Optimum mapping between tesselated parallelograms and tesselated rectangles?
I have a lattice whose points are the vertices of many tessellated parallelograms. Each point is located at $mathbf{x}=alpha mathbf a + beta mathbf b$ where $alpha$ and $beta$ are integers and $mathbf {a,b}$ are linearly independent. I would like to (bijectively) map this on to a rectangular lattice $mathbf{x}'=lambda mathbf x + mu mathbf y$, where $lambda,mu$ are again integers but $mathbf{x,y}$ are not only linearly independent but also are orthogonal. Except that they are orthogonal, I have freedom to choose any $mathbf x$ and $mathbf y$ that I want. Each point should move as little as possible when moved from its place on the parallel lattice to its new place on the rectangular lattice. How can this be done?
analytic-geometry transformation
asked Nov 27 '18 at 15:46
Display Name
1988
add a comment |
I have a lattice whose points are the vertices of many tessellated parallelograms. Each point is located at $mathbf{x}=alpha mathbf a + beta mathbf b$ where $alpha$ and $beta$ are integers and $mathbf {a,b}$ are linearly independent. I would like to (bijectively) map this on to a rectangular lattice $mathbf{x}'=lambda mathbf x + mu mathbf y$, where $lambda,mu$ are again integers but $mathbf{x,y}$ are not only linearly independent but also are orthogonal. Except that they are orthogonal, I have freedom to choose any $mathbf x$ and $mathbf y$ that I want. Each point should move as little as possible when moved from its place on the parallel lattice to its new place on the rectangular lattice. How can this be done?
analytic-geometry transformation
asked Nov 27 '18 at 15:46
Display Name
1988
The restrictions of your problem are not clear. If you choose very short basis vectors $|mathbf x|=|mathbf y| = varepsilon ll |mathbf a|, |mathbf b|$, then all points will need to move not more than $varepsilon$, which can be arbitrarily small
– Vasily Mitch
Nov 27 '18 at 16:13
That's a good catch. I forgot to mention the requirement that the mapping between the two lattices be 1-1. If the basis vectors are made very small, too many unfilled points would open up on the rectangular lattice. (The question has been edited to reflect this.)
– Display Name
Nov 27 '18 at 16:41
How about choosing $mathbf x$ and $mathbf y$ so that points along the longer diagonals remain fixed?
– amd
Nov 27 '18 at 23:57
That sounds like a good idea, but I am not sure how to prove that it is optimal.
– Display Name
Nov 29 '18 at 14:53
add a comment |
I have a lattice whose points are the vertices of many tessellated parallelograms. Each point is located at $mathbf{x}=alpha mathbf a + beta mathbf b$ where $alpha$ and $beta$ are integers and $mathbf {a,b}$ are linearly independent. I would like to (bijectively) map this on to a rectangular lattice $mathbf{x}'=lambda mathbf x + mu mathbf y$, where $lambda,mu$ are again integers but $mathbf{x,y}$ are not only linearly independent but also are orthogonal. Except that they are orthogonal, I have freedom to choose any $mathbf x$ and $mathbf y$ that I want. Each point should move as little as possible when moved from its place on the parallel lattice to its new place on the rectangular lattice. How can this be done?
analytic-geometry transformation
asked Nov 27 '18 at 15:46
Display Name
1988
I have a lattice whose points are the vertices of many tessellated parallelograms. Each point is located at $mathbf{x}=alpha mathbf a + beta mathbf b$ where $alpha$ and $beta$ are integers and $mathbf {a,b}$ are linearly independent. I would like to (bijectively) map this on to a rectangular lattice $mathbf{x}'=lambda mathbf x + mu mathbf y$, where $lambda,mu$ are again integers but $mathbf{x,y}$ are not only linearly independent but also are orthogonal. Except that they are orthogonal, I have freedom to choose any $mathbf x$ and $mathbf y$ that I want. Each point should move as little as possible when moved from its place on the parallel lattice to its new place on the rectangular lattice. How can this be done?
analytic-geometry transformation
analytic-geometry transformation
asked Nov 27 '18 at 15:46
Display Name
1988
asked Nov 27 '18 at 15:46
Display Name
1988
edited Nov 27 '18 at 16:42
asked Nov 27 '18 at 15:46
Display Name
1988
asked Nov 27 '18 at 15:46
Display Name
1988
asked Nov 27 '18 at 15:46
Display Name
1988
1988
The restrictions of your problem are not clear. If you choose very short basis vectors $|mathbf x|=|mathbf y| = varepsilon ll |mathbf a|, |mathbf b|$, then all points will need to move not more than $varepsilon$, which can be arbitrarily small
– Vasily Mitch
Nov 27 '18 at 16:13
That's a good catch. I forgot to mention the requirement that the mapping between the two lattices be 1-1. If the basis vectors are made very small, too many unfilled points would open up on the rectangular lattice. (The question has been edited to reflect this.)
– Display Name
Nov 27 '18 at 16:41
How about choosing $mathbf x$ and $mathbf y$ so that points along the longer diagonals remain fixed?
– amd
Nov 27 '18 at 23:57
That sounds like a good idea, but I am not sure how to prove that it is optimal.
– Display Name
Nov 29 '18 at 14:53
add a comment |
The restrictions of your problem are not clear. If you choose very short basis vectors $|mathbf x|=|mathbf y| = varepsilon ll |mathbf a|, |mathbf b|$, then all points will need to move not more than $varepsilon$, which can be arbitrarily small
– Vasily Mitch
Nov 27 '18 at 16:13
That's a good catch. I forgot to mention the requirement that the mapping between the two lattices be 1-1. If the basis vectors are made very small, too many unfilled points would open up on the rectangular lattice. (The question has been edited to reflect this.)
– Display Name
Nov 27 '18 at 16:41
How about choosing $mathbf x$ and $mathbf y$ so that points along the longer diagonals remain fixed?
– amd
Nov 27 '18 at 23:57
That sounds like a good idea, but I am not sure how to prove that it is optimal.
– Display Name
Nov 29 '18 at 14:53
The restrictions of your problem are not clear. If you choose very short basis vectors $|mathbf x|=|mathbf y| = varepsilon ll |mathbf a|, |mathbf b|$, then all points will need to move not more than $varepsilon$, which can be arbitrarily small
– Vasily Mitch
Nov 27 '18 at 16:13
The restrictions of your problem are not clear. If you choose very short basis vectors $|mathbf x|=|mathbf y| = varepsilon ll |mathbf a|, |mathbf b|$, then all points will need to move not more than $varepsilon$, which can be arbitrarily small
– Vasily Mitch
Nov 27 '18 at 16:13
That's a good catch. I forgot to mention the requirement that the mapping between the two lattices be 1-1. If the basis vectors are made very small, too many unfilled points would open up on the rectangular lattice. (The question has been edited to reflect this.)
– Display Name
Nov 27 '18 at 16:41
That's a good catch. I forgot to mention the requirement that the mapping between the two lattices be 1-1. If the basis vectors are made very small, too many unfilled points would open up on the rectangular lattice. (The question has been edited to reflect this.)
– Display Name
Nov 27 '18 at 16:41
How about choosing $mathbf x$ and $mathbf y$ so that points along the longer diagonals remain fixed?
– amd
Nov 27 '18 at 23:57
How about choosing $mathbf x$ and $mathbf y$ so that points along the longer diagonals remain fixed?
– amd
Nov 27 '18 at 23:57
That sounds like a good idea, but I am not sure how to prove that it is optimal.
– Display Name
Nov 29 '18 at 14:53
That sounds like a good idea, but I am not sure how to prove that it is optimal.
– Display Name
Nov 29 '18 at 14:53
add a comment |
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The restrictions of your problem are not clear. If you choose very short basis vectors $|mathbf x|=|mathbf y| = varepsilon ll |mathbf a|, |mathbf b|$, then all points will need to move not more than $varepsilon$, which can be arbitrarily small
– Vasily Mitch
Nov 27 '18 at 16:13
That's a good catch. I forgot to mention the requirement that the mapping between the two lattices be 1-1. If the basis vectors are made very small, too many unfilled points would open up on the rectangular lattice. (The question has been edited to reflect this.)
– Display Name
Nov 27 '18 at 16:41
How about choosing $mathbf x$ and $mathbf y$ so that points along the longer diagonals remain fixed?
– amd
Nov 27 '18 at 23:57
That sounds like a good idea, but I am not sure how to prove that it is optimal.
– Display Name
Nov 29 '18 at 14:53