So I'm programming a recursive program that is supposed to draw Koch's snowflake using OpenGL, and I've got the program basically working except one tiny issue. The deeper the recursion, the weirder 2 particular vertices get. Pictures at the bottom.
EDIT: I don't really care about the OpenGL aspect, I've got that part down. If you don't know OpenGL, all that the glVertex does is draw a line between the two vertices specified in the 2 method calls. Pretend its drawLine(v1,v2). Same difference.
I suspect that my method for finding points is to blame, but I can't find anything that looks incorrect.
I'm following the basically standard drawing method, here are the relevant code snips
(V is for vertex V1 is the bottom left corner, v2 is the bottom right corner, v3 is the top corner):
double dir = Math.PI;
recurse(V2,V1,n);
dir=Math.PI/3;
recurse(V1,V3,n);
dir= (5./3.)* Math.PI ;
recurse(V3,V2,n);
Recursive method:
public void recurse(Point2D v1, Point2D v2, int n){
double newLength = v1.distance(v2)/3.;
if(n == 0){
gl.glVertex2d(v1.getX(),v1.getY());
gl.glVertex2d(v2.getX(),v2.getY());
}else{
Point2D p1 = getPointViaRotation(v1, dir, newLength);
recurse(v1,p1,n-1);
dir+=(Math.PI/3.);
Point2D p2 = getPointViaRotation(p1,dir,newLength);
recurse(p1,p2,n-1);
dir-=(Math.PI*(2./3.));
Point2D p3 = getPointViaRotation(p2, dir, newLength);
recurse(p2,p3,n-1);
dir+=(Math.PI/3.);
recurse(p3,v2,n-1);
}
}
I really suspect my math is the problem, but this looks correct to me:
public static Point2D getPointViaRotation(Point2D p1, double rotation, double length){
double xLength = length * Math.cos(rotation);
double yLength = length * Math.sin(rotation);
return new Point2D.Double(xLength + p1.getX(), yLength + p1.getY());
}
N = 0 (All is well):
N = 1 (Perhaps a little bendy, maybe)
N = 5 (WAT)
I can't see any obvious problem code-wise. I do however have a theory about what happens.
It seems like all points in the graph are based on the locations of the points that came before it. As such, any rounding errors that occurs during this process eventually start accumulating, eventually ending with it going haywire and being way off.
What I would do for starters is calculating the start and end points of each segment before recursing, as to limit the impact of the rounding errors of the inner calls.
One thing about Koch's snowflake is, that the algorithm will lead to a rounding issue one time (it is recursive and all rounding errors add up). The trick is, to keep it going as long as possible. There're three things you can do:
If you want to get more detailed, the only way is to expand the possibilities of Double. You will need to use your own range of coordinates and transform them, every time you actually paint on the screen, to screen coordinates. Your own coordinates should zoom and show the last recursion step (the last triangle) in a coordination system of e.g. 100x100. Then calculate the three new triangles on top of that, transform into screen coordinates and paint.
The line dir=Math.PI/3; divides by 3 instead of (double) 3. Add the . after the 3
Make sure you use Point2D.Double anywhere. Your code should do so, but I would explicitely write it everywhere.
You won the game, when you still have a nice snowflake but get a Stackoverflow.
So, it turns out I am the dumbest man alive.
Thanks everyone for trying, I appreciate the help.
This code is meant to handle an equilateral triangle, its very specific about that (You can tell by the angles).
I put in a triangle with the height equal to the base (not equilateral). When I fixed the input triangle, everything works great.
Related
ive been looking for a while and cannot find solution
so I am trying to find a point on a polyline that is either a certain distance toward the opposite point or away from the main point.(I have pictures) if you are familiar with celestial navigation this is called the intercept method. so here is the almost working code.
double distance = ap.distanceToAsDouble(gp)*0.00053996;
double convert = nauticalMiles;
double t = convert / distance;
double actlon = ((1 - t)*ap.getLongitude() + t*gp.getLongitude());
double actlat = ((1 - t)*ap.getLatitude() + t*gp.getLatitude());
bear = ap.bearingTo(new GeoPoint(actlat,actlon));
actual = ap.destinationPoint(nauticalMiles/0.00053996,bear);
so "ap" is the point that the point I am trying to find will be (nauticalMiles) "away" or "toward" the "gp" and they are correct so "bear" is the angle kind of from ap to gp that almost works completely. "actual" is the point I am trying to find. here is an image of how this works out.
as you can see th actual point is toward the gp point from the ap point denoted with the yellow polyline. BUT!!!! if you look at the picture the yellow line is not exactly on the red line so I think that it is off a bit. so I tried to do
bear = ap.bearingTo(gp);
actual = ap.destinationPoint(nauticalMiles/0.00053996,bear);
instead doing that whole calculation that was in the code sample previously which should work but I am getting this
which is totally off it is not on the line. so for right now I am sticking with the first approach because it is relatively accurate but that is why I am asking this question. What calculation could I do to make the yellow line exactly on the red line? I am trying to get the most accurate calculation that I can get. thank you for your time.
I'm trying to make some sort of 3D editor, and I'm trying to make an "orbit" tool, much like the one in Blender.
Now I'd like to know what the angle is. I used the code provided by José Pereda that can be found here, but I need the angles to be in the range from 0 to 2π. I cannot retrieve them from the acquired angles since the output values don't range from -π to π and are different for every output.
Also, the extracted angles need to be relative to each other, much like if you put 3 different Rotate transformations where every following one is relative to the preceding ones, preferably in YXZ order, since that's the order I use everywhere else, as well as the model format the editor is going to export into.
As requested, I have uploaded the part of the code that is relevant to the question on Pastebin. As far as I could tell from the printed values, the angles are relative to each other in XYZ order. For angle calculation, I use the snippet from the linked question:
public static Vec3d getAngle(Node n) {
Transform T = n.getLocalToSceneTransform();
double roll = Math.atan2(-T.getMyx(), T.getMxx());
double pitch = Math.atan2(-T.getMzy(), T.getMzz());
double yaw = Math.atan2(T.getMzx(), Math.sqrt(T.getMzy() * T.getMzy() + T.getMzz() * T.getMzz()));
return new Vec3d(roll, pitch, yaw);
}
Just a quick notice, I have very little knowledge of matrices and their transformations, since that's something we'll be taking in our school next year, which makes it harder for me to understand.
I am created two bodies in Box2d, they are: Player and Platform;
I wanted to create game like Doodle Jump, but I don't know how to create "infinite world with generating platforms";
There is my code where I am creating Array:
buckets = new Array<Bucket>();
for(int i=1;i<BUCKET_COUNT;i++){
buckets.add(new Bucket(W/2,BUCKET_MARGIN*i, world));
}
And this code where I am "Trying" to change position of each platform when camera position is change:
for(Bucket bucket : buckets){
if(cam.position.y - (cam.viewportHeight/2) > bucket.getBody().getPosition().y + 22/PPM){
bucket.repos(W/2,bucket.getBody().getPosition().y + BUCKET_MARGIN);
}
}
It works! But it changes last platform position to very far bottom:
Regarding how to create an infinite world?
Use the world instance shift origin method. In the Box2D 2.3.2 C++ library code, this is the b2World::ShiftOrigin(const b2Vec2& newOrigin) method. Here's an excerpt of this method's declaration along with its documentation:
/// Shift the world origin. Useful for large worlds.
/// The body shift formula is: position -= newOrigin
/// #param newOrigin the new origin with respect to the old origin
void ShiftOrigin(const b2Vec2& newOrigin);
In Java, you should be able to find a similar interface method.
With shift origin you keep the viewport to the physics world centered on (or near) the physical world origin (of 0, 0). This is basically the practical means of accomplishing what Yevhen Danchenko suggested in the comments.
A reason for using this is that the implementation of floating-point arithmetic which Box2D relies on, is itself not infinitely wide ranging nor infinitely accurate. So shifting things helps keep things closer to the origin where the floating-point values are more accurate and keeps things from going off the range of practically usable values assuming that you'll only ever be showing a limited range of x and y values.
I have been trying to solve this for a few hours, and the internet is pretty unfruitful on the subject.
I need help detecting and solving collisions between rectangles, and not just detecting, but note I mentioned solving as well.
These are two boxes, with x/y width/heights. I simply need to detect when they are overlapping, and push one of the boxes out of the other one smoothly.
Also, note that one box is stationary - and the other is moving.
Does anyone have anything on this (or can give me an example?) I'd really appreciate it.
I need the boxes to be able to rest on top of each other as well.
Thank you!
I'm not sure what the context here is (Are these boxes moving or stationary? Are you looking for a physically accurate resolution, or simply a geometrically correct one?), but it seems like you could accomplish this in the following way:
1) Determine if there is a box collision
2) Determine the intersection of the two boxes, which would produce a third box. The width and height of the box is your penetration depth.
3) move the center of one of the boxes by the penetration depth, (x - width, y - height).
This should cause the boxes to become disjoint.
FYI: Intersection of two boxes can be computed by taking the max of the mins and the mins of the maxes from both boxes.
Here is some code from my engine for box intersection:
bool Bounds::IntersectsBounds(const Bounds &other) const
{
return !(min.x > other.max.x || max.x < other.min.x
|| min.y > other.max.y || max.y < other.min.y);
}
bool Bounds::Intersection(const Bounds &other, Bounds &outBounds) const
{
if (!this->IntersectsBounds(other)) {
return false;
}
outBounds.min.x = std::max(min.x, other.min.x);
outBounds.min.y = std::max(min.y, other.min.y);
outBounds.max.x = std::min(max.x, other.max.x);
outBounds.max.y = std::min(max.y, other.max.y);
return true;
}
In this case, the "outBounds" variable is the intersection of the two boxes (which in this case is your penetration depth). You can use the width/height of this box to perform your collision resolution.
Yeah! This is a pretty common problem! You may want to check out the gamedev portion of the stack exchange network!
Detection
bool collide(float x1,float y1,float sx1,float sy1, float x2, float y2, float sx2, float sy2){
if (x1+sx1 <= x2)
return false;
if (x2+sx2 <= x1)
return false;
if (y1+sy1 <= y2)
return false;
if (y2+sy2 <= y1)
return false;
return true;
}
Resolution
As far as an answer, this depends on the type of application you are going for. Is it a sidescroller, top-down, tile based? The answer depends on the response to this question. I'll assume something dynamic like a sidescroller or top-down action game.
The code is not difficult, but the implementation can be. If you have few objects moving on the screen you can use a similar system to mine, which goes something like the following:
Get a list of objects you are currently colliding with, in order of distance from the current object.
Iterate through the objects, and resolve collisions using the following method
Check if the object has some special collision type (teleporter, etc) by sending that object a message, and checking on the return value (a teleporter will take care of the collision resolution)
check if the previous bottom position of our current object (A) was above the top side of the object in question(B), if so that means you have had a bottom collision. Resolve by setting the y position of A to the y position of B minus the height of A
(IF THE PREVIOUS FAILED) check if the previous right side of A was to the left of the left side of B, if so that means you have had a right side collision. Resolve by setting the x position of A to B's position minus A's width
(IF THE PREVIOUS FAILED) check if the previous left side of A was to the right of the right side of B, if so that means you have had a left side collision. Resolve by setting the x position of A to B's x position plus B's width
(IF THE PREVIOUS FAILED) check if the previous top side of A was below the bottom side of B, if so you have had a top side collision. Resolve by setting the y position of A to the y position of B plus B's height
Whew. It is important that you have the objects sorted according to distance, it will catch on edges if you check collisions with an object that is farther away!
I hope that makes sense!
Edit: Apparently doesn't work in Android.
https://stackoverflow.com/a/15515114/3492994
Using the available classes from the 2D Graphics API.
Rectangle r1 = new Rectangle(100, 100, 100, 100);
Line2D l1 = new Line2D.Float(0, 200, 200, 0);
System.out.println("l1.intsects(r1) = " + l1.intersects(r1));
What this doesn't tell you, is where...
I've just tried to write "line" code to visualize a simple math;
Here it is
Ploygon polygon=new Ploygon();
int x,y;
ploygon.addPoint(0,0);
polygon.addPoint(width,height);
g.drawPolygon(polygon);
The code gives y=x effect;
OK... it is quite simple code; But the thing I am interested to get is points each N pixels during the statement period as {x0,y0}{0,0} and {x1,y1} {width,height} and that is the problem :(
The polygon xpoints array is not handy because it may contain just the same points which were added when addPoint(x,y) method was invoked; so in my case there just two added points which are connected by Polygon but what about all the rest points which stay between these points {x0,y0}{0,0} and {x1,y1} {width,height} ? How to get them?
For example. Coming back to the previous snippet how to find out what point x,y value is when (height%N)=0 etc?
Is there the most optimal way?
Thanks
What you have to realise here is that you are no longer working with pixels/coordinates per se, but you are working with vectors. You'd get much the same image from a polygon contained the coordinates (-500,-500) and (500,500) which is drawn onto a Graphics object which represents the (clipped) area from (0,0) in the bottom left to (100,100) in the bottom right. (ignoring for now that the actual coordinate system of Graphics has an inverted y-axis).
Therefore you have to solve this in a more back-to-basic's Math way rather than a “read the pixels” way. Unless you just want to determine if a given point is in the shape (for which the Shape interface offers a built-in method), you would be looking at calculating the slope of a line and determining functions which represent your line. For instance continuing from the example you have two points (-500,-500) and (500,500) which gives a slope of 1000/1000 = 1. So you could rewrite that function in terms of your x-coordinates as f(x) = -500 + (x + 500). Then if you want to know if the point (100,200) is on that line all you need to do is calculate f(100) and see that it isn't.
Getting back to your example, finding points which match a predicate (height%N =0), we'd be looking for f(x) == 0 mod N and so 'all' you'd need to do is solve the equation for x.