James Gregson's Website

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.

Reordering Matrix Products

Here are a few operations for dealing with objective functions of matrix-valued variables:

Starting Definitions

Let $$A$$ be a $$N\times M$$ matrix with entry $$A_{ij}$$ being the entry at row $$i$$ and column $$j$$. The $$i$$'th row of $$A$$ is then $$A_{i*}$$ and the $$j$$'th column is $$A_{*j}$$.

Row and Column Vectorization

Define the column vectorization operator which returns a single column vector containing the columns of $$A$$ stacked

\begin{equation*} \newcommand{\cvec}{{\bf\mbox{cvec}}} \cvec(A) = \begin{bmatrix} A_{*1} \\ A_{*2} \\ \vdots \\ A_{*M} \end{bmatrix} \end{equation*}

likewise the row vectorization operator returns a single column vector containing the (transposes of) rows of $$A$$ stacked:

\begin{equation*} \newcommand{\rvec}{{\bf\mbox{rvec}}} \rvec(A) = \cvec(A^T) = \begin{bmatrix} A_{1*}^T \\ A_{2*}^T \\ \vdots \\ A_{N*} \end{bmatrix} \end{equation*}

In numpy, matrices are stored in row-major order, so the Python definition of these operaions is inverted as below:

def cvec( A ):
"""Stack columns of A into single column vector"""
return rvec(A.T)

def rvec( A ):
"""Stack rows of A into a single column vector"""
return A.ravel()


Similarly, the inverses of $$\cvec(A)$$ and $$\rvec(A)$$ unpack the vectorized values back into a matrix of the original shape:


with corresponding python definitions:

def rmat( v, nr ):
"""Reshape vector v into matrix with nr rows"""
return v.reshape((nr,-1))

def cmat( v, nc ):
"""Reshape vector v into matrix with nc columns"""
return v.reshape((cols,-1)).T


Finally, define two additional operators. The first of these is the spread operator, which takes the Kronecker product between an identity matrix and the input matrix, resulting in copies of the input matrix along the diagonal:

\begin{equation*} \newcommand{\spread}[2]{{\bf\mbox{spread}}_{#2}\left(#1\right)} \spread{A}{r} = I_{r\times r} \otimes A = \begin{bmatrix} A & 0 & \dots & 0 \\ 0 & A & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & A \end{bmatrix} \end{equation*}

The second of these is the split operator which reverses the order of the arguments to the Kronecker product, basically replacing each entry $$A_{ij}$$ with $$A_{i,j} I_{r\times r}$$:

\begin{equation*} \newcommand{\split}[2]{{\bf\mbox{split}}_{#2}\left(#1\right)} \split{A}{r} = A \otimes I_{r\times r} = \begin{bmatrix} A_{11} I_{r\times r} & A_{12} I_{r\times r} & \dots & A_{1M} I_{r\times r} \\ A_{21} I_{r\times r} & A_{22} I_{r\times r} & \dots & A_{2M} I_{r\times r} \\ \vdots & \vdots & \ddots & \vdots \\ A_{N1} I_{r\times r} & A_{N2} I_{r\times r} & \dots & A_{NM} I_{r\times r} \end{bmatrix} \end{equation*}

Python implementation of these are one-liners that just call the existing numpy Kronecker product:

def spread( M, n ):
"""Make n copies of M along the matrix diagonal"""
return np.kron( np.eye(n), M )

def split( M, n ):
"""Replace each entry of M with a n-by-n identity matrix scaled by the entry value"""
return np.kron( M, np.eye(n) )


With these, it is possible to define matrix products. There are also some fun properties that come from the Kronecker product:

Matrix Products

Let $$X$$ be a $$N\times K$$ matrix and $$Y$$ be a $$K\times M$$ matrix, then their product $$W = XY$$ will be a $$N\times M$$ matrix and the following are true:

\begin{align*} \rvec(W) &=& \split{X}{M} \rvec(Y) \\ \cvec(W) &=& \spread{X}{M} \cvec(Y) \\ \rvec(W) &=& \spread{Y^T}{N} \rvec(X) \\ \cvec(W) &=& \split{Y^T}{N} \cvec(X) \end{align*}

What these let you do is unbury any (single) term that's in the middle of some horrendous expression. For example, say there's an expression $$W = AXB$$ with $$A \in \mathcal{R}^{N\times R}$$, $$X \in \mathcal{R}^{R\times S}$$ and $$B \in \mathcal{R}^{S \times M}$$ and you want to isolate $$X$$ as a column vector. Then you have:

\begin{align*} \cvec(W) &=& \spread{A}{M} \split{B^T}{R} \cvec(X) \\ &=& \left( I_{M\times M} \otimes A \right) \left( B^T \otimes I_{R\times R} \right) \cvec(X) \\ &=& \left( I_{M\times M} B^T \right) \otimes \left( A I_{R\times R} \right) \cvec(X) \\ &=& \left( B^T \otimes A \right) \cvec(X) \end{align*}

The step in the third line is the Kronecker product identity:

\begin{equation*} \left( A \otimes B \right) \left( C \otimes D \right ) = \left( A C \right) \otimes \left( B D \right) \end{equation*}

The same steps can also be done if you want row-major storage of $$X$$:

\begin{align*} \rvec(W) &=& \split{A}{M} \spread{B^T}{R} \rvec(X) \\ &=& \left( A \otimes I_{M\times M} \right) \left( I_{R\times R} \otimes B^T \right) \rvec(X) \\ &=& \left( A \otimes B^T \right) \rvec(X) \end{align*}

In these cases the dimensions work out correctly but you don't need to worry too much about them since the dimensions of $$X$$ are constrained by the column could in $$A$$ and the row count in $$B$$. This only applies when operating with them symbolically of course.

Frobenius Norm Problems

With this information, it becomes pretty easy to deal with Frobenius norm problems like:

\begin{equation*} X = \mbox{argmin} \frac{1}{2} \| A X B - Y \|_F^2 \end{equation*}

I have found these frustrating in the past since I invariably end up in indexing hell. But it becomes much easier with the identities above. Applying them first of all isolates the X term to the right-hand-side of the expression and also converts the Frobenius norm to an $$L_2$$ norm. Making the substitutions $$\tilde{x} = \cvec(X)$$ and $$\tilde{y} = \cvec(y)$$,

\begin{align*} \tilde{x}_* &=& \mbox{argmin} \frac{1}{2} \| \left( B^T \otimes A \right) \tilde{x} - \tilde{y} \|_2^2 \\ &=& \mbox{argmin} \frac{1}{2} \tilde{x}^T \left( B \otimes A^T \right) \left( B^T \otimes A \right) \tilde{x} - \tilde{x}^T \left( B \otimes A^T \right) \tilde{y} + \mbox{constants} \\ &=& \mbox{argmin} \frac{1}{2} \tilde{x}^T \left( B B^T \otimes A^T A \right) \tilde{x} - \tilde{x}^T \left( B \otimes A^T \right) \tilde{y} \end{align*}

This now looks like a standard least-squares problem, albeit one with this Kronecker product mashed into it. Setting the gradient to zero gives the solution as:

\begin{align*} \tilde{x}_* &=& \left( B B^T \otimes A^T A \right)^{-1} \left( B \otimes A^T \right ) \tilde{y} \\ &=& \left( (B B^T)^{-1} \otimes (A^T A)^{-1} \right) \left( B \otimes A^T \right) \tilde{y} \\ &=& \left( (B B^T)^{-1} B \otimes (A^T A)^{-1} A^T \right) \tilde{y} \end{align*}

This result has a similar form to $$AXB = \left( B^T \otimes A\right)\cvec(X)$$, which can be used to get back to the matrix form of the expression:

\begin{align*} X &=& (A^T A)^{-1} A^T Y \left( (B B^T)^{-1} B\right)^T \\ &=& (A^T A)^{-1} A^T Y B^T (B B^T)^{-1} \end{align*}

It's also possible to come at this from a different direction

\begin{align*} X &=& \mbox{argmin} \frac{1}{2} \| A X B - Y \|_F^2 \\ &=& \mbox{argmin} \frac{1}{2} \mbox{Tr}\left( (AXB - Y)(AXB - Y)^T \right) \end{align*}

The gradient of this is given in the Matrix Cookbook.

\begin{equation*} \frac{\partial}{\partial X} \frac{1}{2} \mbox{Tr}\left( (AXB - Y)(AXB - Y)^T \right) = A^T \left( A X B - Y \right) B^T \end{equation*}

Setting the gradient of this to zero and isolating $$X$$ gives the following:

\begin{align*} A^T A X B B^T - A^T Y B^T = 0 \\ X = (A^T A)^{-1} A^T Y B^T (B B^T)^{-1} \end{align*}

which matches the previous result.

Nov 10, 2018

\begin{equation*} \newcommand{pd}[2]{\frac{\partial #1}{\partial #2}} \end{equation*}

Let $$x = \left[ x_1, x_2, x_3 \right]^T$$ and $$y_1 = f(x)$$ be a scalar function. Then $$\nabla y_1$$ is:

\begin{equation*} \nabla y_1 = \left[ \pd{f}{x_1}, \pd{f}{x_2}, \pd{f}{x_3} \right] \end{equation*}

i.e., a row-vector. This allows it to be compatible with the corresponding definition of the Jacobian when, i.e, when

\begin{equation*} y = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} f_1(x) \\ f_2(x) \end{bmatrix} \end{equation*}

then the Jacobian of $$y$$ is:

\begin{equation*} J_y = \begin{bmatrix} \pd{f_1}{x_1} & \pd{f_1}{x_2} & \pd{f_1}{x_3} \\ \pd{f_2}{x_1} & \pd{f_2}{x_2} & \pd{f_2}{x_3} \end{bmatrix} = \begin{bmatrix} \nabla f_1 \\ \nabla f_2 \end{bmatrix} \end{equation*}

This also allows linearization of the scalar or vector function to be consistent. Here $$\delta x$$

\begin{align*} y_1(x+\delta x) \approx y_1(x) + \nabla y_1 \delta x \\ y (x+\delta x) \approx y(x) + J_y \delta x \end{align*}

That's great, but where it is useful is in differentiating more complex expressions. For example in the context of optimization or regression $$E = \frac{1}{2} \| y \|^2$$. $$E$$ is a scalar function of a vector function of a vector and it's a pain to differentiate this symbolically with respect to the parameters $$x$$.

What helps here is this:

\begin{equation*} \pd{E}{x} = \pd{E}{y} \pd{y}{x} = y^T J_y \end{equation*}

The transpose on the $$y$$ factor is included because it it fits the definition that the gradient of a scalar function taking a vector argument is a row-vector. The above expression expands to:

\begin{equation*} \pd{E}{x} = \left[ y_1, y_2 \right] \begin{bmatrix} \pd{f_1}{x_1} & \pd{f_1}{x_2} & \pd{f_1}{x_3} \\ \pd{f_2}{x_1} & \pd{f_2}{x_2} & \pd{f_2}{x_3} \end{bmatrix} = \begin{bmatrix} y_1 \nabla y_1 + y_2 \nabla y_2 \end{bmatrix} \end{equation*}

which shows that the terms involving $$y_1$$ and $$y_2$$ are correctly decoupled and only combine through the summation implied by the 2-norm.

What about when $$y = A x - b$$? Well, it's pretty easy to substitute in:

\begin{align*} \pd{E}{y} = y^T \\ \pd{y}{x} = A \\ \pd{E}{x} = (A x - b)^T A \end{align*}

Wait! What's up with that? Isn't it supposed to be $$A^T(A x -b)$$? Well, yes, but that's when gradients are column vectors. Which they're not here. And if you transpose the row vector result you get the expected column vector result.

\begin{equation*} \pd{E}{x}^T = \left( (A x - b)^T A \right)^T = A^T (A x -b) \end{equation*}

So what happens if you decide to make gradients column vectors? Well, for starters, you have to redefine the del operator from $$\nabla = \left[ \pd{}{x_1}, \pd{}{x_2}, \dots, \pd{}{x_N} \right]$$ to:

\begin{equation*} \nabla = \begin{bmatrix} \pd{}{x_1} \\ \pd{}{x_2} \\ \vdots \\ \pd{}{x_3} \end{bmatrix} \end{equation*}

Redefining fundamental mathematical things should be the first clue this is wrong but pressing on using the same definitions, you get:

\begin{equation*} \nabla y_1 = \begin{bmatrix} \pd{y_1}{x_1} \\ \pd{y_1}{x_2} \\ \pd{y_1}{x_3} \end{bmatrix} \end{equation*}

Stacking these gradients column-wise gives something resembling the Jacobian but which is actually its transpose:

\begin{equation*} \pd{y}{x} = J_y^T = \begin{bmatrix} \pd{y_1}{x_1} & \pd{y_2}{x_1} \\ \pd{y_1}{x_2} & \pd{y_2}{x_2} \\ \pd{y_1}{x_3} & \pd{y_2}{x_3} \end{bmatrix} \end{equation*}

The chain rule then gives:

\begin{equation*} \pd{E}{x} = \pd{E}{y} \pd{y}{x} \end{equation*}

which leads to total garbage since the array dimensions are not compatible for matrix multiplication anymore.

\begin{equation*} \pd{E}{x} \neq y J^T = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \begin{bmatrix} \pd{y_1}{x_1} & \pd{y_2}{x_1} \\ \pd{y_1}{x_2} & \pd{y_2}{x_2} \\ \pd{y_1}{x_3} & \pd{y_2}{x_3} \end{bmatrix} \hspace{0.25cm}{\bf\mbox{Incompatible!}} \end{equation*}

But this can be fixed by changing the order of the chain rule from $$\pd{E}{x} = \pd{E}{y} \pd{y}{x}$$ to $$\pd{E}{x} = \pd{y}{x} \pd{E}{y}$$ to get:

\begin{equation*} \pd{E}{x} = J^T y \end{equation*}

which for the example of $$y = A x - b$$ gives the expected column oriented gradient of $$\nabla E = A^T (A x - b)$$. The difference here is that matrix composition left multiplies new transformations onto an existing one while the chain rule is typically written with right multiplication. This is really the only difference between the column-oriented and row-oriented gradients, other than the discrepancy between the definitions of the gradient and Jacobian matrix. When the Jacobian and gradient are defined correctly (i.e. row-oriented) this does not occur.

What this ends up meaning is that in order for the results to work out correctly with column-oriented gradients you have to carry the Jacobian transpose throughout your work and remember to re-order the chain rule, otherwise you need to forensically reconstruct what the dimensions should be from the scalar terms of your objective function (in optimization anway). I feel this is too high a price to pay.

On the other hand, defining points in one domain as column-vectors and gradients as row-vectors means that expressions involving both need some added boilerplate, e.g.:

\begin{equation*} v = x - \delta \nabla y \hspace{0.25cm}{\bf\mbox{Incompatible!}} \end{equation*}

with $$\delta$$ a scalar no longer have compatible dimensions. Instead the equation above would need to be expressed as:

\begin{equation*} v = x - \delta (\nabla y)^T \end{equation*}

This is awkward but tends not to be a big problem since many programming langages have good array broadcasting that allows the transpose to be omitted when programming.

Some Random Example

Let:

\begin{equation*} A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix} \end{equation*}

and:

\begin{equation*} B = \begin{bmatrix} b_{11} & b_{12} & b_{13} & b_{14} \\ b_{21} & b_{22} & b_{23} & b_{24} \end{bmatrix} \end{equation*}

and let $$A$$ & $$B$$ be functions of $$\theta \in \mathcal{R}^5$$. Now try to minimize the Frobenius norm of $$Y = AB$$:

\begin{equation*} F(\theta) = \frac{1}{2} \| Y \|_F^2 = \mbox{Tr}\left(Y Y^T\right) \end{equation*}

$$F(\theta)$$ is a scalar and $$\theta$$ is a vector so the gradient should be a row-vector of size $$1\times5$$ in the end. Starting with the chain rule an issue is evident:

\begin{equation*} \nabla F(\theta) = Y \pd{Y}{\theta} \end{equation*}

$$Y$$ is the wrong shape $$3\times4$$ and $$\pd{Y}{\theta}$$ is a $$3 \times 4 \times 5$$ array where the final index (k) holds $$\pd{Y}{\theta_k}$$. In order to make this work, we can concatenate the rows $$y_{i*}$$ of $$Y$$ to get a $$1\times 12$$ matrix $$\tilde{Y}$$ i.e.

\begin{equation*} \tilde{Y} = \begin{bmatrix} y_{1*} & y_{2*} & y_{3*} \end{bmatrix} = \mbox{vec}(Y^T)^T \end{equation*}

where $$\mbox{vec}(P)$$ stacks the columns of $$P$$ into a column vector. We can likewise stack the columns $$y_{*j}$$ of $$Y$$ to get a $$12x1$$ column vector $$\hat{Y}=\mbox{vec}(Y)$$. The derivative of this with respect to $$\theta$$ is just the Jacobian of each column of $$Y$$, stacked row-wise:

\begin{equation*} \pd{\hat{Y}}{\theta} = \begin{bmatrix} \pd{y_{*1}}{\theta} \\ \pd{y_{*2}}{\theta} \\ \pd{y_{*3}}{\theta} \\ \pd{y_{*4}}{\theta} \end{bmatrix} \end{equation*}

By doing this, the sum over all matrix entries implied by the Frobenius norm is absorbed as an inner product of $$\tilde{Y} \in \mathcal{R}^{1\times12}$$ and $$\pd{\hat{Y}}{\theta} \in \mathcal{R}^{12\times5}$$. This gives the correct dimensions for the gradient of $$1\times 5$$. Note that everything works pretty cleanly by adopting the row-vector form for gradients.

We still need to compute the $$\pd{Y_{*i}}{\theta}$$ terms. Starting from the definition of $$Y = AB$$ it's pretty easy to see that $$y_{*i} = A b_{*i}$$. We want to get $$\pd{y_{*i}}{\theta} \in \mathcal{R}^{3\times5}$$ and so go to the chain rule:

\begin{equation*} \pd{y_{*i}}{\theta} = A \pd{b_{*i}}{\theta} + \pd{A}{\theta} b_{*i} \end{equation*}

The first term is no problem, it is simply a transform ($$A$$) applied to the Jacobian of $$b_{*i}$$. The shapes work out correctly. The second term is more problematic, $$\pd{A}{\theta}$$ is a $$3\times 2 \times 5$$ array while $$b_{*i}$$ is $$2 \times 1$$. One way to address this is to stack rows of $$A$$ into a column vector using the vectorization operator and splay $$b_{*i}$$ using the Kronecker Product:

\begin{equation*} A b_{*i} = \begin{bmatrix} b_{*i}^T & {\bf 0} & {\bf 0} \\ {\bf 0} & b_{*i}^T & {\bf 0} \\ {\bf 0} & {\bf 0} & b_{*i}^T \end{bmatrix} \begin{bmatrix} a_{1*}^T \\ a_{2*}^T \\ a_{3*}^T \end{bmatrix} = \left( {\bf I_{3\times3}} \otimes b_{*i}^T \right) \mbox{vec}(A^T) \end{equation*}

Rotation Matrix Euler Angles

This page describes a basic approach to extracting Euler angles from rotation matrices. This follows what OpenCV does, (which is where I got the process) but I also wanted to understand why this approach was used.

Here I'm considering rotation performed first around x-axis ($$\alpha$$), then y-axis ($$\beta$$) and finally z-axis ($$\gamma$$). For this, the rotation matrix is shown below. Other rotation orders are analogous.

\begin{equation*} R(\alpha,\beta,\gamma) = \left[\begin{matrix}\cos{\left (\beta \right )} \cos{\left (\gamma \right )} & \sin{\left (\alpha \right )} \sin{\left (\beta \right )} \cos{\left (\gamma \right )} - \sin{\left (\gamma \right )} \cos{\left (\alpha \right )} & \sin{\left (\alpha \right )} \sin{\left (\gamma \right )} + \sin{\left (\beta \right )} \cos{\left (\alpha \right )} \cos{\left (\gamma \right )} & 0\\\sin{\left (\gamma \right )} \cos{\left (\beta \right )} & \sin{\left (\alpha\right )} \sin{\left (\beta \right )} \sin{\left (\gamma \right )} + \cos{\left (\alpha \right )} \cos{\left (\gamma \right )} & - \sin{\left (\alpha \right )} \cos{\left (\gamma \right )} + \sin{\left (\beta \right )} \sin{\left (\gamma \right )} \cos{\left (\alpha \right )} & 0\\- \sin{\left (\beta \right )} & \sin{\left (\alpha \right )} \cos{\left (\beta \right )} & \cos{\left (\alpha \right )} \cos{\left (\beta \right )} & 0\\0 & 0 & 0 & 1\end{matrix}\right] \end{equation*}

To solve for the Euler angles, we can use a little bit of trigonometry. The best way, in my opinion, is to find pairs of entries that are each products of two factors. Using the identity $$\sin^2 + \cos^2 = 1$$ can be used to isolate specific angles, e.g. $$R_{0,0}$$ and $$R_{0,1}$$ can be used to find $$\cos(\beta)$$ which unlocks the rest of the angles:

\begin{align*} (-1)^k \cos(\beta + k \pi) = \sqrt{ \cos^2(\beta) ( \sin^2(\gamma) + \cos^2(\gamma)) } = \sqrt{ R_{0,0}^2 + R_{1,0}^2 } \\ \cos(\beta) \approx \sqrt{ R_{0,0}^2 + R_{1,0}^2 } \end{align*}

Pay attention to the first line: For any true value of $$\beta$$, this formula will return the same value for $$\beta + k\pi$$ where $$k$$ is an arbitrary integer. This will in turn determine the angles $$\alpha$$ and $$\gamma$$ in order to be compatible. The approximate equality will return $$\beta \in [-\pi/2, \pi/2]$$.

Once $$\cos(\beta)$$ is found we can find the angles directly using atan2:

\begin{align*} \beta = \mbox{atan2}( -R_{2,0}, \cos(\beta) ) \\ \alpha = \mbox{atan2}\left( \frac{R_{2,1}}{\cos(\beta)}, \frac{R_{2,2}}{\cos(\beta)} \right) \\ \gamma = \mbox{atan2}\left( \frac{R_{1,0}}{\cos(\beta)}, \frac{R_{0,0}}{\cos(\beta)} \right) \end{align*}

The other issue occurs when $$|\cos(\beta)| = 0$$, which causes division by zero in the equations for $$\alpha,\gamma$$ ($$\beta$$ is still well defined). In this case there are a number of options. What OpenCV does is to arbitrarily decide that $$\gamma = 0$$, which means that $$\sin(\gamma) = 0$$ and $$\cos(\gamma) = 1$$. The formulas in this case are:

\begin{align*} \beta = \mbox{atan2}( -R_{2,0}, \cos(\beta) ) \\ \alpha = \mbox{atan2}( -R_{1,2}, R_{1,1} ) \\ \gamma = 0 \end{align*}

There are, of course, other options. Python code implementing these operations is below, along with unit testing for the degenerate and non-degerate cases.

import unittest
import numpy

def x_rotation( theta ):
"""Generate a 4x4 homogeneous rotation matrix about x-axis"""
c = numpy.cos(theta)
s = numpy.sin(theta)
return numpy.array([
[ 1, 0, 0, 0],
[ 0, c,-s, 0],
[ 0, s, c, 0],
[ 0, 0, 0, 1]
])

def y_rotation( theta ):
"""Generate a 4x4 homogeneous rotation matrix about y-axis"""
c = numpy.cos(theta)
s = numpy.sin(theta)
return numpy.array([
[ c, 0, s, 0],
[ 0, 1, 0, 0],
[-s, 0, c, 0],
[ 0, 0, 0, 1]
])

def z_rotation( theta ):
"""Generate a 4x4 homogeneous rotation matrix about z-axis"""
c = numpy.cos(theta)
s = numpy.sin(theta)
return numpy.array([
[ c,-s, 0, 0],
[ s, c, 0, 0],
[ 0, 0, 1, 0],
[ 0, 0, 0, 1]
])

def xyz_rotation( angles ):
"""Generate 4x4 homogeneous rotation in x then y then z order"""
return numpy.dot( z_rotation(angles[2]), numpy.dot( y_rotation(angles[1]), x_rotation(angles[0]) ) )

def xyz_rotation_angles( R, eps=1e-8, check=False ):
"""Back out the Euler angles that would lead to the given matrix"""

if check and numpy.linalg.norm( numpy.dot( R.transpose(), R ) - numpy.eye(4) ) > eps:
raise ValueError('Input matrix is not a pure rotation, R\'R != I')

cos_beta = numpy.sqrt( R[0,0]**2 + R[1,0]**2 )
if numpy.abs(cos_beta) > eps:
alpha = numpy.arctan2( R[2,1]/cos_beta, R[2,2]/cos_beta )
beta  = numpy.arctan2(-R[2,0], cos_beta )
gamma = numpy.arctan2( R[1,0]/cos_beta, R[0,0]/cos_beta )
return numpy.array((alpha,beta,gamma))
else:
alpha = numpy.arctan2(-R[1,2], R[1,1] )
beta  = numpy.arctan2(-R[2,0], cos_beta )
gamma = 0
return numpy.array((alpha,beta,gamma))

class test_euler_angles(unittest.TestCase):
def test_angles(self):
"""Do fuzz testing on arbitrary rotations, limiting beta to valid range"""
for i in range(1000):
alpha = (numpy.random.rand()-0.5)*numpy.pi*2
beta  = (numpy.random.rand()-0.5)*numpy.pi*0.9999
gamma = (numpy.random.rand()-0.5)*numpy.pi*2

ang = xyz_rotation_angles( xyz_rotation([alpha,beta,gamma]))
self.assertAlmostEqual( alpha, ang[0] )
self.assertAlmostEqual( beta,  ang[1] )
self.assertAlmostEqual( gamma, ang[2] )

def test_degeneracies(self):
"""Do fuzz testing on the degenerate condition of beta = +- pi/2"""
for i in range(1000):
alpha = (numpy.random.rand()-0.5)*numpy.pi*2
beta  = numpy.sign(numpy.random.randn())*numpy.pi/2
gamma = (numpy.random.rand()-0.5)*numpy.pi*2
R     = xyz_rotation((alpha,beta,gamma))
ang   = xyz_rotation_angles( R )
R2    = xyz_rotation( ang )
# check that beta is recovered, gamma is set to zero and
# the rotation matrices match to high precision
self.assertAlmostEqual( beta, ang[1] )
self.assertAlmostEqual( ang[2], 0 )
for j in range(16):
self.assertAlmostEqual( R.ravel()[j], R2.ravel()[j] )

if __name__ == '__main__':
unittest.main()


Transformation Matrix Jacobians

Starting with the $$4\times4$$ homogeneous transform matrix with parameters $$\beta = [ \theta_x, \theta_y, \theta_z, t_x, t_y, t_z ]^T$$ where rotations are performed in XYZ order and using the following substitutions:

\begin{align*} c_x = \cos(\theta_x) \\ s_x = \sin(\theta_x) \\ c_y = \cos(\theta_y) \\ s_y = \sin(\theta_y) \\ c_z = \cos(\theta_z) \\ s_z = \sin(\theta_z) \\ \end{align*}

the vector function mapping a point $$p = [p_x, p_y, p_z, 1]^T$$ in the body coordinate system to a point in the world coordinate system $$w = [w_x, w_y, w_z, 1]^T$$ is:

\begin{equation*} \begin{bmatrix} w_x \\ w_y \\ w_z \\ 1 \end{bmatrix} = F( p, \beta ) = \left[\begin{matrix}c_{y} c_{z} & - c_{x} s_{z} + c_{z} s_{x} s_{y} & c_{x} c_{z} s_{y} + s_{x} s_{z} & t_{x}\\c_{y} s_{z} & c_{x} c_{z} + s_{x} s_{y} s_{z} & c_{x} s_{y} s_{z} - c_{z} s_{x} & t_{y}\\- s_{y} & c_{y} s_{x} & c_{x} c_{y} & t_{z}\\0 & 0 & 0 & 1\end{matrix}\right]\begin{bmatrix} p_x \\ p_y \\ p_z \\ 1 \end{bmatrix} \end{equation*}

and the Jacobian with respect to the parameters $$\beta$$ is:

\begin{equation*} J_F = \left[\begin{matrix}p_{y} \left(c_{x} c_{z} s_{y} + s_{x} s_{z}\right) + p_{z} \left(c_{x} s_{z} - c_{z} s_{x} s_{y}\right) & c_{x} c_{y} c_{z} p_{z} + c_{y} c_{z} p_{y} s_{x} - c_{z}p_{x} s_{y} & - c_{y} p_{x} s_{z} + p_{y} \left(- c_{x} c_{z} - s_{x} s_{y} s_{z}\right) + p_{z} \left(- c_{x} s_{y} s_{z} + c_{z} s_{x}\right) & 1 & 0 & 0\\p_{y} \left(c_{x} s_{y} s_{z} - c_{z} s_{x}\right) + p_{z} \left(- c_{x} c_{z} - s_{x} s_{y} s_{z}\right) & c_{x} c_{y} p_{z} s_{z} + c_{y} p_{y} s_{x} s_{z} - p_{x} s_{y} s_{z} & c_{y} c_{z} p_{x} + p_{y} \left(- c_{x} s_{z} + c_{z} s_{x} s_{y}\right) + p_{z} \left(c_{x} c_{z} s_{y} + s_{x} s_{z}\right) & 0 & 1 & 0\\c_{x} c_{y} p_{y} - c_{y} p_{z} s_{x} & - c_{x} p_{z} s_{y} - c_{y} p_{x} -p_{y} s_{x} s_{y} & 0 & 0 & 0 & 1\\0 & 0 & 0 & 0 & 0 & 0\end{matrix}\right] \end{equation*}

Python code for these respective operations is below:

def make_transform( beta ):
c_x = numpy.cos(beta[0])
s_x = numpy.sin(beta[0])
c_y = numpy.cos(beta[1])
s_y = numpy.sin(beta[1])
c_z = numpy.cos(beta[2])
s_z = numpy.sin(beta[2])
t_x = beta[3]
t_y = beta[4]
t_z = beta[5]
return numpy.array([
[c_y*c_z, -c_x*s_z + c_z*s_x*s_y, c_x*c_z*s_y + s_x*s_z, t_x],
[c_y*s_z, c_x*c_z + s_x*s_y*s_z, c_x*s_y*s_z - c_z*s_x, t_y],
[-s_y, c_y*s_x, c_x*c_y, t_z],
[0, 0, 0, 1]
])

def make_transform_jacobian( beta, p ):
c_x = numpy.cos(beta[0])
s_x = numpy.sin(beta[0])
c_y = numpy.cos(beta[1])
s_y = numpy.sin(beta[1])
c_z = numpy.cos(beta[2])
s_z = numpy.sin(beta[2])
t_x = beta[3]
t_y = beta[4]
t_z = beta[5]
p_x = p[0]
p_y = p[1]
p_z = p[2]
return numpy.array([
[p_y*(c_x*c_z*s_y + s_x*s_z) + p_z*(c_x*s_z - c_z*s_x*s_y), c_x*c_y*c_z*p_z + c_y*c_z*p_y*s_x - c_z*p_x*s_y, -c_y*p_x*s_z + p_y*(-c_x*c_z - s_x*s_y*s_z) + p_z*(-c_x*s_y*s_z + c_z*s_x), 1, 0, 0],
[p_y*(c_x*s_y*s_z - c_z*s_x) + p_z*(-c_x*c_z - s_x*s_y*s_z), c_x*c_y*p_z*s_z + c_y*p_y*s_x*s_z - p_x*s_y*s_z, c_y*c_z*p_x + p_y*(-c_x*s_z + c_z*s_x*s_y) + p_z*(c_x*c_z*s_y + s_x*s_z), 0, 1, 0],
[c_x*c_y*p_y - c_y*p_z*s_x, -c_x*p_z*s_y - c_y*p_x - p_y*s_x*s_y, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0]
])


I generated these using sympy to build the transformations and used a finite-difference Jacobian function to check the output.