A Local-Global Principle for Vertex-Isoperimetric Problems

A Local-Global Principle for Vertex-Isoperimetric Problems Sergei L. Bezrukov∗ Department of Mathematics and Computer Science University of Wisconsin - Superior Superior, WI 54880-4500, USA

Oriol Serra Departament de Matem`atica Aplicada i Telem`atica Modul C3, Campus Nord, Universitat Polit`ecnica de Catalunya Jordi Girona 1, 08034 Barcelona, Spain

Dedicated to Daniel J. Kleitman on his 65th birthday Abstract We consider the vertex-isoperimetric problem for cartesian powers of a graph G. A total order  on the vertex set of G is called isoperimetric if the boundary of sets of a given size k is minimum for any initial segment of , and the ball around any initial segment is again an initial segment of . We prove a local-global principle with respect to the so-called simplicial order on Gn (see Section 2 for the definition). Namely, we show that the simplicial order n is isoperimetric for each n ≥ 1 iff it is so for n = 1, 2. Some structural properties of graphs that admit simplicial isoperimetric orderings are presented. We also discuss new relations between the vertex-isoperimetric problems and Macaulay posets.

1

Introduction

Discrete isoperimetric problems have been widely studied in the literature due to their theoretical interest and numerous applications (see [3] for a survey). Although all these problems share some common features, each specific version of the problem requires particular techniques and gives rise to specific results. In this paper we consider the vertex-isoperimetric problem on graphs. Let G = (V, E) be a finite simple graph, and let A ⊆ V . Consider the (vertex) boundary of A, and the ball around ∗

Partially supported by the Spanish Research Council under project TIC97-0963.

1

A respectively defined as Γ(A) = {v ∈ V \ A | distG (v, A) = 1} B(A) = A ∪ Γ(A). Our objective is to find for any given m a subset A ⊆ V such that |A| = m and |Γ(A)| ≤ |Γ(X)| for any X ⊆ V with |X| = m. We call such subsets optimal . The result of Harper [11] for the n-cube is one of the first and most celebrated solutions of a vertex-isoperimetric problem. A generalization of this result for finite grids, i.e., the cartesian products of chains, was obtained by Bollob´as and Leader [8] and Bezrukov [5]. Wang and Wang [15, 16] gave the solution for infinite grids (with respect to Γ(·)), and finite tori with a slightly different definition of the boundary. A solution for finite even tori, i.e., the cartesian products of even cycles, with respect to the boundary operator Γ(·) defined above was considered by Bollob´as and Leader [9] and definitely solved by Karachanjan [13] and Riordan [14]. All the above mentioned results share the two following common features. First, there exists a sequence of nested optimal subsets A0 , A1 , . . . , A|V | , i.e. ∅ = A0 ⊂ A1 ⊂ · · · ⊂ A|V | = V,

|Am | = m,

for m = 0, 1, . . . , |V |

Second, the ball B(Am ) for any Am of this sequence is again a subset of the sequence. It turns out that similar properties are important and also appear in many other discrete extremal problems [7]. This fact suggests the following definition. Let  be a total order on the vertex set V of a graph G = (V, E). The initial segment of length m in order  is denoted by F(m). The order  is called isoperimetric if for any m, 1 ≤ m ≤ |V |, it satisfies the following two conditions: Nestedness: Continuity:

|Γ(F(m))| = min{|Γ(A)| | A ⊂ V, |A| = m} B(F(m)) is an initial segment of  .

(1) (2)

Thus, the above results can be stated as saying that products of chains and products of even cycles admit isoperimetric orderings. For cartesian products of chains, the known isoperimetric orderings happen to be simplicial (cf. Section 2 for precise definition). In addition to the vertex-isoperimetric problems discussed above, substantial research was done on edge-isoperimetric problems where, instead of minimizing the vertex boundary, the objective was to minimize the number of edges connecting a set A ⊆ V with V \ A. Representation of solutions to the edge-isoperimetric problem on a graph G = (V, E) and its cartesian powers in terms of a total order on the vertex set, i.e. the nestedness of the solutions, is a common practice in this area (see survey [4] for more information). In [2] Ahlswede and Cai proved the following remarkable result, which they call the local-global principle: if the lexicographic order provides nestedness in the edge-isoperimetric problem for n = 2, then it is so for any n ≥ 3. It is hard to overestimate the importance of this result. One has to check just a finite number of cases for n = 2 in order to make sure that the lexicographic order provides nestedness of solutions of the edge-isoperimetric problem for any 2

larger dimension. It turns out that a similar local-global principle for the edge-isoperimetric problem is valid also with respect to some other total orders [4]. The present authors obtained a local-global principle for the shadow minimization problem on cartesian products of Macaulay posets [7] with respect to the lexicographic order. It turns out that posets, in contradistinction to graphs in the edge-isoperimetric problem, have to satisfy a number of properties for the validity of this principle. Although these are quite different problems, a well developed technique for solving both is based on a set compression and works well starting with three graphs (resp. posets) in the product. The case n = 2 under this approach requires a special investigation. It seems that the local-global principle is a general phenomena for various discrete extremal problems, and for many orders. The effect takes place mostly due to “cartesianess” of the product, and some nice properties of the considered total orders. Presently, however, it seems difficult to specify the orders for which a local-global principle can hold for a specific extremal problem on cartesian products of corresponding structures, particularly if we just know the two above-mentioned instances of such problems. This is an interesting and important question for future research. There are many connections between the three mentioned problems: shadow minimization on posets and both edge-isoperimetric and vertex-isoperimetric problems on graphs (cf. [6]). However, the differences between the resulting extremal sets and between the involved orders require a specific treatment of each one, and the techniques can not be directly translated from one problem to another. For example, the local-global principle fails to hold for the vertexisoperimetric problem with respect to the lexicographic order, as can be easily checked in the case of the complete graph K2 . An isoperimetric inequality for the powers of arbitrary complete graphs is derived in [12]. In this paper we prove the validity of the local-global principle for the vertex-isoperimetric problem on cartesian products of graphs with respect to the simplicial order defined in the next section. Our main result is the following theorem (K2,2 and K4 − e are shown in Fig. 2(a)). Theorem 1 Let G = (V, E) be a connected graph different from K2,2 and K4 −e. The simplicial ordering n is isoperimetric in Gn for any n ≥ 1 if and only if it is so for n = 1, 2. The paper is organized as follows. Section 2 introduces the required technical definitions and preliminary results. It turns out that isoperimetricity of a simplicial order implies some structural properties of the base graph G. These properties include the existence of a pair of antipodal vertices and a certain distribution of the boundaries around them. The precise statements and their proofs are given in Section 3. Section 4 is devoted to the proof of our main result, Theorem 1. Some examples of graphs to which the local-global principle for simplicial orders can be applied are considered in Section 5. There we also present new relations between the vertex-isoperimetric problem and Macaulay posets. Final remarks in Section 6 conclude the paper. We present there a new strong necessary condition for the simplicial order to be isoperimetric and discuss the ways of generating the graphs for which Theorem 1 is applicable.

3

2

Simplicial orders and isoperimetry

Throughout the paper G = (V, E) denotes a finite simple connected graph. We consider the cartesian powers Gn = (V n , E n ) of G, where two vertices x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) are adjacent iff (xi , yi ) ∈ E for exactly one index i and xj = yj for j 6= i. G

Let ≤ be a total order of V , and let 0 be the smallest vertex of V in this order. For x ∈ V denote by kxk the distance between x and 0 in G. Now for n ≥ 1 and x = (x1 , . . . , xn ) ∈ V n P define kxk = ni=1 kxi k. The set Vtn = {x ∈ V n | kxk = t} is called the t-th level of Gn . We denote by h + 1 the number of levels of G. Thus, the number of levels in Gn is hn + 1. G

The simplicial order n on the set V n associated with ≤ is defined as follows. For n ≥ 1, we write x = (x1 , . . . , xn ) ≺n y = (y1 , . . . , yn ) iff (i) kxk < kyk, or G

(ii) kxk = kyk and there is s, 1 ≤ s ≤ n, such that xi = yi for i < s and xs > ys . G

In particular, for n = 1, we have x  y if either kxk < kyk or kxk = kyk and x ≥ y. The initial segment of length m in order n is denoted by F(m, n ). The ordinal of a vertex x ∈ V n , is denoted by #(x, n ). We often write , F n (m) and #(x) for n , F(m, n ) and #(x, n ) respectively, assuming that the reference to  and n is clear from the context. For x ∈ V n denote F n (x) = {z ∈ V n | z n x}. Recall that a total order on Gn is called isoperimetric if the nestedness and continuity conditions (1) and (2) are satisfied. We often denote the vertices of G by their cardinal numbers in the G

range from 0 to k = |V | − 1 according to the order ≤. Examples of isoperimetric orders for the paths P2 and P3 and their cartesian powers are shown in Fig. 1. The upper shadow (resp. lower shadow ) of a subset A ⊂ Vt , is defined as ∇(A) = Γ(A) ∩ Vt+1 ∆(A) = Γ(A) ∩ Vt−1 . We write ∇(x) and ∆(x) for ∇({x}) and ∆({x}) respectively. By the definition of Vt we have ∇(Vt ) = Vt+1

for 0 ≤ t < h,

(3)

and ∆(x) 6= ∅ unless x = 0. However, we may have ∆(Vt ) 6= Vt−1 , and ∇(x) = ∅ for some x ∈ Vt .

4

s1

s0

s0

@ s 21 @6 @ @ s 11 @s20 s 01 @4 @5 3 @ @ @ s 01 @s10 @2 1 @ @s00

s111 @7 @ s011 s101@s110 @6 @5 4 @ @ s001@s010@s100 @3 2 1 @ @s000

s2 s1

s 12 @7 @

b.

a.

s 22 @8 @

0

c.

0

Figure 1: Isoperimetric orders on P2 and P3 (a), on P23 (b), and on P32 (c)

G

As it is shown in Section 3, the following property of a total order ≤ plays a fundamental G

G

role for the isoperimetricity of simplicial orders associated to ≤. The total order ≤ is called level-greedy if for any x ∈ V with ∇(x) 6= ∅ the following two conditions are satisfied y ∈ ∇(x) G

y < min ∇(x) and kyk > kxk

implies implies

G

y>x

(4)

G

y<x

(5)

G

≤

We conclude the section with the following remark. Remark 1 Let G be a graph with |V | = p and h = 1 for whose cartesian powers the simplicial order is isoperimetric. Then, for each vertex x ∈ V1 , we must have |Γ(x)| ≥ |Γ(0)| = |V | − 1, so that G = Kp . Since for the cartesian powers of complete graphs with p > 2 no nested solutions exist in general [12], in the sequel we always assume that either G = K2 or h > 1.

3

Structural properties G

Throughout this section we assume that G is a connected graph with a total order ≤ on its G

vertex set. We present here some structural properties of G and the order ≤ which will be used for the proof of the main result (Theorem 1) in the next section. G

Let n be the simplicial order associated with ≤ on V n , and let x ∈ V n . We introduce the marginal boundary of a vertex y ∈ V n with respect to F n (x): ∂(x, y) = B({y}) \ B(F n (x) \ {x}). In other words, |∂(x, y)| measures the increment in the size of the ball around the initial segment F n (x) \ {x} as y is added to this set. 5

For 0 ≤ t ≤ h, let at and bt be, respectively, the minimum and maximum vertices in Vt with G

respect to the order ≤. In other terms, at and bt are, respectively, the maximum and minimum vertices in Vt with respect to the order . In particular, a0 = b0 = 0. The first property is just an auxiliary one, and is almost straightforward. Lemma 1 Let n ≥ 1 and let the order n be isoperimetric. Furthermore, let x, y ∈ V n with x ≺n y. Then (a.) |∂(x, x)| ≤ |∂(x, y)|. (b.) If kxk = kyk then minn ∆(x)  minn ∆(y). (c.) If kxk = kyk then maxn ∇(x)  maxn ∇(y). Proof. The first assertion is an immediate consequence of the nestedness property (1) of n . To show the second one, denote x0 = minn ∆(x), y0 = minn ∆(y), and assume y0 ≺n x0 . Then B(F n (y0 )) contains y and does not contain x. Since x n y, then B(F n (y0 )) is not an initial segment. This contradicts the continuity of n (cf. (2)). To show the third assertion, let x0 = maxn ∇(x) and y0 = maxn ∇(y). Suppose on the contrary that y0 ≺n x0 and let z = minn ∆(x0 ). We have z n x ≺n y. Let A = F n (z) and A0 = (A \ {z}) ∪ y. By the continuity property, F n (x0 ) ⊆ B(A), whereas the largest element in B(A0 ) is strictly smaller (in order n ) than x0 . Therefore, |B(A0 )| < |B(A)| and A0 is not an initial segment, contradicting the nestedness of n . 2 Lemma 2 Assume that the simplicial orders  and 2 are isoperimetric in G and G2 respectively. Let x ∈ V such that ∇(x) = ∅. Then kxk = h. Moreover, |Vh | = 1. Proof. Let x with kxk = t be the minimum element (in order ) such that ∇(x) = ∅. Now 0 = |∂(bt , x)| ≥ |∂(bt , bt )| implies ∇(bt ) = ∅ (cf. Lemma 1(a)). By the minimality of x, we have x = bt . Since G is connected, then t ≥ 1 and ∇(bt0 ) 6= ∅ for each t0 with 0 ≤ t0 < t. Assume |Vt | ≥ 2. We show that this assumption leads to a contradiction. Let (η, ζ) be the first vertex in order 2 such that (at , bt ) ∈ ∇((η, ζ)). Denote A = F 2 ((η, ζ)). We have B((bt , bt )) ⊆ F 2 ((at , bt )) ⊆ B(A). (6) First assume ζ = bt (hence η ∈ Vt−1 ). Let (z 0 , z 00 ) be the successor of (η, bt ) in order 2 . Since |Vt | ≥ 2 then z 0 = η and z 00 is the successor of bt in order . Note that (at , z 00 ) ∈ ∂((η, z 00 ), (η, z 00 )). Therefore, (6) implies |∂((η, z 00 ), (η, z 00 ))| ≥ 1 > 0 = |∂((η, z 00 ), (bt , bt ))|, contradicting Lemma 1(a). 6

Now assume η = at (hence ζ = bt−1 ). If |∇(bt−1 )| ≥ 2, then |∂((at , bt−1 ), (at , bt−1 ))| ≥ |∇(bt−1 )| ≥ 2 > 1 ≥ |∂((at , bt−1 ), (bt , bt ))|, that contradicts Lemma 1(a). Finally, if ∇(bt−1 ) = {bt }, then (at , at ) 6∈ B(A). Therefore, there is a vertex (z 0 , z 00 ) 2 (at , bt−1 ) such that |∂((z 0 , z 00 ), (z 0 , z 00 ))| ≥ 1. But then, using (6), |∂((z 0 , z 00 ), (z 0 , z 00 ))| ≥ 1 > 0 = |∂((z 0 , z 00 ), (bt , bt ))|, 2

which contradicts Lemma 1(a).

Lemma 3 Let G 6∈ {K2,2 , K4 − e} be a connected graph. Assume the simplicial orders  and G

G

2 in G and G2 associated with the order ≤ are isoperimetric. Then ≤ is level-greedy. G

Proof. First suppose that there exist x, y ∈ V such that x ∈ ∆(y) and y < x. We call (x, y) an inverted pair. We show that the assumption of the existence of an inverted pair leads to a contradiction. Denote t = min{kxk | (x, y) is an inverted pair}. Since V0 = {0} we conclude that t ≥ 1. Let (u, v) be the smallest (in order 2 ) inverted pair in Vt × Vt+1 . Fact 1 ∇(x) = Vi+1 for all x ∈ Vi and i, 0 ≤ i < h. Fix i < h, and let z = min ∆(ai+1 ). Since (u, ai+1 ) ≺2 (v, bi ), Lemma 1(b) implies (u, z) = min ∆(u, ai+1 ) 2 min ∆(v, bi ) = (u, bi ), 2 2 



which, in turn, implies z = bi . Applying this lemma again, for any y ∈ Vi+1 with y ≺ ai+1 we have min ∆(y)  min ∆(ai+1 ) = bi . Therefore, bi ∈ ∆(y) and thus ∇(bi ) = Vi+1 . Now applying Lemma 1(a) with x ∈ Vi , we have |Vi+1 | = |∂(bi , bi )| ≤ |∂(bi , x)| ≤ |Vi+1 |. Hence, ∇(x) = ∇(bi ) = Vi+1 . 2 Fact 2 (u, v) = (bt , bt+1 ). G

G

Since v ∈ ∇(u) ∩ ∇(bt ) by Fact 1, and v < u ≤ bt , then (bt , v) is an inverted pair too. Since (u, v) ≺2 (bt , v), then, taking into account the minimality of (u, v), we get u = bt . Now if h = 2 G

G

the assertion is true due to Lemma 2, so we assume h ≥ 3. Note that bt+1 ≥ v and u = bt > v. G

G

G

Assume bt+1 > v. Let V + = {x ∈ Vt+1 | bt < x} and V − = {x ∈ Vt+1 | bt > x}. Note that bt+1 ∈ V + by the minimality of (u, v) = (bt , v), and v ∈ V − . Therefore, V + × {bi−1 } ⊆ F 2 ((bt , bi )) for i = 1, 2, . . . , h. This implies |∂((bt , bi ), (v, bi−1 ))| = |Vi |. Applying Lemma 1(a), we get 1 + |Vi+1 | ≤ |V − | + |Vi+1 | = |∂((bt , bi ), (bt , bi ))| ≤ |∂((bt , bi ), (v, bi−1 ))| = |Vi |. Summing these inequalities for i = 1, . . . , h − 1 yields (h − 2) + |Vh−1 | ≤ |V1 |. Since h ≥ 3 then |Vh−1 | < |V1 |. But then, using Lemma 2, 1 + |V1 | = |∂(0, 0)| > |∂(0, bh )| = 1 + |Vh−1 |, that contradicts Lemma 1(a). This contradiction implies v = bt+1 . 2 7

G

Fact 3 bt0 > bt00 for all t0 , t00 with t ≤ t0 < t00 ≤ h. G

First suppose that bt is not the maximum element in V in order ≤. Let x be the closest element G

to 0 such that x > bt . By Fact 2, we have t0 = kxk ≥ t + 2. Since (x, 0) 2 (bt , bt0 −t ), then, G

by Lemma 1(b), (bt0 −1 , 0) 2 (bt , bt0 −t−1 ). This implies bt0 −1 > bt , contradicting the choice of x. G

Hence, bt is the maximum element of V in order ≤. Now assume the assertion is not true, and let (t0 , t00 ) be a pair such that t ≤ t0 < t00 ≤ h G

and bt0 < bt00 . We choose this pair so that t0 is minimum, and among all such pairs let t00 be minimum. Since bt is the maximum element by above, then t0 ≥ t + 1. G

G

G

Note that bt00 −1 < bt00 , since otherwise bt00 −1 > bt00 > bt0 , contradicting the choice of the pair (t0 , t00 ). Now since (bt00 , b1 ) ≺2 (bt0 , bt00 −t0 +1 ), then (bt00 , 0) 2 (bt0 −1 , bt00 −t0 +1 ) by Lemma 1(b). G

2

This implies bt00 > bt0 −1 , contradicting the choice of (t0 , t00 ). G

Fact 4 bt−1 < bh . G

G

Let t0 ≥ t be the largest integer such that bt−1 < bt0 . Then t0 exists because bt−1 < bt by G

the choice of the inverted pair (bt , bt+1 ). Now if t0 < h, then bt−1 > bt0 +1 , which implies (bt−1 , bt0 +1 ) ≺2 (bt0 +1 , bt−1 ). Applying Lemma 1(b), we get (bt−1 , bt0 ) 2 (bt0 , bt−1 ). This implies G

2

bt−1 > bt0 , a contradiction. Fact 5 |Vi | = 1 for 0 ≤ i ≤ h. G

G

If ai 6= bi for some i > t, then (bi , ai ) ≺2 (ai , bi ). By Fact 3, bi−1 > bi > ai , hence (bi−1 , ai ) = min2 ∆((bi , ai )) and (bi−1 , bi ) = min2 ∆((ai , bi )). Now Lemma 1(b) implies (bi−1 , ai ) 2 (bi−1 , bi ). This, in turn, implies ai = bi , i.e., |Vi | = 1 for t < i ≤ h. It follows from Facts 2 - 4 that G

G

G

G

G

G

b0 < · · · < bt−1 < bh < bh−1 < · · · < bt .

(7)

This, in turn, implies (bt , bh−t ) is the smallest element of Vh2 . By Fact 1, |∇(bt )| = |Vt+1 | ≥ 1, and similarly |Vh−t+1 | ≥ 1. Since (bt , bh−t ) ≺2 (bh , 0), applying Lemma 1(a) one has 2 ≤ |Vt+1 | + |Vh−t+1 | = |∂((bt , bh−t ), (bt , bh−t ))| ≤ |∂((bt , bh−t ), (bh , 0))| = |V1 |. Assume t < h − 1. It follows from above that |Vh−1 | = |Vh | = 1. Therefore, |∂(0, bh )| = |Vh−1 | + 1 = 2 < 3 ≤ |V1 | + 1 = |∂(0, 0)|, contradicting Lemma 1(a). Suppose t = h − 1. Then by (7), (bh−1 , bj ) ≺2 (bj−1 , bh ) for j = 1, 2, . . . , h − 2. This and Lemma 1(a) imply 1 + |Vj+1 | = |∂((bh−1 , bj ), (bh−1 , bj ))| ≤ |∂((bh−1 , bj ), (bj−1 , bh ))| = |Vj |. 8

Summing these equalities for j = 1, 2, . . . , h − 2 yields (h − 2) + |Vh−1 | ≤ |V1 |. But then for h ≥ 3 one has |∂(0, bh )| = 1 + |Vh−1 | < 2 + |Vh−1 | ≤ 1 + |V1 | = |∂(0, 0)|, which contradicts Lemma 1(a). G

Assume h = 2. Since G 6∈ {K2,2 , K4 − e}, then |V1 | ≥ 3. First note that if x > b2 for some x ∈ V1 then (x, b1 ) 2 (b2 , 0). Lemma 1(b) implies (x, 0) 2 (b1 , 0), from where x = b1 follows. G

G

G

G

G

G

Thus, 0 < a1 < p < · · · < q < b2 < b1 for some p, q ∈ V1 (if |V1 | = 3 then p = q). Therefore (0, b1 ) is the successor of (a1 , 0) in V12 (with respect to 2 ), and (p, 0) is its predecessor. Note that ∇(a1 , 0) ∩ B(F 2 ((p, 0))) = {(b2 , 0)}. Thus, |∂((a1 , 0), (a1 , 0))| = |∇(a1 , 0)| − 1 = |V1 |. On the other hand ∇(0, b1 ) also consists of |V1 | + 1 elements (0, b2 ), (a1 , b1 ), . . . , (b1 , b1 ), but at least the last two of them are already in B(F 2 ((p, 0))). Hence, |∂((a1 , 0), (0, b1 ))| ≤ |V1 | − 1 < |V1 | = |∂((a1 , 0), (a1 , 0))|, 2

contradicting Lemma 1(a).

It follows from all the above facts that, if there is an inverted pair, then G is a chain and (7) 2 , hence holds. Therefore, (bt , bh−1 ) is the smallest element of Vt+h−1 ∂((bt , bh−1 ), (bt , bh−1 )) = ∇((bt , bh−1 )) = {(bt , bh ), (bt+1 , bh−1 )}. On the other hand, (bh , bt−1 ) ≺2 (bt , bh−1 ), and ∇((bh , bt−1 )) = {(bh , bt )}. Therefore, |∂((bt , bh−1 ), (bt , bh−1 ))| = 2 > 1 = |∂((bt , bh−1 ), (bh , bt−1 ))|, contradicting Lemma 1(a). This shows that there are no inverted pairs. To prove the second relation (5) in the definition of the level–greediness, assume the contrary, i.e. that G

G

x < y < x0 = min ∇(x)

and kyk > kxk.

(8)

G

≤

Let q = kyk − kxk > 0. Then (y, bh−q ) ≺2 (x, bh ). Denote z = max ∇(bh−q ). Since there are no inverted pairs, then max2 ∇((y, bh−q )) = (y, z). Now Lemma 1(c) implies (y, z) 2 G

max2 ∇(x, bh ) = (x0 , bh ). Hence, y ≥ x0 contradicting (8). This concludes the proof.

2

Remark 2 Note that Lemma 3 is not true for the graphs K2,2 and K4 − e. Fig. 2(a) shows a G

total ordering ≤ of the vertices of K2,2 such that the simplicial orders  and 2 on K2,2 and G

2 respectively are isoperimetric but ≤ fails to be level-greedy. K2,2

The same order also provides a counterexample for K4 − e that can be obtained from K2,2 by adding the dotted edge in Fig. 2(a). 9

3

s22  @  @   s32 s21 @ s 12  s23 @  @ @   @  @ @   s@ s s s  02 11 13  31@ s33@ s20   @ @  @   @  @ @   @s@ @ s s  01 03 10 s30  @   @  @s 00

s2 @ @ qs q q q q q q q q@ q s1 @ @ @s 0

a.

b.

2 Figure 2: Counterexample to Lemma 3 for K2,2 (a) and K2,2 (b).

Now we introduce a very important property of the graph G. This property, explored in Lemma 4, and Lemma 10 are the key points of our technique. For x ∈ Vt define the apex of x as α(x) = F(x) ∩ Vt . The order  is called upper-monotone if, for any x, y ∈ V with x  y, the following condition holds |Γ(F(m))| + |Γ(F(m0 ))| ≥ |Γ(F(m − δ))| + |Γ(F(m0 + δ))|,

(9)

where m = #(x), m0 = #(y) and δ = δ(m, m0 ) = min{|α(x)|, |Vkyk | − |α(y)|}. In other words, if we take two copies G1 and G2 of G, and initial segments F(m) and F(m0 ) with m ≤ m0 , we should be able to move some vertices from the smaller initial segment to the larger one without increasing the sum of their boundaries (see Fig. 3). We denote this transformation by T (F(m), F(m0 )). u $ u $' u $ u $' u $ u $ ' ' ' '

y

zs ys

s

xs

y

s

xs

⇒

xs zs

or

u % u %& u % u %& u % u % & & & &

a.

b.

c.

Figure 3: The upper-monotonicity: the original configuration (a), and the resulting sets after T (F(m), F(m0 )) if δ = |α(x)| (b) and δ < |α(x)| (c)

G

Lemma 4 Let G 6∈ {K2,2 , K4 −e} be a graph with a total order ≤ on its vertex set. Suppose that G

the simplicial orders  and 2 associated with ≤ are isoperimetric in G and G2 respectively. Then the order  in G is upper-monotone. 10

G

Proof. By Lemma 3, the order ≤ is level-greedy. Let 0  x  y. Denote m = #(x), m0 = #(y) with m ≤ m0 , and let δ = δ(m, m0 ) as in (9). Furthermore, denote Ux = F(m) − F(m − δ), Uy = F(m0 + δ) − F(m0 ),

Ux = B(F(m)) \ B(F(m − δ)), Uy = B(F(m0 + δ)) \ B(F(m0 )).

One has Ux ⊆ Vkxk , Uy ⊆ Vkyk , and |Ux | = |Uy | = δ. Furthermore, Ux ⊆ Vkxk+1 and Uy ⊆ Vkyk+1 . Our objective is to show |Ux | ≥ |Uy |, which is equivalent to (9). Avoiding trivial cases we assume kyk − kxk ≤ h − 1. Case 1. Assume δ = |α(x)|. Let z ∈ V with #(z) = m0 + δ (cf. Fig. 3(b)), and denote t = h − (kyk − kxk) − 1 and A = F 2 ((bt , z)). For any (u, v) ∈ B(A) one has k(u, v)k ≤ t + kyk + 1 = h + kxk. Let A0 = (A \ ({bt } × Uy )) ∪ ({bh } × Ux ). Since A0 is not an initial segment of 2 , then |B(A)| ≤ |B(A0 )|. We show |B(A0 )| ≤ |B(A)| − |{bt } × Uy | + |{bh } × Ux |.

(10)

This, in turn, will follow from {bt } × Uy ⊆ B(A) \ B(A0 ) and B(A0 ) \ B(A) ⊆ {bh } × Ux .

(11)

Let us show (11). Obviously, {bt } × Uy ⊆ B(A). Assume (bt , v) ∈ ({bt } × Uy ) ∩ B(A0 ). Then there exists (u, w) ∈ B((bt , v))∩A0 such that either u = bt or w = v. Since u = bt contradicts the definition of Uy , then w = v. This implies u ∈ ∆(bt ), so k(u, v)k = (t − 1) + (kyk + 1) = t + kyk. G

By the level-greediness we have u < bt , which implies (u, v) 2 (bt , y). Since t ≤ h − 1, then (u, v) 6∈ {bh } × Ux , so (u, v) 6∈ A0 , a contradiction. This proves the first inclusion in (11). To show the second inclusion in (11) let (u, v) ∈ B(A0 ) \ B(A). Then either u = bh or u ∈ Vh−1 . If the later holds, then k(u, v)k = h + kxk − 1 = t + kyk, so (u, v) ∈ B(A). On the other hand, if u = bh then, either (u, v) = (bh , v) ≺2 min2 ({bt } × Uy ) ∈ B(A) or v ∈ Ux . Hence, (11) and thus (10) is established. This in combination with |B(A)| ≤ |B(A0 )|, implies |Ux | ≥ |Uy |. Case 2. Assume δ < |α(x)|. Let z ∈ V with #(z) = m − δ (cf. Fig. 3(c)). Denote t = kyk − kxk + 1, and let A = F 2 ((bt , z)) as above. For any (u, v) ∈ B(A) one has k(u, v)k ≤ t + kxk + 1. Let A0 = (A \ ({b0 } × Uy )) ∪ ({bt } × Ux ). Since A0 is not an initial segment, then |B(A)| ≤ |B(A0 )|. Similarly as above one can show {b0 } × Uy ⊆ B(A) \ B(A0 ) and B(A0 ) \ B(A) ⊆ {bt } × Ux . This implies |B(A)| ≤ |B(A0 )| ≤ |B(A)| − |{b0 } × Uy | + |{bt } × Ux |, and thus |Ux | ≥ |Uy |.

11

2

Remark 3 Note that if the order  is upper-monotone and kxk = kyk for some x, y ∈ V then the transformation T (F(x), F(y)) does not increase the sum of the boundaries regardless whether x  y or not. This is true because the sum of the boundaries for the resulting sets is the same in both cases x ≺ y and x  y. We will use this remark in the proof of Lemma 11. We call a pair (x, y) of vertices of G antipodal pair if distG (x, y) = diam(G), and there are no other diametrically opposite vertices for x or y. For instance, any pair of complementary vertices form an antipodal pair in the n-cube, whereas an odd cycle has no antipodal pairs. With this terminology, the above lemmas can be put together into the following theorem. G

Theorem 2 Let G 6∈ {K2,2 , K4 − e} be a connected graph with a total order ≤ on its vertex G

set. Furthermore, let the simplicial orders  and 2 associated with ≤ be isoperimetric in G and G2 respectively. Then G

(a.) The order ≤ is level-greedy. G

(b.) The first and last vertices in order ≤ form an antipodal pair. (c.) The order  is upper-monotone in G. Proof. The first property is proved in Lemma 3. To show the second one note that Lemma 2 G

implies ∇(v) 6= ∅ for any v ∈ Vt and any t < h. Therefore, if k is the last vertex in order ≤ then k ∈ Vh . Moreover, |Vh | = 1. Hence, distG (0, k) = h and no other vertex is at distance h from 0 or k. Furthermore, for any two vertices x ∈ Vt and y ∈ Vt0 there is a (x, y) path P1 that passes through 0, and there is a (x, y) path P2 that passes through k. Therefore, distG (x, y) ≤ min{|P1 |, |P2 |} = min{t + t0 , 2h − (t + t0 )} ≤ h, so h = diam(G). Finally, the last property is provided by Lemma 4.

4

2

The main result

In this Section we prove Theorem 1. The only if part of this theorem is trivial. For the if part, we first show that the simplicial order n in Gn satisfies the continuity property (cf. (2)) G

provided that it is so for n = 1 and the order ≤ in G is level-greedy. Then we reduce the problem to compressed sets by showing that their boundary is not smaller than the one of initial segments. G

Again we assume that a total order ≤ of the vertex set of a connected graph G = (V, E) is G

fixed and the simplicial order n on Gn is associated with ≤. The following property that we call consistency is straightforward. 12

Lemma 5 Let x, y ∈ V n for n ≥ 2 and xi = yi for some i, 1 ≤ i ≤ n. Furthermore, let x0 , y0 ∈ V n−1 be obtained from x and y respectively by omitting their ith entries. Then x n y iff x0 n−1 y0 . We will use this assertion throughout the proof without explicit references. The following lemmas are aimed to prove the if part of Theorem 1. G

Lemma 6 Let ≤ be a level-greedy total order on a graph G. Assume the simplicial order  satisfies the continuity property in G and ∇(bi ) 6= ∅ for i < h. Then n satisfies the continuity property in Gn for any n ≥ 1. Proof. Assume the contrary, i.e., for some n ≥ 2 there is an initial segment A ⊂ V n such that B(A) is not an initial segment. For some t ≥ 0 one has A=

[

Vin ∪ (A ∩ Vtn ) and B(A) =

i zr . Furthermore, let z0 = (z10 , . . . , zn0 ) be the largest element in ∆(z) ∩ A and let zs0 ∈ ∆(zs ) for some s ∈ {1, . . . , n}. Therefore, zi = zi0 for i 6= s. We construct a vertex x0 ∈ ∆(x) satisfying x0 n z0 . Then z0 ∈ A and A is an initial segment will imply x0 ∈ A. This, in turn, implies x ∈ B(A), a contradiction. If r > s then the vector x0 ∈ ∆(x) obtained from x by replacing xs = zs with zs0 satisfies x0 ≺n z0 and we have a contradiction, since now x0 ∈ A. Assume r ≤ s. If xi 6= 0 for some i > r then consider the vertex x0 obtained from x by replacing G

G

xi with x0i ∈ ∆(xi ). One has x0r = xr > zr ≥ zr0 , where the last inequality follows from the G

level-greediness of ≤ if s = r and it is an equality if s > r. Hence, x0 ≺n z0 and we have again a contradiction. Therefore, either xi = 0 for each i > r or s = r = n. Since kxk = kzk = t + 1, we have P kxr k = i≥r kzi k ≥ kzr k. Suppose kxr k = kzr k. Then, zi = 0 for i > r and, therefore, s = r. By the continuity of  in G, zr ∈ B(F(zr0 )) and xr ≺ zr imply xr ∈ B(F(zr0 )). Therefore, there is x0r ∈ ∆(xr ) ∩ F(zr0 ). Then, the vertex x0 obtained from x by replacing xr with x0r satisfies x0 ≺n z0 and we have a contradiction. Hence, we have t0 = kxr k > kzr k and r < n. Since z is the successor of x and xi = 0 for i > r, then xr = at0 . We have x = (x1 , . . . , xr−1 , at0 , 0, . . . , 0) and z = (x1 , . . . , xr−1 , bt0 −1 , b1 , 0, . . . , 0). 13

Thus, r ≤ s ≤ r + 1, which implies z00 = (x1 , . . . , xr−1 , bt0 −1 , 0 . . . , 0) n z0 . Since z0 ∈ A and A is an initial segment, then z00 ∈ A. The proof will be completed if we show that z00 ∈ ∆(x), because in this case one can choose x0 = z00 . In other words, we have to show bt0 −1 ∈ ∆(at0 ) or, equivalently, at0 ∈ ∇(bt0 −1 ). G

Indeed, if it is not the case, then ∇(bt0 −1 ) 6= ∅ implies at0 < min G ∇(bt0 −1 ). Since kat0 k > kbt0 −1 k ≤

G

G

n

then at0 < bt0 −1 by (5). On the other hand, x ≺ z implies at0 > bt0 −1 . This contradiction completes the proof. 2 G

Remark 4 Note that the assumption on the level-greediness on ≤ in the above Lemma can not be omitted. As an example, the order on the 3-cube Q3 shown in Fig. 4(a) is isoperimetric, but the corresponding simplicial order in (Q3 )2 does not satisfy the continuity property: the ball around the initial segment {(0, 0), (3, 0)} is not an initial segment (cf. Fig. 4(b)).

s7 @ @ s6 s 5 @s 4 @ @ @ @ s 3 @s 2 @s 1 @ @ @s 0

23 ···

e

31

32

33

40

50

60

sP s s e s s PP HH @  PPH  H@ PP H  s10 s20PP @v H 30 ··· @ @ @v00

a.

b.

Figure 4: A non level-greedy isoperimetric order on G (a) that is not isoperimetric in G2 (b)

Remark 5 The assumption ∇(bi ) 6= ∅ for i < h in Lemma 6 can not be omitted. For example, G

for the graph G and the level-greedy order ≤ shown in Fig. 5(a) the simplicial order  satisfies the continuity, but the ball around {(0, 0), (2, 0)} in G2 is not an initial segment (cf. Fig. 5(b)).

s2 @ @

s4 @ @

21

s3

22

30

40

s s e e HH @  H  H@  H H s01 s02 s v 10 @ Q  20 Q  A
Q A
 Q  Q A
 v00

···

@s 1

@s 0

a.

b.

Figure 5: Violating the condition ∇(bi ) 6= ∅ for i < h in G (a) can break the continuity of 2 in G2 (b)

14

We proceed by introducing the compression operator. Let A ⊂ V n , n > 1 and a ∈ V . For 1 ≤ i ≤ n, denote A(i, a) = {z = (z1 , . . . , zn ) ∈ A | zi = a}, ˆ a) = {z = (z1 , . . . , zn−1 ) ∈ V n−1 | (z1 , . . . , zi−1 , a, zi , . . . , zn−1 ) ∈ A}. A(i, ˆ a) consists of the vertices of V n−1 that are obtained from the vertices of A(i, a) by That is, A(i, ˆ a) for any a ∈ V deleting the i-th entry. Denote by C = Ci (A) the subset of V n such that, C(i, n−1 is the initial segment of size |A(i, a)| of order  . We call Ci (A) the i-compression of A. We say that A is compressed if Ci (A) = A for any i, 1 ≤ i ≤ n. Lemma 7 Let the simplicial order n−1 be isoperimetric in Gn−1 for some n > 1. Then for any A ⊂ V n , and any i, 1 ≤ i ≤ n, |Γ(Ci (A))| ≤ |Γ(A)|. Proof. Denote H = Gn−1 , and let C = Ci (A) be the i-compression of A for some fixed i. We show |B(C)| ≤ |B(A)| from where the lemma follows. One has |B(A)| =

P

u∈V

|(B(A))(i, u)|, and

ˆ u))|, max{|A(i, v)| | v ∈ ΓG (u)}}. |(B(A))(i, u)| ≥ max{|BH (A(i,

(13)

ˆ u) is an initial segment of length |A(i, u)| in V n−1 . By the assumpBy the definition of C, C(i, ˆ u))| ≤ |BH (A(i, ˆ u))| tion of the lemma, the order n−1 is isoperimetric in H. Hence, |BH (C(i, n−1 ˆ due to the nestedness property, and BH (C(i, u)) is an initial segment in V due to the continuity property. Therefore, ˆ u))|, max{|C(i, v)| | v ∈ ΓG (u)}}. |(B(C))(i, u)| = max{|BH (C(i,

(14) 2

Now (13) and (14) imply |B(C)| ≤ |B(A)|.

For a subset A ⊂ V n , denote #(A) = x∈A #(x). Note that #(Ci (A)) ≤ #(A) and equality holds iff Ci (A) = A. Therefore, by applying the compression for i = 1, . . . , n sufficiently many times we can transform A into a compressed set C such that |C| = |A| and |Γ(C)| ≤ |Γ(A)|. P

Any initial segment in V n for n > 1 is a compressed set, however, the converse is not true. The next lemmas explore some structural properties of compressed sets when n ≥ 3. G

In the sequel we always assume that ≤ is a level-greedy ordering, and that the simplicial orders  and 2 are isoperimetric in G and G2 respectively. G

Lemma 8 Let ≤ be a level-greedy order on V and let ∇(x) 6= ∅ for all x ∈ V with kxk < h. Then, G

at ≤ at+1

and

G

bt ≤ bt+1

15

for all t = 0, . . . , h − 1.

(15)

G

Proof. The assertion is obvious for t = 0. For t > 0 let y ∈ ∆(at+1 ). Since y ∈ Vt , then at ≤ y G

by the definition of at . On the other hand, y < at+1 follows from (4). These properties imply the first inequality in (15). The second inequality can be established similarly by taking into G

G

2

account bt < x ≤ bt+1 for any x ∈ ∇(bt ).

Denote by ftn and lnt the first and last vectors of Vtn , 0 ≤ t ≤ nh. Obviously, f0n = ln0 = (0, . . . , 0) n and, by Lemma 2, fnh = lnnh = (k, . . . , k). G

Let 0 < t < nh, and represent t in the form t = qh + r with 0 ≤ r < h. Since ≤ is level-greedy, Lemma 8 and Lemma 2 imply ftn = (k, . . . , k , br , 0, . . . , 0) |

lnt

{z

}

q

|

{z

n−q−1

= (0, . . . , 0, ar , k, . . . , k ). |

{z

n−q−1

}

|

{z q

(16)

}

(17)

}

Let A ⊆ V n be compressed, and let y ∈ A, x ∈ V n , and x ≺n y. Note that, if xi = yi for some 1 ≤ i ≤ n, then x ∈ A because A is compressed. We often use this remark and even introduce ↓

the special notation x ≺ y to indicate that x ≺ y and the two vectors agree in some entry. n Let n ≥ 2. Note that, if 0 < t ≤ (n − 1)h, then the last entry of ftn and of ft−1 is 0, whereas if t ≥ (n − 1)h then the first entry of both vectors is k. Similarly, if t ≥ h then the last entry of lnt and of lnt−1 is k and for t ≤ h the first entry of both vectors is 0. Hence, ↓

↓

n ft−1 ≺ ftn

and lnt−1 ≺ lnt

for n ≥ 2 and t ≥ 1.

(18)

By using the definition of the simplicial order it is easy to check that ↓

↓

n−1 n−1 (ft−1 , u)  (ftn−1 , v) and (ln−1 , v) t−1 , u)  (lt

(19)

for any u ∈ Vt0 , v ∈ Vt0 −1 , n > 2 and t, t0 ≥ 1. By (18), if ftn (resp. lnt ) belongs to a compressed set A, then ftn0 ∈ A (resp. lnt0 ∈ A) for all 0 ≤ t0 ≤ t. Let y = (y1 , . . . , yn ) be the largest vertex of a compressed set A ⊂ V n , and let x = (x1 , . . . , xn ) be the smallest vector of V n \ A. If A is not an initial segment, then x ≺ y. The next Lemma explores (19) and provides some additional information concerning the structure of A. It turns out that A is not very far away from being an initial segment. Lemma 9 Let A ⊆ V n be a compressed set and n ≥ 3. (a.) If A ∩ Vtn 6= ∅, then ftn ∈ A. (b.) If lnt ∈ A, then z ∈ A for all z ∈ V n with kzk ≤ t. 16

(c.) kxk ≥ kyk − 1. Proof. Let z = (z1 , . . . , zn ) ∈ A ∩ Vtn , and assume z  ftn = (f1 , . . . , fn ). Let t = qh + r with 0 ≤ r < h. From (16) we have fn = 0 if q < n − 1 and fn = br otherwise. In either case kzn k ≥ kfn k. Applying (19) one has ↓

↓

↓

↓

n−1 n−1 n−1 , zn )  (ft−kz , bkzn k )  (ft−kz , bkzn k−1 )  · · · z  (ft−kz nk nk n k+1 ↓

n−1 n−1  (ft−kf , bkfn k ) = (ft−kf , fn ) = ftn . nk nk

Thus, ftn ∈ A. This proves the first assertion. To prove the second assertion let z ∈ Vtn and lnt = (l1 , . . . , ln ). We have ln = k if t ≥ h and ln = at otherwise. Hence, kln k ≥ kzn k. Applying (19) one has ↓

↓

↓

↓

n−1 n−1 z  (ln−1 t−kzn k , zn )  (lt−kzn k , akzn k )  (lt−kzn k−1 , akzn k+1 )  · · · ↓

n−1 n  (ln−1 t−kln k , akln k ) = (lt−kln k , ln ) = lt .

Thus, z ∈ A. Moreover, by (18), lnt0 ∈ A for all 0 ≤ t0 ≤ t. Applying the above arguments for t0 we prove that kzk ≤ t implies z ∈ A. Finally, for the last assertion let t = kyk = qh + r with 0 ≤ r < h. By part (a) of this lemma we have ftn ∈ A. Furthermore, by part (b), it suffices to show that lnt−2 ∈ A. One has ftn = (k, . . . , k , br , 0, . . . , 0), |

{z

}

q

|

{z

n−q−1

}

lnt−2 = (0, . . . , 0, ar0 , k, . . . , k ), |

{z

}

|

n−q 0 −1

{z

}

q0

where t−2 = q 0 h+r0 . We can assume that these two vectors do not have a common entry, since otherwise we trivially have lnt−2 ∈ A due to the compression. In particular, h > 1. Moreover, as n ≥ 3, we have q ≥ 1 and n − q 0 − 1 ≥ 1. Therefore, ↓

↓

ftn  (0, k, . . . , k , br , bh−1 , 0, . . . , 0)  lnt−2 , |

{z

q−1

}

|

{z

n−q−2

}

2

which implies lnt−2 ∈ A.

The next lemma is one of the most important for our technique. The proof of the main result (Theorem 1) is based on this lemma. G

Lemma 10 Let V be ordered by a level-greedy order ≤, and let A ⊂ V n for some n ≥ 3 be a compressed set. Furthermore, let ∇(z) 6= ∅ for any z ∈ V with kzk < h. Then kxn k ≥ kyn k. Proof. Denote m1 = ky1 k − kx1 k and mn = kyn k − kxn k. Assume the contrary, i.e., that mn > 0. We show that this assumption leads to a contradiction. 17

Since A is not an initial segment, we have x  y, and since x 6∈ A and A is compressed, we have x1 6= y1 . Let x = (x1 , x0 , xn ) and y = (y1 , y0 , yn ), where x0 , y0 are vectors of dimension n − 2. G

By Lemma 9(c), either kxk = kyk, which implies x1 > y1 , or kxk = kyk − 1. We consider two cases. Case 1. Assume m1 > 0. Then, kx0 k − mn ≥ ky0 k + m1 − 1 ≥ 0, where the first inequality follows from kxk ≥ kyk − 1. One has ↓

↓

↓

↓

n−2 0 y = (y1 , y0 , yn )  (x1 , ln−2 kx0 k−mn , yn )  · · ·  (x1 , lkx0 k , xn )  (x1 , x , xn ) = x.

The first inequality in this chain is obvious if kyk > kxk, since k(y1 , y0 , yn )k = kyk and G

n−2 k(x1 , lkx If kyk = kxk it follows from x1 > y1 . The last inequality is a 0 k−m , yn )k = kxk. n consequence of the definition of the simplicial order. The remaining inequalities follow from (19). These relations imply x ∈ A, i.e., a contradiction.

Case 2. Assume now m1 ≤ 0, and suppose first that m1 + mn ≤ 0. This implies, in particular, ky1 k + mn ≤ kx1 k. Denote Z = ∇(∇(· · · ∇(y1 ) · · ·) (mn times). For any z ∈ Z one has kzk = ky1 k + mn , and, G

G

since ≤ is level-greedy, then (4) implies y1 < z. Moreover, if kyk = kxk then it follows from G

G

G

(5) and y1 < x1 that there exists z ∈ Z such that y1 < z ≤ x1 . One has ↓

↓

↓

↓

n−2 n−2 n−2 y = (y1 , y0 , yn )  (y1 , fky 0 k , yn )  (z, lky0 k , xn )  (x1 , lkx0 k , xn )  x.

The first inequality in this chain follows directly from the definition of the simplicial order. The G

second one is implied by the fact z > y1 . The third inequality is obvious if kyk > kxk, because G

n−2 k(z, ln−2 ky0 k , xn )k = kyk and k(x1 , lkx0 k , xn )k = kxk. If kyk = kxk it follows from x1 ≥ z.

Finally, assume m1 + mn > 0. Then ky0 k ≤ kx0 k. One has ↓

↓

↓

↓

↓

n−2 n−2 n−2 y = (y1 , y0 , yn )  (y1 , fky 0 k , yn )  · · ·  (y1 , fky0 k+m , xn )  (x1 , lkx0 k , xn )  x. n

The first inequality here is similar to one above. The last inequality is obvious if kyk > kxk, G

and follows from x1 > y1 if kyk = kxk. The remaining inequalities follow from (19). In all cases we conclude that x ∈ A. This completes the proof.

2

Proof of Theorem 1. G As mentioned above, the necessity of the conditions is obvious. For the sufficiency, let ≤ be a total order on the vertex set of G = (V, E) such that the associated simplicial orders 1 and 2 are isoperimetric in G and G2 respectively. By Lemma 6, the order n satisfies the continuity property for all n ≥ 1. Therefore it remains to prove that this order satisfies the nestedness property. G

By Theorem 2, the order ≤ is level-greedy and  is upper-monotone. Moreover, the vertices 0 and k = |V | − 1 in the first order form an antipodal pair in G. 18

'$

'$

V n (x0 ) xc ' s 0

V n (y0 )

t XX y XX XXX XX

y

Xs

s & %

(x1 , . . . , xn−1 , 0)

t0

s & %

(y1 , . . . , yn−1 , 0)

&

$

%

Figure 6: Usage of the upper-monotonicity property in the proof of Theorem 1 Let n ≥ 3 and A ⊂ V n . Let C ⊂ V n be an initial segment of size |A|. Our objective is to show that |B(A)| ≥ |B(C)|. By Lemma 7, we may assume that A is a compressed set. If A is not an initial segment, let y = (y1 , . . . , yn ) be the maximum element of A and x = (x1 , . . . , xn ) be the minimum element of V n \ A. Then x ≺n y and, by Lemma 10, we have kxn k ≥ kyn k. Let t = kxn k and t0 = kyn k (cf. Fig. 6). For z ∈ V n−1 , denote V n (z) = {(z, u) | u ∈ V }, A(z) = A ∩ V n (z). Let y0 = (y1 , . . . , yn−1 ) and x0 = (x1 , . . . , xn−1 ). Note that xn 6= 0 since otherwise yn = 0 ↓

(Lemma 10), and thus x ≺ y, contradicting x 6∈ A. Denote U = {(x1 , . . . , xn−1 , u) | xn  u  at }, W = {(y1 , . . . , yn−1 , w) | bt0  w  yn }. Since A is a compressed set, then U ⊂ V n \ A and W ⊂ A. We use the upper-monotonicity property of order  to move some vectors from W to U without increasing the boundary. Let δ = min{|U |, |W |}, and let U 0 be the set of the first δ vertices of U , and W 0 be the set of the last δ vertices of W (in order ). Then for the set A0 = (A \ W 0 ) ∪ U 0 we have |A0 | = |A|. First let us show that B(A0 ) \ B(A) ⊆ (B(A0 (x0 )) \ B(A(x0 ))) ∩ V n (x0 ) B(A) \ B(A0 ) ⊇ (B(A(y0 )) \ B(A0 (y0 ))) ∩ V n (y0 ),

(20) (21)

Let A0 = {z ∈ A | z ≺ x}. By the definition of x and y, A0 is an initial segment and 19

A0 ∩ W = ∅. Since A0 \ A = U 0 ⊆ U , then (20) will follow if we prove B(U ) \ V n (x0 ) ⊂ B(A0 ).

(22)

Let u = (x1 , . . . , xn−1 , u) ∈ U and v = (v1 , . . . , vn ) ∈ B(u) \ V n (x0 ). The vectors u and v differ in one entry which is not the last one. Assume xj 6= vj for some j ∈ {1, . . . , n − 1}, ˆ = (v0 , uˆ), where v0 = (v1 , . . . , vn−1 ). We have thus −1 ≤ kuk − kvk ≤ 1. Let uˆ ∈ ∆(u) and v ˆ ∈ A0 . v ∈ B(ˆ v), and will show v G

G

Indeed, if kvj k = kxj k + 1 then by the level-greediness of order ≤, we have vj > xj . Therefore, ˆ ≺ x, which in turn implies v ˆ ∈ A0 . Similarly, if since in this case kˆ vk = kxk, we have v ˆ ∈ A0 . In both cases, v ∈ B(A0 ). This kvj k ≤ kxj k, then kˆ vk < kxk, which again implies v proves (22), which implies (20). To prove (21) suppose that u = (y1 , . . . , yn−1 , u) ∈ (B(A(y0 )) \ B(A0 (y0 ))) ∩ V n (y0 ). Note that kuk > kyk. Let v ∈ B(u) \ V n (y0 ). There is j ∈ {1, . . . , n − 1} such that u and v differ only G

G

in the j-th entry. If kvj k = kyj k − 1 then the level-greediness of ≤ implies vj < yj . Now if kvj k ≥ kyj k then kvk > kyk. In both cases v  y, and, thus v 6∈ A0 . This proves (21). Let (x0 , xn ) be the last element in A(x0 ), and let m = #(xn ) and m0 = #(yn ) in G. By (20) and (21) respectively, |B(A0 ) \ B(A)| ≤ |(B(A0 (x0 )) \ B(A(x0 ))) ∩ V n (x0 )| = |B(F(m + δ))| − |B(F(m))|, |B(A) \ B(A0 )| ≥ |(B(A(y0 )) \ B(A0 (y0 ))) ∩ V n (y0 )| = |B(F(m0 ))| − |B(F(m0 − δ))|. Hence, by the upper-monotonicity of order  in G one has |B(A0 ) \ B(A)| ≤ |B(A) \ B(A0 )|. Therefore, |B(A0 )| ≤ |B(A)| and #(A0 ) < #(A). Note that #(C) ≤ #(A), and equality holds iff C = A. Therefore, by successively applying the above arguments we can transform A into the initial segment without increasing the boundary. This completes the proof. 2

5

Applications

In order to apply Theorem 1 to the cartesian powers Gn of a graph G one has to solve the vertexisoperimetric problem (VIP) for n = 1 and n = 2. For many (simple) graphs the case n = 1 is not so difficult, but even for them the case n = 2 requires certain attempts and particular methods that are hardly extendible to other graphs. As it is mentioned in the introduction, presently the existence of isoperimetric orders is known just for a few families of graphs, and for a long time the theory of VIP was presumably the theory of these examples. Here we propose a new and unified technique for solving the case n = 2 for bipartite graphs, which is based on an auxiliary (and simpler than VIP, in a sense) extremal poset problem. This result along with the local-global principle provides infinite series of graphs with isoperimetric orders, as we show in the next section. 20

Let G be a bipartite graph. Consider G as a ranked poset PG with levels given by {Vt }ht=0 whose Hasse diagram is isomorphic to G. Thus, PG has h + 1 levels and we denote by PG2 the second cartesian power of PG . Now, for given t and m consider the shadow minimization problem (SMP) that consists in finding a set A ⊆ Vt so that |A| = m and |∇(A)| ≤ |∇(B)| for any B ⊆ Vt with |B| = m. The poset PG is called Macaulay if there exists a total order  on its element set (called a Macaulay order ), so that (i) for any t and m the set A represented as the initial segment of Vt with respect to the order  is a solution to the SMP, and (ii) ∇(A) is an initial segment of Vt+1 . These conditions are analogs of the nestedness and continuity that we use in this paper. This definition of a Macaulay poset differs slightly from the standard one, but is equivalent to it. For more information on Macaulay posets the reader is referred to [10]. G

Theorem 3 Let G 6= K2,2 be a bipartite graph with a level-greedy order ≤ on its vertex set, G

and let  and 2 be simplicial orders associated with ≤ in G and G2 respectively. Assume the order  is upper-monotone and isoperimetric in G and 2 is a Macaulay order for PG2 . Then the order 2 is isoperimetric in G2 . To prove the theorem let A ⊆ V 2 be an optimal set. We show that |B(F 2 (|A|))| ≤ |B(A)|. By Lemma 7 we can assume A is compressed. We also assume #(A) is minimum among all optimal compressed sets of the same size. Let y = (y1 , y2 ) be the largest vertex of A and let x = (x1 , x2 ) be the smallest vertex of V 2 \ A and suppose that x ≺2 y. Thus, kxk ≤ kyk and without loss of generality we may assume that x1 6= y1 and x2 6= y2 . We need the following technical lemma. Lemma 11 Let the conditions of Theorem 3 be satisfied and let A 6= F 2 (|A|) be a compressed optimal set with minimum #(A) among all optimal compressed sets of the same size. Then for x and y specified above one has kxk = kyk − 1. Proof. If kxk = kyk then the Macaulayness of the order 2 and the assumptions on A imply A = F 2 (|A|). Therefore, we can assume kxk ≤ kyk − 1. For the sake of contradiction assume kxk ≤ kyk − 2. We show that there exists an optimal set with a smaller value of #(·). ↓

↓

Case 1. Assume kx1 k < ky1 k. Now, if kx2 k < ky2 k then (y1 , y2 )  (x1 , y2 )  (x1 , x2 ). This implies x ∈ A, i.e. a contradiction. Therefore, kx2 k ≥ ky2 k. Hence, the statement of Lemma 10 is fulfilled, and we can apply the arguments of the proof of Theorem 1 to obtain an optimal set with a smaller value of #(·). Case 2. Assume t = kx1 k ≥ ky1 k. Since |V0 | = 1, then t ≥ 1. We will use the transformation of Lemma 4 to construct an optimal set C with #(C) < #(A). Let (z1 , z2 ) ∈ V 2 with kz1 k > 0. First we show that if R = A ∩ (Vkz1 k × Vkz2 k+2 ) 6= ∅, then {z1 } × ∇(z2 ) ⊆ B(A \ Vkyk ).

21

(23)

↓

↓

Indeed, if (u1 , u2 ) ∈ R then for any (z10 , z20 ) ∈ S = ∆(z1 ) × ∇(z2 ) we have (u1 , u2 )  (u1 , z20 )  (z10 , z20 ). Since (u1 , u2 ) ∈ A and A is compressed, then (z10 , z20 ) ∈ A, so S ⊆ A \ Vkyk . Now (23) follows from {z1 } × ∇(z2 ) ⊆ ∇(S). Let i the largest index such that ky1 k ≤ i ≤ t and W = A∩(Vi ×Vkyk−i ) 6= ∅. By the maximality of i, for any (w1 , w2 ) ∈ W we have A ∩ (∇(w1 ) × ∆(w2 )) = ∅. Therefore, (∇(w1 ) × {w2 }) ∩ B(A \ {(w1 , w2 )}) = ∅.

(24)

Let w = (w1 , w2 ) be the maximum element in W . Now let j be the smallest index such that i ≤ j ≤ t and U = (Vj × Vkxk−j ) \ A 6= ∅, and let u = (u1 , u2 ) be the smallest element in U . Now kwk = kyk, kuk = kxk, and kyk ≥ kxk + 2 imply kw2 k ≥ ku2 k + 2. First assume i = j. Since A is compressed, then i > 0. For z ∈ V denote V 2 (z) = {(z1 , z2 ) ∈ V 2 | z2 = z} and consider the sets A0 = A ∩ V 2 (u2 ) and A00 = A ∩ V 2 (w2 ). Then A0 and A00 correspond to initial segments of order  in V 2 (u2 ) and V 2 (w2 ) respectively. Now we apply the transformation T (A00 , A0 ). Let C be the resulting set. By Remark 3 and (24), the set of boundary elements in V 2 (u2 ) ∪ V 2 (w2 ) has not increased. Furthermore, for any (z1 , z2 ) ∈ C the conditions kw1 k = ku1 k, kw2 k ≥ ku2 k + 2, and (23) imply {z1 } × ∇(z2 ) ⊂ B(A ∩ C). 2 Finally, since C \ A ⊂ Vkyk , then ∆(C \ A) ⊆ A ∩ C. Hence C is an optimal set, and obviously #(C) < #(A). Now assume i < j. This means that Vj−1 × Vkxk−j+1 ⊆ A, which, in turn, implies (23) for any (z1 , z2 ) ∈ U . Since (24) holds for any (w1 , w2 ) ∈ W , then similarly to the above the transformation T (A00 , A0 ) leads to an optimal set C with #(C) < #(A). 2 Proof of Theorem 3. It follows from Lemma 6 that the order 2 satisfies the continuity property. Therefore, it remains to prove that this order satisfies the nestedness property. 2 By Lemma 11, kxk = kyk − 1. Since the order 2 is Macaulay, then replacing the sets A ∩ Vkxk 2 2 2 2 and A ∩ Vkyk with initial segments of Vkxk and Vkyk (in order  ) leads to an optimal set C. It is easily seen that #(C) ≤ #(A).

Case 1. Assume kx1 k < ky1 k. Then we can obtain an optimal set with a smaller value of #(·) as described in Lemma 11. Case 2. Assume t = kx1 k = ky1 k. Since ky2 k = kx2 k + 1, then A ∩ (Vt+1 × V ) 6= ∅. Therefore, Vt × {0} ⊆ A, in particular, kx1 k > 0. Now on one hand, (z, y2 ) ∈ A for any z ∈ Vt−1 because (y1 , y2 ) ∈ A and A is compressed. However, on the other hand k(z, y2 )k = kxk and 2 (z, y2 ) 2 (x1 , x2 ) = x. Thus, (z, y2 ) 6∈ A because A ∩ Vkxk is an initial segment of order 2 in 2 Vkxk . The obtained contradiction implies this case is impossible. 2 Case 3. Assume kx1 k > ky1 k. Since A ∩ Vkxk is an initial segment, then for any z = (z1 , z2 ) such that kzk = kxk and kz1 k < kx1 k one has z 6∈ A. In particular, (y1 , z2 ) 6∈ A for any z2 with ky1 k + kz2 k = kxk = kyk − 1. On the other hand, (y1 , z2 ) ≺2 (y1 , y2 ), hence (y1 , z2 ) ∈ A

22

1

3

5

2

4

6

s s s 0 @ @ s @ @ HH Hs @s @s

··· ···

2h − 3

s s HH 2h − 1 @ Hs @   s @s 2h − 2

G

Figure 7: The graph G(h) with a total order ≤ because A is compressed. The obtained contradiction also implies this case is impossible and completes the proof. 2 Theorems 1 and 3 imply a solution to VIP for the n-cube [11] and the grid [8] as special cases. Now we introduce a new family of bipartite graphs with a nested series of solutions to VIP. The graph G(h) for h ≥ 3 consists of 2h vertices so that the subgraph induced by the vertex set Vt ∪ Vt+1 for 1 ≤ t ≤ h − 2 is isomorphic to K2,2 (cf. Fig. 7). It is easily seen that the simplicial order  is isoperimetric for G(h). In our paper [7] we 2 proved that the poset PG(h) is Macaulay. Thus, according to Theorem 3, the order 2 is isoperimetric for G2 (h), and by Theorem 1 it is so for any n ≥ 3. Note that if the simplicial order is isoperimetric for some bipartite graph G then the poset PG is Macaulay (the converse, however, is not necessarily true). Applying this observation to the graph Gn (h) we get another n proof for a result in [7] that the poset PG(h) is Macaulay. This, in turn, implies a solution [2] to an edge-isoperimetric problem on the cartesian powers of complete bipartite graphs Kh,h , as it is shown in [7].

6

Concluding remarks

The results obtained so far in this paper concentrate on the isoperimetricity of simplicial orders in graphs. The existence of an isoperimetric order in a graph G is itself a strong property of the graph. For example, Theorem 2 provides some necessary conditions for G implied by the fact that the simplicial order is isoperimetric. In order to study the existence of isoperimetric orders, it is of major interest to know more about the structure of the graphs which would admit them. Here we present a nontrivial necessary condition in this direction which is based on the following result. Proposition 1 Let O be a total order on the vertex set of a graph G satisfying the continuity property. If each initial segment of the order O is an optimal set, then any final segment is an optimal set as well. Proof. Denote by L(m, O) ⊆ V the set consisting of the last m elements in the order O. Furthermore, let C be an optimal set with |C| = m. We show |B(L(m, O))| ≤ |B(C)|.

23

Consider A = V \ B(C). Then B(A) ⊆ V \ C, hence, |B(A)| ≤ |V | − |C|. Therefore, for the optimal set F(|A|, O) one has |B(F(|A|, O))| ≤ |B(A)| ≤ |V | − |C| = |V | − m.

(25)

Now consider D = V \ B(F(|A|, O)), and denote m0 = |D|. Then D = L(m0 , O) by the continuity, and m0 ≥ m by (25). Furthermore, B(D) ⊆ V \ F(|A|, O), hence, |B(L(m0 , O))| ≤ |V | − |F(|A|, O)| = |V | − |A| = |B(C)|.

(26)

Since m ≤ m0 , then L(m, O) ⊆ L(m0 , O), and hence B(L(m, O)) ⊆ B(L(m0 , O)). Therefore, by (26), |B(L(m, O))| ≤ |B(L(m0 , O))| ≤ |B(C)|. 2 Therefore, the complement of any set within a nested family of optimal subsets satisfying the continuity property is an optimal set too. A similar property holds also with respect to a related isoperimetric problem (cf. [4]), that consist of finding a set A ⊆ V with minimum size of Φ(A) = {x ∈ A | B(x) 6⊆ A} among all the sets of the same cardinality. This isoperimetric problem is equivalent to one studied in this paper due to the identity Φ(A) = Γ(V \ A). Being applied with respect to the simplicial order, Proposition 1 and Theorem 2 imply |Vi | = |Vh−i | for i = 0, 1, . . . , h. G

In other words, balls of radius i centered in the first and the last vertices in order ≤ are optimal sets, and, moreover, have the same size. This can be considered as a kind of a symmetry in G, and provides a necessary condition for the simplicial order (and any other level-by-level total order) to generate nested families of optimal sets. Concerning the sufficient conditions: due to Theorem 3 we can reduce a solution to VIP to the theory of Macaulay posets. In particular, as it follows from [7], the following operation can be used to generate infinite families of (bipartite) graphs for which Theorem 1 is applicable. Let G

G be such a graph along with a total order ≤. Then the poset PG satisfies the conditions of Theorem 3. For v ∈ Vt with t < h let u ∈ Vt+1 be defined as F(u) = B(F(v)). Now assume G

there exists a vertex w ∈ Vt+1 such that u < w and v, w are not adjacent. Add (v, w) to the edge set of G. It follows from [7] (cf. Section 7) that the simplicial order is isoperimetric for powers of the new graph. Similarly, by connecting the vertices of the same level, this techniques can be adopted to construct even non-bipartite graphs for whose cartesian powers the simplicial order is isoperimetric. A natural extension of our results would be to find an analog of Theorem 3 for non-bipartite graphs. It can be shown that Lemma 11 is valid in this case too, however, replacing the sets 2 2 A ∩ Vkxk and A ∩ Vkyk with the initial segments may not lead to an optimal set. Another interesting direction for further research is the study of the local-global principle with respect to other total orders.

Acknowledgments The authors are grateful to two anonymous referees for their constructive comments that significantly improved the quality of the paper. 24

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