Line-Clipping

A few days ago, I discovered I had at least one, maybe two mathematical bugs. The first bug is in the line clipping algorithm. Here’s a quick reminder of what’s going on there:

There are six points of interest: the four intersection points, and the two endpoints of the line. We sort the six points according to their parameter, and take the two innermost points as the endpoints of the clipped line. To check if the line is outside the box we make sure that the result points’ parameters are between 0 and 1 (which correspond to the original endpoints).

Turns out this is not always true – here’s a counter example:

Counter Example for the Line Clipping Algorithm

In the counter example you can see a line, whose endpoints lie between the intersections. The algorithm described above will say this line lies completely inside the box. Clearly this is not true. How do we fix this? Simple actually. We check for each intersection point whether or not it is on the edges of the clipping rectangle. If at most one point is on – our line is outside. I hope this is the end of it. It will require more rigorous testing.

The second bug was in my solution to the random selection challenge. I wasn’t too sure of my solution, but I really liked it, as its code is rather elegant. So when meeting someone I used to work with, I told her about it. After thinking about my solution a bit, she said it is probably wrong, and gave me the reason: While the probability of a point x passing the first stage is indeed (proportional to) p(x), the probability of it passing the second stage is not 1 (or constant). I still need to think about this one, so no easy fix yet. I guess this reopens the challenge, so if anyone feels like writing a good proof or a counter example for my solution, I’d be happy to see it. I’ll be even happier to see a completely different solution.

I actually wrote about that a long time ago, in the wiki that once was algorithm.co.il. Since the wiki is long gone, and I figured this is a very elegant algorithm, I decided I should write about it.

First, some background – while writing Ascii-Plotter, I had to plot lines. To plot lines correctly I had to clip them. Now since this was a fun project, I decided I wanted to figure out how to do that myself, even though the problem is really solved. So I sat down with pen and paper, and did a little thinking, and I came up this nice algorithm.

The idea is simply this: consider a line, that is not parallel to any of the axes. This line intersects all the four lines that make up the limits of our screen. Here is a diagram:

An intersection of a line with a bounding box

Now notice that there are six points of interest: the four intersection points, and the two endpoints of the line. For each point, we find its parameter in the parametric representation of the line, where a and b are the endpoints:

x(t) = a+t(b-a)

To find out the actual endpoints of the clipped line, we sort the six points according to their parameter, and take the two innermost points (with zero based indexing, those would be 2 and 3). What if the line is outside the box you say? We check that the result points’ parameters are between 0 and 1 (which correspond to the original endpoints).

Note that for a line parallel to one of the axes, a similar method applies, but without two of the intersection points. However, it is probably easier to just calculate the clipping directly in that case.

Here is some sample code, without all the surrounding checks (for parallelism with the axes, trivial clipping etc…).
The full version is available in the code of Ascii-Plotter.

points_t = [0.0,1.0]
 
points_t.append( float(rect_bottom_left[0]-line_pt_1[0])/(line_pt_2[0]-line_pt_1[0]) )
points_t.append( float(rect_top_right[0]-line_pt_1[0])/(line_pt_2[0]-line_pt_1[0]) )
points_t.append( float(rect_bottom_left[1]-line_pt_1[1])/(line_pt_2[1]-line_pt_1[1]) )
points_t.append( float(rect_top_right[1]-line_pt_1[1])/(line_pt_2[1]-line_pt_1[1]) )
 
points_t.sort()
if points_t[2] < 0 or points_t[2] >= 1 or points_t[3] < 0 or points_t[2]>= 1:
	return None
result = [(pt_1 + t*(pt_2-pt_1)) for t in (points_t[2],points_t[3]) for (pt_1, pt_2) in zip(line_pt_1, line_pt_2)]
return (result[0],result[1]), (result[2], result[3])

Line-Clipping

Two Mathematical Bugs

Elegant Line Clipping Algorithm