Comments (6)
- cosine similarity is used to filter points that close to each other in Euclidean space but actually not close on surface. Assume one point on left leg and the other on right left -- they can be close , but their normal will be opposite.
- it is just parameter you can tune. It depends on your sampling density.
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Hi! Thanks for your answer. I still have some doubts.
1)
a)
cosine similarity is used to filter points that close to each other in Euclidean space but actually not close on surface. Assume one point on left leg and the other on right left -- they can be close , but their normal will be opposite.
Then, what you do is compute the cosine similarity between the normal vector of the current point and the normals of each neighbour, isn't it?
b)
About that I have a doubt here:
nn_p = nn_p[cos_similar > -0.5]
Why do you just take the values larger than -0.5?
it is just parameter you can tune. It depends on your sampling density.
Do you have any idea of the factor you will use if you increment the sampling density? Imagine you increase the number of points to 6k. Which value should be used?
Thank you so much for your attention!
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1 yes I am computing the cosine similarity. -0.5 (meaning the angling between normals are smaller than 120 degree) is also a parameter that you can tune. This is based on our previous experiments.
2. i would say the best way to determine this is by visualization. Check whether non-geodesic neighbors are picked by a radius. Note there won't be a perfect value -- even 1, when legs overlap, incorrect neighbors will still be picked.
If you increase to 6K, you can still use 5. That means you are being conservative -- only trust 5 closest ones. Dijkstra will take care of long-range distance.
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About the 5 closest vertices. I'm assuming that you are doing that because you are trying to create the initial paths for the Dijkstra algorithm, is that right? I assume that all that process is to make the algorithm more efficient, isn't it? Because you could use the dijkstra algorithm directly to the mesh.
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yes
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Thanks!
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