This is a Heyacrazy puzzle.

Rules of Heyacrazy:

  • Shade some cells of the grid.

  • Shaded cells cannot be orthogonally adjacent; unshaded cells must be orthogonally connected.

  • When the puzzle is solved, you must not be able to draw a line segment that passes through two borders, and does not pass through any shaded cells or grid vertices.

For an example puzzle and its solution, see this question.

enter image description here

This puzzle is the most difficult Heyacrazy I have made. But it has a solution without "case-bashing" -- there is a clean logical path, one that does not involve any deep hypotheticals. Any sort of "what-if" logic should be very easy to do in your head, without ever marking any cells you are unsure of (even if only for the purpose of arriving at a contradiction).


1 Answer 1




I'm not entirely sure this is the logical path Deusovi is looking for, but it should be logically sound nonetheless. Let's start off in the lower left:

The cells marked $A$ and $B$ cannot both be shaded, since that would leave $C$ as a closed off unshaded cell (violating rule #2). If we shade $C$, that doesn't really do us any good - it doesn't impact any of the walls, so it might as well be unshaded. For that reason we can assume one of three situations: none of them are shaded, $A$ is shaded, or $B$ is shaded.

Regardless of which of those it is, we need an opening in the line of walls above it so it can be connected to the unshaded cells in the top half of the puzzle. This is a problem if $B$ is unshaded, because $B$ can use all paths to openings (except two) as a vision line - and the two possible paths are impossible due to rule #2:

From this we can conclude that $B$ must be shaded, and that $A$ therefore is unshaded. From here it's mostly a matter of finding a way to put the opening in a place that $A$ does not have vision of. This opening cannot be in the left half, which means that we can deduce the lower-left quarter of the puzzle:

For the lower-right, we have two possible ways to fill it in without having a vision line from $A$. One of those options leads to a new vision line violating rule #3 though:

From here it is more or less straight forward. I have drawn in the vision lines that led to each of the below steps:

  • 3
    $\begingroup$ That's correct! There's a cleaner way to get the break-in, but you got the gist of it: There's no 3x3 area available, so the path from A/B to the "escape route" of the staircase wall must be at most 2-wide (in one of the two possible dimensions). In any 2-wide hallway with at most one kink (e.g. right-right-right-up-right-right), the ends of the hallway can see each other. So there must be at least two kinks in the hallway before the escape, and the only way to do that in the given space is to make a sideways ʔ from R7C1 to R7C4. That gets you directly to your "lower-left quarter" image. $\endgroup$
    – Deusovi
    Commented Aug 27, 2019 at 17:27

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