0
$\begingroup$

I have seen people using pooling and subsampling synonymously. I have also seen people use them as different processes. I am not sure though if I have correctly inferred what they mean, when they use the terms with distinct meanings. I think these people mean that the pooling part is selecting a submatrix from an input matrix and the subsampling part is selecting yet another submatrix that satisfies some condition from the first submatrix.

So say we have a $100 \times 100$ image. We do $10 \times 10$ non-overlapping pooling with $5 \times 5$ max subsampling. That would mean we slide the $10 \times 10$ "pool" across the image in strides of 10 and at every step we select the $5 \times 5$ submatrix inside the pool that has the maximum sum. That $5 \times 5$ matrix is what comes out of the $10 \times 10$ pool at the current potion. So in the end we have a $50 \times 50$ image.

Can you confirm that this usage of the terms pooling and subsampling exists?

I inferred this definition, as I cannot make sense of how some people use the two terms otherwise. For example in this video, or rather the people from the paper he is talking about (which I can't find because he only has the author name on his slide and no year).

$\endgroup$
  • 1
    $\begingroup$ I think this is asking the same question: stats.stackexchange.com/questions/354944/… $\endgroup$ – Recessive Jul 29 '19 at 2:03
  • $\begingroup$ @Recessive I had read that one. It does not cover the case where people use the terms not as synonyms. It does not explain what a "$10 \times 10$ pool $5 \times 5$ subsampling" is supposed to be. $\endgroup$ – lo tolmencre Jul 29 '19 at 12:12
  • $\begingroup$ From what I know, a maxpooling layer is a method for subsampling. So in that case, the terms are somewhat interchangable, but other types of pooling layers may also be a form of subsampling. $\endgroup$ – Recessive Jul 30 '19 at 4:09
  • $\begingroup$ @Recessive The person from the video answered me now, and he says in this case by subsampling the authors meant the stride of the pooling. $\endgroup$ – lo tolmencre Jul 30 '19 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.