Mesh Parameterization

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This tour explores 2-D parameterization of 3D surfaces using linear methods.

A review paper for mesh parameterization can be found in:

M.S. Floater and K. Hormann, Surface Parameterization: a Tutorial and Survey in Advances in multiresolution for geometric modelling, p. 157-186, 2005.

See also:

K. Hormann, K. Polthier and A. Sheffer Mesh parameterization: theory and practice, Siggraph Asia Course Notes

In [1]:
from __future__ import division

import numpy as np
import scipy as scp
import pylab as pyl
import matplotlib.pyplot as plt

from nt_toolbox.general import *
from nt_toolbox.signal import *
#from nt_solutions import meshdeform_1_parameterization as solutions

import warnings
warnings.filterwarnings('ignore')

%matplotlib inline
%load_ext autoreload
%autoreload 2

Conformal Laplacian

The conformal Laplacian uses the cotan weights to obtain an accurate discretization of the Laplace Beltrami Laplacian.

They where first introduces as a linear finite element approximation of the Laplace-Beltrami operator in:

U. Pinkall and K. Polthier, Computing discrete minimal surfaces and their conjugates Experimental Mathematics, 2(1):15-36, 1993.

First load a mesh. The faces are stored in a matrix $F = (f_j)_{j=1}^m \in \RR^{3 \times m}$ of $m$ faces $f_j \in \{1,\ldots,n\}^3$. The position of the vertices are stored in a matrix $X = (x_i)_{i=1}^n \in \RR^{3 \times n}$ of $n$ triplets of points $x_k \in \RR^3$

In [2]:
from nt_toolbox.read_mesh import * 

[X, F] = read_mesh("nt_toolbox/data/nefertiti.off")
n = np.shape(X)[1]

In order to perform mesh parameterization, it is important that this mesh has the topology of a disk, i.e. it should have a single B.

First we compute the boundary $B = (i_1,\ldots,i_p)$ of the mesh. By definition, for the edges $(i_k,i_{k+1})$, there is a single adjacent face $ (i_k,i_{k+1},\ell) $.

In [3]:
from nt_toolbox.compute_boundary import *
B = compute_boundary(F)

Length of the boundary.

In [4]:
p = len(B)

Display the boundary.

In [5]:
from nt_toolbox.plot_mesh import * 

plt.figure(figsize=(10,10))
#plot mesh
plot_mesh(X, F, el=80, az=-100, lwdt=.6, dist=6)

#plot boundary
for i in range(len(B)):
    plt.plot(X[0,B], X[1,B], X[2,B], color="red", lw=3)

The conformal Laplacian weight matrix $W \in \RR^{n \times n}$ is defined as $$ W_{i,j} = \choice{ \text{cotan}(\al_{i,j}) + \text{cotan}(\be_{i,j}) \qifq i \sim j \\ \quad 0 \quad \text{otherwise}. } $$ Here, $i \times j$ means that there exists two faces $ (i,j,k) $ and $ (i,j,\ell) $ in the mesh (note that for B faces, one has $k=\ell$).

The angles are the angles centered as $k$ and $\ell$, i.e. $$ \al_{i,j} = \widehat{x_i x_k x_j } \qandq \be_{i,j} = \widehat{x_i x_\ell x_j }. $$

Compute the conformal 'cotan' weights. Note that each angle $\alpha$ in the mesh contributes with $1/\text{tan}(\alpha)$ to the weight of the opposite edge. We compute $\alpha$ as $$ \alpha = \text{acos}\pa{ \frac{\dotp{u}{v}}{\norm{u}\norm{v}} } $$ where $u \in \RR^3, v \in \RR^3$ are the edges of the adjacent vertices that defines $\al$.

In [6]:
W = sparse.coo_matrix(np.zeros([n,n]))

for i in range(3):
    i2 = (i+1)%3
    i3 = (i+2)%3
    u = X[:,F[i2,:]] - X[:,F[i,:]]
    v = X[:,F[i3,:]] - X[:,F[i,:]]
    # normalize the vectors   
    u = u/np.tile(np.sqrt(np.sum(u**2,0)), (3,1))
    v = v/np.tile(np.sqrt(np.sum(v**2,0)), (3,1))
    # compute angles
    alpha = 1/np.tan(np.arccos(np.sum(u*v, 0)))
    alpha = np.maximum(alpha, 1e-2*np.ones(len(alpha))) #avoid degeneracy
    W = W + sparse.coo_matrix((alpha,(F[i2,:],F[i3,:])),(n,n))
    W = W + sparse.coo_matrix((alpha,(F[i3,:],F[i2,:])),(n,n))

Compute the symmetric Laplacian matrix $L = D-W$ where $D = \mathrm{Diag}_i\pa{\sum_j W_{i,j}}$

In [7]:
d = W.sum(0)
D = sparse.diags(np.ravel(d),0)
L = D - W

Fixed Boundary Harmonic Parameterization

The problem of mesh parameterization corresponds to finding 2-D locations $(y_i = (y_i^1,y_i^2) \in \RR^2$ for each original vertex, where $ Y = (y_i)_{i=1}^n \in \RR^{2 \times n} $ denotes the flattened positions.

The goal is for this parameterization to be valid, i.e. the 2-D mesh obtained by replacing $X$ by $Y$ but keeping the same face connectivity $F$ should not contained flipped faces (all face should have the same orientation in the plane).

We consider here a linear methods, that finds the parameterization, that impose that the coordinates are harmonic inside the domain, and have fixed position on the boundary (Dirichlet conditions) $$ \forall s=1,2, \quad \forall i \notin B, \quad (L y^s)_i = 0, \qandq \forall j \in B, \quad y^s_j = z_j^s. $$

In order for this method to define a valid parameterization, it is necessary that the fixed position $ z_j = (z^1_j,z^2_j) \in \RR^2 $ are consecutive points along a convex polygon.

Compute the fixed positions $Z=(z_j)_j$ for the vertices on $B$. Here we use a circle.

In [8]:
p = len(B)
t = np.linspace(0,2*np.pi,p+1)
t = np.delete(t,p)
Z = np.vstack((np.cos(t),np.sin(t)))

Computing the parameterization requires to solve two independent linear system $$ \forall s=1,2, \quad L_1 y^s = r^s $$ where $L_1$ is a modified Laplacian, the is obtained from $L$ by $$ \choice{ \forall i \notin B, \quad (L_0)_{i,j} = L_{i,j} \\ \forall i \in B, \quad (L_0)_{i,i}=1, \\ \forall i \in B, \forall j \neq i, \quad (L_0)_{i,i}=0, } $$ i.e. replacing each row indexed by $B$ by a 1 on the diagonal.

In [9]:
L1 = np.copy(L.toarray())
L1[B,:] = 0
for i in range(len(B)):
    L1[B[i], B[i]] = 1
L1 = sparse.coo_matrix(L1)

Set up the right hand size $R$ with the fixed position.

In [10]:
R = np.zeros([2,n])
R[:,B] = Z

Solve the two linear systems.

In [11]:
from scipy.sparse import linalg

Y = np.zeros([2,n])
Y[0,:] = linalg.spsolve(L1, R[0,:])
Y[1,:] = linalg.spsolve(L1, R[1,:])

Display the parameterization.

In [12]:
plt.figure(figsize=(10,10))
plot_mesh(np.vstack((Y,np.zeros(n))),F, lwdt=1, c="lightgrey")

Mesh Parameterization on a Square

One can perform a fixed B parameterization on a square. This is useful to compute a geometry image (a color image storring the position of the vertices).

Exercise 1

Compute the fixed positions $Z$ of the points indexed by $B$ that are along a square. Warning: $p$ is not divisible by 4.

In [13]:
run -i nt_solutions/meshdeform_1_parameterization/exo1
In [14]:
## Insert your code here.

Exercise 2

Compute the parameterization $Y$ on a square.

In [15]:
run -i nt_solutions/meshdeform_1_parameterization/exo2
In [16]:
## Insert your code here.

Exercise 3

Shift the $B$ positions so that the eyes of the model are approximately horizontal.

In [17]:
run -i nt_solutions/meshdeform_1_parameterization/exo3
In [18]:
## Insert your code here.

Re-align the Texture

To map correctly a real image on the surface, the texture needs to be aligned. We use here a simple affine mapping to map the eye and mouth of the image on the corresponding location on the surface.

Load a texture image $T$.

In [19]:
from numpy import random

C = random.rand(1000,3)
C = C/np.repeat(np.sum(C,1)[:,np.newaxis],3,1)
C[np.lexsort((C[:,2],C[:,1],C[:,0]))]

lambd=np.transpose(C)
In [20]:
n1 = 256
T = load_image("nt_toolbox/data/lena.bmp", n1)
T = T[::-1,:]

Display the texture on the mesh, using the parametrization of the mesh as texture coordinates.

In [21]:
from mpl_toolkits.mplot3d import Axes3D
from mpl_toolkits.mplot3d import art3d
from scipy import interpolate

fig = plt.figure(figsize = (10,10))
ax = fig.add_subplot(111, projection='3d')

x = np.linspace(0, 1, np.shape(T)[0])
f = interpolate.RectBivariateSpline(x, x, np.transpose(T))

for i in range(np.shape(F)[1]):
    
    px = Y[0,F[:,i]]
    py = Y[1,F[:,i]]
    
    points_x = np.dot(np.transpose(lambd),px[:,np.newaxis])
    points_y = np.dot(np.transpose(lambd),py[:,np.newaxis])
    points = np.hstack((points_x,points_y))
    
    col = rescale(f.ev(points_x, points_y))
    col = np.repeat(col,3,1)
    
    Px = X[0,F[:,i]]
    Py = X[1,F[:,i]]
    Pz = X[2,F[:,i]]
    
    Points_x = np.dot(np.transpose(lambd),Px[:,np.newaxis])
    Points_y = np.dot(np.transpose(lambd),Py[:,np.newaxis])
    Points_z = np.dot(np.transpose(lambd),Pz[:,np.newaxis])

    ax.scatter(Points_x,Points_y,Points_z,c=col,s=5,lw=0,alpha=1)
    ax.view_init(elev=80, azim=-100)
    ax.axis("off")
    ax.dist=5