What does "convolve k filters" mean in the AlphaGo paper?

On page 27 of the DeepMind AlphaGo paper appears the following sentence:

The first hidden layer zero pads the input into a $$23 \times 23$$ image, then convolves $$k$$ filters of kernel size $$5 \times 5$$ with stride $$1$$ with the input image and applies a rectifier nonlinearity.

What does "convolves $$k$$ filters" mean here?

Does it mean the following:

The first hidden layer is a convolutional layer with $$k$$ groups of $$(19 \times 19)$$ neurons, where there is a kernel of $$(5 \times 5 \times numChannels + 1)$$ parameters (input weights plus a bias term) used by all the neurons of each group. $$numChannels$$ is 48 (the number of feature planes in the input image stack).

All $$(19 \times 19 \times k)$$ neurons' outputs are available to the second hidden layer (which happens to be another convolutional layer, but could in principle be fully connected).

?

• I believe it means they perform convolution with $k$ filters in the given layer (this is if I'm not getting my terminology confused) Aug 21, 2020 at 11:11
• Regarding your first question, I agree with @DavidIreland. I am pretty sure that they convolve $k$ different filters with the same $23 \times 23$ image, where each of the filters has the same depth as the input images, provided you're performing a 2D convolution. The $k$ is a hyper-parameter there. I am not so sure about the second paragraph. I would need to think about it a little more.
– nbro
Aug 21, 2020 at 11:35