zeo, i.e., each watchman emains stationay, and we have the at galley poblem. At the othe limit, if m = 1, we have the watchman oute poblem. Anothe con

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1 Optimum Guad Coves and m-watchmen Routes fo Resticted Polygons Svante Calsson y Bengt J. Nilsson yz Simeon Ntafos x Abstact A watchman, in the teminology of at galleies, is a mobile guad. We conside seveal watchman and guad poblems fo dieent classes of polygons. We intoduce the notion of vision spans along a path (oute) which povide a natual connection between the At Galley poblem, the m-watchmen poblem and the watchman oute poblem. We pove that nding the minimum numbe of vision points, i.e., static guads, along a shotest watchman oute is NP-had. We povide a linea time algoithm to compute the best set of static guads in a histogam polygon. The m-watchmen poblem, minimize total length of outes fo m watchmen, is NP-had fo simple polygons. We give a (n 3 + n 2 m 2 )-time algoithm to compute the best set of m watchmen in a histogam. 1 Intoduction The poblem of placing guads in an at galley so that evey point in the galley is visible to at least one guad has been consideed by seveal eseaches. If the galley is epesented by a polygon, having n vetices, and the guads ae points in the polygon, then visibility poblems can be equivalently stated as poblems of coveing the galley with stashaped polygons. Chvatal [5] and Fisk [8] poved that bn=3c guads ae always sucient and sometimes necessay to guad a polygonal at galley. Many othe esults on at galley poblems and geneal visibility poblems can be found in [12]. Chin and Ntafos [3, 4] conside the poblem of nding shotest watchman outes fo dieent classes of polygons. The oute is the path taced by a moving guad who in total sees the whole polygon. They pove that the poblem is NP-had fo polygons with holes and fo 3-dimensional simple polyheda. They also give a (n)-time algoithm that solves the poblem fo simple ectilinea polygons. In addition they povide a solution to the slightly dieent question of nding a shotest watchman oute given a \doo", a point on the bounday that the oute must pass though. They give an O(n 4 )-time algoithm that solves this poblem fo simple polygons without holes. Recently Tan, Hiata, and Inagaki [14] impoved this bound to O(n 3 ). Both these bounds ely on the linea time tiangulation algoithm by Chazelle [2]. The stationay at galley poblem and the watchman oute poblem can be thought of as special cases of the m-watchmen poblem. Suppose that we have m mobile watchmen and we want to design outes fo them so that each point in a polygon is visible to some point along a oute. The objective is to minimize the total length of the m outes. Clealy, the total length of an optimum solution inceases as the numbe of watchmen deceases and vice vesa. At one The wok of the second autho was suppoted by the Deutsche Foschungsgemeinschaft unde Gant No. Ot 64/5{4. y Depatment of Compute Science, Lund Univesity, Box 118, Lund, Sweden. z Institut fu Infomatik, Univesitat Feibug, Rheinst. 10{12, D-7800 Feibug, Fed. Rep. of Gemany. x Compute Science Pogam, Univesity of Texas at Dallas, Richadson, TX 75083{0688, U.S.A. 1

2 zeo, i.e., each watchman emains stationay, and we have the at galley poblem. At the othe limit, if m = 1, we have the watchman oute poblem. Anothe connection between static and mobile guad poblems is obtained by consideing a single mobile obot equipped with a vision system. We can think of the obot as pefoming two types of activity, motion and visual data acquisition. To minimize the distance taveled so that it sees all of the polygon, the obot taces a shotest watchman oute. Howeve, the obot's vision system does not need to be active thoughout the oute. Also, it is likely that the vision system pefoms bette if the obot is not moving. We efe to points on the oute at which the obot engages in vision activity as vision points. A continuous potion of the oute whee the obot engages in vision activity is efeed to as a vision span. Then we have the at galley poblem, minimize the numbe of vision points, with the estiction that the solution must lie along a path, e.g., the shotest watchman oute. In this pape we show that nding the best set of vision points along a path is NP-had fo simple polygons. Also, the m-watchmen poblem fo simple polygons is NP-had. In view of these esults, we conside the same poblems fo esticted classes of polygons. Nilsson and Wood [11] teat the class of spial polygons. They give a linea time algoithm fo the minimum stationay guad poblem and a (n 2 ) time algoithm fo the multiple watchmen outes poblem. We conside the same poblem fo histogams whee a (vetical) histogam is a ectilinea polygon with a hoizontal edge, the base, equal in length to the sum of the lengths of all the othe hoizontal edges. An inteesting popety of histogams is that the m-watchmen poblem and the m-vision span poblem, along a shotest watchman oute, have a common optimum solution. Histogams ae the most complex of the esticted polygons that have been studied in the liteatue to have this popety. In the next section we show that nding the best set of vision points on a path is NP-had fo simple polygons. In Section 3 we show that a minimum guad cove fo histogam polygons can be found in linea time. In Section 4 we develop the multiple watchmen outes idea on histogams and pove that an optimum set of watchmen lies on the base of the histogam. This enables us to solve the m-watchmen poblem with dynamic pogamming in (n 3 +n 2 m 2 ) time. 2 Minimum Guad Coves on Paths In this section we study the poblem of positioning stationay guads in simple polygons. We look at guads that ae esticted to lie on a given cuve, the watchman oute. These guading points ae called the vision points of the oute, and we pove that, in geneal, it is NP-had to nd the minimum numbe of vision points on a shotest watchman oute. Definition 2.1 A vision point is a point along a given path in which the watchman will engage in visual data acquisition activity. We take the watchman oute poblem to be the poblem whee a stating point on the bounday is specied. We efe to the watchman oute poblem when no stating point is specied as the oating watchman oute poblem. We have the following esult. Theoem 1 Finding the minimum numbe of vision points along a shotest oating watchman oute in a simple polygon is NP-had. Poof: The poof is by eduction fom 3-SAT. Let a 3-SAT instance have k clauses and n vaiables. We modify the poof of Lee and Lin [9] and Aggawal [1] that nding a minimum 2

3 clause stuctue u 1 _ u 2 _ u 3 u 1 _ u 2 _ u 3 x watchman oute u 1 u 1 u 2 u 2 u 3 u 3 well stuctues Figue 1: Five caves ae added to each clause stuctue, one cave to each well stuctue, and thee caves moe, at x, to the left of the well sequence, and to the ight of it. The coesponding 3-SAT fomula fo this example is (u 1 _ u 2 _ u 3 ) ^ (u 1 _ u 2 _ u 3 ): guad cove fo simple polygons is NP-had. We efe to [9] o [12] fo the complete constuction of the poof and show hee only the modications that have to be made. The poof of Lee and Lin species 6k + 2n + 1 distinguished points such that a minimum vetex guad cove uses 3k+n+1 of them as guad locations and this is a minimum vetex guad cove if and only if the coesponding 3-SAT fomula is satisable. We add a numbe of \caves" to the polygon in such a way that the shotest watchman oute will be foced to visit all the distinguished points in the constuction of Lee and Lin. The caves have two bends so that any watchman oute must ente each cave, one vision point is needed in it and this vision point can not see too much outside the cave. We use one cave fo each liteal stuctue plus one cave on eithe side of a clause stuctue to foce the shotest watchman oute to visit the distinguished points in the clause stuctues giving 5k caves. We put one cave at the bottom of each well stuctue, 2n in total, one cave on eithe side of the well sequence, to tie the oute down, and one cave at the uppe left cone, to foce the oute to visit the point x. This gives a total of 5k+2n+3 caves, each of which needs exactly one vision point. The shotest watchman oute visits all the distinguished points so the polygon can be seen with (3k + n + 1) + (5k + 2n + 3) = 8k + 3n + 4 vision points if and only if the fomula is satisable; see Figue 1 fo an example. Note that the vision points ae not esticted to lie on vetices, eithe of the polygon o of the oute, but this does not pose any poblem, since the spikes of the well stuctues can only be seen fom the distinguished points which ae also isolated points of the oute. 2 Anothe question to look at is the following. Is thee a elation between the minimum numbe of vision points and the minimum numbe of guads fo a polygon? The answe is that, in geneal, thee exists no such elation and examples can be easily constucted. Instead, we tun the question aound and tighten the equiement: Fo what types of polygons is it tue that thee is always a shotest watchman oute with the popety that an optimum set of vision points along the oute is also an optimum solution to the minimum guad cove poblem fo the 3

4 Fo convex and sta polygons the answe to the question is tivial fo the oating watchman oute poblem. The best oute is a single point in the kenel. We have a single vision point which is also a solution to the minimum guad cove poblem. Howeve, fo sta polygons, the coespondence does not hold fo the xed watchman oute poblem. A counteexample is constucted by taking a sta polygon whee the watchman oute does not intesect the kenel. In this case we need at least two vision points on the oute while one guad is enough fo the minimum guad cove poblem. A simila point was made by Toussaint, who investigated diagonal guads in sta polygons, as epoted in [12]. Now conside monotone polygons. Hee the coespondence between vision points and solutions to the minimum guad cove poblem does not hold even fo oating shotest watchman outes. Fo example, conside a monotone polygon in which the shotest oute is a staight line connecting two limit lines. We can make the notches be such that two of them can be seen fom a single point in the polygon but not fom a watchman oute. This also holds fo ectilinea monotone polygons; see Figue 2. Fo spial polygons it can be poven that the st and last links of the oating shotest watchman oute connect to the bounday chain consisting of eex vetices (We disegad the special case when the shotest oute consists of only one link). Hence, the end points of the oute must be vision points and also evey second vetex of the oute, othewise thee is some edge on the chain of eex vetices which is not seen by a vision point. In the following we will look close at a special class of polygons fo which the vision point poblem and the minimum guad cove poblem ae equivalent in the sense that the minimum guad cove lies on one, of many possible, shotest watchman outes. 3 Minimum Guad Coves of Histogams The poblem of nding minimum guad coves, o equivalently minimum coves of stashaped polygons, fo simple polygons has been poven to be NP-had by Lee and Lin [9] and Aggawal [1]. The poblem is still open fo ectilinea polygons but is NP-had when minimum coves with convex ectilinea polygons (ectangles) in ectilinea polygons ae consideed; see Culbeson and Reckhow [6]. These esults indicate that the poblem of nding the best guad cove is NP-had even fo ectilinea polygons. Theefoe one may ask the following question: Fo what classes of polygons ae thee ecient solutions to the optimum guad coveing poblem? Fo example, fo stashaped polygons thee is a linea time algoithm to nd the kenel of the polygon; see [10]. Fo spial polygons an algoithm exists which uns in linea time; see [11]. In this section we show that fo histogam polygons thee is a linea time algoithm to nd the best guad set. We begin the section by dening this class of polygons. Definition 3.1 A histogam is a simple ectilinea polygon that is monotone with espect to the hoizontal axis, and such that one of the monotone chains foming its bounday is a single staight line segment. We will hencefoth assume that the lowe monotone chain is a staight line segment, see Figue 3, and call this segment the base of the histogam. Edelsbunne et al [7] use an equivalent denition and dene a vetical histogam as an othogonal polygon with a hoizontal edge, the base, equal in length to the sum of the lengths of the othe hoizontal edges. The uppe monotone chain will hencefoth be efeed to as the top chain. 4

5 S S S S S a a 0 S a 00 S S b S S b 0 Figue 2: Optimum cove, the guads b and b 0, is not on any shotest watchman oute contay to the vision points a, a 0, and a 00. Note that the top chain consists of a sequence of pais of staicase pattens, one up staicase followed by a down staicase. We call the pat of the polygon unde such a pai a pyamid. Sack [13] denes a pyamid to be hoizontally convex vetical histogam. Hoizontal convexity is the same as monotonicity along a vetical line. As shown in Figue 3 two consecutive pyamids have an ovelapping ectangula egion unde the hoizontal edge that stats the up staicase of one pyamid and ends the down staicase of the othe. The two staicases of a pyamid ae connected by the top edge; see Figue 3. We will also use the tem west edge to denote a vetical edge that has the inteio to the ight of the edge. An east edge is dened similaly as a vetical edge having the inteio to the left. Note that a pyamid can be viewed as a stashaped polygon with the kenel unde the top edge. Conside the visibility polygon fom an abitay point s lying on the base. The bounday of this visibility polygon consists of segments that ae also segments of the histogam and segments that only intesect a discete and bounded numbe of points (usually two) on the top chain. Conside the lines collinea with these last segments. These lines intesect the point s, at least one vetex on the top chain of the histogam, and some othe point on the top chain in the ode as we move though the inteio of the histogam fom s; see Figue 3. We call these lines suppoting lines. The intesection of the suppoting line and the histogam consists of a numbe of connected components, in most cases, howeve, only one. We ae only inteested in the component containing s and wheneve we efe to the suppoting line we will mean this component. Each suppoting line is associated to the pyamid containing the segment end point of the suppoting line that is not s. It is woth noticeing that not all pyamids need to have a suppoting line fom a point associated to it. A suppoting line divides a pyamid into one section which is visible fom s and anothe section whee thee may exist points that ae not seen fom s, since the suppoting line passes though a vetex. Fist, we pove some simple popeties of histogams. The following lemma povides the main basis fo ou algoith. Lemma 3.1 Let s and t be two points on the base of a histogam, s being to the ight of t, then s sees at least as much of the pat of the polygon to the ight of s as t does. Poof: Take a point x inteio to the polygon and to the ight of s, such that x is seen by t. We must pove that s also sees x. Poject x othogonally onto the base giving point x 0. the tiangle having cones at t, x, and x 0 contains s and is convex. Hence, s sees x. 2 5

6 west edge ` east edge s Figue 3: Histogam with thee pyamids. ` is the suppoting line of point s with espect to pyamid 2. Lemma 3.2 Fo any histogam, a shotest oating watchman oute and a minimum guad cove is positioned on the base of the histogam between the leftmost east edge and the ightmost west edge. Poof: To see that a shotest oating watchman oute lies on the base between the two edges, note that in ode fo the watchman to see both the leftmost and the ightmost pyamid it must intesect the kenels of these pyamids, i.e., coss the extensions of the leftmost east edge and the ightmost west edge. Also, since the two extended edges ae paallel, all hoizontal segments between the two extended edges have the same length, thus we can pick the segment along the base to be ou shotest oute. To see that a minimum guad cove lies on the base, take any guad cove and poject the guads othogonally onto the base. This can be done since the polygon is monotone. Take a point x visible fom the oiginal guad g. Let g 0 be the pojection of the guad g. Poject the point x onto the base giving point x 0. The fou points g 0, g, x, and x 0 dene a convex aea, actually a tapezoidal aea, and hence, x is seen by g 0. Any guad to the left of the leftmost east edge o to the ight of the ightmost west edge can be moved, without aecting visibility, onto the intesection points of the extensions of the two edges and the base since they ae still in the kenels of the \outemost" pyamids. 2 Since an optimal set of guads lies on a shotest oating watchman oute, the guading points ae also vision points on the oute. We state an algoithm to nd a minimum guad cove of a histogam. The algoithm uses a geedy appoach, walking fom left to ight. The st guad is placed on the base unde the leftmost east edge. The algoithm then ties to place the next guad as fa to the ight of the pevious one as possible. We dene the function patne(s; 4) of a point s on the base and a pyamid 4 to be the futhest point fom s, on the base, whee s and the patne collectively see all of 4. If s aleady sees all of 4 we let the patne be undened. The following lemma shows how to compute the patne of a given point with espect to some pyamid. Lemma 3.3 Let s be a guading point not unde the top edge of a pyamid 4. One of the following holds: 1. The suppoting line of patne(s; 4) w..t. 4 meets the suppoting line of s on the top edge of 4 o 6

7 Poof: Assume w.l.o.g. that s is to the left of the top edge of 4. We need to pove that if 2. does not hold then 1. holds. So, assume that patne(s; 4) does not lie unde the top edge of 4. Suppose at most one of the suppoting lines intesects the top edge, then one of s and patne(s; 4) does not see any pat of the top edge. Thus, the othe guading point, in ode to see all of the top edge, must lie below the top edge. This is a contadiction to the assumption that neithe s no patne(s; 4) lies below the top edge. Suppose the suppoting lines both intesect the top edge but at dieent points, and the two lines intesect outside the polygon, then the segment of the top edge between the intesection points is visible fom both s and the patne so we can move the patne and incease the distance to s. We have a contadiction to the assumption that the patne is the futhest point fom s etaining visibility. 1 Suppose nally that the suppoting lines intesect the top edge and each othe but inside the histogam, then the segment of the top edge between the intesection points is not seen fom any of the points. Again we have a contadiction to the visibility assumption. 2 Given a point s on the base and a pyamid 4 the suppoting line can be computed easily. Fist detemine whethe s lies to the left o ight of the top edge of 4. Assume it lies to the left of 4, then follow the left staicase down, maintaining the inteio line though s and the vetex while emembeing the line having minimum slope. Once the suppoting line is given the point patne(s; 4) can be computed by taking the intesection point of the suppoting line of s and the top edge of 4, then by following the ight staicase down, maintaining the inteio line though the intesection point and the vetex of the staicase. The line having minimum absolute slope is the suppoting line of the patne, and hence, the patne is the intesection of this suppoting line with the base. If the suppoting line of s does not intesect the top edge of 4, the patne is the point on the base unde the leftmost east edge of 4. If s lies to the ight of 4, a simila scheme can be used to compute the suppoting line and the patne. We dene lp(s), the limit point to the ight of s to be the closest patne to the ight of s. Moe fomally lp(s) = min (patne(s; 4 i)); (1) 4 i s whee the value of a point is its distance fom the leftmost edge of the histogam; 4 i, 1 i k; ae the pyamids of the histogam; and means \not to the left of." Similaly we dene the limit point to the left of s, denoted llp(s). Next we show that the limit point to the ight of a given point s is always the futhest point on the base which etains visibility between the two points. Lemma 3.4 Let s be a point on the base of a histogam. Thee does not exist any point to the ight of lp(s), on the base, which togethe with s sees all of the top monotone chain between the two points. Poof: By the denition of limit point we know that lp(s) is the patne of s with espect to some pyamid 4. Thus, fo any point v to the ight of lp(s) the suppoting lines of s and v, with espect to 4, will intesect the bounday of the polygon at distinct points and the potion of the bounday between the intesection points is neithe visible fom s no fom v. 2 1 By etaining visibility we mean that all of the histogam between a point and its limit point is visible fom the two points. 7

8 to the ight of s. This is done fom left to ight. Fo each pyamid the patne to s is computed and the leftmost patne found so fa is emembeed. The pocess stops when the top edge of the next pyamid to be scanned is to the ight of the leftmost patne found, since all the patne points of the pyamids to the ight of this point have to be to the ight of the leftmost patne. This gives us the following algoithm fo the minimum cove. Algoithm Histogam-Minimum-Cove Input: A histogam polygon P Output: A minimum guad cove fo P Step 1 Divide P into its pyamids, 4 1 ; 4 2 ; : : : ; 4 k Step 2 Place the guad g 1 at the intesection of the base and the extension of the leftmost east edge of P Step 3 while not all pyamids ae guaded /* g i was the last guad to be placed */ do Step 3.1 g i+1 := lp(g i ) Step 3.2 i := i + 1 endwhile End Histogam-Minimum-Cove Suppose we have two sets of guads that each cove a histogam. Index the guads in the two sets fom left to ight as G = fg 1 ; g 2 ; : : :g and F = ff 1 ; f 2 ; : : :g. Dene the < elation between two guads as a < b if and only if ( guad a is to the left of guad b a and b coincide and a has smalle index Theoem 2 The algoithm Histogam-Minimum-Cove nds an optimum guad cove fo a histogam polygon in linea time. Poof: The subdivision into pyamids takes linea time since the histogam is monotone. Each pyamid is then scanned only once when the patne of the peviously placed guad is computed. Thus, the algoithm above uses (n) time. To see that thee does not exist a bette guad cove, assume the contadiction. Suppose thee is a guad cove having fewe guads. Let G be the solution obtained by ou algoithm and F the solution, having less guads, that most closely matches G going left to ight. Find the st position whee the solutions die, i.e., g i = f i but g i+1 6= f i+1. If f i+1 < g i+1, then f i+1 can be moved to the ight limit point of f i, lp(f i ) = lp(g i ) = g i+1, without aecting visibility. This contadicts ou assumption that F matched G the best. On the othe hand, g i+1 < f i+1 is not possible because that implies that F is not a guad cove since g i+1 = lp(g i ), and by Lemma 3.4 this is the ightmost point which togethe with g i sees all of the top monotone chain between the two guads. 2 An example of an optimum cove fo the histogam in Figue 3 is given in Figue 4. Suppose now that we have two histogams, one with the uppe chain as base and the othe with the lowe chain as base. We place them so that the bases coincide. What we have is a monotone ectilinea polygon in which the shotest watchman oute is a staight line. It is no longe tue that a minimum guad cove will also be a minimum set of vision points on a shotest watchman oute; see Figue 2 and the discussion of the peceding section. Howeve, if we estict the guads to lie on the oute, then the same appoach we used fo histogams will give an optimum set of vision points. 8

9 g 1 g 2 g 3 Figue 4: A simple example of the optimum cove poduced by the algoithm. The only dieence is that we have to conside both monotone chains that make up the bounday, i.e., pyamids ae on both sides of the base and the next guad point has to be computed with espect to both chains. Thus, a minimum set of vision points on a given staight line which is also the shotest watchman oute fo a monotone ectilinea polygon can be found in (n) time. 4 Multiple Watchmen Routes The poblem of detemining visibility popeties using moving guads, fom now on denoted watchmen, has eceived signicantly less attention than the equivalent poblems with static guads. Howeve thee ae some esults in this, hade, setting. If a watchman is allowed to \patol" a line segment in the polygon it can be shown that bn=4c watchmen ae necessay and sucient to guad the polygon; see [12]. In this section we show that the poblem of computing the m shotest watchmen outes is NP-had. Fo histogam polygons we ae able to show that the poblem can be computed in polynomial time, and give a (n 3 + n 2 m 2 ) time algoithm based on dynamic pogamming. The algoithm computes (n 2 m) candidate outes and chooses the m outes that give the shotest total solution. It should be noted that, although histogams ae polygons with a vey simple stuctue, the watchmen outes poblem becomes vey had even fo this \simple" class of polygons. Definition 4.1 The m-watchmen outes ae a set of m oating outes which togethe sees a polygon. We dene the length of the m-watchmen outes as the sum of the lengths of the outes. The m-watchmen outes poblem is the poblem of nding the m-watchmen outes with shotest total length. We pove that the m-watchmen outes poblem is NP-had. Theoem 3 The m-watchmen outes poblem in simple polygons is NP-had. Poof: It follows fom the intactability of nding the minimum guad cove fo a simple polygon; see [1, 9]. Suppose thee exists an algoithm to compute the best m watchmen outes in polynomial time, then by unning it fo m = 1; 2; : : : ; bn=3c and, fo each m, check whethe the esulting outes have total length 0, we have a polynomial time algoithm fo the minimum guad cove poblem. Thus, we have poved the NP-hadness of the m-watchmen outes poblem. 2 9

10 In this section we will look at how to compute the best set of outes, fo any m moving watchmen in a histogam. It is inteesting to note that this poblem, although solvable in polynomial time, is much hade than nding the best set of static guads. To see this intuitively, conside the static guad poblem of the pevious section. Each guad is allowed to move only along the base, which gives each guad one degee of feedom. We will show that the watchmen outes also lie on the base but each watchman's oute still has two degees of feedom since both of the end points can move. We sketch the algoithm we ae going to apply. Instead of concening ouselves with the watchmen outes, which lie on the base, we will teat the poblem of computing the gaps between each pai of consecutive watchmen, i.e., compute the m? 1 longest gaps such that the visibility equiements ae not violated. Ou st task is to dene a bounded set of possible gaps between watchmen that povably always contains the solution. We then apply dynamic pogamming to choose the m? 1 gaps with longest total length. The main poblem lies in poving that the specied set of gaps always contains the solution. We st pove that the optimum set of watchmen outes always lies on the base. Lemma 4.1 Thee is an optimum solution in which all of the m oating outes ae disjoint spans along the single watchman oute on the base of the histogam. Poof: Take any solution and poject it othogonally onto the base. Let point x be visible fom some oiginal watchman w. We need to show that the pojected watchman w 0 also sees x. Since w sees x thee is a point s w on w fo which the segment between x and s w is inteio to the polygon. By Lemma 3.2 the pojected point s 0 w also sees x and since w 0 is the pojected watchman, s 0 w lies on w 0. The two end points of the single watchman oute see the two extemal pyamids, so any watchman of the solution which lies outside the oute can be moved to one of the end points without aecting visibility. Also, this can neve incease the total length of the solution. 2 We will hencefoth efe to the single shotest watchman oute on the base as the W-oute. Futhemoe, we denote the left end point of the W-oute as p and the ight end point as q. The next thing to show is that nding the gaps is equivalent to nding the watchmen outes. Lemma 4.2 Finding the m? 1 gaps with the lagest total length, such that visibility equiements ae fullled, is equivalent to the m-watchmen outes poblem. Poof: Take an optimum solution to the watchmen outes poblem. The leftmost watchman has its left end point at p, the left end point of the W-oute, since if the point is to the left of p, it can be moved to the ight without violating the visibility equiement. If the point is to the ight of p, it does violate visibility. This follows since othewise the W-oute would not be a shotest watchman oute. By symmety, the ight end point of the ightmost watchman has its ight end point at the point q, ight end point of the W-oute. Now, since the total length of the m watchmen and the m? 1 gaps between the watchmen is xed, i.e., equals the length of the W-oute, the lemma follows. 2 In ode to make the poblem tactable we wish to discetize it and nd a bounded numbe of points on the base which can be end points of the watchmen outes. To nd these points we begin by making the following denition. Definition 4.2 The concavity chains of the W-oute and a pyamid 4 ae two eex chains, called the left and ight concavity chains. The left concavity chain is the shotest path between 10

11 ??? s p? J J J J J J J@ s@ XXXX s s s s J Figue 5: Pyamid with extended concavity chains and pimay event points. q p, the left end point of the W-oute and the left end point of the top edge of 4. The ight concavity chain is dened similaly as the shotest path between q and the ight end point of the top edge of 4; see Figue 5. The left concavity chains ae easily computed by a scan of the top chain fom left to ight given the point p. The point p and the st vetex on the top chain to the ight of p ae pushed in this ode onto a stack. Fo each vetex, until the top edge of the pyamid is found, a check is made that the cuent vetex and the two top elements on the stack do not fom a ight tun. If that is the case, the stack is popped and the check is pefomed again until the points fom a left tun. The cuent vetex is then pushed onto the stack and the next vetex of the top chain becomes the new cuent vetex. The ight concavity chains can be computed in a simila way by a scan in the evese diection. Definition 4.3 The pimay event points of a pyamid 4 ae the intesection points of the W-oute and the extended links of the concavity chain that also intesect the top edge of 4; see Figue 5. The set of pimay event points ae the pimay event points of all the pyamids excluding the st and the last pyamids plus the two points p and q. Definition 4.4 The seconday event points ae all the points s on the W-oute such that patne(s; 4 i ) = patne(s; 4 j ), fo any two pyamids 4 i and 4 j ; see Figue 6. The pimay event points can be computed by following the concavity chains, extending the links, and computing the intesection with the W-oute if the top edge is intesected. Since each vetex of a pyamid can allow at most one pimay event point, thee ae at most a linea numbe of these points. The seconday event points can be computed by looking at each pai of concavity chains. Conside a vetex on a concavity chain. The two pimay event points, which have suppoting lines passing though the vetex, specify the end points of an inteval on the base. Not all vetices intesect the suppoting lines of two event points but in this case we take the inteval to be empty. In this way evey vetex on a concavity chain has an inteval associated to it. Now, fo each pai of vetices lying on left concavity chains fo pyamids 4 i and 4 j, with i 6= j, compute the intesection of the associated intevals. If the intesection is empty, we ae done. If the intesection is non-empty, let I be the intesection inteval. Compute the patnes of the 11

12 4 a a j 4 i A A aaaaaaaaaaaaaaaaaa A A A A A A s As s patne(s; 4 i ) = patne(s; 4 j ) Figue 6: Seconday event points dene local maxima of the distance function between guading point and patne. end points of I fo both pyamids 4 i and 4 j. These fou points dene two patne intevals to I, one fo 4 i and one fo 4 j, denoted J i and J j. If one of J i and J j is contained in the othe, then thee is a seconday event point in I with a patne in J i and J j. If J i and J j ovelap o ae disjoint, then thee is no seconday event point in I (coesponding to this pai of vetices). To see this note that the patne of a point s in I depends linealy on s; see Lemma 4.3. This means that if s = a + (b? a)t, then lp(s) = a i + (b i? a i )t, whee the points a and b ae the end points of I and the points a i and b i ae the end points of J i. Thus, the seconday event point can be computed by solving the equation 2 a i + (b i? a i )t = a j + (b j? a j )t: (2) We can poceed similaly with the ight concavity chains. Fo the seconday event points, note that a pai of vetices fom two dieent pyamids can allow a seconday event point. Hence, O(n 2 ) is an uppe bound on the numbe of seconday event points. We can, in fact, constuct an example of a polygon which has n = 8k + 8 vetices and at least k(k? 1)=2 seconday event points, giving (n 2 ) seconday event points. Hence, the bound is asymptotically tight. The constuction is depicted in Figue 7. The polygon consists of k + 2 pyamids whee the k middle ones ae used to geneate seconday event points. The left concavity chain of each pyamid consists of 4 segments and 5 vetices. The segments adjacent to the middle vetex of a concavity chain ae extended to intesect the base at the end points of an inteval I. We denote the left end point of I with l and the ight one with. Futhemoe we let a i = patne(l; 4 i+1 ) and b i = patne(; 4 i+1 ), with J i = [a i ; b i ]. The pyamids ae then dawn in such a way that a i+1 = a i + and b i+1 = b i? =2 i, whee is a small value, with < (b 1? a 1 )=k. Using this constuction and Equation 2 it is then a simple matte of computing the seconday event points and veifying that thee is one seconday event point pe pai of pyamids, i.e., k(k? 1)=2 seconday event points in total. In the following we denote the pimay and seconday event points simply as event points when we do not cae of what type they ae. Each event point e has coesponding limit points lp(e) and llp(e), as dened in Section 3, which dene maximum visibility gaps fom the event point. This means that two watchmen spanning the W-oute, except the pat between the event point and one of the limit points, see the whole polygon and the limit point is also the futhest such point fom the event point; see Lemma 3.4. The limit points can be computed using the same technique as in the algoithm Histogam-Minimum-Cove pesented in Section 3. 2 With a slight abuse of notation, since we ae actually only concened with the x-coodinates of these points. 12

13 4 j 4 i I J j J i Figue 7: Example of histogam whee each pai of pyamids geneates a seconday event point. The next lemma shows that these gaps ae bette than any othe set of gaps. The idea of the poof is simply that any gap with end points that ae not event points can be slided slightly to the ight o left and since thee is a linea elation between the point and its limit point the size of the gap inceases in one of the diections. Lemma 4.3 If [x; lp(x)] is a gap on the base of a histogam, such that x does not coincide with an event point o a limit point, then thee exists a gap [y; y 0 ] with y eithe slightly to the left of x o to the ight of x, such that the gap [y; y 0 ] is at least as long as the gap [x; lp(x)]. Poof: Let x 0 = lp(x). By the denition of ight limit point we know that x 0 is the patne of x with espect to some pyamid 4. Hence, by Lemma 3.3, the suppoting lines of x and x 0 intesect on the top edge of 4, and pass though the vetices v 4 and v4 0 of 4. Let a be a point such that it is eithe an event point with a 0 = lp(a) o a left limit point with a = llp(a 0 ) and a being the closest such point to the left of x. Similaly let b be the closest such point to the ight of x. Take a point z in the inteval [a; b]. We need to ensue that z and lp(z) have suppoting lines with espect to 4 which intesect on the top edge of 4 and also pass though the vetices v 4 and v4 0 of 4. Suppose this is not tue. Eithe the suppoting lines of z and lp(z) intesect at the top of anothe pyamid o the lines pass though dieent vetices than v 4 and v4 0 of 4. Hence, we have two cases to take cae of: 1. The case when the suppoting lines of z passes though an othe vetex than v 4, say u 4. Take the uppe of the two vetices (w.l.o.g. assume this to be v 4 ) and its adjacent vetex w lying in between v 4 and u 4 on the concavity chain (w = u 4 is possible). Extending the link [v 4 ; w] gives a pimay event point in the inteval between x and z which is a contadiction to the assumption that a and b wee the closest such points to x. Similaly you can pove this fo lp(z). 13

14 !!!!!!!!!!!!!!!!!!!!!!? L u LL a 4 v 4 v??? L LL?! aaaaaaaaaaaaaaaa s s s s s s a x b a 0 x 0 b 0 t b t a Figue 8: Gap [b; b 0 ] is bette than any gap stating between event points a and b. The points a 0, b 0, and x 0 ae limit points to points a, b, and x, espectively. 2. The case when the suppoting lines of z and lp(z) intesect on the top edge of anothe pyamid 4 +. If the suppoting lines of z and lp(z) also intesect on the top edge of 4, then we immediately have a contadiction since then z is a seconday event point. Hence thee is a point s between x and z whee the intesection point of the suppoting lines switches fom pyamid 4 to pyamid 4 +. We can assume that thee is no point between x and z fo which the coesponding suppoting lines intesect on the top edge of a thid pyamid. If the suppoting lines of s and lp(s) intesect on the top edges of both 4 and 4 +, we have again a contadiction since s is then a seconday event point. Hence we can assume that the suppoting lines of s and lp(s) intesect on the top edge of only one pyamid, w.l.o.g. assume this to be 4, wheeas all points to one side of s have suppoting lines which togethe with the suppoting lines of thei ight limit points intesect on the top edge of 4 +. We must pove that s o lp(s) is a pimay event point, so assume the contadiction. Since neithe s no lp(s) is an event point thee exists a neighbouhood aound s fo which the suppoting lines of all points and thei ight limit points intesect on the top edge of 4, but this is not tue fo s so eithe s o lp(s) is a pimay event point. This poves ou st claim, that the suppoting lines of z and lp(z), with espect to 4, intesect on the top edge of 4 and also pass though the vetices v 4 and v4 0. Event points mak the points whee suppoting lines move fom one vetex to anothe; see Figue 8, whee suppoting lines to the left of a intesect vetex u 4 and suppoting lines to the ight of a intesect vetex v 4. Let t s be the point of intesection of the top edge of 4 and the suppoting line of point s on the base. Now, since the tiangles with cones at a; x; v 4 and t a ; t x ; v 4, as well as tiangles with cones at a 0 ; x 0 ; v4 0 and t a; t x ; v4 0, ae simila, we have j[a; x]j j[t a ; t x ]j = j[a; v 4]j j[t a ; v 4 ]j and j[a 0 ; x 0 ]j j[t a ; t x ]j = j[a0 ; v4 0 ]j j[t a ; v4 0 ]j and, hence, j[a; x]j j[a 0 ; x 0 ]j = j[a; v 4]j j[a 0 ; v4 0 ]j j[t a; v4 0 ]j = c; a constant. j[t a ; v 4 ]j 14

15 j[y; lp(y)]j = j[x + ; lp(x + )]j = j[x; lp(x)]j + (c? 1): (3) When c 1 we can chose < 0, i.e., y is to the left of x. Similaly, when c > 1 we chose > 0 giving y to the ight of x. In both cases [y; lp(y)] is at least as long as [x; lp(x)]. 2 The constant c obtained in the poof we call the sliding constant of gap g = [x; lp(x)] o in shot notation c g. One might think that the gaps dened by the event points and the limit points, in this way, would be a sucient set to dene the best m watchmen outes. Howeve, this is not the case. The eason fo this is that two gaps in the set might ovelap, and by placing a static guad somewhee on the ovelap the esulting gap (containing the static guad) could be lage than the combined oiginal gaps. We pove that it is always pefeable to place the static guad at an end point of such an ovelap, and thus fo each gap it is only necessay to compute new gaps dened by m? 2 static guads to the left o ight of any initially given gap. We dene the ight popagation points to be a sequence of points p 0 ; p 1 ; : : : ; p m?1 such that p 0 is an event point and p i = lp(p i?1 ). Similaly we dene the left popagation points popagating to the left of p 0. Such a sequence of points dene gaps with static guads at the end points. Fo each event point we need to conside only m? 1 popagation points since we have at most m watchmen in total. Each event point coesponds to two sequences of popagation points, one popagating to the left and one popagating to the ight. This gives us 2(m? 1) O(n 2 ) = O(n 2 m) gaps in total. We let the set of gaps dened in this way be G and pove that the m? 1 gaps giving an optimal solution ae in this set, G. Lemma 4.4 The set G contains m? 1 non-ovelapping gaps that have total length no smalle than an optimum set of gaps. Poof: We make a poof by compaing a solution contained in G with an allegedly optimal solution with at least one gap not in G. Let F = (f 1 ; f 2 ; : : : ; f m?1 ) be an optimal solution whee f 1 ; f 2 ; : : : ; f m?1 ae gaps odeed fom left to ight. We will show that thee is a solution no smalle than the solution F having all its gaps in G. Let f i be the leftmost gap of F not in G. Let g j and g j+1 be the gaps in G that have thei left end points closest to the left end point of f i. We assume that the left end point of g j is to the left of the left end point of f i and thet the left end point of g j+1 is to the ight of the left end point of f i. The following cases aise. f i contains g j+1. f i?1 F By Lemma 3.4, the gap g j+1 is the lagest gap etaining visibility. Gap f i theefoe violates the visibility equiement. f i is contained in g j. f i g j+1 f i?1 F 15 f i g j

16 1 i?1 j i+1 m?1 j i+1 a static guad can be placed anywhee in the ovelap. f i and g j ovelap. This case divides into two subcases. The st subcase has two vaiants as depicted in the gue below. f i?1 F f i?1 F If f i does not have an end point in common with neithe f i?1 no f i+1, then by Equation 3 in Lemma 4.3, thee is a gap f 0 i with jf 0 i j = jf ij + (c fi? 1) jf i j. If c fi = 1 the size of f 0 i has not inceased but we can then choose f 0 i = g j+1 o join f 0 i to f i+1 with a static guad between them, whicheve occus st. In the latte case we have tansfomed the solution to the second subcase whee f i and f i+1 have a static guad between them. The geneal vesion is depicted in the following gue. g j f i f i g j+1 f i?1 f i f i+`?1 f i+` F It is clea that f i and f i?1 do not have a static guad between them since f i?1 2 G implies that the end points of f i ae popagation points of some event points. Hence, assume that the set ff i ; f i+1 ; : : : ; f i+`?1 g is a maximal set of consecutive gaps with only static guading points between them. By epeated application of Equation 3 in Lemma 4.3 we can choose ff 0 i ; f 0 i+1 ; : : : ; f 0 i+`?1 g to be a sequence of gaps such that jf 0 i j + jf 0 i+1 j + + jf 0 i+`?1 j = jf i j + jf i+1 j + + jf i+`?1 j + (c fi c fi+1 c f i+`?1? 1) jf ij + jf i+1 j + + jf i+`?1 j and incease the total size of the solution. If c fi c c fi+1 f = 1, then again we can choose f 0 i+`?1 i = g j+1 o join f i+`?1 to f i+` with a static guad in between, whicheve occus st. Repeated application of the agument poves the lemma in the case when f i and g j ovelap. f i and g j ae disjoint. This case can neve occu since the union of the gaps in G span the complete W-oute. Theefoe some gap in G will intesect f i. We get the following algoithm Algoithm m-watchmen-routes Input: A histogam polygon P and a positive intege m Output: The m shotest watchmen outes in P Step 1 Divide P into its pyamids, 4 1 ; 4 2 ; : : : ; 4 k Step 2 Compute the event points fo pyamids 4 1 to 4 k Step 3 fo each event point do Step 3.1 Compute the m? 1 gaps popagating to the ight fom the event point endfo Step 4 Redo step 3 fo the gaps popagating to the left Step 5 Sot the gap end points on x-coodinate Step 6 Apply dynamic pogamming to choose among the set of gaps End m-watchmen-routes 16 2

17 mate of s. In Steps 3 and 4 we take cae of the possibility of a sequence of popagation points. These points ae computed using the same technique as in Section 3. The dynamic pogamming scheme will scan the gap end points fom left to ight, which is why these points ae soted in Step 5. In Step 6 we assume a matix M of size m O(n 2 m). Item M[i; j] contains the best solution of i watchmen fom the left initial point p, the left end point of the W-oute, to the j th gap end point. The best solution to item M[i; j] is achieved by taking the best of two items: 1. item M[i? 1; k], whee k < j and k is the index of the mate of the gap end point having index j. This coesponds to placing a static guad at the gap end point with index j o 2. item M[i; j? 1] + fthe length between the gap end points having indices j? 1 and jg. This coesponds to a watchman walking the distance between the j? 1 st and the j th gap end points. Lemma 4.5 The algoithm uns in (n 3 + n 2 m 2 ) time and uses (n 2 m 2 ) stoage. Poof: Steps 1 and 2 take O(n 2 ) time by the discussion following Denitions 4.3 and 4.4. Steps 3 and 4 can take up to O(n 3 ) time since one scan of the polygon is pefomed to compute each sequence of popagation points and thee ae O(n 2 ) of these sequences. Soting of the O(n 2 m) gap end points takes O(n 2 m log n) time in Step 5. By the pevious discussion Step 6 takes O(n 2 m 2 ) time since the matix has at most n 2 m 2 enties and each enty can be computed in constant time. This gives a time complexity of O(n 2 (n + m 2 + m log n)) = O(n 2 (n + m 2 )). The stoage bound follows fom the fact that the matix has m O(n 2 m) enties. Again, the example, having (n 2 ) seconday event points, mentioned in the beginning of this section, uses (n 3 + n 2 m 2 ) time and (n 2 m 2 ) stoage which shows the tightness of the complexity bounds fo ou algoithm. 2 Lemmas 4.4 and 4.5 give the following theoem. Theoem 4 The m-watchmen-routes algoithm computes an optimum set of m watchmen outes in (n 3 + n 2 m 2 ) time using (n 2 m 2 ) stoage. 5 Conclusions We have pesented a numbe of esults in this pape. The st is that computing the estiction of a minimal guad set on a watchman oute is NP-had. We have also pesented a linea time algoithm to compute the minimal guad set in a histogam polygon, which is also a minimal set of vision points on the watchman oute along the base of the histogam. In the last section we have shown an algoithm to compute the best set of m watchmen in histogam polygons. Ou algoithm uns in (n 3 + n 2 m 2 ) time and uses (n 2 m 2 ) stoage. Othe poblems elated to the poblem of nding the best watchmen fo any value of watchmen (which is NP-had in the geneal case) is that of nding best watchmen fo a xed numbe of watchmen, e.g., does thee exist a polynomial time algoithm to compute the best 2-watchmen outes fo simple polygons? Even fo the single watchman oute case the poblem has not been completely answeed, although, as stated in the intoduction, a (n) time algoithm exists fo simple ectilinea polygons. 17

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