Place pieces (red) on the board, so every piece is adjacent to 4, 5 or 6 other pieces.
I've created this puzzle myself playing around with cellular automatons.
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$\begingroup$ Welcome to PSE (Puzzling Stack Exchange)! $\endgroup$– Will.Octagon.GibsonCommented Apr 6 at 7:26
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1$\begingroup$ Adjacent includes diagonally? $\endgroup$– quaragueCommented Apr 6 at 8:41
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1$\begingroup$ Nice puzzle! Simple yet not trivial to solve. $\endgroup$– Kris Van BaelCommented Apr 6 at 10:39
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5$\begingroup$ @quarague including diagonal (otherwise 5 or 6 neighbors weren't possible). $\endgroup$– DromliusCommented Apr 6 at 12:50
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2 Answers
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A symmetric solution (with pieces represented by Os):
X O O X X O O O O X O O X O O X O O O O X X O O X
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$\begingroup$ Welcome to PSE (Puzzling Stack Exchange)! $\endgroup$ Commented Apr 6 at 7:24
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Why the accepted solution is unique:
No corner can have a piece in it, because corners are only next to three squares.
Corollary: As a consequence, every piece next to a corner (like the starting piece) must be next to a piece in every non-corner neighbor. In particular, it forces this configuration:
_ _ _ _ _
_ _ _ _ _
_ _ X O O
_ _ O O O
_ _ O O X
I could proceed with some case-bashing, but instead I'll ask:
How many pieces can be on an edge?
An isolated piece can't, because it only has three non-edge neighbors, but three on an edge is also impossible, because it forces this configuration via Corollary:
_ _ _ _ _
_ _ _ _ _
O O _ O O
O O 7 O O
X O O O X
This means that every edge must have a domino on it (or nothing at all, but any shape fitting in a 4-wide box is impossible by a variation of the above)
How can this be arranged?
Dominoes automatically form pairs by the Corollary, so we have this:
X O O X X
O O O _ X
O O X O O
X _ O O O
X X O O X
Both unknown squares must have pieces to satisfy the edge cells, yielding the accepted solution.
