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Without more details about the nature of the dataset, it is impossible to know for sure. However, here are a few likely causes: You were calling predict on training data, not testing data. The network will be a lot more sure about images that it trained on than on images it has never seen before. Your model overfit the data. This can happen when you use an ...

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This is not necessarily the only way to do this but it would be the approach I'd take. Assuming your agents position is a vector in $\mathbb{R}^d$, then I would have the network take as input this position vector and pass it through a fully connected layer. I would also take as input the matrix and pass it through a convolutional layer(s) and flatten the ...

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in this tutorial, it taught us to intentionally provide false labels to "fool" the discriminator, does it make discriminator actually inaccurate? When training GANs, the training steps for the generator and discriminator are separate: There is a training stage for the discriminator, where it is presented with a mix of generated and real data, all ...

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Not sure I have understood well your second question but I am gonna try to see if I can help you. Question 1 Yes, the authors meant just that. I see were the confusion might come. So: The authors say: "[...] yields a field of 512-dimensional descriptor vectors" You say: "[...] i get back 512 individual image-patches of size 14x14" You ...

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Hi for me this worked perfectly. I encoded with conv2d and dense and then I flatten I and reshape in the decoder after the dense layer so the encoder and decoder are symmetrical. The only difference is that in my case I use images (224,224,1) # create encoder # 28,28 -> 1st conv2d (filter 3x3,relu activation, padding, strides == 'jumps') ...

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Don't know if you have this doubt anymore, but this would be helpful for those who are facing similar problems- You will need to find the correct weights with which you add these two loses by hyperparameter search. That is, find the best $\lambda$ for the loss- $$L = Loss_1 + \lambda(Loss_2)$$ Here $Loss_1$ and $Loss_2$ can be any losses. Here, we take ...

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