I have brown filled svg paths and i want to detect and alert my user if there is any shape behind or above another shape. I know intersection list gets if they intersect at the edges but what happens if i want to detect a shape that is behind another shape but doesnt intersect at the edges?
The encoluseList method seems to be dealing with bounding boxes and not this.
Any ideas?
To detect if a path/shape overlaps another
1. Calculating the area covered by the final shape achieved
2. Calculating the sum of areas of all the shapes independently(since this is SVG and the details of each path element is known, this can be done)
3. Comparing the 2 areas.If the 2 areas are the same, then there is no overlapping otherwise at least 2 shapes overlap.
The tricky step is step 1 which can be approximately calculated using pixel painting algorithm(my preference). For other methods, you can go through the following stackoverflow question concerning area of overlapping circles
Related
Given a number of points on a 2d surface and radiuses for these points I can easily paint circles for them. What I need is an algorithm that only paints the envelope (right word for what I am looking for?) or outer bound of these combined circles. Additionally a second set of circles can 'encroach' on these circles, resulting in a kind of 'border'.
Image of what I am looking for
A quick way to draw the outline of the union of the disks is to
fill all disks in yellow, then
fill all disks in white with a smaller radius.
This can be adapted to the "encroached" circles, provided you only fill the remaining portions of the disks. Unfortunately, in a general setting finding the remaining portions can be an uneasy geometric problem.
There is an alternative approach which can work in all cases:
fill an image with zeroes, then for all disks fill every pixel with the value of the distance to the circumference (maximum at the center), but only keep the highest value so far.
while you do this, fill a second image with the color of the disk that achieved that highest value. (Initialize the image with the background color.)
At the end of this process, the first image will represent a "terrain" made of intersecting cones; and for every point of the terrain, you will know the color.
finally, color the pixels that have a height smaller than the desired stroke width, using the color map.
You can do the drawing in two steps.
1) Draw the outline using the following method: For each point, draw a circle using your favorite circle-drawing method, but before drawing a pixel, ensure that it is not contained inside any other circle. Do this for every point and you will get your outline.
2) Draw the borders between different sets using the following method: For each pair of points from different sets, calculate the two intersection points of the circles. If there is an intersection, the border can be drawn as a segment joining these two points. However, you have to make two lines, one for circle A, and another for circle B. To draw the line for circle A, slightly offset the segment towards point A. Then, use your favorite line-drawing method, but before drawing a pixel, ensure that it is closer to point A that any other point of the opposite set. After drawing the line, repeat the process for circle B. Note that both segment are not guaranteed to be the same length since the asymmetry of the points of the different sets. It will, however, always form a closed shape when all outlines and borders are drawn.
I have a 2d grid where pixel centers are at the intersection of two half-grid lines, as shown below.
I also have a shape that is drawn on this grid. In my case the shape is a glyph, and is described by segments. Each segment has a start point, end point and a number of off-curve points. These segments can be quadratic curves or lines. What's important is that I can know the points and functions that make up the outline of the shape.
The rule for deciding which pixels should be turned on is simple: if the center of the pixel falls within the shape outline, turn that pixel on. The following image shows an example of applying this rule.
Now the problem I'm facing has to do with anti aliasing. What I'd like to do is to calculate what percentage of the area of a given pixel falls within the outline. As an example, in the image above, I've drawn a red square around a pixel that would be about 15% inside the shape.
The purpose of this would be so that I can then turn that pixel on only by 15% and thus get some cleaner edges for the final raster image.
While I was able to find algorithms for determining if a given point falls within a polygon (ray casting), I wasn't able to find anything about this type of problem.
Can someone can point me toward some algorithms to achieve this? Also let me know if I'm going about this problem in the wrong way!
This sounds like an X, Y problem.
You are asking for a way to calculate the perecentage of pixel coverage, but based on your question, it sounds that what you want to do is anti alias a polygon.
If you are working only with single color 2D shapes (i.e red, blue, magenta... squares, lines, curves...) A very simple solution is to create your image and blur the result afterwards.
This will automatically give you a smooth outline and is simple to implement in many languages.
I'm doing a triangle rasteriser and it all works but now I want to use parallel computing to draw a bunch of triangles.
This means that the triangles can be drawn in any random order each frame, the problem is this disordered drawing is causing artifacts.
As a test case, let's consider two triangles that share two vertices, like so:
If we look at a grid of pixels the vertices will look like this:
The filling convention I'm using is just ceil()
So if we draw the blue triangle it will look like this
The problem that now becomes apparent is that if we now draw the yellow triangle, because they share the two vertices, the yellow triangle will be drawn over the blue one:
This isn't a problem on it's own, the problem is IF we draw them in the reverse order (yellow first, blue second) then the blue one will be drawn over instead:
This causes obvious artifacts because in any frame the order can be random so you will see a flickering line as one triangle is drawn over another and vise versa.
Is there any thing that can be done to ensure that the final image will look identical no matter the order the triangles are drawn in?
Always truncate to integer device pixels, and always do it the same way. This could be floor or ceiling or even rounding (if your sure both triangles get numerically identical inputs then they should round the same).
And define the boundaries of the triangle to include the lowest endpoint but exclude the highest, ie., min(x0,x1) <= x < max(x0,x1) and similarly for the y range.
I have a set of 2D points, unorganized, and I want to find the "contour" of this set (not the convex hull). I can't use alpha shapes because I have a speed objective (less than 10ms on an average computer).
My first approach was to compute a grid and find the outline squares (squares which have an empty square as a neighbor). So I think I downsized efficiently my numbers of points (from 22000 to 3000 roughly). But I still need to refine this new set.
My question is : how do I find the real outlines points among my green points ?
After a weekend full of reflexions, I may have found a convenient solution.
So we need a grid, we need to fill it with our points, no difficulty here.
We have to decide which squares are considered as "Contour". Our criteria is : at least one empty neighbor and at least 3 non empty neighbors.
We lack connectivity information. So we choose a "Contour" square which as 2 "Contour" neighbors or less. We then pick one of the neighbor. From that, we can start the expansion. We just circle around the current square to find the next "Contour" square, knowing the previous "Contour" squares. Our contour criteria prevent us from a dead end.
We now have vectors of connected squares, and normally if our shape doesn't have a hole, only one vector of connected squares !
Now for each square, we need to find the best point for the contour. We select the one which is farther from the barycenter of our plane. It works for most of the shapes. Another technique is to compute the barycenter of the empty neighbors of the selected square and choose the nearest point.
The red points are the contour of the green one. The technique used is the plane barycenter one.
For a set of 28000 points, this techniques take 8 ms. CGAL's Alpha shapes would take an average 125 ms for 28000 points.
PS : I hope I made myself clear, English is not my mothertongue :s
You really should use the alpha shapes. Maybe use only green points as inputs of the alpha alpha algorithm.
In an infinite 2D space there are a set of lines, each line having a start and end point, and a time of creation: Line(p0, p1, t).
I want to find the lines that should be rendered in a top-down view of this 2D space (higher values of t show up closer to the viewport, not that it should be relevant.)
The intuitive answer is "check if either point is within the viewport coordinates," but this falls down when the points are further apart than the viewport area covers.
The other idea I had was using something like geohash, this would limit precision i.e. maximum zoom level of the viewport. The idea is enumerating the hashes of the cells intersected and storing them. This way querying is a matter of asking the right question.
Are there any ideal solutions? Has this been solved before?
I think you need to check two conditions: one that the rectangle of viewport overlaps the rectangle with corners (p0,p1) and the second that some corners of viewport rectangle are on the different sides of the whole line which contains line segment (p0,p1).
The task of finding rectangle overlap can be solved very effectively for very large number of rectangles using R-trees (http://en.wikipedia.org/wiki/R-tree).
The second task can be reduced to checking signs of the cross product of (p1-p0) x (corner_coordinate-p0)
(all three quantities taken as 3-d vectors with third coordinate equal to zero, the result will be vector along the perpendicular direction). There should be corners with the opposite sign of this cross product.