message: simplify: Clamp grid size to current search bounds
  This makes sure grid_size doesn't go completely outside of the bounds
  which may confuse the search and reset it to something unexpected.
diff: @@ -1238,8 +1238,6 @@ size_t meshopt_simplifySloppy(unsigned int* destination, const unsigned int* ind
  unsigned int* vertex_cells = allocator.allocate<unsigned int>(vertex_count);
  unsigned int* vertex_cells_temp = allocator.allocate<unsigned int>(vertex_count);
  - // TODO: we don't need a table this large if we're limiting the number of unique elements we're adding to the table
  - // Especially if we can alloc this *after* the counting pass, with an approx *2 size or whatever?
  size_t table_size = hashBuckets2(vertex_count);
  HashCell* table = allocator.allocate<HashCell>(table_size);
  @@ -1259,15 +1257,13 @@ size_t meshopt_simplifySloppy(unsigned int* destination, const unsigned int* ind
  {
  // instead of starting in the middle, let's guess as to what the answer might be! triangle count usually grows as a square of grid size...
  grid_size = int(sqrtf(float(target_cell_count)));
  - grid_size = (grid_size < 1) ? 1 : (grid_size > 1024) ? 1024 : grid_size;
  + grid_size = (grid_size < min_grid) ? min_grid : (grid_size > max_grid) ? max_grid : grid_size;
  }
  else if (pass <= SLOP_INTERPOLATION)
  {
  float k = (float(target_index_count / 3) - float(min_triangles)) / (float(max_triangles) - float(min_triangles));
  - float kl = 0.0f; // TODO: should we even clamp since we fall back to binary? to what?
  - k = (k < kl) ? kl : (k > 1 - kl) ? 1 - kl : k;
  grid_size = int(float(min_grid) * (1 - k) + float(max_grid) * k);
  - grid_size = (grid_size < 1) ? 1 : (grid_size > 1024) ? 1024 : grid_size;
  + grid_size = (grid_size < min_grid) ? min_grid : (grid_size > max_grid) ? max_grid : grid_size;
  }
  else
  {
  