3D segmentation - 2D kernels?? #29
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Perhaps, in 3D, instead of using large 3D kernels of size n^3, you could instead use sets of three orthogonal 2D kernels = 3n^2.
n=3 : 3D=27, 32D=39=27
n=5 : 3D=125, 32D=325=75
n=7 : 3D=343, 32D=147
n=33 : 3D=36000, 32D=3267
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