We Keep Coding

How to find curve corner points using JTS or NTS?

I have a curve (say JTS edge):
How to find all curve direction change points that surpasses given angle using JTS (Java) or NTS (C#):

I did some research and made some tests on JTS, and the best way I found is:
Create polygons and use the function union
Then iterate over the Coordinates, and create a sub-array on each "hard angle" (negative scalar product) and when the sum of angle reaches 180 (don't take the last angle to avoid non-function issues)
Then I change the base to an orthonormal base with x(firstElemOfSubArray, lastElemOfSubArray) by computing the base-changing matrix, and I then recompute the sub-array in a new coordinate system
I then create a function using org.apache.commons.math3.analysis.interpolation.SplineInterpolator to interpolate the function of the course, and then I get the derivative and search the extrema (don't take elements with an ordinate that is too low). With its absysse you can find which point is an inflexion point
So the point you search for are first elements of each sub array, and its inflections points (if there are any)

Find the length of a vector based on a rule

I am creating vertex array for a mesh with given points.So far I was able to create a continuous mesh with thickness.However there is a problem at the intersection of two line segments, the vectors between those segment's sizes needs to be bigger or smaller depending on the situation in order to have a continuous look.
What I have now:
With the given angles theta1 and theta2, how can I calculate the length of red vectors?
What I want:
How I structured my mesh:

You're probably making it more complicated than it needs to be.
Let's start by calculating the red arrows. For any line segment (p_i, p_j), we can calculate the segment's normal with:
dir = normalize(p_j - p_i)
normal = (-dir.y, dir.x) //negate if you want the other direction
At the connection point between two segments, we can just average (and re-normalize) the incident normals. This gives us the red arrows.
The only question that remains is how much you need to shift. The resulting offset for the line segment o_l given an offset of the vertex o_v is:
o_l = o_v * dot(normal_l, normal_v)
This means the following: Both normals are unit vectors. Hence, their dot product is at most one. This is the case when both line segments are parallel. Then, the entire offset of the vertex is transferred to the line. The smaller the angle becomes, the smaller becomes the transferred offset. E.g. if the angle between two consecutive line segments is 120°, then the dot product of normals is 0.5. If you shift the vertex by 1 unit along its normal, both line segment will have a thickness of 0.5.
So, in order to produce a specific line thickness (o_l), we need to shift the vertex by o_v:
o_v = o_l / dot(normal_l, normal_v)
The construction with averaging the line segments' normal for the vertex normal ensures that dot(normal_l1, normal_ v) = dot(normal_l2, normal_v), i.e. the resulting line thickness is equal for both lines in any case.

Fix collision detection penetration

I am implementing collision detection in my game, and am having a bit of trouble understanding how to calculate the vector to fix my shape overlap upon collision.
Say for example, I have two squares. squareA and squareB. For both of them, I know their xCo, yCo, width and height. squareA is moving however, so he has a velocity magnitude, and a velocity angle. Let's pretend I update the game once a second. I have illustrated the situation below.
Now, I need a formula to get the vector to fix the overlap. If I apply this vector onto the red square (squareA), they should not be overlapping anymore. This is what I am looking to achieve.
Can anyone help me figure out the formula to calculate the vector?
Bonus points if constructed in Java.
Bonus bonus points if you type out the answer instead of linking to a collision detection tutorial.
Thanks guys!
Also, how do I calculate the new velocity magnitude and angle? I would like sqaureA to continue moving along the x axis (sliding along the top of the blue square)

I had an function that looked something like this:
Position calculateValidPosition(Position start, Position end)
Position middlePoint = (start + end) /2
if (middlePoint == start || middlePoint == end)
return start
if( isColliding(middlePont) )
return calculateValidPosition(start, middlePoint)
else
return calculate(middlePoint, end)
I just made this code on the fly, so there would be a lot of room for improvements... starting by not making it recursive.
This function would be called when a collision is detected, passing as a parameter the last valid position of the object, and the current invalid position.
On each iteration, the first parameter is always valid (no collition), and the second one is invalid (there is collition).
But I think this can give you an idea of a possible solution, so you can adapt it to your needs.

Your question as stated requires an answer that is your entire application. But a simple answer is easy enough to provide.
You need to partition your space with quad-trees
You seem to indicate that only one object will be displaced when a collision is detected. (For the case of multiple interpenetrating objects, simply correct each pair of objects in the set until none are overlapping.)
Neither an elastic nor an inelastic collision is the desired behavior. You want a simple projection of the horizontal velocity (of square A), so Vx(t-1) = Vx(t+1), Vy(t-1) is irrelevant and Vy(t+1)=0. Here t is the time of collision.
The repositioning of Square A is simple.
Define Cn as the vector from the centroid of A to the vertex n (where the labeling of the vertices is arbitrary).
Define A(t-1) as the former direction of A.
Define Dn as the dot product of A(t-1) and the vector Cn
Define Rn as the width of A measured along Cn (and extending past the centroid in the opposite direction).
Define Sn as the dilation of B by a radius of Rn.
Let j be the vertex of B with highest y-value.
Let k be the vertex that is most nearly the front corner of A [in the intuitive sense, where the value of Dn indicates that Ck is most nearly parallel to A(t-1)].
Let K be the antipodal edge or vertex of A, relative to k.
Finally, translate A so that k and j are coincident and K is coincident with Sk.

Java library for 2d path comparison

I have one 2d line (it can be a curved line, with loops and so on), and multiple similar paths. I want to compare the first path with the rest, and determine which one is the most similar (in percentage if possible).
I was thinking maybe transforming the paths into bitmaps and then using a library to compare the bitmaps, but that seems like overkill. In my case, I have only an uninterrupted path, made of points, and no different colors or anything.
Can anyone help me?
Edit:
So the first line is the black one. I compare all other lines to it. I want a library or algorithm that can say: the red line is 90% accurate (because it has almost the same shape, and is close to the black one); the blue line is 5% accurate - this percentage is made up for this example... - because it has a similar shape, but it's smaller and not close to the black path.
So the criterion of similarity would be:
how close the lines are one to another
what shape do they have
how big they are
(color doesn't matter)
I know it's impossible to find a library that considers all this. But the most important comparisons should be: are they the same shape and size? The distance I can calculate on my own.

I can think of two measures to express similarity between two lines N (defined as straight line segments between points p0, p1... pr)
M (with straight line segments between q0, q1, ...qs). I assume that p0 and q0 are always closer than p0 and qs.
1) Area
Use the sum of the areas enclosed between N and M, where N and M are more different as the area gets larger.
To get N and M to form a closed shape you should connect p0 and q0 and pr and qs with straight line segments.
To be able to calculate the surface of the enclosed areas, introduce new points at the intersections between segments of N and M, so that you get one or more simple polygons without holes or self-intersections. The area of such a polygon is relatively straightforward to compute (search for "polygon area calculation" around on the web), sum the areas and you have your measure of (dis)similarity.
2) Sampling
Take a predefined number (say, 1000) of sample points O that lie on N (either evenly spaced with respect to the entire line, or evenly spaced
over each line segment of N). For each sample point o in O, we'll calculate the distance to the closest corresponding point on M: the result is the sum of these distances.
Next, reverse the roles: take the sample points from M and calculate each closest corresponding point on N, and sum their distances.
Whichever of these two produces the smallest sum (they're likely not the same!) is the measure of (dis)similarity.
Note: To locate the closest corresponding point on M, locate the closest point for every straight line segment in M (which is simple algebra, google for "shortest distance between a point and a straight line segment"). Use the result from the segment that has the smallest distance to o.
Comparison
Method 1 requires several geometric primitives (point, line segment, polygon) and operations on them (such as calculating intersection points and polygon areas),
in order to implement. This is more work, but produces a more robust result and is easier to optimize for lines consisting of lots of line segments.
Method 2 requires picking a "correct" number of sample points, which can be hard if the lines have alternating parts with little detail
and parts with lots of detail (i.e. a lot of line segments close together), and its implementation is likely to quickly get (very) slow
with a large number of sample points (matching every sample point against every line segment is a quadratic operation).
On the upside, it doesn't require a lot of geometric operations and is relatively easy to implement.

How to find if a point exists in which polygon

How to find if a point exists in which given set of polygons ?
I have coordinates like
polygonA = 1(0,0),2(0,5),3(3,4),4(3,5),5( 2,2)
polygonB = 1(10,10),2(10,15),3(13,14),4(13,15),5(12,12)
I have a point as (6,4) now want to search if this point is in any of this polygon or in both or nearest to which polygon.
How to store such data (polygon) ? is there a system / database / algorithm to do this search ?
Update : Thanks all for such fast response...I think i need to be more specific...
How to search = Yes...got list of algorithms and library for the same.
How to store = based on my research SQL and NoSQL db have their solutions.
NoSQL = MongoDb seems closest what i needed. But issue is I can query like "db.places.find({ "loc" : { "$within" : { "$polygon" : polygonB } } })" But cant make query like db.places.find({ "loc" : { "$within" : { } } })
SQL checked postgre and openGIS for some help. But colud not figureout if its possible.
If someone can help me with that...Thanks in advance.

The basic method (if you have a small number of polygons) is to store all polygons in a collection and loop over the elements to check if a point is inside a polygon.
On the other hand, if you have a considerable number of polygons, I would recommend using an R-tree data structure, which is not available in the standard library. You should check this project, if you want to go with R-tree option: http://sourceforge.net/projects/jsi/.
R-tree allows you to index rectangles (bounding boxes of the polygons in this case). So you can find a small number of candidate polygons very fast using R-tree. Then you can loop over the candidate list to get the final result.

You can use the GeneralPath class to help you with deciding if a point intersects a polygon. First, create a GeneralPath with your coordinates added:
GeneralPath gp = new GeneralPath();
double[] x = ...
double[] y = ...
gp.moveTo(x[0], y[0]);
for (int i =1; i < x.length; i++) {
gp.lineTo(x[i], y[i]);
}
gp.closePath();
if (gp.contains(pointX, pointY)) {
...
}
For the problem of which polygon a point is nearer to, this depends a little on how accurately you need a solution.
For an accurate solution., this amounts (without optimisation) to:
take the shortest distance between the point and each of the lines (segments) connecting the vertices of each of the polygons (Java2D apparently doesn't provide a method for this, but the shortest distance from a point to a line is a fairly simple calculation)
which polygon has the line with the shortest distance to the point?
In practice, you can approximate this process for some applications. For example, you could much more efficiently do this:
take the centre point of the bounding rectangle of each polygon (GeneralPath.getBounds() will give you this)
take the distance between the query point and each of these centre points, and see which is closest.
If you do need an accurate answer, then you can combine these techniques to optimise your search among all the vertices. For example, you could order the polygons by the distance to their "centrepoint" (defined as above). Search from minimum to maximum distance. If the minimum distance to a segment that you have found so far is d, then you can automatically rule out any polygon P where the distance from your query point to its "centrepoint" is d + r, where r is half the length of the diagonal of P's bounding rectangle (in other words, for simplicity, you imagine a bounding circle around that bounding box and check that the distance to that bounding circle is further than the nearest point found so far on other polygons).
I don't quite understand the bit about the database. Your polygons are just defined as a series of points. How you decide to store these in memory/file doesn't essentially make any difference to the algorithm.