Sweeping costs of planar domains
SSWEEPING COSTS OF PLANAR DOMAINS
BROOKS ADAMS, HENRY ADAMS, AND COLIN ROBERTS
Abstract.
Let D be a Jordan domain in the plane. We consider a pursuit-evasion, contamination clearing,or sensor sweep problem in which the pursuer at each point in time is modeled by a continuous curve,called the sensor curve . Both time and space are continuous, and the intruders are invisible to the pursuer.Given D , what is the shortest length of a sensor curve necessary to provide a sweep of domain D , so thatno continuously-moving intruder in D can avoid being hit by the curve? We define this length to be the sweeping cost of D . We provide an analytic formula for the sweeping cost of any Jordan domain in terms ofthe geodesic Fr´echet distance between two curves on the boundary of D with non-equal winding numbers.As a consequence, we show that the sweeping cost of any convex domain is equal to its width, and that aconvex domain of unit area with maximal sweeping cost is the equilateral triangle. Introduction
Let D be a Jordan domain, i.e. the homeomorphic image of a disk in the plane. Suppose that continuously-moving intruders wander in D . You and a friend are each given one end of a rope, and your task is to dragthis rope through the domain D in such a way so that every intruder is eventually intersected or caught bythe rope. What is the shortest rope length you need in order to catch every possible intruder? We refer tosuch a continuous rope motion as a sweep of D (Figure 1), and we refer to the length of the shortest suchpossible rope as the sweeping cost of D . t = 0 t = t = t = t = t = 1 Figure 1.
An example sweep of a domain in the plane. A time t = 0 the entire domain iscontaminated, and at time t = 1 the clearing sweep is complete.The problem we consider is only one example of a wide variety of interesting pursuit-evasion problems; seeSection 2 for a brief introduction or [14], for example, for a survey. It is a pursuit-evasion problem in which a r X i v : . [ m a t h . GN ] J u l oth space and time are continuous, the pursuer is modeled at each point in time by a continuous curve, theintruder has complete information about the pursuer’s location and its planned future movements, and thepursuer has no knowledge of the intruder’s movements. Our problem can also be phrased as a contamination-clearing task, in which one must find the shortest rope necessary to clear domain D of a contaminant which,when otherwise unrestricted by the rope, moves at infinite speed to fill its region.As first examples, the sweeping cost of a disk is equal to its diameter, and the sweeping cost of an ellipseis equal to the length of its minor axis. These computations follow from Theorem 5.4, in which we provethat the sweeping cost of domain D is at least as large as the shortest area-bisecting curve in D .One motivation for considering a pursuer which is a rope, or a continuous curve at each point in time,is the context of mobile sensor networks. Suppose there is a large collection of disk-shaped sensors movinginside a planar domain, as considered in [17, 1]. What is the minimal number of sensors needed to clear thisdomain of all possible intruders? If n is the number of sensors, and n is the diameter of each sensor, thenas n → ∞ an upper bound for the number of sensors needed is given by the sweeping cost.As our main result, in Theorem 7.1 we provide an analytic formula for the sweeping cost of an arbitraryJordan domain D . Indeed, the sweeping cost of D is equal to the infimum, taken over all pairs of curves inthe boundary of D whose concatenation wraps a nontrivial number of times around the boundary, of thegeodesic Fr´echet distance between the two curves. The geodesic Fr´echet distance differs from the standardFr´echet distance in that the distance between two points in D is not their Euclidean distance, but insteadthe length of the shortest path between them in D . Our Theorem 7.1 is also closely related to the geodesicwidth between two curves [18].Using our main result, we prove in Theorem 8.1 that the sweeping cost of a convex domain D is equalto the width of D . As a consequence, it follows that the sweeping cost of a polygonal convex domainwith n vertices can be computed in time O ( n ) and space O ( n ) using the rotating calipers technique [28].Furthermore, it follows from [24, 11] that a convex domain of unit area with the maximal possible sweepingcost—i.e. the most expensive convex domain to clear with a rope—is the equilateral triangle.An intriguing open question motivated by our work is the following (Question 7.5). Given a Jordandomain D and two continuous injective curves α, β with image in the boundary ∂D , is the weak geodesicFr´echet distance between α and β equal to the strong geodesic Fr´echet distance? The weak version of theFr´echet distance allows α and β to be reprarametrized non-injectively, whereas the strong version does not.We review related work in Section 2, state our problem of interest in Section 3, and describe some basicproperties of the sweeping cost in Section 4. In Section 5 we provide a lower bound on the sweeping cost interms of shortest area-bisecting curves. We provide analytic formulas for the sweeping costs of Jordan andconvex domains in Sections 7 and 8, and in Section 9 we deduce that the sweeping cost of a unit-area convexdomain is maximized by the equilateral triangle. The conclusion describes related problems of interest, andthe appendix contains two technical lemmas and their proofs.2. Related work
A wide variety of pursuit-evasion problems have appeared in the mathematics, computer science, engi-neering, and robotics literature; see [14] for a survey. Space can modeled in a discrete fashion, for exampleby a graph [3, 7], or as a continuous domain in Euclidean space as we consider here. Time can similarly bediscrete (turn-based) or continuous, as is our case. See [2, 6, 13, 16, 23, 25] for a selection of such problems. further important distinction in a pursuit-evasion problem is whether information is complete (pursuersand intruders know each others’ locations), incomplete (pursuers and intruders are invisible to each other),or somewhere in-between. Our problem can be considered as one in which the pursuer has no knowledge ofthe intruders’ movements, whereas the intruders have complete knowledge of the pursuer’s current positionand future movements. In other words, the pursuer must catch every possible intruder. Evasion problemsin which the pursuer has no information and the intruders have complete information can equivalently becast as contamination-clearing problems, see for example [4, 17, 1]. Indeed, the contaminated region of thedomain at a particular time includes all locations where an intruder could currently be located, and theuncontaminated region is necessarily free of intruders. It is the task of the pursuer to clear the entire domainof contamination, so that no possible intruders could remain undetected.The paper [18] introduces the geodesic width between two polylines, a notion that is very relevant for ourproblem. The geodesic width between two curves α , β is the same as the strong geodesic Fr´echet distancebetween them (Definition 3.3) when domain D is chosen to be a region with boundary consisting of curves α , β , and the two shortest paths connecting the endpoints of α and β . If curves α and β are polylines with n vertices in total, then [18] gives an O ( n log n ) algorithm for computing the geodesic width between them.The goal of our paper is instead to rigorously prove an analytic formula for the sweeping cost of a domain,Theorem 7.1, that is closely related to geodesic widths. Indeed, the right hand side of (6) in Theorem 7.1is unchanged if we replace the geodesic distance between two curves (Definition 3.1) with the weak geodesicFr´echet distance between them (Definition 3.3). The paper [18] also studies sweeps of planar domains bypiecewise linear curves in which the cost of a sweep is not equal to a length, but instead to the number ofvertices or joints in the curve.Related notions to the geodesic width include the isotopic Fr´echet distance [12] and the minimum defor-mation area [29] between two curves α and β . Whereas the geodesic width considers deformations between α and β such that no two intermediate curves intersect, this restriction is not present for the isotopic Fr´echetdistance, which can therefore be defined between intersecting curves. The paper [29] considers a distancebetween two curves on a 2-manifold which is instead an area: the minimal total surface area swept outby any deformation between the two curves. If the curves are piecewise linear in the plane, have n totalvertices, and have I intersection points, then [29] gives an O ( n + I log n ) algorithm to compute the minimumdeformation area between them. 3. Preliminaries and notation
Let d : R × R → R denote the Euclidean metric on R . The distance between two subsets X, Y ⊆ R isdefined as d ( X, Y ) = inf { d ( x, y ) | x ∈ X and y ∈ Y } . We denote the closure of a set X ⊆ R by X . Jordan domains and geodesics.
Let D ⊆ R be a Jordan domain, i.e. the homeomorphic image of aclosed disk in R . It follows that D is compact and simply-connected, and its boundary ∂D is a topologicalcircle. Given a point x ∈ D , we let B ( x, (cid:15) ) = { y ∈ D | d ( x, y ) < (cid:15) } denote the open ball about x in D .We denote the (cid:15) -offset of a set X ⊆ D by B ( X, (cid:15) ) = ∪ x ∈ X B ( x, (cid:15) ). Given a subset X ⊆ D , we define itsboundary as ∂X = X ∩ R \ X .We refer the reader to [10] for the basics of geodesic curves and distances. The length of a continuouspath γ : [ a, b ] → D is defined as in [10, Defintion 2.3.1]; we denote this length by L ( γ ). Curve γ is said to be rectifiable if L ( γ ) < ∞ . Domain D has a length structure ([10, Section 2.1]) in which all continuous pathsare admissible, and the length is given by the function L . The associated geodesic metric d L : D × D → R ,also known as a path-length or intrinsic metric , is d L ( x, y ) = inf { L ( γ ) | γ : [ a, b ] → D is continuous with γ ( a ) = x, γ ( b ) = y } . The precise definition of a geodesic, or length-minimizing curve in D , is given in [10, Definition 2.5.27]. Since D is a Jordan domain, it follows from [9, 8] that each pair of points in D is joined by a unique shortestgeodesic in D . Definition 3.1.
Let D ⊆ R be a Jordan domain. We define the geodesic distance between two curves α, β : [ a, b ] → D to be d L ( α, β ) = max t ∈ [ a,b ] d L ( α ( t ) , β ( t )) . r´echet and geodesic Fr´echet distances. The
Fr´echet distance is a measure of similarity between twocurves α, β : [0 , → R . One application of the Fr´echet distance is in handwriting input recognition for acomputer [27]: in order to properly tell which letters a user has written, the machine must determine whichcurves (representing letters) are the most similar. Other notions of distance, such as the Hausdorff distancebetween the images of the curves, are not necessarily sensitive enough for this task.The intuition behind the Fr´echet distance is that you are walking along path α , your dog is walking alongpath β , and you want to know how long of a leash you need. There are two notions, namely the weak Fr´echetdistance and the strong Fr´echet distance . In the weak case, you and your dog are allowed to backtrack alongyour respective paths, but in the strong case backtracking is forbidden. In general these two distances neednot be equal. Definition 3.2.
Let α, β : [0 , → R be continuous curves. Then the weak (resp. strong) Fr´echet distance between α and β is d Fr´echet( α, β ) = inf a,b max t ∈ [0 , { d ( α ( a ( t ) , β ( b ( t )) } , where the infimum is taken over all continuous a, b : [0 , → [0 ,
1] which are surjective (resp. bijective).If D ⊆ R is a Jordan domain and α, β : [0 , → D are two curves, then we can consider a variant of theFr´echet distance in which the Euclidean metric d is replaced with the geodesic metric d L . Definition 3.3.
Let D ⊆ R be a Jordan domain, and let α, β : [0 , → D be continuous curves. Then the weak (resp. strong) geodesic Fr´echet distance between α and β is d geodesic Fr´echet( α, β ) = inf a,b max t ∈ [0 , { d L ( α ( a ( t ) , β ( b ( t )) } , where the infimum is taken over all continuous a, b : [0 , → [0 ,
1] which are surjective (resp. bijective).
Sensor curves.
Let I = [0 ,
1] be the unit interval. We define a sensor curve to be a time-varying rectifiablecurve in D . Definition 3.4. A sensor curve is a continuous map f : I × I → D such that(i) each curve f ( · , t ) : I → D is rectifiable and injective for t ∈ (0 , f ( I,
0) and f ( I,
1) are each (possibly distinct) single points in ∂D , and(iii) f ( s, t ) ∈ ∂D implies s ∈ { , } or t ∈ { , } .We think of the first input s as a spatial variable and of the second t as a temporal variable; in particular f ( I, t ) is the region covered by the curve of sensors at time t . Assumption (ii) states that the images of thesensor curve at times 0 and 1 are single points, and assumption (iii) implies that (apart from times 0 and 1)only the boundary of the sensor curve intersects ∂D . We define the length of a sensor curve to be L ( f ) = max t ∈ I L ( f ( · , t )) . An intruder is a continuous path γ : I → D . We say that an intruder is caught by a sensor curve f attime t if γ ( t ) ∈ f ( I, t ). A path γ : [0 , t ] → D such that γ ( t (cid:48) ) / ∈ f ( I, t (cid:48) ) for all t (cid:48) ∈ [0 , t ] is called an evasionpath . Sensor curve f is a sweep if every continuously moving intruder γ : I → D is necessarily caught atsome time t , or equivalently, if no evasion path over the full time interval I exists. The following notation will prove convenient. Fix a sensor curve f . We let C ( t ) ⊆ D be the contaminatedregion at time t , and we let U ( t ) ⊆ D be the uncontaminated region at time t . More precisely, C ( t ) = { x ∈ D | ∃ γ : [0 , t ] → D with γ ( t ) = x and γ ( t (cid:48) ) / ∈ f ( I, t (cid:48) ) ∀ t (cid:48) ∈ [0 , t ] } and U ( t ) = D \ C ( t ) . Note that sensor curve f is a sweep if and only if C (1) = ∅ , or equivalently U (1) = D . Definition 3.5.
Let F ( D ) be the set of all sensor curve sweeps of D . The sweeping cost of D isSC( D ) = inf f ∈F ( D ) L ( f ) . Remark 3.6.
The results of Sections 4–5 hold even if assumptions (ii) and (iii) in Definition 3.4 are removed. Our definition is similar to the graph-based definition in [3, Definition 2.1]. . Properties of sensor sweeps
We now prove some basic properties of sensor sweeps and the contaminated and uncontaminated regions.
Lemma 4.1. If x and x (cid:48) are in the same path-connected component of D \ f ( I, t ) , then x ∈ U ( t ) if and onlyif x (cid:48) ∈ U ( t ) .Proof. Suppose for a contradiction that x ∈ U ( t ) but x (cid:48) / ∈ U ( t ). Since x (cid:48) ∈ C ( t ), there exists an evasionpath γ : [0 , t ] → D with γ ( t ) = x (cid:48) . Since x and x (cid:48) are in the same path-connected component, there exists apath β : I → D \ f ( I, t ) with β (0) = x and β (1) = x (cid:48) .Note that β ( I ) and f ( I, t ) are compact, since they are each a continuous image of the compact set I .As any metric space is normal, there exist disjoint neighborhoods containing β ( I ) and f ( I, t ). Because f is uniformly continuous (it is a continuous function on a compact set), we can choose δ > f ( I, [ t − δ , t ]) remains in this open neighborhood disjoint from β ( I ), giving(1) β ( I ) ∩ f ( I, [ t − δ , t ]) = ∅ . Since metric space D is normal, there exist disjoint neighborhoods containing γ ( t ) = x (cid:48) and f ( I, t ). Since γ is continuous and f is uniformly continuous, we can choose δ > γ ([ t − δ , t ]) and f ( I, [ t − δ , t ])remain in these disjoint neighborhoods, giving(2) γ ([ t − δ , t ]) ∩ f ( I, [ t − δ , t ]) = ∅ Let δ = min { δ , δ } . Using (1) and (2) we can define an evasion path γ : [0 , t ] → D with γ ( t ) = x . Indeed,let γ ( t (cid:48) ) = γ ( t (cid:48) ) if t (cid:48) ≤ t − δγ (2 t (cid:48) − t + δ ) if t − δ < t (cid:48) ≤ t − δ β ( δ ( t (cid:48) − t + δ )) if t − δ < t (cid:48) ≤ t. This contradicts the fact that x ∈ U ( t ). (cid:3) Lemma 4.2.
For all t ∈ I , the set U ( t ) is closed and the set C ( t ) is open in D .Proof. Suppose x ∈ C ( t ). Since f ( I, t ) is closed and x / ∈ f ( I, t ), there exists some ε > B ( x, ε ) ∩ f ( I, t ) = ∅ . Note all x (cid:48) ∈ B ( x, ε ) are in the same path-connected component of D \ f ( I, t ) as x via astraight line path. Hence Lemma 4.1 implies B ( x, ε ) ⊆ C ( t ), showing C ( t ) is open in D . It follows that U ( t ) = D \ C ( t ) is closed in D . (cid:3) Lemma 4.3. If h : D → h ( D ) is a homeomorphism onto its image h ( D ) ⊆ R , then a sensor curve f : I × I → D is a sweep of D if and only if sensor curve hf : I × I → h ( D ) is a sweep of h ( D ) . Figure 2.
A homeomorphism h : D → h ( D ). Proof.
Note that if γ : I → D is an evasion path for f , then hγ : I → h ( D ) is an evasion path for hf .Conversely, if γ : I → h ( D ) is an evasion path for hf , then h − γ : I → D is an evasion path for f . (cid:3) . A lower bound on the sweeping cost
In this section we prove that the sweeping cost of a Jordan domain is at least as large as the length of theshortest area-bisecting curve. The first lemma is a version of the intermediate value theorem with slightlyrelaxed hypotheses.
Lemma 5.1. If f : [ a, b ] → R is upper semi-continuous and left continuous and if f ( a ) < u < f ( b ) , thenthere exists some c ∈ ( a, b ) with f ( c ) = u .Proof. Let S be the set of all x ∈ ( a, b ) with f ( x ) < u . Then S is nonempty since a ∈ S , and S is boundedabove by b . Hence by the completeness of R , the supremum c = sup S exists. We claim that f ( c ) = u .Let (cid:15) >
0. Since f is left-continuous, there is some δ > | f ( x ) − f ( c ) | < (cid:15) whenever x ∈ ( c − δ, c ]. By the definition of supremum, there exists some y ∈ ( c − δ, c ] that is contained in S , giving f ( c ) < f ( y ) + (cid:15) < u + (cid:15) . Since this is true for all (cid:15) >
0, it follows that f ( c ) ≤ u .It remains to show f ( c ) ≥ u . Let (cid:15) >
0. Since f is upper semi-continuous, there exists a δ > f ( c ) > f ( x ) − (cid:15) whenever x ∈ ( c − δ, c + δ ). Let y ∈ ( c, c + δ ) and note that y / ∈ S , giving f ( c ) > f ( y ) − (cid:15) ≥ u − (cid:15) .It follows that f ( c ) ≥ f ( u ). (cid:3) If S ⊆ R is a measurable set, then we let area( S ) denote its area. Lemma 5.2.
The function area( U ( t )) is upper semi-continuous and left continuous.Proof. Let t ∈ I . We will show that area( U ( t )) is right upper semi-continuous and left continuous at t ,which implies the function is both upper semi-continuous and left continuous.For right upper semi-continuity, note for t ≥ t we have(3) U ( t ) ⊆ U ( t ) ∪ f ( I, [ t , t ]) . Since sensor curve f : I × I → D is a continuous function on a compact domain, it is also uniformly continuous.Hence for all (cid:15) > δ such that(4) f ( I, [ t , t + δ ]) ⊆ B ( f ( I, t ) , (cid:15) ) . It follows that for all t ∈ [ t , t + δ ] we havearea( U ( t )) − area( U ( t )) ≤ area( f ( I, [ t , t ])) by (3) ≤ area( B ( f ( I, t ) , (cid:15) )) by (4) ≤ L ( f ( I, t )) (cid:15) + π(cid:15) , where the last inequality is by a result of Hotelling (see for example [21, Equation (2.1)]). Hence area( t ) isright upper semi-continuous.To see that area( U ( t )) is left continuous at t ∈ I , we must show that for all sequences { s i } with 0 ≤ s i ≤ t and lim i s i = t , we have lim i area( U ( s i )) = area( U ( t )). We claim(5) U ( t ) \ f ( I, t ) ⊆ lim inf i U ( s i ) ⊆ lim sup i U ( s i ) ⊆ U ( t ) , where the middle containment is by definition. We now justify the first and last containment.To prove U ( t ) \ f ( I, t ) ⊆ lim inf i U ( s i ) it suffices to show that for any x ∈ U ( t ) \ f ( I, t ) there existsan (cid:15) > x ∈ U ( t − δ ) for all δ ∈ [0 , (cid:15) ). Fix (cid:15) such that x / ∈ f ( I, t ) for t ∈ ( t − (cid:15), t ]. Suppose fora contradiction that x ∈ C ( t − δ ) for some δ ∈ [0 , (cid:15) ). Hence there exists an evasion path γ : [0 , t − δ ] → D with γ ( t − δ ) = x . It is possible to extend γ to an evasion path ˜ γ : [0 , t ] → D defined by˜ γ ( t ) = (cid:40) γ ( t ) if t ∈ [0 , t − δ ] x if t ∈ ( t − δ, t ] . This contradicts the fact x ∈ U ( t ), thus giving the first containment.We now show lim sup i U ( s i ) ⊆ U ( t ). If x / ∈ U ( t ), then there exists an evasion path γ : [0 , t ] → D with γ ( t ) = x . Since C ( t ) is open by Lemma 4.2, there exists some δ > B ( x, δ ) ⊆ C ( t ), andhence B ( x, δ ) ∩ f ( I, t ) = ∅ . Since f is uniformly continuous, there is some (cid:15) > ( x, δ ) ∩ f ( I, [ t − (cid:15) , t ]) = ∅ , and since γ is continuous there is some (cid:15) > γ ([ t − (cid:15) , t ]) ⊆ B ( x, δ ).Let (cid:15) = min { (cid:15) , (cid:15) } . Reparametrize γ to get a continuous curve ˜ γ : [0 , t ] → D with˜ γ ( t ) = γ ( t ) if x ∈ [0 , t − (cid:15) ) γ ( t ) ∈ B ( x, δ ) if t ∈ [ t − (cid:15), t − (cid:15)/ γ ( t ) = x if t ∈ [ t − (cid:15)/ , t ] . The evasion path ˜ γ shows x / ∈ lim sup i U ( s i ), giving the third containment and finishing the proof of (5).Set f ( I, t ) has Lebesgue measure zero since curve f ( · , t ) is rectifiable, giving area( U ( t ) \ f ( I, t )) =area( U ( t )). Thus (5) impliesarea(lim sup i U ( s i )) = area( U ( t )) = area(lim inf i U ( s i )) . Since area( D ) is finite, Lemma A.1 implies lim sup i area( U ( s i )) ≤ area(lim sup i U ( s i )) and area(lim inf i U ( s i )) ≤ lim inf i area( U ( s i )), givinglim sup i area( U ( s i )) ≤ area( U ( t )) ≤ lim inf i area( U ( s i )) . Hence lim i area( U ( s i )) = area( U ( t )) as required. (cid:3) Remark 5.3.
The function area( U ( t )) need not be right continuous. Indeed, consider a sensor curve asshown below in Figure 3, where area( U ( t )) >
0, and where there is some (cid:15) > < (cid:15) < (cid:15) ,only one point on the sensor curve at time t + (cid:15) intersects ∂D and area( U ( t + (cid:15) )) = 0. tarea ( U ( t )) t = t t = t + (cid:15) t Figure 3.
An example sensor curve where the function area( U ( t )) is not right continuous.The shaded region is U ( t ) and the unshaded region is C ( t ).As a consequence we obtain the following lower bound on the sweeping cost. Theorem 5.4. If D is a Jordan domain, then the sweeping cost SC( D ) is at least as large as the length ofthe shortest area-bisecting curve in D .Proof. Suppose that f is a sweep of D . Note that area( U (0)) = 0 and area( U (1)) = area( D ). By Lemmas 5.1and 5.2, there exists some time t (cid:48) ∈ I with area( U ( t (cid:48) )) = area( D ). So L ( f ( · , t (cid:48) )) and hence SC( D ) is atleast as large as the shortest area-bisecting curve in D . (cid:3) Example 5.5. If D = { ( x, y ) ∈ R | x + y ≤ } is the unit disk, then SC( D ) = 2. Proof.
To see SC( D ) ≤
2, consider the sweep f : [ − , × I → D defined by f ( s, t ) = (2 s √ t − t , t − D is a diameter [19]. Hence weapply Theorem 5.4 to get SC( D ) ≥ (cid:3) Example 5.6.
Let a, b >
0. If D = { ( x, y ) ∈ R | ( x/a ) + ( y/b ) ≤ } is the convex hull of an ellipse, thenSC( D ) = min { a, b } . = 0 t = t = t = 1 Figure 4.
A sweep of the unit disk. The shaded region is U ( t ) and the unshaded region is C ( t ). Proof.
To see SC( D ) ≤ min { a, b } , construct a sweep much like in Example 5.5.For the reverse direction, the solution to [26, Chapter X, Problem 33] states that because the ellipsehas a center of symmetry, the shortest area-bisecting curve is a straight line. All area-bisecting lines passthrough the center of the ellipse, and hence have length at least min { a, b } . It follows from Theorem 5.4that SC( D ) ≥ min { a, b } . (cid:3) A lemma of no progress
In Sections 7–9 we will restrict attention to sensor curves f with boundary points f (0 , t ) , f (1 , t ) ∈ ∂D for all t ∈ I . The motivation behind this assumption is Lemma 6.2, which states that if f (0 , t ) / ∈ ∂D or f (1 , t ) / ∈ ∂D , then the uncontaminated region at time t is as small as possible, namely U ( t ) = f ( I, t ).The following lemma is from [30]; see also its statement in [22, page 164].
Lemma 6.1 (Zoretti) . If K is a bounded maximal connected subset of a plane closed set M and (cid:15) > , thenthere exists a simple closed curve J enclosing K such that J ∩ M = ∅ and J ⊆ B ( K, (cid:15) ) . Lemma 6.2.
Let f : I × I → D be a sensor curve. If f (0 , t ) / ∈ ∂D or f (1 , t ) / ∈ ∂D and U ( t ) (cid:54) = D , then U ( t ) = f ( I, t ) .Proof. Without loss of generality suppose f (0 , t ) / ∈ ∂D . It suffices to show that D \ f ( I, t ) is a single path-connected component, because then Lemma 4.1 and the fact that U ( t ) (cid:54) = D will imply C ( t ) = D \ f ( I, t )and hence U ( t ) = f ( I, t ). Let x, x (cid:48) ∈ D \ f ( I, t ); we must find a path in D \ f ( I, t ) connecting x and x (cid:48) .There are two cases: when f (1 , t ) / ∈ ∂D , and when f (1 , t ) ∈ ∂D .In the first case f (1 , t ) / ∈ ∂D , note that f ( I, t ) is disjoint from ∂D . Hence by compactness there existssome (cid:15) > d ( f ( I, t ) , ∂D ∪ { x, x (cid:48) } ) < (cid:15) . By Lemma 6.1 (with K = f ( I, t ) and M = ∂D ∪ { x, x (cid:48) } ),there exists a simple closed curve J in D enclosing f ( I, t ) but not enclosing x or x (cid:48) . We may thereforeconnect x and x (cid:48) by a path in D \ f ( I, t ) consisting of three pieces: a path in D from x to J , a path in D from x (cid:48) to J , and a path in J connecting these two endpoints (Figure 5). x x (cid:48) x x (cid:48) γ Figure 5.
The first case in the proof of Lemma 6.2, with J drawn in red. n the second case f (1 , t ) ∈ ∂D , pick some point y ∈ D \ ( f ( I, t ) ∪ { x, x (cid:48) } ). By translating D in theplane we may assume that y = (cid:126)
0. Define the inversion function i : R \ { (cid:126) } → R \ { (cid:126) } by i ( r cos θ, r sin θ ) =( r cos θ, r sin θ ); note i is the identity map. Let (cid:15) > d ( i ( f ( I, t ) ∪ ∂D ) , { i ( x ) , i ( x (cid:48) ) } ) < (cid:15). By Lemma 6.1 (with K = i ( f ( I, t ) ∪ ∂D ) and M = { i ( x ) , i ( x (cid:48) ) } ), there exists a simple closed curve J in R enclosing i ( f ( I, t ) ∪ ∂D ) but not enclosing i ( x ) or i ( x (cid:48) ). By the Jordan curve theorem, i ( x ) and i ( x (cid:48) ) are inthe same (exterior) connected component E of R \ J . Since E is open it is also path-connected, and hencewe can connect i ( x ) and i ( x (cid:48) ) by a path γ in E . The path i ( γ ) is therefore a path in D \ f ( I, t ) connecting x and x (cid:48) (Figure 6). x x (cid:48) i ( x ) i ( x (cid:48) ) i ( ) x x (cid:48) γ i − ( γ ) i i − Figure 6.
The second case in the proof of Lemma 6.2, with J and i ( J ) drawn in red. (cid:3) Sweeping cost of a Jordan domain
As motivated by Lemma 6.2, for the remainder of the paper we restrict attention to sensor curves satisfying f ( s, t ) ∈ ∂D if and only if s ∈ { , } or t ∈ { , } .Given curves α, β : I → ∂D with α (1) = β (0) (see Figure 7), we define the concatenated curve α · β : I → ∂D by α · β ( t ) = (cid:40) α (2 t ) if 0 ≤ t ≤ β (2 t −
1) if < t ≤ . We define the inverse curve β − : I → ∂D by β − ( t ) = β (1 − t ). αβ wn( γ ) = − γ ) = 0 wn( γ ) = 1 wn( γ ) = 2 Figure 7. (Left) Two curves α, β : I → ∂D with α (1) = β (0). (Right) Example winding numbers.Since ∂D is homeomorphic to the circle, given a loop γ : I → ∂D (with γ (0) = γ (1)) we can denotethe winding number of γ , i.e. the number of times γ wraps around ∂D , by wn( γ ). The winding number is ositive (resp. negative) for loops that wrap around in the counterclockwise (resp. clockwise) direction. Notethat if α and β are paths in ∂D with α (0) = β (0) and α (1) = β (1), then α · β − is a loop.Our main result is an analytic formula for the sweeping cost of a Jordan domain. Theorem 7.1.
The sweeping cost of a Jordan domain D is (6) SC( D ) = inf { d L ( α, β ) | α, β : I → ∂D, α (0) = β (0) , α (1) = β (1) , wn( α · β − ) (cid:54) = 0 } . Equation (6) is closely related to the geodesic width between two polylines [18], and also the isotopicFr´echet distance between two curves [12]. Indeed, note that the right hand side of (6) is unchanged if wereplace d L ( α, β ) with the weak geodesic Fr´echet distance between α and β (see Definition 3.3). Remark 7.2.
The value of the right hand side of (6) is unchanged if we replace wn( α · β − ) (cid:54) = 0 withwn( α · β − ) ∈ {− , } . Proof of Remark 7.2.
Let α, β : I → ∂D with α (0) = β (0) and α (1) = β (1), and suppose | wn( α · β − ) | ≥ < t < α ( t ) = β ( t ) and wn( α | [0 ,t ] · β | − ,t ] ) ∈ {− , } . The claim followssince d L ( α | [0 ,t ] , β | [0 ,t ] ) ≤ d L ( α, β ) . (cid:3) The following lemma will be used to prove the ≤ direction in (6). Lemma 7.3.
Let D be a Jordan domain. Suppose α, β : I → ∂D with α (0) = β (0) , α (1) = β (1) , and wn( α · β − ) (cid:54) = 0 . If f : I × I → D is any sensor curve with f (0 , t ) = α ( t ) and f (1 , t ) = β ( t ) , then f is asweep of D .Proof. Let D = { ( x, y ) ∈ R | x + y ≤ } be the unit disk. We first prove this claim in the case when D = D .By Lemma A.2, there exists a point p ∈ R \ D and two continuous families of curves g α , g β : I × I → R such that • g α (0 , t ) = p = g β (0 , t ), • g α (1 , t ) = α ( t ), • g β (1 , t ) = β ( t ), and • g α ( s, t ) , g β ( s, t ) / ∈ D for s < S be the circle of unit circumference, i.e. [0 ,
1] with endpoints 0 and 1 identified. Define a continuousmap g : S × I → D via g ( s, t ) = g α (3 st, t ) if 0 ≤ s < f (3 s − , t ) if ≤ s < g β (3 t (1 − s ) , t ) if ≤ s ≤ . Note g ( · , t ) is indeed a (possibly non-simple) map from the circle since g α (0 , t ) = p = g β (0 , t ) for all t . Definea continuous signed distance d ± : R × I → R by d ± ( x, t ) = (cid:40) d ( x, g ( S , t )) if x ∈ g ( S , t ) or wn( g ( · , t ) , x ) = 0 − d ( x, g ( S , t )) if x / ∈ g ( S , t ) and wn( g ( · , t ) , x ) (cid:54) = 0 . Here wn( g ( · , t ) , x ) denotes (for x / ∈ g ( S , t )) the winding number of the map g ( · , t ) : S → R \ { x } (cid:39) S .Note that wn( g ( · , t ) , x ) is constant on each connected component of R \ g ( S , t ), and that d ± is continuous.Given any intruder path γ : I → D , the continuous function d ± ( γ ( t ) , t ) : I → R satisfies d ± ( γ (0) , ≥ f ( I,
0) is a single point in ∂D ) and d ± ( γ (1) , ≤ f ( I,
1) is a single point in ∂D andwn( α · β − ) (cid:54) = 0). By the intermediate value theorem there exists some t (cid:48) ∈ I with d ± ( γ ( t (cid:48) ) , t (cid:48) ) = 0, andhence γ ( t (cid:48) ) ∈ g ( S , t (cid:48) ) ∩ D = f ( I, t (cid:48) ). So γ is not an evasion path, and f is a sweep of D .We now handle the case when D is an arbitrary Jordan domain. By definition there exists a homeo-morphism h : D → D . Note that h − α, h − β : I → ∂ D with h − α (0) = h − β (0), h − α (1) = h − β (1), andwn( h − α · h − β − ) (cid:54) = 0. Since h − f : I × I → D is a sensor curve, it follows from our proof in the case ofthe disk that h − f is a sweep of D . Hence f is a sweep of D by Lemma 4.3. (cid:3) roof of Theorem 7.1. Let c = inf { d L ( α, β ) | α, β : I → ∂D, α (0) = β (0) , α (1) = β (1) , wn( α · β − ) (cid:54) = 0 } . We first prove the ≤ direction of (6). Let (cid:15) > α, β : I → ∂D with α (0) = β (0), α (1) = β (1), wn( α · β − ) (cid:54) = 0, and d L ( α, β ) ≤ c + (cid:15) . Define f : I × I → D byletting f ( · , t ) : I → D be the unique constant-speed geodesic in D between f (0 , t ) = α ( t ) and f (1 , t ) = β ( t ),which exists by [9, 8]. Lemma 7.3 implies that f is a sweep, and hence we haveSC( D ) ≤ L ( f ) = d L ( α, β ) ≤ c + (cid:15). Since this is true for all (cid:15) >
0, we have SC( D ) ≤ c .For the ≥ direction of (6), suppose that f is a sensor curve with L ( f ) < c . For notational convenience,define α, β : I → ∂D by α ( t ) = f (0 , t ) and β ( t ) = f (1 , t ). Then necessarily wn( α · β − ) = 0, and furthermore(7) α ( t ) = β ( t ) implies wn( α | [0 ,t ] · β | − ,t ] ) = 0 , since otherwise we’d have L ( f ) ≥ d L ( α, β ) ≥ d L ( α | [0 ,t ] , β | [0 ,t ] ) ≥ c, a contradiction. We will show that f is not a sweep of D by showing the existence of an evasion path γ : I → D whose image furthermore lives in ∂D .Indeed, consider the 1-dimensional evasion problem in ∂D where the region covered by the sensors attime t is { α ( t ) , β ( t ) } . In this 1-dimensional problem, it is clear that the uncontaminated region in ∂D iseither (i) a single point α ( t ) = β ( t ), (ii) a closed interval in ∂D with endpoints α ( t ) and β ( t ), or (iii) all of ∂D . Equation (7), however, rules out the possibility of (iii). It follows that the contaminated region in ∂D is always a nonempty open interval in ∂D with continuously varying endpoints α ( t ) and β ( t ). Therefore wecan define an evasion path γ : I → ∂D , for example by letting γ ( t ) be the midpoint of the open interval ofthe uncontaminated region in ∂D . This evasion path γ is also an evasion path for our original 2-dimensionalproblem in D , as γ : I → ∂D ⊆ D satisfies γ ( t ) / ∈ f ( I, t ) for all t . This gives the ≥ direction of (6). (cid:3) Question 7.4.
Does Theorem 7.1 hold even if assumption (iii) in Definition 3.4 is removed, i.e. if the interiorof a sensor curve is also allowed to touch ∂D ? Question 7.5.
For any Jordan domain D in the plane and injective curves α, β : I → ∂D , we conjecturethat the weak geodesic Fr´echet distance between α and β is equal to their strong geodesic Fr´echet geodesicdistance.There are simple counterexamples to Question 7.5 when α and β are not injective, or when they do notmap to ∂D . Closely related is following question: is the value of (6) unchanged if we require α and β to beinjective? 8. Sweeping cost of a convex domain
Given a convex Jordan domain D ⊆ R , its width w ( D ) is defined as w ( D ) = min (cid:107) v (cid:107) =1 max x ∈ R L ( D ∩ { x + tv | t ∈ R } ) , where v is a unit direction vector in R . Alternatively, the width w ( D ) is the smallest distance between twoparallel supporting lines on opposite sides of D (Figure 8). Theorem 8.1. If D is a convex Jordan domain, then SC( D ) = w ( D ) .Proof. We first show SC( D ) ≤ w ( D ). By Theorem 7.1, it suffices to showinf { d L ( α, β ) | α, β : I → ∂D, α (0) = β (0) , α (1) = β (1) , wn( α · β − ) (cid:54) = 0 } ≤ w ( D ) . Let v some direction vector realizing the width, i.e. w ( D ) = max x ∈ R L ( D ∩ { x + tv | t ∈ R } ). Considersweeping through all lines in R parallel to v ; the intersection of these lines with ∂D traces out two continuous (D) Figure 8.
The width w ( D ) of a domain D .curves α, β : I → ∂D with α (0) = β (0), α (1) = β (1), and wn( α · β − ) = ± We have d L ( α, β ) ≤ w ( D ),giving SC( D ) ≤ w ( D ).To finish the proof, we need some background on planar convex domains. A point x ∈ ∂D is smooth if ithas a unique supporting hyperplane, and otherwise x is a vertex containing a range of angles [ θ , θ ] ⊆ S (with θ (cid:54) = θ ) which are the outward normal directions of supporting hyperplanes of D at x . Away from thevertices, the unique supporting hyperplane of x ∈ ∂D varies continuously with x . By [5, Proposition 11.6.2],the set of vertices of the closed convex domain D is countable.We now show SC( D ) ≥ w ( D ). Given (cid:15) >
0, let α and β be (cid:15) -close to realizing the infimum in (6), meaningSC( D ) + (cid:15) ≥ d L ( α, β ). For notational convenience we assume that α (0) = β (0) and α (1) = β (1) are notvertices of ∂D (our same proof technique works regardless). Let T = { t , t , t , . . . } ⊆ I be a countablesubset such that t ∈ T if either α ( t ) or β ( t ) is a vertex of ∂D . Let t = 0, and if | T | is finite, then let t | T | +1 = 1. For i = 1 , , . . . , | T | , choose weights w i > (cid:80) i w i = w < ∞ ; this is possible since T iscountable. Let p : ∂D × S → ∂D and p : ∂D × S → S be the projection maps. It is possible to define a continuous map g α : [0 , w ] → ∂D × S satisfying the following properties. • Each g α ( s ) is equal to a point ( α ( t ) , v ) ∈ ∂D × S with t ∈ I such that v is the outward normalvector to a supporting hyperplane of D at α ( t ). • If t i ≤ t ≤ t i +1 , then α ( t ) = p g α ( t + (cid:80) ij =1 w j ). • For all 0 ≤ s ≤ w i , we have p g α ( s + t i + (cid:80) i − j =1 w j ) = α ( t i ). • As s varies from 0 to w i , angle p g α ( s + t i + (cid:80) i − j =1 w j ) varies over the range of supporting hyperplanesof D at α ( t i ) (which may be a single angle if α ( t i ) is not a vertex of ∂D ).Define g β : [0 , w ] → ∂D × S similarly (with α replaced everywhere by β ). Note that g α (0) = g β (0), that g α (1 + w ) = g β (1 + w ), and that p g α and p g β wrap in opposite directions around S . Hence for some s ∈ [0 , w ] the supporting hyperplanes corresponding to g α ( s ) and g β ( s ) will be parallel and on oppositesides of D . It follows that SC( D ) + (cid:15) ≥ d L ( α, β ) ≥ d ( p g α ( s ) , p g β ( s )) ≥ w ( D ) . Since this is true for all (cid:15) >
0, we have SC( D ) ≥ w ( D ). (cid:3) The paper [20] shows that for D ⊆ R a convex polygonal domain with n vertices, the width and hencethe sweeping cost of D can be computed in time O ( n ) and space O ( n ) using the rotating calipers technique. This is not quite precise if ∂D contains a straight line segment of non-zero length parallel to v (there are at most two suchsegments). In this case, pick an arbitrary point on each such line segment; each such point will be either the starting point orthe ending point for both α and β . . Extremal shapes
Which convex shape of unit area has the largest sweeping cost? The papers [11, Theorem 4.3] and [24]state that if D is a bounded planar convex domain, then area( D ) ≥ w ( D ) / √
3, where equality is achievedif D is an equilateral triangle. The next corollary follows immediately from Theorem 8.1. Corollary 9.1.
Let D be a convex Jordan domain. Then area( D ) ≥ SC( D ) √ , where equality is achieved if D is an equilateral triangle. Hence the equilateral triangle has the maximalsweeping cost over all planar convex domains of the same area. The next example shows that there is no extremal shape for non-convex Jordan domains.
Example 9.2.
A (non-convex) domain D of unit area may have arbitrarily large sweeping cost. Proof.
Consider a deformation of an equilateral triangle with unit side lengths where we deform each edgetowards the center of the triangle (Figure 9). Note that as the sweeping cost converges to √ (the distancefrom the center to a vertex) from above, the area of the shape tends zero. Rescaling each shape in thisdeformation to have area one shows that a non-convex domain of unit area may have arbitrarily largesweeping cost. Figure 9.
We deform each edge of the triangle towards the center of the triangle, producinga three-pronged shape. As the sweeping costs of the shapes converge to a fixed constant,the areas converge to zero. (cid:3)
Conclusion
Given a Jordan domain D in the plane, we show that the sweeping cost of D is at least as large as theshortest area-bisecting curve in D , and we give a formula for the sweeping cost in terms of the geodesicFr´echet distance between two curves on the boundary of D with non-equal winding numbers. We show thatthe sweeping cost of any convex domain is equal to its width. Therefore, the sweeping cost of a polygonalconvex domain with n vertices can be computed in time and space O ( n ), and a convex domain of unit areawith maximal sweeping cost is the equilateral triangle.We end by mentioning two related settings of interest. First, let D be a compact region in the plane,perhaps not simply-connected. Suppose the pursuer is now a union of curves. What can one say about thesweeping cost of D , measured as the sum of the curve lengths? Second, let D ⊂ R n be the homemorphicimage of the closed n -dimensional ball. What are the properties of the sweeping cost of D , when swept byan ( n − Acknowledgements
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Appendix A. Additional lemmas and proofs
Lemma A.1. If ( X, µ ) is a measure space and U i is a sequence of measurable sets in X , then (1) µ (lim inf i U i ) ≤ lim inf i µ ( U i ) , and (2) µ (lim sup i U i ) ≥ lim sup i µ ( U i ) if µ ( X ) < ∞ . roof. Recall lim sup i U i = ∩ ∞ i =1 ( ∪ ∞ j = i U i ). Since the ∪ ∞ j = i U i are a decreasing sequence of sets, and since µ ( X ) < ∞ , [15, Proposition 1.2.3] implies µ (lim sup i U i ) = lim i µ ( ∪ ∞ j = i U i ). Since U i ⊆ ∪ ∞ j = i U j , we have µ ( U i ) ≤ µ ( ∪ ∞ j = i U j ), and hence µ (lim sup i U i ) = lim i µ ( ∪ ∞ j = i U i ) ≥ lim sup i µ ( U i ) , giving (2). The proof of (1) is similar except that the finiteness assumption is unnecessary. (cid:3) Lemma A.2.
Let D = { ( x, y ) ∈ R | x + y ≤ } be the unit disk. Given any point p ∈ R \ D and anycurve α : I → ∂ D , there exists a continuous function g : I × I → R such that • g (0 , t ) = p for all t ∈ I , • g (1 , t ) = α ( t ) for all t ∈ I , and • g ( s, t ) / ∈ D for s < .Proof. Given r ≥
0, let α r ( t ) : I → R be defined by α r ( t ) = (1 + r ) α ( t ). Fix some (cid:15) >
0. Pick a single curve γ : I → R \ D with γ (0) = p and γ (1) = α (cid:15) (0); this is possible since R \ D is connected by the Jordan curvetheorem. We define the function g : I × I → R as follows: g ( s, t ) = γ (3 s ) if 0 ≤ s < α (cid:15) ((3 s − t ) if ≤ s < α (3 − s ) (cid:15) ( t ) if ≤ s ≤ . Note that g is continuous and satisfies all of the required conditions. (cid:3) E-mail address : [email protected] E-mail address : [email protected] E-mail address : [email protected]@rams.colostate.edu