Abstract and Applied Analysis

Volume 2014, Article ID 341910, 14 pages

http://dx.doi.org/10.1155/2014/341910

## Nonasymptotic Densities for Shape Reconstruction

^{1}Department of Mathematics, Washington State University, Pullman, WA 99164-3113, USA^{2}Department of Mathematics, University of Tennessee Knoxville, Knoxville, TN 37996-1320, USA

Received 17 August 2013; Revised 25 February 2014; Accepted 25 February 2014; Published 20 May 2014

Academic Editor: Victor Kovtunenko

Copyright © 2014 Sharif Ibrahim et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

In this work, we study the problem of reconstructing shapes from simple nonasymptotic densities measured only along shape boundaries. The particular density we study is also known as the integral area invariant and corresponds to the area of a disk centered on the boundary that is also inside the shape. It is easy to show uniqueness when these densities are known for all radii in a neighborhood of , but much less straightforward when we assume that we only know the area invariant and its derivatives for only one . We present variations of uniqueness results for reconstruction (modulo translation and rotation) of polygons and (a dense set of) smooth curves under certain regularity conditions.

#### 1. Introduction

This work discusses the integral area invariant introduced by Manay et al. [1], particularly with regard to reconstructability of shapes. This topic has been considered previously by Fidler et al. [2, 3] for the case of star-shaped regions. Recent results have shown local injectivity in the neighborhood of a circle [4] and for graphs in a neighborhood of constant functions [5].

The present work does not assume a star-shaped condition but does make use of a tangent-cone graph-like condition which is local to the integral area circle. We also present an interpretation of the integral area invariant as a nonasymptotic density. This is based on a poster presented by the authors [6].

Our tangentially graph-like and tangent-cone graph-like conditions (Definitions 5 and 7 in Section 2) restrict our attention to shapes with boundaries that can locally (i.e., within radius ) be viewed as graphs of functions in a Cartesian plane in one particular orientation (in the case of tangentially graph-like) or a particular set of orientations (for tangent-cone graph-like). Intuitively, these conditions guarantee that the boundary does not turn too sharply within the given radius and that working locally in Euclidean space is the same as working locally on the boundary of our shapes (i.e., the shape boundary does not pass through any given invariant circle multiple times, Section 2.2). These simplifying assumptions allow us to explicitly analyze what happens when we move along the boundary and to work locally without worrying about global effects.

We show that the tangent-cone graph-like property can be preserved when approximating a shape with a polygon (Section 3) and discuss what the derivatives of these nonasymptotic densities represent (Section 4) and show that all tangentially graph-like boundaries can be reconstructed (modulo translations and rotations) given sufficient information about the nonasymptotic density and its derivatives (Section 5 and Appendix A).

The main contribution of this paper is to show (under our tangent-cone graph-like condition) that all polygons (Theorem 29 in Section 6) and a -dense set of boundaries (Theorem 30 in Section 7) are reconstructible (modulo translations and rotations). We briefly discuss and sketch the proofs of these two theorems.

Theorem 1. *For a polygon which is tangent-cone graph-like with radius , suppose that one has the integral area invariant where is parameterized by arc length. Suppose that for all one knows and its first derivatives with respect to (disk radius) and (position along the boundary). This information is sufficient to completely determine up to translation and rotation; that is, one can recover the side lengths and angles of .*

The proof of this theorem uses the discontinuities in the derivative to determine the locations of vertices (and thus the side lengths between them). We combine the derivative and the one-sided derivative information when centered on a vertex to recover the angles at which the polygon enters and exits the circle (which might not be the polygon vertex angle if the circle contains another vertex). Doing this with the other one-sided derivative gives the same thing but using the orientation determined by the other polygon side incident to the vertex. The combination of these yields the polygon's angle at each vertex.

Theorem 2. *Define is a simple closed curve and tangentially graph-like for . Suppose that, for , for all , and for each , one knows the first-, second-, and third-order partial derivatives of . Then the set of reconstructible is dense in where reconstructability is modulo reparametrization, translation, and rotation.*

The first part of the proof shows that the derivative information can be used to obtain the curvature. However, it is not the curvature at the boundary point where the circle is centered but rather the curvature at each of the points where the boundary enters and exits the circle. Although the Euclidean distance to these points is known, the arc length distances are not and can vary from point to point. Thus the sequences of curvatures we obtain also lose the arc length parameterization of our area invariant. The rest of the proof is concerned with finding the arc length distance from the center to the entry and exit points which effectively recovers the curvature for all points. This relies on matching up the unique features of exit angle sequences with each other which in turn relies on the existence of unique maxima and minima in these sequences. While this is not true in general, it can be arranged to be so by a suitable small perturbation of the boundary (which is why our result is one of density rather than for all shapes).

This is a theoretical paper about a measure that is useful in applications: we do not pretend that the reconstruction techniques in our proofs are practically useful. In fact, the reconstructions we use to show uniqueness would be seriously disturbed by the noise that any practical application would encounter. We do, however, comment on some possible approaches to reconstruction (Section 8) using the ORTHOMADS direct search algorithm [7] to successfully reconstruct shapes which are not predicted by our theory.

#### 2. Notation and Preliminaries

Unless otherwise specified, we will be assuming throughout this paper that is a compact set with simple closed, piecewise continuously differentiable boundary of length . Let be a continuous arc-length parameterization of (see Figure 1). We will adopt the convention that traverses in a counterclockwise direction so it always keeps the interior of on the left (there is no compelling reason for this particular choice, but adopting a consistent convention allows us to avoid some ambiguities later). Note that and that restricted to is a bijection. Denote by the closed disk and the circle of radius centered at the point .

In geometric measure theory, the -dimensional density of a set at a point is given by where is the -dimensional Hausdorff measure and is the volume of the unit ball in [8]. In the current context, the -dimensional density of at is simply While we can evaluate this for all , just knowing the density at every point along the boundary is generally insufficient to reconstruct the original shape. If exists, then is approximated arbitrarily well for sufficiently small by replacing with its tangent line (which gives us an area of exactly ). Hence, we have at any point where is differentiable. That is, just knowing (i.e., the limit) is insufficient to distinguish any two shapes with boundary.

Contrast this with the situation where we know for every and (i.e., we have all of the values needed to compute the limit as well). This added information is sufficient to uniquely identify curves by recovering their curvature at every point (see Appendix A).

One natural question to ask (and the focus of the present work) is whether failing to pass to the limit (i.e., using some fixed radius instead of the limit or all ) and collecting the values for all points along the boundary preserves enough information to reconstruct the original shape. That is, can a nonasymptotic density (perhaps along with information about its derivatives) be used as a signature for shapes?

##### 2.1. Definitions

*Definition 3. *In the current context, the integral area invariant [1] is denoted by and given by

*Remark 4. *Note the lack of the normalizing factor in the definition of . Since we presume that is fixed and known for the situations we study, it is trivial to convert data between the forms and ; we choose to leave out the normalizing factor in the definition of as it is the integral area invariant of Manay et al. [1] and this form proves useful when computing derivatives in Section 4.

We introduce the tangentially graph-like condition as a simplifying assumption for the shapes we consider.

*Definition 5. *For a fixed radius , one says that is graph-like (GL) at a point (or graph-like on ) if it is possible to impose a Cartesian coordinate system such that the set of points is the graph of some function in this coordinate system. Without loss of generality, one adopts the convention that is the origin so that . One defines tangentially graph-like (TGL) in the same way but further requires that be continuously differentiable and (noting that is because is). This is illustrated in Figure 2(a). Without loss of generality (and in keeping with our convention that traverses counterclockwise), one assumes that the interior of is “up” in the circle (i.e., for sufficiently small ). If is (tangentially) graph-like on for all , one says that is (tangentially) graph-like for radius .

It is instructive to consider what is* not* considered graph-like or tangentially graph-like. Violations of the graph-like condition are generally due to a radius that is too large (certainly, choosing a radius so large that all of is in the disk will do it). For example, a unit side length square is not graph-like with radius for any (position the circle at the center of a side; see Figure 3(a)). Notice that the same square is graph-like with any radius or below. A shape can fail to be tangentially graph-like while still being graph-like if it fails to be a graph in the required orientation but works in some other (see Figure 3(b)).

We would like to consider shapes with corners but our tangentially graph-like condition requires that the boundary be differentiable everywhere. The following definitions allow us to generalize the tangentially graph-like condition to this situation by using one-sided derivatives.

*Definition 6. *Given a piecewise function , one defines the tangent cone of at a point (which is denoted by ) in terms of the one-sided derivatives. In particular, one lets where and .

*Definition 7. *One extends the tangentially graph-like notion to boundaries that are piecewise by defining to be tangent-cone graph-like (TCGL) at a point if it is graph-like at for every orientation in the tangent cone of at . More precisely, for every and every pair of distinct points , one has . See Figure 2(b).

*Remark 8. *It is clear that in Definition 6 is a convex cone. The tangent cone is dependent on the direction in which traverses (which by convention was counterclockwise) since an arc-length traversal would have different tangent cones (namely, iff ). However, these differences are irrelevant to the application of Definition 7.

*Remark 9. *Note that when is , there is only one direction in for each (i.e., the tangent to at ). Thus, the definitions of tangentially graph-like and tangent-cone graph-like coincide when is and every tangentially graph-like boundary is tangent-cone graph-like.

##### 2.2. Two-Arc Property

The graph-like condition implies (in proof of the following lemma) that will never be entirely contained in the disk, no matter where on the boundary we center it. That is, some part of lies outside of for every .

Lemma 10. *Let and . If is graph-like on , then .*

*Proof. *Suppose by way of contradiction that . Since is a simple closed curve, we have . As is graph-like at with radius , there exists some orientation for which is the graph of a well-defined function. However, is a simple closed curve so it is not the graph of a function in any orientation, yielding a contradiction.

The next result is the reason we find the tangent-cone graph-like condition useful. It says that if is tangent-cone graph-like with radius , then, for every , the disk has only two points of intersection with and these are transverse. In other words, this means that when working locally in the disk , we need only to consider a single piece of .

Theorem 11. *If is tangent-cone graph-like with radius at , then and crosses transversely at these points. As a result, for every , there is a unique arc along between them in .*

*Proof. *By Lemma 10, we have that . Note that contains an interior point () and at least two boundary points of the disk (since ). As is connected and simply closed, there must exist an arc of within the disk going from some point on through to another point on .

Suppose ; that is, there are other points of intersection. Letting denote one of these, there are two cases to consider (illustrated in Figure 4).*(a) ** Does Not Cross ** at *. As is tangent-cone graph-like at , then is a graph in every orientation in the tangent cone of at . In particular, note that the tangent line to at is in this cone. However, the line from to is normal to this line and thus is not graph-like in this orientation, a contradiction. Therefore, this case cannot occur. This argument applies to all points in so we immediately have the result that always crosses transversely.*(b) ** Crosses ** at *. There exists such that there is a path along in from to . That is, there exist (without loss of generality, ) such that , , and the image of under is contained in (but does not include , since it is on another arc and is simple). Thus, enters at and exits at .

If we can find and in the tangent cone of at satisfying , we will contradict that is tangent-cone graph-like.

Define by
Note that is in the tangent cone of at so that is graph-like using the orientation given by .

Define by . Note that from both and are directions pointing into the circle so . Similarly, points out and points in so that .

Observe that (and therefore ) is piecewise continuous since is piecewise . By a piecewise continuous analogue of the intermediate value theorem, there exists such that
By continuity of the inner product and , we have
Similarly, .

If is differentiable at , then and we have our contradiction. Otherwise, let and . As both and are in the convex tangent cone of at , any positive linear combination of them is as well. Letting , we have
Noting that is continuous in , we apply the intermediate value theorem to obtain such that . Letting , we obtain our contradiction.

Therefore, there are no other points of intersection and .

*Definition 12. *One says that has the two-arc property for a given radius if, for every point , one has that divides into two connected arcs: and . Instead of considering how divides , one can equivalently frame the definition in terms of how divides . That is, has the two-arc property if the circle is divided into two connected arcs by for every .

Corollary 13. *If is tangent-cone graph-like for some radius , then it has the two-arc property.*

*Proof. *This is a trivial consequence of Theorem 11.

Corollary 14. *If is tangentially graph-like for some radius , then it has the two-arc property for radius .*

*Remark 15. *While the assumption of the two-arc property for disks of radius does not imply the two-arc property for all (see Figure 5), it is the case that TGL for does imply that is TGL for all . The fact that is TGL for all follows easily from the definition of TGL and the fact that .

##### 2.3. Notation

Suppose that is tangent-cone graph-like with radius and we have some such that is tangentially graph-like at with radius . Since is TGL at , it has two points of intersection with by Theorem 11. In the orientation forced by the TGL condition, one of these points of intersection must be on the right side of the circle and one must be on the left side.

With reference to Figure 6 we define and so that is the point of intersection on the right and is the point of intersection on the left. The notation is motivated by the fact that in general due to our convention that traverses counterclockwise. The only case where this is not true is when is in the disk but even then it will hold for a suitably shifted that starts at some point outside the current disk.

The quantities and are the angles that the rays from the origin to the right and left points of intersection, respectively, make with the positive axis. We can assume and .

We define as the angle between the vector and the vector , the one-sided tangent to at the point of intersection on the right.

That is, we are measuring the angle between the outward normal to the disk at the point of intersection and the actual direction is going as it exits the disk. We define similarly. We have due to the fact that all circle crossings are transverse by Theorem 11.

When the proper to use is implied by context, we will often simply write , , , , , and in place of , and so forth.

##### 2.4. Calculus on Tangent Cones

The following result is a version of the intermediate value theorem for elements of the tangent cones.

Lemma 16. *Suppose that is tangent-cone graph-like on and such that . Further suppose that , , and , and let . Then, there exists such that either or is in .*

*Proof. *Let be a unit vector in with . We have . It suffices to consider only as the argument is identical in the other case. Note that since for some nonnegative constants not both zero, at least one of the inner products on the right is nonnegative. Using the notation of Definition 6, we define and have . We similarly define with respect to such that .

Define
and . Since , the argument proceeds as in Theorem 11 to yield and such that . Thus, for some . In particular, so either or (depending on the sign of ).

In addition to the intermediate value theorem, we have an analogous mean value theorem for tangent cone elements.

Lemma 17. *Suppose that is a simple, arc-length parameterized curve with piecewise continuous derivative defined on except possibly on finitely many points. Further suppose that the image of has no cusps. Then there exists in such that either or is in .*

*Proof. *Let be a unit vector with . Consider and note that is defined wherever is differentiable. We have . Thus, either everywhere it is defined or it takes on both positive and negative values. In particular, there exists a point such that either or .

If , then we have so that for some . As , we have which gives us our conclusion.

If , there exists such that . Note that and for some and let .

By the convexity of , we have with which follows as in the previous case.

The following lemma tells us that the tangent-cone graph-like condition is sufficient to apply Lemma 17.

Lemma 18. *If is tangent-cone graph-like for some radius , then has no cusps.*

*Proof. *Suppose has a cusp at . Then, using the terminology of Definition 6 and the fact that is arc length parameterized, we have . We let and note that . Letting with , we have , contradicting the fact that is tangent-cone graph-like. Therefore, has no cusps.

##### 2.5. TCGL Boundary Properties

The following technical lemmas allow us to bound various distances and areas encountered in tangent-cone graph-like boundaries.

Lemma 19. *Suppose that is tangent-cone graph-like with radius and points with . Then one of the arcs (call it ) along between and is such that, for any two points , one has .*

*Proof. *Note that so that there is an arc along from to which is fully contained in the interior of by Theorem 11. We will call this arc .

For all on , let denote the subpath of from to (so ). We claim that is contained in for all on (thus, is contained in ). Indeed, if this were not the case, then there must be some on such that is contained in but is nonempty (i.e., we can move the disk along until some part of the subpath hits the boundary). That is, the subpath has a tangency with the disk which is impossible because of Theorem 11.

Let and note that since is contained in for all on , we have that is contained in . Therefore, for all as desired.

Lemma 20. *If , where is as in the previous lemma, then the arc length between and along is at most .*

*Proof. *Since is tangentially graph-like, for any , , the angle between and is at most . Since this is true for all , there is a point and such that the angle between and tangent vectors for any other point is at most .

This means that is the graph of a Lipschitz function of rank 1 in the orientation defined by . This does not necessarily imply that , , or is the graph of a Lipschitz function; we explore a Lipschitz condition for the disks in Section 3. Let with , . Then the arclength from to is given by

Lemma 21. *If is tangent-cone graph-like with radius and with , then the image of together with the straight line from to encloses a region with area.*

*Proof. *By Lemma 20, we have that the image of under has arc length . Therefore, the region of interest has perimeter at most so by the isoperimetric inequality it has area at most from which the conclusion follows.

#### 3. TCGL Polygonal Approximations

If is tangent-cone graph-like with radius , it can sometimes be nice to know that there is an approximating polygon to which is also tangent-cone graph-like. The following lemmas explore this idea.

Lemma 22. *If is TCGL with radius , then, for each , there exists a polygonal approximation to that is TCGL with radius and such that every point on is within distance of the polygon.*

*Proof. *First, choose a finite number of points along the boundary such that the arc length along between any two neighboring points is no more than . These will be the vertices of our polygon. Similarly to , we let be an arclength parameterization of this polygon so that they both encounter their common points in the same order.

The fine spacing between vertices guarantees that we obtain the bound. Indeed, given any point and its neighboring vertices and , the arc length along from to plus that from to is at most by assumption. Since Euclidean distance is bounded above by arc length, we have . This bound in turn implies that at least one of and is bounded above by .

Consider a point on a side of the polygon (i.e., not a vertex) and its neighboring vertices and (chosen with and ). By Lemma 17, there exists such that . Note that this is the only member of up to positive scalar multiplication.

Combining the arcs along and between and , we obtain a closed curve with total length at most , so that the distance between any two points on the curve is at most . That is, for any and , we have .

Let . Then so that is contained in .

Let be distinct points on the polygon in and consider the line connecting them. This line also intersects on such that we have , , and so that . As for some scalar , we have
since is TCGL at with radius and . Thus, is TCGL at with radius .

The case where is a vertex is similar but we must consider an arbitrary vector in the inner product. We wish to show that, for every , there is a such that either or and , after which the proof follows as in the first case with (or ) in place of . We let and let and be the neighboring vertices (so and ).

As above, there exist such that , , and . Note that is exactly the set of positive linear combinations of these vectors. By Lemma 16, for every , there is a such that . As , the proof is complete.

*Definition 23. *One says that is tangentially graph-like and Lipschitz (TGLL) with radius if is tangentially graph-like with radius and there is some constant such that, for every , the arc is the graph of a Lipschitz function (in the same orientation used by the tangentially graph-like definition) and that the Lipschitz constant is at most .

*Remark 24. *Note that tangentially graph-like does not imply tangentially graph-like and Lipschitz: taking to be a square with side length whose corners are replaced by quarter circles of radius and then considering disks of radius centered on yields one example.

Because is arclength parameterized by , for all . Since is assumed on its compact domain , is uniformly continuous: for any , there is a such that if , then .

We will use the fact that always crosses transversely to prove that is in fact TGLL on slightly bigger disks of radius as long as one takes a somewhat bigger Lipschitz constant . It is then an immediate result of Lemma 22 that we can find an approximating polygon that is TCGL with radius .

Lemma 25. *If is TGLL with radius , then it is TGLL with radius for some and there is an approximating polygon which is TCGL with radius .*

*Proof. **Step 1.* Show that the quantities and are continuous as a function of (see Figure 6).

Define . Taking the derivative, we get
Because and are both less than and is graph-like in the disk, we have that both elements of this derivative are nowhere zero. By the implicit function theorem, we get that and are continuous functions of . From this it follows that and are continuous on .*Step 2*. From the previous step and the compactness of , we get that and are both bounded by . We define . Fix a . Define by . Then where , the external normal to at (see Figure 7). On any interval in where we have that is one to one and strictly increasing. Define and . We showed above that .

For , and will have to have together turned by at least radians. And until they have turned this far, . But for some . (Choosing works.) And is uniformly continuous on . Therefore, there is a such that on , , and both turn by less than . Therefore, for , we have that and intersects once for each , where .

A completely analogous argument works to show that intersects once for each .

Define to be the distance from to . Since is TGL, is greater than zero for all and is continuous in . Therefore, there is a smallest distance such that for all . Define .

Therefore, intersects exactly twice for for any .

A similar argument shows that intersects exactly twice for for any . Defining , we get that intersects exactly twice for , with the additional fact that at all those intersections.*Step 3.* TGLL implies that there is a constant such that is the graph of a function whose* x*-axis direction is parallel to and this function is Lipschitz with Lipschitz constant .

Since is uniformly continuous, there will be a such that if , then . Define . Define . Then is the graph of a Lipschitz function with Lipschitz constant at most when is used as the* x*-axis direction. That is, for all , is TGLL with Lipschitz constant for disks of radius . The result follows by Lemma 22.

#### 4. Derivatives of

Lemma 26. *Using the notation of Figure 6, we have . That is, the derivative exists and equals the length of the curve .*

*Proof. *We have the following (see Figure 8):
This difference of areas can be modeled by the difference in the circular sectors of and with angle . The actual area depends on the image of outside of , but this correction will be a subset of the circular segment of which is tangent to at the point exits. This has area by Lemma 21.

Thus we have

Lemma 27. *Using the notation of Figures 6 and 9, one has .*

*Proof. *We haveThe situation is illustrated in Figure 9 where we can see that the area being added as we go from to is the shaded region on the right with height and, considering first-order terms only, uniform width so has area . Similarly, we are subtracting the area on the left. Therefore, we have

#### 5. Reconstructing Shapes from T-Like Data

In this section, we consider the case where nonasymptotic densities and first derivatives are known along a T-shaped set (i.e., for all with a fixed radius and for all with a fixed ). We show that this information is sufficient to guarantee reconstructability modulo reparametrizations, translations, and rotations.

Lemma 28. *Assume that is TGL for (and thus all ). Then if one knows , , and for , one can reconstruct for all modulo reparametrizations, translation, and rotations (see Figure 10).*

*Proof. *As was shown in Section 4, gives us the length of the arc and tells us precisely what position this arc is along with respect to the direction . The assumption of TGL for implies TGL for (see Remark 15) and this implies that has the 2-arc property and transverse intersections with for all disks corresponding to . Since we care only about reconstructing a curve isometric to the original curve, we choose and . Taken together, and locate both points in for all . This yields . Now, simply increase , sliding the center of a disk of radius along , using to find the element of outside , and using the fact that the other element of is inside and known. This process can be continued until the entire curve is traced out in .

#### 6. TCGL Polygon Is Reconstructible from and without Tail

Theorem 29. *For a tangent-cone graph-like polygon , knowing , , and for all and a particular for which is tangent-cone graph-like is sufficient to completely determine up to translation and rotation; that is, one can recover the side lengths and angles of .*

*Proof. *For a given and where and exist, we can use them to obtain as the length of the circular arc between the entry and exit points by Lemma 26 and as the difference in heights of the entry and exit points by Lemma 27.

We wish to recover and from these quantities. Note that if is one possible solution, then so is and so solutions always come in pairs.

We can imagine placing a circular arc with angle on our circle and sliding it around until the endpoints have the appropriate height difference, yielding and . Note that since is tangent-cone graph-like, one endpoint must be on the left side of the circle and the other must be on the right and we cannot slide either endpoint to or beyond the vertical line through the center of the circle.

Therefore, as we slide the right endpoint down, the left endpoint slides up so that the height difference as a function of the slide is strictly monotonic. Therefore, the slide that gives us and is unique for a given starting arc placement. However, there are two starting arc placements: the first calls the angle for the right endpoint and the left endpoint (so the interior of is “up” in the circle) and the second swaps these (so the interior of is “down”). Since we have adopted the convention that is traversed in a counterclockwise direction (so the interior of is up in the circles), we therefore pick the first option; this gives us a unique solution for and .

This procedure works whenever and exist which is certainly true whenever the density disk does not touch a vertex of either at its center or on its boundary because if we avoid these cases, then there is only one graph-like orientation to deal with and is for all the points that enter into the computation. In fact, with a moment's thought, we can make a stronger statement than this: always exists and exists as long as the center of the density disk is not a vertex of the polygon.

We can identify the values at which does not exist to obtain the arc length positions of the vertices (and therefore obtain side lengths). For a given corresponding to a vertex, we can find and the one-sided derivatives and . These correspond to the graph-like orientations required by the polygon sides adjacent to the current vertex.

Referring to Figure 11, the one-sided derivatives along with the argument at the beginning of the proof yield the angles , , , and . Thus, we can calculate which means that the polygon vertex at has angle .

Doing this for all corresponding to vertices, we can determine all of the angles of the polygon. With the side lengths identified earlier, this completely determines the polygon up to translation and rotation.

#### 7. Simple Closed Curves Are Generically Reconstructible Using Fixed Radius Data

We will assume that is TGL for the radius . We will also assume that we know the first, second, and third derivatives of for . Under these assumptions, is generically reconstructible. By generic we mean the* admittedly weak* condition of density—reconstructible curves are dense in the space of simple closed curves.

Theorem 30. *Define as a simple closed curve and TGL for . Suppose that, for , for all , and for each , one knows the first-, second-, and third-order partial derivatives of . Then the set of reconstructible is dense in where reconstructability is modulo reparametrization, translation, and rotation.*

*Proof. *In Section 4 we showed that and , where the notation is as in Figure 6. Because is TGL, we can solve for and from these two derivatives as in the proof of Theorem 29.

*Claim 1. *The following equations hold: and .

*Proof of Claim 1. *Simply differentiate the expressions we already have for and .

We wish to express this in terms of and . Note that if we expand the circle radius by , the right exit point moves approximately (i.e., considering first-order terms only) a distance of (so , a fact we will use later to compute curvature). Therefore, Straightforward techniques yield and a similar calculation shows that .

Therefore, rewriting the second derivatives of in terms of and , we get Using these 2 derivatives, together with the previous two, we can solve for and whenever . Since we are assuming that the curve is a simple closed curve, is always true.

*Claim 2. *Knowing and gives us and , the curvatures of at and .

*Proof of Claim 2. *Computing, we get
Since and are the only unknowns, we end up having to invert
again and this is always nonsingular, giving us and as a function of , the coordinate of the center of the disk.

Relative to the horizontal, the angle of the curve at is so the rate of change in angle as we expand the circle is . Recalling that rate of movement of this exit point as we expand the circle is given by , we have that the curvature is given by . Similarly, .

*Claim 3. *Generically, we can deduce from knowledge of , , , and .

*Proof. *We outline the proof without some of the explicit constructions that follow without much trouble from the outline. We have that and . All four of these quantities (the left- and right-hand sides of each of the 2 equations) are the turning angles between the tangent to the curve at the center of the disk and the tangent to the curve at a point away from the center of the disk.

Now we use this correspondence between the curves to solve for and . But these curves can differ by a homeomorphism of the domain. Thus, we can only find the correspondence if there is a distinguished point on those curves as well as no places where the values attained are constant. The turning angle curves, having isolated critical points and a unique maximum or minimum, are enough.

To get isolated extrema, start by approximating the curve with another one that agrees in at a large but finite number of points (i.e., agrees in tangent direction as well as position) and has isolated critical points in the derivative of the tangent direction. Now perturb to one that is close (but not close) by using oscillations about the curve so that the 2nd and 3rd derivatives are never simultaneously below the bounds on the 2nd and 3rd derivatives of the curve we started with. We do this in a way that alternates around the curve. See Figure 12. In a bit more detail, suppose that . Choose a starting point on the curve; works. Now begin perturbing at the point in the positive direction such that . We name the newly perturbed curve and we keep . We continue perturbing until we have reached defined by . We begin perturbing again when we reach . Continue in this fashion around . The last piece, shown in green in the figure, will require a perturbation that is distinct in size due to the fact that it will interact with the perturbation that starts at . On this last piece, we enforce . All these perturbations can be chosen with isolated singularities in derivatives, thus giving us curves that are monotonic between isolated singularities. (In fact, we might as well choose all perturbations to be piecewise polynomial perturbations. This immediately gives us the isolated singularities and monotonicity that we want.)

Finally, if there is not a distinct maximum, we can choose one of the maxima and add a small twist to the curve at that point. See Figure 13. The idea is that a small twist, applied to leading edge of the tangents we are comparing to get the turning angle, will increase the angle most at the center of the twist. If this corresponds to a nonunique global maximum, we end up with a unique global maximum.

Now the correspondence scheme works. That is, we know that the global maximums must match, and because the turning angle curves are monotonic between isolated critical points, we can find the homeomorphisms in that move the turning angle curves into correspondence.

Taken together, the last two claims give us the curvature as a function of arclength. This determines up to translations and rotations.

#### 8. Numerical Experiments

In this section, we consider a numerical curve reconstruction for the situation in which is known for a given radius but no derivative information is available. This reconstruction is more strict than the scenarios of Sections 5–7. Our motivation is to explore whether any can be uniquely and practically reconstructed with this limited information.

We consider , the set of* simple* polygons of ordered vertices parameterized by the set with , as
for some coefficients . In this way, the polygon is a discrete approximation of a curve. The sides of are not necessarily of equal length.

We take the vector signature to be the discrete area densities of computed at each vertex. Given such a signature for fixed radius and fixed partition , we seek satisfying Equation (21) represents a nonlinearly constrained optimization problem with continuous nonsmooth objective. The constraint ensures that polygons are simple though any optimal reconstruction is not expected to lie on the feasible region boundary except in cases of noisy signatures. This approach to reconstructing curves seeks a polygon that matches a given discrete signature, rather than an analytic sequential point construction procedure.

We use the direct search ORTHOMADS algorithm [7] to solve this problem. MADS class algorithms do not require objective derivative information [7, 9] and converge to second-order stationary points under reasonable conditions on nonsmooth functions [10]. We implement our constraint using the extreme barrier method [11] in which the objective value is set to infinity whenever constraints are not satisfied. We utilize the standard implementation with partial polling and minimal spanning sets of directions.

We performed a series of numerical tests using the synthetic shamrock curve shown in black in the upper portion of Figure 14. This curve is given as a polygon in with discretization coefficients (). A sequence of reconstructions was performed with all integer values . The reconstruction begins with initial coefficients, , which determine a regular -gon with approximately the same interior area as the shamrock (as determined by the signature . In particular, the va