Well, I'm developing a application using Java3D, which can solve Rubik's cube showing the user step-step solution.
I did my Algorithm part perfectly, and now the issue is with 3D cube,
Till now, i've created 27 cubes as in real one.
The problem is that, inorder to rotate a face clockwise or anti-clockwise, I should add those individual cubes to a group which in turn can be used to rotate the face anti-clockwise or clockwise direction as i said, but indeed the nodes in edges have to be in multiple groups, so that they can rotate either on X axis or Y axis or else we can take two axis of rotation.
so, my question now is, how to add a node to multiple groups ?
Or if it is not possible, Well then there must be a way to construct a Rubiks cube and to rotate its faces, How to do that!
Can you help me with this!!
Adding the cubes to BranchGroups and dynamically updating them or managing multiple groups sounds complicated.
Perhaps consider keeping the cubes independent. Create three transforms: one to rotate clockwise/counter-clockwise 90 degrees about each of the three axes. To rotate a face, apply the same transform to each of the cubes in that face -- since they are all rotating about the same axis anyways.
Related
What are the approaches to color-filling a curved concave shape in libgdx?
Let us assume that:
1) The shape-to-be-rendered is built from an array of vertices that are close to each other.
2) Edges between the vertices are known.
3) The vertices' positions might dynamically change over time. We are guaranteed that no self-intersections will occur (and that the shape will not be hollow).
The right-most picture is what I'm trying to render in libgdx (with-or-without the outline).
From what I've read, triangulation is a popular approach to non-curved shapes, but in order for it look any good for shapes with curvature, I imagine we would need a huge number of vertices (so that the many lines "zoomed out" resemble a smooth curve).
Triangulatin is not the only one way to go with this. I am not familiar with your gfx lib but I would do this in low level:
Subdivide your polyline to set of convex ones
you need to know which segment is line and which is curve and for the curve you also need to know the control points from neighboring patch so the connection is smooth. The concave boundary can be detected by the fact that angle change swaps sign so if you got consequent points p(0)...p(n) with consistent widing rule then
d(i ) = p(i+1)-p(i )
d(i-1) = p(i )-p(i-1)
n(i) = cross( d(i) , d(i-1) )
n(i).z * n(i-1).z < 0.0
so the cross product will give you normal and if the direction of the normal swaps it mean the winding is changed... The equatio assumes points are in xy plane or parallel to it. If not the case the last line should be
dot( n(i) , n(i-1) ) < 0.0
if true p(i) is your concave boundary and you should split your shape by it
fill the polylines
you can use the same approach as for triangles or convex polygons see:
how to rasterize rotated rectangle
Algorithm to fill triangle
render polyline outline
Of coarse if you do not have fast pixel access or horizontal line rendering available is this not a good way to go. There is one simple but not as fast option:
render outline
flood fill inside
this step is slow which might cause problems for bigger resolutions.
I've recently been looking into LibGDX and seem to have hit a wall, seen in the picture, the blue dot represents the users finger, the map generation it self is where i seem to get stuck, does LibGDX provide a method of dynamically drawing curved objects? I could simply generate them myself as images but then the image is hugely stretched to the point of the gap for the finger can fit 3! But also would need to be 1000's of PX tall to accommodate the whole level design.
Is it such that i should be drawing hundreds of polygons close together to make a curved line?
On a side not i'll need a way of determining when the object has from bottom to top so i can generate another 'chunk' of map.
You don't need hundreds of polygons to make a curve like you drew. You could get away with 40 quads on the left, and 40 on the right, and it would look pretty smooth. Raise that to 100 on each side and it will look almost perfectly smooth, and no modern device is going to have any trouble running that at 60fps.
You could use the Mesh class to generate a procedural mesh for each side. You can make the mesh stay in one spot, locked to the camera, and modify it's vertices and UVs to make it look like you are panning down an infinitely long corridor. This will take a fair amount of math up front but should be smooth sailing once you have that down.
Basically, your level design could be based on some kind of equation that takes Y offset as an input. Or it could be a long array of offsets, and you could use a spline equation or linear equation to interpolate between them. The output would be the UV and X coordinates which can be used to update each of the vertices of your two meshes.
You can use the vertex shader to efficiently update the UV coordinates, using a constant offset uniform parameter that you update each frame. That way you don't have to move UV data to the GPU every frame.
For the vertex positions, use your Mesh's underlying float[] and call setVertices() each frame to update it. Info here.
Actually, it might look better if you leave the UV's and the X positions alone, and just scroll the Y positions up. Keep a couple quads of padding off top and bottom of screen, and just move the top quad to the bottom after it scrolls off screen.
How about creating a set of curved forms that can be put together variably. Like the gap in the middle will at the top and bottom of each image be in the middle (with the same curvature at end and beginning points)...
And inbetween the start and end points you can go crazy on the shape.
And finally, you can randomly put those images together and get an endless world.
If you don't want to stop in the middle each time, you could also have like three entry and exit points (left, middle, right)... and after an image that ends left, you of course need to add an image that starts left, but might end somewhere else...
i want to find a circular object(Iris of eye, i have used Haar Cascase with viola Jones algorithm). so i found that hough circle would be the correct way to do it. can anybody explain me how to implement Hough circle in Java or any other easy implementation to find iris with Java.
Thanks,
Duda and Hart (1971) has a pretty clear explanation of the Hough transform and a worked example. It's not difficult to produce an implementation directly from that paper, so it's a good place for you to start.
ImageJ provides a Hough Circle plugin. I've been playing around with it several times in the past.
You could take a look at the source code if you want or need to modify it.
If you want to find an iris you should be straightforward about this. The part of the iris you are after is actually called a limbus. Also note that the contrast of the limbus is much lower than the one of the pupil so if image resolution permits pupil is a better target. Java is not a good option as programming language here since 1. It is slow while processing is intense; 2. Since classic Hough circle requires 3D accumulator and Java probably means using a cell phone the memory requirements will be tough.
What you can do is to use a fact that there is probably a single (or only a few) Limbuses in the image. First thing to do is to reduce the dimensionality of the problem from 3 to 2 by using oriented edges: extract horizontal and vertical edges that together represent edge orientation (they can be considered as horizontal and vertical components of edge vector). The simple idea is that the dominant intersection of edge vectors is the center of your limbus. To find the intersection you only need two oriented edges instead of three points that define a circle. Hence dimensionality reduction from 3 to 2.
You also don’t need to use a classical Hough circle transform with a huge accumulator and numerous calculations to find this intersection. A Randomized Hough will be much faster. Here is how it works (~ to RANSAC): you select a minimum number of oriented edges at random (in your case 2), find the intersection, then find all the edges that intersect at approximately the same location. These are inliers. You just iterate 10-30 times choosing a different random sample of 2 edges to settle in a set with maximum number of inliers. Hopefully, these inliers lie on the limbus. The median of inlier ray intersections will give you the center of the circle and the median distance to the inliers from the center is the radius.
In the picture below bright colors correspond to inliers and orientation is shown with little line segment. The set of original edges is shown in the middle (horizontal only). While original edges lie along an ellipse, Hough edges were transformed by an Affine transform to make those belonging to limbus to lie on a circle. Also note that edge orientations are pretty noisy.
This is sort of a homework question, however no expectations for code or whatever just an idea or hint towards the following problem.
I have a set of cubes in 3D world coordinates and i have to display them using two projections in two separate areas, parallel and perspective. The parallel went fine, no problems there, however displaying the same scene using perspective projection is becoming a nuisance for me.
The world to screen coordinates seemed like a good idea, but i don't know on which coordinates to apply them to, the original real coordinates, the new coordinates.
Thank you for your time.
PS: we are only allowed Java2D Api.
To perform a perspective projection, you need two additional things: the perspective point (where the "eye" is) and the projection plane. With a parallel projection, the perspective point/eye and plane can be any arbitrary distance from the objects (e.g., the cubes). But it is a little more complex with perspective projection.
Once you establish your eye and projection plane, you will need to iterate over your cubes. Ideally, you would iterate over them from the farthest cube to the eye to the nearest - that way the nearer cubes will overwrite the farther ones.
For each cube, determine the distance from the eye for each point. Then for each face (again in order of decreasing distance), calculate the projected points for each vertex. You can skip those faces with occluded points (the farthest vertex for each cube).
To calculate the projected point for a particular vertex, you need to find the point on the projection plane. This point will be where the line from the eye to the vertex intersects the projection plane. This will require some math, but should not be too difficult.
I'm currently writing an application that actually acts as a "cut" tool for 3D meshes. Well, I had some problems with it now which I am clueless on how to solve, since it is my first application.
I have loaded a model from an object file onto the canvas, then on the same canvas, I use the mouse drag event to draw lines to define the cutting point.
Let us say I want to cut a ball into half and I draw the line in the middle. How do I detect the vertices of the ball under the line.
Secondly, if I rotate/translate the ball, would all the the vertices information change?
Think of what you'd do in the real world: You can't cut a ball with a line, you must use a knife (a line has no volume). To cut the ball, you must move the knife through the ball.
So what you're looking after is a plane, not a line. To get such a plane, you must use some 3D math. What you have is the canvas orientation and the "side view" of the plane (which looks like a line).
So the plane you're looking for is perpendicular to the canvas. A simple way to get such a plane is to take the canvas orientation and create a plane which has the same orientation and then rotate the plane around the line by 90°.
After that, you can visit all edges of your model and determine on which side of the plane they are. For this, determine on which side of the plane the end points of the edge are. Use the cross product. If they are on the same side (both results of the cross products will have the same sign), you can ignore the edge. Otherwise, you need to determine the intersection point of the edge and plane. Create new edges and connect them accordingly.
See this page for some background on the math. But you should find some helper methods for all this in your opengl library.
if I rotate / translate the ball, would all the the vertices information change
Of course.
It's not going to be that easy.
I assume the line you are drawing induces a plane which then cuts the sphere.
To do so, you have to calculate the intersecting area of the sphere and the plane.
This is not a trivial task and I suggest using an existing framework for this or if you really want to do this yourself, read about basic intersection problems to get a feeling for this kind of problem. This paper offers a good introduction to various intersection tests.
In general boundary represended volumes, as in your case, are difficult to handle when it comes to more advanced manipulations. Cutting a sphere in half is easy compared to burring a small hole into it. Sometimes it's better to use a volume representation, like tetrahedral meshes or CSG.
Regarding your second question, you shouldn't rotate or translate the sphere, rotate and translate the camera.