Outline Building Parametrizations The Users Perspective Results - - PDF document
A Manifold-Based Construction for Surface Fitting Lecture 6 - February 5, 2009 - 2-3 PM Outline Building Parametrizations The Users Perspective Results Conclusions Suggested Readings 2 Building Parametrizations Recall
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M ⊆ Rm Ω2 Rn Ω1 θ2 θ1 ϕ12 ϕ21 Ω12 Ω21
Recall the big picture...
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GIVEN
M ⊆ Rm Ω2 Rn Ω1 θ2 θ1 ϕ12 ϕ21 Ω12 Ω21
Our goal now is to define a family, (θv)v∈I, of parametriza- tions.
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To define (θv)v∈I, we specify a family of shape functions and a family of weight functions, both of which are described as follows:
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For each u ∈ I, we define the shape function, ψu : u ⊂ R2 → R3 , associated with Ωu, as the B´ ezier surface patch of bi-degree (m, n), ψu(p) =
bu
j,k · Bm j (x) · Bn k (y) ,
where
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cos(π/mu), (x, y) are the coordinates of p, ψu(p) =
bu
j,k · Bm j (x) · Bn k (y)
Ωu
L = cos(π/mu) (2 · l(u), 0) [−L + 2 · l(u), −L] [L + 2 · l(u), −L] [L + 2 · l(u), L] [−L + 2 · l(u), L]
p = (x, y)
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j,k) ⊂ R3 are the control points, with 0 ≤ j ≤ m and
0 ≤ k ≤ n, R3
b0,0 b1,0 b2,0 b0,1 b1,1 b2,1 b0,2 b1,2 b2,2
Ex: m = 2 and n = 2
p1,0 p2,0 p0,1 p1,1 p2,1 p0,2 p1,2 p2,2 = (L + 2 · l(u), L) p0,0 = (−L + 2 · l(u), L)
R2 ψu(p) =
bu
j,k · Bm j (x) · Bn k (y)
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ψu(p) =
bu
j,k · Bm j (x) · Bn k (y)
Bl
i(t) =
l i
r − t r − s l−i · t − s r − s i is the i-th Bernstein polynomial of degree l over the affine frame [s, r], where [s, r] = [−L+2·l(u), L+2·l(u)] (or [s, r] = [−L, L]), for every i ∈ {0, . . . , l}. Note that Bl
i is a scalar function, and recall that the following
holds:
Bm
j (x) · Bn k (y) = 1, ∀x, y ∈ [0, 1] .
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So, the convex hull property holds for p ∈ u.
p1,0 p2,0 p0,1 p2,1 p0,2 p1,2
R2
p1,1 p0,0 p2,2 1 2 3 4 x 0.5 1 1.5 2 y 1 2 3 4 z 1 1 2 3 4 z
b00 b01 b10 b20 b11 b21 b22
ψu(p1,1) = ψu((2 · l(u), 0))
R3 u
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Very nice, but how do we pick the control points? We use a least squares fitting procedure. The idea is to create a large collection, (pj, p
j)j∈J, of pairs
p
j ∈ R3.
We can think of p
j as the image of pj under an unknown
function, f : R2 → R3, we wish to locally approximate using
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We can create (pj, p
j)j∈J by placing the pj’s in the Tu’s and
the p
j’s in ST .
But, since ST is piecewise linear, it is better (for the sake of visual quality) to place the p
j’s on a “curved” surface that
approximates |ST |. Which “curved” surface?
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Good choices:
PN triangle surfaces are cheap to construct and evaluate, while subdivision approaches usually yield pretty good-looking sur- faces. We now describe the computation of the parameter and sam- ple points in a way that is (almost) independent of the choice
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The idea is to assume that we have a surface, S ⊂ R3, which is the union of finitely many parametric patches, bσ : R2 → R3, each of which is associated with a triangle, σ, of ST , and all of them are the defined in the same affine frame, say ⊂ R2: S =
bσ() .
σ bσ()
bσ()
R3
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S |ST | It is desirable to define S continuous (at least!)
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How can we compute the control points? First, sample the affine frame of ψu in a uniform manner: We choose m = n = mu + 1 and sample the above frame with 4 · m2 points, where mu is the degree of vertex u in ST . HEURISTIC!
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su(σ)
Second, for each triangle su(σ) of Tu, map the sample points in su(σ) to . st(u, ST )
σ u
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To map from su(σ) to , we simply use a barycentric map- ping: Choose an order, a, b, c, for the vertices, a, b, and c of su(σ). Now, every point p ∈ su(σ) can be expressed as p = λ · a + µ · b + ν · c , where λ, µ, and ν are the barycentric coordinates of p with respect to su(σ) and associated with the vertices a, b, and c, respectively.
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Finally, the point q ∈ corresponding to p ∈ su(σ) is given by q = λ · a + µ · b + ν · c , where a = (0, 0), b = (1, 0), and c = (1/2, √ 3/2) are the vertices of . Assume that = [(0, 0), (1, 0), (1/2, √ 3/2)].
su(σ) a b
c
b c
p q
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Third, we compute p = bσ(q).
b c
q
σ bσ()
p = bσ(q)
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su(σ) a b
c
p
Sampling su(σ)
b c
q
Barycentric Mapping Evaluation of bσ
σ bσ()
p = bσ(q)
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Now, we have a collection of pairs, (p, p), of parameter and sample points associated with each vertex, u, of ST , as shown below:
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Let M be the number of elements of the collection of pairs, (p, p), associated with u. We assume that M is larger than m × n = (mu + 1)2. We can assemble three linear equation systems, AX = Bl, with l = 1, 2, 3, each of which has exactly M × (mu + 1)2
the form ψu(p) = p
l =
⇒
bu
j,k · Bm j (x) · Bn k (y) = p l ,
where p
l the l-th coordinate of p. Since M m, the systems
are overdetermined!
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The unknows are the m×n = (mu+1)2 control points, (bu
j,k),
We have a typical setup for using least squares, and that’s what we do! Once we compute the control points, we have ψu.
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Ultimately, we want to compute θu: GIVEN
M ⊆ Rm Ω2 Rn Ω1 θ2 θ1 ϕ12 ϕ21 Ω12 Ω21
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Why not let θu = ψu? First, the domain of θu is Ωu, while the domain of ψu is R2.
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Second, θu(p) and θv(q) must be the same point in R3 if q = ϕvu(p). GIVEN
M ⊆ Rm Ω2 Rn Ω1 θ2 θ1 ϕ12 ϕ21 Ω12 Ω21
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It is EXTREMELY unlikely that ψu fullfils the second condi- tion. So, what can we do? Answer: use cutoff functions!
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Ωu
(2 · l(u), 0)
p p − (2 · l(u), 0)
For each u ∈ I, we define a weight function, γu : R2 → R , associated with Ωu is given by γu(p) = ξ(p − (2 · l(u), 0)) , for every p ∈ R2, where · denotes Euclidean distance, and
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ξ : R → R is such that, for every t ∈ R, ξ(t) = 1 if t ≤ H1 if t ≥ H2 1/(1 + e2·s)
where H1, H2 are constant, with 0 < H1 < H2 < 1, and s =
√ 1 − H
√ 1 − H
H = t − H1 H2 − H1
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0.2 0.4 0.6 0.8 1 t 0.2 0.4 0.6 0.8 1 G1
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L2 = cos(π/mu) L1 = 0.25 · L2
Ωu
HEURISTIC!
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“θu(p) and θv(q) must be the same point in R3 if q = ϕvu(p)” The problem can be solved by defining θu as θu(p) = ψu(p) · γu(p) , but the above does not solve the problem “the domain of θu is Ωu, while the domain of ψu is u”
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Well, let us use another trick: θu(p) =
, where Ju(p) = {v ∈ I | p ∈ Ωuv} . Ju(p) has at least one vertex (i.e., u) and at most three.
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rv,σ Tw Tv rv,σ Tu p
ψu(p)
ϕvu(p) ϕwu(p)
ψv ◦ ϕwu(p) ψw ◦ ϕwu(p)
Ju(p) = {u, v, w}
u v w
θu(p)
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The θu’s are all C∞! In θu(p) =
, the term ψw ◦ ϕwu(p) can be viewed as the contribution of Ωw to θu(p), which is weighted by γw ◦ ϕwu(p) .
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We can show that θu(p) = θv ◦ ϕvu(p) = θw ◦ ϕwu(p) . A key observation for proving the above fact is Ju(p) = Jv(ϕvu(p)) = Jw(ϕwu(p)) . Later, to convince yourself, take a look at the formulas for θv ◦ ϕvu(p) and θw ◦ ϕwu(p) .
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FINALLY, our surface is given by S =
θu(Ωu) . How can we evaluate S? Is the “user” supposed to know about sets of gluing data and parametrizations?
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Let t be a triangle in ST and p be any point in t: t p
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Map p to an equilateral triangle in R2. We can do that by using barycentric coordinates.
u v w p
(0, 0) (1, 0)
2, √ 3 2
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q
(0, 0) (1, 0)
2, √ 3 2
u v w
rv,σ Tw Tv rv,σ Tu
R−1
(v,u) ◦ g−1 v
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It might very well be the case that R−1
(v,u) ◦ g−1 v
is not in Ωv! This can be easily tested... If R−1
(v,u) ◦ g−1 v
∈ Ωv then we consider Ωu (and maybe Ωw). But, there is a subtle detail... The barycentric mapping from t to the canonical triangle changes!
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(0, 0) (1, 0)
q
2, √ 3 2
Tw Tv rv,σ Tu
p u v w
R−1
(u,w) ◦ g−1 u
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rv,σ Tw Tv rv,σ Tu
p u v w
R−1
(w,v) ◦ g−1 w
(0, 0) (1, 0)
q
2, √ 3 2
q
Tu rv,σ Tw Tv rv,σ
p u v w
R−1
(v,u) ◦ g−1 v
p
ϕuv ϕwv
ϕwv = ϕwu ◦ ϕuv
ϕwu
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Is the “user” supposed to know about sets of gluing data and parametrizations? DEFINITELY NOT! This is because we created an explicit homeomorphism be- tween |ST | and S.
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This homeomorphism hides the differentiable structure of S from the user! The user just needs to know about ST , and then provide “us” with the barycentric coordinates of a point p in a triangle σ
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OPEN PROBLEM: Can we make it polynomial?
Unfortunately, the map θv is NOT polynomial.
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We can easily make γv a Ck polynomial, for any finite k. The difficulty really lies in making ϕvu (rational) polynomial. A solution for this problem would make a PPS as simple to evaluate as any polynomial surface (such as B´ ezier and Splines surfaces).
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For a good survey on the existing constructions, see
Parametrization with Manifolds. In ACM SIGGRAPH 2006 Courses (SIGGRAPH’06), pages 1-81, New York, NY, USA, 2006. ACM Press.
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There are many extensions of this work. If you are interested, please talk to Luiz or myself. We will be glad to discuss some
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Surfaces from Triangle Meshes Using Parametric Pseudo-Manifolds, preprint submitted to Computer & Graphics. Download a PDF from the course web page:
http://w3.impa.br/∼lvelho/ppm09
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