xucao-42 / bilateral_normal_integration Goto Github PK
View Code? Open in Web Editor NEWOfficial implementation of "Bilateral Normal Integration" (BiNI), ECCV 2022.
License: GNU General Public License v3.0
Official implementation of "Bilateral Normal Integration" (BiNI), ECCV 2022.
License: GNU General Public License v3.0
Dear author,
Thx for sharing your great work! Although the method is designed for single view normal integration. I wonder if I can apply it to multiview senario? Especially for the case where input views are sparse (leq 8 and distributed 360). If so, cloud you please give a rough instruction about the usage, assuming we have 8 rgbs, masks, normals?
Hi,
thanks for your great work. I run into the above mentioned cuda error when running the cupy script on the following normal map.
I was wondering whether you can reproduce the issue and have an idea where to start looking. Potentially somewhere where the masks are computed or the diagonal_data_term is updated.
It works with the numpy script, but I want to use the cupy script as it is faster for many of my normal_maps.
Looking forward to hearing from you.
Best,
Jan
Hello,
Thanks for this amazing work, but I am confused about one sentence in the paper. In the Section 2.1, you mention "we omit the dependencies of p and n on u for brevity."
Could you please give me more explanations on "dependencies"? I mean what do dependencies of p and n on u mean here? And, when ommitting dependencies, what are the assumptions?
Thank you for your time and patience.
Zhenshan,
Regards
你好,想请教下,已知z关于u,v的梯度,怎么转化成代码中的nx, ny, nz法线图的形式
我想知道MADEs计算的细节,以及是如何进行计算的
In equation 17 of the paper the depth differences are scaled by the normal vector z component. The paper reads:
Here, the depth differences are scaled by nz to measure the difference along the normal direction at the point.
I wondered why this should make sense, and alas, reading the code i do not find that the scaling with nz is used. In the python numpy script line 290 and 291, the weights are constructed as
wu = sigmoid((A2 @ z) ** 2 - (A1 @ z) ** 2, k)
wv = sigmoid((A4 @ z) ** 2 - (A3 @ z) ** 2, k)
,
ie directly applying the sigmoid to the difference of the squared one-sided derivatives, no scaling by nz.
Do i miss something?
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