I have an image like this one
with 3 distinct regions. Using a breath first 4 neighbor queue, I have implemented a basic flood fill that distinguishes between the 3.
Now I need to find the center of mass of these regions with each pixel weighing one unit of weight.
Whats the best way of going about that?
The simplest way is to keep three arrays, sumx, sumy and count, each with one entry per label (3 in your case), and all initialized to 0. Then run through the image once, and for each labeled pixel add the x coordinate to the corresponding bin in sumx, the y coordinate to the corresponding bin in sumy, and 1 to the corresponding bin in count.
At the end, for each label l you can compute sumx[l]/count[l] and sumy[l]/count[l]. These are the unweighted centers of gravity (centroids).
Related
I am learning theory of machine learning and have some confusion about VC dimensions. According to the text book, the VC dimension of 2D axis-aligned rectangles is 4 which means it cannot shatter 5 points.
I found an example here: Cornell
However I still cannot understand this example. What if we use a rectangle like this (the red one)
Then we can classify this point out of them. Why is this incorrect?
We are supposed to draw the rectangle containing +ve examples only for any combination of the given 5 points. Here you see that, for any combination of the given points, a rectangle that corresponds to the points with maximum x-coordinate, minimum x-coordinate, maximum y-coordinate, and minimum y-coordinate will always contain the fifth point also. Hence, the set of 5 points cannot be shattered.
I want to find all pixels in an image (in Cartesian coordinates) which lie within certain polar range, r_min r_max theta_min and theta_max. So in other words I have some annular section defined with the parameters mentioned above and I want to find integer x,y coordinates of the pixels which lie within it. The brute force solution comes to mid offcourse (going through all the pixels of the image and checking if it is within it) but I am wondering if there is some more efficient solution to it.
Thanks
In the brute force solution, you can first determine the tight bounding box of the area, by computing the four vertexes and including the four cardinal extreme points as needed. Then for every pixel, you will have to evaluate two circles (quadratic expressions) and two straight lines (linear expressions). By doing the computation incrementally (X => X+1) the number of operations drops to about nothing.
Inside a circle
f(X,Y) = X²+Y²-2XXc-2YYc+Xc²+Yc²-R² <= 0
Incrementally,
f(X+1,Y) = f(X,Y)+2X+1-2Xc <= 0
If you really want to avoid that overhead, you will resort to scanline conversion techniques. First think of filling a slanted rectangle. Drawing two horizontal lines by the intermediate vertices, you decompose the rectangle in two triangles and a parallelogram. Then for any scanline that crosses one of these shapes, you know beforehand what pair of sides you will intersect. From there, you know what portion of the scanline you need to fill.
You can generalize to any shape, in particular your circle segment. Be prepared to a relatively subtle case analysis, but finding the intersections themselves isn't so hard. It may help to split the domain with a vertical through the center so that any horizontal always meets the outline twice, never four times.
We'll assume the center of the section is at 0,0 for simplicity. If not, it's easy to change by offsetting all the coordinates.
For each possible y coordinate from r_max to -r_max, find the x coordinates of the circle of both radii: -sqrt(r*r-y*y) and sqrt(r*r-y*y). For every point that is inside the r_max circle and outside the r_min circle, it might be part of the section and will need further testing.
Now do the same x coordinate calculations, but this time with the line segments described by the angles. You'll need some conditional logic to determine which side of the line is inside and which is outside, and whether it affects the upper or lower part of the section.
What method would you use to compute a distance that represents the number of "jumps" one has to do to go from one area to another area in a given 2D map?
Let's take the following map for instance:
(source: free.fr)
The end result of the computation would be a triangle like this:
A B C D E F
A
B 1
C 2 1
D 2 1 1
E . . . .
F 3 2 2 1 .
Which means that going from A to D, it takes 2 jumps.
However, to go from anywhere to E, it's impossible because the "gap" is too big, and so the value is "infinite", represented here as a dot for simplification.
As you can see on the example, the polygons may share points, but most often they are simply close together and so a maximum gap should be allowed to consider two polygons to be adjacent.
This, obviously, is a simplified example, but in the real case I'm faced with about 60000 polygons and am only interested by jump values up to 4.
As input data, I have the polygon vertices as an array of coordinates, from which I already know how to calculate the centroid.
My initial approach would be to "paint" the polygons on a white background canvas, each with their own color and then walk the line between two candidate polygons centroid. Counting the colors I encounter could give me the number of jumps.
However, this is not really reliable as it does not take into account concave arrangements where one has to walk around the "notch" to go from one polygon to the other as can be seen when going from A to F.
I have tried looking for reference material on this subject but could not find any because I have a hard time figuring what the proper terms are for describing this kind of problem.
My target language is Delphi XE2, but any example would be most welcome.
You can create inflated polygon with small offset for every initial polygon, then check for intersection with neighbouring (inflated) polygons. Offseting is useful to compensate small gaps between polygons.
Both inflating and intersection problems might be solved with Clipper library.
Solution of the potential neighbours problem depends on real conditions - for example, simple method - divide plane to square cells, and check for neighbours that have vertices in the same cell and in the nearest cells.
Every pair of intersecting polygons gives an edge in (unweighted, undirected) graph. You want to find all the path with length <=4 - just execute depth-limited BFS from every vertice (polygon) - assuming that graph is sparse
You can try a single link clustering or some voronoi diagrams. You can also brute-force or try Density-based spatial clustering of applications with noise (DBSCAN) or K-means clustering.
I would try that:
1) Do a Delaunay triangulation of all the points of all polygons
2) Remove from Delaunay graph all triangles that have their 3 points in the same polygon
Two polygons are neightbor by point if at least one triangle have at least one points in both polygons (or obviously if polygons have a common point)
Two polygons are neightbor by side if each polygon have at least two adjacents points in the same quad = two adjacent triangles (or obviously two common and adjacent points)
Once the gaps are filled with new polygons (triangles eventually combined) use Djikistra Algorithm ponderated with distance from nearest points (or polygons centroid) to compute the pathes.
In connected components labeling, if I see that the pixel to the left and the pixel above the current pixel have the same color but different labels, can't I automatically reassign their labels to be the same (instead of doing with an equivalence table)?
Wikipedia and MathWorks assigns the minimum label to the current pixel but otherwise leave the neighboring pixels the same. Then, they polish the label table with another pass. Unless I'm mistaken my tweak will allow me to label the image uniformly in a single pass. Is there an example in which my little tweak will break the algorithm?
You wouldn't eliminate the second pass. If you did change the labels of the neighboring pixels, what about their neighboring pixels? Basically, if this event happens, you've discovered the two labels are in the same equivalence class; but you'd still have to walk over everything you've examined so far to reassign those labels. You may as well just do that on the second pass and do all the reassigning in one sweep.
Example:
+-+-+-+
|?|?|A|
+-+-+-+
|B|B|x|
+-+-+-+
You examine pixel x, it matches both pixels north and west. Suppose A is the minimum label. So you choose to label the three pixels A, but that won't relabel the other B pixel. You still have to record that A==B, and will still have to sweep through to relabel any B's that remain. Furthermore, you might later find that A itself is equivalent to some other smaller label, and you'd have to relabel all these pixels later.
I have a set of points to define a shape. These points are in order and essentially are my "selection".
I want to be able to contract this selection by an arbitrary amount to get a smaller version of my original shape.
In a basic example with a triangle, the points are simply moved along their normal which is defined by the points to the left and the right of the points in question.
Eventually all 3 points will meet and form one point but until that point they will make a smaller and smaller triangle.
For more complex shapes, when moving the individual points inward, they may pass through the outer edge of the shape resulting in weird artifacts. Obviously I'll need to cull these points and remove them from the array.
Any help in exactly how I can do that would be greatly appreciated.
Thanks!
This is just an idea but couldn't you find the center of mass of the object, create a vector from the center to each point, and move each point along this vector?
To find the center of mass would of course involve averaging each x and y coordinate. Getting a vector is as simple a subtracting the point in question with the center point. Normalizing and scaling are common vector operations that can be found with the Google.
EDIT
Another way to interpret what you're asking is you want to erode your collection of points. As in morphology erosion. This is typically applied to binary images but you can slightly modify the concept to work with a collection of points. Essentially, you need to write a function that, given a point, will return true (black) or false (white) depending on if that point is inside or outside the shape defined by your points. You'd have to look up how to do that for shapes that aren't always concave (it's harder but not impossible).
Now, obviously, every single one of your actual points will return false because they're all on the border (by definition). However, you now have a matrix of points around your point of interest that define where is "inside" and where is "outside". Average all of the "inside" points and move your actual point along the vector between itself and towards this average. You could play with different erosion kernels to see what works best.
You could even work with a kernel with floating point weights instead of either/or values which will affect your average calculation proportional to their weights. With this, you could approximate a circular kernel with a low number of points. Try the simpler method first.
Find the selection center (as suggested by colithium)
Map the selection points to the coordinate system with the selection center at (0,0). For example, if the selection center is at (150,150), and a given selection point is at (125,75), the mapped position of the point becomes (-25,-75).
Scale the mapped points (multiply X and Y by something in the range of 0.0..1.0)
Remap the points back to the original coordinate system
Only simple maths required, no need to muck about normalizing vectors.