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Toroidal Heyacrazy: Rainstorm

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Toroidal Heyacrazy: Rainstorm

Heyacrazy: CrossesStatue Park: Knight's LinesHeyawake: An Introductory PuzzleHeyawacky: Ace of CupsHeyacrazy: CrossesHeyacrazy: ForksHeyacrazy: “LMI”Heyacrazy: CareeningHeyacrazy: No DiagonalsHeyacrazy: Empty SpaceHeyacrazy: Dissection

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3

$begingroup$

This is a Toroidal 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.

Additional rule for Toroidal Heyacrazy:

• 
  The grid 'wraps around' on both sides, with the right side connecting to the left side and the top connecting to the bottom. Red lines mark where the grid wraps around, but otherwise have no effect. (In particular, they do not count as 'borders' for the previous rule). Nine copies of the grid have been provided for your convenience.

^{The vanilla puzzles are done, now on to the variants! This is the first of three variant puzzles I've made - I noticed that in standard Heyacrazy puzzles, you can often figure out that the corner is unshaded simply because there are no adjacent walls, and so nothing can force it to be shaded. This is a puzzle that doesn't suffer from that particular problem.}

logical-deduction grid-deduction

share|improve this question

edited 8 hours ago

Deusovi

asked 9 hours ago

Deusovi♦Deusovi

73.9k77 gold badges255255 silver badges325325 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

This is a Toroidal 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.

Additional rule for Toroidal Heyacrazy:

• 
  The grid 'wraps around' on both sides, with the right side connecting to the left side and the top connecting to the bottom. Red lines mark where the grid wraps around, but otherwise have no effect. (In particular, they do not count as 'borders' for the previous rule). Nine copies of the grid have been provided for your convenience.

^{The vanilla puzzles are done, now on to the variants! This is the first of three variant puzzles I've made - I noticed that in standard Heyacrazy puzzles, you can often figure out that the corner is unshaded simply because there are no adjacent walls, and so nothing can force it to be shaded. This is a puzzle that doesn't suffer from that particular problem.}

logical-deduction grid-deduction

share|improve this question

edited 8 hours ago

Deusovi

asked 9 hours ago

Deusovi♦Deusovi

73.9k77 gold badges255255 silver badges325325 bronze badges

$endgroup$

add a comment | 

$begingroup$

This is a Toroidal 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.

Additional rule for Toroidal Heyacrazy:

• 
  The grid 'wraps around' on both sides, with the right side connecting to the left side and the top connecting to the bottom. Red lines mark where the grid wraps around, but otherwise have no effect. (In particular, they do not count as 'borders' for the previous rule). Nine copies of the grid have been provided for your convenience.

^{The vanilla puzzles are done, now on to the variants! This is the first of three variant puzzles I've made - I noticed that in standard Heyacrazy puzzles, you can often figure out that the corner is unshaded simply because there are no adjacent walls, and so nothing can force it to be shaded. This is a puzzle that doesn't suffer from that particular problem.}

logical-deduction grid-deduction

share|improve this question

edited 8 hours ago

Deusovi

asked 9 hours ago

Deusovi♦Deusovi

73.9k77 gold badges255255 silver badges325325 bronze badges

$endgroup$

share|improve this question

edited 8 hours ago

Deusovi

asked 9 hours ago

Deusovi♦Deusovi

73.9k77 gold badges255255 silver badges325325 bronze badges

share|improve this question

edited 8 hours ago

Deusovi

asked 9 hours ago

Deusovi♦Deusovi

73.9k77 gold badges255255 silver badges325325 bronze badges

asked 9 hours ago

Deusovi♦Deusovi

73.9k77 gold badges255255 silver badges325325 bronze badges

asked 9 hours ago

Deusovi♦Deusovi

73.9k77 gold badges255255 silver badges325325 bronze badges

2 Answers
2

active

oldest

votes

3

$begingroup$

To block both of these lines, either the central cell of the blue rectangle needs to be shaded, or both of its neighbours inside the rectangle.




The latter option does not work. There is no way to block both of these lines now.




So the central cell must be filled in. As a result, a couple of cells emerge which obviously must be shaded.




Need to fill in two cells to block these lines.




Only one way to block this line.




Only one way to block both of these lines without shutting out unshaded cells.




Regardless of how this line is blocked, the circled cell must be empty.




Only one way to block each of these lines now.




Only one legal way to block both these lines.




Only one way to block each of these lines.




Zooming out a little bit, there's only one legal way to block both of these lines.




This cell must be empty to avoid cutting off unshaded cells vertically.




This tricky line in the corner needs to be blocked as well. Only one way to do it.




This one as well (even trickier).




Finally, this one. The last cell needs to be empty to avoid shutting in the unshaded cells.




Final position.

share|improve this answer

answered 7 hours ago

jafejafe

35.7k55 gold badges9999 silver badges359359 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Impressive answer. May I ask what software you are using? I use GIMP but it's not good at drawing straight lines. Also, just a minor nitpick: "avoid cutting off unshaded cells vertically" doesn't seem to be a requirement, since we're on a torus, the unshaded cells may still be connected via the red portal.
  $endgroup$
  – Arnaud Mortier
  6 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArnaudMortier The point is that if you fill in the cell in question, the cells wouldn't be connected. As in, the entire thing would be cut in half vertically because the vertical wall of shaded cells continues across the red border as well. This was all done in Paint, although I do use GIMP from time to time.
  $endgroup$
  – jafe
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean there is a difference between being connected on the torus and being connected in the plane with an infinite number of copies of the square in both directions. I believe that the former is implied here.
  $endgroup$
  – Arnaud Mortier
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Funny we were writing simultaneously. I see what you mean, but still, imagine a donut with a ring on it going around in one direction. The complement of the ring is an annulus, still perfectly connected. In the language of this puzzle, if an entire column was shaded and the rest of the squares wasn't, it would break many rules but not the "unshaded are connected" rule.
  $endgroup$
  – Arnaud Mortier
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, note that this step can be skipped and seen as a direct consequence of the next one where a neighbouring square is shaded.
  $endgroup$
  – Arnaud Mortier
  5 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Solution:

I'm at work and kinda lazy, so I'll just show some stages along the path. From each of these to the next is a fairly easy inference.

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

80.1k33 gold badges202202 silver badges308308 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I think I have seen something that is definitely wrong...
  $endgroup$
  – Gareth McCaughan♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah. Looking at what I had immediately before that error, I think I must just have filled the wrong cell by accident. Revising now.
  $endgroup$
  – Gareth McCaughan♦
  7 hours ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

To block both of these lines, either the central cell of the blue rectangle needs to be shaded, or both of its neighbours inside the rectangle.




The latter option does not work. There is no way to block both of these lines now.




So the central cell must be filled in. As a result, a couple of cells emerge which obviously must be shaded.




Need to fill in two cells to block these lines.




Only one way to block this line.




Only one way to block both of these lines without shutting out unshaded cells.




Regardless of how this line is blocked, the circled cell must be empty.




Only one way to block each of these lines now.




Only one legal way to block both these lines.




Only one way to block each of these lines.




Zooming out a little bit, there's only one legal way to block both of these lines.




This cell must be empty to avoid cutting off unshaded cells vertically.




This tricky line in the corner needs to be blocked as well. Only one way to do it.




This one as well (even trickier).




Finally, this one. The last cell needs to be empty to avoid shutting in the unshaded cells.




Final position.

share|improve this answer

answered 7 hours ago

jafejafe

35.7k55 gold badges9999 silver badges359359 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Impressive answer. May I ask what software you are using? I use GIMP but it's not good at drawing straight lines. Also, just a minor nitpick: "avoid cutting off unshaded cells vertically" doesn't seem to be a requirement, since we're on a torus, the unshaded cells may still be connected via the red portal.
  $endgroup$
  – Arnaud Mortier
  6 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArnaudMortier The point is that if you fill in the cell in question, the cells wouldn't be connected. As in, the entire thing would be cut in half vertically because the vertical wall of shaded cells continues across the red border as well. This was all done in Paint, although I do use GIMP from time to time.
  $endgroup$
  – jafe
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean there is a difference between being connected on the torus and being connected in the plane with an infinite number of copies of the square in both directions. I believe that the former is implied here.
  $endgroup$
  – Arnaud Mortier
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Funny we were writing simultaneously. I see what you mean, but still, imagine a donut with a ring on it going around in one direction. The complement of the ring is an annulus, still perfectly connected. In the language of this puzzle, if an entire column was shaded and the rest of the squares wasn't, it would break many rules but not the "unshaded are connected" rule.
  $endgroup$
  – Arnaud Mortier
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, note that this step can be skipped and seen as a direct consequence of the next one where a neighbouring square is shaded.
  $endgroup$
  – Arnaud Mortier
  5 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

To block both of these lines, either the central cell of the blue rectangle needs to be shaded, or both of its neighbours inside the rectangle.




The latter option does not work. There is no way to block both of these lines now.




So the central cell must be filled in. As a result, a couple of cells emerge which obviously must be shaded.




Need to fill in two cells to block these lines.




Only one way to block this line.




Only one way to block both of these lines without shutting out unshaded cells.




Regardless of how this line is blocked, the circled cell must be empty.




Only one way to block each of these lines now.




Only one legal way to block both these lines.




Only one way to block each of these lines.




Zooming out a little bit, there's only one legal way to block both of these lines.




This cell must be empty to avoid cutting off unshaded cells vertically.




This tricky line in the corner needs to be blocked as well. Only one way to do it.




This one as well (even trickier).




Finally, this one. The last cell needs to be empty to avoid shutting in the unshaded cells.




Final position.

share|improve this answer

answered 7 hours ago

jafejafe

35.7k55 gold badges9999 silver badges359359 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Impressive answer. May I ask what software you are using? I use GIMP but it's not good at drawing straight lines. Also, just a minor nitpick: "avoid cutting off unshaded cells vertically" doesn't seem to be a requirement, since we're on a torus, the unshaded cells may still be connected via the red portal.
  $endgroup$
  – Arnaud Mortier
  6 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArnaudMortier The point is that if you fill in the cell in question, the cells wouldn't be connected. As in, the entire thing would be cut in half vertically because the vertical wall of shaded cells continues across the red border as well. This was all done in Paint, although I do use GIMP from time to time.
  $endgroup$
  – jafe
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean there is a difference between being connected on the torus and being connected in the plane with an infinite number of copies of the square in both directions. I believe that the former is implied here.
  $endgroup$
  – Arnaud Mortier
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Funny we were writing simultaneously. I see what you mean, but still, imagine a donut with a ring on it going around in one direction. The complement of the ring is an annulus, still perfectly connected. In the language of this puzzle, if an entire column was shaded and the rest of the squares wasn't, it would break many rules but not the "unshaded are connected" rule.
  $endgroup$
  – Arnaud Mortier
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, note that this step can be skipped and seen as a direct consequence of the next one where a neighbouring square is shaded.
  $endgroup$
  – Arnaud Mortier
  5 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

To block both of these lines, either the central cell of the blue rectangle needs to be shaded, or both of its neighbours inside the rectangle.




The latter option does not work. There is no way to block both of these lines now.




So the central cell must be filled in. As a result, a couple of cells emerge which obviously must be shaded.




Need to fill in two cells to block these lines.




Only one way to block this line.




Only one way to block both of these lines without shutting out unshaded cells.




Regardless of how this line is blocked, the circled cell must be empty.




Only one way to block each of these lines now.




Only one legal way to block both these lines.




Only one way to block each of these lines.




Zooming out a little bit, there's only one legal way to block both of these lines.




This cell must be empty to avoid cutting off unshaded cells vertically.




This tricky line in the corner needs to be blocked as well. Only one way to do it.




This one as well (even trickier).




Finally, this one. The last cell needs to be empty to avoid shutting in the unshaded cells.




Final position.

share|improve this answer

answered 7 hours ago

jafejafe

35.7k55 gold badges9999 silver badges359359 bronze badges

$endgroup$

share|improve this answer

answered 7 hours ago

jafejafe

35.7k55 gold badges9999 silver badges359359 bronze badges

answered 7 hours ago

jafejafe

35.7k55 gold badges9999 silver badges359359 bronze badges

answered 7 hours ago

jafejafe

35.7k55 gold badges9999 silver badges359359 bronze badges

2

$begingroup$

Solution:

I'm at work and kinda lazy, so I'll just show some stages along the path. From each of these to the next is a fairly easy inference.

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

80.1k33 gold badges202202 silver badges308308 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I think I have seen something that is definitely wrong...
  $endgroup$
  – Gareth McCaughan♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah. Looking at what I had immediately before that error, I think I must just have filled the wrong cell by accident. Revising now.
  $endgroup$
  – Gareth McCaughan♦
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Solution:

I'm at work and kinda lazy, so I'll just show some stages along the path. From each of these to the next is a fairly easy inference.

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

80.1k33 gold badges202202 silver badges308308 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I think I have seen something that is definitely wrong...
  $endgroup$
  – Gareth McCaughan♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah. Looking at what I had immediately before that error, I think I must just have filled the wrong cell by accident. Revising now.
  $endgroup$
  – Gareth McCaughan♦
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Solution:

I'm at work and kinda lazy, so I'll just show some stages along the path. From each of these to the next is a fairly easy inference.

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

80.1k33 gold badges202202 silver badges308308 bronze badges

$endgroup$

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

80.1k33 gold badges202202 silver badges308308 bronze badges

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

80.1k33 gold badges202202 silver badges308308 bronze badges

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

80.1k33 gold badges202202 silver badges308308 bronze badges