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Feb 21 '19 at 15:31 history edited nbro CC BY-SA 4.0
added 33 characters in body
Mar 24 '18 at 10:01 comment added M.kazem Akhgary I think the answers are same. in my answer I can assume number of weights w = ij + jk + kl. basically sum of n * n_i between layers as you noted.
Mar 20 '18 at 13:50 comment added maaartinus @DuttaA No. There's a constant amount of work per weight, which gets repeated e times for each of m examples. I didn't bother to compute the number of weights, I guess, that's the difference.
Mar 20 '18 at 11:00 comment added user9947 From the above answer don't you think itdepends on more factors?
Mar 20 '18 at 6:16 history answered maaartinus CC BY-SA 3.0