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
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$begingroup$
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).
logical-deduction grid-deduction three-dimensional
$endgroup$
add a comment |
$begingroup$
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).
logical-deduction grid-deduction three-dimensional
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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).
logical-deduction grid-deduction three-dimensional
$endgroup$
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).
logical-deduction grid-deduction three-dimensional
logical-deduction grid-deduction three-dimensional
asked 8 hours ago
Deusovi♦Deusovi
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I've been really enjoying these puzzles. This was the best one yet.
Reasoning:
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:
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.
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.
Another unshaded cell comes from the observation that one of those two must be shaded.
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.
Once more, two diagonally-adjacent shaded cells. No contradiction this time; we merely realize that a couple nearby cells must be unshaded.
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.
Here, I hit a wall. Couldn't figure out the next step for a good ten minutes. But eventually, I saw it:
Sneaky, sneaky isometric stuff! Now, the rest of the puzzle falls into place.
Only one way to block both of these without cutting anything off:
Nice.
New contributor
$endgroup$
3
$begingroup$
That's correct - nicely done! (And I'm glad you've been enjoying them!)
$endgroup$
– Deusovi♦
4 hours ago
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I've been really enjoying these puzzles. This was the best one yet.
Reasoning:
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:
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.
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.
Another unshaded cell comes from the observation that one of those two must be shaded.
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.
Once more, two diagonally-adjacent shaded cells. No contradiction this time; we merely realize that a couple nearby cells must be unshaded.
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.
Here, I hit a wall. Couldn't figure out the next step for a good ten minutes. But eventually, I saw it:
Sneaky, sneaky isometric stuff! Now, the rest of the puzzle falls into place.
Only one way to block both of these without cutting anything off:
Nice.
New contributor
$endgroup$
3
$begingroup$
That's correct - nicely done! (And I'm glad you've been enjoying them!)
$endgroup$
– Deusovi♦
4 hours ago
add a comment |
$begingroup$
I've been really enjoying these puzzles. This was the best one yet.
Reasoning:
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:
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.
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.
Another unshaded cell comes from the observation that one of those two must be shaded.
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.
Once more, two diagonally-adjacent shaded cells. No contradiction this time; we merely realize that a couple nearby cells must be unshaded.
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.
Here, I hit a wall. Couldn't figure out the next step for a good ten minutes. But eventually, I saw it:
Sneaky, sneaky isometric stuff! Now, the rest of the puzzle falls into place.
Only one way to block both of these without cutting anything off:
Nice.
New contributor
$endgroup$
3
$begingroup$
That's correct - nicely done! (And I'm glad you've been enjoying them!)
$endgroup$
– Deusovi♦
4 hours ago
add a comment |
$begingroup$
I've been really enjoying these puzzles. This was the best one yet.
Reasoning:
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:
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.
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.
Another unshaded cell comes from the observation that one of those two must be shaded.
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.
Once more, two diagonally-adjacent shaded cells. No contradiction this time; we merely realize that a couple nearby cells must be unshaded.
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.
Here, I hit a wall. Couldn't figure out the next step for a good ten minutes. But eventually, I saw it:
Sneaky, sneaky isometric stuff! Now, the rest of the puzzle falls into place.
Only one way to block both of these without cutting anything off:
Nice.
New contributor
$endgroup$
I've been really enjoying these puzzles. This was the best one yet.
Reasoning:
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:
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.
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.
Another unshaded cell comes from the observation that one of those two must be shaded.
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.
Once more, two diagonally-adjacent shaded cells. No contradiction this time; we merely realize that a couple nearby cells must be unshaded.
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.
Here, I hit a wall. Couldn't figure out the next step for a good ten minutes. But eventually, I saw it:
Sneaky, sneaky isometric stuff! Now, the rest of the puzzle falls into place.
Only one way to block both of these without cutting anything off:
Nice.
New contributor
New contributor
answered 5 hours ago
Khuldraeseth na'BaryaKhuldraeseth na'Barya
1715 bronze badges
1715 bronze badges
New contributor
New contributor
3
$begingroup$
That's correct - nicely done! (And I'm glad you've been enjoying them!)
$endgroup$
– Deusovi♦
4 hours ago
add a comment |
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
add a comment |
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$begingroup$
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