rank_hard_loss_layer.cpp

[Robert0812]

How much amount of contribution does tloss2 have in the forward process? I notice that you only have the term D(x, x-) rather than D(x+, x-) in your ICCV paper.

[kli-casia]

I have the same problem

[xiaolonw]

Just see it as an easy way to increase training samples.

[kli-casia]

@xiaolonw that's make sense, thank you xiaolong

[icodingc]

@xiaolonw
rank_hard_loss.cpp第228行。 如果loss = max(0,||x - x1||^2 - ||x-x2||^2 + margin)的话， 我的理解是梯度应该这样计算： dloss/dx = 2(x2-x1) dloss/dx1 = -2(x-x1) dloss/dx2 = 2(x-x2) 但是代码中跟我理解的不一样，请问是我理解错么，还请解释一下。

[xiaolonw]

@icodingc

that is because we are using cosine distance, which means after normalization layer, ||x|| = 1, ||x1|| = 1. Thus loss = max(0, (2-2x_x1) - (2-2x_x2) + margin )

[icodingc]

thank you xiaolong.