Develop Reference

How to overlay one MATRIX over another? [duplicate]

Say you have a collection of points with coordinates on a Cartesian coordinate system.
You want to plot another point, and you know its coordinates in the same Cartesian coordinate system.
However, the plot you're drawing on is distorted from the original. Imagine taking the original plane, printing it on a rubber sheet, and stretching it in some places and pinching it in others, in an asymmetrical way (no overlapping or anything complex).
(source)
You know the stretched and unstretched coordinates of each of your set of points, but not the underlying stretch function. You know the unstretched coordinates of a new point.
How can you estimate where to plot the new point in the stretched coordinates based on the stretched positions of nearby points? It doesn't need to be exact, since you can't determine the actual stretch function from a set of remapped points unless you have more information.
other possible keywords: warped distorted grid mesh plane coordinate unwarp

Ok, so this sounds like image warping. This is what you should do:
Create a Delaunay triangulation of your unwarped grid and use your knowledge of the correspondences between the warped and unwarped grid to create the triangulation for the warped grid. Now you know the corresponding triangles in each image and since there is no overlapping, you should be able to perform the next step without much difficulty.
Now, to find the corresponding point A, in the warped image:
Find the triangle A lies in and use the transformation between the triangle in the unwarped grid and the warped grid to figure out the new position.
This is explained explicitly in detail here.
Another (much more complicated) method is the Thin Plate Spline (which is also explained in the slides above).

I understood that you have one-to-one correspondence between the wrapped and unwrapped grid points. And I assume that the deformation is not so extreme that you might have intersecting grid lines (like the image you show).
The strategy is exactly what Jacob suggests: Triangulate the two grids such that there is a one-to-one correspondence between triangles, locate the point to be mapped in the triangulation and then use barycentric coordinates in the corresponding triangle to compute the new point location.
Preprocess
Generate the Delaunay triangulation of the points of the wrapped grid, let's call it WT.
For every triangle in WT add a triangle between the corresponding vertices in the unwrapped grid. This gives a triangulation UWT of the unwrapped points.
Map a point p into the wrapped grid
Find the triangle T(p1,p2,p3) in the UWT which contains p.
Compute the barycentric coordinates (b1,b2,b3) of p in T(p1,p2,p3)
Let Tw(q1,q2,q3) be the triangle in WT corresponding to T(p1,p2,p3). The new position is b1 * q1 + b2 * q2 + b3 * q3.
Remarks
This gives a deformation function as a linear spline. For smoother behavior one could use the same triangulation but do higher order approximation which would lead to a bit more complicated computation instead of the barycentric coordinates.

The other answers are great. The only thing I'd add is that you might want to take a look at Free form deformation as a way of describing the deformations.
If that's useful, then it's quite possible to fit a deformation grid/lattice to your known pairs, and then you have a very fast method of deforming future points.

A lot depends on how many existing points you have. If you have only one, there's not really much you can do with it -- you can offset the second point by the same amount in the same direction, but you don't have enough data to really do any better than that.
If you have a fair number of existing points, you can do a surface fit through those points, and use that to approximate the proper position of the new point. Given N points, you can always get a perfect fit using an order N polynomial, but you rarely want to do that -- instead, you usually guess that the stretch function is a fairly low-order function (e.g. quadratic or cubic) and fit a surface to the points on that basis. You then place your new point based on the function for your fitted surface.

Calculate shape orientation in Java (Image analysis)

I have an image such as this:
and I need to calculate the orientation of it. In this case the shape is pointing towards the top left of the screen. Accuracy isn't hugely important as long as 3 or 4 calculations average out to within 5 degrees or so of the actual orientation (it will be moving slightly).
Can anyone point me towards an algorithm to do this? I don't mind if the orientation is returned as a double or as a vector.

If the image is always T-shaped, you can simply get the furthest pair of pixels, then find the furthest pair from either both of those (the edges of the T), find which is further from the other two, draw a line from that one to the middle point of those two.
You can further refine it by then finding the base of the T by comparing the middle line with the edges of the base, and adjusting the angle and offset until it is actually in the middle.

The definitive solution is impossible I guess, since requires image recognition. I would project the 2D image onto axis, i.e. obtain the width and height of the image and get direction vector from these values taking them as components.

First, a couple of assumptions:
The center and centroid are "close"
The descending bar of the T is longer than the cross-bar
First, determine the bounding rectangle of the image and find the points of the image that lie along this rectangle. For points that lie along the line and are a certain distance from one another (say 5 pixels to pick a value) you'll need to only take 1 point from that cluster. At the end of this you should have 3 points, i.e. a triangle. The shortest side of the triangle should be the cross-bar (from assumption 2), i.e. find the two points closest to each other. The line that is perpendicular to the line crossing those two points is then your orientation line, i.e. find the angle between it and the horizontal axis.

I would try morphological skeletonization to simplify the image, followed by some straightforward algorithm to determine the orientation of the longer leg of the skeleton.

The solution in the end was to use a Convex Hull Algorithm, which finds the minimum number of points needed to enclose a shape with a bound.

How to implement Hough Circle in Java

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.

How does Affine Transform really work in Java?

I have been using Affine Transform to rotate a String in my java project, and I am not an experienced programmer yet, so it has taking me a long time to do a seemingly small task.. To rotate a string.
Now I have finally gotten it to work more or less as I had hoped, except it is not as precisely done as I want... yet.
Since it took a lot of trial and error and reading the description of the affine transform I am still not quite sure what it really does. What I think I know at the moment, is that I take a string, and define the center of the string (or the point which I want to rotate around), but where does matrices come into this? (Apparently I do not know that hehe)
Could anyone try and explain to me how affine transform works, in other words than the java doc? Maybe it can help me tweak my implementation and also, I just would really like to know :)
Thanks in advance.

To understand what is affine transform and how it works see the wikipedia article.
In general, it is a linear transformation (like scaling or reflecting) which can be implemented as a multiplication by specific matrix, and then followed by translation (moving) which is done by adding a vector. So to calculate for each pixel [x,y] its new location you need to multiply it by specific matrix (do the linear transform) and then add then add a specific vector (do the translation).

In addition to the other answers, a higher level view:
Points on the screen have a x and a y coordinate, i.e. can be written as a vector (x,y). More complex geometric objects can be thought of being described by a collection of points.
Vectors (point) can be multiplied by a matrix and the result is another vector (point).
There are special (ie cleverly constructed) matrices that when multiplied with a vector have the effect that the resulting vector is equivalent to a rotation, scaling, skewing or with a bit of trickery translation of the input point.
That's all there is to it, basically. There are a few more fancy features of this approach:
If you multiply 2 matrices you get a matrix again (at least in this case; stop nit-picking ;-) ).
If you multiply 2 matrices that are equivalent to 2 geometric transformations, the resulting matrix is equivalent to doing the 2 geometric transformations one after the other (the order matters btw).
This means you can encode an arbitrary chain of these geometric transformations in a single matrix. And you can create this matrix by multiplying the individual matrices.
Btw this also works in 3D.
For more details see the other answers.

Apart from the answers already given by other I want to show a practical tip namely a pattern I usually apply when rotating strings or other objects:
move the point of rotation (x,y) to the origin of space by applying translate(-x,-y).
do the rotation rotate(angle) (possible also scaling will be done here)
move everything back to the original point by translate(x,y).
Remember that you have to apply these steps in reverse order (see answer of trashgod).
For strings with the first translation I normally move the center of the bounding box to the origin and with the last translate move the string to the actual point on screen where the center should appear. Then I can simply draw the string at whatever position I like.
Rectangle2D r = g.getFontMetrics().getStringBounds(text, g);
g.translate(final_x, final_y);
g.rotate(-angle);
g.translate(-r.getCenterX(), -r.getCenterY());
g.drawString(text, 0, 0);
or alternatively
Rectangle2D r = g.getFontMetrics().getStringBounds(text, g);
AffineTransform trans = AffineTransform.getTranslateInstance(final_x, final_y);
trans.concatenate(AffineTransform.getRotateInstance(-angle));
trans.concatenate(AffineTransform.getTranslateInstance(-r.getCenterX(), -r.getCenterY()));
g.setTransform(trans);
g.drawString(text, 0, 0);

As a practical matter, I found two things helpful in understanding AffineTransform:
You can transform either a graphics context, Graphics2D, or any class that implements the Shape interface, as discussed here.
Concatenated transformations have an apparent last-specified-first-applied order, also mentioned here.

Here is purely mathematical video guide how to design a transformation matrix for your needs http://www.khanacademy.org/video/linear-transformation-examples--scaling-and-reflections?topic=linear-algebra
You will probably have to watch previous videos to understand how and why this matrices work though. Anyhow, it's a good resource to learn linear algebra if you have enough patience.

Two Viewports using Java2D

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.