Gallai-Ramsey numbers of odd cycles
Zhao Wang, Yaping Mao, Colton Magnant, Ingo Sciermeyer, Jinyu Zou
aa r X i v : . [ m a t h . C O ] A ug Gallai-Ramsey numbers of odd cycles ∗ Zhao Wang † , Yaping Mao ‡ , Colton Magnant § ,Ingo Schiermeyer ¶ , Jinyu Zou k .August 29, 2018 Abstract
Given two graphs G and H and a positive integer k , the k -colorGallai-Ramsey number, denoted by gr k ( G : H ), is the minimum in-teger N such that for all n ≥ N , every k -coloring of the edges of K n contains either a rainbow copy of G or a monochromatic copy of H .We prove that gr k ( K : C ℓ +1 ) = ℓ · k + 1 for all k ≥ ℓ ≥ All graphs considered in this work are simple, with no loops or multiple edges.By a coloring of a graph, we mean a coloring of the edges of the graph. A ∗ Supported by the National Science Foundation of China (Nos. 11601254, 11551001,11161037, 61763041, 11661068, and 11461054) and the Science Found of Qinghai Province(Nos. 2016-ZJ-948Q, and 2014-ZJ-907) and the Qinghai Key Laboratory of Internet ofThings Project (2017-ZJ-Y21). † College of Science, China Jiliang University, Hangzhou 310018, China. [email protected] ‡ School of Mathematics and Statistis, Qinghai Normal University, Xining, Qinghai810008, China. [email protected] § Department of Mathematics, Clayton State University, Morrow, GA, 30260, USA. [email protected] ¶ Technische Universit¨at Bergakademie Freiberg, Institut f¨ur Diskrete Mathematik undAlgebra, 09596 Freiberg, Germany.
[email protected] k School of Computer, Qinghai Normal University, Xining, Qinghai 810008, China. [email protected] rainbow if every edge receives a distinct color. Allstandard notation used here comes from [5].Given two graphs G and H and a positive integer k , the k -color Gallai-Ramsey number, denoted by gr k ( G : H ), is the minimum integer N suchthat for all n ≥ N , every k -coloring of K n contains either a rainbow copy of G or a monochromatic copy of H .The theory of Gallai-Ramsey numbers has grown by leaps and boundsin recent years, especially for the case where G = K . Many individualgraphs have been solved for any number of colors but classes of graphs havegenerally only been bounded from above and below. We refer the interestedreader to the survey [11] (with an updated version at [12]) for more generalinformation. In particular, the following has been shown for odd cycles. Theorem 1 ([10, 16]) . Given integers n ≥ and k ≥ , n k + 1 ≤ gr k ( K : C n +1 ) ≤ (2 k +3 − n log n. For the triangle C = K , several authors obtained the following result. Theorem 2 ([1, 6, 14]) . For k ≥ , gr k ( K : K ) = ( k/ + 1 if k is even, · ( k − / + 1 if k is odd . For the C , Fujita and Magnant obtained the following. Theorem 3 ([10]) . For any positive integer k ≥ , we have gr k ( K : C ) = 2 k +1 + 1 . More recently, other authors have extended these results to longer oddcycles in a sequence of papers.
Theorem 4 ([2, 3]) . For integers ℓ ≤ and k ≥ , we have gr k ( K : C ℓ +1 ) = ℓ · k + 1 . Based on these results, it has been widely believed that the lower boundof Theorem 1 is indeed the correct value. In this paper, we confirm thatbelief with the following, our main result.
Theorem 5.
For integers ℓ ≥ and k ≥ , we have gr k ( K : C ℓ +1 ) = ℓ · k + 1 . Preliminaries
We begin this section with the fundamental tool in the study of coloredcomplete graphs containing no rainbow triangle. Here a colored completegraph is called
Gallai colored if it contains no rainbow triangle.
Theorem 6 ([4, 13, 15]) . In any Gallai colored complete graph, there existsa partition of the vertices (called a
Gallai partition ) such that there are atmost two colors on the edges between the parts and only one color on edgesbetween each pair of parts.
We will use this result as a matter of fact. Since a subgraph consistingof an arbitrarily selected vertex from each part of a Gallai partition is a 2-colored complete graph, we call this subgraph the reduced graph of the Gallaipartition.
Lemma 1.
Let k ≥ , H be a graph with | H | = m , and let G be a Gallaicoloring of the complete graph K n containing no monochromatic copy of H .If G = A ∪ B ∪ B ∪ · · · ∪ B k − where A uses at most k colors (say from [ k ] ), | B i | ≤ m − for all i , and all edges between A and B i have color i , then n ≤ gr k ( K : H ) − . Note that this lemma uses the assumed structure to provide a bound on | G | even if G itself uses more than k colors. Proof.
For i = j , all edges between B i and B j must have either color i orcolor j to avoid a rainbow triangle. Since | B i | ≤ m − < | H | , changingall edges within B i that are not a color in [ k −
1] to color k cannot createa monochromatic copy of H . The result of this modification is a k -coloredcopy of K n with no rainbow triangle and no monochromatic copy of H so n ≤ gr k ( K : H ) − Theorem 7 ([7]) . Let G be a graph of order n ≥ . If δ ( G ) ≥ n , then G ishamiltonian. Lemma 2.
If there are at least ℓ + 1 vertices in a Gallai colored completegraph G with only one color appearing on edges between parts of the Gallaipartition and all parts of order at most ℓ , then G contains a monochromaticcopy of C ℓ +1 . roof. Let blue be the color of the edges between parts. Since each partof the Gallai partition has order at most ℓ , every vertex has degree at least2 ℓ − ( ℓ − ≥ ℓ + 1 in blue. By Theorem 7, G contains a blue copy of C ℓ +1 .Next a helpful result on the existence of paths. Theorem 8 ([8]) . Let G be a graph on n vertices and let k ≥ be an integer.If e ( G ) > k − n , then G contains a path with k edges (i.e. a copy of P k +1 ). We use Theorem 8 to prove the following colored version. Let d R ( v ) and d B ( v ) denote the red and blue degrees of the vertex v respectively. Lemma 3.
Let G be a graph of order n with edges colored by red and blue.If for every vertex v ∈ V ( G ) and for some non-negative integers a and b with a + b ≥ , we have d R ( v ) + d B ( v ) ≥ a + b − , then G contains either a redcopy of P a or a blue copy of P b .Proof. Let ¯ d ( G ), ¯ d R ( G ), and ¯ d B ( G ) denote the average degree, average reddegree, and average blue degree of G respectively. Then ¯ d ( G ) = ¯ d R ( G ) +¯ d B ( G ). Since d R ( v ) + d B ( v ) ≥ a + b − v ∈ V ( G ), we have¯ d R ( G ) + ¯ d B ( G ) ≥ a + b − d R ( G ) > a − d B ( G ) > b − d R ( G ) > a −
2, then there are more than n ( a − red edges in G , so byTheorem 8, G contains a red copy of P a . On the other hand, if ¯ d B ( G ) > b − n ( b − blue edges in G , so by Theorem 8, G containsa blue copy of P b , completing the proof.Finally, we state some other known results that will be used in the proof. Theorem 9 ([9, 17, 18]) . R ( C m , C n ) = n − for ≤ m ≤ n , m odd, ( m, n ) = (3 , , n − m/ for ≤ m ≤ n , m and n even, ( m, n ) = (4 , , max { n − m/ , m − } for ≤ m < n , m even and n odd. Theorem 10 ([10, 16]) . Given integers n ≥ and k ≥ , ( n − k + n + 1 ≤ gr k ( K : C n ) ≤ ( n − k + 3 n.
4n particular, we mostly use the immediate corollary of Theorem 10 that gr ( K : C ℓ − ) ≤ ℓ − Proof.
Let ℓ ≥ G be a Gallai colored complete graph K n containingno monochromatic copy of C ℓ +1 . Suppose, for a contradiction, that n = ℓ · k + 1 . This proof is by induction on k . Since the cases k = 1 and k = 2 are eithertrivial or follow from Theorem 9 respectively, we assume k ≥ T be a maximal set of vertices T = T ∪ T ∪· · ·∪ T k where each subset T i has all edges to G \ T in color i and | G \ T | ≥ ℓ constructed iteratively byadding at most 2 ℓ vertices at a time, with at most ℓ vertices being added toeach of two sets T i at a time. We first claim that | T i | is small for all i . Claim 1.
For each i with ≤ i ≤ k , we have | T i | ≤ ℓ and furthermore, T i = ∅ for some i .Proof. By the iterative definition of T , we may assume that this is the firststep in the iterative construction where T violates one of these assumptions.That is, assume that | T i | ≤ ℓ for all i and either • at most two sets T i and T j have | T i | , | T j | > ℓ , or • no set is empty and at most one set T i has | T i | > ℓ .In either case, we have | T | ≤ ( k + 1) ℓ .We first show that T i = ∅ for some i so suppose the latter item above. If k ≥
4, then | G \ T | ≥ [ ℓ k + 1] − [( k + 1) ℓ ] ≥ ( ℓ − k + 3 ℓ. By Theorem 10, there is a monochromatic copy of C ℓ contained within G \ T .Since T i = ∅ for all i , this cycle can easily be extended to a monochromaticcopy of C ℓ +1 , for a contradiction. We may therefore assume k = 3. If any set T i has | T i | ≥ ℓ + 1, then G \ T contains no edges of color i and | G \ T | ≥ ℓ + 1so there is a monochromatic copy of C ℓ +1 in one of the other colors. We maytherefore assume that | T i | ≤ ℓ for all i ∈ { , , } and so | G \ T | ≥ ℓ + 1.5ithin G ′ = G \ T , consider a Gallai partition and let r be the numberof parts in this partition of order at least ℓ and suppose red and blue are thecolors appearing on edges between the parts of the partition. If r ≥
2, thentwo of these large parts, say with red edges between, along with any vertexof T , produces a red copy of C ℓ +1 . This means we may immediately assume r ≤
1. If r = 1 say with H as the large part, then let G R and G B be thesets of vertices in G ′ \ H H .If | G R ∪ T | ≥ ℓ + 1 or | G B ∪ T | ≥ ℓ + 1, then there is a red or blue copy of C ℓ +1 respectively, meaning that | G R ∪ T | , | G B ∪ T | ≤ ℓ . Then G R and G B can be added to T to produce a larger set than T with the same properties,contradicting the maximality of T . We may therefore assume r = 0. Within G ′ , every vertex has degree at least | G ′ | − ( ℓ −
1) when restricted to the redand blue edges. Since | G ′ | ≥ ℓ + 1, we have d ( v ) ≥ (5 ℓ + 1) − ( ℓ −
1) = 4 ℓ + 2 > [2 ℓ ] + [2 ℓ ] − ℓ . Thispath along with a vertex of T with appropriately colored edges producesa monochromatic copy of C ℓ +1 , a contradiction, completing the proof that T i = ∅ for some i .Next we show that | T i | ≤ ℓ for all i . First suppose there is only oneset, T i with | T i | > ℓ . Any edge of color i within G \ T would produce amonochromatic copy of C ℓ +1 so G \ T contains no edge of color i . Then byLemma 1, we have | G \ T i | ≤ gr k − ( K : C ℓ +1 ) − n = | T i | + | G \ T i |≤ ℓ + gr k − ( K : C ℓ +1 ) − < gr k ( K : C ℓ +1 ) , a contradiction.Finally suppose there are two sets T i and T j with ℓ + 1 ≤ | T i | ≤ ℓ and ℓ + 1 ≤ | T i | ≤ ℓ so G \ T contains no edge of color i or j . Then by Lemma 1,we have | G \ ( T i ∪ T j ) | ≤ gr k − ( K : C ℓ +1 ) − n = | T i | + | T j | + | G \ ( T i ∪ T j ) |≤ ℓ + 2 ℓ + gr k − ( K : C ℓ +1 ) − < gr k ( K : C ℓ +1 ) , a contradiction, completing the proof of Claim 1.6onsider a Gallai partition of G \ T with the minimum number of parts t and let H , . . . , H t be the parts of the partition where | H | ≥ | H | ≥ · · · ≥| H t | , say with | H | = s . Certainly s ≥ G \ T would be a2-coloring of order at least n ≥ . ℓ , producing the desired monochromaticcycle by Theorem 9. Without loss of generality, suppose red (color 1) and blue(color 2) are the two colors appearing on edges between parts in the Gallaipartition. Note that the choice of t to be minimum implies that both redand blue are either connected or absent in the reduced graph so in particular,every part has red edges to at least one other part and blue edges to at leastone other part. Claim 2.
We may assume | H | ≤ ℓ .Proof. Suppose not, so | H | ≥ ℓ + 1. Let r be the number of parts with H i with | H i | ≥ ℓ so | H | ≥ | H | ≥ · · · ≥ | H r | ≥ ℓ + 1 and call these parts large . Certainly r ≤ C ℓ +1 .We break the remainder of the proof into cases based on the value of r . Case 1. r = 5 . Since there can be no monochromatic triangle in the reduced graph re-stricted to the 5 large parts, the reduced graph must be the unique 2-coloringof K containing two complementary copies of C . Either one of these cyclesyields a monochromatic copy of C ℓ +1 , completing the proof of this case. Case 2. r = 4 . If t = 4, then there are no vertices in G \ ( H ∪ · · · ∪ H ), so since | H i | ≥ ℓ + 1, no part can contain any red or blue edges as such an edge wouldyield a monochromatic copy of C ℓ +1 . This also means that T = T = ∅ .Define H ′ = H ∪ T , H ′ = H ∪ T ∪ T ∪ · · · ∪ T k and let H ′ i = H i for i ≥ | H ′ i | ≤ gr k − ( K : C ℓ +1 ) − i ∈ { , } and by Lemma 1, we have | H ′ i | ≤ gr k − ( K : C ℓ +1 ) − i ∈ { , } . We therefore get that n = X i =1 | H ′ i |≤ gr k − ( K : C ℓ +1 ) − < gr k ( K : C ℓ +1 ) , t ≥ v in G \ ( H ∪ · · · ∪ H ). The reduced graph restricted to the large parts couldeither be two complementary copies of P or a matching in one color with allother edges between parts in the other color.First suppose the reduced graph consists of a red matching, say H H and H H with all blue edges otherwise in between the parts. In order toavoid creating a red copy of C ℓ +1 using edges between H and H along with v , the vertex v must have all blue edges to one of H or H and similarly toone of H or H , say H and H . Then the blue edges between H and H along with v yield a blue copy of C ℓ +1 , for a contradiction.Finally suppose the reduced graph is two complementary copies of P ,say H H H H in red and the remaining edges in blue. To avoid creating ablue copy of C ℓ +1 using the blue edges between H and H along with v , thevertex v must have all red edges to either H or H , suppose H . In orderto avoid creating a red copy of C ℓ +1 using the red edges between H and H along with v , the vertex v must have all blue edges to H . In order toavoid creating a blue copy of C ℓ +1 using the blue edges between H and H along with v , the vertex v must have all red edges to H . Then vH H H H v induces a red copy of C in the reduced graph, yielding a red copy of C ℓ +1 in G for a contradiction, completing the proof in this case. Case 3. r = 3 . The cycle among the 3 large parts cannot be monochromatic so suppose,without loss of generality, that the edges from H to H are blue and allother edges between these parts are red. Let A be the set of vertices in G \ ( H ∪ H ∪ H ) with blue edges to H and H and red edges to H , let B be the set with red edges to H and H and blue edges to H , and let C be theset with blue edges to H and H and red edges to H . Note that any or allof these sets of vertices may be empty and G = H ∪ H ∪ H ∪ A ∪ B ∪ C ∪ T .Also note that T = T = ∅ .Both A and C cannot be nonempty since the blue edges between H and H along with a blue path of the form H CH AH yields a blue copy of C ℓ +1 . Without loss of generality, suppose C = ∅ . The parts H and H eachcontain no red or blue edges and the set H ∪ A ∪ B contains no red edges.Let H ′ = H ∪ T and let H ′ = H ∪ T ∪ T ∪ · · · ∪ T k so by Lemma 1, we8ave | H ′ i | ≤ gr k − ( K : C ℓ +1 ) − i ∈ { , } . This yields n = | H ′ | + | H ′ | + | H ∪ A ∪ B |≤ gr k − ( K : C ℓ +1 ) −
1] + [ gr k − ( K : C ℓ +1 ) − < gr k ( K : C ℓ +1 ) , a contradiction, completing the proof of this case. Case 4. r = 2 . Without loss of generality, suppose the edges between H and H are red.Let A be the set of vertices in G \ ( H ∪ H ) with red edges to H and blueedges to H , let B be the set with blue edges to H ∪ H , and let C be theset with blue edges to H and red edges to H . No vertex b ∈ B can havered edges to both a ∈ A and c ∈ C since the red edges between H and H along with a red path of the form H abcH would yield a red copy of C ℓ +1 in G . Note that T = ∅ . If | A ∪ B | ≥ ℓ + 1 (or similarly | B ∪ C | ≥ ℓ + 1),then neither A ∪ B nor H (respectively neither B ∪ C nor H ) can containany blue edges.If both | A ∪ B | ≥ ℓ + 1 and | B ∪ C | ≥ ℓ + 1, then either B = ∅ or | B | > ℓ and {| A | , | C |} = { , } . Assuming the first case, with | A | ≥ ℓ + 1and | C | ≥ ℓ + 1, A and C must also contain no red edges. By the samecalculations as in Case 2, this is a contradiction. In the latter case, we have T = T = ∅ and the calculations are again easy.Next suppose one of | A ∪ B | or | B ∪ C | is at least ℓ + 1 and the other isnot, say | A ∪ B | ≥ ℓ + 1. Suppose further that | A | ≥ ℓ + 1 so A contains nored or blue edges. By Lemma 1, we get n = | H ∪ B ∪ C ∪ T | + | H ∪ T k | + | A ∪ T ∪ · · · ∪ T k − |≤ [ gr k − ( K : C ℓ +1 ) −
1] + 2[ gr k − ( K : C ℓ +1 ) − < gr k ( K : C ℓ +1 ) , a contradiction, meaning that we may assume | A | ≤ ℓ . Then again usingLemma 1, we get n = | H ∪ B ∪ C ∪ T k | + | H ∪ A ∪ T ∪ T ∪ · · · ∪ T k − |≤ gr k − ( K : C ℓ +1 ) − < gr k ( K : C ℓ +1 ) , a contradiction. In fact, the same analysis as this last subcase also applieswhen both | A ∪ B | ≤ ℓ and | B ∪ C | ≤ ℓ , completing the proof of this case.9 ase 5. r = 1 . Let A be the set of vertices in G \ H with red edges to H and let B bethe set with blue edges to H . If | A | ≤ ℓ and | B | ≤ ℓ , we move both A and B to T , contradicting the maximality of T . If one of these sets is large and theother is not, say | A | ≥ ℓ + 1 and | B | ≤ ℓ , then neither H nor A can containany red edge (and T = ∅ ). This means that | A | ≤ gr k − ( K : C ℓ +1 ) − | A ∪ T ∪ T ∪ · · · ∪ T k − | ≤ gr k − ( K : C ℓ +1 ) − | H ∪ B ∪ T k | ≤ gr k − ( K : C ℓ +1 ) − n = | A ∪ T ∪ T ∪ · · · ∪ T k − | + | H ∪ B ∪ T k |≤ gr k − ( K : C ℓ +1 ) − < gr k ( K : C ℓ +1 ) , a contradiction.Finally suppose | A | ≥ ℓ + 1 and | B | ≥ ℓ + 1 so H contains neitherred nor blue edges, A contains no red edge, B contains no blue edge, and T = T = ∅ . Note that A and B consist of parts of the Gallai partition withorder at most ℓ . By Lemma 2, we know that | A | ≤ ℓ and | B | ≤ ℓ . ByLemma 1, we have | H ∪ T ∪ T ∪ · · · ∪ T k − | ≤ gr k − ( K : C ℓ +1 ) − | A ∪ T k | ≤ ℓ . Putting these together, we get n = | H ∪ T ∪ T ∪ · · · ∪ T k − | + | A ∪ T k | + | B |≤ [ gr k − ( K : C ℓ +1 ) −
1] + 3 ℓ + 2 ℓ< gr k ( K : C ℓ +1 ) , a contradiction. This completes the proof of Case 5 and Claim 2.By Claim 2, we know that all parts in the Gallai partition of G have atmost ℓ vertices. Since no such part can contain a copy of C ℓ +1 this meanswe may assume k = 3 so n = 8 ℓ + 1. Claim 3. | T | ≤ ℓ .Proof. Suppose, for a contradiction, that | T | > ℓ and call color 3 green. Let m = | T | so by Claim 1, this means that ℓ < m ≤ ℓ . There can be at most2 ℓ − m disjoint copies of P in green in G \ T since otherwise the vertices of T could be used along with these copies of P to create a green copy of C ℓ +1 .This means we can delete a set F of at most 6 ℓ − m vertices from G \ T
10o leave behind at most a matching in green. Let m i = | T i | for i ∈ { , } sosince m > ℓ , we get | G \ ( T ∪ F ) | ≥ (8 ℓ + 1) − (6 ℓ − m ) − m − m − m > (6 ℓ + 1) − m − m . By Claim 1, it is never the case that m > i , so suppose m > m = 0, and so we wish to find either a red copy of C ℓ − m +1) or a bluecopy of C ℓ +1 . There are (6 ℓ + 1) − m vertices remaining in G \ ( T ∪ F ). ByTheorem 9, we have R ( C ℓ +1 , C ℓ − m +1) ) = max { ℓ − m + 1 , ℓ − m + 3 } . First consider the subcase that 3 ℓ − m + 1 > ℓ − m + 3, so there are atmost 3 ℓ − m parts in the Gallai partition of G \ ( T ∪ F ) to avoid the desiredmomochromatic cycle. With at least 6 ℓ + 1 − m vertices, this means thereare at least [6 ℓ − − m ] − [3 ℓ − m ] = 3 ℓ − ℓ − ℓ − − m vertices available, this means that m = 1. Conversely, theassumption that 3 ℓ − m + 1 > ℓ − m + 3 with m = 1 yields 3 ℓ > ℓ − ℓ ≥
3. We may therefore assume that 3 ℓ − m + 1 ≤ ℓ − m + 3. This means there are at most 4 ℓ − m + 2 parts in the Gallaipartition of G \ ( T ∪ F ). We must therefore have at least[6 ℓ − − m ] − [4 ℓ − m + 2] = 2 ℓ + 3 m − C ℓ +1 or C ℓ +2 (whichever is odd) or a red even cycle C ℓ − m +1 or C ℓ − m +2 (whichever iseven). With the given assumptions, Theorem 9 implies that there can be atmost 2 ℓ − m + 2 parts of order 2. Putting these last observations together,we arrive at 5 m ≤
3, a contradiction.Finally suppose m = m = 0. Then | G \ ( T ∪ F ) | ≥ ℓ + 2. Let r bethe number of parts of order 2 in the Gallai partition of G \ ( T ∪ F ). ByTheorem 9, we know that r ≤ ℓ + 2 since otherwise there would exist amonochromatic odd cycle of length ℓ + 1 or ℓ + 2 (whichever is odd) in thereduced graph, making a monochromatic copy of C ℓ +1 in G . Since there areat least 6 ℓ + 2 vertices, this means there are actually at least(6 ℓ + 2) − ℓ + 2) = 2 ℓ − ℓ ≥
3, there is at least one such part, say X . If r = 2 ℓ + 2, then the reduced graph on these parts along with x must producea monochromatic odd cycle of length ℓ + 1 or ℓ + 2, making a monochromaticcopy of C ℓ +1 in G . This means r ≤ ℓ + 1 so there are at least(6 ℓ + 2) − ℓ + 1) = 2 ℓ parts of order 1. Therefore, there must be at least (2 ℓ + 1) + 2 ℓ = 4 ℓ + 1parts in the Gallai partition. By Theorem 9, there is a monochromatic copyof C ℓ +1 in the reduced graph, and therefore in G .Our final claim shows that the largest part H is even smaller than pre-viously claimed. Claim 4. | H | ≤ ℓ .Proof. Let G R (and G B ) be the sets of vertices in G \ ( H ∪ T ) with red (orblue respectively) edges to H , say with | G R | ≥ | G B | . We first show that s = | H | ≤ ℓ so suppose that ℓ +13 ≤ s ≤ ℓ . By Claim 3, we know that T ≤ ℓ so G R ≥ (8 ℓ + 1) − ℓ − s . If there are at least 2 ℓ +1 − s disjoint red edges within G R , then there is a redcopy of C ℓ +1 using a red complete bipartite graph between H and G R andincluding these disjoint edges. We may therefore delete at most 2(2 ℓ − s )vertices from G R to produce a subgraph G ′ R containing no red edges with | G ′ R | ≥ ℓ + 1 − ℓ − s − ℓ + 4 s = 14 s − ℓ + 24 ≥ ℓ + 2012 > ℓ + 1so by Lemma 2, there is a blue copy of C ℓ +1 within G ′ R , a contradiction. Wemay therefore assume that ℓ +12 ≤ | H | ≤ ℓ .Next we show that | G B | ≥ ℓ . Otherwise | G B | ≤ ℓ − | G R | ≥ ℓ + 1 − ℓ − s − (2 ℓ − P be a longest red path within G R so | P | < ℓ − s ) + 1 to avoid creating a red copy of C ℓ +1 . Note that | P | ≥ P is a longest red path, its end vertices u and v , say12ith u ∈ H u and v ∈ H v , must have no red edges to F = G R \ ( P ∪ H u ∪ H v ).Then | F | ≥ ℓ + 1 − ℓ − s − (2 ℓ − − | P | − s − . For every vertex x ∈ F , the degree of x within F in red or blue is at least8 ℓ + 1 − ℓ − s − (2 ℓ − − | P | − s − − s ≥ [2( ℓ − s ) + 3 − | P | ] + [2 ℓ − − . By Lemma 3, there is either a red copy of P ℓ − s )+3 −| P | or a blue copy of P ℓ − within F , either case resulting in a monochromatic copy of C ℓ +1 . Thismeans we may assume | G B | ≥ ℓ .Let P be a longest red path within G R and let P be a longest bluepath within G B . Note that | P i | ≤ ℓ − s ) + 1 ≤ ℓ for i ∈ { , } to avoidcreating a monochromatic copy of C ℓ +1 and | P | , | P | ≥ v ∈ T (if T = ∅ ), v , v ∈ H , v ∈ G R \ P , v ∈ G B \ P , and v v be any green edge within G \ T (note that such an edge must existsince otherwise G \ T is 2-colored with more than 4 ℓ + 1 vertices). Let G ′ = G \ ( P ∪ P ∪ { v , v , . . . , v } ) so | G ′ | ≥ (8 ℓ + 1) − (2 ℓ + 7) > ℓ − . By Theorem 10, there is a monochromatic copy of C ℓ − in G ′ . Regardlessof the color (red, blue, or green) and location with respect to the sets (i.e. H , G R , G B , . . . ), we may construct a monochromatic copy of C ℓ +1 usingvertices of G \ G ′ , a contradiction.Let G R and G B denote the sets of vertices in G \ H with red and blueedges respectively to H . Without loss of generality, suppose | G R | ≥ | G B | .Next we show that | G B | ≥ ℓ + 6. Otherwise | G B | ≤ ℓ + 5 so | G R | ≥ ℓ + 1 − ℓ − s − (cid:0) ℓ + 5 (cid:1) . Let P be a longest red path within G R so | P | < ℓ − s ) + 1 to avoid creating a red copy of C ℓ +1 . Note that | P | ≥ P is a longest red path, its end vertices u and v , saywith u ∈ H u and v ∈ H v , must have no red edges to F = G R \ ( P ∪ H u ∪ H v ).Then | F | ≥ ℓ + 1 − ℓ − s − (cid:18) ℓ (cid:19) − | P | − s − . For every vertex x ∈ F , the degree of x within F restricted to red and blueis at least8 ℓ +1 − ℓ − s − (cid:18) ℓ (cid:19) −| P |− s − − s ≥ [2( ℓ − s )+3 −| P | ]+[2 ℓ − − .
13y Lemma 3, there is either a red copy of P ℓ − s )+3 −| P | or a blue copy of P ℓ − within F , either case resulting in a monochromatic copy of C ℓ +1 . Thismeans we may assume | G B | ≥ ℓ/ P be a longest red path within G R and let P be a longest blue pathwithin G B . Note that | P i | ≤ ℓ − s ) + 1 for i ∈ { , } to avoid creatinga monochromatic copy of C ℓ +1 and | P | , | P | ≥ H R and H R be the parts of the Gallai partition of G \ T which contain the endvertices of P (where it is possible that H R = H R ) and similarly let H B and H B be the parts containing the end vertices of P . Let u u . . . u | P | bethe vertices of P and let v v . . . v | P | be the vertices of P .Let G ′ B = G B \ ( P ∪ H B ∪ H B ) so | G ′ B | ≥ ( ℓ + 6) − | P | − s + 2 =[2( ℓ − s ) + 1 − | P | ] + [ ℓ + 10] −
3. By Lemma 3, there is either a blue pathof order 2( ℓ − s ) + 1 − | P | or a red path of order at least ℓ + 10. The bluepath, in combination with P and H , would produce a blue copy of C ℓ +1 so we may assume there is a red path in G B of order at least ℓ + 10. Bythe same argument, there is a blue path of order at least ℓ + 10 in G R since | G R | ≥ | G B | . In particular, these observations mean that there is a red edge x y in G B \ ( P ∪ H B ∪ H B ) and a blue edge x y in G R \ ( P ∪ H R ∪ H R ).Let w ∈ T (if T = ∅ ), w w be any green edge within G \ T (note thatsuch an edge must exist since otherwise G \ T is 2-colored with more than4 ℓ + 1 vertices), w , w ∈ H , and let x y and x y be defined as above.Let Q = { u , u , . . . , u ℓ − , u | P | } and let Q = { v , v , . . . , v ℓ − , v | P | } wheresome vertices in these sets may not exist if | P i | < ℓ . Finally let G ′ = G \ ( { w , w , w , w , w , x , y , x , y } ∪ Q ∪ Q )so | G ′ | ≥ | G |− (9+2 ℓ ) > ℓ −
9. By Theorem 10, there exists a monochromaticcopy of C ℓ − within G ′ , say C .If C is green, then since all parts of the Gallai partition of G \ T haveorder at most ℓ , C must use edges from T to G ′ \ T . Let e = uv be onesuch edge with u ∈ T and v / ∈ T . Then replacing the edge uv with the path uw w w v produces a green copy of C ℓ +1 .Since we are no longer applying the assumption that | G R | ≥ | G B | , we mayassume, without loss of generality, that C is red and claim that a symmetricargument would hold if C was blue. If C contains two vertices in G R , say u and v , at distance (along C ) at most 2, then replacing this path with thered path uw u w v or uw u u w v (of the appropriate length) produces ared copy of C ℓ +1 . Thus, there can be no two vertices in C ∩ G R at distance14t most 2 along C . More generally, if P is longer then there can be no twovertices u and v in G R ∩ C at distance at most | P | along C .Similarly, if there is an edge of C from H to G R , say uv with u ∈ H and v ∈ G R , then we replace this edge with the red path uu u w v to producea red copy of C ℓ +1 . This means there is no edge from H to G R on C andtherefore, C ∩ H = ∅ .If C contains two vertices u and v with u, v ∈ G B \ ( P ∪ H B ∪ H B ) atdistance at most 2 on C , then replacing the edge with a red path uv x v | P | v or uv x y v | P | v of the appropriate length produces a red copy of C ℓ +1 . Thismeans C cannot have two vertices at distance at most 2 in G B \ ( P ∪ H B ∪ H B ). If C contains any vertex u ∈ H B i for i ∈ { , } , then since C cannotbe contained entirely within H B i , there is an edge of C from H B i to G \ H B i ,say uv where v / ∈ H B i . Supposing i = 1 without loss of generality, we canreplace the edge uv with the red path ux y v v to produce a red copy of C ℓ +1 . We therefore know that C ∩ ( H B ∪ H B ) = ∅ .If | P | ≥
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