Problem with CNN architecture for small images of size 6x6

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Muhammad Jaleed Khan 2017-5-22

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评论： Javier Pinzón 2017-10-25

I'm trying to develop a CNN classifier for a large dataset of small images of size 6x6. I have modified the original code of an example (Train a Convolutional Neural Network Using Data in ImageDatastore) from MATLAB Help. I need to increase number of layers to the network more deep to get better results.

Original Code of CNN layers:

layers = [imageInputLayer([28 28 1]);
          convolution2dLayer(5,20);
          reluLayer();
          maxPooling2dLayer(2,'Stride',2);
          fullyConnectedLayer(5);
          softmaxLayer();
          classificationLayer()];

Modified Code of CNN layers:

% Define the convolutional neural network architecture. 
  layers = [imageInputLayer([6 6 1]);
            convolution2dLayer(5,20,'Padding',3);
            reluLayer();
            maxPooling2dLayer(2,'Stride',2);
            convolution2dLayer(5,20);
            reluLayer();
            maxPooling2dLayer(5,'Stride',2);
            fullyConnectedLayer(5);
            softmaxLayer();
            classificationLayer()];

Now I am getting this error:

Error using nnet.cnn.layer.Layer>iInferSize (line 261)
Layer 7 is expected to have a different size.

It is expecting the pooling layer to have different size. I have tried 2x2 and 3x3 sizes of pooling layer but it gives the same errors. Please help me fix this issue so that I can add more number of layers to the CNN.

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Answer 1

Javier Pinzón 2017-6-1

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https://ww2.mathworks.cn/matlabcentral/answers/341312-problem-with-cnn-architecture-for-small-images-of-size-6x6#answer_269219

编辑：Javier Pinzón 2017-6-1

在 MATLAB Online 中打开

Hello Muhammad,

First of all, you have errors when calculating the output volumen of each layer. Lets check:

Convolution 1:

OutV1 = (6 - 5 + 3*2)/1 + 1 = 8

Maxpooling:

OutV2 = 8 / 2 = 4

Convolution 2:

OutV3 = (4 - 5)/1 + 1 = 0

And then... you dont have any output volume from convolution 2 ownwards... so you need to recalculate your filter sizes... I really recommend you to use sizes of 2 or 3 in the convolutions, and also add some padding of the size of the filter - 1 in each convolution layer, to keep a considerably volumen, i.e., if you use a filter of size of 3, use "padding" = 1, so you will have:

Out volume = (In - 3 + 2*1)/1 + 1 = In

Remember:

 Output Volume = [("Input Volumen" - "Filter Size" + 2 * "Padding")/"Stride"] + 1

hope it helps if it is not too late =)

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shubham gupta 2017-10-21

why +1 is added to Output Volume

Javier Pinzón 2017-10-25

+1 is related to the "Bias", each layer has the activation neuron, for that reason there is a +1 in the formula

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