[CST-2] Advanced Graphics

[Alvin]

I think it's a misprint in the question, sorry Chris!  Anyway, the logic
goes: 

if there are 2k-1 basis functions, then there must be (k-1) at each end of
the knot vector, plus 1 which is the uniform knot vector.  I.e.

0,0...0,0 | 1,2....,k-1 | k,k,...,k

    k           k-1            k

So in total, there must be >= 3k - 1 knots

By definition, there are n+k+1 knots so

3k - 1 <= n+k+1

N >= 2k - 2 for there to be 2k - 1 basis functions.

Or N < 2k - 2 for there to be (k-1)(k-2)/2, which we're not expected to
prove, thankfully.

Well the supervisor seems quite happy with this explanation...

Alvin

> 2001/8/4
> 
> Parts (a) and (b) :
> Says how many distinct basis functions are there for an open uniform
> B-spline with n>=2k-3.
> Well the number of distinct basis functions is n+1 for n up to *and
> including 2k-2*. Then it just stays the same.
> 
> So what's wrong with this - cos in my solution the boundary should be
> n>2k-3, not n>=2k-3. I guess I got the maths wrong but my brain is fried
> so help ...
> 
> --
> Matej Pfajfar
> St John's College, University of Cambridge, UK
> 
> GPG Public Keys @ http://matejpfajfar.co.uk/keys
> WARNING: THIS E-MAIL ACCOUNT WILL BE DELETED ON
>        15/07/2002. PLEASE USE mp@cantab.net.
> 
> 
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