Mixing
mapping a cube into a tetrahedron
Clash Royale CLAN TAG#URR8PPP
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I have a cube with vertex at (0,0,0),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,1,0),(1,0,1),(1,1,1), and a tetrahedron with vertex at (0,0,0),(0,0,1),(0,1,0),(1,0,0).
How do I map the cube's coordinates to the tetrahedron bijectively? Are there an elegant function that would do it?
Note: on 2-dimension, to map a square to a triangle I can do f(x,y) = (x-xy/2,y-xy/2); I'm struggling to extend this to 3-dimension
coordinate-systems
asked Jul 26 at 2:05
More water plz
1,331320
add a comment |Â
up vote
0
down vote
favorite
I have a cube with vertex at (0,0,0),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,1,0),(1,0,1),(1,1,1), and a tetrahedron with vertex at (0,0,0),(0,0,1),(0,1,0),(1,0,0).
How do I map the cube's coordinates to the tetrahedron bijectively? Are there an elegant function that would do it?
Note: on 2-dimension, to map a square to a triangle I can do f(x,y) = (x-xy/2,y-xy/2); I'm struggling to extend this to 3-dimension
coordinate-systems
asked Jul 26 at 2:05
More water plz
1,331320
It’s certainly possible: just scale each ray from the origin linearly to take the edge of the cube to the edge of the tetrahedron. I’m interested to see whether anyone can do it more elegantly than that.....
– G Tony Jacobs
Jul 26 at 2:10
that's what I was thinking. But I'm not sure how to express that process by a function
– More water plz
Jul 26 at 2:26
You would have to write down separate functions depending which face of the cube and the tetrahedron each ray intersects. Like I said, not very elegant.
– G Tony Jacobs
Jul 26 at 2:57
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a cube with vertex at (0,0,0),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,1,0),(1,0,1),(1,1,1), and a tetrahedron with vertex at (0,0,0),(0,0,1),(0,1,0),(1,0,0).
How do I map the cube's coordinates to the tetrahedron bijectively? Are there an elegant function that would do it?
Note: on 2-dimension, to map a square to a triangle I can do f(x,y) = (x-xy/2,y-xy/2); I'm struggling to extend this to 3-dimension
coordinate-systems
asked Jul 26 at 2:05
More water plz
1,331320
I have a cube with vertex at (0,0,0),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,1,0),(1,0,1),(1,1,1), and a tetrahedron with vertex at (0,0,0),(0,0,1),(0,1,0),(1,0,0).
How do I map the cube's coordinates to the tetrahedron bijectively? Are there an elegant function that would do it?
Note: on 2-dimension, to map a square to a triangle I can do f(x,y) = (x-xy/2,y-xy/2); I'm struggling to extend this to 3-dimension
coordinate-systems
asked Jul 26 at 2:05
More water plz
1,331320
edited Jul 26 at 2:25
asked Jul 26 at 2:05
More water plz
1,331320
asked Jul 26 at 2:05
More water plz
1,331320
asked Jul 26 at 2:05
More water plz
1,331320
1,331320
It’s certainly possible: just scale each ray from the origin linearly to take the edge of the cube to the edge of the tetrahedron. I’m interested to see whether anyone can do it more elegantly than that.....
– G Tony Jacobs
Jul 26 at 2:10
that's what I was thinking. But I'm not sure how to express that process by a function
– More water plz
Jul 26 at 2:26
You would have to write down separate functions depending which face of the cube and the tetrahedron each ray intersects. Like I said, not very elegant.
– G Tony Jacobs
Jul 26 at 2:57
add a comment |Â
It’s certainly possible: just scale each ray from the origin linearly to take the edge of the cube to the edge of the tetrahedron. I’m interested to see whether anyone can do it more elegantly than that.....
– G Tony Jacobs
Jul 26 at 2:10
that's what I was thinking. But I'm not sure how to express that process by a function
– More water plz
Jul 26 at 2:26
You would have to write down separate functions depending which face of the cube and the tetrahedron each ray intersects. Like I said, not very elegant.
– G Tony Jacobs
Jul 26 at 2:57
It’s certainly possible: just scale each ray from the origin linearly to take the edge of the cube to the edge of the tetrahedron. I’m interested to see whether anyone can do it more elegantly than that.....
– G Tony Jacobs
Jul 26 at 2:10
It’s certainly possible: just scale each ray from the origin linearly to take the edge of the cube to the edge of the tetrahedron. I’m interested to see whether anyone can do it more elegantly than that.....
– G Tony Jacobs
Jul 26 at 2:10
that's what I was thinking. But I'm not sure how to express that process by a function
– More water plz
Jul 26 at 2:26
that's what I was thinking. But I'm not sure how to express that process by a function
– More water plz
Jul 26 at 2:26
You would have to write down separate functions depending which face of the cube and the tetrahedron each ray intersects. Like I said, not very elegant.
– G Tony Jacobs
Jul 26 at 2:57
You would have to write down separate functions depending which face of the cube and the tetrahedron each ray intersects. Like I said, not very elegant.
– G Tony Jacobs
Jul 26 at 2:57
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
A potentially useful observation is that the unit cube and unit simplex in any number of dimensions can be expressed in terms of the maximum norm $||vec x||_M=max(|x_i|)$ and the taxicab norm $||vec x||_T=sum_i|x_i|$, allowing us to make use of the convenient properties of norms:
$$
textUnit Cube: mathcal C=
$$
$$
textUnit Simplex: mathcal S=vec x
$$
Note that your cube and tetrahedron are the $d=3$ case.
One way to transform between these shapes, then, is to map each vector $vec x$ in the cube to a vector $vec y$ in the same direction such that $||vec x||_M=||vec y||_T$. This gives us a simple function, which multiplies $vec x$ by some positive constant $c(vec x)$, for which we can find an expression.
$$beginalign
||vec x||_M & = ||cvec x||_T \
||vec x||_M & = |c| ||vec x||_T \
c & = fracvec x
endalign$$
This allows us to write down a function, taking care of the problematic zero case:
$$
f:mathcal Ctomathcal S, f(vec x) = begincases
fracvec xvec x & vec x neq vec 0 \
vec 0 & vec x = vec 0
endcases
$$
From here, showing that this is a bijection should be straightforward.
answered Jul 26 at 3:00
Kajelad
1,848619
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
A potentially useful observation is that the unit cube and unit simplex in any number of dimensions can be expressed in terms of the maximum norm $||vec x||_M=max(|x_i|)$ and the taxicab norm $||vec x||_T=sum_i|x_i|$, allowing us to make use of the convenient properties of norms:
$$
textUnit Cube: mathcal C=
$$
$$
textUnit Simplex: mathcal S=vec x
$$
Note that your cube and tetrahedron are the $d=3$ case.
One way to transform between these shapes, then, is to map each vector $vec x$ in the cube to a vector $vec y$ in the same direction such that $||vec x||_M=||vec y||_T$. This gives us a simple function, which multiplies $vec x$ by some positive constant $c(vec x)$, for which we can find an expression.
$$beginalign
||vec x||_M & = ||cvec x||_T \
||vec x||_M & = |c| ||vec x||_T \
c & = fracvec x
endalign$$
This allows us to write down a function, taking care of the problematic zero case:
$$
f:mathcal Ctomathcal S, f(vec x) = begincases
fracvec xvec x & vec x neq vec 0 \
vec 0 & vec x = vec 0
endcases
$$
From here, showing that this is a bijection should be straightforward.
answered Jul 26 at 3:00
Kajelad
1,848619
add a comment |Â
up vote
2
down vote
A potentially useful observation is that the unit cube and unit simplex in any number of dimensions can be expressed in terms of the maximum norm $||vec x||_M=max(|x_i|)$ and the taxicab norm $||vec x||_T=sum_i|x_i|$, allowing us to make use of the convenient properties of norms:
$$
textUnit Cube: mathcal C=
$$
$$
textUnit Simplex: mathcal S=vec x
$$
Note that your cube and tetrahedron are the $d=3$ case.
One way to transform between these shapes, then, is to map each vector $vec x$ in the cube to a vector $vec y$ in the same direction such that $||vec x||_M=||vec y||_T$. This gives us a simple function, which multiplies $vec x$ by some positive constant $c(vec x)$, for which we can find an expression.
$$beginalign
||vec x||_M & = ||cvec x||_T \
||vec x||_M & = |c| ||vec x||_T \
c & = fracvec x
endalign$$
This allows us to write down a function, taking care of the problematic zero case:
$$
f:mathcal Ctomathcal S, f(vec x) = begincases
fracvec xvec x & vec x neq vec 0 \
vec 0 & vec x = vec 0
endcases
$$
From here, showing that this is a bijection should be straightforward.
answered Jul 26 at 3:00
Kajelad
1,848619
add a comment |Â
up vote
2
down vote
up vote
2
down vote
A potentially useful observation is that the unit cube and unit simplex in any number of dimensions can be expressed in terms of the maximum norm $||vec x||_M=max(|x_i|)$ and the taxicab norm $||vec x||_T=sum_i|x_i|$, allowing us to make use of the convenient properties of norms:
$$
textUnit Cube: mathcal C=
$$
$$
textUnit Simplex: mathcal S=vec x
$$
Note that your cube and tetrahedron are the $d=3$ case.
One way to transform between these shapes, then, is to map each vector $vec x$ in the cube to a vector $vec y$ in the same direction such that $||vec x||_M=||vec y||_T$. This gives us a simple function, which multiplies $vec x$ by some positive constant $c(vec x)$, for which we can find an expression.
$$beginalign
||vec x||_M & = ||cvec x||_T \
||vec x||_M & = |c| ||vec x||_T \
c & = fracvec x
endalign$$
This allows us to write down a function, taking care of the problematic zero case:
$$
f:mathcal Ctomathcal S, f(vec x) = begincases
fracvec xvec x & vec x neq vec 0 \
vec 0 & vec x = vec 0
endcases
$$
From here, showing that this is a bijection should be straightforward.
answered Jul 26 at 3:00
Kajelad
1,848619
A potentially useful observation is that the unit cube and unit simplex in any number of dimensions can be expressed in terms of the maximum norm $||vec x||_M=max(|x_i|)$ and the taxicab norm $||vec x||_T=sum_i|x_i|$, allowing us to make use of the convenient properties of norms:
$$
textUnit Cube: mathcal C=
$$
$$
textUnit Simplex: mathcal S=vec x
$$
Note that your cube and tetrahedron are the $d=3$ case.
One way to transform between these shapes, then, is to map each vector $vec x$ in the cube to a vector $vec y$ in the same direction such that $||vec x||_M=||vec y||_T$. This gives us a simple function, which multiplies $vec x$ by some positive constant $c(vec x)$, for which we can find an expression.
$$beginalign
||vec x||_M & = ||cvec x||_T \
||vec x||_M & = |c| ||vec x||_T \
c & = fracvec x
endalign$$
This allows us to write down a function, taking care of the problematic zero case:
$$
f:mathcal Ctomathcal S, f(vec x) = begincases
fracvec xvec x & vec x neq vec 0 \
vec 0 & vec x = vec 0
endcases
$$
From here, showing that this is a bijection should be straightforward.
answered Jul 26 at 3:00
Kajelad
1,848619
edited Jul 26 at 3:06
answered Jul 26 at 3:00
Kajelad
1,848619
answered Jul 26 at 3:00
Kajelad
1,848619
answered Jul 26 at 3:00
Kajelad
1,848619
1,848619
add a comment |Â
add a comment |Â
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It’s certainly possible: just scale each ray from the origin linearly to take the edge of the cube to the edge of the tetrahedron. I’m interested to see whether anyone can do it more elegantly than that.....
– G Tony Jacobs
Jul 26 at 2:10
that's what I was thinking. But I'm not sure how to express that process by a function
– More water plz
Jul 26 at 2:26
You would have to write down separate functions depending which face of the cube and the tetrahedron each ray intersects. Like I said, not very elegant.
– G Tony Jacobs
Jul 26 at 2:57