Simple algorithm for drawing filled ellipse in C/C++

Question

On SO, found the following simple algorithm for drawing filled circles:

for(int y=-radius; y<=radius; y++)
    for(int x=-radius; x<=radius; x++)
        i         


        

Answer 1

Simpler, with no double and no division (but be careful of integer overflow):

for(int y=-height; y<=height; y++) {
    for(int x=-width; x<=width; x++) {
        if(x*x*height*height+y*y*width*width <= height*height*width*width)
            setpixel(origin.x+x, origin.y+y);
    }
}

We can take advantage of two facts to optimize this significantly:

• Ellipses have vertical and horizontal symmetry;
• As you progress away from an axis, the contour of the ellipse slopes more and more.

The first fact saves three-quarters of the work (almost); the second fact tremendously reduces the number of tests (we test only along the edge of the ellipse, and even there we don't have to test every point).

int hh = height * height;
int ww = width * width;
int hhww = hh * ww;
int x0 = width;
int dx = 0;

// do the horizontal diameter
for (int x = -width; x <= width; x++)
    setpixel(origin.x + x, origin.y);

// now do both halves at the same time, away from the diameter
for (int y = 1; y <= height; y++)
{
    int x1 = x0 - (dx - 1);  // try slopes of dx - 1 or more
    for ( ; x1 > 0; x1--)
        if (x1*x1*hh + y*y*ww <= hhww)
            break;
    dx = x0 - x1;  // current approximation of the slope
    x0 = x1;

    for (int x = -x0; x <= x0; x++)
    {
        setpixel(origin.x + x, origin.y - y);
        setpixel(origin.x + x, origin.y + y);
    }
}

This works because each scan line is shorter than the previous one, by at least as much as that one was shorter than the one before it. Because of rounding to integer pixel coordinates, that's not perfectly accurate -- the line can be shorter by one pixel less that that.

In other words, starting from the longest scan line (the horizontal diameter), the amount by which each line is shorter than the previous one, denoted dx in the code, decreases by at most one, stays the same, or increases. The first inner for finds the exact amount by which the current scan line is shorter than the previous one, starting at dx - 1 and up, until we land just inside the ellipse.

                       |         x1 dx x0
                       |######    |<-->|
 current scan line --> |###########    |<>|previous dx
previous scan line --> |################  |
two scan lines ago --> |###################
                       |##################### 
                       |###################### 
                       |######################
                       +------------------------

To compare the number of inside-ellipse tests, each asterisk is one pair of coordinates tested in the naive version:

 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************
 *********************************************

... and in the improved version:

                        *
                             **
                                  ****
                                       ***
                                          ***
                                            ***
                                             **
                                             **