Isometries and Equivalences Between Point Configurations, Extended To $\varepsilon$-diffeomorphisms

TL;DR

This paper extends the Orthogonal Procrustes Problem to approximate mappings called $\varepsilon$-diffeomorphisms, providing examples and an algorithm for partitioning point configurations to facilitate their construction.

Contribution

It introduces the concept of $\varepsilon$-diffeomorphisms to the Orthogonal Procrustes Problem and proposes an algorithm for partitioning point configurations for easier map construction.

Findings

Examples where complete maps cannot be constructed despite matching distributions

An algorithm for partitioning configurations into polygons

Extension of the Orthogonal Procrustes Problem to $\varepsilon$-diffeomorphisms

Abstract

In this announcement, we deal with the Orthogonal Procrustes Problem, in which two point configurations are compared in order to construct a map to optimally align the two sets. This extends this to $\varepsilon$-diffeomorphisms, introduced by [1] Damelin and Fefferman. Examples will be given for when complete maps can not be constructed, for if the distributions do match, and finally an algorithm for partitioning the configurations into polygons for convenient construction of the maps. A revision of this announcement is the memoir preprint: arXiv: 2103.09748, [0], submitted for consideration for publication.

Figures

Equations

$$x\mapsto Ax+\vec{t}$$

$$\varphi \cdot \{i,j\}\cap \varphi \cdot \{i,k\}\ne \varnothing .$$

$$A_{ijk}=\sqrt{\frac{s(\frac{s}{2}-d{p}_{ij})(\frac{s}{2}-d{p}_{ik})(\frac{s}{2}-d{p}_{jk})}{2}}\text{where }s:=d{p}_{ij}+d{p}_{ik}+d{p}_{jk}$$

$$B_{i^{\prime }{j}^{\prime }{k}^{\prime }}=\sqrt{\frac{s^{\prime }(\frac{s^{\prime }}{2}-d{q}_{i^{\prime }{j}^{\prime }})(\frac{s^{\prime }}{2}-d{q}_{i^{\prime }{k}^{\prime }})(\frac{s^{\prime }}{2}-d{q}_{j^{\prime }{k}^{\prime }})}{2}}\text{where }{s}^{\prime }:=d{q}_{i^{\prime }{j}^{\prime }}+d{q}_{i^{\prime }{k}^{\prime }}+d{q}_{j^{\prime }{k}^{\prime }}$$

$$\mathcal{A}=\big{\{}A_{ijk}|\forall\{i,j,k\}\subseteq\{1,...,n\}\big{\}}$$

$$\mathcal{B}=\big{\{}B_{i^{\prime}j^{\prime}k^{\prime}}|\forall\{i^{\prime},j^{\prime},k^{\prime}\}\subseteq\{1,...,n\}\big{\}}\}$$

$$\mathcal{A}_{1}=\big{\{}A_{ijk}|A_{ijk}\in\mathcal{A}\text{ and }\exists B_{i^{\prime}j^{\prime}k^{\prime}}\in\mathcal{B}\text{ s.t. }A_{ijk}=B_{i^{\prime}j^{\prime}k^{\prime}}\big{\}}$$

$$\mathcal{B}_{1}=\big{\{}B_{i^{\prime}j^{\prime}k^{\prime}}|B_{i^{\prime}j^{\prime}k^{\prime}}\in\mathcal{B}\text{ and }\exists A_{ijk}\in\mathcal{A}\text{ s.t. }B_{i^{\prime}j^{\prime}k^{\prime}}=A_{ijk}\big{\}}$$

$${\mathcal{A}}_2=\mathcal{A}\backslash {\mathcal{A}}_1{\mathcal{B}}_2=\mathcal{B}\backslash {\mathcal{B}}_1$$

$$\bigl{(}\begin{smallmatrix}i&j&k&\cdots\\ i^{\prime}&j^{\prime}&k^{\prime}&\cdots\end{smallmatrix}\bigr{)}$$

$$r:=\text{Diagonal}\{{\mathcal{DP}}_{ijkl}\}s:=\text{Diagonal}\{{\mathcal{DP}}_{ijkl}\backslash \{r\}\}$$

$$\{a,b,c,d\}:={\mathcal{DP}}_{ijkl}\backslash \{r,s\}$$

$$A_{ijkl}=\frac{1}{4}\sqrt{4r^2{s}^2-(a^2+c^2-b^2-d^2{)}^2}$$

$$r^{\prime }:=\text{Diagonal}\{{\mathcal{D}\mathcal{Q}}_{i^{\prime }{j}^{\prime }{k}^{\prime }{l}^{\prime }}\}{s}^{\prime }:=\text{Diagonal}\{{\mathcal{D}\mathcal{Q}}_{i^{\prime }{j}^{\prime }{k}^{\prime }{l}^{\prime }}\backslash \{r\}\}$$

$$\{a^{\prime },b^{\prime },c^{\prime },d^{\prime }\}:={\mathcal{D}\mathcal{Q}}_{i^{\prime }{j}^{\prime }{k}^{\prime }{l}^{\prime }}\backslash \{r^{\prime },s^{\prime }\}$$

$$B_{i^{\prime }{j}^{\prime }{k}^{\prime }{l}^{\prime }}=\frac{1}{4}\sqrt{4r^{\prime 2}{s}^{\prime 2}-(a^{\prime 2}+c^{\prime 2}-b^{\prime 2}-d^{\prime 2}{)}^2}$$

$$\mathcal{A}=\big{\{}A_{ijkl}|\forall\{i,j,k,l\}\subseteq\{1,...,n\}\big{\}}$$

$$\mathcal{B}=\big{\{}B_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}|\forall\{i^{\prime},j^{\prime},k^{\prime},l^{\prime}\}\subseteq\{1,...,n\}\big{\}}$$

$$\mathcal{A}_{1}=\big{\{}A_{ijkl}|A_{ijkl}\in\mathcal{A}\text{ and }\exists B_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\in\mathcal{B}\text{ s.t. }A_{ijkl}=B_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\big{\}}$$

$$\mathcal{B}_{1}=\big{\{}B_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}|B_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\in\mathcal{B}\text{ and }\exists A_{ijkl}\in\mathcal{A}\text{ s.t. }B_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}=A_{ijkl}\big{\}}$$

$${\mathcal{A}}_2=\mathcal{A}\backslash {\mathcal{A}}_1{\mathcal{B}}_2=\mathcal{B}\backslash {\mathcal{B}}_1$$

$$(1-{\epsilon }_{ij})\le \frac{d{p}_{ij}}{d{q}_{i^{\prime }{j}^{\prime }}}=\frac{||p_i-p_j||}{||q_i^{\prime }-q_j^{\prime }||}\le (1+{\epsilon }_{ij})\text{, }\forall \{i,j\}\subseteq \{1,...,n\}\text{, given }i\ne j$$

$$\sqrt{(B_{i^{\prime }{j}^{\prime }{k}^{\prime }}{)}^2-\frac{1}{16}\cdot H_1}\le A_{ijk}\le \sqrt{(B_{i^{\prime }{j}^{\prime }{k}^{\prime }}{)}^2+\frac{1}{16}\cdot H_2}$$

$$d{q}_{i^{\prime }{j}^{\prime }}\cdot {\epsilon }_{ij-}\le d{p}_{ij}\le d{q}_{i^{\prime }{j}^{\prime }}\cdot {\epsilon }_{ij+}$$

$$d{q}_{i^{\prime }{k}^{\prime }}\cdot {\epsilon }_{ik-}\le d{p}_{ik}\le d{q}_{i^{\prime }{k}^{\prime }}\cdot {\epsilon }_{ik+}$$

$$d{q}_{j^{\prime }{k}^{\prime }}\cdot {\epsilon }_{jk-}\le d{p}_{jk}\le d{q}_{j^{\prime }{k}^{\prime }}\cdot {\epsilon }_{jk+}$$

$$E:=max\big{\{}\varepsilon_{st}|\{s,t\}\subseteq\{i,j,k\}\big{\}}$$

$$E_{-}:=(1-E)E_{+}:=(1+E)$$

$$⟹d{q}_{s^{\prime }{t}^{\prime }}\cdot E_{-}\le d{p}_{st}\le d{q}_{s^{\prime }{t}^{\prime }}\cdot E_{+}\text{ , for all pairs {s,t}}\subseteq \{i,j,k\}$$

$$s:=d{p}_{ij}+d{p}_{ik}+d{p}_{jk}$$

Full text

1. 1 Introduction

2. 2 Procrustes Problem

3. 3 Non-Reconstructible Configurations

4. 3.1 Example

5. 4 Partition Into Polygons

6. 4.1 Considering Areas Of Triangles - 10-step algorithm

7. 4.2 Considering Areas Of Quadrilaterals

8. 5 Partition Into Polygons For $\varepsilon$-distortions

9. 5.1 Areas Of Triangles For $\varepsilon$-distortions

10. 5.2 Considering Areas Of Triangles - Part 2

11. 5.3 Areas Of Quadrilaterals For $\varepsilon$-distortions

12. 5.4 Considering Areas Of Quadrilaterals - Part 2

13. 6 Reconstruction From Distances

14. 6.1 One-Sided Error Algorithm

15. 6.2 Generalizing The Results Of [3] For $\varepsilon$-distortions

16. 6.3 Construction Of Points Given Distance Distribution

17. 6.4 Kabsch’s Algorithm - Find Rotation And Translation [4],[5]

18. 6.5 Example Results And Visualization

19. 6.6 Alternative Way For Constructing Rotation After Relabelling And Shifting

20. 6.7 SVD Approach After Relabelling

21. 6.8 On Matching Point Configurations

22. 6.9 Kabsch’s Algorithm On $\varepsilon$-diffeomorphisms

23. 6.10 Difference In Rotations On $\varepsilon$-diffeomorphisms Using Kabsch’s Algorithm

1 Introduction

2 Procrustes Problem

Procrustes Problem: Let $D,n\in\mathbb{N}$, $O(D)$ the set of orthogonal transformations, and $A(D)$ the set of affine transformations in $\mathbb{R}^{D}$. Considering the following action:

[TABLE]

where $A\in O(D)$, $\vec{t}\in\mathbb{R}^{D}$, $d(\cdot)$ the Euclidean distance in $\mathbb{R}^{D}$. $X=\{x_{1},\ldots,x_{n}\}$ and $Y=\{y_{1},\ldots,y_{n}\}$ are collections of distinct points in $\mathbb{R}^{D}$. If $d(x_{i},x_{i})=d(y_{i},y_{j})$ for all $1\leq i,j\leq n$, then $\exists\varphi\in A(D)$ such that $\varphi(x_{i})=y_{i}$ for all $i,j\in\{1,...,n\}$. Given the collections $X$ and $Y$, find $A\in O(D)$ and $\vec{t}\in\mathbb{R}^{D}$, or $\phi\in A(D)$.

There are several proofs of this problem, which will be omitted.

3 Non-Reconstructible Configurations

In [2] we see an example of two set of points, where the set of their distances match but the point configurations do not, for a total number of points $n=4$. [2] Proposition 2.1 gives a simple way to check such cases.

Definition 3.1**.**

By relabelling, we mean that if there’s an initial labelling of ordered points in two incongruent configurations, we reorder them in such a way that there’s a correspondence between the points.

An example is if there’s an initial labelling of ordered points $\{a,b,c,d,e\}$ in $X$ and $\{\alpha,\beta,\epsilon,\delta,\gamma\}$ in $Y$, where $a$ corresponds to $\alpha$, $b$ to $\beta$, $c$ to $\gamma$, $d$ to $\delta$ and $e$ to $\epsilon$, one such relabelling will come from the permutation \bigl{(}\begin{smallmatrix}1&2&3&4&5\\ 1&2&5&4&3\end{smallmatrix}\bigr{)}.

Proposition 3.1**.**

$[2]$ *Suppose * $n\neq 4.$ A permutation $\phi\in S_{\binom{n}{2}}$ is a relabelling if and only if for all pairwise distinct indices $i,j,k\in\{1,...,n\}$ we have:

[TABLE]

In other words, we need to take the edges of equal length between the two configurations we are considering and check if there’s a mutual vertex between all such pairs for a given permutation $\phi\in S_{\binom{n}{2}}$. This permutation is what will give us the labelling if it does exist.

In the context of our problem, we consider the given n-point configurations $\{p_{1},...,p_{n}\}$ and $\{q_{1},...,q_{n}\}$ with their corresponding pairwise distances $D_{P}=\{dp_{ij}|dp_{ij}=d(p_{i},p_{j}),1\leq i,j\leq n\}$ and $D_{Q}=\{dq_{ij}|dq_{ij}=d(q_{i},q_{j}),1\leq i,j\leq n\}$ with $D_{P}=D_{Q}$ up to some reordering and $|D_{P}|=|D_{Q}|=\binom{n}{2}$.

We then want to find if $\exists\{i,k\},\{j,l\}$ such that $d(p_{i},p_{k})=d(q_{j},q_{l})\Leftrightarrow dp_{ik}=dq_{jl}\text{ }\forall i,j,k,l\in\{1,...,n\}$ for a permutation $\phi\in S_{\binom{n}{2}}$. In the case where this isn’t true, we need to disregard a certain number of bad points from both configurations in order to achieve this.

3.1 Example

Below is an example with two different 4-point configurations in $\mathbb{R}^{2}$ which have the same distribution of distances. The corresponding equal distances between the 2 configurations are represented in the same color, and the we have two edges with distances $1,2\text{ and }\sqrt{5}$, but it’s obvious that there doesn’t exist a Euclidean transformation between the two.

From this example we can construct infinitely many sets of 2 different configurations with the same distribution of distances. This can be done by simply adding as many points as desired on the same location across the dashed line in both the configurations of figure 2.

In both the above example and the one mentioned in [2], it suffices to exclude one point from the two configurations and you will be able to get a Euclidean motion to move from one configuration to the other.

Conjecture 3.1**.**

For two n-point configurations $P,Q\in\mathbb{R}^{2}$ with $D_{P}=D_{Q}$ for which $\nexists A\in O(2)$, $\vec{t}\in\mathbb{R}^{2}$ such that Q=\big{(}AP+\vec{t}\big{)} assuming the points have been labelled appropriately, then $\exists p_{i}\in P$ and $\exists q_{i}\in Q$ such that Q\backslash q_{i}=\big{(}A(P\backslash p_{i})+\vec{t}\big{)}, for some $A\in O(2)$, $\vec{t}\in\mathbb{R}^{2}$.

If the above conjecture holds, it suffices to exclude a single bad-point from both $P$ and $Q$, such that $P$ and $Q$ differ only by a Euclidean motion. Iterating through the potential pairs of bad-points will take $\mathcal{O}(n^{2})$, the issue still arises in determining whether the points we excluded results in two congruent configurations.

4 Partition Into Polygons

One approach we can take in order to see which points should be excluded from our 2 configurations $P$ and $Q$, is to partition the entire configurations into smaller polygons and compare polygons of the same area, in order to determine existing point correspondences between $P$ and $Q$. For any subsets $\{i,j,...\}\subseteq\{1,...,n\}$ or $\{s,t\}\subseteq\{1,...,k\}$ we consider in the upcoming sections, the elements of each subset will be distinct.

4.1 Considering Areas Of Triangles - 10-step algorithm

Considering our two n-point configurations $P=\{p_{1},...,p_{n}\}$ and $Q=\{q_{1},...,q_{n}\}$, we partition them into a total of $\binom{n}{3}\textit{ triangles }$ and considering the distance between our 3 points in each case, let’s say indexed $i,j,k$, we have the distances $dp_{ij},dp_{ik},dp_{jk}$ and analogously $dq_{i^{\prime}j^{\prime}},dq_{i^{\prime}k^{\prime}},dq_{j^{\prime}k^{\prime}}$.

We now compute the areas as follows:

[TABLE]

[TABLE]

and consider the sets of areas

[TABLE]

[TABLE]

where $|\mathcal{A}|=|\mathcal{B}|=\binom{n}{3}=\frac{n(n-1)(n-2)}{6}$.

We further partition the above sets as follows:

[TABLE]

[TABLE]

[TABLE]

Note that it may not be true that $|\mathcal{A}_{1}|=|\mathcal{B}_{1}|$, as the areas need not all be distinct.We essentially want the shapes formed by our points in the two sets which are “identical”. Let’s assume our sets $\mathcal{A}_{1}$ and $\mathcal{B}_{1}$ are in ascending order with respect to the modes of the areas. We undertake the following steps in order to check which points to disregard and permutations are valid:

Disregard all points from from $P$ and $Q$ which are vertices of triangles in $\mathcal{A}_{2}$ and $\mathcal{B}_{2}$ respectively, but at the same time not vertices of any triangle in $\mathcal{A}_{1}$ and $\mathcal{B}_{1}$. 2. 2.

In order, we take $A_{ijk}\in\mathcal{A}_{1}$ and the corresponding triangles in $\mathcal{B}_{1}$, with $A_{ijk}=B_{i^{\prime}j^{\prime}k^{\prime}}$. 3. 3.

If the distances of the sides of the triangles corresponding to $A_{ijk}$ and $B_{i^{\prime}j^{\prime}k^{\prime}}$ don’t match, disregard the triangles with area $B_{i^{\prime}j^{\prime}k^{\prime}}$. 4. 4.

If the distances match up, we assign the points the corresponding points from $P$ to $Q$ and essentially start constructing our permutation, so thus far we have:

[TABLE]

Alternatively, we can match the points between $A_{ijk}$ and $B_{i^{\prime}j^{\prime}k^{\prime}}$ which have the same corresponding angles. 5. 5.

Note that we might have more than 1 possible permutation, so for now we keep track of all of them and list them as \alpha_{s}^{(t)}=\bigl{(}\begin{smallmatrix}i&j&k&\cdots\\ i^{\prime}&j^{\prime}&k^{\prime}&\cdots\end{smallmatrix}\bigr{)} for $s$ being the indicator of the triangle we take from $\mathcal{A}_{1}$, and $t$ being the indicator of the corresponding triangle in $\mathcal{B}_{1}$ in order (so if 3 triangles correspond, we have $t\in\{1,2,3\}$).

• •

For triangles with 3 distinct inner angles we will have 1 permutation, for isosceles triangles 2 permutations, and for equilateral triangles 3!=6 permutations.

• •

In the case of squares when considering quadrilaterals, we will have 4!=24 permutations (will be discussed in section 3.3).

• •

This can be thought of as matching angles between equidistant edges of our polygons. 6. 6.

Go to the next triangle in $\mathcal{A}_{1}$ (which might have the same area as our previous triangle), and repeat steps 2-4

• •

If the distances of our current triangle match with those of our previous triangle, simply take all previous permutations and “concatenate” them. So for example \big{(}\alpha_{1}^{(1)}\big{)}_{2}=\bigl{(}\begin{smallmatrix}\alpha_{1}^{(1)}\alpha_{1}^{(2)}\end{smallmatrix}\bigr{)}, where the index $v$ in \big{(}\alpha_{s}^{(t)}\big{)}_{v} indicates the combination we have with $\alpha_{s}^{(t)}$ being the first element of the permutation as above. We therefore get a total of $\prod_{\iota=0}^{\nu-1}(t-\iota)$ permutations we are currently keeping track of, where $\nu$ is the number of elements in the constructions thus far. Note that in the above procedure we assume no common points, and the case where mutual points exists is described below. 7. 7.

If our current and previous triangles share points, we consider the combination of two triangles in $\mathcal{B}_{1}$ with the same corresponding areas and shapes and matching points as the combination of the two triangles taken from $\mathcal{A}_{1}$, so we’ll either get a quadrilateral (if they share 2 points) or two triangles sharing an vertex (not a pentagon), and check whether all $\binom{4}{2}$ or $\binom{5}{2}$ distances between our 2 shapes match up. If they do, we replace or extend the permutations we are keeping track of, and disregard any permutations from before which don’t satisfy the conditions of this bullet-point. 8. 8.

If our current and previous triangles don’t share points, we essentially repeat steps 2-5 and extend the permutations we are keeping track of, in a similar manner to that shown in step 6. 9. 9.

At this point we have traversed through all triangles in both $\mathcal{A}_{1}$ and $\mathcal{B}_{1}$ with the same area, and have constructed permutations (not necessarily all of the same size) which can be considered as sub-correspondence of points between $P$ and $Q$ (meaning that more points may be included to the correspondences). We are now going to be considering the triangles with area of the next lowest mode and repeat steps 2-8, while keeping track of the permutations we have thus far. Some steps though will be slightly modified as now we are considering various shapes (corresponding to our permutations), and in the above steps when referring to our ”previous triangle”, we will now be considering our ”previous shapes”. 10. 10.

Repeating the above until we traverse through all triangles in $\mathcal{A}_{1}$ and $\mathcal{B}_{1}$ will give us a certain number of permutations, and for our problem we can simply take the permutations of the largest size (might have multiple) and the points which aren’t included in that permutation can be considered as bad points for the problem. Note that certain points might be considered as bad for certain permutations and not for others, which depends entirely on $P$ and $Q$.

Brief Explanation On The Above Approach

The idea of the above approach is to disregard non-identical shapes and configurations of the point sets $P$ and $Q$, while simultaneously constructing the desired permutations of sub-configurations which have the same shape. Note that we start of with the triangle areas which have the smallest mode in order to simplify the implementation of this algorithm. There will exist a diffeomorphism between $P$ and $Q$ if and only if the maximum permutations constructed have size $n$, where all points will be included. A drawback of this approach, is that we keep track of a relatively large number of permutations through out this process, but when going through each set of triangles of the same area, a lot of them are disregarded in step 7.

4.2 Considering Areas Of Quadrilaterals

Alternatively, we can partition $P=\{p_{1},...,p_{n}\}$ and $Q=\{q_{1},...,q_{n}\}$, by partitioning them into a total of $\binom{n}{4}\textit{ quadrilaterals }$, and consider the $\binom{4}{2}=6$ distances between our 4 points in each case. If we take 4 distinct points indexed $i,j,k,l$, we have the set of distances $\mathcal{DP}_{ijkl}=\{dp_{ij},dp_{ik},dp_{il},dp_{jk},dp_{jl},dp_{kl}\}$ and analogously $\mathcal{DQ}_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}=\{dq_{i^{\prime}j^{\prime}},dq_{i^{\prime}k^{\prime}},dq_{i^{\prime}l^{\prime}},dq_{j^{\prime}k^{\prime}},dq_{j^{\prime}l^{\prime}},dq_{k^{\prime}l^{\prime}}\}$.

We now compute the areas as follows:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and consider the sets of areas

[TABLE]

[TABLE]

where $|\mathcal{A}|=|\mathcal{B}|=\binom{n}{4}=\frac{n(n-1)(n-2)(n-3)}{24}$.

We further partition the above sets as follows:

[TABLE]

[TABLE]

[TABLE]

We can now follow the same algorithm described in section 3.1, with the exception that now we’ll be considering 4 points at a time, rather than 3. Depending on the point-configurations $P$ and $Q$, either this approach or the previous approach might be more efficient, but this cannot be determined a priori.

5 Partition Into Polygons For $\varepsilon$-distortions

We extend our previous work to $\varepsilon$-distortions.

5.1 Areas Of Triangles For $\varepsilon$-distortions

For notational convenience we will be using the same notation used in section 3.1, as well as the fact that our sets will have the following property:

[TABLE]

rather than $dp_{ij}=dq_{i^{\prime}j^{\prime}}\Leftrightarrow||p_{i}-p_{j}||=||q_{i}^{\prime}-q_{j}^{\prime}||$.

Theorem 5.1**.**

For our usual setup, it holds that for three points in our two point configurations $P$ and $Q$ with indices and areas $\{i,j,k\}$, $A_{ijk}$ and $\{i^{\prime},j^{\prime},k^{\prime}\}$, $B_{i^{\prime}j^{\prime}k^{\prime}}$ respectively, the points can be mapped from $P$ to $Q$ through an $E$-distorted diffeomorphism if and only if

[TABLE]

where $H_{1},H_{2}$ depend on E:=max\big{\{}\varepsilon_{st}|\{s,t\}\subseteq\{i,j,k\}\big{\}}, and the elements of the distribution of distances of $B_{i^{\prime}j^{\prime}k^{\prime}}$.

Proof.

Considering the area of the triangles defined by the point $p_{i},p_{j},p_{k}$ and the corresponding points $q_{i}^{\prime},q_{j}^{\prime},q_{k}^{\prime}$, we define $\varepsilon_{ij-}:=(1-\varepsilon_{ij})$, $\varepsilon_{ij+}:=(1+\varepsilon_{ij})$, and get the following 3 inequalities for each triangle:

[TABLE]

[TABLE]

[TABLE]

In order to simplify our computations we define:

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

It then follows that for all pairs $\{s,t\}\subseteq\{i,j,k\}$, that

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

[TABLE]

Taking advantage of the triangle inequality, $s^{\prime}\geq 2dq_{i^{\prime}j^{\prime}}$, we get the following bounds:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

We know that the area of the triangle defined by the points in the configurations $P$ and $Q$ are respectively:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

In order to be as precise as possible we don’t undertake any simplifications, and from the above inequalities considering the indices $\{i,j,k\}$ and $\{i^{\prime},j^{\prime},k^{\prime}\}$, we get:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Comparing the areas of two corresponding triangles from the 2 point-configurations we then get:

[TABLE]

[TABLE]

[TABLE]

∎

5.2 Considering Areas Of Triangles - Part 2

For areas of triangles for $\varepsilon$-distortions, we construct the sets of areas of the partitioned triangles as follows:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

We then follow the exact same 10-step algorithm to get the desired result for $\varepsilon-$distortions, although now it is very unlikely that 2 or more triangles will have the exact same area.

5.3 Areas Of Quadrilaterals For $\varepsilon$-distortions

Just as above, for notational convenience we will be using the same notation used in section 3.1, as well as the fact that our sets will have the following property:

[TABLE]

rather than $dp_{ij}=dq_{i^{\prime}j^{\prime}}\Leftrightarrow||p_{i}-p_{j}||=||q_{i}^{\prime}-q_{j}^{\prime}||$.

Theorem 5.2**.**

For our usual setup, it holds that for four points in our two point configurations $P$ and $Q$ with indices and areas $\{i,j,k,l\}$, $A_{ijkl}$ and $\{i^{\prime},j^{\prime},k^{\prime},l^{\prime}\}$, $B_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}$ respectively, the points can be mapped from $P$ to $Q$ through an $E$-distorted diffeomorphism if and only if

[TABLE]

where $\hat{H}_{1},\hat{H}_{2}$ depend on E:=max\big{\{}\varepsilon_{st}|\{s,t\}\subseteq\{i,j,k,l\}\big{\}}, and the elements of the distribution of distances of $B_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}$.

Proof.

We consider our two n-point configurations $P=\{p_{1},...,p_{n}\}$ and $Q=\{q_{1},...,q_{n}\}$, and partition them into a total of $\binom{n}{4}\textit{ quadrilaterals }$, and take into account $\binom{4}{2}=6$ distances between our 4 points in each case. If we take the 4 points indexed $i,j,k,l$, we have the set of distances $\mathcal{DP}_{ijkl}=\{dp_{ij},dp_{ik},dp_{il},dp_{jk},dp_{jl},dp_{kl}\}$ and analogously $\mathcal{DQ}_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}=\{dq_{i^{\prime}j^{\prime}},dq_{i^{\prime}k^{\prime}},dq_{i^{\prime}l^{\prime}},dq_{j^{\prime}k^{\prime}},dq_{j^{\prime}l^{\prime}},dq_{k^{\prime}l^{\prime}}\}$ for our 2nd configuration. We also define $\varepsilon_{ij-}:=(1-\varepsilon_{ij})$, $\varepsilon_{ij+}:=(1+\varepsilon_{ij})$, and get the following 6 inequalities for each triangle:

[TABLE]

[TABLE]

[TABLE]

Following a similar approach to what was shown previously, we define the following parameters and compute the areas:

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

It then follows that for all pairs $\{s,t\}\subseteq\{i,j,k\}$

[TABLE]

[TABLE]

which imply that

[TABLE]

and

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Comparing the areas of two corresponding quadrilaterals from the 2 point-configurations we then get:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

∎

5.4 Considering Areas Of Quadrilaterals - Part 2

For areas of quadrilaterals for $\varepsilon$-distortions, we construct the sets of areas of the partitioned quadrilaterals as follows:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

We then follow the exact same 10-step algorithm to get the desired result for $\varepsilon-$distortions, although now it is very unlikely that 2 or more quadrilaterals will have the exact same area.

6 Reconstruction From Distances

6.1 One-Sided Error Algorithm

We want to see how likely it is to Construct Point Configurations, given the distance distributions. See [8] for more on measuring the distance between 2D point sets.

Using [3] Theorem 1.3 and considering $n$-point configurations, we can select $\binom{n}{3}$ different sets for $\{i_{0},i_{1},i_{2}\}$, $\binom{n-3}{2}$ for $\{j_{1},j_{2}\}$, $\binom{n-5}{2}$ for $\{k_{1},k_{2}\}$, $\binom{n-7}{2}$ for $\{l_{1},l_{2}\}$ and $\binom{n-9}{2}$ for $\{m_{1},m_{2}\}$.

The total number of possible such $collections$ is:

[TABLE]

Define $N:=\frac{n!}{(n-11)!}\cdot\frac{1}{96}$.

As shown above, the number of 11-tuples one has to check is very large and not practical even though it’s relatively easy to implement. An alternative way to check if our configuration is in fact reconstructible from distances is to run a one sided-error algorithm.

We make use of the polynomial defined in [3] which takes as inputs 6 distances:

g(U,V,W,X,Y,Z):=\Big{[}2U^{2}Z+2UZ^{2}+2V^{2}Y+2VY^{2}+2X^{2}W+2XW^{2}+2UVX+2UYW+2VWZ+2XYZ\Big{]}+\Big{[}-2UVY-2UVZ-2UXW-2UXZ-2UYZ-2UWZ-2VXY-2VXW-2VYW-2VYZ-2XYW-2XWZ\Big{]}

and the following facts:

• •

$g(U,V,W,X,Y,Z)=0$ if and only if the inputs are the sides and diagonals of a well-defined quadrilateral

• •

if $g(d_{\{i_{0},i_{1}\}},d_{\{i_{0},i_{2}\}},d_{\{j_{1},j_{2}\}},d_{\{k_{1},k_{2}\}},d_{\{l_{1},l_{2}\}},d_{\{m_{1},m_{2}\}})\neq 0$ for all such 11-tuples in $P$, then $P$ is reconstructible from distances

For the one-sided error algorithm we assume that $P$ is reconstructible from distances, and select at random an 11-tuple from $P$. If for the given 11-tuple we get that $g(d_{\{i_{0},i_{1}\}},...,d_{\{m_{1},m_{2}\}})=0$, we conclude that $P$ is in fact not reconstructible from distances. The issue here is that we may falsely conclude that $P$ is reconstructible from distances, with an error of $\frac{|\mathcal{K}_{1}|}{N}$, where:

[TABLE]

[TABLE]

[TABLE]

This obvious depends purely on $P$ and the cardinality of $\mathcal{K}_{2}$, and there’s nothing that can be said about it a priori. So depending on how many such bad 11-tuples exist in $P$, we will either have a large or a small error. In order to reduce this we can randomly select $x$ such tuples and check all of if them, and if at least one satisfies $g(d_{\{i_{0},i_{1}\}},...,d_{\{m_{1},m_{2}\}})=0$, we safely conclude that $P$ is not reconstructible from distances. In the case where all the tuples we selected lie in $\mathcal{K}_{1}$ we will have a false conclusion, where the error will be \Big{(}\frac{|\mathcal{K}_{1}|}{N}\Big{)}^{x}\ll\frac{|\mathcal{K}_{1}|}{N}.

6.2 Generalizing The Results Of [3] For $\varepsilon$-distortions

Theorem 6.1**.**

For a generic $P,Q\subset\mathbb{R}^{2}$, where the following conditions hold:

• •

if $dist(P)=dist(Q)$, where $dist(R)=\{dist(r_{i},r_{j})|r_{i},r_{j}\in R\subset\mathbb{R}^{2}\}$

• •

all $\binom{n}{2}$ distances $dist(p_{i},p_{j})$ are distinct and $g(U,V,W,X,Y,Z)=0$

• •

if $\{e_{1},...,e_{6}\}\in E$ for $E:=\{(p_{i},p_{j})|i,j\in\{i,...,n\}\}$, are not the diagonals of a quadrilateral then $g(e_{1},...,e_{6})\neq 0$

then $P\cong Q$.

Definition 6.1**.**

By $dist(P)\approx dist(Q)$, we mean that for each element in $dist(P)$ there exists only one element in $dist(Q)$, such that $(1-\varepsilon)\leq\frac{dp_{ij}}{dq_{i^{\prime}j^{\prime}}}\leq(1+\varepsilon)$ and vice versa.

For our case, we generalize the theorem as follows:

Theorem 6.2**.**

For a generic $P,Q\subset\mathbb{R}^{2}$, if $\exists$ $T\in A(D)$ such that $|T(P)-Q|<\varepsilon$, given $\varepsilon>0$, then the following inequalities hold:

• •

if $dist(P)\approx dist(Q)$, where $dist(R)=\{dist(r_{i},r_{j})|r_{i},r_{j}\in R\subset\mathbb{R}^{2}\}$.

• •

all $\binom{n}{2}$ distances $dist(p_{i},p_{j})$ are distinct and

[TABLE]

where $H$ depends on $\varepsilon$, the polynomial $g(\cdot)$, and the distances $\{U^{\prime},...,Z^{\prime}\}$.

• •

if $\{e_{1},...,e_{6}\}\in E$ for $E:=\{(p_{i},p_{j})|i,j\in\{i,...,n\}\}$ are not the diagonals of a quadrilateral, then

[TABLE]

Proof.

We show the derivation of the analogous conditions for the $\varepsilon$-distortions

• •

The first bullet-point is essentially what we want for the $\varepsilon$-distortions.

• •

We have:

[TABLE]

and we know that select our 6-distance collections, in order to satisfy

[TABLE]

[TABLE]

W.L.O.G., we label and reorder our $6-distance$ collections in the following way:

[TABLE]

[TABLE]

We then get the following inequalities:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

• •

Follow the same steps shown in the proof of the second inequality, with the only difference that the inequality signs are switched, since we want not almost equality. So if in the proof above we had $\beta_{1}\leq\alpha\leq\beta_{2}$, we would now use : $\alpha<\beta_{1}\text{ or }\alpha>\beta_{2}$ $\Leftrightarrow\alpha\notin[\beta_{1},\beta_{2}].$

∎

6.3 Construction Of Points Given Distance Distribution

Question : *Given $dist(P)$ and the total volume $V$ of the convex set of points $P\subset\mathbb{R}^{3}$, with $P$ unique, how do we reconstruct $P\subset\mathbb{R}^{3}$ upto a rigid motion?

In order to solve this problem, we can take the following steps, which directly relate to the 10-step algorithm:

List the distances in increasing order, and call this ordered list $D$, with $E:=|D|=\binom{n}{2}$. All $\binom{n}{2}$ edges correspond to a side of $\binom{n}{3}$ triangles of the $P$ configuration. 2. 2.

Find all possible triangles for $d_{i}\in D$ in increasing order, with $d_{i}$ being the smallest edge of the triangle. So essentially:

$i\leftarrow 1$;

$h\leftarrow 1$;

while $i\leq(E-3)$ do

       take all $\binom{n-i}{2}$ ordered 3-tuples $\{d_{i},d_{j>i},d_{E\geq k>j}\}$;

       if $(d_{k}-d_{j})\geq d_{i}$ then

             $\mathcal{T}_{h}^{(i)}\leftarrow\{d_{i},d_{j},d_{k}\}$;

             $\mathcal{T}^{(i)}=\{\mathcal{T}^{(i)},\mathcal{T}_{h}^{(i)}\}$;

             $h\leftarrow h+1$;

            

      else

             disregard 3-tuple $\{d_{i},d_{j},d_{k}\}$;

            

       end if

      $i\leftarrow i+1$;

      

end while

$\mathcal{T}^{(i)}\leftarrow\bigcup_{i=1}^{E-3}\mathcal{T}^{(i)}$;

$T\leftarrow|\mathcal{T}^{(i)}|$;

So $\mathcal{T}$ would look like \mathcal{T}=\big{\{}\{d_{1},d_{2},d_{3}\},\{d_{1},d_{4},d_{5}\},\{d_{2},d_{4},d_{5}\},...\big{\}} for instance, which is still ordered in ascending order of the first element of the 3-tuple, then the second element and then the third element. 3. 3.

Again in ascending order of $d_{i}\in D$ take all triangle 3-tuples $\mathcal{T}_{\iota}\in\mathcal{T}$ for $\iota\in\{1,...,T\}$, and then construct all possible 6-tuples of edges which form a tetrahedron with $d_{i}$ being the minimum length of its edges, by essentially combining the triangles. This can be done as follows:

take $\mathcal{T}^{(i)}$ for all $i\in\{1,...,E-3\}$ from above;

$i\leftarrow 1;\qquad\qquad$ % index of set $\mathcal{T}^{(i)}$, consisting of all triangles with $d_{i}$ as its smallest side

$j\leftarrow 2;\qquad\qquad$ % index of $\mathcal{T}^{(j)}$, for $E>j>i$

while $i\leq(E-3)$ do

       $h\leftarrow 1;\qquad\qquad$ % index for all elements of $\mathcal{T}^{(i)}$

       $\delta\leftarrow 1;\qquad\qquad$ % index of 6-tuple tetrahedron considering $\mathcal{T}^{(i)}$

       $d^{\prime}_{1}\leftarrow 1^{st}$ element of $\mathcal{T}_{h}^{(i)}$; $h_{1}\leftarrow d^{\prime}_{1}$’s original index;

       $d^{\prime}_{2}\leftarrow 2^{nd}$ element of $\mathcal{T}_{h}^{(i)}$; $h_{2}\leftarrow d^{\prime}_{2}$’s original index;

       $d^{\prime}_{3}\leftarrow 3^{rd}$ element of $\mathcal{T}_{h}^{(i)}$; $h_{3}\leftarrow d^{\prime}_{3}$’s original index;

       $h\leftarrow h+1$;

      

      while $h\leq|\mathcal{T}^{(i)}|$ do

             $\tilde{d^{\prime}}_{1}\leftarrow 1^{st}$ element of $\mathcal{T}_{h}^{(j)}$; $h_{4}\leftarrow\tilde{d}^{\prime}_{1}$’s original index;

             $\tilde{d^{\prime}}_{2}\leftarrow 2^{nd}$ element of $\mathcal{T}_{h}^{(j)}$; $h_{5}\leftarrow\tilde{d}^{\prime}_{2}$’s original index;

             $\tilde{d^{\prime}}_{3}\leftarrow 3^{rd}$ element of $\mathcal{T}_{h}^{(j)}$; $h_{6}\leftarrow\tilde{d}^{\prime}_{3}$’s original index;

             if $\{d_{h_{2}},d_{h_{3}}\}\cap\{d_{h_{4}},d_{h_{5}}\}=\varnothing$ then

                   find the triangles which have one of the following combinations of sides (both can’t occur simultaneously, and we only need to search in $\bigcup_{\eta\in\hat{H}}\mathcal{T}^{\eta}$ for $H:=\{h_{2},h_{3},h_{4}h_{5}\}$, $\hat{H}:=H\backslash max\{H\}$), and their corresponding $3^{rd}$ side is assigned below to $x_{\iota\in\{1,2\}}$:

                   $\qquad\text{(i)}\{d_{h_{2}},d_{h_{4}},x_{1}\}$ & $\{d_{h_{3}},d_{h_{5}},x_{2}\}$

                   $\qquad\text{(ii)}\{d_{h_{2}},d_{h_{5}},x_{1}\}$ & $\{d_{h_{3}},d_{h_{4}},x_{2}\}$

                   if $x_{1}=x_{2}$ then

                         $h_{6}\leftarrow x_{1}$;

                         $\Delta^{(i)}_{\delta}\leftarrow\{d_{h_{1}},d_{h_{2}},d_{h_{3}},d_{h_{4}},d_{h_{5}},d_{h_{6}}\}$;

                         $\Delta^{(i)}\leftarrow\{\Delta^{(i)},\Delta^{(i)}_{\delta}\}$;

                         $\mathcal{V}^{(i)}_{\delta}\leftarrow|\frac{1}{288}\sqrt{det\begin{pmatrix}0&1&1&1&1\\ 1&0&(d_{h_{2}})^{2}&(d_{h_{3}})^{2}&(d_{h_{6}})^{2}\\ 1&(d_{h_{2}})^{2}&0&(d_{h_{1}})^{2}&(d_{h_{4}})^{2}\\ 1&(d_{h_{3}})^{2}&(d_{h_{1}})^{2}&0&(d_{h_{5}})^{2}\\ 1&(d_{h_{6}})^{2}&(d_{h_{4}})^{2}&(d_{h_{5}})^{2}&0\end{pmatrix}}|$;

                        

                        $\mathcal{V}^{(i)}\leftarrow\{\mathcal{V}^{(i)},\mathcal{V}^{(i)}_{\delta}\}$;

                         $\delta\leftarrow\delta+1$;

                        

                  else

                         disregard current 6-tuple;

                   end if

                  

             end if

            $h\leftarrow h+1$;

            

       end while

      

$i\leftarrow i+1$;

      

end while

$\mathcal{V}\leftarrow\bigcup_{\iota=1}^{E-3}\mathcal{V}^{(\iota)};\qquad\qquad$ % Set with all tetrahedron 6-tuples

$\Delta\leftarrow\bigcup_{\iota=1}^{E-3}\Delta^{(\iota)};\qquad\qquad$ % Set with the corresponding volumes of the tetrahedrons 4. 4.

Now we want to find the collection of tetrahedrons which have a sum of volume equal to $V$. In order to do this, we re-arrange out sets $\mathcal{V}$ and $\Delta$ to $\hat{\mathcal{V}}$ and $\hat{\Delta}$, in ascending order with respect to the mode of the $\mathit{faces}$ of the tetrahedrons. We do this because it will be a lot faster to identify whether a tetrahedron is not part of the overall n-point configurations, and we will disregard it. We then go through the following steps:

1. (a)

Take in order the tetrahedron $\delta_{i}\in\hat{\Delta}$ and it’s corresponding volume $\nu_{i}\in\hat{\mathcal{V}}$, and then it’s 3-tuple ”triangle face” with the smallest mode, $t^{(i)}_{1}$. 2. (b)

Find the next tetrahedron in the ordered set $\hat{\Delta}$ which has $t^{(i)}_{1}$ as a face, and combine the two potential tetrahedrons, to get a hexahedron and its volume. 3. (c)

Considering the 6 3-tuple faces of the hexahedron we repeat step (b) and this is done until we exceed the total volume an n-tuple or we have an n-tuple with a total volume less than V. 4. (d)

In this case, we take out the last tetrahedron which was added and add the next possible candidate and check the conditions from step (c) and repeat until all possible candidate tetrahedrons have been checked. If we are not successful, we then go ”2 steps back” and take the next possible candidate for the our $2^{nd}$ most recent selection of a tetrahedron. 5. (e)

We then repeat steps (b)-(d) considering the appropriate shape we have after each iteration. 6. (f)

If we are not successful and have back-traced back to $\delta_{i}$, we consider the next appropriate tetrahedron from $\hat{\Delta}$ and repeat steps (b)-(e). 7. (g)

Given the conditions on our set $dist(P)$, this algorithm will terminate once it successfully finds the tuple set of the exterior faces of the overall convex shape formed by the configuration $P$, and the set of tetrahedrons $\Delta_{final}$ which form it. 5. 5.

At this point, we know the triangles on the exterior of the overall convex shape we are looking for, as well as the which distances from our initial set $dist(P)$ correspond to which point. From here it is therefore only a matter of selecting an arbitrary tetrahedron from $\Delta_{final}$ and placing it in $\mathbb{R}^{3}$, and based on this initialization we construct 3 points $P$, consider the tetrahedrons in $\Delta_{final}$ adjacent to the one we initially constructed and then construct 4 more points, repeat this process until we have exhausted all elements of $\Delta_{final}$. This will therefore give us an orthogonal transformation of $P$.

The construction of $P$ is unique upto a rigid motion, by the setup of the problem.

The point matching problem is further explored in [9].

Open Question : Given $dist(P)$ for $P\subset\mathbb{R}^{D}$, how do we reconstruct $P$ upto a rigid motion?

6.4 Kabsch’s Algorithm - Find Rotation And Translation [4],[5]

There’s a well known algorithm for finding the optimal rotation which minimizes the root mean squared deviation between two paired sets of points. In our case there exists an exact rotation, so by simply running this algorithm we get the exact rotation. This algorithm is broken up into the following three steps:

• •

Shift by Center of Mass

[TABLE]

[TABLE]

• •

**Find the Optimal Rotation

**Here we use the singular value decomposition of the covariance matrix $H=\tilde{P}^{T}\tilde{Q}$.

[TABLE]

[TABLE]

The $D$ matrix is used to take care of any reflections which might have taken place.

• •

Find the Translation $\vec{t}$

[TABLE]

6.5 Example Results And Visualization

Below is an example where construct a random 60-point configuration $P$. Using $P$ we then construct $Q$, such that $P\sim Q$. We then use the method described in section 6.3 to label the points, and that in section 6.4 to confirm that the configurations align. Indeed, we got that $(P-P_{\pi}^{-1}QP_{\pi})=0$.

6.6 Alternative Way For Constructing Rotation After Relabelling And Shifting

After shifting our two configurations as shown in section 6.4, we constructed the matrices $\tilde{P}$ and $\tilde{Q}$ with entries the coordinates of all points of $P$ and $Q$ with center of mass being the origin. Assuming that $n\geq D$, we can randomly select $D$ rows of $\tilde{P}$ and the corresponding rows in $\tilde{Q}$ (we have already relabelled them), and construct a change of basis. In order to do this we need to first confirm that the vectors are linearly independent which is pretty simple using software, as you simply need to make sure that the $D\times D$ matrices consisting of these vectors in their columns have a nonzero determinant, or are full-rank. So if we have selected $D$ linearly independent vector from $\tilde{P}$, which we denote as $\mathfrak{B}_{P}=\{\vec{p_{1}},...,\vec{p_{D}}\}$, and their corresponding vectors in $\tilde{Q}$, $\mathfrak{B}_{Q}=\{\vec{q_{1}},...,\vec{q_{D}}\}$, these form a basis for $\mathbb{R}^{d}$, and the rotation we are looking for is the change of basis between $\mathfrak{B}_{P}$ and $\mathfrak{B}_{Q}$. So we essentially do the following to construct the rotation $R$:

[TABLE]

[TABLE]

6.7 SVD Approach After Relabelling

An alternative way of computing the permutation $R$ is using the $SVD$ and the fact that the singular values of a matrix are unique. We are considering the case where $n\geq D$ and $U\in\mathbb{R}^{D\times D},\Sigma\in\mathbb{R}^{D\times n},V\in\mathbb{R}^{n\times n}$, so we can truncate the right singular vectors matrix $V$, as the first $D$ right-singular vectors for $\mathcal{P}=\begin{pmatrix}\vec{p_{1}}&\ldots&\vec{p_{n}}\end{pmatrix}\in\mathbb{R}^{D\times n}$ and $\mathcal{Q}=\begin{pmatrix}\vec{q_{1}}&\ldots&\vec{q_{n}}\end{pmatrix}\in\mathbb{R}^{D\times n}$ match up.

Notationally we use $A_{D}=A(1:D,1:D)$ to denote the truncation of matrix $A$, by taking the submatrix consisting of the columns and rows 1 through $D$ of $A$.

For $\mathcal{P}=U_{P}\Sigma_{P}{V_{P}}^{T}$ and $\mathcal{Q}=U_{Q}\Sigma_{Q}{V_{Q}}^{T}$, we know that $\Sigma_{P}=\Sigma_{Q}$ and that $\exists R\in O(D)$ s.t. $R\mathcal{P}=\mathcal{Q}$. It then follows that for:

[TABLE]

[TABLE]

If the singular values of $\mathcal{P}$ and $\mathcal{Q}$ are all distinct, then the $SVD$ of the two matrices are unique. If they are not distinct, then $\exists U_{P},U_{Q}$ and $M$ a permutation matrix s.t. $MU_{P_{D}}=U_{Q_{D}}$, which implies that $V_{P_{D}}=V_{Q_{D}}$. So for simplicity let’s consider this singular value decomposition, where $M=I_{D\times D}$. It then follows that for all point configurations which are congruent upto a rigid motion, there exist a singular value decomposition and $R\in O(D)$, s.t.:

[TABLE]

6.8 On Matching Point Configurations

Lemma 6.1**.**

$[6]$ Let $\mathcal{P},\mathcal{Q}\in\mathbb{R}^{D\times n}$ as defined above. Then $\mathcal{P}^{T}\mathcal{P}=\mathcal{Q}^{T}\mathcal{Q}$ if and only if $\exists A\in O(D)$ such that $A\mathcal{P}=\mathcal{Q}$.

The above Lemma implies that a necessary and sufficient condition for two configurations to be equivalent, is that their Gramian-matrices are equal, after translating them such that their center of mass is at the origin.

6.9 Kabsch’s Algorithm On $\varepsilon$-diffeomorphisms

Kabsch’s algorithm also finds the rotation in order to align point configuration $P$ with a point configuration $Q$, where $\forall i\in\{1,...,n\}\text{ }\exists i^{\prime}\in\{1,...,n\}$ such that $||p_{i}-q_{i^{\prime}}||<\frac{\varepsilon}{2}$, for $p_{i}\in P$, and $q_{i^{\prime}}\in Q$, with a given error. Note that for $\epsilon<1$ and the property required for $\varepsilon$-diffeomorphisms $(1-\varepsilon)\leq\frac{||p_{i}-p_{j}||}{||q_{i^{\prime}}-q_{j^{\prime}}||}\leq(1+\varepsilon)$, we get that \big{|}|p_{i}-p_{j}|-|q_{i^{\prime}}-q_{j^{\prime}}|\big{|}\leq\varepsilon. Below we justify that the condition on the points $||p_{i}-q_{i^{\prime}}||<\frac{\varepsilon}{2}$, satisfy the stated equivalent condition for $\varepsilon$-diffeomorphisms:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Since we have exact bounds on the difference $|p_{i}-q_{i^{\prime}}|$, we can easily implement simulation to confirm that Kabsch’s algorithm works on random n-point configurations $P$ with points in $Q$ which satisfy the above condition. The alignment will obviously not be exact. An alternate solution to the point cloud registration problem can be seen in [7], for both the rigid and non-rigid cases.

6.10 Difference In Rotations On $\varepsilon$-diffeomorphisms Using Kabsch’s Algorithm

A simulation over multiple random $P$ and corresponding $Q$ configurations and their alignment was implemented, which resulted in the error-plots provided on the next page. These were over $\varepsilon\in\{0.01,0.02,0.04,.06,0.08,0.1\}$ and $n\in\{10,12,12,...,150\}$. The error was calculated in means of sum of squared differences for each coordinate over 30 averaged simulations, and then averaged the coordinates. What the plots reveal is simply that for greater $\varepsilon$ we have a greater averaged error for the same number of points, while there is no obvious trend. This is also justified by figure 10, where a Least-Squared Fit was used, and it’s obvious that as $n$ increases, the error also tends to increase.

An alternative, more detailed approach to such configurations, is described in [6].

Acknowledgements: Neophytos Charalambides was supported by the University of Michigan Math Department and Research Experience for Undergraduates program, Brad Schwartz was supported by the University of Michigan Undergraduate Research Opportunity Program, and Dr. Steven Damelin by the American Mathematical Society. Thanks goes to all departments and programs for the continued support.