How to represent reflection of a particle inside a standard simplex?
$begingroup$
I am trying to simulate the trajectory of an evolutionary system represented by a vector of probabilities $vec p = [p_1, p_2,...] $. Values are restricted between 0 and 1. As a result we can think the system is contained in a standard simplex.
So far, so good. However, I would like to have the system to reflect every time it reach one of the borders of the simplex, i.e. $p_i =0$ for any $i$. By reflection I mean something similar to physical reflection of a light beam: to leave the border in the same angle and the same distance as the source, but in a different direction.
I found a somewhat simple rule in wikipedia:
${displaystyle operatorname {Ref} _{a}(v)=v-2{frac {vcdot a}{acdot a}}a,}$
However, it seems it requires the plane $a$ passing by origin, plus its describe the behavior of vector instead of a point. I searched for other formulas, but I must admit, got a little lost in the diversity of solutions.
Assuming there I have all positions of the system was up to the point it reaches the border, is there a more direct formula I can use?
linear-algebra geometry simplex reflection
asked Dec 17 '18 at 20:39
JMenezesJMenezes
1336
$endgroup$
add a comment |
$begingroup$
I am trying to simulate the trajectory of an evolutionary system represented by a vector of probabilities $vec p = [p_1, p_2,...] $. Values are restricted between 0 and 1. As a result we can think the system is contained in a standard simplex.
So far, so good. However, I would like to have the system to reflect every time it reach one of the borders of the simplex, i.e. $p_i =0$ for any $i$. By reflection I mean something similar to physical reflection of a light beam: to leave the border in the same angle and the same distance as the source, but in a different direction.
I found a somewhat simple rule in wikipedia:
${displaystyle operatorname {Ref} _{a}(v)=v-2{frac {vcdot a}{acdot a}}a,}$
However, it seems it requires the plane $a$ passing by origin, plus its describe the behavior of vector instead of a point. I searched for other formulas, but I must admit, got a little lost in the diversity of solutions.
Assuming there I have all positions of the system was up to the point it reaches the border, is there a more direct formula I can use?
linear-algebra geometry simplex reflection
asked Dec 17 '18 at 20:39
JMenezesJMenezes
1336
$endgroup$
$begingroup$
To clarify, you want the vector of reflection ?
$endgroup$
– T. Fo
Dec 17 '18 at 20:45
$begingroup$
Also, the values are restricted between $0$ and $1$, and their sum is $1$, or their sum is $le 1$? Just trying to work out exactly what simplex you're referring to.
$endgroup$
– jgon
Dec 17 '18 at 20:47
$begingroup$
I want to know the position the system (particle) will be after it reflects. However, if it is easier I can work with just the vector of reflection. And jgon is correct. It is restricted to 0 and 1 and their sum to 1.
$endgroup$
– JMenezes
Dec 18 '18 at 15:46
add a comment |
$begingroup$
I am trying to simulate the trajectory of an evolutionary system represented by a vector of probabilities $vec p = [p_1, p_2,...] $. Values are restricted between 0 and 1. As a result we can think the system is contained in a standard simplex.
So far, so good. However, I would like to have the system to reflect every time it reach one of the borders of the simplex, i.e. $p_i =0$ for any $i$. By reflection I mean something similar to physical reflection of a light beam: to leave the border in the same angle and the same distance as the source, but in a different direction.
I found a somewhat simple rule in wikipedia:
${displaystyle operatorname {Ref} _{a}(v)=v-2{frac {vcdot a}{acdot a}}a,}$
However, it seems it requires the plane $a$ passing by origin, plus its describe the behavior of vector instead of a point. I searched for other formulas, but I must admit, got a little lost in the diversity of solutions.
Assuming there I have all positions of the system was up to the point it reaches the border, is there a more direct formula I can use?
linear-algebra geometry simplex reflection
asked Dec 17 '18 at 20:39
JMenezesJMenezes
1336
$endgroup$
I am trying to simulate the trajectory of an evolutionary system represented by a vector of probabilities $vec p = [p_1, p_2,...] $. Values are restricted between 0 and 1. As a result we can think the system is contained in a standard simplex.
So far, so good. However, I would like to have the system to reflect every time it reach one of the borders of the simplex, i.e. $p_i =0$ for any $i$. By reflection I mean something similar to physical reflection of a light beam: to leave the border in the same angle and the same distance as the source, but in a different direction.
I found a somewhat simple rule in wikipedia:
${displaystyle operatorname {Ref} _{a}(v)=v-2{frac {vcdot a}{acdot a}}a,}$
However, it seems it requires the plane $a$ passing by origin, plus its describe the behavior of vector instead of a point. I searched for other formulas, but I must admit, got a little lost in the diversity of solutions.
Assuming there I have all positions of the system was up to the point it reaches the border, is there a more direct formula I can use?
linear-algebra geometry simplex reflection
linear-algebra geometry simplex reflection
asked Dec 17 '18 at 20:39
JMenezesJMenezes
1336
asked Dec 17 '18 at 20:39
JMenezesJMenezes
1336
asked Dec 17 '18 at 20:39
JMenezesJMenezes
1336
asked Dec 17 '18 at 20:39
JMenezesJMenezes
1336
asked Dec 17 '18 at 20:39
JMenezesJMenezes
1336
1336
$begingroup$
To clarify, you want the vector of reflection ?
$endgroup$
– T. Fo
Dec 17 '18 at 20:45
$begingroup$
Also, the values are restricted between $0$ and $1$, and their sum is $1$, or their sum is $le 1$? Just trying to work out exactly what simplex you're referring to.
$endgroup$
– jgon
Dec 17 '18 at 20:47
$begingroup$
I want to know the position the system (particle) will be after it reflects. However, if it is easier I can work with just the vector of reflection. And jgon is correct. It is restricted to 0 and 1 and their sum to 1.
$endgroup$
– JMenezes
Dec 18 '18 at 15:46
add a comment |
$begingroup$
To clarify, you want the vector of reflection ?
$endgroup$
– T. Fo
Dec 17 '18 at 20:45
$begingroup$
Also, the values are restricted between $0$ and $1$, and their sum is $1$, or their sum is $le 1$? Just trying to work out exactly what simplex you're referring to.
$endgroup$
– jgon
Dec 17 '18 at 20:47
$begingroup$
I want to know the position the system (particle) will be after it reflects. However, if it is easier I can work with just the vector of reflection. And jgon is correct. It is restricted to 0 and 1 and their sum to 1.
$endgroup$
– JMenezes
Dec 18 '18 at 15:46
$begingroup$
To clarify, you want the vector of reflection ?
$endgroup$
– T. Fo
Dec 17 '18 at 20:45
$begingroup$
To clarify, you want the vector of reflection ?
$endgroup$
– T. Fo
Dec 17 '18 at 20:45
$begingroup$
Also, the values are restricted between $0$ and $1$, and their sum is $1$, or their sum is $le 1$? Just trying to work out exactly what simplex you're referring to.
$endgroup$
– jgon
Dec 17 '18 at 20:47
$begingroup$
Also, the values are restricted between $0$ and $1$, and their sum is $1$, or their sum is $le 1$? Just trying to work out exactly what simplex you're referring to.
$endgroup$
– jgon
Dec 17 '18 at 20:47
$begingroup$
I want to know the position the system (particle) will be after it reflects. However, if it is easier I can work with just the vector of reflection. And jgon is correct. It is restricted to 0 and 1 and their sum to 1.
$endgroup$
– JMenezes
Dec 18 '18 at 15:46
$begingroup$
I want to know the position the system (particle) will be after it reflects. However, if it is easier I can work with just the vector of reflection. And jgon is correct. It is restricted to 0 and 1 and their sum to 1.
$endgroup$
– JMenezes
Dec 18 '18 at 15:46
add a comment |
1 Answer
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oldest
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$begingroup$
Angle measurement requires you to introduce metrics, which can be done differently resulting in different embeddings of the simplex into $mathbf R^{k}$ ($k$ is the size of probability vector). However, let's discuss the most symmetric one (the one that creates a regular simplex as an embedding).
We will embed the simplex into $mathbf R^{k}$ with vertices at the end of basis unit vectors. Thus, our simplex lies on a hyperplane $sum p_i=1$. Its normal is:
$$n=frac1{sqrt k}(1,ldots,1)^T$$
To get the normal to the border lying at the hyperplane, we project the normal, for example, $e_1=(1,0,ldots,0)$ to the hyperplane:
$$v_1'=e_1-n(e_1,v),qquad v_1=frac{v_1'}{|v_1'|}=frac1{sqrt{k(k-1)}}Big((k-1),-1,ldots,-1Big)^T$$
This vector is what we use in formula you have found.
answered Dec 17 '18 at 21:28
Vasily MitchVasily Mitch
2,3141311
$endgroup$
$begingroup$
I am still a little confused with the $n$ in this formulae. In a 3D case, from your first equation I get it $n={1,1,1}$, since $n*sqrt{n} = {1,1,1}$. But I am not quite getting how it becomes a function on $v_1'$ part or what this part is representing. Can you clarify, please? Thanks.
$endgroup$
– JMenezes
Dec 18 '18 at 18:05
$begingroup$
Sorry, it was a letter collision. I have changed the number of $p$-s to be $k$.
$endgroup$
– Vasily Mitch
Dec 18 '18 at 18:13
add a comment |
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$begingroup$
Angle measurement requires you to introduce metrics, which can be done differently resulting in different embeddings of the simplex into $mathbf R^{k}$ ($k$ is the size of probability vector). However, let's discuss the most symmetric one (the one that creates a regular simplex as an embedding).
We will embed the simplex into $mathbf R^{k}$ with vertices at the end of basis unit vectors. Thus, our simplex lies on a hyperplane $sum p_i=1$. Its normal is:
$$n=frac1{sqrt k}(1,ldots,1)^T$$
To get the normal to the border lying at the hyperplane, we project the normal, for example, $e_1=(1,0,ldots,0)$ to the hyperplane:
$$v_1'=e_1-n(e_1,v),qquad v_1=frac{v_1'}{|v_1'|}=frac1{sqrt{k(k-1)}}Big((k-1),-1,ldots,-1Big)^T$$
This vector is what we use in formula you have found.
answered Dec 17 '18 at 21:28
Vasily MitchVasily Mitch
2,3141311
$endgroup$
$begingroup$
I am still a little confused with the $n$ in this formulae. In a 3D case, from your first equation I get it $n={1,1,1}$, since $n*sqrt{n} = {1,1,1}$. But I am not quite getting how it becomes a function on $v_1'$ part or what this part is representing. Can you clarify, please? Thanks.
$endgroup$
– JMenezes
Dec 18 '18 at 18:05
$begingroup$
Sorry, it was a letter collision. I have changed the number of $p$-s to be $k$.
$endgroup$
– Vasily Mitch
Dec 18 '18 at 18:13
add a comment |
$begingroup$
Angle measurement requires you to introduce metrics, which can be done differently resulting in different embeddings of the simplex into $mathbf R^{k}$ ($k$ is the size of probability vector). However, let's discuss the most symmetric one (the one that creates a regular simplex as an embedding).
We will embed the simplex into $mathbf R^{k}$ with vertices at the end of basis unit vectors. Thus, our simplex lies on a hyperplane $sum p_i=1$. Its normal is:
$$n=frac1{sqrt k}(1,ldots,1)^T$$
To get the normal to the border lying at the hyperplane, we project the normal, for example, $e_1=(1,0,ldots,0)$ to the hyperplane:
$$v_1'=e_1-n(e_1,v),qquad v_1=frac{v_1'}{|v_1'|}=frac1{sqrt{k(k-1)}}Big((k-1),-1,ldots,-1Big)^T$$
This vector is what we use in formula you have found.
answered Dec 17 '18 at 21:28
Vasily MitchVasily Mitch
2,3141311
$endgroup$
$begingroup$
I am still a little confused with the $n$ in this formulae. In a 3D case, from your first equation I get it $n={1,1,1}$, since $n*sqrt{n} = {1,1,1}$. But I am not quite getting how it becomes a function on $v_1'$ part or what this part is representing. Can you clarify, please? Thanks.
$endgroup$
– JMenezes
Dec 18 '18 at 18:05
$begingroup$
Sorry, it was a letter collision. I have changed the number of $p$-s to be $k$.
$endgroup$
– Vasily Mitch
Dec 18 '18 at 18:13
add a comment |
$begingroup$
Angle measurement requires you to introduce metrics, which can be done differently resulting in different embeddings of the simplex into $mathbf R^{k}$ ($k$ is the size of probability vector). However, let's discuss the most symmetric one (the one that creates a regular simplex as an embedding).
We will embed the simplex into $mathbf R^{k}$ with vertices at the end of basis unit vectors. Thus, our simplex lies on a hyperplane $sum p_i=1$. Its normal is:
$$n=frac1{sqrt k}(1,ldots,1)^T$$
To get the normal to the border lying at the hyperplane, we project the normal, for example, $e_1=(1,0,ldots,0)$ to the hyperplane:
$$v_1'=e_1-n(e_1,v),qquad v_1=frac{v_1'}{|v_1'|}=frac1{sqrt{k(k-1)}}Big((k-1),-1,ldots,-1Big)^T$$
This vector is what we use in formula you have found.
answered Dec 17 '18 at 21:28
Vasily MitchVasily Mitch
2,3141311
$endgroup$
Angle measurement requires you to introduce metrics, which can be done differently resulting in different embeddings of the simplex into $mathbf R^{k}$ ($k$ is the size of probability vector). However, let's discuss the most symmetric one (the one that creates a regular simplex as an embedding).
We will embed the simplex into $mathbf R^{k}$ with vertices at the end of basis unit vectors. Thus, our simplex lies on a hyperplane $sum p_i=1$. Its normal is:
$$n=frac1{sqrt k}(1,ldots,1)^T$$
To get the normal to the border lying at the hyperplane, we project the normal, for example, $e_1=(1,0,ldots,0)$ to the hyperplane:
$$v_1'=e_1-n(e_1,v),qquad v_1=frac{v_1'}{|v_1'|}=frac1{sqrt{k(k-1)}}Big((k-1),-1,ldots,-1Big)^T$$
This vector is what we use in formula you have found.
answered Dec 17 '18 at 21:28
Vasily MitchVasily Mitch
2,3141311
edited Dec 18 '18 at 18:11
answered Dec 17 '18 at 21:28
Vasily MitchVasily Mitch
2,3141311
answered Dec 17 '18 at 21:28
Vasily MitchVasily Mitch
2,3141311
answered Dec 17 '18 at 21:28
Vasily MitchVasily Mitch
2,3141311
2,3141311
$begingroup$
I am still a little confused with the $n$ in this formulae. In a 3D case, from your first equation I get it $n={1,1,1}$, since $n*sqrt{n} = {1,1,1}$. But I am not quite getting how it becomes a function on $v_1'$ part or what this part is representing. Can you clarify, please? Thanks.
$endgroup$
– JMenezes
Dec 18 '18 at 18:05
$begingroup$
Sorry, it was a letter collision. I have changed the number of $p$-s to be $k$.
$endgroup$
– Vasily Mitch
Dec 18 '18 at 18:13
add a comment |
$begingroup$
I am still a little confused with the $n$ in this formulae. In a 3D case, from your first equation I get it $n={1,1,1}$, since $n*sqrt{n} = {1,1,1}$. But I am not quite getting how it becomes a function on $v_1'$ part or what this part is representing. Can you clarify, please? Thanks.
$endgroup$
– JMenezes
Dec 18 '18 at 18:05
$begingroup$
Sorry, it was a letter collision. I have changed the number of $p$-s to be $k$.
$endgroup$
– Vasily Mitch
Dec 18 '18 at 18:13
$begingroup$
I am still a little confused with the $n$ in this formulae. In a 3D case, from your first equation I get it $n={1,1,1}$, since $n*sqrt{n} = {1,1,1}$. But I am not quite getting how it becomes a function on $v_1'$ part or what this part is representing. Can you clarify, please? Thanks.
$endgroup$
– JMenezes
Dec 18 '18 at 18:05
$begingroup$
I am still a little confused with the $n$ in this formulae. In a 3D case, from your first equation I get it $n={1,1,1}$, since $n*sqrt{n} = {1,1,1}$. But I am not quite getting how it becomes a function on $v_1'$ part or what this part is representing. Can you clarify, please? Thanks.
$endgroup$
– JMenezes
Dec 18 '18 at 18:05
$begingroup$
Sorry, it was a letter collision. I have changed the number of $p$-s to be $k$.
$endgroup$
– Vasily Mitch
Dec 18 '18 at 18:13
$begingroup$
Sorry, it was a letter collision. I have changed the number of $p$-s to be $k$.
$endgroup$
– Vasily Mitch
Dec 18 '18 at 18:13
add a comment |
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$begingroup$
To clarify, you want the vector of reflection ?
$endgroup$
– T. Fo
Dec 17 '18 at 20:45
$begingroup$
Also, the values are restricted between $0$ and $1$, and their sum is $1$, or their sum is $le 1$? Just trying to work out exactly what simplex you're referring to.
$endgroup$
– jgon
Dec 17 '18 at 20:47
$begingroup$
I want to know the position the system (particle) will be after it reflects. However, if it is easier I can work with just the vector of reflection. And jgon is correct. It is restricted to 0 and 1 and their sum to 1.
$endgroup$
– JMenezes
Dec 18 '18 at 15:46