 Convolution dilation

Convolution dilation specifies the spacing between neighboring kernel sample points in a convolutional layer. Dilation one samples adjacent positions; larger dilation separates those positions while keeping the filter's learned coefficient count fixed. Convolution stride instead controls the spacing between successive filter origins.^[ref]

For an own real one-dimensional example, takeand coefficients $(1,1,1)$, with stride one, no padding and no bias. Use the unflipped cross-correlation convention. Dilation one sums three adjacent values and gives $(7,14,28,56,112)$. Dilation two instead uses offsets $0,2,4$:The second filter evaluates a wider input span while still using three coefficients.

For $k$ sampled coefficients separated by $d$ positions, the first-to-last displacement is $d(k-1)$. Including the first position gives the effective convolution kernel sizeThus $k=3,d=2$ spans five positions. At each origin only three of those five positions are sampled directly.

The same linear operation can be represented by a dense width-five kernel $(1,0,1,0,1)$. The inserted zeros are fixed structure, not additional trainable coefficients. Dilation changes the footprint without making the original three-parameter filter a general five-parameter one.^[ref]

For input length $L$, symmetric zero padding $p$ and stride $s$, the usual per-axis size formula becomeswhen the padded input accommodates the footprint. The example gives $\lfloor (7-4-1)/1\rfloor +1=3$ outputs. The floor function counts origins whose final sample remains in range.^[ref]

Dilation can enlarge a unit's spatial neural network receptive field without evaluating every intervening input position. Whether later layers fill those gaps depends on the architecture. A wider span alone does not establish greater numerical influence at every position or better performance on a learning task.