How to do: for looping + increasing of radius each loop? - loops
How do I create a clean and simple code that creates a circle of point/dots within the larger one? Or something similar (I can't post an image of what I want sorry). I was told to try using a for loop around the outside of my code and have the radius increase slightly each iteration of the loop. However, i don't know how to increase the radius?
This is the code I've been experimenting with so far:
size (400, 400);
background(255);
noStroke();
fill(0);
smooth();
translate(width/2, height/2);
int numpoints = 10;
float angleinc = 2 * PI / numpoints;
int radius = 100;
for (int i = 0; i < numpoints; i++) {
float x = cos(angleinc * i) * radius;
float y = sin(angleinc * i) * radius;
ellipse(x, y, 4, 4);
}
Please, any quick help would be appreciated. Also, I'm fairly new to processing and coding, so I'm not the best...
You'll have better luck if you break your problem down into smaller steps. Step one is creating a function that draws a single "ring" of smaller circles. You already have that step done, all you need to do is separate it into its own function:
void drawCircle(int outerRadius, int innerRadius) {
int numpoints = 10;
float angleinc = 2 * PI / numpoints;
for (int i = 0; i < numpoints; i++) {
float x = cos(angleinc * i) * outerRadius;
float y = sin(angleinc * i) * outerRadius;
ellipse(x, y, innerRadius, innerRadius);
}
}
Then, to draw a set of rings of increasing size, you simply call the function multiple times:
drawCircle(50, 8);
drawCircle(75, 12);
drawCircle(100, 16);
Which you can condense into a for loop:
for(int i = 2; i <= 4; i++){
drawCircle(25*i, 4*i);
}
The whole thing would look something like this:
void setup() {
size (400, 400);
}
void draw() {
background(255);
noStroke();
fill(0);
smooth();
translate(width/2, height/2);
for(int i = 2; i <= 4; i++){
drawCircle(25*i, 4*i);
}
}
void drawCircle(int outerRadius, int innerRadius) {
int numpoints = 10;
float angleinc = 2 * PI / numpoints;
for (int i = 0; i < numpoints; i++) {
float x = cos(angleinc * i) * outerRadius;
float y = sin(angleinc * i) * outerRadius;
ellipse(x, y, innerRadius, innerRadius);
}
}
This is just an example, and you'll have to play around with the numbers to make it look exactly like what you want, but the process is the same: break your problem down into smaller steps, isolate those steps into functions that do one thing, and then call those functions to accomplish your overall goal.
I hope i got your question right-
The formula of a circle around the origin is x=Rcos(angle) y=Rsin(angle) where angel is going between 0 to 2*pi
if you want to draw the circle around point lets say around (x',y'), the formula will be x= x' + Rcos(angle) y= y' + rsin(angle)
The code:
float epsilon = 0.0001f;
float R = 5.5.f;
for (float angle = 0.0; angle < 2*PI; angle += epsilon ) {
float x = x' + R*cos(angle);
float y = y' + R*sin(angle);
drawPoint(x,y);
if( /*condition for changing the radius*/ )
{
R = R*2; // or any change you want to do for R
}
}
It's probably easiest if you use two for loops: one for loop to draw the circle at a certain radius and another for loop which has the previous for loop in it which increases the radius.
int numCircles = 3;
//This for loop increases the radius and draws the circle with another for loop
for (int j = 0; j < numCircles; j++)
{
//This for loop draws the actual circle
for (int i = 0; i < numpoints; i++)
{
float x = cos(angleinc * i) * radius;
float y = sin(angleinc * i) * radius;
ellipse(x, y, 4, 4);
}
//(add code here that increases the radius)
}
Related
Radial Waves in Processing
I am currently a bit stuck! Lets say, have a grid of shapes (nested For-Loop) and I want to use a wave to animate it. The wave should have an offset. So far, i can achieve it. Currently the offset affects the Y-axis … But how can I manage to have a RADIAL offset – you know – like the clock hand, or a radar line… I really would like the offset to start from (width/2, height/2) – and then walks around clockwise. Here is my code and the point where I am stuck: void setup() { size(600, 600); } void draw () { background(255); float tiles = 60; float tileSize = width/tiles; for (int x = 0; x < tiles; x++) { for (int y = 0; y < tiles; y++) { float waveOffset = map(y, 0, 60, 0, 300); float sin = sin(radians(frameCount + waveOffset)); float wave = map(sin, -1, 1, 0, tileSize); fill(0); noStroke(); pushMatrix(); translate(tileSize/2, tileSize/2); ellipse(x*tileSize, y*tileSize, wave, wave); popMatrix(); } } } I tried different things – like the rotate(); function etc. but I can't manage to combine it! Thank you for any kind of help!
Right now, you're defining the size of the ellipses based on a transformation of sin(y). A transformation means it looks like a * sin(b * y + c) + d, and in this case you have a = tileSize / 2 b = 300 / 60 = 5 c = frameCount d = tileSize / 2 If you want to do a different pattern, you need to use a transformation of sin(theta) where theta is the "angle" of the dot (I put "angle" in quotes because it's really the angle from the vector from the center to the dot and some reference vector). I suggest using the atan2() function. Solution: float waveOffset = 2*(atan2(y - tiles/2, x - tiles/2)); float sin = sin((frameCount/20.0 + waveOffset));
void setup() { size(600, 600); } void draw () { background(255); float tiles = 60; float tileSize = width/tiles; for (int x = 0; x < tiles; x++) { for (int y = 0; y < tiles; y++) { float waveOffset = atan2(y - tiles/2, x - tiles/2)*0.5; float sin = sin((frameCount*0.05 + waveOffset)); float wave = map(sin, -1, 1, 0, tileSize); fill(0); noStroke(); pushMatrix(); translate(tileSize/2, tileSize/2); ellipse(x*tileSize, y*tileSize, wave, wave); popMatrix(); } } }
Animating a trajectory of Cannon Shells C-program
I am unable to see what I did wrong with my code. I am supposed to determine the trajectory that two cannon shells will take when fired at 500 m/s at an angle of 50 degrees above the horizontal. One of the shells is assumed to experience no drag and the other experiences a drag coefficient of 0.0001. I was wondering as to why my code wasn't plotting correctly. Any help would be great! #include <stdio.h> #include <math.h> #include <stdlib.h> #include "philsplot.h" // Number of shells #define N 2 // Defining the value of PI #define PI 3.1415926535 int main() { double xmin, xmax, ymin, ymax, theta = (50 * (PI/180)); int color, style; double expand; // philsplot function to open an x-window open_plot("800x600"); int i,j; double vx[N], vy[N], x[N], y[N], h, ax[N], ay[N]; h = 0.01; // philsplot function erases the canvas after flushpoint is called erase_plot(); // plot using km for distance xmin = 0; xmax = 1; ymin = 0; ymax = 1; expand = 1.1; color = 1; // philsplot function to set limits of the plotting canvas box_plot(xmin,xmax,ymin,ymax,expand,color,"Distance (km)","Height (km)","RRRR","TTTT"); // philsplot function that maps the plotting canvas onto the screen flush_plot(); style = 2; color = 3; expand = 1.0; // The initial values for(i=0; i<N; i++) { // want the initial x-position to be this x[i] = 0; // want the initial y-position to be this y[i] = 0; // want the initial x-velocity to be this vx[i] = 0.5*cos(theta); // want the initial y-velocity to be this vy[i] = 0.5*sin(theta); } for(j=0; j<N-1; j++) { // Using Euler's method you get this: ax[j] = (-0.0001) * (0.5) * vx[j]; ay[j] = (-0.00981) - (0.0001) * (0.5) * vy[j]; vx[j] = vx[j] - ax[j] * h; vy[j] = vy[j] - ay[j] * h; x[j] = x[j] + vx[j] * h; y[j] = y[j] + vy[j] * h; putpoint_plot(x[j],y[j],10, style, color, expand, 1); flush_plot(); delay_plot(1000); } printf("hit enter for next plot: "); getchar(); return 0; }
Mandelbrot using Pthreads, stepping through array
I am working on a project to rewrite a sequential C-code algorithm for creating a mandelbrot set into a parallel one using pthreads. I've gone up against a wall so to speak, as my version simply outputs a more or less black picture (and nothing to what the original program results into), and I can't really see where I'm going wrong. Simply put, I could use a second pair of eyes on this one. Here is the sequential code snippet that matters: void mandelbrot(float width, float height, unsigned int *pixmap) { int i, j; float xmin = -1.6f; float xmax = 1.6f; float ymin = -1.6f; float ymax = 1.6f; for (i = 0; i < height; i++) { for (j = 0; j < width; j++) { float b = xmin + j * (xmax - xmin) / width; float a = ymin + i * (ymax - ymin) / height; float sx = 0.0f; float sy = 0.0f; int ii = 0; while (sx + sy <= 64.0f) { float xn = sx * sx - sy * sy + b; float yn = 2 * sx * sy + a; sx = xn; sy = yn; ii++; if (ii == 1500) { break; } } if (ii == 1500) { pixmap[j+i*(int)width] = 0; } else { int c = (int)((ii / 32.0f) * 256.0f); pixmap[j + i *(int)width] = pal[c%256]; } } } } Here is my sequential version of the code: void* Mandel(void* threadId) { int x = *(int*)threadId; float xmin = -1.6f; float xmax = 1.6f; float ymin = -1.6f; float ymax = 1.6f; float b = xmin + x * (xmax - xmin) / WIDTH; for (int y = 0; y < 1024; y++) { float a = ymin + y * (ymax - ymin) / WIDTH; float sx = 0.0f; float sy = 0.0f; int ii = 0; while (sx + sy <= 64.0f) { float xn = sx * sx - sy * sy + b; float yn = 2 * sx * sy + a; sx = xn; sy = yn; ii++; if (ii == 1500) { break; } } if (ii == 1500) { pixmap[x+y*(int)WIDTH] = 0; } else { int c = (int)((ii / 32.0f) * 256.0f); pixmap[x + y *(int)WIDTH] = pal[c%256]; } } } Explanation of my thought process: I create 1024 threads in the main function, and then call on the function above with each thread. They're supposed to a column each (since the x is a constant between 0 and 1023, while the y value changes from 0 to 1023 within the function). As you can see, most of the mathematical meat in the function itself is the same in both the sequential and my parallel versions of the code. Because of this, I think the problem comes from how I'm stepping through the array, but I cannot see the problem with my own eyes. Regardless, the value that ii eventually receives is used to calculate c, which in turn is used to decide the color value that's to be saved in the corresponding position in pixmap. (pal is basically just a large array filled with color values). This function is the only piece of the code that I've actually touched to any major degree. The only difference in the main function is that I've created threads in it with the instruction to carry out the function Mandel. I assume that anyone willing to help will want more information, and please let me know of any improvements to this post in case I have posted too little information.
I found an answer shortly after posting this code. Silly me. Anywho, the problem wasn't stepping through the array within the function, but actually how I created the threads. This is how it looked in main when the problem occurred: for(int k = 0; k < 1024; k++) { pthread_create(&threads[k], NULL, (void*)Mandel, (void*) k); } The problem with doing it this way was that the loop continued, changing the value of k for the next thread, meaning that the last thread that we just created suddenly got an erronous value for k. This was solved by using an int array which saves the values of k as we go through each k value and create each thread, like shown below: int id[1024]; for(int k = 0; k < 1024; k++) { pthread_create(&threads[k], NULL, (void*)Mandel, (void*)(id+k)); } I would like to point towards the post in the following link, for helping me answer this problem: Pass integer value through pthread_create Just goes to show that you can often find the answers you're looking for if you search long enough, for the most of the time.
How to generate a set of points that are equidistant from each other and lie on a circle
I am trying to generate an array of n points that are equidistant from each other and lie on a circle in C. Basically, I need to be able to pass a function the number of points that I would like to generate and get back an array of points.
It's been a really long time since I've done C/C++, so I've had a stab at this more to see how I got on with it, but here's some code that will calculate the points for you. (It's a VS2010 console application) // CirclePoints.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include "stdio.h" #include "math.h" int _tmain() { int points = 8; double radius = 100; double step = ((3.14159265 * 2) / points); double x, y, current = 0; for (int i = 0; i < points; i++) { x = sin(current) * radius; y = cos(current) * radius; printf("point: %d x:%lf y:%lf\n", i, x, y); current += step; } return 0; }
Try something like this: void make_circle(float *output, size_t num, float radius) { size_t i; for(i = 0; i < num; i++) { const float angle = 2 * M_PI * i / num; *output++ = radius * cos(angle); *output++ = radius * sin(angle); } } This is untested, there might be an off-by-one hiding in the angle step calculation but it should be close. This assumes I understood the question correctly, of course. UPDATE: Redid the angle computation to not be incrementing, to reduce float precision loss due to repeated addition.
Here's a solution, somewhat optimized, untested. Error can accumulate, but using double rather than float probably more than makes up for it except with extremely large values of n. void make_circle(double *dest, size_t n, double r) { double x0 = cos(2*M_PI/n), y0 = sin(2*M_PI/n), x=x0, y=y0, tmp; for (;;) { *dest++ = r*x; *dest++ = r*y; if (!--n) break; tmp = x*x0 - y*y0; y = x*y0 + y*x0; x = tmp; } }
You have to solve this in c language: In an x-y Cartesian coordinate system, the circle with centre coordinates (a, b) and radius r is the set of all points (x, y) such that (x - a)^2 + (y - b)^2 = r^2
Here's a javascript implementation that also takes an optional center point. function circlePoints (radius, numPoints, centerX, centerY) { centerX = centerX || 0; centerY = centerY || 0; var step = (Math.PI * 2) / numPoints, current = 0, i = 0, results = [], x, y; for (; i < numPoints; i += 1) { x = centerX + Math.sin(current) * radius; y = centerY + Math.cos(current) * radius; results.push([x,y]); console.log('point %d # x:%d, y:%d', i, x, y); current += step; } return results; }
fast algorithm for drawing filled circles?
I am using Bresenham's circle algorithm for fast circle drawing. However, I also want to (at the request of the user) draw a filled circle. Is there a fast and efficient way of doing this? Something along the same lines of Bresenham? The language I am using is C.
Having read the Wikipedia page on Bresenham's (also 'Midpoint') circle algorithm, it would appear that the easiest thing to do would be to modify its actions, such that instead of setPixel(x0 + x, y0 + y); setPixel(x0 - x, y0 + y); and similar, each time you instead do lineFrom(x0 - x, y0 + y, x0 + x, y0 + y); That is, for each pair of points (with the same y) that Bresenham would you have you plot, you instead connect with a line.
Just use brute force. This method iterates over a few too many pixels, but it only uses integer multiplications and additions. You completely avoid the complexity of Bresenham and the possible bottleneck of sqrt. for(int y=-radius; y<=radius; y++) for(int x=-radius; x<=radius; x++) if(x*x+y*y <= radius*radius) setpixel(origin.x+x, origin.y+y);
Here's a C# rough guide (shouldn't be that hard to get the right idea for C) - this is the "raw" form without using Bresenham to eliminate repeated square-roots. Bitmap bmp = new Bitmap(200, 200); int r = 50; // radius int ox = 100, oy = 100; // origin for (int x = -r; x < r ; x++) { int height = (int)Math.Sqrt(r * r - x * x); for (int y = -height; y < height; y++) bmp.SetPixel(x + ox, y + oy, Color.Red); } bmp.Save(#"c:\users\dearwicker\Desktop\circle.bmp");
You can use this: void DrawFilledCircle(int x0, int y0, int radius) { int x = radius; int y = 0; int xChange = 1 - (radius << 1); int yChange = 0; int radiusError = 0; while (x >= y) { for (int i = x0 - x; i <= x0 + x; i++) { SetPixel(i, y0 + y); SetPixel(i, y0 - y); } for (int i = x0 - y; i <= x0 + y; i++) { SetPixel(i, y0 + x); SetPixel(i, y0 - x); } y++; radiusError += yChange; yChange += 2; if (((radiusError << 1) + xChange) > 0) { x--; radiusError += xChange; xChange += 2; } } }
Great ideas here! Since I'm at a project that requires many thousands of circles to be drawn, I have evaluated all suggestions here (and improved a few by precomputing the square of the radius): http://quick-bench.com/mwTOodNOI81k1ddaTCGH_Cmn_Ag The Rev variants just have x and y swapped because consecutive access along the y axis are faster with the way my grid/canvas structure works. The clear winner is Daniel Earwicker's method ( DrawCircleBruteforcePrecalc ) that precomputes the Y value to avoid unnecessary radius checks. Somewhat surprisingly that negates the additional computation caused by the sqrt call. Some comments suggest that kmillen's variant (DrawCircleSingleLoop) that works with a single loop should be very fast, but it's the slowest here. I assume that is because of all the divisions. But perhaps I have adapted it wrong to the global variables in that code. Would be great if someone takes a look. EDIT: After looking for the first time since college years at some assembler code, I managed find that the final additions of the circle's origin are a culprit. Precomputing those, I improved the fastest method by a factor of another 3.7-3.9 according to the bench! http://quick-bench.com/7ZYitwJIUgF_OkDUgnyMJY4lGlA Amazing. This being my code: for (int x = -radius; x < radius ; x++) { int hh = (int)std::sqrt(radius_sqr - x * x); int rx = center_x + x; int ph = center_y + hh; for (int y = center_y-hh; y < ph; y++) canvas[rx][y] = 1; }
I like palm3D's answer. For being brute force, this is an amazingly fast solution. There are no square root or trigonometric functions to slow it down. Its one weakness is the nested loop. Converting this to a single loop makes this function almost twice as fast. int r2 = r * r; int area = r2 << 2; int rr = r << 1; for (int i = 0; i < area; i++) { int tx = (i % rr) - r; int ty = (i / rr) - r; if (tx * tx + ty * ty <= r2) SetPixel(x + tx, y + ty, c); } This single loop solution rivals the efficiency of a line drawing solution. int r2 = r * r; for (int cy = -r; cy <= r; cy++) { int cx = (int)(Math.Sqrt(r2 - cy * cy) + 0.5); int cyy = cy + y; lineDDA(x - cx, cyy, x + cx, cyy, c); }
palm3D's brute-force algorithm I found to be a good starting point. This method uses the same premise, however it includes a couple of ways to skip checking most of the pixels. First, here's the code: int largestX = circle.radius; for (int y = 0; y <= radius; ++y) { for (int x = largestX; x >= 0; --x) { if ((x * x) + (y * y) <= (circle.radius * circle.radius)) { drawLine(circle.center.x - x, circle.center.x + x, circle.center.y + y); drawLine(circle.center.x - x, circle.center.x + x, circle.center.y - y); largestX = x; break; // go to next y coordinate } } } Next, the explanation. The first thing to note is that if you find the minimum x coordinate that is within the circle for a given horizontal line, you immediately know the maximum x coordinate. This is due to the symmetry of the circle. If the minimum x coordinate is 10 pixels ahead of the left of the bounding box of the circle, then the maximum x is 10 pixels behind the right of the bounding box of the circle. The reason to iterate from high x values to low x values, is that the minimum x value will be found with less iterations. This is because the minimum x value is closer to the left of the bounding box than the centre x coordinate of the circle for most lines, due to the circle being curved outwards, as seen on this image The next thing to note is that since the circle is also symmetric vertically, each line you find gives you a free second line to draw, each time you find a line in the top half of the circle, you get one on the bottom half at the radius-y y coordinate. Therefore, when any line is found, two can be drawn and only the top half of the y values needs to be iterated over. The last thing to note is that is that if you start from a y value that is at the centre of the circle and then move towards the top for y, then the minimum x value for each next line must be closer to the centre x coordinate of the circle than the last line. This is also due to the circle curving closer towards the centre x value as you go up the circle. Here is a visual on how that is the case. In summary: If you find the minimum x coordinate of a line, you get the maximum x coordinate for free. Every line you find to draw on the top half of the circle gives you a line on the bottom half of the circle for free. Every minimum x coordinate has to be closer to the centre of the circle than the previous x coordinate for each line when iterating from the centre y coordinate to the top. You can also store the value of (radius * radius), and also (y * y) instead of calculating them multiple times.
Here's how I'm doing it: I'm using fixed point values with two bits precision (we have to manage half points and square values of half points) As mentionned in a previous answer, I'm also using square values instead of square roots. First, I'm detecting border limit of my circle in a 1/8th portion of the circle. I'm using symetric of these points to draw the 4 "borders" of the circle. Then I'm drawing the square inside the circle. Unlike the midpoint circle algorith, this one will work with even diameters (and with real numbers diameters too, with some little changes). Please forgive me if my explanations were not clear, I'm french ;) void DrawFilledCircle(int circleDiameter, int circlePosX, int circlePosY) { const int FULL = (1 << 2); const int HALF = (FULL >> 1); int size = (circleDiameter << 2);// fixed point value for size int ray = (size >> 1); int dY2; int ray2 = ray * ray; int posmin,posmax; int Y,X; int x = ((circleDiameter&1)==1) ? ray : ray - HALF; int y = HALF; circlePosX -= (circleDiameter>>1); circlePosY -= (circleDiameter>>1); for (;; y+=FULL) { dY2 = (ray - y) * (ray - y); for (;; x-=FULL) { if (dY2 + (ray - x) * (ray - x) <= ray2) continue; if (x < y) { Y = (y >> 2); posmin = Y; posmax = circleDiameter - Y; // Draw inside square and leave while (Y < posmax) { for (X = posmin; X < posmax; X++) setPixel(circlePosX+X, circlePosY+Y); Y++; } // Just for a better understanding, the while loop does the same thing as: // DrawSquare(circlePosX+Y, circlePosY+Y, circleDiameter - 2*Y); return; } // Draw the 4 borders X = (x >> 2) + 1; Y = y >> 2; posmax = circleDiameter - X; int mirrorY = circleDiameter - Y - 1; while (X < posmax) { setPixel(circlePosX+X, circlePosY+Y); setPixel(circlePosX+X, circlePosY+mirrorY); setPixel(circlePosX+Y, circlePosY+X); setPixel(circlePosX+mirrorY, circlePosY+X); X++; } // Just for a better understanding, the while loop does the same thing as: // int lineSize = circleDiameter - X*2; // Upper border: // DrawHorizontalLine(circlePosX+X, circlePosY+Y, lineSize); // Lower border: // DrawHorizontalLine(circlePosX+X, circlePosY+mirrorY, lineSize); // Left border: // DrawVerticalLine(circlePosX+Y, circlePosY+X, lineSize); // Right border: // DrawVerticalLine(circlePosX+mirrorY, circlePosY+X, lineSize); break; } } } void DrawSquare(int x, int y, int size) { for( int i=0 ; i<size ; i++ ) DrawHorizontalLine(x, y+i, size); } void DrawHorizontalLine(int x, int y, int width) { for(int i=0 ; i<width ; i++ ) SetPixel(x+i, y); } void DrawVerticalLine(int x, int y, int height) { for(int i=0 ; i<height ; i++ ) SetPixel(x, y+i); } To use non-integer diameter, you can increase precision of fixed point or use double values. It should even be possible to make a sort of anti-alias depending on the difference between dY2 + (ray - x) * (ray - x) and ray2 (dx² + dy² and r²)
If you want a fast algorithm, consider drawing a polygon with N sides, the higher is N, the more precise will be the circle.
I would just generate a list of points and then use a polygon draw function for the rendering.
It may not be the algorithm yo are looking for and not the most performant one, but I always do something like this: void fillCircle(int x, int y, int radius){ // fill a circle for(int rad = radius; rad >= 0; rad--){ // stroke a circle for(double i = 0; i <= PI * 2; i+=0.01){ int pX = x + rad * cos(i); int pY = y + rad * sin(i); drawPoint(pX, pY); } } }
The following two methods avoid the repeated square root calculation by drawing multiple parts of the circle at once and should therefore be quite fast: void circleFill(const size_t centerX, const size_t centerY, const size_t radius, color fill) { if (centerX < radius || centerY < radius || centerX + radius > width || centerY + radius > height) return; const size_t signedRadius = radius * radius; for (size_t y = 0; y < radius; y++) { const size_t up = (centerY - y) * width; const size_t down = (centerY + y) * width; const size_t halfWidth = roundf(sqrtf(signedRadius - y * y)); for (size_t x = 0; x < halfWidth; x++) { const size_t left = centerX - x; const size_t right = centerX + x; pixels[left + up] = fill; pixels[right + up] = fill; pixels[left + down] = fill; pixels[right + down] = fill; } } } void circleContour(const size_t centerX, const size_t centerY, const size_t radius, color stroke) { if (centerX < radius || centerY < radius || centerX + radius > width || centerY + radius > height) return; const size_t signedRadius = radius * radius; const size_t maxSlopePoint = ceilf(radius * 0.707106781f); //ceilf(radius * cosf(TWO_PI/8)); for (size_t i = 0; i < maxSlopePoint; i++) { const size_t depth = roundf(sqrtf(signedRadius - i * i)); size_t left = centerX - depth; size_t right = centerX + depth; size_t up = (centerY - i) * width; size_t down = (centerY + i) * width; pixels[left + up] = stroke; pixels[right + up] = stroke; pixels[left + down] = stroke; pixels[right + down] = stroke; left = centerX - i; right = centerX + i; up = (centerY - depth) * width; down = (centerY + depth) * width; pixels[left + up] = stroke; pixels[right + up] = stroke; pixels[left + down] = stroke; pixels[right + down] = stroke; } }
This was used in my new 3D printer Firmware, and it is proven the fastest way for filled circle of a diameter from 1 to 43 pixel. If larger is needed, the following memory block(or array) should be extended following a structure I wont waste my time explaining... If you have questions, or need larger diameter than 43, contact me, I will help you drawing the fastest and perfect filled circles... or Bresenham's circle drawing algorithm can be used above those diameters, but having to fill the circle after, or incorporating the fill into Bresenham's circle drawing algorithm, will only result in slower fill circle than my code. I already benchmarked the different codes, my solution is 4 to 5 times faster. As a test I have been able to draw hundreds of filled circles of different size and colors on a BigTreeTech tft24 1.1 running on a 1-core 72 Mhz cortex-m4 https://www.youtube.com/watch?v=7_Wp5yn3ADI // this must be declared anywhere, as static or global // as long as the function can access it ! uint8_t Rset[252]={ 0,1,1,2,2,1,2,3,3,1,3,3,4,4,2,3,4,5,5,5,2,4,5,5, 6,6,6,2,4,5,6,6,7,7,7,2,4,5,6,7,7,8,8,8,2,5,6,7, 8,8,8,9,9,9,3,5,6,7,8,9,9,10,10,10,10,3,5,7,8,9, 9,10,10,11,11,11,11,3,5,7,8,9,10,10,11,11,12,12, 12,12,3,6,7,9,10,10,11,12,12,12,13,13,13,13,3,6, 8,9,10,11,12,12,13,13,13,14,14,14,14,3,6,8,9,10, 11,12,13,13,14,14,14,15,15,15,15,3,6,8,10,11,12, 13,13,14,14,15,15,15,16,16,16,16,4,7,8,10,11,12, 13,14,14,15,16,16,16,17,17,17,17,17,4,7,9,10,12, 13,14,14,15,16,16,17,17,17,18,18,18,18,18,4,7,9, 11,12,13,14,15,16,16,17,17,18,18,18,19,19,19,19, 19,7,9,11,12,13,15,15,16,17,18,18,19,19,20,20,20, 20,20,20,20,20,7,9,11,12,14,15,16,17,17,18,19,19 20,20,21,21,21,21,21,21,21,21}; // SOLUTION 1: (the fastest) void FillCircle_v1(uint16_t x, uint16_t y, uint16_t r) { // all needed variables are created and set to their value... uint16_t radius=(r<1) ? 1 : r ; if (radius>21 ) {radius=21; } uint16_t diam=(radius*2)+1; uint16_t ymir=0, cur_y=0; radius--; uint16_t target=(radius*radius+3*radius)/2; radius++; // this part draws directly into the ILI94xx TFT buffer mem. // using pointers..2 versions where you can draw // pixels and lines with coordinates will follow for (uint16_t yy=0; yy<diam; yy++) { ymir= (yy<=radius) ? yy+target : target+diam-(yy+1); cur_y=y-radius+yy; uint16_t *pixel=buffer_start_addr+x-Rset[ymir]+cur_y*buffer_width; for (uint16_t xx= 0; xx<=(2*Rset[ymir]); xx++) { *pixel++ = CANVAS::draw_color; }}} // SOLUTION 2: adaptable to any system that can // add a pixel at a time: (drawpixel or add_pixel,etc_) void FillCircle_v2(uint16_t x, uint16_t y, uint16_t r) { // all needed variables are created and set to their value... uint16_t radius=(r<1) ? 1 : r ; if (radius>21 ) {radius=21; } uint16_t diam=(radius*2)+1; uint16_t ymir=0, cur_y=0; radius--; uint16_t target=(radius*radius+3*radius)/2; radius++; for (uint16_t yy=0; yy<diam; yy++) { ymir= (yy<=radius) ? yy+target : target+diam-(yy+1); cur_y=y-radius+yy; uint16_t Pixel_x=x-Rset[ymir]; for (uint16_t xx= 0; xx<=(2*Rset[ymir]); xx++) { //use your add_pixel or draw_pixel here // using those coordinates: // X position will be... (Pixel_x+xx) // Y position will be... (cur_y) // and add those 3 brackets at the end }}} // SOLUTION 3: adaptable to any system that can draw fast // horizontal lines void FillCircle_v3(uint16_t x, uint16_t y, uint16_t r) { // all needed variables are created and set to their value... uint16_t radius=(r<1) ? 1 : r ; if (radius>21 ) {radius=21; } uint16_t diam=(radius*2)+1; uint16_t ymir=0, cur_y=0; radius--; uint16_t target=(radius*radius+3*radius)/2; radius++; for (uint16_t yy=0; yy<diam; yy++) { ymir= (yy<=radius) ? yy+target : target+diam-(yy+1); cur_y=y-radius+yy; uint16_t start_x=x-Rset[ymir]; uint16_t width_x=2*Rset[ymir]; // ... then use your best drawline function using those values: // start_x: position X of the start of the line // cur_y: position Y of the current line // width_x: length of the line // if you need a 2nd coordinate then :end_x=start_x+width_x // and add those 2 brackets after !!! }}
I did pretty much what AlegGeorge did but I changed three lines. I thought that this is faster but these are the results am I doing anything wrong? my function is called DrawBruteforcePrecalcV4. here's the code: for (int x = 0; x < radius ; x++) // Instead of looping from -radius to radius I loop from 0 to radius { int hh = (int)std::sqrt(radius_sqr - x * x); int rx = center_x + x; int cmx = center_x - x; int ph = center_y+hh; for (int y = center_y-hh; y < ph; y++) { canvas[rx][y] = 1; canvas[cmx][y] = 1; } }