Gallai-Ramsey numbers for graphs with chromatic number three
aa r X i v : . [ m a t h . C O ] J un Gallai-Ramsey numbers for graphswith chromatic number three
Qinghong Zhao ∗ and Bing Wei † Department of Mathematics, University of Mississippi, University, MS 38677, USA
Abstract
Given a graph H and an integer k ≥
1, the Gallai-Ramsey number GR k ( H ) isdefined to be the minimum integer n such that every k -edge coloring of K n containseither a rainbow (all different colored) triangle or a monochromatic copy of H . Inthis paper, we study Gallai-Ramsey numbers for graphs with chromatic numberthree such as b K m for m ≥
2, where b K m is a kipas with m +1 vertices obtained fromthe join of K and P m , and a class of graphs with five vertices, denoted by H . Wefirst study the general lower bound of such graphs and propose a conjecture for theexact value of GR k ( b K m ). Then we give a unified proof to determine the Gallai-Ramsey numbers for many graphs in H and obtain the exact value of GR k ( b K )for k ≥
1. Our outcomes not only indicate that the conjecture on GR k ( b K m ) istrue for m ≤
4, but also imply several results on GR k ( H ) for some H ∈ H whichare proved individually in different papers. Key words : Gallai coloring, Gallai-Ramsey number, Rainbow triangle : 05C55; 05D10; 05C15
In this paper, we only deal with finite, simple and undirected graphs. Given a graph G and the vertex set V ( G ), let | G | denote the order of G and G [ W ] denote the subgraph of G induced by a set W ⊆ V ( G ). Given disjoint vertex sets X, Y ⊆ V ( G ), if each vertexin X is adjacent to all vertices in Y and all the edges between X and Y are colored withthe same color, then we say that X is mc -adjacent to Y , that is, X is blue-adjacent to Y if all the edges between X and Y are colored with blue. We use P n , C n and K n to denote the path, cycle and complete graph on n vertices, respectively; and S n todenote the star on n + 1 vertices. A kipas b K m is obtained by deleting one edge on therim of a wheel with m + 1 vertices. For an integer k ≥
1, we define [ k ] = { , . . . , k } . ∗ Corresponding Author. Supported by the Summer Graduate Research Assistantship Program ofGraduate School. E-mail address: [email protected] (Q. Zhao). † Supported in part by the summer faculty research grant from CLA at the University of Mississippi.E-mail address: [email protected] (B. Wei).
Theorem 1.1 ([5],[7]) . For a complete graph G under any edge coloring without rain-bow triangle, there exists a partition of vertices (called a Gallai-partition) with parts V , V , . . . , V ℓ , ℓ ≥ , such that there are at most two colors on the edges between theparts and only one color on the edges between each pair of parts. We define G = G [ { v , . . . , v ℓ } ] = K ℓ as a reduced graph of G , where v i ∈ V i for i ∈ [ ℓ ]. Obviously, there exists a monochromatic copy of H in G if ℓ ≥ R ( H ), whichleads to a monochromatic copy of H in G . In honor of Gallai’s result, the edge coloringof a complete graph without rainbow triangle is called Gallai coloring . Given graphs H , . . . , H k and an integer k ≥
1, the
Ramsey number R ( H , . . . , H k ) is defined to bethe minimum integer n such that every k -edge coloring of K n contains a monochro-matic copy of H i in color i ∈ [ k ], and the Gallai-Ramsey number GR ( H , . . . , H k )is defined to be the minimum integer n such that every Gallai k -coloring of K n con-tains a monochromatic copy of H i in color i ∈ [ k ]. We simply write R k ( H ) and GR k ( H ) when H = · · · = H k , and R (( k − s ) H s +1 , sH ) and GR k (( k − s ) H s +1 , sH )when H = · · · = H s and H s +1 = · · · = H k . Clearly, GR ( F, H ) = R ( F, H ) and GR k ( H ) ≤ R k ( H ) for k ≥
1. However, determining GR k ( H ) for H is far from trivial,even for a small graph. The general behavior of GR k ( H ) for H was established in [6]. Theorem 1.2 ([6]) . Let H be a fixed graph with no isolated vertices. Then GR k ( H ) isexponential in k if H is not bipartite, linear in k if H is bipartite but not a star, andconstant (does not depend on k ) when H is a star. In 2015, Fox, Grinshpun and Pach [3] posed a conjecture for GR k ( K t ). The cases t = 3 , t = 5 in [11] while there is no any results for the cases t ≥
6. For more information onthis topic, we refer the readers to [3, 4]. Let H denote a class of graphs of order fivewith chromatic number three. It is known that H contains twelve graphs (see Fig. 1). H H H H H H H H H H H H Fig. 1: The graphs with five vertices of chromatic number three.2n this paper, we study Gallai-Ramsey numbers for some graphs in H and b K m for m ≥
2. Recently, Gallai-Ramsey numbers for many graphs above have been determinedin [15] for H and H , [9] for the graphs from H to H , [16] for H , [14],[17] for H and [12] for H . However, the Gallai-Ramsey numbers for the remaining graphs H , H and H have not been determined so far. In this paper, we gain the exact valuesof GR k ( H ) for H ∈ { H , H , H } . Let F = { H , H , H , H , H , H , H , H } . Wefirst study the general lower bound of Gallai-Ramsey numbers for all graphs in F and b K m for m ≥ GR k ( H ) for H ∈ F under the enlightenment from [14] in Section 3. Theorem 1.3.
For k ≥ , if H ∈ F \ H , then GR k ( H ) = ( ( R ( H ) − · ( k − / + 1 , if k is even,4 · ( k − / + 1 , if k is odd and if H = H , then GR ( H ) = 5 , GR ( H ) = 7 and GR k ( H ) = ( k/ + 1 , if k ≥ · ( k − / + 1 , if k ≥ K ∼ = b K , K \ { e } ∼ = b K and H ∼ = b K . In this paper, we come up witha new approach to construct the general lower bound of GR k ( H ) for some H , whichwill provide a direction for us to study Gallai-Ramsey numbers of b K m for m ≥
2. Theexact values of GR k ( b K ) and GR k ( b K ) have been determined as follows: Theorem 1.4 ([1],[6]) . For k ≥ , GR k ( b K ) = ( k/ + 1 if k is even,2 · ( k − / + 1 if k is odd. Theorem 1.5 ([16]) . For k ≥ , GR ( b K ) = 4 and GR k ( b K ) = ( · ( k − / + 1 , if k ≥ · ( k − / + 1 , if k ≥ GR k ( b K ). Theorem 1.6.
Let k ≥ and s be an integer with ≤ s ≤ k . Then GR k (( k − s ) P , s b K ) = · s/ + 1 , if s is even and s < k ,2 · s/ , if s is even and s = k ,4 · ( s − / + 1 , if s is odd.3ne can obtain the Gallai-Ramsey number of b K from Theorem 1.6 by setting s = k . With the support of the general lower bound of GR k ( b K m ) given in Section 2and the exact values of GR k ( b K m ) for 2 ≤ m ≤
4, we propose a conjecture as follows:
Conjecture 1.7.
For k ≥ and m ≥ , GR ( b K m ) = m + 1 and GR k ( b K m ) = ( R ( b K m ) − · ( k − / + 1 , if k is even and m is odd, R ( b K m ) + ( m/ · (5 k/ − , if k is even and m is even, max { R ( b K m ) − , m } · ( k − / + 1 , if k ≥ Theorem 1.8 ([2]) . GR k ( P ) = 3 for k ≥ . Theorem 1.9 ([8]) . R ( H ) = 7 , R ( H ′ ) = 9 and R ( H ′′ ) = 10 , where H ′ ∈{ H , H , H , H } and H ′′ ∈ { H , H , H , H } . Theorem 1.10 ([13]) . R ( P , b K ) = 5 . In this section, we first give the general lower bound of GR k ( H ) for H ∈ F or H = b K m with m ≥ | H | = 5. Suppose H = H and k ≥
1. Let g (1) = | H | − g (2) = R ( H ) − R ( H ) = 7 by Theorem 1.9 and let g ( k ) = ( k/ , if k ≥ · ( k − / , if k ≥ GR ( H ) = | H | , GR ( H ) = R ( H ) and K is a subgraph of H , GR k ( H ) ≥ g ( k ) + 1 for k ≥
1. Suppose H ∈ F \ H and k ≥
1. Let g ( k ) = ( ( R ( H ) − · ( k − / , if k is even,4 · ( k − / , if k is odd.Suppose H = b K m for m ≥ k ≥
1. Let g (1) = | H | − g ( k ) = ( R ( H ) − · ( k − / , if k is even and m is odd, R ( H ) + ( m/ · (5 k/ − − , if k is even and m is even, max { R ( H ) − , | H | − } · ( k − / , if k ≥ G, c ) to denote a complete graph G under the Gallai coloring c .For all k ≥
1, we now construct the complete graph ( G k , c k ) on g ( k ) vertices recursivelywhich contains no monochromatic copy of H ∈ b K m ∪ F \ H , where c k : E ( G k ) −→ [ k ] is4 Gallai k -coloring. Let K be colored with colors k − k without monochromatictriangle and ( G k − , c k − ) be the construction on g ( k −
2) vertices with colors in [ k − Case 1. k is even and m is odd for all H ∈ b K m ∪ F \ H . For k = 2, let G be 2-colored complete graph on g (2) = R ( H ) − H under c : E ( G ) −→ { , } . For all even k ≥ G k , c k ) be the construction obtained by replacing each vertex of K with a copyof ( G k − , c k − ) such that the colors on all edges between the corresponding copies of( G k − , c k − ) are consistent with K . Case 2. k is odd for all H ∈ b K m ∪ F \ H . For k = 1, let G be 1-colored complete graph on g (1) = | H | − H under c : E ( G ) −→ { } . For all odd k ≥
3, let( G k − , c k − ) be obtained from Case 1. Note that ( G k − , c k − ) is also obtained by samemethod as in Case 1 for even m . If 2( R ( H ) − ≥ | H | − G k , c k ) be theconstruction obtained by joining two copies of ( G k − , c k − ) such that all edges betweentwo copies are colored with color k . If 2( R ( H ) − < | H | − G k , c k ) bethe construction obtained from the copies of ( G k − , c k − ) by the same method as inCase 1 under K . Case 3. k is even and m is even for H = b K m . For k = 2, let G be 2-colored complete graph on g (2) = R ( b K m ) − b K m under c : E ( G ) −→ { , } . For all even k ≥
4, let ( G k , c k ) be the construction obtained from the copies of ( G k − , c k − ) andthe following copies of G ′ k − and G ′′ k − by the same method as in Case 1 under K . Let X i , X j be any two corresponding copies such that the color on the edges between X i and X j is color i ∈ { k − , k } . Since there can not be monochromatic copy of b K m in( G k , c k ), the following properties must hold. • There is neither S m/ nor P m in color i within either X i or X j . • If X i contains S m/ − or P m − in color i , then X j has no edge in color i .We first construct the Gallai ( k − G ′ k − ( res . G ′′ k − ) foreven k ≥ k −
1] ( res . [ k ] \ { k − } ) recursively which contains neithermonochromatic copy of b K m in color i ∈ [ k −
2] nor S m/ in color k − res . k ). Let G ′ ( res . G ′′ ) be a complete graph on m/ k − res . k ).Clearly, G ′ ( res . G ′′ ) contains no S m/ in color k − res . k ). For all even k ≥
4, let G ′ k − ( res . G ′′ k − ) be the construction with colors in [ k − ∪ { k − } ( res . [ k − ∪ { k } )such that G ′ = G ′ ( res . G ′′ = G ′′ ) if k = 4. Let K (a complete graph of order five)be colored with colors k − k − G ′ k − ( res . G ′′ k − ) be obtained from the copies of G ′ k − ( res . G ′′ k − ) by the same method as5n Case 1 under K . Since G ′ k − ( res . G ′′ k − ) is a Gallai ( k − b K m in color i ∈ [ k −
4] nor S m/ in color k − res . k ), G ′ k − ( res . G ′′ k − ) is a desired construction on ( m/ · ( k − / verticeswithout rainbow triangle.By the properties above, there is at least one copy of ( G k − , c k − ), at most twocopies of G ′ k − and at most two copies of G ′′ k − embedding in K . For all even k ≥ G k , c k ) (see Fig. 2. color k is on the edges of outer five cycle and color k − K with one copy of ( G k − , c k − ), two vertices of K with two copies of G ′ k − and two vertices of K with two copies of G ′′ k − . G ′′ k − G ′′ k − G ′ k − G k − G ′ k − G k − G k Fig. 2: An example of the construction for all even k ≥ m ≥ Then | G k | = | G k − | + 2 | G ′ k − | + 2 | G ′′ k − | for all even k ≥ m ≥
2. Thus weget the following recurrence equation: ( g ( k ) = g ( k −
2) + 2 m · ( k − / , k ≥ g (2) = R ( b K m ) − . Note that g ( k ) = g ( k −
2) + 2 m · ( k − / is a nonhomogeneous equation. Since 5 isnot a solution of characteristic equation r − g ( k ) − g ( k −
2) = 0, the particular solution of the nonhomogeneous equationis (2 m · ( k − / ) · a . Then(2 m · ( k − / ) · a = (2 m · ( k − / ) · a + 2 m · ( k − / and thus a = 5 /
4. Furthermore, the general solution of the homogeneous equation is g ′ ( k ) = b · ( − k + b · (1) k . So the general solution of nonhomogeneous equation is g ( k ) = g ′ ( k ) + (2 m · ( k − / ) · (5 /
4) = b + b + ( m/ · k/ k is even. Since g (2) = R ( b K m ) − b + b = R ( b K m ) − − (5 m/ g ( k ) = R ( b K m ) + ( m/ · (5 k/ − − k ≥ m ≥ G k − and G k − have no monochromatic copy of H under c k − : E ( G k − ) −→ [ k −
1] and c k − : E ( G k − ) −→ [ k −
2] respectively in the correspondingcases, ( G k , c k ) is a desired construction on g ( k ) vertices for all k ≥ GR k ( H ) ≥ g ( k ) + 1 for all k ≥ H ∈ b K m ∪ F \ H .For H = H , we have g ( k −
1) = ( ( k − / , if k ≥ · ( k − / , if k ≥ g ( k −
2) = ( ( k − / , if k ≥ · ( k − / , if k ≥ g ( k −
2) = 6 if k = 4.Let t = ( R ( H ) − H ∈ F \ H , we have g ( k −
1) = ( t · ( k − / , if k ≥ · ( k − / , if k ≥ g ( k −
2) = ( t · ( k − / , if k ≥ · ( k − / , if k ≥ g ( k − ≥ g ( k − g ( k − ≥ g ( k − k +3 ≥ g ( k − k ≥ H ∈ F \ H and k ≥ H . Furthermore, g ( k ) + 1 > g ( k −
1) + k + 1 , if k ≥ H ∈ F ,2 g ( k − , if k ≥ H ∈ F \ H and k ≥ H .2 g ( k −
1) + 2 , if k ≥ H ∈ { H , H , H } ,5 g ( k − , if k ≥ H ∈ F \ H and k ≥ H . ( ∗ )We next give the general lower bound of GR k (( k − s ) P , s b K ) for k ≥ s with0 ≤ s ≤ k . For each integer k ≥ s with 0 ≤ s ≤ k , let w ( k, s ) = · s/ , if s is even and s < k ,2 · s/ − , if s is even and s = k ,4 · ( s − / , if s is odd.For all k ≥ s with 0 ≤ s ≤ k , we now construct the Gallai k -colored completegraph G sk on w ( k, s ) vertices recursively which contains neither monochromatic copy of b K in color i ∈ [ s ] nor P in color j ∈ [ k ] \ [ s ].Assume that s = 0. By Theorem 1.8, let G k be the Gallai k -colored complete graphon w ( k,
0) = GR k ( P ) − P in color j ∈ [ k ]. Assume s = k .By Theorem 1.9, we see that 2( R ( b K ) − < | b K | − G kk can be obtained7y Case 2 when s is odd and by Case 3 when s is even.Now it remains to consider the cases 1 ≤ s < k . By Theorem 1.8, let G ′ be aGallai ( k − s )-colored complete graph on 2 vertices with colors in [ k ] \ [ s ] containingno monochromatic copy of P . Then we construct G ′′ by joining two copies of G ′ suchthat all the edges between two copies are colored with color 1. Clearly, G ′′ is a Gallai( k − s + 1)-colored complete graph on 4 vertices which contains neither monochromaticcopy of b K in color 1 nor P in color [ k ] \ [ s ]. For s = 1, let G k = G ′′ . For all s ≥ G s − k − be the construction on w ( k − , s −
2) vertices with colors in [ k ] \ { s − , s } such that G s − k − = G ′ if s = 2. Let K (a complete graph of order five) be coloredwith colors s − s without monochromatic triangle. Let G sk be the constructionobtained from the copies of G s − k − by the same method as in Case 1 under K . Since G s − k − is a Gallai ( k − b K in color i ∈ [ s −
2] nor P in color j ∈ [ k ] \ [ s ], G sk is a desired constructionon | G sk | = 4 · ( s − / vertices for odd s ≥ | G sk | = 2 · s/ vertices for even s ≥ GR k (( k − s ) P , s b K ) ≥ w ( k, s ) + 1 for all k ≥ s with 0 ≤ s ≤ k .For k ≥ s with 1 ≤ s ≤ k , we have w ( k, s −
1) = ( · ( s − / , if s is even,2 · ( s − / , if s is odd,and w ( k − , s −
1) = · ( s − / , if s is even,2 · ( s − / − , if s is odd and s = k ,2 · ( s − / , if s is odd and s < k .Thus w ( k, s − ≥ w ( k − , s − ≥ s + 1. For k ≥ s with 2 ≤ s ≤ k , we have w ( k − , s −
2) = w ( k, s −
2) = ( · ( s − / , if s is even,4 · ( s − / , if s is odd,and w ( k − , s −
2) = · ( s − / , if s is even and s < k ,2 · ( s − / − , if s is even and s = k ,4 · ( s − / , if s is odd.Thus w ( k, s −
2) = w ( k − , s − ≥ w ( k − , s − ≥
2. Furthermore, w ( k, s ) + 1 > ( w ( k, s − , if k ≥ ≤ s ≤ k ,4 w ( k − , s −
2) + w ( k − , s − , if k ≥ ≤ s ≤ k . ( ⋆ )8 Proof of main results
By constructions in Section 2, it suffices to show that GR k ( H ) ≤ g ( k ) + 1 with H ∈ F and GR k (( k − s ) P , s b K , ) ≤ w ( k, s )+ 1 for k ≥ s with 0 ≤ s ≤ k . We proceed theproof by induction on k for GR k ( H ) with H ∈ F and k + s for GR k (( k − s ) P , s b K ).The case for k = 1 is trivial. By Theorems 1.9 and 1.10, GR ( H ) = R ( H ) = g (2) + 1, GR ( b K ) = R ( b K ) = w (2 , GR ( P , b K ) = R ( P , b K ) = w (2 , s = 0 is Theorem 1.8. Therefore, we may assume that k ≥ s with 1 ≤ s ≤ k .Suppose GR k ( H ) with H ∈ F holds for all k ′ < k and GR k (( k − s ) P , s b K ) holds forall k ′ + s ′ < k + s . Let G = K g ( k )+1 for GR k ( H ) with H ∈ F and G = K w ( k,s )+1 for GR k (( k − s ) P , s b K ). Let c : E ( G ) −→ [ k ] be any Gallai k -coloring of G . Suppose ( G, c )contains no monochromatic copy of H for GR k ( H ) with H ∈ F and ( G, c ) containsneither monochromatic copy of b K in color i ∈ [ s ] nor P in color j ∈ [ k ] \ [ s ] for GR k (( k − s ) P , s b K ). Choose ( G, c ) with k minimum.Let A ⊆ V ( G ) and let p, q ∈ [ k ] be two distinct colors. By induction for GR k ( H )with H ∈ F , if ( G [ A ] , c ) contains no edge in color p ( res . p and q ), then | A | ≤ g ( k − res . | A | ≤ g ( k − p, q ∈ [ s ] if s ≥
2. By induction for GR k (( k − s ) P , s b K ),if ( G [ A ] , c ) contains no edge in color p ( res . p and q ), then | A | ≤ w ( k − , s −
1) ( res . | A | ≤ w ( k − , s − G [ A ] , c ) contains edges in color p ( res . p and q )but does not contain P in color p ( res . p and q ), then | A | ≤ w ( k, s −
1) as ( G [ A ] , c )contains neither monochromatic copy of b K in color i ∈ [ s ] \ { p } nor P in color j ∈ ([ k ] \ [ s ]) ∪ { p } ( res . | A | ≤ w ( k, s −
2) as ( G [ A ] , c ) contains neither monochromaticcopy of b K in color i ∈ [ s ] \ { p, q } nor P in color j ∈ ([ k ] \ [ s ]) ∪ { p, q } ). Furthermore, if( G [ A ] , c ) contains edges in color p but does not contain P in color p and edge in color q , then | A | ≤ w ( k − , s − u , u , . . . , u t ∈ V ( G ) be a maximum sequence of vertices chosen as follows: foreach j ∈ [ t ], all edges between u j and V ( G ) \ { u , u , . . . , u j } are colored the same colorunder c . Let U = { u , u , . . . , u t } . Notice that U is possibly empty. For each u j ∈ U ,let c ( u j ) be the unique color on the edges between u j and V ( G ) \ { u , u , . . . , u j } . Claim 1. c ( u i ) = c ( u j ) for all i, j ∈ [ t ] with i = j .Proof. Suppose that c ( u i ) = c ( u j ) for some i, j ∈ [ t ] with i = j . We may assume that u j is the first vertex in the sequence u , . . . , u t such that c ( u j ) = c ( u i ) for some i ∈ [ t ]with i < j . We may further assume that the color c ( u i ) is red. Thus the edge u i u j iscolored with red under c . Let W = V ( G ) \ { u , u , . . . , u j } . Then all the edges between { u i , u j } and W are colored with red under c . We first consider the proof for GR k ( H )with H ∈ F . By the pigeonhole principle, j ≤ k + 1. Note that ( G [ W ] , c ) containsno red edge, otherwise we obtain a red H ∈ F . By induction, | W | ≤ g ( k − ∗ ), | G | ≤ g ( k −
1) + k + 1 < g ( k ) + 1, which is a contradiction. We next considerthe proof for GR k (( k − s ) P , s b K ). Note that the color on the edges between u j and9 ( G ) \ { u , u , . . . , u j } belongs to [ s ], otherwise we obtain a monochromatic copy of P with color in [ k ] \ [ s ] as | G | ≥ w ( k, k ≥ s with 1 ≤ s ≤ k . Clearly, redbelongs to [ s ]. By the pigeonhole principle, j ≤ s +1. By induction, | W | ≤ w ( k − , s − G [ W ] , c ) contains no red edge. Recall that w ( k, s − ≥ w ( k − , s − ≥ s + 1. By( ⋆ ), | G | ≤ w ( k − , s −
1) + s + 1 ≤ w ( k, s − < w ( k, s ) + 1, which is impossible. (cid:4) By Claim 1, | U | ≤ k for GR k ( H ) with H ∈ F and | U | ≤ s for GR k (( k − s ) P , s b K ).Consider a Gallai-partition of G \ U with parts V , V , . . . , V ℓ such that ℓ ≥ | V | ≥ | V | ≥ . . . ≥ | V ℓ | . Let G be the reduced graph of G \ U with vertices v , . . . , v ℓ . By Theorem 1.1, we may further assume that the edges of G are colored with red or blue. It is obvious that any monochromatic copy of H in G would yield a monochromatic copy of H in G \ U for GR k ( H ) with H ∈ F and anymonochromatic copy of b K or P in G would yield a monochromatic copy of b K or P in G \ U for GR k (( k − s ) P , s b K ). Let V r = { V i | V i is red-adjacent to V under c , i ∈ { , . . . , ℓ }} and V b = { V i | V i is blue-adjacent to V under c , i ∈ { , . . . , ℓ }} .Let R = S V i ∈V r V i and B = S V i ∈V b V i . Then | G | = | V | + | R | + | B | + | U | . Without lossof generality, we may assume that | B | ≤ | R | . Obviously, | R | ≥
2, otherwise the vertexin R or B can be added to U , contrary to the maximality of t in U . As ( G [ V ∪ R ] , c )contains a red P , red belongs to [ s ] for GR k (( k − s ) P , s b K ). It is worth noting that if | V | ≥
2, then ( G [ V ∪ R ] , c ) contains a red C and thus no vertex in U is red-adjacentto V ( G ) \ U under c . It follows that ( G [ U ] , c ) contains no red edge. Furthermore,if ( G [ V ] , c ) doesn’t contain red edge or red P , and neither does ( G [ V ∪ U ] , c ) or( G [ V ∪ B ∪ U ] , c ) when | B | ≤ Claim 2. B = ∅ for GR k (( k − s ) P , s b K ) and thus ≤ s ≤ k and blue belongs to [ s ] .Proof. Suppose B = ∅ . Assume ℓ ≥
3. Since R is red-adjacent to V under c , wecan actually find a Gallai-partition with only two parts, contrary to the fact that ℓ is as small as possible. Thus ℓ = 2. Since | V | = | R | ≥ | V | ≥
2. Thus novertex in U is red-adjacent to V ( G ) \ U under c and so | U | ≤ s −
1. Clearly, thereis no red P within either G [ V ] or G [ R ] under c . Note that | V | ≥
3, otherwise | G | = | V | + | R | + | U | ≤ s − < w ( k, s ) + 1. If there exist red edges within ( G [ V ] , c ),then ( G [ R ] , c ) doesn’t contain red edge, and neither does ( G [ R ∪ U ] , c ). By induction, | V | ≤ w ( k, s −
1) and | R ∪ U | ≤ w ( k − , s − w ( k − , s − ≤ w ( k, s − ⋆ ), | G | = | V | + | R ∪ U | ≤ w ( k, s −
1) + w ( k − , s − ≤ w ( k, s − < w ( k, s ) + 1,which is impossible. So there is no red edge within ( G [ V ] , c ). By induction, | V ∪ U | ≤ w ( k − , s − ⋆ ), | G | = | V ∪ U | + | V | ≤ w ( k − , s − ≤ w ( k, s − < w ( k, s )+1,contrary to the fact that | G | = w ( k, s ) + 1. Thus B = ∅ . Suppose s = 1 for k ≥ s ]. Note that | V | = 1, otherwise there is a blue P within ( G [ V ∪ B ] , c ). Then ( G, c ) only contains red and blue edges, contrary to our10ssumption that k ≥
3. Thus 2 ≤ s ≤ k . Suppose blue is in [ k ] \ [ s ]. By Theorem 1.10, R ( P , b K ) = 5 and so ℓ ≤
4. Thus | V | ≥ | G \ U | ≥ w ( k, − ≥
8. But there willbe a blue P in ( G [ V ∪ B ] , c ), which is a contradiction. (cid:4) Claim 3.
Let
C, D be two disjoint sets of V ( G ) such that | C | ≥ , | D | = 2 and C is redor blue-adjacent to D under c . For GR k ( H ) with H ∈ F , there is no red or blue edgewithin ( G [ C ] , c ) if H ∈ F \ H ( | C | ≥ for H = H ) or ( G [ D ] , c ) if H ∈ { H , H } .Proof. Suppose not. Then there will be a red or blue H ∈ F in ( G [ C ∪ D ] , c ), contraryto our assumption that ( G, c ) contains no monochromactic copy of H ∈ F . (cid:4) Claim 4. | V | ≥ for GR k ( H ) with H ∈ F and | V | ≥ for GR k (( k − s ) P , s b K ) .Proof. We first consider the proof for GR k ( H ) with H ∈ F . By Theorem 1.9, ℓ ≤ H = H and ℓ ≤ H ∈ F \ H . Then | V | ≥ | G \ U | ≥ f (3) − H = H and | G \ U | ≥ g (3) − H ∈ F \ H . Since | R | ≥
2, no vertex in U isred-adjacent to V ( G ) \ U under τ and so | U | ≤ k −
1. Thus | V | ≥ H ∈ F \ H as | G \ U | ≥ g (3) − H = H and k ≥
4, then | V | ≥ | G \ U | ≥ g (4) − H = H and k = 3. Suppose | V | = 2. Then | R | ≥ | G \ U | ≥ | R | ≥ | B | . By Claim 3, ( G [ R ] , c ) doesn’tcontain red edge and neither does ( G [ R ∪ U ] , c ). By induction, | R ∪ U | ≤ g (2) = 6.It follows that | B | ≥ | G | = g (3) + 1 = 11. Thus no vertex in U is blue-adjacentto V ( G ) \ U under c and so | U | ≤
1. By Claim 3 again, ( G [ B ] , c ) contains no blueedge. Now, we see that the color on the edges between any pairs of { V , . . . , V ℓ } in( G [ R ] , c ) and ( G [ B ] , c ) is blue and red, respectively. Note that there is no blue K in( G [ R ] , c ) as ( G [ V ∪ B ] , c ) contains blue edges. Then there are at most two parts of { V , . . . , V ℓ } in ( G [ R ] , c ), which means that | B | ≤ | R | ≤
4. Thus | U | = 1 as | G | = 11.Since | V | = 2, ( G [ V ] , c ) only contains one color. Such color is different from red andblue, otherwise there is a red K in ( G [ V ∪ R ] , c ) or a blue K in ( G [ V ∪ B ] , c ), whichtogether with a red edge in ( G [ B ] , c ) or a blue edge in ( G [ R ] , c ) yields a monochromaticcopy of H . Let green denote such color. Then ( G [ V ∪ U ] , c ) contains a green K ,which forbids green edge in ( G [ R ∪ B ] , c ). By induction, | R ∪ B | ≤ g (2) = 6. Then | G | = | V ∪ U | + | R ∪ B | ≤ < g (3) + 1, contrary to the fact that | G | = g ( k ) + 1.We next consider the proof for GR k (( k − s ) P , s b K ). By Claim 2, we only need toassume that 3 ≤ s ≤ k for k ≥ G \ U, c ) when s = 2, which forces | V | ≥
2. By Theorem 1.9, R ( b K ) = 10. By Claim2, ℓ ≤ | V | ≥ | G \ U | ≥ w ( k, − ≥
18 for k ≥ ≤ s ≤ k . (cid:4) For the convenience of proofs, we now show that the desired result holds for H = H and k = 3. By Claims 3 and 4, ( G [ V ] , c ) contains no red edge. Suppose | B | ≤
1. Thenthere is no red edge in ( G [ V ∪ B ∪ U ] , c ). By induction, | V ∪ B ∪ U | ≤ g (2) = 6and so | R | ≥ | G | = 11. By Claims 3 and 4, there is no red edge in ( G [ R ] , c ). Byinduction, | R | ≤ g (2) = 6. Since R ( K ) = 6, | V ∪ B ∪ U | ≤ | R | ≤
5. Then11 G | = | V ∪ B ∪ U | + | R | ≤ < g (3) + 1, which is impossible. Suppose | B | ≥
2. ByClaims 3 and 4, there is no blue edge in ( G [ V ] , c ) and so ( G [ V ∪ U ] , c ) contains only onecolor which is different from red and blue, say green. It follows that 3 ≤ | V ∪ U | ≤ G [ V ∪ U ] , c ) contains a green K , which forbids green edge in ( G [ R ∪ B ] , c ).By induction, | R ∪ B | ≤ g (2) = 6. Then | G | = | V ∪ U | + | R ∪ B | ≤ < g (3) + 1,which is a contradiction. Thus we may assume that k ≥ H = H in later proofs. Claim 5. | B | ≥ for GR k ( H ) with H ∈ F and | B | ≥ for GR k (( k − s ) P , s b K ) .Proof. We first suppose | B | ≤ | R | ≥
2. By Claim 4,there is no red P within either ( G [ R ] , c ) or ( G [ V ∪ B ∪ U ] , c ) for GR k (( k − s ) P , s b K ).By induction, | R | ≤ w ( k, s −
1) and | V ∪ B ∪ U | ≤ w ( k, s − ⋆ ), | G | = | V ∪ B ∪ U | + | R | ≤ w ( k, s − < w ( k, s ) + 1, which is a contradiction. We nextconsider the case for GR k ( H ) with H ∈ F . Assume that | R | = 2. Let H ∈ F \ H . Notethat | V | ≥ | G | ≥
21. By Claims 3, there is no red edge within ( G [ V ∪ B ∪ U ] , c ),By induction, | V ∪ B ∪ U | ≤ g ( k − ∗ ), | G | = | V ∪ B ∪ U | + | R | ≤ g ( k −
1) + 2 ≤ g ( k −
1) + k + 1 < g ( k ) + 1, which is impossible. Let H = H . Clearly, there is no red P in ( G [ V ∪ B ∪ U ] , c ). Then all the red edges in ( G [ V ∪ B ∪ U ] , c ) induce a matching,say M . Let X , X be two disjoint sets of V ∪ B ∪ U such that X ∪ X = V ∪ B ∪ U and M is part of the edges between X and X under c . Then there is no red edgewithin either ( G [ X ] , c ) or ( G [ X ] , c ). By induction, | V ∪ B ∪ U | ≤ g ( k − ∗ ), | G | = | V ∪ B ∪ U | + | R | ≤ g ( k −
1) + 2 < g ( k ) + 1, which is a contradiction. Assumethat | R | ≥
3. By Claims 3 and 4, there is no red edge within either ( G [ V ∪ B ∪ U ] , c )or ( G [ R ] , c ). By induction, | V ∪ B ∪ U | ≤ g ( k −
1) and | R | ≤ g ( k − ∗ ), | G | = | V ∪ B ∪ U | + | R | ≤ g ( k − < g ( k ) + 1, contrary to the fact that | G | = g ( k ) + 1.We next suppose | B | = 2 for GR k ( H ) with H ∈ F . Then no vertex in U is red orblue-adjacent to V ( G ) \ U under c and so | U | ≤ k −
2. Recall that 2 g ( k − ≥ g ( k −
1) + g ( k − ≥ g ( k − k +3 ≥ g ( k − | R | = 2. Let H ∈ F \ H . ByClaims 3 and 4, there is neither red nor blue edge within ( G [ V ∪ U ] , c ). By induction, | V ∪ U | ≤ g ( k − ∗ ), | G | = | V ∪ U | + | R | + | B | ≤ g ( k − ≤ g ( k − < g ( k )+1,which is a contradiction. Let H = H . Obviously, there is neither red P nor blue P within ( G [ V ∪ U ] , c ), Similar to the arguments for | B | ≤
1, Since (
G, c ) has no rainbowtriangle, all the red and blue edges in ( G [ V ∪ U ] , c ) induce a matching. Then there isneither red nor blue edge within either ( G [ X ] , c ) or ( G [ X ] , c ). By induction, | V ∪ U | ≤ g ( k − ∗ ), | G | = | V ∪ U | + | R | + | B | ≤ g ( k − ≤ g ( k − < g ( k )+ 1, whichis impossible. Assume that | R | ≥
3. By Claims 3 and 4, there is no red edge withineither ( G [ R ] , c ) or ( G [ V ∪ U ] , c ). By induction, | R | ≤ g ( k −
1) and | V ∪ U | ≤ g ( k − H ∈ { H , H , H } . By ( ∗ ), | G | = | V ∪ U | + | R | + | B | ≤ g ( k −
1) + 2 < g ( k ) + 1,which is a contradiction. Let H ∈ F \ { H , H , H } . Note that | V | ≥
4, otherwise | G | = | V | + | R | + | B | + | U | ≤ g ( k −
1) + k + 3 ≤ g ( k − < g ( k ) + 1. By Claims 3,there is no blue edge within ( G [ V ∪ U ] , c ). By induction, | V ∪ U | ≤ g ( k − ∗ ),12 G | = | V ∪ U | + | R | + | B | ≤ g ( k −
2) + g ( k −
1) + 2 ≤ g ( k − < g ( k ) + 1, which yieldsa contradiction. (cid:4) By Claims 3-5, there is no red edge ( res . P ) within ( G [ R ] , c ) and no blue edge( res . P ) within ( G [ B ] , c ) for GR k ( H ) with H ∈ F ( res . GR k (( k − s ) P , s b K )).Define Y = { V i : | V i | = 1 , i ∈ { , . . . , ℓ }} and Y = { V i : | V i | ≥ , i ∈ { , . . . , ℓ }} . Then | Y ∪ Y | = | R ∪ B | . Let | R ∩ Y t | and | B ∩ Y t | be the number of the common parts in { V , . . . , V } , where t = 1 ,
2. We see that | R ∩ Y | ≤ | B ∩ Y | ≤
2, otherwise thereis a blue H ∈ F ∪ b K in ( G [ R ] , c ) or a red H ∈ F ∪ b K in ( G [ B ] , c ). Thus we have thefollowing three facts:1. If | R ∩ Y | = 2 or | B ∩ Y | = 2, then | R ∩ Y | = 0 or | B ∩ Y | = 0.2. If | R ∩ Y | = 1 or | B ∩ Y | = 1, then | R ∩ Y | ≤ | B ∩ Y | ≤ | R ∩ Y | = 0 or | B ∩ Y | = 0, then | R | ≤ | B | ≤ Proof.
We only consider the proof for R . The proof for B is similar. Suppose | R ∩ Y | ≥ Y is blue-adjacent to Y and so there is a blue H ∈ F ∪ b K in( G [ R ] , c ), which is impossible. Suppose | R ∩ Y | ≥ V i , V j , V k ∈ Y andlet V s ∈ Y . Then V i , V j and V k are blue-adjacent to V s and there is at most one red edgein ( G [ V i ∪ V j ∪ V k ] , c ). Thus there is a blue H ∈ F ∪ b K in ( G [ V s ∪ V i ∪ V j ∪ V k ] , c ), whichis impossible. Suppose | R | ≥ G [ R ] , c ) must contain red and blueedges, contrary to the fact that there is no red edge within ( G [ R ] , c ) for GR k ( H ) with H ∈ F . For GR k (( k − s ) P , s b K ), since ( G [ R ] , c ) contains no red P and R ( P , b K ) = 5,there is a blue b K in ( G [ R ] , c ), which is a contradiction. (cid:4) By facts 1-3, it is easily seen that | R | ≤ | V | as 4 ≤ | V | + 2 ≤ | V | . Then | B | ≤ | R | ≤ | V | . By Claims 4 and 5, no vertex in U is red or blue-adjacent to V ( G ) \ U under c and thus ( G [ U ] , c ) contains neither red nor blue edge.We first consider the proof for GR k ( H ) with H ∈ F . By Claims 3-5, ( G [ V ∪ U ] , c )contains neither red nor blue edge. By induction, | V ∪ U | ≤ g ( k − k ≥ H = H and k ≥ H ∈ F \ H . As | R | ≤ | V | , | R | + | B | ≤ g ( k − ∗ ), | G | = | V ∪ U | + | R | + | B | ≤ g ( k − < g ( k ) + 1, which is a contradiction.Assume that H = H and k = 4. Then | V ∪ U | ≤
5, otherwise ( G [ V ∪ U ] , c ) contains amonochromatic K , say green, as R ( K ) = 6, which forbids green edge in ( G [ R ∪ B ] , c ).By induction, | R ∪ B | ≤ g (3) = 10. Then | G | = | V ∪ U | + | R ∪ B | ≤ < g (4)+ 1, whichis impossible. Now, | G | = | V ∪ U | + | R | + | B | ≤ < g (4) + 1, contrary to the factthat | G | = g ( k ) + 1. We next turn to the proof for GR k (( k − s ) P , s b K ). We know that | R | ≥
2. By Claim 5, ( G [ V ∪ U ] , c ) contains neither red P nor blue P . By induction, | V ∪ U | ≤ w ( k, s − w ( k, s −
2) = w ( k − , s − ≥ w ( k − , s − ≥ | R ∩ Y | ≤ | B ∩ Y | ≤
2. Suppose | R ∩ Y | ≤
1. By facts 2 and3, | B | ≤ | R | ≤ | V | + 2. By ( ⋆ ), | G | = | V ∪ U | + | R | + | B | ≤ w ( k, s −
2) + 4 ≤ w ( k − , s −
2) + w ( k − , s − < w ( k, s ) + 1, which is impossible. Thus | R ∩ Y | = 2.By facts 2 and 3 again, | B | ≤ | V | + 2 if | B ∩ Y | ≤
1. By ( ⋆ ), | G | = | V ∪ U | + | R | + | B | ≤ w ( k, s − ≤ w ( k − , s − w ( k − , s − < w ( k, s )+1, which is a contradiction.Thus | B ∩ Y | = 2. By fact 1, | R ∩ Y | = 0 and | B ∩ Y | = 0. Then ℓ = 5. Since R ( b K ) = 10, there is no K in G \ U whose edges are colored with red and blue under c , which implies that at least one part of { V , . . . , V } contains neither red nor blueedge. Note that | V ∪ U | ≤ w ( k, s −
2) = w ( k − , s −
2) and | V | ≥ | V | ≥ . . . ≥ | V | .By ( ⋆ ), | G | = | V ∪ U | + | R | + | B | ≤ w ( k − , s −
2) + w ( k − , s − < w ( k, s ) + 1,contrary to the fact that | G | = w ( k, s ) + 1.This completes the proof of Theorems 1.3 and 1.6. (cid:4) References [1] F. R. K. Chung, R. Graham, Edge-colored complete graphs with precisely coloredsubgraphs, Combinatorica 3 (1983) 315–324.[2] R.J. Faudree, R. Gould, M. Jacobson and C. Magnant, Ramsey numbers in rain-bow triangle free colorings, Australas. J. Combin. 46 (2010) 269–284.[3] J. Fox, A. Grinshpun, J. Pach, The Erd˝os-Hajnal conjecture for rainbow triangles,J. Combin. Theory Ser. B 111 (2015) 75–125.[4] S. Fujita, C. Magnant, K. Ozeki, Rainbow generalizations of Ramsey theory: asurvey, Graphs Combin. 26 (2010) 1–30.[5] T. Gallai, Transitiv orientierbare Graphen, Acta Math. Acad. Sci. Hung. 18 (1967)25–66.[6] A. Gy´arf´as, G. S´ark¨ozy, A. Seb˝o, S. Selkow, Ramsey-type results for Gallai color-ings, J. Graph Theory 64 (2010) 233–243.[7] A. Gy´arf´as, G. Simonyi, Edge colorings of complete graphs without tricoloredtriangles, J. Graph Theory 46 (2004) 211–216.[8] G.R.T. Hendry, Ramsey numbers for graphs with five vertices, J. Graph Theory,13 (1989) 245–248.[9] X-H. Li, L-G. Wang, Gallai-Ramsey numbers for a class of graphs with five vertices,preprint, arXiv:1811.06134.[10] H. Liu, C. Magnant, A. Saito, I. Schiermeyer, Y. Shi. Gallai-Ramsey number for K , J. Graph Theory 94 (2020) 192–205.[11] C. Magnant, I. Schiermeyer, Gallai-Ramsey number for K5