The correction says that p=0 means valid conv while when applying the formula Andrew mentioned, it doesn’t make a difference to set p=0 or not. I mean p=0 still results in “same convolution”
From my understanding, the formula: (n + 2p - f / s) + 1 == (39 + 2*0 - 3 / 1) + 1 == (39 + 0 - 3 / 1) + 1 == (39 - 3 / 1) + 1 == (36 / 1) + 1 == 37.
So, how p=0 affects padding?
For same padding,
\begin{aligned}
(n + 2p -f) / s + 1 &= n \Leftrightarrow \\
n + 2p - f &= (n - 1) * s \Leftrightarrow \\
p &= ((n - 1) * s + f - n) / 2.
\end{aligned}
With your example:
\begin{aligned}
p &= ((39 - 1) * 1 + 3 - 39) / 2 \Rightarrow \\
p &= (38 + 3 - 39) /2 \Rightarrow \\
p &= 1.
\end{aligned}
Thus, if the padding is 1, you will have the same output size as the input size, hence it is called same convolution.
To double check
(n + 2p -f) / s + 1 = (39 + 2 * 1 - 3) / 1 + 1 = 39,
which is equal to the input size.
Valid convolution is no padding, hence p = 0.
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