    # Supporting Ruled Polygons

We explore several problems related to ruled polygons. Given a ruling of a polygon P, we consider the Reeb graph of P induced by the ruling. We define the Reeb complexity of P, which roughly equates to the minimum number of points necessary to support P. We give asymptotically tight bounds on the Reeb complexity that are also tight up to a small additive constant. When restricted to the set of parallel rulings, we show that the Reeb complexity can be computed in polynomial time.

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## 1 Introduction

Gauss’s Theorema Egregium states that any isometric embedding of a surface preserves the (Gaussian) curvature everywhere on the surface . A particularly important example of this is the case of flat, or rectifiable, surfaces. The Gaussian curvature is the product of the so-called principal curvatures, and so, zero-curvature implies that at every point, some direction lies in a straight line (a principal curvature of zero). Ruled surfaces are one such example, but the most well-known example is pizza. If one rolls a (flat) triangular piece of pizza in one direction, the curvature in that direction is non-zero and thus, unless the pizza stretches or tears, the curvature in the orthogonal direction must be zero. This is what keeps the tip of the pizza from flopping downward.

In this paper, we attempt to establish a theoretical foundation for algorithmic problems on isometric embeddings of rectifiable polygons in . These are planar polygons that have embedded in so that the curvature stays zero everywhere locally, but the embedding may not lie in a plane. Our original motivation came from the question of how many single points of contact where necessary to support a polygon so that none of the corners can “flop”. A similar problem was studied in robotics for holding cloth , in which the researchers rediscovered the Art Gallery problem . To correctly state such problems in the zero-curvature case, we first establish a vocabulary for rulings and give a description of the intrinsic topology. The ruling is the set of lines of zero curvature in some isometric embedding. This allows us to abstract away issues of physics (gravity, for example) and embeddings (self-intersection). Along the way, we connect these rulings to a generalization of Reeb graphs that allows us to connect ruled polygons to an art gallery-type theorem, phrased in terms of topological simplification.

For a ruled polygon with no holes, cutting along a ruling line divides it into two pieces. The ruling is supported by a set of points if every line of the ruling has a point of on both pieces of . For example, two points suffice to support a triangle (slice of pizza). We give the formal definition of the Reeb complexity of a polygon in Section 4, but roughly, it corresponds to the minimum size support set over all possible rulings. In Section 4, we prove that all -gons have Reeb complexity at most , and we give a family of polygons with Reeb complexity .

We conclude with a collection of open problems and research directions that we believe could be of interest as they provide connections between classic problems in computational geometry such as monotone polygons, Hamiltonian triangulations, and art galleries and growing new areas such as Reeb graphs and topological simplification.

## 2 Definitions

Let be a simple polygon in the plane. Let be an isometric embedding of into . This is one for which the distance between any pair of points on the embedded surface is the same as in the plane. Every point on will have a principle curvature of zero in some direction. We will limit ourselves to nondegenerate embeddings, in which there is a unique such direction at each point. These correspond to line segments covering the polygon. Rather than working directly with embeddings of , we will look at the patterns induced by these line segments on itself. Thus, we use Gauss’s theorem to make statements about nondegenerate isometric embeddings by reasoning directly about planar polygons.

A ruling of is a set of line segments in with both endpoints on the boundary, whose interiors partition the interior of . Moreover, we require that no two distinct segments in a ruling are collinear and intersect. This last condition may seem strange at first, but is a fundamental issue in rulings, particularly in defining a topology on the rulings. According to this definition, the segment through the reflex vertex of the polygon cannot be replaced by two shorter line segments. Segments that contain a reflex vertex in their relative interior are called branch segments.

A degenerate embedding of does not have a corresponding ruling. This happens both for the extreme cases where the entire embedding lies in a plane, but also in other more interesting cases.

We say that a ruling is simple if it has no branch segments. A ruling is parallel if every pair of segments are parallel. A ruling is Morse if no branch segment contains more than one reflex vertex in its relative interior.

## 3 The Topology of Rulings

Given a simple polygon with or without holes, the Reeb graph  with respect to a continuous function is the quotient space where if and only if and are in the same connected component of , where . For an accessible introduction to Reeb graphs, see . We can construct a similar space from the ruling lines of . That is we define the Reeb graph of a ruling of , , to be the quotient space , where if and only if for some segment . The branch segments of decompose the polygons into pieces which correspond to edges in the Reeb graph, glued together at internal nodes corresponding to the branch segments themselves. has a natural graph metric (i.e., is a 1-dimensional stratified metric space) where the distance between two equivalence classes is the Hausdorff distance between the line segments containing and . Recall that the Hausdorff distance between two compact subsets and of a metric space is defined as

The constructions of a Reeb graph of a ruling and the Reeb graph constructions align for particularly nice rulings, hence the naming convention. If we have a parallel ruling , then the Reeb graph of the ruling is equivalent to the Reeb graph formed from the height function orthogonal to the ruling. Alternatively, if we have a simple ruling on , then we may consider the midpoint of each line segment , which collectively trace out a path . We can then define a continuous function by , where if is on line segment , , yielding .

For completeness’ sake, the remainder of this section will explore the relations between the homotopy type and homology of a polygon and its Reeb graph of a ruling , . A known result in  states that if a space is locally path connected and is partitioned into connected equivalence classes by and is semilocally simply connected, then is surjective, where is the topological quotient map. Since each are path-connected, and all other conditions are satisfied, we have that is a surjection. Note that we use the fact that is homotopy equivalent to .

With regards to homotopy, if has holes then it is homotopy equivalent to , where is the wedge sum formed by taking the disjoint union of copies of and adjoining them each at a single point. The fundamental group of , , is then equal to , the free product on generators. If has no holes, then it is contractible, and we have that and are homotopy equivalent.

With regards to homology, the Hurewicz Theorem states there is an isomorphism between and , where the former is the abelianization of . Since , then as expected.

The following theorem due to Vietoris  allows us to provide an isomorphism between the homology of and that of .

[Vietoris-Begle Mapping Theorem] Given compact metric spaces and and surjective map , if for all , for all , , then is an isomorphism for and a surjection for .

Consider the quotient map , which is surjective by definition. Given , each fiber is the line segment corresponding to the equivalence class , thus contractible, so is acyclic for all dimensions . Theorem 3 then implies that .

## 4 Asymptotically Tight Bounds

Let be a simple polygon with vertices and holes. We define the Reeb complexity of as the minimum number of leaves in over all possible rulings of . In this section we show that the Reeb complexity of is upper bounded by and can be as large as . To show the upper bound we consider the special case of parallel rulings, whereas to show the lower bound we construct a family of polygon for which any ruling must induce a Reeb graph with leaves.

### 4.1 Upper Bound

For any vector

, there is a parallel ruling defined by sweeping the line orthogonal to across . We think of as the “height” direction, and the function maps to its “height”, , in the direction . Here denotes the dot product. The Reeb graph is the quotient space constructed by contracting the connected components of to single points as sweeps through in the direction . We denote by the number of branch (internal) nodes in and by the number of leaves.

A reflex vertex of a polygon is a vertex whose interior angle is strictly greater than ; see Figure 1. We denote by the set of reflex vertices of and define . Note that a reflex vertex cannot be on the convex hull of . The reflex vertices play an important role in determining the number of leaves in a Reeb graph induced by a parallel ruling. Each reflex vertex bears witness to a closed set of vectors which, if the rulings are induced by any vector in the set, eliminate from the Reeb graph. That is is not a critical point of and does not correspond to a node of .

Let be a reflex vertex of . Denote by and the normals to the edges and adjacent to , respectively. We will consider two double cones defined by the normals at apex . Consider the vector and notice that . Define the closed double cone ; see Figure 1. Figure 1: A polygon with a reflex vertex at p. The cone Cp is shown in green and is defined by the lines with directions n1 and n2 and intersecting p

Let be a simple polygon with or without holes and be an arbitrary direction. Consider an arbitrary leaf node . Then for all , . Suppose, for the sake of contradiction, that there exists a reflex vertex . Consider the ruling line at , and divide into two rays with base point . Since is a leaf node, the ruling line at locally intersects the interior of in a single connected component. Thus one of the two rays, say , points into the exterior of between the two edges adjacent to . However, this implies that we can perturb in directions and while still locally intersecting the interior of . This contradicts that fact that is a leaf node.

Let be a simple polygon with or without holes, be a reflex vertex, and be an arbitrary direction. Then creates a branch node in if and only if . Note that, as a consequence of Lemma 4.1, can only create a branch node in . Suppose that creates a branch node in . For this to occur, the ruling line at locally intersects the interior of in two connected components. This happens if and only if the vector orthogonal to is not in the set .

From Lemmas 4.1 and 4.1, we see that defines precisely the set of vectors such that the ruling eliminates from . In Section 5, we will use the cones to compute the Reeb complexity of a polygon when restricted to the set of parallel rulings.

When is Morse, every branch node of has degree 3. In this case we have that , where is the total number of edges in . Since the holes are disjoint, each hole creates a cycle in adding one edge to the total number of edges, giving . Combining the expressions we get the relation . Note that when we recover the relationship between the number of internal nodes and the number of leaves in a tree. Furthermore when is not Morse, the equality becomes the inequality .

Let be a simple polygon with holes, be the number of reflex vertices, and be an arbitrary direction. If is Morse, then has at most leaves. In the worst case, the vector is not in for any . By Lemma 4.1 every reflex vertex creates a branch node in . Since is Morse, we have the relationship .

This result is tight in that there exists a polygon and a direction such that has exactly leaves; see Figure 2. However it is easy to construct polygons where all but a constant number of vertices are reflex vertices. In such cases we can bound the number of branch nodes in the Reeb graph much more tightly. Figure 2: A polygon with 6 reflex vertices and a parallel ruling for which the Reeb graph has 8 leaves. However had we taken v to be the x-direction, the Reeb graph would only have 2 leaves, and the ruling would be simple.

Let be a simple polygon with holes and vertices, and let be an arbitrary direction. Then number of branch nodes . Let denote the number of non-reflex vertices and note that . By Lemma 4.1, a reflex vertex either forms a branch node or is eliminated from ; it follows that . Similarly, as a consequence of Lemma 4.1, every leaf node is created by the ruling passing over one or more non-reflex vertices and it follows that . Combining these inequalities with the inequality , we have that

 b+2−2h ≤l ≤m 2b+2−2h ≤m+b ≤m+k =n b ≤n−2+2h2.

When is Morse, we can use the relation to bound the Reeb complexity of any ruling as . Notice that this bound holds for any arbitrary ruling, not just parallel rulings, as increasing the set of rulings considered only decreases the Reeb complexity.

Let be a simple polygon with holes and vertices. Let be any ruling of . Then the Reeb complexity is at most .

### 4.2 Lower Bound

Consider once again the example shown in Figure 2. Had we chosen the direction orthogonal to , the Reeb graph would have only 2 leaves. To establish a lower bound, and thus show that our result is asymptotically tight, we construct a family of polygons whose Reeb complexity is .

Consider two concentric circles centered at the origin and with radii respectively, where . We parameterize and . For each , we construct a set of vertices, of which will be placed on , with the remaining vertices being placed on . The first set of vertices are placed on at for . The second set of vertices are placed on at for . The edges of the polygon are constructed by connecting the th vertex of to the and th vertices of . See Figure 3. Notice that every vertex on is a reflex vertex of the polygon.

Now consider a vertex on , and suppose that some (not necessarily parallel) ruling eliminates from . Then there exists some line segment of the ruling with endpoint . The other endpoint of can be contained in one of only two (when

is odd) or three (when

is even) other spikes of the polygon, due to the limited visibility at . We prove this statement in the following paragraphs. Crucial to this argument is the fact that must be on the boundary of in a spike different than that of for to be a valid ruling.

Let be the vertices on that are adjacent to . The length of the segment , the length of the arc connecting and . The affine hulls of the edges and each intersect in two points. Let and be the intersections not equal to and ; see Figure 4. We will show that the length of the segment , which, for an appropriate choice of , covers at most intervals when is odd and intervals when is even. Figure 4: For the spike at p we consider the affine hulls of the edges adjacent to p. The intersection of these affine hulls with C2 define p′1 and p′2. The point q can lie in any of the spikes spanned by the segment p′1p′2.

Let and be the midpoints of the segments and respectively. Notice that the triangles and are similar. Define the lengths and . Since and are similar we have the relationship . We can write the length for some . Here is the distance between and . Similarly we can write ; see Figure 4. Then we have that

 d′ =d|pm′||pm| =dr1+r2sinφr1−r2+δ ≤πnr1+r2sinφr1−r2+δ ≤πnr1+r2r1−r2+δ ≤πnr1+r2r1−r2,

where the first inequality follows from , the second from , and the third from . Then taking we have that

 |p′1p′2| =2d′ ≤2πnr1+1r1−1.

The vertices on are spaced so that each interval has arc-length . All that remains is to compute the number of intervals a segment of length can cover. When , this amounts to computing the angle , shown in Figure 4, when is at its maximum value. The angle . Then the maximum number of intervals spanned by the segment is

 2arcsin(πnr1+1r1−1)2πn=nπarcsin(πnr1+1r1−1),

which approaches as . For the purposes of this construction we take . When , for all , the segment can span at most intervals when is odd and at most when is even.

Suppose that corresponds to the th spike. When is odd must intersect one of only two possible intervals, those corresponding to the spikes and . The th and th spike have overlapping regions of a single interval. However if a ruling line attempts to eliminate the vertex on the th spike, the resulting line segment must intersect . This is impossible in any valid ruling, and it follows that at most spikes can be eliminated from . Thus the Reeb complexity of this polygon is at least . Note that the polygon has vertices in total, so this matches our upper bound. In conjunction with Theorem 4.1, we’ve established the following theorem.

Let be a simple polygon with holes and with vertices. Let be any ruling of . Then the Reeb complexity of is upper bounded by . Furthermore there exists simple polygons for which the Reeb complexity is at least .

While our bound is asymptotically tight, the additive difference between the example used to establish the lower bound and the upper bound proved in Theorem 4.1 is . It remains open whether there exists a polygon for which every direction induces a Reeb graph with exactly leaves.

## 5 Computing the Reeb Complexity for Parallel Rulings

Given a simple polygon with holes and vertices we wish to compute the Reeb complexity of . In Section 6, we conjecture that this problem is NP-complete for general rulings. In the special case of parallel rulings, we show that the problem can be solved in time.

By Lemma 4.1, finding a parallel ruling of minimum Reeb complexity is equivalent to finding a vector that is contained in the maximum number of cones . We use the standard duality transform that maps a point to the line . In the dual plane, a parallel ruling dualizes to a vertical line, because each line in has the same slope. Similarly, the two lines that bound the cone dualize to two points where are the slopes of and are the -intercepts. Figure 5: The set of cones for each reflex vertex translated to the origin. The boundaries of these cones dualize to points in the dual plane. We disregard the y-coordinate of the dualized points and consider the resulting list of slopes on the x-axis. Then finding a vector in the maximum number of cones is equivalent to finding a line in the maximum number of intervals. Dualizing the entire dotted line as a set of points, gives the desired ruling.

The algorithm begins by computing the cone for each reflex vertex . The duality transform is applied to the set of lines , giving the set of points . Notice that the -intercept values can be disregarded, as we are interested in a vector based at the origin that lies in the maximum number of cones translated to the origin. The set of slopes define a set of intervals. Sort the endpoints of the intervals and call the resulting list . Finding a vector that is contained in the maximum number of cones is equivalent to finding the vertical line that lies in the maximum number of intervals in .

There is one remaining caveat. Notice that traversing the list of slopes in increasing order corresponds, in the primal plane, to traversing the cones in rotary order starting with the vector and performing a rotation of . The vector may already lie in a subset of the cones . Consider and its corresponding pair . In this case, the first endpoint of the interval that the traversal encounters in is an exit event, not an entry event. The pair corresponds to the interval . To account for this, the algorithm first computes , and labels the elements of with the correct entry/exit labels. The algorithm keeps a counter , initialized with the value , and traverses incrementing on each entry event, and decrementing on each exit event. The maximum value of this counter gives the minimum number of leaves , where is the number of reflex vertices.

The runtime of the algorithm is dominated by sorting , which takes time. The other steps of the algorithm – computing the cones , dualizing the boundary lines , computing the subset , assigning the correct labels to the intervals, and computing – can all be done in time. The correctness of the algorithm follows from Lemma 4.1. We have established the following theorem.

Let be a simple polygon with holes and vertices. The Reeb complexity of , restricted to the set of parallel rulings, can be computed in time.

Note that the algorithm presented in this section is equivalent to an algorithm in the primal space where a vector is rotated once around the origin. As the vector rotates around the origin, the algorithm keeps track of entry and exit events defined by each cone. We chose to present the algorithm in the dual space because future extensions to more general classes of rulings will likely operate in the dual space. As shown in Figure 5, a parallel ruling corresponds to a vertical line in the dual space, since each line segment of the ruling has identical slope. More general rulings correspond to curves in the dual space. Characterizing the set of curves that correspond to valid rulings of a polygon is likely to be an important first step to settling algorithmic questions related to Reeb complexity.

## 6 Conclusions and Open Problems

Many problems on Reeb complexity of polygons and rulings remain open.

1. Give an algorithm to determine if the Reeb complexity of a polygon is at most a given bound . We conjecture that this problem is NP-Complete.

2. A special case of the problem above is to test if a polygon admits a simple ruling. When this problem was posed at the open problem session of CCCG 2016, David Eppstein observed that a polygon admits a simple ruling if and only if some subdivision of the edges results in a polygon that admits a Hamiltonian triangulation . It may be possible to adapt the algorithm in that paper to this problem. There is likely also a connection to sweepable polygons , 2-walkable  polygons, and algorithms for detecting them.

3. What rulings correspond to physically realizable rulings? A similar problem is to characterize the rulings that result from a given support set under the effects of gravity.

4. Is every Reeb graph of a ruling on also the Reeb graph of a continuous function on ?

5. A ruling is called proper if no two line segments share an endpoint. Can the Reeb complexity of a polygon change if we only permit proper rulings?

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