# James Gregson's Website

## Reordering Matrix Products II

Given two matrices $$A \in \mathcal{R}^{M\times S}$$ and $$B \in \mathcal{R}^{S \times N}$$ it is helpful to be able to reorder their product $$A B \in \mathcal{R}^{M\times N}$$ in order to compute Jacobians for optimization. For this it is common to define the $$\mbox{vec}(.)$$ operator which flattens a matrix into a vector by concatenating columns:

\begin{equation*} \newcommand{\vec}[1]{\mbox{vec}{\left(#1\right)}} \vec{A} = \begin{bmatrix} A_{11} \\ \vdots \\ A_{M1} \\ \vdots \\ A_{1S} \\ \vdots \\ A_{MS}\end{bmatrix} \end{equation*}

Using this operator, the (flattened) product $$AB$$ can be written as one of:

\begin{align*} \vec{AB} &=& (B^T \otimes I_A)\vec{A} \\ &=& (I_B \otimes A)\vec{B} \end{align*}

where $$\otimes$$ is the Kronecker Product and $$I_A$$ is an identity matrix with the same number of rows as $$A$$ and $$I_B$$ is an idenity matrix with the same number of columns as $$B$$.

From the two options for expressing the product, it's clear that $$(B^T \otimes I_A)$$ is the (vectorized) Jacobian of $$AB$$ with respect to $$A$$ and $$(I_B \otimes A)$$ is the corresponding vectorized Jacobian with respect to $$B$$.

That's great but if our linear algebra library happens to represent matrices in row-major order it will be necessary to sprinkle a large number of transposes in to get things in the right order to apply these formulas. That will be tedious, error-prone and inefficient. Fortunately, equivalent formulas are available for row-major ordering by defining a $$\mbox{ver}(.)$$ operator that flattens a matrix by concatenating rows to form a column vector:

\begin{equation*} \newcommand{\ver}[1]{\mbox{ver}{(#1)}} \ver{A} = \begin{bmatrix} A_{11} \\ \vdots \\ A_{1S} \\ \vdots \\ A_{M1} \\ \vdots \\ A_{MS} \end{bmatrix} \end{equation*}

Given this it is fairly quick to work out that the equivalent to the first two formulas using the $$\ver{.}$$ operator are:

\begin{align*} \ver{AB} &=& (I_A \otimes B^T)\ver{A} \\ &=& ( A \otimes I_B)\ver{B} \end{align*}

Using the same logic it can be concluded that $$(I_A \otimes B^T)$$ is the vectorized Jacobian of $$AB$$ with respect to $$A$$ and $$(A \otimes I_B)$$ is the corresponding Jacobian with respect to $$B$$.

It's also worth noting that these are simplifications of a more general case for the product $$AXB$$ that brings $$X$$ to the right of the expression. The corresponding $$\vec{.}$$ and $$\ver{.}$$ versions are:

\begin{align*} \vec{AXB} &=& (B^T \otimes A) \vec{X} \\ \ver{AXB} &=& (A \otimes B^T) \ver{X} \end{align*}

which makes it clear where the identity matrices $$I_A$$ and $$I_B$$ in the previous expressions originate.

All these can be demonstrated by the following python code sample:

from typing import Tuple
import numpy as np

def vec( M: np.ndarray ):
'''Flatten argument by columns'''
return M.flatten('F')

def ivec( v: np.ndarray, shape: Tuple[int,int] ):
'''Inverse of vec'''
return np.reshape( v, shape, 'F' )

def ver( M: np.ndarray ):
'''Flatten argument by rows'''
return M.flatten('C')

def iver( v: np.ndarray, shape: Tuple[int,int] ):
'''Inverse of ver'''
return np.reshape( v, shape, 'C' )

def test_vec_ivec():
A = np.random.standard_normal((3,4))
assert np.allclose( A, ivec( vec( A ), A.shape ) )

def test_ver_iver():
A = np.random.standard_normal((3,4))
assert np.allclose( A, iver( ver(A), A.shape ) )

def test_products():
A = np.random.standard_normal((3,4))
X = np.random.standard_normal((4,4))
B = np.random.standard_normal((4,5))

AB1 = A@B

Ia = np.eye( A.shape[0] )
Ib = np.eye( B.shape[1] )

AB2a = ivec( np.kron( B.T, Ia )@vec(A), (A.shape[0],B.shape[1]) )
AB2b = ivec( np.kron(  Ib,  A )@vec(B), (A.shape[0],B.shape[1]) )

assert np.allclose( AB1, AB2a )
assert np.allclose( AB1, AB2b )

AB3a = iver( np.kron( A,   Ib )@ver(B), (A.shape[0], B.shape[1]) )
AB3b = iver( np.kron( Ia, B.T )@ver(A), (A.shape[0], B.shape[1]) )

assert np.allclose( AB1, AB3a )
assert np.allclose( AB1, AB3b )

AXB1 = A@X@B
AXB2 = ivec( np.kron(B.T,A)@vec(X), (A.shape[0],B.shape[1]) )
AXB3 = iver( np.kron(A,B.T)@ver(X), (A.shape[0],B.shape[1]) )

assert np.allclose( AXB1, AXB2 )
assert np.allclose( AXB1, AXB3 )

if __name__ == '__main__':
test_vec_ivec()
test_ver_iver()
test_products()


Hopefully this is helpful to someone.

## Quaternion Notes

These are some helpful quaternion identities, all summarized from Kok et al., Using Inertial Sensors for Position and Orientation Estimation although I have put the scalar component last.

A quaternion $$q$$ is represented as a 4-tuple, or an imaginary vector portion $$q_v=[q_x,q_y,q_z]^T$$ and a scalar value $$q_s$$ giving $$q = [q_v^T, q_s]^T$$. The norm of a quaternion is $$\|q\|_2 = (q_v^T q_v + q_s^2)^{\frac{1}{2}}$$, the conjugate of a quaternion is $$q^* = [-q_v^T, q_s]^T$$ and the inverse of a quaternion is $$q^{-1} = \frac{q^*}{\|q\|_2^2}$$.

Unit quaternions have $$\|q\|_2 = 1$$, in which case $$q^* = q^{-1}$$. Unit quaternions also represent spatial rotations/orientations of $$\theta$$ radians around a unit axis $$a$$ via $$q = [a^T \sin\left(\frac{\theta}{2}\right), \cos\left(\frac{\theta}{2}\right) ]^T$$. Both $$q$$ and $$-q = [-q_v, -q_s]^T$$ represent the same orientation, although the one with positive scalar component takes the shortest rotation to achieve that orientation since $$\cos^{-1}(|x|) \leq \cos^{-1}(-|x|)$$.

The multiplication of two quaternions is $$p \otimes q = [ (p_s q_v + q_s p_v + p_v \times q_v)^T, p_s q_s - p_v \times q_v ]^T$$. The presence of the cross-product means that $$p\otimes q \neq q \otimes p$$. When $$p, q$$ are unit quaternions, $$p\otimes q$$ is equivalent to chaining the two rotations. The rotation matrix corresponding to a unit quaternion is:

\begin{equation*} R(q) = \begin{bmatrix} 1 - 2 q_y^2 - 2*q_z^2 & 2 q_x q_y - 2 q_z q_s & 2 q_x q_z + 2 q_y q_s \\ 2 q_x q_y + 2 q_z q_s & 1 - 2 q_x^2 - 2 q_z^2 & 2 q_y q_z - 2 q_x q_s \\ 2 q_x q_z - 2 q_y q_s & 2 q_y q_z + 2 q_x q_s & 1 - 2 q_x^2 - 2 q_y^2 \end{bmatrix} \end{equation*}

or in python:

def quat2mat( q: np.ndarray ) -> np.ndarray:
assert q.ndim == 1 and q.size == 4
qx,qy,qx,qs = q
return np.ndarray((
( 1 - 2*qy**2 - 2*qz**2,     2*qx*qy - 2*qz*qs,     2*qx*qz + 2*qy*qs ),
(     2*qx*qy + 2*qz*qs, 1 - 2*qx**2 - 2*qz**2,     2*qy*qz - 2*qx*qs ),
(     2*qx*qz - 2*qy*qs,     2*qy*qz + 2*qx*qs, 1 - 2*qx**2 - 2*qy**2 )
))


The, usually non-unit, quaternion representation of a vector $$v^\wedge = [ v^T, 0]^T$$. The vector can be rotated by the unit-quaternion $$q$$ via $$R(q) v = (q\otimes v^\wedge \otimes q^*)_v$$ where only the vector portion is retained.

The quaternion-quaternion product can be represented by matrix multiplication in either the same, or reversed order, i.e.:

\begin{align*} \newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} p \otimes q &= p^L q = q^R p \hspace{1cm} & \mbox{where} \\ p^L &= \begin{bmatrix} p_s I + \lfloor p_v \rfloor_\times & p_v \\ -p_v^T & p_s \end{bmatrix} & \\ q^R &= \begin{bmatrix} q_s I - \lfloor q_v \rfloor_\times & q_v \\ -q_v^T & q_s \end{bmatrix} & \end{align*}

where $$\lfloor m \rfloor_\times$$ is the matrix representation of the cross-product. This is extremely useful when differentiating quaternion expressions since $$p^L = \pd{}{q}(p\otimes q)$$ and $$q^R = \pd{}{p}(p\otimes q)$$.

A Rodrigues rotation vector $$\eta = \theta a$$ where $$\|a\|_2 = 1$$ can be mapped to a quaternion via the exponential map:

\begin{align*} \exp_q(\eta) =& q = \begin{pmatrix} a \sin\left(\frac{\theta}{2}\right) \\ \cos\left(\frac{\theta}{2}\right) \end{pmatrix} \\ \log_q(q) =& \eta = 2 q_v \end{align*}

Under the assumption that $$\theta << 1$$, the following approximations can be made:

\begin{align*} \exp_q(\eta) \approx& \begin{pmatrix} \frac{\eta}{2} \\ 1 \end{pmatrix} \\ \pd{\exp_q(\eta)}{\eta} \approx& \begin{pmatrix} \frac{1}{2} I_{3\times3} \\ 0 \end{pmatrix} \\ \pd{\log_q(q)}{q} \approx& \begin{pmatrix} 2 I_{3\times3} & 0_{3\times1} \end{pmatrix} \end{align*}

The above are also very helpful in manifold optimization over quaternion states where the Newton update is generally small which results in linear Jacobians with respect to the update variables.

## QR Decomposition Notes

I find this process easier to reason through by writing out a small 3-column version explicitly:

\begin{equation*} \begin{bmatrix} A_{*0} & A_{*1} & A_{*2} \end{bmatrix} = \begin{bmatrix} Q_{*0} & Q_{*1} & Q_{*2} \end{bmatrix} \begin{bmatrix} R_{00} & R_{01} & R_{02} \\ & R_{11} & R_{12} \\ & & R_{22} \end{bmatrix} \end{equation*}

which gives:

\begin{align*} &A_{*0} = Q_{*0} R_{00} \\ &A_{*1} = Q_{*0} R_{01} + Q_{*1} R_{11} \\ &A_{*2} = Q_{*0} R_{02} + Q_{*1} R_{12} + Q_{*2} R_{22} \end{align*}

From this and the orthonormal properties of $$Q$$, the meaning of the $$R_{ji}$$ values are clear, the projection of the $$i$$'th column of $$A$$ onto the $$j$$'th column of $$Q$$. These allows fixing the values of $$R_{ji}$$ and norms of $$Q_{*i}$$:

• $$Q_{*i}^T Q_{*j} = 0, \forall i \neq j$$: columns of $$Q$$ are orthogonal so each $$R_{j*}$$ value must account for entire parallel component of $$A_{*j}$$ onto each column of $$Q$$.
• $$\|Q_{*i}\|_2 = 1, \forall i$$: constrains $$\|Q_{*i}\|$$ and corresponding $$R_{ji}$$ values.

Therefore:

• $$R_{00} = \|A_{*0}\|_2$$ and $$Q_{*0} = A_{*0}/R_{00}$$
• $$R_{01} = Q_{*0}^T A_{*1}$$, $$s_1 = A_{*1} - R_{01} Q_{*0}$$, $$R_{11} = \|s_1\|_2$$ $$Q_{*1} = s_1/R_{11}$$
• $$R_{02} = Q_{*0}^T A_{*2},~R_{12} = Q_{*1}^T A_{*2}, s_2 = A_{*2} - R_{02} Q_{*0} - R_{12} Q_{*1}, R_{22} = \| s_2 \|_2, Q_{*2} = s_2/R_{22}$$

or in general:

• $$R_{ji} = Q_{*j}^T A_{*i}, \forall j < i$$
• $$s_{i} = A_{*i} - \sum_j R_{ji} Q_{*j}$$
• $$R_{ii} = \| s_{i} \|_2, Q_{*i} = s_i/R_{ii}$$

and in python:

def qr_gs( A, inplace=True ):
'''Decompose A into Q*R with Q orthonormal and R upper triangular using classical Gram-Schmidt (unstable)'''
A = A if inplace else A.copy()
R = np.zeros((A.shape[1],A.shape[1]))
for i in range( A.shape[1] ):
for j in range(i):
R[j,i] = np.sum(A[:,j]*A[:,i])
A[:,i] -= R[j,i]*A[:,j]
R[i,i] = np.linalg.norm(A[:,i])
A[:,i] /= R[i,i]
return A,R


The form above generates $$Q$$ column by column but has stability issues due to the use of classical Gram-Schmidt. It can be improved by replacing classical Gram-Schmidt with modified Gram-Schmidt:

def qr_mgs( A, inplace=True ):
'''Decompose A into Q*R with Q orthonormal and R upper triangular using modified Gram-Schmidt

Assumes columns of A are linearly independent.
'''
A = A if inplace else A.copy()
R = np.zeros((A.shape[1],A.shape[1]))
for i in range( A.shape[1] ):
R[i,i] = np.linalg.norm(A[:,i])
A[:,i] /= R[i,i]
for j in range(i+1,A.shape[1]):
R[i,j] = np.dot(A[:,j],A[:,i])
A[:,j] -= R[i,j]*A[:,i]
return A,R


The inner loop of this can be replaced entirely with vectorization in python but the above form gives a reasonable starting point for porting to another language like C or C++. These implementations should also check for $$R_{ii} = 0$$ which indicate a rank deficient system.

## Least Squares & MLE

Given a linear model $$y_i = x_i^T \beta$$, we collect $$N$$ observations $$\tilde{y_i} = y_i + \epsilon_i$$ where $$\epsilon_i \sim \mathcal{N}(0,\sigma^2)$$. The likelihood of any one of our observations $$\tilde{y_i}$$ given current estimates of our parameters $$\beta$$ is then:


and the likelihood of all of them occuring is:

\begin{equation*} P(\tilde{y} | \beta) = \prod_{i=1}^N P(\tilde{y}_i | \beta ) \end{equation*}

The parameters most likely to explain the observations, given only the observations and definition of the model, are those that maximize this likelihood. Equivalently, we can find the parameters that minimize the negative log-likelihood since the two have the same optima:

\begin{align*} \beta^* &=& \argmax_\beta P(\tilde{y} | \beta) \\ &=& \argmin_\beta - \ln( P(\tilde{y} | \beta ) \\ &=& \argmin_\beta - N\log{\frac{1}{\sqrt{2\pi\sigma^2}}} - \log{ \prod_{i=1}^N P(\tilde{y}_i | \beta ) } \end{align*}

The benefit of this is that the terms decouple in the case of normally distributed residuals. Ditching constant factors and collecting the product terms into the exponent the above becomes:

\begin{align*} \beta^* &=& \argmin_\beta -\ln\left( \exp{-\frac{1}{2}\sum_{i=1}^N \left(\frac{\tilde{y}_i - M_i^T \beta}{\sigma}\right)^2} \right) \\ &=& \argmin_\beta \frac{1}{2\sigma^2} \sum_{i=1}^N \left( \tilde{y}_i - M_i^T \beta \right)^2 \end{align*}

which gives the least-squares version.

## Least Squares & MAP

We can also add a prior for the parameters $$\beta$$, e.g. $$\beta_i \sim \mathcal{N}(0,\sigma_\beta)$$. In this case, each component of $$\beta$$ is independent. The MAP estimate for $$\beta$$ is then:

\begin{align*} \beta^* &=& \argmax_\beta P(\tilde{y} | \beta) P(\beta) \\ &=& \argmax_\beta \prod_i P(\tilde{y}_i | \beta) \prod_j P(\beta_j) \\ &=& \argmax_\beta \prod_i \frac{1}{\sqrt{2\pi\sigma^2}}\exp{ -\frac{1}{2}\left(\frac{e_i}{\sigma}\right)^2} \prod_j \frac{1}{\sqrt{2\pi\sigma_\beta^2}}\exp{ -\frac{1}{2}\left(\frac{\beta_j}{\sigma_\beta}\right)^2 } \end{align*}

The same negative log trick can then be used to convert products to sums, also ditching constants, to get:

\begin{align*} \beta^* &=& \argmin \frac{1}{2\sigma^2} \sum_i e_i^2 + \frac{1}{2\sigma_\beta^2} \sum_j \beta_j^2 \\ &=& \argmin \frac{1}{2\sigma^2} \sum_i \left( \tilde{y}_i - x_i^T \beta \right)^2 + \frac{1}{2\sigma_\beta^2} \sum_j \beta_j^2 \end{align*}

Using different priors will of course give different forms for the second term, e.g. choosing components of $$\beta$$ to follow a Laplace distribution $$\beta_j \sim \frac{\lambda}{2} \exp{-\lambda |\beta_j|}$$, gives:

\begin{equation*} \beta^* = \argmin \frac{1}{2\sigma^2} \sum_i \left( \tilde{y}_i - x_i^T \beta \right)^2 + \lambda \sum_j |\beta_j| \end{equation*}

## Essential Matrix

The essential matrix relates normalized coordinates in one view with those in another view of the same scene and is widely used in stereo reconstruction algorithms. It is a 3x3 matrix with rank 2, having only 8 degrees of freedom.

Given $$p$$ as a normalized point in image P and $$q$$ as the corresponding normalied point in another image Q of the same scene, the essential matrix provides the constraint:

\begin{equation*} q^T E p = 0 \end{equation*}

It's important that $$p$$ and $$q$$ be normalized homogeneous points, i.e. the following where $$[u_p,v_p]$$ are the pixel coordinates of point p:

\begin{equation*} p = K^{-1} \begin{bmatrix} u_p \\ v_p \\ 1 \end{bmatrix} \end{equation*}

By fixing the extrinsics of image P as $$R_P = I$$ and $$t_P = [0,0,0]^T$$ the essential matrix relating P and Q can defined by the relationship:

\begin{equation*} E = t_\times R \end{equation*}

where $$R$$ and $$t$$ are the relative transforms of camera Q with respect to camera P and $$t_\times$$ is the matrix representation of the the cross-product:

\begin{equation*} t_\times = \begin{bmatrix} 0 & -t_z & t_y \\ t_z & 0 & -t_x \\ -t_y & t_x & 0 \end{bmatrix} \end{equation*}

## Load/Save Wavefront .OBJ Files in Python

It's pretty useful to be able to load meshes when doing graphics. One format that is nearly ubiquitous are Wavefront .OBJ files, due largely to their good features-to-complexity ratio. Sure there are better formats, but when you just want to get a mesh into a program it's hard to beat .obj which can be imported and exported from just about anyhwere.

The code below loads and saves .obj files, including material assignments (but not material parameters), normals, texture coordinates and faces of any number of vertices. It supports faces with/without either/both of normals and texture coordinates and (if you can accept 6D vertices) also suports per-vertex colors (a non-standard extension sometimes used for debugging in mesh processing). Faces without either normals or tex-coords assign the invalid value -1 for these entries.

From the returned object, it is pretty easy to do things like sort faces by material indices, split vertices based on differing normals or along texture boundaries and so on. It also has the option to tesselate non-triangular faces, although this only works for convex faces (you should only be using convex faces anyway though!).

# wavefront.py
import numpy as np

class WavefrontOBJ:
def __init__( self, default_mtl='default_mtl' ):
self.path      = None               # path of loaded object
self.mtllibs   = []                 # .mtl files references via mtllib
self.mtls      = [ default_mtl ]    # materials referenced
self.mtlid     = []                 # indices into self.mtls for each polygon
self.vertices  = []                 # vertices as an Nx3 or Nx6 array (per vtx colors)
self.normals   = []                 # normals
self.texcoords = []                 # texture coordinates
self.polygons  = []                 # M*Nv*3 array, Nv=# of vertices, stored as vid,tid,nid (-1 for N/A)

def load_obj( filename: str, default_mtl='default_mtl', triangulate=False ) -> WavefrontOBJ:
"""Reads a .obj file from disk and returns a WavefrontOBJ instance

Handles only very rudimentary reading and contains no error handling!

Does not handle:
- relative indexing
- subobjects or groups
- lines, splines, beziers, etc.
"""
# parses a vertex record as either vid, vid/tid, vid//nid or vid/tid/nid
# and returns a 3-tuple where unparsed values are replaced with -1
def parse_vertex( vstr ):
vals = vstr.split('/')
vid = int(vals[0])-1
tid = int(vals[1])-1 if len(vals) > 1 and vals[1] else -1
nid = int(vals[2])-1 if len(vals) > 2 else -1
return (vid,tid,nid)

with open( filename, 'r' ) as objf:
obj = WavefrontOBJ(default_mtl=default_mtl)
obj.path = filename
cur_mat = obj.mtls.index(default_mtl)
for line in objf:
toks = line.split()
if not toks:
continue
if toks[0] == 'v':
obj.vertices.append( [ float(v) for v in toks[1:]] )
elif toks[0] == 'vn':
obj.normals.append( [ float(v) for v in toks[1:]] )
elif toks[0] == 'vt':
obj.texcoords.append( [ float(v) for v in toks[1:]] )
elif toks[0] == 'f':
poly = [ parse_vertex(vstr) for vstr in toks[1:] ]
if triangulate:
for i in range(2,len(poly)):
obj.mtlid.append( cur_mat )
obj.polygons.append( (poly[0], poly[i-1], poly[i] ) )
else:
obj.mtlid.append(cur_mat)
obj.polygons.append( poly )
elif toks[0] == 'mtllib':
obj.mtllibs.append( toks[1] )
elif toks[0] == 'usemtl':
if toks[1] not in obj.mtls:
obj.mtls.append(toks[1])
cur_mat = obj.mtls.index( toks[1] )
return obj

def save_obj( obj: WavefrontOBJ, filename: str ):
"""Saves a WavefrontOBJ object to a file

Warning: Contains no error checking!

"""
with open( filename, 'w' ) as ofile:
for mlib in obj.mtllibs:
ofile.write('mtllib {}\n'.format(mlib))
for vtx in obj.vertices:
ofile.write('v '+' '.join(['{}'.format(v) for v in vtx])+'\n')
for tex in obj.texcoords:
ofile.write('vt '+' '.join(['{}'.format(vt) for vt in tex])+'\n')
for nrm in obj.normals:
ofile.write('vn '+' '.join(['{}'.format(vn) for vn in nrm])+'\n')
if not obj.mtlid:
obj.mtlid = [-1] * len(obj.polygons)
poly_idx = np.argsort( np.array( obj.mtlid ) )
cur_mat = -1
for pid in poly_idx:
if obj.mtlid[pid] != cur_mat:
cur_mat = obj.mtlid[pid]
ofile.write('usemtl {}\n'.format(obj.mtls[cur_mat]))
pstr = 'f '
for v in obj.polygons[pid]:
# UGLY!
vstr = '{}/{}/{} '.format(v[0]+1,v[1]+1 if v[1] >= 0 else 'X', v[2]+1 if v[2] >= 0 else 'X' )
vstr = vstr.replace('/X/','//').replace('/X ', ' ')
pstr += vstr
ofile.write( pstr+'\n')


A function that round-trips the loader is shown below, the resulting mesh dump.obj can be loaded into Blender and show the same material ids and texture coordinates. If you inspect it (and remove comments) you will see that the arrays match the string in the function:

from wavefront import *

data = '''
# slightly edited blender cube
mtllib cube.mtl
v 1.000000 1.000000 -1.000000
v 1.000000 -1.000000 -1.000000
v 1.000000 1.000000 1.000000
v 1.000000 -1.000000 1.000000
v -1.000000 1.000000 -1.000000
v -1.000000 -1.000000 -1.000000
v -1.000000 1.000000 1.000000
v -1.000000 -1.000000 1.000000
vt 0.625000 0.500000
vt 0.875000 0.500000
vt 0.875000 0.750000
vt 0.625000 0.750000
vt 0.375000 0.000000
vt 0.625000 0.000000
vt 0.625000 0.250000
vt 0.375000 0.250000
vt 0.375000 0.250000
vt 0.625000 0.250000
vt 0.625000 0.500000
vt 0.375000 0.500000
vt 0.625000 0.750000
vt 0.375000 0.750000
vt 0.125000 0.500000
vt 0.375000 0.500000
vt 0.375000 0.750000
vt 0.125000 0.750000
vt 0.625000 1.000000
vt 0.375000 1.000000
vn 0.0000 0.0000 -1.0000
vn 0.0000 1.0000 0.0000
vn 0.0000 0.0000 1.0000
vn -1.0000 0.0000 0.0000
vn 1.0000 0.0000 0.0000
vn 0.0000 -1.0000 0.0000
usemtl mat1
# no texture coordinates
f 6//1 5//1 1//1 2//1
usemtl mat2
f 1/5/2 5/6/2 7/7/2 3/8/2
usemtl mat3
f 4/9/3 3/10/3 7/11/3 8/12/3
usemtl mat4
f 8/12/4 7/11/4 5/13/4 6/14/4
usemtl mat5
f 2/15/5 1/16/5 3/17/5 4/18/5
usemtl mat6
# no normals
f 6/14 2/4 4/19 8/20
'''
with open('test_cube.obj','w') as obj:
obj.write( data )
save_obj( obj, 'dump.obj' )

if __name__ == '__main__':


Hope it's useful!

## Useful Mesh Functions in Python

These are collections of functions that I end up re-implementing frequently.

import numpy as np

def manifold_mesh_neighbors( tris: np.ndarray ) -> np.ndarray:
"""Returns an array of triangles neighboring each triangle

Args:
tris (Nx3 int array): triangle vertex indices

Returns:
nbrs (Nx3 int array): neighbor triangle indices,
or -1 if boundary edge
"""
if tris.shape[1] != 3:
raise ValueError('Expected a Nx3 array of triangle vertex indices')

e2t = {}
for idx,(a,b,c) in enumerate(tris):
e2t[(b,a)] = idx
e2t[(c,b)] = idx
e2t[(a,c)] = idx

nbr = np.full(tris.shape,-1,int)
for idx,(a,b,c) in enumerate(tris):
nbr[idx,0] = e2t[(a,b)] if (a,b) in e2t else -1
nbr[idx,1] = e2t[(b,c)] if (b,c) in e2t else -1
nbr[idx,2] = e2t[(c,a)] if (c,a) in e2t else -1
return nbr

if __name__ == '__main__':
tris = np.array((
(0,1,2),
(0,2,3)
),int)
nbrs = manifold_mesh_neighbors( tris )

tar = np.array((
(-1,-1,1),
(0,-1,-1)
),int)
if not np.allclose(nbrs,tar):
raise ValueError('uh oh.')


## Useful Transformation Functions in Python

These are collections of functions that I end up re-implementing frequently.

from typing import *
import numpy as np

def closest_rotation( M: np.ndarray ):
U,s,Vt = np.linalg.svd( M[:3,:3] )
if np.linalg.det(U@Vt) < 0:
Vt[2] *= -1.0
return U@Vt

def mat2quat( M: np.ndarray ):
if np.abs(np.linalg.det(M[:3,:3])-1.0) > 1e-5:
raise ValueError('Matrix determinant is not 1')
if np.abs( np.linalg.norm( M[:3,:3].T@M[:3,:3] - np.eye(3)) ) > 1e-5:
raise ValueError('Matrix is not orthogonal')
w = np.sqrt( 1.0 + M[0,0]+M[1,1]+M[2,2])/2.0
x = (M[2,1]-M[1,2])/(4*w)
y = (M[0,2]-M[2,0])/(4*w)
z = (M[1,0]-M[0,1])/(4*w)
return np.array((x,y,z,w),float)

def quat2mat( q: np.ndarray ):
qx,qy,qz,qw = q/np.linalg.norm(q)
return np.array((
(1 - 2*qy*qy - 2*qz*qz,     2*qx*qy - 2*qz*qw,     2*qx*qz + 2*qy*qw),
(    2*qx*qy + 2*qz*qw, 1 - 2*qx*qx - 2*qz*qz,     2*qy*qz - 2*qx*qw),
(    2*qx*qz - 2*qy*qw,         2*qy*qz + 2*qx*qw, 1 - 2*qx*qx - 2*qy*qy),
))

if __name__ == '__main__':
for i in range( 1000 ):
q = np.random.standard_normal(4)
q /= np.linalg.norm(q)
A = quat2mat( q )
s = mat2quat( A )
if np.linalg.norm( s-q ) > 1e-6 and np.linalg.norm( s+q ) > 1e-6:
raise ValueError('uh oh.')

R = closest_rotation( A+np.random.standard_normal((3,3))*1e-6 )
if np.linalg.norm(A-R) > 1e-5:
print( np.linalg.norm( A-R ) )
raise ValueError('uh oh.')


## Mesh Processing in Python: Implementing ARAP deformation

Python is a pretty nice language for prototyping and (with numpy and scipy) can be quite fast for numerics but is not great for algorithms with frequent tight loops. One place I run into this frequently is in graphics, particularly applications involving meshes.

Not much can be done if you need to frequently modify mesh structure. Luckily, many algorithms operate on meshes with fixed structure. In this case, expressing algorithms in terms of block operations can be much faster. Actually implementing methods in this way can be tricky though.

To gain more experience in this, I decided to try implementing the As-Rigid-As-Possible Surface Modeling mesh deformation algorithm in Python. This involves solving moderately sized Laplace systems where the right-hand-side involves some non-trivial computation (per-vertex SVDs embedded within evaluation of the Laplace operator). For details on the algorithm, see the link.

As it turns out, the entire implementation ended up being just 99 lines and solves 1600 vertex problems in just under 20ms (more on this later). Full code is available.

## Building the Laplace operator

ARAP uses cotangent weights for the Laplace operator in order to be relatively independent of mesh grading. These weights are functions of the angles opposite the edges emanating from each vertex. Building this operator using loops is slow but can be sped up considerably by realizing that every triangle has exactly 3 vertices, 3 angles and 3 edges. Operations can be evaluated on triangles and accumulated to edges.

Here's the Python code for a cotangent Laplace operator:

from typing import *
import numpy as np
import scipy.sparse as sparse
import scipy.sparse.linalg as spla

def build_cotan_laplacian( points: np.ndarray, tris: np.ndarray ):
a,b,c = (tris[:,0],tris[:,1],tris[:,2])
A = np.take( points, a, axis=1 )
B = np.take( points, b, axis=1 )
C = np.take( points, c, axis=1 )

eab,ebc,eca = (B-A, C-B, A-C)
eab = eab/np.linalg.norm(eab,axis=0)[None,:]
ebc = ebc/np.linalg.norm(ebc,axis=0)[None,:]
eca = eca/np.linalg.norm(eca,axis=0)[None,:]

alpha = np.arccos( -np.sum(eca*eab,axis=0) )
beta  = np.arccos( -np.sum(eab*ebc,axis=0) )
gamma = np.arccos( -np.sum(ebc*eca,axis=0) )

wab,wbc,wca = ( 1.0/np.tan(gamma), 1.0/np.tan(alpha), 1.0/np.tan(beta) )
rows = np.concatenate((   a,   b,   a,   b,   b,   c,   b,   c,   c,   a,   c,   a ), axis=0 )
cols = np.concatenate((   a,   b,   b,   a,   b,   c,   c,   b,   c,   a,   a,   c ), axis=0 )
vals = np.concatenate(( wab, wab,-wab,-wab, wbc, wbc,-wbc,-wbc, wca, wca,-wca, -wca), axis=0 )
L = sparse.coo_matrix((vals,(rows,cols)),shape=(points.shape[1],points.shape[1]), dtype=float).tocsc()
return L


Inputs are float $$3 \times N_{verts}$$ and integer $$N_{tris} \times 3$$ arrays for vertices and triangles respectively. The code works by making arrays containing vertex indices and positions for all triangles, then computes and normalizes edges and finally computes angles and cotangents. Once the cotangents are available, initialization arrays for a scipy.sparse.coo_matrix are constructed and the matrix built. The coo_matrix is intended for finite-element matrices so it accumulates rather than overwrites duplicate values. The code is definitely faster and most likely fewer lines than a loop based implementation.

## Evaluating the Right-Hand-side

The Laplace operator is just a warm-up exercise compared to the right-hand-side computation. This involves:

• For every vertex, collecting the edge-vectors in its one-ring in both original and deformed coordinates
• Computing the rotation matrices that optimally aligns these neighborhoods, weighted by the Laplacian edge weights, per-vertex. This involves a SVD per-vertex.
• Rotating the undeformed neighborhoods by the per-vertex rotation and summing the contributions.

In order to vectorize this, I constructed neighbor and weight arrays for the Laplace operator. These store the neighboring vertex indices and the corresponding (positive) weights from the Laplace operator. A catch is that vertices have different numbers of neighbors. To address this I use fixed sized arrays based on the maximum valence in the mesh. These are initialized with default values of the vertex ids and weights of zero so that unused vertices contribute nothing and do not affect matrix structure.

To compute the rotation matrices, I used some index magic to accumulate the weighted neighborhoods and rely on numpy's batch SVD operation to vectorize over all vertices. Finally, I used the same index magic to rotate the neighborhoods and accumulate the right-hand-side.

The code to accumulate the neighbor and weight arrays is:

def build_weights_and_adjacency( points: np.ndarray, tris: np.ndarray, L: Optional[sparse.csc_matrix]=None ):
L = L if L is not None else build_cotan_laplacian( points, tris )
n_pnts, n_nbrs = (points.shape[1], L.getnnz(axis=0).max()-1)
nbrs = np.ones((n_pnts,n_nbrs),dtype=int)*np.arange(n_pnts,dtype=int)[:,None]
wgts = np.zeros((n_pnts,n_nbrs),dtype=float)

for idx,col in enumerate(L):
msk = col.indices != idx
indices = col.indices[msk]
values  = col.data[msk]
nbrs[idx,:len(indices)] = indices
wgts[idx,:len(indices)] = -values


Rather than break down all the other steps individually, here is the code for an ARAP class:

class ARAP:
def __init__( self, points: np.ndarray, tris: np.ndarray, anchors: List[int], anchor_weight: Optional[float]=10.0, L: Optional[sparse.csc_matrix]=None ):
self._pnts    = points.copy()
self._tris    = tris.copy()
self._nbrs, self._wgts, self._L = build_weights_and_adjacency( self._pnts, self._tris, L )

self._anchors = list(anchors)
self._anc_wgt = anchor_weight
E = sparse.dok_matrix((self.n_pnts,self.n_pnts),dtype=float)
for i in anchors:
E[i,i] = 1.0
E = E.tocsc()
self._solver = spla.factorized( ( self._L.T@self._L + self._anc_wgt*E.T@E).tocsc() )

@property
def n_pnts( self ):
return self._pnts.shape[1]

@property
def n_dims( self ):
return self._pnts.shape[0]

def __call__( self, anchors: Dict[int,Tuple[float,float,float]], num_iters: Optional[int]=4 ):
con_rhs = self._build_constraint_rhs(anchors)
R = np.array([np.eye(self.n_dims) for _ in range(self.n_pnts)])
def_points = self._solver( self._L.T@self._build_rhs(R) + self._anc_wgt*con_rhs )
for i in range(num_iters):
R = self._estimate_rotations( def_points.T )
def_points = self._solver( self._L.T@self._build_rhs(R) + self._anc_wgt*con_rhs )
return def_points.T

def _estimate_rotations( self, def_pnts: np.ndarray ):
tru_hood = (np.take( self._pnts, self._nbrs, axis=1 ).transpose((1,0,2)) - self._pnts.T[...,None])*self._wgts[:,None,:]
rot_hood = (np.take( def_pnts,   self._nbrs, axis=1 ).transpose((1,0,2)) - def_pnts.T[...,None])

U,s,Vt = np.linalg.svd( rot_hood@tru_hood.transpose((0,2,1)) )
R = U@Vt
dets = np.linalg.det(R)
Vt[:,self.n_dims-1,:] *= dets[:,None]
R = U@Vt
return R

def _build_rhs( self, rotations: np.ndarray ):
R = (np.take( rotations, self._nbrs, axis=0 )+rotations[:,None])*0.5
tru_hood = (self._pnts.T[...,None]-np.take( self._pnts, self._nbrs, axis=1 ).transpose((1,0,2)))*self._wgts[:,None,:]
rhs = np.sum( (R@tru_hood.transpose((0,2,1))[...,None]).squeeze(), axis=1 )
return rhs

def _build_constraint_rhs( self, anchors: Dict[int,Tuple[float,float,float]] ):
f = np.zeros((self.n_pnts,self.n_dims),dtype=float)
f[self._anchors,:] = np.take( self._pnts, self._anchors, axis=1 ).T
for i,v in anchors.items():
if i not in self._anchors:
raise ValueError('Supplied anchor was not included in list provided at construction!')
f[i,:] = v
return f


The indexing was tricky to figure out. Error handling and input checking can obviously be improved. Something I'm not going into are the handling of anchor vertices and the overall quadratic forms that are being solved.

## So does it Work?

Yes. Here is a 2D equivalent to the bar example from the paper:

import time
import numpy as np
import matplotlib.pyplot as plt

import arap

def grid_mesh_2d( nx, ny, h ):
x,y = np.meshgrid( np.linspace(0.0,(nx-1)*h,nx), np.linspace(0.0,(ny-1)*h,ny))
idx = np.arange(nx*ny,dtype=int).reshape((ny,nx))
quads = np.column_stack(( idx[:-1,:-1].flat, idx[1:,:-1].flat, idx[1:,1:].flat, idx[:-1,1:].flat ))
return np.row_stack((x.flat,y.flat)), tris, idx

nx,ny,h = (21,5,0.1)
pnts, tris, ix = grid_mesh_2d( nx,ny,h )

anchors = {}
for i in range(ny):
anchors[ix[i,0]]    = (       0.0,           i*h)
anchors[ix[i,1]]    = (         h,           i*h)
anchors[ix[i,nx-2]] = (h*nx*0.5-h,  h*nx*0.5+i*h)
anchors[ix[i,nx-1]] = (  h*nx*0.5,  h*nx*0.5+i*h)

deformer = arap.ARAP( pnts, tris, anchors.keys(), anchor_weight=1000 )

start = time.time()
def_pnts = deformer( anchors, num_iters=2 )
end = time.time()

plt.triplot( pnts[0], pnts[1], tris, 'k-', label='original' )
plt.triplot( def_pnts[0], def_pnts[1], tris, 'r-', label='deformed' )
plt.legend()
plt.title('deformed in {:0.2f}ms'.format((end-start)*1000.0))
plt.show()


The result of this script is the following:

This is qualitatively quite close to the result for two iterations in Figure 3 of the ARAP paper. Note that the timings may be a bit unreliable for such a small problem size.

A corresponding result for 3D deformations is here (note that this corrects an earlier bug in rotation computations for 3D):

## Performance

By using SciPy's sparse matrix functions and pre-factoring the systems, the solves are very quick. This is because SciPy uses optimized libraries for the factorization and backsolves. Prefactoring the system matrices for efficiency was a key point of the paper and it's nice to be able to take advantage of this from Python.

To look at the breakdown of time taken for right-hand-side computation and solves, I increased the problem size to 100x25, corresponding to a 2500 vertex mesh. Total time for two iterations was roughly 35ms, of which 5ms were the linear solves and 30ms were the right-hand-side computations. So the right-hand-side computation completely dominates the overall time. That said, using a looping approach, I could not even build the Laplace matrix in 30ms to say nothing of building the right-hand-side three times over, so the vectorization has paid off. More profiling would be needed to say if the most time-consuming portion is the SVDs or the indexing but overall I consider it a success to implement a graphics paper that has been cited more than 700 times in 99 lines of Python.

## Tsai Camera Calibration

Projection of known 3D points into a pinhole camera can be modeled by the following equation:

\begin{align*} \begin{bmatrix} \hat{u}_c \\ \hat{v}_c \\ \hat{w}_c \end{bmatrix} = \begin{bmatrix} f & 0 & u_0 \\ 0 & f & v_0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} r_{xx} & r_{xy} & r_{xz} & t_x \\ r_{yx} & r_{yy} & r_{yz} & t_y \\ r_{zx} & r_{zy} & r_{zz} & t_z \end{bmatrix} \begin{bmatrix} x_w \\ y_w \\ z_w \\ 1 \end{bmatrix} \\ = K E x \end{align*}

The matrix $$K$$ contains the camera intrinsics and the matrix $$E$$ contains the camera extrinsics. The final image coordinates are obtained by perspective division: $$u_c = \hat{u}_c/\hat{w}_c$$ and $$v_c = \hat{v}_c/\hat{w}_c$$.

Estimating $$K$$ and $$E$$ jointly can be difficult because the equations are non-linear and may converge slowly or to a poor solution. Tsai's algorithm provides a way to obtain initial estimates of the parameters for subsequent refinement. It requires a known calibration target that can be either planar or three-dimensional.

The key step is in estimating the rotation and x-y translation of the extrinsics $$E$$ in such a way that the unknown focal length is eliminated. Once the rotation is known, the focal-length and z translation can be computed.

This process requires the optical center $$[u_0, v_0]$$ which is also unknown. However, given reasonable quality optics this is often near the image center so the center can be used as an initial estimate.

This is the starting point for the estimation.

## Estimating (most of) the Extrinsics

This is divided into two cases, one for planar targets and one for 3D targets.

Case 1: 3D Non-planar Target:

For calibration targets with 8+ points that span 3 dimensions. Points in the camera frame prior to projection are:

\begin{equation*} \begin{bmatrix} x_c \\ y_c \\ z_c \end{bmatrix} = \begin{bmatrix} r_{xx} x_w + r_{xy} y_w + r_{xz} z_w + t_x \\ r_{yx} x_w + r_{yy} y_w + r_{yz} z_w + t_y \\ r_{zx} x_w + r_{zy} y_w + r_{zz} z_w + t_z \end{bmatrix} \end{equation*}

The image points relative to the optical center are then:

\begin{equation*} \begin{bmatrix} u_c - u_0 \\ v_c - v_0 \end{bmatrix} = \begin{bmatrix} u_c' \\ v_c' \end{bmatrix} = \begin{bmatrix} (f x_c + u_0 z_c)/z_c - u_0 \\ (f y_c + v_0 z_c)/z_c - v_0 \end{bmatrix} = \begin{bmatrix} f x_c/z_c \\ f y_c/z_c \end{bmatrix} \end{equation*}

The ratio of these two quantities gives the equation:

\begin{equation*} \frac{u_c'}{v_c'} = \frac{x_c}{y_c} \end{equation*}

or, after cross-multiplying and collecting terms to one side:

\begin{align*} v_c' x_c - u_c' y_c = 0 \\ [ v_c' x_w, v_c' y_w, v_c' z_w, -u_c' x_w, -u_c' y_w, -u_c' z_w, 1, -1] \begin{bmatrix} r_{xx} \\ r_{xy} \\ r_{xz} \\ r_{yx} \\ r_{yy} \\ r_{yz} \\ t_x \\ t_y \end{bmatrix} = 0 \end{align*}

Each point produces a single equation like the above which can be stacked to form a homogeneous system. With a zero right-hand-side, scalar multiples of the solution still satisfy the equations and a trivial all-zero solution exists. Calling the system matrix $$M$$, the solution to this equation system is the eigenvector associated with the smallest eigenvalue of $$M^T M$$. The computed solution will be proportional to the desired solution. It is also possible to fix one value, e.g. $$t_y=1$$ and solve an inhomogeneous system but this may give problems if $$t_y \neq 0$$ for the camera configuration.

The recovered $$r_x = [ r_{xx}, r_{xy}, r_{xz} ]$$ and $$r_x = [ r_{yx}, r_{yy}, r_{yz} ]$$ should be unit length. The scale of the solution can be recovered as the mean length of these vectors and used to inversely scale the translations $$t_x, t_y$$.

Case 2: 2D Planar Targets:

For 2D targets, arbitrarily set $$z_w=0$$, leading to:

\begin{equation*} \begin{bmatrix} x_c \\ y_c \\ z_c \end{bmatrix} = \begin{bmatrix} r_{xx} x_w + r_{xy} y_w + t_x \\ r_{yx} x_w + r_{yy} y_w + t_y \\ r_{zx} x_w + r_{zy} y_w + t_z \end{bmatrix} \end{equation*}

Use the same process as before, a homogeneous system can be formed. It is identical to the previous one, except the $$r_{xz}$$ and $$r_{yz}$$ terms are dropped since $$z_w=0$$. Each equation ends up being:

\begin{equation*} [ v_c' x_w, v_c' y_w, -u_c' x_w, -u_c' y_w, 1, -1] \begin{bmatrix} r_{xx} \\ r_{xy} \\ r_{yx} \\ r_{yy} \\ t_x \\ t_y \end{bmatrix} = 0 \end{equation*}

Solving this gives $$r_{xx}, r_{xy}, r_{zx}, r_{zy}, t_x, t_y$$ up to a scalar $$k$$. It does not provide $$r_{xz}, r_{yz}$$. In addition, we know that the rows of the rotation submatrix are unit-norm and orthogonal, giving three equations:

\begin{align*} r_{xx}^2 + r_{xy}^2 + r_{xz}^2 = k^2 \\ r_{yz}^2 + r_{yy}^2 + r_{yz}^2 = k^2 \\ r_{xx} r_{yx} + r_{xy} r_{yy} + r_{xz} r_{yz} = 0 \end{align*}

After some algebraic manipulations the factor $$k^2$$ can be found as:

\begin{equation*} k^2 = \frac{1}{2}\left( r_{xx}^2 + r_{xy}^2 + r_{yx}^2 + r_{yy}^2 + \sqrt{\left( (r_{xx} - r_{yy})^2 + (r_{xy} + r_{yx})^2\right)\left( (r_{xx}+r_{yy})^2 + (r_{xy} - r_{yx})^2 \right) } \right) \end{equation*}

This allows $$r_{xz}, r_{yz}$$ to be found as:

\begin{align*} r_{xz} = \pm\sqrt{ k^2 - r_{xx}^2 - r_{xy}^2 } \\ r_{yz} = \pm\sqrt{ k^2 - r_{yz}^2 - r_{yy}^2 } \end{align*}

## Estimate Remaining Intrinsics and Focal Length

Estimating the focal length and z-translation follows a similar process. Using the definitions above we have:

\begin{align*} u_c' = f \frac{x_c}{z_c} \\ v_c' = f \frac{y_c}{z_c} \end{align*}

Cross-multiplying and collecting gives:

\begin{align*} (r_{xx} x_w + r_{xy} y_w + r_{xz} z_w + t_x) f - u_c' t_z = u_c' (r_{zx} x_w + r_{zy} y_w + r_{zz} z_w) \\ (r_{yx} x_w + r_{yy} y_w + r_{yz} z_w + t_y) f - v_c' t_z = v_c' (r_{zx} x_w + r_{zy} y_w + r_{zz} z_w) \end{align*}

Solving this system gives the desired $$f, t_x$$ values.

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