Linear interpolation along a "Bresenham line"

Question

I'm using linear interpolation for animating an object between two 2d coordinates on the screen. This is pretty close to what I want, but because of rounding, I get a jagged motion. In ASCII art:

ooo
  ooo
    ooo
      oo

Notice how it walks in a Manhattan grid, instead of taking 45 degree turns. What I'd like is linear interpolation along the line which Bresenham's algorithm would have created:

oo
  oo
    oo
      oo

For each x there is only one corresponding y. (And swap x/y for a line that is steep)

So why don't I just use Bresenham's algorithm? I certainly could, but that algorithm is iterative, and I'd like to know just one coordinate along the line.

I am going to try solving this by linearly interpolating the x coordinate, round it to the pixel grid, and then finding the corresponding y. (Again, swap x/y for steep lines). No matter how that solution pans out, though, I'd be interested in other suggestion and maybe previous experience.

Answer 1

Bresenham's algorithm for lines was introduced to draw a complete line a bit faster than usual approaches. It has two major advantages:

• It works on integer variables
• It works iteratively, which is fast, when drawing the complete line

The first advantage is not a great deal, if you calculate only some coordinates. The second advantage turns out as a disadvantage when calculating only some coordinates. So after all, there is no need to use Bresenham's algorithm.

Instead, you can use a different algorithm, which results in the same line. For example the DDA (digital differential analyzer). This is basically, the same approach you mentioned.

First step: Calculate the slope.

m = (y_end - y_start) / (x_end - x_start)

Second step: Calculate the iteration step, which is simply:

i = x - x_start

Third step: Calculate the coresponding y-value:

y = y_start + i * m
  = y_start + (x - x_start) * (y_end - y_start) / (x_end - x_start)