Mixing
![List comprehension iterating over two lists is not working as expected [duplicate]](https://lh3.googleusercontent.com/-3J_iVpOejjY/AAAAAAAAAAI/AAAAAAAAAIU/zxbGBjr8bRw/s72-c/photo.jpg?sz=32)
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 Jan 16 at 9:03
KilianKilian
1183
$endgroup$
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 Jan 16 at 9:03
KilianKilian
1183
$endgroup$
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 Jan 16 at 9:03
KilianKilian
1183
$endgroup$
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 Jan 16 at 9:03
KilianKilian
1183
asked Jan 16 at 9:03
KilianKilian
1183
asked Jan 16 at 9:03
KilianKilian
1183
asked Jan 16 at 9:03
KilianKilian
1183
asked Jan 16 at 9:03
KilianKilian
1183
1183
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 Jan 16 at 10:35
VinceVince
47926
$endgroup$
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
Jan 16 at 11:03
1
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
Jan 16 at 11:26
$begingroup$
It is more or less the same but it allows vectorial computation. Do not forget that when I say "sort", this can be done implicetly as both horizontal list and vertical list are already sorted. Bresenham is more interesting if you care on where exactly is the crossing point.
$endgroup$
– Vince
Jan 16 at 13:26
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 Jan 16 at 11:25
David RicherbyDavid Richerby
67.5k15102193
$endgroup$
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 Jan 16 at 10:35
VinceVince
47926
$endgroup$
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
Jan 16 at 11:03
1
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
Jan 16 at 11:26
$begingroup$
It is more or less the same but it allows vectorial computation. Do not forget that when I say "sort", this can be done implicetly as both horizontal list and vertical list are already sorted. Bresenham is more interesting if you care on where exactly is the crossing point.
$endgroup$
– Vince
Jan 16 at 13:26
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 Jan 16 at 10:35
VinceVince
47926
$endgroup$
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
Jan 16 at 11:03
1
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
Jan 16 at 11:26
$begingroup$
It is more or less the same but it allows vectorial computation. Do not forget that when I say "sort", this can be done implicetly as both horizontal list and vertical list are already sorted. Bresenham is more interesting if you care on where exactly is the crossing point.
$endgroup$
– Vince
Jan 16 at 13:26
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 Jan 16 at 10:35
VinceVince
47926
$endgroup$
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 Jan 16 at 10:35
VinceVince
47926
answered Jan 16 at 10:35
VinceVince
47926
answered Jan 16 at 10:35
VinceVince
47926
answered Jan 16 at 10:35
VinceVince
47926
47926
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
Jan 16 at 11:03
1
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
Jan 16 at 11:26
$begingroup$
It is more or less the same but it allows vectorial computation. Do not forget that when I say "sort", this can be done implicetly as both horizontal list and vertical list are already sorted. Bresenham is more interesting if you care on where exactly is the crossing point.
$endgroup$
– Vince
Jan 16 at 13:26
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
Jan 16 at 11:03
1
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
Jan 16 at 11:26
$begingroup$
It is more or less the same but it allows vectorial computation. Do not forget that when I say "sort", this can be done implicetly as both horizontal list and vertical list are already sorted. Bresenham is more interesting if you care on where exactly is the crossing point.
$endgroup$
– Vince
Jan 16 at 13:26
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
Jan 16 at 11:03
$begingroup$
Thanks. Let me take a closer look at it and I'll accept your answer once I got it translated into code
$endgroup$
– Kilian
Jan 16 at 11:03
1
1
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
Jan 16 at 11:26
$begingroup$
Ingenious but this seems much less efficient than Bresenham.
$endgroup$
– David Richerby
Jan 16 at 11:26
$begingroup$
It is more or less the same but it allows vectorial computation. Do not forget that when I say "sort", this can be done implicetly as both horizontal list and vertical list are already sorted. Bresenham is more interesting if you care on where exactly is the crossing point.
$endgroup$
– Vince
Jan 16 at 13:26
$begingroup$
It is more or less the same but it allows vectorial computation. Do not forget that when I say "sort", this can be done implicetly as both horizontal list and vertical list are already sorted. Bresenham is more interesting if you care on where exactly is the crossing point.
$endgroup$
– Vince
Jan 16 at 13:26
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 Jan 16 at 11:25
David RicherbyDavid Richerby
67.5k15102193
$endgroup$
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 Jan 16 at 11:25
David RicherbyDavid Richerby
67.5k15102193
$endgroup$
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 Jan 16 at 11:25
David RicherbyDavid Richerby
67.5k15102193
$endgroup$
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 Jan 16 at 11:25
David RicherbyDavid Richerby
67.5k15102193
answered Jan 16 at 11:25
David RicherbyDavid Richerby
67.5k15102193
answered Jan 16 at 11:25
David RicherbyDavid Richerby
67.5k15102193
answered Jan 16 at 11:25
David RicherbyDavid Richerby
67.5k15102193
67.5k15102193
add a comment |
add a comment |
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