What would be the convolutional layer output by keras.layers.Conv2D when conv output is fractional?

Question

I have input ($n=224$), strides ($s=4$), filter size ($k=11$) and no padding which gives me a fractional conv output:

$$\texttt{conv output} = (n-k+2p)/s + 1 = 54.25$$

My question is how keras.layers.Conv2D is generating conv output when the above formula results in a fraction?

For the above parameters keras.layers.Conv2D is giving conv output as $54$. Why is it taking the floor of the fractional conv output? I read that when it is in fraction we need to adjust $n$, $s$, $k$ to make that output an integer.

Answer 1

In a given dimension, for a given input size ($n$), filter size ($k$), stride ($s$) and padding ($p$), the formula for the number of convolutional windows is

$$\lfloor (n - k + p + s)/s\rfloor,$$

which in your case is

$$\lfloor (224 - 11 + 0 + 4)/4\rfloor = \lfloor 54.25\rfloor = 54.$$

The question is why the floor is taken (as opposed to the ceiling for example). The reason is that the stride of $s = 4$ forces a distance of 4 pixels between each convolutional window. Without padding ($p = 0$), there is no space left for a 55th convolutional window at the right-most part while maintaining a distance of 4 pixels.

This is illustrated well here,$^\dagger$ in a 2D convolutional example where (horizontally) $n = 5$, $k = 2$, $s = 2$, $p = 0$:

Here the lack of padding means only two steps can be taken in the horizontal dimension.

Filling in the formula gives the same answer:

$$\lfloor (5 - 2 + 0 + 2)/2\rfloor = \lfloor 2.5\rfloor = 2.$$

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$^\dagger$: Zhang, A., Lipton, Z., Li, M., & Smola, A. (2023). Dive into Deep Learning. Cambridge University Press. https://D2L.ai