Logic Masters Deutschland e.V.

Polyominous

(Eingestellt am 25. November 2022, 20:10 Uhr von 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:

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

Zuletzt geändert am 23. Dezember 2022, 22:43 Uhr

Gelöst von ONeill, Oripy, apothycus, Myxo, MagnusJosefsson, amarins, 85392, Jaych, jkuo7, SudokuExplorer, uniquicity, Doziam, wooferzfg, Jakhob, h5663454, cair, Jesper, Mark Sweep, farodin64, Franjo, Raistlen, ... moeve, Bellsita, misko, Deathranger999, jlaitio, Uhu, ManuH, matiasv5, TJReds, smartmagpie, NIGHTCRAULER, OGRussHood, tobymgk, indolentfool, Xavi 17, MicroStudy, dogfarts, cameleon, Vebby, GexTed
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Kommentare

am 23. Dezember 2022, 22:43 Uhr von mathpesto
Added penpa+ link

Zuletzt geändert am 23. Dezember 2022, 22:42 Uhr

am 23. Dezember 2022, 21:45 Uhr von 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 :)

am 30. November 2022, 06:06 Uhr von 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 :)

am 26. November 2022, 11:04 Uhr von farodin64
Very smooth solve flow, thanks for the fun!

am 26. November 2022, 07:12 Uhr von wooferzfg
Beautiful!

am 25. November 2022, 21:16 Uhr von Myxo
Lovely!

am 25. November 2022, 20:57 Uhr von ONeill
Very nice! Thanks :)

Schwierigkeit:2
Bewertung:93 %
Gelöst:64 mal
Beobachtet:1 mal
ID:000C4O

Variantenkombination

Lösung abgeben

Lösungscode:

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