On 22 March 2010 21:57, Andrea Aime wrote:
> Maybe by adding missing points outside of the specified range
> to make the cubic algorithm work?
> By the way I read the spec we can imagine repeating the first
> and the last point indefinitely below and above what was
> specified without really changing the outcome?
> So I guess we could add one copy of the first point below
> the first point and one copy of the last point above the
> last point just to handle the corner case of X falling
> in the area we cannot cover with the given points.
> Wondering, would doubling the point in place work?
> Like, we assume the first point is there two times in
> the expression? And same for the last?
> Would that provide a continuous output?

Here's a test I did with the cubic algorithm...
http://imagebin.org/89895

The first and last points are added such that they like outside the
range of the "real" interpolation points but have the same y-value as
their neighbour.  For the cubic algorithm it is actually a requirement
that no points have tied x values - I'll have to add a check for that
to the function.

So there could be this scheme:

N points   Algorithm
0             any           throw an Exception
1             any           evaluate point's value and return for all inputs
>=2         lin / cos     ok
2             cubic         fall back to linear (since 4 collinear
points would give ~linear results)
3             cubic         add dummy point before or after to put
lookup value into
                                  middle interpolation interval
>=4         cubic         ok

Plus, for 2 or more points, treat lookup values outside the range of
the points as per the spec.

What do you think ?

Michael

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