aa r X i v : . [ m a t h . C O ] A ug Symmetric punctured intervals tile Z Stijn Cambie ∗ Abstract
Extending the methods of Metrebian (2018), we prove that any symmetric puncturedinterval tiles Z . This solves a question of Gruslys, Leader and Tan (2016). Given n , let T be a tile in Z n , i.e. a finite subset of Z n . Recently, confirming a conjecture ofChalcraft that was posed on MathOverflow, Gruslys, Leader and Tan [2] showed that T tiles Z d for some d . This is an existence result and they wondered about better bounds in termsof the dimension n and the size | T | . They conjectured the following for the case n = 1 . Conjecture 1.1 (Gruslys, Leader, Tan [2]) . For any positive integer t there is a number d such that any tile T in Z with | T | = t tiles Z d . Let us note that Adler and Holroyd [1] had earlier investigated which tiles in Z can tile Z .In general, for any fixed d , there are one-dimensional tiles which cannot tile Z d , see Section 4.When dealing with one-dimensional tiles, we find it convenient to use the same notation asin [1]: a tile T in Z which is the union of n intervals I up to I n , such that the length of interval I i is a i and the gap between I i and I i +1 is b i , will be denoted by a ( b ) a ( b ) a . . . ( b n − ) a n .With this notation, we can state the concrete question that motivated this work [2, Qu. 21].It asks for the optimal tiling dimension for symmetric punctured intervals. Question 1.2 (Gruslys, Leader, Tan [2]) . What is the least d for which T = k (1) k tiles Z d ? Very recently, Metrebian [4] showed that d ≤ suffices and that d = 3 is optimal when k isodd or k ≡ . He noted that for k ≥ one has d ≥ , while for k ∈ { , } the optimal d equals k . We extend Metrebian’s methods to solve Question 1.2 in its entirety. Theorem 1.3.
The least d for which T = k (1) k tiles Z d equals min { k, } . This will be a corollary of a slightly more general result, Proposition 3.4 below.The organization of the paper is as follows. In Section 2 we prove a lemma implying thatit is enough to find some structured partial tilings of Z . We exhibit such constructions inSection 3. After having constructed upper bounds on the dimension d such that T tiles Z d ,we then in Section 4 examine lower bounds that are related to Conjecture 1.1. We concludein Section 5 with some speculative thoughts towards Conjecture 1.1. ∗ Department of Mathematics, Radboud University Nijmegen, Postbus 9010, 6500 GL Nijmegen, TheNetherlands. Email: [email protected]. This work has been supported by a Vidi Grant of the NetherlandsOrganization for Scientific Research (NWO), grant number . . . From partial to complete tilings
To tile Z d with a one-dimensional tile T for some minimal d > , one clearly needs tiles directedin each of the d orthogonal directions. Choosing one particular direction and removing alltiles in the tiling of Z d in this direction, one has partial tilings of spaces isomorphic to Z d − which are orthogonal to the chosen direction. Noting that T has some gap, one can derive acontradiction assuming there are only two partial tilings.In this section, we will prove that three different partial tilings can be enough when T contains only one gap, i.e. when T is the union of two intervals. This is done in Lemma 2.1which is a generalization of Lemma in [4]. Lemma 2.1.
Let T be the one-dimensional tile k ( m ) ℓ . Suppose there are three disjoint subsets A, B, C of Z d with the same cardinality such that one can tile Z d \ ( A ∪ B ) , Z d \ ( A ∪ C ) and Z d \ ( B ∪ C ) with T . Then T tiles Z d +1 .Proof. First assume m < min { k, ℓ } . We construct a subset Y ⊂ Z × { , , } such that | Y ∩ ( { z } × { , , } ) | = 2 for every z ∈ Z and such that T tiles Y. Let ( x, i ) ∈ Y for some x ∈ Z and i ∈ { , , } if and only if x − i ( k + l ) ≡ , , . . . , k ; k + m + 1 , k + m + 2 . . . , k + m + ℓ (mod 3 k + 3 ℓ ) or ≡ k + ℓ + 1 , . . . , k + 2 ℓ ; 2 k + 2 ℓ + m + 1 , . . . , k + 2 ℓ + m (mod 3 k + 3 ℓ ) . The construction has been sketched in Figure 1 for { , , . . . , k + ℓ ) }×{ , , } . By gluinginfinitely many copies of that picture together, one gets the full construction of Y . k m ℓ k − m mℓ k k + ℓ k + ℓ )012 π π π Figure 1: Construction of Y .Now we explain why this construction meets the conditions we need. Let S = { , , . . . , k } , S = { k + m + 1 , k + m + 2 . . . , k + m + ℓ } , S = { k + ℓ + 1 , . . . , k + 2 ℓ } and S = { k + 2 ℓ + m + 1 , . . . , k + 2 ℓ + m } . Let S o = S + S and S e = S + S . Then both S o ∪ (( k + ℓ ) + S o ) ∪ (2( k + ℓ ) + S o ) and S e ∪ (( k + ℓ ) + S e ) ∪ (2( k + ℓ ) + S e ) cover all elementsin Z k + ℓ ) Z exactly once, from which the result follows.The elements of A ∪ B ∪ C can be partitioned into triples { a i , b i , c i } since A, B, C havethe same cardinality. Every set Z × { a i , b i , c i } has a subset Y i ∼ = Y which can be tiled by T in the same manner, i.e. there exists a partition { Z , Z , Z } of Z such that for every i we have Y i ∩ ( { z } × { a i , b i , c i } ) = { a i , b i } for every z ∈ Z , Y i ∩ ( { z } × { a i , b i , c i } ) = { a i , c i } for every z ∈ Z and Y i ∩ ( { z } × { a i , b i , c i } ) = { b i , c i } for every z ∈ Z . Now Z d +1 \ ( ∪ i Y i ) can be written as Z × (cid:0) Z d \ ( A ∪ B ) (cid:1) ∪ Z × (cid:0) Z d \ ( A ∪ C ) (cid:1) ∪ Z × (cid:0) Z d \ ( B ∪ C ) (cid:1) and bythe assumptions this can be tiled by T as well, so T tiles Z d +1 . Looking at Figure 1, everyhyperplane π i will be covered by the intersections with ∪ i Y i and a partial tiling isomorphic toone of Z d \ ( A ∪ B ) , Z d \ ( A ∪ C ) or Z d \ ( B ∪ C ) . m ≥ min { k, ℓ } , where we assume without loss of generality k = min { k, ℓ } , onecan glue two copies T , T of T together to a tile T ′ with k ′ = ℓ ′ = k + ℓ and m ′ = m − k by taking T = {− k, − k + 1 , . . . , − } ∪ { m, m + 1 , . . . m + ℓ − } and T = {− k − ℓ, − k − ℓ + 1 , . . . , − k − } ∪ { m − k, m − k + 1 , . . . , m − } . See Figure 2 for a depiction. When m ′ ≥ k ′ , one can glue ⌊ m ′ /k ′ + 1 ⌋ copies of T ′ together, which are translates of T ′ with initialpoint at , k ′ , . . . , ⌊ m ′ /k ′ ⌋ k ′ . Hence we have reduced this to the case which has been provenalready. ℓ m k ℓ ′ k ′ m ′ T T T ′ Figure 2: Gluing T and T and copies T ′ . Z Rather than focusing solely on Question 1.2, we consider a slightly more general setting.Throughout this section, we let T be a punctured interval tile, which is the union of aninterval of length k and an interval of length ℓ with a gap of size . So T = k (1) ℓ equalsa translate of {− k, − k + 1 , . . . , − , , , . . . , ℓ } as a subset of Z . By applying Lemma 2.1, weprove that in most cases T tiles Z . When tiles do not tile Z , the partial tilings cannot beonly horizontal (similarly not only vertical). Hence it is natural to try to combine partialvertical tilings with partial horizontal tilings up to a set of the desired form.As a warm up, we construct three partial tilings of the plane satisfying the conditions ofLemma 2.1 when T is the symmetric punctured interval k (1) k with k ≡ . Proposition 3.1. If k ≡ , then T = k (1) k tiles Z .Proof. Let X be a set of diagonals which are a distance k + 1 apart, e.g. X = { ( x, y ) ∈ Z | x − y ≡ k +1) } . Let A = { ( x, y ) ∈ X | x ≡ , and B = { ( x, y ) ∈ X | x ≡ , } . Furthermore, choose C = B + (0 ,
1) = { ( x, y ) ∈ Z | y − x ≡ k + 1) , x ≡ , } . The construction is shown in Figure 3. Then Z \ ( A ∪ B ) = Z \ X can be tiledby T in many ways. Note that one can tile Z \ ( A ∪ C ) easily by vertical copies of T , i.e. forevery x ∈ Z one can tile ( { x } × Z ) \ ( A ∪ C ) straightforwardly with copies of T .One can also see that Z \ ( B ∪ C ) can be tiled, by placing copies of T horizontally. Onecan check that for every y ∈ Z the set ( Z × { y } ) \ ( B ∪ C ) is periodic (with period k + 4 ) andits period can be covered with two copies of T , which have one edge in common (i.e. whichare translates of each other with distance k + 1 ).By Lemma 2.1, we know T tiles Z as the conditions of the lemma are satisfied.In the not-necessarily-symmetric case, we start with two constructions that work for certainpunctured intervals and then proceed to Proposition 3.4, which implies the upperbound inTheorem 1.3. Proposition 3.2. If ∤ k + ℓ , then T = k (1) ℓ tiles Z . A B Ckk k
Figure 3: Construction of
A, B, C for T = k (1) k where k ≡ . Proof. If k + ℓ is odd, then K = k + ℓ +1 is even. Take A = { ( x, y ) ∈ Z | x ≡ y (mod K ) , x ≡ } , B = { ( x, y ) ∈ Z | x ≡ y (mod K ) , x ≡ } and C = { ( x, y ) ∈ Z | x ≡ y − K ) , x ≡ } .Now one can tile Z \ ( A ∪ B ) and Z \ ( A ∪ C ) by placing copies of T horizontal, i.e. byplacing copies of T from ( x − k, y ) to ( x + ℓ, y ) for any ( x, y ) ∈ A ∪ B , resp. A ∪ C. Similarly onecan tiles Z \ ( B ∪ C ) with vertical copies of T from ( x, y − k ) to ( x, y + ℓ ) for any ( x, y ) ∈ B ∪ C .Hence the result follows from Lemma 2.1. Proposition 3.3. If ∤ k, ℓ , then T = k (1) ℓ tiles Z . Proof.
Let K = k + ℓ + 2 . Take A = { ( x, y ) ∈ Z | x ≡ y (mod K ) , x ≡ } and A = A + ( k + 1 ,
0) = { ( x, y ) ∈ Z | x ≡ y + k + 1 (mod K ) , x ≡ } . Let B i = (1 ,
1) + A i and C i = (0 ,
1) + A i for every i ∈ { , } . Set A = A ∪ A , B = B ∪ B and C = C ∪ C . Now one can tile Z \ ( A ∪ B ) and Z \ ( A ∪ C ) by placing copies of T horizontal, i.e. byplacing copies of T from ( x − k, y ) to ( x + ℓ, y ) for any ( x, y ) ∈ A ∪ B , resp. A ∪ C . Similarly one can tile Z \ ( B ∪ C ) with vertical copies of T from ( x, y − k ) to ( x, y + ℓ ) forany ( x, y ) ∈ B ∪ C .Hence the result follows from Lemma 2.1. Proposition 3.4. If v ( k ) = v ( ℓ ) , then T = k (1) ℓ tiles Z . Proof.
Let v ( k ) = v ( ℓ ) = n and q = 2 n . When n = 0 , the result follows from Proposition 3.3.So from now on, we assume n ≥ .Let A ⊂ Z be the sets containing the elements ( x, y ) if and only if x − y [( k + ℓ + 2)( q −
1) + 1] ≡ i ( ℓ + k + 2) + j (mod 2( ℓ + k + 2) q ) for some ≤ i ≤ q − and j ∈ { , k + 1 } . Let B = ( q ( ℓ + k + 2) ,
0) + A . Let C ⊂ Z be thesets containing the elements ( x, y ) if and only if x − y [( k + ℓ + 2)( q −
1) + 1] ≡ i ( ℓ + k + 2) + j (mod 2( ℓ + k + 2) q ) for some q ≤ i ≤ q − and j ∈ { k, ℓ + k + 1 } . One can see a depiction of this in Figure 4 inthe case q = 2 , n = 1 . ℓ legend: A B C ... ...
Figure 4: Construction of partial planar tilings when v ( k ) = v ( ℓ ) .Now one can tile Z \ ( A ∪ B ) with T as A ∪ B is the union of diagonals which are alternatelydistance k + 1 and ℓ + 1 apart. One can tile Z \ ( A ∪ C ) horizontally. For this, it is enough totile one horizontal line as every horizontal line is a translate of that one and due to periodicityin particular the set (cid:0) Z \ ( A ∪ C ) (cid:1) ∩ ( { , , . . . , ℓ + k + 2) q − } × { } )= ( { , , . . . , ℓ + k + 2) q − } × { } ) \ ( A ∪ C ) . For this, use translates of T starting at (1 + i ( ℓ + k + 2) , for ≤ i ≤ q − and at ( i ( ℓ + k + 2) , for q ≤ i ≤ q − . To finish, we note that we can tile Z \ ( B ∪ C ) vertically. For this, we only have to check ( { } × Z ) \ ( B ∪ C ) , since gcd { ℓ + k + 2) q, ( k + ℓ + 2)( q −
1) + 1 } = 1 and hence every verticalline is up to some translation identical to every other vertical line. By noting that B and C are subsets of some diagonals on the plane, one checks that ( { } × Z ) ∩ B = { }×{ y | y ≡ i ( ℓ + k +2)+ j (mod 2( ℓ + k +2) q ) , q ≤ i ≤ q − , j ∈ { , ℓ +1 }} . − [ i ( ℓ + k + 2) + j ] · [( k + ℓ + 2)( q −
1) + 1] ≡ − [ i ( k + ℓ + 2)( q −
1) + i + j ( q − k + ℓ + 2) − j (mod 2 q ( k + ℓ + 2)) ≡ ( i − j ( q − k + ℓ + 2) − j (mod 2 q ( k + ℓ + 2)) since q | l + k. When j = 0 , we get i ( k + ℓ + 2) (mod 2 q ( k + ℓ + 2)) for q ≤ i ≤ q − . When j = ℓ + 1 , we get i ( k + ℓ + 2) + k + 1 (mod 2 q ( k + ℓ + 2)) for q ≤ i ≤ q − , since ℓ ≡ q (mod 2 q ) and | q , so ( ℓ + 1)( q − ≡ − q ) . Similarly one has ( { } × Z ) ∩ C = { }× { y | y ≡ i ( ℓ + k + 2)+ j (mod 2( ℓ + k + 2) q ) , ≤ i ≤ q − , j ∈ {− k, }} . Hence one can tile ( { } × Z ) \ ( B ∪ C ) by putting vertical tiles starting at (0 , i ( ℓ + k +2) − k +1) for every i ≡ , , . . . , q − q ) and (0 , i ( ℓ + k + 2) − k ) for every i ≡ q, q + 1 , . . . , q − q ) .Hence the result follows from Lemma 2.1.We can handle a good fraction more cases as follows, but for the remaining cases for T = k (1) ℓ , we suspect that an additional idea may be needed. Proposition 3.5. If k ≡ , | ℓ (or vice versa) then T = k (1) ℓ tiles Z . Proof.
Let K = k + ℓ + 2 and note that K is even. We will choose A, B, C again such thatwe can apply Lemma 2.1.Let j = k , g = gcd { j, K } and K ′ = Kg . We will construct a permutation a , a . . . , a K of { , . . . , K } satisfying a i − a i − ≡ j, j +1 (mod K ) for every ≤ i < K . Write i − q i K ′ + r ,where q i and ≤ r ≤ K ′ − are integers. For every ≤ i ≤ K , choose ≤ a i ≤ K suchthat a i ≡ j ( i −
1) + q i (mod K ) . In particular, one can note that a i − a i − ≡ j + 1 (mod K ) exactly when i is a multiple of K ′ and otherwise a i − a i − ≡ j (mod K ) . To check that this isa permutation, note that if a i ≡ a h (mod K ) then q i ≡ q h (mod g ) . Since ≤ q i , q h ≤ g − , this implies q i = q h . Hence ji ≡ jh (mod K ) ⇒ i ≡ h (mod K ′ ) and so combining with q i = q h we conclude i = h. For every ≤ i ≤ K , let U i = { ( x, y ) ∈ Z | ( x, y ) ≡ (2 i, i ) , (2 i + 1 , i ) , (2 i, i + k + 1) , (2 i + 1 , i + k + 1) (mod 2 K ) } . For every ≤ i ≤ K • if a i − a i − ≡ j (mod K ) , let X i = { ( x, y ) ∈ Z | x ≡ a i − , a i +1 (mod 2 K ) , y or y − k − ≡ a i − , a i (mod 2 K ) } , • if a i − a i − ≡ j + 1 (mod K ) , let X i = { ( x, y ) ∈ Z | x ≡ a i − +1 , a i (mod 2 K ) , y or y − k − ≡ a i − , a i (mod 2 K ) } . Let A ∪ B = U = K [ i =1 U i , A ∪ C = X = K [ i =1 X i and B ∪ C = U △ X. A = U ∩ X, B = U \ X and C = X \ U. Now Z \ ( A ∪ B ) = Z \ U can be tiled vertically by copies of T , by placing those copiesstarting in all ( x, y ) for x ≡ i, i + 1 (mod 2 K ) , y ≡ i + 1 (mod 2 K ) for every ≤ i ≤ K .The set Z \ ( A ∪ C ) = Z \ X can be tiled horizontally by copies of T. Put those copiesstarting at ( x, y ) for some x ≡ a i − + 1 , y or y − k − ≡ a i − , a i (mod 2 K ) when a i − a i − ≡ j (mod K ) and otherwise for some x ≡ a i − + 2 , y or y − k − ≡ a i − , a i (mod 2 K ) .For the last case, we will check that Z \ ( B ∪ C ) can be tiled vertically by copies of T again.For every ≤ i ≤ K , the set (cid:0) U a i − ∪ U a i (cid:1) △ X i equals { ( x, y ) ∈ Z | ( x, y ) or ( x, y − k − ≡ v (mod 2 K ) , for some v ∈ V } where V = { (2 a i − , a i ) , (2 a i − + 1 , a i − ) , (2 a i , a i ) , (2 a i + 1 , a i − ) } if a i − a i − ≡ j (mod K ) , or V = { (2 a i − , a i − ) , (2 a i − + 1 , a i ) , (2 a i , a i − ) , (2 a i + 1 , a i ) } if a i − a i − ≡ j + 1 (mod K ) . So clearly (cid:0) U a i − ∪ U a i (cid:1) △ X i can be tiled vertically.The conclusion now follows easily as U △ X = K/ [ i =1 (cid:0) U a i − ∪ U a i (cid:1) △ X i . Hence the result follows from Lemma 2.1.
Corollary 3.6.
More than of the punctured intervals tile Z . Proof.
For this, we combine Proposition 3.2 (half of the cases), Proposition 3.4 (one third ofthe cases) and Proposition 3.5 (one eighth of the cases).
Conjecture 1.1 should be a substantially more difficult problem than Question 1.2. In this sec-tion, to give an indication of the subtleties, we collect two classes of (known) one-dimensionaltiles that do not tile Z d for a given d .Let T k be the tile k ( k − k − . . . ( k − k − | {z } k times ( k − k as considered in [2].Let D n be the tile . . . (1)2 | {z } n times , as considered in [3].The following proposition shows that for every d , one can find k and n such that neither D n nor T k tiles Z d . The reason behind this is slightly different for the two tiles. The first usessparseness of tiles put in one direction. The other considers the intersection of the tiles withsubdivisions of Z d . Proposition 4.1. T k does not tile Z d for d < k +2 k − k − and D n does not tile Z d for n > d − .Proof. In the case of T k , one looks to the maximum volume covered by tiles in one of the d orthogonal directions in a hypercube [ N ] d . When N → ∞ , the ratio of the volume covered by7hese tiles will have a limsup which is at most k − k +2 k − from which the result follows as thesum of the ratios over the d directions should sum to .We assume D n tiles Z d and look to the intersection of this fixed tiling with a hypercube [ N ] d .Look to the d possible partitions of Z d in hypercubes with side length . Call a nonemptyintersection of [ N ] d with a hypercube of side length for a given partition a subregion. Wenow count the total number D of intersections of a subregion of a partition and a D n whichare of size , in two different ways.For each of the d partitions, there are less than (cid:0) N + 2 (cid:1) d subregions. Each subregionwill contain at most d − intersections with a D n of size . Hence D < d (cid:0) N + 2 (cid:1) d d − = d − ( N + 6) d .On the other hand, there are at least ( N − n ) d n D n ’s completely inside the hypercube. Every D n of these, intersects n subregions in exactly places for each of · d − partitions. For d − partitions, these D n intersects n − small hypercubes in exactly places and smallhypercubes in exactly one place. This implies that D ≥ ( N − n ) d n · d − (3 n − .Hence d − ( N + 6) d > ( N − n ) d n · d − (3 n − for all N , in particular one finds that theleading coefficients satisfy d − ≥ d − · n − n ⇒ n ≤ d − . In the case of D n , this generalizes the ‘only if’ part of Proposition in [3]. Let us remarkthat this also follows from a straightforward generalization of Theorem in [3], which concerns‘convolutions’ of tiles. In case it might be of use to others, we use the notation of [3] to statethe generalization (and leave the proof to the reader). Theorem in [3] is n = 2 and d = 2 . Proposition 4.2 ([3]) . Suppose T ⊂ Z n is a tile. Suppose that S ⊂ Z d is a symmetric tile(i.e. no matter how the tile is oriented, it is a translate of itself ). Then if for some m ∈ N onehas | S ⋆ m T | < | S || T | , or if | S ⋆ m T | ∞ < | T | and | S | 6 = 0 , then T does not tile Z d . Since the examples we presented in Section 4 contain many gaps, it is natural to wonder ifthe following is true. If so, it would prove Conjecture 1.1.
Question 5.1.
Does there exist a function f : N → N such that any tile T ⊂ Z with at most N gaps, i.e. T is the union of at most N + 1 intervals, tiles Z f ( N ) ? This question naturally leads to a number of subproblems, which if solved could lead toprogress in Conjecture 1.1. • Does any punctured interval k (1) ℓ tile Z ? • Does any one-dimensional tile k ( m ) ℓ tile Z d for some small (uniform) choice of d ? • Find the smallest d such that D n tiles Z d .Answering the first subquestion affirmatively would improve upon Corollary 3.6 and wouldconfirm Question in [4]. By the work in this paper, the remaining open cases are ≤ v ( k ) < v ( ℓ ) , the smallest case being the tile T = 4(1)8 . For the second subquestion, by the reduction used at the end of the proof of Lemma 2.1,one knows that it is enough to do this for m < k, ℓ.
When d = gcd { m, k, ℓ } > , one can form8he union of a k × d × d and a ℓ × d × d cuboid with a m × d × d gap. So one only needsto consider the case with gcd { k, m, ℓ } = 1 . In the case m | k + ℓ + m, one can extend theconstruction in Proposition 3.2 by choosing K = k + ℓm + 1 and replacing every element in theconstruction of A, B and C by squares of size m × m. For this, it could be helpful to find sometiling where m ∤ k + ℓ, e.g. T = 3(2)4 and then find a general construction for these cases. References [1] A. Adler and F. C. Holroyd. Some results on one-dimensional tilings.
Geom. Dedicata ,10(1-4):49–58, 1981.[2] V. Gruslys, I. Leader, and T. S. Tan. Tiling with arbitrary tiles.
Proc. Lond. Math. Soc.(3) , 112(6):1019–1039, 2016.[3] A. U. O. Kisisel. Polyomino convolutions and tiling problems.
J. Combin. Theory Ser. A ,95(2):373–380, 2001.[4] H. Metrebian. Tiling with punctured intervals.