Why is the robust loss function \rho applied to the residual's squared norm instead of its norm?

[Chi Zhang]

Hi,

I have some confusion about the loss function \rho in the overall objective function (Equation 1) that ceres is optimizing. ( http://ceres-solver.org/nnls_tutorial.html#introduction )  My question is:

• Shouldn't the robust loss function \rho_i be applied to the the norm of the i-th residual, instead of to its squared norm?
  

For example, if I choose \rho to be a Huber loss function in a bundle adjustment problem. And there was an outlier of wrong correspondences which causes a reprojection_error=200 pixel. Before the error is passed to the Huber loss function, it would have been magnified to 200*200, which will surely bias the whole optimization. Even with the Huber loss function applied, the error would still be 200*200 - \epsilon, which doesn't seem to be to robust. 

I'm rather confused how the robust loss functions work in ceres. Am I getting the whole idea wrong?

Best,

Chi

[Sameer Agarwal]

Chi,

You can apply it to either, its just a matter of how you code the loss function. We assume that we apply it to the squared norm, therefore the identity/null loss function when applied to the cost function gives you the squared norm.

And the SoftL1 is equivalent to (except for a small region) taking the squareroot of the squared norm.

Sameer

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[Chi Zhang]

Hi Sameer,

Ah I see. Should have had checked the API references more carefully;-) Thank you very much!

Chi