Reindexing two lists / flattening indices
$begingroup$
If I have two lists/indiced that run from -N to N, call them A and B. How can I create a third index, C such that C goes from 0 to $(2N+1)^2-1$?
combinatorics
asked Jan 17 at 20:15
yankeefan11yankeefan11
1,0041920
$endgroup$
add a comment |
$begingroup$
If I have two lists/indiced that run from -N to N, call them A and B. How can I create a third index, C such that C goes from 0 to $(2N+1)^2-1$?
combinatorics
asked Jan 17 at 20:15
yankeefan11yankeefan11
1,0041920
$endgroup$
add a comment |
$begingroup$
If I have two lists/indiced that run from -N to N, call them A and B. How can I create a third index, C such that C goes from 0 to $(2N+1)^2-1$?
combinatorics
asked Jan 17 at 20:15
yankeefan11yankeefan11
1,0041920
$endgroup$
If I have two lists/indiced that run from -N to N, call them A and B. How can I create a third index, C such that C goes from 0 to $(2N+1)^2-1$?
combinatorics
combinatorics
asked Jan 17 at 20:15
yankeefan11yankeefan11
1,0041920
asked Jan 17 at 20:15
yankeefan11yankeefan11
1,0041920
asked Jan 17 at 20:15
yankeefan11yankeefan11
1,0041920
asked Jan 17 at 20:15
yankeefan11yankeefan11
1,0041920
asked Jan 17 at 20:15
yankeefan11yankeefan11
1,0041920
1,0041920
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Call $i = -N, -N + 1, cdots, N$ the index of $A$, and $j = -N, -N + 1, cdots, N$ the index of $B$. The index
$$ bbox[5px,border:2px solid blue]
{
k = (i + N)(2N + 1) + (j + N)
}
$$
goes from $k = 0$ for $(i, j) = (-N, -N)$ to $k = 2N(2N + 1) + 2N = (2N + 1)^2 - 1$ for $(i, j) = (N, N)$
answered Jan 17 at 20:34
caveraccaverac
14.8k31130
$endgroup$
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
$begingroup$
Call $i = -N, -N + 1, cdots, N$ the index of $A$, and $j = -N, -N + 1, cdots, N$ the index of $B$. The index
$$ bbox[5px,border:2px solid blue]
{
k = (i + N)(2N + 1) + (j + N)
}
$$
goes from $k = 0$ for $(i, j) = (-N, -N)$ to $k = 2N(2N + 1) + 2N = (2N + 1)^2 - 1$ for $(i, j) = (N, N)$
answered Jan 17 at 20:34
caveraccaverac
14.8k31130
$endgroup$
add a comment |
$begingroup$
Call $i = -N, -N + 1, cdots, N$ the index of $A$, and $j = -N, -N + 1, cdots, N$ the index of $B$. The index
$$ bbox[5px,border:2px solid blue]
{
k = (i + N)(2N + 1) + (j + N)
}
$$
goes from $k = 0$ for $(i, j) = (-N, -N)$ to $k = 2N(2N + 1) + 2N = (2N + 1)^2 - 1$ for $(i, j) = (N, N)$
answered Jan 17 at 20:34
caveraccaverac
14.8k31130
$endgroup$
add a comment |
$begingroup$
Call $i = -N, -N + 1, cdots, N$ the index of $A$, and $j = -N, -N + 1, cdots, N$ the index of $B$. The index
$$ bbox[5px,border:2px solid blue]
{
k = (i + N)(2N + 1) + (j + N)
}
$$
goes from $k = 0$ for $(i, j) = (-N, -N)$ to $k = 2N(2N + 1) + 2N = (2N + 1)^2 - 1$ for $(i, j) = (N, N)$
answered Jan 17 at 20:34
caveraccaverac
14.8k31130
$endgroup$
Call $i = -N, -N + 1, cdots, N$ the index of $A$, and $j = -N, -N + 1, cdots, N$ the index of $B$. The index
$$ bbox[5px,border:2px solid blue]
{
k = (i + N)(2N + 1) + (j + N)
}
$$
goes from $k = 0$ for $(i, j) = (-N, -N)$ to $k = 2N(2N + 1) + 2N = (2N + 1)^2 - 1$ for $(i, j) = (N, N)$
answered Jan 17 at 20:34
caveraccaverac
14.8k31130
answered Jan 17 at 20:34
caveraccaverac
14.8k31130
answered Jan 17 at 20:34
caveraccaverac
14.8k31130
answered Jan 17 at 20:34
caveraccaverac
14.8k31130
14.8k31130
add a comment |
add a comment |
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