Non-Existence of Annular Separators in Geometric Graphs

Non-existence of annular separators in geometric graphs

Farzam Ebrahimnejad∗ James R. Lee†

Paul G. Allen School of Computer Science & Engineering University of Washington

Abstract Benjamini and Papasoglou (2011) showed that planar graphs with uniform polynomial volume growth admit 1-dimensional annular separators: The vertices at graph distance ' from any vertex can be separated from those at distance 2' by removing at most $ ' vertices. They asked whether geometric 3-dimensional graphs with uniform polynomial volume( ) growth similarly admit 3 1 -dimensional annular separators when 3 > 2. We show that this fails in a strong sense: For( any− 3) > 3 and every B > 1, there is a collection of interior-disjoint spheres in ℝ3 whose tangency graph has uniform polynomial growth, but such that all annular separators in have cardinality at least 'B.

1 Introduction

The well-known Lipton-Tarjan separator theorem [LT79] asserts that any =-vertex planar graph has a balanced separator with $ √= vertices. By the Koebe-Andreev-Thurston circle packing ( ) theorem, every planar graph can be realized as the tangency graph of interior-disjoint circles in the plane. One can deﬁne 3-dimensional geometric graphs by analogy: Take a collection of “almost 3 non-overlapping” bodies (E ℝ : E + , where each (E is “almost round,” and the associated { ⊆ ∈ } geometric graph contains an edge D,E if (D and (E “almost touch.” { } As a prototypical example, suppose we require that every point G ℝ3 is contained in at most : ∈ of the bodies (E , each (E is a Euclidean ball, and two bodies are considered adjacent whenever { } (D (E ≠ . These are precisely the intersection graphs of :-ply systems of balls, studied by ∩ ∅ Miller, Teng, Thurston, and Vavasis [MTTV97]. Those authors also provide a generalization of the Lipton-Tarjan separator theorem: For : = $ 1 , such an intersection graph contains a balanced

arXiv:2107.09790v1 [math.CO] 20 Jul 2021 ( ) separator of size $ =1 1 3 . ( − / ) Similarly, ﬁnite-element graphs associated to simplicial complexes with bounded aspect ratio can be viewed as subgraphs of geometric overlap graphs [MTTV98], and one again obtains balanced separators of size $ =1 1 3 . This covers a number of scenarios commonly arising in applications of ( − / ) the ﬁnite-element method; we refer to the discussion of well-shaped meshes in [ST07, §6.2].

∗[email protected] †[email protected]

1 We will not be too concerned with the particular notion of geometric graph used since our construction satisﬁes all these commonly employed sets of assumptions. Indeed, it can be cast as the tangency graph of a sphere packing, where adjacent spheres have uniformly comparable radii. It can also be cast as the 1-skeleton of a 3-dimensional simplicial complex whose simplices have uniformly bounded aspect ratio as studied. Such graphs were studied by, for instance, by Plotkin, Rao, and Smith [PRS94] in their work on shallow minors (see also the followup work [Ten98]). Annular separators. Note that the preceding results deal with global separators that separate the entire graph into two roughly equal pieces. In many settings, especially those arising in physical simulation, it useful to consider local separators. Let = +, be an undirected graph with path ( ) metric 3, and deﬁne graph balls and graph spheres, respectively, by G,' := H + : 3 G,H 6 ' ( ) ∈ ( ) ( G,' := H + : 3 G,H = ' . ( ) ∈ ( )

Suppose that for some > 1, we we want to separate ( G,' from ( G, ' by removing a ( ) ( ) small set of nodes * G, ' G,' . We refer to * as an annular separator. See Figure 1. ⊆ ( )\ ( )

(a) Separating ( G,' from ( G, 2' (b) A random triangulation1 of S2 ( ) ( ) Figure 1: Annular separators

It is easy to see that even if is a planar graph, small annular separators don’t necessary exist. But in many cases, one can ﬁnd annular separators of size $ ' . For instance, this is true for most ( ) vertices at most scales in a uniformly random triangulation of the 2-dimensional sphere [Kri05] (a fact which extends experimentally to a variety of other models of random planar maps, e.g., those studied in [GHS20]). These models also have the properties that the cardinality of graph balls tends : to grow asymptotically like G,' ' (as ' , up to lower-order ﬂuctuations), where the | ( )| ∼ → ∞ exponent : depends on the model. For random triangulations, one has : = 4 [Ang03].

1Depiction of a random triangulation is due to Nicolas Curien.

2 Benjamini and Papasoglou [BP11] give an explanation of this phenomenon as follows. Suppose that is an inﬁnite planar graph and we assume, additionally, that has uniform polynomial growth of degree :: There exist numbers , : > 1 such that

1 : : − ' 6 G,' 6 ' , G +, ' > 1. | ( )| ∀ ∈

Then for every G + and ' > 1, there is a set * G, 2' G,' whose removal ∈ ( ) ⊆ ( )\ ( ) disconnects ( G,' from ( G, 2' in , and such that * 6 $ ' . This applies equally well to ( ) ( ) | | ( ) ﬁnite graphs: Indeed, the authors actually show that if the graph metric restricted to G, 2' has ( ) doubling constant , then one can ﬁnd an annular separator of size at most '. We remark that there are rich families of planar graphs with uniform polynomial growth arising in a variety of contexts; see [BS01, BK02, EL20]. Indeed, one can obtain planar graphs with uniform polynomial growth of degree : for all real degrees : > 1. Moreover, many models of random planar graphs have an almost sure asymptotic version of this property [DG20, GHS20]. The authors of [BP11] asked whether an analog of this phenomenon holds in higher dimensions. For instance, if is a graph with uniform polynomial growth that can be geometrically represented in ℝ3, does it hold that has annular separators of size $ '2 ? We give examples showing that for ( ) 3 > 3, this phenomenon fails in a strong way. Say that a graph is sphere-packed in ℝ3 if is the tangency graph of a collection of interior- disjoint spheres in ℝ3. Say that is "-uniformly sphere-packed in ℝ3 if the collection of spheres can be taken such that the radii of any two tangent spheres lies in the interval " 1," , and that is [ − ] uniformly sphere-packed in ℝ3 if this holds for some " < . ∞ Theorem 1.1 (Arbitrarily large annular separators). For every 3 > 3 and B > 1, there is a number 2 > 0 and a graph satisfying:

1. has uniform polynomial growth of degree $ B . ( ) 2. is uniformly sphere-packed in ℝ3.

B 3. For every G + , there are at least 2' vertex-disjoint paths from ( G,' to ( G,' for any ∈ ( ) ( ) ( 0) '0 > ' > 1.

Clearly if has uniform polynomial growth of degree 3, then one of the '-many spheres ( G,' 1 ,( G,' 2 ...,( G, 2' must be an annular separator of size $ '3 1 . We show that ( + ) ( + ) ( ) ( − ) the moment the growth degree exceeds 3, there are graphs sphere-packed in ℝ3 that don’t have 3 1 -dimensional annular separators. ( − ) Theorem 1.2 (Nearly-dimensional growth rate). For every 3 > 3 and > 0, there are numbers 2, > 0 and a graph satisfying:

1. has uniform polynomial growth of degree at most 3 . + 2. is uniformly sphere-packed in ℝ3.

3 1 3. For every G + , there are at least 2' vertex-disjoint paths from ( G,' to ( G,' for ∈ ( ) ( − )+ ( ) ( 0) any '0 > ' > 1.

3 Note that the two preceding theorems refer to infinite graphs. A version for families of finite graphs appears in Theorem 2.7. Preliminaries. We will consider primarily connected, undirected graphs = +, , which we ( ) equip with the associated path metric 3. We write + and , respectively, for the vertex and ( ) ( ) edge sets of . If * + , we write * for the subgraph induced on *. ⊆ ( ) [ ] For E +, let deg E denote the degree of E in . Let diam := sup 3 G,H denote ∈ ( ) ( ) G,H + ( ) the diameter of (which is only finite for finite and connected), and for a subset∈ ( +, denote ⊆ diam ( := supG,H ( 3 G,H . For E + and A > 0, we use E,A = D + : 3 D,E 6 A to ( ) ∈ ( ) ∈ ( ) { ∈ ( ) } denote the closed ball in . For subsets (,) +, we write 3 (,) := inf 3 B,C : B (,C ) . ⊆ ( ) { ( ) ∈ ∈ } For two expressions and , we use the notation . to denote that 6 for some universal constant . The notation . ,, denotes that 6 , , where , , ··· ( ···) ( ···) denotes a number depending only on the parameters , , etc. We write for the conjunction . . . ∧

2 Tilings of the unit cube

Fix the dimension 3 > 2. Our constructions are based on tilings of subsets of ℝ3 by axis-parallel hyperrectangles, a generalization of the planar constructions in [EL20]. A 3-dimensional tile is an 3 axis-parallel closed hyperrectangle ℝ , i.e., a set of the form = 01, 11 02, 12 0 , 1 ⊆ [ ] × [ ] × · · · × [ 3 3] for numbers satisfying 08 < 18 for each 8 = 1, 2, . . . , 3. We will encode such a tile as a 3 1 -tuple ? , ℓ1 , ℓ2 , . . . , ℓ , where ? := ( + ) ( ( ) ( ) ( ) 3( )) ( ) 01, 02, . . . , 03 and ℓ8 := 18 08 is the length of the projection of along the 8th axis. ( ) ( ) − Ð 3 A tiling Z is a finite collection of interior-disjoint tiles. Denote Z := Z . If ' ℝ , we say ∈ ⊆ that Z is a tiling of ' if Z = '. We associate to a tiling its tangencyJ graphK Z with vertex set Z and ( ) with an edge betweenJ twoK tiles , Z whenever has non-zero 3 1 -dimensional volume. ∈ ∩ ( − ) Denote by the set of all tilings of the unit 3-dimensional cube 0, 1 3. See Figure 2(a) for a T3 [ ] tiling of 0, 1 3 and Figure 2(b) for a representation of its tangency graph. For the remainder of the [ ] paper, we will consider only tilings Z for which Z is connected. ( ) Definition 2.1 (Tiling product). For Y, Z , define the product Y Z as the tiling formed by ∈ T3 ◦ ∈ T3 replacing every tile in Y by an (appropriately scaled) copy of Z. More precisely: For every Y ∈ and Z, there is a tile ' Y Z with ℓ8 ' := ℓ8 ℓ8 , and ∈ ∈ ◦ ( ) ( ) ( )

?8 ' := ?8 ?8 ℓ8 , ( ) ( ) + ( ) ( ) for each 8 = 1, 2, . . . , 3.

If Z and = > 0, we use Z = := Z Z to denote the =-fold tile product of Z with itself, ∈ T3 ◦ · · · ◦ where Z 0 := O and O := 0, 1 3 is the identity tiling. The following observation shows that 3 3 {[ ] } ∈ T3 this is well-deﬁned.

Observation 2.2. The tiling product is associative: Y Z [ = Y Z [ for all Y, Z , [ 3. 3 ( ◦ ) ◦ ◦ ( ◦ ) ∈ T Moreover, if Od consists of the single tile 0, 1 , then Z Od = Od Z for all Z . ∈ T3 [ ] ◦ ◦ ∈ T3

4 3 (a) Z( ) (b) The tangency graph 3,6,3 h i Figure 2: A tiling and its tangency graph

2.1 The construction

3 Given a sequence = 1,..., 1 with 8 ℕ, we deﬁne an associated tiling Z( ) 3 as follows: 3h 1 i ∈ 3 1 3 1 ∈ T For 1 6 8 6 1, ﬁll 0, 1 − 8 1 1, 8 1 with 8 − copies of 0, 1 8 − 0, 1 1 formed into a [ ] × [( − )/ 3 / ] [ / ] × [ / ] ( ) 8 8 grid. For example, see Z 3,6,3 in Figure 2(a). The following observation will be useful. | ×{z · · · × } h i 3 1 times − 3 3 3 = = ( ) ( ) = ( ) Observation 2.3. For 1,..., 0 , 0 10 ,..., 10 , we have Z Z Z where h i ◦ 0 ⊗ 0

0 = 10 ,..., 10 ,..., 0 0 ,..., 0 0 . ⊗ 1 1 1 1 = 3 3 = In particular, for = > 1 it holds that Z( ) = Z ( )= , where ⊗ := is the =-fold tensor ⊗ ⊗ ⊗ · · · ⊗ 3 product of with itself. See Figure 3(a) for a representation of Z ( ) 2 . We will use these two 3,6,3 ⊗ representations interchangeably throughout the paper. h i

We use the notations := 1 and | | 3 3 3 1 3 1 ( ) := Z ( ) = − − . | | | | 1 + · · · + 1 Note that 3 = 3 3 , hence | ⊗ 0|( ) | |( )| 0|( ) 3 3 = Z ( )= = ( ) . (2.1) ⊗ | | Further, denote 3 3 log ( ) k( ) := | | . ( ) log | |

5 3 3 3 3 (a) Z( ) Z( ) = Z( ) (b) Z( ) 3,6,3 3,6,3 3,6,3 2 ˜ 3,6,3 2 h i ◦ h i h i⊗ h i⊗ Figure 3: A tile product and its cube packing

We now state three key lemmas proved that are proved in subsequent sections, and then use them to prove our main theorems. The ﬁrst establishes uniform polynomial volume growth for the 3 graphs Z ( )= . It is proved in Section 3. ⊗

Lemma 2.4 (Volume growth). Consider = 1,..., 1 with min = 1 = 1 = 1. Then for all h i ( n) 3 o 3 > 2, there is a number = 3, such that the family of graphs = Z ( )= : = > 0 has uniform ( ) ℱ ⊗ polynomial growth of degree : = k 3 in the sense that ( )( ) 1 : : − ' 6 G,' 6 ' , ,G + , 1 6 ' 6 diam . | ( )| ∀ ∈ ℱ ∈ ( ) ( ) The second lemma, proved in Section 4 establishes a lower bound on the size of annular separators.

Lemma 2.5 (Separator size). For every 3 > 2 and 1 > 1, there is a number 2 = 2 3, 1 such that for 3 1 ( ) any = 1,..., 1 with min = 1, the following holds. Denote : := k( − ) . For any = > 1, if h3 i ( ) : ( ) = Z ( )= , and E + , then there are at least 2' disjoint paths from ( E,' to ( E,'0 , for any ( ⊗ ) ∈ ( ) ( ) ( ) 1 6 ' < ' 6 diam 3 3 1 . 0 ( )/( ( + ))

If is an undirected graph, let us write < for the <-subdivision of , where each edge of [ ] is replaced by a path of length <. We will sometimes consider + + < via the obvious ( ) ⊆ ([ ] ) identiﬁcation. The next lemma is proved in Section 5.

Lemma 2.6. Consider a sequence = 1,..., with 1 = , and such that h 1i 1 8 1 8 max + , ℕ, 1 6 8 < 1. (2.2) 8 8 1 ∈ ∀ +

6 Then for every 3 > 2, there are numbers < = < 3, and " = " 3, such that for every = > 0: If 3 ( ) 3 ( ) = Z ( )= , then < is "-uniformly sphere-packed in ℝ . ( ⊗ ) [ ] With these results in our hand, let us first prove a finitary version of our main theorems. The corresponding infinite version appears in Section 2.2. Define

1 B 3, : := 3 1 : 3 1 . ( ) − + ( − ) − 3 1 − This represents our basic tradeoﬀ: One can construct 3-dimensional geometric graphs with uniform polynomial growth of degree : and such that every annular separator has size Ω 'B 3,: . + ( ( )) Theorem 2.7 (Finite graph families). For every 3 > 3, : > 3, and > 0, there is a family of ﬁnite ℱ graphs satisfying:

1. For some : 6 : and every , ˜ + ∈ ℱ : G,' ' ˜ , G + , 1 6 ' 6 diam . | ( )| 3, ∀ ∈ ( ) ( ) 3 2. Each is "-uniformly sphere-packed in ℝ for some " .3, 1. 3. There is a number 2 & 1 such that for every and G + , there are at least 2'B 3,: 3, ∈ ℱ ∈ ( ) ( ) vertex-disjoint paths from ( G,' to ( G,' for every 1 6 ' < ' 6 2diam . ( ) ( 0) 0 ( ) n 3 o In light of Lemma 2.4–Lemma 2.6, we can take = Z ( )= < : = > 0 as long as we can ﬁnd ℱ [ ( ⊗ )] a sequence suited to the parameters. To this end, consider 3 > 2 and parameters ℎ, ?, @ ℕ such ∈ that ? > @3. Deﬁne the sequence

?,@,ℎ ( ) := 1,C1,...,C1, 1 , h | {z } i 1 2 copies − @ 3 1 ? 3@ ?,@,ℎ where 1 := ℎ ( − ) and C := ℎ − . Note that C > 1 since ? > @3. Moreover, ( ) satisﬁes (2.2) by construction.

Lemma 2.8. Let 3 > 2 and : := ? @ be as above. Then for all > 0, there is some ℎ0 ℕ so that for all / ∈ ℎ > ℎ0 the following statements hold:

3 ?,@,ℎ 1. k( ) ( ) : < . ( ) − 3 1 ?,@,ℎ 2. k( − ) ( ) B 3, : < . ( ) − ( ) Proof. It holds that

3 ?,@,ℎ ?,@,ℎ 3 3 1 3 1 k( ) ( ) = log ( ) ( ) = log 1 − 2 1 2 C − ( ) 1(| | ) 1 ( + ( − ) ) 3 1 3 1 ? 3@ = log 1 − 2 1 2 ℎ( − )( − ) 1 ( + ( − ) ) 3 1 : 3 = log 1 − 2 1 2 1 − . 1 ( + ( − ) )

7 Furthermore, we have ?,@,ℎ 3 = 3 1 : 3 = lim log1 ( ) ( ) lim log1 1 − 2 1 2 1( − ) :. ℎ (| | ) 1 ( ( + ( − ) )) →∞ →∞ Therefore for all > 0, for all suﬃciently large values of ℎ, it holds that k 3 ?,@,ℎ : 6 . | ( )( ( )) − | Similarly, we have ?,@,ℎ 3 1 = lim log1 ( ) ( − ) B 3, : , 1 (| | ) ( ) →∞ hence we can choose ℎ suﬃciently large to as to satisfy the second condition as well.

2.2 Construction of the inﬁnite graphs

For this section alone, we consider tilings of the nonnegative orthant 0, 3. 3 [ ∞) Note that for any sequence = 1,..., , one can view Z ( ) as the tangency graph of a h 1i ( ) packing of cubes by changing the height of cubes in layer 8 from 1 1 to 1 8. See Figure 3(b) for an 3 / / example. Let Z ( ) denote this rescaled tiling. By convention, we insist that one corner of the tiling ˜ still lies at the origin, implying that

3 3 1 Z( ) = 0, 1 − 0, , J ˜ K [ ] × [ ( )] where 1 1 := 1 . ( ) 1 + · · · + 1

If we now assume that 1 = 1 = 1, then we have the chain of inclusions:

3 3 3 3 ( ) ( ) ( ) ( ) Z 0 Z = 1 Z = Z = 1 ˜⊗ ⊆ · · · ⊆ ˜⊗( − ) ⊆ ˜ ⊗ ⊆ ˜⊗( + ) ⊆ · · · 3 3 3 This gives rise, in a straightforward way, to the infinite tiling Z ( ℕ) , with Z ( ℕ) = 0, . We define ˜⊗ ˜⊗ [ ∞) 3 3 J K the infinite tangency graph ( ) := Z ( ℕ) . ˆ ( ⊗ ) = = = = 3 Theorem 2.9. If 1,..., 1 has 1 1 1 min , then ˆ ( ) has uniform polynomial growth of h i ( ) 3 1 degree k 3 , and for every G + and ' > ' > 1, there are at least 2'k( − ) vertex-disjoint paths ( )( ) ∈ ( ) 0 ( ) from ( G,' to ( G,' , where 2 & 1. ( ) ( 0) ,3 3 3 Proof. For = > 1, denote = := Z ( )= and := ( ). We can think of = as an induced subgraph ( ⊗ ) ˆ ˆ of in the obvious way. Consider a vertex E + and radii ' > ' > 1. Then there is some = > 1 ˆ ∈ ( ˆ ) 0 such that E, 2'0 + = . ˆ ( ) ⊆ ( ) In this case, the it holds that 3 G,H = 3= G,H for all G,H E,'0 , since any path originating ˆ ( ) ( ) ∈ ˆ ( ) in E,'0 and leaving + = must have length at least 2'0. Therefore E,'0 = = E,'0 , ˆ ( ) ( ) ˆ ( ) ( ) and the uniform polynomial growth and seperator size assertions then follow immediately from Lemma 2.4 and Lemma 2.5.

The following theorem is proved in Section 5.

Theorem 2.10. If = 1,..., 1 satisﬁes 1 = 1 and (2.2), then there is some number < = < 3, 3 h i 3 ( ) such that ( ) < is uniformly sphere-packed in ℝ . [ ˆ ]

8 3 Volume growth analysis

Our goal is now to prove Lemma 2.4. The next section provides a few key lemmas about the size of balls in products of tilings, which are mostly straightforward generalizations of the bounds in [EL20] (for the case 3 = 2). With these in hand, we prove Lemma 2.4 in Section 3.2.

3.1 Volume growth in tile products

The next lemma is straightforward.

Lemma 3.1. Consider Y, Z and = Y Z . If - Y Z, then -, diam Z > Z . ∈ T3 ( ◦ ) ∈ ◦ | ( ( ( )))| | | 3 Let 41, . . . , 4 denote the standard basis of ℝ . If Z , we write 8 Z Z for the set { 3} ∈ T3 ( ) ⊆ ( ( )) of edges in the 8th direction, i.e., those edges , Z where is orthogonal to 48. { } ∈ ( ( )) ∩ Thus we have a partition Z = 1 Z Z . ( ( )) ( ) ∪ · · · ∪ 3( ) For Z and 1 6 8 6 3, denote ∈ #Z , 8 = 0 Z : , 0 8 Z , ( ) { ∈ { } ∈ ( )} and #Z := #Z , 1 #Z , 3 . Moreover, we deﬁne ( ) ( ) ∪ · · · ∪ ( ) ℓ9 Z , 8 := max ( ) : #Z , 8 , 1 6 9 6 3 ( ) ℓ9 ∈ ( ) ( ) Z 8 := max Z , 8 : Z ( ) { ( ) ∈ } Z := max Z 8 : 1 6 8 6 3 { ( ) } !Z := max ℓ8 : Z , 1 6 8 6 3 . { ( ) ∈ }

We will take Z := 1 if Z contains a single tile. It is now straightforward to check that Z bounds the degrees in Z . ( ) Lemma 3.2. For a tiling Z and Z, it holds that ∈ T3 ∈ 3 23 deg Z 6 3 Z . ( )( ) · Proof. Denote

3 := G1,...,G ℝ : ?8 Zℓ8 6 G8 6 ?8 1 Z ℓ8 , ˜ ( 3) ∈ ( ) − ( ) ( ) + ( + ) ( ) where ?8 denotes the 8th coordinate of ? . Clearly ( ) ( ) 3 Ö 3 vol = 1 2 Z ℓ8 = 1 2 Z vol . 3( ˜) (( + ) ( )) ( + ) 3( ) 8=1

Furthermore, by the deﬁnition of Z , it holds that 0 for 0 #Z . And for any such 0, it 3 ⊆ ˜ ∈ ( ) holds that vol > − vol . Hence, 3( 0) Z 3( ) 3 3 #Z − vol 6 1 2 Z vol . | ( )| Z 3( ) ( + ) 3( ) 3 23 Using Z > 1, this yields deg Z 6 3 Z . ( )( ) 9 Lemma 3.3. Consider Y, Z and let = Y Z . Then for any - Y Z, it holds that ∈ T3 ( ◦ ) ∈ ◦ 23 2 32 -, 1 !Z 6 3 3 1 Z . (3.1) | ( /( Y ))| ( Y) ( + )| |

Proof. For . Y Z, let . Y denote the unique tile for which . .. Let ΓY - denote the set of ∈ ◦ ˆ ∈ ⊆ ˆ ( ˆ ) paths in Y that originate from - and such that 8 Y 6 1 for each 8 = 1, . . . , 3. In other ( ) ˆ | ∩ ( )| words, the set of paths in Y starting at - and containing at most one edge in every direction. ( ) ˆ Denote by # - := Ð + the set of vertices reachable via such paths. Y ΓY - ˜ ( ˆ ) ∈ ( ˆ ) ( ) Note that because we allow one edge in every direction,

3 3 3 3 - ℓ1 - , ℓ1 - ℓ - , ℓ - r#Y - z , (3.2) ˆ + [− ( ˆ )/ Y ( ˆ )/ Y] × · · · × [− 3( ˆ )/ Y 3( ˆ )/ Y] ⊆ ˜ ( ˆ ) where ’ ’ here is the Minkowski sum ' ( := A B : A ',B ( . + + { + ∈ ∈ } We will now show that 23 r -, 1 !Z z r#Y - z . (3.3) ( /( Y )) ⊆ ˜ ( ˆ ) It will follow that 3 3 1 -, 1 Y+ !Z 6 Z #Y - 6 Z 3 1 max deg Y , | ( /( ))| | | · | ˜ ( ˆ )| | | · ( + ) Y ( )( ) ∈ and then (3.1) follows from Lemma 3.2.

To establish (3.3), consider any path - = -0,-1,-2,...,- in with - ∉ #Y - . Let : 6 ℎ h ℎi ˆ ℎ ˜ ( ˆ ) be the smallest index for which - ∉ #Y - . Then: ˆ : ˜ ( ˆ )

-0,-1,...,-: 1 r#Y - z (3.4) − ⊆ ˜ ( ˆ ) 3 -: 1 % r#Y - z 0, 1 ≠ . (3.5) − ∩ ˜ ( ˆ ) ∩ ( ) ∅

3 Now (3.4) implies that for 9 6 : 1, we have -9 #Y - , which implies ℓ8 -9 6 ℓ8 - since -9 − ˆ ∈ ˜ ( ˆ ) ( ˆ ) Y ( ˆ ) ˆ can be reached from - by a path of length at most 3 in Y . It follows that ˆ ( ) 3 ℓ8 -9 6 !Zℓ8 -9 6 !Z ℓ8 - , 9 6 : 1, 1 6 8 6 3. (3.6) ( ) ( ˆ ) Y ( ˆ ) − And (3.5) together with (3.2) shows that

: 1 Õ− @ ℓ8 -9 > ℓ8 - , 1 6 8 6 3. (3.7) ( ) ( ˆ )/ Y 9=0

Combining (3.6) and (3.7) now gives

1 ℎ 1 > : 1 > , − − 23 Y !Z verifying (3.3) and completing the proof.

10 3.2 Volume growth in iterated products

Our goal is now to prove Lemma 2.4. To this end, ﬁx 3 > 2.

Observation 3.4. For 1 6 8 6 3 1, we have 3 8 = 1. It further holds that − Z( ) ( ) 8 3 = 3 1 = max : 1 6 8, 9 6 1, 8 9 = 1 . Z ( ) Z ( ) ( ) 9 | − |

Given Observation 2.3 and Observation 3.4, the next lemma is straightforward.

Lemma 3.5. Consider = 1,..., 1 and 0 = 0 ,..., 1 . If 0 = 0 , then h i 1 0 1 10

3 3 = 3 6 max 3 , 3 . Z ( ) Z ( ) Z ( ) Z ( ) Z ( ) ( ) ◦ 0 ⊗ 0 0

Proof. Note that, by deﬁnition for 1 6 9 6 10 and 0 6 8 6 1 1, we have 0 81 9 = 8 0. Hence, − ( ⊗ ) 0+ 9

09 1  + 9 < 1 0 810 9 1  0 0 ( ⊗ ) + + = 9 8 1 10 8 1 0 810 9  + = + 9 = 1 , ( ⊗ ) +  8 0 8 0  10 where we used 0 = 0 in the second case. 1 10

Corollary 3.6. Let = 1,..., . If 1 = , then for = > 1 we have h 1i 1

3 6 3 . Z ( )= Z( ) ⊗

Lemma 3.7. Consider = 1,..., with 1 = = 1. For every = > 0, it holds that h 1i 1 = 3 = 1 1 6 diam Z ( )= 6 3 1 1 . (3.8) − ( ⊗ ) ( + )

3 3 1 = Proof. Let denote the “bottom” layer of tiles in Z ( )= , i.e., those contained in 0, 1 − 0, 1 1 . C ⊗ [ ] × [ / ] 3 3 = Take := Z ( )= . For every Z ( )= , it holds that ℓ3 = 1− . In particular, this yields the lower ( ⊗ ) ∈ ⊗ ( ) bound in (3.8) since any path from the bottom to the top (in dimension 3) requires 1= tiles. 3 = This also implies that for Z ( )= we have 3 , 6 1 . Furthermore, as 1 = 1 by assumption, ⊗ = = ∈ ( C) = = = we have ⊗ 1 = 1 , and therefore by construction is a 3 1 -dimensional 1 1 1 ( ) = [C] ( − ) × × · · · × grid. Hence we have diam 6 3 1 1 , and now the upper bound in (3.8) follows by the (C) ( − ) triangle inequality.

3 3 3 3 := ( ) ( ) ( ) = ( ) Proof of Lemma 2.4. Denote Z = and ﬁx Z = . For any 0 6 9 6 =, write Z = Z = 9 ( ⊗ ) ∈ ⊗ ⊗ ⊗( − ) ◦ 3 3 9 Z ( )9 and let denote the copy of Z ( )9 containing . By Lemma 3.7, we have diam 6 3 1 1 , ⊗ ˆ ⊗ ( ˆ) ( + ) 9 and thus , 3 1 1 . ( ( + ) ) ⊇ ˆ Employing Lemma 3.1 therefore gives

9 9 3 3 3 = ( ) = ( ) , 3 1 1 > Z 9 ( ) , Z = , 9 0, 1, . . . , = . ( ( + ) ) | ˆ| ⊗ | | ∀ ∈ ⊗ ∈ { }

11 The desired lower bound now follows using monotonicity of , A with respect to A. | ( )| To prove the upper bound, ﬁrst note that by Corollary 3.6 we have

3 6 3 . 1. Z ( )= Z( ) ⊗

9 3 Moreover, as min = 1, we have ! 3 = 1 , hence applying Lemma 3.3 with ( = Z ( ) and Z ( ) − = 9 ( ) 9 ⊗( − ) ⊗ 3 ) = Z ( )9 gives ⊗

9 9 3 3 3 ( ) = ( ) , 1 .3, Z 9 ( ) , Z = , 9 0, 1, . . . , = , ( / ) | ⊗ | | | ∀ ∈ ⊗ ∈ { } for some .,3 1, completing the proof.

4 The size of annular separators

3 We now prove that the graphs Z ( )= do not have small annular separators. ( ⊗ ) 3 3 1 Deﬁnition 4.1 (Projection of tiles). For 3 > 2 and a tile ℝ , we write Π3 1 ℝ − for the ⊆ − ( ) ⊆ projection of onto the last 3 1 coordinates. Furthermore, for a tiling Z , we deﬁne − ∈ T3 Π3 1 Z := Π3 1 Z , − ( ) { − ( ) | ∈ } and for a tile Π3 1 Z , ∈ − ( ) Π 1 = Π = −3 1 ; Z : Z : 3 1 . − ( ) { ∈ − ( ) } Observation 4.2. For any sequence = 1,..., , the following hold: h 1i 3 3 1 (a) Π3 1 Z( ) = Z( − ). − ( ) 3 1 3 Π ( ) Π ( ) (b) For all 3 1 Z , the tangency graph −3 1 ; Z is a path. ∈ − ( ) ( − ( )) 3 1 3 1 3 ≠ Π ( ) Π ( ) Π ( ) (c) For all 0 3 1 Z , the sets −3 1 ; Z and −3 1 0; Z are disjoint. ∈ − ( ) − ( ) − ( ) See Figure 4.

1 3 3 We will now use the family Π− ; Z( ) : Π3 1 Z( ) of pairwise disjoint paths to locate 3 1 − 3 { − ( ) ∈ ( )} paths across annuli in Z ( )= . We ﬁrst remark that diameter of each such path is long in . ( ⊗ ) 3 Lemma 4.3. For every Π3 1 Z( ) , ∈ − ( ) 1 3 1 diam 3 Π− ; Z ( ) > 1, Z ( ) 3 1 ( )( − ( )) ! 3 − Z( )

3 Proof. Let Π1 : ℝ ℝ denote the projection onto the ﬁrst coordinate. Then the Euclidean diameter 1 3 → 3 Π Π ( ) ( ) of 1 −3 1 ; Z is precisely 1. On the other hand, for any path % in Z , the Euclidean ( − ( )) ( ) diameter of Π1 % is at most len % 1 ! 3 , and the result follows. ( ) ( ( ) + ) Z( )

12 1 3 3 Π ( ) Π ( ) Figure 4: Projection to the last 3 1 coordinates and −3 1 ; Z for one 3 1 Z . − − ( ) ∈ − ( )

With this in hand, we can now exhibit many disjoint paths across annuli in .

= 3 = 3 Proof of Lemma 2.5. Deﬁne : Z ( )= and ℎ : log1 ' 3 1 . Let Z ( )= be arbitrary. Write ( ⊗ ) b ( /( + ))c ∈ ⊗ 3 3 3 3 3 ( ) = ( ) ( ) ( ) ( ) Z = Z = ℎ Z ℎ , and let Z = be the copy of Z ℎ that contains . By Lemma 3.7, we have ⊗ ⊗( − ) ◦ ⊗ ˆ ∈ ⊗ ⊗ ℎ diam 6 3 1 1 6 ', where the latter inequality follows from our deﬁnition of ℎ. Thus ( ˆ) ( + ) ˆ , ' . ⊆ ( ) 3 3 But as is a translation of Z ( )ℎ , by Observation 4.2(a), it holds that Π3 1 Z ( )ℎ is a translation of ˆ ⊗ − ( ⊗ ) 3 1 Z ( −ℎ ). This yields ⊗ ℎ 3 1 3 1 Π3 1 , ' > Π3 1 = Z ( −ℎ ) = ( − ) . | − ( ( )) | | − ( ˆ)| | ⊗ | | |

1 3 Using Observation 4.2(b)–(c), the sets Π− ; Z ( )= : Π3 1 , ' form a collection of { 3 1( ⊗ ) ∈ − ( ( ))} vertex-disjoint paths in , and Lemma 4.3 gives− a lower bound on the diameter of every such path in . It follows that for 1 1 '0 < © 1ª , (4.1) 2 ! 3 − ® Z ( )= « ⊗ ¬ 3 1 ℎ there are at least vertex-disjoint paths originating in , ' and leaving , ' . (| |( − )) ( ) ( 0) Finally, note that, by assumption, we have min > 1, and therefore min = > 1=, implying ( ) ( ⊗ ) that = ! 3 = 1− . Z ( )= ⊗ Since diam 6 3 1 1= by Lemma 3.7, the constraint (4.1) is implied by ( ) ( + ) 1 diam '0 < ( ) 1 . 2 3 1 − + 13 We may assume that diam > 6 3 1 , otherwise the statement of the lemma is vaccuous (since ( ) ( + ) we may choose the constant 2 suﬃciently small depending on 3), in which case this is implied by

diam '0 6 ( ) , 3 3 1 ( + ) completing the proof.

5 Sphere-packing representations

We will now prove Lemma 2.6 and Theorem 2.10. To this end, we require some regularity from our cube packings. Say that two closed, axis-parallel cubes , ℝ3 are neatly tangent if: ⊆ (i) and are interior-disjoint.

(ii) If and intersect along 3 1 -dimensional faces and , then either ( − ) ⊆ ⊆ ⊆ ∩ or . ⊆ ∩ A collection of closed, axis-parallel cubes is a neat cube packing if every pair ≠ is either C ∈ C neatly tangent or else disjoint. The aspect ratio of the packing is deﬁned by C ℓ := max ( ) : , , ≠ , (C) ℓ ∈ C ∩ ∅ ( ) where ℓ denotes the sidelength of a cube . Say that a graph is admits an -uniform neat cube ( ) packing in ℝ3 if is the tangency graph of a neat cube packing with 6 . C (C) 3 Lemma 5.1. If = 1,..., satisﬁes 1 = = 1 = min and (2.2) holds, then the graphs Z ( ) h 1i 1 ( ) ( ) 3 3 and ( ) admit an -uniform neat cube packing in ℝ with = 3 . ˆ Z( )

3 Proof. First take := Z ( ), as deﬁned in Section 2.2. Under the integrality assumptions on , it holds C ˜ that is a neat packing, since one of the ratios 8 1 8 or 8 8 1 is an integer for every 1 6 8 < 1 C + / / + (see Figure 3(b) for an illustration). Moreover, we have

= 3 = 3 . Z ( ) Z ( ) (C) ˜

3 For the second assertion, we take := Z ( ) (as deﬁned in Section 2.2). Corollary 3.6 asserts that C ⊗∞

3 6 3 , = > 0, Z ( )= Z( ) ⊗ ∀ hence 3 < , as desired. Z ( ) ⊗∞ ∞ Given the preceding lemma, the next result suﬃces to prove Lemma 2.6 and Theorem 2.10.

Lemma 5.2. If admits an -uniform neat cube packing in ℝ3, then there are numbers < 6 $ 3 and 2 3 ( ) " 6 $ 3 such that the subdivision < is "-uniformly sphere-packed in ℝ . ( ) [ ]

14 To prove this, we need a simple result on sphere packings that satisfy prescribed tangencies. For a general closed axis-parallel cube ℝ3 and > 0, we define % to denote the boundary of ⊆ in ℝ3, and we use for the standard Euclidean distance. For a point G ℝ3 and a set ( ℝ3, k · k ∈ ⊆ define dist2 G,( := inf G H : H ( . Let us also define ( ) {k − k ∈ }

% := G % :# : dist2 G, > ℓ = 23 1 , { ∈ { ∈ ℱ ( ) ( )} − } where ℓ is the sidelength of , and is the collection of the 23 facets of (i.e., the 3 1 - ( ) ℱ ( − ) dimensional faces of ). These are the boundary points that lie on exactly once face of and are ℓ -far from every other. ( ) Lemma 5.3. For every 3 > 2 and 0, 1 2 , the following holds. Consider a closed axis-parallel cube ∈ ( / ) ℝ3 of sidelength ℓ, and a set of points % % such that G H > ℓ for every G ≠ H %. Then ⊆ ⊆ k − k ∈ there is a finite collection of interior-disjoint spheres contained in such that: S 1. The radius of every sphere in is at least ℓ 603 . S /( ) 2. For every ? %, there is a sphere ( ? tangent to ?. ∈ ( ) ∈ S 3. The tangency graph of is the <-subdivision of a star graph whose leaves are the spheres ( ? : ? % , S { ( ) ∈ } with < = 2 63 . + d e 3 Proof. By scaling and translation, it suffices to prove the lemma for = 0, 1 . Denote 20 := [ ] 1 , 1 ,..., 1 , and define the sphere ( 2 2 2 ) 3 (0 := G ℝ : G 20 = 1 4 . { ∈ k − k / } For each ? %, define ( ? to be the unique sphere of radius 8 that is contained in 0, 1 3 and ∈ ( ) / [ ] such that ( ? % 0, 1 3 = ? . Such a sphere exists because ? % 0, 1 3. ( ) ∩ [ ] { } ∈ [ ] Let 2? denote the center of ( ? , and let 20, 2? denote the line segment from 20 to 2?. Define I? ( ) [ ] to be the point where 20, 2? intersects (0, and define I to be the point where 20, 2? intersects [ ] 0? [ ] ( ? . Let I? ,I 20, 2? denote the line segment connecting I? to I . ( ) [ 0?] ⊆ [ ] 0? As G H > for all G ≠ H %, it holds that I I > 4 8 > 2. Note that k − k ∈ k 0G − 0H k − ( / ) / 3 0, 1 2 20, √3 2 , where the latter object is the Euclidean ball of radius √3 2 about 20. [ ] ⊆ ( / ) / Let IG denote the point where the line through 20, 2? intersects %2 20, √3 2 . Then we have ˜ [ ] ( / ) IG IH > I I , and by similarity of the triangles defined by 20, IG , IH and 20,I ,I , it k ˜ − ˜ k k 0G − 0H k { ˜ ˜ } { 0G 0H } holds that IG IH I0G I0H > k ˜ − ˜ k > . k − k √3 2√3 It follows that min 0 1 : 0 IG ,I0G , 1 IH ,I0H > IG IH > . (5.1) {k − k ∈ [ ] ∈ [ ]} k − k 2√3 Note also that for every ? %, ∈ 1 1 1 3 6 6 20 2? 6 I? I0 6 20 2? 6 diam2 0, 1 6 √3, (5.2) 8 4 − 4 k − k − 4 + 8 − ? k − k ([ ] ) where we have used < 1 2. /

15 63 Let ? : 0, I? I I? ,I be a parameterization of I? ,I by arclength. Define : := [ k − 0? k] → [ 0?] [ 0?] d e I? I and A := k − 0? k , and let ( be the collecion of interior-disjoint spheres of radius A centered at the ? 2: ˜? ? points ? A? , ? 2A? , ? 4A? ,..., ? 2 : 1 A? . ( ) ( ) ( ) ( ( − ) ) Note that the tangency graph of (˜? is a path, and that the first sphere is tangent to (0, while the last is tangent to ( ? . ( ) h i By (5.2), for each ? %, we have A% , . In particular, (5.1) implies that if ( (? and ∈ ∈ 603 6√3 ∈ ˜ ( (? for ? ≠ ? %, then ( and ( are disjoint. 0 ∈ ˜ 0 0 ∈ 0 Thus the collection of spheres Ø := (0 ( ? : ? % (? S { } ∪ { ( ) ∈ } ∪ ˜ ? % ∈ satisfies the conditions of the lemma.

Proof of Lemma 5.2. Suppose that is a neat cube packing whose tangency graph is and such that C 6 . For every pair , with ≠ , let 2 , be the center of mass of . (C) ∈ C ∩ ∅ ( ) ∩ Deﬁne the set of points % := 2 , : , ≠ . Since is a neat cube packing and { ( ) ∈ C ∩ ∅} C we have ℓ > ℓ , it follows that with := 1 2 , we have % % , and ( ) ( )/ /( ) ⊆

G H > ℓ , G ≠ H %. k − k ( ) ∀ ∈

Therefore we can use Lemma 5.3 to replace each cube by a corresponding collection of ∈ C S spheres whose tangency graph is the <-subdivision of a star, the leaves of which are tangent to the points in %. Note that Lemma 5.3 gives < = 2 12 3 . + d e Moreover, any two adjacent spheres have their ratio of radii contained in the interval " 1," [ − ] for 603 " := = 120 23. Ð Thus if we deﬁne := , then 2< is the tangency graph of , and the corresponding S ∈C S [ ] S sphere-packing is "-uniform.

Acknowledgements

This research was partially supported by NSF CCF-2007079 and a Simons Investigator Award.

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