A new 4 x 4 arrow puzzle
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Can anyone help me with solving this puzzle:
Draw arrows in all fields around the diagram in a way that every arrow is pointing at least one number inside. The numbers inside the boxes equal the number of arrows pointing at them. The arrows can point horizontally, vertically or diagonally.
Here is an example showing how to solve this type of puzzle.
![arrows puzzle[1]](https://i.stack.imgur.com/EzcXY.jpg)
This is from a job interview so I have no source.
grid-deduction
edited Dec 4 '18 at 16:53
GentlePurpleRain♦
16.9k568136
asked Nov 13 '18 at 14:29
TeditedutuTeditedutu
562
$endgroup$
add a comment |
$begingroup$
Can anyone help me with solving this puzzle:
Draw arrows in all fields around the diagram in a way that every arrow is pointing at least one number inside. The numbers inside the boxes equal the number of arrows pointing at them. The arrows can point horizontally, vertically or diagonally.
Here is an example showing how to solve this type of puzzle.
This is from a job interview so I have no source.
grid-deduction
edited Dec 4 '18 at 16:53
GentlePurpleRain♦
16.9k568136
asked Nov 13 '18 at 14:29
TeditedutuTeditedutu
562
$endgroup$
$begingroup$
I believe there are 2^8 + (3^8) = 6817 combinations of puzzle solutions, so in principle this can be bruteforced.
$endgroup$
– Parseltongue
Nov 13 '18 at 16:18
2
$begingroup$
@Parseltongue I think your math is off; it should actually be 2^8 times 3^8 = 1679616, shouldn't it? Still trivial for a modern computer, but obviously it scales far too quickly to be a feasible approach at higher size squares. This makes me curious whether this is a NP problem; I can see a way of mapping it to a Boolean satisfiability problem, which would be NP (though that's certainly not the most efficient approach), and the sums can probably provide more information, perhaps revealing an algorithm in P. But it reminds me of Sudoku, which is in NP... I wonder.
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– Graham
Nov 13 '18 at 20:58
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You're right! My mistake. And those were my thoughts as well... I immediately started trying to code up an algorithm to solve this, but decided Brute Force was easiest.
$endgroup$
– Parseltongue
Nov 13 '18 at 21:04
$begingroup$
Here is an "assisted solver" I whipped up to play around with this puzzle. It's just a bare-bones set of<button>and<span>elements in a grid. The values in the grid update live as you rotate the arrows. I see that the answer has already been posted, and a tool like this probably already exists, but perhaps someone will find this one useful.
$endgroup$
– benj2240
Nov 13 '18 at 23:15
$begingroup$
@benj2240 - that is absolutely amazing! Learned a lot reading the code.
$endgroup$
– Parseltongue
Nov 14 '18 at 18:57
add a comment |
$begingroup$
Can anyone help me with solving this puzzle:
Draw arrows in all fields around the diagram in a way that every arrow is pointing at least one number inside. The numbers inside the boxes equal the number of arrows pointing at them. The arrows can point horizontally, vertically or diagonally.
Here is an example showing how to solve this type of puzzle.
This is from a job interview so I have no source.
grid-deduction
edited Dec 4 '18 at 16:53
GentlePurpleRain♦
16.9k568136
asked Nov 13 '18 at 14:29
TeditedutuTeditedutu
562
$endgroup$
Can anyone help me with solving this puzzle:
Draw arrows in all fields around the diagram in a way that every arrow is pointing at least one number inside. The numbers inside the boxes equal the number of arrows pointing at them. The arrows can point horizontally, vertically or diagonally.
Here is an example showing how to solve this type of puzzle.
This is from a job interview so I have no source.
grid-deduction
grid-deduction
edited Dec 4 '18 at 16:53
GentlePurpleRain♦
16.9k568136
asked Nov 13 '18 at 14:29
TeditedutuTeditedutu
562
edited Dec 4 '18 at 16:53
GentlePurpleRain♦
16.9k568136
asked Nov 13 '18 at 14:29
TeditedutuTeditedutu
562
edited Dec 4 '18 at 16:53
GentlePurpleRain♦
16.9k568136
edited Dec 4 '18 at 16:53
GentlePurpleRain♦
16.9k568136
edited Dec 4 '18 at 16:53
GentlePurpleRain♦
16.9k568136
16.9k568136
asked Nov 13 '18 at 14:29
TeditedutuTeditedutu
562
asked Nov 13 '18 at 14:29
TeditedutuTeditedutu
562
asked Nov 13 '18 at 14:29
TeditedutuTeditedutu
562
562
$begingroup$
I believe there are 2^8 + (3^8) = 6817 combinations of puzzle solutions, so in principle this can be bruteforced.
$endgroup$
– Parseltongue
Nov 13 '18 at 16:18
2
$begingroup$
@Parseltongue I think your math is off; it should actually be 2^8 times 3^8 = 1679616, shouldn't it? Still trivial for a modern computer, but obviously it scales far too quickly to be a feasible approach at higher size squares. This makes me curious whether this is a NP problem; I can see a way of mapping it to a Boolean satisfiability problem, which would be NP (though that's certainly not the most efficient approach), and the sums can probably provide more information, perhaps revealing an algorithm in P. But it reminds me of Sudoku, which is in NP... I wonder.
$endgroup$
– Graham
Nov 13 '18 at 20:58
$begingroup$
You're right! My mistake. And those were my thoughts as well... I immediately started trying to code up an algorithm to solve this, but decided Brute Force was easiest.
$endgroup$
– Parseltongue
Nov 13 '18 at 21:04
$begingroup$
Here is an "assisted solver" I whipped up to play around with this puzzle. It's just a bare-bones set of<button>and<span>elements in a grid. The values in the grid update live as you rotate the arrows. I see that the answer has already been posted, and a tool like this probably already exists, but perhaps someone will find this one useful.
$endgroup$
– benj2240
Nov 13 '18 at 23:15
$begingroup$
@benj2240 - that is absolutely amazing! Learned a lot reading the code.
$endgroup$
– Parseltongue
Nov 14 '18 at 18:57
add a comment |
$begingroup$
I believe there are 2^8 + (3^8) = 6817 combinations of puzzle solutions, so in principle this can be bruteforced.
$endgroup$
– Parseltongue
Nov 13 '18 at 16:18
2
$begingroup$
@Parseltongue I think your math is off; it should actually be 2^8 times 3^8 = 1679616, shouldn't it? Still trivial for a modern computer, but obviously it scales far too quickly to be a feasible approach at higher size squares. This makes me curious whether this is a NP problem; I can see a way of mapping it to a Boolean satisfiability problem, which would be NP (though that's certainly not the most efficient approach), and the sums can probably provide more information, perhaps revealing an algorithm in P. But it reminds me of Sudoku, which is in NP... I wonder.
$endgroup$
– Graham
Nov 13 '18 at 20:58
$begingroup$
You're right! My mistake. And those were my thoughts as well... I immediately started trying to code up an algorithm to solve this, but decided Brute Force was easiest.
$endgroup$
– Parseltongue
Nov 13 '18 at 21:04
$begingroup$
Here is an "assisted solver" I whipped up to play around with this puzzle. It's just a bare-bones set of<button>and<span>elements in a grid. The values in the grid update live as you rotate the arrows. I see that the answer has already been posted, and a tool like this probably already exists, but perhaps someone will find this one useful.
$endgroup$
– benj2240
Nov 13 '18 at 23:15
$begingroup$
@benj2240 - that is absolutely amazing! Learned a lot reading the code.
$endgroup$
– Parseltongue
Nov 14 '18 at 18:57
$begingroup$
I believe there are 2^8 + (3^8) = 6817 combinations of puzzle solutions, so in principle this can be bruteforced.
$endgroup$
– Parseltongue
Nov 13 '18 at 16:18
$begingroup$
I believe there are 2^8 + (3^8) = 6817 combinations of puzzle solutions, so in principle this can be bruteforced.
$endgroup$
– Parseltongue
Nov 13 '18 at 16:18
2
2
$begingroup$
@Parseltongue I think your math is off; it should actually be 2^8 times 3^8 = 1679616, shouldn't it? Still trivial for a modern computer, but obviously it scales far too quickly to be a feasible approach at higher size squares. This makes me curious whether this is a NP problem; I can see a way of mapping it to a Boolean satisfiability problem, which would be NP (though that's certainly not the most efficient approach), and the sums can probably provide more information, perhaps revealing an algorithm in P. But it reminds me of Sudoku, which is in NP... I wonder.
$endgroup$
– Graham
Nov 13 '18 at 20:58
$begingroup$
@Parseltongue I think your math is off; it should actually be 2^8 times 3^8 = 1679616, shouldn't it? Still trivial for a modern computer, but obviously it scales far too quickly to be a feasible approach at higher size squares. This makes me curious whether this is a NP problem; I can see a way of mapping it to a Boolean satisfiability problem, which would be NP (though that's certainly not the most efficient approach), and the sums can probably provide more information, perhaps revealing an algorithm in P. But it reminds me of Sudoku, which is in NP... I wonder.
$endgroup$
– Graham
Nov 13 '18 at 20:58
$begingroup$
You're right! My mistake. And those were my thoughts as well... I immediately started trying to code up an algorithm to solve this, but decided Brute Force was easiest.
$endgroup$
– Parseltongue
Nov 13 '18 at 21:04
$begingroup$
You're right! My mistake. And those were my thoughts as well... I immediately started trying to code up an algorithm to solve this, but decided Brute Force was easiest.
$endgroup$
– Parseltongue
Nov 13 '18 at 21:04
$begingroup$
Here is an "assisted solver" I whipped up to play around with this puzzle. It's just a bare-bones set of
<button> and <span> elements in a grid. The values in the grid update live as you rotate the arrows. I see that the answer has already been posted, and a tool like this probably already exists, but perhaps someone will find this one useful.$endgroup$
– benj2240
Nov 13 '18 at 23:15
$begingroup$
Here is an "assisted solver" I whipped up to play around with this puzzle. It's just a bare-bones set of
<button> and <span> elements in a grid. The values in the grid update live as you rotate the arrows. I see that the answer has already been posted, and a tool like this probably already exists, but perhaps someone will find this one useful.$endgroup$
– benj2240
Nov 13 '18 at 23:15
$begingroup$
@benj2240 - that is absolutely amazing! Learned a lot reading the code.
$endgroup$
– Parseltongue
Nov 14 '18 at 18:57
$begingroup$
@benj2240 - that is absolutely amazing! Learned a lot reading the code.
$endgroup$
– Parseltongue
Nov 14 '18 at 18:57
add a comment |
4 Answers
4
active
oldest
votes
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The first clue I spotted was:
The second column has exactly one vertical arrow.
This can be proven as:
It can't have two vertical arrows because of the 1 in it. If it has zero vertical arrows, then the 4 at (2,2) is forced, and then the 4 at (2,4) is forced. But then the 4 at (4,4) can't be achieved, because of it's six arrow squares, three have already been used: (4,0), (0,4) and (3,5).
Secondly you can spot that:
As the 1 in the second column is already covered, the arrows on row 1 must point diagonally downwards.
answered Nov 13 '18 at 18:31
JonMark PerryJonMark Perry
18.6k63888
$endgroup$
add a comment |
$begingroup$
I think this is the answer desired:
I started by
Assuming at least one arrows each in an inverted A shape, based on the prevalence of 5s and 4s, in columns 1 and 4, and rows 2 and 4. After that, it was primarily guesswork, placing lines, then working backwards to determine what arrows would cause those lines.
answered Nov 13 '18 at 16:27
SconibulusSconibulus
14.4k128100
$endgroup$
$begingroup$
This is indeed the correct answer. bravo!
$endgroup$
– ABcDexter
Nov 13 '18 at 19:03
add a comment |
$begingroup$
This solution:
- places an arrow in every field around the diagram;
- each arrow points at one or more numbers inside;
- satisfies the condition of having the numbers in the boxes equaling the numbers of arrows pointing at them!
That being said, the person who is giving you the interview may not like it.
:P
answered Nov 13 '18 at 15:18
Excited RaichuExcited Raichu
6,43521166
$endgroup$
$begingroup$
These are a standard puzzle form: puzzlepicnic.com/puzzles... but something about this puzzle makes me believe it's unsolvable. I wonder if it's just one of those task-persistence measures given in job interviews to see how long you'll try before you give up. Or, if it is solvable, it doesn't have characteristics of other arrow puzzles, which generally have one square with only one logical set of arrows possible.
$endgroup$
– Parseltongue
Nov 13 '18 at 15:32
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I too thought about odd angles for arrows, but you sir are taking it to another level x)
$endgroup$
– Cashbee
Nov 13 '18 at 15:35
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Bottom left corner?
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– Greg
Nov 13 '18 at 15:49
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@Greg all of the outer squares it touches have an arrow pointing at it
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– Excited Raichu
Nov 13 '18 at 15:51
$begingroup$
Wow, that is next level!
$endgroup$
– Greg
Nov 13 '18 at 15:58
|
show 1 more comment
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This is a possibility:
I'm sure there are others.
edited Dec 4 '18 at 16:59
GentlePurpleRain♦
16.9k568136
answered Nov 14 '18 at 10:45
AnonAnon
111
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The first clue I spotted was:
The second column has exactly one vertical arrow.
This can be proven as:
It can't have two vertical arrows because of the 1 in it. If it has zero vertical arrows, then the 4 at (2,2) is forced, and then the 4 at (2,4) is forced. But then the 4 at (4,4) can't be achieved, because of it's six arrow squares, three have already been used: (4,0), (0,4) and (3,5).
Secondly you can spot that:
As the 1 in the second column is already covered, the arrows on row 1 must point diagonally downwards.
answered Nov 13 '18 at 18:31
JonMark PerryJonMark Perry
18.6k63888
$endgroup$
add a comment |
$begingroup$
The first clue I spotted was:
The second column has exactly one vertical arrow.
This can be proven as:
It can't have two vertical arrows because of the 1 in it. If it has zero vertical arrows, then the 4 at (2,2) is forced, and then the 4 at (2,4) is forced. But then the 4 at (4,4) can't be achieved, because of it's six arrow squares, three have already been used: (4,0), (0,4) and (3,5).
Secondly you can spot that:
As the 1 in the second column is already covered, the arrows on row 1 must point diagonally downwards.
answered Nov 13 '18 at 18:31
JonMark PerryJonMark Perry
18.6k63888
$endgroup$
add a comment |
$begingroup$
The first clue I spotted was:
The second column has exactly one vertical arrow.
This can be proven as:
It can't have two vertical arrows because of the 1 in it. If it has zero vertical arrows, then the 4 at (2,2) is forced, and then the 4 at (2,4) is forced. But then the 4 at (4,4) can't be achieved, because of it's six arrow squares, three have already been used: (4,0), (0,4) and (3,5).
Secondly you can spot that:
As the 1 in the second column is already covered, the arrows on row 1 must point diagonally downwards.
answered Nov 13 '18 at 18:31
JonMark PerryJonMark Perry
18.6k63888
$endgroup$
The first clue I spotted was:
The second column has exactly one vertical arrow.
This can be proven as:
It can't have two vertical arrows because of the 1 in it. If it has zero vertical arrows, then the 4 at (2,2) is forced, and then the 4 at (2,4) is forced. But then the 4 at (4,4) can't be achieved, because of it's six arrow squares, three have already been used: (4,0), (0,4) and (3,5).
Secondly you can spot that:
As the 1 in the second column is already covered, the arrows on row 1 must point diagonally downwards.
answered Nov 13 '18 at 18:31
JonMark PerryJonMark Perry
18.6k63888
answered Nov 13 '18 at 18:31
JonMark PerryJonMark Perry
18.6k63888
answered Nov 13 '18 at 18:31
JonMark PerryJonMark Perry
18.6k63888
answered Nov 13 '18 at 18:31
JonMark PerryJonMark Perry
18.6k63888
18.6k63888
add a comment |
add a comment |
$begingroup$
I think this is the answer desired:
I started by
Assuming at least one arrows each in an inverted A shape, based on the prevalence of 5s and 4s, in columns 1 and 4, and rows 2 and 4. After that, it was primarily guesswork, placing lines, then working backwards to determine what arrows would cause those lines.
answered Nov 13 '18 at 16:27
SconibulusSconibulus
14.4k128100
$endgroup$
$begingroup$
This is indeed the correct answer. bravo!
$endgroup$
– ABcDexter
Nov 13 '18 at 19:03
add a comment |
$begingroup$
I think this is the answer desired:
I started by
Assuming at least one arrows each in an inverted A shape, based on the prevalence of 5s and 4s, in columns 1 and 4, and rows 2 and 4. After that, it was primarily guesswork, placing lines, then working backwards to determine what arrows would cause those lines.
answered Nov 13 '18 at 16:27
SconibulusSconibulus
14.4k128100
$endgroup$
$begingroup$
This is indeed the correct answer. bravo!
$endgroup$
– ABcDexter
Nov 13 '18 at 19:03
add a comment |
$begingroup$
I think this is the answer desired:
I started by
Assuming at least one arrows each in an inverted A shape, based on the prevalence of 5s and 4s, in columns 1 and 4, and rows 2 and 4. After that, it was primarily guesswork, placing lines, then working backwards to determine what arrows would cause those lines.
answered Nov 13 '18 at 16:27
SconibulusSconibulus
14.4k128100
$endgroup$
I think this is the answer desired:
I started by
Assuming at least one arrows each in an inverted A shape, based on the prevalence of 5s and 4s, in columns 1 and 4, and rows 2 and 4. After that, it was primarily guesswork, placing lines, then working backwards to determine what arrows would cause those lines.
answered Nov 13 '18 at 16:27
SconibulusSconibulus
14.4k128100
answered Nov 13 '18 at 16:27
SconibulusSconibulus
14.4k128100
answered Nov 13 '18 at 16:27
SconibulusSconibulus
14.4k128100
answered Nov 13 '18 at 16:27
SconibulusSconibulus
14.4k128100
14.4k128100
$begingroup$
This is indeed the correct answer. bravo!
$endgroup$
– ABcDexter
Nov 13 '18 at 19:03
add a comment |
$begingroup$
This is indeed the correct answer. bravo!
$endgroup$
– ABcDexter
Nov 13 '18 at 19:03
$begingroup$
This is indeed the correct answer. bravo!
$endgroup$
– ABcDexter
Nov 13 '18 at 19:03
$begingroup$
This is indeed the correct answer. bravo!
$endgroup$
– ABcDexter
Nov 13 '18 at 19:03
add a comment |
$begingroup$
This solution:
- places an arrow in every field around the diagram;
- each arrow points at one or more numbers inside;
- satisfies the condition of having the numbers in the boxes equaling the numbers of arrows pointing at them!
That being said, the person who is giving you the interview may not like it.
:P
answered Nov 13 '18 at 15:18
Excited RaichuExcited Raichu
6,43521166
$endgroup$
$begingroup$
These are a standard puzzle form: puzzlepicnic.com/puzzles... but something about this puzzle makes me believe it's unsolvable. I wonder if it's just one of those task-persistence measures given in job interviews to see how long you'll try before you give up. Or, if it is solvable, it doesn't have characteristics of other arrow puzzles, which generally have one square with only one logical set of arrows possible.
$endgroup$
– Parseltongue
Nov 13 '18 at 15:32
$begingroup$
I too thought about odd angles for arrows, but you sir are taking it to another level x)
$endgroup$
– Cashbee
Nov 13 '18 at 15:35
$begingroup$
Bottom left corner?
$endgroup$
– Greg
Nov 13 '18 at 15:49
$begingroup$
@Greg all of the outer squares it touches have an arrow pointing at it
$endgroup$
– Excited Raichu
Nov 13 '18 at 15:51
$begingroup$
Wow, that is next level!
$endgroup$
– Greg
Nov 13 '18 at 15:58
|
show 1 more comment
$begingroup$
This solution:
- places an arrow in every field around the diagram;
- each arrow points at one or more numbers inside;
- satisfies the condition of having the numbers in the boxes equaling the numbers of arrows pointing at them!
That being said, the person who is giving you the interview may not like it.
:P
answered Nov 13 '18 at 15:18
Excited RaichuExcited Raichu
6,43521166
$endgroup$
$begingroup$
These are a standard puzzle form: puzzlepicnic.com/puzzles... but something about this puzzle makes me believe it's unsolvable. I wonder if it's just one of those task-persistence measures given in job interviews to see how long you'll try before you give up. Or, if it is solvable, it doesn't have characteristics of other arrow puzzles, which generally have one square with only one logical set of arrows possible.
$endgroup$
– Parseltongue
Nov 13 '18 at 15:32
$begingroup$
I too thought about odd angles for arrows, but you sir are taking it to another level x)
$endgroup$
– Cashbee
Nov 13 '18 at 15:35
$begingroup$
Bottom left corner?
$endgroup$
– Greg
Nov 13 '18 at 15:49
$begingroup$
@Greg all of the outer squares it touches have an arrow pointing at it
$endgroup$
– Excited Raichu
Nov 13 '18 at 15:51
$begingroup$
Wow, that is next level!
$endgroup$
– Greg
Nov 13 '18 at 15:58
|
show 1 more comment
$begingroup$
This solution:
- places an arrow in every field around the diagram;
- each arrow points at one or more numbers inside;
- satisfies the condition of having the numbers in the boxes equaling the numbers of arrows pointing at them!
That being said, the person who is giving you the interview may not like it.
:P
answered Nov 13 '18 at 15:18
Excited RaichuExcited Raichu
6,43521166
$endgroup$
This solution:
- places an arrow in every field around the diagram;
- each arrow points at one or more numbers inside;
- satisfies the condition of having the numbers in the boxes equaling the numbers of arrows pointing at them!
That being said, the person who is giving you the interview may not like it.
:P
answered Nov 13 '18 at 15:18
Excited RaichuExcited Raichu
6,43521166
edited Nov 13 '18 at 15:25
answered Nov 13 '18 at 15:18
Excited RaichuExcited Raichu
6,43521166
answered Nov 13 '18 at 15:18
Excited RaichuExcited Raichu
6,43521166
answered Nov 13 '18 at 15:18
Excited RaichuExcited Raichu
6,43521166
6,43521166
$begingroup$
These are a standard puzzle form: puzzlepicnic.com/puzzles... but something about this puzzle makes me believe it's unsolvable. I wonder if it's just one of those task-persistence measures given in job interviews to see how long you'll try before you give up. Or, if it is solvable, it doesn't have characteristics of other arrow puzzles, which generally have one square with only one logical set of arrows possible.
$endgroup$
– Parseltongue
Nov 13 '18 at 15:32
$begingroup$
I too thought about odd angles for arrows, but you sir are taking it to another level x)
$endgroup$
– Cashbee
Nov 13 '18 at 15:35
$begingroup$
Bottom left corner?
$endgroup$
– Greg
Nov 13 '18 at 15:49
$begingroup$
@Greg all of the outer squares it touches have an arrow pointing at it
$endgroup$
– Excited Raichu
Nov 13 '18 at 15:51
$begingroup$
Wow, that is next level!
$endgroup$
– Greg
Nov 13 '18 at 15:58
|
show 1 more comment
$begingroup$
These are a standard puzzle form: puzzlepicnic.com/puzzles... but something about this puzzle makes me believe it's unsolvable. I wonder if it's just one of those task-persistence measures given in job interviews to see how long you'll try before you give up. Or, if it is solvable, it doesn't have characteristics of other arrow puzzles, which generally have one square with only one logical set of arrows possible.
$endgroup$
– Parseltongue
Nov 13 '18 at 15:32
$begingroup$
I too thought about odd angles for arrows, but you sir are taking it to another level x)
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– Cashbee
Nov 13 '18 at 15:35
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Bottom left corner?
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– Greg
Nov 13 '18 at 15:49
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@Greg all of the outer squares it touches have an arrow pointing at it
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– Excited Raichu
Nov 13 '18 at 15:51
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Wow, that is next level!
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– Greg
Nov 13 '18 at 15:58
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These are a standard puzzle form: puzzlepicnic.com/puzzles... but something about this puzzle makes me believe it's unsolvable. I wonder if it's just one of those task-persistence measures given in job interviews to see how long you'll try before you give up. Or, if it is solvable, it doesn't have characteristics of other arrow puzzles, which generally have one square with only one logical set of arrows possible.
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– Parseltongue
Nov 13 '18 at 15:32
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These are a standard puzzle form: puzzlepicnic.com/puzzles... but something about this puzzle makes me believe it's unsolvable. I wonder if it's just one of those task-persistence measures given in job interviews to see how long you'll try before you give up. Or, if it is solvable, it doesn't have characteristics of other arrow puzzles, which generally have one square with only one logical set of arrows possible.
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– Parseltongue
Nov 13 '18 at 15:32
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I too thought about odd angles for arrows, but you sir are taking it to another level x)
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– Cashbee
Nov 13 '18 at 15:35
$begingroup$
I too thought about odd angles for arrows, but you sir are taking it to another level x)
$endgroup$
– Cashbee
Nov 13 '18 at 15:35
$begingroup$
Bottom left corner?
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– Greg
Nov 13 '18 at 15:49
$begingroup$
Bottom left corner?
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– Greg
Nov 13 '18 at 15:49
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@Greg all of the outer squares it touches have an arrow pointing at it
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– Excited Raichu
Nov 13 '18 at 15:51
$begingroup$
@Greg all of the outer squares it touches have an arrow pointing at it
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– Excited Raichu
Nov 13 '18 at 15:51
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Wow, that is next level!
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– Greg
Nov 13 '18 at 15:58
$begingroup$
Wow, that is next level!
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– Greg
Nov 13 '18 at 15:58
|
show 1 more comment
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This is a possibility:
I'm sure there are others.
edited Dec 4 '18 at 16:59
GentlePurpleRain♦
16.9k568136
answered Nov 14 '18 at 10:45
AnonAnon
111
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add a comment |
$begingroup$
This is a possibility:
I'm sure there are others.
edited Dec 4 '18 at 16:59
GentlePurpleRain♦
16.9k568136
answered Nov 14 '18 at 10:45
AnonAnon
111
$endgroup$
add a comment |
$begingroup$
This is a possibility:
I'm sure there are others.
edited Dec 4 '18 at 16:59
GentlePurpleRain♦
16.9k568136
answered Nov 14 '18 at 10:45
AnonAnon
111
$endgroup$
This is a possibility:
I'm sure there are others.
edited Dec 4 '18 at 16:59
GentlePurpleRain♦
16.9k568136
answered Nov 14 '18 at 10:45
AnonAnon
111
edited Dec 4 '18 at 16:59
GentlePurpleRain♦
16.9k568136
edited Dec 4 '18 at 16:59
GentlePurpleRain♦
16.9k568136
edited Dec 4 '18 at 16:59
GentlePurpleRain♦
16.9k568136
16.9k568136
answered Nov 14 '18 at 10:45
AnonAnon
111
answered Nov 14 '18 at 10:45
AnonAnon
111
answered Nov 14 '18 at 10:45
AnonAnon
111
111
add a comment |
add a comment |
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I believe there are 2^8 + (3^8) = 6817 combinations of puzzle solutions, so in principle this can be bruteforced.
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– Parseltongue
Nov 13 '18 at 16:18
2
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@Parseltongue I think your math is off; it should actually be 2^8 times 3^8 = 1679616, shouldn't it? Still trivial for a modern computer, but obviously it scales far too quickly to be a feasible approach at higher size squares. This makes me curious whether this is a NP problem; I can see a way of mapping it to a Boolean satisfiability problem, which would be NP (though that's certainly not the most efficient approach), and the sums can probably provide more information, perhaps revealing an algorithm in P. But it reminds me of Sudoku, which is in NP... I wonder.
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– Graham
Nov 13 '18 at 20:58
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You're right! My mistake. And those were my thoughts as well... I immediately started trying to code up an algorithm to solve this, but decided Brute Force was easiest.
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– Parseltongue
Nov 13 '18 at 21:04
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Here is an "assisted solver" I whipped up to play around with this puzzle. It's just a bare-bones set of
<button>and<span>elements in a grid. The values in the grid update live as you rotate the arrows. I see that the answer has already been posted, and a tool like this probably already exists, but perhaps someone will find this one useful.$endgroup$
– benj2240
Nov 13 '18 at 23:15
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@benj2240 - that is absolutely amazing! Learned a lot reading the code.
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– Parseltongue
Nov 14 '18 at 18:57