How does num_deformable_groups work?

[victorhcm]

First off, thank you for your amazing port, @1zb!

How does num_deformable_groups work? I was wondering if it is splitting the offsets, thus each group of input channels will have a given set of offsets (which may be different from other groups). For instance, if we have an input of 6 channels (1, 6, H, W), and num_deformable_groups=2, then we would have our input split between two tensors of (1, 3, H, W) and (1, 3, H, W), each with its respective instance of offset mask (1, 3, kH, kW). Is that correct?

[1zb]

Yes. You are right.

inC must be divisible by num_deformable_groups. Then each group has their own offset.

[victorhcm]

Thank you for your attention! 😊

[victorhcm]

Sorry for bothering you again with this, @1zb, I have a follow up question. In the mxnet repository, the authors provide two options, num_group and num_deformable_group. According to their definition, num_group splits the input data, while num_deformable_group splits the output tensor and applies the i-th offset group to the i-th output group.

Thus, here in this repository, is num_deformable_group actually equivalent to num_group in the original implementation, as it is splitting the input data instead of output?

[1zb]

Thanks for your response.

According to these lines and these lines, the parameter num_deformable_group splits the input tensor along channel axis, which contradicts the definition. I will take a look at the code again.

The parameter num_group or groups exists in vanilla CNN. You might find useful discussions here.

[victorhcm]

I think you're right, it does seem to be splitting the input tensor. I should check their model to see how they are being used, to make sure.

Thank you for the num_group links, by the way!

[victorhcm]

I was checking their code, it seems they really got the definitions reversed. In this line, they divide the number of output channels by the number of groups, while here they convolve the i-th output with the i-th weight.