Logic Masters Deutschland e.V.

Polyominous

(Published on 25. November 2022, 20:10 by mathpesto)

This puzzle is inspired by Josh Johnson's Killomino, so be sure to check that out!


Rules:

Divide the grid into polyominoes so that no two polyominoes of the same size share an edge. Each cell contains a number equal to the size of its polyomino. Numbers within a cage may not repeat and must sum to the total, if given.

(Note: A polyomino's size may exceed 9. Suggested notation in the cells for such regions is a corner mark for the tens place and a center mark for the ones place.)


Solve:

Solution code: Numbers in Rows 7 and 9 (left to right, no spaces)

Last changed on on 23. December 2022, 22:43

Solved by ONeill, Oripy, apothycus, Myxo, MagnusJosefsson, amarins, 85392, Jaych, jkuo7, SudokuExplorer, uniquicity, Doziam, wooferzfg, Jakhob, h5663454, cair, Jesper, Mark Sweep, farodin64, Franjo, Raistlen, ... Just me, tuturitu, zhall12570, Xendari, Felis_Timon, Steven R, Chefofdeath, widjo, sth, rubbeng, zhantyzgz, Bibet, kamkam, PURB97, zuzanina, P12345, nottabird, moss, mdjvz, woz, StephenR, GTLSE
Full list

Comments

on 23. December 2022, 22:43 by mathpesto
Added penpa+ link

Last changed on 23. December 2022, 22:42

on 23. December 2022, 21:45 by TJReds
penpa link: https://tinyurl.com/2ob3vev4
(for those who may prefer penpa like myself)

Absolutely gorgeous puzzle!

--------

Thanks so much, I will add the link :)

on 30. November 2022, 06:06 by Emckee2006
I stared for a moment, thought I had it, then had to restart (it was broken) and got almost to the end, only to think I'd broken it again. Luckily the light bulbs went off and everything worked like a charm. This was a delightful way to spend some time :)

on 26. November 2022, 11:04 by farodin64
Very smooth solve flow, thanks for the fun!

on 26. November 2022, 07:12 by wooferzfg
Beautiful!

on 25. November 2022, 21:16 by Myxo
Lovely!

on 25. November 2022, 20:57 by ONeill
Very nice! Thanks :)

Difficulty:2
Rating:93 %
Solved:86 times
Observed:2 times
ID:000C4O

Variant combination

Enter solution

Solution code:

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