two problem about the paper and source code

[yanyiss]

hello, i've read your paper: QuadriFlow: A Scalable and Robust Method fo Quadrangulation. but i got puzzled with this two problems:


so how i can get (7) by splitting? i thought (7) should be the 1-norm of difference between the two variables, but why it's the sum?
Another question is that in your source code, you seemed to give up transferring (7) to MCF but select the 1-norm of d as your optimization objective restricting a small change with d-star. so i want to know if i took my wrong way or it's truth.
glad to your answer.

[jrsjrsjrsjrs]

so how i can get (7) by splitting? i thought (7) should be the 1-norm of difference between the two variables, but why it's the sum?

Hi Yanyiss, ∂d_e^+ and ∂d_e^- are not the same as d_e and d*_e. Rather: if d_e - d_e^* is positive, the we set ∂d_e^+ = d_e - d_e^* and ∂d_e^- = 0. Whereas if d_e - d_e^* is negative, the we set ∂d_e^+ = 0 and ∂d_e^- = d_e^* - d_e. That way, we obtain the property that ∂d_e^+ + ∂d_e^- = ||d_e - d*_e||_1, while also ensuring that ∂d_e^+ and ∂d_e^- are always nonnegative. I hope this helps, Jonathan

[yanyiss]

so how i can get (7) by splitting? i thought (7) should be the 1-norm of difference between the two variables, but why it's the sum?
Hi Yanyiss, ∂d_e^+ and ∂d_e^- are not the same as d_e and d*_e. Rather: if d_e - d_e^* is positive, the we set ∂d_e^+ = d_e - d_e^* and ∂d_e^- = 0. Whereas if d_e - d_e^* is negative, the we set ∂d_e^+ = 0 and ∂d_e^- = d_e^* - d_e. That way, we obtain the property that ∂d_e^+ + ∂d_e^- = ||d_e - d*_e||_1, while also ensuring that ∂d_e^+ and ∂d_e^- are always nonnegative. I hope this helps, Jonathan

Thanks a lot and it does help me. By the way, I found I got mistakes about the second question while reading the codes. So I've understood them. Thanks again for your time.

[jrsjrsjrsjrs]

Thanks a lot and it does help me.

Great!

By the way, I found I got mistakes about the second question while reading the codes. So I've understood them.

I'm glad to hear that, because I didn't work on the codes and I can't help with that.

Thanks again for your time.

My pleasure. Jonathan