Any formula to get point in Bresenham's 2D Line without loop?

Question

Let say we have (x1,y1) and (x2,y2) which might be faraway

Can we calculate the point in line that has N distance from (x1,y1)?

My simple idea is add count in for loop but this will be expensive if distance is very large

*note: floating point can't be use

Answer 1

A = (x₁, y₁)
B = (x₂, y₂)

C = B - A = (x₂ - x₁, y₂ - y₁) = (x_c, y_c)
C² = C . C = x_c² + y_c²
|C| = sqrt(C²)
k = C/|C| = (x_c/|C|, y_c/|C|)

The point you're looking for is D = A + Nk

Answer 2

In the first octant (0 <= y2-y1 <= x2-x1), the formula is:

y(x) = y1 + round( (x-x1)(y2-y1) / (x2-x1) )

If you can't use floating point, then you probably need to user flooring division. You can do the rounding like this:

y(x) = y1 + floor( ( 2(x-x1)(y2-y1) + (x2-x1) ) / 2(x2-x1) )

The whole point of Bresenham's algorithm is to calculate this formula incrementally to avoid the multiplication and division, which were a lot more expensive at the time than they usually are now.

NOTE: I'm assuming you want a distance along x or y, because you want to clip the line. If you really want a Euclidean distance, then see @Beta answer

Answer 3

I'm having the same issue, Bresenham's algorithm is drawing a nice line, but I just can't pick any point without iterating through the coordinates, because of its nature, to avoid the rounding and other "glitches" it decides to place a pixel on-the-fly, which is very handy during rendering, but it makes geometric calculations quite expensive (line crossing/intersection finding for example).

I can't use floating point types either, because I'm developing to an MCU right now which doesn't have FPU, but the following code from my project works quite well, very fast, and I haven't noticed any faults yet.

inline static int16_t GetYOfLineAtX(int16_t x0, int16_t y0, int16_t x1, int16_t y1, int16_t x) {
    return (y1 - y0) * (x - x0) / (x1 - x0) + y0;
}

inline static int16_t GetXOfLineAtY(int16_t x0, int16_t y0, int16_t x1, int16_t y1, int16_t y) {
    return (x1 - x0) * (y - y0) / (y1 - y0) + x0;
}

It is straightforward: x0, y0 is your line starting point, x1, y1 is the ending point (slope may be negative or positive, direction doesn't matter), the x or y parameter on the two functions are the required coordinate, it returns the other axis value.