Re: Quadratic Bezier intersection (a special case)

I do not believe this is a correct conclusion. Consider the curves
(x(s),y(s)) = (s,s*s) for s >= 0 and (x(t),y(t)) = (t,t) for t >= 0.
Both are monotone in the x- and y-components. However, the curves
intersect in two points.

You are right, it was my mistake.
Then, I also need to estimate whether the curves can potentially
intersect to get rid of expensive operations in most cases
(intersections are rare). This estimation must be computationally
cheap. I feel the criterion is any of line segments p0-p1 and p1-p2
have intersection (it needs to check 4 pair of line segments) then the
curves may potentially intersect (false positives are OK, but not false
negatives!). But I'm not quite sure in it. The 100% robust criterion is
to check the triangles p0-p1-p2 for overlapping.

Still, is there any simplification for the monotone cases? There can be
no more than 2 intersection points, so, looks like a quadratic equation
:)
Will you enlighten me?

.

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