All–
I’ve encountered what I believe to be a bug in the behavior of itk::ConnectedComponentImageFilter. In the below example, I create a 2x2x2 “Checkerboard” image, where there are 4 foreground pixels, none of which share a face. I then use the itk::ConnectedComponentImageFilter to label the pixels, with FullyConnected set to Off. In the result, three pixels are correctly labeled, but the pixel at [1, 0, 1] is incorrectly labeled as being “connected” to the pixel at [1, 1, 0], although they do not share a face.
Here is a MCE:
#include <itkConnectedComponentImageFilter.h>
using TImage = itk::Image<unsigned char, 3>;
using TConnected = itk::ConnectedComponentImageFilter< TImage, TImage >;
int main() {
std::bitset<8> bits(105); // 3D Checkerboard
std::cout << bits << std::endl; // 01101001
const auto image = TImage::New();
TImage::RegionType region({{0,0,0}}, {{2,2,2}});
image->SetRegions( region );
image->Allocate();
image->FillBuffer( 0 );
for (size_t i = 0; i < 8; ++i) {
std::bitset<3> idx(i);
image->SetPixel( {{idx[0], idx[1], idx[2]}}, bits[i] );
}
const auto connected = TConnected::New();
connected->SetInput( image );
connected->FullyConnectedOff();
connected->Update();
itk::ImageRegionConstIterator<TImage> it(connected->GetOutput(), connected->GetOutput()->GetLargestPossibleRegion());
while (!it.IsAtEnd()) {
std::cout << "Index : Label => " << it.GetIndex() << " : " << static_cast<int>(it.Get()) << std::endl;
++it;
}
// Index : Label => [0, 0, 0] : 1
// Index : Label => [1, 0, 0] : 0
// Index : Label => [0, 1, 0] : 0
// Index : Label => [1, 1, 0] : 2
// Index : Label => [0, 0, 1] : 0
// Index : Label => [1, 0, 1] : 2 <= Incorrect Label
// Index : Label => [0, 1, 1] : 3
// Index : Label => [1, 1, 1] : 0
return EXIT_SUCCESS;
}
And here is a screenshot of the output, where I’ve extracted the labeled image as a mesh where each box (corresponding to one pixel) is colored according to its pixel value. Note that none of the four foreground pixels share a face, and yet two pixels have the same label:
Thanks in advance, and apologies if there is an error somewhere in my logic.
Best,
–Davis
