Body: <p>On page 27 of the <a href="http://airesearch.com/wp-content/uploads/2016/01/deepmind-mastering-go.pdf" rel="nofollow noreferrer">DeepMind AlphaGo paper</a> appears the following sentence:</p>
  <blockquote>
  <p>The first hidden layer zero pads the input into a <span class="math-container">$23 \times 23$</span> image, then convolves <span class="math-container">$k$</span> filters of kernel size <span class="math-container">$5 \times 5$</span> with stride <span class="math-container">$1$</span> with the input image and applies a rectifier nonlinearity.</p>
  </blockquote>
  <p>What does &quot;convolves <span class="math-container">$k$</span> filters&quot; mean here?</p>
  <p>Does it mean the following:</p>
  <blockquote>
  <p>The first hidden layer is a convolutional layer with <span class="math-container">$k$</span> groups of <span class="math-container">$(19 \times 19)$</span> neurons, where there is a kernel of <span class="math-container">$(5 \times 5 \times numChannels + 1)$</span> parameters (input weights plus a bias term) used by all the neurons of each group. <span class="math-container">$numChannels$</span> is 48 (the number of feature planes in the input image stack).</p>
  <p>All <span class="math-container">$(19 \times 19 \times k)$</span> neurons' outputs are available to the second hidden layer (which happens to be another convolutional layer, but could in principle be fully connected).</p>
  </blockquote>
  <p>?</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-08-21T11:11:19.233Z
Comments.Text: I believe it means they perform convolution with $k$ filters in the given layer (this is if I'm not getting my terminology confused)
Comments.UserDisplayName: David
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-08-21T11:35:57.153Z
Comments.Text: 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.
Comments.UserDisplayName: nbro
Tags: <convolutional-neural-networks><convolution><alphago><filters><convolutional-layers>
Title: What does "convolve k filters" mean in the AlphaGo paper?
UserDisplayName: William Ehlhardt
ThreadText: # What does "convolve k filters" mean in the AlphaGo paper?
  
  (Asked by: username_0 on 2020-08-19)
  
  On page 27 of the [DeepMind AlphaGo paper](http://airesearch.com/wp-content/uploads/2016/01/deepmind-mastering-go.pdf) 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).
  
  ?
  
  **Tags:** convolutional-neural-networks, convolution, alphago, filters, convolutional-layers
  
  > **Comments:**
  > **username_1**: I believe it means they perform convolution with $k$ filters in the given layer (this is if I'm not getting my terminology confused)  
  > **username_2**: 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.  
  
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  ## 0 answers
  
  