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Proof for max. distance between two points inside a circle
Clash Royale CLAN TAG#URR8PPP
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I have an algorithm in mind which can solve one of my algorithm's problems.
The algorithm links Blue-Dot to any of the Red-dot such that their mutual-distance is maximum compared to the distance between blue-dot and any other red-dot. (see the image below for a better understanding).
I am pretty much clear how to do that but I need more mathematical proves. I have done something on my part as well, but any help which makes it more evident mathematically will be highly appreciated. (I know the intuitive part well, but I need mathematical proof)
- The central-node (black-dot in the centre, knows the cartesian coordinates of all dots and easily gets polar coordinates.
- The circle is divided into four quadrants. This way Angle $angle POR$ can easily be found.
- It is for sure that the red-dot in the same quadrant is out-of-question since it will definitely have a smaller distance compared to other red-dots.
- Using the law of cosines, length $r$ can be found for the rest of the three quadrants.
- But, the ideal candidate is the exact opposite quadrant as following the law of cosine $r = sqrta^2 + b^2 - 2abcos(angle POR)$ and in the opposite quadrant $angle POR$ is very close to 180, hence $cos(angle POR) = -1$. This will make $r$ to be maximum. But I need mathematical proof or is this proof enough.
geometry euclidean-geometry circle polar-coordinates
asked Aug 2 at 12:01
Kashan
341110
 |Â
show 3 more comments
up vote
0
down vote
favorite
I have an algorithm in mind which can solve one of my algorithm's problems.
The algorithm links Blue-Dot to any of the Red-dot such that their mutual-distance is maximum compared to the distance between blue-dot and any other red-dot. (see the image below for a better understanding).
I am pretty much clear how to do that but I need more mathematical proves. I have done something on my part as well, but any help which makes it more evident mathematically will be highly appreciated. (I know the intuitive part well, but I need mathematical proof)
- The central-node (black-dot in the centre, knows the cartesian coordinates of all dots and easily gets polar coordinates.
- The circle is divided into four quadrants. This way Angle $angle POR$ can easily be found.
- It is for sure that the red-dot in the same quadrant is out-of-question since it will definitely have a smaller distance compared to other red-dots.
- Using the law of cosines, length $r$ can be found for the rest of the three quadrants.
- But, the ideal candidate is the exact opposite quadrant as following the law of cosine $r = sqrta^2 + b^2 - 2abcos(angle POR)$ and in the opposite quadrant $angle POR$ is very close to 180, hence $cos(angle POR) = -1$. This will make $r$ to be maximum. But I need mathematical proof or is this proof enough.
geometry euclidean-geometry circle polar-coordinates
asked Aug 2 at 12:01
Kashan
341110
4
I do not understand the problem. You need a mathematical proof of what ?
– nicomezi
Aug 2 at 12:09
If red dots are not fixed (meaning your image is just one of the infinite possible cases) then the third assumption is wrong. I think your approach isn't bullet-proof. I’ll suggest a different way: given $N$ is the number of red dots then $textargmax_i in [1,N] left|vecP-vecR_iright|$ return the index of the red dot you’re interested into.
– Giulio Scattolin
Aug 2 at 12:37
$Giulio Scattolin: red-dots are uniformaly distributed over the entire 2D plane, I have tried to show only those that are closer to the black dot, since this is one of the consideration that whatever the location of Blue -dot, red dot must always be closer to the center.
– Kashan
Aug 2 at 19:44
@Giulio Scattolin again, I know I can find that using what you have said, but I have a large number of nodes, and I don't want to use Brute-force algorithm for my design, for this purpose, I have divided the circle in 4-quadrants. Once I have proven that the nodes on the opposite quadrant are most probable to have largest distance with blue-dot, I can then apply your said formula
– Kashan
Aug 2 at 20:59
1
@Kashan I was thinking about a "2D binary tree". Then you can determine the location of the point which is the farthest from the blue one if it existed: finally search from that area.
– Giulio Scattolin
Aug 2 at 22:56
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have an algorithm in mind which can solve one of my algorithm's problems.
The algorithm links Blue-Dot to any of the Red-dot such that their mutual-distance is maximum compared to the distance between blue-dot and any other red-dot. (see the image below for a better understanding).
I am pretty much clear how to do that but I need more mathematical proves. I have done something on my part as well, but any help which makes it more evident mathematically will be highly appreciated. (I know the intuitive part well, but I need mathematical proof)
- The central-node (black-dot in the centre, knows the cartesian coordinates of all dots and easily gets polar coordinates.
- The circle is divided into four quadrants. This way Angle $angle POR$ can easily be found.
- It is for sure that the red-dot in the same quadrant is out-of-question since it will definitely have a smaller distance compared to other red-dots.
- Using the law of cosines, length $r$ can be found for the rest of the three quadrants.
- But, the ideal candidate is the exact opposite quadrant as following the law of cosine $r = sqrta^2 + b^2 - 2abcos(angle POR)$ and in the opposite quadrant $angle POR$ is very close to 180, hence $cos(angle POR) = -1$. This will make $r$ to be maximum. But I need mathematical proof or is this proof enough.
geometry euclidean-geometry circle polar-coordinates
asked Aug 2 at 12:01
Kashan
341110
I have an algorithm in mind which can solve one of my algorithm's problems.
The algorithm links Blue-Dot to any of the Red-dot such that their mutual-distance is maximum compared to the distance between blue-dot and any other red-dot. (see the image below for a better understanding).
I am pretty much clear how to do that but I need more mathematical proves. I have done something on my part as well, but any help which makes it more evident mathematically will be highly appreciated. (I know the intuitive part well, but I need mathematical proof)
- The central-node (black-dot in the centre, knows the cartesian coordinates of all dots and easily gets polar coordinates.
- The circle is divided into four quadrants. This way Angle $angle POR$ can easily be found.
- It is for sure that the red-dot in the same quadrant is out-of-question since it will definitely have a smaller distance compared to other red-dots.
- Using the law of cosines, length $r$ can be found for the rest of the three quadrants.
- But, the ideal candidate is the exact opposite quadrant as following the law of cosine $r = sqrta^2 + b^2 - 2abcos(angle POR)$ and in the opposite quadrant $angle POR$ is very close to 180, hence $cos(angle POR) = -1$. This will make $r$ to be maximum. But I need mathematical proof or is this proof enough.
geometry euclidean-geometry circle polar-coordinates
asked Aug 2 at 12:01
Kashan
341110
asked Aug 2 at 12:01
Kashan
341110
asked Aug 2 at 12:01
Kashan
341110
asked Aug 2 at 12:01
Kashan
341110
341110
4
I do not understand the problem. You need a mathematical proof of what ?
– nicomezi
Aug 2 at 12:09
If red dots are not fixed (meaning your image is just one of the infinite possible cases) then the third assumption is wrong. I think your approach isn't bullet-proof. I’ll suggest a different way: given $N$ is the number of red dots then $textargmax_i in [1,N] left|vecP-vecR_iright|$ return the index of the red dot you’re interested into.
– Giulio Scattolin
Aug 2 at 12:37
$Giulio Scattolin: red-dots are uniformaly distributed over the entire 2D plane, I have tried to show only those that are closer to the black dot, since this is one of the consideration that whatever the location of Blue -dot, red dot must always be closer to the center.
– Kashan
Aug 2 at 19:44
@Giulio Scattolin again, I know I can find that using what you have said, but I have a large number of nodes, and I don't want to use Brute-force algorithm for my design, for this purpose, I have divided the circle in 4-quadrants. Once I have proven that the nodes on the opposite quadrant are most probable to have largest distance with blue-dot, I can then apply your said formula
– Kashan
Aug 2 at 20:59
1
@Kashan I was thinking about a "2D binary tree". Then you can determine the location of the point which is the farthest from the blue one if it existed: finally search from that area.
– Giulio Scattolin
Aug 2 at 22:56
 |Â
show 3 more comments
4
I do not understand the problem. You need a mathematical proof of what ?
– nicomezi
Aug 2 at 12:09
If red dots are not fixed (meaning your image is just one of the infinite possible cases) then the third assumption is wrong. I think your approach isn't bullet-proof. I’ll suggest a different way: given $N$ is the number of red dots then $textargmax_i in [1,N] left|vecP-vecR_iright|$ return the index of the red dot you’re interested into.
– Giulio Scattolin
Aug 2 at 12:37
$Giulio Scattolin: red-dots are uniformaly distributed over the entire 2D plane, I have tried to show only those that are closer to the black dot, since this is one of the consideration that whatever the location of Blue -dot, red dot must always be closer to the center.
– Kashan
Aug 2 at 19:44
@Giulio Scattolin again, I know I can find that using what you have said, but I have a large number of nodes, and I don't want to use Brute-force algorithm for my design, for this purpose, I have divided the circle in 4-quadrants. Once I have proven that the nodes on the opposite quadrant are most probable to have largest distance with blue-dot, I can then apply your said formula
– Kashan
Aug 2 at 20:59
1
@Kashan I was thinking about a "2D binary tree". Then you can determine the location of the point which is the farthest from the blue one if it existed: finally search from that area.
– Giulio Scattolin
Aug 2 at 22:56
4
4
I do not understand the problem. You need a mathematical proof of what ?
– nicomezi
Aug 2 at 12:09
I do not understand the problem. You need a mathematical proof of what ?
– nicomezi
Aug 2 at 12:09
If red dots are not fixed (meaning your image is just one of the infinite possible cases) then the third assumption is wrong. I think your approach isn't bullet-proof. I’ll suggest a different way: given $N$ is the number of red dots then $textargmax_i in [1,N] left|vecP-vecR_iright|$ return the index of the red dot you’re interested into.
– Giulio Scattolin
Aug 2 at 12:37
If red dots are not fixed (meaning your image is just one of the infinite possible cases) then the third assumption is wrong. I think your approach isn't bullet-proof. I’ll suggest a different way: given $N$ is the number of red dots then $textargmax_i in [1,N] left|vecP-vecR_iright|$ return the index of the red dot you’re interested into.
– Giulio Scattolin
Aug 2 at 12:37
$Giulio Scattolin: red-dots are uniformaly distributed over the entire 2D plane, I have tried to show only those that are closer to the black dot, since this is one of the consideration that whatever the location of Blue -dot, red dot must always be closer to the center.
– Kashan
Aug 2 at 19:44
$Giulio Scattolin: red-dots are uniformaly distributed over the entire 2D plane, I have tried to show only those that are closer to the black dot, since this is one of the consideration that whatever the location of Blue -dot, red dot must always be closer to the center.
– Kashan
Aug 2 at 19:44
@Giulio Scattolin again, I know I can find that using what you have said, but I have a large number of nodes, and I don't want to use Brute-force algorithm for my design, for this purpose, I have divided the circle in 4-quadrants. Once I have proven that the nodes on the opposite quadrant are most probable to have largest distance with blue-dot, I can then apply your said formula
– Kashan
Aug 2 at 20:59
@Giulio Scattolin again, I know I can find that using what you have said, but I have a large number of nodes, and I don't want to use Brute-force algorithm for my design, for this purpose, I have divided the circle in 4-quadrants. Once I have proven that the nodes on the opposite quadrant are most probable to have largest distance with blue-dot, I can then apply your said formula
– Kashan
Aug 2 at 20:59
1
1
@Kashan I was thinking about a "2D binary tree". Then you can determine the location of the point which is the farthest from the blue one if it existed: finally search from that area.
– Giulio Scattolin
Aug 2 at 22:56
@Kashan I was thinking about a "2D binary tree". Then you can determine the location of the point which is the farthest from the blue one if it existed: finally search from that area.
– Giulio Scattolin
Aug 2 at 22:56
 |Â
show 3 more comments
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4
I do not understand the problem. You need a mathematical proof of what ?
– nicomezi
Aug 2 at 12:09
If red dots are not fixed (meaning your image is just one of the infinite possible cases) then the third assumption is wrong. I think your approach isn't bullet-proof. I’ll suggest a different way: given $N$ is the number of red dots then $textargmax_i in [1,N] left|vecP-vecR_iright|$ return the index of the red dot you’re interested into.
– Giulio Scattolin
Aug 2 at 12:37
$Giulio Scattolin: red-dots are uniformaly distributed over the entire 2D plane, I have tried to show only those that are closer to the black dot, since this is one of the consideration that whatever the location of Blue -dot, red dot must always be closer to the center.
– Kashan
Aug 2 at 19:44
@Giulio Scattolin again, I know I can find that using what you have said, but I have a large number of nodes, and I don't want to use Brute-force algorithm for my design, for this purpose, I have divided the circle in 4-quadrants. Once I have proven that the nodes on the opposite quadrant are most probable to have largest distance with blue-dot, I can then apply your said formula
– Kashan
Aug 2 at 20:59
1
@Kashan I was thinking about a "2D binary tree". Then you can determine the location of the point which is the farthest from the blue one if it existed: finally search from that area.
– Giulio Scattolin
Aug 2 at 22:56