James Gregson's Website

Nov 03, 2019

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 ))
    tris  = np.vstack((quads[:,(0,1,2)],quads[:,(0,2,3)]))
    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:

As Rigid As Possible deformation

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):

As Rigid As Possible deformation in 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.

Mar 09, 2019

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.

Feb 09, 2019

Extract Image Patches in Python

There are a number of imaging algorithms that work with patch representations. Non-local means, dictionary learning and Gaussian Mixture Models are some examples.

In order to use these, you need to be able to extract image patches and reconstruct images efficiently. There are probably faster methods but the following does the trick for me without much in the way of dependencies:

import numpy

def extract_grayscale_patches( img, shape, offset=(0,0), stride=(1,1) ):
    """Extracts (typically) overlapping regular patches from a grayscale image

    Changing the offset and stride parameters will result in images
    reconstructed by reconstruct_from_grayscale_patches having different
    dimensions! Callers should pad and unpad as necessary!

    Args:
        img (HxW ndarray): input image from which to extract patches

        shape (2-element arraylike): shape of that patches as (h,w)

        offset (2-element arraylike): offset of the initial point as (y,x)

        stride (2-element arraylike): vertical and horizontal strides

    Returns:
        patches (ndarray): output image patches as (N,shape[0],shape[1]) array

        origin (2-tuple): array of top and array of left coordinates
    """
    px, py = numpy.meshgrid( numpy.arange(shape[1]),numpy.arange(shape[0]))
    l, t = numpy.meshgrid(
        numpy.arange(offset[1],img.shape[1]-shape[1]+1,stride[1]),
        numpy.arange(offset[0],img.shape[0]-shape[0]+1,stride[0]) )
    l = l.ravel()
    t = t.ravel()
    x = numpy.tile( px[None,:,:], (t.size,1,1)) + numpy.tile( l[:,None,None], (1,shape[0],shape[1]))
    y = numpy.tile( py[None,:,:], (t.size,1,1)) + numpy.tile( t[:,None,None], (1,shape[0],shape[1]))
    return img[y.ravel(),x.ravel()].reshape((t.size,shape[0],shape[1])), (t,l)

def reconstruct_from_grayscale_patches( patches, origin, epsilon=1e-12 ):
    """Rebuild an image from a set of patches by averaging

    The reconstructed image will have different dimensions than the
    original image if the strides and offsets of the patches were changed
    from the defaults!

    Args:
        patches (ndarray): input patches as (N,patch_height,patch_width) array

        origin (2-tuple): top and left coordinates of each patch

        epsilon (scalar): regularization term for averaging when patches
            some image pixels are not covered by any patch

    Returns:
        image (ndarray): output image reconstructed from patches of
            size ( max(origin[0])+patches.shape[1], max(origin[1])+patches.shape[2])

        weight (ndarray): output weight matrix consisting of the count
            of patches covering each pixel
    """
    patch_width  = patches.shape[2]
    patch_height = patches.shape[1]
    img_width    = numpy.max( origin[1] ) + patch_width
    img_height   = numpy.max( origin[0] ) + patch_height

    out = numpy.zeros( (img_height,img_width) )
    wgt = numpy.zeros( (img_height,img_width) )
    for i in range(patch_height):
        for j in range(patch_width):
            out[origin[0]+i,origin[1]+j] += patches[:,i,j]
            wgt[origin[0]+i,origin[1]+j] += 1.0

    return out/numpy.maximum( wgt, epsilon ), wgt

if __name__ == '__main__':
    import cv2
    import time
    import matplotlib.pyplot as plt

    img = cv2.imread('lena.png')[:,:,::-1]

    start = time.time()
    p, origin = extract_grayscale_patches( img[:,:,2], (8,8), stride=(1,1) )
    end = time.time()
    print( 'Patch extraction took: {}s'.format(numpy.round(end-start,2)) )
    start = time.time()
    r, w = reconstruct_from_grayscale_patches( p, origin )
    end = time.time()
    print('Image reconstruction took: {}s'.format(numpy.round(end-start,2)) )
    print( 'Reconstruction error is: {}'.format( numpy.linalg.norm( img[:r.shape[0],:r.shape[1],2]-r ) ) )

    plt.subplot( 131 )
    plt.imshow( img[:,:,2] )
    plt.title('Input image')
    plt.subplot( 132 )
    plt.imshow( p[p.shape[0]//2] )
    plt.title('Central patch')
    plt.subplot( 133 )
    plt.imshow( r )
    plt.title('Reconstructed image')
    plt.show()

Remove the end bit to have only numpy as a dependency.

Feb 01, 2019

Assignment Problems

Assignment problems occur in many fields that deal with resource allocation. There are many variants, however the simplest involves \(N\) agents indexed by \(i\) which must perform \(N\) tasks indexed by \(j\). The cost for agent \(i\) to perform task \(j\) is encoded in a cost matrix \(C\) that is weighted component-wise by assignments from an assignment matrix \(W_{ij}\) to form the objective.

It's common to have the requirements that each task must be completed and that no agent can perform more than one task. Often it is required that each agent perform a single task to completion but I'm not considering that case here.

\begin{align*} \newcommand{\argmin}{{\bf\mbox{argmin}}} \newcommand{\st}{{\bf\mbox{subject to:}}} \newcommand{\gap}{\hspace{0.25cm}} W = \argmin\gap& &\sum_{ij} C_{ij} W_{ij} \\ \st\gap& & \sum_i W_{ij} = 1\gap\forall j\\ & & \sum_j W_{ij} = 1\gap\forall i\\ & & W_{ij} \geq 0 \gap\forall i,j \end{align*}

The problem can be solved using the Hungarian algorithm but I thought I'd give it a try with ADMM to see how well it works.

The probability simplex constraints are a pain but it's pretty easy to solve this by introducing splitting variables \(A, B\) each equal to \(W\). This gives the following modified problem:

\begin{align*} W = \argmin\gap& &\sum_{ij} C_{ij} W_{ij} + \sum_i \mathcal{I}_\Delta(A_{i*}) + \sum_j \mathcal{I}_\Delta(B_{*j}) \\ \st\gap& & W - A = 0 \\ & & W - B = 0 \end{align*}

The function \(\mathcal{I}_\Delta(v)\) is the indicator function for probability simplex constraints on the rows of \(V\):

\begin{equation*} \mathcal{I}_\Delta(v) = \begin{cases} 0 & \gap\mbox{if } v_i \geq 0 \mbox{ and } \sum_j v_i = 1 \gap\forall i \\ \infty &\gap\mbox{otherwise} \end{cases} \end{equation*}

The proximal operator for this functions requires Euclidean projection onto the non-negativity and sum contraints. It's possible to alternately project on the non-negativity and sum contraints but this converges slowly. However, an \(O(N \log N)\) algorithm exists and can be implemented in Python as below:

def prox_probability_simplex( v ):
    """Algorithm of https://eng.ucmerced.edu/people/wwang5/papers/SimplexProj.pdf"""
    j = numpy.arange(1,len(v)+1)
    u = numpy.sort(v)[::-1]
    y = numpy.cumsum( u )
    rho = numpy.argmax((u+(1.0-y)/j>0)*j)+1
    off = (1-y[rho-1])/rho
    ret = numpy.maximum( v+off, 0.0 )
    return ret

Neglecting constants w.r.t \(W, A, B\), the augmented Lagrangian of this problem is:

\begin{equation*} \mathcal{L}(W,A,U_A,B,U_B) = \sum_{ij} C_{ij} W_{ij} + \sum_i \mathcal{I}_\Delta(A_{i*}) + \sum_j \mathcal{I}_\Delta(B_{*i}) + \frac{\rho}{2}\| W - A + U_A \|_2^2 + \frac{\rho}{2}\| W - A + U_A \|_2^2 \end{equation*}

ADMM consists of alternating minimizations for the solution \(W\) and the splitting variables \(A\) and \(B\). After each set of updates, the (scaled) Lagrange multipliers \(U_A, U_B\) are updated. The objective terms parallelize really cleanly, the first and last two terms parallelize per-component of \(W\). The \(\mathcal{I}_\Delta(.)\) terms each parallelize independently over rows and columns of the splitting variables.

The resulting algorithm is as follows. I have not parallelized the solves although the \(W\) solve is vectorized at least:

def assignment_problem( C, W=None, rho=0.1, thresh=0.55, early_stop=True, num_iterations=10000 ):
    """Applies an ADMM algorithm to the assignment problem

    This algorithm will only solve the problem at the limit of iterations. Early
    stopping or running too few iterations could give a non-optimal or invalid
    matching (i.e. not a permutation of agents to tasks)

    That said, it often does okay!

    Inputs:
        C (NxN matrix): Input cost matric mapping agents (row indices) to tasks
            (column indices). Invalid assignments should be indicated with very
            large positive entries

        W (NxN matrix): Initial assignments to warm-start the process or None
            to start from all zeros

        rho (scalar): Lagrangian augmentation weight

        thresh (scalar): Threshold above which assignments are treated as hard
            for the purposes of returned assignments

        early_stop (boolean): enable early stopping in the case that thresholding
            the assignment matrix by thresh produces a valid assignment. This
            does not necessarily produce an optimal solution!

        num_iterations (int): maximum number of iterations to perform, may be
            fewer if early_stop == True

    Returns:

        assignment (N-element integer array): assignments from agents to task
            where each element assignment[i]=v maps agent i to task v

        assignment matrix (NxN matrix): solution variables to warm-start if
            a valid solution was not found

        loss (scalar): the achieved loss/cost of the solution

        iter (integer): number of iterations performed
    """
    if W is None:
        W = numpy.zeros_like(C)

    A = W.copy()
    B = W.copy()
    Ua = W-A
    Ub = W-B

    loss = numpy.sum(C*W)
    for iter in range( num_iterations ):
        # update primary variables, trivially parallelizes per component
        W = numpy.maximum( (rho*(A-Ua)+rho*(B-Ub)-C)/(2.0*rho), 0.0 )

        # update row constraints, trivially parallizes by row
        for i in range(A.shape[0]):
            A[i,:] = prox_probability_simplex( W[i,:] + Ua[i,:] )
        Ua += (W - A)

        # update column constraints, trivially parallelizes by column
        for i in range(A.shape[1]):
            B[:,i] = prox_probability_simplex( W[:,i] + Ub[:,i] )
        Ub += (W - B)

        tloss = numpy.sum(C*W)
        rowsum = numpy.sum(W > 0.55, axis=0)
        colsum = numpy.sum(W > 0.55, axis=1)
        if early_stop and numpy.max(rowsum) == 1 and numpy.min(rowsum) == 1 and numpy.max(colsum) == 1 and numpy.min(colsum) == 1:
            return numpy.argmax( W > thresh, axis=1 ), W, tloss, iter

    return numpy.argmax( W > thresh, axis=1 ), W, tloss, iter

def check_validity( x ):
    """Checks that a solution from the above is a permutation"""
    return numpy.linalg.norm( numpy.sort(x) - numpy.arange(len(x)) ) > 1e-8

It's worth noting a few things:

  • With the splitting and ADMM, this algorithm will only exactly solve the problem in the limit of iteration count.
  • Early stopping can reduce the iterations needed to achieve a valid, low-cost matching by 10-50X depending on the problem but this matching may not be optimal. For some test problems with 10x10 cost matrices having standard normal entries this happened about 1% of the time. The achieved loss values are typically within 1-2% of each other.
  • Early stopping does always generate identical output running the algorithm for many more iterations. Subsets of agents/tasks where multiple matchings produce close results tend to be hard to resolve.

Jan 28, 2019

Dictionary Learning

Dictionary learning consists of jointly solving for both a dictionary \(D\), the columns of which store features and weights \(X\) which store coefficients applied to \(D\) required to best approximate a collection of input datapoints stored as columns of \(Y\). The weights \(X\) are required to be sparse, i.e. have relatively few non-zero entries:

\begin{equation*} \newcommand{\gap}{\hspace{0.25cm}} \newcommand{\argmin}{{\bf\mbox{argmin}}} \newcommand{\st}{{\bf\mbox{subject to:}}} D, X = \argmin \frac{1}{2} \| Y - DX \|_F^2 + \lambda \| X \|_1 \end{equation*}

An additional constraint must be added to \(D\) because, without one, the 1-norm in the objective can be made arbitrarily small by allowing \(D\) to grow arbitrarily large. Typically this constraint is that columne of \(D\) have unit 2-norm.

This is a non-convex problem, but it can be solved relatively well by iterating weight and dictionary updates, leaving the other fixed. Ignoring the norm constraint, the dictionary update is a linear least-squares problem and can be solved as:

\begin{equation*} D = \left( \left( X X^T + \epsilon I \right)^{-1} X Y^T \right)^T \end{equation*}

After this, the columns can be normalized.

The weight update is a sparse-coding problem and can be solved using ADMM. The first step is to introduce a splitting variable that allows the Frobenius-norm and 1-norm to be handled separately:

\begin{align*} X = \argmin & & \frac{1}{2} \| Y - D X \|_F^2 + \lambda \| Z \|_1 \\ \st & & X - Z = 0 \end{align*}

Once in this form, the update algorithm for the weights can be written from inspection based on the link above using \(F(x) = \argmin \frac{1}{2}\| Y - D X \|_F^2\) and \(G(Z) = \lambda \| Z \|_1\).

\begin{align*} \newcommand{\prox}[2]{{\bf\mbox{prox}_{#1}}\left(#2\right)} X^{k+1} &=& \prox{F}{Z^k - U^k} \\ Z^{k+1} &=& \prox{\lambda G}{X^{k+1} + U^k} \\ U^{k+1} &=& U^k + X^{k+1} - Z^{k+1} \end{align*}

All that remains is to determine the proximal operators and initialization. The proximal operator for \(X\) is for a pretty conventional least-squares problem:

\begin{align*} \prox{F}{V} &=& \argmin\gap F(X) + \frac{1}{2}\| X - V \|_F^2 \\ &=& \argmin\gap \frac{1}{2} \| Y - D X \|_F^2 + \frac{1}{2} \| X - V \|_F^2 \\ &=& \left( D^T D + I \right)^{-1} \left( D^T Y + V \right) \end{align*}

The corresponding proximal operator for \(Z\) is for 'pure' 1-norm, which is just the soft-thresholding operator:

\begin{align*} \prox{\lambda G}{v} &=& \argmin\gap G(Z) + \frac{1}{2\lambda} \| Z - V \|_1 \\ &=& ( |V| - \lambda )_+ {\bf\mbox{sgn}}( V ) \end{align*}

If you're in doubt where either of these come from, consult the ADMM link above.

This leads to some interesting choices. Solving the split problem has two sets of variables, \(X\) and \(Z\). \(X\) minimizes the data term but \(Z\) enforces the constraints. Nominally, both of these should be equal but this only holds at the limit of iteration count, assuming the overall algorithm works at all (the problem is non-convex after all). Which should be used when updating \(D\)?

For a test problem involving around 5.5M points of 48 dof data, I found that using \(Z\) is vastly preferable. Using a dictionary with only 100 atoms (~2X overcomplete) yields reconstruction errors around 1% or less but the percentage of non-zeros in the coding weights is only around 25%. It settled relatively quickly. Using \(X\) on the other hand oscillated quite a bit and did not reach these levels of accuracy/sparsity.

I conclude that it's better to use a looser fit than slacker constraint satisfaction, for this problem at least. Presumably the dictionary/codebooks adapt to each other better when they are consistent. More 'compressible' problems may yield sparser solutions.

import numpy

def sparse_coding( Y, D, X, rho, num_iterations, Z=None, U=None ):
    if Z is None:
        Z = X.copy()
    if U is None:
        U = X - Z

    # precompute solve and part of RHS
    iDtD = numpy.linalg.inv( numpy.dot(D.transpose(),D) + numpy.eye(D.shape[1]) )
    DtY  = numpy.dot( D.transpose(), Y )

    for iter in range(num_iterations):
        print('    Sparse coding iteration [{}/{}]'.format(iter+1,num_iterations) )
        # primary update
        X = numpy.dot( iDtD, DtY + Z - U )

        # splitting variable update
        T = X + U
        Z = numpy.maximum( numpy.abs(T) - rho, 0.0)*numpy.sign(T)

        # lagrange multiplier update
        U = T - Z
    return X, Z, U

def dictionary_learning( Y, num_atoms, rho=0.001, num_outer_iterations=10, num_inner_iterations=10, epsilon=1e-8 ):
    # initialize the dictionary and weights
    X = numpy.random.standard_normal( (num_atoms, Y.shape[1]) )
    D = numpy.random.standard_normal( (Y.shape[0], num_atoms) )

    # outer loop to fit dictionary and weights
    Z = X.copy()
    U = X - Z
    for outer in range(num_outer_iterations):
        print( '  Outer iteration [{}/{}]'.format(outer+1,num_outer_iterations) )

        # dictionary update
        D = numpy.linalg.solve(
            numpy.dot(Z,Z.transpose())+epsilon*numpy.eye(X.shape[0]),
            numpy.dot(Z,Y.transpose())
        ).transpose()
        for i in range(D.shape[1]):
            D[:,i] /= numpy.linalg.norm(D[:,i])

        # sparse coding weight update
        X, Z, U = sparse_coding( Y, D, X, rho, num_inner_iterations, Z, U )

        # print some stats
        print( '    ||Y-DX|| RMS error: {}'.format( numpy.sqrt(numpy.mean(numpy.square(Y - numpy.dot(D,Z)))) ) )
        print( '          mean(nnz(X)): {}'.format( numpy.mean( numpy.sum(numpy.abs(Z)>1e-4, axis=0) ) ) )

    # return dictionary and solution variables
    return D, Z

Jan 12, 2019

PyQt5 + PyOpenGL Example

It took a little bit for me to find working example code for PyOpenGL+PyQt5. The following exposes a GLUT-like interface where you can provide callback functions for basic window, mouse and keyboard events.

# simple_viewer.py

from PyQt5 import QtWidgets
from PyQt5 import QtCore
from PyQt5 import QtGui
from PyQt5 import QtOpenGL

class SimpleViewer(QtOpenGL.QGLWidget):

    initialize_cb    = QtCore.pyqtSignal()
    resize_cb        = QtCore.pyqtSignal(int,int)
    idle_cb          = QtCore.pyqtSignal()
    render_cb        = QtCore.pyqtSignal()

    mouse_move_cb    = QtCore.pyqtSignal( QtGui.QMouseEvent )
    mouse_press_cb   = QtCore.pyqtSignal( QtGui.QMouseEvent )
    mouse_release_cb = QtCore.pyqtSignal( QtGui.QMouseEvent )
    mouse_wheel_cb   = QtCore.pyqtSignal( QtGui.QWheelEvent )

    key_press_cb     = QtCore.pyqtSignal( QtGui.QKeyEvent )
    key_release_cb   = QtCore.pyqtSignal( QtGui.QKeyEvent )

    def __init__(self, parent=None):
        self.parent = parent
        QtOpenGL.QGLWidget.__init__(self, parent)
        self.setMouseTracking(True)

    def mouseMoveEvent( self, evt ):
        self.mouse_move_cb.emit( evt )

    def mousePressEvent( self, evt ):
        self.mouse_press_cb.emit( evt )

    def mouseReleaseEvent( self, evt ):
        self.mouse_release_cb.emit( evt )

    def keyPressEvent( self, evt ):
        self.key_press_cb.emit(evt)

    def keyReleaseEvent( self, evt ):
        self.key_release_cb.emit(evt)

    def initializeGL(self):
        self.initialize_cb.emit()

    def resizeGL(self, width, height):
        if height == 0: height = 1
        self.resize_cb.emit(width,height)

    def paintGL(self):
        self.render_cb.emit()

And here is a basic example using it:

"""About the simplest PyQt OpenGL example with decent interaction"""
import sys

from OpenGL.GL import *
from OpenGL.GLU import *

from simple_viewer import SimpleViewer

width = 800
height = 600
aspect = width/height

def resize( w, h ):
    width = w
    height = h
    aspect = w/h
    glViewport( 0, 0, width, height )

def initialize():
    glEnable(GL_DEPTH_TEST)
    glClearColor( 0.7, 0.7, 1.0, 0.0 )

def render():
    glClear( GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT )

    glMatrixMode( GL_PROJECTION )
    glLoadIdentity()
    gluPerspective( 45.0, aspect, 0.1, 10.0 )

    glMatrixMode( GL_MODELVIEW )
    glLoadIdentity()
    gluLookAt( 0.0, 2.0, 5.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0 )

    glPointSize(5.0)
    glLineWidth(5.0)
    glBegin(GL_LINES)
    glColor3f(  1.0, 0.0, 0.0 )
    glVertex3f( 1.0, 0.0, 0.0 )
    glVertex3f( 0.0, 0.0, 0.0 )
    glColor3f(  0.0, 1.0, 0.0 )
    glVertex3f( 0.0, 1.0, 0.0 )
    glVertex3f( 0.0, 0.0, 0.0 )
    glColor3f(  0.0, 0.0, 1.0 )
    glVertex3f( 0.0, 0.0, 1.0 )
    glVertex3f( 0.0, 0.0, 0.0 )
    glEnd()

def mouse_move( evt ):
    print('Mouse move {}: [{},{}]'.format(evt.button(),evt.x(),evt.y()) )

def mouse_press( evt ):
    print('Mouse press {}: [{},{}]'.format(evt.button(),evt.x(),evt.y()) )

def mouse_release( evt ):
    print('Mouse release {}: [{},{}]'.format(evt.button(),evt.x(),evt.y()) )

def key_press( evt ):
    print('Key press {}'.format(evt.key()) )

def key_release( evt ):
    print('Key release {}'.format(evt.key()) )

# create the QApplication
app = SimpleViewer.application()

# set up the display
viewer = SimpleViewer()
viewer.resize_cb.connect( resize )
viewer.initialize_cb.connect( initialize )
viewer.render_cb.connect( render )

# keyboard & mouse interactions
viewer.key_press_cb.connect( key_press )
viewer.key_release_cb.connect( key_release )
viewer.mouse_press_cb.connect( mouse_press )
viewer.mouse_release_cb.connect( mouse_release )
viewer.mouse_move_cb.connect( mouse_move )

# resize the window
viewer.resize( width, height )
viewer.show()

# main loop
try:
    sys.exit(app.exec_())
except SystemExit:
    pass

Nov 15, 2018

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:

\begin{align*} \newcommand{\cmat}{{\bf\mbox{cmat}}} \newcommand{\rmat}{{\bf\mbox{rmat}}} \cmat\left( \cvec(A), M \right) = A \\ \rmat\left( \rvec(A), N \right) = A \\ \end{align*}

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:

\begin{align*} \spread{M}{n}^T &=& \spread{M^T}{n} \\ \spread{M}{n}^{-1} &=& \spread{M^{-1}}{n} \\ \split{M}{n}^T &=& \split{M^T}{n} \\ \split{M}{n}^{-1} &=& \split{M^{-1}}{n} \end{align*}

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 02, 2018

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()

Nov 02, 2018

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.

Oct 25, 2018

Create Docker Image

This process follows this link . First update apt and install docker.io

$ sudo apt-get update
$ sudo apt install docker.io

Next start the service and enable it at boot time:

$ sudo systemctl start docker
$ sudo systemctl enable docker

Finally add your current user to the docker group:

$ sudo usermod -a -G docker $USER

and log out of your account and back in.

Create Dockerfile

Create a blank Dockerfile

$ touch Dockerfile

and then edit it to look like the following:

# start with ubuntu
FROM ubuntu:16.04

# update the software repository
RUN apt-get update
RUN apt-get install -y python3-dev

# add user appuser and switch to it
RUN useradd -ms /bin/bash appuser
USER appuser
WORKDIR /home/appuser

# copy over a test message and python
# script to print it to the screen
COPY data/ ./data/
COPY print_message.py .

# run the script at container startup
CMD ["python3", "print_message.py"]

Next make a directory to share with the image and put a simple message file into it.

$ mkdir data
$ echo "Hello from docker" > data/message.txt

Then create a python script to print the message:

if __name__ == '__main__':

    with open('data/message.txt') as f:
        data = f.read()
        print( data )

Build the image and run it:

$ docker build -t test_image
$ docker run -it test_image
Hello from docker

And that's all there is to it.

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