Given a 30 x 30 image of red and green pixels, stored as an array of 0s and 1s with 1 being red and 0 being green.
The image starts out as green and random patterns of red gets drawn on top.
All of the outer most pixels of the image are also painted red.
The question is how to fill every single pockets of green that's not connected to the biggest pocket of green with red?
Get/write a flood-fill algorithm that, starting at some pixel, fills all connected pixels with a different value and, while doing that, counts the pixels.
Have an int biggestSize = 0 and a Point biggestStartPoint = null variable with the given initial values.
Scan the image.
When you get at a green pixel, flood-fill with blue.
If the count of that flood-fill is bigger than biggestSize, flood-fill the old biggest area (from biggestStartPoint) with red. Store the new count and start pixel in biggestSize and biggestStartPoint.
If the count isn't bigger, flood-fill the (now blue) region with red, and leave the variables unchanged.
Finally, flood-fill the biggest area with green.
Related
Given a 2D pixel array where any pixel can be either 0 or 1, what algorithm would output a new 2D pixel array, where each pixel with a value of 1 would get a new value based on the minimum amount of line segments required to reach a specific "light source" pixel, while only crossing pixels that have an input value of 1? Input values of 0 would not change.
Example of input array, magenta cross represents the "light source" pixel:
https://cdn3.imggmi.com/uploads/2019/1/8/2a5f6dd0ebdc9c72115f9ce93af3337a-full.png
Output array with output values 1 and 2 (Photoshopped, not a pixel perfect image):
https://cdn3.imggmi.com/uploads/2019/1/8/0025709aaa826c26ee0a8e17476419cb-full.png
Red region = 1 line segment away from source
Yellow region = 2 line segments away from source
White region = 3 or more line segments away from source
(The algorithm wouldn't stop at 3, it would continue until every pixel is evaluated.)
EDIT: I'm not sure whether StackOverflow is the right Stack Exchange site to post this in, if not please let me know!
Make your point of origin the origin of a polar coordinate system. Convert the block corners to polar coordinates.
Now, treat your point source as a searchlight, sweeping from 0 to 2*PI. The beam continues until it hits the frame edge or a black box. This defines a polygon that you fill with magenta (1 line segment, direct lighting).
That's the easy part. Now you get to repeat this for every pixel that lies on a magenta-white (1-0) boundary of the polygon. This defines a finite set of secondary polygons; fill those with yellow (code 2).
Repeat this process with the yellow-white (2-0) boundaries to identify the 3 pixels; iterate until you run out of pixels.
In other paradigms, I've applied interval algebra to blocking segments (e.g. where one block partially shadows another), but I think that the polar polygon attack will get you to a solution in fewer hours of coding.
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 have a pictures of goods. There is a white border around some images. White color is not uniform and has some shades (poor quality etc.). I need to cut this border. To remove white color I use:
bd.threshold(bd, rect, pt, ">", threshold, color, maskColor);
There are some none transparent pixels after threshold because threshold color is unique for every image. BitmapData.getColorBoundsRect return region include none transparent pixels. I need region without this pixels(only image). Check each pixel is bad for big pictures. What is the most economical way to do this(find green region on picture below)? Sorry for my bad english and thanks for any help.
There are four edges of the image: left, right, top, down. Check each of them starting from the edge and moving towards inside of the image.
For example, let's take top edge (y = 0).
choose any horisontal position in the edge, for example, x = 10.
check pixel at (x, y).
if it is transparent, move the edge down: y++;
goto 2 and repeat until the pixel is not transparent.
choose different horisontal position and goto 2.
Repeat several times with different x. If there are only occasional non-transparent pixels, repeating the process 5-10 times will give you a new top edge which will most probably be 100% precise. It doesn't matter if the image is big, you only check several places in the edge. Do the same for left, right and bottom edges. Then copy the image defined by these edges.
If the edge quality is really bad then it's better to edit all images manually.
As part of more complex algorithm I need following:
let say I have a circle with radius R1 drawn on discrete grid (image) (green on image below)
I want to draw circle that have radius R2 that is bigger then R1 with one pixel (red on image below).
At each algorithm step to draw circles with increasing radius in a way that each time I have a filled circle.
How can I find the points to fill at each step so at the end of each step I have fully filed circle?
I have thinking of some circle rasterization algorithm, but this will lead to some gaps in filling. Another way is to use some mathematical morphology operation like dilation but this seems to be computationally expensive to do.
I am generally looking for way to do this on arbitrary shape but initially circle algorithm will be enough.
Your best option is to draw and fill a slightly larger red circle, and then draw and fill the green circle. Then redo on next iteration.
To only draw the 1px border is quite tricky. Your sample image is not even quite consistent. At some places a white pixel occurs diagonally to a green pixel, and in other places that pixel is red.
Edit:
borderPixels = emptySet
For each green pixel, p
For each neighbor n to p
If n is white
Add n to *borderPixels`
Do whatever you like with borderPixels (such as color them red)
My current solution for circle.
Based on well known Midpoint circle algorithm
create set of points for 1 octant for R1 radius (light green pixels)
create set of points for 1 octant for R2 radius (dark orange pixels)
for each row in image compare X coordinate for orange and green pixels and get 0 or 1 (or whatever) number of pixels in-between (light orange).
repeat for each octant (where for some octants columns instead of rows have to be compared)
This algorithm can be applied for other types of parametric shapes (Bezier curve based for example)
For non-parametric shapes (pixel based) image convolution (dilation) with kernel with central symmetry (circle). In other words for each pixel in shape looking for neighbors in circle with small radius and setting them to be part of the set. (expensive computation)
Another option is to draw a circle/shape with a 2pixel wide red border, and then draw a green filled circle/shape with NO border. Which should leave an approximately 1px wide edge.
It depends on how whatever technique you use resolves lines to pixels.
Circle algorithms tend to be optimised for drawing circles.....See the link here
I want to extract the points from given image. the image is shown below..
The points I want are the green upper point and the red point. I tried pixel by pixel comparison but it is too slow. I need a better algorithm. What are your suggestions ?
If the points are always going to be at a known radius from the center then you can just check the points that lie on the circumference.
Pixel by pixel comparison is going to be hard to beat. You can improve your search speed on the green line considerably by using a divide-and-conquer method.
If the image width is x and its height is y, search all of the pixels located at x={0...x},y={y/4,3*y/4} for a green pixel. If none is found, search all the pixels along x={x/4,3*x/4},y={0...y}. As soon as you find a green pixel p at coordinates px,py, search that pixel's two-pixel neighborhood which is farther from the center of the image (that is, {px,py+1},{px+1,py} if p is in the upper right, {px,py-1},{px-1,py} if p is in the lower left, {px,py+1},{px-1,py} if p is in the upper left, or {px,py-1},{px+1,py} if p is in the lower right quadrant. Update p to be the first green neighbor you find. Iterate until p has no more green neighbors. Worst-case this algorithm is ~O(2*(x+y)+(1/2)*max(x,y)) ~= O(2.5*max(x,y)) ~= O(x), which is a lot better than O(x*y) if you simply check the color value of every {x,y} pair.
Finding the red dot is going to be expensive, though, no less expensive than O(x*y) since the only way to improve the cost of searching for a single red pixel will be by subsampling the image (O(x*y)) and then searching the whole image (now O(sqrt(x*y))) for the red pixel.
I like par's idea, though, if the two dots are always the same distance from the center of the image then you can just search the pixels that fall along the circumference of that radius!