The layer number of grids
aa r X i v : . [ m a t h . M G ] S e p THE LAYER NUMBER OF GRIDS
GERGELY AMBRUS, ALEXANDER HSU, BO PENG, AND SHIYU YAN
Abstract.
The peeling process is defined as follows: starting with a finite point set X ⊂ R d , werepeatedly remove the set of vertices of the convex hull of the current set of points. The number ofpeeling steps needed to completely delete the set X is called the layer number of X . In this paper,we study the layer number of the d -dimensional integer grid [ n ] d . We prove that for every d ≥ n ] d is at least Ω (cid:16) n dd +1 (cid:17) . On the other hand, we show that for every d ≥
3, ittakes at most O ( n d − / ) steps to fully remove [ n ] d . Our approach is based on an enhancement ofthe method used by Har-Peled and Lidick´y [9] for solving the 2-dimensional case. Introduction
History.
Consider a finite point set X ⊂ R d with d ≥
1. Define the peeling process as follows:in every step, we remove the set of vertices of the convex hull of the previous iteration. The sets ofpoints removed in each step form the convex layers of X , while the total number of steps neededto completely delete X is the layer number of X , which we denote by τ ( X ).The convex layer decomposition of planar sets was first studied by Eddy [6] and Chazelle [5]from the algorithmic point of view. The latter article gave an O ( n log n ) running time algorithmfor computing the convex layers of an n -element planar point set. Therefore, layer numbers maybe computed quickly and efficiently.Almost 20 years later, Dalal [7] determined the expected layer number of random point sets. Heproved that if X is a set of n random points chosen independently from the d -dimensional unitball, then E ( τ ( X )) = Θ( n / ( d +1) ).Let [ n ] d = { , . . . , n } d be the n × n × . . . × n d -dimensional integer grid. Har-Peled and Lidick´y [9]studied the peeling process of the planar set [ n ] . They proved the asymptotically sharp bound τ ([ n ] ) = Θ( n / ), which provides an example when random points and lattice points behavesimilarly (see the survey article of B´ar´any [4] for such phenomena). It is natural to believe thatthis analogy also holds for higher dimensional cases. Conjecture 1.
The layer number of the grid [ n ] d satisfies τ ([ n ] d ) = Θ( n d/ ( d +1) ) for every d ≥ . If true, the above asymptotic estimate would match the result of Dalal [7] on random pointsets. Conjecture 1 initiated our current research project. We study the layer number of the higherdimensional grids [ n ] d , with particular focus on the 3-dimensional case. Although we are not ableto reach the asymptotically sharp estimate of Conjecture 1, our results provide the first non-trivialestimates for the layer number of [ n ] d with d ≥ n × n grid seem toconverge to circles. This observation has been given an experimental verification by Eppstein, Har-Peled, and Nivasch [8], who established an interesting connection between the planar grid peelingprocess and the affine curve-shortening flow. Further algorithmic applications of the peeling processwere given in [1]. Mathematics Subject Classification.
Key words and phrases.
Layer numbers, peeling process, convex hull, integer points, geometric processes.Research of the first author was supported by NKFIH grants PD-125502 and KKP-133819. .2. Definitions.
First, we rigorously define the peeling process. Starting with a finite point set X = X of R d , let us recursively define X i = X i − \ ext( X i − ) for each i ≥
1, where ext( Y ) standsfor the set of extreme points of Y (that is, the set of vertices of the convex hull of Y ). The smallestindex i for which X i = ∅ is called the layer number of X , denoted by τ ( X ) (this is sometimes alsoreferred to as the convex depth of X ). The point sets removed in each step are called the convexlayers of X .In the article, we are going to study the peeling process of [ n ] d . For a given d and n , set X = X to be [ n ] d , and denote by P i the convex hull of X i introduced above. Then P i is a convex latticepolytope, which is going to be referred to as the i th convex layer polytope of [ n ] d . Naturally, thepolytopes P i form a nested sequence, starting from the cube [1 , n ] d , and shrinking to the emptyset. We also note that since the initial set [ n ] d is symmetric, each P i is symmetric as well.The subsequent arguments are based on lattice geometric observations. A core notion is thefollowing. Definition 1 (Primitive vector) . An integer vector ( x , x , . . . , x d ) ∈ Z d is primitive if not allcoordinates are and all coordinates are coprime, meaning the greatest common divisor of all x i is . Geometrically, this translates to the condition that the segment connecting the origin and theprimitive point does not contain any other integer points. For each positive integer µ , define V µ to be the set of primitive vectors with each coordinatecontained in the interval [0 , µ ]. Definition 2 (Direction of category k) . Fix a step of the peeling process of [ n ] d with the corre-sponding convex layer polytope P . Assume that P is non-degenerate. For each vector v ∈ V µ , thereexist two supporting hyperplanes to P that are orthogonal to v . By symmetry, these two hyperplaneshave isomorphic intersections with P . If these intersections are k -dimensional faces of P , where ≤ k ≤ d − , we define v to be of category k in this given peeling step. By the standard notation, for a convex polytope P ⊂ R d , let f k ( P ) denote the number of k -dimensional faces of P , for each k ∈ [0 , d ]. For all other common definitions regarding convex setswe refer to the monograph of Schneider [12].All asymptotic notations in the article are meant for a fixed d while n converges to ∞ , with theimplied constants depending on d .2. Bounds on the layer number
A lower bound on τ ([ n ] d ) .Theorem 1. The layer number of a d -dimensional grid [ n ] d is bounded below by Ω (cid:16) n dd +1 (cid:17) .Proof. G.E. Andrews [2] proved that for any convex lattice polytope P , f ( P ) ≤ O (cid:16) Vol( P ) d − d +1 (cid:17) , and this bound is sharp.Consider the peeling process of [ n ] d with the corresponding convex layer polytopes P i . For each i ,the volume of P i is at most n d . Thus, the upper bound on the number of vertices of each layeris ( n d ) d − d +1 . Hence, the number of vertices removed in each peeling step is at most O (cid:0) n d − dd +1 (cid:1) , whichyields the lower bound on the layer number τ (cid:16) [ n ] d (cid:17) ≥ n d O (cid:18) n d − dd +1 (cid:19) = Ω (cid:16) n dd +1 (cid:17) . (cid:3) onjecture 1 states that this lower bound is tight.2.2. Upper bounds on τ ([ n ] ) . The subsequent arguments are based on the approach of Har-Peled and Lidick´y [9] for studying the planar case of Conjecture 1.
Lemma 1.
Let H be a hyperplane determined by d affinely independent points of [ n ] d . There existsa primitive vector, with each coordinate bounded by O ( n d − ) , normal to H .Proof. Let the d lattice points be denoted by w , . . . , w d . Consider the vectors v i = w i − w d foreach 1 ≤ i ≤ d −
1. We may find a vector u normal to H by computing the generalized crossproduct of the vectors v , . . . , v d − : u = ^ ( v , . . . , v d − ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) v · · · v d ... . . . ... v d − · · · v dd − e · · · e d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) where ( e , . . . , e d ) is the standard basis of R d , and each v i is of the form v i = ( v i , . . . , v di ). Theabove formula implies that the coordinates of u are all integers. Let ˜ u be the unique primitivevector contained in the segment 0 u . Then ˜ u is also normal to H . In order to estimate thecoordinates of ˜ u , we calculate the above determinant through Laplace expansion. The coefficientof each e i is the determinant of the ( d − × ( d −
1) matrix obtained by deleting the last row and i th column, whose absolute value is at most ( d − n d − . Thus, the norm of u is bounded aboveby √ d ( d − n d − = O ( n d − ). (cid:3) For any nonzero vector v ∈ Z d , let H v denote the set of hyperplanes orthogonal to v whichcontain at least one point of [ n ] d . Lemma 2.
Given a primitive vector v ∈ V µ , the number of hyperplanes with normal vector v thatintersect [ n ] d is bounded by | H v | ≤ dnµ .Proof. Fix a primitive vector v = ( a , a , ..., a d ) ∈ V µ . Consider a hyperplane H orthogonal to v with its defining equation a x + ... + a d x d = c . Assume that H contains a point ( x , . . . , x d ) ∈ [ n ] d .Since x i ∈ Z and 1 ≤ x i ≤ n for every i , and 0 ≤ a i ≤ µ for every i , the value of c is bounded fromabove by d n max( a i ) ≤ dnµ , while being strictly positive. Since c is an integer, this implies thedesired bound. (cid:3) Lemma 3.
For every d ≥ , | V m | = Θ( m d ) holds with the implied constant depending on d .Proof. Avoiding any conflicts with the rest of the paper, for this proof exclusively, let µ denote theM¨obius function defined in [10, pp. 234]. Introduce Jordan’s totient function J r ( k ) which countsthe r -tuples of positive integers all less than or equal to k that form a coprime ( r + 1)-tuple togetherwith k . This is a generalization of Euler’s totient function, which is given by J .Results from Andrica and Piticari [3] give us J r ( k ) = k r X a | k µ ( a ) a r = X a | k (cid:18) ka (cid:19) r µ ( a ) = X aa ′ = k ( a ′ ) r µ ( a ) . ow, | V m | = m X i =1 J d − ( i ) = X aa ′ ≤ m ( a ′ ) d − µ ( a )= m X a =1 µ ( a ) ⌊ m/a ⌋ X a ′ =1 ( a ′ ) d − = m X a =1 µ ( a ) ⌊ m/a ⌋ d d + 12 j ma k d − + d − X i =2 B i i ! p i − j ma k ( d − − i +1 ! = ( ∗ )by Faulhaber’s Formula, where B i denotes the i th Bernoulli number, and p i − = p !( p − i +1)! . Contin-uing the above chain of equalities,( ∗ ) = 1 d m X a =1 µ ( a ) (cid:18)j ma k d + O (cid:18)j ma k d − (cid:19)(cid:19) ≥ d d m X a =1 µ ( a ) (cid:18)(cid:16) ma (cid:17) d + O (cid:18)(cid:16) ma (cid:17) d − (cid:19)(cid:19) = m d d d m X a =1 µ ( a ) a d ! + O m d − ∞ X a =1 a d − ! = m d d d ∞ X a =1 µ ( a ) a d ! − m d d d ∞ X a = m +1 µ ( a ) a d + O (cid:16) m d − (cid:17) ≥ m d d d ζ ( d ) − m d d d Z ∞ m x d d x + O (cid:16) m d − (cid:17) = m d d d ζ ( d ) − O ( m ) + O (cid:16) m d − (cid:17) = m d dζ ( d ) + O ( m d − ) . (cid:3) After the necessary preparations, we are now able to prove a nontrivial upper bound on the layernumber of higher dimensional grids. Here comes our first estimate for the d = 3 case. Theorem 2.
The number of steps needed to peel away [ n ] is at most O ( n / ) .Proof. Consider a given step of the peeling process of [ n ] with the corresponding convex layerpolyhedron P . Each vector v ∈ V µ may be of category 0 , , or 2, as introduced in Definition 2.If v is of category 0, then the intersection of P with either of its supporting planes of normal vector v is just one vertex, which is deleted in the next step of the peeling process. Thus, the number ofplanes orthogonal to v intersecting the remaining set decreases by at least 2. By Lemma 2, initiallythere are at most 3 nµ such planes, hence v may be of category 0 in at most nµ steps.If v is of category 1, we know that the intersection of P with its supporting planes orthogonal to v are edges of P . Being contained in [ n ] , these may not contain more than n grid points. In eachsubsequent step, the two endpoints of the remaining edge is deleted, so it takes at most ( n + 1) / P with its supporting plane. Therefore,by Lemma 2 again, any given v may be of category 1 in at most n ( n + 1) µ peeling steps. et M = 2 n µ . Since the above arguments show that v may be of category 0 or 1 in at most n µ peeling steps, v must be of category 2 in at least M/ M iterationsof the peeling process.Denote by c k,i the number of category k directions in V µ in the i th peeling step.By Euler’s polyhedron formula, f ( P i ) − f ( P i ) + f ( P i ) = 2 holds. Moreover, since f ( P i ) ≥ / f ( P i ), we also have f ( P i ) ≥ f ( P i ) /
2. Since for each category 2 primitive vector v containedin V µ , there exists a pair of opposites facets of P perpendicular to v , f ( P i ) ≥ c ,i and hence f ( P i ) ≥ c ,i . By Lemma 3, we reach the following chain of inequalities:(1) M X i =1 f ( P i ) ≥ M X i =1 c ,i ≥ n µ | V µ | ≥ γn µ with some positive constant γ . Setting µ = γ − / n / , we find that the total number of verticesof the convex hulls in the first M iterations is at least n , which is the number of all grid pointsin [ n ] . This implies that the peeling process must terminate in at most M = 2 n µ = O ( n / )iterations. (cid:3) It should be noted that the above argument may not be applied to the higher dimensional cases,since the number of facets may be much more than the number of vertices (see the relevant UpperBound Theorem by McMullen [11]).The main restricting factor on our bound is the worst case for the number of steps a primitivevector can be of category 1, in that it is rare that some, if any, edges will contain n vertices andtake n/ O ( nµ ) upper bound for the total number of steps in which a given direction may be ofcategory 1. If that was true, M could be set to be Θ( nµ ) in the above proof, leading to the desiredtight upper bound Θ( n / ) for the layer number τ ([ n ] ). This sketches a possible line of attack forConjecture 1.Although we could not reach the upper bound of Conjecture 1, we are still able to improve onthe result of Theorem 2. In the remaining part of the section, we present this strengthened bound.The main tool is the following general lemma. Lemma 4.
For all d ≥ there exists a positive constant α d > such that the number of primitivevectors in V µ which are not perpendicular to any non-zero lattice vector of norm at most α d µ /d isat least | V µ | .Proof. For a given ν > v ∈ V µ suchthat the hyperplane v ⊥ does not contain any non-zero lattice point of Z d with norm less than ν .The set of these vectors in V µ may be obtained by going over all short non-zero vectors, anddropping out all primitive vectors in V µ that are perpendicular to it. For a given w ∈ Z d with | w | < ν , we claim that w ⊥ contains at most µ d − vectors in V µ . This follows from selecting abasis vector e i not contained in w ⊥ , and considering the projection of w ⊥ ∩ V µ to e ⊥ i along e i . Theprojection mapping restricted to w ⊥ is one-to-one, it maps lattice points to lattice points, and theimage is contained in the ( d − µ . Therefore, the number ofimage points, and thus, the number of points in w ⊥ ∩ V µ , is at most µ d − .Hence, for each short vector, the number of vectors deleted from V µ in the above process is at most µ d − . On the other hand, the number of vectors w ∈ Z d with | w | ≤ ν is at most (2 ν + 1) d = Θ( ν d ).Thus, the total number of vectors deleted from V µ is at most O ( µ d − ν d ). Since, by Lemma 3, | V µ | = Θ( µ d ), setting ν = α d µ /d with an appropriate constant α d (depending on d ) guaranteesthat at most half of the vectors of V µ are dropped out. (cid:3) tilizing Lemma 4 we are able to improve our upper bound on the layer number of the 3-dimensional grid. Theorem 3.
The layer number of [ n ] is at most O ( n / ) .Proof. Let V ′ µ be the set of vectors in V µ specified in Lemma 4. Select v ∈ V ′ µ arbitrary. Then,any line ℓ perpendicular to v may contain at most s = √ n/ ( α µ / ) + 1 lattice points of [ n ] : thelength of the segment ℓ ∩ [0 , n ] is at most √ n , while the shortest nonzero lattice vector parallelto ℓ must be of norm at least α µ / .Therefore, if v is of category 1 in a given step of the peeling process with the correspondingconvex layer polytope P , it takes at most ( s + 1) / P and its supporting hyperplane perpendicular to v . Thus, by Lemma 2, thenumber of steps in which v is of category 1 is at most N = 6 /α · n µ / . The number of steps inwhich v is of category 0 is at most nµ .Set M ′ = 2( N + nµ ), and apply the same argument as in the proof of Theorem 2 for the first M ′ steps of the peeling process, but replacing the set V µ of considered normal vectors by V ′ µ . Let c ′ ,i denote the number of category 2 directions in V ′ µ in the i th step of the peeling process of [ n ] .Similarly to (1), we arrive at the following chain of inequalities: M ′ X i =1 f ( P i ) ≥ M ′ X i =1 c ′ ,i ≥ M ′ | V ′ µ | = Θ( n µ / )Θ( µ ) = γ ′ n µ / with some positive constant γ ′ . We finish the proof by setting µ = ( γ ′ ) − / n / , which by thesame argument as before results in the upper bound τ ([ n ] ) ≤ O ( n / ). (cid:3) Upper bound for higher dimensional cases.
We conclude the paper by the extension ofthe bound of Theorem 3 to higher dimensions using a simple recursive argument.
Theorem 4.
For every d ≥ , the layer number of the d -dimensional grid satisfies τ ([ n ] d ) ≤ O ( n d − / ) .Proof. We will apply induction on d . The d = 3 case is provided by Theorem 3. We shall provethe d -dimensional case, assuming that the estimate is valid for [ n ] d − .Consider any set A ⊂ [ n ] d . Let H be the hyperplane defined by the equation x = 1, which istangent to the cube [1 , n ] d . The restriction of the ( d -dimensional) peeling process of A to H is thenequivalent to the ( d − A ∩ H . Since A ∩ H is contained in a copyof [ n ] d − , this must terminate in at most O ( n ( d − − / ) steps, by the inductive hypothesis. Thesame reasoning may be applied to each of the 2 d boundary hyperplanes of the cube [1 , n ] d . Thus,we obtain that after O ( n ( d − − / ) peeling steps of A , the remaining set will contain no points onthe boundary of the cube [1 , n ] d .Now, we note that the grid [ n ] d may be written as the union of ⌈ n/ ⌉ cubic shells. Applying theabove argument for each of these shells results in the upper bound O ( n d − / ) for the layer numberof [ n ] d . (cid:3) Acknowledgements
This research was done under the auspices of the Budapest Semesters in Mathematics program.
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E-mail address , G. Ambrus: [email protected]
Alexander Hsu, Department of Mathematics, Reed College, Portland, OR, USA
E-mail address , A. Hsu: [email protected]
Bo Peng, Department of Mathematics, Carleton College, Northfield, MN, USA
E-mail address , B. Peng: [email protected]
Shiyu Yan, Department of Mathematics, Carleton College, Northfield, MN, USA
E-mail address , S. Yan: [email protected]@carleton.edu