Bucket computation, cutting array with lines
$begingroup$
Given an NxN array, drawing a line from the edge's midpoint to the opposite field how can the N buckets be found covering the majority of the line's path?
A visual aid:
Is there a better way to this than computing the linear equation and iteratively advancing in tiny steps to check which bucket we fall on? If not what is the lower limit step size to choose?
algorithms computational-geometry
asked 3 hours ago
KilianKilian
1162
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Kilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
Given an NxN array, drawing a line from the edge's midpoint to the opposite field how can the N buckets be found covering the majority of the line's path?
A visual aid:
Is there a better way to this than computing the linear equation and iteratively advancing in tiny steps to check which bucket we fall on? If not what is the lower limit step size to choose?
algorithms computational-geometry
asked 3 hours ago
KilianKilian
1162
New contributor
Kilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Given an NxN array, drawing a line from the edge's midpoint to the opposite field how can the N buckets be found covering the majority of the line's path?
A visual aid:
Is there a better way to this than computing the linear equation and iteratively advancing in tiny steps to check which bucket we fall on? If not what is the lower limit step size to choose?
algorithms computational-geometry
asked 3 hours ago
KilianKilian
1162
New contributor
Kilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Given an NxN array, drawing a line from the edge's midpoint to the opposite field how can the N buckets be found covering the majority of the line's path?
A visual aid:
Is there a better way to this than computing the linear equation and iteratively advancing in tiny steps to check which bucket we fall on? If not what is the lower limit step size to choose?
algorithms computational-geometry
algorithms computational-geometry
asked 3 hours ago
KilianKilian
1162
New contributor
Kilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
KilianKilian
1162
New contributor
Kilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
KilianKilian
1162
New contributor
Kilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
KilianKilian
1162
asked 3 hours ago
KilianKilian
1162
1162
New contributor
Kilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kilian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For the considered line you know the start position (xs, ys) and the end (xe, ye). For instance on the N=7 visual you provide, let's take the first upper left line:
xs = 0
ys = 4.5
xe = 7
ye = 2.5
(I use coordinates right, down starting from upper left corner).
Let's take a parameter t of the linear position on the line with t=0 on start and t=1 on end. You know how many verticals or horizontals your line crosses and at which t value. In our exemple:
- 6 verticals crossed at t = v/7 with v=1 to 6 (not counting outlines).
- 2 horizontal crossed at t = 0.25 and t = 0.75.
Now just sort t increasing all these crossings:
(1/7, V), (0.25, H) (2/7, V), (3/7, V), (4/7, V), (5/7, V), (0.75, H), (6/7, V).
They correspond to cell change starting in cell (1, 5) to cell (7, 3). So you can compute the relative length of the line in each cell. For instance:
cell (1, 5) L = 1/7-0 = 1/7
cell (2, 5) L = 0.25-1/7
...
And you know how to select the X pixels now.
answered 1 hour ago
VinceVince
1844
$endgroup$
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
1 hour ago
1
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
51 mins ago
add a comment |
$begingroup$
You probably want to look at Bresenham's line drawing algorithm. Rather than moving along the line itself, you move along the $x$-coordinates a "pixel" (i.e., array entry) at a time and compute which $y$-value puts the greatest part of the line in the pixel $(x,y)$.
answered 51 mins ago
David RicherbyDavid Richerby
66.4k15101190
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the considered line you know the start position (xs, ys) and the end (xe, ye). For instance on the N=7 visual you provide, let's take the first upper left line:
xs = 0
ys = 4.5
xe = 7
ye = 2.5
(I use coordinates right, down starting from upper left corner).
Let's take a parameter t of the linear position on the line with t=0 on start and t=1 on end. You know how many verticals or horizontals your line crosses and at which t value. In our exemple:
- 6 verticals crossed at t = v/7 with v=1 to 6 (not counting outlines).
- 2 horizontal crossed at t = 0.25 and t = 0.75.
Now just sort t increasing all these crossings:
(1/7, V), (0.25, H) (2/7, V), (3/7, V), (4/7, V), (5/7, V), (0.75, H), (6/7, V).
They correspond to cell change starting in cell (1, 5) to cell (7, 3). So you can compute the relative length of the line in each cell. For instance:
cell (1, 5) L = 1/7-0 = 1/7
cell (2, 5) L = 0.25-1/7
...
And you know how to select the X pixels now.
answered 1 hour ago
VinceVince
1844
$endgroup$
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
1 hour ago
1
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
51 mins ago
add a comment |
$begingroup$
For the considered line you know the start position (xs, ys) and the end (xe, ye). For instance on the N=7 visual you provide, let's take the first upper left line:
xs = 0
ys = 4.5
xe = 7
ye = 2.5
(I use coordinates right, down starting from upper left corner).
Let's take a parameter t of the linear position on the line with t=0 on start and t=1 on end. You know how many verticals or horizontals your line crosses and at which t value. In our exemple:
- 6 verticals crossed at t = v/7 with v=1 to 6 (not counting outlines).
- 2 horizontal crossed at t = 0.25 and t = 0.75.
Now just sort t increasing all these crossings:
(1/7, V), (0.25, H) (2/7, V), (3/7, V), (4/7, V), (5/7, V), (0.75, H), (6/7, V).
They correspond to cell change starting in cell (1, 5) to cell (7, 3). So you can compute the relative length of the line in each cell. For instance:
cell (1, 5) L = 1/7-0 = 1/7
cell (2, 5) L = 0.25-1/7
...
And you know how to select the X pixels now.
answered 1 hour ago
VinceVince
1844
$endgroup$
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
1 hour ago
1
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
51 mins ago
add a comment |
$begingroup$
For the considered line you know the start position (xs, ys) and the end (xe, ye). For instance on the N=7 visual you provide, let's take the first upper left line:
xs = 0
ys = 4.5
xe = 7
ye = 2.5
(I use coordinates right, down starting from upper left corner).
Let's take a parameter t of the linear position on the line with t=0 on start and t=1 on end. You know how many verticals or horizontals your line crosses and at which t value. In our exemple:
- 6 verticals crossed at t = v/7 with v=1 to 6 (not counting outlines).
- 2 horizontal crossed at t = 0.25 and t = 0.75.
Now just sort t increasing all these crossings:
(1/7, V), (0.25, H) (2/7, V), (3/7, V), (4/7, V), (5/7, V), (0.75, H), (6/7, V).
They correspond to cell change starting in cell (1, 5) to cell (7, 3). So you can compute the relative length of the line in each cell. For instance:
cell (1, 5) L = 1/7-0 = 1/7
cell (2, 5) L = 0.25-1/7
...
And you know how to select the X pixels now.
answered 1 hour ago
VinceVince
1844
$endgroup$
For the considered line you know the start position (xs, ys) and the end (xe, ye). For instance on the N=7 visual you provide, let's take the first upper left line:
xs = 0
ys = 4.5
xe = 7
ye = 2.5
(I use coordinates right, down starting from upper left corner).
Let's take a parameter t of the linear position on the line with t=0 on start and t=1 on end. You know how many verticals or horizontals your line crosses and at which t value. In our exemple:
- 6 verticals crossed at t = v/7 with v=1 to 6 (not counting outlines).
- 2 horizontal crossed at t = 0.25 and t = 0.75.
Now just sort t increasing all these crossings:
(1/7, V), (0.25, H) (2/7, V), (3/7, V), (4/7, V), (5/7, V), (0.75, H), (6/7, V).
They correspond to cell change starting in cell (1, 5) to cell (7, 3). So you can compute the relative length of the line in each cell. For instance:
cell (1, 5) L = 1/7-0 = 1/7
cell (2, 5) L = 0.25-1/7
...
And you know how to select the X pixels now.
answered 1 hour ago
VinceVince
1844
answered 1 hour ago
VinceVince
1844
answered 1 hour ago
VinceVince
1844
answered 1 hour ago
VinceVince
1844
1844
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
1 hour ago
1
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
51 mins ago
add a comment |
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
1 hour ago
1
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
51 mins ago
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
1 hour ago
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
1 hour ago
1
1
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
51 mins ago
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
51 mins ago
add a comment |
$begingroup$
You probably want to look at Bresenham's line drawing algorithm. Rather than moving along the line itself, you move along the $x$-coordinates a "pixel" (i.e., array entry) at a time and compute which $y$-value puts the greatest part of the line in the pixel $(x,y)$.
answered 51 mins ago
David RicherbyDavid Richerby
66.4k15101190
$endgroup$
add a comment |
$begingroup$
You probably want to look at Bresenham's line drawing algorithm. Rather than moving along the line itself, you move along the $x$-coordinates a "pixel" (i.e., array entry) at a time and compute which $y$-value puts the greatest part of the line in the pixel $(x,y)$.
answered 51 mins ago
David RicherbyDavid Richerby
66.4k15101190
$endgroup$
add a comment |
$begingroup$
You probably want to look at Bresenham's line drawing algorithm. Rather than moving along the line itself, you move along the $x$-coordinates a "pixel" (i.e., array entry) at a time and compute which $y$-value puts the greatest part of the line in the pixel $(x,y)$.
answered 51 mins ago
David RicherbyDavid Richerby
66.4k15101190
$endgroup$
You probably want to look at Bresenham's line drawing algorithm. Rather than moving along the line itself, you move along the $x$-coordinates a "pixel" (i.e., array entry) at a time and compute which $y$-value puts the greatest part of the line in the pixel $(x,y)$.
answered 51 mins ago
David RicherbyDavid Richerby
66.4k15101190
answered 51 mins ago
David RicherbyDavid Richerby
66.4k15101190
answered 51 mins ago
David RicherbyDavid Richerby
66.4k15101190
answered 51 mins ago
David RicherbyDavid Richerby
66.4k15101190
66.4k15101190
add a comment |
add a comment |
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