How can I draw a major pieslice in C, using the function pieslice()?
pieslice(X-centre,Y-centre,StrtAngle,EndAngle,Radius).
I am trying to draw a major sector or pieslice in C, using the pieslice function; I want the start angle to be 135 degrees and end angle to be 235 degrees, but at the same time it should be the major sector, not the minor sector.
I tried all the four combinations
pieslice(100,100,135,-135,20)
pieslice(200,200,225,135,30)
pieslice(300,300,225,360+135,30)
pieslice(400,400,135,225,20)
pieslice(50,50,0,135,30);
pieslice(50,50,225,0,30);
But all of them draw the corresponding minor sector not the major sector. Can someone please advise me how to do that?
Here is a screenshot of the output:
Thanks for your effort and time.
Now, I could not make the pieslice to work my way. However with the following tweak, I am able to get around the problem and get the desired output. I wrote a user defined function slice(int x-centre, int y-centre,int sangle, int eangle, int radius) similar to pieslice.
I hope it is useful for those who get stuck in a similar kind of situation:
void slice(int x, int y, int sangle, int eangle, int rad)
{
int i,j,color;
if(sangle>eangle){
color=getcolor();
setcolor(getcolor()) ;
setfillstyle(SOLID_FILL,color);
circle(x,y,rad);
floodfill(x,y,color);
setcolor(getbkcolor());
setfillstyle(SOLID_FILL,getbkcolor());
pieslice(x,y,eangle,sangle,rad);
setcolor(color);
}
}
Draw two pie slices with the same centre and radius, one from 0 to 135 degrees, and one from 225 to 0 degrees. It seems that the function is forcing the pie slices to be always less than 180 degrees, so this should work around that.
See also: http://electrosofts.com/cgraphics/
here is my game loop using pieslice . pacman moves to and fro.
for (int dx = 10, dy = 0, dt = 100; c != 'q';) {
if ((x + rx + dx) > getmaxx() || (x + rx + dx) < 0) {
dx = -dx;
rx = -rx;
}
if ((y + ry + dy) > getmaxy() || (y + ry + dy) < 0) {
dy = -dy;
ry = -ry;
}
delay(dt);
cleardevice();
//gotoxy(1,1);
//cout << x+rx << " " << y+ry << " " << m;
pieslice(x, y, (sa + m), (ea - m), RADIUS);
//floodfill(x,y,getcolor());
if (m + dm < 0 || m + dm > ea)
dm = -dm;
m += dm;
x += dx;
y += dy;
if (kbhit())
c = getch();
}
i need to draw a polygon of "n" sides given 2 points (the center and 1 of his vertex) just that i suck in math. I have been reading a lot and all this is what i have been able to figure it (i dont know if it is correct):
Ok, i take the distance between the 2 points (radius) with the theorem of Pythagoras:
sqrt(pow(abs(x - xc), 2) + pow(abs(y - yc), 2));
And the angle between this 2 points with atan2, like this:
atan2(abs(y - yc), abs(x - xc));
Where xc, yc is the center point and x, y is the only vertex know.
And with that data i do:
void polygon(int xc, int yc, int radius, double angle, int sides)
{
int i;
double ang = 360/sides; //Every vertex is about "ang" degrees from each other
radian = 180/M_PI;
int points_x[7]; //Here i store the calculated vertexs
int points_y[7]; //Here i store the calculated vertexs
/*Here i calculate the vertexs of the polygon*/
for(i=0; i<sides; i++)
{
points_x[i] = xc + ceil(radius * cos(angle/radian));
points_y[i] = yc + ceil(radius * sin(angle/radian));
angle = angle+ang;
}
/*Here i draw the polygon with the know vertexs just calculated*/
for(i=0; i<sides-1; i++)
line(points_x[i], points_y[i], points_x[i+1], points_y[i+1]);
line(points_y[i], points_x[i], points_x[0], points_y[0]);
}
The problem is that the program dont work correctly because it draw the lines not like a polygon.
Someone how know enough of math to give a hand? im working in this graphics primitives with C and turbo C.
Edit: i dont want to fill the polygon, just draw it.
Consider what 360/sides actually returns if sides is not a factor of 360 (this is integer division - see what 360/7 actually returns).
There is no need to use degrees at all - use 2*Math_PI/(double)nsides and work throughout in radians.
also you can omit the final line by using the modulus function (module nsides).
If you have more than 7 sides you will not be able to store all the points. You don't need to store all the points if you are simply drawing the polygon rather than storing it - just the last point and the current one.
You should be using radians in all your calculations. Here's a complete program that illustrates how best to do this:
#include <stdio.h>
#define PI 3.141592653589
static void line (int x1, int y1, int x2, int y2) {
printf ("Line from (%3d,%3d) - (%3d,%3d)\n", x1, y1, x2, y2);
}
static void polygon (int xc, int yc, int x, int y, int n) {
int lastx, lasty;
double r = sqrt ((x - xc) * (x - xc) + (y - yc) * (y - yc));
double a = atan2 (y - yc, x - xc);
int i;
for (i = 1; i <= n; i++) {
lastx = x; lasty = y;
a = a + PI * 2 / n;
x = round ((double)xc + (double)r * cos(a));
y = round ((double)yc + (double)r * sin(a));
line (lastx, lasty, x, y);
}
}
int main(int argc, char* argv[]) {
polygon (0,0,0,10,4); // A diamond.
polygon (0,0,10,10,4); // A square.
polygon (0,0,0,10,8); // An octagon.
return 0;
}
which outputs (no fancy graphics here, but you should get the idea):
===
Line from ( 0, 10) - (-10, 0)
Line from (-10, 0) - ( 0,-10)
Line from ( 0,-10) - ( 10, 0)
Line from ( 10, 0) - ( 0, 10)
===
Line from ( 10, 10) - (-10, 10)
Line from (-10, 10) - (-10,-10)
Line from (-10,-10) - ( 10,-10)
Line from ( 10,-10) - ( 10, 10)
===
Line from ( 0, 10) - ( -7, 7)
Line from ( -7, 7) - (-10, 0)
Line from (-10, 0) - ( -7, -7)
Line from ( -7, -7) - ( 0,-10)
Line from ( 0,-10) - ( 7, -7)
Line from ( 7, -7) - ( 10, 0)
Line from ( 10, 0) - ( 7, 7)
Line from ( 7, 7) - ( 0, 10)
I've written the polygon function as per your original specification, passing in just the two co-ordinates. As an aside, you don't want those abs calls in your calculations for radius and angle because:
they're useless for radius (since -n2 = n2 for all n).
they're bad for angle since that will force you into a specific quadrant (wrong starting point).
I'm not going to just give you the answer, but I have some advice. First, learn how line drawing works INSIDE AND OUT. When you have this down, try to write a filled triangle renderer.
Generally, filled polygons are drawn 1 horizontal scan line at a time, top to bottom. You're job is to determine the starting and stopping x coordinate for every scan line. Note that the edge of a polygon follows a straight line (hint, hint)... :)
You're trying to draw a filled poly I guess?
If you're going to try to draw the polys using a line primitive, you're going to have a lot of pain coming to you. dicroce actually gave you some very good advice on that front.
Your best bet is to find a primitive that fills for you and supply it a coordinates list. It's up to you to determine the coordinates to give it.
I think the main trouble is: atan2(abs(y - yc), abs(x - xc)); is giving you radians, not degrees, just convert it to degrees and try.
/* all angles in radians */
double ainc = PI*2 / sides;
int x1, y1;
for (i = 0; i <= sides; i++){
double a = angle + ainc * i;
int x = xc + radius * cos(a);
int y = yc + radius * sin(a);
if (i > 0) line(x1, y1, x, y);
x1 = x; y1 = y;
}
Or, you could save the points in an array and call the DrawPoly routine if you've got one.
If you want a filled polygon, call FillPoly if you've got one.
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;
}
}
I created this function that draws a simple polygon with n number of vertexes:
void polygon (int n)
{
double pI = 3.141592653589;
double area = min(width / 2, height / 2);
int X = 0, Y = area - 1;
double offset = Y;
int lastx, lasty;
double radius = sqrt(X * X + Y * Y);
double quadrant = atan2(Y, X);
int i;
for (i = 1; i <= n; i++)
{
lastx = X; lasty = Y;
quadrant = quadrant + pI * 2.0 / n;
X = round((double)radius * cos(quadrant));
Y = round((double)radius * sin(quadrant));
setpen((i * 255) / n, 0, 0, 0.0, 1); // r(interval) g b, a, size
moveto(offset + lastx, offset + lasty); // Moves line offset
lineto(offset + X, offset + Y); // Draws a line from offset
}
}
How can I fill it with a solid color?
I have no idea how can I modify my code in order to draw it filled.
The common approach to fill shapes is to find where the edges of the polygon cross either each x or each y coordinate. Usually, y coordinates are used, so that the filling can be done using horizontal lines. (On framebuffer devices like VGA, horizontal lines are faster than vertical lines, because they use consecutive memory/framebuffer addresses.)
In that vein,
void fill_regular_polygon(int center_x, int center_y, int vertices, int radius)
{
const double a = 2.0 * 3.14159265358979323846 / (double)vertices;
int i = 1;
int y, px, py, nx, ny;
if (vertices < 3 || radius < 1)
return;
px = 0;
py = -radius;
nx = (int)(0.5 + radius * sin(a));
ny = (int)(0.5 - radius * cos(a));
y = -radius;
while (y <= ny || ny > py) {
const int x = px + (nx - px) * (y - py) / (ny - py);
if (center_y + y >= 0 && center_y + y < height) {
if (center_x - x >= 0)
moveto(center_x - x, center_y + y);
else
moveto(0, center_y + y);
if (center_x + x < width)
lineto(center_x + x, center_y + y);
else
lineto(width - 1, center_y + y);
}
y++;
while (y > ny) {
if (nx < 0)
return;
i++;
px = nx;
py = ny;
nx = (int)(0.5 + radius * sin(a * (double)i));
ny = (int)(0.5 - radius * cos(a * (double)i));
}
}
}
Note that I only tested the above with a simple SVG generator, and compared the drawn lines to the polygon. Seems to work correctly, but use at your own risk; no guarantees.
For general shapes, use your favourite search engine to look for "polygon filling" algorithms. For example, this, this, this, and this.
There are 2 different ways to implement a solution:
Scan-line
Starting at the coordinate that is at the top (smallest y value), continue to scan down line by line (incrementing y) and see which edges intersect the line.
For convex polygons you find 2 points, (x1,y) and (x2,y). Simply draw a line between those on each scan-line.
For concave polygons this can also be a multiple of 2. Simply draw lines between each pair. After one pair, go to the next 2 coordinates. This will create a filled/unfilled/filled/unfilled pattern on that scan line which resolves to the correct overall solution.
In case you have self-intersecting polygons, you would also find coordinates that are equal to some of the polygon points, and you have to filter them out. After that, you should be in one of the cases above.
If you filtered out the polygon points during scan-lining, don't forget to draw them as well.
Flood-fill
The other option is to use flood-filling. It has to perform more work evaluating the border cases at every step per pixel, so this tends to turn out as a slower version. The idea is to pick a seed point within the polygon, and basically recursively extend up/down/left/right pixel by pixel until you hit a border.
The algorithm has to read and write the entire surface of the polygon, and does not cross self-intersection points. There can be considerable stack-buildup (for naive implementations at least) for large surfaces, and the reduced flexibility you have for the border condition is pixel-based (e.g. flooding into gaps when other things are drawn on top of the polygon). In this sense, this is not a mathematically correct solution, but it works well for many applications.
The most efficient solution is by decomposing the regular polygon in trapezoids (and one or two triangles).
By symmetry, the vertexes are vertically aligned and it is an easy matter to find the limiting abscissas (X + R cos(2πn/N) and X + R cos(2π(+1)N)).
You also have the ordinates (Y + R sin(2πn/N) and Y + R sin(2π(+1)N)) and it suffices to interpolate linearly between two vertexes by Y = Y0 + (Y1 - Y0) (X - X0) / (X1 - X0).
Filling in horizontal runs is a little more complex, as the vertices may not be aligned horizontally and there are more trapezoids.
Anyway, it seems that I / solved / this myself again, when not relying on assistance (or any attempt for it)
void polygon (int n)
{
double pI = 3.141592653589;
double area = min(width / 2, height / 2);
int X = 0, Y = area - 1;
double offset = Y;
int lastx, lasty;
while(Y-->0) {
double radius = sqrt(X * X + Y * Y);
double quadrant = atan2(Y, X);
int i;
for (i = 1; i <= n; i++)
{
lastx = X; lasty = Y;
quadrant = quadrant + pI * 2.0 / n;
X = round((double)radius * cos(quadrant));
Y = round((double)radius * sin(quadrant));
//setpen((i * 255) / n, 0, 0, 0.0, 1);
setpen(255, 0, 0, 0.0, 1); // just red
moveto(offset + lastx, offset + lasty);
lineto(offset + X, offset + Y);
} }
}
As you can see, it isn't very complex, which means it might not be the most efficient solution either.. but it is close enough.
It decrements radius and fills it by virtue of its smaller version with smaller radius.
On that way, precision plays an important role and the higher n is the less accuracy it will be filled with.