Other correct solutions were sent in by Jennifer Hatch, David Aberle, Ray Kremer, Shawn Lord. Further solutions were submitted by Burkart Venzke, Robert T. McQuaid, Jim Benstead, Jens Voss, Al Zimmermann, Dane Brook, Jan Siwanowicz, Michael Lynch, Al Langen, Denis Borris, Tim Kelley, Philippe Fondanaiche, and a partial solution by Lucas Thiessen.

Because the neighborhoods of both a and b are odd, their difference, c + g, is even.  Symmetry requires all such "dominoes" to be even.  The neighborhood of f  is the sum of the neighborhood of a, two "dominoes", and the square labeled k.  The neighborhood of a is odd, and the dominoes are even; as the neighborhood of f  must be odd, k must then be even, so k = 0.  By symmetry all four center squares are zero.  Then each of the "dominoes" must be all zero, too.  That leaves the corner squares, which must be ones.

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