triangles leaves a smaller spiral polgon. Repeating this process covers S with at most dt=e lights. Generaliing the polgon shown in Fig. 1 establishes

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1 Verte -Lights for Monotone Mountains Joseph O'Rourke Abstract It is established that dt=e = dn=e?1 verte -lights suce to cover a monotone mountain polgon of t = n? triangles. A monotone mountain is a monotone polgon one of whose chains is a single segment, and a verte -light is a oodlight of aperture whose ape is a verte. Kewords. art galler theorems, oodlights, monotone polgons. 1 Introduction It was established in [ECOUX9] that for an <, there is a polgon that cannot be illuminated with an -oodlight at each verte. An -oodlight (or -light) is a light with aperture no more than. A verte -light is one whose ape is placed at a verte, aiming a cone of light of up to into the polgon. Each verte ma be assigned at most one light. The result of [ECOUX9] is then that n verte -lights do not alwas suce when <. Let a polgon P have t triangles in an triangulation, t = n? ; we will phrase bounds in terms of t. For =, an eas argument shows that t verte -lights alwas suce: place a light at an ear tip, cut o the ear, and recurse. This raises the question of nding a better upper bound. Urritia phrased the problem this wa [Urr97]: is there a c < 1 such that cn verte -lights alwas suce? The largest lower bound is c = via an eample of F. Santos. In this paper we pursue this question, but onl in special cases. In particular, we show that c = 1 for spirals and, more interestingl, for monotone mountains. A monotone mountain is a monotone polgon one of whose chains is a single segment. More precisel, a monotone chain is a polgonal chain whose Department of Computer Science, Smith College, Northampton, MA 0106, USA. orourke@cs.smith.edu. Supported b NSF grant CCR intersection with an vertical line is at most one point. A monotone mountain consists of one monotone chain, whose etreme (left and right) vertices are connected b a single segment. Note this base edge need not be horiontal. 1 Fig. shows a monotone mountain with base edge. Although this is a severel restricted class of polgons, it deserves attention for three reasons: the eamples establishing the results of [ECOUX9] (and [OX94]) are \nearl" monotone mountains; the problem is alread not completel trivial for monotone mountains; and there is some reason to hope similar techniques will appl to the unrestricted problem. We start with a result on spiral polgons, where the problem is trivial. Spiral Polgons A spiral polgon consists of two joined polgonal chains: a chain of ree vertices, and a chain of conve vertices. Theorem 1 A spiral polgon S of t = n? triangles ma be covered b dt=e = dn=e? 1 verte -lights; some spirals require this man. Proof: If S has no ree vertices, S is conve and can be covered with one verte -light at an verte. So assume S has at least one ree verte. Let,, and be three consecutive vertices of S, with ree, conve, and conve. Such a triple alwas eists, because an polgon has at least three conve vertices. The segment must be an internal diagonal of the polgon. Therefore at least two triangles are incident to in an triangulation of S. Placing a light at, as shown in Fig. 1, therefore covers at least two triangles; because is conve, the light covers the entire angle at. Removing the covered 1 This denition diers in this respect from that introduced in [OX94], which demanded a horiontal base edge. Page 1

2 triangles leaves a smaller spiral polgon. Repeating this process covers S with at most dt=e lights. Generaliing the polgon shown in Fig. 1 establishes that the bound is tight. Figure : A wrong orienting decision at can lead to suboptimal coverage. Figure 1: Placing a -light at covers at least two triangles. The light is shown as a full -light, although onl the angle interior to the polgon is relevant. Notice that the procedure implied b this proof places lights onl on conve vertices. One reason spiral polgons are so eas is that lights never need be placed on ree vertices, and so the potentiall dif- cult decision of how to orient a -light at a ree verte need not be confronted. 4 Worst Case It is clear that if the number of triangles incident to in Fig. from the left is k and from right is also k, then a lower bound of c = 1 is attained: t = k + 1, and k + 1 = dt=e lights are necessar, one at and k on the opposite ree chain. The same bound is acheived b the shape shown in Fig.. In this polgon, the etension of v 1 v meets v v 6 ; the etension of v v meets v 4 v ; and so on. 4 Non-Localit Monotone mountains are more dicult than spirals for two reasons: ree vertices cannot be avoided, and the decision of how of orient a light at ree verte cannot be made locall. Man art galler theorems can be proved inductivel as follows: cut o a small piece, illuminate that piece, and recurse on the remainder [O'R87]. The reason this paradigm works is that decisions can be made locall: what happens in the small piece is independent of the shape of the remainder of the polgon. This is not the case with the verte -light problem, even for monotone mountains. Consider the polgon shown in Fig., and imagine tring to decide whether to shine the light at left or right, basing the decision onl on the portion of the polgon to the left of. One can see that no c < 1 can be achieved without looking at the structure of the right portion: if the \wrong" decision is made at (as illustrated), then an arbitraril large fraction of all remaining vertices will need lights. Although the decision is obvious in this case, as it can be based on the number of triangles incident to, the eect might be more subtle. 0 6 Figure : dt=e lights are necessar: t = and d=e = are needed. We prove this simple fact for later reference: Lemma 1 The generaliation of the polgon M in Fig. requires dt=e = dn=e? 1 verte -lights. Proof: Each verte on the left chain can onl see two vertices on the right chain, and vice versa: v can see v and v, because the etensions of v 1 v and v v straddle v ; etc. Thus at most (in fact eactl) three triangles are incident to v in a triangulation of M. A -light at v can onl full cover two of these three triangles, because v is ree. So each light covers at most two triangles, and dt=e are needed overall. 1 Page

3 Dualit One wa to view the phenomenon illustrated in Fig. is as follows: the polgon naturall partitions into two monotone mountain subpolgons at. If at light is placed at and aimed left, then in the right subpolgon, placing a light at is forbidden (as that would place two lights at one verte). Moreover, that eample shows that a (sub)polgon with one verte forbidden a light could in fact require one light per triangle. However, there is an interesting \dualit" at pla here, in the following sense: if a polgon with one forbidden verte requires man lights, then placing a light at the forbidden verte permits it to be covered with few lights. In other words, there is no polgon structure that is both bad with a forbidden verte and bad without that verte forbidden. If M is a monotone mountain with etreme left and right vertices and, let L 10 (M) be the number of verte -lights needed to cover M when verte is assigned a light and is forbidden to have a light; and let L 01 (M) be the number needed when is assigned a light and is forbidden. Note that, in these denitions, not onl is one verte forbidden a light, but the other etreme verte must be assigned a light. The precise statement of dualit is captured in the following lemma: Lemma is the ke to the main theorem in the net section. We now prove it via induction. Proof: Let M be a monotone mountain of t triangles. The induction hpothesis is that L 10 (M 0 ) + L 01 (M 0 ) t for an monotone mountain M 0 of t 0 < t triangles. The base case is a single triangle T, t = 1, when L 10 (T ) = L 01 (T ) = 1, and so L 10 (T ) + L 01 (T ) = = t + 1. Let the base edge of M be, and let be the verte rst encountered b sweeping the line containing verticall; see Fig.. It must be the case that both and are internal diagonals. This provides a natural partition of M into three pieces: 4, a subpolgon A sharing diagonal, and a subpolgon B sharing diagonal. Note that it ma well be that either A or B is the empt polgon ;; if both are empt, t = 1 and we fall into the base case of the induction. A B Lemma For an monotone mountain M of t triangles, L 10 (M) + L 01 (M) t + 1. The generaliation of Fig. 4 establishes that the sum is sometimes as large as t+1: here L 10 (M) = 1 (v 0 assigned) and L 01 (M) = t (v 0 forbidden, as illustrated). Figure : Induction partition of M into A, B, and It is clear that A and B are monotone mountains. In particular, the angle at in A is conve, as is the angle at in B: for the monotone chain enters from the left and leaves it from the right (Fig. 6), as do the diagonals and respectivel. We prove the lemma in two cases. Case 1: Neither A nor B is empt. We compute a bound on L 10 (M), which places a light at but forbids a light at. Because the angle at in M is conve, the light at covers 4. This light also serves as a light at in A. It makes sense in this situation to place a light at and aim it into B. Doing this gives us an upper bound on L 10 (M), upper because this sensible light placement and orientation at might not optimal. This strateg ields Figure 4: Dualit: L 10 (M) + L 01 (M) = t + 1. L 10 (M) L 10 (A) + L 10 (B) : (1) Page

4 B Figure 6: The monotone chain enters each verte from the left halfplane and leaves in the right halfplane. Figure 7: A = ;. Analogous reasoning (again the light at (illustrated in Fig. ) covers 4) ields L 01 (M) L 01 (A) + L 01 (B) : () Adding Eqs. 1 and ields L 10 (M)+L 01 (M) [L 10 (A)+L 01 (A)]+[L 10 (B)+L 01 (B)] : Suppose A contains a triangles and B contains b triangles, so that t = a + b + 1. Then appling the induction hpothesis to each ields L 10 (M) + L 01 (M) [a + 1] + [b + 1] This is the claim to be proved. It onl remains to handle the case where one of A or B is empt. Case : A = ; but B is not empt. This case is illustrated in Fig. 7; the case with B = ; is smmetric and need not be considered. If a light is placed at, it serves to cover 4, and the reasoning is just as before: L 10 (M) 1 + L 10 (B) : If a light is placed at, then it covers 4 (as illustrated in Fig. 7), and there is no need to an additional light to cover the empt A: Adding ields L 01 (M) L 01 (B) : L 10 (M) + L 01 (M) 1 + [L 10 (B) + L 01 (B)] L 10 (M) + L 01 (M) 1 + [b + 1] 6 Main Result With Lemma in hand, the nal step is eas: Theorem A monotone mountain polgon M of t = n? triangles ma be covered b dt=e = dn=e?1 verte -lights; some monotone mountains require this man. Proof: We know from Lemma that Let L(M) = minfl 10 (M); L 01 (M)g : B the pigeonhole principle, L(M) b(t + 1)=c = dt=e : Lemma 1 established that this bound can be attained (Fig. ). The proofs of Lemma and Theorem impl a simple algorithm: compute a bound on L 10 (M) b placing lights at the left corners of A and of B and recursing, and compute a bound on L 01 (M) similarl. Use the light placement of whichever is smaller. The algorithm is easil seen to be O(n log n): spend linear time nding, and recursivel process the pieces. This leads to the familiar divide-and-conquer recurrence. An eample is shown in Fig. 8. Here M has t = 14 triangles, and L 10 (M) + L 01 (M) + 10 = t + 1. This eample illustrates a number of features of the light placements implied b the bound computation on L 10 and L 01 : 1. Ever verte that is not a local maimum is assigned a light in either the L 10 or L 01 computation. (Some vertices are assigned a light in both.). All the lights in the L 10 placement aim to the right; and all those in the L 01 placement aim to the left. Page 4

5 . The sum L 10 (M) + L 01 (M) achieved is alwas eactl t + 1, because blindl following the procedure places lights even if the might not be needed (e.g., when M is conve). 4. Lights at ree vertices are turned either full counterclockwise (in L 10 ) or clockwise (in L 10 ): intermediate positions are never needed. 7 Discussion Man of the features present in monotone mountains hold for the problem for general simple polgons as well: for eample, non-localit. For other features, it remains unclear: for eample, whether ever light ma be full turned (observation 4 above). In an case, I believe that a version of the dualit described in Lemma holds and will be a ke to solving Urrutia's problem. I conjecture c = is achievable. References [ECOUX9] V. Estivill-Castro, J. O'Rourke, J. Urrutia, and D. Xu. Illumination of polgons with verte oodlights. Inform. Process. Lett., 6:9{1, 199. [O'R87] J. O'Rourke. Art Galler Theorems and Algorithms. Oford Universit Press, New York, NY, [OX94] J. O'Rourke and D. Xu. Illumination of polgons with 90 verte lights. In Snapshots of Computational and Discrete Geometr, volume, pages 108{ 117. School Comput. Sci., McGill Univ., Montreal, PQ, Jul Technical Report SOCS [Urr97] J. Urrutia. Art galler and illumination problems. In J.-R. Sack and J. Urrutia, editors, Handbook on Computational Geometr. North-Holland, To appear. Figure 8: Eample: t = 14, L 10 = (top), L 01 = 10 (bottom). Page