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What does the KL_loss mean? #4

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zlyx525 opened this issue Jan 25, 2018 · 2 comments
Open

What does the KL_loss mean? #4

zlyx525 opened this issue Jan 25, 2018 · 2 comments

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@zlyx525
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zlyx525 commented Jan 25, 2018

Thanks very much for sharing the code and I think this is very great job. But I have some questions in understanding the code.
Firstly, what does the KL_loss mean? According to my understand, the SSE_loss in the code corresponds to the MSE_loss in the paper, but you didn't mentioned the KL_loss in the paper?
Secondly, you did't use the KL_loss and SSE_loss when updating the params, why?
Thirdly, z_x = tf.add(z_x_mean, tf.multiply(tf.sqrt(tf.exp(z_x_log_sigma_sq)), eps)) # grab our actual z, what does "grab actual z "mean?
Finally, What's the role of the function "average_gradients" @htconquer

@zoeyuchao
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Hi, I have the same questions as yours. Have you find the answers?

@ozen
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ozen commented Jan 28, 2019

I came across this during research.

What was implemented is a variational autoencoder (VAE). So the output of the encoder is a probability distribution instead of a latent vector. Learning more about VAEs would let you answer your own questions.

I suggest this blog post: https://lilianweng.github.io/lil-log/2018/08/12/from-autoencoder-to-beta-vae.html

"Loss Function" section explains the KL divergence etc.

"Reparameterization Trick" section explains "grabbing actual z".

I haven't read the paper fully yet, haven't the authors mentioned that this is a VAE?

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