Cover Decomposability of Convex Polygons and Octants

Transcription

1 Cover Decomposability of Convex Polygons and Octants Sayan Bandyapadhyay Department of Computer Science University of Iowa Iowa City, USA September 7, Introduction Imagine a universe, which is basically a set of points (that may be infinite), and a collection of sensors. Each sensor has a specified covering region in the universe, i.e, a subset of the universe which it covers (monitors). Moreover, the sensors are powered by battery and they have two alternating modes of action, active and passive. In active mode a sensor covers its region and in passive mode the battery is charged and the sensor cannot work. The duration of each mode is one unit time, and once a sensor is turned on it is alternated between the two modes. The goal is to schedule the sensors in a way so that they cover the universe for all the time. One possible approach to achieve this goal is to decompose the collection of sensors into two disjoint sets so that each of these sets covers all the points of the universe. Then we can schedule the sensors in the following way. First, all the sensors of a set are turned on at the same time, say t. Then all the sensors of the other set are turned on at time t + 1. Note that due to the one unit interval between the starting times, when the sensors of one set are charged the sensors of the other set remain active. Hence the universe gets covered for all the time. Now the fundamental problem in this approach is the decomposition of the set of sensors with such a property. Intuitively the decomposition is possible in that way if each point in the universe is covered by many sensors. In this technical report we discuss the decomposability of such sets with respect to specific types of covering region, for example intervals in R or convex polygons in R 2. To define the problem formally we need some definitions at first. A family of geometric sets P in R d is an m-fold covering of a set S if every point in S is contained in at least m sets of P. A 1-fold covering is simply called a covering. The simplest geometric sets to study are intervals in R. In Figure 1(a) a 2-fold covering of a portion of R with intervals is shown. Note that each point is covered by at least 2 intervals. Moreover, the two collections of intervals shown by dotted segments and bold segments individually covers the portion of R. Thus the whole collection of intervals is decomposable into two subcollections (dotted and bold) each of which is a sayan-bandyapadhyay@uiowa.edu 1

2 covering. In fact, one can show that any 2-fold covering of a portion of R with intervals can be decomposed into two coverings. Figure 1(b) shows a 2-fold covering of a set of three points with squares. Note that here the collection of three squares cannot be decomposed into two coverings, as no single square contains all of the three points. With all these discussions one might be interested in the following question: Given an m-fold covering P of a subset S of R d with a family of geometric sets, is there two disjoint subcollections of P such that each subcollection is a covering of S? Based on the family of geometric sets considered there are different notions of decomposability. In this technical report we study the families formed by translates or homothets (a homothet of an object is obtained by applying translation and scaling to it) of a given geometric set. (a) (b) Figure 1: (a) Decomposition of a 2-fold covering. (b) An indecomposable covering. 1.1 Covering with Translates The most straightforward family to consider is the one formed by translates of a geometric set. Before moving on we introduce the following notion of decomposability. A geometric set P in R d is cover-decomposable if there exists a constant m (that depends on P ) such that every m-fold covering of any subset of R d with a finite number of translates of P can be decomposed into two coverings of the same subset. The minimum constant m for which P is cover-decomposable, is called the cover-decomposability constant of P. The cover-decomposability problem we consider in this report is as follows. Given a geometric set P in R d, determine whether P is cover-decomposable. As mentioned earlier the cover-decomposability constant of any interval in R is 2. On the other hand the cover-decomposability constant of squares in R 2 is at least 3. In 1980 Pach [12] proposed the problem of determining all the cover-decomposable sets in the plane. He conjectured that every convex object is coverdecomposable. Pach [9] showed that centrally-symmetric convex polygons are cover-decomposable. Tardos et al. [16] have proved that cover-decomposability constant of triangles is at most 43. Recently, Keszegh et al. [5] have reduced the constant to 12. Finally, Palvolgyi and Toth [15] have showed that convex polygons are cover-decomposable. However, in a very recent result Palvolgyi prove that unit discs are not coverdecomposable [14]. On the other hand, Pach et al. [10] and Palvolgyi [13] have independently showed that convexity is a necessary condition and there exists non-convex polygons which are not cover-decomposable. Partly motivated by scheduling sensors, an extension of cover-decomposability has also been studied. The problem is to decompose a k-fold covering into as many disjoint coverings as possible. The proof of cover-decomposability of triangles given by Tardos et al. [16] implies that any k-fold covering with 2

3 translates of a triangle can be decomposed into Ω(log k) coverings. Pach et al. [11] showed that any k-fold covering with translates of a centrally-symmetric convex polygon can be decomposed into Ω( k) coverings. Aloupis et al. [1] improved their result by showing a decomposition into Ω(k) coverings. More recently, Gibson and Varadarajan [4] have extended the latter result for general convex polygons. Comparing to the planar version of cover-decomposability problem very few works have been done for the version in R 3. Recently, Keszegh et al. [5] have considered the problem for octants. They show that octants are cover-decomposable for the constant 12. More recently, Keszegh et al. have shown that any O(k 4.53 )-fold covering with translates of an octant can be decomposed into k coverings [6] (see also [2],[3],[7]). 1.2 Covering with Homothets A variant of cover-decomposability has recently caught much attention. In this variant in addition to translation, scaling is allowed to construct the family of sets which covers the points. Indeed, this is the most natural extension of the problem of covering with translates. We note that the cover decomposability with translates of an octant in R 3 implies the cover decomposability with homothets of a triangle in R 2. The intersections of a set of octants with a plane in R 3 that is not parallel to any axis form a set of homothetic triangles (see Figure 2). To be precise the intersection of the translates of the octant containing (-,-,- ) with the x + y + z = 0 plane generates homothetic copies of an equilateral triangle. Given a set of homothets of a triangle in a plane we can transform them to homothets of the equilateral triangle using affine transformation. We rotate the plane so that it coincides with the plane x + y + z = 0. Then from each homothet of equilateral triangle we get an unique translate of the octant containing (-,-,- ). For any point set lying on x + y + z = 0 plane we can show that this set of translates of the octant is cover-decomposable. Thus any 12-fold covering with a set of homothets of a triangle is cover-decomposable [5]. Similarly, any O(k 4.53 )-fold covering with homothets of a triangle can be decomposed into k coverings [6]. These results motivated the study of cover-decomposability for homothets of a convex polygon. Recently, Kovacs [8] show that for any convex polygon P with at least four sides and k > 0, there exists a k-fold covering of the plane with homothetic copies of P that cannot be decomposed into two coverings. a c b Figure 2: The intersection of an octant and a plane forms the triangle abc 3

4 1.3 Other Related Works Pach et al. [10] considered a variant of cover-decomposability problem where the family of geometric sets is any arbitrary collection of a single class of objects. They mainly studied three classes of objects: strips, straight lines, and axis-parallel rectangles. A strip is an unbounded set between two parallel lines. They showed that the family of strips, the family of straight lines and the family of axis-parallel rectangles are not cover-decomposable. 1.4 Our Results and Techniques In this technical report we partly discuss the works of Gibson and Varadarajan [4] and Keszegh and Palvolgyi [5]. Gibson and Varadarajan [4] consider the problem of decomposing k-fold covering with translates of a convex polygon into as many disjoint coverings as possible. They show that any k-fold covering with translates of a convex polygon can be decomposed into Ω(k) coverings. For simplicity they first consider the case where the convex polygon is centrally symmetric. A convex polygon P with its centroid at the origin is called centrally-symmetric if for any point (p x, p y ) belong to P, the point ( p x, p y ) also belong to P. They transform their problem to a coloring problem with wedges (a wedge is an unbounded set between two nonparallel lines). In this problem the points needs to be colored with two colors in a way so that any wedge containing at least k points contains points of both colors. They design a simple and elegant iterative coloring algorithm which colors all the points with two colors satisfying the above constraint. Then they extend their solution for general convex polygons. In this report we discuss the cover-decomposability problem for only centrally-symmetric convex polygon. The reader is referred to the original paper for the general case. Keszegh and Palvolgyi [5] consider the cover-decomposability problem for translates of an octant in R 3. They show that any 12-fold covering with translates of an octant can be decomposed into 2 coverings. An overview of their technique is as follows. They first transform the problem into a more tractable planar coloring problem with axis parallel wedges. Then they design an algorithm to construct a bipartite graph with special properties where the set of vertices is a set of planar points. The beauty of this algorithm is that any coloring of the vertices of this special bipartite graph with two colors also solves the coloring problem with wedges. In this report we discuss the transformation of the 3D-problem into the planar problem, and the algorithm which solves the latter problem. Organization of this report. The organization of this technical report is as follows. In Section 2 we discuss the cover-decomposability of any k-fold covering with translates of a centrally-symmetric convex polygon into Ω(k) coverings. In Section 3 we discuss the convex decomposabilty of any 12-fold covering with translates of an octant. 2 Optimally Decomposing Covering with Translates of a Centrally Symmetric Convex Polygon A convex polygon P with its centroid at the origin is called centrally-symmetric if for any point (p x, p y ) belong to P, the point ( p x, p y ) also belong to P. We show that any k-fold covering of any subset of 4

5 P (p j ) p j p j O i O i P (O i ) p k P (O i ) p k P (p k ) (a) (b) Figure 3: (a) P (O i ) contains p j, but not p k. (b) O i is contained in P (p j ), but not in P (p k ) the plane with translates of a centrally-symmetric convex polygon can be decomposed into Ω(k) coverings. Pach [9] showed a basic trick to solve any planar cover-decomposability problem. He show that using a dual transformation a planar cover-decomposability problem can be reduced to a coloring problem. We use this trick to reduce our problem to a coloring problem which is more convenient to deal with than solving the cover-decomposability problem directly. 2.1 Dualization et P = {P i } be a set of translates of a finite convex polygon P in R 2 and S be a set of planar points. Also, let O i be the centroid of P i. For a planar set T and a point x, let T (x) be the translate of T with centroid at x. We consider the geometric dual of the cover-decomposability problem of P. In dual version a translate P i of the convex polygon is transformed to the point O i and a point p S is transformed to P (p), where P is the reflection through p of P i (p). Note that for a centrally symmetric polygon P, P and P are same. It is obvious that p P i (x) iff x P (p) (see Figure 3). Thus if P = {P i } is a k-fold covering of S, then for each p S, P (p) contains at least k points from the set O = {O i }. Moreover, P is decomposable into f(k) coverings if and only if the points in O can be colored with f(k) colors such that for each p S, P (p) contains at least one point of each color, where f(k) is a function of k. Now, P is cover-decomposable if and only if P is. Hence we have the following lemma. emma 1. Any k-fold covering with translates of a convex polygon P is decomposable into Ω(k) coverings iff there exists a constant α 1 so that for any collection of points Q in the plane, it is possible to assign each point of Q a color from {1, 2,..., k α }, such that every translate of P containing at least k points of Q contains at least one point of each color. Henceforth, we focus on the dual coloring problem. We are given a centrally-symmetric convex polygon P and a set of points which we need to color. In the next subsection we reduce the problem involving the the polygon P to a problem involving wedges which is easier to deal with. 5

6 W P (a) (b) Figure 4: (a) A polygon P placed on the grid. (b) A wedge W of P containing a constant fraction of the points inside P, in a gridcell 2.2 Polygon to Wedges et p 0, p 1,..., p n 1 be the vertices of P in counter-clockwise order. et d be equal to half of the minimum among the distances between any two points on non-consecutive edges of P. Consider a sqaure grid of side d on the plane. For x R 2, P (x) intersects at most β grid cells, where β is a constant which depends on the number of sides of P. Thus if P (x) contains at least k points of a set S, there exists a grid cell which contains at least k β points of S (see Figure 4). For each p i consider an i-wedge obtained by extending the two sides of P incident on p i. The way the grid is laid ensures that any grid cell intersects at most two consective sides of a translate. Thus the intersection of P (x) and a grid cell is same as the intersection of the grid cell and a translate of an i-wedge. Hence for any grid cell containing at least k β points of S P (x) there exists an i, such that a translate of the i-wedge contains the same set of points. Because of these observations it is sufficient to prove the following theorem. Theorem 1. There exists a constant α 1 so that for any k 1000n.(5n) n and any collection Y of points in the plane, it is possible to assign each point in Y a color from {1, 2,..., k α }, such that every translate of an i-wedge of P that contains at least k points from Y contains at least one point of each color. From now onwards we focus on the problem involving wedges. 2.3 Coloring Algorithm for Wedges Consider any arbitrary set Y of points which we need to color. Without loss of generality we assume the points are in general position. We denote an i-wedge with apex at x by W i (x). et the load load Y (W i (x)) of the wedge W i (x) is the number of points of Y that it contains. We define a boundary for an i-wedge and a positive integer r, where i {1, 2,..., n 1}. This boundary has the property that any i-wedge with its apex inside or on the boundary has load at least r and any i-wedge with its apex outside the boundary has load strictly less than r. This boundary is called level curve (see Figure 5). et W j i = {x} be the set of apices of W i such that load Y (W i (x)) = j. The level curve C i (r) is the boundary of the region W r i = j r W j i for each i {1, 2,..., n 1}. 6

7 h i τ i W i Figure 5: An example of a level curve C i (r) with r = 2 Note that C i (r) is a monotone staircase path with edges that are parallel to the edges of an i-wedge. We have the following observations. Observation 1. For any x C i (r), r load Y (W i (x)) r + 1. Observation 2. Any i-wedge W such that load Y (W ) r contains an i-wedge whose apex belongs to C i (r). Note that two extreme edges of C i (r) are semi-infinite. One is a ray parallel to the edge p i 1 p i. We denote its origin by h i. For any point y on this ray, W i (y) Y = W i (h i ) Y. The other extreme edge is a ray parallel to the edge p i p i+1. et τ i be the origin of this ray. Similarly, for any point on this ray, W i (y) Y = W i (τ i ) Y (see Figure 5). Because of Observation 2 it is sufficient to prove Theorem 1 for the i-wedges with apex on C i (k), for each i {1, 2,..., n 1}. Now we discuss a routine color(i, Q, t) which takes as input a level curve C i (k), a subset Q of Y and a positive integer t. Also the input has the guarantee that for any i-wedge W with apex on C i (k), W Q 2t. This routine outputs a partial coloring of Q with colors {1, 2,..., t} such that any i-wedge W with apex on C i (k) (a) contains a point colored j, for 1 j t, and (b) contains at most 2t colored points. ater we ll use this routine to color all the points of Y. For any q Q, consider the set I(q) = {u C i (k) q W i (u)} of apexes of i-wedges containing q which is an interval of C i (k). The color(i, Q, t) routine chooses a subset of these intervals in each iteration. The intervals are chosen in such an order that if interval I properly contains interval I, then I is chosen before I (see Figure 6 for an illustration). In this order an interval is chosen if it covers an uncovered point of C i (k). After selecting a cover of C i (k) we repeatedly throw out intervals which are redundant, i.e, even after its deletion the current set of intervals covers all points of C i (k). The final non-reduntant set of intervals has the property that no more than two intervals cover any point of C i (k). The points of Q correspond to the chosen intervals are given color 1. This process is repeated t 1 more times after deleting the colored points from Q. Consider any i-wedge W i (x) with its apex x on C i (k). Recall that W i (x) Q 2t. As in each iteration at most two intervals cover x, at most two points of W i (x) Q can be colored in one iteration. Thus after t iterations W i (x) contains at most 2t colored points and a point colored j, for 1 j t. Thus we have the following observation. 7

8 I q q I W i Figure 6: The interval I (shown by dotted lines) corresponding to q is chosen before the interval I (shown by dashed lines) corresponding to q Observation 3. The partial cover computed by color(i, Q, t) has the property that any i-wedge with apex at C i (k) has at most 2t colored points. Moreover, if q and q are points in Q such that q W i (q ) (I(q) properly contains I(q )), then q is colored only if q is colored. Algorithm 1 1: Y Y 2: for i 0 to n 1 do 3: min{load Y (W j (x)) : x C j (k) and j = i, i + 1,..., n 1} 4: X i Y { j k W j i } 5: Run color(i, X i, 64n ). et Y i X i be the points assigned a color during this call. 6: Y Y \ Y i. The algorithm that achieves the coloring claimed in Theorem 1 is Algorithm 1. Y is the set of uncolored points which is initially set to be Y. For each i = 0 to n 1, this algorithm computes the minimum load among the loads of j-wedges with apex on C j (k), for j i. X i is the set of uncolored points of Y which are lying outside or on the level curve C i (k). The algorithm then calls the routine color(i, X i, 64n ). The points which are colored by this call are deleted from Y. After the i th iteration any i-wedge with apex on C i (k) contains points colored 1, 2,..., 64n. This is Ω(k) colors if Ω(k). In the remaining of this section we establish that. We show that, which is at least k before 0 th iteration drops by at most a constant factor (O(n)) with each iteration. For two distinct points q and q, if W i (q) W i (q ), we say that q dominates q. If q and q both are uncolored before iteration i, then by Observation 3 q is colored in iteration i only if q is already colored. Recall that Y i is the set of points which have been assigned a color in iteration i. From Observation 3 and the fact that in each iteration only 64n colors are used we have the following observation. Observation 4. For any x C i (k), load Yi (W i (x)) 32n. Consider any centrally-symmetric convex polygon P. A line through a vertex of P is a tangent if it intersects P only at that vertex. Now consider any two vertices p i and p j of P. If two parallel tangents can 8

9 p i p j Figure 7: p j is the antipodal vertex of p i in the polygon. All other vertices are non-antipodal with respect to p i be drawn through p i and p j so that P lies in between those two tangents, then p i and p j are called antipodal vertices. Otherwise they are called non-antipodal. As P is centrally-symmetric for each vertex v = (x, y) there is exactly one antipodal vertex ( x, y). The other vertices are non-antipodal with respect to v (see Figure 7). Consider a j-wedge W with apex at x. If p i and p j are antipodal, (1) every line that is tangent to W i (x) is also a tangent to W and (2) W and W i (x) are on opposite sides of these tangents (see Figure 8). W is called antipodal with respect to an i-wedge if p i and p j are antipodal. Otherwise W is called non-antipodal with respect to an i-wedge. W W i (x) Figure 8: A line tangent to W i (x) must lie in the shaded region (e.g., the dotted line). By definition W is antipodal with respect to the i-wedge W i (x) Intuitively if in iteration i < j too many points get colored, then in iteration j some j-wedge with apex on C j (k) might not contain Ω(k) points of X j. We show that this cannot be the case. To be precise we show that each wedge W of j-type (j i) have at most a constant factor of its uncolored points assigned a color in iteration i of Algorithm 1. Note that there are two cases: W is non-antipodal with respect to an i-wedge or antipodal with respect to an i-wedge. Corresponding to these two cases we have two lemmas. The first lemma handles the former case. emma 2. Suppose W is a non-antipodal wedge with respect to an i-wedge and load Xi (W ) 6. At the end of i th iteration of Algorithm 1, W has load at least 5n from points in Y. The following lemma consider the case that W is antipodal with respect to an i-wedge. emma 3. Suppose W is an antipodal wedge with respect to an i-wedge and load Xi (W ) 6. At the end of i th iteration of Algorithm 1, W has load at least 5n from points in Y. 9

10 For the time being we assume that these two lemmas hold. We use these lemmas to complete the proof of Theorem Proof of Theorem 1 Proof. In iteration i, we color the points in X i so that each i-wedge contains points of 64n different colors. This gives Ω(k) different colors as long as is Ω(k) prior to iteration i of the algorithm. Consider any j-wedge W, for j > i. We show that in every iteration the load of W of uncolored points, which is at least, falls by at most a constant (5n) factor. Suppose load Xi (W ) < 6 then, W contains at least 6 uncolored points after iteration i. Therefore consider the case where load Xi (W ) 6. If W is non-antipodal with respect to an i-wedge we invoke emma 2. If W is antipodal with respect to an i-wedge we invoke emma 3. In both of the cases the load fall by at most a factor of 5n in iteration i. As k 1000n.(5n) n, when we reach iteration j of the algorithm W still contains Ω(k) uncolored points. Hence after iteration j, W contains points of all the Ω(k) colors which completes the proof of the theorem. Now we prove emma Proof of emma 2 Proof. et x be the apex of W. Now there could be two cases (a) W C i (k) = φ, and (b) W C i (k) φ. For the former case to be true one of the halflines of W must be parallel with a halfline of an i-wedge as W is non-antipodal with respect to an i-wedge. Without loss of generality we assume that W has a halfline parallel to the side p i p i+1 of P (see Figure 9). Then W X i W i (τ i ) by the definition of τ i. We thus have 32n load Yi (W ) load Yi (W i (τ i )) by Observation 4. Hence, the load of uncolored points in W after iteration i is at least 6 32n > 5n. So consider the second case W C i (k) φ. Say W is a j-wedge. Here we have two subcases to consider: when walking counter-clockwise around P (i) we encounter p j after p i and before the vertices antipodal to p i, and (ii) we encounter p j after the vertices antipodal to p i and before p i. C i (k) W i (τ i ) x τ i W Figure 9: Illustration for the proof of emma 2 (Case (a)): the shaded region does not contain any point We focus on the first case, since the other is symmetric. et z be the point where W intersects with C i (k) (to be precise the last point of intersection as one walks clockwise around the boundary of W ). If W does 10

11 not contain τ i, then W Y i W i (z) Y i (see Figure 10(a)). Thus load Yi (W ) load Yi (W i (z)) Thus the load of uncolored points in W after iteration i is at least 6 32n > 5n. et us therefore assume that W contains τ i in its interior (see Figure 10(b)). et a denote the intersection point of the boundaries of W i (z) and W i (τ i ). If load Xi (W i (a)) 8n, then since load Y i (W i (a)) load Yi (W i (τ i )) 32n, there are uncolored points in W i(a) after iteration i. Since any point in W i (a) dominates points in W Y i that are not contained in W i (z) W i (τ i ), by Observation 3 W Y i (W i (z) W i (τ i )) Y i. Thus, load Yi (W ) load Yi (W i (z)) + load Yi (W i (τ i )) Thus there must be at least 6 16n > 5n uncolored points left in W. C i (k) 16n 32n. W i (z) z τ i x z W i (z) x a τ i C i (k) W W W i (τ i ) (a) (b) Figure 10: (a) W does not contain τ i. (b) W contains τ i in its interior et us therefore consider the case where load Xi (W i (a)) < 8n. Now the way is defined, load X i (W i (y)), for y C i (k). Thus load Xi (W i (τ i ) \ W i (a)) > 8n > 2. Also, load Y i (W i (τ i ) \ W i (a)) load Yi (W i (τ i ) 32n. This means the load of the points in W i(τ i ) \ W i (a) that are uncolored after iteration i is at least 2 32n > 5n. As W i(τ i ) \ W i (a) W the load of the uncolored points left in W is at least, which completes the proof of this lemma. 5n astly, we prove emma Proof of emma 3 Proof. et x be the apex of W. We start with the case when both the halflines of W intersect C i (k). Consider any point z W C i (k). et a z (resp. b z ) be the leftmost (resp. rightmost ) point where the boundaries of W and W i (z) intersect (see Figure 11(a)). Suppose R z be the quadrilateral with vertices a z, x, b z, z, i.e, R z = W W i (z). Suppose that load Xi (R z ) 5n + 32n. By Observation 4, load Y i (W i (z)) and since R z W i (z), R z contains load of at least 5n of uncolored points. Since R z W, same holds for W also n,

12 z 1 W i (z) a z x R z z b z τ i W W i (z ) a z x z b z z 2 W (a) (b) Figure 11: (a) The region R z. (b) The constructed point z et us therefore consider the case when load Xi (R z ) < 5n + 32n for all z W C i(k). Since load Xi (W i (z)), we have that load Xi (W i (a z )) load Xi (W i (b z )) 5n 32n > 8. et z 1 and z 2 be the leftmost and rightmost points on C i (k) W respectively. et us define the points a z1 and b z2 in a way similar to how we defined a z and b z before. Notably, in this case R z1 and R z2 are degenerate rectangles (line segments). Notice that a z1 = z 1 and thus load Xi (W i (a z1 )). Similarly, load Xi (W i (b z2 )). Thus for each point z on C i (k) while walking from z 1 to z 2 the value load Xi (W i (a z )) decreases monotonically by 1 at a time. et z be the last point on C i (k), while traversing from z 1 to z 2, such that load Xi (W i (a z )) 16 (see Figure 11(b)). We note that load Xi (W i (a z )) Thus load Xi (W i (b z )) load Xi (W i (a z ) W i (b z )) load Xi (W i (a z )) > 8 ( ) 16 Now consider any point z W \W i (z ). It must be that W i (z ) contains either W i (a z ) or W i (b z ) both of which have load at least 16. Without loss of generality we assume W i(z ) contains W i (a z ). The points in W i (a z ) dominates z and z can be colored after coloring of all these points. But since load Yi (W i (a z )) load Yi (W i (z )) 32n < 16 load X i (W i (a z )), z is not colored in iteration i. It follows that W Y i W i (z ) Y i, and thus load Yi (W ) load Yi (W i (z )) 32n. And so the load of the uncolored points in W is at least 5n. We now consider the case when at least one halfline of W do not intersect C i (k). We show that we can find a wedge W such that both of its halflines intersect C i (k), W Y = W Y, and W is antipodal with respect to an i-wedge. Then we can use the previous argument to show W contains Ω() uncolored points after iteration i, which implies that W also contains Ω() uncolored points. We place a wedge identical to W just behind of it so that the new wedge contains exactly same set of points from Y like W (see Figure 12). We bend in the halflines of the wedge just enough so that the wedge still contains the same set of points as W and the halflines are no longer parallel to the edges of C i (k). This is our wedge W. Both halflines now intersect C i (k) and W is antipodal with respect to an i-wedge. This completes the proof of the lemma. 12

13 h i τ i C i (k) x W 3 Cover Decomposability of Octants Figure 12: The first step in constructing W et W 1 be the octant with apex at the origin which contains the point (,, ). We prove that W 1 is cover-decomposable, i.e., any 12-fold covering of any subset of R 3 with a finite number of translates of W 1 can be decomposed into two coverings. Note that as any octant can be transformed to W 1 using rotation and translation this proves that any octant is cover-decomposable. Instead of proving this statement directly we consider the dual setting. Each point (x, y, z) in the primal space is transformed to the octant with apex (x, y, z) containing the point (,, ) in the dual space. Each translate of W 1 with apex (p, q, r) in the primal space is transformed to the point (p, q, r) in the dual space. Suppose a translate of W 1 with apex (p, q, r) in the primal space contains a point (x, y, z). Then in the dual space the transformed octant with apex (x, y, z) containing (,, ) contains the transformed point (p, q, r). Thus if in the primal space a point is covered by k octants then in dual space the corresponding octant contains k points. Now suppose in dual space the points can be colored with two colors such that any octant with at least 12 points contains points of both colors. Then in primal space it is possible to color the octants using two colors such that any point covered by at least 12 octants is contained in octants of both colors. Thus based on the color assigned to the octants we can decompose the set of octants into two parts each of which is a 12-fold covering. Thus it is sufficient to prove the following theorem. Theorem 2. Any finite set of points in R 3 can be colored using two colors such that any translate of a given octant with at least 12 points contains points of both colors. We consider the octant W with apex at origin containing (,, ). Without loss of generality we ll prove Theorem 2 for translates of W. To be precise we show that the problem in R 3 can be transformed into a more tractable planar problem. et P be a finite set of points in R 3. For simplicity, assume that all the z-coordinates of these points are different (otherwise, it is possible to perturb the points). et p i be the point in P with i th smallest z- coordinates. Also let P i = {p 1,..., p i }. Denote the projection of P on the z = 0 plane by P. Similarly denote the projection of the point p i by p i, the projection of W by W and let P i = {p 1,..., p i }. Now we prove the following lemma. emma 4. The following two statements are equivalent: i. P can be colored using two colors so that for any i and any translate of W containing at least 12 points of P i, it is true that the intersection of the translate and P i contains points of both colors 13

14 ii. P can be colored using two colors so that any translate of W with at least 12 points of P contains points of both colors. Proof. Suppose such a coloring of P is possible. We assign the same color to the corresponding points in P. Now consider any translate of W containing at least 12 points of P. et p t be the point in the translate with maximum z-coordinate. Then the points contained in W is a subset of P t. Thus the quadrant corresponding to the translate of W contains at least 12 points of P t and the coloring of P ensures that the intersection of the quadrant and P t contains both colors. So the intersection of the translate of W and P t contains both colors. Hence the translate of W also contains points of both colors. Now say such a coloring of P exists. We use the same color for the corresponding points in P. Consider any P i and a translate of W such that the translate contains at least 12 points of P i. Also consider an octant such that its projection on z = 0 plane is the translate of W, it contains same points of P i like the translate of W, and it does not contain any point of P \P i. The existence of such an octant follows from the existence of the translate of W which contains at least 12 points of P i. Note that as the octant contains at least 12 points of P i P the coloring of P ensures that the octant contains points of both colors. Thus the corresponding translate of W also contains points of both colors which completes the proof of the lemma. Henceforth we call such a coloring of planar point set P a good coloring. By emma 4 it follows that to prove Theorem 2 it is sufficent to show the existence of a good coloring of P. To avoid intricate notations we drop the apostrophe from all the symbols. Thus we want to find a good coloring of a planar point set P so that for any i and for any translate of the quadrant W (with apex at origin and contains (, )) with at least 12 points of P i contains points of both colors. One way to visualise this problem is to consider an online setting. In every iteration i we have a point set P i, and we already have a good coloring of P i. The goal is to color the next coming point p i+1 in a way so that it gives a good coloring of P i+1. To get a good coloring of P we construct a bipartite graph recursively with some special properties. As we know that any bipartite graph is 2-colorable we ll show that any such two coloring is a good coloring of P. Now we introduce some notations to set up the stage. For each i we construct the bipartite graph G i recursively on the point set P i, starting with the graph containing just one vertex p 1. We also maintain a staircase S i, a set of pairwise incomparable points in every step i. Thus, before the i th step we have a bipartite graph G i 1 on P i 1 with some special properties and a staircase S i 1. In i th step we construct a new graph G i having those special properties and a new staircase S i using G i 1 and S i 1. We ensure that the edge set of G i 1 is a subset of edge set of G i. Now we introduce some defintions. We refer to a translate of the quadrant W as a wedge. At any step i, a point p is good if any wedge containing p also contains an edge of G i. Thus any wedge containing p contains two points of different colors in the final coloring. At step i, consider the staircase S i. We order the points in S i with respect to their x-coordinates. Two points of S i are called neighbors if they are consecutive in this order. A point p in S i is called almost good if any wedge containing p and its neighbors contains two points of P i which are connected by an edge of G i. If p has only one neighbor p, then p to be almost good, any wedge containing p and p must contain an edge of G i. Note that the good points and neighbor of a good point are always almost good. For any planar point p, let p x and p y be its x and y-coordinate respectively. Consider two points p and q. If p x < q x but p y > q y we say p is NW (North-West) from q and q is SE (South-East) from p. In this case 14

15 the pair of points is called incomparable. Similarly, if p x < q x and p y < q y we say p is SW from q and q is NE from p. In this case we say q dominates p. We say a point p P i is above the staircase S i if there exists a point q in S i such that p dominates q. If p is not above or on the staircase, then we say p is below the staircase. Now we state the properties which are maintained in each step. At any step j: 1. Every point above the staircase is good. 2. Every point of the staircase is almost good. 3. All the points below the staircase are pairwise incomparable. 4. If a wedge only contains the points below the staircase then it contains at most 3 points. et E i be the edge set of G i. For i = 1, S 1 = {p 1 } and E 1 = φ. Hence these properties are vacuously satisfied. Suppose these properties hold at time i 1 1. Now we process the point p i using the following algorithm maintaining all the properties. Algorithm Step i Set G = G i 1, E = E i 1 and S = S i 1 Step (a) If p i is above the staircase S i 1, do the following. Otherwise, skip to Step (b). Set S i = S i 1. As p i is above S i there is a point q S i so that p i dominates q. Add an edge between p i and q. Set E i = E i 1 {(p i, q)}. Thus any wedge containing p i always contains the edge (p i, q) and by induction all the properties are maintained. Terminate the algorithm. Note that we proceed further if and only if p i is below S i 1. Step (b) If there are two points p and q below S so that p dominates q, do the following. Otherwise, skip to Step (c). Because of Property 3 one of p or q is p i and there is no other point r below the staircase so that r dominates p. et N be the set of points of S which dominates p. Now define the new staircase S = (S \ N) {p}. Thus all the points in the staircase remain pairwise incomparable. Add an edge between p and q. Thus E = E {(p, q)}. By induction all the points above the old staircase are good. The only other points above S are the points in N. Now any wedge which contain a point p of N must contain p as p dominates p (see Figure 13). Thus that wedge also contains the edge (p, q) and p is good. Hence Property 1 is satisfied. Now the points in S except p and its neighbor(s) are almost good by induction as their neighbor(s) remain unchanged. Now as p is a good point, p and its neighbor(s) are almost good. Thus Property 2 is also satisfied. Repeat Step (b) until Property 3 is satisfied, then proceed to Step (c). 15

16 p p p i q q Figure 13: Repeated applications of Step (b) of the algorithm Step (c) If there exists 4 points (pairwise incomparable) below the staircase S such that there is a wedge V such that V contains these 4 points but no points of S, do the following. Otherwise, skip to Step (d). Denote these 4 points by q 1, q 2, q 3 and q 4 in increasing order of their x-coordinate. et M be the set of points in S so that each of them dominates either q 2 or q 3. Update S to (S \ M) {q 2, q 3 }. Add the edges between q 1 and q 2, and q 3 and q 4. Thus the new set of edges E = E {(q 1, q 2 ), (q 3, q 4 )}. By induction all the points above the old staircase are good. The only other points above S are the points in M. Now any wedge V which contain a point q of M must contain either q 2 or q 3 as q dominates q 2 or q 3 (see Figure 14 (a)). Now we claim that V also contains q 1 or q 4. If not then as V contains all the 4 points V must contain V (see Figure 14 (b)). Thus V contains the point q of old staircase which is a contradiction. Thus V contains at least one of the edges {(q 1, q 2 ), (q 3, q 4 )} and thus q is good. Hence Property 1 holds. Property 2 is true by induction for all the points in S except q 2, q 3 and their neighbor(s). Now the wedge containing q 2 and its neighbor in S other than q 3 contains the edge (q 1, q 2 ) and thus they are almost good. Similarly, q 3 and its neighbor in S other than q 2 are also almost good. Thus Property 2 also holds. Note that Property 3 is already satisfied in Step (b). Repeat Step (c) until Property 4 holds, then proceed to Step (d). q 1 q 2 q 1 q 2 q 1 q 2 V V q 3 q 4 q 3 q 4 q 3 q 4 (a) (b) Figure 14: Application of Step (c) of the algorithm. Step (d) Set S i = S, G i = G. Terminate the algorithm. It s not hard to see that after step i also all the four properties are maintained. Note that G n might be disconnected. For example if P consists of just three incomparable points, G n is a graph with three isolated vertices as no edge is added by the algorithm in this case. Now we prove that G n is a bipartite graph and to be precise we show that it is a collection of trees. 16

17 V 1 V 2 V V 3 Figure 15: Any monochromatic wedge can contain at most 11 points. emma 5. The final graph G n is a forest. Proof. Note that it is sufficient to show that G n does not contain any cycle. For the sake of contradiction suppose that there is a cycle C in G n. et e = (p, q) be the edge of C which is added at first before adding any other edge of C. et G be the graph obtained by addition of e at t th iteration. Then G does not contain any other edge of C. Now there could be three cases, e is added at Step (a) or (b) or (c) of the algorithm. Say e is added at Step (a). Then one of p or q has to be p t, the point added at iteration t. Without loss of generality let p = p t. As p is added for the first time in this iteration there is no existing edge between p and any other vertex of G. Now note that p is a point above the current staircase S and thus no further edge is added which has p as an endpoint. Thus p and hence (p, q) can never be part of any cycle which is a contradiction. Now say e is added at Step (b). Then before addition of e, p and q are below the current staircase and both are comparable. Without loss of generality say p dominates q and thus p is added to the staircase. Now the only case where an edge is added such that one of its endpoints is p is that the other endpoint is a new added point, say p l which lies above the current staircase. But, as no other edges can be added such that p l is one of its endpoints p cannot be part of a new cycle and thus (p, q) also cannot be a member of any cycle which is a contradiction. astly, assume that e is added at Step (c). Then without loss of generality let p is added to the staircase. Thus in this case also no further edges can be added such that one of its endpoints is p and it is a part of a cycle. Hence (p, q) cannot be part of any cycle which is again a contradiction. Thus in all of the cases we are getting contradiction which completes the proof of the lemma. From emma 5 it follows that G n is two colorable. Now we have the following lemma. emma 6. Any two coloring of G n is a good coloring of P. Proof. Consider any two coloring of vertices in G n. Also consider any wedge V that contains at least 12 points of P i. If V contains a point above S i, then Property 1 ensures that it contains points of both colors. Now say it does not contain such a point. If it contains at least three points from the staircase (consecutive), by applying Property 2 to the middle point it follows that V contains points of both colors. Then say V contains at most two points of S i. Then we can found three wedges whose union covers all the points in P i V such that no wedge contains a point of S i (see Figure 15). By Property 4 each such wedge can contain 17

18 at most three points of P i. Thus in total V contains at most = 11 points which is a contradiction. Thus the latter case cannot occur and this completes the proof of the lemma. emma 6 completes the proof of Theorem 2. 4 Conclusion In this technical report we summarize the results of two papers on cover-decomposability: one by Gibson and Varadarajan [4] and the other by Keszegh and Palvolgyi [5]. The main result of the first paper is an algorithm which decomposes any k-fold covering of any subset of R d, with translates of a given geometric set, into Ω(k) coverings of the same subset. By duality, this also shows that any set of points in R d can be colored by Ω(k) colors such that any translate of a given geometric set containing at least k points contains at least one point of each color. Recently, Kovacs [8] show that for any convex polygon P with at least four sides and k > 0, there exists a k-fold covering of the plane with homothetic copies of P that cannot be decomposed into two coverings. But as in case of homothets the dual problem is not equivalent to the primal one this doesn t say anything about the dual version of the problem. Hence the following problem still remains unsolved. Given a geometric set P and any set of points S in R d, is it possible to color the points in S using two colors such that any homothet of P containing at least m points contains at least one point of each color, for some constant m? The second paper show that the cover-decomposability constant for octants in R 3 is at most 12. On the other hand, they show a counterexample proving that this constant is at least 4. Thus it is interesting to study up to what extent this gap can be reduced. References [1] Greg Aloupis, Jean Cardinal, Sébastien Collette, Stefan angerman, David Orden, and Pedro Ramos. Decomposition of multiple coverings into more parts. Discrete & Computational Geometry, 44(3): , [2] Jean Cardinal, Kolja B. Knauer, Piotr Micek, and Torsten Ueckerdt. Making triangles colorful. JoCG, 4(1): , [3] Jean Cardinal, Kolja B. Knauer, Piotr Micek, and Torsten Ueckerdt. Making octants colorful and related covering decomposition problems. In SODA, pages , [4] Matt Gibson and Kasturi R. Varadarajan. Optimally decomposing coverings with translates of a convex polygon. Discrete & Computational Geometry, 46(2): , [5] Balázs Keszegh and Dömötör Pálvölgyi. Octants are cover-decomposable. Discrete & Computational Geometry, 47(3): , [6] Balázs Keszegh and Dömötör Pálvölgyi. Convex polygons are self-coverable. CoRR, abs/ ,

19 [7] Balázs Keszegh and Dömötör Pálvölgyi. Octants are cover-decomposable into many coverings. Comput. Geom., 47(5): , [8] István Kovács. Indecomposable coverings with homothetic polygons. CoRR, abs/ , [9] János Pach. Covering the plane with convex polygons. Discrete & Computational Geometry, 1:73 81, [10] János Pach, Gábor Tardos, and Géza Tóth. Indecomposable coverings. In CJCDGCGT, pages , [11] János Pach and Géza Tóth. Decomposition of multiple coverings into many parts. Comput. Geom., 42(2): , [12] János Pach. Decomposition of multiple packing and covering. In 2. Kolloquium über Diskrete Geometrie, pages , Salzburg, Institut für Mathematik der Universität Salzburg. Professor Pach s number: [009]. [13] Dömötör Pálvölgyi. Indecomposable coverings with concave polygons. Discrete & Computational Geometry, 44(3): , [14] Dömötör Pálvölgyi. Indecomposable coverings with unit discs. CoRR, abs/ , [15] Dömötör Pálvölgyi and Géza Tóth. Convex polygons are cover-decomposable. Discrete & Computational Geometry, 43(3): , [16] Gábor Tardos and Géza Tóth. Multiple coverings of the plane with triangles. Discrete & Computational Geometry, 38(2): ,