Comments (9)
mkazhdan
commented on September 4, 2024
1. I'm not sure. Though it shouldn't make a difference. Scaling the B-splines by different values does not change the function space they span. From the perspective of iterative solvers, convergence may be different depending on the scale, but only if functions within the same level are scaled by different amounts. (Which is not the case with or without per-level normalization.)
What is important is that when splatting a normal into the tree the integral of the splat is proportional to the estimated area of the sample. (This does require scaling coarser nodes by 1/w^3.)
2. Yes, that's how evaluation is done.
…On March 5, 2018 2:41:41 AM EST, antibes0415 ***@***.***> wrote:
Hi. I have two questions.
1. In Poisson Surface Reconstruction (2006), the value of the base
function is sacled by (1 / o.w^3).
Do you implement this feature in screened Poisson, i.e. is the B-spline
function scaled by (1 / o.w^3) in 3D or (1 / o.w) in 1D whenever it is
called ?
2. Is the implicit function value of an arbitary point computed as the
linear combination of the solution of related nodes ( x(p) =
sum(xiBi(p) ) ? Except for B-spline function, there is no other scaling
factor to adjust the weight of solution value of different depth nodes,
right ?
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mkazhdan
commented on September 4, 2024
More precisely:
When scaling the different levels by different amounts, you need to be careful that the prolongation (and restriction) operators are adjusted accordingly. I believe the later paper uses the standard B-splines prolongation stencil, which means no per-level re-scaling.
…On March 5, 2018 2:41:41 AM EST, antibes0415 ***@***.***> wrote:
Hi. I have two questions.
1. In Poisson Surface Reconstruction (2006), the value of the base
function is sacled by (1 / o.w^3).
Do you implement this feature in screened Poisson, i.e. is the B-spline
function scaled by (1 / o.w^3) in 3D or (1 / o.w) in 1D whenever it is
called ?
2. Is the implicit function value of an arbitary point computed as the
linear combination of the solution of related nodes ( x(p) =
sum(xiBi(p) ) ? Except for B-spline function, there is no other scaling
factor to adjust the weight of solution value of different depth nodes,
right ?
--
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#49
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antibes0415
commented on September 4, 2024
Thanks. I've implemented unscreened Poisson by myself. Maybe due to different ways of implementation, I have to use an extra factor to adjust the weight among different levels ( x(p) = sum(factor * xi * Bi(p) ). Right now I try to implement the screened Poisson, however the adaptive weight of the screen term ( 2^d ) described in your paper just doesn't work for my code. Anyway I will be trying. Thanks.
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antibes0415
commented on September 4, 2024
Hi. Where can I find the standard B-spline prolongation stencil ? Could you give me some help ?
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mkazhdan
commented on September 4, 2024
In 1D, I believe that the prolongation stencil for B-splines of degree d has entries:
{ Choose(d+1,0) , ... , Choose(d+1,d+1) } / 2^d
where Choose(a,b) is "a choose b". That is:
Choose(a,b) = a! / ( (a-b)! * b! )
In dimension k, it's the k-th tensor product of the 1D stencil.
Note that even- (resp. odd-) degree B-splines have an even (resp. odd) number of entries in their stencil. This is because even- (resp. odd-) degree B-splines are centered at cell centers (resp. corners).
…-- Misha
On March 8, 2018 12:11:31 AM EST, antibes0415 ***@***.***> wrote:
Hi. Where can I find the standard B-spline prolongation stencil ? Could
you give me some help ?
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antibes0415
commented on September 4, 2024
Much thanks. I've found this 1D stencil described in your V4 Poisson code comments. But I was not so sure.
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antibes0415
commented on September 4, 2024
Hi, I have another question about the prolongation. I use the 2 degree B-spline and the prolongation stencil is {1/4, 3/4, 3/4, 1/4}. For example, in 1D, if the solution of a node T in depth D-1 is b(T), and the B-spline range of the node is [-1.5w, 1.5w], then the prolonged solutions of related 6 nodes in depth D are {0,1/4, 3/4, 3/4, 1/4,0} * b(T) ? To my understanding, it's like this:
node T in depth D - 1:
|-------------|
b(T)
nodes in depth D covered by node T:
|------|------|------|------|------|------|
0 0.75b(T) 0.25b(T)
0.25b(T) 0.75b(T) 0
But when there are some neighbor nodes T1,T2,... in depth D-1, how to compute the prolonged solution in overlap nodes in depth D ?
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antibes0415
commented on September 4, 2024
Hi. I just accumulate those prolonged solutions of depth D - 1 nodes for overlapped depth D nodes. It works. Much thanks. Finally I get the same quality and accuracy mesh compared with mesh generated by your codes.
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mkazhdan
commented on September 4, 2024
Great. Glad to hear things worked out.
From: antibes0415 <[email protected]>
Sent: Wednesday, March 14, 2018 10:39 PM
To: mkazhdan/PoissonRecon <[email protected]>
Cc: mkazhdan <[email protected]>; Comment <[email protected]>
Subject: Re: [mkazhdan/PoissonRecon] Question about the B-spline function (#49)
Hi. I just accumulate those prolonged solutions of depth D - 1 nodes for overlapped depth D nodes. It works. Much thanks. Finally I get the same quality and accuracy mesh compared with mesh generated by your codes.
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