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This is a Heyacrazy puzzle, originally constructed for a test for Logic Masters India.

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.

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  • $\begingroup$ Brilliant concept! I'd love to see some circles next. $\endgroup$ – greenturtle3141 Aug 23 at 23:30
  • $\begingroup$ @greenturtle3141 Thank you! Unfortunately, circles don't work in an interesting way, because any arc in a cell immediately tells you that that cell is shaded. $\endgroup$ – Deusovi Aug 24 at 1:03
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First of all, these four squares must have two shaded cells in order to prevent all of these three lines. There are two ways to fill them.

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If we try to shade them like this, we'll soon reach a contradiction. Let's see how that happens.

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First of all, the fourth cell on row 4 must be shaded to prevent a line passing through it. That means the fourth cell of line 2 must be empty to prevent cutting off empty cells.

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We must now shade two cells on the first row to block these lines.

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But now there is already an unblockable line crossing row 2 on the right. So this is the contradiction – the starting cells have to be shaded the other way.

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Starting out the other way, again we need to shade a cell on row 4 to prevent this line. Again the corresponding cell on row 2 needs to be empty to avoid shutting out empty cells.

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Only one way to block both these lines.

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Fill in the last cell of row 4 to block this line. This means cell 3 of row 2 must be empty because filling it would cut off empty cells.

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Now, only one way to block this line.

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Some cells can be marked as empty as blocking a nearby line in any possible way would always force that cell to be empty.

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Cell 3 of the second-to-last row must be filled, because leaving it empty would make it impossible to block both of these lines without leaving an empty cell stranded.

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The bottom of the "L" needs to either have its central cell (row 4 cell 2) shaded, or the cells to the right to it and below it. The latter way makes it impossible to complete the bottom left corner without shutting out empty cells.

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So it has to be the central cell that blocks the bottom of the L.

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Only one way to block this line now. This forces the top left corner cell to be empty, as well as the first cell of row 5.

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One way to block this line. This solves the bottom left corner.

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This line can only be blocked one way, forcing the fourth cell of the bottom row to be empty as well.

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From here, there are two ways to block both of these lines: the two diagonals of the 2x2 square. Only one of them leaves us with all empty cells connected.

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Leading to the final solution.

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    $\begingroup$ That's correct - nicely done (and very thorough explanation)! The intended path at the start is slightly cleaner, not using contradiction: one of the two diagonals of the M must be shaded, and so the other must be unshaded. That gives the two shaded cells in the top row either way, and then the open border in the top right resolves the diagonals. $\endgroup$ – Deusovi Aug 21 at 21:49
  • $\begingroup$ nice logic :) wasn't able to solve during the contest $\endgroup$ – ABcDexter Sep 3 at 20:22
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The following appears to be the unique solution

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but I feel like I have failed to spot some useful principles, because I was only able to prove this by doing a painful amount of case-splitting ("there must be a filled square either here or here, so let's try both and see what happens..."). So I'm not going to try to describe my reasoning; it would be pages long.

If someone else wants to post something that solves this with nice elegant explanations, I will not be in any way upset if they get the glorious green checkmark rather than me.

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