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I am trying to implement CNN using python Numpy.
I searched so much, but all I found was for one filter with one channel for Convolution.

Suppose we have an X as Image with this shape: (N_Height, N_Width, N_Channel) = (5,5,3)

And Let's say I have 16 filters with this shape: (F_Height, F_Width, N_Channel) = (3,3,3) , stride=1 and padding=0

Forward:

Output shape after conv2d will be

(
math.floor((N_Height - F_Height + 2*padding)/stride + 1 )),
math.floor((N_Width- F_Width + 2*padding)/stride + 1 )),
filter_count
)

So the output of this layer will be an array with this shape: (Height, Width, Channel) = (3, 3, 16)

BackPropagation:

Suppose $dL/dh$ is the input for my layer in backpropagation with this shape: (3,3,16)

Now I must find $dL/dw$ and $dL/dx$: $dL/dw$ to update my filters params and $dL/dx$ to pass it as input to the previous layer as Loss respect to the input X.

From this answer Error respect to filters weights I found how to calculate $dL/dw$.

The problem I have in BackPropagation is I don't know how to calculate $dL/dx$ having this shape:(5,5,3) and pass it to the prev layer.

I read lots of articles in Medium and other sites but I don't get how to calculate it:

How Backpropagation works in a CNN
The best explanation of Convolutional Neural Networks on the Internet!
Backpropagation In Convolutional Neural Networks
How to propagate error back to previous layer in CNN?

Thanks in advance :)

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    $\begingroup$ Hi and welcome to AI SE! Please, format your symbols of the derivates with MathJax, i.e. wrap them with two $ symbols. $\endgroup$ – nbro May 16 at 13:11
  • $\begingroup$ Thanks, I did . $\endgroup$ – A.Najafi May 16 at 13:26
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While this may not be the answer you were looking for, I hope this explanation will help you to understand applying backpropagation to a CNN. Fundamentally, convolutional layers are no different than dense layers, however there are restrictions. The key one is weight-sharing which allows a CNN to be much more efficient than a regular dense layer (as well as it being sparse due to locality). Imagine we are transforming a 4x4 image into a 2x2 image. Since we are inputting a 16-vector, and outputting a 4-vector, we need a weights matrix of 4x16:

enter image description here

This has 64 parameters. In a convolutional layer, we can accomplish this by convolving a 3x3 kernel over the image:

$$ K= \begin{bmatrix} k_{1,1} & k_{1,2} & k_{1,3} \\ k_{2,1} & k_{2,2} & k_{2,3} \\ k_{3,1} & k_{3,2} & k_{3,3} \end{bmatrix} $$

This convolution is equivalent to multiplying by the weights matrix:

enter image description here

As you can see, this only requires 9 parameters and backpropagation can be applied to update these parameters.

Image Source: https://towardsdatascience.com/intuitively-understanding-convolutions-for-deep-learning-1f6f42faee1

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  • $\begingroup$ thanks but as you said it is not my question's answer. I finally find $dL/dX$ for each channel in my example 16 then I add all of them and I get my $dL/dX$ $\endgroup$ – A.Najafi May 27 at 8:15

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