Logic Masters Deutschland e.V.

Selfish Cages

(Published on 13. November 2022, 21:34 by Ennead)



Normal sudoku rules apply.

Digits in cages cannot repeat and must sum to the small corner total, where given.

Pairs of cages that share an edge must share a single digit.




f-puzzles CTC


Solution code: Column 3 followed by row 8 (18 digits, no spaces)

Last changed on on 13. November 2022, 23:46

Solved by AnalyticalNinja , yttrio, davidagg, SKORP17, PixelPlucker, dumediat, Myxo, polar, glum_hippo
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Comments

Last changed on 21. June 2024, 07:36

on 20. June 2024, 21:30 by glum_hippo
Surprising in moments, and very hard
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Thanks for ploughing through it! I appreciate your tenacity. This puzzle is much too hard really. I'm not sure I could solve it myself without looking at the notes!

Last changed on 6. December 2022, 09:03

on 6. December 2022, 05:51 by PixelPlucker
This was a proper challenge and also very gripping. Thanks, Ennead!

- You are very welcome! Well done for getting through it. I appreciate your persistence!

Last changed on 17. November 2022, 16:31

on 17. November 2022, 16:18 by SKORP17
Dürfen sich die Cages nur jeweils 1 Ziffer teilen oder auch mehr als 1?

- Only 1 digit. (Nur eine Ziffer.)

Last changed on 16. November 2022, 16:06

on 16. November 2022, 15:25 by davidagg
It was certainly a challenge, and I'm not sure I always followed the intended path. Thanks for setting it.

Last changed on 16. November 2022, 15:28

on 16. November 2022, 15:21 by davidagg
Really hard!

- Yes. It's the hardest puzzle I have made, both for me and you! Thanks for solving. I hope you enjoyed the challenge.

Last changed on 16. November 2022, 15:23

on 14. November 2022, 02:47 by yttrio
An interesting rule set with some cool logic. This was quite difficult, but felt very rewarding to finally solve.

- Thanks for solving (and for your help taming the original version.) Glad you liked it!

on 13. November 2022, 23:46 by Ennead
Fixed broken solution code.

Difficulty:5
Rating:N/A
Solved:9 times
Observed:8 times
ID:000C01

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