Isometric Heyacrazy - Now In 3D!Heyacrazy: CrossesHeyawake: An Introductory PuzzleHeyacrazy: CrossesHeyacrazy: ForksHeyacrazy: “LMI”Heyacrazy: CareeningHeyacrazy: No DiagonalsHeyacrazy: Empty SpaceHeyacrazy: DissectionToroidal Heyacrazy: RainstormHeyacrazier: Big Three

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Isometric Heyacrazy - Now In 3D!


Heyacrazy: CrossesHeyawake: An Introductory PuzzleHeyacrazy: CrossesHeyacrazy: ForksHeyacrazy: “LMI”Heyacrazy: CareeningHeyacrazy: No DiagonalsHeyacrazy: Empty SpaceHeyacrazy: DissectionToroidal Heyacrazy: RainstormHeyacrazier: Big Three






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








9












$begingroup$


This is a Heyacrazy puzzle on an isometric grid.



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.




Special rules for the isometric grid:




  • This grid is a three-dimensional structure, not a hexagon. Lines should be interpreted as paths on that structure. So in the below image, the blue path is a straight line (going diagonally across a 1x2 rectangle), and the red path is not (because it bends about a third of the way down).
    enter image description here



enter image description here










share|improve this question









$endgroup$













  • $begingroup$
    Pretty clear, but just to be 100%: "lines [are] paths on that structure" means that the lines run along the 3D surface, not through the 3D structure itself, yeah? So the blue line is "straight" on the unfolded mesh, not straight in 3D space and we don't need to consider lines leaving the surface.
    $endgroup$
    – Alconja
    6 hours ago










  • $begingroup$
    @Alconja Yes, that's correct.
    $endgroup$
    – Deusovi
    6 hours ago

















9












$begingroup$


This is a Heyacrazy puzzle on an isometric grid.



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.




Special rules for the isometric grid:




  • This grid is a three-dimensional structure, not a hexagon. Lines should be interpreted as paths on that structure. So in the below image, the blue path is a straight line (going diagonally across a 1x2 rectangle), and the red path is not (because it bends about a third of the way down).
    enter image description here



enter image description here










share|improve this question









$endgroup$













  • $begingroup$
    Pretty clear, but just to be 100%: "lines [are] paths on that structure" means that the lines run along the 3D surface, not through the 3D structure itself, yeah? So the blue line is "straight" on the unfolded mesh, not straight in 3D space and we don't need to consider lines leaving the surface.
    $endgroup$
    – Alconja
    6 hours ago










  • $begingroup$
    @Alconja Yes, that's correct.
    $endgroup$
    – Deusovi
    6 hours ago













9












9








9





$begingroup$


This is a Heyacrazy puzzle on an isometric grid.



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.




Special rules for the isometric grid:




  • This grid is a three-dimensional structure, not a hexagon. Lines should be interpreted as paths on that structure. So in the below image, the blue path is a straight line (going diagonally across a 1x2 rectangle), and the red path is not (because it bends about a third of the way down).
    enter image description here



enter image description here










share|improve this question









$endgroup$




This is a Heyacrazy puzzle on an isometric grid.



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.




Special rules for the isometric grid:




  • This grid is a three-dimensional structure, not a hexagon. Lines should be interpreted as paths on that structure. So in the below image, the blue path is a straight line (going diagonally across a 1x2 rectangle), and the red path is not (because it bends about a third of the way down).
    enter image description here



enter image description here







logical-deduction grid-deduction three-dimensional






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 8 hours ago









DeusoviDeusovi

74k7 gold badges255 silver badges325 bronze badges




74k7 gold badges255 silver badges325 bronze badges














  • $begingroup$
    Pretty clear, but just to be 100%: "lines [are] paths on that structure" means that the lines run along the 3D surface, not through the 3D structure itself, yeah? So the blue line is "straight" on the unfolded mesh, not straight in 3D space and we don't need to consider lines leaving the surface.
    $endgroup$
    – Alconja
    6 hours ago










  • $begingroup$
    @Alconja Yes, that's correct.
    $endgroup$
    – Deusovi
    6 hours ago
















  • $begingroup$
    Pretty clear, but just to be 100%: "lines [are] paths on that structure" means that the lines run along the 3D surface, not through the 3D structure itself, yeah? So the blue line is "straight" on the unfolded mesh, not straight in 3D space and we don't need to consider lines leaving the surface.
    $endgroup$
    – Alconja
    6 hours ago










  • $begingroup$
    @Alconja Yes, that's correct.
    $endgroup$
    – Deusovi
    6 hours ago















$begingroup$
Pretty clear, but just to be 100%: "lines [are] paths on that structure" means that the lines run along the 3D surface, not through the 3D structure itself, yeah? So the blue line is "straight" on the unfolded mesh, not straight in 3D space and we don't need to consider lines leaving the surface.
$endgroup$
– Alconja
6 hours ago




$begingroup$
Pretty clear, but just to be 100%: "lines [are] paths on that structure" means that the lines run along the 3D surface, not through the 3D structure itself, yeah? So the blue line is "straight" on the unfolded mesh, not straight in 3D space and we don't need to consider lines leaving the surface.
$endgroup$
– Alconja
6 hours ago












$begingroup$
@Alconja Yes, that's correct.
$endgroup$
– Deusovi
6 hours ago




$begingroup$
@Alconja Yes, that's correct.
$endgroup$
– Deusovi
6 hours ago










1 Answer
1






active

oldest

votes


















7













$begingroup$

I've been really enjoying these puzzles. This was the best one yet.




Done!




Reasoning:




Step 1
The first thing we notice is that we have a ton of opposite-color pairs. This observation gives us one unshaded cell immediately, the lowest colored cell in this picture:
Step 2
Now, let's suppose that middle cell is shaded. We can fill in a little neighborhood, but before long, we find ourselves unable to both both of the blue lines. So our supposition was wrong.
Step 3
Another little neighborhood. There's another cell that must be shaded. Also, examining both possible configurations of the bottom part gives us a few more unshaded cells.
Step 4
Another unshaded cell comes from the observation that one of those two must be shaded.
Step 5
We have two options for filling in those two center cells to block that horizontal line through them. Let's give one of them a try. In the upper left, there must be two diagonally-adjacent shaded cells, but either option cuts some unshaded ones off from the rest. So of those center cells, it's the left one that must be shaded.
Step 6
Once more, two diagonally-adjacent shaded cells. No contradiction this time; we merely realize that a couple nearby cells must be unshaded.
Step 7
Consider a line through those blue cells. Clearly, one must be shaded. If it's the leftmost one, then its lower-right neighbor must be as well—any of these three will block off the top. The unshaded cells must therefore connect around the bottom. Also, no two of those three cells can be shaded.
Step 8
Here, I hit a wall. Couldn't figure out the next step for a good ten minutes. But eventually, I saw it:
Step 9
Sneaky, sneaky isometric stuff! Now, the rest of the puzzle falls into place.
Step 10
Only one way to block both of these without cutting anything off:
Done!
Nice.







share|improve this answer








New contributor



Khuldraeseth na'Barya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





$endgroup$










  • 3




    $begingroup$
    That's correct - nicely done! (And I'm glad you've been enjoying them!)
    $endgroup$
    – Deusovi
    4 hours ago














Your Answer








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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7













$begingroup$

I've been really enjoying these puzzles. This was the best one yet.




Done!




Reasoning:




Step 1
The first thing we notice is that we have a ton of opposite-color pairs. This observation gives us one unshaded cell immediately, the lowest colored cell in this picture:
Step 2
Now, let's suppose that middle cell is shaded. We can fill in a little neighborhood, but before long, we find ourselves unable to both both of the blue lines. So our supposition was wrong.
Step 3
Another little neighborhood. There's another cell that must be shaded. Also, examining both possible configurations of the bottom part gives us a few more unshaded cells.
Step 4
Another unshaded cell comes from the observation that one of those two must be shaded.
Step 5
We have two options for filling in those two center cells to block that horizontal line through them. Let's give one of them a try. In the upper left, there must be two diagonally-adjacent shaded cells, but either option cuts some unshaded ones off from the rest. So of those center cells, it's the left one that must be shaded.
Step 6
Once more, two diagonally-adjacent shaded cells. No contradiction this time; we merely realize that a couple nearby cells must be unshaded.
Step 7
Consider a line through those blue cells. Clearly, one must be shaded. If it's the leftmost one, then its lower-right neighbor must be as well—any of these three will block off the top. The unshaded cells must therefore connect around the bottom. Also, no two of those three cells can be shaded.
Step 8
Here, I hit a wall. Couldn't figure out the next step for a good ten minutes. But eventually, I saw it:
Step 9
Sneaky, sneaky isometric stuff! Now, the rest of the puzzle falls into place.
Step 10
Only one way to block both of these without cutting anything off:
Done!
Nice.







share|improve this answer








New contributor



Khuldraeseth na'Barya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





$endgroup$










  • 3




    $begingroup$
    That's correct - nicely done! (And I'm glad you've been enjoying them!)
    $endgroup$
    – Deusovi
    4 hours ago
















7













$begingroup$

I've been really enjoying these puzzles. This was the best one yet.




Done!




Reasoning:




Step 1
The first thing we notice is that we have a ton of opposite-color pairs. This observation gives us one unshaded cell immediately, the lowest colored cell in this picture:
Step 2
Now, let's suppose that middle cell is shaded. We can fill in a little neighborhood, but before long, we find ourselves unable to both both of the blue lines. So our supposition was wrong.
Step 3
Another little neighborhood. There's another cell that must be shaded. Also, examining both possible configurations of the bottom part gives us a few more unshaded cells.
Step 4
Another unshaded cell comes from the observation that one of those two must be shaded.
Step 5
We have two options for filling in those two center cells to block that horizontal line through them. Let's give one of them a try. In the upper left, there must be two diagonally-adjacent shaded cells, but either option cuts some unshaded ones off from the rest. So of those center cells, it's the left one that must be shaded.
Step 6
Once more, two diagonally-adjacent shaded cells. No contradiction this time; we merely realize that a couple nearby cells must be unshaded.
Step 7
Consider a line through those blue cells. Clearly, one must be shaded. If it's the leftmost one, then its lower-right neighbor must be as well—any of these three will block off the top. The unshaded cells must therefore connect around the bottom. Also, no two of those three cells can be shaded.
Step 8
Here, I hit a wall. Couldn't figure out the next step for a good ten minutes. But eventually, I saw it:
Step 9
Sneaky, sneaky isometric stuff! Now, the rest of the puzzle falls into place.
Step 10
Only one way to block both of these without cutting anything off:
Done!
Nice.







share|improve this answer








New contributor



Khuldraeseth na'Barya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





$endgroup$










  • 3




    $begingroup$
    That's correct - nicely done! (And I'm glad you've been enjoying them!)
    $endgroup$
    – Deusovi
    4 hours ago














7














7










7







$begingroup$

I've been really enjoying these puzzles. This was the best one yet.




Done!




Reasoning:




Step 1
The first thing we notice is that we have a ton of opposite-color pairs. This observation gives us one unshaded cell immediately, the lowest colored cell in this picture:
Step 2
Now, let's suppose that middle cell is shaded. We can fill in a little neighborhood, but before long, we find ourselves unable to both both of the blue lines. So our supposition was wrong.
Step 3
Another little neighborhood. There's another cell that must be shaded. Also, examining both possible configurations of the bottom part gives us a few more unshaded cells.
Step 4
Another unshaded cell comes from the observation that one of those two must be shaded.
Step 5
We have two options for filling in those two center cells to block that horizontal line through them. Let's give one of them a try. In the upper left, there must be two diagonally-adjacent shaded cells, but either option cuts some unshaded ones off from the rest. So of those center cells, it's the left one that must be shaded.
Step 6
Once more, two diagonally-adjacent shaded cells. No contradiction this time; we merely realize that a couple nearby cells must be unshaded.
Step 7
Consider a line through those blue cells. Clearly, one must be shaded. If it's the leftmost one, then its lower-right neighbor must be as well—any of these three will block off the top. The unshaded cells must therefore connect around the bottom. Also, no two of those three cells can be shaded.
Step 8
Here, I hit a wall. Couldn't figure out the next step for a good ten minutes. But eventually, I saw it:
Step 9
Sneaky, sneaky isometric stuff! Now, the rest of the puzzle falls into place.
Step 10
Only one way to block both of these without cutting anything off:
Done!
Nice.







share|improve this answer








New contributor



Khuldraeseth na'Barya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





$endgroup$



I've been really enjoying these puzzles. This was the best one yet.




Done!




Reasoning:




Step 1
The first thing we notice is that we have a ton of opposite-color pairs. This observation gives us one unshaded cell immediately, the lowest colored cell in this picture:
Step 2
Now, let's suppose that middle cell is shaded. We can fill in a little neighborhood, but before long, we find ourselves unable to both both of the blue lines. So our supposition was wrong.
Step 3
Another little neighborhood. There's another cell that must be shaded. Also, examining both possible configurations of the bottom part gives us a few more unshaded cells.
Step 4
Another unshaded cell comes from the observation that one of those two must be shaded.
Step 5
We have two options for filling in those two center cells to block that horizontal line through them. Let's give one of them a try. In the upper left, there must be two diagonally-adjacent shaded cells, but either option cuts some unshaded ones off from the rest. So of those center cells, it's the left one that must be shaded.
Step 6
Once more, two diagonally-adjacent shaded cells. No contradiction this time; we merely realize that a couple nearby cells must be unshaded.
Step 7
Consider a line through those blue cells. Clearly, one must be shaded. If it's the leftmost one, then its lower-right neighbor must be as well—any of these three will block off the top. The unshaded cells must therefore connect around the bottom. Also, no two of those three cells can be shaded.
Step 8
Here, I hit a wall. Couldn't figure out the next step for a good ten minutes. But eventually, I saw it:
Step 9
Sneaky, sneaky isometric stuff! Now, the rest of the puzzle falls into place.
Step 10
Only one way to block both of these without cutting anything off:
Done!
Nice.








share|improve this answer








New contributor



Khuldraeseth na'Barya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share|improve this answer



share|improve this answer






New contributor



Khuldraeseth na'Barya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








answered 5 hours ago









Khuldraeseth na'BaryaKhuldraeseth na'Barya

1715 bronze badges




1715 bronze badges




New contributor



Khuldraeseth na'Barya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




New contributor




Khuldraeseth na'Barya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • 3




    $begingroup$
    That's correct - nicely done! (And I'm glad you've been enjoying them!)
    $endgroup$
    – Deusovi
    4 hours ago













  • 3




    $begingroup$
    That's correct - nicely done! (And I'm glad you've been enjoying them!)
    $endgroup$
    – Deusovi
    4 hours ago








3




3




$begingroup$
That's correct - nicely done! (And I'm glad you've been enjoying them!)
$endgroup$
– Deusovi
4 hours ago





$begingroup$
That's correct - nicely done! (And I'm glad you've been enjoying them!)
$endgroup$
– Deusovi
4 hours ago


















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