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.
Related
I'm looking for a way to match a list of 2D coordinates (which are in drawing order) to a shape from a list of shapes, or to find if it doesn't match any shape.
I've had a look at this, but I'm not too sure how exactly I'd relate that to things like straight/bent lines.
One idea I've thought of is to determine all the corners of each shape from the list (in order), and then check if the points from the list of coordinates that are found to be corners match one of these shapes (in the same order), using something like this. This might be a bit too strict for what I want to do however, as I want there to be a decent amount of leeway in the detection.
I'm also not sure what exactly I should use to pull corners (or lines etc.) from the list of coordinates - would it be worth turning it into an image or should I just try to detect things straight from the coordinate list?
For example, a possible input could be:
And some shapes to match could be: or or
And in this case, it would match the L shape.
For context, the reason I'm doing this is for a game where you have to draw symbols in the air to cast a spell.
If there aren't too many distinct shapes, here is what I woud try:
turn the curves to a (length, angle) representation; the length can be estimated as the sums of segment lengths, and the angle as the direction of the segments;
to be able to compare two representations, normalize the total length to unit (this will make the comparison size-independent), and sample at regular distances; said differently, obtain a vector of N angles at points equally spaced along the curve;
now to recognize a curve, compute the sum of absolute differences with similar vectors obtained from reference model, and keep the smallest sum.
There is no need for strict accuracy.
Note that this representation is not rotation-invariant, on purpose. It is also sensitive to the drawing order (from one endpoint to the other, or conversely), this can be unwanted. Finally, beware of the closed paths (circles), that people might start anywhere.
I have a collection of java.awt.Shape objects covering a two-dimensional plane with no overlap. These are from a data set of U.S. Counties, at a fairly low resolution. For an (x,y) latitude/longitude point, I want a quick way to identify the shape which county contains that point. What's an optimal way to index this?
Brute force would look like:
for (Shape eachShape : countyShapes) {
if (eachShape.contains(x, y)) {
return eachShape;
}
}
To optimize this, I can store the min/max bounds of the (possibly complex) shapes, and only call contains(x, y) on shapes whose rectangular bounds encompass a given x,y coordinate. What's the best way to build this index? A SortedMultiset would work for indexing on the x minima and maxima, but how to also include the y coordinates in the index?
For this specific implementation, doing a few seconds of up-front work to index the shapes is not a problem.
If possible you could try a bitmap with each shape in a different color. Then simply query the point and the color and lookup the shape.
This question is outside the scope of Stackoverflow but the answer is probably Binary Space Partitioning.
Roughly:
Divide the space in two either on the x coordinate or y coordinate using the mid-point of the range.
Create a list of counties on the two sides of that line (and divided by that line).
On each side of that line divide the collections again by the other dimension.
Continue recursively building a tree dividing alternately by x and y until you reach a satisfactory set of objects to examine by brute force.
The conventional algorithm actually divides the shapes lying across the boundary but that might not be necessary here.
A smart implementation might look for the most efficient line to divide on which is the one where the longest of the two lists is the smallest.
That involves more up front calculation but a more efficient and consistently performing partition.
You could use an existing GIS library like GeoTools and then all the hard work is already done.
Simply load your shapefile of counties and execute queries like
"the_geom" contains 'POINT(x y)
The quickstart tutorial will show you how to load the shapes, and the query tutorial will show you how to query them.
Having min an max values of coordinates of the bounds not guarantee that you can determine if one point is in or out in any situations. If you want achieve this by yourself you should implement some algorithm. There's a good one that is called "radial algorithm", I recommend that uses this, and it isn't so complicated to implement, there are sufficient bibliography and examples.
Hope this help.
I'm currently working on my master's thesis where I get:
A Delaunay triangulation drawn for me with given n points in (x, y, z) form.
My task is to use this triangulation and make contour lines at given z values.
I have been nearly successful at doing this by implementing the wikipedia spline interpolation : https://en.wikipedia.org/wiki/Spline_interpolation
My problem is that I get contour lines crossing each other when implementing the splines while of course the linear drawings doesn't cross.
Parametric cubic spline interpolated contour lines
If you look at the bottom part of the screen you see two contour lines crossing, I don't have enough reputation points to show that the linear drawings doesn't. You can also see that from point to point that the edges are way too rounded.
What I've tried is to interpolate more points between any pair of points to make more knot points along the lines, this to restrict them further, but to get non-crossing lines the splines look too much like a linear drawing which isn't satisfactory to the eye.
What I'd like to know, is not actual code implementation of the how, but maybe a pointer to how, readings and so forth.
(NB, I'm going to make this from scratch, no libraries).
Question: How to make a higher degree polynomial function which doesn't curve too much outside its linear counterpart. By too much I mean that a given contour at let's say 50 meters, that it doesn't cross a contour at 60 meters.
Any help is highly appreciated.
You can try a weighted delaunay triangulation. It's defined as the euklidian distance minus the weight.
Couples of years ago I solved similar task.
Here are some of my working notes. Probably it would help you.
Refer to XoomCode AcidMaps plugin, on github:
https://github.com/XoomCode/AcidMaps/tree/master/examples/isolines
Here is a demo:
http://ams.xoomcode.com/flex/index.html
Set, for example, the renderer type "Sparse" and interpolation strategy as "Linear", then press the "Update" button.
Refer to VividSolutions JTS Java library:
http://www.vividsolutions.com/jts/download.htm
http://mike.teczno.com/notes/curves-through-points.html
http://blog.csdn.net/xsolver/article/details/8913390
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.
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.