Convex hulls of bounded curvature

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Convex hulls of bounded curvature

Jean-Daniel Boissonnat, Sylvain Lazard

To cite this version:

Jean-Daniel Boissonnat, Sylvain Lazard. Convex hulls of bounded curvature. Canadian Conference on Computational Geometry (CCCG’96), 1996, Ottawa, Canada. pp.14-19. �inria-00442802�

Jean-Daniel Boissonnat

?

Sylvain Lazard

?

Abstract

In this paper, we consider the problem of computing a convex hull of bounded curvature of a set^S of points in the plane, i.e. a set containing^S and whose boundary is a curve of bounded curvature of minimal length. We prove that, if the radius of the smallest disk that contains

S is greater than¹, such a hull is unique. We show that the computation of a convex hull of bounded curvature reduces to convex programming or to solving a set of algebraic systems.

1 Introduction

The convex hull of a set of points in the plane is dened as the smallest set, or equivalently, the set of smallest perimeter that contains all the points. We consider in this paper convex hulls of bounded curvature. A curve is said of bounded curvature if it is^C1and if its curvature is upper bounded by¹everywhere it is dened. We dene a convex hull of bounded curvature of a set^S of points in the plane as a set containing^S and whose boundary is a curve of bounded curvature of minimal length.

Convex sets of bounded curvature have been con

sidered in the context of non-holonomic motion plan

ning [ART95, BL96] but we are not aware of any previous work devoted to the construction of such hulls.

In the sequel, the boundary of a region^Rwill be de

noted by^@R. A polygon whose vertices are^M1;:::;Mn such that ^M1;:::;Mn appear in this order on the boundary of the polygon will simply be called polygon

M

1 :::M

n. When necessary, the suxⁱof a vertex^Miof a polygon^M1:::Mn will be considered modulo ⁿ. Two polygons are said to be geometrically equal if they dene the same region; notice that two polygons that are geo

metrically equal have the same non-at vertices but may have dierent at vertices.

In the sequel,^S denotes a set of points in the plane,^T a convex hull of bounded curvature of^S. ^P is the usual convex hull of ^S and ^P1;:::;Pn denote the vertices of

P. ^Di denotes the closed disk of unit radius centered at ^Pi, for any ^{i 2 f1;:::;ng}. ^Q is a polygonal region whose perimeter is minimal and that intersects all the disks^D1;:::;Dn(observe that it is not required that the boundary of^Q intersects all the disks ^D1;:::;Dn). As we will see,^Qplays a central role in the characterization and in the computation of^T.

?INRIA, BP 93

06902 Sophia-Antipolis Cedex, France E-mail : ^boissonn, lazard@sophia.inria.fr

If the radius^rof the smallest disk that contains^S is smaller or equal to¹, any disk of unit radius containing

S is plainly a convex hull of bounded curvature of ^S. Notice that if ^{r < 1}there exists an innite number of such disks, and, if ^{r = 1} such a disk is unique. We assume in the sequel that ^r>1.

2 Properties of

^T

Lemma 1

^T is convex and contains ^P.

Proof:

^T is convex because otherwise its convex hull has shorter perimeter and its boundary is a curve of bounded curvature. Thus,^T contains^Pbecause^Pis the smallest

convex that contains^S. ²

We easily deduce from this lemma that the convex hull of bounded curvature of ^S is equal to the convex hull of bounded curvature of the vertices of the convex hull of

S.

Lemma 2

^@T consists of line segments and arcs of unit circles passing through the vertices of ^P.

Proof:

Since the radius of the smallest disk that con

tains ^S is strictly greater than ¹, ^@T is not reduced to a unit circle and passes through some points of ^S, which, by the remark above, are vertices of^P. Let^S0be the set of vertices of ^P. Clearly, any arc of^@T joining two oriented points ^(A;) and^(B;)in ^IR2nS0 is a lo

cally shortest path of bounded curvature¹ joining these two oriented points. Then, according to [BCL94] and [PBGM62, Theorem.25], any arc of ^@T in ^IR2nS0 is a curve ^C1 of one of the two types^CSC or^CCC where

C denotes a unit circular arc and^Sa line segment. The paths of type ^CCC cannot appear in ^@T because ^@T is convex. Thus, any circular arc that appears in ^@T is followed and preceded by a line segment and must pass through a vertex of^P.

2

Notice that not all the vertices of^Pnecessarily belong to^@T : Figure 1 shows^@T when^P is a square; when we add to^Pa fth vertex^Athat belongs to^T,^T is still the convex hull of bounded curvature of these ve vertices yet not all ve vertices belong to^@T.

Now, we transform the problem of computing^T into a more standard problem in Euclidean geometry (see Fig

ure 1) :

1A curve^Cis a locally shortest path of bounded curvature join

ing^(A;) and^(B;)if and only if any curve of bounded curva

ture joining^(A;)and^(B;)and contained in a suciently small neighborhood of^Cis longer than^C.

Q P

@T A

Figure 1: Example where not all the vertices of^Pbelong to^@T

Proposition 3

^T is the Minkowski sum of the disk of unit radius centered at the origin and of a polygonal re

gion ^Q which is, among the regions that intersect all the disks^D1;:::;Dn, one whose perimeter is minimal.

Proof:

First, notice that, since ^T is convex, the sum of the lengths of the circular arcs of^@T is equal to ². Hence, the perimeter of^T is equal to²plus the sum of the lengths of the line segments of^@T.

We recall that the eroded region of ^T by the unit disk ^D centered at the origin is ^(TcD)c where "^:c"

denotes the complementation and the Minkowski sum. In other words, the eroded region of ^T by ^D is

T n[

P2@T

D(P)where^D(P)is the translated of^Dcen

tered at ^P. Let ^Q denote the eroded region of ^T by

D.

As^@T is convex and of bounded curvature,^Q is con

vex, non empty and the Minkowski sum of^Qand^D is equal to ^T. Moreover, as ^T contains ^P, ^Q intersects all the disks^D1;:::;Dn. The perimeter of^T is equal to

2plus the perimeter of^Q. Thus,^Q is, among the re

gions that intersect all the disks^D1;:::;Dn, one whose

perimeter is minimal. ²

Let^Qdenote a polygonal region of minimal perimeter that intersects all the disks ^D1;:::;Dn. We will prove some properties of^Qin Section 3 and show in Section 4 that^Qis unique and therefore equal to^Q.

3 Properties of

^Q

Lemma 4

^Qis convex.

Proof:

^Q is convex because, otherwise, its convex hull has a perimeter strictly smaller than the one of^Q and its convex hull still intersects all the disks ^D1;:::;Dn and still has a bounded curvature; that contradicts the

denition of^Q. ²

Lemma 5

^QP.

Proof:

Assume for a contradiction that ^{Q 6 P}. The idea of the proof is to project the part of ^Q outside ^P

onto^P. Notice that we cannot simply replace the part of ^Q that is outside ^P by an arc of ^@P because the resulting polygon may possibly not intersect all the disks

D

1

;:::;D

n (see Figure 2a).

Precisely, each point of^Qoutside^P is projected onto the closest point of ^@P (see Figure 2b). That transfor

mation shortens^@Q. Moreover, each point of^Qthat be

longs to a disk^Diis projected onto a point that belongs to the same disk because ^P is convex. Thus, the trans

formed polygon still intersects all the disks ^D1;:::;Dn. As the perimeter of ^Q is minimal, we have a contradic

tion. ²

P Q

A

B

P

(a)

(b)

Figure 2: For the proof of Lemma 5

Lemma 6

^@Qintersects all the disks ^D1;:::;Dn.

Proof:

Assume for a contradiction that ^@Q does not intersect a disk ^Di0. As, by hypothesis, ^Q\Di0 is not empty, ^Pi0 belongs to the interior of^Q. Thus, by Lemma 5,^Pi0 belongs to the interior of^P which contra

dicts the hypothesis that ^Pi0 is a vertex of^P. ²

Proposition 7

There exists ^{M1 2 D1;:::;Mn 2 Dn} such that the polygon ^M1:::Mn is geometrically equal to the polygon ^Q.

Proof:

By Lemma 6, there exists ^{Mi 2Di\@Q}, ⁸ⁱ²

f1;:::;ng.

If the points^M1;:::;Mn appear in this order on ^@Q, we take ^{Mi = Mi} (^{8i 2 f1;:::;ng}). The polygon

M

1 :::M

n is geometrically equal to^Q. Indeed, as^Q is convex and^Mi2@Q, the polygon^M1:::Mn is convex and included in ^Q. Thus, the perimeter of the polygon

M

1 :::M

n is not greater than the perimeter of^Q, and it intersects all the disks ^D1;:::;Dn (because^{Mi 2Di}).

As, by denition,^Qis a polygon intersecting all the disks

D

1

;:::;D

n whose perimeter is minimal, ^Q is geometri

cally equal to the polygon^M1:::Mn.

If the points^M1;:::;Mn do not appear in this order on ^@Q, let ^Mi0 be the intersection point between ^@Q and the line segment^PiMi which is the closest from^Pi (see Figure 3). By construction, ^{Mi0 2 Di} and the line segment^PiMi0 does not intersect the interior of^Q.

2

P

Q

Mi M

0

i Pi

M

j

=M 0

j

Pj

Dj

Figure 3: For the proof of Lemma 7

If the points^M10;:::;Mn0 appear in this order on^@Q, by the same argument as above, the polygon^M10:::Mn0 is geometrically equal to^Q. Otherwise, there exist two consecutive points^Mi0 and^Mj0 on^@Qsuch that the line segments ^PiMi0 and ^PjMj0 intersect², because the seg

ments ^{P1M10;:::;PnMn0} belong to ^P (^P is convex and contains^Q) and do not intersect the interior of^Q. The two segments^PiMi0and^PjMj0 can intersect only if^Mi0or

M 0

j belongs to the intersection between the two disks^Di and^Dj(see Figure 4a). We assume, without loss of gen

erality, that^{Mi0 2Di\Dj}. We then dene^{Mj00 =Mi0}. The number of intersection points between the segments

P

1 M

0

1

;:::;P

n M

0

ndecreases by¹when we replace^Mj0 by

M 00

j

=M 0

i (see Figure 4b) : actually, on one hand, the segments^PiMi0 and ^PjMj00 do not intersect contrary to the segments ^PiMi0 and ^PjMj0; on the other hand, the

"new" segment^PjMj00 can only be intersected by a seg

ment intersecting the "old" segment^PjMj0 (because^Mi0 and^Mj0 are consecutive on^@Q). Furthermore, we claim that the line segment^PjMj00 does not intersect the inte

rior of^Q. Indeed (see Figure 4c), let^Hi be the union of the two half-planes limited by the edges of^Qincident to

M 0

i that do not contain ^Q. Let ^Hic be the complemen

tary of ^Hi. By construction, ^PiMi0 does not intersect the interior of ^Q, thus ^{PiMi0 Hi}. If ^PjMj00 intersects the interior of ^Q, ^{Pj 2 Hci} and so ^{PjMj0 Hci}. Then

P

i M

0

i

\P

j M

0

j

= ;, which contradicts our assumption and proves the claim.

Repeating this procedure for all the pairs of con

secutive points on ^@Q such that the corresponding line segments intersect, we dene a list of points

M

1

;:::;M

n that belong to^@Q such that the segments

P

1 M

1

;:::;P

n M

n do not pairwise intersect and do not intersect the interior of^Q.

The fact that the segments ^{P1M1;:::;PnMn} do not pairwise intersect, are included in^Pand do not intersect the interior of ^Q, implies that the points ^M1;:::;Mn appear in this order on ^@Q. By the same argument as above, the polygon^M1:::Mn is geometrically equal to

Q. ²

Remark 8

The points^M1;:::;Mnmay not be unique.

2We say that two line segments intersect if their relative interior intersect.

Q

H c

i

H

i M 0

i

=M 00

j

M 0

j

Pi Pj Pj

P

i M

0

i

=M 00

j M

0

j

Dj

D

i

P

j

M 0

j M

0

i

=M 00

j Q

P

Pi

(a) (b)

(c)

Figure 4: For the proof of Proposition 7

For example in Figure 1, the number of non-at vertices of^Qis smaller than the number of vertices of^P. In that case, one of the^Miis a at vertex of^Qand moving^Mi inside ^Di along^@Qdoes not modify^@Q.

Proposition 9

If ^{M12 D1;:::;Mn 2Dn} are the ver

tices of a polygon geometrically equal to^Q, then, for any non-at³vertex ^Mi,

M

i belongs to the boundary of ^Di,

the segments^{Mi?1 Mi}and^MiMi+1 do not intersect the interior of^Di,

the line ^PiMi is the bisector of the two lines

M

i?1 M

i and ^MiMi+1 that separates ^Mi?1 and

M

i+1.

Proof:

The rst claim of the proposition is a direct consequence of the second one.

Let ^Mi be a non-at vertex of ^Q. As ^Mi is not at, the line segment ^{Mi?1 Mi+1} does not intersect

D

i. As the perimeter of ^Q is minimal, the polygo

nal line ^{Mi?1 MiMi+1} is, among the polygonal lines

M

i?1 MM

i+1 such that^M2Di, one of smallest length.

The set of points ^M such that the length of the polyg

onal line^{Mi?1 MMi+1} is equal to a given^lis an ellipse whose focuses are ^Mi?1 and ^Mi+1 . It follows that ^Mi is the common point of^Di and the ellipse whose focuses are^Mi?1 and^Mi+1 that is tangent to ^Di and does not enclose^Di (see Figure 5). This proves the second claim of the proposition.

3If ^Q is reduced to a point, ^{Mi = Mj} for all ^(i;j) and we consider the vertices^Mias at, by convention.

A well known property of the ellipses is that the nor

mal line to an ellipse at a point ^M is the bisector of the two lines^{Mi?1 M} and^MMi+1 that separates the fo

cuses. In our case, the normal to the ellipse at the point

M

i is also normal to the boundary of ^Di and so passes through^Pi. Therefore, ^PiMi is the bisector of the two lines ^{Mi?1 Mi} and ^MiMi+1 that separates ^Mi?1 and

M

i+1. ²

M

i?1

M

i+1 M

i P

i D

i

Figure 5: ^PiMi is the bisector of the two lines^{Mi?1 Mi} and^MiMi+1 that separates^Mi?1 and^Mi+1

4 Uniqueness of

^Q

and of

^T

We dene the following function^f :

f : D

1

:::D

n

! IR

M

1

;:::;M

n

7! kM

1 M

2 k+kM

2 M

3 k+:::

+kM

n?1 M

n k+kM

n M

1 k

where ^kMiMi+1k denotes the Euclidean distance be

tween the points^Mi and^Mi+1.

Proposition 10

^f is a convex function.

Proof:

A simple computation, omitted here, yields the

proposition. ²

Proposition 11

^f(M1;:::;Mn)is the minimum of^f if and only if the polygon^M1:::Mn is, among the regions that intersect all the disks ^D1;:::;Dn, one of minimal perimeter.

Proof:

Let ^M1:::Mn be such that ^f(M1;:::;Mn) is minimum. Let ^Q be a polygon of minimal perimeter that intersects all the disks ^D1;:::;Dn. Clearly, the perimeter of ^Q is smaller or equal to the perimeter of the polygon^M1:::Mn. By Proposition 7, ^Qis geomet

rically equal to a polygon ^M1?:::Mn? where ^{Mi? 2 Di}. As^{f(M1?;:::;Mn?)}cannot be smaller than the minimum of^f,^{f(M1?;:::;Mn?)}, which is equal to the perimeter of

Q, is also equal to^f(M1;:::;Mn), which is equal to the perimeter of the polygon^M1:::Mn.

Conversely, if^f(M1;:::;Mn)is not the minimum of^f, the perimeter of the polygon^M1:::Mn is not minimum

among the regions that intersect all the disks^D1;:::;Dn.

2

Lemma 12

Let ⁱ be the angle ^{6 (P??i?P?i?1!?P?i?M!i)} where

M

i is a point of the boundary of ^Di and let^{Ui [0;2]}

be the set of the ⁱ such that ^Mi2P (¹ⁱⁿ). Let^g be

g : U

1

:::U

n

! IR

1

;:::;

n

7! kM

1 M

2 k+kM

2 M

3 k+:::

+kM

n?1 M

n k+kM

n M

1 k

and let ^{D [0;2]n} be the open set of the ⁼

(

1

;:::;

n ) 2 U

1

::: U

n such that the polygon

M

1 :::M

n does not intersect the interior of the disks

D

1

;:::;D

n.

Then, ^g is locally strictly convex on ^D, i.e. for any

2 D, there exists an open neighborhood of such that the restriction of^gon this neighborhood is a strictly convex function.

Remark 13

^g is not convex on ^D because it can be shown that^Dis not a convex set. Notice that^2Dif and only if the polygon^M1:::Mndoes not intersect the interior of the disks^D1;:::;Dn and if the edge^MiMi+1 is neither tangent to^Di nor to^Di+1,^8i2f1;:::;ng.

Proof:

We consider the function

~ g

i : U

i U

i+1

! IR

i

;

i+1

7! kM

i M

i+1 k:

We show by computing the Hessian matrix of^g~i that ^~gi is locally strictly convex at any point^(i;i+1)such that the relative interior of the line segment^MiMi+1does not intersect (and is not tangent to)^Di and^Di+1. It follows that ^~gi is locally strictly convex on the projection^Di of

D onto ^UiUi+1. We omit here these computations.

Let^gi be the function

g

i : U

1

:::U

n

! IR

1

;:::;

n

7! kM

i M

i+1 k:

Since ^{g =P1ingi}, the fact that ^~gi is locally strictly convex on^Di implies that^g is locally strictly convex on

D. ²

Proposition 14

^Qis unique.

Proof:

Let^Q be a polygon of minimal perimeter that intersects all the disks ^D1;:::;Dn. Since the radius of the smallest disk that contains^S is strictly greater than

1,^\1inDi=;; therefore,^Qis not reduced to a point.

By Proposition 7,^Qis geometrically equal to a polygon

M

1 :::M

n such that ^Mi2Di,^8i2f1;:::;ng.

Let^Mi1;:::;Miq be the non-at vertices of^Qand let

U be the set of all the polygons that intersect the^qdisks

D

i1

;:::;D

iq.

The proof consists of three steps : rst, we show that the polygon ^Mi1:::Miq is, among the polygons of ^U,

4