# The receptive field of a stack of dilated convolution layers with exponentially increasing dilation

In the paper that describes the multi-scale context aggregation by dilated convolutions, the authors state that their proposed architecture is motivated by the fact that dilated convolutions support exponentially expanding receptive fields without losing resolution or coverage, and use an example to illustrate the same:

Let $$F_0,F_1,..., F_{n−1} : \mathbb{Z}^2 → \mathbb{R}$$ be discrete functions and let $$k_0, k_1 , ..., k_{n−2} : Ω_1 → \mathbb{R}$$ be discrete $$3×3$$ filters. Consider applying the filters with exponentially increasing dilation : $$F_{i+1} = F_i ∗_{2^i} k_i$$ for $$i = 0,1,...,n−2$$.

Define the receptive field of an element $$\textbf{p}$$ in $$F_{i+1}$$ as the set of elements in $$F_0$$ that modify the value of $$F_{i+1}(\textbf{p})$$. Let the size of the receptive field of $$\textbf{p}$$ in $$F_{i+1}$$ be the number of these elements. It is easy to see that the size of the receptive field of each element in $$F_{i+1}$$ is $$(2^{i+2}−1)×(2^{i+2}−1)$$. The receptive field is a square of exponentially increasing size. It would be of great help if I could get a derivation of the last formula $$(2^{i+2}−1)×(2^{i+2}−1)$$ !

I read here that when using dilated convolutions, the filter size, initially $$k$$ would be expanded to $$k + (k-1)(r-1)$$, $$r$$ being the dilation factor.

I also read in this answer that the layer wise receptive field size could be calculated using the formula $$s_{l_{i+1}} = s_{l_i} + (kernelsize−1) * dilationfactor$$, where $$l_i$$ denotes the different network layers and $$s$$ the receptive field size, $$s_{l_0} = 1$$. I don't know how this formula was derived either.

I can't seem to figure out how these formulas can be used to calculate the resulting generalized receptive field size in an inductive manner.

I mean, applying the formula $$s_{l_{i+1}} = s_{l_i} + (kernelsize−1) * dilationfactor$$, the receptive field size will be $$s_{l_{i+1}} = s_{l_{i}} + 2 * 2^i$$ since $$2^i$$ will be the dilation factor when calculating the receptive field size for layer $$i+1$$, and $$kernelsize - 1$$ would be $$3 - 1 = 2$$.

So that would be the same as $$s_{l_{i}} + 2^{i+1}$$

$$\implies s_{l_{i-1}} + (s^{i}) + 2^{i+1}$$

$$\implies s_{l_{i-1}} + (2^i) + 2^{i+1}$$ using the same formula for layer $$i$$.

Expanding backwards this way, we get :

$$\implies 1 + 2^1 + 2^2 + ... + 2^{i+1}$$

$$\implies 2^{(i+1) + 1} - 1$$ using the formula to calculate sum of powers of 2.

$$\implies 2^{i+2} - 1$$

OH MY GOD!

I was able to derive it correctly, but only because I had the formula $$s_{l_{i+1}} = s_{l_i} + (kernelsize−1) * dilationfactor$$ at hand.
I would appreciate it if somebody could explain the logic behind the formula $$s_{l_{i+1}} = s_{l_i} + (kernelsize−1) * dilationfactor$$. 