Timeline for What are deconvolutional layers?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 1, 2019 at 0:10 | comment | added | X.X | @andriys thanks for the illustration, very informative. I was initially thinking that the upsampled input image (with interleaving rows and columns of 0) is literally "convolved" with the filter kernel. But the real operation is more like what you said: the kernel is multiplied by the input elements then "tiled" into the output with potential overlap (if kernel size is bigger than stride). I think the underlying math and gradient computing would be also different from the "conventional" Conv2D operation. | |
| Nov 1, 2019 at 0:05 | comment | added | X.X | @JamesBond I think this is what the padding parameter in the Conv2DTransposed() function in the tensorflow.keras controls. Sometime it is desirable to have the output strictly multiple (double, triple, etc) of the original size. In this case, from the input size of 2x2 (wxh) to a 4x4. | |
| Jun 25, 2018 at 13:29 | comment | added | James Bond | @andriys In the image that you've shown, why is the final result cropped? | |
| Nov 19, 2017 at 9:49 | comment | added | andriys | Sorry, @Alex, but I fail to understand why intermediate output is 7. Can you please elaborate? | |
| Nov 19, 2017 at 9:45 | history | edited | andriys | CC BY-SA 3.0 | added code sample |
| Nov 14, 2017 at 14:59 | comment | added | Alex | I think your calculation is wrong here. The intermediate output should be 3+ 2*2=7, then for a 3x3 kernel the final output should be 7-3+1 = 5x5 | |
| Jul 4, 2017 at 22:12 | review | First posts | |||
| Jul 5, 2017 at 2:22 | |||||
| Jul 4, 2017 at 22:09 | history | answered | andriys | CC BY-SA 3.0 |
