Estimating Rigid Transformations of Noisy Point Clouds Using the Universal Manifold Embedding

Abstract

We present a closed form solution to the problem of registration of fully overlapping 3D point clouds undergoing unknown rigid transformations, as well as for detection and registration of sub-parts undergoing unknown rigid transformations. The solution is obtained by adapting the general framework of the universal manifold embedding (UME) to the case where the transformations the object may undergo are rigid. The UME nonlinearly maps functions related by certain types of geometric transformations of coordinates to the same linear subspace of some Euclidean space while retaining the information required to recover the transformation. Therefore registration, matching and classification can be solved as linear problems in a low dimensional linear space. In this paper, we extend the UME framework to the special case where it is a priori known that the geometric transformations are rigid. While a variety of methods exist for point cloud registration, the method proposed in this paper is notably different as registration is achieved by a closed form solution that employs the UME low dimensional representation of the shapes to be registered.

Appendices

Proof for proposition 1

Proof

First, notice that

(55)

(56)

Adding and subtracting $\int \limits _{X_{miss}^{(i)}}\mathbf{x}' d\mathbf{x}$ to (56), we obtain

$$\begin{aligned}&\int \limits _{{\mathbb {R}}^n}\mathbf{x}' \mathbb {1}_{[l_i,h_i]}(\widetilde{h}(\mathbf{x}))d\mathbf{x}\nonumber \\&\quad = \int \limits _{X_{in}^{(i)}}\mathbf{x}' d\mathbf{x}+ \int \limits _{X_{miss}^{(i)}}\mathbf{x}' d\mathbf{x}- \int \limits _{X_{miss}^{(i)}}\mathbf{x}'d\mathbf{x}+\int \limits _{X_{out}^{(i)}}\mathbf{x}'d\mathbf{x}\end{aligned}$$

(57)

substituting (55) in (57)

$$\begin{aligned}&\int \limits _{{\mathbb {R}}^n}\mathbf{x}' \mathbb {1}_{[l_i,h_i]}(\widetilde{h}(\mathbf{x}))d\mathbf{x}= \int \limits _{{\mathbb {R}}^n}\mathbf{x}' \mathbb {1}_{[l_i,h_i]}(h(\mathbf{x}))d\mathbf{x}\nonumber \\&-\int \limits _{X_{miss}^{(i)}}\mathbf{x}' d\mathbf{x}+\int \limits _{X_{out}^{(i)}}\mathbf{x}'d\mathbf{x}\end{aligned}$$

(58)

Hence,

$$\begin{aligned}&\int \limits _{{\mathbb {R}}^n}\mathbf{x}' \mathbb {1}_{[l_i,h_i]}(\widetilde{h}(\mathbf{x}))d\mathbf{x}- \int \limits _{{\mathbb {R}}^2}\mathbf{x}' \mathbb {1}_{[l_i,h_i]}(h(\mathbf{x}))d\mathbf{x}=\nonumber \\&\int \limits _{X_{out}^{(i)}}\mathbf{x}'d\mathbf{x}- \int \limits _{X_{miss}^{(i)}}\mathbf{x}'d\mathbf{x}\end{aligned}$$

(59)

$\square$

Mean UME Error Matrix Detailed Derivation

Following definition 1, we notice that

$$\begin{aligned} \int \limits _{X_{out}^{(i)}}\mathbf{x}'d\mathbf{x}= \int \limits _{X_{above}^{(i)}\cup X_{below}^{(i)}}\mathbf{x}'\mathbb {1}_{[l_i-h(\mathbf{x}),h_i-h(\mathbf{x})]}(n(\mathbf{x}))d\mathbf{x}\end{aligned}$$

(60)

and similarly

$$\begin{aligned}&\int \limits _{X_{miss}^{(i)}}\mathbf{x}'d\mathbf{x}=\nonumber \\&\int \limits _{X_{within}^{(i)}}\mathbf{x}'\mathbb {1}_{[h_i-h(\mathbf{x}),\infty ]}(n(\mathbf{x})) d\mathbf{x}+ \mathbf{x}'\mathbb {1}_{[-\infty ,l_i-h(\mathbf{x})]}(n(\mathbf{x}))d\mathbf{x}\nonumber \\ \end{aligned}$$

(61)

Therefore,

(62)

similarly it is found that

$$\begin{aligned} E\left( \int \limits _{X_{miss}^{(i)}}\mathbf{x}'d\mathbf{x}\right) = \int \limits _{X_{within}^{(i)}}\mathbf{x}'P_{miss}^{(i)}(\mathbf{x};h(\mathbf{x}))d\mathbf{x}\end{aligned}$$

(63)

Substituting (63) and (62) into (32) and taking the expectation

(64)

Proof of Proposition 2

Proof

The proof is a two step procedure. In the first step, the rotation between two sets of points is found by minimizing the sum of squared errors independently of the translation, showing that $E({\hat{\mathbf{R}}}) = {\bar{\mathbf{R}}}$. Next, using the result for the rotation, it is shown that $E({\hat{\mathbf{t}}}) ={\bar{\mathbf{t}}}$. Define $\widetilde{\mathbf{h}}^c_i$ and $\mathbf{g}^c_i$, as in (19). From (21) and (22), the solutions for ${\hat{\mathbf{R}}}$ and ${\bar{\mathbf{R}}}$ in (41) and (42) can be written as

$$\begin{aligned} {\hat{\mathbf{R}}}&= \underset{\mathbf{R}}{{\text {argmin}}}\sum _{i=1}^Pm_i^2||\widetilde{\mathbf{h}}_{i}^c-{\mathbf{g}^c}_{i}\mathbf{R}||^2 \end{aligned}$$

(65)

$$\begin{aligned} {\bar{\mathbf{R}}}&= \underset{\mathbf{R}}{{\text {argmin}}}\sum _{i=1}^Pm_i^2||E(\widetilde{\mathbf{h}}_{i}^c)-{\mathbf{g}^c}_{i}\mathbf{R}||^2 \end{aligned}$$

(66)

Thus, the estimated rotation is determined by the centered UME matrices, and hence, it can be obtained independently of the estimated translation. In addition, from (22) we have that the translation is selected such that the estimation error of (41) and (42) is determined completely by (65) and (66). Define the estimation errors

$$\begin{aligned} \mathbf{e}_{res,i}&= \widetilde{\mathbf{h}}_{i}^c-{\mathbf{g}^c}_{i}{\hat{\mathbf{R}}} \end{aligned}$$

(67)

$$\begin{aligned} {\bar{\mathbf{e}}}_{res,i}&= E(\widetilde{\mathbf{h}}_{i}^c)-{\mathbf{g}^c}_{i}{\bar{\mathbf{R}}} \end{aligned}$$

(68)

Taking the expectation of both sides of (67), and substituting (67) into (68), we have

$$\begin{aligned} \mathbf{g}_i^c{\bar{\mathbf{R}}} +{\bar{\mathbf{e}}}_{res,i} = \mathbf{g}_i^cE({\hat{\mathbf{R}}}) +E(\mathbf{e}_{res,i}) \end{aligned}$$

(69)

Taking the $m_i^2$ weighted average of (69), we have

$$\begin{aligned}&\frac{\sum _{i=1}^P m_i^2 \mathbf{g}_i^c}{\sum _{i=1}^P m_i^2}E({\hat{\mathbf{R}}})+ \frac{\sum _{i=1}^P m_i^2E(\mathbf{e}_{res,i})}{\sum _{i=1}^P m_i^2}= \frac{\sum _{i=1}^P m_i^2 \mathbf{g}_i^c}{\sum _{i=1}^P m_i^2}{\bar{\mathbf{R}}} \nonumber \\&\quad +\frac{\sum _{i=1}^P m_i^2{\bar{\mathbf{e}}}_{res,i}}{\sum _{i=1}^P m_i^2} \end{aligned}$$

(70)

Since the solution is obtained using the weighted-and-centered versions $\widetilde{\mathbf{h}}_{i}^c, {\mathbf{g}^c}_{i}$, it can be easily shown using the definition of the centering process in (19) that $\frac{\sum _{i=1}^P m_i^2\mathbf{e}_{res,i}}{\sum _{i=1}^P m_i^2}=0$ and $\frac{\sum _{i=1}^P m_i^2{\bar{\mathbf{e}}}_{res,i}}{\sum _{i=1}^P m_i^2}=0$. Thus, (70) becomes

$$\begin{aligned} \frac{\sum _{i=1}^P m_i^2 \mathbf{g}_i^c}{\sum _{i=1}^P m_i^2}E({\hat{\mathbf{R}}})= \frac{\sum _{i=1}^P m_i^2 \mathbf{g}_i^c}{\sum _{i=1}^P m_i^2}{\bar{\mathbf{R}}} \end{aligned}$$

(71)

Assuming $\mathbf{g}_i^c$ are not all zeros

$$\begin{aligned} E({\hat{\mathbf{R}}}) = {\bar{\mathbf{R}}} \end{aligned}$$

(72)

As for the translation, let us evaluate the mean translation from (41) by substituting (22):

$$\begin{aligned}&E({\hat{\mathbf{t}}}) = E\left[ \frac{\sum _{j=1}^Pm_j\widetilde{\mathbf{h}}_j}{\sum _{j=1}^Pm_j^2} -\frac{\sum _{j=1}^Pm_j\mathbf{g}_j}{\sum _{j=1}^Pm_j^2}{\hat{\mathbf{R}}}\right] ^T= \nonumber \\&\quad = \left[ \frac{\sum _{j=1}^Pm_jE(\widetilde{\mathbf{h}}_j)}{\sum _{j=1}^Pm_j^2} -\frac{\sum _{j=1}^Pm_j\mathbf{g}_j}{\sum _{j=1}^Pm_j^2}E({\hat{\mathbf{R}}})\right] ^T=\nonumber \\&\quad = \left[ \frac{\sum _{j=1}^Pm_jE(\widetilde{\mathbf{h}}_j)}{\sum _{j=1}^Pm_j^2} -\frac{\sum _{j=1}^Pm_j\mathbf{g}_j}{\sum _{j=1}^Pm_j^2}{\bar{\mathbf{R}}}\right] ^T = {\bar{\mathbf{t}}} \end{aligned}$$

(73)

where the last equality is due to the definition of (42). $\square$

Real Data Registration Results