How to chose the threshold for LocLoss?

[dangweili]

Usually the [theta_1_1 theta_1_2; theta_2_1 theta_2_2] is mall (less than 1). The parameters [theta_1_3 theta_1_3] will be much bigger (because this is the offset of cropped feature map or image part). So how to set the threshold of LocLoss for those two different types of parameters? Or I misunderstand the implement? Any suggestions? @daerduoCarey

[daerduoCarey]

I think you misunderstand the definition for theta_1_3 and theta_2_3. They should not be too large and should be of the same scale with other thetas. All thetas are defined to map a location from the source image to the target image in [-1, 1] scale, instead of the actual pixel space scale, e.g. [0, 224], [0, 256]. For example, any location in the source image or the target image is defined to be in [-1, 1], which is the relative location within the whole image.

Hope my reply answer your question. Thank you.

[dangweili]

Got it. I misunderstand it. Thanks a lot. @daerduoCarey

[dangweili]

Another question about the details. Consider the feature map is CHW. C is the channel. H, W are the height and width of the feature maps. Are the parameters [theta_1_1 theta_1_2 theta_1_3] correspond with the transform of features' width (denoted as x in original paper)? Or I misunderstand the order of x and y in the original paper? Thanks. @daerduoCarey

[daerduoCarey]

Yes. You're right.

[dangweili]

OK. Thanks a lot. @daerduoCarey

[shaoming20798]

Yes. You're right.

Hi, author。In your code, I find that x_t and y_t are reversed, here is your source code

for(int i=0; i<output_H_ * output_W_; ++i) {
    data[3 * i] = (i / output_W_) * 1.0 / output_H_ * 2 - 1;               // this is y_t, I think, cause it is scale by 'output_H'
    data[3 * i + 1] = (i % output_W_) * 1.0 / output_W_ * 2 - 1;     // x_t
    data[3 * i + 2] = 1;
}

And based on these codes, theta is also reversed, [theta_1_1 theta_1_2 theta_1_3] correspond with [y_t x_t translation]。
Emm.....it looks weird. Is there something I miss? I will appreciate it if you could give me a reply. Thanks!