Approximate Guarding of Monotone and Rectilinear Polygons


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1 Aroximate Guarding of Monotone and Rectilinear Polygons Erik Krohn Bengt J. Nilsson Abstract We show that vertex guarding a monotone olygon is NPhard and construct a constant factor aroximation algorithm for interior guarding monotone olygons. Using this algorithm we obtain an aroximation algorithm for interior guarding rectilinear olygons that has an aroximation factor indeendent of the number of vertices of the olygon. If the size of the smallest interior guard cover is OPT for a rectilinear olygon, our algorithm roduces a guard set of size O(OPT 2 ). Comutational geometry Art gallery roblems Monotone olygons Rectilinear olygons Aroximation algorithms 1 Introduction The art gallery roblem is erhas the best known roblem in comutational geometry. It asks for the minimum number of guards to guard a sace having obstacles. Originally, the obstacles were considered to be walls mutually connected to form a closed Jordan curve, hence, a simle olygon. Tight bounds for the number of guards necessary and sufficient were found by Chvátal [7] and Fisk [17]. Subsequently, other obstacle saces, both more general and more restricted than simle olygons have also been considered for guarding roblems, most notably, olygons with holes and simle rectilinear olygons [21, 32]. Art gallery roblems are motivated by alications such as lineofsight transmission networks in terrains, such as, signal communications and broadcasting, cellular telehony systems and other telecommunication technologies as well as lacement of motion detectors and security cameras. We distinguish between two tyes of guarding roblems in simle olygons. Vertex guarding considers only guards ositioned at vertices of the olygon, whereas interior guarding allows the guards to be laced anywhere in the interior of the olygon. The comutational comlexity question of guarding simle olygons was settled by Aggarwal [1] and Lee and Lin [26] indeendently when they showed that the roblem is NPhard for both vertex guards and interior guards. Further results have shown that already for very restricted subclasses of olygons the roblem is still NPhard [2, 30]. Chen et al. [5] claim that vertex guarding a monotone olygon is NPhard, however the details of their roof are omitted and still to be verified. We resent a new roof that vertex guarding a monotone olygon is NPhard. The aroximation comlexity of guarding olygons has been studied by Eidenbenz and others. Eidenbenz [14] shows that olygons with holes cannot be efficiently guarded by fewer than Ω(log n) times the otimal number of interior or vertex guards, unless P=NP, where n is the number of vertices of the olygon. Brodén et al. and Eidenbenz [2, 13] indeendently rove that interior guarding simle olygons is APXhard. Deartment of Comuter Science, University of Wisconsin Oshkosh, Oshkosh, WI, 54901, USA. Deartment of Comuter Science, Malmö University, SE Malmö, Sweden. 1
2 s t monotone rectilinear Figure 1: Illustrating the olygon classes. Any olygon (with or without holes) can be efficiently vertex guarded with logarithmic aroximation factor in n, the number of vertices of the olygon. The algorithm is a simle reduction to SET COVER and goes as follows [19]: comute the arrangement roduced by the visibility olygons of the vertices. Next, let each vertex v corresond to a set in the set cover instance consisting of elements corresonding to the faces of the arrangement that lie in the visibility olygon of v. The greedy algorithm for SET COVER will then roduce a guard cover having logarithmic aroximation factor. The above result can be imroved for simle olygons using randomization, giving an algorithm with exected running time O(nOPT 2 v log4 n) that roduces a vertex guard cover with aroximation factor O(log OPT v ) with high robability, where OPT v is the smallest vertex guard cover for the olygon [12]. Taking the same aroach one ste further, Deshande et al. [11] resent a seudoolynomial randomized algorithm for finding a guard cover (without any restriction on lacement) with aroximation factor O(log OPT). We rove olynomial time deterministic aroximation algorithms for interior guarding of monotone and rectilinear olygons. As we have already mentioned, vertex guarding of monotone olygons is NPhard, and furthermore, otimally guarding rectilinear olygons is also NPhard [23]. This rovides the basis for our interest in aroximation algorithms for these roblems. The art gallery roblem concerns itself with covering olygons using star shaed ieces, the visibility olygons of the guards. Covering olygons with other tye of objects, e.g., convex olygons, etc., remains NPhard in general; [8, 9, 16, 20, 29, 32, 33, 34]. The next section contains some useful definitions. Section 3 contains our NPhardness roof for monotone olygons and in Sections 4 and 5 we describe the aroximation algorithms for guarding monotone and rectilinear olygons resectively. 2 Definitions A olygon P is lmonotone if there is a line of monotonicity l such that any line orthogonal to l has a simly connected intersection with P. When we talk about monotone olygons, we will henceforth assume that they are xmonotone, i.e., the xaxis is the line of monotonicity for the olygons we consider; see Figure 1. The boundary of a monotone olygon P can be subdivided into two chains, the uer chain U and the lower chain D. Let s and t be the leftmost and rightmost vertices of P resectively. The chain U consists of the boundary ath followed from s to t in clockwise direction, whereas D is the boundary ath followed from s to t in counterclockwise direction. A olygon P is rectilinear if the boundary of P consists of axis arallel line segments. Hence, at each vertex, the interior angle between the two connecting boundary edges is either 90 or 270 degrees; see Figure 1. 2
3 q r r r t SP(,t) r Figure 2: Illustrating the roof of Lemma 2.1. Let VP() denote the visibility olygon of P from the oint, i.e, the set of oints in P that can be connected with a line segment to without intersecting the outside of P. Consider a artial set of guard oints g 1,..., g m inpand the union of their visibility olygons m i=1 VP(g i), the set P \ m i=1 VP(g i) is the region of P not seen by the oints g 1,..., g m. This region consists of a set of simly connected olygonal regions called ockets bounded by either the olygon boundary or the edges of the visibility olygons. The following definitions are useful for monotone olygons. Since the xaxis is the line of monotonicity it makes sense to say that an object A in the olygon is to the left or to the right of some other object B if there is vertical line that searates the two objects. We will occasionally use A B (A B) to denote that A is to the right (to the left) of B. Let q be a oint in VP(). We denote by VP R (, q) the art of VP() that lies to the right of q. Similarly, VP L (, q) is the art of VP() to the left of q. Hence, VP() = VP L (, q) VP R (, q) for all oints q P. We also denote VP R () = VP R (, ) and VP L () = VP L (, ). In the sequel, we will also let SP(, q) denote the shortest (Euclidean) ath between oints and q inside P. LEMMA 2.1 If q is a oint on SP(, t) inside a monotone olygon P, then VP R (, q) VP R (q). PROOF: Let r be a oint to the right of q in P that is visible from. To rove that r is seen from q consider the vertical line through r and its intersection oint r with SP(, t). The three oints, r, and r define a olygon in P having three convex vertices and ossibly some reflex vertices on the ath SP(, r ). Since r sees both and r, r sees all of the ath SP(, r ) and hence also the oint q; see Figure 2. 3 NPHardness of Vertex Guarding Monotone Polygons In this section, we will show that vertex guarding a monotone olygon is NPhard. The reduction is from Monotone 3SAT (M3SAT) [18, age 259 (roblem L02)]. An M3SAT instance (X, C) consists of a air of sets, a set of Boolean variables, X = {x 1, x 2,..., x n } and a set of clauses, C = {c 1, c 2,..., c m }. Each clause contains three literals, c i = x j x k x l, a ositive clause, or c i = x j x k x l, a negative clause, for 1 j, k, l n. An M3SAT instance is satisfiable, if a satisfying truth assignment for C exists such that all clauses c i are true. An ordinary 3SAT instance can easily be transformed to an M3SAT instance by taking each nonmonotone clause and relacing it by three monotone ones as follows. c i = x j x k x l ( z i1 z i2 x l ) (z i1 x j x k ) (z i2 x j x k ) c i = x j x k x l (z i1 z i2 x l ) ( z i1 x j x k ) ( z i2 x j x k ) 3
4 x x d( x) x x x x d(x) b(x) Starting attern b(x) d(x) d( x) b(x) Negative Positive Variable atterns Figure 3: The different tyes of variable atterns. where z i1 and z i2 are new variables used only in these three clauses. It is easy to verify that a truth assignment makes clause c i true if and only if the truth assignment makes all the three monotone relacement clauses true as well. By aroriately dulicating clauses, we can assume that the instance has m clauses where m is odd and the instance has (m + 1)/2 ositive clauses and (m 1)/2 negative clauses. Also, let K = n (m + 1). We show that any M3SAT instance is olynomially transformable to an instance of vertex guarding a monotone olygon. We construct a monotone olygon P from the M3SAT instance such that P is guardable by K or fewer guards if and only if the M3SAT instance is satisfiable. We first resent some basic gadgets to show how the olygon is constructed. We then connect these gadgets together to create a olygon. Starting Pattern: The lower boundary of the olygon is divided into two arts, the left and the right sides. The first gadgets on the left side are the starting atterns. The starting atterns are shown to the left in Figure 3. In each attern, the bottom of the downward sike b(x) is the distinguished vertex of the attern. This area is only seen by vertices x and x and must be guarded by one of these two vertices. This attern aears along the left side of the lower boundary of the monotone olygon a total of n times, one corresonding to each variable. Variable Pattern: On the left and the right side of the lower boundary we have variable atterns that verify the assigned truth value of each variable. This attern is shown to the right in Figure 3. Once again, the bottom of the sike at b(x) must be guarded by either x or x. The attern has additional distinguished vertices that we call ledges d(x) and d( x) that must both be seen and this is what forces the choice of guard lacement at either x or x. Figure 4 shows how the starting atterns are connected to variable atterns. If we choose x j in the starting attern, we are forced to continuing to choose x j in each of the subsequent variable atterns. If we at some variable attern would choose x j instead of x j, the ledge d( x j ) is not seen. Similarly, if we in the starting attern choose x j, we are, by the same argument, forced to continuing to choose x j in each of subsequent variable atterns. Clauses: For each clause c in the boolean formula, there is a sequence of variable atterns x 1,...,x n along either the left or the right side of the lower boundary and a clause attern along the uer boundary of the olygon. On the left side of the lower boundary, the variable attern sequence corresonds to negative clauses, on the right side, to ositive clauses. The clause attern on the uer boundary consists of three vertices in an uward sike such that the to vertex of the sike is only seen by the variable atterns corresonding to the literals in the clause; see 4
5 x j x j x k x k. x j x j x j Starting atterns x j x k x k x k x k Positive variable atterns Negative variable atterns Figure 4: Variable atterns transferring logical values. Figure 5. We denote the to vertex of the sike by c to corresond to the clause. We choose our truth value for each variable in the starting variable atterns. The truth values are then mirrored in turn between variable atterns on the right side, corresonding to ositive clauses, and variable atterns on the left side, corresonding to negative clauses, of the lower boundary. Truth values do not change in the mirroring rocess since a variable x j in clause c i only sees the ledge d(x j ) in the next variable attern and none of the other ledges. Similarly x j only sees ledge d( x j ) in the next variable attern; see Figure 4. In the examle of Figure 5 the M3SAT clause corresonds to c = x 1 x 3 x 5. Hence, a vertex guard lacement that corresonds to a truth assignment that makes c true, will have at least one guard on x 1, x 3 or x 5 and can therefore see vertex c without additional guards. We still have variables x 2 and x 4 in the clause, however none of them or their negations see the vertex c. They are there simly to transfer their truth values in case these variables are needed in later clauses. The monotone olygon we construct consists of 4n +(6n + 4)m + 2 vertices. Each starting variable attern having four vertices, each variable attern six vertices, the clause sike consists of three vertices lus one blocking vertex at the start of each clause sequence on the lower boundary and the two leftmost and rightmost oints of the olygon. Consider an M3SAT instance (x 1 x 2 x 3 ) ( x 1 x 3 x 5 ) (x 3 x 4 x 5 ). Figure 6 shows how this instance is transformed into a monotone olygon and a lacement of guards corresonding to the satisfying truth assignment x 1 = x 2 = x 4 = x 5 = false, x 3 = true. Exactly K = n(m + 1) guards are required to guard the olygon since there are K bottom vertices b(x j ) at downward sikes and no vertex in the olygon can see more than one such b(x j ) vertex. If the M3SAT instance is satisfiable, then we lace guards at vertices in accordance to whether the variable is true or false in each of the sequences of variable atterns. Each clause vertex is seen since one of the literals in the associated clause is true and the corresonding vertex has a guard. 5
6 clause c x 1 x 3 x 1 x 2 x 3 x 4 x 5 x 5 Figure 5: A variable attern sequence with its clause sike. Figure 6: Examle reduction of (x 1 x 2 x 3) ( x 1 x 3 x 5) (x 3 x 4 x 5). Points with white centers mark the guards. 6
7 ke(r) R Figure 7: Illustrating the concet of a kernel exansion. The darker shaded areas are the comonents of R and the lighter shaded area is the kernel exansion of R. Suose we have a vertex guard cover of size exactly K. Since each bottom sike b(x j ) is guarded there is a guard at one of x j, x j, or b(x j ) itself. They together make u K guards so there can be no other guards. Since each clause vertex c i is also seen, we can establish which of the guards see this vertex and deduce a satisfying truth assignment from this guard lacement. We have roved the following theorem. THEOREM 1 Finding the smallest vertex guard cover for a monotone olygon is NPhard. Note that our roof does not immediately generalize to interior guards. In the next section, we show how to aroximate the minimum number of interior guards in a monotone olygon. 4 Interior Guarding Monotone Polygons 4.1 The Guarding Algorithm Our algorithm for guarding a monotone olygon P incrementally guards P starting from the left, moving right. Hence, we are interested in the structure of the ockets that occur when guarding is done in this way. We define the main region that we will be interested in, the sear, that will guide the lacement of our guards. Let R be a, ossibly disconnected, set of oints in P and let q be a oint in P. We denote by R L (q) the oints of R to the left of q. Let v R be a leftmost oint of R, hence, if q is to the left of v R, then R L (q) =. DEFINITION 4.1 The kernel exansion of R, denoted ke(r), is formally defined as the set of oints ke(r) def = {q v R R L (q) VP(q)}, i.e., all the oints q in P to the right of v R that see everything in R L (q); see Figure 7. kernel exansion of a region consisting of two connected comonents. In Section we describe how to comute kernel exansions efficiently. Figure 7 shows the LEMMA 4.1 A kernel exansion ke(r) is a monotone olygon. 7
8 (a) uer ocket (b) uer ocket lower ocket boundary ocket (c) (d) Figure 8: Illustrating ockets. Points with white centers are guards. The shaded areas are the visibility olygons. PROOF: Let q and q be two oints in ke(r) having the same xcoordinate, hence, R L (q) = R L (q ) and assume that R L (q). Let be any oint in R L (q) seen by both q and q. Since the three oints, q, and q form a (ossibly degenerate) triangle in P, any oint between q and q will also see. This means that any vertical line has a simly connected intersection with ke(r) so the region is monotone. Assume that we have a artial guard cover in P with the roerty that all guards are to the left of any ockets remaining in P. Consider such a ocket. We say that is a boundary ocket if it is adjacent to both the uer and lower boundaries U and D of P, an uer ocket if it is adjacent only to the uer boundary U, a lower ocket if it is adjacent only to the lower boundary D and a middle ocket if it is adjacent to neither U nor D; see Figures 8(a) (c). We show in Section that our incremental guarding algorithm never roduces any middle ockets, so we can disregard them for now. Since we assume that the guards all lie to the left of all ockets, it is easy to see that the cover can only generate one boundary ocket. Let be such a boundary ocket. In Lemma 4.7 of Section we show that the kernel exansion of a olygonal region is comletely determined by the vertices of the region. We subdivide the vertices of into three sets, V M, V U and V D, where V M are the vertices interior to P, V U are the vertices coinciding with the uer boundary U of P and V D are the vertices coinciding with the lower boundary D of P. We let P U () be the straight line ath visiting the vertices in V M V U in order from left to right, i.e., all vertices of V U are visited along the uer boundary U. Similarly, we let P D () be the straight line ath visiting the vertices in V M V D in order from left to right. Consider the uer ockets resulting from a artial guard cover and enumerate them U 1, U 2,... from left to right. Similarly enumerate the lower ockets D 1,D 2,... from left to right and denote the boundary ocket B. Define two sets as follows: Q U Q D def = U 1 U 2 P U ( B ), def = D 1 D 2 P D( B ), with which we can establish the main regions of interest in the resentation. 8
9 uer sear s U U s D t uer sear ti u U Figure 9: The dark shaded area is the uer sear. DEFINITION 4.2 For X being one of U or D, let the kernel exansion of Q X be the sear with resect to X, i.e., def s X = ke(q X ). Hence, we have the two sear tyes s U and s D corresonding to the sequences of ockets we are currently considering; see Figure 9 for an examle of the uer sear s U. Since the intersection of monotone olygons is also monotone, a sear can be comuted by a lane swee algorithm going from left to right, maintaining the uer and lower boundaries of the kernel exansions; see Section The rightmost intersection oint between the uer and lower boundary of a sear is called the sear ti and we denote sear tis by u U and u D, corresonding to the two tyes of sears; see Figure 9. By the definition of these oints, u X has the roerty of being the rightmost oint that sees all the oints of the ockets of tye X to the left of it. The sears are deendent on the lacement of the reviously laced guards so we will henceforth refer to them as s U (G ) and s D (G ) given the artial guard set G. For each sear s X (G ), we similarly arameterize the sear ti u X (G ). If G =, the uer and lower sears s U ( ) and s D ( ) together with the uer and lower sear tis u U ( ) and u D ( ) are well defined since all of P is considered a boundary ocket. We can now give the details of our guarding algorithm, dislayed in Figure 10. Each iteration of the algorithm begins by comuting the sears and the sear tis, which we show how to do efficiently in Section Ste 4 selects the leftmost of u U (G) and u D (G), lacing a guard g at this oint in Ste 5. Ste 6 results in the addition of g to the guard set only if s X(G) actually intersects l g, where X denotes the remaining ocket tye different from X. We show how to erform Ste 7 efficiently in Section Since all uer and lower ockets are guarded after the algorithm has concluded, we know that the comlete boundary of P is seen by the guards laced. In fact, we rove in Lemma 4.3 that also the interior of P is seen. We claim the following theorem and dedicate Sections 4.2 and 4.3 to roving it. THEOREM 2 The algorithm GUARDMONOTONEPOLYGON comutes a guard cover of size at most 30OPT for a monotone olygon P in olynomial time, where OPT is the size of the smallest guard cover for P. 9
10 Algorithm Inut: GUARDMONOTONEPOLYGON A monotone olygon P Outut: A guard cover for P 1 G := while not all ockets are guarded do 2 Comute s U (G) and u U(G) if there are uer ockets 3 Comute s D (G) and u D(G) if there are lower ockets 4 if u U(G) is to the left of u D(G) then X := U, X := D else X := D, X := U endif 5 Place a guard g at u X(G) and let l g be the vertical line segment through g 6 Place a guard g at an intersection oint of l g and s X(G), if they intersect 7 Place a guard ĝ on l g so that u X(G {g, g, ĝ}) lies as far to the right as ossible 8 G := G {g,g, ĝ} endwhile return G End GUARDMONOTONEPOLYGON Figure 10: The algorithm for guarding monotone olygons. To hel the reader, we rovide a table of the notation we introduce. Symbol Name Exlanation VP() the visibility olygon of the oint VP L (, q), the art of VP() to the left of q VP R (, q) the art of VP() to the right of q VP B (, q, r) the art of VP() between q and r R L (q) the set of oints of R to the left of q ke(r) kernel exansion the set of oints that see all of R L () Q X the union of tye X ockets s X sear the kernel exansion of Q X, ke(q X ) u X sear ti the rightmost oint of a sear, s X l the line or segment through the oint v X base v X lies on the boundary of VP(u X ) and u X lies on the boundary of VP(u X ) 4.2 Correctness and Aroximation Factor Correctness We know from the construction of algorithm GUARDMONOTONEPOLYGON that it will guard the boundary of the olygon. However, we need to rove that it will also guard the interior of the olygon. To rove this, it is sufficient to show that our algorithm never roduces a middle ocket. We do this in two stes. The first ste is to show that all guards cannot be on one side of a middle ocket, i.e., a middle ocket can never be generated to the right of the guards as they are laced by the algorithm. The second ste is to show that when the algorithm 10
11 g u q l gd r e u l q q u q q d e d r g d Figure 11: Illustrating the roof of Lemma 4.2. laces new guards, a middle ocket to the left of these guards can never be generated. LEMMA 4.2 Consider a middle ocket of a artial guard set in a monotone olygon. Let r be the leftmost oint in. Not all guards of the artial guard set can be to the left of r. PROOF: Assume for a contradiction, that all the guards are to the left of r. Let r be the rightmost oint of and let e u and e d be the uer and lower edges resectively of that are also adjacent to r. Neither e u nor e d can be vertical since this would immediately give a contradiction, requiring a guard to the right of r. Let q be a oint interior to, below e u and above e d, let l q be the vertical line through q and let q u and q d be the intersection oints of l q with e u and e d resectively. Since all guards are to the left of r, no single guard can see both oints q u and q d, so there are at least two different guards g u and g d that see these oints. Furthermore, the lines of sight from g u to q u and from g d to q d cannot cross, otherwise, either g u or g d will see oints inside the middle ocket giving us a contradiction. Assume without loss of generality that g u is further to the left than g d, otherwise we reverse the roles of g u and g d in the following argument. Let l gd be the vertical line through g d. It intersects the line of sight from g u to q u at q. The four oints g d, q d, q u and q form a convex olygon with the guard g d in one corner. Hence, q is seen by g d, contradicting our assumtion that q is in a middle ocket; see Figure 11. Lemma 4.2 shows that as algorithm GUARDMONOTONEPOLYGON laces guards incrementally in the olygon, it can never generate a middle ocket to the right of the rightmost guard laced so far. Next, we show that the algorithm will not generate a middle olygon to the left of this guard either, thus giving us the following lemma. LEMMA 4.3 The algorithm GUARDMONOTONEPOLYGON never introduces middle ockets, and hence, roduces a comlete guard cover. PROOF: We make a roof by contradiction and assume that in iteration i of the algorithm, as guards g i, g i and ĝ i are ositioned in the olygon, a middle ocket i is generated between guard triles g j, g j, ĝ j and g j+1, g j+1, ĝ j+1, where j < i. We assume furthermore that i is the first iteration index that generates a middle ocket and hence that i is generated in this iteration. Let G i denote the set of guards laced by the algorithm from iteration 1 until iteration i has comleted. 11
12 Let r be a oint in i and consider the situation just after iteration i 1. The oint r belongs to a ocket i 1 that is either an uer, a lower or a boundary ocket after this iteration. Consider the situation as the algorithm laces guards g i, g i and ĝ i during iteration i. By Lemma 4.2, r lies to the left of g i since there are no other guards to the right of g i. Without loss of generality, we can assume that g i is laced at u U (G i ) in Ste 5 of the algorithm, as the argument when g i is laced at u D (G i ) is comletely analogous. We make a case analysis on whether i 1 is an uer, a lower or a boundary ocket. Assume first that i 1 is an uer ocket, then we have an immediate contradiction since g i = u U (G i 1 ) and u U (G i 1 ), by definition, sees all oints in the uer ockets to the left of itself, and hence, also the oint r. Assume next that i 1 is a lower ocket, then either the vertical line l gi through g i intersects s D (G i 1 ), in which case g i sees r, giving us a contradiction, or l g i does not intersect s D (G i 1 ). In this case, u D (G i 1 ) is to the left of u U (G i 1 ), contradicting the selection in Ste 4. If we assume that i 1 is a boundary ocket, it lies to the right of g i 1. Assume for a contradiction that the oint r, to the left of g i, is not seen by g i or g i. This means that some art of the olygon boundary hides r from g i and g i. Assume first that this is U, i.e., the shortest ath SP(r, g i) touches U at some vertex v. This means that the vertex v on U to the left of v is not seen, contradicting that g i = u U (G i 1 ), since v P U ( i 1 ); see the definition of Q U in Section 4.1. On the other hand, if SP(r, g i ) touches D at some vertex v, then the vertex v on D to the left of v is not seen, giving us that u D (G i 1 ) is to the left of u U (G i 1 ), since v P D ( i 1 ), contradicting the selection in Ste 4. Therefore, algorithm GUARDMONOTONEPOLYGON roduces a guard cover that sees all the boundary of the olygon and it never generates a middle ocket. Hence it roduces a comlete guard cover for the monotone olygon Bases and Shadows We continue with a discussion that becomes fairly technical. We associate a secific region, a shadow, to the right of a sear and show that if two sears of the same tye, i.e., uer or lower sears, generated by a artial guard set that obeys certain conditions, then the associated shadows do not intersect. We use this information in the next section to bound the number of guards that our algorithm will roduce. We begin by defining two concets. Fix a artial guard cover G and let s X be the sears with resect to G, for the ocket tye X being U or D. To each sear s X we associate a oint called a base of the sear, denoted v X, with X being U or D. Let l ux be the vertical line through the sear ti u X of the sear s X and let Q X be the region, the set of ockets, such that s X = ke(q X ); see Definition 4.2. DEFINITION 4.3 A base of s X is the rightmost oint v X in Q X and on the boundary X of P such that the sear ti u X lies on the boundary of VP(v X ) and v X lies on the boundary of VP(u X ). A oint v U is called an uer base and a oint v D is called a lower base. Note that a base can lie on a boundary edge infinitely close to a vertex without being on the vertex. See Figure 12 for an examle of an uer base. The second concet that we define is that of a shadow. DEFINITION 4.4 For a sear s X with X being one of U or D, define the shadow of s X, denotedshd X, to be the art of the visibility olygon of the base v X strictly to the right of u X. Hence, shd X = VP R (v X, u X )\l ux, where l ux is the vertical line through u X ; see Figure
13 base v U shadow shd U Figure 12: Examle of a base and shadow for an uer sear. We arameterize the shadows, in the same way as the sears, to be deendent on the lacement of the reviously laced guards and refer to them as shd U (G ) and shd D (G ) given the artial guard set G. We rove a technical lemma that will be useful to bound the number of guards roduced by our algorithm. LEMMA 4.4 If G and G + are two artial guard covers of P such that G G + and u X (G ) G +, for X being one of U or D, then shd X (G ) shd X (G + ) =. PROOF: Since the sear tye is fixed in each case, we can simlify our notation and let u = u X (G ), u + = u X (G + ), v = v X (G ), v + = v X (G + ), shd = shd X (G ), and shd + = shd X (G + ). We denote by X the oosite boundary of X in P, i.e., if X is U then X is D and vice versa. Let Q and Q + be the ocket regions of the two guard sets with resect to X. If u + lies on the boundary X (excet for the degenerate case when u + lies on a reflex vertex of X), then we immediately have that shd + = roving the lemma. Assume now that u + does not lie on X (or that the degenerate case has occurred), then the sear ti u + is adjacent to two boundary edges e and e of the sear, where e is art of the line segment [v +, u + ]. Consider the extension of e, the other edge, from u + towards the left until it reaches the exterior of P at. We claim that the line segment [, u + ] must touch the boundary X at some oint. Assume that it does not, then there is a oint on the extension of [v +, u + ] towards the right that sees as much of Q + as u + does, thus contradicting that u + is a sear ti. This also shows that [, u + ] touches the boundary of some ocket in Q +. Let q denote the leftmost oint on X that intersects the segment [, u + ]. The line segment [, q] artitions P into two subolygons, P + and P, where P + contains u +. This gives us two cases; see Figure 13. v lies in P. If the extension of [v, u ] towards the right does not cross [, q], then all of shd also lies in P and cannot intersect shd + which lies in P +. If the extension of [v, u ] towards the right does cross [, q], then extend [q, u + ] towards the right until it reaches the exterior of P at. In this case, all of shd in P + lies on the same side of [q, ] as v +, 13
14 v + P + u shd q u + P + P v u shd q v + u + v P (a) shd + (b) shd + u v + v P + P q u + g shd (c) shd + Figure 13: Illustrating the roof of Lemma 4.4 with X = U. since v + lies on X and q lies on X. Hence, shd + lies on the oosite side of [q, ] so shd and shd + cannot intersect; see Figure 13(a). v lies in P +. If u lies in P, then all of shd lies in P and the two shadows cannot intersect. Assume now that u lies in P + and that v sees oints in shd +. We have two subcases. If u also sees oints in shd +, we have an immediate contradiction since then u sees oints in some ocket of Q + or oints of the ath P X () of Q +, if is a boundary ocket; see Figure 13(b). If u does not see oints in shd +, there is a art of the boundary X blocking vision between u and shd +. Note that u cannot see q since otherwise it would also see oints of. Therefore, there is second guard g in G + seing q. The guard g must lie in P otherwise it sees oints in. Furthermore, visibility from g into must be blocked by the boundary X, which must then cross the line segment [, q], a contradiction; see Figure 13(c). This concludes the roof Serial Guard Covers We define a secial tye of guard set that will hel us rove the aroximation factor of our algorithm. DEFINITION 4.5 We define restricted guards as follows: For a region R, a guard g is Rrestricted, if we only consider the restricted visibility olygon of g to be VP(g) def = VP(g) R. A guard g is a left (or right) guard, if g is VP L (g)restricted (or VP R (g)restricted). Next, we define a stri subdivision of P. 14
15 G yaxis Ḡ P P Figure 14: The reflected olygon P. DEFINITION 4.6 A stri subdivision of a monotone olygon P is a subdivision of the olygon by the introduction of vertical segments connecting the uer and lower boundary. Each stri is a subolygon of P bounded by a left vertical edge (ossibly degenerating to the left end oint s), a ortion of the uer boundary U, a right vertical edge (ossibly degenerating to the right end oint t) and a ortion of the lower boundary D. We are now in a osition to define serial guard sets. DEFINITION 4.7 A guard set is serial, if P is subdivided into K stris s 0,...,s K 1 ordered from left to right so that the left edge of stri s 0 has zero or one s 0 restricted guard, the left edge of every other stri s i, 0 < i < K, has exactly one s i restricted guard and the right edge of each stri s i, 0 i < K, contains zero or more left guards. No other guards are laced in the olygon. We say that a serial guard set is an uer serial guard cover, if the s i restricted guard on the left edge of each stri s i, together with the left guards on the right edges of the stris, sees the uer boundary of s i. Similarly, a guard set is a lower serial guard cover, if the s i restricted guard on the left edge of each stri s i, together with the left guards on the right edges of the stris, sees the lower boundary of s i. LEMMA 4.5 A monotone olygon has an uer serial guard cover with at most 2OPT s i restricted guards and at most 3OPT left guards. PROOF: Let G be a guard cover for P. Reflect P along the yaxis to get the reversed olygon P having the guard cover Ḡ, the set G reflected along the yaxis; see Figure 14. Our roof is constructive and iteratively laces restricted guards in the olygon P. Do a lane swee from left to right on P. Initially, let H 0 be the emty guard set and let l 0 be the vertical line through the leftmost oint of P. Iteratively, given the artial guard set H j and the line l j, we construct the next artial guard set H j+1 and the next vertical line l j+1 as follows: Obtain the uer sear ti u = u U (H j ) and let l j+1 be the maximal vertical line segment through u interior to P. Let H j+1 include the guards in H j and let s j be the stri in P bounded by l j and l j+1. We lace the following additional guards on l j+1, 1. an s j restricted guard ḡ j at u, 2. a right guard at u, 3. each guard from Ḡ in s j is moved along its shortest ath to s, the rightmost oint of P, until it reaches l j+1. At this oints lace a right guard. Each of the guards thus laced is added to H j+1 ; see Figure 15. According to Lemma 2.1, any right guard ḡ R on l j+1 will see at least as much to the right of l j+1 in P as the original guard ḡ in Ḡ. Hence, we have the 15
16 Ḡ l j+1 VP(H j) ḡ j VP(H j+1) ḡ j 1 s j VP(ḡ j) l j Figure 15: Illustrating the roof of Lemma 4.5. invariant that the guard set H j+1 sees at least as much of the uer boundary of P as the guards to the left of u in ḠṪhe rocess terminates after K iterations when the lane swee reaches the rightmost oint of P. In the last iteration we have two ossibilities. Either, the right guards in H K 1 together see the uer boundary of s K 1, in which case we do not have to add any s K 1 restricted guard at the right edge of s K 1, or s K 1 must by necessity contain guards from Ḡ. We differentiate between these cases when we count the number of guards laced. The s j restricted guard ḡ j laced at u U (H j ) will see the uer boundary of s j together with the right guards in H j. Hence, the restricted guard set H K sees the uer boundary of P. We count the number of guards laced according to their tye, 1, 2, or 3, above. The number of Tye 3 right guards in H K is Ḡ = G since each of these right guards corresonds to a guard in G. The number of Tye 2 right guards is the same as the number of Tye 1 s j restricted guards since both tyes are laced at uer sear tis u. It remains to count the Tye 1 s j restricted guards. If s j contains guards from Ḡ, we can associate ḡ j to such a guard ḡ in Ḡ. In articular, if the rocess laces guard ḡ K 1 in the last iteration, there is a guard from Ḡ in s K 1 and we can associate ḡ K 1 to this guard. On the other hand, if s j contains no guards from Ḡ, then an uer base v U(H j ) of s U (H j ) is not seen by the guards in Ḡ to the left of l j, i.e., in s 0,..., s j 1, since no right guard in H j sees the base. Therefore, the base v U (H j ) must be seen by a guard ḡ in Ḡ lying in the uer shadow shd U(H j ). Since there is a Tye 2 right guard at the osition of ḡ j, the rerequisites of Lemma 4.4 are fulfilled and we know that no two uer shadows shd U (H j ) and shd U (H j ) intersect for j j, and hence, ḡ can only see one base. We can therefore associate ḡ j to a guard ḡ from Ḡ in the uer shadow shd U(H j ). Note that, if s K 1 contains no guard from Ḡ, then this stri is comletely seen by the guards in revious stris and the rocess laces no s K 1 restricted guard in s K 1. In this way, any guard ḡ in Ḡ can be associated to at most two s jrestricted guards. Hence, the number of Tye 1 s j restricted guards, and thus also the number of Tye 2 right guards, is at most 2 Ḡ = 2 G. Next, reflect the set H K back along the yaxis to become a guard set U of P. We claim that U is uer serial with at most 2 G s i restricted guards, for stris s i, 0 i < K, and at most 3 G left guards. This follows since 16
17 a stri s j in P when reflected back becomes a stri s i in P, with i = K j 1. Each s j restricted guard ḡ j in H K lies on the right edge of s j and sees the uer boundary of s j so the corresonding s i restricted guard g i in U lies on the left edge of s i and sees the uer boundary of s i. Finally, the right guards of H K on the left edges of stris in P corresond to left guards of U in right edges of stris in P. The number of guards has not changed so U is uer serial as claimed. By choosing G to be an otimal guard cover for P, we have that G = OPT, thus roving the lemma. We can, using the same roof technique, show a corresonding lemma for lower serial guards Aroximation Factor Next, we establish the aroximation factor of the algorithm. LEMMA 4.6 The algorithm GUARDMONOTONEPOLYGON laces at most 30OPT guards in P, where OPT is the size of the smallest guard cover for P. PROOF: To bound the total number of guards, we establish the number of guards laced by Stes 5 7 throughout the iterations of algorithm GUARDMONOTONEPOLYGON. To do so, we comare the number of guards laced in each ste with the size of an uer and a lower serial guard cover. Let G U be the set of guards assigned in Stes 5 7 in the iterations of the algorithm when the selection in Ste 4 makes X = U and X = D; see Figure 10. Similarly, let G D be the guards assigned when X = D and X = U. Consider first the set G U and order the guard triles in this set from left to right, G U = {g 1, g 1, ĝ 1, g 2, g 2, ĝ 2,...}. In iteration i of the loo, our algorithm erforms Stes 5 7 with X = U and laces guards g j, g j and ĝ j, all having the same xcoordinate, with j i being the roer index in G U. The next time the algorithm erforms Stes 5 7 with X = U it laces guards g j+1, g j+1 and ĝ j+1. Let G i be the set of guards laced in iterations 1 to i by the algorithm. The guards g j, g j and ĝ j are the rightmost guards in G i. We comare the number of guards in G U with the size of an uer serial guard cover U and show that G U contains at most 3 U guards. To do this, we construct a secondary guard set H incrementally starting with the emty guard set H 0. For every index j > 0 we go through the guard triles in G U as follows: If U has an srestricted guard g in the interval between g j and g j+1 (in the case of j = 0 we consider the leftmost end oint s of P to be the imaginary guard g 0 ), for some stri s, then we let H j := {g j, g j, ĝ j} H j 1, if j > 0. If U has no srestricted guard in the interval between g j and g j+1, then this whole interval is contained in a stri s associated to U. This means that the uer base v U (G i ) is either seen by the srestricted guard g s to the left of g j or by a left guard in U in the uer shadow shd U (G i ). If g s sees v U (G i ) then the shortest ath from g s to t, the rightmost oint of P, crosses the vertical line through g j at a oint. Let H j := {g j, g j, } H j 1, i.e., we exchange the guard ĝ j for a guard at. We claim that g s does not see v U (H j ) but this follows immediately by Lemma 2.1. Furthermore, u U (H j ) is not to the right of u U (G i ), since by our algorithm ĝ j is laced so that u U (G i ) is as far to the right as ossible. If g s does not see v U (G i ) then we let H j := {g j, g j, ĝ j} H j 1. We have that v U (G i ) = v U (H j ). The set H constructed according to the rules given above obeys three imortant criteria: 17
18 1. H = G U, 2. for each j, either there is an srestricted guard between g j and g j+1 or no srestricted guard g s in U sees the uer base v U (H j ), 3. u U (H j ) is not to the right of u U (G i ), where i is the iteration index when guards g j, g j and ĝ j are laced. Let us count the number of guards in H. If two subsequent triles {g j, g j, j} and {g j+1, g j+1, j+1} in H have an srestricted guard g s in the interval between them, then we associate the trile {g j, g j, j} to g s. We call such an association an αassociation. By construction, an srestricted guard g s in U can only be αassociated to a guard trile in H once. If two subsequent triles {g j, g j, j} and {g j+1, g j+1, j+1} in H do not have any srestricted guard g s in the interval between them, then we know that the uer base v U (H j ) is seen by a left guard g in U in the uer shadow shd U (H j ) and we associate the trile {g j, g j, j} to g. We call such an association a βassociation. Since the uer sear s U (H j ) and s U (H j ) obey the rerequisites of Lemma 4.4, for any j j, the two uer shadows shd U (H j ) and shd U (H j ) do not intersect. This means that a left guard g can only be β associated to a guard trile in H once. From this we can deduce that the number of guard triles in H is at most the number of srestricted guards and left guards in U together. From Lemma 4.5, we know that this is at most 5OPT. By a comletely symmetrical argument we can construct a set of guard triles H of the same size as the set G D and deduce that the number of guard triles in this set is also bounded by 5OPT. The total number of guards constructed by our algorithm is therefore bounded by G U + G D = H + H 3 U + 3 D 30OPT, as claimed. 4.3 Comutation Comuting Kernel Exansions Let, q and r be three oints in P. We let VP B (, q, r) denote the art of the visibility olygonvp() between the oints q and r. Let R be a ossibly disconnected olygonal region in P having m vertices and assume that the vertices are ordered v 1,...,v m from left to right. We claim the following lemma. LEMMA 4.7 ke(r) = m 1 i=1 i VP B (v j, v i, v i+1 ) m VP R (v j, v m ). j=1 j=1 PROOF: From Definition 4.1 we have that ke(r) = { v 1 R L () VP()} where R L () is the art of R to the left of. Let be a oint between v i and v i+1. We show that is in ke(r) if and only if i j=1 VP(v j). Assume first that i j=1 VP(v j). In this case, there is a vertex v j of R such that does not see v j. Hence, there is a oint in R to the left of, the vertex v j, not seen by, so is not in ke(r) roving the first imlication of the equivalence. 18
19 q l q v j l q R q q r Figure 16: Illustrating the roof of Lemma 4.7. Assume next that ke(r). In this case, there is a oint q R L () not seen from. Consider the shortest ath SP(, q). Let [, q] be the last segment and let [r, ] be the enultimate segment of SP(, q). Since does not see q, SP(, q) consists of at least two segments and because of the monotonicity of P, the oint is a vertex of P to the left of. Let l q be the maximal line segment interior to P from through q to. The segment l q contains a maximal subsegment l q comletely contained in R. Let q and q be the two end oints of l q, with q to the left of q ; see Figure 16. Assume that [r,, ] forms a right turn. Follow the boundary of R in counterclockwise order from q to q, above the segment [, ], until the first vertex v j of R is encountered. Since v j is above [, ] and to the left of, the oint does not see v j and j i. If [r,, ] forms a left turn we can make a symmetric argument to show that there is a vertex of R to the left of not seen by. We have thus roved both directions of the equivalence. Lemma 4.7 gives us a method to comute the kernel exansion of a region. We begin by ordering the vertices of the region from left to right. For each vertex, in order, we comute the visibility olygon [15, 22, 25] and establish the aroriate intersections in successive order; see Figure 17. The comlexity of the algorithm is O(m log m + mn), where m is the number of vertices of R and n is the number of vertices of P. If R is monotone, the comlexity reduces to O(mn) since the sorting of the vertices of R can be done in linear time. The algorithm reeatedly comutes intersections between two monotone olygons and combines the result with the left art established in revious iterations. The intersection between two monotone olygons having n and n vertices resectively can be comuted in O(n + n ) time with a lane swee algorithm. Using the linear time intersection algorithm we can successively comute the intersections between visibility olygons, obtaining the kernel exansions of the aroriate ocket regions, i.e., the sears of each tye. Since the number of ocket vertices is at most linear in total, we have the following lemma. LEMMA 4.8 A sear in a monotone olygon can be comuted in quadratic time Comuting the Next Guard Consider Ste 7 of algorithm GUARDMONOTONEPOLYGON. In an iteration, just before we reach Ste 7, we have a artial guard set G with the rightmost guards at g and g and we are suosed to lace a third guard ĝ on the same vertical line in such a way that the sear ti of tye X with resect to G {ĝ} is as far to the 19
20 Algorithm Inut: COMPUTEKERNELEXPANSION A region R in a monotone olygon P Outut: The kernel exansion ke(r) 1 Order the vertices of R from left to right; v 1,..., v m 2 Let K := VP R(v 1) 3 for each vertex v i in order from v 2,..., v m do 3.1 Comute VP R(v i) 3.2 Let K L(v i) and K R(v i) be the two arts of K to the left and right of v i resectively 3.3 Let K := K L(v i) (K R(v i) VP R(v i)) endfor 4 return K End COMPUTEKERNELEXPANSION Figure 17: The algorithm for comuting kernel exansions. l g r r ê y = ax + b q q e u X(G { }) u X(G {}) Figure 18: Illustrating the movement of u X(G {}). right as ossible. Let l g be the vertical line through g and g. The line l g intersects U at U and D at D. The algorithm emulates a sliding rocess whereby a oint slides along l g from U to D and we maintain the sear ti u X (G {}) as a function of, continuously udating the sear ti as moves along l g. To accurately detect for which oint that the oint u X (G {}) is rightmost, we let the xcoordinate of u X (G {}) be a function of the ycoordinate of. We denote this function by x X (y), where y corresonds to the arameter of the vertical line l g = (1 y) U + y D. The sear ti u X (G {}) is adjacent to two edges e and e of the corresonding sear and, if u X (G {}) moves when moves along l g, at least one of these edges must move as moves. One of the two edges, say e, extends towards the left, reaching a oint r on a boundary edge ê of X. The edge e, on the other hand, can either coincide with the oosite boundary X (when u X (G {}) lies on X) or it extends towards the left, touching a vertex of the boundary X before it reaches a vertex v of a tye X ocket. In the most general case, both r and v move as moves. Consider first the movement of r on ê, where ê is a segment on the line y = ax + b. The suorting segment [, r] touches X at a oint q and the other suorting segment [u X (G {}), r] touches X at a oint q ; see Figure 18. We want to establish the equation of the line coinciding with [u X (G {}), r] in terms of the ycoordinate 20
21 r r q q q q q q q q (a) (b) (c) Figure 19: The cases for arameter change. of. If we let r be a function of, we have that r is the intersection oint between the lines y = y() y(q) x() x(q) x + y(q) y() y(q) x() x(q) x(q) and y = ax + b. So, by setting the two linear functions equal, we obtain the coordinates of the oint r. The two coordinates are each the ratio between two affine functions in y(), i.e., ( c y() + d r = y() + h, c y() + d ), y() + h where c, d, h, c, and d, are constants deendent on a, b and q. Hence, the line through r and q can be established to be y = g(y())x + k(y()), where g(y) = (αy + β)/(y + γ) and k(y) = (α y + β )/(y + γ), for constants α, β, γ, α and β. With similar calculations we can establish the other suorting line that intersects the vertex v of a tye X ocket to have the equation y = g (y())x + k (y()) as a function of y(). The function x X (y()) is the xcoordinate of the intersection oint between the two suorting lines, i.e., giving us g(y)x X (y) + k(y) = g (y)x X (y) + k (y), x X (y) = k (y) k(y) g(y) g (y) = Ay2 + By + C A y 2 + B y + C, where the constants A, B, C, A, B and C only deend on the oints of contact that the four suorting lines corresonding to visibility olygon edges make with the boundary. The constant arameters A, B, C, A, B and C can change value as the suorting lines make contact on different vertices and edges of the olygon and ocket boundaries. We are interested in comuting these oints of arameter change to be able to udate the function x X (y) aroriately; see Figure 19. These occur when: the convex vertex of an edge of VP(G {}) adjacent to a ocket becomes incident to two vertices on the olygon boundary of P; see Figures 19(a) and (b). the convex vertex of an edge of VP(u X (G {})) adjacent to a ocket becomes incident to two vertices on the olygon boundary of P or to one vertex of the boundary of P and one vertex of a tye X ocket of G ; see Figures 19(a) and (c). We can establish a suerset of these oints on l g that we call the rimary event oints by comuting the visibility olygon of each boundary and ocket vertex to the right of l g and obtaining the, at most two, intersection 21
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