Counting Blanks in Polygonal Arrangements
CCounting Blanks in Polygonal Arrangements
Arseniy Akopyan ∗ Erel Segal-Halevi † September 24, 2018
Abstract
Inside a two dimensional region (“cake”), there are m non-overlapping tiles of acertain kind (“toppings”). We want to expand the toppings while keeping them non-overlapping, and possibly add some blank pieces of the same “certain kind”, such thatthe entire cake is covered. How many blanks must we add?We study this question in several cases: (1) The cake and toppings are general poly-gons. (2) The cake and toppings are convex figures. (3) The cake and toppings areaxes-parallel rectangles. (4) The cake is an axes-parallel rectilinear polygon and thetoppings are axes-parallel rectangles. In all four cases, we provide tight bounds on thenumber of blanks. Introduction
Consider a two-dimensional cake C with m non-overlapping toppings Z , Z , . . . , Z m . We wantto cut the cake without harming the toppings. I.e, we want to partition the entire cake tonon-overlapping pieces Z (cid:48) , Z (cid:48) , . . . , such that each topping is contained in a piece and eachpiece contains at most a single topping. There are some geometric constraints on the pieces,e.g, they should be polygonal or convex or rectangular. This might require us to add some“blanks” — pieces with no topping. For example, in the rectangular cake at the left of Figure1 there are m = 4 rectangular toppings. If the pieces must be axes-parallel rectangles, thenwe will need to add at least one blank, denoted at the right by Z (cid:48) . Given m and the geometricconstraint on the pieces, how many blanks must we add in the worst case (for the worst initialarrangement of toppings)?Besides cutting cakes, an additional application of this question is for re-division of land[6]. There are some small lots on a piece of land. The owners have built on their initial lotsand need to keep them. They would like to expand their lots and then fill the land withadditional lots — but not too many additional lots. All lots are required to have a “nice”geometric shape. How may additional lots do they need to add in the worst case?This paper answers this question under three different geometric constraints on the top-pings and pieces: polygonality, convexity and rectangularity. In all these cases, we prove thatfor any initial arrangement of toppings there exists a maximal expansion — where the top-pings expand inside the cake until they cannot expand any further while keeping the geometric ∗ Institute of Science and Technology Austria (IST Austria), Am Campus 1, 3400 Klosterneuburg, [email protected] † Ariel University (Ariel 40700) and Bar-Ilan University (Ramat-Gan 52900), Israel. [email protected] a r X i v : . [ c s . C G ] A p r Z Z Z Z Z (cid:48) Z Z (cid:48) Z Z (cid:48) Z Z (cid:48) Z (cid:48) Figure 1:
Left: a rectangular cake with 4 rectangular toppings.
Right: an expansion of the toppings to 4 larger pieces and a fifth blank piece.constraints. Since we are interested in a worst-case upper bound, it is sufficient to considersuch maximal arrangements, since the initial arrangement might already be maximal. So ourquestion becomes: how many blanks can there be in a maximal arrangement of pieces? Weanswer this question in four cases. The cake and toppings are polygons. Then in any maximal arrangement, the entirecake is covered — there are no blanks. The cake and toppings are convex figures. Then in any maximal arrangement, theuncovered spaces are all convex, and there are at most 2 m − The cake and toppings are axes-parallel rectangles, as in Figure 1. Then in any maximalarrangement, all uncovered spaces are axes-parallel rectangles, and there are at most m −(cid:100) √ m − (cid:101) of them The toppings are still axes-parallel rectangles, but the cake may be an arbitrary simply-connected axes-parallel polygon. Let T be the number of reflex vertices (270-degree interiorangles) in the cake. Then in any maximal arrangement, the remaining uncovered spaces canbe partitioned into b axes-parallel rectangles, where b ≤ m − (cid:100) √ m − (cid:101) + T .All the results are tight in the following sense: In each case, for every m (and T ), there isan explicit construction where the number of blanks equals the bound.In addition, for cases 2 and 3 we consider a related question. Suppose the toppings liein the unbounded plane R , and we do not expand them. We define a hole as a connectedcomponent of the plane that remains outside of toppings (a hole can be unbounded). Howmany holes can there be? We prove the following answers. In any arrangement of m pairwise-disjoint convex figures in the plane, the number ofholes is at most 2 m − In any arrangement of m pairwise-disjoint axes-parallel rectangles in the plane, thenumber of holes is at most m − k ∈ { , , , } , case k is handled in Section k .In each case, the proof consists of two steps. First, we show that any initial arrangementof toppings has a maximal expansion. Then, we prove that in any maximal arrangement, thenumber of blanks is bounded as claimed. This method works in cases 1–4, but it may notwork in other cases; some counter-examples are discussed in Section 5.2igure 2: Left: a cake with three simply-connected topping.
Right:
Tubes connecting the toppings to each other and to the cake boundary.
In this warm-up section, the cake and each of the m toppings is a polygon. Initially, we allowpolygons to be not simply-connected, in contrast to the classic definition. We prove that inthis case it is possible to expand the toppings such that there will be no blanks. Theorem 1.1.
Let C (“cake”) be a polygon and Z , . . . , Z m (“toppings”) be pairwise-disjointpolygons contained in C . Then there exists a partition of C into polygons, C = Z (cid:48) (cid:116) · · · (cid:116) Z (cid:48) m , where Z i ⊆ Z (cid:48) i for all i ≤ m .Proof. The new partition can be created by the following procedure.Let C ∗ be the cake outside the toppings, C ∗ := C \ (cid:83) mi =1 Z i . Since C and the Z i are allpolygons, C ∗ is a union of a finite number of polygons, say: C ∗ = H (cid:116) · · · (cid:116) H n , where thesides of each H j are made of subsets of the sides of C and the sides of the toppings. Moreover,every H j must have at least one side that overlaps with a side of some topping Z i (since thecake itself is connected). So Z i ∪ H j is a polygon. Replace Z i with Z i ∪ H j and repeat theprocedure for the remaining components of C ∗ .Note: Theorem 1.1 implies that, if the initial arrangement of toppings is any maximalarrangement, then it has no blanks.While Theorem 1.1 is easy to prove, it is not so easy to extend to toppings that areconnected but not polygonal. A counter-example is given in Section 5.On the other hand, Theorem 1.1 can be extended to the case where the cake and thetoppings are simply-connected polygons . Initially, add to each topping thin polygonal tubesthat connect it to the cake boundary and to each of the other toppings, as long is it is possibleto do so without overlapping other toppings (see Figure 2). Then proceed as in Theorem 1.1:add each component of C ∗ to an adjacent topping. The tubes guarantee that the toppingsremain simply-connected. In this section, the cake and each of the m toppings is a convex figure . The symbol (cid:116) denotes union of pairwise-interior-disjoint sets.
Left: a tiling of the plane by hexagons.
Right: a modification of the tiling where near each vertex there is a blank.
Theorem 2.1.
Let C (“cake”) be a convex figure and Z , . . . , Z m (“toppings”) be pairwise-disjoint convex figures in C , where m ≥ . There exists a partition of C into m + b convexfigures, C = Z (cid:48) (cid:116) · · · (cid:116) Z (cid:48) m + b , where Z i ⊆ Z (cid:48) i for all i ≤ m , and b ≤ m − . Moreover, forevery m there exists an arrangement of m toppings where in every such partition, b = 2 m − . The techniques and example in the proof are very similar to [1]. We first prove the secondpart of the theorem by showing an arrangement for which b = 2 m − To get intuition for the lower bound construction, consider the tiling of the plane withhexagons, shown in Figure 3/Left. Each point in which 3 hexagons meet is called a ver-tex . By slightly moving the hexagons, it is possible to create, near each vertex, a triangularblank, as is shown in Figure 3/Right. No hexagon can be expanded towards an adjacentblank while remaining convex and disjoint from the other hexagons (i.e, the arrangement ismaximal). Each blank touches three hexagons and each hexagon touches six blanks. Hence,the number of blanks is asymptotically twice the number of hexagons, which gives an initialapproximation b ≈ m . In a finite tiling, there are boundary conditions. For example, whenthe tiling is contained in a square cake, there are Θ( √ m ) triangles near the boundary of C .These can be discarded or joined with nearby toppings, so the total number of blanks is only2 m − Θ( √ m ).Figure 4 shows a more sophisticated arrangement of toppings that has only a constant num-ber of blanks in the boundary. The arrangement is based on a construction of Edelsbrunner,Robison and Shen [1]. For every integer k ≥
0, the toppings are: • • k medium intermediate hexagons; • m mod 3 is 0 or 1 or 2.Near the vertices of the toppings, there are triangular blanks. Note that the arrangement is maximal — there is no way to expand any topping into any blank. Therefore the final partitionconsists exactly of the m toppings and b blanks shown in the figure. We now calculate b . Eachblank is triangular so it is adjacent to 3 toppings, so below each blank is counted three times: • Each external hexagon is adjacent to 3 blanks, for a total of 9;4
Each intermediate hexagon is adjacent to 6 blanks, for a total of 18 k ; • Regarding the inner polygons: – When m mod 3 = 0 — there are three inner quadrangles, which are adjacent to3 · – When m mod 3 = 1 — there are one triangle and three pentagons, which areadjacent to 3 + 3 · – When m mod 3 = 2 — there are a triangle, a quadrangle, a pentagon and 2hexagons, which are adjacent to 3 + 4 + 5 + 2 · ∀ k ≥ • m = 6 + 3 k and b = (9 + 18 k + 12) / k = 2 m − • m = 7 + 3 k and b = (9 + 18 k + 18) / k = 2 m − • m = 8 + 3 k and b = (9 + 18 k + 24) / k = 2 m − • m = 3 and b = 1 = 2 m − • m = 4 and b = 3 = 2 m − • m = 5 and b = 5 = 2 m − m −
5, as claimed.
For the upper bound we first prove a theorem for an unbounded cake.
Theorem 2.2.
Let Z , . . . , Z m , with m ≥ , be pairwise-interior-disjoint convex figures in theplane. Define a hole as a connected component of R \ ∪ mi =1 Z i . Then there are at most m − holes. This bound is tight.Proof. Define a graph G ( V, E ) where the vertices are the m toppings and there is an edgebetween each two toppings whose boundaries meet. Since the toppings are convex, every twotoppings meet at most once (in a single point or segment), so G is planar and simple. Eachhole is a face of G . Therefore, it is sufficient to prove that | F | ≤ | V | −
4, where F is the setof faces of the graph.We can assume w.l.o.g. that G is connected; otherwise we can just add edges betweenconnected components of G , since this does not change V or F . Therefore, by Euler’s formula: | V | − | E | + | F | = 2. Every edge touches at most 2 faces. Every interior face touches at least3 edges. In a connected planar graph with at least 3 vertices, the exterior face too touches atleast 3 edges. Therefore: 2 | E | ≥ | F | . Substituting in Euler’s formula gives the result.Tightness is proved by the same example of the previous subsection, adding 1 for theunbounded region outside the cake. 5igure 4: Top:
A maximal arrangement of hexagonal toppings in a triangle cake.
Bottom: m mod 3.
6e now return to proving the upper bound in Theorem 2.1. First, we expand the toppingsinside the cake C by the following procedure. Pick an arbitrary topping, say Z . Among allconvex figures in C containing Z and not overlapping any other topping, choose one thatis inclusion-maximal. Replace Z with this maximal element. Repeat the procedure for theremaining toppings.This procedure results in a maximal arrangement — an arrangement where no topping canbe expanded further while remaining convex and disjoint from the others. Pinchasi [5, Claim2] proved that, in an maximal arrangement, every hole inside the cake is a convex polygonand does not have a common boundary with the cake. By Theorem 2.2, the total number of holes inside the cake, plus the entire unboundedregion outside the cake, is at most 2 m −
4. Therefore, the total number of bounded holesinside the cake is at most 2 m −
5. Since each hole is convex, we can make each hole a singleblank piece in the final partition, and get b ≤ m − In this section, the cake and each of the m toppings is an axes-parallel rectangle. Theorem 3.1.
Let C (“cake”) be an axes-parallel rectangle and Z , . . . , Z m (“toppings”) bepairwise-disjoint axes-parallel rectangles in C . There exists a partition of C into m + b axes-parallel rectangles, C = Z (cid:48) (cid:116) · · · (cid:116) Z (cid:48) m + b , where Z i ⊆ Z (cid:48) i for all i ≤ m , and b ≤ m − (cid:100) √ m − (cid:101) .Moreover, for every m there exists an arrangement where in all partitions b = m − (cid:100) √ m − (cid:101) . In the proofs below we will use the following property of the function (cid:100) √ m − (cid:101) : (cid:100) √ m − (cid:101) = (cid:40) k k < m ≤ k + k k + 1 k + k < m ≤ ( k + 1) . Consider the tiling in Figure 5/Left. Here, four squares meet near each vertex. By movingthem slightly, we get the arrangement in Figure 5/Middle, where near each vertex there is asquare blank. Each blank touches four squares and each square touches four blanks. Hence,the number of squares and blanks is asymptotically the same.When the tiling is finite, there are boundary conditions. Suppose the cake is a square andit is tiled by m = ( k + 1) smaller squares in a grid of k + 1 times k + 1. Then, the number ofinner vertices, that can be converted to blanks, is k = m − (2 k + 1) = m − (cid:100) √ m − (cid:101) . SeeFigure 5/Right.When m is not a square number, e.g. m = k + t for some t >
0, the lower bound can beconstructed by gluing k squares at the right and then k + 1 squares at the top of a k × k tiling.Every glued square adds a 4-vertex that can be converted to a blank, except the first one at the The existence of such maximal element can be proved based on the Kuratowski–Zorn lemma. The proofis straightforward and we omit it. See Lemma 3.3 in [4]. An alternative proof of this fact follows from the proof for the case of rectangle cake and rectangle piecesin Subsection 3.2. It is proved there that every hole must be inner and have only convex vertices; the sameproof is valid for the case of convex cake and convex pieces. We are grateful to an anonymous referee for suggesting the form (cid:100) √ m − (cid:101) for this function. Left: a tiling of the plane by squares.
Middle: a modification of the tiling where near each vertex there is a blank.
Right: an arrangement contained in a rectangular cake. There are m = 16 toppings and m − (cid:100) √ m − (cid:101) = 9 blanks. Z Z Z Z Z Z HB B A A A A Figure 6: A rectangular cake C , rectangular toppings Z j , and a hole H . Note the configurationis not maximal.right and the first one at the top. These are exactly the points where the function (cid:100) √ m − (cid:101) increases by 1. Therefore the number of blanks always remains m − (cid:100) √ m − (cid:101) . We first expand the toppings as follows. For the topping Z consider a rectangle Z (cid:48) withmaximal area among all axis-parallel rectangles contained in C , containing Z and avoidingall other toppings. Substitute Z by Z (cid:48) . Repeat for the remaining toppings.We now have a maximal arrangement of toppings. Again we define a hole as a connectedcomponent of C \ ∪ mi =1 Z i . It is clear that in a maximal arrangement every hole is simply-connected, since a hole that is not simply-connected contains a topping which can be expanded.We will now prove that in any maximal arrangement, every hole is an axes-parallel rectangleand is not adjacent to the cake boundary. We need several definitions regarding a hole H (the8xamples refer to Figure 6): Definition 3.1. A hole-vertex of H is a point on the boundary of H that is a vertex of atopping or of the cake. A hole-vertex is called: • convex — if the internal angle adjacent to it is less than ◦ (i.e, ◦ ; like A , A , B , B ); • nonconvex — if the internal angle adjacent to it is at least ◦ (like. A , A ). Definition 3.2. A hole-edge of H is a line-segment between adjacent hole-vertices of H . Ahole-edge is called: • inner-edge — if it is contained in the interior of C ; • boundary-edge — if it is contained in the boundary of C . Definition 3.3.
A hole-vertex is called: • inner-vertex — if it is contained in the interior of C (links two inner-edges; like A i ); • boundary-vertex — if it is on the boundary of C (touches a boundary-edge; like B i ). Definition 3.4.
A hole is called: • inner-hole — if it is contained in the interior of C (so it has only inner-edges); • boundary-hole — if it has a common boundary with C (so it has some boundary-edges). Any inner-edge e of a hole H represents an opportunity to expand some topping into H .Such opportunity can only be blocked by an adjacent topping with an edge e (cid:48) perpendicular to e . Let v be the vertex that connects e and e (cid:48) . We call v a blocking-vertex of e . For example, A is a blocking-vertex of the edge A A . It is easy to see that, if v is a blocking-vertex of e : • v is a convex vertex — it connects two perpendicular edges of different toppings. • v is an inner vertex — it connects two edges of toppings. • v blocks only e — it cannot simultaneously block e (cid:48) .In a maximal arrangement, each inner-edge must have at least one blocking-vertex. Therefore:For every hole H : H ) ≥ H ) (1)We now consider inner-holes and boundary-holes separately.An inner hole has only inner-vertices and inner-edges, and their number must be equal,so: For every inner-hole H : H ) = H ) (2)Combining (1) and (2) implies that, in an inner-hole, all vertices are convex, so:Every inner-hole is a rectangle. (3)9 boundary-hole ’s boundary contains sequences of adjacent boundary-edges and sequencesof adjacent inner-edges. Each boundary sequence contains an alternating sequence of boundary-vertex — boundary-edge — boundary-vertex — boundary-edge — ... boundary-vertex. So:For every boundary-hole H : H ) ≥ H ) + 1 (4)Summing (1) and (4) gives:For every boundary-hole H : H ) ≥ H ) + 1But this is impossible, since in every hole the total number of vertices and edges must beequal. Therefore: There are no boundary holes. (5)(3) and (5) imply that, in a maximal arrangement, all holes look like Figure 5: each hole is arectangle adjacent to four toppings. Our upper bound matches the lower bound of Subsection 3.1: m − (cid:100) √ m − (cid:101) . The boundrelies on the following lemmas: Lemma 3.1.
For all positive integers k , k , t : ( k − k − − t ≤ k k − t − (cid:100) (cid:112) k k − t − (cid:101) . Proof.
First, we prove the inequality for t = 0. It is sufficient to show that (cid:100) √ k k − (cid:101) ≤ k + k −
1. Indeed, for a fixed sum k + k , the product k k is maximized when k = k orwhen k +1 = k . In the former case (cid:100) √ k k − (cid:101) = 2 k −
1, in the latter case (cid:100) √ k k − (cid:101) = (cid:100) (cid:112) k + k (cid:101) − k + 1) − k . In both cases the inequality holds with equality.For t >
0, the difference (RHS minus LHS) is even larger, so the inequality remainstrue.Suppose a rectangle is partitioned into m smaller rectangles (with no holes). Define a3 -vertex as a point in which three rectangles meet and a 4 -vertex as a point in which fourrectangles meet. Lemma 3.2.
Suppose a rectangular cake C is partitioned into m rectangles. Then, the numberof -vertices is at most m − (cid:100) √ m − (cid:101) .Proof. First, suppose that there are only 4-vertices (no 3-vertices). Then C is partitioned, bythe lines containing all sides of all rectangles, to a grid of k by k smaller rectangles. So thetotal number of 4-vertices is ( k − k −
1) and the total number of rectangles is m = k k .Then the lemma follows from Lemma 3.1 setting t = 0.Next, suppose that there are 3-vertices (refer to Figure 7 for the examples). Choose one3-vertex (e.g. A ). Add a segment that separates the rectangle that does not have vertex at v to two parts (e.g. the segment A A ). This converts the 3-vertex to 4-vertex. Additionally,there are several cases depending on where the other end of the additional segment lands:10 A A B B C C D D Figure 7: Bounding the number of 4-vertices in a rectangle tiling.(a) If it lands on the boundary of C (like with D D and C C ), then no further action isrequired; the number of rectangles grows by 1 and the number of 4-vertices grows by 1.(b) If it lands on the boundary of another rectangle (like with A A ), then a new 3-vertexis created, and can be handled in the same way by continuing the segment. This must stopbecause eventually the line hits the boundary of C .(c) If it lands on a 3-vertex (like with B B ), then an additional 4-vertex is created; thenumber of rectangles grows by 1 and the number of 4-vertices grows by 2.In all cases, the number of 4-vertices grows at least as much as the number of rectangles.Continue this procedure until all 3-vertices disappear. Then, the partition is a grid formedby the lines containing the sides of the all original rectangles. Suppose the grid has k × k rectangles (e.g. in Figure 7, k = 3 and k = 4). So, the total number of rectangles is k k and the total number of 4-vertices is ( k − k − t removal steps. So,the original number of 4-vertices was at most ( k − k − − t and the original number ofrectangles m was exactly k k − t . From Lemma 3.1 it follows that the number of original4-vertices is at most m − (cid:100) √ m − (cid:101) .We now return to the proof of Theorem 3.1. Recall from Subsection 3.2 that we have amaximal arrangement of toppings, where each hole is a rectangle bounded by four toppings.Each of these toppings is blocked from expanding into the hole, either by the next toppingcounter-clockwise (as in Figure 5) or by the next topping clockwise. Our goal is to show thatthe total number of such holes is at most m − (cid:100) √ m − (cid:101) . We will remove the holes one byone, by contracting each hole to a 4-vertex without changing the number of toppings.Each hole is contracted in two steps: vertical and horizontal . The vertical contraction isillustrated in Figure 8. Consider the hole [ x , x ] × [ y , y ]. There are two cases regarding thetoppings surrounding the hole: either the top has a side at x = x and the bottom has a sideat x = x (in the clockwise case), or vice versa (in the counter-clockwise case). These casesare entirely analogous. We assume the former case, as in Figure 8/Top.Let y be the bottom coordinate of the topping below the hole. Transform the arrangementof rectangles in the following way. Every point ( x, y ), where x > x and y ∈ [ y , y ], istransformed to ( x, y (cid:48) ), where: y (cid:48) := y + y − y y − y · ( y − y ) , y y x x y y y x x Figure 8:
Top:
An arrangement with 5 holes.
Bottom:
One of the holes is contractedvertically. The other rectangles and holes change their size but remain in the picture.so the ray ( x > x , y = y ) goes to ( x > x , y = y ) and the ray ( x > x , y = y ) remains inits place. Now, the rectangle on top of the hole can be extended down, so that it covers thehole (see 8/Bottom).The horizontal contraction is very similar. Let x be the left coordinate of the rectangleto the left of the hole. Every point ( x, y ), where y < y and x ∈ [ x , x ], is transformed to( x (cid:48) , y ), where: x (cid:48) := x + x − x x − x · ( x − x ) , so the ray ( y < y , x = x ) goes to ( y < y , x = x ) and the ray ( y < y , x = x ) remains inits place. Now, instead of the hole, there is a single 4-vertex at the point ( x , y ).The shrinking does not change the combinatorics of the arrangement: the number oftoppings does not change, no new holes are created, and no 4-vertices disappear.After all holes are contracted, the situation is as in Lemma 3.2, where the number of4-vertices is upper-bounded. Therefore, the number of holes in the original configuration isupper-bounded by the same expression. For completeness, we formulate the analogue of Theorem 2.2 for axis-parallel rectangles.12igure 9: An arrangement of m = 9 rectangles with m − Theorem 3.2.
Let Z , . . . , Z m , with m ≥ , be pairwise-interior-disjoint axes-parallel rectan-gles in the plane. Define a hole as a connected component of R \ ∪ mi =1 Z i . Then there are atmost m − holes. This bound is tight.Proof. The proof of the upper bound is similar to Theorem 2.2; the only difference is that here,every face of the planar graph touches at least four edges (instead of three). Therefore, we have2 | E | ≥ | F | (instead of 2 | E | ≥ | F | ). Substituting this in Euler’s formula | V | − | E | + | F | = 2gives that | F | ≤ | V | −
2, so the number of holes is at most m − In this section, the toppings are still axes-parallel rectangles, but now the cake C can be anysimply-connected axes-parallel rectilinear polygon. The “complexity” of a rectilinear polygonis characterized by the number of its reflex vertices — vertices with internal angle 270 ◦ . It isknown that a rectilinear polygon with T reflex vertices can always be partitioned to at most T + 1 rectangles [3, 2], and this bound is tight when the vertices of C are in general position.Since our goal is to bound the number of blank rectangles, we expect the bound to depend on T , in addition to m (the number of toppings). Theorem 4.1.
Let C (“cake”) be a simply-connected axes-parallel polygon with T reflex ver-tices, and Z , . . . , Z m (“toppings”) be pairwise-disjoint axes-parallel rectangles in C . Thereexists a partition of C into m + b axes-parallel rectangles, C = Z (cid:48) (cid:116) · · · (cid:116) Z (cid:48) m + b , where Z i ⊆ Z (cid:48) i for all i ≤ m , and b ≤ m + T − (cid:100) √ m − (cid:101) . Moreover, for every m and T there exists anarrangement where in every such partition, b = m + T − (cid:100) √ m − (cid:101) . Take a worst-case arrangement of rectangular toppings in a rectangular cake, such as the oneshown in Figure 5/Right. For any T ≥
1, it is possible to convert the rectangular cake to arectilinear polygon with T reflex vertices by adding a narrow rectilinear “staircase” with T “steps”, as shown in Figure 10.The number of blanks in the rectangle is m − (cid:100) √ m − (cid:101) . The additional rectilinear partis a single hole , but it contains T blanks (its rectangular components). All in all, we need m + T − (cid:100) √ m − (cid:101) blanks. 13igure 10: A rectilinear cake made of a union of a rectangle and a rectilinear “staircase”with T = 4 steps. The rectangle part contains a maximal arrangement of m = 16 toppingsand m − (cid:100) √ m − (cid:101) = 9 blanks. The rectilinear part adds T reflex vertices (circled) and T rectangular blanks. A A B B Z Z Z Z Z Z H H H Figure 11: A rectilinear cake with six rectangular topping Z , . . . , Z , and three rectilinearholes H , . . . , H . Note the configuration is not maximal since Z can be extended upwardsinto H . The dashed lines inside the holes indicate a possible partitioning of the holes to eightrectangular blanks. 14 .2 Upper bound We first expand the arrangement to a maximal arrangement of toppings as in Subsection 3.2.In the case of a rectangle cake, we proved that all holes are inner-holes and rectangular.This is no longer true here: Figure 11 shows that we can have boundary-holes that are notrectangular. It is still true that in a maximal arrangement every hole is simply-connected,since a hole that is not simply-connected contains a topping which can be expanded.Moreover, it is still true that every inner-edge must be blocked by a hole-edge perpendicularto it, and the blocking vertex must be a convex-vertex. However, now the blocking-vertex canbe either an inner-convex-vertex, or a boundary-vertex that is a reflex-vertex of C (like B in Figure 11). We call such vertex connection-reflex- C -vertex since it connects an inner-edgeand a boundary-edge. So now, for every hole H : C -vertices( H ) + H ) ≥ H ) (6)In an inner-hole there are only inner-vertices. Therefore, conclusion (3) is still true — everyinner-hole is a rectangle.In a boundary-hole, inequality (4) is still true, since in each sequence of adjacent boundary-edges, there is a vertex for each edge plus one additional vertex:For every boundary-hole H : H ) ≥ H ) + 1 (4)Summing (6) and (4) gives: C -vertices( H ) + H ) − H ) (7) ≥ H ) + 1Now, in every hole, H ) = H ). Combining this with (7) gives: H ) +1 ≤ C -vertices( H ) (8)In addition, the boundary of H contains some boundary-reflex-vertices, all of which are reflex-vertices of C that are not connection-vertices: H ) = C -vertices( H ) (9)Adding the latter two inequalities gives, for every boundary-hole H : H ) +1 ≤ C -vertices( H ) (10)A rectilinear polygon with x reflex vertices can always be partitioned to at most x + 1rectangles [3, 2]. Therefore, each boundary-hole H can be partitioned into rectangles whosenumber is at most the number of reflex-cake-vertices in the boundary of H . Moreover, everyreflex-cake-vertex is a vertex of at most one boundary-hole. Therefore, summing over allboundary-holes gives that all boundary-holes can be partitioned into rectangles whose totalnumber is at most T — the number of reflex-vertices of C .Adding the (at most) m − (cid:100) √ m − (cid:101) rectangular inner-holes gives that the total numberof rectangular blanks is at most: T + m − (cid:100) √ m − (cid:101) .15igure 12: A rectangular cake with two simply-connected toppings. The circular blankscannot be attached to any topping without crossing the other one. Therefore there is anunbounded number of blanks (4 of them are shown). We considered three families of pieces: polygons, convex figures, and axes-parallel rectangles.For each of these families, we proved that: (a) any arrangement of toppings in the familycan be expanded to a maximal arrangement, and (b) the number of blanks in any maximalarrangement is bounded. These two facts are not necessarily true for other families. Weprovide several examples below. Suppose that all pieces must be axes-parallel rectangles whose lengths are irrational numbers,while the cake is the unit square. Then, even a single topping cannot be expanded to amaximal arrangement. However, it may still be possible to find a partition with a boundednumber of blanks. For example, with a single topping it is sufficient to add 3 blanks. We didnot calculate how many blanks may be needed in general.
Suppose that all pieces must be path-connected. This very natural constraint is not as simpleas it seems: the number of blanks in a maximal arrangement might be unbounded. Anexample is shown in Figure 12. The cake is a rectangle. There are two toppings; they are notonly path-connected but also simply-connected. There are four circular blanks which cannotbe attached to any topping, since any path from a blank to a topping must cross the othertopping. Therefore the arrangement is maximal. It is straightforward to extend this exampleto contain any number of blanks.If the pieces are required to be connected but not path-connected, then the discs can beattached to any of the toppings, so in a maximal expansion of this example there are no blanks.We do not know if this is always the case when the cake and pieces are connected.
One way to overcome the problem in the previous section is to require that the toppings andpieces be topologically closed, in addition to being path-connected. However, in this case amaximal arrangement might not exist. These examples are based on suggestions by anonymous referees. Z Z Z Figure 13: A configuration of toppings that can be extended in two ways: optimal ( Z leftwards) and sub-optimal ( Z downwards).For example, suppose the cake is a union of:1. the curve ( x, sin(1 /x )) for x ∈ (0 , , − , (0 , , sin(1)).This cake is closed and path-connected. The two toppings are any two points on curve (3).They too are closed and path-connected. But there is no maximal expansion to pieces thatare closed and path-connected: if an expanded piece intersects both curve (1) and segment(2) then it is not path-connected, and if a piece contains only curve (1) then it is not closed. Suppose we are given a cake with m toppings and we want to expand the toppings such thatthe number of blanks is as small as possible. Suppose we use the following greedy algorithm:Find a maximal expansion of each topping in turn, in an arbitrary order.The results in this paper imply that, in the four cases studied, the number of blanksattained by this greedy algorithm is upper-bounded by some function of m . However, ourresults do not imply that the number of blanks is minimal: in specific cases it may possibleto extend the toppings in a way that attains a smaller number of blanks — less than theworst-case minimum. For example, in Figure 13, if Z is extended downwards, then we getone blank, which is the worst-case minimum for m = 4 toppings. However, if Z is extendedleftwards, then we can get zero blanks.This opens an interesting algorithmic question: given a specific arrangement of toppings,how can we find an expansion with a minimum number of blanks? A related natural questionis how to minimize, instead of the number of blanks, their total perimeter or area.17 .5 Restricted toppings This paper only considered cases where the geometric constraint on the initial toppings is thesame as on the final pieces. But when the geometric constraints on the toppings are stricter,the upper bound might be lower.For example, suppose the toppings must be points. Then, in both the convex case andthe rectangular case the upper bound is b = 0. In the convex case we can take the Voronoitesseletion (it works for disks as well; in this case we get weighted Voronoi). In the rectangularcase, we can just separate the points by vertical lines, and if two points have same x-coordinate,separate them by horizontal lines. Acknowledgements
This paper started as a question in the Math Overflow website [7]. Also participated inthe discussion: Steven Stadnicki, Fedor Petrov and Mikhail Ostrovskii. We are grateful toReuven Cohen, David Peleg, Herbert Edelsbrunner, Rom Pinchasi, Ashkan Mohammadi, JorgeFernandez and N. Bach for discussions about the problem studied in this paper.We would like to thank the editorial board of SIDMA and three anonymous referees forthe careful reading of our paper and several useful suggestions that significantly improved itsreadability and correctness.A. A. is supported by People Programme (Marie Curie Actions) of the European Union’sSeventh Framework Programme (FP7/2007-2013) under REA grant agreement n ◦ [291734].E. S.-H. was supported by the Doctoral Fellowships of Excellence at Bar-Ilan University,the Mordecai and Monique Katz Graduate Fellowship Program at Bar-Ilan University, andISF grant 1083/13. References [1] H. Edelsbrunner, A. D. Robison, and X.-J. Shen. Covering convex sets with non-overlapping polygons.
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