about the loss

[JoyHuYY1412]

Thank you for your work. I am not very familiar with MATLAB coding, but according to your paper, I think the loss you used for instance segmentation is:
the margin loss after every step of mean-shift from 0 to T, and each time step t you use the updated data to get the loss lt
Also, since the mean-shift kernel is set (according to your paper), all you train is the embedding from pixel to the 64-dimension feature. Am I right?

[aimerykong]

Yes. You can think in this way. The mean-shift behaves more like a loss rather than parametric module. But here are two things to notice -- 1. In the toy dataset (mnist), I set T as large as 5, but in practice, I use T=1 because the GPU memory is not enough to hold multiple copies of the transformed data. Of course this can be improved in different ways but I didn't push this in this paper. 2. My remark (1) also indicates a different thing from what you say about the mean-shift kernel. Actually, the kernel is computed in each loop based on the new data, not the original 64-dim raw features. Hope this helps. Thanks for your interest.

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On Sun, Dec 2, 2018 at 1:46 AM JoyHuYY1412 ***@***.***> wrote: Thank you for your work. I am not very familiar with MATLAB coding, but according to your paper, I think the loss you used for instance segmentation is: the margin loss after every step of mean-shift from 0 to T, and each time step t you use the updated data to get the loss lt Also, since the mean-shift kernel is set (according to your paper), all you train is the embedding from pixel to the 64-dimension feature. Am I right? — You are receiving this because you are subscribed to this thread. Reply to this email directly, view it on GitHub <#5>, or mute the thread <https://github.com/notifications/unsubscribe-auth/AGKZJKrLwOUxsioWnfPAg8nBR7fsmsUXks5u06FegaJpZM4Y9Uwp> .

[JoyHuYY1412]

Yes, thank you for your help. I meant that the kernel's form is fixed since you choose a fixed bandwidth, while the data is updated each mean-shift step.