Minimum Weight Convex Steiner Partitions

Minimum weight convex Steiner partitions∗ Adrian Dumitrescu† Csaba D. Tóth‡ May 24, 2009 Abstract New tight bounds are presented on the minimum length of planar straight line graphs connecting n given points in the plane and having convex faces. Specifically, we show that the minimum length of a convex Steiner partition for n points in the plane is at most O(log n/ log log n) times longer than a Euclidean minimum spanning tree (EMST), and this bound is the best possible. Without Steiner points, the corresponding bound is known to be Θ(log n), attained for n vertices of a pseudo-triangle. We also show that the minimum length convex Steiner partition of n points along a pseudo-triangle is at most O(log log n) times longer than an EMST, and this bound is also the best possible. Our methods are constructive and lead to O(n log n) time algorithms for computing convex Steiner partitions having O(n) Steiner points and weight within the above worst-case bounds in both cases. 1 Introduction Geometric spanner networks for n given points in Euclidean plane have been studied extensively, particularly in the last 20 years [15, 35]. Some desirable properties of such spanners are constant stretch factor, constant degree, weight proportional to the Euclidean minimum spanning tree, and others. A convex Steiner partition for a set S of points in the plane is a planar straight line graph G where the vertex set of G contains S and the boundary of every face is a convex polygon; specifically, every bounded face is convex, and the unbounded face is the complement of a convex polygon. A convex (non-Steiner) partition has the additional property that every vertex of G is a point in S (i.e., there are no Steiner vertices). Similarly, one can consider Steiner triangulations and triangulations for a point set S. Clearly, every triangulation for S is a convex partition for S, and every Steiner triangulation for S is a Steiner convex partition for S. The weight of a planar straight line graph or network is the total Euclidean length of its edges. We denote by W = W (S) the weight of a Euclidean minimum spanning tree (EMST) for a finite point set S. Every spanning network for S is at least as heavy as the Euclidean minimum Steiner tree of S, whose weight is known to be at least W/2 [19].1 A convex (non-Steiner) partition or triangulation for n points is a planar graph on n vertices, hence it has O(n) edges. Since the weight of each edge is at most W , the weight of a convex partition or triangulation is trivially bounded by O(W n). This naı̈ve bound is tight apart from a constant factor: Kirkpatrick [24] ∗ A preliminary version of this paper appeared in the Proceedings of the 19th ACM-SIAM Symposium on Discrete Algorithms, ACM Press, 2008, pp. 581–590. † Department of Computer Science, University of Wisconsin-Milwaukee, WI 53201-0784, USA. E-mail: [email protected]. Supported in part by NSF CAREER grant CCF-0444188. ‡ Department of Mathematics, University of Calgary, AB, T2N 1N4, Canada. E-mail: [email protected] 1 Gilbert and Pollak √ [19] conjectured that the Steiner ratio, i.e., the ratio of the weights of the EMST and the minimum Steiner tree, is at most 2/ 3. The validity of the proof of this conjecture given by Du and Hwang [11] has recently been questioned by O. de Wet [41]. 1 exhibited sets of n points, shown in Figure 1(a), where the minimum weight of any convex partition is Ω(W n). Clarkson [10] proved that any set of n points in the plane admits a Steiner triangulation of weight O(W log n), and Eppstein [14] showed that this bound is the best possible. His construction consists of 4 vertices of a square and n − 4 points evenly distributed along a circle placed in the interior of the square, see Figure 1(b). Both lower bound constructions consist of points along a reflex chain and a few other points off this chain. These constructions can be realized with a vertex set of O(1) pseudo-triangles, that is, polygons with exactly three convex vertices called corners, see Figure 1(c). t s (a) (b) (c) (d) Figure 1: (a) n points for which any triangulation or convex partition has weight Ω(W n). (b) n + 4 points for which any Steiner triangulation has weight Ω(W log n), and any convex Steiner partition has length Ω(W log log n). (c) A pseudo-triangle with the bisectors of its three corner-angles. (d) A nonconvex face where compass routing fails to route from s to t. Our contribution. (i) Given n points along a pseudo-triangle, we prove that they admit a convex Steiner partition of weight O(W log log n), and this bound is the best possible (Theorem 1). Recall that the weight of a minimum Steiner triangulation may be as large as Ω(W log n) for n vertices of a pseudo-triangle. (ii) Given n points in the plane, we prove that they admit a convex Steiner partition of weight O(W log n/ log log n), and this bound is the best possible (Theorem 2). This is a (log log n)-factor improvement over the corresponding bound for minimum weight Steiner triangulations. We prove our upper bounds constructively, using O(n) Steiner points in both cases. Our methods lead to O(n log n) time algorithms in the real RAM model for computing convex Steiner partitions within these bounds. Networks with constant stretch factor. A spanner network for set S of points in the plane is a connected planar straight line graph G whose vertex set contains S. The vertex dilation (also known as stretch factor) of G is the maximum ratio between the length of the shortest path in G and the Euclidean distance for any pair of points in S. In contrast, the geometric dilation of G is the maximum ratio between the length of the shortest path in G and the Euclidean distance for any two points, at vertices or on edges of G. For the Ω(W log n/ log log n) lower bound, listed above in (ii), on the length of a minimum Steiner partition of a general point set, it is crucial that the unbounded face in a convex partition is required to be the complement of a convex set. If we drop this condition (but still require that all bounded faces be convex), then the minimum weight network would be the minimum Steiner tree, which has only a single, unbounded face. We show that the Ω(W log n/ log log n) lower bound holds even if we replace the convexity condition on the unbounded face by another condition on the stretch factor. We present (Theorem 3) n-element point sets for which any spanner network whose stretch factor is o(n) and whose bounded faces are convex, has 2 weight Ω(W log n/ log log n). The condition on the stretch factor rules out the EMST in some cases, for instance, for n equally spaced points along a circle. It is known that given n points in the plane, there is a plane spanner network that (i) is short, namely, it has weight O(W ); (ii) is small, namely, it has O(n) vertices and edges; and (iii) has constant vertex dilation [4, 31]. Recently, we have shown [12] that property (iii) can be strengthened by requiring constant geometric dilation rather than constant vertex dilation: There is a network plane spanner network with properties (i), (ii), and with constant geometric dilation. We observed that some of the faces of such networks are often non-convex. Theorem 3 shows that non-convex faces are in general unavoidable in a spanner network with properties (i)–(iii). Motivation. Location-based routing, also known as geographic routing, has been studied extensively and is considered one of the most promising routing protocols in interconnection networks, sensor networks, and mobile networks [16, 23]. Even abstract networks are often embedded in Euclidean space (with “virtual coordinates”) to be able to use location-based routing [28, 37, 39]. In geographic routing, each node stores its geographic coordinates, and makes routing decisions based on the coordinates of the packet destination, eliminating expensive routing tables. Variants differ in the number of bits carried along with a packet and in the amount of information stored at each node. A minimalist model is compass routing [27], where (i) the packets carry no routing information other than the target’s coordinates, (ii) each node knows only its own coordinates and the directions of the adjacent edges, and (iii) they route each packet on a link whose direction is closest to the current direction of the target. Compass routing works on the Delaunay triangulation of a point set, as it was shown in [6]. As mentioned earlier, the Delaunay triangulation of n points in the plane may be Ω(n) times heavier than an EMST [24]. However, Bern et al. [2] showed that for every set S of n points in the plane, there is a super set S ′ , S ⊆ S ′ , with |S ′ | = O(n), such that the weight of the Delaunay triangulation of S ′ is O(W log n). This O(W log n) bound is the best possible due to Eppstein’s lower bound on minimum weight Steiner triangulations [14]. A randomized version of compass routing guarantees delivery on any convex network [3]. The deterministic compass routing, however, can easily run into a loop: it fails for some source-destinations pairs in nonplanar graphs, and in planar straight line graphs with some bounded faces non-convex (Figure 1(d)). Moreover, any deterministic memoryless routing protocol fails on some convex network [5]. It is worth noting that many competing protocols, such as face routing [7], allow O(1) memory carried along with each packet and they guarantee delivery on any plane graph; with O(1) memory one can also guarantee that each packet is routed along a path at most constant-times longer than the Euclidean distance between source and destination [6, 21]. Related results. In this paper, we prove tight worst-case upper bounds on the minimum weight convex Steiner partition for n points in the plane in terms of n and W , and present polynomial-time algorithms for computing convex Steiner partitions within these bounds. No polynomial exact or approximation algorithm is known for computing the minimum convex Steiner partition for n given points. The problem is not known to be NP-hard, either, although we suspect that this is the case. The related minimum weight triangulation problem has seen many developments in the last few years. Mulzer and Rote [34] proved that computing the minimum weight triangulation is NP-hard. Remy and Steger [40] gave a quasi-polynomial time approximation scheme for minimum weight triangulation. Levcopoulos and Krznaric [29] proposed O(n log n) time algorithms for constant-factor approximation of the minimum weight triangulation and the minimum weight convex partition, based on earlier work by Plaisted and Hong [38]. Eppstein [14] gave an O(n log n) time algorithm for computing a 316-factor approximation of the minimum weight Steiner triangulation of n points. Gudmundsson and Levcopoulos recently gave a 3 tight O(W log n) upper bound on the minimum weight pseudo-triangulation of n points in the plane [20]. In a minimal pseudo-triangulation, every vertex is incident to a reflex angle, and so this is, in some sense, the opposite of convex partition. The minimum weight convex partition problem restricted to the interior of a simple polygon can be solved exactly. Gilbert [18] and Klincsek [25] independently gave O(n3 ) time algorithms for computing a minimum weight convex partition of a simple polygon with n vertices using dynamic programming. Levcopoulos and Lingas [30] showed that the minimum weight of a convex Steiner partition of the interior of a simple polygon P with n vertices, m of which are reflex, is always O(|P | log n) and sometimes Ω(|P | log m/ log log m). Here |P | denotes the perimeter of P . Our Theorem 4 in Section 3 shows that this lower bound is tight: Every simple polygon P with m reflex vertices admits a convex Steiner partition of weight O(|P | log m/ log log m). The minimum number convex partition problem asks for the minimum number of faces in a convex partition of n points. Knauer and Spillner [26] recently showed that any planar n-element point set admits faces (improving an earlier bound of 10n−18 by Neumann-Lara et a convex partition with at most 15n−24 11 7 12n al. [36]); Garcı́a-López and Nicolás [17] gave a lower bound construction of 11 − 2 for n ≥ 4. Knauer 30 and Spillner [26] also gave a polynomial time 11 -approximation for the minimum number convex partition problem. No corresponding results are known for the minimum number convex Steiner partition problem. Restricted to simple polygons, these problems have efficient solutions. Keil and Snoeyink [22] gave an O(n3 ) time algorithm for computing the minimum number convex partition of a simple polygon with n vertices; this problem is NP-hard for polygons with holes [32]. Chazelle and Dobkin [9] gave an O(n3 ) time algorithm for the minimum number convex Steiner partition of a simple polygon with n vertices. Definitions and notations. If A is a finite set, let #A denote the cardinality of A. For a polygonal curve γ, let |γ| denote the length (or weight) of γ. For a polygon P , let |P | denote the perimeter of P . A convex chain is a polygonal chain whose vertices are consecutive vertices of some convex polygon. A reflex chain is a convex chain appearing on the boundary of a nonconvex polygon with all internal angles of the chain larger than π. Let γ be a convex (or reflex) chain with endpoints a and b; see Figure 2(a). The turning angle of γ is the angle in [0, 2π] of the rays along the first and last segments of γ. The width of γ is the width of the smallest parallel strip that contains γ and is parallel to the segment ab connecting the endpoints of γ. Denote by h(γ) the half-plane containing γ determined by the line other than ab bounding this strip. A pseudo-triangle is a simple polygon with exactly three vertices with interior angles less than π, called corners; see Figure 1(c). 2 Vertex sets of pseudo-triangles and reflex chains In this section, we prove a tight bound on the minimum weight convex Steiner partition for a vertex set of a pseudo-triangle. The corners partition the pseudo-triangle into three reflex chains (some of which may consist of a single segment). Each reflex chain is a polygonal path of reflex internal angles and total turning angle of at most π. If all three reflex chains of a pseudo-triangle P are straight line segments then P is a triangle; if two reflex chains are straight line segments then P is a one-chain pseudo-triangle; see Figure 1(a). Theorem 1 (i) For every set S of n points lying along a pseudo-triangle, a convex Steiner partition of weight √ O(W log log n) with n + O( n) Steiner points can be computed in O(n) time. 4 (ii) For every n, there is a set S of n points along a pseudo-triangle such that any convex Steiner partition for S has weight Ω(W log log n). Upper bound. We construct a convex Steiner partition for S by augmenting the vertex set S with new (Steiner) vertices and new edges as follows. Refer to Figure 1(c). Add the 3 edges of the convex hull conv(S) of S to the network. Each edge has weight at most W , hence their total weight is O(W ). Since the three polygons between the convex hull of the pseudo-triangle and the three reflex chains are convex, it suffices to partition the interior of the pseudo-triangle. First we reduce the problem to one-chain pseudotriangles. Note that the weight of a reflex chain of P is less than twice the diameter of P , hence |P | = O(W ). Lemma 1 The interior of every pseudo-triangle P with n vertices can be partitioned into three one-chain pseudo-triangles and a (possibly degenerate) triangle along three line segments of total weight O(|P |). Proof. The bisector of an interior angle of P at a corner separates the two adjacent reflex chains of P . The bisectors of the three corners bound a (possibly degenerate) triangle ∆ lying in the interior of P ; see Figure 1(c). Consider the line segment along each bisector between a corner and its intersections with the other two bisectors. These segments jointly partition the interior of P into three one-chain pseudo-triangles (each adjacent to a reflex chain of P ) and ∆. Each segment is shorter than a diagonal of P , hence its weight is at most |P |/2. The total weight of the three segments is O(|P |). ✷ Lemma 2 The width of a convex chain γ of turning angle α, 0 < α ≤ π2 , is at most α4 |γ|. Proof. Draw a circle in which ab is a chord of inscribed angle π − α. Note that γ must lie in the circular disk cap bounded by the chord ab, otherwise its turning angle is more than α. The width of this cap is (|ab|/2) sin(α/2) < α|γ|/4, hence this is also an upper bound on the width of γ. ✷ c c h(γ) c w b α λ3 λi B3 γ B2 a Ai λ2 B1 λ0 a λ1 a (a) b b (b) (c) Figure 2: (a) The a reflex chain of width w and turning angle α. (b) A one-chain pseudo-triangle with corners a, b, and c, and with reflex chains λ0 , λ1 , λ2 , and λ3 . (c) In step i, we choose a vertex set Ai , halfplanes h(γ) for γ ∈ Γi . Lemma 3 The interior of every one-chain pseudo-triangle P with n vertices has a convex Steiner partition √ of weight at most O(|P | log log n) with at most n+O( n) Steiner points. Such a partition can be computed in O(n) time. 5 Proof. Denote the corners of a one-chain pseudo-triangle P by a, b, and c such that the reflex chain λ0 between a and b has n − 3 internal vertices, and the other two reflex chains are the line segments ac and bc; see Figure 2(b). We partition the interior int(P) of P by pairwise non-crossing reflex chains λi , i = 1, 2, . . . , t = O(log log n). Each reflex chain connects segments ac and bc, and λi lies in the interior of the one-chain pseudo-triangle bounded by λi−1 , ac, and bc. It is clear that the weight of each reflex chain is at most |ac| + |bc|. The last reflex chain, λt , will be a single segment, and so the portion of int(P) bounded by λt , ac, and bc is a triangle. Once these polygonal chains are constructed, we still need to subdivide the faces Bi ⊂ int(P) between consecutive reflex chains, λi−1 and λi , into convex faces. Note that the vertices of λi are convex vertices of Bi , and the internal vertices of λi−1 are reflex vertices of Bi . Partition Bi by rays emanating from each reflex vertex of Bi and subdividing the reflex angles into convex angles. We will choose the chains λi and the rays that partition Bi such that the portion of the rays lying in Bi have total weight proportional to the weight of λi for every i = 1, 2, . . . t. We define the reflex chains λi , i = 0, 1, . . . , t, inductively. λ0 is the given reflex chain of the one-chain pseudo-triangle P . Assume that λi is given, and we need to construct λi+1 . We will partition λi into a set Γi of subchains ordered along λi such that the width of each subchain γ ⊂ λi , where γ ∈ Γi , is at most the average weight of the segments in γ. For every i, let mi denote the number of internal vertices of λi , and let αi < π denote its turning angle. Initially, we have i = 0; λ0 is the reflex chain of P , and m0 = n − 3. As long as mi > 0, construct the chain λi+1 as follows. We choose a subset Ai of vertices of λi , and partition λi into a set Γi of subchains, each lying between consecutive vertices in Ai . Choose the splitting points Ai by the following simple algorithm: Put the endpoint λi ∩ bc of λi into Ai . Traverse λi from λi ∩ bc to λi ∩ ac. Denoting by γv ⊂ λi the subchain between the previous vertex of Ai and a vertex v ∈ λi , put v into Ai if γv is the maximal √ √ subchain with at most ⌊ mi ⌋ segments and turning angle of at most αi / mi . Finally, put the endpoint λi ∩ ac into Ai (this process is schematically shown in Figure 2(c)). The number of subchains of λi created √ in this way is |Γi | ≤ 2 mi . Let Ci be the intersection of the halfplanes h(γ) for γ ∈ Γi . Let λi+1 be the portion of the boundary of Ci lying in P . Finally, for each vertex v of each subchain γ ∈ Γi , partition Bi by the ray emanating from v and perpendicular to h(γ). In particular, we draw two rays at the common endpoint of any two consecutive subchains in Γi . This completes the construction of the chain λi+1 and the partition of Bi into convex faces. The weight of each ray drawn at a vertex of a subchain γ ∈ Γi is at most the width of γ, which is at √ √ √ most αi |γ|/(4 mi ) ≤ π4 |γ|/ mi by Lemma 2. By construction, each γ ∈ Γi has at most mi vertices. Hence the total weight of the rays that partition Bi+1 into convex faces is X √ π X π |γ| π π ≤ · (⌊ mi ⌋ + 1) · · √ |γ| = |λi | ≤ |P |. 4 mi 2 2 2 γ∈ Γi γ∈ Γi √ Since mi+1 ≤ |Γi | ≤ 2 mi , we also have t = O(log log n). Summing over all i = 0, 1, . . . , t − 1, the total weight of the resulting convex Steiner partition of P is O(|P | · t) = O(|P | log log n), as required. The number of Steiner points is upper bounded by t t t X X X √ i i 21−1/2 · (n − 3)1/2 = n + O( n). mi ≤ O(log log n) + (2 + mi ) = 2(t + 1) + i=0 i=0 i=0 Each polygonal chain λi+1 is constructed in a single traversal of λi , in O(mi ) time. Between the chains λi and λi+1 , every segment splitting a reflex angle at an internal vertex of λi hits ac, bc, or one of two possible edges of λi+1 ,P so each ray can be computed in O(1) time. Hence, the total runtime of the partitioning algorithm is O( ti=0 mi ) = O(n). ✷ 6 Lemma 4 The interior of every pseudo-triangle P with n vertices has a convex Steiner partition of weight √ O(|P | log log n) with n + O( n) Steiner points. Such a partition can be computed in O(n) time. Proof. Partition the pseudo-triangle P by Lemma 1 into three one-chain pseudo-triangles and a (possibly degenerate) small interior triangle, as in Figure 1(c). This partition has weight O(|P |) and at most 3 Steiner points. By Lemma 4, the interior of each pseudo-triangle can be partitioned into convex faces using √ ✷ n + O( n) Steiner points and edges of total weight O(|P | log log n) in O(n) time. This completes the proof of the upper bound in part (i) of Theorem 1. Lower bound construction with points along a reflex chain. We prove that the weight of every convex Steiner partition for a set S of n + 4 points arranged as indicated in Figure 1(b) is at least Ω(W log log n). Since this construction can be tiled with 5 convex faces and 4 congruent pseudo-triangles as shown in the figure, we also obtain an Ω(W log log n) lower bound for the minimum Steiner partition of n vertices of a pseudo-triangle. Consider an integer n such that log log n is even. Let S0 be a set of n evenly spaced points on a circle of unit radius centered at the origin o; let S be the union of S0 and the 4 vertices of a square Q at points (±2, ±2); see Figure 1(b). Clearly Q = conv(S). Denoting by Wn the length of an EMST for S, observe √ For sufficiently large n, an EMST consists of n − 1 edges that limn→∞ Wn = 2π + 4(2 2 − 1) ≈ 13.60. √ of conv(S0 ) and 4 edges, each of length 2 2 − 1, connecting the vertices of Q to conv(S0 ). Let G be a convex Steiner partition for S. For r > 0, denote by C(r) (resp. D(r)) the circle (resp. disk) of radius r centered at o. For 0 < r < R, let K(r, R) = D(R) \ D(r) be the annulus between the circles C(R) and C(r). We construct inductively a sequence of k = (log log n)/2 concentric circles of radii 1 = r0 < r1 < . . . < rk ≤ 3/2 and show (in Lemma 7) that the length of the portion of G lying in each annulus K(ri , ri+1 ) is Ω(1). This immediately implies |G| = Ω(log log n). We say that a set of m points on a circle C is dense if every arc of C of size (measured by the angle of apex at the center of C) at least 4π/m contains at least one of the points. Along each circle C(ri ), we choose a dense set Ai ⊂ G ∩ C(ri ), which consists of some intersection points of the circle C(ri ) with vertices or edges of G. Initially, let A0 = S0 be the a dense set of m0 = n points of S along C(r0 ). We next describe how we choose the radius ri and the point set Ai for i = 1, 2, . . . , k. We will choose the widths εi = ri+1 − ri of the annuli to satisfy the following two conditions: 1. εi should be small enough so that the circle C(ri+1 ) intersects G in a large dense set of points. 2. εi should be large enough so that the length of the portion of G in K(ri , ri+1 ) is at least Ω(1). We choose the radii ri so that they satisfy the recurrence ri+1 = ri + 1/(9mi ) for i = 1, 2, . . . , k. The √ following lemma ensures that we find a dense set of at least mi points in C(ri+1 ) ∩ G for every i = 0, 1, . . . , k − 1, if ri+1 < 3/2. This shows that −i mi ≥ n 2 , (1) and we repeat the argument k = (log log n)/2 times, where mi ≥ 10 holds for every 0 ≤ i ≤ k. √ Lemma 5 For any two points p, q ∈ C(ri ) with ∠poq ≥ 1/ mi , the chords of the circle C(ri+1 ) whose midpoints are p and q, respectively, are disjoint. 1 Proof. Consider two points p, q ∈ C(ri ) and let α = ∠poq. Recall that ri+1 = ri + 9m . If the chords of i the circle C(ri+1 ) with midpoints p and q, respectively, share a common endpoint, then ri ri cos α = = 1 . ri+1 ri + 9m i 7 G C(ri) γ D C(ri+1) γ′ s q1 p q2 Figure 4: The graph G contains a portion of length at least εi in each half-disk lying in the annulus K(ri , ri+1 ). Figure 3: A half-disk D whose boundary (black) consists of a segment s and a half-circle γ, and a convex partition G (grey).  From the Taylor expansion of cosine, for α ∈ 0, π2 we have 1− α2 α4 α2 α2 < cos α < 1 − + ≤1− . 2 2 4! 3 (2) Hence, the chords with midpoints at p and q are disjoint if 1− α2 ri ≤ 1 . 3 ri + 9m i √ The inequality holds for α = 1/ mi and ri ≥ 1. (3) ✷ The following simple lemma is crucial for enforcing the convexity of faces in the partition. Lemma 6 Let D be a closed half-disk of radius r centered at p and bounded by a diameter segment s and a half-circle γ; see Figure 3. Let G be network such that p lies at a vertex or on an edge of G, and the half-disk D is covered by convex faces of G. Then the weight of the portion of G lying in D is at least r. Proof. It is enough to show that every halfcircle centered at p and lying in D intersects the network G, since then the lower bound on the weight follows by integration. Assume to the contrary, that there is a halfcircle γ ′ centered at p lying in D but disjoint from G. Denote the endpoints of γ ′ by q1 , q2 ∈ s. The points q1 and q2 are on opposite sides of p and lie in the interior of some faces of G. They must lie in distinct faces of G, otherwise any point in the segment q1 q2 , including p, would be in the interior of the same face by convexity. However, they must lie in the same face of G, since they can be connected by curve γ ′ disjoint from G. This is a contradiction, hence every halfcircle centered at p and lying in D intersects G. ✷ pWe next choose a set of points Ai ⊆ C(ri ) ∩ G inductively for every i = 1, 2, . . . k. Partition C(ri ) into 2⌊ mi /2⌋2 ≤ mi /2 congruent arcs. By the induction hypothesis, Ai is a dense set along C(r pi ), and so ′ ⊂ A consist of exactly one point of A from every ⌊ m /2⌋-th there is a point of Ai in each arc. Let A i i i pi √ arc. The cardinality of A′i is #A′i = 2⌊ mi /2⌋ ≥ mi if mi ≥ 2. Draw the chords of circle C(ri+1 ) with midpoints at points in A′i . Note that each chord is tangent to C(ri ). The size of an arc between consecutive 8 points in A′i is at least p ⌊ mi /2⌋ − 1 1 ≥√ , 2π p 2 mi 2⌊ mi /2⌋ if mi > 2. By Lemma 5, these chords of C(ri+1 ) are pairwise disjoint, and determine disjoint caps of C(ri+1 ). Recall that we assumed that disk D(ri+1 ) is covered by bounded faces whenever ri+1 < 3/2. By Lemma 6, every cap contains a point in G ∩ C(ri+1 ). Let Ai+1 consists of one point of G ∩ C(ri+1 ) from √ each cap. The cardinality of Ai+1 is mi+1 = #A′i ≥ mi . Moreover, the point set Ai+1 is dense, since every arc of size least 4π/mi+1 contains at least one entire cap, hence at least one point of Ai+1 . Lemma 7 The total weight of the portion of G that lies in K(ri , ri+1 ) is at least 1/50 for every i = 0, 1, . . . , k − 1. 1 Proof. Recall that the width of K(ri , ri+1 ) is ri+1 − ri = 9m . Partition C(ri ) into ⌊mi /2⌋ congruent i arcs. Since Ai is dense, each arc of size 4π/mi contains a point of Ai . Pick one point of Ai from every other arc, and let A′′i denote the resulting set of at least ⌊mi /4⌋ points. Any two consecutive points in A′′i are separated by an arc of size at least 8π/mi . For every point p ∈ A′′i , construct a half-disk centered at p with radius εi and bounded by the tangent to C(ri ) at p; see Figure 4. These half-disks are pairwise disjoint and lie in the annulus K(ri , ri+1 ). By Lemma 6, the weight of the portion of G lying in each half-disk of radius εi is at least εi . Hence, the length of G ∩ K(ri , ri+1 ) is at least #A′′i · εi ≥ jm k i 4 · 1 1 ≥ . 9mi 50 ✷ A standard calculation using (1) gives that mk ≥ log n holds for sufficiently large n. Note also that we assumed in Lemma 7 that each annulus K(ri , ri+1 ) is covered by bounded (convex) faces of G. Since the sequence of mi ’s is non-decreasing, we have k−1 X i=0 k−1 εi ≤ X k 3 1 εi ≤ . ≤ , therefore rk = 1 + 9 log n 2 2 i=0 The disk of radius 3/2 centered at the origin lies in Q. Since the annuli K(ri , ri+1 ), for 0 ≤ i ≤ k − 1, have pairwise disjoint interiors, and each annulus is covered by bounded faces of G, we can apply Lemma 7, and the total weight of G is at least k/50 = Ω(W log log n). This completes the proof of part (ii) of Theorem 1. 3 General point sets in the plane In this section we derive asymptotically tight bounds on the minimum length of a convex Steiner partition for n points in the plane. Theorem 2 (i) For n points in the plane, there is a convex Steiner partition with O(W log n/ log log n) weight and O(n) Steiner points. Such a partition can be computed in O(n log n) time. (ii) For every n, there is a set S of n points in the plane such that any convex Steiner partition for S has weight Ω(W log n/ log log n). 9 We also extend the lower bound construction and show that there are n-element point sets for which any spanner network with all bounded faces convex and stretch factor o(n) has weight Ω(W log n/ log log n). Theorem 3 For every n, there is an n-element point set S in the plane such that any spanner network for S, whose bounded faces are convex and whose stretch factor is o(n), must have weight Ω(W log n/ log log n). Our upper bound in Theorem 2 is based on first partitioning the convex hull of an n-element point set into polygons, and then partitioning the polygons into convex faces. A convex Steiner partition of a polygon P is a planar straight line graph G, where the boundary of every bounded face of G is convex and all edges of P are covered by edges of G (in particular, there is no constraint on the unbounded face of G). For convex Steiner partitions of polygons, we prove the following. Theorem 4 Every polygon P with n vertices, m of which are reflex, admits a convex Steiner partition with O(|P | log m/ log log m) weight and O(n) Steiner points. Such a partition can be computed in O(n log n) time. The upper bound for the weight is the best possible, as it was shown by Levcopoulos and Lingas [30]. Interestingly, in our lower bound construction in Theorem 2, the EMST is a path which can be completed to a simple polygon by adding one edge. The resulting simple polygon P is, in fact, somewhat reminiscent of the lower bound construction presented in [30]. 3.1 Lower bound construction Proof of Theorem 2(ii). For every n > 4, we describe a set of at most n points in the plane for which any convex Steiner partition has weight Ω(W log n/ log log n). Refer to Figure 5. Let k ∈ N be the maximal integer such that kk < n. Since n > 4, we have k ≥ 2 and k = Θ(log n/ log log n). Consider a circular arc subtended by an inscribed angle α = π/(4k); that is, the central angle is 2α = π/(2k). Let β be a polygonal path connecting k + 1 points evenly distributed along this arc. We construct inductively, in k steps, a polygonal path γk , whose vertex set will be the point set S. The initial polygonal path is a straight line segment γ0 = ab. In step i = 1, 2, . . . , k, construct a polygonal path γi by replacing each segment s of γi−1 by a scaled copy of the polygonal path β above segment s, as illustrated in Figure 5. The polygonal path γi consists of ki segments of equal length. Let Si denote the set of vertices of γi , for i = 0, 1, . . . , k. Since γi is a refinement of γi−1 , we have Si−1 ⊂ Si . By induction, we have #Si = ki + 1. Our point set for the lower bound construction is S = Sk . Observe that by the definition of k, #S = #Sk ≤ n. Every edge of γi makes an angle of at most iα with the x-axis. So every edge of the final curve γk makes an angle at most kα ≤ π/4 with the x-axis. Hence, an EMST of Si is the path γi for each i (e.g., by Prim’s algorithm). We give an upper bound on the weight of γk . In each step, the increase in length of the path is bounded by a constant factor: 2α α 1 24k π 1 |γi | < ≤ < = =1+ <1+ , 3 |γi−1 | 2 sin α α − α /6 1 − α/6 24k − π 24k − π k for k ≥ 2. The weight of γk is |γk | ≤ |γ0 | · (1 + 1/k)k < |ab| · e = O(|ab|), where e stands for the base of the natural logarithm. Next, we show that the weight of any convex Steiner partition of S is Ω(|γk |·k) = Ω(W log n/ log log n). By Lemma 6, it is enough to construct a set of pairwise disjoint half-disks of total radii Ω(|ab| · k) such that each half-disk is centered at a point of S and is contained in the convex hull of S. Specifically, we 10 construct disjoint half-disks of radius |ab|/(32ki ) centered at at least half of the points in Si \ Si−1 , for i = 1, 2, . . . , k − 1 (but not for the last level, i = k). So the total radii of these disks will be k−1 X #(Si \ Si−1 ) i=1 2 k−1 |ab| |ab| X ki − ki−1 · = = Ω(|ab| · k) = Ω 32ki 64 ki i=1  W log n log log n  . The length of each segment of the path γi is at least |ab|/ki because in each step, we replace a segment by a sequence of k segments of larger total length. If the diameter of the polygonal path β is ℓ, then its width is 2 sin2 α2 α π ℓ α ℓ 1 − cos α ·ℓ = ·ℓ≥ . · ℓ = tan · ≥ · ℓ = 2 sin α 4 sin α2 cos α2 2 2 4 16k 8k Since the vertices of a scaled copy of the path β ⊂ γi lie along a circular arc, at least k/2 of its k − 1 internal vertices lie at a distance at least ℓ/(16k) from the diameter of β (that is, at least half-width distance from the chord of the circular arc). For each copy of the path β along the paths γi for i = 1, 2, . . . , k − 1 (except for the last level, i = k), at each such vertex v ∈ β, v ∈ Si \ Si−1 , place a half-disk of radius |ab|/(32ki ) ≤ ℓ/(32k) centered at v and bounded by the tangent line of the circular arc at v. For a fixed i, let Di denote this set of congruent half-disks. width a b Figure 5: Lower bound construction for k = 2 : S = S3 . Each segment is replaced with a sequence of 4 segments. We place half-disks of radius |ab|/(32k i ) centered at at least half of the points in Si \ Si−1 , for i = 1, 2, . . . , k − 1. k−1 Di lies in the convex hull of S, and that the half-disks in It remains to show that any half-disk in ∪i=1 are pairwise disjoint. Note that the half-disks in Di do not lie entirely below γi , and so γi does not separate half-disks in Di and Di+1 . We show that the half-disks in Di (i) are pairwise disjoint; (ii) lie above curve γi−1 , at distance at least |ab|/(32ki ) from γi−1 ; (iii) lie below γi+1 and their portions above γi remains within distance at most |ab|/(64ki+2 ) from γi . (i) It is easy to see that the half-disks in Di , centered at vertices of γi , are pairwise disjoint: The half-disks in Di have radius |ab|/(32ki ). Consecutive vertices of γi in the same copy of β are at least |ab|/ki distance apart, and so the x-coordinates of any two vertices of γi (even in different copies of β) are at least cos(π/4)|ab|/ki distance apart. (ii) By construction, the centers of the half-disks in Di are above γi−1 , at distance at least |ab|/(16ki ) from γi−1 . Hence every half disk in Di is above γi−1 , at distance at least |ab|/(32ki ) from γi−1 . (iii) Consider a circle passing through the vertices of a copy of β along γi . Each edge of β subtends an inscribed angle of α/k = π/(4k2 ); and the diameter of a half-disk centered at a vertex of β is tangent to this circle. Hence, at the center of each half-disk in Di , the incident edges of γi make an angle of π/(4k2 ) with the diameter of the half-disk. The portions of the half-disk above γi remain within distance at most sin(π/(4k2 )) · |ab|/(32ki ) ≤ |ab|/(64ki+2 ) from γi . Therefore the half-disks in Di lie below γi+1 , yet they are disjoint from any half-disk in Di+1 . This completes the proof of Theorem 2(ii). ✷ k−1 ∪i=1 Di Proof of Theorem 3. We modify the previous construction as follows. Let S be a set of at most 2n points: a set S1 of n points distributed evenly on a circle of radius 2 centered at the origin o, and a set S2 of at most n points making our construction in Figure 5, starting from a horizontal segment γ0 = ab centered at o, where |ab| = 2. In particular, S2 is contained in a circle of unit radius centered at o. 11 An EMST of S consists of a polygonal path of n − 1 edges connecting consecutive points along the circle of radius 2, a polygonal path of n − 1 edges connecting consecutive points along γk , and a unit segment connecting the two. Denoting by Wn the length of an EMST for S, we have limn→∞ Wn ≤ 2e + 4π + 1 = O(1). Consider a spanner network G for S with stretch factor o(n) and all bounded faces convex. By the condition on the stretch factor, the shortest path between any two consecutive vertices of conv(S) has length o(n) · 4π/n = o(1). Hence the unbounded face is disjoint from the unit disk if n is sufficiently large. That is, all faces of the network that cover the unit disk are convex, and our previous argument shows that the network has weight Ω(log n/ log log n) = Ω(W log n/ log log n), as claimed. ✷ Remark. Consider again the point set S for the lower bound construction in Figure 5. Let P be the simple polygon obtained by connecting the two endpoints a and b of γk . Note that P is x-monotone, and |P | ≤ 2W . We have given a lower bound on the total weight of the portions of a convex Steiner partition of S that lie in some pairwise disjoint half-disks. Observe that these half-disks are disjoint from the edges of P and lie in the interior of P . It follows that any Steiner partition of P has weight Ω(|P | log n/ log log n). In our construction, the convex vertices of P are a, b, and the vertices Sk \ Sk−1 introduced in the last level. The number of reflex vertices is m = Θ(n/k) = Θ(n log log n/ log n), and so log m/ log log m = Θ(log n/ log log n). This provides an alternative construction for the lower bound of Ω(|P | log m/ log log m) on the minimum weight Steiner partition of a simple polygon P with m reflex vertices, first established by Levcopoulos and Lingas [30]. 3.2 Upper bound—reduction to flat poygons Let S be a set of n points in the plane. We show that S admits a convex Steiner partition of weight O(W log n/ log log n) by reducing the partition problem for points to the corresponding partition problem for polygons. We proceed as follows. Compute the convex hull and an EMST of S. The EMST partitions the interior of the convex hull into weakly simple polygons [33, Section 10.2]. The total perimeter of these polygons is at most 4W . Since the maximum degree of an EMST is at most 6, the total number of vertices of these polygons is at most 6n. We construct a convex Steiner partition for each of these polygons, and then Theorem 4 completes the proof of Theorem 2(i). We now present the proof of Theorem 4. Given a simple polygon P with n vertices, m of which are reflex, we construct a convex Steiner partition of P in three stages. The first stage (Lemma 8) partitions the interior of P into convex polygons and monotone polygons (defined below) by introducing new edges of total weight O(|P |). The second stage (Lemma 10) partitions the interior of each x-monotone polygon M into convex polygons and monotone 1-flat polygons (defined below) introducing new edges of total weight O(|M |). The third stage (Lemma 11) partitions every monotone 1-flat polygon F into convex faces by adding new edges of total weight O(|F | log m/ log log m). We proceed with the details. γ a P p P b D(P, s) s s (a) q (b) Figure 6: (a) A 1-flat polygon P . (b) A polygon P and a domain D(P, s) spanned by a side s of P . 12 Definitions. A diagonal of a polygon P is a line segment connecting two vertices of P through the interior of P . A chord of a polygon P is a line segment that connects two points on the boundary of P (at vertices or on edges of P ) and whose relative interior is disjoint from the exterior of P . For a parameter ε > 0, a polygon P is ε-flat if it is bounded by a base side s and a polygonal curve γ connecting the endpoints of s, and lying in one of the closed halfplanes determined by s, such that for any chord ab of P with a, b ∈ γ, the portion of γ between a and b has weight at most (1 + ε)|ab|; see Figure 6 (i). In particular, we have |γ| ≤ (1 + ε)|s|. Let d be a direction, represented by a directed or undirected line or line segment (e.g., a coordinate axis). A polygonal path γ is d-monotone if the intersection of γ with every line orthogonal to d is connected (that is, the intersection is a point, a line segment, or the empty set). A polygon P is d-monotone if it is bounded by a base side s parallel to d and a d-monotone path γ connecting the endpoints of s. We sometimes omit the direction d, and call a polygon monotone if it is d-monotone in some direction d. (Note that our definition of monotone polygon is slightly different from the standard one [1]). Denote by P the closed polygonal domain bounded by P ; that is, P is the closure of int(P). For a polygon P and a side s, let D(P, s) be the set of all points p ∈ P such that the line segment pq lies in P , q ∈ s and pq is orthogonal to s; refer to Fig. 6(ii). Let the polygon M (P, s) be the boundary of the domain D(P, s). Note that M (P, s) is s-monotone. Lemma 8 Every polygon P with n vertices admits a partition into convex faces and monotone polygons, such that the partition has O(|P |) weight, O(n) Steiner points, and can be computed in O(n log n) time. Furthermore, every reflex vertex of a monotone face is a reflex vertex of P . Proof. We partition P recursively. The input of each recursive step is a pair (Q, s), where Q is a simple polygon and s is a side of Q. For a polygon Q, let v(Q) denote the number of vertices of Q. Initially, we start with the pair (P, s0 ), where s0 is an arbitrary side of P , and v(P ) = n. One recursion step works as follows (refer to Figure 7): We are given a pair (Q, s). Stop if Q is s-monotone or convex. Otherwise compute the s-monotone polygon M (Q, s) spanned by s in Q. The edges of M (Q, s) partition Q into polygonal faces. Denote the resulting polygons outside of M (Q, s) by Q1 , Q2 , . . . , Qt , for some t ≥ 1. The polygon M (Q, s) has exactly one side adjacent to each Qi , which we denote by si . We have v(Qi ) ≤ v(Q) for i = 1, 2, . . . , t. Case 1 If [t = 1 and v(Q1 ) < v(Q)] or [t ≥ 2], then partition Q along all segments si and recurse on (Qi , si ) for i = 1, . . . , t. Case 2 If t = 1 and v(Q1 ) = v(Q), then find a chord f of Q1 parallel to s1 such that f 6⊆ s1 , the endpoints of f lie on the edges of Q1 adjacent to s1 , and f contains a vertex of Q1 ; see Figure 7(c). Partition Q along f into a convex quadrilateral Q0 adjacent to s; and polygons Q′1 , Q′2 , . . . , Q′t′ , for some t′ ≥ 1. Each Q′i has exactly one side s′i along f . Recurse on (Q′i , s′i ) for i = 1, . . . , t′ . First, we estimate the weight of the partition. Consider one step of the recursion. In Case 1, Q is partitioned along the sides of M (Q, s) perpendicular to s, and the monotone polygon M (Q, s) is discarded from further consideration. The partitioning edges si , i = 1, 2, . . . , t, become the base sides in the subproblems (Qi , si ). Charge the weight of each si to the common boundary of M (Qi , si ) and the input polygon P . The weight of this portion of the boundary of M (Qi , si ) is at least si , and will not be charged again—since M (Qi , si ) is discarded in the next step of the recursion. In Case 2, Q is partitioned along a chord f perpendicular to s, and a convex quadrilateral Q0 , which is strictly larger than M (Q, s), is discarded from further consideration. The edges s′i , i = 1, 2, . . . , t′ , along f become the base sides of the subproblems. Charge the weight of each s′i to the common boundary of M (Q′i , s′i ) and the input polygon P as above. Over all, 13 each portion of the boundary of P is charged at most once. Hence the total weight of the new edges is at most |P |, and the weight of the entire network (including P ) is at most 2|P |. Q2 Q3 s2 Q1 Q s1 M (Q, s) s s (a) Q1 Q′1 s1 Q1 Q′2 Q2 s2 f s′ s′ 2 1 s1 Q1 Q2 s1 s′1 Q′1 f s1 s1 Q1 (c) (d) Q1 s1 Q1 s1 s2 s1 Q1 s3 (b) Q1 Q2 (e) (f) Figure 7: Partitioning a polygon P into convex and monotone polygons. Next, we estimate the number of Steiner points. For a polygon Q, let count(Q) = 2v(Q) − 6. Consider a step of the recursion that produces t ≥ 1 subproblems (Qi , si ), i = 1, 2, . . . , t. In Case 1, at most t new Steiner points are created (at most one endpoint of each si ). In Case 2, at most 2 Steiner points are created (namely, the endpoints of the chord f ). So in both cases, at most 2t Steiner points are created. We claim that t X count(Qi ) ≤ count(Q) − t. i=1 That is, the total count decreases by at least t. Initially, count(P ) = 2n−6. We have count(Q) ≥ 2·3−6 = 0 for any subproblem (Q, s) throughout the recursion. This gives an upper bound of 2(2n − 6) = 4n − 12 on the number of new Steiner points created altogether. We now justify the above claim. For a subproblem (Q, s), let V (Q, s) denote the set of vertices of Q with the exception of the endpoints of s. Clearly, #V (Q, s) = v(Q) − 2. As above, consider a step of the recursion that produces subproblems (Qi , si ), i = 1, 2, . . . , t. The sets V (Qi , si ) are pairwise disjoint 14 S P P and ti=1 V (Qi , si ) ⊆ V (Q, s). Therefore, ti=1 (v(Qi ) − 2) ≤ v(Q) − 2, and ti=1 count(Qi ) ≤ count(Q) − 2(t − 1). For t ≥ 2, we have t ≤ 2(t − 1), so our claim is established in this case. Now assume that t = 1 (i.e., a single subproblem (Q1 , s1 ) is produced). In Case 1, we assumed that v(Q1 ) < v(Q). In Case 2, a vertex of Q incident to the chord f belongs to V (Q, s) but lies at the base side of the subproblem, hence v(Q1 ) < v(Q). This implies that for t = 1, we also have count(Q1 ) ≤ count(Q) − 2 in both cases. It is easy to implement the algorithm in O(n log n) time. Assume w.l.o.g. that s0 is horizontal. Then every partitioning segment is axis-parallel, and is incident to a vertex of P or a Steiner point. Using a ray-shooting data structure for the polygon P [8], compute all axis-aligned rays from every vertex of P in advance; and sort the heads of rays along each edge of P . During the partition algorithm, we insert new rays from every new Steiner point. Shoot a ray from the endpoints of si orthogonally to si into the interior of Qi (in Case 1) and from the endpoints of f orthogonally to f in the interior of Q′i (in Case 2). The rays allow computing each polygon M (Q, s) in time proportional to its number of vertices. The sorted list of ray heads along the edges allows finding the segment f incident to a vertex of P . ✷ Consider now a x-monotone polygon P with a horizontal base s. Let C denote the set of all chords of P whose endpoints are not in the interior of s; and let H ⊂ C denote the set of horizontal chords. For every ab ∈ C, let µ(ab) denote the portion of the boundary of P between a and b that does not contain s. Lemma 9 Let P be a x-monotone polygon with a horizontal base s. If there is a constant κ > 0 such √ that |µ(ab)| ≤ (1 + κ)|ab| for all horizontal chords ab ∈ H, then |µ(ab)| ≤ (1 + κ) 2|ab| for all chords ab ∈ C. b µ(ab) c d π(ab) P a s Figure 8: A chord ab in a monotone polygon P with a horizontal base s, the boundary path µ(ab) (bold), and the x-monotone path π(ab) (dashed). Proof. Let ab ∈ C be a chord of P ; see Figure 8. Assume that ab is not horizontal, otherwise µ(ab) ≤ (1 + κ)|ab|. Assume without loss of generality that a has smaller y-coordinate than b. Let π(ab) be an x-monotone path between a and b such that every segment along π(ab) is either horizontal or lies along an edge of P . Since the√weight of π(ab) is at most that of an axis-aligned staircase path between a and b, we have |π(ab)| ≤ 2|ab|. For each horizontal portion √ of π(ab), say, cd ⊂ π(ab), we have ✷ µ(cd) ≤ (1 + κ)|cd|. Hence |µ(ab)| ≤ (1 + κ)|π(ab)| ≤ (1 + κ) 2|ab|. Lemma 10 Every x-monotone polygon P with a horizontal base s and n vertices admits a Steiner partition into convex polygons and 1-flat x-monotone faces such that the partition has O(|P |) weight, O(n) Steiner points, and it can be computed in O(n log n) time. Furthermore, every reflex vertex of a face is a reflex vertex of P . Proof. We sweep a horizontal line ℓ top-down over P , and insert horizontal chords along ℓ when certain events occur. Let Q denote the polygonal face in the current partition adjacent to the base s. Initially, let 15 ℓ ℓ ℓ a b Start a Event 2 ℓ b Event 1 a b Event 2 ℓ ab ℓ Event 2 Stop Figure 9: Partitioning an x-monotone polygon with horizontal base s into convex faces and 1-flat x-monotone faces. Q = P . The face Q is always an x-monotone polygon with base s, and each insertion of a horizontal chord along ℓ cuts off a polygon from Q. The algorithm is designed so that each polygon cut off from Q is either convex or 1-flat with a horizontal base along the current position of ℓ. For every chord ab ∈ H lying along the sweep-line ℓ, we define µQ (ab) to be the portion of the boundary of Q between a and b that does not contain s. Sweep a horizontal line ℓ from the top vertex down until ℓ reaches the base s. If any of the following two events occurs for a chord ab ∈ H lying along ℓ, partition Q along ab into two faces, and let Q be the face below ab. √ Event 1. µQ (uv) contains at least three vertices of Q in its interior and |µQ (ab)| = 2|ab|. √ Event 2. a or b is a reflex vertex of Q and |µQ (ab)| ≥ 2|ab|. In each step, we inserted a chord ab and partitioned Q along chords ab into√two faces. The face above ab is either convex and its common boundary with the new face Q is√at least 2-times heavier than |ab|; or it is 1-flat and x-monotone with base ab by Lemma 9, with κ = 2 − 1. This face is discarded from further consideration, and the remaining face (adjacent to s) is partitioned recursively. In each step, we √ discard a portion of weight at least 2|ab| from the boundary of Q, and introduce a new boundary segment of weight |ab|. If we charge the weight of each chord ab uniformly toP the portion of the polygon P that it ∞ √ −i 1 times. So the replaces, then each point along the boundary of P is charged at most i=1 ( 2) = √2−1 1 total length of the partition (including the weight of P ) is at most (1 + √2−1 )|P | = O(|P |). Next, we estimate the number of Steiner points. Let count(Q) denote the number of vertices of Q plus the number of vertices of Q with an acute interior angle. Initially, when Q = P , we have count(Q) ≤ 2n, since there are no more than n acute angles. We claim that every partition step decreases count(Q) by at least one, and introduces at most two Steiner points. This implies that the number of new (Steiner) vertices cannot exceed 4n. In Event 2, at least three vertices are removed from Q, and at most two new (Steiner) vertices are created, neither of which may have an acute interior angle. In Event 1, one or two convex vertices are removed and at most one new (Steiner) vertex is created, which does not have acute interior angle. If only one vertex is removed from Q, then the part of Q above ab is a triangle, and µQ (ab) ≥ |ab| is possible only if the triangle has an acute angle at the vertex opposite to ab. This confirms our claim that count(Q) strictly decreases in each step, and concludes the proof of the lemma. ✷ 16 3.3 Upper bound—partitioning a flat polygon In this section, we partition a monotone 1-flat polygon recursively. The intermediate polygons in the recursion steps are not necessarily monotone or flat, however, they have a very special structure: A polygon P is a clamp polygon with spine ab if it is bounded by a convex chain (path) σ and a d-monotone polygonal path γ (for some direction d), with both paths connecting the same two points a and b; see Figure 10. The width of the clamp polygon is the width of the minimum strip that contains the polygon and is parallel to ab. Observe that every s-monotone polygon is a clamp polygon, where the convex chain is the side s. γ d w b a P σ Figure 10: A clamp polygon P , bounded by a convex chain σ (in bold) and a d-monotone polygonal path γ. Lemma 11 Every 1-flat x-monotone polygon F with a horizontal base and n vertices, m of which are reflex, admits a convex Steiner partition with O(|F | log m/ log log m) weight and O(m) Steiner points. Such a partition can be computed in O(n + m log n) time. Proof. Consider a 1-flat x-monotone polygon F bounded by a horizontal side s0 and an x-monotone polygonal path γ0 . Assume that F has n vertices, m of which are reflex. We partition F recursively. We describe a generic step of the recursion, where we are given a clamp polygon Q with r reflex vertices, bounded by a convex chain and an x-monotone polygonal path γ ⊆ γ0 , and we want to partition Q into convex faces and some clamp polygons Qi , each having at most r/2 reflex vertices. We say that the weight of the problem associated with Q is |γ|. The polygonal path γ of Q will be partitioned among the subproblems Qi ; and so the total weight of the subproblems in each level of the recursion is at most |γ0 |. In a recursion step, we will introduce new edges of total weight O(|γ0 |). Intuitively, it is enough to show that the “average depth” of the recursion is O(log m/ log log m), in order to establish a O(W log m/ log log m) bound on the weight of the partition. One step of the recursion. We are given a clamp polygon Q with r ≥ 1 reflex vertices, a spine s, where Q is bounded by a convex chain σ and a polygonal path γ ⊆ γ0 . By rotating Q, if necessary, we may assume that the spine s is horizontal, and γ is d-monotone for some direction d (which is not necessarily horizontal). It is clear that any vertex of Q with maximal y-coordinate is a vertex of γ. We also assume that one of the vertices of Q with minimal y-coordinate is a vertex of γ. This property holds for the initial clamp polygon F , and we maintain the property for every clamp polygon Q during the recursion. It follows that |γ| ≥ 2w, where w is the width of Q. Let k = ⌈|γ|/w⌉ ≥ 2. We partition the polygonal path γ into at least k subpaths such that (i) each subpath passes through at most r/k reflex vertices of Q; (ii) each subpath passing through a reflex vertex of Q has weight at most |γ|/k; and (iii) a furthest point of γ from the supporting line of the spine of Q is the endpoint of a subpath. 17 The partition is done by successively selecting the elements of a set A ⊂ γ of spliting points. Refer to Figure 11. The splitting points are selected as follows: Place one endpoint of γ in A. Move a point p along γ continuously from this endpoint to the other. If the subpath of γ between the previous splitting point of A and p ∈ γ satisfies any of the following conditions, then insert p into A: (1) p is the (⌊r/k⌋ + 1)th reflex vertex of γ along the subpath; (2) the weight of the subpath is at least |γ|/k and it passes through at least one reflex vertex; (3) the weight of the subpath exceeds |γ|/k and p is its first reflex vertex; (4) p is a the furthest point of γ from the supporting line of the spine of Q, which may be either above or below the spine; or (5) p is the right endpoint of γ. The cardinality of A is at most 2k + 1, with the two endpoints of γ, at most k − 1 additional splitting points of type (1), at most k − 1 additional splitting points of types (2) and (3), and the furthest point from the spine (4). Next we partition the interior of Q into a set F of at least k polygons by using the points A ⊂ γ and by drawing some new segments. If γ is x-monotone then drop a vertical segment from each point a ∈ A to the convex chain σ. If γ is not x-monotone, however, additional splitting points may be necessary. Let γ∗ be the lower envelope of γ (that is, the set of points p ∈ γ such that there is no other point of γ with the same x-coordinate and a smaller y-coordinate). If γ is x-monotone, then γ = γ∗ , otherwise γ∗ ⊂ γ consists of several components. If pq is a vertical segment that connects distinct components of γ∗ , then the portion of γ between p and q is called a pocket of γ bounded by pq. For every point a ∈ γ in a pocket bounded by pq, there is a line segment orthogonal to d that connects a to pq, since γ is d-monotone. Now, connect every splitting point a ∈ A to the convex chain σ as follows. If a ∈ γ∗ , then drop a vertical segment from a to the convex chain σ. If a is part of a pocket bounded by some vertical segment pq, then insert p and q as new splitting points, extend the vertical segment pq to σ, and connect a to pq by a segment orthogonal to d. Let A′ denote the union of A and the sets {p, q} for each a ∈ A lying in a pocket bounded by pq. There are at most 2k + 1 splitting points in A. Each vertical edge between γ and the convex chain σ has weight at most w. The segments orthogonal to d have weight at most 2w since F is 1-flat and the weight of a pocket bounded by pq is at most 2|pq|, where |pq| ≤ w. So the total weight of the partition edges is at most   |γ| + w |γ| w ≤ 12|γ|. w<8 (2k + 1)(1 + 2)w = (6k + 3)w ≤ 8kw = 8 w w d a γ q γ′ s(β) β w ℓ(β) s p Q σ Figure 11: One level of partitions for a clamp polygon Q. The splitting points in A′ partition the path γ into a set Γ of subpaths, satisfying conditions (i)–(iii). By connecting consecutive points of A′ along γ, we obtain a polygonal path γ ′ (dashed path in Figure 11). So far, we have partitioned Q into a set F of polygons, each of which is bounded by a convex chain and a path in Γ. For each path β ∈ Γ, let s(β) ⊂ γ ′ denote the segment connecting the endpoints of β. Shift segment s(β) continuously to a position ℓ(β) such that it partitions the corresponding face in F into a convex face and a clamp polygon with spine s(β); see Figure 11. Since |ℓ(β)| ≤ |s(β)|, the total weight of the segments ℓ(β), for all β ∈ Γ, is |γ ′ | ≤ |γ|. The total weight of all new edges introduced in one 18 recursion step is at most 12|γ| + |γ| = 13|γ|. Partition all non-convex clamp polygons recursively until all faces are convex. This completes the description of our partition algorithm. Analysis. Assume F is the initial 1-flat clamp polygon with n vertices, m of which are reflex, a base s0 and a polygonal path γ0 . (We can assume m is large enough, when needed.) Obviously, we have |γ0 | ≤ |F |. At level j of the recursion, j ≥ 0, we construct a polygonal path λj connecting the two endpoints of the spine s0 as follows. At the root level, λ0 is a straight line segment λ0 = s0 . At level j, we construct λj by replacing some segments of λj−1 with polygonal paths: If a segment s ⊂ λj−1 is the spine of a subproblem with at least one reflex vertex, then s is replaced by the polygonal path γ ′ (defined previously); if a segment s ⊂ λj−1 is the spine of a convex clamp polygon in a subproblem, then s is replaced by the (convex) chain γ of that clamp polygon. We can establish a piecewise linear homeomorphism Hj : λj−1 → λj . If a segment s ⊆ λj−1 is replaced by a polygonal path γ ′ ⊆ λj of weight |γ ′ |, then Hj maps s to γ ′ while increasing each portion of s with the same factor. By transitivity, the composition Hj ◦ Hj−1 ◦ · · · ◦ H1 is a piecewise linear homeomorphism between s0 and λj . We next show that the weights of the approximations λj monotonically increase (|λj−1 | ≤ |λj |, for j ≥ 1) but remain in the range |s0 | ≤ |λj | ≤ |γ0 | ≤ 2|s0 |. Consider the recursion tree T . In T , the subproblem at a terminal node at level j is a convex clamp polygon, and so the corresponding portion of λj is a portion of the initial path γ0 . At the next level, j + 1, only the clamp polygons adjacent to λj are further partitioned. Each non-terminal node v in T corresponds to a subproblem, where a spine sv is replaced by a path γv′ and v has at least kv ≥ 2 children (recall that one step of the recursion partitions γv ⊆ γ0 into at least kv subpaths). By the definition of kv , we have kv ≥ |γv |/wv ≥ |sv |/wv . Observe that the path γv′ reaches a vertex of γv at distance at least wv /2 from the spine sv , so its weight is at least p p p |γv′ | ≥ 2 (|sv |/2)2 + (wv /2)2 = |sv |2 + wv2 ≥ |sv | 1 + 1/kv2 . Let nodes v(i), i = 0, 1, . . . , t, in T form a path from the root to a leaf, where v(i) lies at level i of the recursion. If we follow a path v(0), v(1), . . ., v(t) in the tree of the recursion, then the number of reflex vertices in the subproblems decreases by a factor of at least kv(i) for i = 0, 1, 2 . . . , t − 2. We have zero reflex vertices at a terminal node, and we may have fewer than kv(t−1) reflex vertices at the parent of a terminal node. At any other node v(i) ∈ V (T ), the number of reflex vertices decreases by a factor of at least kv(i) . Since at the root of T corresponds to the initial problem with m reflex vertices, the product of the kv(i) values along the path is bounded by m: t−2 Y i=0 kv(i) ≤ m. In any chain v(0), v(1), . . . , v(t) from the root to a leaf in T , there are at most 4 log m/ log log m nodes with kv(i) ≥ (log m)1/4 , since ((log m)1/4 )4 log m/ log log m = 2(log log m)(log m/ log log m) = 2log m = m. We next establish lower bounds on the weight of the approximation paths pλj . In each step with kv ≤ 1/4 ′ (log m) , a segment sv is replaced by a path γv of weight at least |sv | · 1 + 1/(log m)1/2 . In each step with k > (log m)1/4 , we use the trivial lower bound that a segment sv is replaced by a path γv′ of weight at least |sv |. Let h ≥ 0 be an integer. We show that the total weight of those portions of the initial segment p s0 that undergo at least h⌈log m/ log log m⌉ steps that each expand the weight by a factor of at least 1 + 1/(log m)1/2 is at most |s0 |/2h . This is immediate for h = 0; and if it did not hold for h ≥ 1, 19 then the weight of the final approximation path γ0 would be  h log m 2 log log m 1 1+ √ |γ0 | ≥ log m √ m √log m ! 2hloglog  log m 1 |s0 | = 1+ √ 2h log m h  √ log m 2 log log m ≥ 2h |s0 | > 2|s0 |, ≥ |s0 | 2 |s0 | 2h  for m ≥ 2256 , using the fact that (1 + 1/x)x ≥ 2 for x ≥ 1. This contradicts our assumption that F is 1-flat and so |γ0 | ≤ 2|s0 |. For any h ≥ 4 and j = h⌈log m/ log log m⌉, we have |λj ∩ γ0 | ≥ (1 − 24−h )|γ0 |, that is, the common portion of λj and γ0 has weight at least (1 − 24−h )|γ0 |. This implies that for any h ≥ 4, at level j = h⌈log m/ log log m⌉, we are left with subproblems of total weight at most |γ0 |/2h . At each level of the recursion, we introduce new edges whose total weight is proportional to the weight of the current subproblems in that level. Therefore, for h ≥ 4, the total weight of the new edges introduced between levels h⌈log m/ log log m⌉ and (h+1)⌈log m/ log log m⌉ is at most (1/2)(h−4) O(|γ0 |)·⌈log m/ log log m⌉. This also holds for h = 0, 1, 2, 3. The weight of the final convex Steiner partition of F is bounded by ∞  h−4 X 1 h=0 2       log m log m log m O(|γ0 |) · = O |γ0 | · = O |F | · . log log m log log m log log m The number of Steiner points in level j of the recursion is proportional to the total number splitting points created along γj−1 . Each splitting point in a set A lies at a reflex vertex or between two reflex vertices of γ0 ; and there are at most one splitting point of A between two consecutive reflex vertices of γ0 . Hence, there are at most 2m − 1 splitting points along paths λj for all j ≥ 0, and the number of Steiner points is O(m). The runtime of constructing the partition is O(n + m log n). In an O(n)-time preprocessing step, traverse γ0 and for each convex chain β ⊂ γ0 between consecutive reflex vertices, compute the weight and build a binary search structure, which can report for a query value q > 0 a point p ∈ β in O(log n) time such that the weight of the portion of β between its left endpoint and p equals q. In order to find the splitting points in a clamp polygon Q, we traverse the corresponding path γ ⊆ γ0 once. We have seen that the number of reflex vertices traversed more than h⌈log m/ log log m⌉ times is O(m/2h ), and so we traverse reflex vertices O(m log m/ log log m) times in total. We can skip convex vertices, since we have computed the weight of γ0 between consecutive reflex vertices. There is at most one splitting point in A between any two reflex vertices, each of which can be located in O(log n) time based on a binary search structure. We spend O(m log n) total time on finding points of A between reflex vertices. For connecting the splitting points to the convex chain σ of the corresponding clamp polygon Q, we use a ray shooting data structure [8] for F , which can be constructed in O(n) time and permits O(log n) query time. Since each segment s(β) either connects the endpoints of β or is tangent to a reflex vertex of β, we can compute s(β) in a single traversal of the reflex vertices of β. The total time for partitioning F is O(m log n). This completes the proof of Lemma 11. ✷ Lemma 11 was the last in the reduction chain, so the proof of Theorem 4 is now complete. 20 4 Conclusion We deduced tight bounds on the weight of a minimum convex partition of a point set in terms of their Euclidean minimum spanning trees. The worst-case ratio of the minimum Steiner partition and an EMST is Θ(log n/ log log n) in general, and Θ(log log n) for the special case of pseudo-triangles. We conclude with some remaining unanswered questions. 1. Our partition for n points has O(n) Steiner points. Does every set of n points admit a convex Steiner partition of weight O(W log n/ log log n) and with only o(n) Steiner points? 2. Does every set of n points admit a convex Steiner partition of weight O(W log n/ log log n) and with O(n) Steiner points, such that every face is fat (that is, the ratio of the radii of the minimum enclosing and maximum inscribed circles over all faces is bounded by a constant)? 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