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

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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).

?

William Ehlhardt

Posted 2020-08-19T04:28:46.577

Reputation: 31

I believe it means they perform convolution with $k$ filters in the given layer (this is if I'm not getting my terminology confused) – David Ireland – 2020-08-21T11:11:19.233

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 – 2020-08-21T11:35:57.153

No answers