Method ComputeDivisions

Common/DataModel/vtkBoundingBox.cxx:472–555

------------------------------------------------------------------------------ Compute the number of divisions given the current bounding box and a target number of buckets/bins. Note that degenerate bounding boxes (i.e., one or more of the edges are zero length) are handled properly.

// target number of buckets/bins. Note that degenerate bounding boxes (i.e.,
// one or more of the edges are zero length) are handled properly.
vtkIdType vtkBoundingBox::ComputeDivisions(vtkIdType totalBins, double bounds[6], int divs[3]) const
{
  // This will always produce at least one bin
  totalBins = (totalBins <= 0 ? 1 : totalBins);

  // First determine the maximum length of the side of the bounds. Keep track
  // of zero width sides of the bounding box.
  int numNonZero = 0, nonZero[3], maxIdx = (-1);
  double max = 0.0, lengths[3];
  this->GetLengths(lengths);

  // Use a finite tolerance when detecting zero width sides to ensure that
  // numerical noise doesn't cause an explosion later on. We'll consider any
  // length that's less than 0.1% of the average length to be zero:
  double totLen = lengths[0] + lengths[1] + lengths[2];
  const double zeroDetectionTolerance = totLen * (0.001 / 3.);

  for (int i = 0; i < 3; ++i)
  {
    if (lengths[i] > max)
    {
      maxIdx = i;
      max = lengths[i];
    }
    if (lengths[i] > zeroDetectionTolerance)
    {
      nonZero[i] = 1;
      numNonZero++;
    }
    else
    {
      nonZero[i] = 0;
    }
  }

  // If the bounding box is degenerate, then one bin of arbitrary size
  if (numNonZero < 1)
  {
    divs[0] = divs[1] = divs[2] = 1;
    bounds[0] = this->MinPnt[0] - 0.5;
    bounds[1] = this->MaxPnt[0] + 0.5;
    bounds[2] = this->MinPnt[1] - 0.5;
    bounds[3] = this->MaxPnt[1] + 0.5;
    bounds[4] = this->MinPnt[2] - 0.5;
    bounds[5] = this->MaxPnt[2] + 0.5;
    return 1;
  }

  // Okay we need to compute the divisions roughly in proportion to the
  // bounding box edge lengths.  The idea is to make the bins as close to a
  // cube as possible. Ensure that the number of divisions is valid.
  double f = static_cast<double>(totalBins);
  f /= (nonZero[0] ? (lengths[0] / totLen) : 1.0);
  f /= (nonZero[1] ? (lengths[1] / totLen) : 1.0);
  f /= (nonZero[2] ? (lengths[2] / totLen) : 1.0);
  f = pow(f, (1.0 / static_cast<double>(numNonZero)));

  for (int i = 0; i < 3; ++i)