10
$\begingroup$

This is a Heyacrazy puzzle, designed at request of testsolvers of previous puzzles who wanted to see how longer diagonals would work.

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

$\endgroup$
7
$\begingroup$

Solution:

There are lots of quick and easy deductions at the bottom, starting with the two cells that on their own contain two edges and leading to this:
enter image description here
One of the two cells on the top edge where two lines meet must be shaded. If we try letting it be the right-hand one, we quickly get to this impossible situation
enter image description here
and so instead it must be the left-hand one:
enter image description here
One of the two currently-blank cells in column 3 must be shaded. Can it be the lower one? If so, then we have this
enter image description here
which is impossible because now lines emanating SE-ish from the third cell on the first row require two adjacent cells to be shaded. So in fact it was the upper of the two cells in column 3 that's shaded:
enter image description here
requiring the cell one left and one down from the top right corner to be shaded, which immediately requires all the other cells near it to be unshaded for connectivity
enter image description here
leading us rapidly to
enter image description here
and then
enter image description here
and we're done.

$\endgroup$
  • 1
    $\begingroup$ Beaten by half an hour? Dang. $\endgroup$ – Rand al'Thor Aug 23 at 16:43
  • $\begingroup$ Maybe I just saw it half an hour before you did. Or maybe I'm quicker :-). $\endgroup$ – Gareth McCaughan Aug 23 at 16:45
  • $\begingroup$ Probably the latter. It's my first go at this type of puzzle. $\endgroup$ – Rand al'Thor Aug 23 at 16:46
  • $\begingroup$ That's correct, and exactly the intended path - nicely done! (Well, you can use more advanced logic to get rid of the hypotheticals, but it's roughly the same thing anyway.) $\endgroup$ – Deusovi Aug 23 at 17:12
3
$\begingroup$

Labelling the rows 1 (top) to 7 (bottom) and the columns A (left) to F (right), our first step is that

A5 and F6 are clearly shaded, which means A4, A6, B5, F5, F7, E6 are unshaded. Then B6 is also shaded, which means C6 and B7 are unshaded. For connectivity of unshaded squares, now A7, C7, and E7 must be unshaded.

first steps

Next target to be shaded is

E5, with unshaded around it, and then D6, with unshaded around it. For connectivity, now C5 must be unshaded. Then D4 must be shaded, with unshaded around it.

next steps

Now we have a few sets of squares of which at least one must be shaded:

D1 and E1, C1 and C2 and C3, D1 and E2. That means if E1 is shaded, then D1 is not shaded and so E2 is shaded, contradiction. So D1 is shaded, which means C1, E1, and D2 are unshaded, which means exactly one of C2 and C3 is shaded.

and again

Trying a similar deduction technique again:

At least one of C2 and E3 must be shaded. So either C2 is shaded, or C3 and E3 both are. In the latter case, we get the following which is impossible (draw a line from A1 to B6):

can't be solved!

So in fact

C2 is shaded, which means B2 and C3 are unshaded, and B1 for connectivity. Now E2 must be shaded (I think), and then everything else in the top right corner must be unshaded.

almost done

Finally,

B3 must be shaded, so B4 and A3 are unshaded, and A2 for connectivity. Then A1 must be shaded and we're done:

final solution

$\endgroup$
  • $\begingroup$ Your configuration with "no unique solution" has in fact no solution because it already has a line (roughly horizontal, near the top) crossing two diagonals and no shaded squares. $\endgroup$ – Gareth McCaughan Aug 23 at 17:02
  • $\begingroup$ (This doesn't make your answer wrong, of course. In fact it makes it righter :-).) $\endgroup$ – Gareth McCaughan Aug 23 at 17:02
  • $\begingroup$ @Gareth Are you sure? I found this one, but that seems to pass through a square corner, and adjusting it in any direction would make it pass through a shaded square. $\endgroup$ – Rand al'Thor Aug 23 at 17:07
  • $\begingroup$ @GarethMcCaughan True, but if it did happen to be nonunique there, that isn't license to say it must be the other possibility! The deduction "either A or B; A leads to multiple solutions, so it must be B" is definitely invalid, because if the puzzle has multiple solutions you can't just pick one arbitrarily. (A different split might have led to a different 'unique' solution!) $\endgroup$ – Deusovi Aug 23 at 17:08
  • $\begingroup$ @Randal'Thor The top edge there is the top left, not R1C3. $\endgroup$ – Deusovi Aug 23 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.