 18:07

on 9. June 2024, 18:03 by OutOfMyMindBRB
What a brilliant puzzle! Severe "damage" to the usual scanning... had to restart 5 times until i got the hang of it, quite a challenging and fun solve!

BTW, the solution check does not work for this one.

on 9. June 2024, 16:42 by Al Fresco
Updated the wording in the rules.
Appologies for the unclear wording.

on 9. June 2024, 15:54 by sujoyku
No Briks, I agree that the wording of the rules could be more clear. But I guess it should be possible that in row 1 the 23 and the 21 cage share a repeated digit. But inside of one cage no digit may repeat.

on 9. June 2024, 14:53 by Briks
I any case there are double digits in the two cages which isn't allowed by the rules.

on 9. June 2024, 14:34 by sujoyku
I think it is meant differently, Briks. If I understand the rules correctly, the repeater cell may be in a cage. But then the other instance of that digit in the row/column may not be contained in that same cage.

on 9. June 2024, 13:37 by Briks
It is not possible to fill the cages in the first row (23, 21) with digits when there is no repaeater cell in one of the cages.

Difficulty:3
Rating:N/A
Solved:3 times
Observed:5 times
ID:000IFP

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