Nurikabe: SixesHexagonal KurokuronStatue Park: FiveStatue Park: Knight's LinesA Sincere Nurikabe Puzzle3D Statue Park: U shapesStatue Park: Apollo 11Heyawake: An Introductory PuzzleQuick Nurikabe Puzzle: $4$sNurikabe minicubes: the Headache, the Panache, the ApacheQuick Kurodoko Puzzle: Threes and Triples
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Nurikabe: Sixes
Hexagonal KurokuronStatue Park: FiveStatue Park: Knight's LinesA Sincere Nurikabe Puzzle3D Statue Park: U shapesStatue Park: Apollo 11Heyawake: An Introductory PuzzleQuick Nurikabe Puzzle: $4$sNurikabe minicubes: the Headache, the Panache, the ApacheQuick Kurodoko Puzzle: Threes and Triples
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$begingroup$
This is a Nurikabe puzzle. The goal is to paint some cells black so that the resulting grid satisfies the rules1 of Nurikabe:
- Numbered cells are white.
- White cells are divided into regions, all of which contain exactly one number. The number indicates how many white cells there are in that region.
- Regions of white cells cannot be adjacent to one another, but they may touch at a corner.
- Black cells must all be orthogonally connected.
- There are no groups of black cells that form a $2times2$ square anywhere in the grid.
1 Paraphrased from the original rules on Nikoli
grid-deduction
asked 9 hours ago
jafejafe
36.3k5 gold badges101 silver badges361 bronze badges
$endgroup$
add a comment
|
$begingroup$
This is a Nurikabe puzzle. The goal is to paint some cells black so that the resulting grid satisfies the rules1 of Nurikabe:
- Numbered cells are white.
- White cells are divided into regions, all of which contain exactly one number. The number indicates how many white cells there are in that region.
- Regions of white cells cannot be adjacent to one another, but they may touch at a corner.
- Black cells must all be orthogonally connected.
- There are no groups of black cells that form a $2times2$ square anywhere in the grid.
1 Paraphrased from the original rules on Nikoli
grid-deduction
asked 9 hours ago
jafejafe
36.3k5 gold badges101 silver badges361 bronze badges
$endgroup$
add a comment
|
$begingroup$
This is a Nurikabe puzzle. The goal is to paint some cells black so that the resulting grid satisfies the rules1 of Nurikabe:
- Numbered cells are white.
- White cells are divided into regions, all of which contain exactly one number. The number indicates how many white cells there are in that region.
- Regions of white cells cannot be adjacent to one another, but they may touch at a corner.
- Black cells must all be orthogonally connected.
- There are no groups of black cells that form a $2times2$ square anywhere in the grid.
1 Paraphrased from the original rules on Nikoli
grid-deduction
asked 9 hours ago
jafejafe
36.3k5 gold badges101 silver badges361 bronze badges
$endgroup$
This is a Nurikabe puzzle. The goal is to paint some cells black so that the resulting grid satisfies the rules1 of Nurikabe:
- Numbered cells are white.
- White cells are divided into regions, all of which contain exactly one number. The number indicates how many white cells there are in that region.
- Regions of white cells cannot be adjacent to one another, but they may touch at a corner.
- Black cells must all be orthogonally connected.
- There are no groups of black cells that form a $2times2$ square anywhere in the grid.
1 Paraphrased from the original rules on Nikoli
grid-deduction
grid-deduction
asked 9 hours ago
jafejafe
36.3k5 gold badges101 silver badges361 bronze badges
asked 9 hours ago
jafejafe
36.3k5 gold badges101 silver badges361 bronze badges
asked 9 hours ago
jafejafe
36.3k5 gold badges101 silver badges361 bronze badges
asked 9 hours ago
jafejafe
36.3k5 gold badges101 silver badges361 bronze badges
asked 9 hours ago
jafejafe
36.3k5 gold badges101 silver badges361 bronze badges
36.3k5 gold badges101 silver badges361 bronze badges
add a comment
|
add a comment
|
1 Answer
1
active
oldest
votes
$begingroup$
First, any two numbered cells that are one space apart must be separated by a black cell, which gives us the top-right, and given the "all black cells must be connected" condition one straightforwardly (and stair-wise) gets to
Again, from here the continuation still follows from the same rule applied to the bottom-right, until
Now note that the orange square below cannot be white, it's too far away from any numbered cell, and therefore the red one nearby cannot be black, because of the 2x2 rule. Moreover, there is no way the red cell can be connected to another 6 than the one on its right, because the one on its right would not have 6 friends.
We therefore get this, right after a second application of the 2x2 rule:
From here, there is only one way to connect the black pieces while leaving enough space for the top 6 to have friends:
answered 8 hours ago
Arnaud MortierArnaud Mortier
5,95314 silver badges50 bronze badges
$endgroup$
2
$begingroup$
You beat me to it. I've played quite a few Nurikabes, and this one was not very hard but good fun. I must confess that at the point in your explanation with the orange square, I just saw the rest of solution without reasoning it through exactly, just by knowing the black parts had to be connected together around the remaining islands.
$endgroup$
– Jaap Scherphuis
8 hours ago
3
$begingroup$
@JaapScherphuis It was quite fun indeed! I forbid myself to "see" things, as I was scolded by Deusovi once for doing that :)
$endgroup$
– Arnaud Mortier
8 hours ago
add a comment
|
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1 Answer
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1 Answer
1
active
oldest
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votes
$begingroup$
First, any two numbered cells that are one space apart must be separated by a black cell, which gives us the top-right, and given the "all black cells must be connected" condition one straightforwardly (and stair-wise) gets to
Again, from here the continuation still follows from the same rule applied to the bottom-right, until
Now note that the orange square below cannot be white, it's too far away from any numbered cell, and therefore the red one nearby cannot be black, because of the 2x2 rule. Moreover, there is no way the red cell can be connected to another 6 than the one on its right, because the one on its right would not have 6 friends.
We therefore get this, right after a second application of the 2x2 rule:
From here, there is only one way to connect the black pieces while leaving enough space for the top 6 to have friends:
answered 8 hours ago
Arnaud MortierArnaud Mortier
5,95314 silver badges50 bronze badges
$endgroup$
2
$begingroup$
You beat me to it. I've played quite a few Nurikabes, and this one was not very hard but good fun. I must confess that at the point in your explanation with the orange square, I just saw the rest of solution without reasoning it through exactly, just by knowing the black parts had to be connected together around the remaining islands.
$endgroup$
– Jaap Scherphuis
8 hours ago
3
$begingroup$
@JaapScherphuis It was quite fun indeed! I forbid myself to "see" things, as I was scolded by Deusovi once for doing that :)
$endgroup$
– Arnaud Mortier
8 hours ago
add a comment
|
$begingroup$
First, any two numbered cells that are one space apart must be separated by a black cell, which gives us the top-right, and given the "all black cells must be connected" condition one straightforwardly (and stair-wise) gets to
Again, from here the continuation still follows from the same rule applied to the bottom-right, until
Now note that the orange square below cannot be white, it's too far away from any numbered cell, and therefore the red one nearby cannot be black, because of the 2x2 rule. Moreover, there is no way the red cell can be connected to another 6 than the one on its right, because the one on its right would not have 6 friends.
We therefore get this, right after a second application of the 2x2 rule:
From here, there is only one way to connect the black pieces while leaving enough space for the top 6 to have friends:
answered 8 hours ago
Arnaud MortierArnaud Mortier
5,95314 silver badges50 bronze badges
$endgroup$
2
$begingroup$
You beat me to it. I've played quite a few Nurikabes, and this one was not very hard but good fun. I must confess that at the point in your explanation with the orange square, I just saw the rest of solution without reasoning it through exactly, just by knowing the black parts had to be connected together around the remaining islands.
$endgroup$
– Jaap Scherphuis
8 hours ago
3
$begingroup$
@JaapScherphuis It was quite fun indeed! I forbid myself to "see" things, as I was scolded by Deusovi once for doing that :)
$endgroup$
– Arnaud Mortier
8 hours ago
add a comment
|
$begingroup$
First, any two numbered cells that are one space apart must be separated by a black cell, which gives us the top-right, and given the "all black cells must be connected" condition one straightforwardly (and stair-wise) gets to
Again, from here the continuation still follows from the same rule applied to the bottom-right, until
Now note that the orange square below cannot be white, it's too far away from any numbered cell, and therefore the red one nearby cannot be black, because of the 2x2 rule. Moreover, there is no way the red cell can be connected to another 6 than the one on its right, because the one on its right would not have 6 friends.
We therefore get this, right after a second application of the 2x2 rule:
From here, there is only one way to connect the black pieces while leaving enough space for the top 6 to have friends:
answered 8 hours ago
Arnaud MortierArnaud Mortier
5,95314 silver badges50 bronze badges
$endgroup$
First, any two numbered cells that are one space apart must be separated by a black cell, which gives us the top-right, and given the "all black cells must be connected" condition one straightforwardly (and stair-wise) gets to
Again, from here the continuation still follows from the same rule applied to the bottom-right, until
Now note that the orange square below cannot be white, it's too far away from any numbered cell, and therefore the red one nearby cannot be black, because of the 2x2 rule. Moreover, there is no way the red cell can be connected to another 6 than the one on its right, because the one on its right would not have 6 friends.
We therefore get this, right after a second application of the 2x2 rule:
From here, there is only one way to connect the black pieces while leaving enough space for the top 6 to have friends:
answered 8 hours ago
Arnaud MortierArnaud Mortier
5,95314 silver badges50 bronze badges
answered 8 hours ago
Arnaud MortierArnaud Mortier
5,95314 silver badges50 bronze badges
answered 8 hours ago
Arnaud MortierArnaud Mortier
5,95314 silver badges50 bronze badges
answered 8 hours ago
Arnaud MortierArnaud Mortier
5,95314 silver badges50 bronze badges
5,95314 silver badges50 bronze badges
2
$begingroup$
You beat me to it. I've played quite a few Nurikabes, and this one was not very hard but good fun. I must confess that at the point in your explanation with the orange square, I just saw the rest of solution without reasoning it through exactly, just by knowing the black parts had to be connected together around the remaining islands.
$endgroup$
– Jaap Scherphuis
8 hours ago
3
$begingroup$
@JaapScherphuis It was quite fun indeed! I forbid myself to "see" things, as I was scolded by Deusovi once for doing that :)
$endgroup$
– Arnaud Mortier
8 hours ago
add a comment
|
2
$begingroup$
You beat me to it. I've played quite a few Nurikabes, and this one was not very hard but good fun. I must confess that at the point in your explanation with the orange square, I just saw the rest of solution without reasoning it through exactly, just by knowing the black parts had to be connected together around the remaining islands.
$endgroup$
– Jaap Scherphuis
8 hours ago
3
$begingroup$
@JaapScherphuis It was quite fun indeed! I forbid myself to "see" things, as I was scolded by Deusovi once for doing that :)
$endgroup$
– Arnaud Mortier
8 hours ago
2
2
$begingroup$
You beat me to it. I've played quite a few Nurikabes, and this one was not very hard but good fun. I must confess that at the point in your explanation with the orange square, I just saw the rest of solution without reasoning it through exactly, just by knowing the black parts had to be connected together around the remaining islands.
$endgroup$
– Jaap Scherphuis
8 hours ago
$begingroup$
You beat me to it. I've played quite a few Nurikabes, and this one was not very hard but good fun. I must confess that at the point in your explanation with the orange square, I just saw the rest of solution without reasoning it through exactly, just by knowing the black parts had to be connected together around the remaining islands.
$endgroup$
– Jaap Scherphuis
8 hours ago
3
3
$begingroup$
@JaapScherphuis It was quite fun indeed! I forbid myself to "see" things, as I was scolded by Deusovi once for doing that :)
$endgroup$
– Arnaud Mortier
8 hours ago
$begingroup$
@JaapScherphuis It was quite fun indeed! I forbid myself to "see" things, as I was scolded by Deusovi once for doing that :)
$endgroup$
– Arnaud Mortier
8 hours ago
add a comment
|
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