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What makes a convolution kernel separable? How would I be able to tell what those separable parts were in order to do two 1D convolutions instead of a 2D convolution>

Thanks

Paul R
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Derek
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2 Answers2

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If the 2D filter kernel has a rank of 1 then it is separable. You can test this in e.g. Matlab or Octave:

octave-3.2.3:1>     sobel = [-1 0 1 ; -2 0 2 ; -1 0 1];
octave-3.2.3:2>     rank(sobel)
ans =  1
octave-3.2.3:3>

See also: http://blogs.mathworks.com/steve/2006/11/28/separable-convolution-part-2/ - this covers using SVD (Singular Value Decomposition) to extract the two 1D kernels from a separable 2D kernel.

See also this question on DSP.stackexchange.com: Fast/efficient way to decompose separable integer 2D filter coefficients

Community
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Paul R
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    SVD is the way to go here. Separable (ie. rank 1) kernels are very specific, and SVD allows you to approximate your kernel by a (small) sum of separable ones. – Alexandre C. May 04 '11 at 21:19
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you can also split the matrix into symmetric and skew parts and separate each part, which can be effective for larger 2d convolutions.

Philip Oakley
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  • can you give example to clarify? – mrgloom Feb 11 '15 at 15:52
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    If you can arrange your 2d matrix as the vector product `x.y' +u.v'` etc you can do a set of 1d convolutions of the rows and columns in stead of the 2d convolution, needing only 4N multiply/adds rather than N^2. If the `u.v'` has a smaller size then the reduction is greater. This usually assumes you have prior knowledge of the matrix's structure to ease the separation. It will also depend on your compute engine - a GPU may favour another structure. – Philip Oakley Feb 11 '15 at 23:36