monotonicCubicInterpolant, Implementation Question

[Maik Figura]

Hello Magnus, 

when I was looking at the source code of the monotonicCubicInterpolant, I was wondering how the case (which is described in the original paper) delta_k != 0 and signum(d_k) == signum(d_k+1) == signum(delta_k) is checked against? I could not yet figure out, how you make sure that when delta_k != 0 all signums are the same. From what I understand, the case delta_k =! 0 is implicit by checking for if(delta_k == 0) but this does not include the signums condition, does it? 

template<class T>

T monotonicCubicInterpolant(const T &f1, const T &f2, const T &f3, const T &f4, 

                            double t)

{

  T d_k = T(.5) * (f3 - f1);

  T d_k1 = T(.5) * (f4 - f2);

  T delta_k = f3 - f2;

  if (delta_k == static_cast<T>(0)) {

    d_k = static_cast<T>(0);

    d_k1 = static_cast<T>(0);

  }

  T a0 = f2;

  T a1 = d_k;

  T a2 = (T(3) * delta_k) - (T(2) * d_k) - d_k1;

  T a3 = d_k + d_k1 - (T(2) * delta_k);

  T t1 = t;

  T t2 = t1 * t1;

  T t3 = t2 * t1;

  return a3 * t3 + a2 * t2 + a1 * t1 + a0;

}

Cheers 

Maik 

[Magnus Wrenninge]

Hi Maik,

Sorry for the late reply, I was out at Siggraph last week.

I think you're right, we don't correctly check for that.

 if (delta_k == static_cast<T>(0)) {

    d_k = static_cast<T>(0);

    d_k1 = static_cast<T>(0);

  }

should really be

 if (delta_k == static_cast<T>(0) ||

    (sign(d_k) != sign(delta_k) || sign(d_k1) != sign(delta_k))) {

    d_k = static_cast<T>(0);

    d_k1 = static_cast<T>(0);

 } 

I'll submit a ticket for it. Thanks for finding that.

Magnus

Maik 

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