In one of mine applications I am dealing with graphics objects. I am using open source GPC library to clip/merge two shapes. To improve accuracy I am sampling (adding multiple points between two edges) existing shapes. But before displaying back the merged shape I need to remove all the points between two edges.
But I am not able to find an efficient algorithm that will remove all points between two edges which has same slope with minimum CPU utilization. Currently all points are of type
PointF
I am calculating slope using following function
private float Slope(PointF point1, PointF point2)
{
return (point2.Y - point1.Y) / (point2.X - point1.X);
}
Any pointer on this will be a great help.
What algorithm are you currently using? I can think only of going through all point and for each 3 to check wherher middle point is on vector (or close to) defined by 2 other points.
Do you need math for that operation?
Just to be clear, you have three points A = (a,b), C = (c,d), and E = (e,f), and are wondering if the segment AE goes through C and thus you can replace the pair of segments AC and CE with the single segment AE?
slope AC = (d-b)/(c-a) = slope CE = (f-d)/(e-c)
multiply through by the denominators, you get
(d-b)(e-c) = (f-d)(c-a)
that's just four subtracts, two multiplies, and a compare. You'll need to do the comparison with some error tolerance due to the use of floating point.
Well.. I found the solution for my question. Instead of using Sampling method provided by SDK, I created my own sampling method which insert a point between two points at a fixed distance. This reduces the number of point that I need to process and in turn reducing processor usage.
Related
Lets assume we have a bezier curve with a start p0 of (0, 0) and an end p4 of (100, 0). Right now it would basically be a line with no curve yet. Now lets assume I want to calculate the two missing control points (p1 p2) based on a given angle. What is the best way to achieve this?
Lets assume I wanted something like this:
https://1.bp.blogspot.com/_W3ZUYKgeEpk/SDcAerq1xkI/AAAAAAAAAAc/W9OnovkzgPI/s400/RectanglularControlPoly.jpg
I mean depending on the position of the control points it forms a triangle of some sort, that is why I was wondering if its possible.
Controls points that go through a Bezier point with a given angle, lie on the tangent with that angle.
The resulting bending will be the softer the farther away the control point is chosen, so there are many different solutions with the same angle and different curvature..
To find control points with equally soft curvatures for two Bezier points simply find the crossing of the two tangents! Use the crossing as the common control point for both segments, i.e. have C1 = C2.
For any sort of symmetrical curve you need to keep the deviations from the crossing symmetrical, i.e. 50%, 10% etc..
Note that for optimizing the overall shape one also needs to look at the neighbouring points; in general the provided GDI function does a good job; so it is worth considering simply adding more Bezier points for controlling the shape; but of course using the perfect set of control points is the most economic solution.
Update: I have added an example of how well a circle (orange) gets approximated by the math in this interesting post.
Short version: An exact solution isn't really possible but the best fit for a quarter circle is to move the control point to ~0.55% of the crossing point. (d=r*4*(sqrt(2)-1)/3). Sometimes instead of using a 4 segment solution an 8 segment solution is used for even closer approximation..
private void button_Click(object sender, EventArgs e)
{
int w = Math.Abs(P2.Left - P1.Left);
int h = Math.Abs(P2.Top - P1.Top);
C2.Left = (int) (P2.Left + w * 0.5523f);
C2.Top = P2.Top;
C1.Left = P1.Left;
C1.Top = (int) (P1.Top + h * 0.5523f);
C1.Parent.Invalidate();
}
The code uses Labels for the points and control points..
Btw: Adding ellipses/circles to a GraphicsPath will create bezier curves that seem to be approximated just like that.
We have 3d system with the source is sending signal and four receivers are receiving it. We know the coordinate location of these four receivers. We have the Time difference of arrival. Could you tell us how to calculate the location of the source.
Your problem is actually almost the same as the GPS navigation system. There you have four transmitters at known locations and one receiver in an unknown location, but the principle is exactly the same.
You get a system of four equations:
|p_1 - p_x| = c * (t + t_1)
|p_2 - p_x| = c * (t + t_2)
|p_3 - p_x| = c * (t + t_3)
|p_4 - p_x| = c * (t + t_4)
where p_x is the position vector of the transmitter and p_n the position vectors of the receivers. t is an unknown (even though it can be solved), t_n are the time differences to each receiver. |v| is the norm of vactor v (length, i.e. distance between stations). c is the speed of light.
There are different ways of solving this group of equations. You might want to have a look at any textbook or web resource of GPS mathematics.
If you know the locations of all four receivers, then you can draw circles around them, with each circle having a radius proportional to the time it took the signal to reach that receiver. Where those four circles intersect is the position of the transmitter.
It's a little bit more complicated than that because two circles will intersect in two positions. So you'll need the ability to discard one of the two points from the first circle intersection.
If the circles don't exactly intersect at one common point, you'll need some fuzzy matching to get the intersection point.
See https://math.stackexchange.com/questions/39561/finding-the-intersecting-points-on-two-circles for information on how to compute circle intersection.
I'm trying to find the fastest and easiest way in a C# program to calculate the intersections of two circles. From what I can tell there are two possible methods, and you'll have to forgive me for not knowing the official names for them.
We're assuming you know the center points for both circles and their exact radii, from which you can calculate the distance between them, so all that is missing are the point(s) of intersection. This is taking place on a standard x-y plot.
The first is a kind of substitution method like the one described here where you combine the two circle formulas and isolate either x or y, then sub it back in to an original formula to end up with a quadratic equation that can be solved for two (or possibly one or none) coordinates for an axis, which then lets you find the corresponding coordinates on the other axis.
The second I have seen a reference to is using a Law of Cosines method to determine the angles, which would then let you plot a line for each side on the grid, and put in your radius to find the actual intersection point.
I have written out the steps for the first method, and it seems rather lengthy. The second one is going to take some research/learning to write out but sounds simpler. What I've never done is translate processes like this into code, so I don't know ultimately which one will be the easiest for that application. Does anyone have advice on that? Or am I perhaps going about the the complete wrong way? Is there a library already out there that I can use for it instead of reinventing the wheel?
Some context: I'm worried mainly about the cost to the CPU to do these calculations. I plan on the application doing a heck of a lot of them at once, repeatedly, hence why I want the simplest way to accomplish it.
Computational geometry is almost always a pain to implement. It's also almost always quite CPU-intensive. That said, this problem is just algebra if you set it up right.
Compute d = hypot(x2-x1, y2-y1), the distance between the two centres. If r1 + r2 < d, there is no intersection. If r1+r2 == d, the intersection is at (x1, y1) + r1/(r1+r2) * (x2-x1,y2-y1). If d < abs(r1-r2), one circle is contained in the other and there is no intersection. You can work out the case where the two circles are tangent and one is contained in the other. I will only deal with the remaining case.
You want to find distances h orthogonal to (x2-x1,y2-y1) and p parallel to (x2-x1,y2-y1) so that p^2 + h^2 = r1^2 and (d-p)^2 + h^2 = r2^2. Subtract the two equations to get a linear equation in p: d^2-2dp = r2^2-r1^2. Solve this linear equation for p. Then h = sqrt(r1^2 - p^2).
The coordinates of the two points are (x1,y1) + p (x2-x1,y2-y1) / d +/- h (y2-y1, x1-x2) / d. If you work through the derivation above and solve for p/d and h/d instead, you may get something that does fewer operations.
I am trying to write an algorithm (in c#) that will stitch two or more unrelated heightmaps together so there is no visible seam between the maps. Basically I want to mimic the functionality found on this page :
http://www.bundysoft.com/wiki/doku.php?id=tutorials:l3dt:stitching_heightmaps
(You can just look at the pictures to get the gist of what I'm talking about)
I also want to be able to take a single heightmap and alter it so it can be tiled, in order to create an endless world (All of this is for use in Unity3d). However, if I can stitch multiple heightmaps together, I should be able to easily modify the algorithm to act on a single heightmap, so I am not worried about this part.
Any kind of guidance would be appreciated, as I have searched and searched for a solution without success. Just a simple nudge in the right direction would be greatly appreciated! I understand that many image manipulation techniques can be applied to heightmaps, but have been unable to find a image processing algorithm that produces the results I'm looking for. For instance, image stitching appears to only work for images that have overlapping fields of view, which is not the case with unrelated heightmaps.
Would utilizing a FFT low pass filter in some way work, or would that only be useful in generating a single tileable heightmap?
Because the algorithm is to be used in Unit3d, any c# code will have to be confined to .Net 3.5, as I believe that's the latest version Unity uses.
Thanks for any help!
Okay, seems I was on the right track with my previous attempts at solving this problem. My initial attemp at stitching the heightmaps together involved the following steps for each point on the heightmap:
1) Find the average between a point on the heightmap and its opposite point. The opposite point is simply the first point reflected across either the x axis (if stitching horizontal edges) or the z axis (for the vertical edges).
2) Find the new height for the point using the following formula:
newHeight = oldHeight + (average - oldHeight)*((maxDistance-distance)/maxDistance);
Where distance is the distance from the point on the heightmap to the nearest horizontal or vertical edge (depending on which edge you want to stitch). Any point with a distance less than maxDistance (which is an adjustable value that effects how much of the terrain is altered) is adjusted based on this formula.
That was the old formula, and while it produced really nice results for most of the terrain, it was creating noticeable lines in the areas between the region of altered heightmap points and the region of unaltered heightmap points. I realized almost immediately that this was occurring because the slope of the altered regions was too steep in comparison to the unaltered regions, thus creating a noticeable contrast between the two. Unfortunately, I went about solving this issue the wrong way, looking for solutions on how to blur or smooth the contrasting regions together to remove the line.
After very little success with smoothing techniques, I decided to try and reduce the slope of the altered region, in the hope that it would better blend with the slope of the unaltered region. I am happy to report that this has improved my stitching algorithm greatly, removing 99% of the lines reported above.
The main culprit from the old formula was this part:
(maxDistance-distance)/maxDistance
which was producing a value between 0 and 1 linearly based on the distance of the point to the nearest edge. As the distance between the heightmap points and the edge increased, the heightmap points would utilize less and less of the average (as defined above), and shift more and more towards their original values. This linear interpolation was the cause of the too step slope, but luckily I found a built in method in the Mathf class of Unity's API that allows for quadratic (I believe cubic) interpolation. This is the SmoothStep Method.
Using this method (I believe a similar method can be found in the Xna framework found here), the change in how much of the average is used in determining a heightmap value becomes very severe in middle distances, but that severity lessens exponentially the closer the distance gets to maxDistance, creating a less severe slope that better blends with the slope of the unaltered region. The new forumla looks something like this:
//Using Mathf - Unity only?
float weight = Mathf.SmoothStep(1f, 0f, distance/maxDistance);
//Using XNA
float weight = MathHelper.SmoothStep(1f, 0f, distance/maxDistance);
//If you can't use either of the two methods above
float input = distance/maxDistance;
float weight = 1f + (-1f)*(3f*(float)Math.Pow(input, 2f) - 2f*(float)Math.Pow(input, 3f));
//Then calculate the new height using this weight
newHeight = oldHeight + (average - oldHeight)*weight;
There may be even better interpolation methods that produce better stitching. I will certainly update this question if I find such a method, so anyone else looking to do heightmap stitching can find the information they need. Kudos to rincewound for being on the right track with linear interpolation!
What is done in the images you posted looks a lot like simple linear interpolation to me.
So basically: You take two images (Left, Right) and define a stitching region. For linear interpolation you could take the leftmost pixel of the left image (in the stitching region) and the rightmost pixel of the right image (also in the stitching region). Then you fill the space in between with interpolated values.
Take this example - I'm using a single line here to show the idea:
Left = [11,11,11,10,10,10,10]
Right= [01,01,01,01,02,02,02]
Lets say our overlap is 4 pixels wide:
Left = [11,11,11,10,10,10,10]
Right= [01,01,01,01,02,02,02]
^ ^ ^ ^ overlap/stitiching region.
The leftmost value of the left image would be 10
The rightmost value of the right image would be 1.
Now we interpolate linearly between 10 and 1 in 2 steps, our new stitching region looks as follows
stitch = [10, 07, 04, 01]
We end up with the following stitched line:
line = [11,11,11,10,07,04,01,02,02,02]
If you apply this to two complete images you should get a result similar to what you posted before.
I have to be able to set a random location for a waypoint for a flight sim. The maths challenge is straightforward:
"To find a single random location within a quadrangle, where there's an equal chance of the point being at any location."
Visually like this:
An example ABCD quadrangle is:
A:[21417.78 37105.97]
B:[38197.32 24009.74]
C:[1364.19 2455.54]
D:[1227.77 37378.81]
Thanks in advance for any help you can provide. :-)
EDIT
Thanks all for your replies. I'll be taking a look at this at the weekend and will award the accepted answer then. BTW I should have mentioned that the quadrangle can be CONVEX OR CONCAVE. Sry 'bout dat.
Split your quadrangle into two triangles and then use this excellent SO answer to quickly find a random point in one of them.
Update:
Borrowing this great link from Akusete on picking a random point in a triangle.
(from MathWorld - A Wolfram Web Resource: wolfram.com)
Given a triangle with one vertex at
the origin and the others at positions v1
and v2, pick
(from MathWorld - A Wolfram Web Resource: wolfram.com)
where A1
and A2 are uniform
variates in the interval [0,1] , which gives
points uniformly distributed in a
quadrilateral (left figure). The
points not in the triangle interior
can then either be discarded, or
transformed into the corresponding
point inside the triangle (right
figure).
I believe there are two suitable ways to solve this problem.
The first mentioned by other posters is to find the smallest bounding box that encloses the rectangle, then generate points in that box until you find a point which lies inside the rectangle.
Find Bounding box (x,y,width, height)
Pick Random Point x1,y1 with ranges [x to x+width] and [y to y+height]
while (x1 or y1 is no inside the quadrangle){
Select new x1,y1
}
Assuming your quadrangle area is Q and the bounding box is A, the probability that you would need to generate N pairs of points is 1-(Q/A)^N, which approaches 0 inverse exponentially.
I would reccommend the above approach, espesially in two dimensions. It is very fast to generate the points and test.
If you wanted a gaurentee of termination, then you can create an algorithm to only generate points within the quadrangle (easy) but you must ensure the probablity distribution of the points are even thoughout the quadrangle.
http://mathworld.wolfram.com/TrianglePointPicking.html
Gives a very good explination
The "brute force" approach is simply to loop through until you have a valid coordinate. In pseudocode:
left = min(pa.x, pb.x, pc.x, pd.x)
right = max(pa.x, pb.x, pc.x, pd.x)
bottom = min(pa.y, pb.y, pc.y, pd.y)
top = max(pa.y, pb.y, pc.y, pd.y)
do {
x = left + fmod(rand, right-left)
y = bottom + fmod(rand, top-bottom)
} while (!isin(x, y, pa, pb, pc, pd));
You can use a stock function pulled from the net for "isin". I realize that this isn't the fastest-executing thing in the world, but I think it'll work.
So, this time tackling how to figure out if a point is within the quad:
The four edges can be expressed as lines in y = mx + b form. Check if the point is above or below each of the four lines, and taken together you can figure out if it's inside or outside.
Are you allowed to just repeatedly try anywhere within the rectangle which bounds the quadrangle, until you get something within the quad? Might this even be faster than some fancy algorithm to ensure that you pick something within the quad?
Incidentally, in that problem statement, I think the use of the word "find" is confusing. You can't really find a random value that satisfies a condition; the randomizer just gives it to you. What you're trying to do is set parameters on the randomizer to give you values matching certain criteria.
I would divide your quadrangle into multiple figures, where each figure is a regular polygon with one side (or both sides) parallel to one of the axes. For eg, for the figure above, I would first find the maximum rectangle that fits inside the quadrangle, the rectangle has to be parallel to the X/Y axes. Then in the remaining area, I would fit triangles, such triangles will be adjacent to each side of the rectangle.
then it is simple to write a function:
1) get a figure at random.
2) find a random point in the figure.
If the figure chosen in #1 is a rectangle, it should be pretty easy to find a random point in it. The tricky part is to write a routine which can find a random point inside the triangle
You may randomly create points in a bound-in-box only stopping after you find one that it's inside your polygon.
So:
Find the box that contains all the points of your polygon.
Create a random point inside the bounds of the previously box found. Use random functions to generate x and y values.
Check if that point is inside the polygon (See how here or here)
If that point is inside the polygon stop, you're done, if not go to step 2
So, it depends on how you want your distribution.
If you want the points randomly sampled in your 2d view space, then Jacob's answer is great. If you want the points to be sort of like a perspective view (in your example image, more density in top right than bottom left), then you can use bilinear interpolation.
Bilinear interpolation is pretty easy. Generate two random numbers s and t in the range [0..1]. Then if your input points are p0,p1,p2,p3 the bilinear interpolation is:
bilerp(s,t) = t*(s*p3+(1-s)*p2) + (1-t)*(s*p1+(1-s)*p0)
The main difference is whether you want your distribution to be uniform in your 2d space (Jacob's method) or uniform in parameter space.
This is an interesting problem and there's probably as really interesting answer, but in case you just want it to work, let me offer you something simple.
Here's the algorithm:
Pick a random point that is within the rectangle that bounds the quadrangle.
If it is not within the quadrangle (or whatever shape), repeat.
Profit!
edit
I updated the first step to mention the bounding box, per Bart K.'s suggestion.