Large Monochromatic Components in Two-colored Grids

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1 Large Monochromatic Components in Two-colore Gris Jiří Matoušek 1 Aleš Přívětivý 2 Department of Applie Mathematics 1,2 an Institute of Theoretical Computer Science (ITI) 1 Charles University Malostranské nám. 25, Praha 1 Czech Republic {matousek,privetivy}@kam.mff.cuni.cz Abstract Let D n enote the -imensional gri with iagonals, that is, the graph with vertex set {1, 2,..., n} an with eges connecting every two vertices that iffer by at most 1 in every coorinate. We prove that for an arbitrary coloring of the vertices of D n by two colors there exists a monochromatic connecte subgraph with at least n 1 2 n 2 vertices, an thus the horizontal layer coloring (by the parity of the first coorinate) is almost optimal. We also consier a -imensional triangulate gri; this is the graph of a triangulation of the soli cube [1, n] that refines the subivision of [1, n] into the gri of unit cubes. Here every 2-coloring has a monochromatic connecte subgraph with Ω(n 1 / ) vertices. These results are prove by combining combinatorial an topological arguments with suitable isoperimetric inequalities, an they can be viewe as -imensional generalizations of the planar HEX lemma. The secon author s research is supporte by grant from Czech Science Founation GD201/05/H014. 1

2 Figure 1: A triangulate square gri. 1 Introuction The HEX lemma. Let Tn 2 be the graph of an n n triangulate square gri in the plane as in Fig. 1. Here n counts the number of rows an columns of vertices; so the picture has n = 6. The well-known planar HEX lemma asserts that if the vertices of Tn 2 are colore re an blue in an arbitrary way, then there exists a path in Tn 2 connecting the top an bottom sies an having all vertices re, or a path connecting the left an right sies an having all vertices blue. In particular, uner any two-coloring of the vertices of Tn 2 there is a monochromatic connecte subgraph with at least n vertices. Triangulate gris. Our results can be regare as -imensional generalizations of the HEX lemma. First we consier a -imensional triangulate gri Tn, which is efine geometrically as follows. We begin with the -imensional soli cube [1, n] an we subivie it into the gri of unit cubes; each unit cube is of the form [i 1, i 1 +1] [i 2, i 2 +1] [i, i +1], 1 i 1, i 2,...,i n 1. Then we triangulate each of the unit cubes in such a way that the simplices of all of these triangulations taken together form a triangulation of [1, n] (here a triangulation is a simplicial complex in the sense of algebraic topology; see Section 2). Then Tn is the graph of such a triangulation; that is, the vertex set is [n] (where we use the shorthan [n] for the set {1, 2,..., n}), an the eges of Tn are the eges of the triangulation. Thus, for given n an, Tn is not efine uniquely, but rather it stans for an arbitrary graph from a (finite) family. We consier an arbitrary coloring of [n] by two colors an we ask, what 2

3 can be sai about the number of vertices of the largest monochromatic connecte subgraph of Tn? The obvious horizontal layer coloring, by the parity of the first coorinate, shows an upper boun of n 1 for this quantity in the worst case, an one might suspect this coloring to be optimal. However, it turns out that for 3, there are better colorings at least for some of the possible triangulate gris. Namely, let us efine the ith iagonal layer L n(i) by L n(i) = {x [n] : j=1 x j = i}, an the iagonallayer coloring by coloring all L n(i) with even i re an all L n(i) with o i blue. As we will verify in Section 6, there are triangulate gris Tn with no eges connecting L n(i) an L n(i ) with i i 2, an for these, the largest monochromatic connecte component is the largest iagonal layer. For = 3, for example, max i L n (i) is approximately 3 4 n2. We prove that the iagonal-layer coloring is not far from optimal in the worst case. To state the result precisely, let us efine, for α [0, 1] an given n an, It can be checke that i 1/2 = { i α = min i : L n(j) > αn }. j i (n+1) 2 an that L n(i 1/2 ) is either the single largest iagonal layer or one of the two largest iagonal layers. We prove the following. Theorem 1.1 For an arbitrary 2-coloring of the vertices of a -imensional triangulate gri Tn, n 3, there exists a monochromatic connecte subgraph with at least L n (i 2/3) Ln 1(i 1/2) vertices. The last quantity is of orer n 1. We conjecture that the right answer is L n (i 1/2) instea of L n (i 2/3) Ln 1 (i 1/2 ) ; that is, the iagonal-layer coloring is optimal in the worst case. This woul follow from a natural an very plausible-looking conjecture state in Section 7. The numerical values of L n (i 1/2) an L n (i 2/3) Ln 1(i 1/2) are not too far from each other, as is illustrate by the following table (showing approximate values for small fixe an n ): L n(i 1/2 ) L n(i 2/3 ) L 1 n (i 1/2 ) = n n 2 = n n 3 = n n 4 3

4 Let us remark that the proof of Theorem 1.1 works for graphs G somewhat more general than Tn. However, since we on t see any goo use of such greater generality at the moment, we stick to the concrete geometric formulation as above. If neee, the properties of the graph actually require can be reconstructe from the proof. Gri with all iagonals. Next, we consier the -imensional gri with all iagonals of the unit cubes ae. That is, we efine the graph D n, the -imensional gri with iagonals, as the graph with vertex set [n] an ege set {{u, v} : u, v [n], u v 1}, where u v = max i u i v i. In this case we show that the horizontal-layer coloring is almost optimal (the remaining lower-orer term is probably an artifact of the proof metho): Theorem 1.2 For an arbitrary 2-coloring of the vertices of the gri with iagonals Dn, there exists a monochromatic connecte subgraph with at least n 1 2 n 2 vertices. Relate work. There is a ifferent an well-known -imensional generalization of the planar HEX lemma appearing e.g. in Gale [G79] (also see Linial an Saks [LS93] for a ifferent proof an an application in computer science). It asserts that if the vertices of T n are colore by colors, then there is a monochromatic path connecting two opposite facets of the cube [1, n]. In particular, there is a monochromatic connecte component with at least n vertices. The problem consiere in this paper fits in the following general context. For an arbitrary graph G an an integer k, let us efine ξ k (G) as the smallest m such that there exists a coloring of the vertices of G by k colors with no monochromatic connecte subgraph having more than m vertices. (So the usual chromatic number χ(g) equals min{k : ξ k (G) = 1}.) The quantity ξ k (G) has been stuie extensively for graphs of boune egree, mainly for k = 2. It is easy to see that any graph G of maximum egree 3 has ξ 2 (G) 2. Alon, Ding, Oporowski, an Vertigan [ADO+03] prove that every graph G of maximum egree 4 satisfies ξ 2 (G) 57. Haxell, Szabó, an Taros [HST03] improve this to ξ 2 (G) 6 an prove that ξ 2 (G) for every graph G of maximum egree 5. On the other han, Alon et al. [ADO+03] constructe 6-regular graphs G with ξ 2 (G) arbitrarily large. For graphs G of maximum egree 3 it was also shown in [BS05] that they amit two-coloring where one color inuces inepenent an set, while the other color inuces components of size at most 189. Earlier 4

5 work on this subject [DOS+96], [JW96] mainly focuse on more specific questions concerning line graphs of 3-regular graphs. These investigations culminate in [Tho99] showing that the eges of every 3-regular graph can be 2-colore so that each monochromatic component is a path of length at most 5. Outline of the proofs of Theorems 1.1 an 1.2. To prove Theorem 1.1, we consier a re-blue coloring of [n]. Assuming that no monochromatic connecte component in Tn is very large, we prove that Tn has a connecte monochromatic separator, where a separator in a graph G is a subset S V (G) whose removal isconnects G into components of size at most 1 2 V (G). The only property of T n that we use for this part is that it is the graph of a simply connecte simplicial complex. Theorem 1.1 then follows from a known result about the vertex expansion of the orinary gri graph (not triangulate). Theorem 1.2 follows along the same lines, only with a boun on the vertex expansion of Dn applie in the en. We erive this boun from the ege-isoperimetric inequality for the orinary gri. 2 Topological preliminaries Here we present some material from elementary algebraic topology. We briefly recall even very stanar things in orer to make the paper more accessible; see, e.g., [Hat01] or [Mun84] for more etails an backgroun. Simplicial complexes. A (geometric) simplex σ is the convex hull of a finite affinely inepenent set A in some R. The points of A are the vertices of σ. The imension of σ is imσ := A 1. Thus a k-simplex (k-imensional simplex) has k + 1 vertices. The convex hull of an arbitrary subset of vertices of a simplex σ is a face of σ (every face is itself a simplex). A nonempty family K of simplices is a (geometric) simplicial complex if the following two conitions hol: 1. Each face of any simplex σ K is also a simplex of K. 2. The intersection σ 1 σ 2 of any two simplices σ 1, σ 2 K is a face of both σ 1 an σ 2. In this paper we consier only simplicial complexes with finitely many simplices. A subcomplex of a simplicial complex K is a subset L of the simplices of K that constitutes a simplicial complex. 5

6 The union of all simplices in a simplicial complex K is the polyheron of K an it is enote by K. In this situation K is calle a triangulation of the topological space K. The vertex set of K, enote by V (K), is the union of the vertex sets of all simplices of K. The graph of K, enote by G(K), has vertex set V (K) an two vertices are connecte by an ege if they are containe in a common 1-simplex (ege) of K. Simplicial maps an simplicial approximation. Let K an L be simplicial complexes. A simplicial map of K into L is a continuous map f : K L such that the image of every simplex σ K is a simplex of L (an in particular, every vertex of K is mappe to a vertex of L), an moreover, f restricte to each simplex σ K is an affine map. Let K an K be simplicial complexes. We call K a refinement of K if K = K an every simplex of K is containe in some simplex of K (an thus for every nonempty simplex σ of K there is a subcomplex of K that is a triangulation of σ). We will nee the following proposition, which is a variation of the stanar simplicial approximation theorem. Proposition 2.1 Let K an L be simplicial complexes an let f : K L be an arbitrary continuous map. Then there is a refinement K of K an a simplicial map f of K into L such that ( ) If σ K is a simplex such that f(σ) is completely containe in some simplex τ L, then f maps all vertices of K lying in σ to vertices of τ. In particular, if f(v) V (L) for some v V (K), then f(v) = f(v). The usual simplicial approximation theorem also yiels f homotopic to f (but we on t nee this part). On the other han, conition ( ) oesn t appear in the formulations of the simplicial approximation theorem in the literature, but it immeiately follows from the stanar proof; see, for example, [Hat01] or [Mun84]. A variant of the HEX lemma. Let T be a 2-imensional simplicial complex with T homeomorphic to B 2, the unit isk in the plane, an let C T be the cycle in the graph G(T) corresponing to the bounary of B 2. (So G(T) can be rawn in the plane with the outer face boune by C T an with all inner faces being triangles.) We nee the following variant of the HEX lemma: 6

7 A u v B Figure 2: A variant of the HEX lemma Lemma 2.2 Let T an C T be as above, an let the vertices of T be colore re an blue. Let the vertices of be partitione into four sets, forming consecutive segments along C T (see Fig. 2), as follows: a set A of re an blue vertices, a single re vertex u, a set B of re an blue vertices, an a single re vertex v. If there is no blue path (path consisting of blue vertices) from A to B, then u an v are connecte by a re path. See Fig. 2 Proof. The case where C has four vertices (an thus A an B are singletons) an A an B are blue is Claim in [MN98]. The case of A, B arbitrary reuces to the previous case by aing two new vertices a an b in the outer face (Fig. 3), coloring them blue, connecting a to all vertices of A {u, v} an b to all vertices of B {u, v}. 3 Monochromatic connecte separators We recall that a topological space X is simply connecte if each continuous map f : S 1 X can be extene to a continuous map f : B 2 X (S 1 enotes the unit circle, the bounary of B 2 ). A simplicial complex K is simply connecte if the topological space K is simply connecte. Here is the main result of this section. Proposition 3.1 Let G = G(K) be the graph of a (finite) simply connecte simplicial complex K. For an arbitrary coloring of V (G) by two colors there 7

8 a A u v B b Figure 3: A reuction to the 4-cycle case. exists a monochromatic connecte separator, that is, a subset S V (G) such that all vertices in S have the same color, the subgraph inuce by S is connecte, an each component of G \ S has at most 1 2 V (G) vertices. We nee the following lemma. Lemma 3.2 Let G be a graph as in Proposition 3.1 with V (G) colore re an blue. Let C be a cycle in G, let u, v V (C) be two istinct re vertices, an let A an B be the two connecte components (arcs) of C \ {u, v}. If there is no blue path from a vertex of A to a vertex of B in G, then u an v are connecte by a re path. Proof. The goal is to apply Lemma 2.2. Let us choose a triangulation T with T homeomorphic to the unit isk an with the bounary cycle C T such that C T has the same length as C. Let f 0 : V (C T ) V (C) be an isomorphism of the two cycles, an let f : C T C be the (unique) extension of f 0 to a simplicial map. (So here we also regar C T as a 1- imensional subcomplex of T an C as a 1-imensional subcomplex of K.) Since T is homeomorphic to B 2 an since K is simply connecte, f can be extene to a continuous map f : T K. Applying our version of the simplicial approximation theorem (Proposition 2.1) to the mapping f, we obtain a refinement T of T an a simplicial 8

9 map f of T into K. Let C T be the subcomplex of T that refines C T (so C T = C T is the bounary of the isk T ). Accoring to conition ( ) in Proposition 2.1, we have f(v) = f(v) for every vertex v of C T, an for every ege (1-simplex) e of C T with vertices u an v, all vertices of C T lying in e are mappe by f to either f(u) or f(v). Each vertex of T is mappe to a vertex f(v) V (G) since f is a simplicial map, so we can efine the color of each v V ( T) as the color of f(v). We note that if x 1, x 2,...,x k is a monochromatic path in T, then f(x 1 ), f(x 2 ),..., f(x k ) is a monochromatic walk in G. A monochromatic walk can be shortcut to a monochromatic path. Now we check that T, CT, an the two re vertices u T = f 1 (u) an v T = f 1 (v) satisfy the conitions of Lemma 2.2. Since the two components A an B of C \ {u, v} are not connecte by a blue path in G, it follows that the two components of C \ {u T, v T } are not connecte by a blue path in T either. Therefore, by Lemma 2.2, u T an v T are connecte by a re path, an u an v are connecte by a re path too, as claime. Proof of Proposition 3.1. If the vertices of G are colore with only one color, we can take V (G) for S an we are one. Otherwise, we choose S as the vertex set of a monochromatic connecte subgraph of G that is inclusion-maximal (no vertex of the same color can be ae), an such that the largest component G 1 of G \S has the smallest possible number of vertices. Without loss of generality, let S be colore blue. We claim that G 1 has at most than 1 2 V (G) vertices. For contraiction, we thus assume V (G 1 ) > 1 2 V (G). We efine S as the union of all inclusion-maximal re connecte subgraphs of G 1 that have at least one vertex connecte to S (so S is the re foam on the blue surface of S facing G 1 ). We will show that S is better than S; namely, that S inuces an inclusion-maximal connecte re subgraph of G an the largest component of G \ S is smaller than G 1. The set S is re by efinition. Next, we show that S inuces a connecte subgraph (then it will also be clear that it is inclusion-maximal, by the choice of S ). It suffices us to show that any two vertices u an v of S are connecte by a re path in G. Moreover, it suffices us to show this only for u an v that are both connecte to S by an ege. Let u S be a blue neighbor of u an let v S be a blue neighbor of v. Since S is a blue connecte subgraph, there 9

10 is a blue path P 1 connecting u an v. Since G 1 is connecte, u an v are connecte by a path P 2 in G 1, which is is vertex-isjoint from P 1. The path P 2 may consist of vertices of both colors, but there is no blue path from a vertex of P 2 to P 1 (by the inclusion-maximality of S). The paths P 1 an P 2 together form a cycle C in G. The graph G an the cycle C satisfy all conitions of Lemma 3.2, an so we have a re path connecting u an v. Hence S inee inuces a connecte subgraph. Each component of G\S is containe either in G 1 \S, or in G\G 1. We have V (G 1 ) \ S < V (G 1 ) since S, an V (G) \ V (G 1 ) < 1 2 V (G). Thus the largest component of G \ S is strictly smaller than G 1, an this contraicts the choice of S. 4 Isoperimetric inequalities For a graph G the an a set A V (G), the vertex bounary of A in G is the set vert-b G (A) = {v V (G) : v A, {u, v} E(G) for some u A}. An isoperimetric inequality for G (or more precisely, a vertex-isoperimetric inequality) bouns from below the quantity min{vert-b G (A) : A V (G), A = s} as a function of s. The orinary gri. Let G n be the graph of the orinary gri (not triangulate). Explicitly, V (G n ) = [n] an u, v V (G n ) are connecte by an ege if u v 1 = i=1 u i v i = 1. A vertex-isoperimetric inequality for G n was establishe by Bollobás an Leaer [BL91a]. For stating it, we efine the simplicial orer on [n] by setting x < simpl y if either i=1 x i < i=1 y i, or if i=1 x i = i=1 y i an x precees y lexicographically; that is, for some i [] we have x j = y j for j = 1, 2,..., i 1 an x i < y i. Theorem 4.1 ([BL91a]) For every A [n ] we have vert-b G n (A) vert-b G n (I simpl ( A )), where I simpl (s) enotes the set of the first s elements of [n] in the orering < simpl. To boun from below the size of the vertex bounary of arbitrary set A [n] we use the following lemma: 10

11 Lemma 4.2 Let A [n] with A = βn. Then we have vert-b G n (A) L n (i β) L 1 n (i 1/2). Proof. By Theorem 4.1 it suffices to prove the statement only for A of the form I simpl ( A ). Then A is a subset of the union of the first i β iagonal layers an the first i β 1 layers are fully containe in A. Let us write D for the intersection of A with L n (i β). If D =, then vert-b G n (A) = L n(i β ) an we are one. If D, then vert-b G n (A) is a isjoint union of two sets B 1 an B 2, where B 1 L n (i β) an B 2 L n(i β + 1). It is easy to check that B 1 = L n(i β ) \ D an if we show that B 2 D Ln 1(i 1/2), we are one. Let D be D shifte by 1 in the first component; that is, D = D + (1, 0,...,0). Similarly, let L be L n(i β ) + (1, 0,...,0). Since D L, we have D \ [n] L \ [n] an from the fact D [n] B 2 we arrive at B 2 D L \ [n]. The rest follows from the fact that L \ [n] is exactly a copy of the set Ln 1(i β n), which has always smaller size than the mile layer Ln 1(i 1/2) as will be shown in Lemma 4.3. Let us remark that some error term in the previous estimation of vert-b G n (A) is neee. For example, vert-b G n (I simpl (21)) = 5 for the case n = 3, = 3, but L 3 3 (i 21/27) = 6. Lemma 4.3 For any n, N the following hol: ( ) ( (a) L (n+1) L n 2 i = (n+1) n 2 + i) for every i = 0, 1, 2,..., an therefore, i 1/2 = (n+1) 2. (b) The function i L n (i) is non-ecreasing on integers i (n+1) 2 = i 1/2 an non-increasing on integers i (n+1) 2. Proof. (a) The function f ( (x 1,...,x ) ( ) ( ) = (n + 1 x 1,...,n + 1 x ) maps L (n+1) n 2 i to L (n+1) n 2 + i an it is easy to check from efinition that i 1/2 = (n+1) 2. (b) The proof is by inuction on. For = 1, 2 an any n 1 the statement trivially hols. Fix some n an > 2. We can partition arbitrary 11

12 set L n(i) into n sets L 1,...,L n accoring to the last component of the vertices; that is L t = {x L n (i) : x = t} for all t [n]. It is easy to check that each L t is isomorphic to Ln 1 (i t) an thus L n (i + 1) L n (i) = n L 1 n (i + 1 t) n L 1 n (i t) t=1 t=1 = L 1 n (i) L 1 n (i n). We istinguish four cases: 1. If i (n+1)( 1) 2 then by the inuctive hypothesis we have L 1 n (i n) L 1 n (i). 2. If (n+1)( 1) 2 < i < (n+1) 2 then by (a) we have L 1 n (i) ( ) = L 1 (n + 1)( 1) i an since n (n + 1)( 1) (i n) (n + 1)( 1) i 2 we have by the inuctive hypothesis (i n) L 1 ( ) (n + 1)( 1) i = Ln 1 (i). L 1 n n 3. If (n+1) 2 i < (n+1)(+1) 2 + n then by (a) we have L 1 n (i n) = L 1 n ((n + 1)( 1) (i n)) an since (n + 1)( 1) i (n + 1)( 1) (i n) 2 we have by the inuctive hypothesis L 1 n (i) ( L 1 n (n + 1)( 1) (i n)) = L 1 n (i n). 12

13 4. If i (n+1)(+1) 2 + n then by the inuctive hypothesis we have L 1 n (i n) L 1 n (i). Proposition 4.4 For every α (0, 1 2 ) an every A [n] with αn A (1 α)n we have vert-b G n (A) L n(i 1 α ) L 1 n (i 1/2 ). Proof. Let A = βn. By Lemma 4.2 we have vert-b G n (A) L n (i β) Ln 1 (i 1/2 ). If α β 1 2, then also i α i β i 1/2 an by Lemma 4.3 we have L n(i β ) L n(i α ). Similarly, for 1 2 < β (1 α) we have L n(i β ) L n(i 1 α ). Since i 1/2 (n + 1) i α i (1 α), by Lemma 4.3 we have L n(i α ) = L n((n + 1) i α ) L n(i (1 α) ), an thus vert-b G n (A) L n (i (1 α)) Ln 1(i 1/2). The gri with iagonals. Here we erive the following vertex-isoperimetric inequality: Proposition 4.5 For any set A [n] with 1 4 n A 3 4 n we have vert-b D n (A) n 1 2 n 2. We are going to erive this result from an ege-isoperimeric inequality for the orinary gri G n. For A [n], let ege-b G n (A) = {{u, v} E(G n ) : u A, v A} be the ege bounary of A in G n. Theorem 4.6 (Bollobás an Leaer [BL91b]) Let A [n]. Then 4 A /n for A < n /4, ege-b G n (A) n 1 for n /4 A 3n /4, 4(n A )/n for A > 3n /4. First we show, following [BL91a] almost verbatim, that it suffices to prove Proposition 4.5 for sets A that are own-sets. We begin with the necessary efinitions from [BL91a]. A set A [n] is a own-set if x A implies y A for all y [n] satisfying x i y i for all i []. For A [n] an 1 i, let the i-section of A at x [n] 1 be efine as A i (x) = {t [n] : (x 1,..., x i 1, t, x i,...,x 1 ) A}. 13

14 For A [n] an 1 i, let the i-compression of A be efine by giving its i-sections: (C i (A)) i (x) = {1,..., A i (x) } for all x [n] 1. In other wors, C i compresses each i-section of A ownwars. We note that C i (A) = A, an it is also easy to check that if a set A is i-compresse, then so is C j (A). So the set A = C n (C n 1 (... C 1 (A)...)) is i-compresse for every i [], an thus it is a own-set. Lemma 4.7 Let A [n]. Then there is a own-set A [n] with A = A an vert-b D n (A ) vert-b D n (A). Proof. For A [n], let N(A) = A vert-b D n (A) enote the close neighborhoo of A. By the above remarks it suffices to prove that N(C i (A)) N(A) for all A [n] an all i []. Let us write B for C i (A). The case = 1 is easy; we have N(A) A + 1 = N(B) for all A [n] apart from A = an A = [n]. For A = an A = [n] we have N(A) = A = B = N(B). For > 1 it is sufficient to show that for each x [n] 1 we have Fixing an arbitrary x [n] 1, an (N(B)) i (x) (N(A)) i (x). (N(A)) i (x) = (N(B)) i (x) = y [n] 1 : y x 1 y [n] 1 : y x 1 N ( A i (y) ) N ( B i (y) ). The sets N(B i (y)) are initial segments of [n], hence they are neste, an thus (N(B)) i (x) max N ( B i (y) ). y [n] 1 : y x 1 From the one-imensional case we know N ( B i (y) ) N ( A i (y) ), an therefore, (N(B)) i (x) (N(A)) i (x). 14

15 Proposition 4.5 now follows from Theorem 4.6, the previous lemma, an the next one: Lemma 4.8 Let A [n] be a own-set. Then vert-b D n (A) ege-b G n (A) 2 n 2. To prove this lemma we nee the following claim: Claim 4.9 Let A [n] be a own-set. Then ege-b G n (A) n 1. Proof. For any ege {u, v} ege-b G n (A) there is a unique i-section A i (x) that contains both of the vertices u an v. Since A is a own-set, there is at most one ege e ege-b G n (A) in every i-section, an we have n 1 ifferent i-sections in [n]. Proof of Lemma 4.8. Let E = ege-b G n (A) \ {{u, v} E(G n) : u i = v i = n for some i}. We show that E can be injectively mappe into vert-b D n (A). Let us construct a mapping f : E [n] in the following way: For arbitrary ege {u, v} E, let i be the number of the component in which u an v iffer by 1, an without loss of generality, let u i v i. Then we put f({u, v}) = u + (1, 1,...,1, 0,...,0). }{{} i times1 It is easy to check that u j < n for j i, an hence z = f({u, v}) [n] inee. We have u A an v / A. If we ha z A, then since A is a own-set an v i z i for all i [], it woul follow that v A as well. Thus z / A, an since u z = 1, we arrive at f({u, v}) vert-b D n (A). Now we suppose for contraiction that there exist two eges {u 1, v 1 }, {u 2, v 2 } E that are mappe by f to the same vertex z. Let i 1 be the number of the component where u 1 an v 1 iffer by 1 an let i 2 be the number of the component where u 2 an v 2 o so. Without loss of generality let i 1 < i 2. Then v 2 = (z 1 1,...,z i2 1 1, z i2, z i2+1,...,z ) 15

16 is in all components less or equal to u 1 = (z 1 1,...,z i1 1 1, z i1 1, z i1+1,..., z ), an since A is a own-set, u 1 A implies v 2 A. This contraiction shows that f is injective as claime. It remains to estimate the size of E. For i [] let us put F i = {{u, v} ege-b G n (A) : u i = v i = n}. It is easy to see that where F i = ege-b G 1(A n i ), A i = {(x 1,..., x 1 ) [n] 1 : (x 1,...,x i 1, n, x i,...,x 1 ) A}. Since A i is a own-set, Claim 4.9 gives F i ( 1)n 2, an thus E ege-b G n (A) F i ege-b G n (A) ( 1)n 2, i=1 which is even slightly better than claime in the lemma. 5 Proofs of Theorems 1.1 an 1.2 We will nee the fol- Monochromatic connecte separators again. lowing stanar consequence of Proposition 3.1. Corollary 5.1 Let G be the graph of a simply connecte simplicial complex, an let V (G) be colore re an blue. Then there exists a partition of V (G) into three isjoint sets A, B an S such that no ege connects a vertex of A to a vertex of B, S is a monochromatic connecte subgraph of G, an A, B 2 3 V (G). Proof. Let S be as in Proposition 3.1, an let V 1, V 2,..., V m be the vertex sets of the components of G \ S orere by size in the escening orer. Let i be the largest inex with V 1 V 2... V i 2 3 V (G). Then 16

17 V V i 2 3 V (G) an V i V m 1 3 V (G). Let us put A = V 1 V 2 V i an B = V \ A \ S. Obviously, A 2 3 V (G), an since V (G) B A V i+1 an B = V i+1 + V i V m V i V (G), we arrive at B 2 3 V (G) too. Proof of Theorem 1.1. We consier a two-colore graph Tn as in the theorem an a partition V (Tn) = A B S as in Corollary 5.1 (S is a monochromatic connecte separator). We may assume S 1 3 n, for otherwise, we woul be one. Then A or B is between 1 3 n an 2 3 n ; let us fix the notation A, B so that 1 3 n A 2 3 n. We now consier the gri graph G n as a subgraph of T n. We have S vert-b T n (A) vert-b G n (A), an vert-b G n (A) L n(i 2/3 ) Ln 1 (i 1/2 ) by Proposition 4.4. The require estimates for L n(i 2/3 ) an Ln 1(i 1/2) are establishe in the appenix. Theorem 1.1 is prove. Proof of Theorem 1.2. We consier a 2-coloring of the graph Dn. By eleting suitable iagonals from Dn we obtain a triangulate gri Tn. Using Corollary 5.1 we again fin a partition V (Tn ) = A B S. We may assume S 1 3 n, for otherwise, we woul be one. We can again assume 1 3 n A 2 3 n, an using Proposition 4.5 we get S vert-b D n (A) n 1 2 n 2. Theorem 1.2 is prove. 6 An upper boun for certain triangulate gris In this section we show that there exists a triangulate gri Tn in which every ege connects either two vertices from the same iagonal layer L n(i) or vertices from two consecutive iagonal layers L n (i) an L n (i+1). Thus, each monochromatic connecte component in the iagonal-layer coloring of such Tn is containe in some L n(i). To see that such triangulation exists, we consier the arrangement A of the following hyperplanes: Hj i = {x : x i = j} for j [n], i = 0,...,, an S j = {x : x k = j} for j =,..., n. The vertices of this arrangement are of two types: the first type are the vertices forme as the intersection of hyperplanes among Hj i, an the secon type are vertices forme as the intersection of one hyperplane S j an 1 of the hyperplanes Hj i. The vertices of the first type are exactly the gri points of [n]. An intersection of 1 hyperplanes Hj i is either empty or a line parallel to one 17

18 of the coorinate axes, an if we intersect this line by some hyperplane S j we get a vertex with integral coorinates. Thus, if we restrict ourselves to the hypercube [1, n], the only possible vertices of our arrangement are the gri points of [n]. The arrangement A is a polyheral complex. Let C be the subcomplex of A consisting of all faces containe in [1, n]. The union of all faces of C is exactly [1, n], C refines the subivision of [1, n] into unit cubes, an it remains to refine C to a triangulation. This can be one, for example, using the bottom-vertex triangulation (or canonical triangulation) of Clarkson [Cla88]. For each face F of C, which is a convex polytope, we efine the bottom vertex v as the vertex of F with the lexicographically smallest coorinate vector. The bottom-vertex triangulation of C is efine inuctively. For k 1, we assume that all faces of C imension at most k have alreay been triangulate (for k = 1 this is satisfie automatically, since the 1-imensional faces are segments). Given a (k + 1)-imensional face F, we let v be the bottom vertex of F, an we triangulate F as follows. For each simplex σ from the union of the triangulations of all proper faces of F, we consier the cone over σ with apex v, that is, the simplex conv(σ {v}). Taking all of these simplices plus all simplices in the triangulation of all proper faces of F yiels the bottom-vertex triangulation of F. It is not har to verify that this inee efines a simplicial complex refining C, an thus the esire triangulate gri T n. 7 Open problems We have consiere colorings of the vertices of [n] by two colors. The case of colors is essentially solve by the -imensional HEX lemma. A natural an, in our opinion, very interesting question is, what happens for colorings by k colors for 3 k 1? An obvious conjecture is that the minimum possible size of the largest monochromatic connecte subgraph shoul be about n +1 k in this case (for an k fixe an n ), but at present we have nothing nontrivial to say about this question. A natural boun one woul expect in Theorem 1.1 (for a triangulate gri) woul be the size of the mile iagonal layer L n(i 1/2 ). However, our metho yiels only L n(i 2/3 ). The improve boun of L n(i 1/2 ) woul follow immeiately by our metho from the following conjecture: If A [n] 18

19 is such that removing it from the gri G n leaves no connecte component larger than 1 2 n, then A L n (i 1/2). This oesn t seem to follow from the vertex-isoperimetric inequality for G n, since in principle, for example, we might have a situation where G n \A has three connecte components of size about 1 3 n each, an A is a common bounary of all three. While this or similar situations seem unlikely, we currently on t have a proof of the conjecture. We note that Kleitman [Kle86] prove a similar conjecture for the hypercube (that is, for the gri G 2), but aapting his metho to larger gris seems nontrivial. In the appenix below, we erive asymptotic bouns for the size of the iagonal layers L n(i 1/2 ) an L n(i 2/3 ), using a quantitative version of the central limit theorem. However, the estimate quantity is of a quite basic nature an it might well be that consierably more precise results are known. Acknowlegments We woul like to thank Nati Linial for iscussions that were the starting point of our research reporte in this paper, an Imre Leaer for useful avice concerning isoperimetric inequalities. References [ADO+03] N. Alon, G. Ding, B. Oporowski, D. Vertigan. Partitioning into graphs with only small components. Journal of Combinatorial Theory (Series B), 87: , [BL91a] [BL91b] [BS05] [Cla88] B. Bollobás an I. Leaer. Compressions an Isoperimetric Inequalities. Journal of Combinatorial Theory (Series A) 56:47-62, B. Bollobás an I. Leaer. Ege-isoperimetric inequalities in the gri. Combinatorica 11(4): , R. Berke an T. Szabó. Relaxe two-coloring of cubic graphs. DMTCS Proceeings of 2005 European Conference on Combinatorics, Graph Theory an Aplications, K. L. Clarkson. A ranomize algorithm for closest-point queries. SIAM J. Comput., 17: ,

20 [DOS+96] G. Ding, B. Oporowski, D. Saners, D. Vertigan, preprint, Louisiana State University, Baton Rouge, [Fel43] [G79] W. Feller. Generalization of a Probability Limit Theorem of Cramer. Transactions of the American Mathematical Society Vol. 54, No. 3, pp , D. Gale. The game of hex an the Brouwer fixe-point theorem. Amer. Math. Monthly 86: , [Hat01] A. Hatcher. Algebraic Topology. Cambrige University Press, Cambrige Electronic version available at [HST03] P. Haxell, T. Szabó, G. Taros. Boune size components partitions an transversals. Journal of Combinatorial Theory (Series B) 88: , [JW96] B. Jackson an N. Wormal. On the linear k-arboricity of cubic graphs. Discrete Math 162: , [Kle86] D. Kleitman. On a problem of Yuzvinsky on separating the n- cube. Discrete Math. 60: , [LS93] [MN98] N. Linial an M. Saks. Low iameter graph ecompositions. Combinatorica 13: , J. Matoušek an J. Nešetřil. Invitation to Discrete Mathematics. Oxfor University Press, Oxfor [Mun84] J. R. Munkres. Elements of Algebraic Topology. Aison-Wesley, Reaing, MA, [Tho99] C. Thomassen. Two-colouring the eges of a cubic graph such that each monochromatic component is a path of length at most 5. Journal of Combinatorial Theory (Series B) 75: , A Appenix: Bouning the size of iagonal layers Let us recall that L n(i) = {x [n] : j=1 x j = i} an i α = min{i : j i L n(j) > αn }. Here we prove the following estimates: 20

21 Proposition A.1 There exist constants C c > 0 such that for all n > 1 an all we have c n 1 L n(i 2/3 ) L n(i 1/2 ) C n 1. We use the following quantitative version of the central limit theorem by Feller [Fel43]. Theorem A.2 Let X 1,..., X be inepenent ranom variables an suppose that their secon moments σ 2 1,..., σ 2 exist. Let S = X X an s 2 = σ σ2, let F (t) = Prob[S t] be the istribution function of S an let Φ(u) be the istribution function of the stanar normal istribution. Let there exist a non-increasing sequence λ 1,...,λ, such that X k λ k s k for k = 1,...,. If 0 < λ u < (3 5)/4, then 1 F (us ) = e (1/2)u2 Q (u) ( (1 Φ(u)) + θλ e (1/2)u2). Here θ has an upper boun inepenent of the sequence {X k } an of ; furthermore Q (u) = q,ν u ν where coefficient q,ν epens only on the ν first moments of X 1,...,X. If, more particularly, 0 < λ u < 1/12 then an ν=1 θ < 9 q,ν < (1/7)(12λ ) ν. Moreover, for any 0 < i < j we have Q j (u) Q i (u) (1/2)(s 2 j s 2 i)/s 2 n. Let us assume that n is o, n = 2k+1; the case of even n is similar. Let us consier the sequence of inepenent ranom variables X 1,..., X, where each X i is uniformly istribute on the set of integers { k, k + 1,...,k}. 21

22 For this sequence X 1,...(, X we efine ) S, F, s as in Feller s theorem. It is easy to see that n F t n+ 2 = u i=0 L n (i) for all t N. In our case k 3 s k, so we can put λ = 3. For 0 u 1 an > 10 6 it is seen that λ u < an thus Q (u) So we arrive at q,ν u ν 1 q,ν 7 (12λ ) ν ν=1 12λ 7 ν=0 ν=1 ν=1 (12λ ) ν 36 7 ν=0 ( e 3 1/2 e 1 2 u2 Q (u) e 3 1/ /2 e 1 2 u2 Q (u) /2. ) ν 6. From 1 F (us ) = e (1/2)u2 Q (u) ( (1 Φ(u)) + θλ e (1/2)u2) we erive 1 F (us ) ( /2)( 1 Φ(u) /2) F (us ) Φ(u) 32 1/2 an also 1 F (us ) (1 3 1/2)( 1 Φ(u) 27 1/2) F (us ) Φ(u) /2. Let q 1 = Φ 1 (0.7) an q 2 = Φ 1 (0.8); both of them satisfy 0 < q 1, q 2 < 1. Then the i-layers for i [ q 1 s + n+ 2, q 2 s + n+ 2 ] contain together at least 1 30 n vertices, since F (q 2 s ) F (q 1 s ) Φ(q 2 ) 32 1/2 Φ(q 1 ) 32 1/2 = 1/ /2 1/30. There are at most (q 2 s q 1 s ) k 1/2 such layers, so the largest of them, L n(q 1 s + n+ 1 2 ), contains at least n 1 vertices. Now the ith layer 15 1/2 22

23 for i = q 1 s + n+ 2 cuts off a set of size at least 2 3 n from the gri, since F (q 1 s ) Φ(q 1 ) 32 1/2 = 7/ / > 2/3. Thus, L n(i 2/3 ) > L n(q 1 s + n+ 2 ) an therefore, L n(i 2/3 ) 1 n /2 Let us remin that uring all the previous steps we still assume that > In the following proposition we show the upper boun on the size of the mile layer. To prove this, we nee to show that two consecutive layers cannot iffer too much in size. Lemma A.3 For any n,, i N the following hol: (a) L n (i 1/2) 1000 n 1, (b) L n (i) L n 2 n (i + 1) L 1 n (i n) Proof. First we show that L n(i) L n(i + 1) Ln 1 (i n) Ln 1(i 1/2). Let L be L n (i) shifte by 1 in the first component, i.e. L = L n(i) + (1, 0,...,0). Since (L [n] ) L n(i + 1), we arrive at L n (i) L n (i + 1) L \ [n]. It is easy to check that L \ [n] is exactly a copy of the set Ln 1 (i n), which has always smaller size than the mile layer Ln 1 (i 1/2 ). The rest of (b) follows from (a). To prove (a) for 10 6, we note that size of any layer L n(i) is boune above by n 1 an for 10 6 we have n n 1. To prove (a) for > 10 6, we estimate the size of the n 6 consecutive layers L n (i) for i [ ] i 1/2, i 1/2 + n 6 : F n ( i 1/2 + n ) F ( ) n i1/2 6 ( ( n Φ 6s Φ ) + 32 ) ( Φ (0) 32 ) ( 1 ) Φ(0) + 64 ( 1 1/2 2π 65 1/2. 0 ) e (1/2)y2 y + 64 Among the n/6 layers the size of the smallest layer L n ( i1/2 + n 6 ) is at most ( L n i 1/2 + n ) 65 1/2 n 400i n 1. 6 n/6 23

24 Using (b) we estimate the size of the mile layer as L n (i 1/2) ( L n i 1/2 + n ) ) n + (2000 n n 1. 24