RAMSEY-TYPE NUMBERS FOR DEGREE SEQUENCES

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1 RAMSEY-TYPE NUMBERS FOR DEGREE SEQUENCES ARTHUR BUSCH UNIVERSITY OF DAYTON DAYTON, OH MICHAEL FERRARA, MICHAEL JACOBSON 1 UNIVERSITY OF COLORADO DENVER DENVER, CO STEPHEN G. HARTKE 2 UNIVERSITY OF NEBRASKA LINCOLN LINCOLN, NE Abstract. A (finite) sequence of nonnegative integers is graphic if it is the degree sequence of some simple graph G. Given graphs G 1 and G 2, we consider the smallest integer n such that for every n-term graphic sequence π, there is some graph G with degree sequence π with G 1 G or with G 2 G. When the phrase some graph in the prior sentence is replaced with all graphs the smallest such integer n is the classical Ramsey number r(g 1, G 2 ) and thus we call this parameter for degree sequences the potential-ramsey number and denote it r pot(g 1, G 2 ). In this paper, we give exact values for r pot(k n, K t), r pot(c n, K t), and r pot(p n, K t) and consider situations where r pot(g 1, G 2 ) = r(g 1, G 2 ). 1. Introduction A (finite) nonnegative integer sequence π = (d 1,..., d n ) is graphic if it is the degree sequence of some simple graph G. In that case, we say that G realizes π or is a realization of π, and we write π = π(g). Given a positive integer n, we let GS n denote the set of n-term graphic sequences. A graphic sequence π is unigraphic if all realizations of π are isomorphic, and we denote π = (d 1,..., d n ) = (n 1 d 1,..., n 1 d n ). When convenient we will reorder the terms in nonincreasing (or another) order. The study of graphic sequences dates to the 1950s and 1960s, and includes the characterization of graphic sequences by Havel [1] and Hakimi [11] and an independent characterization by Erdös and Gallai [5]. Subsequent research sought to describe the graphic sequences which are realized by graphs with certain desired structural properties. Such problems can be broadly classified into two types, first described as forcible problems and potential problems by A.R. Rao in [20]. In a forcible degree sequence problem, a specified graph property must exist in every realization of the degree sequence π, while in a potential degree sequence problem, the structure sought must be found in at least one realization of π. Results on forcible degree sequences are often stated as traditional problems in structural 1 Partially supported by UCD GK12 project, National Science Foundation grant DGE Partially supported by National Science Foundation grant DMS

2 2 graph theory, where a necessary and/or sufficient condition is given in terms of the degrees of the vertices of a given graph (e.g., the proverbial First Theorem of Graph Theory ). Two older, but exceptionally thorough surveys on forcible and potential problems are due to Hakimi and Schmeichel [10] and S.B. Rao [21]. Other work [7, 14, 15] on potentially graphic sequences has focused on the study of potentially H-graphic sequences, or sequences which have realizations containing a specified subgraph H. Here, we consider similar problems for both a degree sequence π and the complementary sequence π simultaneously. Motivated by a number of other degree sequence variants of well known problems in extremal graph theory, including potentially graphic sequence analogues of the Turán problem [4, 7, 25], the Erdős-Sös conjecture [24], Hadwiger s conjecture [2, 22] and the Sauer-Spencer theorem [1], the purpose of this paper is to propose a potential degree sequence variant of Ramsey number r(g 1, G 2 ), i.e., the minimum integer n such that for any graph G of order n, either G 1 G or G 2 G. Specifically, given two graphs G 1 and G 2 and a graphic sequence π, we write that π (G 1, G 2 ) if either π is potentially G 1 -graphic or π is potentially G 2 -graphic. We define the potential-ramsey number of G 1 and G 2, denoted r pot (G 1, G 2 ), to be the smallest non-negative integer n such that π (G 1, G 2 ) for every π GS n. In other words, the problem of determining r pot (G 1, G 2 ) is the potential analogue to the forcible problem of determining the classical Ramsey number r(g 1, G 2 ). In the sections that follow, we consider how known results from Ramsey theory can be utilized to establish some initial results for potential-ramsey numbers, give several bounds on r pot (G 1, G 2 ) and also determine exact values for a number of families of graphs. Before we proceed, we introduce some additional notation and terminology. If u and v are adjacent vertices in G, we write u v and say also that uv is an edge of G. For a subgraph H and a vertex v in G, N H (v) denotes those neighbors of v lying in H and we let d H (v) = N H (v). Similarly, given vertices u and v in H, we let dist H (u, v) denote the distance from u to v in H. A trivial component of a graph G is a component of order one. 2. Bounds The first claim in this section establishes a concrete link between the classical Ramsey number r and the Ramsey-potential number r pot. Claim 1. For any graphs G 1 and G 2, r pot (G 1, G 2 ) r(g 1, G 2 ). This relationship is of limited use, since exact values of r(g 1, G 2 ) are known for very few collections of graphs (for a thorough dynamic survey see [19]). Even when r(g 1, G 2 ) is known, this bound is often far from optimal as a number of our results will subsequently demonstrate. However, there are instances for which the bound is sharp. For instance, in light of [12], we have

3 RAMSEY-TYPE NUMBERS FOR DEGREE SEQUENCES n + t 1, r pot (K 1,n, K 1,t ) = r(k 1,n, K 1,t ) = n + t, if n, t are both even, otherwise. Furthermore, the following lemma demonstrates that in certain cases it is straightforward to determine when equality holds in Claim 1. Lemma 2.1. Let r = r(g 1, G 2 ) and let G be a graph of order r 1 such that G 1 G and G 2 G. If π(g) is unigraphic, then r(g 1, G 2 ) = r pot (G 1, G 2 ). This allows us to obtain several interesting results. For instance, it is well known that r(k, K ) = 6 and that C 5 is the unique graph on five vertices containing no triangle and no independent set of size three. Since π(c 5 ) = (2, 2, 2, 2, 2) is unigraphic, we conclude that r pot (K, K ) = r(k, K ) = 6. In [9], it was shown that for n m 2, r(p n, P m ) = n + m 2 1, where the lower bound was established via consideration of the graph G = K n 1 K m As π(g) is unigraphic, we obtain that m r pot (P n, P m ) = r(p n, P m ) = n Next we give a bound on r pot (G, K n ) for a number of choices of G. The 1- dependence number of a graph G [8], denoted α (1) (G), is the maximum order of an induced subgraph H of G with (H) 1. Theorem 2.2. Let G be a graph of order n with α (1) (G) n 1 and t 2. Then r pot ((G, K t ) 2t + n α (1) (G) 2. Proof. Let l = n α (1) (G) 1 and consider π = π(k l (t 1)K 2 ), which is unigraphic. The result follows from two simple observations. First, π is uniquely realized by (K 2t 2 (t 1)K 2 ) K l which contains no K t. Secondly, any copy of G lying in the unique realization of π requires at least α (1) (G) + 1 vertices from the t 1 independent edges, which is impossible as any such collection of vertices would necessarily induce a subgraph of G with order at least α (1) (G) + 1 and maximum degree at most one. This theorem yields a straightforward corollary, springing from the fact that for any graph G of order n without isolaed vertices, π(k n 1 K t 2 ) (G, K t ). Corollary 2.. Let G be a graph of order n with α (1) (G) n 1 and no isolated vertices, and let t 2. Then r pot ((G, K t ) max{2t + n α (1) (G) 2, n + t 2}. In Sections and 4, we show that equality holds in Corollary 2. when G = K n, C n or P n.

4 4. r pot (K n, K t ) Few exact values of r(k n, K t ) are known, and it is one of the foremost problems in combinatorics to determine even r(k 5, K 5 ). Despite this, we determine r pot (K n, K t ) for n t 2. The following results will be useful as we proceed. Theorem.1 (R. Luo [16]). Let π = (d 1,..., d n ) be a nonincreasing graphic sequence with d 2 and d n 1. Then π is potentially C -graphic if and only if π (2, 2, 2, 2) and π (2, 2, 2, 2, 2). Theorem.2 (J-H Yin and J-S Li [2]). Let π = (d 1,..., d n ) be a nonincreasing graphic sequence and k 1 be an integer. (a) If d k k 1 and d 2k k 2 then π is potentially K k -graphic. (b) If d k k 1 and d i 2(k 1) i for 1 i k 1, then π is potentially K k -graphic. We are now ready to prove the main result of this section. Theorem.. For n t, r pot (K n, K t ) = + t 4 except when n = t =, in which case r pot (K, K ) = 6. Proof. The case where n = t = has already been discussed in Section 2, so we may assume that n 4 and t. The fact that r pot (K n, K t ) + t 4 follows from Theorem 2.2 and/or consideration of π((k 2 (n 1)K 2 ) K t ), which is unigraphic. Therefore, let π = (d 1,..., d k ) be a nonincreasing graphic sequence of length k = + t 4 with complementary sequence π = (d 1,..., d k ), which we will also assume to be nonincreasing. If d n < n 1, then d k n+1 = d n+t n + t 2t. As t, this gives that d 1,..., d t 2t > t 1, so π is potentially K t -graphic by part (b) of Theorem.2. In a similar manner, if d t < t 1 then π is potentially K n - graphic since d n d. Consequently, we may assume that d n n 1 and d t t 1. Additionally, if t =, then k 7 so π is potentially K -graphic by Theorem.1. Hence we may assume that t 4. Now, if d n 2, then part (a) of Theorem.2 implies that π is potentially K n -graphic. Hence, we may assume that d n, so that d k +1 = d t (k 1) (n ) 2t 2. As we have that d t 1 t 1, we may apply part (b) of Theorem.2 unless d t 2 < t or, more specifically since d t t 1, if d t 2 = t 1. In this case, however, then d k (t 2)+1 = d 1 4, so we can apply part (b) of Theorem.2 to π provided d 1. We complete the proof by observing that if this were not the case, we would have d 1 = 4 and thus d k = k 1 ( 4) = t 1. As t 4, k 2t, and so π is potentially K t -graphic by part (a) of Theorem.2.

5 RAMSEY-TYPE NUMBERS FOR DEGREE SEQUENCES 5 4. r pot (C n, K t ) Exact values of r(c n, K t ) are known for all t 7, and the conjecture that r(c n, K t ) = (n 1)(t 1) + 1 for n t has been outstanding since 1976 [, 6]; a recent proof for n 4m + 2, m is given in [17]. In contrast, in this section, we show r pot (C n, K t ) is linear in n and t for all n, t 2. Theorem 4.1. For t 2 and n with t, rpot (C n, K t ) = n + t 2. Proof. The fact that r pot (C n, K t ) n+t 2 follows from consideration of π(k n 1 K t 2 ), which is unigraphic. To show the reverse inequality, let π = (d 1,..., d k ) be a graphic sequence with k = n + t 2. If t = 2, the result is immediate, so we assume next that t = and n = 5 and order π = (d 1, d 2, d, d 4, d 5, d 6 ) to be nonincreasing. By Theorem.1, if d 2 then π is potentially K -graphic. We leave it to the reader to check that if π is a nonincreasing sequence with six terms that does not satisfy this condition, then π must be potentially C 5 -graphic. Hence, we may assume that n 6, t and t. If no realization of π contains a cycle, then every realization of π is bipartite, and hence has an independent set of size at least n+t 2 2. Thus, in the complement of G we have either a clique on at least t vertices, or n+t 2 2 t 1 which implies that t n. Thus, we can assume that some realization of π contains a cycle. Obviously, if G contains a cycle of length n, there is nothing to prove, so we assume that no cycle in any realization of π has length n. We proceed by considering the following cases. Case 1: Some realization G of π contains a cycle of length n + 1. Let C = v 1 v 2 v n v n+1 v 1 be a cycle of length n + 1 in G and note that v i v i+2 for any index i, as no realization of π contains an n-cycle. Furthermore, if v i v i+ E(G) for some index i, then we can exchange the edges v i v i+ and v i+1 v i+2 for the nonedges v i+1 v i+ and v i v i+2 to obtain a realization G in which v i v i+2 v i+ v n+1 v 1 v i is a cycle of length n. Thus, we can assume that for any i, neither v i v i+2 nor v i v i+ is an edge of G. Consider a vertex x V (G) C with d = d C (x) maximum. If d = 0, we have either d G (x) = 0 for all x V (G) C, or we have x, y V (G) C with xy E(G). In the latter case, we can exchange the edges v 1 v 2, v 2 v and xy with the nonedges v 1 x, v y and v 1 v to construct a realization of π containing the n- cycle v 1 v v 4 v n+1 v 1. If, instead, d G (x) = 0 for each vertex in V (G) C, then V (G) C along with v 1, v and v 5 form an independent set of size t unless v 1 v 5 is in E(G). Then, however, we can exchange the edges v 1 v 5, v 2 v and v v 4 for the nonedges v 1 v, v v 5 and v 2 v 4 to construct a realization of π containing the n-cycle v 1 v 2 v 4 v 5... v n+1. Thus, we can assume that there is some vertex x V (G) C such that v i N(x) for some index i. If xv i+ E(G), then G contains a cycle of length n, so there exists an index j such that v j N(x) and v j+1 / N(x). Now, replacing the edges xv j and v j+1 v j+2 with the nonedges xv j+1 and v j v j+2 gives a realization G of π in which v j v j+2 v j+... v n+1 v 1 v 2... v j is a cycle of length n. Case 2: Some realization G of π contains a cycle of length n + 2.

6 6 Let C = v 1 v 2 v n v n+2 v 1 be a cycle of length n + 2 in G, and note that Case 1 and the assumption that no realization of G contains an n-cycle, neither v i v i+2 nor v i v i+ are edges of G for any index i. If C is not an induced cycle, choose a chord v i v j E(G) so that dist C (v j, v i ) > is minimum. Next, we note that by replacing the edges v i v i+1, v i v j and v i+2 v i+ with the nonedges v i v i+2, v i v i+ and v i+1 v j we obtain a realization of π in which v i v i+ v i+4 v n+2 v 1 v 2 v i is a cycle of length n. Therefore, we can assume that C is an induced C n+2 in G. Suppose then that there is a vertex x in V (G) C such that xv i is an edge of G for some i. Then either v 1 v 2 v i xv i+ v i+4 v n+2 v 1 is a cycle of length n + 1, or we can choose an index j such that xv j E(G) but xv j+1 / E(G). Then, by exchanging the edges v j x and v j+1 v j+2 with the nonedges v j v j+2 and xv j+1 we obtain a realization of π which contains a cycle of length n + 1. Furthermore, if there exists any edge xy with x, y / {v 1,..., v n+2 }, then replacing the edges v 1 v 2, v 2 v and xy with the edges v 1 v, v 2 x and v 2 y yields a realization of π where v 1 v v 4 v n+2 v 1 is a cycle of length n + 1. Thus G is isomorphic to C n+2 K t 4, and as n + 2 8, α(g) t. Case : Some realization G of π contains a cycle of length m > n + 2. Choose a realization G containing a cycle v 1 v 2 v m v 1 with m > n + 2 maximum. If v n v 1, then the vertices v 1, v 2,..., v n obviously give the desired subgraph isomorphic to C n. Similarly, since we can assume G contains no cycle of length n+2, v n+1 v m, as this edge would complete a cycle of length n + 2. Thus, replacing the edges v m v 1 and v n v n+1 with the non-edges v n v 1 and v n+1 v m gives a graph G realizing π which contains C n. Case 4: Every realization of π has circumference at most n 1. As above, let G be a realization of π containing a longest cycle C = v 1 v 2... v m with m n 1 and suppose that G has the maximum circumference amongst all realizations of π. Let H be the subgraph of G induced by V (G) V (C). Claim 1. H is acyclic. Assume otherwise, and let x 1 x 2 x p x 1 be a cycle in H. Then, if v i x j E(G), it follows that neither v i+1 x j E(G) nor v i+2 x j+1 E(G). Otherwise, v 1 v 2 v i x j v i+1 v i+2 v m v 1 or v 1 v 2 v i x j x j+1 v i+2 v m v 1 is a cycle of length greater than m in G, a contradiction. Thus we conclude, without loss of generality, that v 1 x 2 and v 2 x 1, and by replacing the edges v 1 v 2 and x 1 x 2 with v 1 x 2 and v 2 x 1, we obtain a realization of π with a cycle v 1 x 2 x x p x 1 v 2 v v m v 1 of length m + p, a contradiction which establishes that H is acyclic. Claim 2. Either (H) 1 or the only non-trivial component of H is a star. Assume first that there exist vertices x and y in H such that d H (x) 2 and d H (y) 2. Clearly, we can choose x and y so that xy E(G). Since H is acyclic, then for any vertices x N H (x) {y} and y N H (y) {x} we have x y and x y. As the circumference of G is m, there exists an index i such that v i x / E(G) and v i+1 y / E(G). By replacing xx, yy and v i v i+1 with v i x, v i+1 y and x y, we obtain a graph G realizing π containing the (m + 2)-cycle v 1 v i xyv i+1 v m v 1, a contradiction. Hence H has at most one vertex of degree at least two.

7 RAMSEY-TYPE NUMBERS FOR DEGREE SEQUENCES 7 Assume then that (H) > 1, and that H has at least two non-trivial components. Choose x V (H) with d H (x) > 1, and choose vertices x and x in N(x), and and edge yy in a component of H different from the component containing x. We may assume, without loss of generality, that v 1 x. If, in addition, v 2 x then v 1 v 2 xx x xv 1 alternates between edges and non-edges in G and can be used to obtain a realization of π containing a cycle of length m + 1 > m, a contradiction. A similar argument holds if v m x so both v 2 x and v m x are edges in G. Now, if v 2 y / E(G), then replacing edges xx, yy and v 1 v 2 with v 1 x, v 2 y and y x yields a realization of π in which v 1 xv 2 v m v 1 is a cycle of length m+1, which contradicts the maximality of m. We can therefore assume v 2 y E(G), and by an identical argument, that v m y E(G). But then v 2 v n y yv 2 is a cycle of length m + 1 in G. This last contradiction establishes the claim. Claim. If the only non-trival component of H is a star, then π (C n, K t ). Let x be the unique vertex in V (H) with d H (x) > 1. Clearly, by the maximality of m, v i x E(G) implies that v i+1 x for every i, and the argument in Claim 2 establishes that we could construct a larger cycle if x has two consecutive nonneighbors on C. We conclude that m is even and, without loss of generality, x v i if and only if i is even. Next, observe that if we can choose an odd index i such that v i y for some y V (H) x, then either v 1 v 2 v i yxv i+1 v i+2 v m v 1 is a cycle of length m + 1 in G or, xy / E(G). In the latter case, we can exchange the edges v i y and xx with the nonedges v i x and x y for any x N H (x) to obtain a realization of π in which v 1 v 2 v i xv i+1 v i+2 v m v 1 is a cycle of length m + 1. Thus, we conclude that no vertex in V (H) is adjacent to v 1 or v. If v 1 v, we have the cycle v 1 v v 2 xv 4 v 5 v m v 1 of length m + 1 in G, and if v 1 v, then V (H) {v 1, v } {x} is a set of at least (k m) = (n m) + t 1 t independent points in G, completing the proof of the claim. As a result of Claims 1 -, we may assume that (H) 1. First, consider the case that (H) = 0. Since V (H) = k m t 1, we can assume equality holds, and conclude that m = n 1. Now, either every vertex v i is incident with at least one vertex of H, or we can choose a vertex v j such that V (H) {v j } is an independent set of size at least t. If every v i has at least one neighbor in V (H), then since n 1 > t 1, we conclude that some vertex x V (H) is adjacent to vertices v i and v j with i < j, and by the maximality of m, i < j 1. We choose x and v i v j so that j i > 1 is as small as possible. We now choose y V (H) N(v i+1 ), and observe that y x. The minimality of j i ensures that v j y and v i+1 x. Thus, the graph obtained by replacing edges v i+1 y and v j x with v i+1 x and v j y is a realization of π which contains a cycle v 1 v 2 v i xv i+1 v i+2 v n v 1 of length m+1 = n. Therefore, it remains to examine the possibility that (H) = 1. First, we assume that H contains exactly two (necessarily adjacent) vertices of degree one, say x and y. We also note that if v i x E(G) for any index i, then the maximality of m guarantees that v i±1 x and v i±2 y. In this case, by exchanging the edges v i+1 v i+2 and xy with the nonedges v i+1 x and v i+2 y we obtain a realization of π with an independent set of size V (H). Since H has exactly one edge and order k m, we conclude that either H has t independent points, or m is one of n 1 or n 2. If m = n 2, then V (H) = t and we can use an argument identical to the case where (H) = 0 to show that there is some vertex on C that has no neighbor

8 8 in H, implying the existence of an appropriate independent set. Alternatively, if m = n 1, then V (H) = t 1 and we can conclude again that there is some vertex v j is adjacent to no vertex of H. Since m = n 1 5, we can then choose an index i j, j 1 so that replacing the edges v i v i+1 and xy with v i x and v i+1 y gives realization of π with t independent points (the vertices in V (H) together with the vertex v j ). Next, assume that H contains 2l 4 vertices of degree one and p isolated vertices and choose xx, yy E(H). If v i x E(G) but v i x, then we can produce a realization of π in which (H) = 2 by replacing edges v i x, xx and yy with the nonedges xy, xy and v i x. The previous claims then imply π (C n, K t ). Furthermore, we note that by replacing the edges xx and yy with the edges xy and x y in H, we obtain another realization of π, and as a result the preceeding argument immediately implies that v i y E(G). Thus, each vertex v i adjacent to any vertex x with d H (x) = 1 is adjacent to every vertex of H with degree one. Let S = {v i v i x for some x with d H (x) = 1}, and let s = S. First, suppose that S = and that x i y i, 1 i l are the (disjoint) edges in H. If l m, then we can exchange x i y i and the edge v i v i+1 for the nonedges xv i and yv i+1. This creates a realization of π in which the vertices of H contain an independent set of size p+l+m. If p+l+m t we are done, so suppose otherwise. Then k (p + l + m) = l n 1, which implies that k 2, a contradiction. Hence we may assume that l < m, which again allows us to exchange edges in H and edges on C to create a realization of π in which H induces an independent set of size p + 2l. As k = n + t 2 and m n 1, this implies that, in fact, p + 2l = t 1 and m = n 1. Now, as above, if every vertex v i has a neighbor in H, we can construct a realization of G containing a longer cycle, so there is some vertex v j with no neighbor in H. As t and l t 1 2, we conclude that l n = m 2, so we can use edges in E(C) that are not incident to v j to break the l edges in H and create a realization of π in which V (H) {v j } forms an independent set of size t, as desired. Thus, we now assume that S. Observe that for every v i S, the maximality of m implies that v i+2 and v i 2 are not adjacent to any vertex of V (H) with degree one. Additionally, the vertices v i+1 and v i 1 are not adjacent to any vertex x V (H), as we would either have a cycle of length m + 1 if d H (x) = 1, or, if d H (x) = 0, we can choose yy E(H) and replace v i±1 x and v i y with v i±1 y and v i x and obtaian a realization in which v 1 v j y yv j+1 v n v 1 is a cycle of length m + 1 for j = i, i 1. Furthermore, we note that each edge v i v j of the graph induced by C S, and any edge xx E(H) can be replaced with the edges v i x and v j x to obtain a realization of π with fewer edges in the graph induced by V (H). Since each edge of the cycle not incident with a vertex of S is in C S and dist C (v i, v j ) for any distinct v i and v j in S, there are at least m 2s edges in this graph. If m 2s > l, then we can eliminate every edge of H without using an edge incident with v i+1 for some v i S. This gives a realization of π in which V (H) {v i+1 } is a collection of at leat t independent vertices. Thus, we can assume that m 2s l, and by using each of these m 2s edges of the cycle to eliminate an edge of H, we conclude that there is a realization of π with an independent set of size l + p + (m 2s). So either

9 RAMSEY-TYPE NUMBERS FOR DEGREE SEQUENCES 9 π (C n, K t ) or (1) l + p + m 2s t 1. Now, let S + = {v i+1 v i S}, and suppose first that there is some v S + and z V (H) such that v z and d H (z) = 0. Then, for any edge xy in H we could exchange the edges vz and xy for the nonedges to vx and zy. This results in x having consecutive neighbors on C, a contradiction to the maximality of m. We also claim that S + forms an independent set in G. If not, then there are adjacent vertices v i+1 and v j+1 in S +, which implies that for any edge xy in H, v i xyv j v j 1 v i+1 v j+1 v j+2 v i is an (m + 2)-cycle in G, a contradiction. Thus, S + along with the p isolates in H and one vertex from each edge in H comprise an independent set in G. Thus, either π (C n, K t ) or (2) l + p + s t 1. Summing (1) and (2) gives 2l + 2p + m s 2t 2. Now, since 2l + p = V (H) = n + t 2 m, we have (n s) + p t. Note that s m n 1, and hence +1 + p t, contradicting t and completing the proof. Theorem 4.2. For t and n 4 with t >, n r pot (C n, K t ) = 2t 2 +. Proof. The lower bound stems from Theorem 2.2 and/or consideration of π(k ( n 1) (t 1)K 2 ), which is unigraphic. To show the reverse inequality, let π = (d 1,..., d k ) be graphic with k = 2t 2 + n. If no realization of π contains a cycle, then every realization of π is bipartite, and hence has an independent set of size at least t. Thus, we can assume that some realization of π contains a cycle. Obviously, if G contains a cycle of length n, there is nothing to prove, so we assume that no cycle in any realization of π has length n. Due to the similarities between this proof and that of Theorem 4.1, we only sketch the following cases, which suffice to complete the proof. Case 1: Some realization G of π contains a cycle of length n + 1. Let C = v 1 v 2 v n v n+1 v 1 be a cycle of length n+1 in G. Just as in our proof of Theorem 4.1, this requires that d(x) = 0 for every x V (G) V (C). Then either v 1 v and v 1 v v 4 v n+1 v 1 is a cycle of length n, or V (G) {v 2, v 4,..., v n+1 } is a set of k (n 1) = 2t 1 t indpendent vertices. Case 2: Some realization G of π contains a cycle v 1 v 2 v n+2 v 1. Again, the aguments in Case 2 of Theorem 4.1 imply that G = C n+2 K k n 2 and since n 4, this graph has an indpendent set of size at least k n + 1 t. Case : Some realization of π has circumference m > n + 2. The proof is identical to the proof of Case of Theorem 4.1. Case 4: Every realization of π has circumference m n 1.

10 10 Let G be a realization of π containing a longest cycle C = v 1 v 2... v m with m n 1 and suppose that G has the maximum circumference amongst all realizations of π. Let H be the subgraph of G induced by V (G) V (C) and note that H has order at least k (n 1) t so that H contains at least one edge. Following the argumentation in the proof of Theorem 4.1 we conclude that H = lk 2 pk 1 for some integers p and l. Proceeding in the same manner, we define the set S = {v i v i x E(G) for some vertex x with deg H (x) = 1}. and note that v i S implies that none of v i±1, v i±2 S. Thus, the graph V (C) S has at least m 2 S edges and each edge of this graph can be paired with an edge of H. Replacing these edges with two vertex disjoint non-edges in the induced 2K 2 gives a realization of π with one fewer edge in the graph induced by V (H). Thus, we have some realization of π with an independent set of size at least l+p+m 2 S, so either π (C n, K t ) or l + p + m 2s t 1. Since V (H) = 2l + p = 2t 2 + m, we must have 2(l + p + m 2 S ) 2t 2 (2t 2 + p + V (H) + p + m 2 S 2t 2 m) + p + m 2 S 2t 2 m S p + An easy case analysis then shows that the left hand side is strictly positive except when p = 0, m = n 1, S = m and n 1 (mod ). However, this in turn requires k = 2t 2 + n+2 +1, and V (H) = 2l = 2t. As n 1 (mod ), +1 is odd, a contradiction which completes the proof. The techniques developed in the proofs of Theorems 4.1 and 4.2 also allow us to determine r pot (P n, K t ). In fact, r pot (P n, K t ) and r pot (C n, K t ) differ by at most one. Note, however, that r(p n, K t ) = (n 1)(t 1) + 1 [18], which is the conjectured value for r(c n, K t ) for all n t. Theorem 4.. For n 6 and t, n + t 2, r pot (P n, K t ) = if t, 2t 2 + n, if t >. Proof. When t, we note that rpot (P n, K t ) r pot (C n, K t ) = n + t 2, and equality is established via Corollary 2. and/or by observing that K n 1 K t 2 contains neither P n nor K t. Therefore, we may assume t >. Theorem 2.2 establishes that rpot (P n, K t ) 2t 2 + n and if n 0 (mod ), we observe that rpot (P n, K t ) r pot (C n, K t ) = 2t 2 + n, establishing the result. Finally, for n 0 (mod ), we note that by Theorem 4.1, every degree sequence π with k 2t 2 + n terms has a realization G which contains either a subgraph isomorphic to C n 2 or an independent set of cardinality t. If the latter, π (P n, K t ), and if the former, we note that there are at least 2t 2 + n (n 2) = 2t t vertices in V (G) V (C n 2). Since we are assuming α(g) < t, there is

11 RAMSEY-TYPE NUMBERS FOR DEGREE SEQUENCES 11 at least one edge xy E(G) with x, y / V (C n 2 ). Now, without loss of generality, either vx E(G) and G contains a subgraph isomorphic to P n, or we can select any edge v i v i+1 E(C n 2 ) and exchange the edges v i v i+1 and xy with the non-edges v i x and v i+1 y to obtain a realization of π which contains a path of order n. 5. Conclusion Our results in this paper deal only with a 2-color Ramsey-type parameter for degree sequences; we are also able to define a multicolored version of r pot. Let π i,..., π k be graphic sequences, with π i = (d (i) 1,..., d(i) n ) for all i (they need not be monotone). Then, as defined in [1], π 1,..., π k pack if there exist edge-disjoint graphs G 1,..., G k, all with vertex set {v 1,..., v n }, such that for each i, and d Gi (v j ) = d (i) j d G1 G k (v j ) = n i=1 d (i) j. In traditional graph packing, the vertex sets of the graphs in question may be permuted prior to their embedding into K n. Note that when packing degree sequences, the ordering of the terms in π 1,..., π k remains fixed. With this definition in mind, we define r pot (G 1,..., G k ) to be the smallest integer n such that for any collection of n-term graphic sequences π 1,..., π k that sum, termwise, to n 1 and pack, there exist edge disjoint graphs F 1,..., F k all with vertex set {v 1,..., v n }, such that d Fi (v j ) = d (i) j for all i, j and also that F i contains G i as a subgraph for some i. As is the case with the two-color version, the multicolor potential-ramsey number is bounded from above by the classical Ramsey number, but the added challenges inherent in degree sequence packing make the determination of r pot (G 1,..., G k ) for k both interesting and significantly more difficult than the k = 2 case. References [1] Arthur H. Busch, Michael J. Ferrara, Stephen G. Hartke, Michael S. Jacobson, Hemanshu Kaul, and Douglas B. West. Packing of graphic n-tuples. Journal of Graph Theory, pages n/a n/a, [2] Z. Dvorak and B. Mohar. Chromatic number and complete graph substructures for degree sequences. Arxiv preprint arxiv: , [] P. Erdős, R. J. Faudree, C. C. Rousseau, and R. H. Schelp. On cycle-complete graph Ramsey numbers. J. Graph Theory, 2(1):5 64, [4] P. Erdös, M. S. Jacobson, and J. Lehel. Graphs realizing the same degree sequences and their respective clique numbers. In Graph theory, combinatorics, and applications, Vol. 1 (Kalamazoo, MI, 1988), Wiley-Intersci. Publ., pages Wiley, New York, [5] Paul Erdős and Tibor Gallai. Graphs with prescribed degrees of vertices (hungarian). Mat. Lapok, 11: , [6] R. J. Faudree and R. H. Schelp. Some problems in Ramsey theory. In Theory and applications of graphs (Proc. Internat. Conf., Western Mich. Univ., Kalamazoo, Mich., 1976), volume 642 of Lecture Notes in Math., pages Springer, Berlin, [7] Michael J. Ferrara and John Schmitt. A general lower bound for potentially H-graphic sequences. SIAM J. Discrete Math., 2(1): , 2008/09.

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