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Can I uniquely identify a cell in an arbitrary “grid†by two numbers?
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Suppose I have a paper on which I draw some kind of "hull" polygon. I divide that polygon into many other polygons which I call "cells". Now I want to uniquely identify each cell with the geometric center and the radius of the circumscribed circle.
Is that possible?
Simple example
Let's say our "hull polygon" is an A4 paper. And the polygons we divide it into are simply the rectanlges printed on it:
Claim: All cells are uniquely identified by their center of mass, if all cells are convex.
Proof: Let $C$ be the centers of mass of two convex polygons $A_P$ and $B_P$. The center of mass of a convex polygon is within the polygon, hence $C in A_P$ and $C in B_P$. As each cell is guaranteed not to overlap with any other cell, $A_P = B_P$. Hence the center of mass uniquely identifies the cell if all cells are convex polygons.
Non-convex cells
Let's imagine we have a non-convex cell (the blue one below). The blue polygon and the red one might have the same center of mass. But at least the radius of their circumscribed circle is different. But I'm not sure if that is enough and I have no idea how to prove it.
geometry
asked Jul 31 at 15:49
Martin Thoma
4,10093695
add a comment |Â
up vote
2
down vote
favorite
Suppose I have a paper on which I draw some kind of "hull" polygon. I divide that polygon into many other polygons which I call "cells". Now I want to uniquely identify each cell with the geometric center and the radius of the circumscribed circle.
Is that possible?
Simple example
Let's say our "hull polygon" is an A4 paper. And the polygons we divide it into are simply the rectanlges printed on it:
Claim: All cells are uniquely identified by their center of mass, if all cells are convex.
Proof: Let $C$ be the centers of mass of two convex polygons $A_P$ and $B_P$. The center of mass of a convex polygon is within the polygon, hence $C in A_P$ and $C in B_P$. As each cell is guaranteed not to overlap with any other cell, $A_P = B_P$. Hence the center of mass uniquely identifies the cell if all cells are convex polygons.
Non-convex cells
Let's imagine we have a non-convex cell (the blue one below). The blue polygon and the red one might have the same center of mass. But at least the radius of their circumscribed circle is different. But I'm not sure if that is enough and I have no idea how to prove it.
geometry
asked Jul 31 at 15:49
Martin Thoma
4,10093695
So the piece of paper is bounded?
– Arnaud Mortier
Jul 31 at 15:57
Yes. And also the hull polygon. I only mentioned the piece of paper to make sure this is very "basic".
– Martin Thoma
Jul 31 at 15:59
Do you mean the center of the polygon, or the center of the circumscribing circle? If the latter, then definitely not; consider a 'wide' obtuse triangle such that the center of the circumscribing circle isn't in the polygon, and then reflect the triangle about that center.
– Steven Stadnicki
Jul 31 at 16:07
Looking at the wikipedia page, I think I should write "Centroid" instead of "center of mass".
– Martin Thoma
Jul 31 at 16:27
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose I have a paper on which I draw some kind of "hull" polygon. I divide that polygon into many other polygons which I call "cells". Now I want to uniquely identify each cell with the geometric center and the radius of the circumscribed circle.
Is that possible?
Simple example
Let's say our "hull polygon" is an A4 paper. And the polygons we divide it into are simply the rectanlges printed on it:
Claim: All cells are uniquely identified by their center of mass, if all cells are convex.
Proof: Let $C$ be the centers of mass of two convex polygons $A_P$ and $B_P$. The center of mass of a convex polygon is within the polygon, hence $C in A_P$ and $C in B_P$. As each cell is guaranteed not to overlap with any other cell, $A_P = B_P$. Hence the center of mass uniquely identifies the cell if all cells are convex polygons.
Non-convex cells
Let's imagine we have a non-convex cell (the blue one below). The blue polygon and the red one might have the same center of mass. But at least the radius of their circumscribed circle is different. But I'm not sure if that is enough and I have no idea how to prove it.
geometry
asked Jul 31 at 15:49
Martin Thoma
4,10093695
Suppose I have a paper on which I draw some kind of "hull" polygon. I divide that polygon into many other polygons which I call "cells". Now I want to uniquely identify each cell with the geometric center and the radius of the circumscribed circle.
Is that possible?
Simple example
Let's say our "hull polygon" is an A4 paper. And the polygons we divide it into are simply the rectanlges printed on it:
Claim: All cells are uniquely identified by their center of mass, if all cells are convex.
Proof: Let $C$ be the centers of mass of two convex polygons $A_P$ and $B_P$. The center of mass of a convex polygon is within the polygon, hence $C in A_P$ and $C in B_P$. As each cell is guaranteed not to overlap with any other cell, $A_P = B_P$. Hence the center of mass uniquely identifies the cell if all cells are convex polygons.
Non-convex cells
Let's imagine we have a non-convex cell (the blue one below). The blue polygon and the red one might have the same center of mass. But at least the radius of their circumscribed circle is different. But I'm not sure if that is enough and I have no idea how to prove it.
geometry
asked Jul 31 at 15:49
Martin Thoma
4,10093695
asked Jul 31 at 15:49
Martin Thoma
4,10093695
asked Jul 31 at 15:49
Martin Thoma
4,10093695
asked Jul 31 at 15:49
Martin Thoma
4,10093695
4,10093695
So the piece of paper is bounded?
– Arnaud Mortier
Jul 31 at 15:57
Yes. And also the hull polygon. I only mentioned the piece of paper to make sure this is very "basic".
– Martin Thoma
Jul 31 at 15:59
Do you mean the center of the polygon, or the center of the circumscribing circle? If the latter, then definitely not; consider a 'wide' obtuse triangle such that the center of the circumscribing circle isn't in the polygon, and then reflect the triangle about that center.
– Steven Stadnicki
Jul 31 at 16:07
Looking at the wikipedia page, I think I should write "Centroid" instead of "center of mass".
– Martin Thoma
Jul 31 at 16:27
add a comment |Â
So the piece of paper is bounded?
– Arnaud Mortier
Jul 31 at 15:57
Yes. And also the hull polygon. I only mentioned the piece of paper to make sure this is very "basic".
– Martin Thoma
Jul 31 at 15:59
Do you mean the center of the polygon, or the center of the circumscribing circle? If the latter, then definitely not; consider a 'wide' obtuse triangle such that the center of the circumscribing circle isn't in the polygon, and then reflect the triangle about that center.
– Steven Stadnicki
Jul 31 at 16:07
Looking at the wikipedia page, I think I should write "Centroid" instead of "center of mass".
– Martin Thoma
Jul 31 at 16:27
So the piece of paper is bounded?
– Arnaud Mortier
Jul 31 at 15:57
So the piece of paper is bounded?
– Arnaud Mortier
Jul 31 at 15:57
Yes. And also the hull polygon. I only mentioned the piece of paper to make sure this is very "basic".
– Martin Thoma
Jul 31 at 15:59
Yes. And also the hull polygon. I only mentioned the piece of paper to make sure this is very "basic".
– Martin Thoma
Jul 31 at 15:59
Do you mean the center of the polygon, or the center of the circumscribing circle? If the latter, then definitely not; consider a 'wide' obtuse triangle such that the center of the circumscribing circle isn't in the polygon, and then reflect the triangle about that center.
– Steven Stadnicki
Jul 31 at 16:07
Do you mean the center of the polygon, or the center of the circumscribing circle? If the latter, then definitely not; consider a 'wide' obtuse triangle such that the center of the circumscribing circle isn't in the polygon, and then reflect the triangle about that center.
– Steven Stadnicki
Jul 31 at 16:07
Looking at the wikipedia page, I think I should write "Centroid" instead of "center of mass".
– Martin Thoma
Jul 31 at 16:27
Looking at the wikipedia page, I think I should write "Centroid" instead of "center of mass".
– Martin Thoma
Jul 31 at 16:27
add a comment |Â
1 Answer
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Theoretically, yes.
Assuming that each cell $A$ contains a disk of some size (meaning that there is a point in $A$ that has positive distance from the complement $A^c$) you can consider the set $$I_A=xin Bbb R_+,exists ain A colon d(a,A^c)=x$$
Since the hull polygon is bounded, each $I_A$ is an interval of the form $(0,M_A)$.
Now consider $$leftain A, d(a,A^c)leq fracM_A2right$$
That is a non-empty compact subset of $A$, and you can consider the two coordinates of its largest element, in the sense of any total order of your choice, for example a lexicographical order: take the element with largest $x$-value, and if there are several with this value, take from them the one with the largest $y$-value.
No two polygons can have the same two coordinates assigned in this way, because of the positive distance from the complement.
answered Jul 31 at 16:04
Arnaud Mortier
18k21958
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Theoretically, yes.
Assuming that each cell $A$ contains a disk of some size (meaning that there is a point in $A$ that has positive distance from the complement $A^c$) you can consider the set $$I_A=xin Bbb R_+,exists ain A colon d(a,A^c)=x$$
Since the hull polygon is bounded, each $I_A$ is an interval of the form $(0,M_A)$.
Now consider $$leftain A, d(a,A^c)leq fracM_A2right$$
That is a non-empty compact subset of $A$, and you can consider the two coordinates of its largest element, in the sense of any total order of your choice, for example a lexicographical order: take the element with largest $x$-value, and if there are several with this value, take from them the one with the largest $y$-value.
No two polygons can have the same two coordinates assigned in this way, because of the positive distance from the complement.
answered Jul 31 at 16:04
Arnaud Mortier
18k21958
add a comment |Â
up vote
0
down vote
Theoretically, yes.
Assuming that each cell $A$ contains a disk of some size (meaning that there is a point in $A$ that has positive distance from the complement $A^c$) you can consider the set $$I_A=xin Bbb R_+,exists ain A colon d(a,A^c)=x$$
Since the hull polygon is bounded, each $I_A$ is an interval of the form $(0,M_A)$.
Now consider $$leftain A, d(a,A^c)leq fracM_A2right$$
That is a non-empty compact subset of $A$, and you can consider the two coordinates of its largest element, in the sense of any total order of your choice, for example a lexicographical order: take the element with largest $x$-value, and if there are several with this value, take from them the one with the largest $y$-value.
No two polygons can have the same two coordinates assigned in this way, because of the positive distance from the complement.
answered Jul 31 at 16:04
Arnaud Mortier
18k21958
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Theoretically, yes.
Assuming that each cell $A$ contains a disk of some size (meaning that there is a point in $A$ that has positive distance from the complement $A^c$) you can consider the set $$I_A=xin Bbb R_+,exists ain A colon d(a,A^c)=x$$
Since the hull polygon is bounded, each $I_A$ is an interval of the form $(0,M_A)$.
Now consider $$leftain A, d(a,A^c)leq fracM_A2right$$
That is a non-empty compact subset of $A$, and you can consider the two coordinates of its largest element, in the sense of any total order of your choice, for example a lexicographical order: take the element with largest $x$-value, and if there are several with this value, take from them the one with the largest $y$-value.
No two polygons can have the same two coordinates assigned in this way, because of the positive distance from the complement.
answered Jul 31 at 16:04
Arnaud Mortier
18k21958
Theoretically, yes.
Assuming that each cell $A$ contains a disk of some size (meaning that there is a point in $A$ that has positive distance from the complement $A^c$) you can consider the set $$I_A=xin Bbb R_+,exists ain A colon d(a,A^c)=x$$
Since the hull polygon is bounded, each $I_A$ is an interval of the form $(0,M_A)$.
Now consider $$leftain A, d(a,A^c)leq fracM_A2right$$
That is a non-empty compact subset of $A$, and you can consider the two coordinates of its largest element, in the sense of any total order of your choice, for example a lexicographical order: take the element with largest $x$-value, and if there are several with this value, take from them the one with the largest $y$-value.
No two polygons can have the same two coordinates assigned in this way, because of the positive distance from the complement.
answered Jul 31 at 16:04
Arnaud Mortier
18k21958
answered Jul 31 at 16:04
Arnaud Mortier
18k21958
answered Jul 31 at 16:04
Arnaud Mortier
18k21958
answered Jul 31 at 16:04
Arnaud Mortier
18k21958
18k21958
add a comment |Â
add a comment |Â
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So the piece of paper is bounded?
– Arnaud Mortier
Jul 31 at 15:57
Yes. And also the hull polygon. I only mentioned the piece of paper to make sure this is very "basic".
– Martin Thoma
Jul 31 at 15:59
Do you mean the center of the polygon, or the center of the circumscribing circle? If the latter, then definitely not; consider a 'wide' obtuse triangle such that the center of the circumscribing circle isn't in the polygon, and then reflect the triangle about that center.
– Steven Stadnicki
Jul 31 at 16:07
Looking at the wikipedia page, I think I should write "Centroid" instead of "center of mass".
– Martin Thoma
Jul 31 at 16:27