Exploring Simple Grid Polygons

Transcription

1 Exploring Simple Grid Polygon Chritian Icking 1, Tom Kamphan 2, Rolf Klein 2, and Elmar Langetepe 2 1 Univerity of Hagen, Praktiche Informatik VI, Hagen, Germany. 2 Univerity of Bonn, Computer Science I, Römertraße 164, Bonn, Germany. Abtract. We invetigate the online exploration problem of a hortighted mobile robot moving in an unknown cellular room without obtacle. The robot ha a very limited enor; it can determine only which of the four cell adjacent to it current poition are free and which are blocked, i. e., unacceible for the robot. Therefore, the robot mut enter a cell in order to explore it. The robot ha to viit each cell and to return to the tart. Our interet i in a hort exploration tour, i. e., in keeping the number of multiple cell viit mall. For abitrary environment without hole we provide a trategy producing tour of length S C + 1 E 3, where C denote the number of cell the area, and 2 E denote the number of boundary edge the perimeter of the given environment. Further, we how that our trategy i competitive with a factor of 4, and give a lower bound of 7 for our problem. Thi leave a 3 6 gap of only 1 between the lower and the upper bound. 6 Key word: Robot navigation, exploration, covering, online algorithm, competitive analyi, lower bound, grid polygon 1 Introduction Exploring an unknown environment and earching for a target in unknown poition are among the baic tak of autonomou mobile robot. Both problem have received a lot of attention in computational geometry and in robotic; ee e. g. [3, 5, 9, 12, 13, 17]. We ue a imple model for the robot and it environment: the robot i hortighted, and the urrounding i ubdivided by a rectangular integer grid, imilar to a cheboard. Eentially, there are two motivation for uing thi model intead of a robot with a full viion ytem: Firt, even a laer canner ha a reliable range of only a few meter. Hence, the robot ha to move toward more ditant area in order to explore them. Second, ervice robot like lawn mower or cleaner need to get cloe to their work area. The robot enor provide the information, which of the four neighbor of the currently occupied cell do not belong to the polygon and which one do. The robot can enter the latter cell. The robot tak i to viit every cell inide the polygon and to return to the tart cell. Sometime, thi tak in alo called covering. Even though our robot doe not know it environment in advance it i intereting to ak how hort a tour can be in the offline ituation, i. e., when the environment i already known. Thi amount to contructing a hortet traveling aleperon tour on the free cell.

2 If the polygonal environment contain obtacle, the problem of finding uch a minimum length tour i known to be NP-hard, ee Itai et al. [14]. There are 1 + ε approximation cheme by Grigni et al. [7], Arora [2], and Mitchell [16], and a 53 approximation by Arkin et al. [1]. 40 In a polygon without obtacle, the complexity of contructing offline a minimum length tour eem to be open. Ntafo [18] and Arkin et al. [1] have hown how to approximate the minimum length tour with factor of 4 3 and 6 5, repectively. Uman and Lenhart [19] have provided an O(C 4 ) algorithm for deciding if there exit a Hamiltonian cycle, i. e., a tour that viit each of the C cell of a polygon exactly once. For the related problem of Hamiltonian path, Everett [4] ha given a polynomial algorithm for certain grid graph. In thi paper our interet i in the online verion of the cell exploration problem. Exploring a grid polygon with hole wa conidered by Icking et al. [10, 11] and independently by Gabriely and Rimon [6]. Icking et al. howed a lower bound of 2 for thi problem and introduced an exploration trategy that need no more than C E+3H +W 2 tep,3 ee [15], where C denote the number of cell, E the number of boundary edge, H the number of hole and W i a meaure for the winding of the polygon. Gabriely and Rimon howed an upper bound of C + B, where B denote the number of boundary cell. We conider the exploration of polygon without hole. Although both problem eem to be cloely related there i an important difference: We have a lower bound of 2 for polygon with hole, but it turn out that we can do much better in imple polygon. An upper bound for our exploration trategy i given in term of the polygon area, C, and the perimeter, E. While C i the number of free cell, E i the number of edge between a free cell and a blocked cell, ee for example Fig. 1. We ue E to ditinguih between kinny and thick environment. For thick environment, E O( C) hold; thu, the number of additional cell viit i ubtantially maller than C. Only in polygon that do not contain any 2 2-quare of free cell, C = 24 E = 40 = 2(C+1) C = 24 E = 20 << 2C Fig. 1. The perimeter, E, to ditinguih between thin and thick environment. E achieve it maximum value of 2(C + 1), and our upper bound i equal to 2C 2, but in thi cae one cannot do better, ince even the optimal offline trategy need that number of tep. Our paper i organized a follow: in Sect. 2 we give more detailled decription of our robot and the environment. We give a lower bound for our problem in Sect. 3. In Sect. 4 we preent an exploration trategy, SmartDFS. The analyi how in Sect. 5 that thi trategy ue no more than C + 1 2E 3 tep and i in fact competitive with a factor of 4 3. SmartDFS wa implemented in a Java-Applet available in the internet, ee [8]. 3 We aume that the cell have unit ize, o the length of the path i equal to the number of tep from cell to cell.

3 2 Definition We conider a imple model for the environment of the robot: the robot move in a urrounding with a grid tructure. More preciely, a cell i a baic block in our environment, defined by a pair (x, y) IN 2. A cell i either free and can be viited by the robot, or blocked, i. e., unacceible for the robot. We call two cell adjacent, if they hare a common edge, and touching, if they hare a common edge or corner. A grid polygon, P, i a connected et of free cell. A polygon without blocked cell inide it boundary i called imple. From it current poition, the robot can find out which of the adjacent cell are free and which are blocked, and it can move in one tep to an adjacent free cell, ee Fig. 1. The robot ha enough memory to tore a map of known cell. 3 A Lower Bound Theorem 1. Every trategy for the exploration of a imple grid polygon with C cell need at leat 7 6 C tep. Proof. We aume that the robot tart in a corner of the polygon, ee Fig. 2(i). W. l. o. g. we aume that the trategy decide to walk one tep to the eat. For the econd tep, the trategy ha two poibilitie: either it leave the wall with a tep to the outh, ee Fig. 2(ii), or it continue to follow the wall with a further tep to the eat, ee Fig. 2(iii). In the firt cae, we cloe the polygon a hown in Fig. 2(iv). The robot need at leat 8 tep to explore thi polygon, but the optimal trategy need only 6 tep yielding a factor of 8 6. In the econd cae we proceed a follow. If the robot leave the boundary, we cloe the polygon a hown in Fig. 2(v) and (vi). The robot need 12 tep, but 10 tep are ufficient. In the mot intereting cae, the robot till follow the wall, ee Fig. 2(vii). In thi cae, the robot need at leat 28 tep to explore thi polygon, wherea an optimal trategy need only 24 tep. Thu, we achieve a factor of 7 6. We can eaily extend thi pattern to polygon of arbitrary ize by repeating the contruction uing the entry and exit cell denoted by the arrow in Fig. 2(iv) (vii). Thi contruction cannot lead to overlapping polygon or polygon with hole, ince the polygon alway extend to the ame direction. (ii) (i) (iv) (v) (vi) (iii) (vii) Fig.2. A lower bound for the exploration of imple polygon. The dahed line how the optimal olution, denote the robot poition.

4 4 An Exploration Strategy A a firt approach, we can apply a imple depth-firt earch algorithm (DFS): The polygon i explored following the left-hand rule, i. e., for every entered cell the robot trie to continue it path to an adjacent and unexplored cell, preferring a tep to the left over a traight tep over a tep to the right. Thi reult in a complete exploration, but take 2C 2 tep. Since the hortet tour need at leat C tep, DFS turn out to be 2-competitive. However, there i no reaon to viit each cell twice jut becaue thi i required in ome pecial ituation like dead end of width 1. In the following, we introduce two improvement to DFS. (i) c 2 (ii) DFS improved DFS c 1 c 1 c 2 Fig.3. Improvement to DFS: (i) optimize return path, (ii) detect polygon plit. The firt improvement i to return directly to thoe cell that have unexplored neighbor. See e. g. Fig. 3(i): DFS walk from c 1 to c 2 through the completely explored corridor. A more efficient trategy walk on a hortet path on cell that are already known from c 1 to c 2. Now, oberve the polygon hown in Fig. 3(ii). With DFS, the robot walk four time through the narrow corridor. A more clever olution explore the right part immediately after the firt viit of c 1, and continue with the left part, reulting in only two viit. The cell c 1 ha the property that the graph of unviited cell plit into two component after c 1 i explored. We call cell like thi plit cell. The econd improvement i to recognize and handle plit cell, ee Sect. 5. The following decription of our trategy, SmartDFS, reume both improvement to DFS, ee Fig. 5 for an example. SmartDFS(P, tart): Chooe direction dir, uch that revere(dir) i a blocked cell; ExploreCell(dir); Walk on the hortet path to tart; ExploreStep(bae, dir): if unexplored(bae, dir) then Walk on hortet path to bae; move(dir); ExploreCell(dir); end if ExploreCell(dir): bae := current poition; if not isplitcell(bae) then ExploreStep(bae, ccw(dir)); ExploreStep(bae, dir); ExploreStep(bae, cw(dir)); ele Chooe different order, ee Sect. 5. end if

5 5 The Analyi of SmartDFS SmartDFS explore the polygon in layer, beginning with the cell along the boundary of P and proceeding toward the interior of P. Definition 2. Let P be a grid polygon. The boundary cell of P uniquely define the firt layer of P. The polygon P without it firt layer i called the 1-offet of P. The l-th layer and the l-offet of P are defined ucceively, ee Fig. 4(i). Lemma 3. The l-offet of a imple grid polygon, P, ha at leat 8l edge le than P. Proof. Firt, we cut off blind alley narrower than 2l, ince thoe part of P do not affect the l-offet. We walk clockwie around the boundary cell of the remaining polygon, ee Fig. 4(i). For every left turn the offet gain at mot 2l edge and for every right turn the offet looe at leat 2l edge. Since, there are four more right turn than left turn, we looe at leat 8l edge. (i) Π cut off l 2l edge gained l 2l edge lot P 1 (iii) Q P 1 (ii) c c P 2 c P 2 P 1 Q c c (iv) P 2 Fig. 4. (i) The 2-offet (haded) of a grid polygon; three example for plit cell, (ii) type (II), (iii) and (iv) type (I). Definition 2 allow u to pecify the handling of a plit cell in SmartDFS. Let u conider the ituation hown in Fig. 5(i): SmartDFS ha jut met the firt plit cell, c, in the fourth layer of P. P divide into three part: P = K 1 K2 { viited cell of P }, where K 1 and K 2 denote the connected component of the unviited cell. In thi cae it i reaonable to explore the component K 2 firt ince the tart cell i cloer to K 1. We ue the layer number to decide which component we have to viit at lat. Whenever a plit cell occur in layer l, every component i one of the following type, ee Fig. 4(ii) (iv): (I) K i i completely urrounded by layer l, 4 (II) K i i not urrounded by layer l, or (III) K i i partially urrounded by layer l. In any cae, it i the bet choice to explore the component of type (III) at lat. Note that it may occur that three component arie at a plit cell, but we can handle thi cae a two ucceive plit occuring at the ame plit cell. 4 More preciely, the part of layer l that urround K i i completely viited. For convenience, we will ue lightly loppy, but horter form.

6 K 1 c Q P 1 Q K 1 Q c c P 2 P K 2 K 2 (i) (ii) Fig.5. A decompoition of P at the plit cell c and it handling in martdfs. For the analyi we conider two polygon, P 1 { and P 2, a follow. Let Q be l, if K2 i of type (I) the quare of width 2q + 1 around c with q := l 1, if K 2 i of type (II), where K 2 denote the component that i explored firt, and l denote the layer in which the plit cell wa found. We chooe P 2 P Q, uch that K 2 {c} i the q-offet of P 2, and P 1 := ((P \P 2 ) Q) P, ee Fig. 5. The interection with P i neceary, ince Q may exceed the boundary of P. The choice of P 1, P 2 and Q enure that the robot path in P 1 \Q and in P 2 \Q do not change compared to the path in P. The part of the robot path that lead from P 1 to P 2 and from P 2 to P 1 are fully contained in the quare Q. Jut the part inide Q are bended to connect the appropriate path inide P 1 and P 2, ee Fig. 5. We want to viit every cell in the polygon and to return to. Every trategy need at leat C(P) tep to fulfill thi tak. Thu, we can plit the overall length of the exploration path Π into two part, C(P) and exce(p), with Π = C(P) + exce(p). Since SmartDFS recurively explore K 2 {c}, we want to apply the upper bound inductively to the component K 2 {c}. The following lemma give u the relation between the path length in P and the path length in the two component. Lemma 4. Let P be a imple grid polygon. Let the robot viit the firt plit cell, c, which plit the unviited cell of P into two component K 1 and K 2, where K 2 i of type (I) or (II). With the preceding notion we have exce(p) exce(p 1 ) + exce(k 2 {c}) + 1. Proof. Since c i the firt plit cell, there i no exce in P 2 \(K 2 {c}) and it uffice to conider exce(k 2 {c}) for thi part. After K 2 {c} i finihed, the robot tart at c and explore K 1. For thi part we take exce(p 1 ) into account. Finally, we add one ingle tep, becaue the plit cell c i viitited twice:

7 once, when SmartDFS detect the plit and once more after the exploration of exce(k 2 {c}) i finihed. Altogether, the given bound i achieved. The following lemma can eaily be hown and allow u to charge the number of edge in P 1 and P 2 againt the number of edge in P and Q. Lemma 5. Let P be a imple grid polygon, and let P 1, P 2 and Q be defined a above. The number of edge atify E(P 1 ) + E(P 2 ) = E(P) + E(Q). Lemma 6. Let Π be the hortet path between two cell in a grid polygon P. The length of Π i bounded by Π 1 2E(P) 2. Proof. The maximal ditance i achieved between two cell in the firt layer, and 1 the hortet path between them i never longer than 2 #(cell in the firt layer). Analogouly to Lemma 3, thi layer ha at mot E(P) 4 cell. Now, we can give an upper bound for the number of tep ued to explore a imple polygon. Theorem 7. Let P be a imple grid polygon with C cell and E edge. P can be explored with S C + 1 2E 3 tep. Thi bound i tight. Proof. C i the number of cell and thu a lower bound on the number of tep that are needed to explore the polygon P. We will how by induction on the number of component that exce(p) 1 2E(P) 3 hold. For the induction bae we conider a polygon without any plit cell, i. e., SmartDFS viit all cell and return on the hortet path to the tart cell. Since there i no polygon plit, all cell of P can be viited by a path of length C 1. By Lemma 6 the hortet path back to the tart cell i not longer than 1 2 E 2 and exce(p) 1 2E(P) 3 hold. Now, we aume that there i more than one component during the application of SmartDFS. Let c be the firt plit cell detected in P. When SmartDFS reache c, two new component, K 1 and K 2, occur. We conider the two polygon P 1 and P 2 defined a above uing the quare Q around c. W. l. o.g. we aume that K 2 i recurively explored firt. After K 2 i completely explored, SmartDFS proceed with the remaining polygon. A hown in Lemma 4 we have exce(p) exce(p 1 )+exce(k 2 {c})+1. Now, we apply the induction hypothei to P 1 and K 2 {c} and get exce(p) 1 2 E(P 1) E(K 2 {c}) By applying Lemma 3 to the q-offet K 2 {c} of P 2 we achieve exce(p) 1 2 E(P 1) (E(P 2) 8q) = 1 2 (E(P 1) + E(P 2 )) 4q 5. From Lemma 5 we conclude E(P 1 ) + E(P 2 ) E(P) + 4(2q + 1). Thu, we get exce(p) 1 2E(P) 3. Thi bound i achieved exactly in polygon that do not contain any 2 2-quare of free cell. So far we have hown an upper bound for the number of tep needed to explore a polygon that depend on the number of cell and edge in the polygon. Now we want to analyze SmartDFS in the competitive framework.

8 Corridor of width 1 or 2 play a crucial role in the following, o we refer to them a narrow paage. 5 It i eay to ee that narrow paage are explored optimally. In paage of width 1 both SmartDFS and the optimal trategy viit every cell twice, and in the other cae both trategie viit every cell exactly once. We need two lemmata to how a competitive factor for SmartDFS. The firt one give u a relation between the number of cell and the number of edge for a pecial cla of polygon. Lemma 8. For a imple grid polygon, P, without any narrow paage or plit cell in the firt layer, E(P) 2 3 C(P) + 6 hold. Proof. Conider uch a polygon, P, ee Fig. 6(i). We ucceively remove an outer row or column of at leat three boundary cell, maintaining our aumption on P. Thee aumption enure that we can alway find uch a row or column. Thu, we remove at leat three cell and at mot two edge. Thi decompoition end with a 3 3 block of cell that fulfill E = 2 3C(P)+6. Now, we revere our decompoition, i. e., we ucceively add all row and column until we end up with P. In every tep, we add at leat three cell and at mot two edge. Thu, E 2 3C(P) + 6 i fulfilled in every tep. P Π c SmartDFS (i) (ii) optimal trategy Fig.6. (i) For polygon without narrow paage or plit cell in the firt layer, E(P) 2 C(P) + 6 hold, and the lat explored cell, 3 c, lie in the 1-offet, P (haded), (ii) In a corridor of width 3 and even length, S(P) = 4 SOpt(P) 2 hold. 3 For the ame cla of polygon, we can how that SmartDFS behave lightly better than the bound in Theorem 7. Lemma 9. A polygon of the ame type a in Lemma 8 can be explored uing no more than S(P) C(P) + 1 2E(P) 5 tep. Proof. We have hown S(P) C(P) + 1 2E(P) 3 in Theorem 7. In the proof, we ued Lemma 6 to bound the return path, but thi lemma bound the path between two cell in the firt layer. By our aumption on P, we can completely explore the firt layer of P before viiting another layer, and the return path, 5 More preciely, a cell, c, belong to a narrow paage, if c can be removed without changing the layer number of any other cell.

9 Π, tart in a cell, c, in the 1-offet, P, ee Fig. 6(i). Let denote the firt viited cell in P. Remark that and are at leat touching each other. Now, Π i bounded by a hortet path, Π, from c to in P and a hortet path from to, i. e., Π Π + 2. Π, in turn, i bounded uing Lemma 6 by Π 1 2 E(P ) 2. With Lemma 3, E(P ) E(P) 8 hold, and altogether we get Π 1 2E(P) 4, which i two tep horter than tated in Lemma 6. Theorem 10. The trategy SmartDFS i 4 3 -competitive. Proof. Let P be a imple grid polygon. Firt, we remove all narrow paage from P and get a equence of (ub-)polygon P i, i = 1,...,k, without narrow paage. For every P i, i = 1,...,k 1, the optimal trategy in P explore the part of P that correpond to P i up to the narrow paage that connect P i with P i+1, enter P i+1, and fully explore every P j with j i. Then it return to P i and continue with the exploration of P i. Further, we already know that narrow paage are explored optimally. Thi allow u to conider every P i eparately without changing the competitive factor of P. Now, we oberve a (ub-)polygon P i. We how by induction on the number of plit cell in the firt layer that S(P i ) 4 3 C(P i) 2 hold. Note that thi i exactly achieved in polygon of ize 3 m with m even, ee Fig. 6(ii). If P i ha no plit cell in the firt layer, we can apply Lemma 9 and Lemma 8: S(P i ) C(P i ) E(P ( i) 5 C(P i ) C(P i) + 6 ) 5 = 4 3 C(P i) 2. Two cae occur if we meet a plit cell, c, in the firt layer, ee Fig. 4(ii) (iv). In the firt cae, the new component wa never viited before (type (II)). Here, we define Q := {c}. The econd cae occur, becaue the robot meet a cell, c, that i in the firt layer and touche the current cell, c, ee for example Fig. 4(iii) and (iv). Let Q be the mallet rectangle that contain both c and c. Similar to the proof of Theorem 7, we plit the polygon P i into two part, both including Q. Let P denote the part that include the component of type (II) or (III), P the other part. For Q = 1, ee Fig. 4(ii), we conclude S(P i ) = S(P )+ S(P ) and C(P i ) = C(P ) + C(P ) 1. Applying the induction hypothei to P and P yield S(P i ) = S(P ) + S(P ) 4 3 C(P i) < 4 3 C(P i) 2. For Q { 2, 4 } we gain ome tep by merging the polygon. If we conider P and P eparately, we count the tep from c to c or vice vera in both polygon, but in P i the path from c to c i replaced by the exploration path in P. Thu, we have S(P i ) = S(P )+S(P ) Q and C(P i ) = C(P )+C(P )+ Q. Thi yield S(P i ) = S(P )+S(P ) Q = 4 3 C(P i)+ 1 3 ( Q 6) 2 < 4 3 C(P i) 2. An optimal trategy need C tep, which, altogether, yield a competitive factor of Summary It turned out that the exploration of imple polygon i eaier than the exploration of polygon with hole in term of competitivity. In contrary to the lower bound of 2 for polygon with hole, we have hown a lower bound of 7 6 and an upper bound of 4 3 for imple polygon, leaving a gap of only 1 6. Additionally, we can alo bound the length of an exploration path by C + 1 2E 3 which i tight.

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