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JavaFX Polygon Translation, Rotation, Scaling and its Points

I ran into a strange issue with the polygon class from javafx (java 8).
When I apply a set translate, rotate or scale on the polygon instance it is correctly moving the polygon around on my shape. The problem is, the points in the getPoints() method stay the same.
I started now to create my own methods and moving around the points and resetting them, the methods do what they should, but is it the right way?
Here an example:
private void translatePoints(double translateX, double translateY) {
List<Double> newPoints = new ArrayList<>();
for (int i = 0; i < getPoints().size(); i += 2) {
newPoints.add(getPoints().get(i) + translateX);
newPoints.add(getPoints().get(i + 1) + translateY);
}
getPoints().clear();
getPoints().addAll(newPoints);
}
Is there a way to get the translated, rotated and scaled points after a couple of operations?
Or do I have to implement them all separatly?

Take a look at the subclasses of Transform (Affine, Rotate, Scale, Shear and Translate). They allow you to transform points stored in a double[] array using the transform2DPoints method.
double[] points = new double[] {
0, 0,
0, 1,
1, 1,
1, 0
};
Rotate rot = new Rotate(45, 0.5, 0.5);
Translate t = new Translate(5, 7);
Scale sc = new Scale(3, 3);
for (Transform transform : Arrays.asList(rot, t, sc)) {
transform.transform2DPoints(points, 0, points, 0, 4);
}
System.out.println(Arrays.toString(points));
this way you need to take care of determining the pivot point of transforms where this is relevant on your own.
You could also get resulting transform for a node using Node.getLocalToParentTransform.
double[] points = polygon.getPoints().stream().mapToDouble(Number::doubleValue).toArray();
polygon.getLocalToParentTransform().transform2DPoints(points, 0, points, 0, points.length/2);
System.out.println(Arrays.toString(points));

Box rotation around multiple axises using Matrix4f

The question change a bit, I figured out how to rotate around a single axis
I want to rotate a box around the Y axis using an angle.
The box has a size, and a Vector3f to signal the rotation.
To rotate the box correctly what I do is rotate the origin position then rotate the origin position plus the size, and use those two references to render the box.
However this rotation does not work correctly and causes rendering artifacts.
This is my code to rotate the positions:
Matrix4f matrix = new Matrix4f();
// Rotate the origin position
Vector3f pos = new Vector3f(new Vector3f(blockX, blockY, blockZ));
matrix.m03 = pos.x;
matrix.m13 = pos.y;
matrix.m23 = pos.z;
Vector3f rot = new Vector3f(new Vector3f(0, 1f, 0f));
Matrix4f.rotate((float) Math.toRadians(45f), rot, matrix, matrix);
Vector3f locationMin = new Vector3f(matrix.m03, matrix.m13, matrix.m23);
// Rotate the position with the size
// Top left back is the position of the block
Vector3f sizeRot = new Vector3f(new Vector3f(blockX + size, blockY + size, blockZ + size));
matrix = new Matrix4f();
matrix.m03 = sizeRot.x;
matrix.m13 = sizeRot.y;
matrix.m23 = sizeRot.z;
rot = new Vector3f(new Vector3f(0, 1f, 0f));
Matrix4f.rotate((float) Math.toRadians(45f), rot, matrix, matrix);
Vector3f locationMax = new Vector3f(matrix.m03, matrix.m13, matrix.m23);
// Then here I use the locationMax and the locationMin to render the cube
What could be wrong with this code? Is the logic I am using to rotate the box correct? as in rotate the origin position then rotate the origin position plus the size..
EDIT: I released that rotating after translating is stupid so instead I just rotated the locationMax which is not translated (it is only the size) then I translated and I still get the same result (Graphical Artifacts).
New Code:
float rx = blockX, ry = blockY, rz = blockZ;
Matrix4f matrix = new Matrix4f();
Vector3f rot = new Vector3f(0, 1f, 0f);
matrix = new Matrix4f();
matrix.m03 = size;
matrix.m13 = size;
matrix.m23 = size;
Matrix4f.rotate((float) Math.toRadians(45f), rot, matrix, matrix);
matrix.translate(new Vector3f(rx, ry, rz), matrix);
float mx = matrix.m03;
float my = matrix.m13;
float mz = matrix.m23;
// Here is use rx, ry, rz and mx, my, mz to render the box
============ * I figured it out (See below)* =============
EDIT:
This is what I ended up doing:
// Origin point
Vector4f a = new Vector4f(blockX, blockY, blockZ, 1);
// Rotate a matrix 45 degrees
Matrix4f mat = new Matrix4f();
mat.rotate((float) Math.toRandians(45f), new Vector3f(
0, 1f, 0), mat);
/* Transform the matrix to each point */
Vector4f c = new Vector4f(size.x, 0, size.z, 1);
Matrix4f.transform(mat, c, c);
Vector4f.add(c, a, c);
Vector4f b = new Vector4f(size.x, 0, 0, 1);
Matrix4f.transform(mat, b, b);
Vector4f.add(b, a, b);
Vector4f d = new Vector4f(0, 0, size.z, 1);
Matrix4f.transform(mat, d, d);
Vector4f.add(d, a, d);
// Here is use a, b, c, and d to render the box.
The problem with this is that I want to rotate around all axises and not only around the Y axis. This makes the code very long and unreadable and There are a lot of bugs when I try to rotate around all axises.
Update Question:
How do I take the above code and make it so I can rotate around all 3 axises. I want to do this so I can have a billboard that will always face the camera.
This is how I calculate the angle between the camera and the object:
Vector3f angle = new Vector3f();
// Calculate the distance between camera and object
Vector3f.sub(game.getCamera().getLocation(),
new Vector3f(blockX, blockY, blockZ), angle);
// Calculate the angle around the Y axis.
float vectorAngle = (float) ((float) Math.atan2(angle.z, angle.x) * -1 + (Math.PI / 2.0f));

Billboards are a very common application of computer graphics (as I'm sure you've noticed, since you're asking the question!)
Ultimately I think you are over complicating the problem, based on:
as in rotate the origin position then rotate the origin position plus the size..
For computer graphics, the most common transformations are Scaling, Translating, and Rotating, and you do these in an order to achieve a desired effect (traditionally you scale, then rotate about the origin, then translate the vertex's position).
Additionally, you will have three main matrices to render a model in 3d: World Matrix, View Matrix, and Projection Matrix. I believe you are having misunderstandings of transforming from Model Space to World Space.
Graphics TRS and Matrix info. If you are having conceptual problems, or this answer is insufficient, I highly recommend looking at this link. I have yet to find a better resource explaining the fundamentals of computer graphics.
So right at the moment, you have your three angles (in degrees, in a Vector3) corresponding to the angle difference in the X,Y, and Z coordinate spaces from your billboard and your camera. With this information, we generate the View matrix by first gathering all of our matrix transformations in one place.
I'm going to assume that you already have your Translation and Scaling matrices, and that they both work. This means that we only need to generate our Rotation matrix, and then transform that matrix with the scaling matrix, and then transforming that matrix by our translation matrix.
X Rotation Matrix
Y Rotation Matrix
Z Rotation Matrix
(Images taken from CodingLabs link above)
So you will generate these three matrices, using the X,Y, and Z angles you calculated earlier, and then transform them to consolidate them into a single matrix, transform that matrix by the scaling matrix, and then transform that matrix by the translation matrix. Now you have your awesome matrix that, when you multiply a a vertex by it, will transform that vertex into the desired size, rotation, and position.
So you transform every single vertex point by this generated matrix.
And then after that, you should be done! Using these techniques will hopefully simplify your code greatly, and set you on the right path :)
So now how about some code?
//I do not guarantee that this code compiles! I did not write it in an IDE nor did I compile it
float angleToRotX = 180f;
float angleToRotY = 90f;
float angleToRotZ = 0f;
// example vertex
Vector4f vertex = new Vector4f(0, 1, 0, 1);
// Rotate vertex's X coordinates by the desired degrees
Matrix4f rotationXMatrix = new Matrix4f();
rotationXMatrix.rotX(angleToRotX);
Matrix4f rotationYMatrix = new Matrix4f();
rotationYMatrix.rotY(angleToRotY);
Matrix4f rotationZMatrix = new Matrix4f();
rotationZMatrix.rotZ(angleToRotZ);
//now let's translate it by 1.5, 1, 1.5 in the X,Y,Z directions
Matrix4f translationMatrix = new Matrix4f();
translationMatrix.setTranslate(new Vector3f(1.5, 1, 1.5));
/*
Now we have our three rotational matrices. So we multiply them (transform them) to get a single matrix to transform all of the points in this model to the desired world coordinates
*/
Matrix4f rotationMatrix = new Matrix4f();
rotationMatrix.mul(rotationXMatrix);
rotationMatrix.mul(rotationYMatrix);
rotationMatrix.mul(rotationZMatrix);
Matrix4f worldMatrix = translationMatrix;
worldMatrix.mul(rotationMatrix);
//now worldMatrix, when applied to a vertex, will rotate it by X,Y,Z degrees about the origin of it's model space, and then translate it by the amount given in translationMatrix
worldMatrix.transform(vertex);
//now vertex should be (1.5, 0, 1.5, 1) with (x,y,z,1)
Now this code could really be simplified, and it is excessively verbose. Try it out! I don't have java downloaded on my machine, but I grabbed the methods from the java documentation Here
Here is an image of what is happening (again, taking from coding labs):
(Advanced Info: Quaternions. These are really cool way of orienting a model in 3d space, however I don't quite understand them to the degree I need to in order to explain it to someone else, and I also believe that your problem is more fundamental)

You could generate the matrix without much hassle. The OpenGL matrix looks like the following:
|lx,ux,vx,px| - lx,ly,lz = the left vector
|ly,uy,vy,py| - ux,uy,uz = the up vector
|lz,uz,vz,pz| - vx,vy,vz = the view vector
|0 ,0 ,0 ,1 | - px,py,pz = the translation
All you need to do, is set px,py,pz to the position of your box in the world,
your view vector to the normalized(camera position - box position), your up comes straight from your camera, and the left is calculated via normalized cross product. It's also good practice to reconstruct the up vector, after left one is derived (by another cross product). That's all there's to it.
My solution aims to save you some time coding, rather than explain everything in detail. Hope that is useful to someone.

How can I check if an object(s) are in front of the camera?

I have got some trees, which are greatly lagging the game, so I would like to check if the trees are in front of the camera or not.
I have had some help from the Mathematics forum, and also had a look at This link to help me convert pitch/yaw to the directional vector needed.
But for some reason, whenever I move the camera to the left, the trees become visible, wheras whenever I move it to the right, they become unvisible (So if camera is pointing at +1 on the Z axis, it seems to be rendering the trees, but -1 on the Z axis and it seems to not render them).
(See http://i.gyazo.com/cdd05dc3f5dbdc07577c6e41fab3a549 for a less-jumpy .mp4)
I am using the following code to check if an object is in front of the camera or not:
Ship you = shipsID.get(UID);
int dis = 300;
Vector3f X = new Vector3f(camera.x(), camera.y(), camera.z());
float x = (float) (Math.cos(Math.toRadians(camera.yaw()))*Math.cos(Math.toRadians(camera.pitch())));
float y = (float) (Math.sin(Math.toRadians(camera.yaw()))*Math.cos(Math.toRadians(camera.pitch())));
float z = (float) Math.sin(Math.toRadians(camera.pitch()));
Vector3f V = new Vector3f(x, y, z);
for (Tree tree : trees){
Vector3f Y = new Vector3f(tree.location.x, tree.location.y, tree.location.z);
Vector3f YMinusX = Y.negate(X);//new Vector3f(Y.x - X.x, Y.y - X.y, Y.z - X.z);
float dot = Vector3f.dot(YMinusX, V);
if (dot > 0){
tree.render();
}
}
Is anyone able to tell me what I have done wrong here? I can't work out if it's the math.. Or the code.. Or what?
Camera translation code:
public void applyTranslations() {
glPushAttrib(GL_TRANSFORM_BIT);
glMatrixMode(GL_MODELVIEW);
glRotatef(pitch, 1, 0, 0);
glRotatef(yaw, 0, 1, 0);
lastYaw = yaw;
glRotatef(roll, 0, 0, 1);
glTranslatef(-x, -y, -z);
glPopAttrib();
}
UPDATE:
It appears to be where the camera is looking. For example, if I look to -Z, nothing happens, but if I look to +Z, they all render.
The if (dot > 0) code appears to somehow being +Z rather than +TheCameraRotation.

Your camera rotations yaw around Y, implying Y is your up vector. However, float z = (float) Math.sin(Math.toRadians(camera.pitch())); gives Z for your up vector. There is an inconsistency. I'd start by swapping y and z here, then print everything out every frame so you can see what happens as you rotate the camera. Also render just one tree and print dot. E.g. you might quickly notice the numbers approach 1.0 only when you look at 90 degrees left of the tree which narrows down the problem. As #DWilches notes, swapping cos/sin will change the phase of the rotation, which would produce such an effect.
You might consider limiting the dot product to the camera's field of view. There are still problems in that trees are not just points. A better way would be to test tree bounding boxes against the camera frustum, as #glampert suggests.
Still, the tree geometry doesn't look that complex. Optimization wise, I'd start trying to draw them faster. Are you using VBOs? Perhaps look at methods to reduce draw calls such as instancing. Perhaps even use a few models for LOD or billboards. Going even further, billboards with multiple trees on them. Occlusion culling methods could be used to ignore trees behind mountains.
[EDIT]
Since your trees are all roughly on a plane, you could limit the problem to the camera's yaw:
float angleToTree = Math.atan2(tree.location.z - camera.z(), tree.location.x - camera.x());
float angleDiff = angleToTree - camera.yaw();
if (angleDiff > Math.PI)
angleDiff -= 2.0f * Math.PI;
if (angleDiff < -Math.PI)
angleDiff += 2.0f * Math.PI;
if (abs(angleDiff) < cameraFOV + 0.1f) //bias as trees are not points
tree.render();

Could you write it like this
Ship you = shipsID.get(UID);
int dis = 300;
Vector3f X = new Vector3f(camera.x(), camera.y(), camera.z());
float x = (float) (Math.cos(Math.toRadians(camera.yaw()))*Math.cos(Math.toRadians(camera.pitch())));
float y = (float) (Math.sin(Math.toRadians(camera.yaw()))*Math.cos(Math.toRadians(camera.pitch())));
float z = (float) Math.sin(Math.toRadians(camera.pitch()));
Vector3f V = new Vector3f(x, y, z);
for (Tree tree : trees){
Vector3f Y = new Vector3f(tree.location.x, tree.location.y, tree.location.z);
Vector3f YMinusX = Y.negate(X);//new Vector3f(Y.x - X.x, Y.y - X.y, Y.z - X.z);
float dot = Vector3f.dot(YMinusX, V);
if (dot > 0){
tree.render();
}
}
As you can see there is far less calculation being performed for each tree.

For what I see here the correct formulas are:
x = Math.sin(pitch) * Math.cos(yaw);
y = Math.sin(pitch) * Math.sin(yaw);
z = Math.cos(pitch);
Could you try them ?

Appearance of a triangle strip. Surface normals? Or windings?

Below is a picture of what my outcome is.
I am using flat shading and have put each vertex in their respectable triangle objects. Then I use these vertices to calculate the surface normals. I have been reading that because my triangles share similar vertices that calculating the normals may be an issue? But to me this looks like a windings problem given that every other one is off.
I provided some of my code below to anyone who wants to look through it and get a better idea what the issue could be.
Triangle currentTri = new Triangle();
int triPointIndex = 0;
List<Triangle> triList = new ArrayList<Triangle>()
GL11.glBegin(GL11.GL_TRIANGLE_STRIP);
int counter1 = 0;
float stripZ = 1.0f;
float randY;
for (float x=0.0f; x<20.0f; x+=2.0f) {
if (stripZ == 1.0f) {
stripZ = -1.0f;
} else { stripZ = 1.0f; }
randY = (Float) randYList.get(counter1);
counter1 += 1;
GL11.glVertex3f(x, randY, stripZ);
Vert currentVert = currentTri.triVerts[triPointIndex];
currentVert.x = x;
currentVert.y = randY;
currentVert.z = stripZ;
triPointIndex++;
System.out.println(triList);
Vector3f normal = new Vector3f();
float Ux = currentTri.triVerts[1].x - currentTri.triVerts[0].x;
float Uy = currentTri.triVerts[1].y - currentTri.triVerts[0].y;
float Uz = currentTri.triVerts[1].z - currentTri.triVerts[0].z;
float Vx = currentTri.triVerts[2].x - currentTri.triVerts[0].x;
float Vy = currentTri.triVerts[2].y - currentTri.triVerts[0].y;
float Vz = currentTri.triVerts[2].z - currentTri.triVerts[0].z;
normal.x = (Uy * Vz) - (Uz * Vy);
normal.y = (Uz * Vx) - (Ux * Vz);
normal.z = (Ux * Vy) - (Uy * Vx);
GL11.glNormal3f(normal.x, normal.y, normal.z);
if (triPointIndex == 3) {
triList.add(currentTri);
Triangle nextTri = new Triangle();
nextTri.triVerts[0] = currentTri.triVerts[1];
nextTri.triVerts[1] = currentTri.triVerts[2];
currentTri = nextTri;
triPointIndex = 2;
}
}
GL11.glEnd();

You should be setting the normal before calling glVertex3f (...). A call to glVertex* is basically what finalizes a vertex, it associates the current color, normal, texture coordinates, etc... with the vertex at the position you pass and emits a new vertex.
glVertex — specify a vertex
Description
glVertex commands are used within glBegin / glEnd pairs to specify point, line, and polygon vertices. The current color, normal, texture coordinates, and fog coordinate are associated with the vertex when glVertex is called.
When only x and y are specified, z defaults to 0.0 and w defaults to 1.0. When x, y, and z are specified, w defaults to 1.0.
Chances are very good that this is a large part of your problem. Triangle strips are designed to workaround implicit winding issues. You have to reverse the winding of every triangle when you use a strip, but the rasterizer compensates for this by flipping the winding order used for front/back internally on each alternate triangle.
Update:
Understand of course that the rasterizer is smart enough to flip the front/back winding for each alternate triangle when using a strip but your code is not (at least not currently). You need to compensate for the alternately reversed winding when you calculate the normals yourself on the CPU side.

Actually it's both in one. The direction of the normal depends on the winding used to calculate it. However ultimately it boils down to a normals problem, since that's what determines lighting calculations.
Winding is also important for OpenGL, but you can't change anything about that in a striped primitive.

Drawing filled polygon with libGDX

I want to draw some (filled) polygons with libGDX. It shoudn't be filled with a graphic/texture. I have only the vertices of the polygon (closed path) and tried to visualize with meshes but at some point this is not the best solution, I think.
My code for an rectangle is:
private Mesh mesh;
#Override
public void create() {
if (mesh == null) {
mesh = new Mesh(
true, 4, 0,
new VertexAttribute(Usage.Position, 3, "a_position")
);
mesh.setVertices(new float[] {
-0.5f, -0.5f, 0
0.5f, -0.5f, 0,
-0.5f, 0.5f, 0,
0.5f, 0.5f, 0
});
}
}
// ...
#Override
public void render() {
Gdx.gl.glClear(GL10.GL_COLOR_BUFFER_BIT);
mesh.render(GL10.GL_TRIANGLE_STRIP, 0, 4);
}
is there a function or something to draw filled polygons in an easier way?

Since recent updates of LibGDX, #Rus answer is using deprecated functions. However, I give him/her credits for the new updated version below:
PolygonSprite poly;
PolygonSpriteBatch polyBatch = new PolygonSpriteBatch(); // To assign at the beginning
Texture textureSolid;
// Creating the color filling (but textures would work the same way)
Pixmap pix = new Pixmap(1, 1, Pixmap.Format.RGBA8888);
pix.setColor(0xDEADBEFF); // DE is red, AD is green and BE is blue.
pix.fill();
textureSolid = new Texture(pix);
PolygonRegion polyReg = new PolygonRegion(new TextureRegion(textureSolid),
new float[] { // Four vertices
0, 0, // Vertex 0 3--2
100, 0, // Vertex 1 | /|
100, 100, // Vertex 2 |/ |
0, 100 // Vertex 3 0--1
}, new short[] {
0, 1, 2, // Two triangles using vertex indices.
0, 2, 3 // Take care of the counter-clockwise direction.
});
poly = new PolygonSprite(polyReg);
poly.setOrigin(a, b);
polyBatch = new PolygonSpriteBatch();
For good triangulating algorithms if your polygon is not convex, see the almost-linear earclipping algorithm from Toussaint (1991)
Efficient triangulation of simple polygons, Godfried Toussaint, 1991

Here is a libGDX example which draws a 2D concave polygon.
Define class members for PolygonSprite PolygonSpriteBatch
PolygonSprite poly;
PolygonSpriteBatch polyBatch;
Texture textureSolid;
Create instances, 1x1 size texture used with red pixel as workaround. An array of coordinates (x, y) is used for initialization of the polygon.
ctor() {
textureSolid = makeTextureBox(1, 0xFFFF0000, 0, 0);
float a = 100;
float b = 100;
PolygonRegion polyReg = new PolygonRegion(new TextureRegion(textureSolid),
new float[] {
a*0, b*0,
a*0, b*2,
a*3, b*2,
a*3, b*0,
a*2, b*0,
a*2, b*1,
a*1, b*1,
a*1, b*0,
});
poly = new PolygonSprite(polyReg);
poly.setOrigin(a, b);
polyBatch = new PolygonSpriteBatch();
}
Draw and rotate polygon
void draw() {
super.draw();
polyBatch.begin();
poly.draw(polyBatch);
polyBatch.end();
poly.rotate(1.1f);
}

I believe the ShapeRenderer class now has a polygon method for vertex defined polygons:
ShapeRenderer.polygon()

You can use the ShapeRenderer API to draw simple, solid-color shapes with Libgdx.
The code you've given is a reasonable way to draw solid color polygons too. Its much more flexible than ShapeRenderer, but is a good bit more complicated. You'll need to use glColor4f to set the color, or add a Usage.Color attribute to each vertex. See the SubMeshColorTest example for more details on the first approach and the MeshColorTexture example for details on the second approach.
Another option to think about is using sprite textures. If you're only interested in simple solid colors objects, you can use very simple 1x1 textures of a single color and let the system stretch that across the sprite. Much of Libgdx and the underlying hardware are really optimized for rendering textures, so you may find it easier to use even if you're not really taking advantage of the texture contents. (You can even use a 1x1 white texture, and then use a SpriteBatch with setColor and draw()
to draw different color rectangles easily.)
You can also mix and match the various approaches, too.

Use triangulation algorithm and then draw all triangles as GL_TRIANGLE_STRIP
http://www.personal.psu.edu/cxc11/AERSP560/DELAUNEY/13_Two_algorithms_Delauney.pdf

just wanted to share my related solution with you, namely for implementing and drawing a walkZone with scene2d. I basically had to put together the different suggestions of the others' posts:
1) The WalkZone:
import com.badlogic.gdx.graphics.Pixmap;
import com.badlogic.gdx.graphics.Texture;
import com.badlogic.gdx.graphics.g2d.PolygonRegion;
import com.badlogic.gdx.graphics.g2d.TextureRegion;
import com.badlogic.gdx.math.EarClippingTriangulator;
import com.badlogic.gdx.math.Polygon;
import com.mygdx.game.MyGame;
public class WalkZone extends Polygon {
private PolygonRegion polygonRegion = null;
public WalkZone(float[] vertices) {
super(vertices);
if (MyGame.DEBUG) {
Pixmap pix = new Pixmap(1, 1, Pixmap.Format.RGBA8888);
pix.setColor(0x00FF00AA);
pix.fill();
polygonRegion = new PolygonRegion(new TextureRegion(new Texture(pix)),
vertices, new EarClippingTriangulator().computeTriangles(vertices).toArray());
}
}
public PolygonRegion getPolygonRegion() {
return polygonRegion;
}
}
2) The Screen:
you can then add a listener in the desired Stage:
myStage.addListener(new InputListener() {
#Override
public boolean touchDown(InputEvent event, float x, float y, int pointer, int button) {
if (walkZone.contains(x, y)) player.walkTo(x, y);
// or even directly: player.addAction(moveTo ...
return super.touchDown(event, x, y, pointer, button);
}
});
3) The implementation:
The array passed to te WZ constructor is a set of x,y,x,y... points. If you put them counter-clockwise, it works (I didn't check the other way, nor know how it exactly works); for example this generates a 100x100 square:
yourScreen.walkZone = new WalkZone(new int[]{0, 0, 100, 0, 100, 100, 0, 100});
In my project it works like a charm, even with very intricated polygons. Hope it helps!!

Most answers suggest triangulation, which is fine, but you can also do it using the stencil buffer. It handles both convex and concave polygons. This may be a better solution if your polygon changes a lot, since otherwise you'd have to do triangulation every frame. Also, this solution properly handles self intersecting polygons, which EarClippingTriangulator does not.
FloatArray vertices = ... // The polygon x,y pairs.
Color color = ... // The color to draw the polygon.
ShapeRenderer shapes = ...
ImmediateModeRenderer renderer = shapes.getRenderer();
Gdx.gl.glClearStencil(0);
Gdx.gl.glClear(GL20.GL_STENCIL_BUFFER_BIT);
Gdx.gl.glEnable(GL20.GL_STENCIL_TEST);
Gdx.gl.glStencilFunc(GL20.GL_NEVER, 0, 1);
Gdx.gl.glStencilOp(GL20.GL_INVERT, GL20.GL_INVERT, GL20.GL_INVERT);
Gdx.gl.glColorMask(false, false, false, false);
renderer.begin(shapes.getProjectionMatrix(), GL20.GL_TRIANGLE_FAN);
renderer.vertex(vertices.get(0), vertices.get(1), 0);
for (int i = 2, n = vertices.size; i < n; i += 2)
renderer.vertex(vertices.get(i), vertices.get(i + 1), 0);
renderer.end();
Gdx.gl.glColorMask(true, true, true, true);
Gdx.gl.glStencilOp(GL20.GL_ZERO, GL20.GL_ZERO, GL20.GL_ZERO);
Gdx.gl.glStencilFunc(GL20.GL_EQUAL, 1, 1);
Gdx.gl.glEnable(GL20.GL_BLEND);
shapes.setColor(color);
shapes.begin(ShapeType.Filled);
shapes.rect(-9999999, -9999999, 9999999 * 2, 9999999 * 2);
shapes.end();
Gdx.gl.glDisable(GL20.GL_STENCIL_TEST);
To use the stencil buffer, you must specify the number of bits for the stencil buffer when your app starts. For example, here is how to do that using the LWJGL2 backend:
LwjglApplicationConfiguration config = new LwjglApplicationConfiguration();
config.stencil = 8;
new LwjglApplication(new YourApp(), config);
For more information on this technique, try one of these links:
http://commaexcess.com/articles/7/concave-polygon-triangulation-shortcut
http://glprogramming.com/red/chapter14.html#name13
http://what-when-how.com/opengl-programming-guide/drawing-filled-concave-polygons-using-the-stencil-buffer-opengl-programming/