Enumerating the image of an integer matrix applied to a lattice
0
$begingroup$
Let $Ainmathbb{Z}^{n times m}$ and $xin mathbb{Z}^m$. Also suppose we are given $l,uinmathbb{Z}^m$. I would like to efficiently enumerate
$${Ax ,|, l_i le x_i le u_i text{ for all } i}$$
The subscript $i$ refers to the $i^text{th}$ element of the vector.
$A$ is not necessarily of full rank, so the problem is nontrivial.
Is there a simple algorithm for this problem? Even just a reference would be appreciated.
Thanks.
discrete-mathematics algorithms integer-lattices
asked Dec 16 '18 at 19:50
KurtKurt
550215
$endgroup$
add a comment |
0
$begingroup$
Let $Ainmathbb{Z}^{n times m}$ and $xin mathbb{Z}^m$. Also suppose we are given $l,uinmathbb{Z}^m$. I would like to efficiently enumerate
$${Ax ,|, l_i le x_i le u_i text{ for all } i}$$
The subscript $i$ refers to the $i^text{th}$ element of the vector.
$A$ is not necessarily of full rank, so the problem is nontrivial.
Is there a simple algorithm for this problem? Even just a reference would be appreciated.
Thanks.
discrete-mathematics algorithms integer-lattices
asked Dec 16 '18 at 19:50
KurtKurt
550215
$endgroup$
$begingroup$
What constitutes efficiency? The set you're hoping to enumerate has, in the general case, $J = pi (u_i - l_i)$ entries, each with $n$ entries, so simply writing down the set requires time $Jn$ in the worst case.
$endgroup$
– John Hughes
Dec 16 '18 at 20:14
$begingroup$
The naiive approach is to enumerate all of the $x$, and compute $Ax$ for each $x$. However, if $A$ is not of full rank, then many of the possible $x$ produce the same value of $Ax$. I would like the complexity of the algorithm to scale with the size of the image and not with the size of the domain.
$endgroup$
– Kurt
Dec 16 '18 at 20:21
$begingroup$
"However, if $A$ is not of full rank, then many of the possible $x$ produce the same value of $Ax$" is false in general, although it may be true for some particular bounds. Example: $A = pmatrix{10 & 1 \ 20 & 2}$, $0 le x le 2, 0 le y le 3$. In view of this, I tend to doubt that there's a simple output-dependent algorithm.
$endgroup$
– John Hughes
Dec 16 '18 at 20:26
$begingroup$
Could you provide more details on the context of the problem? I.e. what sizes are $m$, $n$, and the lattice? Is the rank much smaller than $min(m,n)$? If $A$ is big enough and the rank is small enough there may be heuristic algorithms that work.
$endgroup$
– tch
Dec 21 '18 at 22:01
add a comment |
0
0
0
$begingroup$
Let $Ainmathbb{Z}^{n times m}$ and $xin mathbb{Z}^m$. Also suppose we are given $l,uinmathbb{Z}^m$. I would like to efficiently enumerate
$${Ax ,|, l_i le x_i le u_i text{ for all } i}$$
The subscript $i$ refers to the $i^text{th}$ element of the vector.
$A$ is not necessarily of full rank, so the problem is nontrivial.
Is there a simple algorithm for this problem? Even just a reference would be appreciated.
Thanks.
discrete-mathematics algorithms integer-lattices
asked Dec 16 '18 at 19:50
KurtKurt
550215
$endgroup$
Let $Ainmathbb{Z}^{n times m}$ and $xin mathbb{Z}^m$. Also suppose we are given $l,uinmathbb{Z}^m$. I would like to efficiently enumerate
$${Ax ,|, l_i le x_i le u_i text{ for all } i}$$
The subscript $i$ refers to the $i^text{th}$ element of the vector.
$A$ is not necessarily of full rank, so the problem is nontrivial.
Is there a simple algorithm for this problem? Even just a reference would be appreciated.
Thanks.
discrete-mathematics algorithms integer-lattices
discrete-mathematics algorithms integer-lattices
asked Dec 16 '18 at 19:50
KurtKurt
550215
asked Dec 16 '18 at 19:50
KurtKurt
550215
asked Dec 16 '18 at 19:50
KurtKurt
550215
asked Dec 16 '18 at 19:50
KurtKurt
550215
asked Dec 16 '18 at 19:50
KurtKurt
550215
550215
$begingroup$
What constitutes efficiency? The set you're hoping to enumerate has, in the general case, $J = pi (u_i - l_i)$ entries, each with $n$ entries, so simply writing down the set requires time $Jn$ in the worst case.
$endgroup$
– John Hughes
Dec 16 '18 at 20:14
$begingroup$
The naiive approach is to enumerate all of the $x$, and compute $Ax$ for each $x$. However, if $A$ is not of full rank, then many of the possible $x$ produce the same value of $Ax$. I would like the complexity of the algorithm to scale with the size of the image and not with the size of the domain.
$endgroup$
– Kurt
Dec 16 '18 at 20:21
$begingroup$
"However, if $A$ is not of full rank, then many of the possible $x$ produce the same value of $Ax$" is false in general, although it may be true for some particular bounds. Example: $A = pmatrix{10 & 1 \ 20 & 2}$, $0 le x le 2, 0 le y le 3$. In view of this, I tend to doubt that there's a simple output-dependent algorithm.
$endgroup$
– John Hughes
Dec 16 '18 at 20:26
$begingroup$
Could you provide more details on the context of the problem? I.e. what sizes are $m$, $n$, and the lattice? Is the rank much smaller than $min(m,n)$? If $A$ is big enough and the rank is small enough there may be heuristic algorithms that work.
$endgroup$
– tch
Dec 21 '18 at 22:01
add a comment |
$begingroup$
What constitutes efficiency? The set you're hoping to enumerate has, in the general case, $J = pi (u_i - l_i)$ entries, each with $n$ entries, so simply writing down the set requires time $Jn$ in the worst case.
$endgroup$
– John Hughes
Dec 16 '18 at 20:14
$begingroup$
The naiive approach is to enumerate all of the $x$, and compute $Ax$ for each $x$. However, if $A$ is not of full rank, then many of the possible $x$ produce the same value of $Ax$. I would like the complexity of the algorithm to scale with the size of the image and not with the size of the domain.
$endgroup$
– Kurt
Dec 16 '18 at 20:21
$begingroup$
"However, if $A$ is not of full rank, then many of the possible $x$ produce the same value of $Ax$" is false in general, although it may be true for some particular bounds. Example: $A = pmatrix{10 & 1 \ 20 & 2}$, $0 le x le 2, 0 le y le 3$. In view of this, I tend to doubt that there's a simple output-dependent algorithm.
$endgroup$
– John Hughes
Dec 16 '18 at 20:26
$begingroup$
Could you provide more details on the context of the problem? I.e. what sizes are $m$, $n$, and the lattice? Is the rank much smaller than $min(m,n)$? If $A$ is big enough and the rank is small enough there may be heuristic algorithms that work.
$endgroup$
– tch
Dec 21 '18 at 22:01
$begingroup$
What constitutes efficiency? The set you're hoping to enumerate has, in the general case, $J = pi (u_i - l_i)$ entries, each with $n$ entries, so simply writing down the set requires time $Jn$ in the worst case.
$endgroup$
– John Hughes
Dec 16 '18 at 20:14
$begingroup$
What constitutes efficiency? The set you're hoping to enumerate has, in the general case, $J = pi (u_i - l_i)$ entries, each with $n$ entries, so simply writing down the set requires time $Jn$ in the worst case.
$endgroup$
– John Hughes
Dec 16 '18 at 20:14
$begingroup$
The naiive approach is to enumerate all of the $x$, and compute $Ax$ for each $x$. However, if $A$ is not of full rank, then many of the possible $x$ produce the same value of $Ax$. I would like the complexity of the algorithm to scale with the size of the image and not with the size of the domain.
$endgroup$
– Kurt
Dec 16 '18 at 20:21
$begingroup$
The naiive approach is to enumerate all of the $x$, and compute $Ax$ for each $x$. However, if $A$ is not of full rank, then many of the possible $x$ produce the same value of $Ax$. I would like the complexity of the algorithm to scale with the size of the image and not with the size of the domain.
$endgroup$
– Kurt
Dec 16 '18 at 20:21
$begingroup$
"However, if $A$ is not of full rank, then many of the possible $x$ produce the same value of $Ax$" is false in general, although it may be true for some particular bounds. Example: $A = pmatrix{10 & 1 \ 20 & 2}$, $0 le x le 2, 0 le y le 3$. In view of this, I tend to doubt that there's a simple output-dependent algorithm.
$endgroup$
– John Hughes
Dec 16 '18 at 20:26
$begingroup$
"However, if $A$ is not of full rank, then many of the possible $x$ produce the same value of $Ax$" is false in general, although it may be true for some particular bounds. Example: $A = pmatrix{10 & 1 \ 20 & 2}$, $0 le x le 2, 0 le y le 3$. In view of this, I tend to doubt that there's a simple output-dependent algorithm.
$endgroup$
– John Hughes
Dec 16 '18 at 20:26
$begingroup$
Could you provide more details on the context of the problem? I.e. what sizes are $m$, $n$, and the lattice? Is the rank much smaller than $min(m,n)$? If $A$ is big enough and the rank is small enough there may be heuristic algorithms that work.
$endgroup$
– tch
Dec 21 '18 at 22:01
$begingroup$
Could you provide more details on the context of the problem? I.e. what sizes are $m$, $n$, and the lattice? Is the rank much smaller than $min(m,n)$? If $A$ is big enough and the rank is small enough there may be heuristic algorithms that work.
$endgroup$
– tch
Dec 21 '18 at 22:01
add a comment |
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$begingroup$
What constitutes efficiency? The set you're hoping to enumerate has, in the general case, $J = pi (u_i - l_i)$ entries, each with $n$ entries, so simply writing down the set requires time $Jn$ in the worst case.
$endgroup$
– John Hughes
Dec 16 '18 at 20:14
$begingroup$
The naiive approach is to enumerate all of the $x$, and compute $Ax$ for each $x$. However, if $A$ is not of full rank, then many of the possible $x$ produce the same value of $Ax$. I would like the complexity of the algorithm to scale with the size of the image and not with the size of the domain.
$endgroup$
– Kurt
Dec 16 '18 at 20:21
$begingroup$
"However, if $A$ is not of full rank, then many of the possible $x$ produce the same value of $Ax$" is false in general, although it may be true for some particular bounds. Example: $A = pmatrix{10 & 1 \ 20 & 2}$, $0 le x le 2, 0 le y le 3$. In view of this, I tend to doubt that there's a simple output-dependent algorithm.
$endgroup$
– John Hughes
Dec 16 '18 at 20:26
$begingroup$
Could you provide more details on the context of the problem? I.e. what sizes are $m$, $n$, and the lattice? Is the rank much smaller than $min(m,n)$? If $A$ is big enough and the rank is small enough there may be heuristic algorithms that work.
$endgroup$
– tch
Dec 21 '18 at 22:01