Anagram-free colorings of graphs

Transcription

1 Anagram-free colorings of graphs Nina Kamčev Tomasz Luczak Benny Suakov Abstract A sequence S is calle anagram-free if it contains no consecutive symbols r 1 r... r k r k+1... r k such that r k+1... r k is a permutation of the block r 1 r... r k. Answering a question of Erős an Brown, Keränen constructe an infinite anagram-free sequence on four symbols. Motivate by the work of Alon, Grytczuk, Ha luszczak an Rioran [], we consier a natural generalisation of anagram-free sequences for graph colorings. A coloring of the vertices of a given graph G is calle anagram-free if the sequence of colors on any path in G is anagram-free. We call the minimal number of colors neee for such a coloring the anagram-chromatic number of G. In this paper we stuy the anagram-chromatic number of several classes of graphs like trees, minor-free graphs an boune-egree graphs. Surprisingly, we show that there are bouneegree graphs such as ranom regular graphs) in which anagrams cannot be avoie unless we basically give each vertex a separate color. 1 Introuction The stuy of non-repetitive colorings was conceive by a famous result of Thue [19] from He showe that there exists an infinite sequence S on an alphabet of three symbols in which no two ajacent blocks of any length) are the same. In other wors, S contains no sequence of consecutive symbols r 1 r... r n with r i = r i+n for all i n. Note that it is not a priori obvious that the minimal size of the alphabet necessary for an infinite non-repetitive sequence is even finite. Thue s result is interesting in its own right, but it also has influential an surprising applications, the most famous one probably occurring in a solution to the Burnsie problem for groups by Novikov an Ajan [17]. Thue-type problems lea to the evelopment of Combinatorics on Wors, a new area of research with many interesting connections an applications. Generalisations of Thue s result occurre in two irections. Firstly, the setting has been change from sequences to, e.g., the real line, the lattice Z n, or graphs. Seconly, repetitions as a forbien structure can be replace by anagrams, sums, patterns etc. For a formal treatment an references to these problems, we refer the reaer to the survey of Grytczuk [13]. Here we focus on graph colorings, an the structure we are avoiing are anagrams. A sequence r 1 r... r n r n+1... r n is calle an anagram if the secon block, r n+1... r n, is a permutation of r 1 r... r n. A long staning open question of Erős [9] an Brown [6] was whether there exists a sequence on {0, 1,, 3} containing no anagrams. We call such sequences anagramfree. It is easy to check that no such sequence on three symbols exists. In 1968 Evokimov [10] showe that the goal can be achieve with 5 symbols, which was the first finite upper Department of Mathematics, ETH, 809 Zurich. nina.kamcev@math.ethz.ch. Faculty of Mathematics an Computer Science, Aam Mickiewicz University, Poznań, Polan. tomasz@amu.eu.pl. Research supporte by NCN grant 01/06/A/ST1/0061. Department of Mathematics, ETH, 809 Zurich. benjamin.suakov@math.ethz.ch. 1

2 boun. Later Pleasants [18] an Dekking [8] lowere this number to five. Finally, Keränen [15] constructe arbitrarily long anagram-free sequences on four symbols using Thue s iea given a finite anagram-free sequence S on symbols {0, 1,, 3}, we can replace each symbol by a longer wor on the same alphabet in a way that yiels a new, longer anagram-free sequence S. This answere the question of Erős an Brown, but at the same time opene new avenues for further stuies; some of them can be foun in [13]. Bean, Ehrenfeucht an McNulty [5] have stuie the problem of non-repetitive colorings in a continuous setting. A coloring of the real line is calle square-free if no two ajacent intervals of the same length are colore in the same way. More precisely, for any intervals I = [a, b] an J = [b, c] of the same length L > 0, there exists a point x I whose color is ifferent from x + L. In [5], they showe that there exist square-free two-colorings of the real line. The problem of avoiing anagrams also has a continuous variant. Alon, Grytczuk, Lasoń an Micha lek [3] have prove that there exists a measurable 4-coloring of the real line such that no two ajacent segments contain equal measure of every color. Alon, Grytczuk, Ha luszczak an Rioran [] propose another variation on the non-repetitive theme. Let G be a graph. A vertex coloring c : V G) C is calle non-repetitive if any path in G inuces a non-repetitive sequence. Define the Thue number πg) as the minimal number of colors in a non-repetitive coloring of G. It is easy to see that this number is a strengthening of the classical chromatic number, as well as the star-chromatic number. It turns out that the Thue number is boune for several interesting classes of graphs, e.g. πp n ) 3 for a path P n of length n irectly from Thue s Theorem), an πt ) 4 for any tree T. Using the Lovász Local Lemma, the authors of [] showe that πg) c G), where c is a constant an G) enotes the maximum egree of G. They also foun a graph G with πg) c log. Closing the above gap remains an intriguing open question. Another interesting problem is to ecie if the Thue number of planar graphs is finite. A survey of Grytczuk [1] lays out some progress in this irection, as well as numerous relate questions on non-repetitive graph colorings. The investigation of anagram-free colorings of graphs, which we o here, was suggeste in the concluing remarks of []. Let c : V G) C be a vertex coloring of a graph G. Two vertex sets V 1 an V have the same coloring if they have the same number of occurrences of each color, i.e. c 1 a) V 1 = c 1 a) V for each a C. An anagram is a path v 1 v... v n in G whose two segments v 1... v n an v n+1... v n have the same coloring. We enote the minimum number of colors in an anagram-free coloring of G by π α G), an call it the anagram-chromatic number of G. Clearly π α G) n for any n-vertex graph G. The result of Keränen [15] states that π α P n ) 4 for a path P n of length n, so it is only natural to ask what is π α for other families of graphs. It turns out that as soon as we move on from paths, the situation becomes very ifferent. We first show that the anagram-chromatic number of a binary tree alreay increases with the number of vertices. Proposition 1.1. Let T h be a perfect binary tree of epth h, i.e. every non-leaf has two chilren an there are h leaves, all at istance h from the root. Then h log h π αt h ) h + 1. It follows that the anagram-chromatic number of planar graphs is also unboune, but it is still interesting to etermine how quickly it increases with the number of vertices. We observe that in ealing with a family of graphs which amits small seperators such as H-minor-free graphs), this fact can be use to boun π α G) from above. Proposition 1.. Let h 1 be an integer, an let H be a graph on h vertices. Any n-vertex graph G with no H-minor satisfies π α G) 10h 3/ n 1/.

3 In this paper we are particularly intereste in anagram-free colorings of graphs of boune egree. We show that, surprisingly, there are graphs of boune egree such that to avoi anagrams we essentially nee to give every vertex a separate color. We show this by consiering the ranom regular graph G n,, which is chosen uniformly at ranom from all n-vertex -regular graphs. Here we write G n, for the sample graph as well as the unerlying probability space, an we stuy G n, for a constant an n. We say that an event in this space hols with high probability whp) if its probability tens to 1 as n tens to infinity over the values of n for which n is even so that G n, is non-empty). Then our main result can be state as follows. Theorem 1.3. There exists a constant C such that for sufficiently large, with high probability, the ranom regular graph G n, satisfies 1 C log ) n π α G n, ) 1 log ) n. The rest of this paper is organize as follows. We start with some observations on the anagramchromatic number for trees an minor-free graphs. Then, we give the proof of Theorem 1.3. We conclue the article with some open questions an conjectures on anagram-free colorings. We mostly omit floor an ceiling signs for the sake of clarity. The log will enote the base-e logarithm. We will sometimes use stanar O-notation for the asymptotic behaviour of the relative orer of magnitue of two sequences, epening on a parameter n. Specific families of graphs.1 Bouns for trees A binary tree is a tree in which every vertex has at most two chilren. Let T h be a perfect binary tree of epth h, that is to say that every non-leaf has two chilren an there are h leaves, all at istance h from the root. The root is taken to be at epth 0, so a tree consisting of one vertex has epth 0. Coloring each vertex of T h by its istance from the root shows that π α T h ) h + 1. In the following section, we will argue that actually any n-vertex tree can be anagram-free colore with log n colors. Proposition 1.1 asserts the lower boun π α T h ) h log h, which will be proven in this section. Let T be a vertex-colore binary tree an let U be a subtree of T. The effective vertices of U are its root i.e. the vertex of U of the smallest epth), leaves, an vertices of egree three. The effective epth of U is set to h 1, where h is the minimum number of effective vertices on any path from the root to a leaf that is, the epth of the binary tree obtaine by contracting all the internal egree-two vertices of U). Note that if U has effective epth h 1, then it has at least h1 leaves. We say that U is essentially monochromatic if all its effective vertices carry the same color. We will use a Ramsey-type argument to fin a large essentially monochromatic subtree of a given tree. In the statement below Ha 1, a,..., a ) enotes the minimal number h for which any perfect binary tree T of epth h whose vertices are colore using colors 1,,..., contains an essentially i-colore subtree of effective epth a i, for some i []. Lemma.1. Ha 1, a,..., a ) a a. Proof. We use inuction on i=1 a i. The base case is a 1 = = a = 0, for which the claim clearly hols. Let T be a perfect binary tree of epth a a. Suppose that its root v has the color 1, an call its chilren v L an v R. Consier the subtrees T L an T R of epth at least a 1 + +a 1 3

4 roote at v L an v R respectively. If for some i, T L contains an essentially i-colore subtree of effective epth a i, we are one. The same hols for T R. Otherwise, using the inuction hypothesis, T L an T R contain essentially 1-colore subtrees of effective epth a 1 1. Those two subtrees, together with the root v, form an essentially 1-colore subtree of T, as require. Proof of Proposition 1.1. Let T h be colore using < h log h colors. By Lemma.1, it contains an essentially monochromatic subtree U of epth h/. Let u be the root of U, an suppose U is essentially re. There are at least h/ paths from u to the leaves, an the coloring of each path is a multiset of orer at most h + 1. On the other han, there are at most h such multisets. Since h < h/ for our choice of, there is a multiset which occurs on two ifferent paths, say P 1 an P. Let v be the lowest common vertex of P 1 an P, an let l 1 an l be their respective leaves. By construction of U, the vertices v, l 1 an l are re. Hence the segments from l 1 to v, excluing v, an from v to l, excluing l, have the same coloring. We conclue that the given coloring of T h, even restricte to U, contains an anagram.. Graphs with an exclue minor Planar graphs are of special interest when it comes to coloring problems. The Four Color Theorem is one of the most celebrate results in Graph Theory. Moreover, the question of whether the Thue-chromatic number of planar graphs is finite has attracte a lot of attention an is still open. We use separator sets to show that for a large class of minor-free graphs the anagram-chromatic number is of orer O n). The crucial ingreient of our argument is the separator theorem, prove by Alon, Seymour an Thomas [4]. It states that for a given h-vertex graph H, in any graph G with n vertices an no H-minor, one can fin a set S V G) of orer S h 3 n 1, whose removal partitions G into isjoint subgraphs each of which has at most n 3 vertices. Such a set S is calle a separator in G. Using this theorem, we construct a coloring of any proper minor-close family of graphs. For convenience of the reaer, we restate Proposition 1.. Proposition 1.. Let h 1 be an integer, an let H be a graph on h vertices. Any n-vertex graph G with no H-minor satisfies π α G) 10h 3/ n 1/. Proof. The coloring is inuctive suppose the claim hols for graphs on at most n 1 vertices. Let G be as in the statement, an let S be a separating set of vertices in G of orer at most h 3/ n 1/ given by the Separator Theorem. Then G S consists of two vertex-isjoint subgraphs spanne by A 1 V G) an A V G), with A i n 3. The inuce subgraphs G[A i ] o not contain H as a minor, so by the inuctive hypothesis, we can color them using k = 10h 3/ n/3 colors a 1, a,..., a k. Note that the two subgraphs receive colors from the same set. This coloring guarantees that any path containing only vertices from A 1 or A is anagram-free. Furthermore, we assign to each vertex v i S a separate color b i, making any path passing through S anagram-free. Hence the coloring is inee anagram free. As intene, the number of colors use is at most ) h 3/ n 10 1/ h 3/ n 1/. 4

5 Since planar graphs are characterize as graphs containing neither K 5 nor K 3,3 as a minor, we arrive at the following consequence of the above result note that the constant 150 can be replace by 19 if we use the fact that each planar graph has a separator of orer 1.84 n). Corollary.. Let G be an n-vertex planar graph. Then π α G) 150 n. In fact, any hereitary family of graphs with small separators can be colore using the argument from Proposition 1.. For example, it is easy to see that an n-vertex forest F contains a single vertex which separates it into several forests on at most n/ vertices. The same inuctive argument implies π α F ) log n. As for the lower boun for planar graphs, we only have the following moification of the argument we gave for trees. Proposition.3. There is an n-vertex planar graph F n with π α F n ) 1 4 log n. Proof. Let F n be a perfect binary tree with n leaves, plus extra eges between any two vertices on the same level having the same parent. Suppose it is colore in k = 1 4 log n colors. The number of shortest paths from the root to the vertices corresponing to leaves is n, whereas the number of possible colorings of these paths is ) log n+k k 1 < n; hence some two paths have the same coloring. These two paths, minus the share initial segment, can be mae into an anagram. 3 Boune-egree graphs 3.1 A four-regular graph with a large anagram-chromatic number In this section we stuy the number of colors neee to color a boune-egree graph on n vertices so as to avoi all anagrams. The trivial upper boun is n, so we will mainly be intereste in lower bouns for the anagram-chromatic number. Keränen s result implies that graphs of maximum egree two satisfy π α G) 5. It turns out that there are alreay 4-regular graphs G for which π α G) grows rather quickly with the size of the graph. Proposition 3.1. For infinitely many values of n, there exists a 4-regular n-vertex graph H with π α H) n log n. Proof. Note that for each even k 4, there exists a 3-regular k-vertex graph G which is Hamiltonconnecte, which means that any two vertices of G are joine by a Hamilton path. Inee, it can be easily checke that for any m 1, the Cayley graph of C C m+1 with canonical generators is Hamilton-connecte. For a self-containe proof, we refer the reaer to [7]. Let n = k + 1)k. Take k + 1 copies of such G on vertex sets V 1, V,..., V k+1 with V i = k. Furthermore, take a perfect matching M on V 1 V k+1 such that there exists exactly one ege between any two V i an V j, for i j. To see that such a matching exists, enote V i = {v ij : j [k + 1] \ {i}}, an take M = {{v ij, v ji } : 1 i < j k + 1}. Call the resulting graph H. H is 4-regular - any vertex has three ajacent eges belonging to n its copy of G an one ege belonging to M. Suppose that the vertices of H are colore with log n colors. Consier the subsets of form i S V i for any S [k + 1]. There are k+1 such subsets. The coloring of each n i S V i efines a multiset of orer at most n. Given colors, n log the number of such multisets is at most n n = n < k+1. Thus, by the pigeonhole principle, there are two istinct sets S, T [k + 1] such that i S V i an i T V i have the same number of occurrences of each color. The same hols for sets S = S \ T an T = T \ S, which are in aition isjoint. Without loss of generality assume S = {V 1,..., V s } an T = {V s+1,..., V s }. log n 5

6 By the choice of M, we can fin vertices v 1, u 1, v, u,..., v s, u s such that v i, u i V i for i [s], an u i v i+1 are eges in M for i [s 1]. Moreover, we can fin a Hamilton path in each H[V i ] between u i an v i, using Hamilton-connecteness of G. Concatenating these s paths gives us a path in H n which traverses V 1 V V s in orer. This path forms an anagram in H, so π α H) >. log n 3. Ranom regular graphs Let us start with a simple observation which slightly improves the trivial upper boun n for the anagram-chromatic number of a graph. Proposition 3.. Let G be an n-vertex graph with an inepenent set of orer m. Then π α G) n m + 1. Proof. Let S be an inepenent set insie G of orer m. Give each vertex of S the same color, an each vertex of V G)\S its own color. Any path in G contains at least one vertex of V G)\S, so it cannot contain an anagram. This means that our coloring is inee anagram-free. The above boun is essentially optimal for the ranom regular graph G n,. To recapitulate, Theorem 1.3 states that for sufficiently large, with high probability, G n, satisfies 1 ) 105 log n π α G n, ) 1 log ) n. The upper boun is an immeiate consequence of Proposition 3., an the fact that with high probability, G n, contains an inepenent set of orer asymptotic to log n > log n see, for instance, Frieze an Luczak [11]). We will now outline the proof of the lower boun on π α G n, ), which comprises the remainer of the section. Instea of stuying the ranom -regular graph G n,, we will consier the union of two ranom graphs G n,1 an G n, with = 1 +. The asymptotic properties of G n, are contiguous with such a moel see Lemma 9.4 in [14]). Let G 1 = G n,1, G = G n, an c be a given vertex-coloring of G = G 1 G. The first step is to fin two vertex subsets V 1 an V with the same coloring such that G 1 [V 1 ] an G 1 [V ] have goo expansion properties. Then we use the eges of G to exten paths on V 1 an V, eventually builing Hamilton cycles C 1 in G[V 1 ] an C in G[V ]. Finally, we can fin an ege v 1 v G with v i V i an use it to buil a single path S which traverses first the vertices of C 1 an then the vertices of C. The segments S[V 1 ] an S[V ] give an anagram in c. Before proceeing, let us introuce some notation. For a graph G an v V G), we enote the neighborhoo of v in G by N G v). For a vertex set U V G), N G U) = v U N Gv) \ U. The graph inuce on U is G[U], an its ege set is enote by E G U) = EG[U]). For isjoint sets U an T, E G U, T ) is the set of eges with one enpoint in U an one in T. Finally, the corresponing counts are e G U) = E G U) an e G U, V ) = E G U, V ). We enote the uniform probability measure on the space of ranom regular graphs G n, by P, suppressing the inices. All the inequalities below are suppose to hol only for n large enough Ege istribution in the configuration moel In analysing G n,, we pass to the configuration moel of ranom regular graphs. For n even, we take a set of n points partitione into n cells v 1, v,..., v n, each cell containing points. A perfect matching P on [n] inuces a multigraph MP ) in which the cells are regare as vertices an pairs in P as eges. For a fixe egree an P chosen uniformly from the set of perfect matchings P n,, the probability that MP ) is a simple graph is boune away from zero, an 6

7 each simple graph occurs with equal probability. Therefore, if an event hols whp in MP ), then it hols whp even when we conition on the event that MP ) is a simple graph, an therefore it hols whp in G n, for a formal escription of the configuration moel an its basic properties, see, for instance, Chapter 9 of [14]). We use the configuration moel to get a boun on the ege istribution in G n, analogous to the Erős-Rényi moel. The uniform probability measure on P n, is enote by P P. Both inices n an are kept so that each perfect matching P correspons to a unique -regular multigraph MP ). Lemma 3.3. Let V 1 [n], an let B be a set { of pairs of vertices } from V 1. Let E be another set 1 n of pairs of vertices from [n] with E min 4 V 1 B, 0. For a fixe positive integer an P P n, chosen uniformly at ranom, P P [MP ) E an MP ) B = ] ) E e B 5n. n The lemma also hols for more general configurations of B an E, but we state it in the form which is fit for our purpose. The proof is given after Lemma 3.5, which is a boun on the probability that G n, oes not intersect a given set of eges. A crucial ingreient is the following estimate of Alon an Frielan [1], which is a simple corollary of the Brègman boun on the permanent of a 0, 1)-matrix. Theorem 3.4 [1]). Let H be a graph on [N]. Let r 1, r,..., r N be the egrees of the vertices in H. Furthermore, enote r = 1 N N i=1 r i. Then the number of perfect matchings in H is at most N r i!) 1 r i r!) N r. i=1 Lemma 3.5. For each even number N, let F = F N) be a graph on [N] consisting of at least βn eges. Let G N,1 enote a ranom matching on [N]. Then P [G N,1 F = ] e 8β 9 N. Proof. Let PF ) be the set of perfect matchings on [N] which o not intersect our graph F. Since N! the number of perfect matchings on [N] is exactly, we nee to show that N N )! PF ) e 8β 9 N N! N N ).! Consier the complement F of F. The matchings in PF ) are exactly perfect matchings in F. We apply Theorem 3.4 irectly to the graph F with r = N βn, an use Stirling s formula to reach the final result. Inee, we have PF ) N βn)!) P [G N,1 F = φ] N βn)!) 1 1 β) N 1 1 β) ) 1 β)n = O N e )! N N! ) N e ) N N ) = O N e βn. 7

8 Here we use the fact that N! N )! N = Θ1) ) N N e, as well as the inequality 1 β e β. Absorbing the error term into the constant, we get, for N large enough, P [G N,1 F = ] e 8β 9 N. Proof of Lemma 3.3. We will restate the event {MP ) E an MP ) B = } in terms of P, rather than MP ). For a matching M ) [n], we enote the inuce multigraph on V = {v i } i [n] by MM). To save on notation, we write MM) for both the graph an its ege set. Conversely, if e = {v i, v j } is a pair of vertices from V, we enote its corresponing pairs in ) [n] by ẽ = {{x, y} : x v i, y v j }. Finally, for a set E V ), we put Ẽ = e E ẽ. Assume that MP ) E. Then we can fin a matching M P such that M = E = m an MM) = E. Conitioning over the possible choices of M, we have P P [MP ) E MP ) B = ] P P [P M] P P [P B ] = P M. M We boun the two probabilities separately. Fix a choice M = {{x i, y i } : i [m]}, an let W = [n] \ {x 1,..., x m, y 1,..., y m }. Claim 1. P P [P M] n m) m To show this, we just count perfect matchings. The total number of perfect matchings P is n m)!. The points from W can be paire in ways. Altogether, using Stirling s m n)! n )! n formula, we get n n m)! P P [M P ] n m)! ) n! n)! n m)! m ) n ) m n m n m n =1 + o1)) n e n m =1 + o1)) 1 m ) n ) m e n n m n m) m. Here we use the fact that since 1 x e x, we have 1 m Claim. For B = βn, P P [P B ] = P M e β 5 n. n ) n ) n n m e m. Let W be as before, an enote N = W. Using the assumption m n 0, we get N = n m 9n 10. Let B W = B[W ], that is, the set of pairs containe in W which woul inuce B. By putting the matching M asie, we have lost some pairs from B, namely those touching the vertices of M. Each vertex of M is containe in at most V 1 pairs from B, so the hypothesis m 1 4 V 1 B implies B W B m V 1 = βn 1 m V ) 1 βn 1 βn 1 βn. A ranom matching P conitione on P M correspons to a ranom matching on W, i.e. an element of G N,1. Hence we can apply Lemma 3.5 with B W 1 βn, an N = W 9n 10. P P [P B W = P M] P [EG N,1 ) B W = ] e βn e 5 βn. e ) m 8

9 Claim 1 an hol for any choice of the matching M with MM) = E. Putting them together, an using the fact that there are at most m such matchings M, we get Using m n 0, P P [MP ) E an MP ) B = ] m P P [MP ) E an MP ) B = ] m n m) m β e 5 n. ) m e β 5 n = n ) m e β 5 n. n 3.. Expansion properties an Pósa rotations Recall that we will be working with the union of ranom graphs G n,1 an G n,. First we focus on the expansion properties of G 1 = G n,1, which will allow us to o rotations in G 1. The aim is to ientify large sets of vertex pairs, calle boosters, which coul increase the length of the longest path in G 1. Hence all the lemmas in this section will later be applie with replace by 1. The following lemma says that eges in G n,1 are uniformly istribute. Lemma 3.6. For sufficiently large, with high probability G n, has the following two properties: 30 log P1) any vertex set U with U n satisfies e G U) 100 U log ; 10 log P) for any two isjoint vertex subsets T an U with T n an U have 100 log n, we e G T, U) T U 0n. 1) Proof. We prove Lemma 3.6 for G sample accoring to the configuration moel, i.e. take G = MP ), where P is a ranom element of P n,. We start with P). Take vertex sets T an U in G with T = t an U = u. We nee to boun the probability of the event { D T,U = e G T, U) < } 0n tu. For a fixe set of m eges E with m 0n tu, the probability that E GT, U) = E is at most ) m e 5n tu m). n This boun is a irect application of Lemma 3.3 to the ege set E an its bipartite complement T U) \ E. Taking the union boun over all sets E, we get P P [D T,U ] 0n tu m=0 tu m ) n ) m e 5n tu m) 0n tu m=0 etu m ) m e 5n tu m). n The summan is increasing in m, so we boun it using the largest term, m = M = 0n tu. etu 0n P P [D T,U ] M tu n = Me tu n n) e tu 8n. ) 0n tu e 5n tu M) = M 40e) tu 0n e 5n tu M) 9

10 Finally, we take the union boun over all sets T an U of orer at least t 0 = u 0 = n respectively. 100 log P P [G violates P)] n n t=t 0 u=u 0 n t ) n u ) e tu 8n. We use the boun n t) t = e t log vali for t t 0 an large enough. For t 10 log P P [G violates P)] n, we get u log tu 10n n t=t 0 n 8n. u=u 0 e t log +u log e tu 10 log n an 100 log. Similarly, since u n, it hols that t log tu 100n. P P [G violates P)] On )e ) 1 n log = o1). We euce P1) from the following more general statement. ) A ea Claim 3. Fix the constants A 1 an A satisfying 1 A e. Then with high probability, A1 log any vertex set U V G) with U n satisfies e G U) A U log. Introucing A 1 = 30 an A = 100, which inee satisfy 30e 100) 100 e, gives exactly P1). To prove the claim, we fix a set U of orer u, an use Lemma 3.3 to establish that the probability of some A u log eges occurring in U is at most u ) / A u log n ) Au log eu A log n ) Au log. Let D u enote the event that some subset U with U = u spans more than A u log eges. We have ) ) [ Au log ) ] A log u n eu ne eu P P [D u ]. ) u A n log u A n log The term in square brackets is increasing in u, so for u [ ) ] A log u [ e ea1 P P [D u ] A 1 log A A1 log n, e log e A 1 log ] u < u. ) A ea Here we use the conition 1 A e. For u n we use ) to get a stronger boun P P [D u ] O n 1 1 A log ))) u < n 1, vali for large. Putting the two bouns together, n P P [G violates P1)] P P [D u ] + u=1 A 1 log n u= n P P [D u ] n n 1 + A 1 log n u= n u = o1), completing the proof of Claim 3. To recapitulate, applying the claim for A 1 = 30 an A = 100 gives that G = MP ) satisfies P1) with high probability. Since the ranom graph G n, is contiguous to G, we conclue that for large enough, G n, satisfies P1) an P). 10

11 The next step is to buil subsets of [n] which will later give us the require anagram. In everything that follows, take α = Given a -regular graph G, we say that a subset V 1 [n] is G-ense if α log n V 1 α log α 4 n, an any vertex v V 1 has at least 160 log neighbors in V 1. Lemma 3.7. Suppose we are given a -regular graph ) G on [n] with properties P1) an P), an a vertex coloring c : [n] C with C = 1 α log n colors. For sufficiently large an n, there exist two isjoint G-ense sets of vertices V 1, V [n] which have the same coloring. ) Proof. Let c be a coloring of the vertices of G into 1 α log n colors. Claim 4. There exists a subset Z V G) satisfying α log n Z α log n such that each color appears in Z an even number of times, an for all v Z, N G v) Z α 40 log. We construct Z using the following algorithm. Let V G) = [n], an enote δ = α 40 log. ) Let X contain one vertex from each color class with an o number of colors, so X 1 α log n. We ) assume X = 1 α log n, by iscaring more pairs of vertices of the same color if necessary. Furthermore, set R 0 = ˆR 0 =. Note that from this step onwars, all color classes in [n] \ X R i ˆR i ) will have even orer. For i 0, we form R i+1 := R i {v}, ˆR i+1 = ˆR i {w}, where v is the smallest vertex with fewer than δ neighbors in [n] \ X R i ˆR i ), an w is the smallest vertex with cv) = cw). When there are no such vertices v, set Z = [n] \ X R i ˆR i ) an terminate 10 log the algorithm. We claim that this occurs after at most n steps. Suppose that it is not the case an G satisfies P1) an P), but the algorithm continues 10 log beyon t = n steps. Look at the sets R t an Z t = [n] \ X R t ˆR t ). Each vertex from R t has fewer than δ neighbors in Z t, so e G R t, Z t ) < δ R t = α 40 R t log. 10 log On the other han, since R t = t = P) gives e G R t, Z t ) R t Z t 0n R t α log n an Z t = α log n t α log n, the property 0n = α 40 R t log. We reache a contraiction, so inee we have the esire set Z with Z α log n. We show the existence of the require partition of Z into sets V 1 an V using a probabilistic argument. Partition each color class c 1 a) into orere pairs arbitrarily, an enote the collection of pairs by Q. For each pair v, w) Q, ranomly an inepenently put v into V 1 an w into V or vice versa, with probability 1. This guarantees that V 1 an V have the same coloring. Claim 5. With positive probability, the partition satisfies eg G[Vi]v) an i {1, }. α 160 log for all v V i We use the Local Lemma. Fix a vertex v, wlog v V 1. Let B v be the event that fewer than α 160 log neighbors of v in Z have ene up in V 1. Let S be a set of α 40 log neighbors of v in Z, an let T S be the set of vertices whose match accoring to Q oes not lie in S. Note that S contains exactly y = 1 S T ) pairs of Q. If B v occurs, then S V 1 α 160 log, an therefore 1 S S V 1 α 160 log. But this implies 1 T T V 1 = 1 S y) S V 1 y) α log

12 T V 1 is a ranom variable with istribution B T, ) 1, so Chernoff bouns as state in [14, Remark.5]) give [ 1 P [B v ] = P T T V 1 α ] 160 log e T α 160 log ) e 3 log = 3. Here we use T α 40 log an α = 105. Two events B v an B w are epenent only if v an w share a neighbor, or if some two neighbors of v an w are paire. In such a epenency graph, the event B v has egree at most. Since for sufficiently large, e + 1) 3 < 1, the Lovász Local Lemma grants that there is a splitting avoiing all the ba events B v. This is exactly the require splitting. It conclues the proof of Claim 5 an Lemma 3.7. We say a graph G is a p-expaner if it is connecte, an N G U) U for U p. Lemma 3.8. Let G be a -regular graph on vertex set )[n] satisfying P1) an P), an let V 1 [n] be a G-ense subset of [n]. Then G[V 1 ] is a -expaner. Proof. Denote H = G[V 1 ]. To show expansion, suppose for the sake of contraiction that 10 log N H U) < U, an first assume that U n. We can apply P1) to T = U N H U) V 1, 30 log using the assumption T n. This gives eg[t ]) 100 T log. Counting all the eges with an enpoint in U, which certainly lie insie T, we get 1 α 160 U log eg[t ]). The two inequalities imply T α U = 3000 > 3 U, which contraicts our assumption. Seconly, in case V1 10 log 4 U n an N G U) < U, we have V 1 \ U N H U)) V log n. This puts us in the position to apply P) an claim that G contains eges between U an V 1 \ U N H U)), contraicting the efinition of N H U). Hence sets of orer up to V1 4 inee expan in H. Finally, assume that H is not connecte. Let its smallest component be spanne by S V 1, i.e. S V1 an N H S) =. We alreay showe that certainly S > V1 4. But then the fact that E G S, V 1 \ S) = contraicts P). V1 4 We will use these expansion properties to buil long paths an ultimately a Hamilton cycle in G[V i ]. Our approach is base on the rotation-extension technique originally evelope by Pósa. Given a graph G, enote the length number of eges) of the longest path in G by lg). We say that a non-ege {u, v} / EG) is a booster with respect to G if G + {u, v} is Hamiltonian or lg + {u, v}) > lg). We enote the set of boosters in G by BG). Pósa s rotation technique guarantees that there exist plenty of boosters in G see, for instance, Corollary.10 from [16]). Lemma 3.9. Let p be a positive integer. Let G = V, E) be a p-expaner. Then BG) p Using G to hit boosters in G 1 Now we move on to G = G n,. Recall that we woul like to a its eges to G 1 [V 1 ] an complete a cycle on V 1. However, we have to argue carefully because the choice of a G 1 -ense set V 1 will epen on the given vertex coloring. Lemma Let G 1 be a 1 -regular graph on [n] with properties P1) an P), for sufficiently large 1, an let With high probability, G = G n, has the property that for any G 1 -ense subset V 1 [n], G 1 G )[V 1 ] is Hamiltonian. 1

13 The proof of Lemma 3.10 consists of two parts. First we ientify a eterministic property that is sufficient to make G 1 G )[V 1 ] Hamiltonian, an then, using the configuration moel, we show that G n, possesses this property with high probability. Lemma Let H 1 an H be graphs on vertex set V 1. Suppose that for any ege set E EH ) with E V 1, H 1 E is Hamiltonian, or BH 1 E ) EH ). Then the graph H 1 H is Hamiltonian. Proof. We will buil a subset of EH ) such that its aition to H 1 creates a Hamiltonian graph. Start with E 0 =. Assume that E i is a subset of i eges in EH ). If the graph H 1 E i is Hamiltonian, we are one. Otherwise, by hypothesis, EH ) BH 1 E i ) contains an ege e, so we set E i+1 = E i {e}. In each step i, we have lh 1 E i+1 ) > lh 1 E i ), so the process terminates after at most V 1 steps, with a Hamiltonian graph H 1 E i. Lemma 3.1. Let G 1 be a 1 -regular graph on V with properties P1) an P), where V = n an 1 is sufficiently large. Let G = G n, for We say that G A G1 or A G1 occurs) if there exists a G 1 -ense subset V 1 V, an an ege set E V 1 ), E V1, such that G contains E an oes not intersect BG 1 + E)[V 1 ]). It hols that P [A G1 ] = o1). Proof. We will prove the claim for G sample accoring to the configuration moel, which is contiguous to the uniform moel G n,. This allows us to apply Lemma 3.3, which gives us a precise estimate on the probability of non-)occurrence of certain ege sets. Let P P n, be chosen uniformly at ranom. We will actually boun the probability that the inuce multigraph MP ) is in A G1, enote by P P [A G1 ], with a slight abuse of notation for not renaming the event A G1 itself. Fix a G 1 -ense subset V 1 V with V 1 = ξn, an E V 1 ) with E = m V1. ) Recall that α log 1 since V 1 is G 1 -ense, ξn 4 1 n = 105 log n. Note that the graph G 1 +E is a V1 4 -expaner, so we apply Lemma 3.9, which says that the set of boosters B = BG 1 + E)[V 1 ]) contains at least 5 ξ n eges. Applying Lemma 3.3 to E an B, we get ) m P P [MP ) E an MP ) B = ] e 5 n 5 ξ n. Now we can take the union boun over all choices of E an V 1. We cruely boun the number of ways to choose V 1 by n n ξn). The term ξ n m n P P [A G1 ] n ξn ) n ) ξn ξ n m m=1 ) ) m e 1 5 n 4 ξ n. ) m ) m eξ n m is increasing in m in the given range, an hence ) ξn P P [A G1 ] n ξn eξ 1 eξ e ξ. Introucing the value of ξ, the term in brackets is at most e e 105 log e

14 For [ 1 150, ] the term above is upper-boune by , so as claime. P P [A G1 ] ξn e Ωξn) = e Ωξn), Proof of Lemma Since G 1 satisfies P1) an P), for G = G n, it hols with high probability that G / A G1. Hence, given a G 1 -ense set V 1 V we can apply Lemma 3.11 to G 1 [V 1 ] an G [V 1 ] to fin a Hamilton cycle in G 1 G )[V 1 ], as require. We are now reay to put together the proof. Proof of Theorem 1.3. For a given, set = 300 an 1 =. Let be large enough so that , an for Lemma 3.6 an Lemma 3.10 to hol. Moreover, by choosing to be even, we ensure that whenever n is even so that G n, is non-empty), n 1 an n are also even. Generate G 1 = G n,1 an G = G n, on vertex set V. Suppose that G 1 has properties P1) an P), an G satisfies the conclusion of Lemma By Lemma 3.6 an Lemma) 3.10, this α log 1 hols with high probability. We claim that in this case π α G 1 G ) > 1 1 n, where [ ) ] α = 10 5 α log 1 as before. Let c : V 1 1 n be a given coloring. We first use Lemma 3.7 to fin G 1 -ense sets V 1, V V with the same coloring. Then by Lemma 3.10, we conclue that the graphs G 1 G )[V i ] are Hamiltonian, for i = 1,. Let C 1 an C be Hamilton cycles on V 1 an V. G 1 satisfies P), which implies that it contains an ege between some two vertices v 1 V 1 an v V. We form the require path S by going along C 1, using v 1 v to skip to V an then going along C. The segments S[V 1 ] an S[V ] give an anagram in c, as require. It remains to express the boun in terms of. Note that 1 lies between 1 100) 1 an, so Hence π α G 1 G ) > probability. α log log 105 log log. ) n, an by contiguity, the same hols for G n, with high 4 Concluing Remarks In this paper we stuie anagram-free colorings of graphs, an showe that there are very sparse graphs in which anagrams cannot be avoie unless we basically give each vertex a separate color. Our research suggests several interesting questions, some of which we mention here. The first question concerns the lower boun on the anagram-chromatic number for trees. Is there a family of trees T n) on n vertices for which π α T n)) ε log n for some positive constant ε > 0? We remark that this is the case for the analogous problem of fining the anagram-chromatic inex of a tree. Inee, a simple counting argument cf. Proposition.3) shows that if instea of vertex colorings, we color eges of a graph, then to avoi anagrams in the complete binary tree of epth h, we nee to use at least 1 4h colors. In estimating the anagram-chromatic number of planar graphs we relie only on the fact that they have small separators. It woul be interesting to know a better lower boun on π α G) for such graphs. In particular, we woner if there exists a family H n of planar graphs on n vertices such that π α H n ) n ɛ for some absolute constant ɛ > 0. 14

15 Let Gn, ) enote the graph with the largest anagram-chromatic number among all graphs G on n vertices ) with G). Our main result shows that if is large enough then π α Gn, )) n 1 C log, while for = 4 we can only provie a construction which gives π α Gn, 4)) n log n. We believe that there exist cubic graphs for which the anagram-chromatic number grows linearly with the orer of the graph. It woul be nice to know how fast the function f) = 1 lim sup n π α Gn, ))/n ecreases with. Let us recall that from Proposition 3. an Theorem 1.3 it follows that 1 log ) f) = O. We o not know the correct boun, but we have goo reasons to believe that the upper boun can be improve. Inee, consier a graph which is a union of n/ cliques of size / an a ranom n-vertex /-regular graph. We think that using such a construction one can show that f) log ) 1/+o1) / but the proof looks quite involve an woul probably not be worth the effort since it is by no means clear whether it woul give the right orer of f). Finally, Lemma 3.10 motivates questions on Hamiltonicity of small inuce subgraphs of G n,. Pursuing our proof outline, we can prove the following. Claim 6. There is a constant C such that with high probability, G = G n, has the following log property. For any vertex set V 1 [n] of orer at least C n, if the graph H = G[V 1] has minimum egree at least 10n V 1, then H is Hamiltonian. To see this, take G = G n, for = 0 C log, an G1 = G n,1 for 1 =. Consier G = G n,1 G n,. Let V 1 an H = G[V 1 ] satisfy the hypothesis, an enote V 1 = ξn with log ξ C. Since the graph G[V 1] has minimum egree at least ξ 10, an we ensure ξ 0, G 1 [V 1 ] has minimum egree at least ξ 0. This guarantees that G 1[V 1 ], as well as any graph on V 1 containing it, has Θξ n ) boosters. On the other han, the conition e Ωξ) < 1 implies that G hits those boosters with high probability see the calculation in Lemma 3.1). Hence G[V 1 ] is Hamiltonian for any such V 1, an by contiguity, G n, satisfies the claim. The above iscussion leas to the natural question, what is the smallest possible lower boun log on V 1 in Claim 6? Note that V 1 = C n is the best we can get from our approach. Namely, )) 1 the above-mentione conitions require Ω ξ log 1 ξ = ξ 0, i.e ξ = Ω log ). n We also give a lower boun on V 1. Using inepenent sets in G n,, one can fin an inuce unbalance bipartite subgraph of orer log n with high minimum egree, which is obviously non- Hamiltonian. This observation implies that we nee at least V 1 log n. We woner how tight this estimate is. Note ae in proof. After this paper was written an submitte, we learne that Wilson an Woo [0] inepenently stuie graph colorings avoiing abelian squares, which are equivalent to anagrams. In particular, our Proposition 1.1, which shows that complete binary trees have an unboune anagram-free chromatic number, answers one of their questions. Acknowlegement. This work was carrie out when the secon author visite the Institute for Mathematical Research FIM) of ETH Zürich. He woul like to thank FIM for the hospitality an for creating a stimulating research environment. We woul also like to thank the referee for helpful comments. 15

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