Why does ImageFilter fail with radius 1, but works with radius 2?

Question

Bug introduced in 9.0 and persisting through 13.1.0 [CASE:3968469]

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When thinking on a way to answer this question using ImageFilter I have found (with Jason B's help) that we can find the end points of the skeleton by increasing the radius in ImageFilter from 1 to 2 in the following way:

pic = Thinning@Binarize@Import["https://i.stack.imgur.com/K8dm3.png"];
ep2 = ImageFilter[If[#[[3, 3]] == 1 && Total[#, 2] == 3, 1, 0] &, pic, 2];
ppos = PixelValuePositions[ep2, White]

{{271, 546}, {190, 471}, {694, 382}, {899, 366}}

But I cannot figure out why the original approach with radius 1 fails:

ep1 = ImageFilter[If[#[[2, 2]] == 1 && Total[#, 2] == 2, 1, 0] &, pic, 1];
PixelValuePositions[ep1, White]

{}

This looks unexpected because there is no actual difference between these two approaches as can be clearly seen by visualizing the neighborhood pixels in the thinned image:

ArrayPlot[1 - ImageData[ImageTrim[pic, {# - .5}, 2]], Mesh -> True, 
   ImageSize -> 100] & /@ ppos

Can anyone explain this issue? Is it a bug?

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UPDATE: Additional observations

With the help by Martin Büttner I have discovered that in this particular case the filter function is always supplied with zeros when the radius is 1. Even weirder, there are only 76 (!) evaluations of the filter function:

n = 0; count = 0;

ImageFilter[If[Total[#, 2] == 0, ++n; 0, ++n; ++count; 1] &, pic, 1];

{n, count}

{76, 0}

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All the above output was obtained with version 10.4. With version 8.0.4 ImageFilter behaves as expected:

pic = Thinning@Binarize@Import["https://i.stack.imgur.com/K8dm3.png"];
ep1 = ImageFilter[If[#[[2, 2]] == 1 && Total[#, 2] == 2, 1, 0] &, pic, 1];
Total[ImageData[ep1], 2]

4

So this seems to be a regression bug.

_{Reported as [CASE:3968469]}

Answer 1

One can use MorphologicalTransform[image,f] for filtering binary images using 3×3 pixels window:

ep2 = MorphologicalTransform[pic, If[#[[2, 2]] == 1 && Total[#, 2] == 2, 1, 0] &];
ppos = PixelValuePositions[ep2, White]

{{271, 546}, {190, 471}, {694, 382}, {899, 366}}

This method is 2 times faster than the non-buggy ImageFilter, here are timings for Mathematica 8.0.4 on Windows 7 x64:

Do[i1 = ImageFilter[If[#[[2, 2]] == 1 && Total[#, 2] == 2, 1, 0] &, pic, 1], {10}]; // AbsoluteTiming

Do[i2 = MorphologicalTransform[pic, If[#[[2, 2]] == 1 && Total[#, 2] == 2, 1, 0] &], {10}]; // AbsoluteTiming 

i1 == i2

{0.7150409, Null}

{0.3280187, Null}

True

It is also much faster than alternative workarounds listed below and in the original question.

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Another way is to use ArrayFilter instead of ImageFilter:

ep3 = Image@
  ArrayFilter[If[#[[2, 2]] == 1 && Total[#, 2] == 2, 1, 0] &, ImageData[pic], {1, 1}]
ppos3 = PixelValuePositions[ep2, White]

{{271, 546}, {190, 471}, {694, 382}, {899, 366}}

ImageData[ep3] == ImageData@ep2

True

... or to use BlockMap along with ArrayPad:

ep4 = Image@
  BlockMap[If[#[[2, 2]] == 1 && Total[#, 2] == 2, 1, 0] &, 
   ArrayPad[ImageData[pic], 1, "Fixed"], {3, 3}, 1]
ppos4 = PixelValuePositions[ep2, White]