Timeline for How does a max-pooled layer connect to the second Convolution layer?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 10, 2017 at 23:35 | vote | accept | moondra | ||
| Jun 10, 2017 at 23:34 | comment | added | moondra | Thanks, I was missing the summed into one value part.. So they are essentially taking the dot product it seems. I understand now. I actually was able to find explanation here (just now) 'xrds.acm.org/blog/2016/06/…' Thanks for you patience. I asked this question on quora as well, and everyone was giving me vague answers. They could have just said they are taking the dot product or what you said "What you might be overlooking is that they are summed into one value." | |
| Jun 10, 2017 at 23:20 | comment | added | Frobot | And I know it is confusing. I asked nearly the same question on here almost 2 years ago. There isn't much out there that explains the finer details | |
| Jun 10, 2017 at 23:18 | comment | added | Frobot | Each filter in the first layer will multiply with all three channels in the input image. What you might be overlooking is that they are summed into one value, and that is the value you put for that pixel in the resulting activation map. Think of both your input and your filter as a 3D volume. Say the filter is of dimension (w, h, d). You multiply this with a chunk of your input of the same dimension (w, h, d) and get a new volume of values (w, h, d). now you sum across all whd values to get one single value, put that through your activation func, and this is your activation value for yourmap | |
| Jun 10, 2017 at 22:55 | comment | added | moondra | So each filter will only multiply with one of the RGB planes? If we are in the first layer, Having 6 filters would result in 2 filters x R plane, the other two filters x G plane, 2x B plane? Am I getting this right? This is really confusing. I | |
| Jun 10, 2017 at 22:31 | comment | added | Frobot | Actually there is a paper where they had filters with a depth that was less than the number of activation maps at the given layer and they chose certain maps to be convolved with certain filters. This is by no means the standard though | |
| Jun 10, 2017 at 22:26 | comment | added | Frobot | The filters are 3 dimensional, but the activation maps produced by them are not. Each individual map is 2D. After these 2D maps are produced you can think of them all stacked on top of each other as a 3D volume, and then your 3D filters in the next layer convolve with your entire set of maps, and each filter produces one new 2D activation map. It wouldn't make any sense for your filters to all have a depth of 3 (except the first layer) because then either your activation maps would have to be 3D or you would have to assign certain filters to certain maps. neither is the case though | |
| Jun 10, 2017 at 17:28 | comment | added | moondra | I asked on quora as well and this is the response I got from someone who had a PHD in computer vision: 'Its because each convolutional filter is also three dimensional. In this case every filter is of size 32x32x3." I asked him to expand on this, but I'm assuming that each of the 12 activation maps end of having a depth of 3? | |
| Jun 10, 2017 at 3:11 | comment | added | Frobot | I added some more information to my answer. I know most diagrams you can find don't really explain this part so well. Hope that helps | |
| Jun 10, 2017 at 3:11 | history | edited | Frobot | CC BY-SA 3.0 |
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| Jun 10, 2017 at 2:15 | comment | added | moondra | I'm confused. Wouldn't the depth of each filter be only 3 dimensions? I'm more interested in how many activation maps the conv2 would create if I used 8 filters. Thanks. | |
| Jun 9, 2017 at 23:38 | history | answered | Frobot | CC BY-SA 3.0 |