This notebook is all about ICP and it's different implementations. It should be visual and self - descriptive.



Having two scans $P = \{p_i\}$ and $Q = \{q_i\}$ we want to find a transformation (rotation $R$ and translation $t$) to apply to $P$ to match $Q$ as good as possible. In the remainder of this notebook we will try to define what does "as good as possible mean" as well as ways to find such a transformation.

In [1]:
import sys
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import animation, rc
from math import sin, cos, atan2, pi
from IPython.display import display, Math, Latex, Markdown, HTML

The way we will plot the data

In [2]:
def plot_data(data_1, data_2, label_1, label_2, markersize_1=8, markersize_2=8):
    fig = plt.figure(figsize=(10, 6))
    ax = fig.add_subplot(111)
    if data_1 is not None:
        x_p, y_p = data_1
        ax.plot(x_p, y_p, color='#336699', markersize=markersize_1, marker='o', linestyle=":", label=label_1)
    if data_2 is not None:
        x_q, y_q = data_2
        ax.plot(x_q, y_q, color='orangered', markersize=markersize_2, marker='o', linestyle=":", label=label_2)
    return ax

def plot_values(values, label):
    fig = plt.figure(figsize=(10, 4))
    ax = fig.add_subplot(111)
    ax.plot(values, label=label)
def animate_results(P_values, Q, corresp_values, xlim, ylim):
    """A function used to animate the iterative processes we use."""
    fig = plt.figure(figsize=(10, 6))
    anim_ax = fig.add_subplot(111)
    anim_ax.set(xlim=xlim, ylim=ylim)
    x_q, y_q = Q
    # draw initial correspondeces
    corresp_lines = []
    for i, j in correspondences:
        corresp_lines.append(anim_ax.plot([], [], 'grey')[0])
    # Prepare Q data.
    Q_line, = anim_ax.plot(x_q, y_q, 'o', color='orangered')
    # prepare empty line for moved data
    P_line, = anim_ax.plot([], [], 'o', color='#336699')

    def animate(i):
        P_inc = P_values[i]
        x_p, y_p = P_inc
        P_line.set_data(x_p, y_p)
        draw_inc_corresp(P_inc, Q, corresp_values[i])
        return (P_line,)
    def draw_inc_corresp(points_from, points_to, correspondences):
        for corr_idx, (i, j) in enumerate(correspondences):
            x = [points_from[0, i], points_to[0, j]]
            y = [points_from[1, i], points_to[1, j]]
            corresp_lines[corr_idx].set_data(x, y)
    anim = animation.FuncAnimation(fig, animate,
    return HTML(anim.to_jshtml())

Generate example data

Thoughout this notebook we will be working wigh generated data that looks like this:

In [3]:
# initialize pertrubation rotation
angle = pi / 4
R_true = np.array([[cos(angle), -sin(angle)], 
                   [sin(angle),  cos(angle)]])
t_true = np.array([[-2], [5]])

# Generate data as a list of 2d points
num_points = 30
true_data = np.zeros((2, num_points))
true_data[0, :] = range(0, num_points)
true_data[1, :] = 0.2 * true_data[0, :] * np.sin(0.5 * true_data[0, :]) 
# Move the data
moved_data = R_true.dot(true_data) + t_true

# Assign to variables we use in formulas.
Q = true_data
P = moved_data

plot_data(moved_data, true_data, "P: moved data", "Q: true data")
2021-04-18T23:08:42.663499 image/svg+xml Matplotlib v3.3.4, https://matplotlib.org/

Correspondences computation

We compute correspondences from $P$ to $Q$, i.e. for every $p_i$ we search the closest $q_j$ to it.

In [4]:
def get_correspondence_indices(P, Q):
    """For each point in P find closest one in Q."""
    p_size = P.shape[1]
    q_size = Q.shape[1]
    correspondences = []
    for i in range(p_size):
        p_point = P[:, i]
        min_dist = sys.maxsize
        chosen_idx = -1
        for j in range(q_size):
            q_point = Q[:, j]
            dist = np.linalg.norm(q_point - p_point)
            if dist < min_dist:
                min_dist = dist
                chosen_idx = j
        correspondences.append((i, chosen_idx))
    return correspondences

def draw_correspondeces(P, Q, correspondences, ax):
    label_added = False
    for i, j in correspondences:
        x = [P[0, i], Q[0, j]]
        y = [P[1, i], Q[1, j]]
        if not label_added:
            ax.plot(x, y, color='grey', label='correpondences')
            label_added = True
            ax.plot(x, y, color='grey')

ICP based on SVD

Tldr version. If the scans would match exactly, their cross-covariance would be identity. Therefore, we can iteratively optimize their cross-covariance to be as close as possible to an identity matrix by applying transformations to $P$. Let's dive into details.

Single iteration

In a single iteration we assume that the correspondences are known. We can compute the cross-covariance between the corresponding points. Let $C = \{\{i,j\}:p_i \leftrightarrow q_j\}$ be a set of all correspondences, also $|C| = N$. Then, the cross-covariance $K$ is computed as:

\begin{eqnarray} K &=& E [(q_i - \mu_Q)(p_i - \mu_P)^T] \\ &=& \frac{1}{N}\sum_{\{i,j\} \in C}{(q_i - \mu_Q)(p_i - \mu_P)^T} \\ &\sim& \sum_{\{i,j\} \in C}{(q_i - \mu_Q)(p_i - \mu_P)^T} \end{eqnarray}

Each point has two dimentions, that is $p_i, q_j \in {\rm I\!R}^2$, thus cross-covariance has the form of (we drop indices $i$ and $j$ for notation simplicity):

\begin{equation} K = \begin{bmatrix} cov(p_x, q_x) & cov(p_x, q_y) \\ cov(p_y, q_x) & cov(p_y, q_y) \end{bmatrix} \end{equation}

Intuition: Intuitevely, cross-covariance tells us how a coordinate of point $q$ changes with the change of $p$ coorinate, i.e. $cov(p_x, q_x)$ tells us how the $x$ coordinate of $q$ will change with the change in $x$ coordinate of $p$ given that the points are corresponding. Ideal cross-covariance matrix is an identity matrix, i.e., we want the $x$ coordinates to be ideally correlated between the scans $P$ and $Q$, while there should be no correlation between the $x$ coorinate of points from $P$ to the $y$ coordinate of points in $Q$. In our case, however, the position of $P$ is derived from the position of $Q$ through some rotation $R$ and translation $t$. Therefore, whenever we would move the scan $Q$, scan $P$ would move in a related way, but pertrubed through the rotation and translation applied, making the cross-covariance matrix non-identity.

Knowing the cross-covariance we can compute its SVD decomposition:

\begin{equation} \mathrm{SVD}(K) = USV^T \end{equation}

The SVD decomposition gives us how to rotate our data to align it with its prominent direction with $UV^T$ and how to scale it with its singular values $S$. Therefore:

\begin{eqnarray} R &=& UV^T \\ t &=& \mu_Q - R \mu_P \end{eqnarray}

Let's try this out:

Make data centered

In [5]:
def center_data(data, exclude_indices=[]):
    reduced_data = np.delete(data, exclude_indices, axis=1)
    center = np.array([reduced_data.mean(axis=1)]).T
    return center, data - center

center_of_P, P_centered = center_data(P)
center_of_Q, Q_centered = center_data(Q)
ax = plot_data(P_centered, Q_centered,
               label_1='Moved data centered',
               label_2='True data centered')
2021-04-18T23:08:43.288061 image/svg+xml Matplotlib v3.3.4, https://matplotlib.org/

Compute correspondences

In [6]:
correspondences = get_correspondence_indices(P_centered, Q_centered)
ax = plot_data(P_centered, Q_centered,
               label_1='P centered',
               label_2='Q centered')
draw_correspondeces(P_centered, Q_centered, correspondences, ax)
2021-04-18T23:08:43.777471 image/svg+xml Matplotlib v3.3.4, https://matplotlib.org/

Compute cross covariance

In [7]:
def compute_cross_covariance(P, Q, correspondences, kernel=lambda diff: 1.0):
    cov = np.zeros((2, 2))
    exclude_indices = []
    for i, j in correspondences:
        p_point = P[:, [i]]
        q_point = Q[:, [j]]
        weight = kernel(p_point - q_point)
        if weight < 0.01: exclude_indices.append(i)
        cov += weight * q_point.dot(p_point.T)
    return cov, exclude_indices

cov, _ = compute_cross_covariance(P_centered, Q_centered, correspondences)
[[1113.97274605 1153.71870122]
 [ 367.39948556  478.81890396]]

Find $R$ and $t$ from SVD decomposition

Here we find SVD decomposition of the cross covariance matrix and apply the rotation to $Q$

In [8]:
U, S, V_T = np.linalg.svd(cov)
R_found = U.dot(V_T)
t_found = center_of_Q - R_found.dot(center_of_P)
print("R_found =\n", R_found)
print("t_found =\n", t_found)
[1712.35558954   63.95608054]
R_found =
 [[ 0.89668479  0.44266962]
 [-0.44266962  0.89668479]]
t_found =
 [[  0.4278782 ]

Apply a single correction to $P$ and visualize the result

This is the result after just one iteration. Because our correspondences are not optimal, it is not a complete match.

In [9]:
P_corrected = R_found.dot(P) + t_found
ax = plot_data(P_corrected, Q, label_1='P corrected', label_2='Q')
print("Squared diff: (P_corrected - Q) = ", np.linalg.norm(P_corrected - Q))
[[  0.4278782 ]
[[ 0.89668479  0.44266962]
 [-0.44266962  0.89668479]]
2021-04-18T23:08:44.475427 image/svg+xml Matplotlib v3.3.4, https://matplotlib.org/
Squared diff: (P_corrected - Q) =  16.052894296516953

Let's make it iterative

If we would know the correct correspondences from the start, we would be able to get the optimal solution in a single iteration. This is rarely the case and we need to iterate. That consists of the following steps:

  1. Make data centered by subtracting the mean
  2. Find correspondences for each point in $P$
  3. Perform a single iteration by computing the cross-covariance matrix and performing the SVD
  4. Apply the found rotation to $P$
  5. Repeat until correspondences don't change
  6. Apply the found rotation to the mean vector of $P$ and uncenter $P$ with it.

Working example

As we want to work with centered data and we will be iteratively centering the data, searching for rotation on centered data and uncentering the data at the end of each iteration. It is not the most elegant or efficient way, but it allows us to visualize the clouds nicer.

In [10]:
def icp_svd(P, Q, iterations=10, kernel=lambda diff: 1.0):
    """Perform ICP using SVD."""
    center_of_Q, Q_centered = center_data(Q)
    norm_values = []
    P_values = [P.copy()]
    P_copy = P.copy()
    corresp_values = []
    exclude_indices = []
    for i in range(iterations):
        center_of_P, P_centered = center_data(P_copy, exclude_indices=exclude_indices)
        correspondences = get_correspondence_indices(P_centered, Q_centered)
        norm_values.append(np.linalg.norm(P_centered - Q_centered))
        cov, exclude_indices = compute_cross_covariance(P_centered, Q_centered, correspondences, kernel)
        U, S, V_T = np.linalg.svd(cov)
        R = U.dot(V_T)  
        t = center_of_Q - R.dot(center_of_P)  
        P_copy = R.dot(P_copy) + t
    return P_values, norm_values, corresp_values

P_values, norm_values, corresp_values = icp_svd(P, Q)
plot_values(norm_values, label="Squared diff P->Q")
ax = plot_data(P_values[-1], Q, label_1='P final', label_2='Q', markersize_1=15)
2021-04-18T23:08:44.997894 image/svg+xml Matplotlib v3.3.4, https://matplotlib.org/
2021-04-18T23:08:45.289717 image/svg+xml Matplotlib v3.3.4, https://matplotlib.org/
[37.7609616179743, 16.052894296516953, 4.747691495229575, 0.6307436032154958, 2.6418496970390133e-14, 2.6417913757437442e-14, 2.6105561201727387e-14, 4.026252155082058e-14, 5.0457972328602515e-14, 3.389128540242614e-14]
In [11]:
animate_results(P_values, Q, corresp_values, xlim=(-5, 35), ylim=(-5, 35))

$\newcommand{\b}[1]{\boldsymbol{\mathrm{#1}}}$ $\newcommand{\R}{\boldsymbol{\mathrm{R}}}$ $\newcommand{\x}{\boldsymbol{\mathrm{x}}}$ $\newcommand{\h}{\boldsymbol{\mathrm{h}}}$ $\newcommand{\p}{\boldsymbol{\mathrm{p}}}$ $\newcommand{\q}{\boldsymbol{\mathrm{q}}}$ $\newcommand{\t}{\boldsymbol{\mathrm{t}}}$ $\newcommand{\J}{\boldsymbol{\mathrm{J}}}$ $\newcommand{\H}{\boldsymbol{\mathrm{H}}}$ $\newcommand{\E}{\boldsymbol{\mathrm{E}}}$ $\newcommand{\e}{\boldsymbol{\mathrm{e}}}$ $\newcommand{\n}{\boldsymbol{\mathrm{n}}}$ $\DeclareMathOperator*{\argmin}{arg\,min}$ $\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$

Non-linear Least-squares based ICP

We can alternatively treat every iteration of ICP as a least squares minimization problem. The function we want to minimize is the squared sum of distances between the points of the scans:

\begin{equation} E = \sum_i[\R\p_i + \t - \q_i]^2 \rightarrow \mathrm{min} \end{equation}

To minimize this function we update the pose $\b{R}$, $\b{t}$ (or alternatively represented as a vector $\b{x} = [x, y, \theta]^T$) to which we need to move scan $P$ to overlap it with a query scan $Q$. It is a non-linear function because of the rotation.


We look for correspondeces without moving the data to ensure zero-mean. Therefore the correspondences look worse than in SVD case, where we first ensured that both scans are zero-mean.

In [12]:
correspondences = get_correspondence_indices(P, Q)
ax = plot_data(P, Q, "Moved data", "True data")
draw_correspondeces(P, Q, correspondences, ax)
2021-04-18T23:08:47.265756 image/svg+xml Matplotlib v3.3.4, https://matplotlib.org/


We define $\p_i \in P$ to be points we want to match against $\q_j \in Q$. By "matching" we mean finding pose $\b{x} = [x, y, \theta]^T$ that minimizes the sum of squared lengths of the correspondences. Pose $\x$ can alternatively be resresented by a rotation matrix $\R = \begin{bmatrix} \cos\theta & - \sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$ and a translation vector $\t = [x, y]^T$. We will keep using these representations interchangibly throughtout these notes.

We will further use the following notation: $\h_i(\x) = \R_\theta \p_i + \t$ to denote the points from scan $P$ transformed with $\R$ and $\t$. Additionally, we define error function $\e$ to be:

\begin{eqnarray} \e(\x) &=& \sum_{\{i,j\}\in C}{\e_{i,j}(\x)},\\ \e_{i,j}(\x) &=& \h_i(\x) - \q_j = \R_\theta \p_i + \t - \q_j \end{eqnarray}

This allows us to formulate the minimization problem as follows:

\begin{eqnarray} \b{x}_{query} &=& \argmin_{\x}\{\E(\x)\} \\ &=& \argmin_{\x}\{\sum_{\{i, j\} \in C}{\norm{\e_{i,j}(\x)}^2}\} \\ &=& \argmin_{\x}\{\sum_{\{i, j\} \in C}{\norm{\b{h}_i(\x) - \q_j}^2}\} \end{eqnarray}

Gauss Newton Method

We will be using Gauss Newton method for computing the least squares solution of our non-linear problem. We therefore linearize our function in the vicinity of $\x$. Solving non-linear least squares is equivalent to solving the following system of equations:

\begin{equation} \H \Delta \x = - \E^\prime(\x), \end{equation}

where $\Delta \x$ is the increment of the argument ($[\Delta x, \Delta y, \Delta \theta]$ in our case), $\H$ is the Hessian of $\E$ and $\E^\prime(\x)$ is the derivative over the function we are trying to minimize. We compute the gradient $\E^\prime(\x)$ as follows:

\begin{equation} \E^\prime(\x) = \J(\x) \e(\x) \end{equation}

In Gauss-Newton method we linearize the function around the considered point, which allows us to compute the Hessian as simple as: $\H = \J(\x)^T \J(\x)$


Both the Hessian and the gradient require the computation of a Jacobian. To compute a Jacobian we need a derivative of a rotation matrix:

\begin{equation} \R_\theta^\prime =\frac{\partial}{\partial \theta} \begin{bmatrix} \cos\theta & - \sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} =\begin{bmatrix} -\sin\theta & - \cos\theta \\ \cos\theta & -\sin\theta \end{bmatrix} \end{equation}
In [13]:
def dR(theta):
    return np.array([[-sin(theta), -cos(theta)],
                     [cos(theta),  -sin(theta)]])

def R(theta):
    return np.array([[cos(theta), -sin(theta)],
                     [sin(theta),  cos(theta)]])

Now we have everything to compute the Jacobian $\b{J}$ as follows:

\begin{eqnarray} \b{J} = \frac{\partial \e_{i,j}(\x)}{\partial \x} = \frac{\partial \h_i(\x)}{\partial \x} &=& \Big(\frac{\partial \h_i(\x)}{\partial x}, \frac{\partial \h_i(\x)}{\partial y}, \frac{\partial \h_i(\x)}{\partial \theta}\Big) \\ &=&\Big(\b{I}, \R_\theta^\prime \p_i \Big) \\ &=& \begin{bmatrix} 1 & 0 & -\sin\theta\ p_i^x - \cos\theta\ p_i^y \\ 0 & 1 & \cos\theta\ p_i^x - \sin\theta\ p_i^y \end{bmatrix} \end{eqnarray}
In [14]:
def jacobian(x, p_point):
    theta = x[2]
    J = np.zeros((2, 3))
    J[0:2, 0:2] = np.identity(2)
    J[0:2, [2]] = dR(0).dot(p_point)
    return J

def error(x, p_point, q_point):
    rotation = R(x[2])
    translation = x[0:2]
    prediction = rotation.dot(p_point) + translation
    return prediction - q_point

Solving the Least Squares problem

Now that we know how to compute the Jacobian, we can compute the system of equations, solving which delivers the solution to our problem. We initialize Hessian $\H$ and gradient $\b{g}$ by zeros:

\begin{equation} \b{H} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \ \b{g} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \ \end{equation}

We now need to construct a system of equations solving which would give us the relative pose. For every corresponding pair of points do the following:

\begin{eqnarray} \H &\rightarrow& \b{H} + \J^T \J \\ \b{g} &\rightarrow& \b{g} + \J^T \e \end{eqnarray}

Now that the system of equation is ready, we can find the $\Delta\b{x}$ - the solution to the least squares problem:

\begin{equation} \H \Delta\x = -\b{g} \Longrightarrow \Delta\x = -\b{H}^{-1}\b{g} \end{equation}

This can be solved without actually inverting the matrix in reality.

In [15]:
def prepare_system(x, P, Q, correspondences, kernel=lambda distance: 1.0):
    H = np.zeros((3, 3))
    g = np.zeros((3, 1))
    chi = 0
    for i, j in correspondences:
        p_point = P[:, [i]]
        q_point = Q[:, [j]]
        e = error(x, p_point, q_point)
        weight = kernel(e) # Please ignore this weight until you reach the end of the notebook.
        J = jacobian(x, p_point)
        H += weight * J.T.dot(J)
        g += weight * J.T.dot(e)
        chi += e.T * e
    return H, g, chi

def icp_least_squares(P, Q, iterations=30, kernel=lambda distance: 1.0):
    x = np.zeros((3, 1))
    chi_values = []
    x_values = [x.copy()]  # Initial value for transformation.
    P_values = [P.copy()]
    P_copy = P.copy()
    corresp_values = []
    for i in range(iterations):
        rot = R(x[2])
        t = x[0:2]
        correspondences = get_correspondence_indices(P_copy, Q)
        H, g, chi = prepare_system(x, P, Q, correspondences, kernel)
        dx = np.linalg.lstsq(H, -g, rcond=None)[0]
        x += dx
        x[2] = atan2(sin(x[2]), cos(x[2])) # normalize angle
        rot = R(x[2])
        t = x[0:2]
        P_copy = rot.dot(P.copy()) + t
    return P_values, chi_values, corresp_values

P_values, chi_values, corresp_values = icp_least_squares(P, Q)
plot_values(chi_values, label="chi^2")
2021-04-18T23:08:48.070873 image/svg+xml Matplotlib v3.3.4, https://matplotlib.org/
[279.8595127207585, 135.53670591984252, 148.41156916870492, 138.13600690872124, 164.7628216353993, 162.60937583974695, 172.09657147196538, 161.58393072081944, 125.08676568660141, 74.1879508877521, 54.070226670322775, 41.64904459515214, 19.71413517969046, 9.258599804821097, 8.77264288232226, 4.1720680989042, 1.2769314893000516, 0.16280347501872677, 0.014198093772482398, 0.0012239475378550792, 0.00010514834808173084, 9.024084124479632e-06, 7.742391050343611e-07, 6.64216039416419e-08, 5.69813281717628e-09, 4.888240241032962e-10, 4.193450943848659e-11, 3.597413162964382e-12, 3.0860929402579883e-13, 2.647449094019942e-14]

Animate the result

In [16]:
animate_results(P_values, Q, corresp_values, xlim=(-10, 35), ylim=(-10, 30))

Using point to plane metric with Least Squares ICP

Point to point metric used before is not really the most optimal as can be seen above. It takes quite some iterations for the solution to converge. There is another metric which seems to work better. It is called "point to plane" metric. The idea here is that we still find the closest point, but the error is defined as a projection of the error onto the direction of the normal shot from the found point.

To start, we need to compute the normals of the resulting cloud.

The normal in this 2D case is simple to compute. For a vector $v = [x, y]^\top$ the normal is a vector $n_v = [-y, x]^\top$ as can be shown geometrically.

In [17]:
def compute_normals(points, step=1):
    normals = [np.array([[0, 0]])]
    normals_at_points = []
    for i in range(step, points.shape[1] - step):
        prev_point = points[:, i - step]
        next_point = points[:, i + step]
        curr_point = points[:, i]
        dx = next_point[0] - prev_point[0] 
        dy = next_point[1] - prev_point[1]
        normal = np.array([[0, 0],[-dy, dx]])
        normal = normal / np.linalg.norm(normal)
        normals.append(normal[[1], :])  
        normals_at_points.append(normal + curr_point)
    normals.append(np.array([[0, 0]]))
    return normals, normals_at_points

def plot_normals(normals, ax):
    label_added = False
    for normal in normals:
        if not label_added:
            ax.plot(normal[:,0], normal[:,1], color='grey', label='normals')
            label_added = True
            ax.plot(normal[:,0], normal[:,1], color='grey')
    return ax

Q_normals, Q_normals_to_draw = compute_normals(Q)
ax = plot_data(None, Q, None, 'Q')
ax = plot_normals(Q_normals_to_draw, ax)
2021-04-18T23:08:51.891538 image/svg+xml Matplotlib v3.3.4, https://matplotlib.org/

Point to plane error metric

The error that we minimize is different. Before we minimized the Euclidean error between the 2 points, now we minimize this error projected onto the normal vector of one of the points.

\begin{equation} E = \sum_i (\n_i \cdot (\R_\theta \p_i + \t - \q_j))^2 \rightarrow \mathrm{min}, \end{equation}

here $\n_i$ is the normal direction at the point $\q_i$. This leads to a changed jacobian that is not $2 \times 3$ as before, but $1 \times 3$.

Point to plane Jacobian

In [18]:
from sympy import init_printing, symbols, Matrix, cos as s_cos, sin as s_sin, diff
init_printing(use_unicode = True)

def RotationMatrix(angle):
    return Matrix([[s_cos(angle) , -s_sin(angle)], [s_sin(angle), s_cos(angle)]])

x, y, theta, n_x, n_y, p_x, p_y = symbols('x, y, \\theta, n_x, n_y, p_x, p_y')
t = Matrix([[x], [y]])
X = Matrix([x,y,theta])
n = Matrix([[n_x],[n_y]])
p = Matrix([[p_x], [p_y]])

error_point = RotationMatrix(theta) * p + t
error_normal = n.dot(RotationMatrix(theta) * p + t)

J_point = diff(error_point, X).reshape(3,2).transpose()
J_normal = diff(error_normal, X).reshape(3,1).transpose()
display(Latex("Point to point Jacobian: "), J_point)
display(Latex("Point to plane Jacobian: "), J_normal)
Point to point Jacobian:
$\displaystyle \left[\begin{matrix}1 & 0 & - p_{x} \sin{\left(\theta \right)} - p_{y} \cos{\left(\theta \right)}\\0 & 1 & p_{x} \cos{\left(\theta \right)} - p_{y} \sin{\left(\theta \right)}\end{matrix}\right]$
Point to plane Jacobian:
$\displaystyle \left[\begin{matrix}n_{x} & n_{y} & n_{x} \left(- p_{x} \sin{\left(\theta \right)} - p_{y} \cos{\left(\theta \right)}\right) + n_{y} \left(p_{x} \cos{\left(\theta \right)} - p_{y} \sin{\left(\theta \right)}\right)\end{matrix}\right]$

Faster convergence

This changes very little in the optimization procedure, but the convergence rate is much better.

In [19]:
def prepare_system_normals(x, P, Q, correspondences, Q_normals):
    H = np.zeros((3, 3))
    g = np.zeros((3, 1))
    chi = 0
    for i, j in correspondences:
        p_point = P[:, [i]]
        q_point = Q[:, [j]]
        normal = Q_normals[j]
        e = normal.dot(error(x, p_point, q_point))
        J = normal.dot(jacobian(x, p_point))
        H += J.T.dot(J)
        g += J.T.dot(e)
        chi += e.T * e
    return H, g, chi

def icp_normal(P, Q, Q_normals, iterations=20):
    x = np.zeros((3, 1))
    chi_values = []
    x_values = [x.copy()]  # Initial value for transformation.
    P_values = [P.copy()]
    P_latest = P.copy()
    corresp_values = []
    for i in range(iterations):
        rot = R(x[2])
        t = x[0:2]
        correspondences = get_correspondence_indices(P_latest, Q)
        H, g, chi = prepare_system_normals(x, P, Q, correspondences, Q_normals)
        dx = np.linalg.lstsq(H, -g, rcond=None)[0]
        x += dx
        x[2] = atan2(sin(x[2]), cos(x[2])) # normalize angle
        chi_values.append(chi.item(0)) # add error to list of errors
        rot = R(x[2])
        t = x[0:2]
        P_latest = rot.dot(P.copy()) + t
    return P_values, chi_values, corresp_values

P_values, chi_values, corresp_values = icp_normal(P, Q, Q_normals)
plot_values(chi_values, label="chi^2")
2021-04-18T23:08:53.808042 image/svg+xml Matplotlib v3.3.4, https://matplotlib.org/
In [20]:
animate_results(P_values, Q, corresp_values, xlim=(-10, 35), ylim=(-10, 20))