Logic Masters Deutschland e.V.

Pentomino Minesweeper III

(Published on 2. September 2011, 12:00 by Richard)

Place all twelve pentominos in the empty cells of the grid, so that they don't touch each other, not even diagonally. The numbers indicate how many of eight surrounding cells contain a pentomino part. The pentominos may be mirrored and rotated.

Solution code: Column 2, followed by row 12; The letter of the pentomino or a '-' for an empty cell (24 in total).

Last changed on on 30. October 2011, 06:02
Solved by StefanSch, pokerke, zuzanina, Rollie, Kekes, saskia-daniela, MiR, sandmoppe, CHalb, Statistica, Joe Average, rob, Errorandy, Zzzyxas, Thomster, pin7guin, Saskia, Alex, Luigi, derwolf23, ch1983, ... relzzup, Ute2, pirx, annusia, NikolaZ, mango, Calavera, Joo M.Y, muhorka, EKBM, kiwijam, sf2l, crissu, moss, tuace, Oskama, Matt, Uhu, rubbeng, misko, Nick Smirnov, Krokant, Greg, Echatsum
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Comments

on 16. April 2022, 11:45 by Nick Smirnov
Penpa:
https://tinyurl.com/y4vthgzr

on 16. September 2011, 13:13 by ibag
"I think the measuring system for difficulty is completely out of control..." Yes, I agree!

on 16. September 2011, 13:00 by Richard
@ibag: some TE is necessary in the area middle/bottom half. I was just curious how many clues could be removed and still have a unique solution.
I only don't understand that it has only two stars for difficulty. In my opinion this puzzle is absolutely not suitable for starters. I think the measuring system for difficulty is completely out of control...

on 2. September 2011, 21:03 by Statistica
Sehr sehr schön. Vielen Dank, Richard!

Difficulty:3
Rating:87 %
Solved:74 times
Observed:7 times
ID:00014S
Puzzle variant Pentominoes
Minesweeper

Enter solution

Solution code:

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