[metapost] The strangest issues?
Giuseppe Bilotta
gip.bilotta at iol.it
Fri Mar 18 10:59:55 CET 2005
Thursday, March 17, 2005 Daniel H. Luecking wrote:
> Assuming that t and \tau are differentiable functions of the lambdas,
> for one. But you are also forgetting that \gamma(t) = \omega(\tau) is
> another solution.
Rather than forgetting, I'm just trying to approach the
problem from this point of view: find a finite number of
candidate values, test all of them and select the correct
ones. This allows me to 'a priori' discard certain solutions (like
those having t or \tau equal to 0 or 1), but also to collect
points in other forms (for example, where the t and \tau are
*not* differentiable but still continuous; assuming of
course they can still be expressed as piecewise
differnetiable functions).
> Even if differentiability is satisfied, since you've not done anything
> to restrict t or \tau, \gamma(t) = \omega(\tau) is very likely for the
> generic case. Consider a fixed pair of lambdas, it is likely there will
> be no pair (t,\tau) which maximizes the distance (unless the two cubics
> head off to infinity in parallel) and so any pair (t,\tau) that satifies
> the equations
> (\gamma(t) -\omega(\tau)).\gamma'(t) = 0
> (\gamma(t) -\omega(\tau)).\omega'(\tau) = 0
> will be a minimum (or a saddle point). And minimums happen when the
> curves cross. Minimizing over the lambdas will now likely give you the
> crossing points. Of course, if \gamma is a straight line, the
> alternative is that \omega is a parallel line and (P1-P0) x (P3-P2) = 0.
Well, t and \tau are restricted to the [0,1] range and since
I'm for local, not global, extremal points I really don't
care if they are minima or maxima until I get down to
analyzing them. Which I cannot do if I cannot find them, of
course :)
Plus, the curves \omega and \gamma are not entirely
arbitrary. While I'm trying not to put any constraints on
\gamma, the endpoints of \omega and its initial and final
tangents are in very well fixed relations to \gamma.
Constraints on the lambdas can be also easily created so
that the two curves don't cross in the [0,1]^2 box, unless
the original curve is knotted itself or the radius is higher
than a critical value.
Does this mean I only have to look at the values for which
t and \tau are not differentiable?
--
Giuseppe "Oblomov" Bilotta
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