I read that to compute the derivative of the error with respect to the input of a convolution layer is the same to make of a convolution between deltas of the next layer and the weight matrix rotated by $180°$, i.e. something like
$$\delta^l_{ij}=\delta^{l+1}_{ij} * rot180(W^{l+1})f'(x^l_{ij})$$
with $*$ convolution operator. This is valid with $stride=1$.
However, what happens when stride is greater than $1$? Is it still a convolution with a kernel rotation or I can't make this simplification?
From the paper found from the post linked below:
'We find that max-pooling can simply be replaced by a convolutional layer with increased stride without loss in accuracy on several image recognition benchmarks'
All that means that only values are skipped (=pooling is made) to the matrix, otherwise all works like a convolution should do.
Sources:
https://arxiv.org/pdf/1412.6806.pdf
https://stackoverflow.com/questions/44666390/max-pool-layer-vs-convolution-with-stride-performance
I have just the same problem, and I was trying to derive the backpropagation for the convolutional layer with stride, but it doesn't work.
When you do the striding in the forward propagation, you chose the elements next to each other to convolve with the kernel, then take a step $>1$. This results in the fact that in the backpropagation, in the reverse operation, the delta matrix elements will be multiplied by the kernel elements, (with the rotation) but not as strided, but you are picking elements that are not next to each other, something like $DY_{11} * K_{11} + DY_{13} * K_{12} + DY_{31} * K_{21} + DY_{33} * K_{22}$, which is NOT the equivalent as a convolution with a stride $>1$.
So as far as I am concerned, if I would like to implement the ConvNet by myself to get a better grasp of the concept, I have to implement a different method for the backprop, if I allow strides.
