I have some area X by Y pixels and I need to fill it up pixel by pixel. The problem is that at any given moment the drawn shape should be as round as possible.
I think that this algorithm is subset of Ordered Dithering, when converting grayscale images to one-bit, but I could not find any references nor could I figure it out myself.
I am aware of Bresenham's Circle, but it is used to draw circle of certain radius not area.
I created animation of all filling percents for 10 by 10 pixel grid. As full area is 10x10=100px, then each frame is exactly 1% inc.
A filled disk has the equation
(X - Xc)² + (Y - Yc)² ≤ C.
When you increase C, the number of points that satisfies the equation increases, but because of symmetry it increases in bursts.
To obtain the desired filling effect, you can compute (X - Xc)² + (Y - Yc)² for every pixel, sort on this value, and let the pixels appear one by one (or in a single go if you know the desired number of pixels).
You can break ties in different ways:
keep the original order as when you computed the pixels, by using a stable sort;
shuffle the runs of equal values;
slightly alter the center coordinates so that there are no ties.
Filling with the de-centering trick.
Values:
Order:
Related
I'm trying to plot XY graph in real time using Java. Functions that only rely on X are easy. Just iterate over x0...xn, get value and draw lines between the points. There are a lot of guides on it and it's intuitive.
But there is literally no guide on plotting graphs with x AND y being a variable.
Consider this equation: sin(x^3 * y^2) = cos(x^2 * y^3)
Using online Graph plotter I get this:
While my best result plotting the same function is this:
I just iterate over every pixel on screen and pass pixel positions as parameters to the function. If function's output is close to 0, I color the pixel. As you can see it's bad. It also takes huge amount of processing power. It only redraws once every couple of seconds. And if I try to increase precision, all lines just become thicker. Especially around intersections.
My question is how can I make my program faster and make it produce better looking graphs. Maybe there are some algorithms for that purpose?
The challenge is to chose the correct threshold. Pixels where abs(f(x,y)) is below the threshold should be colored. Pixels above the threshold should be white.
The problem is that if the threshold is too low, gaps appear in places where no pixel is exactly on the line. On the other hand, if the threshold is too high, the lines widen in places where the function is near zero, and the function is changing slowly.
So what's the correct threshold? The answer is the magnitude of the gradient, multiplied by the radius of a pixel. In other words, the pixel should be colored when
abs(f(x,y)) < |g(x,y)| * pixelRadius
The reason is that the magnitude of the gradient is equal to the maximum slope of the surface (at a given point). So a zero crossing occurs within a pixel if the slope is large enough to reduce the function to zero, inside the pixel.
That of course is only rough approximation. It assumes that the gradient doesn't change significantly within the area bounded by the pixel. The function in the question conforms to that assumption reasonably well, except in the upper right corner. Notice that in the graph below, there are Moiré patterns in the upper right. I believe that those are due to the failure in my antialiasing calculation: I don't compensate for a rapidly changing gradient.
In the graph below, pixels are white if
abs(f(x,y)) > |g(x,y)| * pixelRadius
Otherwise the pixel intensity is a number from 0 to 1, with 0 being black and 1 being white:
intensity = abs(f(x,y)) / (|g(x,y)| * pixelRadius)
I don't how exactly the online plotter did, but here are some suggestions.
Simplify your equation, as to this specific one, you can easily have x^2 * y^2 * (x ± y) = (2 * n + 1 / 2) * pi where n for any integer. It's much clearer than the original one.
Draw lines rather than points. Every n here stands for 4 curves, you can now loop over x and figure out y and draw a line between adjacent points.
Hope it helps!
I'm trying to come up with an algorithm to optimize the shape of a polygon (or multiple polygons) to maximize the value contained within that shape.
I have data with 3 columns:
X: the location on the x axis
Y: the location on the y axis
Value: Value of the block which can have positive and negative values.
This data is from a regular grid so the spacing between each x and y value is consistent.
I want to create a bounding polygon that maximizes the contained value with the added condition.
There needs to be a minimum radius maintained at all points of the polygon. This means that we will either lose some positive value blocks or gain some negative value blocks.
The current algorithm I'm using does the following
Finds the maximum block value as a starting point (or user defined)
Finds all blocks within the minimum radius and determines if it is a viable point by checking the overall value is positive
Removes all blocks in the minimum search radius from further value calculations and flags them as part of the final shape
Moves onto the next point determined by a spiraling around the original point. (center is always a grid point so moves by deltaX or deltaY)
This appears to be picking up some cells that aren't needed. I'm sure there are shape algorithms out there but I don't have any idea what to look up to find help.
Below is a picture that hopefully helps outline the question. Positive cells are shown in red (negative cells are not shown). The black outline shows the shape my current routine is returning. I believe the left side should be brought in more. The minimum radius is 100m the bottom left black circle is approximately this.
Right now the code is running in R but I will probably move to something else if I can get the algorithm correct.
In response to the unclear vote the problem I am trying to solve without the background or attempted solution is:
"Create a bounding polygon (or polygons) around a series of points to maximize the contained value, while maintaining a minimum radius of curvature along the polygon"
Edit:
Data
I should have included some data it can be found here.
The file is a csv. 4 columns (X,Y,Z [not used], Value), length is ~25k size is 800kb.
Graphical approach
I would approach this graphically. My intuition tells me that the inside points are fully inside the casted circles with min radius r from all of the footprint points nearby. That means if you cast circle from each footprint point with radius r then all points that are inside at least half of all neighboring circles are inside your polygon. To be less vague if you are deeply inside polygon then you got Pi*r^2 such overlapping circles at any pixel. if you are on edge that you got half of them. This is easily computable.
First I need the dataset. As you did provide just jpg file I do not have the vales just the plot. So I handle this problem like a binary image. First I needed to recolor the image to remove jpg color distortions. After that this is my input:
I choose black background to easily apply additive math on image and also I like it more then white and leave the footprint red (maximally saturated). Now the algorithm:
create temp image
It should be the same size and cleared to black (color=0). Handle its pixels like integer counters of overlapping circles.
cast circles
for each red pixel in source image add +1 to each pixel inside the circle with minimal radius r around the same pixel but in the temp image. The result is like this (Blue are the lower bits of my pixelformat):
As r I used r=24 as that is the bottom left circle radius in your example +/-pixel.
select inside pixels only
so recolor temp image. All the pixels with color < 0.5*pi*r^2 recolor to black and the rest to red. The result is like this:
select polygon circumference points only
Just recolor all red pixels near black pixels to some neutral color blue and the rest to black. Result:
Now just polygonize the result. To compare with the input image you can combine them both (I OR them together):
[Notes]
You can play with the min radius or the area treshold property to achieve different behavior. But I think this is pretty close match to your problem.
Here some C++ source code for this:
//picture pic0,pic1;
// pic0 - source
// pic1 - output/temp
int x,y,xx,yy;
const int r=24; // min radius
const int s=float(1.570796*float(r*r)); // half of min radius area
const DWORD c_foot=0x00FF0000; // red
const DWORD c_poly=0x000000FF; // blue
// resize and clear temp image
pic1=pic0;
pic1.clear(0);
// add min radius circle to temp around any footprint pixel found in input image
for (y=r;y<pic1.ys-r;y++)
for (x=r;x<pic1.xs-r;x++)
if (pic0.p[y][x].dd==c_foot)
for (yy=-r;yy<=r;yy++)
for (xx=-r;xx<=r;xx++)
if ((xx*xx)+(yy*yy)<=r*r)
pic1.p[y+yy][x+xx].dd++;
pic1.save("out0.png");
// select only pixels which are inside footprint with min radius (half of area circles are around)
for (y=0;y<pic1.ys;y++)
for (x=0;x<pic1.xs;x++)
if (pic1.p[y][x].dd>=s) pic1.p[y][x].dd=c_foot;
else pic1.p[y][x].dd=0;
pic1.save("out1.png");
// slect only outside pixels
pic1.growfill(c_foot,0,c_poly);
for (y=0;y<pic1.ys;y++)
for (x=0;x<pic1.xs;x++)
if (pic1.p[y][x].dd==c_foot) pic1.p[y][x].dd=0;
pic1.save("out2.png");
pic1|=pic0; // combine in and out images to compare
pic1.save("out3.png");
I use my own picture class for images so some members are:
xs,ys size of image in pixels
p[y][x].dd is pixel at (x,y) position as 32 bit integer type
clear(color) - clears entire image
resize(xs,ys) - resizes image to new resolution
[Edit1] I got a small bug in source code
I noticed some edges were too sharp so I check the code and I forgot to add the circle condition while filling so it filled squares instead. I repaired the source code above. I really just added line if ((xx*xx)+(yy*yy)<=r*r). The results are slightly changed so I also updated the images with new results
I played with the inside area coefficient ratio and this one:
const int s=float(0.75*1.570796*float(r*r));
Leads to even better match for you. The smaller it is the more the polygon can overlap outside footprint. Result:
If the solution set must be a union of disks of given radius, I would try a greedy approach. (I suspect that the problem might be intractable - exponential running time - if you want an exact solution.)
For all pixels (your "blocks"), compute the sum of values in the disk around it and take the one with the highest sum. Mark this pixel and adjust the sums of all the pixels that are in its disk by deducing its value, because the marked pixel has been "consumed". Then scan all pixels in contact with it by an edge or a corner, and mark the pixel with the highest sum.
Continue this process until all sums are negative. Then the sum cannot increase anymore.
For an efficient implementation, you will need to keep a list of the border pixels, i.e. the unmarked pixels that are neighbors of a marked pixel. After you have picked the border pixel with the largest sum and marked it, you remove it from the list and recompute the sums for the unmarked pixels inside its disk; you also add the unmarked pixels that touch it.
On the picture, the pixels are marked in blue and the border pixels in green. The highlighted pixels are
the one that gets marked,
the ones for which the sum needs to be recomputed.
The computing time will be proportional to the area of the image times the area of a disk (for the initial computation of the sums), plus the area of the shape times the area of a disk (for the updates of the sums), plus the total of the lengths of the successive perimeters of the shape while it grows (to find the largest sum). [As the latter terms might be costly - on the order of the product of the area of the shape by its perimeter length -, it is advisable to use a heap data structure, which will reduce the sum of the lengths to the sum of their logarithm.]
I am writing a function to draw an approximate circle on a square array (in Matlab, but the problem is mainly algorithmic).
The goal is to produce a mask for integrating light that falls on a portion of a CCD sensor from a diffraction-limited point source (whose diameter corresponds to a few pixels on the CCD array). In summary, the CCD sensor sees a pattern with revolution-symmetry, that has of course no obligation to be centered on one particular pixel of the CCD (see example image below).
Here is the algorithm that I currently use to produce my discretized circular mask, and which works partially (Matlab/Octave code):
xt = linspace(-xmax, xmax, npixels_cam); % in physical coordinates (meters)
[X Y] = meshgrid(xt-center(1), xt-center(2)); % shifted coordinate matrices
[Theta R] = cart2pol(X,Y);
R = R'; % cart2pol uses a different convention for lines/columns
mask = (R<=radius);
As you can see, my algorithm selects (sets to 1) all the pixels whose physical distance (in meters) is smaller or equal to a radius, which doesn't need to be an integer.
I feel like my algorithm may not be the best solution to this problem. In particular, I would like it to include the pixel in which the center is present, even when the radius is very small.
Any ideas ?
(See http://i.stack.imgur.com/3mZ5X.png for an example image of a diffraction-limited spot on a CCD camera).
if you like to select pixels if and only if they contain any part of the circle C:
in each pixel place a small circle A with the radius = halv size of the pixel, and another one around it with R=sqrt(2)*half size of the circle (a circumscribed circle)
To test if two circles touch each other you just calculate the center to center distances and subtract the sum of the two radii.
If the test circle C is within A then you select the pixel. If it's within B but not C you need to test all four pixel sides for overlap like this Circle line-segment collision detection algorithm?
A brute force approximate method is to make a much finer grid within each pixel and test each center point in that grid.
This is a well-studied problem. Several levels of optimization are possible:
You can brute-force check if every pixel is inside the circle. (r^2 >= (x-x0)^2 + (y-y0)^2)
You can brute-force check if every pixel in a square bounding the circle is inside the circle. (r^2 >= (x-x0)^2 + (y-y0)^2 where |x-x0| < r and |y-y0| < r)
You can go line-by-line (where |y-y0| < r) and calculate the starting x ending x and fill all the lines in between. (Although square roots aren't cheap.)
There's an infinite possibility of more sophisticated algorithms. Here's a common one: http://en.wikipedia.org/wiki/Midpoint_circle_algorithm (filling in the circle is left as an exercise)
It really depends on how sophisticated you want to be based on how imperative good performance is.
I'm working on a project that requires me to accurately control the number of pixels that are used to draw (roughly) circular stimuli, and although Bresenham's algorithms are great, they don't draw circles of an arbitrary area (to my knowledge). I've tried scripts that interrupt Bresenham's algorithm when the desired area has been plotted, but the results are decidedly hit-or-miss. Does anyone know of a way to plot the "best" circle (somewhat subjective, I know) using a given number of pixels? Many thanks!
A rough way of doing it, for example:
The radius of a circle of area 1000 sq px is sqrt(1000/pi) = 17.8... That circle should then fit in a 35x35 matrix. If you make "indices" for that matrix where the central pixel is (0,0), you can check easily if the pixel falls in the circle or not by substitution into the equation of a circle x^2 + y^2 = r^2. Or you can use the alternative equation for a circle centered at (a,b). If it evaluates to TRUE, it does, if not, it's outside the circle.
As a pseudocode/example, in Python I would do an optimized version of:
import numpy, math
target_area = 1000.0
r = (target_area / math.pi) ** 0.5
m = numpy.zeros((2*r+2,2*r+2))
a, b = r, r
for row in range(0, m.shape[0]):
for col in range(0, m.shape[1]):
if (col-a)**2 + (row-b)**2 <= r**2:
m[row,col] = 1
numpy.sum(m)
#>>> 999
Here is the result when the target area is 100,000 pixels (the actual circle generated is 99988.0):
You could also write a routine to find which areas can be matched more closely than others with this algorithm, and select those values to ensure conformity.
The area of a circle is A=Pi*r2. You're starting from the area and (apparently) want the radius, so we divide both sides by Pi to get: r2=A/pi. Taking the square root of both sides then gives us: r=sqrt(A/pi). Once you have the radius, drawing with most of the normal algorithms should be straightforward.
A simple (but somewhat naive approach) would be to simply count the number of pixels drawn by Bresenham's algorithm for a given radius, and then use binary search to find the radius that produces the desired number of pixels.
My first thought is to use an algorithm with sub-pixel precision. Consider what happens if you're center has irrational coordinates and you gradually increase the radius. This would fill seemingly random pixels around the perimeter as they became included in the circle. You want to avoid symmetry that causes the 4 quadrants of the circle adding pixels at the same time so you get closer to single pixels getting added. How something like this could be implemented I haven't a clue.
I had to solve a single instance of the 3d version once. I needed to a set of lattice points inside a sphere to be less-than or equal to 255. IIRC if r*r = 15 there are 240 points inside the sphere. I was not concerned with getting 255 exactly though.
Supposedly you have 2000 pixels in total that should make up your complete circle. By complete I mean there should be no breakage in pixels and must be connected to each other. Since 2Pi*R = circumference, the running length of diameter of the circle, this is the total amount of pixels you have. Now simply write R = 2000/2*Pi and this will give you the radius. Now you should be able to draw a circle the comprised of 2000 pixels. I hope this is what you wanted.
Let's forget about pixels for a second and let's work through the basic math/geometry.
We all know that
Area of a Circle = Pi * Radius ^2
which is the same as saying
Area of a Circle = Pi * (Diameter / 2) ^2
We all know that
Area of the Square Enclosing the Circle (i.e. each side of the square is tangent to the circle) = Diameter * Diameter
Thus
Ratio of the Circle Area to the Square Area = Circle Area / Square Area = (Pi * (Diameter / 2) ^2) / (Diameter * Diameter) = Pi / 4
Now let's assume that we have a circle and square with a pixel count large enough so that we don't have to worry about the troublesome edge cases around the border of the circle. In fact let's assume for a second that we have a very large diameter (maybe 10,000 or maybe even infinite). With this assumption the following holds:
Number of Pixels in the Circle = (Number of Pixels in the Square) * (Ratio of the Circle Area to the Square Area)
In other words for a sufficiently large number of pixels, the ratio of the areas of a perfectly drawn circle to a perfectly drawn square will approximate the ratio of the number of pixels in a pixelated circle to the number of pixels in the enclosing pixelated square.
Now in a pixelated square, the number of pixels in that square is the number of pixels across times the number of pixels high. Or in other words it is the square's diameter (in pixels) squared. Let's call the square's pixel diameter d. So substituting with the formulas above we have:
Number of Pixels in the Circle = (d * d) * (Pi /4)
So now let's solve for d
d = Sqrt(4 * (Num of Pixels in the Circle) / Pi)
Well we said earlier that d was the diameter of the square. Well it also happens to be the diameter of the circle. So when you want to draw a circle with a certain number of pixels, you draw a circle with the diameter being:
Diameter of Circle = Sqrt(4 * (Desired Number of Pixels in Circle Area) / Pi)
Now obviously you have to make some choices about rounding and so forth (there is no such thing as a fractional pixel), but you get the point. Also, this formula is more accurate as the desired number of pixels for the area of the circle goes up. For small amounts of pixels the roundoff error may not give you exactly the right number of pixels.
Input: a set of rectangles within the area (0, 0) to (1600, 1200).
Output: a point which none of the rectangles contains.
What's an efficient algorithm for this? The only two I can currently think of are:
Create a 1600x1200 array of booleans. Iterate through the area of each rectangle, marking those bits as True. Iterate at the end and find a False bit. Problem is that it wastes memory and can be slow.
Iterate randomly through points. For each point, iterate through the rectangles and see if any of them contain the point. Return the first point that none of the rectangles contain. Problem is that it is really slow for densely populated problem instances.
Why am I doing this? It's not for homework or for a programming competition, although I think that a more complicated version of this question was asked at one (each rectangle had a 'color', and you had to output the color of a few points they gave you). I'm just trying to programmatically disable the second monitor on Windows, and I'm running into problems with a more sane approach. So my goal is to find an unoccupied spot on the desktop, then simulate a right-click, then simulate all the clicks necessary to disable it from the display properties window.
For each rectangle, create a list of runs along the horizontal direction. For example a rectangle of 100x50 will generate 50 runs of 100. Write these with their left-most X coordinate and Y coordinate to a list or map.
Sort the list, Y first then X.
Go through the list. Overlapping runs should be adjacent, so you can merge them.
When you find the first run that doesn't stretch across the whole screen, you're done.
I would allocate an image with my favorite graphics library, and let it do rectangle drawing.
You can try a low res version first (scale down a factor 8), that will work if there is at least a 15x15 area. If it fails, you can try a high res.
Use Windows HRGNs (Region in .net). They were kind of invented for this. But that's not language agnostic no.
Finally you can do rectangle subtraction. Only problem is that you can get up to 4 rectangles each time you subtract one rect from another. If there are lots of small ones, this can get out of hand.
P.S.: Consider optimizing for maximized windows. Then you can tell there are no pixels visible without hit testing.
Sort all X-coordinates (start and ends of rectangles), plus 0 & 1600, remove duplicates. Denote this Xi (0 <= i <= n).
Sort all Y-coordinates (start and ends of rectangles), plus 0 & 1200, remove duplicates. Denote this Yj (0 <= j <= m).
Make a n * m grid with the given Xi and Yj from the previous points, this should be much smaller than the original 1600x1200 one (unless you have a thousand rectangles, in which case this idea doesn't apply). Each point in this grid maps to a rectangle in the original 1600 x 1200 image.
Paint rectangles in this grid: find the coordinates of the rectangles in the sets from the first steps, paint in the grid. Each rectangle will be on the form (Xi1, Yj1, Xi2, Yj2), so you paint in the small grid all points (x, y) such that i1 <= x < i2 && j1 <= y < j2.
Find the first unpainted cell in the grid, take any point from it, the center for example.
Note: Rectangles are assumed to be on the form: (x1, y1, x2, y2), representing all points (x, y) such that x1 <= x < x2 && y1 <= y < y2.
Nore2: The sets of Xi & Yj may be stored in a sorted array or tree for O(log n) access. If the number of rectangles is big.
If you know the minimum x and y dimensions of the rectangles, you can use the first approach (a 2D array of booleans) using fewer pixels.
Take into account that 1600x1200 is less than 2M pixels. Is that really so much memory? If you use a bitvector, you only need 235k.
You first idea is not so bad... you should just change the representation of the data.
You may be interessed in a sparse array of booleans.
A language dependant solution is to use the Area (Java).
If I had to do this myself, I'd probably go for the 2d array of booleans (particularly downscaled as jdv suggests, or using accelerated graphics routines) or the random point approach.
If you really wanted to do a more clever approach, though, you can just consider rectangles. Start with a rectangle with corners (0,0),(1600,1200) = (lx,ly),(rx,ry) and "subtract" the first window (wx1,wy1)(wx2,wy2).
This can generate at most 4 new "still available" rectangles if it is completely contained within the original free rectangle: (eg, all 4 corners of the new window are contained within the old one) they are (lx,ly)-(rx,wy1), (lx,wy1)-(wx1,wy2), (wx2,wy1)-(rx,wy2), and (lx,wy2)-(rx,ry). If just a corner of the window overlaps (only 1 corner is inside the free rectangle), it breaks it into two new rectangles; if a side (2 corners) juts in it breaks it into 3; and if there's no overlap, nothing changes. (If they're all axes aligned, you can't have 3 corners inside).
So then keep looping through the windows, testing for intersection and sub-dividing rectangles, until you have a list (if any) of all remaining free space in terms of rectangles.
This is probably going to be slower than any of the graphics-library powered approaches above, but it'd be more fun to write :)
Keep a list of rectangles that represent uncovered space. Initialize it to the entire area.
For each of the given rectangles
For each rectangle in uncovered space
If they intersect, divide the uncovered space into smaller rectangles around the covering rectangle, and add the smaller rectangles (if any) to your list of uncovered ones.
If your list of uncovered space still has any entries, they contain all points not covered by the given rectangles.
This doesn't depend on the number of pixels in your area, so it will work for large (or infinite) resolution. Each new rectangle in the uncovered list will have corners at unique intersections of pairs of other rectangles, so there will be at most O(n^2) in the list, giving a total runtime of O(n^3). You can make it more efficient by keeping your list of uncovered rectangles an a better structure to check each covering rectangle against.
This is a simple solution with a 1600+1200 space complexity only, it is similar in concept to creating a 1600x1200 matrix but without using a whole matrix:
Start with two boolean arrays W[1600] and H[1200] set to true.
Then for each visible window rectangle with coordinate ranges w1..w2 and h1..h2, mark W[w1..w2] and H[h1..h2] to false.
To check if a point with coordinates (w, h) falls in an empty space just check that
(W[w] && H[h]) == true