On Greedily Packing Anchored Rectangles
Christoph Damerius, Dominik Kaaser, Peter Kling, Florian Schneider
OOn Greedily Packing Anchored Rectangles
A tale of crowns and tiles
Christoph Damerius ! Universität Hamburg, Germany
Dominik Kaaser ! Universität Hamburg, Germany
Peter Kling ! Universität Hamburg, Germany
Florian Schneider ! Universität Hamburg, Germany
Abstract
Consider a set P of points in the unit square U , one of them being the origin. For each point p ∈ P you may draw a rectangle in U with its lower-left corner in p . What is the maximum area suchrectangles can cover without overlapping each other?Freedman [17] posed this problem in 1969, asking whether one can always cover at least 50% of U . Over 40 years later, Dumitrescu and Tóth [11] achieved the first constant coverage of 9 . .
1% might seem low, the authors could not find anyinstance where their algorithm covers less than 50%, nourishing the hope to eventually prove a 50%bound. While we indeed significantly raise the algorithm’s coverage to 39%, we extinguish the hopeof reaching 50% by giving points for which the coverage is below 43 . The
Lower-Left Anchored Rectangle Packing (LLARP) problem considers a finiteset P ⊆ U := [0 , with (0 , ∈ P of input points . The goal is to find a set of non-empty,interior-disjoint rectangles ( r p ) p ∈ P with p being the lower-left corner of r p ⊆ U and suchthat the total area P p ∈ P | r p | is maximized.This problem was first introduced by Freedman [17, Unsolved Problem 11, page 345]in 1969, who asked the question whether for any point set P , the rectangles can always bechosen such that they cover at least 50% of U . It is easy to see that this is the best one canhope for, since putting n equally spaced points along the ascending diagonal of U yields amaximum coverable area of 1 / n → ∞ .Over the years, the LLARP problem reoccurred in form of geometric challenges [13] or inmiscellaneous books and journals about mathematical puzzles [18–20]. Still, it took morethan 40 years until Dumitrescu and Tóth [11] made significant progress towards Freedman’squestion: they considered a natural greedy algorithm and proved that it achieves a coverageof 9 . , and even the question whether a maximumcovering can be found in polynomial time remains elusive.While Dumitrescu and Tóth [11] themselves observed that “ a sizable gap to the conjectured remains ”, they were unable to find instances where it does not reach 50%. This ledthem and others to conjecture a much better quality of their algorithm, making it a naturalcandidate to answer Freedman’s question positively, albeit [11] also mentioned that “ obtainingsubstantial improvements probably requires new ideas ”.Our results indeed attest the greedy algorithm a much better coverage of 39%. However,at the same time we show that there are instances where the coverage stays below 43 . A very recent, still unpublished result slightly raised the greedy algorithm’s coverage to 10 .
39% [12]. a r X i v : . [ c s . C G ] F e b On Greedily Packing Anchored Rectangles
LLARP falls into the class of geometric packing problems, where a typical question is howmuch of a container can be covered using a set of geometric shapes in two or more dimensions.We concentrate on two dimensional packing problems with rectangular containers and shapes.
Complexity of LLARP.
The above mentioned greedy algorithm by Dumitrescu and Tóth[11] considers the input points step by step from top-right to bottom-left, greedily choosingmaximum-area rectangles in each step. It can be shown [11] that this is equivalent topartitioning the unit square into staircase-shaped tiles , one for each input point, and choosingmaximal rectangles within each tile (see Section 2 for the formal algorithm description).While the complexity of LLARP remains unknown, [11] also showed that there is an orderof the input points for which the greedy algorithm achieves an optimal packing (albeit ofunknown value); how to find that ordering remains unclear. Balas and Tóth [6] studied thecombinatorial structure of optimal solutions, proving that the worst-case number of maximalrectangle packings is exponential in the number of input points.
LLARP Variants.
After [11], a series of papers studied special cases and variants of LLARP.Balas et al. [7] allowed rectangles to be anchored in any of the four corners and showed thathere the worst-case coverage lies in [7 / , /
3] and in [5 / , /
23] if the rectangles are restrictedto squares. Akitaya et al. [4] improved the lower bound for such corner-anchored squarepackings to 1 / Further Problems and Applications.
Further related problems include the maximum weightindependent set of rectangles problem [1, 3, 10] (which was used, e.g., in [5] to derive aPTAS for center-anchored rectangle packings) or geometric knapsack [2, 16] and strip packingproblems [14]. All of these differ from LLARP and its variants in that the size of the objectsto be packed is part of the input and the object placement is typically less constrained.Note that LLARP-like problems are not of pure theoretical interest, but have applicationsin, e.g., map labeling. Here, rectangular text labels must be placed under certain constraints(e.g., labels might be scalable but require a fixed ratio and must be placed at a specificanchor) within a given container. We refer to the relatively recent survey [15] for details.
We analyze the greedy
TilePacking from [11] (formally described in Section 2). From ahigh-level view,
TilePacking partitions the unit square into staircase-shaped tiles , eachanchored at an input point, and chooses an area-maximal rectangle in each tile. A naturalway to analyze such an algorithm is to consider tiles’ densities (the ratio between theirarea-maximal rectangles and their own area) and prove a lower bound on the average tiledensity (which immediately yields the covering guarantee). ndeed, intuitively there cannotbe too many low-density tiles, as such tiles “use up” a comparatively large space.Dumitrescu and Tóth [11] follow this approach by defining suitable charging areas C t foreach tile t (suitable trapezoids below/beneath the tile). We also use such a charging scheme,but rely on a much more complex charging area which we refer to as a tile’s crown . Butinstead of directly analyzing a tile’s charging area, we first extract the critical propertiesthat determine the charging scheme’s quality. This general approach (described in Section 3) . Damerius, D. Kaaser, P. Kling, and F. Schneider 3 requires a lower bound function ξ on the tile’s charging ratio | C t | / | t | together with somesimple properties (basically a form of local convexity characterizing the average tile density).We derive such a lower bound and describe simple, symmetric tiles for which it is tight(Figures 8 and 9). We then take an arbitrary tile and show how to gradually transform itinto one of these tiles without increasing its charging ratio. This establishedhat ξ is indeed alower bound and allows us to easily conclude the following theorem. ▶ Theorem 1.
For any nput points,
TilePacking covers at least of the unit square.
While the involved transformations to get from an arbitrary tile to a worst-case tilerequire some care, we showcase the versatility of our approach by first proving a slightlyweaker bound of only 25% (Section 4.2). The analysis of this bound is not only much simplerbut, in fact, takes us halfway to Theorem 1, as the ξ -bound used in this case (Proposition 10)is tight for high-density tiles and all that remains is to refine our bound for low-density tiles(Proposition 14).Our second major result constructs an input instance (depicted in Figure 11) for which TilePacking covers significantly less than 50%. ▶ Theorem 2.
There is a set of input points P for which algorithm TilePacking covers atmost . of the unit square. Let U := [0 , denote the unit square. For a point p ∈ R define x ( p ) and y ( p ) as the x -and y -coordinates of p , respectively. For two points p, p ′ ∈ R we use the notation p ⪯ p ′ toindicate that x ( p ) ≤ x ( p ′ ) and y ( p ) ≤ y ( p ′ ). Similarly, p ≺ p ′ means that x ( p ) < x ( p ′ ) and y ( p ) < y ( p ′ ). The relations “ ⪰ ” and “ ≻ ” are defined analogously. For a set S we denote itsclosure by S . If S is measurable, we use | S | to denote its area.To simplify some geometric constructions and arguments, we use the following line-notation: We define the line ℓ − q ⊆ R as the line through q ∈ R of slope +1. Similarly, wedefine the lines ℓ − q , ℓ − q , and ℓ − q through q with slope 0, −
1, and ∞ , respectively. For linesof type R ∈ { − , − } we write ℓ Rq < ℓ Rq ′ if q ′ = q + ( x,
0) with x > ℓ Rq is left of ℓ Rq ′ .Similarly, for lines of type R ∈ { − , − } we write ℓ Rq < ℓ Rq ′ if q ′ = q + (0 , y ) with y > ℓ Rq is below ℓ Rq ′ . Analogous definitions apply for “ > ”, “ ≤ ”, and “ ≥ ”. Input Sets in General Position.
Remember the problem description from Section 1. Wesay that the input set P is in general position if there are no two (different) points p, p ′ ∈ P with x ( p ) = x ( p ′ ), y ( p ) = y ( p ′ ), or x ( p ) + y ( p ) = x ( p ′ ) + y ( p ′ ). That is, no two points mayshare an x - or y -coordinate and may not lie on the same diagonal of slope −
1. W.l.o.g., werestrict P to be in general position (Lemma 16 in Appendix A explains why this is okay). Tiles and Tile Packings. A tile t ⊆ U is a staircase polygon in U (see Figure 1a). Moreformally, t is defined by its anchor p ∈ U and a set of k upper staircase points Γ t := { q , q , . . . , q k } ⊆ U ordered by increasing x -coordinate and such that q i ≻ p for all q i as well as q i ̸⪯ q j for all q i ̸ = q j . Then t = { q ∈ U | q ⪰ p ∧ ∃ q ′ ∈ Γ t : q ≺ q ′ } . A point p i = (cid:0) x ( q i − ) , y ( q i ) (cid:1) is called a lower staircase point . We define A t ⊆ t as an (arbitrary)area-maximal rectangle in t and ρ t := | A t | / | t | as the tile’s density . For indexed upper staircasepoints q i we often use the shorthands x i := x ( q i ) and y i := y ( q i ).If p and Γ t do not adhere to the above requirements ( q i ≻ p and q i ̸⪯ q j ), but only to theslightly weaker requirements q i ⪰ p q i ̸⪯ q j for all q i ̸ = q j ), then tile t is said to be degenerate .Such tiles have superfluous points in Γ t and play a role in our analysis. On Greedily Packing Anchored Rectangles q q q q q p p p p A t t (a) Staircase points p i , q j andrectangle A t . (b) A packing produced by
TilePacking . tC t q i p q i +1 T p ( q i , q i + ) (c) A tile t , its crown C t , and atower T p ( q i , q i +1 ). Figure 1
Tiles, packings, crowns, and towers. In our figures, tiles are shaded light blue. Upperand lower stair case points are shown in red and black, respectively. A dark blue rectangle representsa (maximal) rectangle of a tile. Crowns are shown in yellow and towers are possibly labeled.
The hyperbola of t is h t := { ( x, y ) ∈ R > | y = p + | A t | /x } . Note that all upper staircasepoints lie between p and h t . Moreover, the points from Γ t ∩ h t span all area-maximalrectangles in t . If, p = (0 ,
0) and | A t | = 1, then t is called normalized .A tile packing of the unit square is a set T of disjoint tiles such that S t ∈T t = U . Inparticular, P t ∈T | t | = |U| = 1. We use A ( T ) := P t ∈T | A t | to denote the area covered bychoosing an area-maximal rectangle A t for each tile t (the area covered by T ). A Greedy Tile Packing Algorithm.
Let us revisit algorithm
TilePacking by Dumitrescuand Tóth [11].
TilePacking processes the points from P from top-right to bottom-left.More formally, it orders P = { p , p , . . . , p n } such that ℓ − p i ≥ ℓ − p i +1 . It then defines for each p i ∈ P the tile t i := { q ∈ U | q ⪰ p i } \ S i − j =1 t j , yielding a tile packing T = { t , t , . . . , t n } .To build its solution to LLARP, TilePacking picks for each p ∈ P the rectangle r p as an(arbitrary) area-maximal rectangle A t ⊆ t in the tile t containing p . Thus, the total areacovered by TilePacking is A ( T ). Figure 1b illustrates the resulting tile packing.Note that by this construction, the lower staircase points of each tile t are input points.Moreover, as already mentioned in Dumitrescu and Tóth [11], for each tile we can define acertain exclusive area that does not contain an input point. ▶ Observation 3.
Consider the tile packing T produced by TilePacking for a set of inputpoints P . Fix a tile t ∈ T and let p ∈ P denote its anchor point. Then the tile’s exclusivearea E t := { q ∈ U | ℓ − q > ℓ − p ∧ ∃ q ′ ∈ Γ t : q ≺ q ′ } does not contain any input point from P .This observation follows by noting that any such input point p ′ ∈ E t would be processedbefore p by TilePacking and “shield” at least one upper staircase point q ′ ∈ Γ t from p ,preventing it from becoming an upper staircase point of tile t . Here we present a general approach to derive lower bounds for the area covered by a giventile packing T . Our approach relies on a suitable charging scheme ( c t ) t ∈T that charges thearea of each tile t ∈ T to a charging area c t >
0. We define c ∗ := P t ∈T c t as the totalcharged area and c t / | t | as the charging ratio of tile t .Assume we are given a piecewise differentiable function ξ : (0 , → R ≥ and a value ρ ∗ ∈ (0 ,
1] with the following properties: . Damerius, D. Kaaser, P. Kling, and F. Schneider 5 ξ is point-convex in ρ ∗ (see Definition 17) with ξ ′ ( ρ ∗ ) < ξ ( ρ ∗ ) is an upper bound on the total charged area c ∗ , andfor any t ∈ T the value ξ ( ρ t ) is a lower bound on t ’s charging ratio c t / | t | .The following lemma then uses ξ to show that ρ ∗ is a lower bound on the area covered by T . ▶ Lemma 4.
Consider a tile packing T with a charging scheme ( c t ) t ∈T together with afunction ξ and a value ρ ∗ as described above. Then T covers an area of at least ρ ∗ . Proof.
Since ξ is point-convex in ρ ∗ , we get for the tangent τ ( ρ ) := ξ ( ρ ∗ ) + ξ ′ ( ρ ∗ ) · ( ρ − ρ ∗ )of ξ in ρ ∗ that τ ( ρ ) ≤ ξ ( ρ ) for all ρ ∈ (0 , A ( T ) = P t ∈T | A t | = P t ∈T | t | · ρ t , thelinearity of τ and the properties of ξ we calculate τ (cid:0) A ( T ) (cid:1) = τ X t ∈T | t | · ρ t ! = X t ∈T | t | · τ ( ρ t ) ≤ X t ∈T | t | · ξ ( ρ t ) ≤ X t ∈T | t | · c t | t | = c ∗ ≤ ξ ( ρ ∗ ) . (1)On the other hand, by definition of τ , we have τ (cid:0) A ( T ) (cid:1) = ξ ( ρ ∗ ) + ξ ′ ( ρ ∗ ) · (cid:0) A ( T ) − ρ ∗ (cid:1) .Combining this with Equation (1) and rearranging yields ξ ′ ( ρ ∗ ) · A ( T ) ≤ ξ ′ ( ρ ∗ ) · ρ ∗ , whichyields the desired result after dividing by ξ ′ ( ρ ∗ ) < ◀ This section introduces the charging scheme we will use to derive our lower bounds for thearea covered by the greedy algorithm
TilePacking (via the approach presented in Section 3).Afterward we derive a first, simple lower bound ξ w : (0 , → R ≥ on the tiles’ charging ratios(as described in Section 3) and prove that it has the properties necessary to apply Lemma 4.While comparatively simple, this already yields that TilePacking covers at least a quarterof the unit square, almost tripling the original guarantee from [11]. Section 5 will refine ξ w in order to derive our main result (Theorem 1). Given a tile packing T from TilePacking , our charging scheme defines an area C t for eachtile t ∈ T and charges t ’s area to c t := | C t | . We first explain how C t is constructed from t .Afterward, we prove useful properties about these areas and their relation to T . Construction of C t . Consider three points p, q = ( x , y ) , q = ( x , y ) ∈ R with q , q ⪰ p , x ≤ x , and y ≥ y . Let p ′ = ( x , y ). The tower T p ( q , q ) with base point p and peak p ′ is the rectangle formed by the lines ℓ − p (the tower’s base ), ℓ − q (the tower’s left side ), ℓ − q i +1 (the tower’s right side ), and ℓ − p i (the tower’s top ). If the subscript p is omitted, the basepoint is assumed to be the origin (0 , t with anchor p and Γ t = { q , q , . . . , q k } orderedby increasing x -coordinate. The charging area of tile t is C t := S k − i =1 T p ( q i , q i +1 ). We referto C t as the crown of tile t . See Figure 1c for an illustration of a tile’s towers and crown.The width and height of a tower T p ( q , q ) correspond to the side lengths of isoscelestriangles (see Figure 2), which yields a formula for | T p ( q , q ) | . By taking derivatives, wealso get formulas for the change when moving q or q horizontally or vertically. ▶ Observation 5.
Consider T p ( q , q ) with q j − p = ( x j , y j ), j ∈ { , } . Let w := x − x ,and h := y − y . Then | T p ( q , q ) | = ( x + y ) · ( w + h ) / On Greedily Packing Anchored Rectangles q q w h p p (cid:48) h h + w √ x + y √ y y x Figure 2 | T p ( q , q ) | is computed viathe catheti of the blue triangles. q (cid:48) q q q (cid:48) T p (cid:48) ( q (cid:48) , q (cid:48) ) T p ( q , q ) p (cid:48) p E t p ∗ p (cid:48)∗ ∆ y Figure 3
Example for Lemma 7. The shown toweroverlap has p ′∗ ∈ E t , violating t ’s exclusive area. ▶ Observation 6.
Consider T p ( q , q ) with q j − p = ( x j , y j ), j ∈ { , } . Let w := x − x ,and h := y − y . Fix α ∈ R and consider the change of | T p ( q , q ) | if either q or q aremoved horizontally or vertically as a linear function of ϵ :(a) If either q ( ϵ ) := q +(0 , α · ϵ ) or q ( ϵ ) := q +( α · ϵ, ∂ | T p ( q , q ) | /∂ϵ = α · ( x + y ) / ∂ | T p ( q , q ) | /∂ϵ = 0.(b) If either q ( ϵ ) := q + ( α · ϵ,
0) or q ( ϵ ) := q + (0 , α · ϵ ), then ∂ | T p ( q , q ) | /∂ϵ = α · (cid:0) w + h − ( x + y ) (cid:1) / ∂ | T p ( q , q ) | /∂ϵ = − α . Properties of the Charging Scheme.
The following results capture basic properties of ourcharging scheme. First, we show that the defined charging areas are disjoint. ▶ Lemma 7.
Consider the tile packing T produced by algorithm TilePacking for a set ofinput points P . For any two different tiles t, t ′ ∈ T , we have C t ∩ C t ′ = ∅ . Proof.
Fix t, t ′ ∈ T and let p, p ′ ∈ P denote their respective anchors. W.l.o.g., assume ℓ − p > ℓ − p ′ , such that TilePacking processes p before p ′ . As crowns consist of towers, it issufficient to show T p ( q , q ) ∩ T p ′ ( q ′ , q ′ ) = ∅ for neighboring q , q ∈ Γ t and q ′ , q ′ ∈ Γ t ′ . Let p ∗ ∈⊆ P and p ′∗ ∈ P denote these towers’ respective peak. W.l.o.g., we assume ℓ − p ∗ < ℓ − p ′∗ ( p ∗ lies left of ℓ − p ′∗ ); the other case follows symmetrically.If ℓ − p ′∗ < ℓ − p , the towers are separated (the top of T p ′ ( q ′ , q ′ ) lies below the base of T p ( q , q ))and cannot intersect. So assume ℓ − p ′∗ > ℓ − p . Then we cannot have p ′∗ ≺ q , since this wouldimply that p ′∗ lies in the exclusive area of t , violating Observation 3.Let ∆ y := q ′ − p ′∗ and note that x (∆ y ) = 0. Define ˜ q := p ∗ − ∆ y and note that p ′∗ ̸≻ ˜ q ,since otherwise q ′ = p ′∗ + ∆ y ≻ ˜ q + ∆ y = p ∗ , which (together with ℓ − p ∗ > ℓ − p > ℓ − p ′ ) wouldmean that p ∗ lies in the exclusive area of t ′ (again violating Observation 3).So ℓ − p ′∗ > ℓ − p ∗ , p ′∗ ̸≺ q , and p ′∗ ̸≻ ˜ q . Together, these imply x ( p ′∗ ) > x ( q ) and y ( p ′∗ ) < y (˜ q ),which in turn imply ℓ − p ′∗ > ℓ − q − ∆ y (see Figure 3). But then, the towers are separated, since ℓ − q ′ = ℓ − p ′∗ +∆ y > ℓ − q − ∆ y +∆ y = ℓ − q ( T p ( q , q )’s right side lies left of T p ′ ( q ′ , q ′ )’s left side). ◀ The next lemma’s proof shows that all crowns lie inside a pentagon formed by U and twoisosceles triangles left and below of U (see Figure 4). With Lemma 7 this implies that thetotal charging area is bounded by the pentagon’s area. . Damerius, D. Kaaser, P. Kling, and F. Schneider 7 (cid:0) − , (cid:1) (0 , (cid:0) , − (cid:1) (1 , , , Figure 4
Pentagon (cid:68) andpoint set P ϵ from Lemma 8. q i +1 q i p i − q i − (cid:96) − q i (cid:96) − q i +1 (cid:96) − q i (cid:96) − p i − T p ( q i − , q i ) T p ( q i , q i + ) p Figure 5
Removing superfluous points from Γ t reduces | C t | by the rectangle between ℓ − q i , ℓ − q i +1 , ℓ − q i , and ℓ − p i − . ▶ Lemma 8.
Consider the tile packing T produced by algorithm TilePacking for a set ofinput points P . The total charging area of T is c ∗ ≤ / . Moreover, this bound is tight, sincethere are arbitrarily small ϵ > and input points P ϵ for which c ∗ ≥ / − ϵ . Proof.
Define the points SW := (0 , , , (cid:68) denote thepentagon enclosed by the lines ℓ − SW , ℓ − NW , ℓ − SE , ℓ − NW , and ℓ − SE (see Figure 4). Since | (cid:68) | = 3 / C t ⊆ (cid:68) for any t ∈ T . For this, in turn, it issufficient to show that any tower T p ( q , q ) of C t lies in (cid:68) .Fix such a tower T p ( q , q ). Since p ≻ SW, we have ℓ − p ≥ ℓ − SW (the base of T p ( q , q ) liesabove the base of (cid:68) ). Similarly, since q , q ∈ U ⊆ (cid:68) , we have ℓ − q ≥ ℓ − NW (the left side of T p ( q , q ) lies right of the left side of (cid:68) ) and ℓ − q ≤ ℓ − SE (the right side of T p ( q , q ) lies left ofthe right side of (cid:68) ). Finally, the topmost point q ∈ U of T p ( q , q ) lies below ℓ − NW and therightmost point q ∈ U of T p ( q , q ) lies to the left of ℓ − SE . Together, we get T p ( q , q ) ⊆ (cid:68) .For the tightness of the bound, choose ϵ > /ϵ ∈ N . Define P ϵ = { SW } ∪{ ( k · ϵ, − k · ϵ ) , (1 − k · ϵ , kϵ ) | k ∈ { , , . . . , /ϵ − } } . As illustrated in Figure 4, thecrown of tile t with anchor SW converges towards (cid:68) as ϵ →
0, such that lim ϵ → | C t | = 3 / ◀ This section proves the following, slightly weaker version of Theorem 1: ▶ Theorem 9.
For any input points,
TilePacking covers at least of the unit square.
Proving this not only serves as a warm-up to illustrate our approach before proving our mainresult but – as we will see in Section 5 – brings us halfway towards proving Theorem 1.So consider a tile packing T produced by TilePacking for some set of input points P .To prove Theorem 9 we follow the approach outlined in Section 3, using the charging schemefrom Section 4.1. That is, the area of t ∈ T is charged to c t = | C t | , where C t represents thecrown of t . To this end, define ρ ∗ := 1 / weak charging ratio bound ξ w : (0 , → R ≥ , ξ w ( ρ ) := 2 · (1 − ρ ) . (2)As a linear function, ξ w is trivially point-convex in ρ ∗ . Moreover, ξ w ( ρ ∗ ) = 3 / ξ w ( ρ ∗ ) ≥ c ∗ . In the remainder of this section we prove the followingproposition, stating that ξ w represents a lower bound on the charging ratio of any t ∈ T . ▶ Proposition 10.
For any tile t we have c t / | t | ≥ ξ w ( ρ t ) . Once this is proven, Theorem 9 follows immediately by applying Lemma 4.
On Greedily Packing Anchored Rectangles q i − q i q i +1 q j − q j q j +1 h t T p ( q j , q j + ) T p ( q j − , q j ) w i h j α j α i T p ( q i − , q i ) T p ( q i , q i + ) p t Figure 6
Moving inner points in Lemma 13.Note that α j > α i < h t q q q q − q − q − R − R − R − R R R R R R q q q Figure 7
Notation for Proposition 10 with k = 3 and m = 5. A Lower Bound on the Charging Ratio.
To prove that ξ w ( ρ t ) lower bounds the chargingratio c t / | t | of any tile t ∈ T , we gradually transform t into a “simpler” tile ˜ t . Our transfor-mations ensure ρ ˜ t = ρ t and c ˜ t / | ˜ t | ≤ c t / | t | . Eventually, ˜ t will be simple enough to directlyprove c ˜ t / | ˜ t | ≥ ξ w ( ρ ˜ t ). The following notation expresses progress via such a transformation:˜ t ⪯ t : ⇔ ρ ˜ t = ρ t and c ˜ t / | ˜ t | ≤ c t / | t | . (3)As a simple example, note that both a tile’s density and charging-ratio are invariantunder translation and concentric scaling w.r.t. its anchor. This gives rise to the followingtransformation, which allows us to restrict our analysis to normalized tiles. ▶ Observation 11.
For a tile t , let ˜ t denote the translation of t such that it is anchored in theorigin and scaled by 1 / | A t | around the origin. Then ˜ t ⪯ t . We call ˜ t the normalization of t .Consider a tile t with anchor p . A transformation may move one of t ’s upper staircasepoints to the same x - or y -coordinate as another point from Γ t ∪{ p } , resulting in a degeneratetile with superfluous points in Γ t (see Section 2). The next lemma states that removing suchsuperfluous points maintains an “equivalent” tile with a smaller crown. ▶ Lemma 12.
Consider a degenerate tile t . The pruned tile ˜ t with the same anchor but with Γ ˜ t := { q ∈ Γ t | ∄ q ′ ∈ Γ t : q ⪯ q ′ } covers the same points, is non-degenerate, and c ˜ t ≤ c t . Proof.
Order Γ t = { q , q , . . . , q k } by non-decreasing x -coordinate and let q = q k +1 = p .W.l.o.g. assume there is some i ∈ { , . . . , k } with y ( q i ) = y ( q i +1 ); the case of identical x -coordinates follows analogously. Let ˜ t denote the (possibly still degenerate) tile with anchor p and Γ ˜ t = Γ t \ { q i } . Note that { q ∈ U | q ⪰ p ∧ q ≺ q i } ⊆ { q ∈ U | q ⪰ p ∧ q ≺ q i +1 } ,which implies ˜ t = t and, thus, ρ t = ρ ˜ t . Removing q i affects the towers T p ( q i − , q i ) with peak p i − and T p ( q i , q i +1 ) with peak q i . Figure 5 illustrates the situation.We now show that c ˜ t ≤ c t , such that ˜ t ⪯ t ; the lemma’s statement then follows byiteration. If i = k , then c ˜ t = c t − | T ( q i − , q i ) | ≤ c t . So assume i < k . Then c ˜ t = c t − | □ | ≤ c t ,where □ is the rectangle enclosed by the lines ℓ − q i , ℓ − q i +1 , ℓ − q i , and ℓ − p i − (see Figure 5). ◀ For the next transformation, remember that the hyperbola h t of a normalized tile t contains exactly those upper staircase points q ∈ Γ t that form maximal rectangles in the tile.We now prove that we can transform t such that at most one q ∈ Γ t lies not on h t . . Damerius, D. Kaaser, P. Kling, and F. Schneider 9 ▶ Lemma 13.
A normalized tile t can be transformed into a tile ˜ t ⪯ t with | Γ ˜ t \ h ˜ t | ≤ . Proof.
Assume | Γ t \ h t | > t = { q , q , . . . , q k } by increasing x -coordinate. Tosimplify border cases, define q = q and q k +1 = q k . Choose q i , q j ∈ Γ t \ h t with i < j .Consider the transformation q i ( ϵ ) := q i + (0 , α i · ϵ ) and q j ( ϵ ) := q j + ( α j · ϵ,
0) with α i , α j ∈ R .Then the tile and crown areas become functions t ( ϵ ) and c t ( ϵ ) of ϵ . We show that thereare non-zero α i , α j such that t ( ϵ ) and, thus, ρ t remain constant and c t ( ϵ ) does not increase.Eventually, this results in a tile ˜ t ⪯ t that has an additional point on h t or that has adegenerate staircase point which we can remove by Lemma 12. In both cases | Γ t \ h t | isreduced and the lemma follows by iteration. Figure 6 illustrates the transformation.The transformation changes only the towers T ( q i − , q i ), T ( q i , q i +1 ), T ( q j − , q j ), and T ( q j , q j +1 ). For l ∈ { , , . . . , k + 1 } let q l = ( x l , y l ) and define w l := x l − x l − for l ̸ = 0and h l := y l − y l +1 for l ̸ = k + 1. Then the transformation changes the tile according to t ′ ( ϵ ) = α i · w i + α j · h j . To keep t constant we make this zero by setting α j = − α i · w i /h j .It remains to find a non-zero α i such that c t ( ϵ ) is non-increasing in ϵ . For this let T i ( ϵ ) := | T ( q i − , q i ) | + | T ( q i , q i +1 ) | and define T j ( ϵ ) analogously. By Observation 6, T ′′ l ( ϵ ) = − α l for l ∈ { i, j } . This yields c ′′ t ( ϵ ) = − α i − α j = − α i · (1 + w i /h j ) < − α i , a negative constant.This allows us to choose α i ∈ { − , +1 } such that c t ( ϵ ) is non-increasing. ◀ With these results, we are ready to prove our first covering guarantee for
TilePacking . Proof of Proposition 10.
Consider an arbitrary tile t . By Observation 11 and Lemma 13,we can assume that t is normalized and that | Γ t \ h t | ≤
1. If | Γ t \ h t | = 1, let q be thatpoint, otherwise choose q ∈ Γ t arbitrarily. Order Γ t = { q − l , . . . , q k } by increasing x -coordinate. To simplify border cases, define q − l − = q l and q k +1 = q k . To ease the notation,for i ∈ { − l − , . . . , k + 1 } we let q i = ( x i , y i ) We furthermore define w i := x i − x i − for i ̸ = l − and h i := y i − y i +1 for i ̸ = k + 1. For i ∈ { − l, . . . , k } inductively define R i := { q ∈ t | q ≺ q i } \ S | j |
2. With these inequalities and since ρ t = | A t | / | t | = 1 / | t | due tothe normalization, the desired statement follows via c t = − X i = − l | T i | + k X i =1 | T i | ≥ k X i = − l | R i | − | t | − | t | · (1 − ρ t ) = | t | · ξ w ( ρ t ) . (4)We now show the above bounds, starting with | T i | ≥ | R i | for | i | >
1. W.l.o.g. we assume i >
1; the case i < − i > q i , q i − ∈ h t and thus(since t is normalized) y j = 1 /x j for j ∈ { i − , i } . This yields x i /y i − = x i − · x i as wellas w i /h i − = ( x i − x i − ) / ( y i − − y i ) = x i − · x i . We use these identities together with | R i | = w i · y i to bound the formula for | T i | from Observation 5: | T i | = 12 · ( x i − + y i ) · ( w i + h i − ) = w i · y i · · (cid:18) x i − y i (cid:19) · (cid:18) h i − w i (cid:19) = | R i | · · (1 + x i − · x i ) · (cid:18) x i − · x i (cid:19) = | R i | · · (1 + x i − · x i ) x i − · x i ≥ | R i | , (5)where the inequality follows since x (1 + x ) /x takes its minimum over [0 , ∞ ) at x = 1.It remains to show that | T − | + | T | ≥ | R − | + 2 | R − | + 2 | R | −
2. Note that by ourdefinition q k +1 = q k , if k = 0 we have | R | = 0 and | T | = 0. Similarly, if l = 0 then | R − | = 0 and | T − | = 0. We assume that not both k and l are zero, as otherwise ξ w ( ρ t ) = ξ w (1) = 0and the proposition becomes trivial. W.l.o.g. let k >
0; the other case follows symmetrically.For α ∈ { − , +1 } (which we fix later) and ϵ ≥ y ( ϵ ) := y + α · ϵ ∈ [ y , /x ] , which moves q either up- or downward, depending on α . Thus, with f ( ϵ ) := | T − | + | T | − | R − | − | R | − | R | + 2 our goal becomes to prove f (0) ≥
0. Tothis end, consider how f ( ϵ ) changes with ϵ . The rectangles | R j | ( j ∈ { − , , } ) changelinearly or remain constant. By Observation 6, ∂ | T | /∂ϵ = 0. Similarly, if l > ∂ | T − | /∂ϵ = − α = − l = 0 we have ∂ | T − | /∂ϵ = 0 (because | T − | remains zero). Thus, in all cases f ′′ ( ϵ ) ≤
0. Then for one of the choices α ∈ { − , +1 } the function f must be non-increasing, meaning its minimum f min lies at one of the borders,where either y = y or y = 1 /x . We consider both possibilities and show that each time f min ≥ f (0) ≥ f min ≥ f min we have y = 1 /x , let t high denote the corresponding tile. Note that q lieson the hyperbola h t . But then | R | = 1 and, thus, f min = | T − | + | T | − | R − | − | R | .Moreover, with q ∈ h t we can apply the calculations for | i | > | T − | ≥ | R − | and | T | ≥ | R | , such that f min ≥ f min we have y = y and let t low denote the corresponding tile.Note that R and R form a rectangle from the origin to the point q on h t , such that | R | + | R | = 1. Thus, f min = | T − | + | T | − | R − | . Define the (degenerate) tile t ′ withΓ t ′ = { q − , q , q } and anchor p , such that its crown area is c t ′ = | T | + | T | . By Lemma 12,for the (non-degenerate) tile ˜ t ′ with Γ ˜ t ′ = Γ t ′ \ { q } we have c ˜ t ′ ≤ c t ′ . The crown c ˜ t ′ consistsof the single tower T ( q − , q ). Since q − , q ∈ h t , we can apply the calculations for | i | > c ˜ t ′ = | T ( q − , q ) | ≥ | R − | . Putting everything together we get f min = | T − | + | T | − | R − | = c t ′ − | R − | ≥ c ˜ t ′ − | R − | ≥ . ◀ This section proves our strong covering guarantee for
TilePacking , namely Theorem 1. Weuse the same approach as for our weak covering guarantee from Section 4.2 but derive astronger bound on the tiles’ charging ratios. More exactly, instead of ξ w we use ξ s : (0 , → R ≥ , ξ s ( ρ ) := ( − ρ · (cid:0) − /ρ ) (cid:1) , if ρ ≤ / ξ w ( ρ ) = 2 · (1 − ρ ) , if ρ > /
2. (6)Most properties required for our approach from Section 3 are easily verified for ξ s (whosefunction graph can be seen in Figure 13 in Appendix A). Indeed, for ρ ∗ := ξ − s (3 / ≈ . ξ ′ s ( ρ ∗ ) ≈ − . <
0. Moreover, ξ s is point-convex in ρ ∗ , since it is convex on (0 , / / ,
1] its tangent t ξ ( ρ ) in ρ ∗ lies below ξ s ( t ξ is steeper and t ξ (1 / ≈ . < ξ w (1 / ρ ∗ and by Lemma 8, we have ξ s ( ρ ∗ ) = 3 / ≥ c ∗ for the totalcharged area c ∗ of a tile packing T produced by TilePacking .The following proposition states the remaining required property from Section 3. ▶ Proposition 14.
For any tile t we have c t / | t | ≥ ξ s ( ρ t ) and this bound is tight. With this, Theorem 1 follows by applying Lemma 4. The remainder of this section outlinesthe analysis of this proposition. Full formal statements and proofs are given in Appendix B. These boundaries ensure that the tile remains valid and normalized. Note that if l = 0, moving q upward also causes the dummy point q − to move upward, such that | R − | and | T − | remain zero. . Damerius, D. Kaaser, P. Kling, and F. Schneider 11 ss h ( x ) = x Figure 8
Low-density tile t h ( s ) for which ξ s is tight. ss Figure 9
High-density tile t l ( s ) for which ξ s is tight. wh ( x ) = x zH r ( x ) z − / z √ Figure 10
Crown contribu-tion of a slide.
Transformation to Worst-case Tiles.
For tiles of density larger than 1 /
2, Proposition 14follows from Proposition 10, since in this regime ξ s ( ρ ) = ξ w ( ρ ). The tightness for such highdensities follows since for any ρ ∈ (1 / ,
1] there is a (symmetric) step tile t = t l ( s ) of size s (depicted in Figure 9) with density ρ t = ρ and c t / | t | = ξ s ( ρ t ). Thus, we restrict our furtherstudy to tiles of density at most 1 /
2. We will show how to gradually transform any suchtile t into a (symmetric) hyperbola tile t h ( s ) ⪯ t of size s (depicted in Figure 8). Again,the tightness follows since for any ρ ∈ (0 , / s such that tile t = t h ( s ) hasdensity ρ t = ρ and c t / | t | = ξ s ( ρ t ).Before we outline the transformation process into such worst-case low-density tiles, weneed to cope with the fact that t h ( s ) is not a staircase polygon and, thus, not captured byour tile definition. However, one can see t h ( s ) as the result of defining Γ t as k equally spacedpoints from the hyperbola { ( x, y ) ∈ [0 , s ) | y = 1 /x } and taking the limit k → ∞ . The nextparagraph formalizes this intuition by introducing generalized tiles and some related notions. Generalized Tiles.
As normal tiles, a generalized tile t ⊆ U is defined by its anchor p ∈ U and Γ t ⊆ U with q ⪯ p and q ̸⪯ q ′ for all q ̸ = q ′ from Γ t . The only difference is that Γ t maybe infinite. All other tile definitions (e.g., point set t , rectangle A t , or density ρ t ) stay intact.From now on the term tile always refers to a generalized tile. We require that the x -coordinates of Γ t can be partitioned into k inclusion-wise maximal, closed intervals I , I , . . . , I k , ordered by increasing x -coordinates. For i ∈ { , , . . . , k } let q − i , q + i ∈ Γ t denote the points realizing the left- and rightmost x -coordinate of I i , respectively. Note that I i may be a point interval, such that q − i = q + i . A section of Γ t is a tuple as follows:a step ( q + i , q − i +1 ), if q + i , q − i +1 ∈ h t ;a slide ( q − i , q + i ), if q − i ̸ = q + i and { q ∈ Γ t | x ( q ) ∈ I i } ⊆ h t ;a double step ( q + i − , q − i , q − i +1 ), if q + i − , q − i +1 ∈ h t and q − i = q + i / ∈ h t ; orthe corners ( q , q ), if q / ∈ h t and q ∈ h t as well as ( q k − , q k ) if q k / ∈ h t and q k − ∈ h t .After applying Lemma 13, all tiles resulting from our transformations can be described as asequence of such sections. Figure 14 illustrates generalized tiles and the different sections. Crown Contribution of Slides.
Our charging scheme from Section 4.1 generalizes naturallyto slides by considering them as the limit of k → ∞ equally spaced upper staircase points.This yields the following complement to Observation 5. ▶ Observation 15.
For a tile t , the contribution of a slide ( q , q ) to the crown C t is H ( q , q ) := (cid:2) ln( z ) + 1 / · ( z − z − ) (cid:3) x x . Proof.
Rotate the hyperbola by π/ t , to obtain the rotatedhyperbola h r ( x ) = √ x + 2. The contribution can then be calculated by integration: Theindefinite integral under h r ( x ) is H r ( x ) := R h r ( x ) dx = x/ · √ x + arsinh ( x/ √ x = ( z − /z ) / √
2, we get H r (( z − /z ) / √
2) = ln( z ) + 1 / · ( z − z − ) (see Figure 10). ◀ Overview of the Transformation Process.
Figure 12 gives an overview of how we graduallytransform an arbitrary tile t with density ρ t ≤ / t h ( s ).Starting with an arbitrary tile t (I), Lemma 23 either enforces Γ t ⊆ h t or exactly one q / ∈ h t ,can be forced to be in a double-step ( q , q, q ) with x ( q ) ≤ ≤ x ( q ) (II), or to be in corner( q, q ′ ) with x ( q ′ ) ≥ q ′ , q ) with x ( q ′ ) ≤ q, q ′ ). Lemma 21 moves steps with an adjacent slide along the hyperbola towards eitherthe left or bottom of the tile, whichever is nearer. Lemma 22 transforms two consecutivesteps left of (1 ,
1) into a step and a slide (or symmetrical on the right side of (1 , t ⊆ h t or t contains no slides which (together with Lemma 27) reduces theremaining cases to those illustrated in (VI) to (IX). For each of these we separately show t h ( s ) ⪯ t (Lemmas 28 and 30 to 32), proving Proposition 14. To show Theorem 2, we construct a point set where
TilePacking fails to cover an area larger than roughly(1 − e − ) /
2. Our goal is to construct a tile ˆ t at the originwhere each maximal rectangle has the same size A . Wetherefore place a large number of k points q i densely on ahyperbola h A centered in the origin. The remaining tilesare constructed such that they have a density close to1 /
2. To this end we add for each point q i on h A a set of(almost) evenly spaced points p i,j with distance roughly ε between each other, see Figure 11. A (0 , f f q − k +1 q − f − (1 , h A q q ... ... ... ... ... ... ... ε ... f k f − k q k − f − k + f k − Figure 11
Our construction.
To get a density close to 1 /
2, the exact coordinates of the points p i,j must be chosencarefully: we place the points on arcs of functions f i described by differential equations,where each f i depends on the two neighboring curves f i − and f i +1 . Formal definitions ofthe point set P ε,k and the functions f i are given in Appendix C.We now sketch the proof of Theorem 2. For simplicity, assume that TilePacking on P ε,k covers half of ˆ U = U \ ˆ t and an area of A of ˆ t . Both assumptions introduce an error; inAppendix C we show that both error terms vanish as k goes to infinity and ε goes to zero.In the full proof, we first show that the f i are well-behaved: their arcs do not intersect eachother, and they only intersect h A at q i . This allow us to show the aforementioned densityclose to 1 / ̸ = ˆ t (unless they are too close to (1 , A . Proof Sketch for Theorem 2.
We analyze the area ρ covered by TilePacking on P ε,k .Let ˆ U = U \ ˆ t . Assume that TilePacking covers half of ˆ U and an area of A of ˆ t . Then ρ ≤ A + | ˆ U| / A + (1 − | ˆ t | ) / ≤ A + (cid:0) − ( A + Z A A/x d x ) (cid:1) / A + A ln A ) / ρ = (1 − e − ) / A = e − . ◀ EFERENCES 13
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Lemma 24Lemma 24Lemma 23 Lemma 26Lemma 25 L e mm a31 L e mm a30 L e mm a28 L e mm a32 Lemma 27
Figure 12
Transforming low-density tiles t with ρ t ≤ / t h ( s ) ⪯ t . For the normalized tiles in Cases (II) and later, the blue dot marks the point (1 , PPENDIX 15
A Auxiliary Notions and Results
The next lemma proves that when analyzing algorithm
TilePacking we can, w.l.o.g., restrictourselves to input point sets P that are in general position. ▶ Lemma 16.
TilePacking achieves the same bound on point sets in general position asfor arbitrary point sets.
Proof.
Assume that
TilePacking chooses the points in some order p , . . . , p n while pro-ducing a tile packing T . Move all points except p n = (0 ,
0) to the top right by at most somesmall ε >
TilePacking stays the same and the point setis in general position.
TilePacking will produce a new tile packing T ′ . The upper staircasepoints of each tile t move by at most ε in either direction and thus A t may increase by atmost O( ε ). Hence A ( T ′ ) ≤ A ( T ) + | P | O( ε ), which proves the statement for ε → ◀ The following definition captures a “local” version of convexity. We use in Section 3 toderive a lower bound on the area covered by a tile packing. ▶ Definition 17.
A function f : I → R is said to be point-convex in x ∈ I if f is differentiablein x and the tangent t of f in x satisfies t ( x ) ≤ f ( x ) for all x ∈ I .An immediate consequence of Definition 17 is that Jensen’s inequality holds for f in x . Thatis, if f is point-convex in x and x = P ki =1 γ i · x i is a convex combination of x i ∈ I (i.e., γ i ≥ P ki =1 γ i = 1), then f ( x ) = t ( x ) = k X i =1 γ i · t ( x i ) ≤ k X i =1 γ i · f ( x i ) . (7) ξ s ( ρ ) ρ . ρ ∗ t ξ . . Figure 13
The plot shows our strong lowerbound function ξ s from Section 5. The greenline is its tangent t ξ in ρ ∗ = ξ − s (3 / ξ s is point-convex in ρ ∗ . I I I I I Figure 14
A generalized tile with four sec-tions formed by the five intervals I to I (ofwhich only I is a proper interval). They form astep (between I and I ), a double-step (between I and I ), a slide ( I ), and a corner ( I , I ). B Details for the Improved Bound
In this section we show the omitted details for the strong covering guarantee of Section 5.Note that we will use [ F ( x )] ba := F ( b ) − F ( a ) as a shorthand notation. Additionally we willuse ⃗x → ⃗x ′ as a notation for substituting ⃗x with ⃗x ′ . Due to Lemma 13 we will always assumethat t satisfies | Γ t \ h t | ≤ ▶ Remark 18.
All transformation Lemmas obviously hold up to mirroring at the ( x = y )-axis. ▶ Observation 19.
In the following we will make often use of very similar arguments to ruleout specific structures of tiles. For this argument we define the area | t ( ⃗x ) | and crown C ( ⃗x ) ofa tile as a function of some (general) parameters. Usually these will be the x i = x ( q i ) ∈ Γ t ,but we leave them as general parameters for now. We then introduce a transformation onthese parameters, i.e., ⃗x ( ε ) with ⃗x (0) = ⃗x as well as defining ⃗X = ⃗x ′ | ε =0 and ⃗Y = ⃗x ′′ | ε =0 .As our transformation needs to leave the area invariant we will enforce that | t ( ⃗x ( ε )) | = | t ( ⃗x ) | for all ε .We then obtain that ∂ ε C ( ⃗x ( ε )) | t ( ⃗x ( ε )) | | ε =0 = ∂ ε C ( ⃗x ( ε )) | ε =0 | t ( ⃗x ) | = ⃗X ⃗ ∇ C ( ⃗x ) | t ( ⃗x ) | Assuming that all parameters can be increased or decreased, the only way we cannotimprove the ratio with a transformation is when: ⃗X ⃗ ∇ C ( ⃗x ) = 0 (8)As it is still possible that our transformation is in a local maximum of, we therefore alsoanalyze the second derivative: ∂ ε C ( ⃗x ( ε )) | t ( ⃗x ( ε )) | | ε =0 = ∂ ε C ( ⃗x ( ε )) | ε =0 | t ( ⃗x ) | = ∂ ε x ′ j ( ε ) C j ( ⃗x ( ε )) | t ( ⃗x ) | | ε =0 = x ′′ j ( ε ) C j ( ⃗x ( ε )) + x ′ j ( ε ) x ′ k ( ε ) C jk ( ⃗x ( ε )) | t ( ⃗x ) | | ε =0 = ⃗Y ⃗ ∇ C ( ⃗x ) + ⃗XH C ⃗X | t ( ⃗x ) | Where C j is shorthand for ∂ x j C ( x i ) and H C is the Hessian of C ( x i ). Thus, we have alocal minimum and can thus not improve the ratio, as long as: ⃗Y ⃗ ∇ C ( ⃗x ) + ⃗XH C ⃗X ≥ | t ( x i ( ε )) | = | t ( x i ) | holds, the only needed information of thetransformation is ⃗X, ⃗Y . As we have shown, the above equations holding is necessary,as otherwise a transformation could improve the tile. Another possibility is, that thetransformation itself is not possible as an involved parameter is on a boundary and cannotbe moved further. We thus can use these equations not holding for specific transformationsto follow that these parameters have to be on a boundary and thus rule out specific tilestructures. ▶ Lemma 20.
Let t be a tile with the leftmost sections being up to one step followed by upto one slide. Then these sections can be replaced by up to one step s = ( q , q ) and up to oneslide h = ( q , q ) such that If s exists, then x x ≥ / √ PPENDIX 17 If h exists, then x x ≤ / √ giving us a tile ˜ t ⪯ t . Proof.
We assume that both leftmost sections, i.e., a step s = ( q , q ), followed by a slide h = ( q , q ) exist. The other cases can be treated similarly.We analyze how | t | and | C t | change as we move q and q along h t . To obtain ˜ t ⪯ t , wemust have | ˜ t | = | t | . This is satisfied if 1 + ( x − x ) /x + ln( x /x ) is left constant. Using thesubstitution x = − x ( α + ln( x )), this term becomes 2 + α + ln( x ), which is independentof x .As we also require that the crown gets smaller, we analyze the contribution C of bothsections to the crown of t . For D = 1 + α + ln( x ) + ( x − x − ): C = | T ( q , q ) | + H ( q , q )= 12 (cid:18) x − x + x − x (cid:19)(cid:18) x + 1 x (cid:19) + (cid:20) ln( z ) + 14 ( z − z − ) (cid:21) x x = D − (cid:18) α + ln( x )) − x + (1 + 2 α (1 + α ) + 2 ln( x )(1 + 2 α + ln( x ))) x (cid:19) ∂C∂x = (1 + α + ln( x )) (cid:0) ( α + ln( x )) − − x (cid:1) x = ( x − x ) (1 − x x )2 x x (Note that the area is only dependent on α and x and not x , so differentiation w.r.t. x isjustified.) The last term is positive iff x x < / √
2. In this case we can decrease x untileither x x = 1 / √ x x > / √
2, we canincrease x until either x x = 1 / √ ◀▶ Lemma 21.
Let t be a tile with a step ( q , q ) and a slide ( q , q ) . If x ≥ /x , then thetwo sections can be replaced by a slide ( q , q ) and a step ( q , q ) , resulting in a tile ˜ t ⪯ t . Proof.
To obtain ˜ t ⪯ t , the tile has to keep its area upon replacing the sections, so ( x − x ) /x + ln ( x /x ) = ln ( x /x ) + ( x − x ) /x . This is fulfilled if we set x = x x x .Let C be the contribution of both sections to C t and ˜ C be the contribution of bothsections to C ˜ t . We are interested in the difference ∆ between both: C = | T ( q , q ) | + H ( q , q )= 12 (cid:18) x − x + x − x (cid:19)(cid:18) x + 1 x (cid:19) + (cid:20) ln( z ) + 14 ( z − z − ) (cid:21) x x ˜ C = H ( q , q ) + T ( q , q )= (cid:20) ln( z ) + 14 ( z − z − ) (cid:21) x x + 12 (cid:18) x − x + x − x (cid:19)(cid:18) x + 1 x (cid:19) = (cid:20) ln( z ) + 14 ( z − z − ) (cid:21) x x x x + 12 (cid:18) x x x − x + x − x x x (cid:19)(cid:18) x x x + 1 x (cid:19) ∆ = ˜ C − C = ( x − x ) ( x − x )(1 − x x )4 x x x (1 , q q q x x x T ( q , q ) T ( q , q ) Lemma 22 (1 , q q q x x x q x T ( q , q ) H ( q , q ) Figure 15
Merging two steps in Lemma 22. ≤ ( x − x ) ( x − x )(1 − x x )4 x x x = 0This immediately gives us a new tile ˜ t ⪯ t . ◀▶ Lemma 22.
Let t be a tile and ( q , q ) , ( q , q ) be steps with x ≤ . Then the two stepscan be replaced with a step ( q , q ) and a slide ( q , q ) resulting in a tile ˜ t ⪯ t . Proof.
See Figure 15 for a depiction. We will look at a generalized tile t ′ with two steps( q , q ) , ( q , q ) and a slide ( q , q ). t can be understood as having q = q , such that theslide vanishes, and ˜ t corresponds to q = q . We will move q to the left on the hyperbolaand let q move such that the transformation is area-preserving. t ′ is then obtained whenboth points coincide. Therefore we show ∂ | C t ′ | /∂x ≥ x := f ( x ) for areapreservation. We first look at a part u of the tile area which has to stay constant. This thenalso implies a formula for the derivative of x = f ( x ) w.r.t. x : u := 1 + ( x − x ) /x + ( x − x ) /x + ln( x /x )= 1 + ( f ( x ) − x ) /f ( x ) + ( x − f ( x )) /x + ln( x /x ) ∂f ( x ) ∂x = f ( x ) ( x − f ( x )) x ( x x − f ( x ) )= x ( x − x ) x ( x x − x )Solving the above area equality for x we obtain the two solutions x ± = 12 ( y ± p y − x x )where y = (3 − u + ln( x /x )) x . Note that at least one of the solutions must be real, so y > x x has to hold. We will show that x = x +2 always yields a smaller crown area. Tosee this, look at the difference ∆ of the contribution C of the sections to the crown area: C = | T ( q , q ) | + | T ( q , q ) | + H ( q , q )= 12 (cid:18) x − x + x − x (cid:19)(cid:18) x + 1 x (cid:19) + 12 (cid:18) x − x + x − x (cid:19)(cid:18) x + 1 x (cid:19) + PPENDIX 19 (cid:20) ln z + 14 ( z + z − ) (cid:21) x x ∆ = C | x = x +2 − C | x = x − = (1 − x x ) p y − x x ( y − x − x )2 x x = (1 − x x ) p y − x x ( x − x )( x − x ) / ( − x )2 x x ≤ < x < x < x ≤ ∂f ( x ) /∂x from above, we get for C := C | x = f ( x ) : ∂C∂x = ( x − x ) X x x x ( x − x x )where X = x x + x x (1 − x x + x x ) + x ( x x − x + 2( x − x − x x ). The term( x − x ) / (2 x x x ( x − x x )) is positive since x = ( y + p y − x x ) / ≥ y/ ≥ √ x x ,so it only remains to show X ≥
0. We apply the substitution x = αβ x and x = βx .The constraints 0 < x < x < x ≤ , x ≥ √ x x then rearrange into 0 < α, β, x ≤ X : X =1 − αβ (2 − β + (1 − α ) β ) − α (1 − β )(2 − α − αβ ) β x ≥ − αβ (2 − β + (1 − α ) β ) − α (1 − β )(2 − α − αβ )=(1 − α )(1 − α (1 − (1 − β ) β )) ≥ (1 − α ) ≥ ◀▶ Lemma 23.
Let t be a tile. Then either there exists a tile ˜ t ⪯ t with Γ ˜ t ⊂ h ˜ t , or t containsa double step ( q , q ) with x ≤ ≤ x or a corner ( q , q ) with x < (if q / ∈ h t , x > if q / ∈ h t ). Proof.
By assumption we only have at most one point q ∈ Γ t which lies not on the hyperbola.If there is no such point, then we are done, as such we assume this point exists.First consider the case where q is part of a corner. (w.l.o.g. we assume it to be of theform ( q, q ), where q / ∈ h t . We then have x < q is then C q = ( x + x )( y − x + x − x ). We will move q to the top left or bottom right such thatthe tile area does not change. This transformation is x → x + x ( x y − x ( y + ε ) − , y → y + ε , whichas can be easily checked leaves the tile area invariant. Using Observation 19 this leads toEquation (9): x − xx − ≥
0, which is easily checked to be false since 0 < x < x < q is on the hyperbola.Now assume that q is part of the double step ( q , q, q ). We denote h = x − y, w = x − x , h = y − x , w = x − x . W.l.o.g. assume that x <
1, we then denote by q and q the two leftmost points. As q ∈ h t , we get that the crown associated with these two pointsis C = ( x − x + y − y )( x + y ). It is easy to check that moving q down decreases the crown as well as | t | .s such, we are only interested in cases where h > h and w > w , asotherwise it is easy to see that we can move q up or right, respectively. This decreases | C t | while increasing | t | . Together with moving q down, we can thus reduce C t further whileleaving | t | invariant until either q degenerates, or q reaches the hyperbola. At the end thisyields a corner, which we have already dealt with.For the remaining cases we move q such that the tile area remains the same, so w h =( w + ε )( h + δ ) (with one of ε , δ being negative), or equally δ = w h / ( w + ε ) − h We thusobtain the transformation h → h − δ , w → w + ε , h → h + δ , w → w − ε .The crown is then given by: C ( ε ) = 12 ( h + w )( x + 1 /x − h ) + 12 ( h + w )( x + w + 1 /x − h − h )For the derivatives w.r.t. ε at ε = 0, we get ∂C ( ε ) ∂ε (cid:12)(cid:12)(cid:12)(cid:12) ε =0 = 12 (cid:18) h ( h − h ) w − w + w (cid:19) ∂ C ( ε ) ∂ε (cid:12)(cid:12)(cid:12)(cid:12) ε =0 = ( h − h ) h w − h < w , as otherwise we get for the first derivative (cid:16) h ( h − h ) w − w + w (cid:17) ≤ (cid:16) w − w − h w (cid:17) < w ≤ h and w < w ≤ h . As wethus find a direction in which our transformation lowers the crown area.Additionally, if the second derivative is negative, C ( ε ) has no minimum and thus theminimum has to be on the boundary. So it remains to handle the case where ( h − h ) h ≥ w , h > h , w > w and w > h . We move q to its new location q ′ = ( x − ( h − γ ) , y +( h − γ )).Since x < q ′ remains below the hyperbola. γ can be chosen such that A = B : For γ = 0,we have A = 0 < ( w − h ) h = B and for γ = h we have A = w h > B . Since A and B change continuously while we change γ in [0 , h ], there must be some γ where A = B andtherefore γw = ( h − γ )( w − ( h − γ )). Note that w ′ > w ′ reduces to γ < h − w − w ,which can be easily checked to be false. As such the resulting tile can be dealt with by theabove method, and it remains to show that crown area is reduced:∆ = 12 ( h ′ + w ′ )( x + 1 /x − h ′ ) + 12 ( h ′ + w ′ )( x + w ′ + 1 /x − h ′ − h ′ ) −
12 ( h + w )( x + 1 /x − h ) −
12 ( h + w )( x + w + 1 /x − h − h )= 12 ( h − w + 3 h )( h − w ) + ( − h + w − h + w ) γ + γ It is easily seen that ∂ ∆ /∂h < γ is independent of h . As we want to show∆ < h to be minimal, i.e., h = w /h + 2 h . We thus obtain:∆ ≤ ( w − w h + 5 h )( h − w )2 h + w γ − w γh + ( − h + w ) γ + γ It is more involved but still straightforward to check that the derivative of this term withrespect to w is also negative. (Note that γ is dependent on w ). Thus choosing w minimal,i.e., w = h we obtain:∆ ≤ γ ( w − w h − h + γ ) ≤ w ≤ w = h and 0 < γ < h . With this we canalways find ˜ t with the required properties. ◀ PPENDIX 21 ▶ Lemma 24.
Consider a tile t with ρ t ≤ / . Then there exists a tile ˜ t ⪯ t , containing atmost one step. Furthermore, if Γ t ⊆ h t then Γ ˜ t ⊆ h ˜ t . Proof. If t only contains one step we are already done. Otherwise, we denote by s = ( q , q )and s ′ = ( q , q ) the leftmost and rightmost step respectively. Assume that no transformationsare available from Lemma 22, Lemma 21 and Lemma 23. We show that ρ t > / q ̸ = q , then at least one section has to exist between s and s ′ . If we assume1 < x < x < x < x (or its mirror), Lemma 22 and Lemma 21 rule out a step or slide,and Lemma 23 exclude a corner/double-step. As such we can assume that x < < x . Wethen differentiate between the cases x < < x (or equivalently its mirror) and x ≤ ≤ x .In the first case, there can only be one slide between s and s ′ . In both cases attaching aslide before s or after s ′ is impossible: Either we get a slide-step-slide sequence, which isimpossible by Lemma 21, or we get a step-slide sequence, which is impossible by Lemma 21.As such s ′ and s are both boundary steps. Since Lemma 20 was inapplicable, we get that x x ≥ / √ y y ≥ / √
2. From x < x and y < y we then get 1 / √ < x and1 / √ < y . With this we bound | t | from above: | t | ≤ x − x ) /x + ln( x /x ) + ( x − x ) /x =3 − x /x − ln( y x ) − y /y ≤ − ( x − / √ x ) − ( y − / √ y )The term z − / √ z is minimized at z = √ /
4. We conclude ρ t = 1 / | t | ≥ / (3 − / ≈ . > / q = q . We get again that x < < x by Lemma 22. This means thatLemma 23 does not allow the existence of a corner/double-step. So s, s ′ are the outermoststeps, and only slides can exist beside s and s ′ . If no slide exists, we end up in the caseabove. By symmetry we can w.l.o.g. choose x < ≤ x , which means that Lemma 21 doesnot permit a slide after s ′ . Therefore a slide h = ( q , q ) exists. By Lemma 21 we know that x > /x . Since s ′ is a boundary step, Lemma 20 gives us 1 / ( x x ) ≥ / √
2. The total areaof | t | is then: | t | =1 + ln( x /x ) + ( x − x ) /x + ( x − x ) /x =3 − x /x − x /x + ln( x /x ) ≤ − x /x − x / √ x x ) ≤ − /x − x / √ x ) ≤ . x being 1 /x − /x > x was maximized. We conclude ρ t = 1 / | t | > / ≤
1. As none of these transformations moves pointsin Γ t away from the hyperbola, this gives the second property. This gives us the required tile˜ t ⪯ t . ◀▶ Lemma 25.
Let t be a tile with a double step ( q , q , q ) and a slide ( q , q ) . We canreplace both sections by a double step ( q , q ) or a sequence of steps and slides between q and q , obtaining a tile t ′ ⪯ t . Proof.
We have that | t R ( ⃗q ) | = 1 + ( x − x ) y + ( x − x ) /x + log( x /x ) and C t R ( ⃗q ) = T ( q , q ) + T ( q , q ) + H ( q , q ) = ( x − x + x − y )( x + y ) + ( x + x )( − x − x + x + y ) + ( − x + x + x − x + 4 log( x x )). Where t R is the part of the tile consistingof the double step/slide. We will transform the tile by moving q up/down while moving q on the hyperbola to leave the tile area invariant. We define this transformation by( y → y + ε, x → − x /W ( − e − x /x x ε − x ε x x )), where W denotes the main branch of theproduct log function. This transformation leaves the rest of the tile invariant, and it is thusstraightforward to check, that | t ( ⃗q ( ε )) | = | t ( ⃗q ) | holds, which as outlined in Observation 19leads to Equation (9)∆ : = 4 x x − x − x x + x (1 + 4 x ) − x x (4 + 2 x − x y ) − x x ( − x ( − x − x ) x + y )) − x (2 x x − x (1 + 4 x )+ x (2 + 2 x − x y )) ≥ ∂ ∆ ∂y = ( x − x ) x > y . Note that x y < q is not onthe hyperbola, thus, inserting y = 1 /x we get that ∆ ′ = x + x − x x − x − x ) x x ≥ x < x < x holds, ∂ ∆ ′ ∂x = 2 x + 4 x ( − x + x ) x > x = x as the maximal possible value for ∆ ′ , we obtain ∆ ′ | x = x = 2 x ( x − x ) < C t , the transformation is thus always possible, unless either the slideis exhausted or q reaches the hyperbola/degenerates. ◀▶ Lemma 26.
Let t be a tile with a slide ( q , q ) and a corner ( q , q ) . We can replace bothsections by a corner ( q , q ′ ) or a sequence of steps and slides between q and q ′ , obtaining atile t ′ ⪯ t . Proof.
We have | t R ( ⃗q ) | = 1 + log( x /x ) + ( x − x ) y and C t R ( ⃗q ) = H ( q , q ) + T ( q , q ) = (1 /x − x − /x + x +4 log( x /x ))+ (1 /x − x + x − y )( x + y ). Where t R is the partof the tile consisting of the corner/slide. We define the transformation x → x + kε , x → x + β , y → y + ε . This transformation leaves the tile besides t R invariant, and as such requiringpreservation of tile area, leads to β = ε + kε +log( x ) − log( x ( x ε − x ε + ky + kε )) / ( y + ε ).Thus | t ( ⃗q ( ε )) | = | t ( ⃗q ) | holds, and we can follow Observation 19, and analyze the resultingequations. Note that the defined transformation is dependent on k , and that the equationshave to be satisfied for every k , as otherwise there is a k defining a valid transformationwhich reduces C t .For k = 0, this results for Equation (8) in x ( x + ( x ( x − x )) /y + 2 y ) = 1, whichreduces to x = ( x + y − x y − x y ) /x . Inserting this into Equation (9), again for k = 0, we obtain y < (1 − x ) / (3 x ). Returning to a general k for Equation (9), we obtaina quadratic polynomial in k . The coefficient of k then has to be positive, as otherwisewe can chose k big enough so that Equation (9) is no longer satisfied. This results in − x + ( x − x − y + 2 x y > y , so only the boundary values (0 < y < (1 − x ) / (3 x )) are ofinterest. For y = 0 the equation does not hold, and inserting y = (1 − x ) / (3 x ) yields − (7 − x + x + 3 x ) / (9 x ) >
0, which has a single maximum for x ≈ .
767 at ≈ − . C t , the transformation is thus always possible forsome k , until either the slide is exhausted or q reaches the hyperbola/degenerates. ◀▶ Lemma 27.
Let t be a tile with ρ t ≤ / , then we can find a tile ˜ t ⪯ t with one of thefollowing properties: PPENDIX 23 Γ ˜ t ⊆ h ˜ t ˜ t consists of only a step and a double step ˜ t consists of only a double step ˜ t consists of only a step and a corner Proof.
By Lemma 24 we first assume one step exists. If | Γ t \ h t | = 0, we are done, so weassume | Γ t \ h t | = 1. t then has to contain a double step or corner, which by Lemma 23 islocated around 1. We can assume that any existing slide is adjacent to the double-step/corner.This is true as otherwise the sections have to be separated by the step, meaning we can useLemma 21 to exchange the step and the slide to make them adjacent, obtaining a tile ˜ t ⪯ t .By this adjacency we can use Lemma 25 or Lemma 26 to either obtain | Γ t \ h t | = 0, or thenonexistence of slides. As such our tile can only consist of a step and a double-step/corner.The possibility of t just consisting of a corner can be excluded, as its density would be largerthan 1 /
2. With this we obtain the result. ◀▶ Lemma 28.
Let t be a tile with ρ t ≤ / with Γ t ⊆ h t . Then | C t | / | t | ≥ ξ s ( ρ t ) . Proof.
By Lemma 24 we can assume that t contains at most one step. Since Γ t ⊆ h t , t can only consist of steps and slides. Using Lemma 21, we can ensure that t has at mostone slide. (Note that a slide must exist, as otherwise ρ t > / | t | ξ s ( ρ t ) − | C t | <
0, since the statement follows by rearranging.First assume that t ’s only section is a slide ( q , q ). Then | t | = 1 + ln( x /x ), and we get:∆ = | t | ξ s ( ρ t ) − H ( q , q )= | t | − | t | − − (cid:2) ln z + ( z − z − ) / (cid:3) x x = (cid:0) x − − x + x − x − (cid:1) / | t | −
1) + | t | − (1 + ln( x /x ))= (cid:0) x − − x + x − x − (cid:1) / x /x ))= 2 sinh((ln( x ) + ln( x )) / sinh(ln( x /x )) < x < x .Now assume that t consists of a step ( q , q ) and a slide ( q , q ) (w.l.o.g. ordered inthis way). In such a case we have | t | = 1 + ( x − x ) /x + ln( x /x ), or rearranged, x = x e x /x + | t |− . Again we calculate∆ = | t | ξ s ( ρ t ) − ( | T ( q , q ) | + H ( q , q ))= | t | − | t | − − (cid:18) x − x + x − x (cid:19)(cid:18) x + 1 x (cid:19) − (cid:20) ln z + z − z − (cid:21) x x Taking the derivative of ∆ w.r.t. | t | (after inserting x = x e x /x + | t |− ), we obtain ∂ ∆ /∂ | t | = − x /x + | t | − x )) <
0. This indicates that ∆ is maximized for smallest | t | .So assume | t | = 2 now, or equivalently, x = x e x /x . By Lemma 20 we can further assumethat x = 1 / ( √ x ). Using the substitution x = 2 − / √ u we get∆ = 12 √ (cid:18) − − √ e + √ e + 1 − e u u + u + u e u (cid:19) ∂ ∆ ∂u = (1 − u )( u + 2 e u − e u (1 + u ))4 √ u e u The derivative above has only one zero, namely u = 1: From 0 < x ≤ x x = 1 / √
2, we candeduce u ∈ (0 ,
2] by the substitution. For the right factor of the derivative’s numerator we then get u + 2 e u − e u (1 + u ) > e u ( e u − (1 + u )) > e u . Hencechecking ∆ at u ’s boundaries and u = 1 is sufficient, where we get lim u → ∆ ≈ − . < − e u ) /u ), ∆ | u =1 ≈ − . < | u =2 ≈ − . < ◀ The following Lemma 29 allows us to restrict ourselves to tiles t with ρ t = 1 / t are on t ’s hyperbola. ▶ Lemma 29.
Let t be a tile with ρ t ≤ / . Then there exists a tile ˜ t with Γ ˜ t ⊆ h ˜ t or ρ ˜ t = 1 / such that, if | C ˜ t | / | ˜ t | ≥ ξ s ( ρ ˜ t ) then also | C t | / | t | ≥ ξ s ( ρ t ) . Proof.
If Γ t ⊆ h t , then the statement is trivial using ˜ t = t . Otherwise let { q i } = Γ t \ h t .W.l.o.g. assume that there is a point in q i − ∈ Γ t with x i − < x i (otherwise look at thesymmetric case; another point has to exist since h t ∩ Γ t ̸ = ∅ ). We construct a new tile ˜ t with ρ ˜ t ≤ / q up by ε . This changes the area, i.e., | ˜ t | = | t | + εw for w = x i − x i − .It is then sufficient to show | C t | − ξ s ( ρ t ) | t | > | C ˜ t | − ξ s ( ρ ˜ t ) | ˜ t | .To see this, consider the term | C t | − ξ s ( ρ t ) | t | = | C t | − ( | t | − | t | − X as wemove q . Since 1 or 2 towers are affected by moving q (depending on whether q is part of adouble step or a corner), the contribution of those towers to the curvature of X is − α = 1. Hence ∂X ∂ε = ∂ | C t | ∂ε − ∂∂ε ( | t | + εw − | t | + εw − − · O(1) − w sinh( | t | − εw )= − · O(1) − w sinh( | ˜ t | ) < ∂X ∂ε < ε ’s sign can be chosen such that X = | C t | − ξ s ( ρ t ) | t | decreases (hence we obtain˜ t with | C t | − ξ s ( ρ t ) | t | > | C ˜ t | − ξ s ( ρ t ) | ˜ t | ). ε ’s magnitude can be chosen such that q i is moved up onto the hyperbola or q i is moveddown and the tile degenerates, which can be transformed into a new tile ˜ t by Lemma 12with ˜ t ⪯ t . We assumed that ρ ˜ t ≤ /
2, so the transformation might stop when the densitybecomes exactly 1 / ◀▶ Lemma 30.
Let t be a tile with ρ t ≤ / consisting only of a double-step ( q , q , q ) . Then | C t | / | t | ≥ ξ s ( ρ t ) . Proof.
By Lemma 29 we may assume ρ t = (or | t | = 2). We can express this in terms of x i , y i to get | t | = 1 + ( x − x ) y + ( y − x ) y = 2, or equivalently, x y = x y + x y . Withthe substitution χ = x y , γ = x /y , α = y /y , β = x /x , or rearranged, x = β √ γχ , x = √ γχ , y = p χ/γ , y = α p χ/γ , this equation becomes χ (1 − α − β ) = 0. We get for C t : | C t | = | T ( q , q ) | + | T ( q , q ) | = 12 (cid:18) x − y + x − x (cid:19) ( x + y ) + 12 (cid:18) y − y + 1 y − x (cid:19) ( x + y )= 1 + 12 (cid:18) γ α + 1 γ β + 1 γ χα + γχβ − γ χα − γχβ − γ χ − γχ (cid:19) + χ (1 − α − β )= 1 + 12 (cid:18) γ α + 1 γ β + 1 γ χα + γχβ − γ χα − γχβ − γ χ − γχ (cid:19) ∂ | C t | ∂χ = 12 (cid:18) γ ( α (1 − α ) −
1) + γ ( β (1 − β ) − (cid:19) < α, β ∈ (0 ,
1) and γ > χ is justified as the area inequality is independent of χ ̸ = 0.) From this we obtain a PPENDIX 25 tile t ′ ⪯ t (namely the one with χ = 1 and same α, β ). Since χ = 1, t ′ is consists of exactlytwo steps. By Lemma 24, we get another tile t ′′ ⪯ t ′ with at most one step and Γ t ′′ ⊆ h t ′′ .The statement then directly follows from Lemma 28. ◀▶ Lemma 31.
Let t be a tile with ρ t ≤ / consisting only of a double step ( q , q , q ) and astep ( q , q ) . Then | C t | / | t | ≥ ξ s ( ρ t ) Proof.
Then | t ( ⃗q ) | = 1 + ( x − x ) y + ( x − x ) /x + ( x − x ) /x and | C t ( ⃗q ) | = T ( q , q ) + T ( q , q ) + T ( q , q ) = 1 / · (1 /x − y + x − x )( x + y ) + 1 / · ( y − /x + x − x )( x +1 /x ) + 1 / · (1 /x − /x + x − x )( x + 1 /x ).We define the transformation ( x → x + ε, x → x + δ ). We then require | t | to be leftinvariant under this transformation, which is true for δ = x ( − x / ( x + x ( x y − ε )).Following Observation 19 for this transformation we obtain for Equation (8): y = 1 /x + ( x ( x − x + x ) x ) / ( − x + x (2 + x x ))Inserting this in Equation (9) we obtain − x + 4 x x − (1 + x ( x − x + x ) ) x − x x + 2 x x + x ( x − x + x ) x − x x ( x − x (2 + x x ) ≥ As the denominator is trivially positive, this is equivalent to: x − x x + (1 + x ( x − x + x ) ) x + 4 x x − x x − x ( x − x + x ) x + x x ≤ The curvature of the term with respect to x is easily checked to be − x x ( x x − < x ( x < x < x ). The resulting term turns out to be the same for x = x and x = x ,namely 4 x − x x + (1 + ( x − x ) x ) x + 4 x x − x x − ( x − x ) x x + x x . Thisterms curvature with regards to a is − x x ( x x − <
0, meaning again only the boundary0 < x < x is relevant. For x = x we obtain ( x − x (2 + x x )) > x = 0 weget 4 x − x x + x + x x + 4 x x − x x − x x + x x . The third derivative of thisw.r.t. x is 12 x (2 x − x )(5 x x − >
0, meaning the second derivative will be minimal forminimal x = x , with this the second derivative is 2(1 + x + 5 x ) >
0, meaning again onlythe boundary terms of x ( x < x < ∞ ) are relevant. For these we get x + 3 x > ∞ > q degenerates/hits the hyperbola, or q degenerates. We thuscan transform t such that either Γ t ⊆ h t , or consists of just a double step. In the former caseLemma 28 gives the required property. In the latter case Lemma 30. ◀▶ Lemma 32.
Let t be a tile with ρ t ≤ / consisting only of a step ( q , q ) where x x ≥ / √ and a corner ( q , q ) . Then there exists a tile t ′ ⪯ t only consisting of two steps. Proof.
Let x := 1 /y . We will first attempt to move q such that | t | = 1 + ( x − x ) /x +( x − x ) /x is invariant, or equivalently, x = ( x − x ) /u for some constant u . In order for | C t | to decrease, it suffices to look at | T ( q , q ) | : | T ( q , q ) | = 12 (cid:18) x − x + x − x (cid:19)(cid:18) x + 1 x (cid:19) = 12 (cid:18) x − ux − x + x − x (cid:19)(cid:18) x + ux − x (cid:19) ∂ | T ( q , q ) | ∂x = x (2 + x x (1 + ( x − x ) x )) − x x ( x − x ) x As we assume that q cannot be moved such that | C t | decreases, we assume that the lastterm is 0, or equally ( x − x ) /x = ( x − x (2 + x x )) / ( x x ). Since x ≥ √ x and x ≥ √ x x ≥ − / , we have | t | = 1 + ( x − x ) /x + ( x − x ) /x ≤ x − / ( √ x )) /x + ( x − x (2 + x x )) / ( x x )= 2 + x − ( x − − − / ) − x − x − − x − < / x − ( x − − − / ) − x − x − = 2 + x − x − (2 / x − − − / x ) ≤ x − x − (2 √ / − < ρ t ≤ /
2, so q must be movable until it lies on thehyperbola, where we get the proclaimed tile t ′ ⪯ t . ◀ We are now ready to show Proposition 14.
Proof of Proposition 14.
By Lemma 4 we already get the result for ρ t > /
2. For tiles with ρ t ≤ / t ′ ⊆ h t ′ . Lemma 28 yields the result. t consists of only a double step. Lemma 30 yields the result. t consists of a step and a double step. Lemma 31 yields the result. t consists of a step and a corner. Lemma 32 yields the result.As such the bound follows.It remains to show the tightness. Note that ξ s exactly corresponds to the tiles t l ( s ) and t h ( s ) shown in Figure 8 and Figure 9: Both tiles are symmetric tiles. First note that for ρ t = 1, the tightness is trivial by choosing an arbitrary tile t with exactly one upper staircasepoint.Let t = t l ( s ) and define Γ t = { q , q } with q = ( v, /v ) and q = (1 /v, v ) for proper v ∈ (0 , | t | = 2 − v , or rearranged v = p − /ρ t and we get | C t | / | t | = | T ( q , q ) | / (2 − v )= 12 (1 /v − v + 1 /v − v )( v + v ) / (cid:0) − v (cid:1) = 2 − ρ t Now let t = t h ( s ). We define Γ t = h t ∩ { ( x, y ) | v ≤ x ≤ /v } . t then consists of exactlyone slide ( q , q ) = (( v, /v ) , (1 /v, v )). With | t | = 1 + ln((1 /v ) /v ) = 1 − v ) ↔ v = e − ρt ,we get | C t | / | t | = | H ( q , q ) | ρ t = ρ t (cid:20) ln( z ) + 14 (cid:0) z − z − (cid:1)(cid:21) e ρt − e − ρt = 1 − ρ t (1 + sinh(1 − /ρ t ))Note that t is only valid in the sense of generalized tiles. However, t can be arbitrarilywell approximated by a non-generalized tile with area and crown size arbitrarily close to | t | and | C t | . ◀ PPENDIX 27 A (0 , q k − f k − f k f f f − k +1 f − k q − k +1 q − f − (cid:0) √ , (cid:1) h A q q ...... ... ... ... ... ... ε ... u xy Figure 16
Construction with k = 4. x ( p ) x ( p ) + εp p (cid:48) tXYf i f i +1 f i − wh Figure 17
Example fora tile created by p ∈ P ε,k .For demonstration purposes,the light blue areas are notshown to scale; in truththey are significantly smallerthan X, Y . C Full Proof of the Upper Bound
In this section, we give the full details for the construction used to show the upper bound inTheorem 2. Our construction is parametrized by k , A , and ε . The main idea is the following(cf. Section 6): we construct a point set where TilePacking fails to cover an area largerthan roughly (1 − e − ) / U clockwise by 45 degrees and let U r denote the unit square U rotated clockwise by 45 degrees around the origin. Observe that in U r , TilePacking processes the points from right to left.Recall that our goal is to construct an L -shaped tile with low density, where each maximalrectangle has roughly the same size A . To this end, we place a large number of points q i on ahyperbola h A . We then add for each point q i on h A a set of points p i,j with x ( p i,j ) = x ( q i )+ jε where j ∈ Z and p i,j ∈ U r . To get a density close to 1 /
2, the y -coordinates of the points p i,j are chosen as follows: we place the points p i,j on arcs of functions f i described by differentialequations (see below), where each f i depends on the two neighboring curves f i − and f i +1 .Observe that TilePacking processes the outer q i before the inner ones. This means,once the tile for some outer point q i has been fixed, the remaining tiles for of all inner pointson f j will be restricted by t q i .We start by formally defining the point set P ε,k and the functions f i . Let h A = { ( x, y ) ∈ U r | x − y = 2 A } be the right branch of a hyperbola centered in the origin that lives in U r . First we define the upper part of the construction with non-negative y coordintes. For i = 0 , . . . , k −
1, densely choose k points q i ∈ h A such that y ( q i ) ≥ √ A = x ( q ) < x ( q ) < · · · < x ( q k ). For i = 0 and i = k , define f ( x ) = 0 and f k ( x ) = √ − x . For 0 < i < k , define f i : [0 , √ → R using f i ( x ( q i )) = y ( q i ) (10)and f ′ i ( x ) = − x ≤ x ( q i )1 − f i ( x ) − f i − ( x ) f i +1 ( x ) − f i − ( x ) for x > x ( q i ) . (11)This means, each f i with 0 < i < k has slope − , x ( q i )), then it intersects h A in q i according to (10), and then it has a slope depending on f i − and f i +1 according to (11).For the symmetric part with negative y coordinates we define q − i = ( x ( q i ) , − y ( q i )) and f − i ( x ) = − f i ( x ) for 0 < i < k . Observe that q − i ∈ h A and the f − i adhere to (11). Finally,define f − k = − f k . We are now ready to define the point set P ε,k for ε > k ∈ N as P ε,k = { (0 , } ∪ [ i,j ∈ Z n (cid:0) jϵ, f i ( jϵ ) (cid:1) (cid:12)(cid:12)(cid:12) − k < i < k, x ( q i ) ≤ jϵ < √ o . We assume that ε and the q i are chosen such that ε divides all x ( q i ). This implies that q i ∈ P ε,k for all − k < i < k . We now show basic properties for the functions f i . ▶ Lemma 33.
Each function f i intersects h A exactly once, namely in q i . Furthermore, f i ( x ) is differentiable for all i = − k, . . . , k , and f − k ( x ) < · · · < f k ( x ) . Proof.
By the choice of the q i , we have f − k (0) < · · · < f k (0). As long as f − k ( x ) < · · · < f k ( x ), the f i +1 ( x ) − f i − ( x ) denominators in the differential equations are positiveand therefore the functions are by definition differentiable and thus continuous. Then wecan also show that each f i cannot have more than one shared point with h A : we have | f ′ i | = | − f i ( x ) − f i − ( x )) / ( f i +1 ( x ) − f i − ( x )) | ≤
1. W.l.o.g. consider the upper branchof h A with the functional form h = √ x − A , then we get | f ′ i | ≤ ≤ x/ √ x − A = h ′ , sothere cannot be more than one shared point.Assume now that the functions are not always ordered, then choose the smallest ˜ x < √ f i (˜ x ) = f i − (˜ x ) for some i . W.l.o.g. we can assume that f i +1 (˜ x ) > f i (˜ x ) (otherwisechoose i + 1 instead or consider the symmetric case). Note that such an i must exist, asotherwise all functions f i meet, which is impossible since the functions f k and f − k only meetat x = √ δ := f i − f i − . There exists a v > f i +1 ( x ) − f i − ( x ) ≥ v holds for all x ≤ ˜ x , where f i is well-defined by its differential equation. When evaluating thefunctions at x , we get δ ′ = f ′ i − f ′ i − = 1 − f ′ i − − f i − f i − f i +1 − f i − ≥ − δv where the last inequality follows from f i +1 ( x ) − f i − ( x ) ≥ v , as well as the properties of thederivative. We therefore see that δ decays (at most) exponentially and can never reach 0,contradicting our assumption of δ (˜ x ) = f i (˜ x ) − f i − (˜ x ) = 0. ◀▶ Lemma 34.
Let u, k > and ˆ U = ( S p ∈ P ε,k ,x ( p ) ≤√ − u t p ) \ ˆ t . Then TilePacking covers | ˆ U| / c ( ε ) area in ˆ U for P ε,k where lim ε → c k ( ε ) = 0 . PPENDIX 29
Proof of Lemma 34.
By Lemma 33, we get f − k ( x ) < · · · < f k ( x ), ensuring that all f i residein U r . Now consider a point p ̸ = (0 , x ( p ) ≤ √ − u that lies on some curve f j and createsthe tile t . Assuming ε < u , there exists another point p ′ = ( x ( p ) + ε, f j ( x ( p ) + ε )) ∈ P ε,k .By Lemma 33, we can assume that f i +1 ( x ) − f i ( x ) > ε for all i = − k, . . . , k − x ≤ √ − u .It follows from | f ′ i | ≤ p ′ is a lower staircase point. (For the same reason there cannotbe points further to the right that are lower staircase points.) t is therefore only restrictedby the tiles from points with x -coordinate x ( p ′ ), giving it an L -shape (see Figure 17). TilePacking will choose the larger one of the two maximal rectangles R and R .Depending on the processing order, TilePacking will process any of the points with the x -coordinate x ( p ) first. This gives rise to areas that can be either covered by t or the tile ofa point on the neighboring curve. As we will see, these areas are negligible, so w.l.o.g. weassume that t covers them.Since all f i are differentiable in [0 , √ g ( x ) with lim x → g ( x ) = 0 such that f i ( x + ε ) = f i ( x ) + f ′ i ( x ) · ε + g ( ε ) · ε .Denote by w, h the dimensions of the rectangle r = R ∩ R . Then w = ( ε + f i ( x ( p ) + ε ) − f i ( x ( p ))) / √ ε + f ′ i ( x ( p )) ε + g ( ε ) ε ) / √ f ′ i ( x ( p )) + g ( ε )) ε/ √ h = (1 − f ′ i ( x ( p )) − g ( ε )) ε/ √ Z be the total area of r plus all areas that could have been covered by neighboringtiles of points with the same x -coordinate x ( p ). It is easy to see that Z = O (cid:0) ε (cid:1) . t also contains two additional rectangles with an area of X = w (( f i ( x ( p )) − f i − ( x ( p ))) / √ − h ) and Y = h (( f i +1 ( x ( p )) − f i ( x ( p ))) / √ − w ). Note that this also holds if x ( q i ± ) > x ( p ),as we extended f i ∓ with lines of slope ±
1. In this case the two rectangles are restricted by q i ∓ ’s tile, respectively.Hence, when evaluating the functions at x ( p ): | X − Y | = | w ( f i − f i − ) / √ − h ( f i +1 − f i ) / √ | = | (1 + f ′ i + g ( ε ))( ε ( f i − f i − )) / − (1 − f ′ i − g ( ε ))( ε ( f i +1 − f i )) / | = | ( f i +1 ( g ( ε ) + f ′ i − − f i − ( g ( ε ) + f ′ i + 1)) + 2 f i | · ε/ | ( g ( ε )( f i +1 − f i − ) + f ′ i ( f i +1 − f i − ) − f i − − f i +1 + 2 f i ) | · ε/ | g ( ε ) || f i +1 − f i − | · ε/ ≤ | g ( ε ) | · ε √ f k ( x ) − f − k ( x ) ≤ √ , √ X > Y . Then for the tile t with area | t | = X + Y + Z , TilePacking covers at most Z + X ≤ Z + X/ Y + | g ( ε ) | · ε √ / | t | / ε ( | g ( ε ) | + ε )). As ˆ U is theunion of such tiles and | P ε,k | = O( k/ε ), we have a total coverage of | ˆ U| / k ( | g ( ε ) | + ε )).This immediately gives us the function c k ( ε ) = O( k ( | g ( ε ) | + ε )) with lim ε → c k ( ε ) = 0. ◀▶ Theorem 35.
TilePacking has no better lower bound than (1 − e − ) / . Proof.
We analyze the area ρ covered by TilePacking on P ε,k for some fixed k and u as ε approaches 0. The bound then follows from letting k go to ∞ and u go to 0.By Lemma 34, TilePacking covers half of ˆ U = ( S p ∈ P ε,k ,x ( p ) ≤√ − u t p ) \ ˆ t (plus c k ( ε )that approaches 0 for ε →
0) for each u >
0. Additionally, at most u area is covered fromall tiles at points p with x ( p ) > √ − u . TilePacking covers at most A + Q area in ˆ t , whereat most Q = max i ( x ( q i ) − x ( q i − )) + max i ( y ( q i ) − y ( q i − )) is additionally covered due tothe q i points only providing an approximation of h A . (Note that all q i lie in P ε,k , so noadditional error is introduced.) In total, using E = Q + c k ( ε ) + u , ρ ≤ A + | ˆ U| / E ≤ A + (1 − | ˆ t | ) / E ≤ A + (1 − ( A + Z A Ax d x )) / E ≤ (1 + A + A ln A ) / E Minimizing the last term leads to ρ = (1 − e − ) / E at A = e − . Q approaches 0 when k goes to ∞ since the q i points lie densely on h A ∩ U r . Therefore the error term E approaches0 for large k and small u, ε ..