the intruer within the room (P; ) before the intruer reaches? The 1-searcher is allowe to move an change the irection of the ashlight arbitrarily an s

Transcription

1 International Journal of Computational Geometry & Applications c Worl Scientic Publishing Company SEARCHING A POLYGONAL ROOM WITH ONE DOOR BY A 1-SEARCHER JAE-HA LEE, SANG-MIN PARK an KYUNG-YONG CHWA Department of Computer Science Korea Avance Institute of Science an Technology Gusong-ong Yusong-gu Taejon Korea fjhlee,smpark,kychwag@jupiter.kaist.ac.kr Receive 14 April 1998 Revise 1 December 1999 Communicate by T. Asano ABSTRACT The 1-searcher is a mobile guar whose visibility is limite to a ray emanating from his position, where the irection of the ray can be change continuously with boune angular rotation spee. Given a polygonal region P with a specie bounary point, is it possible for a 1-searcher to eventually see a mobile intruer that is arbitrarily faster than the searcher within P, before the intruer reaches? We ecie this question in O(n log n)-time for an n-sie polygon. Our main result is a simple characterization of the class of polygons (with a bounary point ) that amits such a search strategy. We also present a simple O(n 2 )-time algorithm for constructing a search scheule, if one exists. Finally, we compare the search capability of a 1-searcher with that of two guars. Keywors: Visibility, motion planning, pursuit-evasion, graph algorithm 1. Introuction Imagine a policeman staning at the unique oor of a ark house into which a robber has sneake. The policeman has a ashlight an shoul n the robber in the ark house. Probably, he will rive the robber to a corner by illuminating every suspicious corner one by one. Can the policeman eventually see the robber, in such a way that the robber couln't escape through the oor? More formally, let us assume that we are given a simple polygon P an a specie bounary point ; we call this conguration (P; ) a (polygonal) room. The 1-searcher is a mobile guar whose visibility is limite to a ray emanating from his position, calle a ashlight, where the irection of the ashlight can be change continuously with boune angular rotation spee. It is assume that the intruer is capable of moving arbitrarily faster than the searcher. Can the 1-searcher, starting at, eventually see Current aress: Jae-Ha Lee, Max-Planck-Institut fur Informatik, Im Statwal, Geb Saarbrucken, Germany, lee@mpi-sb.mpg.e 1

2 the intruer within the room (P; ) before the intruer reaches? The 1-searcher is allowe to move an change the irection of the ashlight arbitrarily an shoul guarantee that the intruer is etecte regarless of his path. A motion control of the 1-searcher subject to these constraints is calle a search scheule. A room (P; ) that amits a search scheule is sai to be 1-searchable. Relate Work Our problem is a `one-oor' variant of the polygon search problem, 9 where a searcher with various visibilities (incluing the 1-searcher consiere in the present paper) has to move to eventually see an intruer in a polygonal region without a oor. Suzuki an Yamashita 9 introuce this polygon search problem an presente some necessary or sucient conitions for a polygon to be searchable by a searcher. Recently Guibas, et al. 3 presente a complete (exponentialtime) algorithm to search a polygon by a searcher. However it is unknown whether a polynomial-time algorithm exists to etermine if a polygon is searchable by a searcher. 9;3 One characterizable variant is a polygonal room with one aitional oor g, written (P; ; g), where an unetecte intruer is assume to move out of an into P through g at any time. This conguration (P; ; g) can be viewe as a corrior with two oors an g, an the goal of the searcher, starting at, is to see or force out the intruer through g before the intruer reaches. This corrior search problem was rst stuie by Icking an Klein. 5 In fact, the searcher type that they consiere is two (bounary) guars (for enition, see Section 6), whose searchability is slightly weaker than that of a 1-searcher. However, the class of corriors searchable by two guars is exactly same as that of 1-searchable corriors (for proof, see Section 6). Some variants of the corrior search problem have also been stuie in Refs. [5,4,1,7,11], but no results are known about searching a room (P; ). Some previous work consiere a group of 1-searchers. The searchlight scheuling problem 8 is that of computing a search scheule of searchlights (stationary 1-searchers) in orer to etect a mobile intruer in a polygon. Recently, the upper an the lower boun on the number of mobile searchers that are neee to search a polygon in group were also investigate in Refs. [12,10]. Our Results We present three necessary conitions for a room (P; ) to be 1-searchable (Section 3) an show that the same conitions are also sucient (Section 4). Our characterization is obtaine by investigating specic patterns of three vertices (calle \orer-inucing triples") that must be cleare in some specic orer. This characterization leas to an O(n log n)-time algorithm for testing the 1-searchability an an O(n 2 )-time algorithm for constructing a search scheule, if one exists, in an n-sie room (P; ) (Section 5). Finally, we show that two guars have the same searchability as a 1-searcher in the corrior search problem but are inferior to a 1-searcher in the room search problem (Section 6). 2

3 2. Denitions an Notations 2.1. Problem Denition A room (P; ) is ene as a simple polygon P an a point on its bounary. The searcher an the intruer are moele as points that can move continuously within P. Let (t) enote the position of the intruer at time t 0. It is assume that : [0; 1)! P is a continuous function, an the intruer is unpreictable in that he is capable of moving arbitrarily fast an his path is unknown to the searcher. Let (t) enote the position of the searcher at time t 0. Let represent a continuous path of the searcher of the form : [0; 1)! P. It is assume that the searcher has a ashlight whose visibility is limite to a ray emanating from (t), where the irection of the ray can be change continuously with boune angular rotation spee. Let (t) enote the enpoint of the ashlight at t. Note that the 1-searcher sees the points on the segment (t)(t) at t. In this paper, it is assume that P contains (t)(t) an bounary points of P may lie in the mile of (t)(t) (see Figure 1). p q Fig. 1. The 1-searcher can see p an q simultaneously. The polygon (P; ) is 1-searchable if it amits a search scheule (; ) satisfying that for every continuous function : [0; 1)! P, there exists a time t 2 [0; 1) such that (t)2(t)(t) an (t 0 )6= for all t 0 <t. This implies the intruer will be seen by the searcher before reaching, regarless of his path. Figure 2 epicts an example of a 1-searchable room an the snapshots of its search scheule. During the search, the 1-searcher traverses the bounary of P monotonically in the clockwise irection, sliing the enpoint of the ashlight in the counterclockwise irection, except that the enpoint of the ashlight backtracks in () Notations enote the bounary of a simple polygon P. For points p; q let C[p; q] enote the connecte bounary chain from p to q in the clockwise irection. As usual, C(p; q) (resp, C(p; q]; C[p; q)) enotes the open chain C[p; q]nfp; qg (resp, C[p; q]nfpg, C[p; q]nfqg). We ene the preceence on the points as follows: for points p; q p q (p `precees' q) if p is encountere before q when we from 3

4 (a) (b) (c) () (e) Fig. 2. Snapshots of a search scheule. (f) clockwise. As the bounary conition, we imagine two points l an r such that l p r for all p an both l an r are `aliases' of (see Figure 3). For three consecutive vertices v i?1 ; v i ; v i+1 (v i?1 v i v i+1 ) of P, Pre(v i ) enotes v i?1 an Succ(v i ) oes v i+1. p q l l p q r Fig. 3. When the preceence between two points is consiere, we say that the one closer to l precees the other. Both l an r are `aliases' of. In this paper, we use the stanar enition of visibility. Two points p an q are visible from each other, if the segment pq is entirely containe in P. Given a point p 2 P, the set of points in P that are visible from p species the visibility polygon of p, enote by VP(p). We ene a cave of a vertex v to be a maximal connecte bounary chain C(p; q) that oes not belong to VP(v). Note that two enpoints p an q of C(p; q) lie on the irecte line?! vp an C(p; q) lies entirely to the left or right of?! vp. It is calle a left cave if it lies to the left of?! vp; a right cave otherwise. A left cave C(p; q) of v is calle the L-cave of v if p=succ(v). Analogously, a right cave C(p; q) of v is calle the R-cave of v if q =Pre(v) (see Figure 4). 3. Necessary Conitions for 1-Searchable Rooms Let (P; ) be a 1-searchable room an (; ) be its search scheule. Accoring to (; ), the searcher sweeps P by the ashlight. Consier a xe time t. Any region that might contain an intruer is sai to be contaminate; otherwise, it is r 4

5 right caves of v R-cave of v VP(v) Pre(v) v left cave of v Fig. 4. Denition of caves. sai to be clear. If a point is contaminate right before t an is clear at t, it is sai to be cleare or becomes clear at t. The following simple observations are helpful. Fact 1 If a point p is cleare at t, p 2 (t)(t). Fact 2 Any region in P that oes not meet (t)(t) is entirely clear or entirely contaminate. Due to the mobility of the intruer, a clear point can become contaminate later. If a point is clear right before t an is contaminate at t, it is sai to be recontaminate at t. This recontamination has an important implication: To avoi a `cycle' of recontaminations, some (three) vertices must be cleare in specic orer. We refer to them as an orer-inucing triple. In Subsection 3.1, we stuy three types of orer-inucing triples. In Subsection 3.2, we escribe three necessary conitions for 1-searchable rooms, each of which forbis a certain combination of orer-inucing triples an caves Orer-inucing Triples Let C = hv 1 ; v 2 ; v 3 i be a triple of vertices such that v 1 v 2 v 3. We call C an s-triple if v 1 has the L-cave an v 2 an v 3 lie in it, an v 3 has the R-cave an v 1 an v 2 lie in it (see Figure 5 (a)). Interestingly, in a room (P; ) containing an s-triple C, the vertex v 2 cannot be cleare last among fv 1 ; v 2 ; v 3 g by any search scheule (Lemma 1 (a)). An s-triple C such that v 2 has the R-cave an v 1 an lie in it is calle an l-triple (see Figure 5 (b)). Note that the R-cave of v 2 entirely contains the chain C[; Pre(v 2 )). Symmetrically, an s-triple C such that v 2 has the L-cave an v 3 an lie in it is calle an r-triple (see Figure 5 (c)). In a room containing an l-triple C (resp, r-triple C), v 3 (resp, v 1 ) is cleare last among fv 1 ; v 2 ; v 3 g. Lemma 1 Suppose that a room (P; ) is 1-searchable an (; ) is its search scheule. (a) If hv 1 ; v 2 ; v 3 i is an s-triple, then (; ) clears v 1 or v 3 last among fv 1 ; v 2 ; v 3 g. (b) If hv 1 ; v 2 ; v 3 i is an l-triple, then (; ) clears v 3 last among fv 1 ; v 2 ; v 3 g. (c) If hv 1 ; v 2 ; v 3 i is an r-triple, then (; ) clears v 1 last among fv 1 ; v 2 ; v 3 g. Proof. (a). Let n(t) enote the number of clear vertices in fv 1 ; v 2 ; v 3 g at t. We emphasize that two vertices in fv 1 ; v 2 ; v 3 g cannot be cleare at the same time, by Fact 1, because they are invisible from each other an so cannot lie on (t)(t) at the same time. Thus n(t) can increase by at most 1 at t. 5

6 v 2 v 2 Pre(v 2) v 2 Succ(v 2) v 1 v 1 v v 1 3 v 3 v 3 (a) (b) Fig. 5. Examples of three orer-inucing triples: (a) s-triple (b) l-triple (c) r-triple. (c) Suppose the searcher is clearing v 1 at t 1 (i.e., v 1 2 (t 1 )(t 1 )). All points in a cave of v 1 oes not meet (t 1 )(t 1 ), so they are all clear or all contaminate at t 1, by Fact 2. Since both v 2 an v 3 are containe in a cave of v 1, they are all clear or all contaminate at t 1. In other wors, n(t 1 ) equals 1 or 3. Since symmetric arguments can be applie to v 3, at the moment v 3 is cleare, n(t) equals 1 or 3. To prove this lemma, we will show the stronger argument that whenever n(t) becomes 3, v 1 or v 3 is cleare (observe it implies this lemma). For contraiction, suppose not at t 00 for the rst time. That is, n(t) becomes 3 an v 2 is cleare at t 00 but never before t 00. Without loss of generality, we assume that n(t) becomes 2 at t 0 (< t 00 ) an oes not change between t 0 an t 00. There are two cases: Case 1. n(t) increases at t 0 (from 1 to 2). Since v 2 is cleare at t 00, the vertex cleare at t 0 is v 1 or v 3. However, it is impossible because whenever v 1 or v 3 is cleare n(t) equals 1 or 3 but n(t 0 )=2. Case 2. n(t) ecreases at t 0 (from 3 to 2). More exactly, after all of fv 1 ; v 2 ; v 3 g is clear, only v 2 is recontaminate at t 0 an later v 2 is cleare at t 00. We show it is impossible. To show this, consier the ynamic shortest path, like rubber ban, from the current position of the searcher to v 1 an v 3, respectively, Clearly, this shortest path ivies the polygon into two regions: upper one an lower one. We assume that when n(t) becomes 3 before t 00, the searcher is clearing v 1 or v 3 (i.e., v 1 or v 3 2(t)(t)). At that moment, the ynamic shortest path lies above (t)(t) an so upper region is all clear or all contaminate, by Fact 1. Since the upper region contains v 2 an thus is clear, the intruer must lie in the lower region, if it is not etecte yet. However, to recontaminate only v 2 at t 0, the intruer must lie in the upper region. Thus the intruer must cross the ynamic shortest path. At the moment that he crosses it, he can sneak into one of v 1 an v 3 along the shortest path. Therefore, v 1 or v 3 is recontaminate before t 0, which is a contraiction. Therefore, we can conclue that any search scheule clears v 1 or v 3 last among fv 1 ; v 2 ; v 3 g. (b). Assume that hv 1 ; v 2 ; v 3 i is an l-triple. Since the chain C[; v 1 ] is entirely containe in a cave of v 2, if v 2 becomes clear while v 1 is contaminate then the intruer at v 1 can escape along C[; v 1 ]. Thus, whenever v 2 becomes clear, v 1 must 6

7 be clear alreay. By combining the fact with the proof of (a), we can easily show that v 3 is cleare last. (c). It is symmetric to (b) Three Necessary Conitions In this subsection, we show three necessary conitions for 1-searchable rooms. We say that a 2 has an s-pair ha 1 ; a 3 i, if ha 1 ; a 2 ; a 3 i is an s-triple. Theorem 1 If a room (P; ) is 1-searchable the following N1; N2; N3 must hol: (N1) There are no vertices a an b such that an b lie in the L-cave or R-cave of a, an an a lie in the L-cave or R-cave of b. (N2) There is no s-triple ha 1 ; a 2 ; a 3 i such that a 1 an a 3 lie in the L-cave or R-cave of a 2. (N3) There are no l-triple ha 1 ; a 2 ; a 3 i an r-triple hb 1 ; b 2 ; b 3 i such that a 2 b 2 an every vertex v lying between a 2 an b 2 has an s-pair hv L ; v R i. Proof. For examples violating N1, N2, N3, see Figure 6. Since N1 an N2 are known necessary conitions for the corrior search problem an the polygon search problem, respectively, proofs of N1 an N2 can be easily euce from those in Refs. [5,9] but we inclue them here for completeness. a2 c1 c 4 a3 c 2 c 3 b1 a b a1 a3 a2 a1 b2 b3 (a) (b) (c) Fig. 6. Examples violating N1, N2, N3. (N1) Suppose two vertices a an b violate N1. The vertices a an b are contaminate at time zero. Consier the time t 2 [0; 1) at which a is cleare for the rst time. It is easily seen that b must alreay be clear at t. Otherwise, the intruer at b will escape along C[b; ] which oes not meet (t)(t). Symmetrically, when b is cleare for the rst time, a must be clear. Therefore, a an b must be cleare simultaneously. However, it is impossible, by Fact 1, since they are mutually invisible an so cannot lie on (t)(t) at the same time. (N2) Suppose that ha 1 ; a 2 ; a 3 i violates N2 but this room is 1-searchable. Consier the time t 2 [0; 1) at which a 2 is cleare for the rst time. By the proof of Lemma 1a, a 1 or a 3 is contaminate at t. Thus the intruer lurking in it can escape through, which is a contraiction. (N3) Before proceeing to a formal proof, which is somewhat technical, we explain basic ieas with the room in Figure 6c. First, let us conrm that this room violates N3: ha 1 ; a 2 ; a 3 i is an l-triple an hb 1 ; b 2 ; b 3 i is an r-triple; every vertex in C(a 2 ; c 2 ] has an s-pair ha 1 ; c 4 i an every vertex in C[c 3 ; b 2 ) has an s-pair hc 1 ; b 3 i. 7

8 Thus this room violates N3. Next, we show this room is not searchable. For contraiction, suppose that this room is searchable an (; ) is a search scheule. Let S be the vertices in C[a 1 ; b 3 ] excluing Succ(a 1 ); Pre(b 3 ). We claim that no vertices are cleare last in S, which is a contraiction because (; ) clears all vertices in S an some vertices last in S. Since ha 1 ; a 2 ; a 3 i is an l-triple an a 3 2S, a 1 cannot be cleare last in S by Lemma 1 (b). Symmetrically, since hb 1 ; b 2 ; b 3 i is an r-triple an b 1 2 S, b 3 cannot be cleare last in S by Lemma 1 (c). Finally, the vertices in C(Succ(a 1 ); Pre(b 3 )) are not cleare last, by Lemma 1 (a), because each of them has an s-pair that are inclue in S. Therefore, no vertices are cleare last, which is a contraiction. Let us give a formal proof. (In the above example, it was clear that a 3 an b 1 belong to S an s-pairs of the vertices in C(Succ(a 1 ); Pre(b 3 )) lie in S. However, it must be shown in the formal proof. See Claim 1 an 2.) For contraiction, suppose that the room violates N3 but is 1-searchable. Since this room violates N3, there exist an l-triple ha 1 ; a 2 ; a 3 i an an r-triple hb 1 ; b 2 ; b 3 i such that a 2 b 2 an every vertex v lying in C(a 2 ; b 2 ) has its s-pair hv L ; v R i. For technical reasons, it is convenient to choose some special ha 1 ; a 2 ; a 3 i an hb 1 ; b 2 ; b 3 i with aitional properties. Without loss of generality, we can assume the following properties. (1) a 1 a 0 1 for any l-triple ha 0 1 ; a 2; a 0 3 i an b0 3 b 3 for any r-triple hb 0 1 ; b 2; b 0 3 i. (2) a 3 is not Pre(b 3 ) an b 1 is not Succ(a 1 ). (Inee otherwise, that is, if a 3 = Pre(b 3 ), ha 1 ; a 2 ; b 3 i is also an l-triple because Pre(b 3 ), being one enpoint of the R-cave of b 3, is a reex vertex an thus the R-cave of b 3 properly contains that of Pre(b 3 ) an the L-cave of a 1 contains b 3. Then we take it as the new ha 1 ; a 2 ; a 3 i.) Let S be the set of vertices in C[a 1 ; b 3 ] excluing Succ(a 1 ) an Pre(b 3 ). Claim 1. a 3 an b 1 belong to S. Proof: First, we note a 3 b 3 (inee otherwise, ha 1 ; b 2 ; a 3 i is an r-triple an it violates the property (1)). Thus, a 3 2 C[a 1 ; b 3 ]. Since a 3 cannot be Pre(b 3 ) by property (2), a 3 belongs to S. The symmetric arguments hol for b 1. We aim to show that no vertices are cleare last in S. Since ha 1 ; a 2 ; a 3 i is an l-triple an a 3 2S, a 1 is not cleare last in S by Lemma 1 (b). Symmetrically,b 3 is not cleare last in S by Lemma 1 (c). Remaining vertices are those in C(Succ(a 1 ); Pre(b 3 )). Claim 2. Each vertex v 2 C(Succ(a 1 ); Pre(b 3 )) has an s-pair hv L ; v R i such that v L ; v R 2 S. Proof: If v belongs to C(Succ(a 1 ); a 2 ] (resp, C[b 2 ; Pre(b 3 ))), ha 1 ; a 3 i (resp, hb 1 ; b 3 i) is an s-pair of v, thus the claim hols. It suces to prove this claim for v 2 C(a 2 ; b 2 ). By N3, every vertex v 2 C(a 2 ; b 2 ) has an s-pair hv L ; v R i. Unless v L belongs to S, either it precees a 1 or it is Succ(a 1 ). In the latter case, it is easily seen that ha 1 ; v R i is also an s-pair of v since Succ(a 1 ) is a reex vertex. If we let v L = a 1, v L 2 S. In the former case, hv L ; v R i is also an s-pair of a 2. Thus hv L ; a 2 ; v R i is an l-triple, which is a contraiction to (1). For v R, symmetric arguments hol. 8

9 Since every vertex in C(Succ(a 1 ); Pre(b 3 )) has an s-pair in S, it cannot be cleare last in S by Lemma 1 (a). Therefore, no vertices are cleare last in S, which is a contraiction Suciency We will show in this section that a room (P; ) is 1-searchable if it satises three necessary conitions in the previous section. Theorem 2 A room (P; ) is 1-searchable, if it satises N1; N2; N3 of Theorem 1. In the rest of this section, we prove Theorem 2. Throughout this section, we assume that the room (P; ) satises N1, N2, N3. Basic iea of the proof is simple: greey algorithms (explaine later) either complete searching successfully or stop to n orer-inucing triples (see Lemma 3,4,5). v R (v L ) v L y x Fig. 7. Currently, V is xy. Suppose that currently, the searcher stans at x an sees y (see Figure 7). Intuitively, we view the segment xy as the cut V that separates the contaminate region from the clear region. We clear P by sweeping it by V in a way that the region below V (i.e., the region that lies on the same sie as with respect to V) is always clear. Now, let us explain how to avance V. Suppose that currently, V is at xy (xy) an the region below V is clear (see Figure 7). First, we n the vertex v L that is rst encountere in the clockwise traversal from x. If y v L, this means x an y lie on one ege. In this case, entire P must have been cleare alreay an so (P; ) is 1-searchable. Thus, we assume v L y. Next, we avance the left enpoint of V to the vertex v L by moving the right enpoint of V back an forth, if necessary. Of course, this movement is not always possible. However, we can always attain it if the room contains no r-triple (Lemma 2). Let (v L ) enote the rst point, encountere in the counterclockwise traversal from, that is visible from v L (see Figure 7). Note that (v L ) always exists. Lemma 2 Suppose the current V is xy (x y) an v L y. If the room (P; ) contains no r-triple, then V can avance to v L (v L ). Proof. We rst ene sub-movements of V. See Figure 8. Suppose that the current V is xy. If some chain C[x; x 0 ] is entirely visible from y, we can avance 9

10 the left enpoint of V continuously to x 0, xing the right enpoint (Figure 8a). We call this movement left-sweep. If there exist two points x 0 ( x) an y 0 ( y) that are mutually visible an the chain C[x; x 0 ] is entirely visible from the intersection point z of two segments xy an x 0 y 0 (Figure 8b), we can move V from xy to x 0 y 0 by performing left-avance-by-rotation, in which we avance the left enpoint to x 0 continuously along the bounary an backtrack the right enpoint until it reaches y 0. Note that uring the execution of left-avance-by-rotation, the right enpoint of V may jump backwars. The right-sweep an the right-avance-by-rotation are ene symmetrically. We emphasize that uring the execution of these sub-movements, the intruer cannot sneak into the region below V. It will turn out that the above four sub-movements suce to search any 1-searchable room. x 0 x (a) y x 0 (b) Fig. 8. Sub-movements of V: (a) left-sweep (b) left-avance-by-rotation. This sub-movement can be implemente by the 1-searcher as follows. If the searcher is at x, he moves along xx 0 looking at z. If he is at y, he looks at z an backtracks from y to y 0 along the segments in orer of numbering. In steps 2 an 5, the 1-searcher moves insie P in the irection parallel to V, looking at two bounary points in V. x 1 y Now, we return to the proof an avance V from xy to v L (v L ). Base on the location of (v L ), there are several cases to consier. If (v L ) lies between y an (as in Figure 8b where v L = x 0 ), the right enpoint of V only has to backtrack, using left-avance-by-rotation. Otherwise, (v L ) y. Even in this case, if the chain C[(v L ); y] is entirely visible from x (as in Figure 7), we can avance V to x(v L ) using right-sweep, an later to v L (v L ) using left-sweep. The above two cases are referre to as Easy cases, in which we can simply avance V to v L (v L ). Thus the remaining case is that (v L ) y an some points in C[(v L ); y] are invisible from x. For convenience we ivie it into two cases. See Figure 9. Case 1. C((v L ); ) lies in a left cave of v L. Case 2. C((v L ); ) lies in the R-cave of v L. (Note that C((v L ); ) cannot lie in a right cave of v L other than the R-cave of v L, since y is visible from x.) In both cases, this lemma can be proven similarly. We avance the right enpoint of V from y to (v L ). During this avance the left enpoint of V may backtrack. For a vertex v in C[(v L ); y], let (v) be the rst point, if it exists, encountere in the counterclockwise traversal of C[; v L ] from v L, that is visible from v (see (v i ) z 6 y 0 10

11 v L x v L (v L ) (v L ) x y y (a) Fig. 9. (a) Case 1 (b) Case 2 (b) in Figure 10). We rst claim that (v) exists for every vertex v in C[(v L ); y]. Inee otherwise, all points in C[; v L ] are invisible from v, which means v L an lie in a cave of v. Since v an lie in a cave of v L, we can n two vertices that violate N1, which is a contraiction. Thus (v) exists. Let v 1 ; v 2 ; ; v m be the vertices in C((v L ); y) represente in counterclockwise orer (see Figure 10). It is easily seen that (v L ) is visible from x; so let v m+1 = (v L ). To prove this lemma, it suces to show that we can avance V from (v i )v i to (v i+1 )v i+1. (The base case of the movement from xy to (v 1 )v 1 is exactly same as the rst or the secon case of the inuction step explaine below.) If (v i+1 ) (v i ) (i = 1 in Figure 10), the left enpoint of V only has to backtrack, which can be one using right-avance-by-rotation. Even when (v i ) (v i+1 ), if the chain C[(v i ); (v i+1 )] is visible from v i+1, V can avance to (v i )v i+1 using right-sweep, an later to (v i+1 )v i+1 using left-sweep. (In fact, the above two cases are symmetric with the Easy cases.) Thus the remaining case is that some points in C[(v i ); (v i+1 )] lie in a cave of v i+1. Let C(p; q) enote a cave of v i+1, rawn with bol lines in Figure 10. v L (v i+1 ) r q p (v i ) v v L i v2 v i v v1 i+1 (v i+1 ) z v i+1 y p q r (v i ) (a) (b) Fig. 10. Avance from (v i )v i to (v i+1)v i+1 (Case 1). v2 v1 y If C(p; q) is a right cave, Pre(q) (r in Figure 10 a) an v i form an s-pair of v L an thus hpre(q); Pre((v L )); v i i is an r-triple (in Case 1) or hpre(q); v L ; v i i violates N2 (in Case 2). If C(p; q) is a left cave, let r be the most clockwise vertex in C(p; q) (Figure 10 b). Note that we only use four sub-movements of V in the proof of this lemma an thus we can assume that only four sub-movements were 11

12 use while V move from l r to xy. Since each enpoint of V moves continuously along the bounary whenever it avances, the left enpoint of V must have passe any point preceing x. Since r (v i+1 )v L an r is a vertex, r x. Thus V must have passe rz for some z. By avancing V from rz to qv i+1, we can make V arrive at qv i+1. The last avance is one by left-avance-by-rotation, because r lies in the left cave of v i+1 an so z precees v i+1. In this way, V can pass any left cave of v i+1 containe in C[(v i ); (v i+1 )] an eventually arrive at (v i+1 )v i+1. Repeating this proceure, we can avance V to (v m+1 )v m+1 = v L (v L ), completing the proof of this lemma. 2 Lemma 2 says that the left enpoint of V can avance locally if the room contains no r-triple. By applying Lemma 2 repeately, we obtain the following lemma. Lemma 3 If the room (P; ) contains no r-triple, it is 1-searchable. Proof. Assume that the room (P; ) contains no r-triple. Let v 0 (= l ), v 1 ; ; v n (= r ) be the vertices of P, represente in clockwise orer. Initially, V is l r. By Lemma 2, V can avance to v 1 (v 1 ). If two points v 1 an (v 1 ) lie on one ege of P, this implies entire P is clear alreay an thus searching is complete. Otherwise, by repeating the same proceure, V can avance through v i (v i ), until two points v i an (v i ) lie on one ege of P. When i becomes n?1, (v n?1 ) = r an so searching is successfully complete. 2 Remark. In the proof of Lemma 2, consier the case that V cannot avance to v L (v L ) ue to some r-triple. There is only one such case (Figure 10 a). The r-triple foun is hr; Pre((v L )); v i i. Note that in this case, hr; v i i is an s-pair of v L, which means Succ(r)v L. Therefore, we can conclue that if the room satises N1, N2, N3 an v Succ(b 0 1) for any r-triple hb 0 1 ; b0 2 ; b0 3i, we can always avance V to v(v) by the proceure in Lemma 2 an 3. This fact will be use in the proof of Lemma 5. Lemma 4 If the room (P; ) contains no l-triple, it is 1-searchable. Proof. It is symmetric to Lemma 3. 2 Finally, we show the following lemma, completing the proof of Theorem 2. Lemma 5 If the room (P; ) contains both l-triples an r-triples, it is 1-searchable. Proof. We rst claim that for any two vertices a an b, if a has the R-cave an b has the L-cave an is containe in both of them, then ab. Its proof is not icult: For contraiction, suppose b a. If b = a, then lies in the R-cave of a an, at the same time, in the L-cave of a(= b), which is impossible. If b=pre(a), then the L-cave of b an the R-cave of a are isjoint each other, so it is impossible that lies in both of them. If b Pre(a), then an b lie in the R-cave of a an an a lie in the L-cave of b, which contraicts N1. Therefore, we have that ab. Let ha 1 ; a 2 ; a 3 i be a rightmost l-triple such that a 0 2 a 2 for every l-triple h; a 0 2 ; i, an let hb 1 ; b 2 ; b 3 i be a leftmost r-triple such that b 2 b 0 2 for every r-triple h; b 0 2 ; i. By the above claim an the enition of the l-triple an r-triple, we have that a 2 b 2. Since this room satises N3, some vertex in C(a 2 ; b 2 ) must have no s-pair; let g enote such a vertex. 12

13 In what follows, we show that the room with a specie vertex g is 1-searchable. We will avance V, as in Lemma 3. Let v 1 ; v 2 ; ; v m (= g) enote the vertices in the chain C(; g] in the clockwise irection. Let (v k ) enote the rst point, encountere in the clockwise traversal from g, that is visible from v k, if such a point exists. Otherwise, let (v k ) = (v k ). (Note that if some point in C(g; ) is visible, is similar to in that they are the 'farthest' points from. If C(g; ) is invisible, is same as. Thus is ene as a combination of an.) Then, we claim: Claim. We can always avance V to v k (v k ), for k =1; 2; ; m. Proof of Claim: The proof procees by inuction on k in increasing orer. As an inuction hypothesis, assume that the current V is v k (v k ). Case 1. If the chain C(g; ) is invisible from v k+1, then v k (v k ) is the same as v k (v k ). Recall from Remark at the en of Lemma 3 that V can avance to v(v) if v Succ(b 0 1) for any r-triple hb 0 1 ; b0 2 ; b0 3i. >From the facts that hb 1 ; b 2 ; b 3 i is a leftmost r-triple an g lies in C(a 2 ; b 2 ) an has no s-pair, g Succ(b 0 1) for any r-triple hb 0 1 ; b0 2 ; b0 3i an v k+1 also oes. Thus V can avance to v k+1 (v k+1 ), by Remark at the en of Lemma 3. Case 2. If some point in the chain C(g; ) is visible from v k+1, there are four cases epening on (v k ) an (v k+1 ). Actually, this case analysis is completely symmetric with that of Case 1 an 2 in Lemma 2. If (v k ) (v k+1 ), we only have to backtrack the right enpoint of V an this is one using left-avance-by-rotation. Even when (v k+1 ) (v k ), if the chain between (v k ) an (v k+1 ) is visible from v k+1, we can move V using left-sweep an right-sweep. The remaining case is that some subchain of C[(v k+1 ); (v k )] forms a cave of v k+1. See Figure 11. Let C(p; q) enote such a cave. If C(p; q) is a left cave (Figure 11a), v k an Succ(p) form an s-pair of g, which contraicts the enition of g. If C(p; q) is a right cave (Figure 11b), let r enote the most counterclockwise vertex in C(p; q). Then by the symmetric argument of Remark at the en of Lemma 3, we have that the right enpoint of V can avance to r if r Pre(a 0 3) for any l-triple ha 0 1 ; a0 2 ; a0 3i. Since ha 1 ; a 2 ; a 3 i is a rightmost l-triple an g lies in C(a 2 ; b 2 ) an has no s-pair, g Pre(a 0 3) for any l-triple ha 0 1 ; a0 2 ; a0 3i, an thus r also oes. Therefore, we can avance V to zr for some z. Since r lies in a right cave of v k+1, v k+1 z. Later we can move V from zr to v k+1 p by right-avance-by-rotation. In this way, V can pass any right cave of v k+1 containe in C[(v k+1 ); (v k )]. Therefore, the claim hols. The last step of the claim implies that we can avance V to gg, which proves this lemma. 2 13

14 g " g " z v k v k+1 (v k+1 ) q p v k v k+1 q (v k+1 ) p r (v k ) (v k ) # (a) Fig. 11. Proof of Claim in Lemma 5. # (b) 5. Algorithms an Their Time Complexities Previously, we have seen a characterization of 1-searchable rooms. However, concrete algorithms have gone unmentione so far. This section corrects the omission an iscusses the eciency of the algorithms. In Subsection 5.1, we give an O(n log n)-time algorithm to test for 1-searchability of an n-sie room (P; ). In Subsection 5.2, we present an O(n 2 )-time algorithm for constructing a search scheule, if one exists, which is worst-case optimal because O(n 2 ) lower boun for the corrior search problem 5 also hols in our problem with small moication. Interestingly (an fortunately), the search scheule construction algorithm is inepenent of the proof-proceures in the previous section an can be unerstoo only on the basis of the fact that a search scheule of any 1-searchable room can be represente as a sequence of sweeps an avance-by-rotations, as observe in the previous section. Let v 0 (= l ), v 1 ; ; v n (= r ) be the vertices represente in clockwise orer, an let e i enote the ege between v i an v i+1. In this section, we assume all eges are open, that is, e i =C(v i ; v i+1 ) Test Algorithm for 1-searchability First of all, we note that N1, N2, N3 are ene as relations on L-caves, R-caves, an vertices. If we view L-caves an R-caves as intervals ene these necessary conitions are relations on intervals. Let l i enote the clockwise enpoint of the L-cave of v i, if it exists, an r i enote the counterclockwise enpoint of the R-cave of v i, if it exists (Figure 12). The L-cave (resp, R-cave) of v i correspons to the interval C[v i+1 ; l i ] (resp, C[r i ; v i?1 ]). The L-cave (resp, R-cave) of a vertex v is maximal if it is not containe in the L-cave (resp, R-cave) of any other vertex. A vertex pair ha 1 ; a 3 i is an s-pair if it is an s-pair of some vertex. An s-pair ha 1 ; a 3 i is maximal if C[a 1 ; a 3 ] is not inclue in C[a 0 1 ; a0 3] for any other s-pair ha 0 1 ; a0 i. 3 As a preprocessing, we compute the sorte lists of L-caves, R-caves, maximal L-caves, maximal R-caves, an maximal s-pairs. Preprocessing is one in three steps: 14

15 v 9 v 13 l 2 v 16 v9 r 22 v 12 v 15 v 7 v 18 v 5 v 20 v 5 v 20 v 22 v 2 l 13 r 12 r 7 Fig. 12. L-caves an R-caves can be viewe as intervals 1. While scanning the bounary, compute L-cave an R-cave of each vertex v i, if they exist. This step can be one in O(n log n) time by computing l i 's an r i 's using ray shooting queries. 2 We assume L-caves an R-caves are sorte with respect to their left enpoint an store in circular lists, respectively. 2. While scanning the list of L-caves elete non-maximal ones. Since the list of L-caves is sorte with respect to the left enpoints in step 1, step 2 runs in linear time. Symmetric proceure computes the list of maximal R-caves. 3. While scanning the list of maximal L-caves an maximal R-caves clockwise, compute maximal s-pairs. Specically, this step can be one as follows: for each maximal L-cave (v i+1 ; l i ) o n the most clockwise maximal R-cave (r j ; v j?1 ) such that r j v i+1 v j?1 l i. Clearly this can be execute in linear time. In the sequel, we briey escribe algorithms for testing N1, N2, N3. First, consier N2. Note that if an s-triple ha 1 ; a 2 ; a 3 i violates N2, some ha 0 1 ; a 2; a 0 3i such that ha 0 1 ; a0 3i is a maximal s-pair also oes. Thus N2 can be teste as follows: While scanning the list of L-caves an maximal s-pairs clockwise, test if there is an L-cave containing two vertices of some s-pair. Next, we execute the same proceure for R-caves. Since maximal s-pairs are sorte, N2 can be teste in linear time after preprocessing. In orer to test N3, we x our attention to l-triples an r-triples with the following special properties. An l-triple ha 1 ; a 2 ; a 3 i is rightmost maximal if a 0 2 a 2 for any l-triple h; a 0 2 ; i an ha 1; a 3 i is a maximal s-pair. An an leftmost maximal r-triple is ene symmetrically. Then we claim that if a room violates N3, it contains a rightmost maximal l-triple ha 1 ; a 2 ; a 3 i an a leftmost maximal r-triple hb 1 ; b 2 ; b 3 i such that every vertex v in C(a 2 ; b 2 ) has its maximal s-pair. This claim hols because the rightmost an leftmost property only make C[a 2 ; b 2 ] narrow. N3 can be teste as follows: First, in orer to n a rightmost maximal l-triple, while scanning the list of R-caves an maximal s-pairs we n the most clockwise R-cave an a maximal s-pair such that the R-cave contains an the rst (but not the secon) vertex of the s-pair. Similarly, we n a leftmost maximal r-triple. If 15

16 both of them exist, while scanning the vertices in C[a 2 ; b 2 ] an the list of maximal s-pairs, test if each vertex in C[a 2 ; b 2 ] has its s-pair. Clearly this can be one in linear time. Testing N1 is simple. Suppose that two vertices a an b violate N1. If the L-cave of a an the L-cave of b violate N1, then the rst an the last maximal L-caves that contain also violate N1; if the R-cave of a an the L-cave of b violate N1, then the rst maximal R-cave containing an the last maximal L-cave containing also violate N1. Analogously, we only have to test N1 for the rst an the last elements containing in the list of maximal L-caves an maximal R-caves. Thus N1 can be teste in constant time after O(n log n) preprocessing. Therefore, we have the following theorem. Theorem 3 To test the 1-searchability of a room, an O(n log n)-time algorithm suces Search Scheule Construction Algorithm Here, we transform the search scheule construction problem to a simple graph search problem. For convenience, the visibility relation is generalize to the weak visibility: An ege e i is visible from an ege e j (resp, a vertex v j ), if some point on e i is visible from some point on e j (resp, v j ). The moie chor system As a ata structure, we use the moie chor system that is a slight moication of the chor system use in Ref. [9]. The moie chor system is represente as a 2n2n array, where each row/column correspons to a vertex or an ege of P. Each entry c ij is associate with a boolean value vis such that c ij.vis is set to 1 if i is visible from j; 0 otherwise. Aitionally, c ij contains many attributes: If i correspons to an ege an c ij :vis is 1, then c ij contains the maximal interval on i that is visible from j. This moie chor system can be constructe in O(n 2 )-time for an n-sie polygon P basically by constructing VP(v) an VP(e), respectively, in linear time from a triangulation of P, for every 2;9 vertex v an ege e of P. Aitional attributes of each cell c ij are four links, namely next-v, next-e, prev-v, prev-e. The link next-v (resp, next-e, prev-v, prev-e) of c ij points the rst vertex (resp, ege, vertex, ege), encountere in the clockwise (resp, clockwise, counterclockwise, counterclockwise) traversal from i, that is visible from j. It is easily seen that these links can be constructe in O(n 2 )-time. Algorithm for constructing a search scheule Recall that a search scheule of any 1-searchable room can be represente as a sequence of sweeps an avance-byrotations. Here, we transform the search scheule construction problem to a graph search problem. In what follows, we refer to graph vertices as noes an graph eges as arcs. Let N(x) enote the vertices an eges ajacent to x (incluing x); N(v i ) = fe i?1 ; v i ; e i g an N(e i ) = fv i ; e i ; v i+1 g. First, we construct a irecte graph G where each noe is an orere pair of vertices an (open) eges of P that are visible from each other, incluing ( l ; r ). There is an arc from a noe (x; y) to (x 0 ; y 0 ) i one of these hols: (1) x 0 2 N(x) an y 0 2 N(y), or (2) y = y 0 an x 0 is 16

17 the rst vertex/ege, encountere in the counterclockwise traversal from x, that is visible from y, or (3) x = x 0 an y 0 is the rst vertex/ege, encountere in the clockwise traversal from y, that is visible from x. The main property of G is as follows: G contains a path from (x; y) to (x 0 ; y 0 ) if an only if V can move from pq to p 0 q 0 where p 2 x; p 0 2 x 0 ; q 2 y; q 0 2 y 0. We briey show this claim by explaining one-to-one corresponence between sub-movements (of sweep an avance-by-rotation) an paths in G. In sub-movements, either both enpoints of V move continuously or one enpoint jumps backwars. The former case correspons to arcs of type (1) in G. The latter case nees some explanation. Observe that an arc of type (2) an (3) is the smallest backwar jump from (x; y) (Figure 13) an so any backwar jump can be represente as a sequence of smallest backwar jumps (arcs of type (2) an (3)) an continuous moves (arcs of type (1)). Thus any sub-movement of V correspons to a path in G. ege x 00 x ege y=y 0 ege x 0 x ege y=y 0 (a) (b) Fig. 13. (a) A backwar jump from (x; y). (b) The smallest backwar jump from (x; y) corresponing to an arc of type (2). What is the maximum out-egree of a noe in G? For example, let us consier the outwar arcs of the noe (x; y). The caniate set of estinations of the arcs ae by the rule (1) is the subset of orere prouct of N(x) an N(y). The number of the estinations ae by the rule (2) is at most 1, because x 0 is uniquely etermine by x an y. Similarly, the number of the estinations ae by the rule (3) is at most 1. Thus the outer egree of each noe is at most 11. Since every noe in G has constant out-egree, the size of G is O(n 2 ). Moreover, it takes constant time to test whether a caniate arc shoul be ae to G, using the moie chor system. Hence, G can be constructe in O(n 2 ) time. To construct a search scheule, we only have to n a path from ( l ; r ) to (v i ; v i ) or (e i ; e i ) in G for some i. If we a a vertex g an connect all noes (v i ; v i ) an (e i ; e i ) to g, a search scheule correspons to a path in G from ( l ; r ) to g. Thus this path can be foun by one reachability test in G, which is one in time linear to the size of the graph. Since the size of G is O(n 2 ), we can construct a search scheule in O(n 2 )-time, which gives the following theorem. Theorem 4 To construct a search scheule, an O(n 2 )-time an O(n 2 )-space algorithm suces. 17

18 6. Comparison of a 1-Searcher an Two Guars In this section, we compare the searchability of a 1-searcher with that of two guars. A corrior is searchable (walkable in Ref. [5]) by two guars if two points L an R can move from to g, along C[; g] an along C[g; ] respectively, in a way that L an R are always mutually visible. If a corrior is searchable by two guars, a 1-searcher can also search it by moving like L an looking at R. However, the 1-searcher has the aitional capability of jumping its one enpoint backwars, as in Figure 8 b. The main question in this section is whether this backwar jump is avantageous for searching a corrior or room. First we show that in the corrior search problem, this jump is not necessary. Theorem 5 Any 1-searchable corrior can be searche by two guars. Proof. Recall from Ref. [5] that a corrior can be searche by two guars if an only if it satises three properties (in our terminology): (1) the chains C[; g] an C[g; ] are weakly visible from each other. (2) there is no hv 1 ; v 3 i that is an s-pair of g. (3) there are no two vertices a 2 C[; g] an b 2 C[g; ] such that the R-cave of a contains b an, an the L-cave of b contains a an,. It suces to show that any 1-searchable corrior also satises above three properties. Since (3) is a special case of N1, (3) is also a necessary conition of a 1-searchable corrior. Next we consier (1) an (2). Suppose a corrior violates (1), that is, some vertex v in C[; g] is not visible from C[g; ]. When a 1-searcher is clearing v (i.e., v 2 (t)(t)), no points in C[g; ] meet (t)(t). Thus the intruer can go from g to along C[g; ], which means (1) is a necessary conition of a 1-searchable corrior. If (2) is violate, that is, hv 1 ; v 3 i is an s-pair of g, hv 1 ; g; v 3 i is an s-triple if we view (P; ) as a room. In orer to prevent the intruer from g to, (t)(t) must ivie the polygon so that an g lie in the opposite sie. By the same arguments as in Lemma 1, when we clear one of v 1 an v 3, the other is contaminate. Therefore, (P; ; g) is not 1-searchable. 2 A room is searchable by two guars if two guars L an R can move from l an r respectively, to some point p (6=) in such a way that L an R are always visible from each other an the position of L always precees that of R. Clearly, if a room is searchable by two guars then it is 1-searchable because the 1-searcher can search it by moving like L an looking at R. However, it is not clear whether the converse is true. Here, we show that two guars are strictly weaker than a 1-searcher in searching a room (P; ). Theorem 6 Some 1-searchable room is not searchable by two guars. Proof. It suces to show an example that is not searchable by two guars but 1-searchable. Such an example is the room in Figure 14. This room is 1-searchable by the following movements of V over time: l r! aa 0! bb 0! bc 0! ae 0! ee 0! ff 0! gg 0! ga! gg. In the rst interval [ l r! aa 0 ], V executes a sequence of left-sweeps an right-sweeps an two enpoints of V move along the bounary of P. Similarly, in all intervals except [ae 0! ee 0 ], two enpoints of V move along the bounary of P. In an exceptional interval [ae 0! ee 0 ], the left enpoint of V jumps backwars. 18

19 g 0 h f 0 e 0 b b 0 e a c 0 f g a 0 Fig. 14. Example of a 1-searchable room that is not searchable by two guars. To show that the room in Figure 14 is not searchable by two guars, we claim that the vertex h cannot be visite by any of L an R. For contraiction, assume that h is visite for the rst time by L at t, that is, l(t) = h for some t. Since L must have passe g 0 at some t 0 (< t), r(t 0 ) lies in C[g; g 0 ] or an ege ajacent to g 0. This implies that R must have alreay passe h before t 0, which is a contraiction. Conversely, assume that h is visite for the rst time by R at t. There must be a time t 0 (<t) such that r(t 0 )=c 0 an that R remains to lie in C(h; c 0 ] uring the time interval [t 0 ; t). Since r(t 0 )=c 0, it must be that b l(t 0 ). Moreover, since l(t)f, L must move from b to f uring the time interval [t 0 ; t). However, since some point in C(e; a) is invisible from any point in C(h; c 0 ], L must jump iscontinuously, which is a contraiction. Therefore, this room is not searchable by two guars, which gives this theorem Concluing Remarks In this paper, we consiere the problem of searching a polygonal room by a 1-searcher, which is an interesting variant of the polygon search problem. 9 In particular, we obtaine a characterization of the class of 1-searchable rooms. Using this characterization, we presente simple algorithms to test the 1-searchabilily of a room an to construct a search scheule. We also showe that the class of rooms searchable by two guars is a proper subset of the class of 1-searchable rooms. However, we cannot obtain a characterization of the class of rooms searchble by two guars an leave it as an open problem. Recently, the results of this paper were extene to consier the searcher with omniirectional visibility in the subsequent paper, 6 which also inclues many variants of the room search problem. Acknowlegements We are grateful to anonymous referees for their helpful comments. This work was supporte by KOSEF(Korea Science an Engineering Founation) uner grant

20 References 1. D. Crass, I. Suzuki, an M. Yamashita. Searching for a mobile intruer in a corrior{ the open ege variant of the polygon search problem. Int. J. of Comp. Geom. an Appl., 5(4):397{412, L. Guibas, J. Hershberger, D. Leven, M. Sharir, an R. Tarjan. Linear-time algorithm for visibility an shortest path problems insie triangulate simple polygons. Algorithmica, 2:209{233, L.J. Guibas, J.C. Latombe, S.M. Lavalle, D. Lin, an R. Motwani. A visibilitybase pursuit-evasion problem. Int. J. of Comp. Geom. an Appl., 9(4-5): , P.J. Heernan. An optimal algorithm for the two-guar problem. Int. J. of Comp. Geom. an Appl., 6(1):15{44, C. Icking an R. Klein. The two guars problem. Int. J. of Comp. Geom. an Appl., 2(3):257{285, J. H. Lee, S. Y. Shin, an K. Y. Chwa. Visibility-base pursuit-evasions in a polygonal room with a oor. In Proc. 15r ACM Symp. Comp. Geom., pages 281{290, G. Narashimhan. On hamiltonian triangulations in simple polygon. In Proc. of Workshop on Algorithm an Data structure, K. Sugihara, I. Suzuki, an M. Yamashita. The searchlight scheuling problem. SIAM J. Comput., 19(6):1024{1040, I. Suzuki an M. Yamashita. Searching for a mobile intruer in a polygonal region. SIAM J. Comput., 21(5):863{888, I. Suzuki, M. Yamashita, H. Umemoto, an T. Kamea. Bushness an a tight worst-case upper boun on the search number of a simple polygon. Information Processing Letters, 66:49{52, L.H. Tseng, P. Heernan, an D.T. Lee. Two-guar walkability of simple polygons. Int. J. of Comp. Geom. an Appl., 8:85{116, M. Yamashita, H. Umemoto, I. Suzuki, an T. Kamea. Searching for mobile intruers in a polygonal region by a group of mobile searchers. In Proc. 13th Annu. ACM Sympos. Comput. Geom., pages 448{450,