1 Introduction
Abstract

Let be a point robot moving in the plane, whose path is constrained to forward motions with curvature at most one, and let be a convex polygon with vertices. Given a starting configuration (a location and a direction of travel) for inside , we characterize the region of all points of that can be reached by , and show that it has complexity . We give an time algorithm to compute this region. We show that a point is reachable only if it can be reached by a path of type CCSCS, where C denotes a unit circle arc and S denotes a line segment.

\title

Reachability by Paths of Bounded Curvature in a Convex Polygon\thanksO.C. was supported by Mid-career Researcher Program through NRF grant funded by the MEST (No. R01-2008-000-11607-0).

1 Introduction

The problem of planning the motion of a robot subject to non-holonomic constraints [16, 25] (for instance, bounds on velocity or acceleration [9, 11, 22], bounds on the turning angle) has received considerable attention in the robotics literature. Theoretical studies of non-holonomic motion planning are far sparser.

In this paper we consider a point robot in the plane whose turning radius is constrained to be at least one and that is not allowed to make reversals. This restriction corresponds naturally to constraints imposed by the steering mechanism found in car-like robots. We assume that the robot is located at a given position (and orientation) inside a convex polygon, and we are interested in the set of points in the polygon that can be reached by the robot. We put no restriction on the orientation with which the robot can reach a point.

The lack of such a restriction distinguishes our work from most of the previous theoretical work on curvature-constrained paths, which usually assumes that not a point, but a configuration (a location with orientation) is given. Dubins [12] was perhaps the first to study curvature-constrained shortest paths. He proved that a curvature-constrained shortest path from a given starting configuration to a given final configuration consists of at most three segments, each of which is either a straight line or an arc of a unit-radius circle. Reeds and Shepp [21] extended this characterization to robots that are allowed to make reversals. Using ideas from control theory, Boissonnat et al. [6] gave an alternative proof for both cases, and Sussmann [26] extended the characterization to the 3-dimensional case.

In the presence of obstacles, Fortune and Wilfong [13] gave a single-exponential decision procedure to verify if two given configurations can be joined by a curvature-constrained path avoiding the polygonal obstacles. On the other hand, computing a shortest bounded-curvature path among polygonal obstacles is NP-hard, as shown by Reif and Wang [23]. Wilfong [27] designed an exact algorithm for the case where the curvature-constrained path is limited to some fixed straight “lanes” and circular arc turns between the lanes. Agarwal et al. [2] considered the case of disjoint convex obstacles whose curvature is also bounded by one, and gave efficient approximation algorithms. Boissonnat and Lazard [8] gave a polynomial-time algorithm for computing the exact shortest paths for the case when the edges of the obstacles are circular arcs of unit radius and straight line segments. Boissonnat et al. [7] gave an algorithm for finding a convex and simple path of bounded curvature within a simple polygon. Agarwal et al. [1] presented an -time algorithm to compute a curvature-constrained shortest path between two given configurations inside a convex polygon. They also showed that there exists an optimal path that consists of at most eight line segments or circular arcs. For general polygonal obstacles, Backer and Kirkpatrick [5] recently gave the first complete approximation algorithm, improving on earlier work that approximated the shortest “robust” path [15, 3].

At least two interesting problems have been studied where not configurations but only locations for the robot are given. The first problem considers a sequence of points in the plane, and asks for the shortest curvature-constrained path that visits the points in this sequence. In the second problem, the Dubins traveling salesman problem, the input is a set of points in the plane, and asks to find a shortest curvature-constrained path visiting all points. Both problems have been studied by researchers in the robotics community, giving heuristics and experimental results [24, 18, 19]. From a theoretical perspective, Lee et al. [17] gave a linear-time, constant-factor approximation algorithm for the first problem. No approximation algorithms are known for the Dubins traveling salesman problem.

Our result is a characterization of the region of points reachable by paths under curvature constraints from a given starting configuration inside a convex polygon . We show that all points reachable from the starting configuration are also reachable by paths of type CCSCS, where C denotes an arc of a unit-radius circle and S denotes a line segment. When has vertices, we show that the reachable region has complexity , and we give an time algorithm to compute this region.

2 Terminology and some lemmas

Let be a convex polygon in the plane. A configuration is a point together with a direction of travel (a unit vector). By a path, we mean a continuously differentiable curve (the image of a -mapping of to ) with average curvature bounded by one in every positive-length interval. Unless stated otherwise, we assume that a path is completely contained in . A configuration on the path is a configuration with on such that is the forward tangent to in . The starting configuration of is the starting point of with its forward tangent. A simple path is a path with no self-intersection; we allow the endpoints of a simple path to coincide, in which case we call it a simple closed path. (Hence, a simple closed path is smooth except possibly at one point.)

Given a configuration , the left disk (right disk ) is the unit disk touching and completely contained in the left (right) halfplane defined by the directed line through . (All unit disks in this paper are unit-radius disks.)

The left directly accessible region is the set of all points in that can be reached by a path with starting configuration consisting of a single (possible zero-length) circular arc on the boundary of followed by a single (possible zero-length) line segment. The right directly accessible region is defined analogously. The directly accessible region is the union of the left and right directly accessible region.

Pestov-Ionin lemma.

The following lemma is perhaps the foundation for all our results. In a slightly less general form, it was proven by Pestov and Ionin [20]. Recently, it has been used for a curve reconstruction problem [14].

Lemma 1 (Pestov-Ionin).

Any simple closed path contains a unit disk in its interior.

For sake of completeness, we sketch a proof analogous to Pestov and Ionin’s.

Lemma 2.

Let be a closed disk, and be a simple path with endpoints such that . Then there is a unit disk touching  that lies within the region bounded by and the exterior arc of

Figure 1: Illustration of Lemma 2. The region is shaded.

(See Figure 1).

Proof.

We proceed by induction on the length of (or, more precisely, by induction on ).

When , we can prove by integration that a unit disk tangent to does not cross . We consider a unit disk tangent to at on the interior side. Since has only one connected component and , clearly is contained in .

Otherwise, let be the point halfway between and on . Let be the largest disk contained in that is tangent to in . If the radius of is larger or equal to , then we are done. If not, must touch in another point . Clearly lies on , and the length of the arc of between and is at most . By the induction hypothesis, a unit disk lies inside the region bounded by and . Since , the lemma follows. ∎

Lemma 1 follows from Lemma 2 by observing that it still holds when degenerates to a point.

Filling.

By we denote the set of all unit-radius disks that are completely contained in , and we let be the union of all the disks in . (See Figure 2.) Both and will be called the filling of .

Figure 2: The filling is shaded in light grey, and the three pockets are shaded in dark grey.

Pockets.

The connected components of are called the pockets of . A pocket of is bounded by a single circular arc (lying on one disk of ) and a connected part of the boundary of . The first and last edge on this connected chain are called the mouth edges of the pocket. Its extremities are called the mouth points. The mouth edges form an angle smaller than  (this is equivalent to observing that the mouth points lie on the same disk of and form an angle smaller than [1]. Agarwal et al. [1] proved the following lemma.

Lemma 3 (Pocket lemma).

A path entering a pocket from cannot leave the pocket anymore.

Proof.

We consider a pocket bounded by a disk . Suppose that there is a path that enters and leaves . There is a subpath of whose endpoints lie on and whose other points are in the interior of . By Lemma 2, there is a unit disk that touches at a point in the interior of , and is contained in . Hence, , and intersects the interior of , a contradiction. ∎

Reachability for a union of disks.

For a set of unit disks, we let denote the set of all unit disks contained in the convex hull of . Equivalently, consists of the unit disks centered at points of the convex hull of the set of all centers of the disks in .

Observation 4.

Let be a set of unit disks in the plane. Then .

Given a convex set , a configuration on the boundary of is a configuration with on the boundary of and such that is tangent to the boundary of in .

Lemma 5.

Let be a set of unit disks, and let be a starting configuration on the boundary of . Then no point in the interior of can be reached by a path starting at and contained in , or even in any convex polygon such that .

Proof.

Assume to the contrary that there is a path with starting configuration on the boundary of and ending point  in the interior of . Assume for the moment that lies completely in .

We extend to infinity using a straight ray, such that the extended path is still . We then extend backwards, by attaching a single loop around the boundary of at . To summarize, the extended path starts at , makes a single loop around the boundary of , then follows the original path , and finally escapes to infinity along a straight line. We can now construct a simple, closed path as follows: starting at infinity, we follow backwards, until we encounter the first intersection of with the part that we have already seen.

Figure 3: Proof of Lemma 5. The set consists of three disks. The convex hull is shaded in light grey, and is in dark grey. The path from to is enlarged into the path .

Such an intersection must exist since the two extensions intersect. We define  to be the part of  between these two self-intersection points. Observe that  lies either on or outside the closed loop .

By the Pestov-Ionin Lemma, contains a unit-disk . Since is contained in , we have . Consequently, the interior of lies in the interior of , a contradiction with the fact that  must lie on or outside .

The lemma still holds when we allow the path to lie inside a convex polygon  with . After all, by Lemma 3, the path cannot return to after it has left it. ∎

The characterization.

Consider a starting configuration on the boundary of the filling . It is easy to see that any point in not in can be reached by a path from . On the other hand, by Lemma 5, no point in can be reached, and so we have a complete characterization of the region reachable from as the complement of .

If the starting configuration lies on the boundary of an arbitrary unit disk contained in , the same characterization holds. For arbitrary starting configurations, however, the situation becomes far more complicated. There is the possibility that no path starting at is tangent to the boundary of , and the filling has no relation to the reachable region.

If there exists a path starting at that is tangent to the boundary, all points outside are reachable, but it is still possible that some points inside are reachable, for instance because they lie in the directly reachable area , or because lies in a pocket with additional maneuvering space (that we would not have been able to exploit if starting inside by Lemma 3). (See Figure 4).

Figure 4: Points in the grey region are reachable from , but are neither in nor in the complement of .

In the rest of this paper, we give a complete characterization of the reachable region, for any starting configuration in . Let us denote the set of points such that is reachable by a path starting from a configuration  by .

3 Paths starting along the boundary

In this section, we assume that the starting configuration is on the boundary of (recall that this means also that the direction is tangent to the boundary). Without loss of generality, we also assume that the direction of is counterclockwise along the boundary, so that points of are reached locally by a left turn from . It turns out that in this situation we can restrict ourselves to paths containing no right-turning arcs.

The forward chain is the longest subchain of the boundary of , that starts counterclockwise from , and that turns by an angle at most . (See Figure 5.)

Figure 5: The forward chain .

In other words, when is directed vertically upward, this chain contains all the edges of that are above its interior, as well as the part of the edge that contains and is above , and, if there is one, the other vertical edge of .

If the forward chain intersects the interior of , then we have the following simple description of the reachable region.

Lemma 6.

If the forward chain intersects the interior of , then .

Proof.

The left directly accessible region can be enlarged to a pocket of the left disk . Thus, no point outside is reachable by the Pocket Lemma (Lemma 3). ∎

We denote by the set of disks contained in that touch the forward chain. Note that always contains the left disk . Let us remark that the set of centers of the disks in need not be connected. For example, Figure 6 shows a situation where consists of just three disks; their centers are marked by black dots (in general, any number of connected components is possible).

Figure 6: Example where the left filling consists of just 3 disks.

For a disk , we define as , where is the first configuration on touching  (by the above, this is well defined). Note that if , then is simply the complement of the interior of . We continue with a lemma on the reachable points outside .

Lemma 7.

Let and be as above, with , and the forward chain not intersecting the interior of . Suppose that there is a path from to , where . Then there exists a disk such that and is not in the interior of , or there exists a disk such that .

Proof.

Without loss of generality, we assume that is directed vertically upward. We first assume that is in the interior of .

Consider the line . Let us first assume that intersects this line top-to-bottom (or tangentially) at . In that case, we extend forward by the semi-infinite ray starting at , and backward by the boundary of . We trace the resulting path backwards from infinity, stopping at the first intersection of the path with the part we have already seen, and thus forming a loop that does not contain  in its interior. We apply the Pestov-Ionin lemma to this loop, and find a unit-radius disk contained in it. (See Figure 7.)

Figure 7: (left) A unit disk lies in the shaded area. (right) is in .

If , then we are done. Otherwise, note that the loop does not cross the boundary of , so . Then we obtain a disk by translating upwards until it touches the forward chain , and we have .

Now we consider the case where intersects the line  bottom-to-top at . Let be the first point of intersection of and the line  (along the path ). If lies between and on the line , then we can apply the argument above to conclude the existence of a disk such that —but then also .

If, finally, lies between and on , then we extend by the semi-infinite ray starting in . This ray must intersect the part of from to , and so again we have found a loop lying in , fulfilling the requirements of the Pestov-Ionin lemma, and containing . As above, there is then a disk with .

We now consider the case where does not lie in the interior of . Then lies in a connected component of different from . Since does not intersect the forward chain, lies entirely below . Let denote the first point on that is on the boundary of . We trace backward from a path along , and then along the lower semi-circle of , until we reach a point that we have already seen. It forms a loop on which we apply the Pestov-Ionin lemma. Thus we find a disk inside that does not contain . If , then we choose and we are done. Otherwise, we obtain by translating upward until it meets the forward chain, and we have . ∎

The lemma above does not give us a complete characterization of the reachable region, as we do not know yet whether we can reach the disks in and tangentially. The following lemma addresses this issue.

Lemma 8.

Assume that the forward chain does not intersect the interior of . (i) If , then there exists a counterclockwise configuration tangent to that can be reached from by a path. (ii) If , then the first configuration on tangent to can be reached from by a path.

Proof.

The lemma is obvious when touches the edge containing , so we assume this is not the case. We first prove (i). If we draw a line segment upward from the leftmost point  of  until we meet the forward chain, we do not intersect . It follows that . We consider a ray that starts at a configuration  tangent to . We start at and move counterclockwise along the boundary of , so that the ray sweeps . Since , this ray must meet  at some point. When the ray first meets , it is tangent to , so we can reach the corresponding configuration by a CS path.

We now prove (ii). We denote by the first configuration on the forward chain that is tangent to . Then , so when we sweep the same ray as in the proof of (i), we meet  tangentially at some configuration . The arc between  and  is inside , so we have a CSC path starting from and going through and . ∎

We are now able to give the following characterization for the reachable region starting from a configuration on the side of . It follows directly from Lemmas 6, 7, and 8.

Proposition 9.

Assume that is a configuration on the boundary of , oriented counterclockwise. Then any point in can be reached by a CSCS path. In addition, we have that:

  • If intersects the interior of , then .

  • If does not intersects the interior of , then .

In the characterization above, it seems that an infinite number of disks could possibly contribute to the boundary of the reachable region. In the following, we show that the contribution of the s along any edge can be reduced to at most two s. We focus on a particular edge . Let denote the counterclockwise direction along this edge. Then we order the counterclockwise configurations along according to direction , that is, for two such configurations and , we say that when .

Lemma 10.

Let and be two counterclockwise configurations on the same edge  of , such that . Let denote the first counterclockwise configuration on such that and intersects . Then .

Proof.

We first assume that intersects the interior of , so that . Then can be enlarged into a pocket, so by Lemma 3, we have .

Otherwise, does not intersect the interior of . Thus, the disk touches . (See Figure 8.)

Figure 8: Proof of Lemma 10.

If , then is a pocket, so by Lemma 3, we have .

Finally, we assume that . Let denote a point in . If  is reached after an arc of with length less than  followed by a line segment , then it is clearly in . (See Figure 8, left). On the other hand, if  is reached by an arc of with length at least , followed by a segment , we claim that . Let be the point of such that the line is tangent to from the left. (See Figure 8, right.) We have to argue that the arc of  from  to  lies in . This follows from the fact that is obtained by translating along  until it touches , that the arc of from to  is in , and that the arc from to  is shorter than the arc from  to . ∎

Now we can show how to construct the reachable region from a configuration on the boundary.

Proposition 11.

Let be a counterclockwise configuration on the boundary of an -sided convex polygon . Then we can compute in time a set of disks, such that .

Proof.

We use the characterization of from Proposition 9. If intersects the interior of , then we just set . So in the remainder of this proof, we assume that does not intersect the interior of .

If , then we first construct the contribution of the disks in . By Observation 4, we only need to find the unit disks whose centers are the vertices of the convex hull of the centers of the disks in . These disks are tangent to at least two edges of , so their centers lie on the medial axis [4, 10]. of . We compute this medial axis in  time using an algorithm by Aggarwal et al. [4], and then check each edge on the medial axis to obtain these disks in  time.

We now observe that if , then we can set and are done. Thus, in the remainder of this proof, we assume that , and explain how to find the contribution of .

Let denote the first point on , starting from in counterclockwise direction, such that . (See Figure 9(left).) Let us call a candidate configuration a configuration  on the counterclockwise boundary of  with the property that and such that is either tangent to two edges of , or is tangent to one edge of  and contains the point .

Consider an arbitrary edge  of . Let denote the first counterclockwise configuration on such that . When is as in Lemma 10, the s of all the disks in that are tangent to are contained in . So we only need to find and to construct the contribution of to .

We observe that and  are candidate configurations: in fact, is the first candidate configuration on , while is the last.

Figure 9: On the left, is shaded. On the right, is shaded, and its medial axis is dashed.

It remains to explain how to compute the candidate configurations efficiently. Denote by a simple polygon obtained by replacing the subchain of that goes counterclockwise from  to  with two or three edges, such that . (See Figure 9(right).) The disk , for a candidate configuration , must lie in  and must touch  in more than one point. It follows that we can find the candidate configurations by first computing the medial axis of  in time using the algorithm by Chin et al. [10], and then checking all edges of the medial axis. ∎

Proposition 11 shows that the reachable region for a configuration on the boundary of  is delimited by  disks. This bound is tight, as shown by the example in Figure 10, where disks of  contribute to the boundary of the reachable region.

Figure 10: Example where disks of contribute to the boundary of the reachable region. Here the disks of are centered along the dashed circle. The reachable region is shaded.

4 Special left-right and right-left paths suffice

Let be a starting configuration. A canonical RL-start from  is a path from  to a configuration  on the boundary of  that begins with a right-turning arc of unit radius and continues with a left-turning arc of unit radius ending at  (and tangent to the boundary of  there) (Figure 11).

Figure 11: A canonical RL-start.

Note that for each edge  of  and for a given , there are at most two canonical RL-starts from  ending on . A canonical LR-start is defined analogously: it begins with a left-turning arc and continues with a right-turning arc.

In this section, we show that for determining the reachability by paths in a convex polygon, it suffices to consider paths of a fairly special form. Namely, we show that a point is reachable if and only if it is directly accessible, or it can be reached by a path that begins with a canonical start.

Dubins [12] showed that the shortest path of bounded curvature between two configurations in the plane is of type CSC or CCC. In the latter case, the middle arc has length more than . (See Figure 12.) We call these paths Dubins paths.

Figure 12: Three types of Dubins paths.
Proposition 12.

Let be a convex polygon, let  be a starting configuration in , and let be reachable from  by a bounded-curvature path. Then lies in the directly accessible region , or it can be reached by a path of one of the following forms: a canonical RL-start followed by a left-turning path (starting on a side of ), or a canonical LR-start followed by a right-turning path (starting on a side of ).

Proof.

Jacobs and Canny [15] showed that, in a polygonal environment, the shortest path of bounded curvature between two configurations is a sequence of Dubins paths. The final configurations of these Dubins paths (except for the last one) all lie on the boundary of the polygonal environment. So, if we denote by a shortest path from to , then can be written as a sequence of  Dubins paths. When , we know that the final configuration of lies on .

We handle three cases separately, according to the type of (see Figure 12). If and are contained in , then , so from now on, we assume that or crosses .

Figure 13: Proof of Proposition 12, case (i).

Case (i).

We assume that is of type . We denote by , , the three circle arcs such that , and recall that has length larger than . If touches we are done, so from now on we assume that does not touch . Let be the disk supporting . Without loss of generality, we assume that and turn counterclockwise and turns clockwise. (See Figure 13(a).) Let denote the center of . We denote by the disk obtained by rotating  by an angle  around . We define as the smallest  such that and touches , or and touches . Since , such a exists. There is a path from to consisting of an arc of , an arc of , a line segment, an arc of  and . (See Figure 13(b).) This means that can be reached by a canonical LR-start and a right-turning path.

Figure 14: Proof of Proposition 12, case (ii).

Case (ii).

We assume that , where is left-turning, is a segment and is right-turning. (See Figure 14.)

Let us first assume that , that is that and that has length less than . In this case, lies in . Indeed, if has length larger than  or if lies to the left of the directed line  defined by , then . (See Figure 14(c)). If has length less than  and lies to the right of , then , since cannot enter because has length less than .

It remains to consider the case where or the length of is at least . Let be the unit disk tangent to and lying to the right of the segment , and denote by the disk obtained by rotating  by an angle  counterclockwise around . (See Figure 14(a).) Let be the clockwise arc of  starting at and going clockwise until the first intersection point with the path , or returning to its starting point if there is no such intersection. Let be the smallest value of such that touches . Since , such a exists. Now can be reached by a path consisting of an arc of , an arc of , a line segment, and an portion of . (See Figure 14(b).) It follows that  can be reached by a canonical LR-start and a right-turning path.

Figure 15: Proof of Proposition 12, case (iii).

Case (iii).

We assume that , where and are right-turning and is a segment. If the length of is less than  and (see Figure 15(a)), then , and therefore . We therefore assume that or the length of is at least  .

We denote by the disk obtained by rotating around the center  of  by an angle . (See Figure 15(b).) When intersects , we denote by the arc of that starts at and goes clockwise until it meets . Otherwise, we denote . As before, let be the smallest value of such that touches . Again, exists since . Now can be reached by a path that consists of an arc of , an arc of , a segment tangent to and , and a subpath of . (See Figure 15(c).) This implies again that  can be reached by a canonical LR-start followed by a right-turning path. ∎

We can give a somewhat different characterization:

Proposition 13.

Let be a convex polygon, let be a starting configuration in , and let be reachable from by a bounded-curvature path. Then is reachable by a path of the form CCSCS. More precisely, is reachable by a path of the form CS, or it is reachable by a path of the form CCSCS, where the two final disks touch the boundary of , and the path goes through these touching points.

5 Putting everything together

In this section, we show how to construct the reachable region when  is an arbitrary configuration in . We obtain it by combining the results in sections 3 and 4. We will prove the following:

Theorem 14.

Let be an -sided, convex polygon, and let be a configuration inside . Then the reachable region from  inside  is delimited by arcs of unit circles, and we can compute in  time.

Proof.

Let be a point in . By Proposition 12, either is in , or it can be reached after a canonical start. We only consider canonical RL-starts; the case of LR-starts can be handled symmetrically.

The directly accessible region is delimited by two circle arcs, which can be computed in time by brute force. We determine the at most  canonical RL-starts by brute force, in time. For each canonical start, by Proposition 11, we compute in time a set of configurations on the side of such that the union of their s form the reachable region after this canonical start.

We have thus obtained a set of  configuration on the boundary of  such that the union of their s with is . By Lemma 10, we only need to keep two such configurations per edge: the first and the last one. As we have only arcs to consider, we can construct by inserting these arcs one by one, and updating the reachable region by brute force. As these arcs are arcs of unit circles, each one of them appears only once along the boundary of . So overall, it takes time. ∎

Acknowledgments

This problem was first posed to us by Hazel Everett. We miss her. We also thank Sylvain Lazard, Ngoc-Minh Lê, and Steve Wismath for discussions on this problem.

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