RAMSEY-TYPE NUMBERS FOR DEGREE SEQUENCES
ARTHUR BUSCH
UNIVERSITY OF DAYTON
DAYTON, OH 45469
MICHAEL FERRARA, MICHAEL JACOBSON1
UNIVERSITY OF COLORADO DENVER
DENVER, CO 80217
STEPHEN G. HARTKE2
UNIVERSITY OF NEBRASKA–LINCOLN
LINCOLN, NE 68588
Abstract. A (finite) sequence of nonnegative integers is graphic if it is the
degree sequence of some simple graph G. Given graphs G1 and G2 , 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 G1 ⊆ G or with G2 ⊆ 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(G1 , G2 )
and thus we call this parameter for degree sequences the potential-Ramsey
number and denote it rpot (G1 , G2 ). In this paper, we give exact values for
rpot (Kn , Kt ), rpot (Cn , Kt ), and rpot (Pn , Kt ) and consider situations where
rpot (G1 , G2 ) = r(G1 , G2 ).
1. Introduction
A (finite) nonnegative integer sequence π = (d1 , . . . , dn ) 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 GSn
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 = (d1 , . . . , dn ) = (n − 1 −
d1 , . . . , n − 1 − dn ). 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 [13] 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
2
Partially supported by UCD GK12 project, National Science Foundation grant DGE-0742434.
Partially supported by National Science Foundation grant DMS-0914815.
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(G1 , G2 ), i.e., the minimum
integer n such that for any graph G of order n, either G1 ⊆ G or G2 ⊆ G.
Specifically, given two graphs G1 and G2 and a graphic sequence π, we write that
π → (G1 , G2 ) if either π is potentially G1 -graphic or π is potentially G2 -graphic.
We define the potential-Ramsey number of G1 and G2 , denoted rpot (G1 , G2 ), to be
the smallest non-negative integer n such that π → (G1 , G2 ) for every π ∈ GSn . In
other words, the problem of determining rpot (G1 , G2 ) is the potential analogue to
the forcible problem of determining the classical Ramsey number r(G1 , G2 ). 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 rpot (G1 , G2 ) 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, NH (v) denotes those neighbors of
v lying in H and we let dH (v) = |NH (v)|. Similarly, given vertices u and v in H,
we let distH (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 rpot .
Claim 1. For any graphs G1 and G2 ,
rpot (G1 , G2 ) ≤ r(G1 , G2 ).
This relationship is of limited use, since exact values of r(G1 , G2 ) are known for
very few collections of graphs (for a thorough dynamic survey see [19]). Even when
r(G1 , G2 ) 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
RAMSEY-TYPE NUMBERS FOR DEGREE SEQUENCES
rpot (K1,n , K1,t ) = r(K1,n , K1,t ) =

 n + t − 1,

n + t,
3
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(G1 , G2 ) and let G be a graph of order r − 1 such that
G1 6⊆ G and G2 6⊆ G. If π(G) is unigraphic, then r(G1 , G2 ) = rpot (G1 , G2 ).
This allows us to obtain several interesting results. For instance, it is well known
that r(K3 , K3 ) = 6 and that C5 is the unique graph on five vertices containing
no triangle and no independent set of size three. Since π(C5 ) = (2, 2, 2, 2, 2) is
unigraphic, we conclude that rpot (K3 , K3 ) = r(K3 , K3 ) = 6.
In [9], it was shown that for n ≥ m ≥ 2, r(Pn , Pm ) = n + m
2 − 1, where the
lower bound was established via consideration of the graph G = Kn−1 ∪ Kb m c−1 .
2
As π(G) is unigraphic, we obtain that
rpot (Pn , Pm ) = r(Pn , Pm ) = n +
jmk
2
− 1.
Next we give a bound on rpot (G, Kn ) for a number of choices of G. The 1dependence 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
rpot ((G, Kt ) ≥ 2t + n − α(1) (G) − 2.
Proof. Let ` = n − α(1) (G) − 1 and consider π = π(K` ∨ (t − 1)K2 ), which is
unigraphic. The result follows from two simple observations. First, π is uniquely
realized by (K2t−2 − (t − 1)K2 ) ∪ K ` which contains no Kt . 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(Kn−1 ∪ K t−2 ) 6→ (G, Kt ).
Corollary 2.3. Let G be a graph of order n with α(1) (G) ≤ n − 1 and no isolated
vertices, and let t ≥ 2. Then
rpot ((G, Kt ) ≥ max{2t + n − α(1) (G) − 2, n + t − 2}.
In Sections 3 and 4, we show that equality holds in Corollary 2.3 when G =
Kn , Cn or Pn .
4
3. rpot (Kn , Kt )
Few exact values of r(Kn , Kt ) are known, and it is one of the foremost problems in combinatorics to determine even r(K5 , K5 ). Despite this, we determine
rpot (Kn , Kt ) for n ≥ t ≥ 2. The following results will be useful as we proceed.
Theorem 3.1 (R. Luo [16]). Let π = (d1 , . . . , dn ) be a nonincreasing graphic
sequence with d3 ≥ 2 and dn ≥ 1. Then π is potentially C3 -graphic if and only if
π 6= (2, 2, 2, 2) and π 6= (2, 2, 2, 2, 2).
Theorem 3.2 (J-H Yin and J-S Li [23]). Let π = (d1 , . . . , dn ) be a nonincreasing
graphic sequence and k ≥ 1 be an integer.
(a) If dk ≥ k − 1 and d2k ≥ k − 2 then π is potentially Kk -graphic.
(b) If dk ≥ k − 1 and di ≥ 2(k − 1) − i for 1 ≤ i ≤ k − 1, then π is potentially
Kk -graphic.
We are now ready to prove the main result of this section.
Theorem 3.3. For n ≥ t ≥ 3,
rpot (Kn , Kt ) = 2n + t − 4
except when n = t = 3, in which case rpot (K3 , K3 ) = 6.
Proof. The case where n = t = 3 has already been discussed in Section 2, so we
may assume that n ≥ 4 and t ≥ 3. The fact that rpot (Kn , Kt ) ≥ 2n + t − 4 follows
from Theorem 2.2 and/or consideration of π((K2n−2 − (n − 1)K2 ) ∪ K t−3 ), which
is unigraphic.
Therefore, let π = (d1 , . . . , dk ) be a nonincreasing graphic sequence of length
k = 2n + t − 4 with complementary sequence π = (d1 , . . . , dk ), which we will also
assume to be nonincreasing. If dn < n − 1, then
dk−n+1 = dn+t−3 ≥ n + t − 3 ≥ 2t − 3.
As t ≥ 3, this gives that d1 , . . . , dt ≥ 2t−3 > t−1, so π is potentially Kt -graphic by
part (b) of Theorem 3.2. In a similar manner, if dt < t − 1 then π is potentially Kn graphic since dn ≥ d2n−3 ≥ 2n − 3. Consequently, we may assume that dn ≥ n − 1
and dt ≥ t − 1. Additionally, if t = 3, then k ≥ 7 so π is potentially K3 -graphic by
Theorem 3.1. Hence we may assume that t ≥ 4.
Now, if d2n ≥ n − 2, then part (a) of Theorem 3.2 implies that π is potentially
Kn -graphic. Hence, we may assume that d2n ≤ n−3, so that dk−2n+1 = dt−3 ≥ (k−
1)−(n−3) ≥ 2t−2. As we have that dt−1 ≥ t−1, we may apply part (b) of Theorem
3.2 unless dt−2 < t or, more specifically since dt ≥ t − 1, if dt−2 = t − 1. In this case,
however, then dk−(t−2)+1 = d2n−1 ≥ 2n − 4, so we can apply part (b) of Theorem
3.2 to π provided d1 ≥ 2n − 3. We complete the proof by observing that if this were
not the case, we would have d1 = 2n − 4 and thus dk = k − 1 − (2n − 4) = t − 1. As
t ≥ 4, k ≥ 2t, and so π is potentially Kt -graphic by part (a) of Theorem 3.2.
RAMSEY-TYPE NUMBERS FOR DEGREE SEQUENCES
5
4. rpot (Cn , Kt )
Exact values of r(Cn , Kt ) are known for all t ≤ 7, and the conjecture that
r(Cn , Kt ) = (n − 1)(t − 1) + 1 for n ≥ t ≥ 3 has been outstanding since 1976 [3, 6];
a recent proof for n ≥ 4m + 2, m ≥ 3 is given in [17]. In contrast, in this section,
we show rpot (Cn , Kt ) is linear in n and t for all n ≥ 3, t ≥ 2.
Theorem 4.1. For t ≥ 2 and n ≥ 3 with t ≤ 2n
3 , rpot (Cn , Kt ) = n + t − 2.
Proof. The fact that rpot (Cn , Kt ) ≥ n+t−2 follows from consideration of π(Kn−1 ∪
Kt−2 ), which is unigraphic.
To show the reverse inequality, let π = (d1 , . . . , dk ) be a graphic sequence with
k = n + t − 2. If t = 2, the result is immediate, so we assume next that t = 3 and
n = 5 and order π = (d1 , d2 , d3 , d4 , d5 , d6 ) to be nonincreasing. By Theorem 3.1, if
d3 ≥ 2 then π is potentially K3 -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 C5 -graphic.
Hence, we may assume that n ≥ 6, t ≥ 3 and t ≤ 2n
3 . 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 d n+t−2
e. Thus, in the complement of G we have
2
≤ t − 1 which implies that t ≥ n.
either a clique on at least t vertices, or n+t−2
2
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 = v1 v2 · · · vn vn+1 v1 be a cycle of length n + 1 in G and note that vi 6∼
vi+2 for any index i, as no realization of π contains an n-cycle. Furthermore,
if vi vi+3 ∈ E(G) for some index i, then we can exchange the edges vi vi+3 and
vi+1 vi+2 for the nonedges vi+1 vi+3 and vi vi+2 to obtain a realization G0 in which
vi vi+2 vi+3 · · · vn+1 v1 · · · vi is a cycle of length n. Thus, we can assume that for any
i, neither vi vi+2 nor vi vi+3 is an edge of G.
Consider a vertex x ∈ V (G) − C with d = dC (x) maximum. If d = 0, we
have either dG (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 v1 v2 , v2 v3 and xy with
the nonedges v1 x, v3 y and v1 v3 to construct a realization of π containing the ncycle v1 v3 v4 · · · vn+1 v1 . If, instead, dG (x) = 0 for each vertex in V (G) − C, then
V (G) − C along with v1 , v3 and v5 form an independent set of size t unless v1 v5
is in E(G). Then, however, we can exchange the edges v1 v5 , v2 v3 and v3 v4 for the
nonedges v1 v3 , v3 v5 and v2 v4 to construct a realization of π containing the n-cycle
v1 v2 v4 v5 . . . vn+1 .
Thus, we can assume that there is some vertex x ∈ V (G)−C such that vi ∈ N (x)
for some index i. If xvi+3 ∈ E(G), then G contains a cycle of length n, so there
exists an index j such that vj ∈ N (x) and vj+1 ∈
/ N (x). Now, replacing the edges
xvj and vj+1 vj+2 with the nonedges xvj+1 and vj vj+2 gives a realization G0 of π
in which vj vj+2 vj+3 . . . vn+1 v1 v2 . . . vj is a cycle of length n.
Case 2: Some realization G of π contains a cycle of length n + 2.
6
Let C = v1 v2 · · · vn vn+2 v1 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 vi vi+2 nor
vi vi+3 are edges of G for any index i. If C is not an induced cycle, choose a chord
vi vj ∈ E(G) so that distC (vj , vi ) > 3 is minimum. Next, we note that by replacing
the edges vi vi+1 , vi vj and vi+2 vi+3 with the nonedges vi vi+2 , vi vi+3 and vi+1 vj we
obtain a realization of π in which vi vi+3 vi+4 · · · vn+2 v1 v2 · · · vi is a cycle of length
n.
Therefore, we can assume that C is an induced Cn+2 in G. Suppose then that
there is a vertex x in V (G) − C such that xvi is an edge of G for some i. Then
either v1 v2 · · · vi xvi+3 vi+4 · · · vn+2 v1 is a cycle of length n + 1, or we can choose an
index j such that xvj ∈ E(G) but xvj+1 ∈
/ E(G). Then, by exchanging the edges
vj x and vj+1 vj+2 with the nonedges vj vj+2 and xvj+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 ∈
/ {v1 , . . . , vn+2 }, then replacing the edges v1 v2 , v2 v3 and xy with the
edges v1 v3 , v2 x and v2 y yields a realization of π where v1 v3 v4 · · · vn+2 v1 is a cycle
of length n + 1. Thus G is isomorphic to Cn+2 ∪ Kt−4 , and as n + 2 ≥ 8, α(G) ≥ t.
Case 3: Some realization G of π contains a cycle of length m > n + 2.
Choose a realization G containing a cycle v1 v2 · · · vm v1 with m > n + 2 maximum. If vn ∼ v1 , then the vertices v1 , v2 , . . . , vn obviously give the desired subgraph
isomorphic to Cn . Similarly, since we can assume G contains no cycle of length n+2,
vn+1 6∼ vm , as this edge would complete a cycle of length n + 2. Thus, replacing
the edges vm v1 and vn vn+1 with the non-edges vn v1 and vn+1 vm gives a graph G0
realizing π which contains Cn .
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 = v1 v2 . . . vm
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 x1 x2 · · · xp x1 be a cycle in H. Then, if vi xj ∈
E(G), it follows that neither vi+1 xj ∈ E(G) nor vi+2 xj+1 ∈ E(G). Otherwise,
v1 v2 · · · vi xj vi+1 vi+2 · · · vm v1 or v1 v2 · · · vi xj xj+1 vi+2 · · · vm v1 is a cycle of length
greater than m in G, a contradiction. Thus we conclude, without loss of generality,
that v1 6∼ x2 and v2 6∼ x1 , and by replacing the edges v1 v2 and x1 x2 with v1 x2 and
v2 x1 , we obtain a realization of π with a cycle v1 x2 x3 · · · xp x1 v2 v3 · · · vm v1 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 dH (x) ≥ 2 and
dH (y) ≥ 2. Clearly, we can choose x and y so that xy ∈ E(G). Since H is acyclic,
then for any vertices x0 ∈ NH (x) − {y} and y 0 ∈ NH (y) − {x} we have x0 6= y 0 and
/ E(G)
x0 6∼ y 0 . As the circumference of G is m, there exists an index i such that vi x ∈
and vi+1 y ∈
/ E(G). By replacing xx0 , yy 0 and vi vi+1 with vi x, vi+1 y and x0 y 0 , we
obtain a graph G0 realizing π containing the (m + 2)-cycle v1 · · · vi xyvi+1 · · · vm v1 ,
a contradiction. Hence H has at most one vertex of degree at least two.
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 dH (x) > 1, and choose vertices x0 and x00 in N (x),
and and edge yy 0 in a component of H different from the component containing
x. We may assume, without loss of generality, that v1 6∼ x. If, in addition, v2 6∼ x
then v1 v2 xx0 x00 xv1 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 vm 6∼ x so both v2 x and vm x are edges in G. Now, if
v2 y ∈
/ E(G), then replacing edges xx0 , yy 0 and v1 v2 with v1 x, v2 y and y 0 x0 yields a
realization of π in which v1 xv2 · · · vm v1 is a cycle of length m + 1, which contradicts
the maximality of m. We can therefore assume v2 y ∈ E(G), and by an identical
argument, that vm y 0 ∈ E(G). But then v2 · · · vn y 0 yv2 is a cycle of length m + 1 in
G. This last contradiction establishes the claim.
Claim 3. If the only non-trival component of H is a star, then π → (Cn , Kt ).
Let x be the unique vertex in V (H) with dH (x) > 1. Clearly, by the maximality
of m, vi x ∈ E(G) implies that vi+1 6∼ 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 ∼ vi
if and only if i is even. Next, observe that if we can choose an odd index i such
that vi ∼ y for some y ∈ V (H) − x, then either v1 v2 · · · vi yxvi+1 vi+2 · · · vm v1 is
a cycle of length m + 1 in G or, xy ∈
/ E(G). In the latter case, we can exchange
the edges vi y and xx0 with the nonedges vi x and x0 y for any x0 ∈ NH (x) to obtain
a realization of π in which v1 v2 · · · vi xvi+1 vi+2 · · · vm v1 is a cycle of length m + 1.
Thus, we conclude that no vertex in V (H) is adjacent to v1 or v3 . If v1 ∼ v3 ,
we have the cycle v1 v3 v2 xv4 v5 · · · vm v1 of length m + 1 in G, and if v1 6∼ v3 , then
V (H) ∪ {v1 , v3 } − {x} is a set of at least (k − m) + 2 − 1 = (n − m) + t − 1 ≥ t
independent points in G, completing the proof of the claim.
As a result of Claims 1 - 3, 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 vi is incident with
at least one vertex of H, or we can choose a vertex vj such that V (H) ∪ {vj } is an
independent set of size at least t. If every vi 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
vi and vj with i < j, and by the maximality of m, i < j − 1. We choose x and vi
vj so that j − i > 1 is as small as possible. We now choose y ∈ V (H) ∩ N (vi+1 ),
and observe that y 6= x. The minimality of j − i ensures that vj 6∼ y and vi+1 6∼ x.
Thus, the graph obtained by replacing edges vi+1 y and vj x with vi+1 x and vj y is a
realization of π which contains a cycle v1 v2 · · · vi xvi+1 vi+2 vn v1 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 vi x ∈ E(G) for any index i, then the maximality
of m guarantees that vi±1 6∼ x and vi±2 6∼ y. In this case, by exchanging the edges
vi+1 vi+2 and xy with the nonedges vi+1 x and vi+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
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 vj is adjacent to no vertex of H. Since m = n − 1 ≥ 5, we can then choose
an index i 6= j, j − 1 so that replacing the edges vi vi+1 and xy with vi x and vi+1 y
gives realization of π with t independent points (the vertices in V (H) together with
the vertex vj ).
Next, assume that H contains 2` ≥ 4 vertices of degree one and p isolated
vertices and choose xx0 , yy 0 ∈ E(H). If vi x ∈ E(G) but vi 6∼ x0 , then we can
produce a realization of π in which ∆(H) = 2 by replacing edges vi x, xx0 and yy 0
with the nonedges xy, xy 0 and vi x0 . The previous claims then imply π → (Cn , Kt ).
Furthermore, we note that by replacing the edges xx0 and yy 0 with the edges xy
and x0 y 0 in H, we obtain another realization of π, and as a result the preceeding
argument immediately implies that vi y ∈ E(G). Thus, each vertex vi adjacent to
any vertex x with dH (x) = 1 is adjacent to every vertex of H with degree one. Let
S = {vi | vi ∼ x for some x with dH (x) = 1}, and let s = |S|.
First, suppose that S = ∅ and that xi yi , 1 ≤ i ≤ ` are the (disjoint) edges in
H. If ` ≥ m, then we can exchange xi yi and the edge vi vi+1 for the nonedges xvi
and yvi+1 . This creates a realization of π in which the vertices of H contain an
independent set of size p+`+m. If p+`+m ≥ t we are done, so suppose otherwise.
Then k − (p + ` + m) = ` ≥ n − 1, which implies that k ≥ 2n − 2, a contradiction.
Hence we may assume that ` < 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 + 2`. As k = n + t − 2 and m ≤ n − 1, this implies that, in fact,
p + 2` = t − 1 and m = n − 1. Now, as above, if every vertex vi has a neighbor in
H, we can construct a realization of G containing
so there is some
a longer cycle,
t−1
and
`
≤
,
we
conclude that
vertex vj with no neighbor in H. As t ≤ 2n
3
2
` ≤ n − 3 = m − 2, so we can use edges in E(C) that are not incident to vj to
“break” the ` edges in H and create a realization of π in which V (H) ∪ {vj } forms
an independent set of size t, as desired.
Thus, we now assume that S 6= ∅. Observe that for every vi ∈ S, the maximality
of m implies that vi+2 and vi−2 are not adjacent to any vertex of V (H) with
degree one. Additionally, the vertices vi+1 and vi−1 are not adjacent to any vertex
x ∈ V (H), as we would either have a cycle of length m + 1 if dH (x) = 1, or, if
dH (x) = 0, we can choose yy 0 ∈ E(H) and replace vi±1 x and vi y with vi±1 y and
vi x and obtaian a realization in which v1 · · · vj y 0 yvj+1 · · · vn v1 is a cycle of length
m + 1 for j = i, i − 1.
Furthermore, we note that each edge vi vj of the graph induced by C − S, and
any edge xx0 ∈ E(H) can be replaced with the edges vi x and vj x0 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 distC (vi , vj ) ≥ 3 for any
distinct vi and vj in S, there are at least m − 2s edges in this graph. If m − 2s > `,
then we can eliminate every edge of H without using an edge incident with vi+1 for
some vi ∈ S. This gives a realization of π in which V (H) ∪ {vi+1 } is a collection of
at leat t independent vertices. Thus, we can assume that m − 2s ≤ `, 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 ` + p + (m − 2s). So either
RAMSEY-TYPE NUMBERS FOR DEGREE SEQUENCES
9
π → (Cn , Kt ) or
(1)
` + p + m − 2s ≤ t − 1.
Now, let S + = {vi+1 | vi ∈ S}, and suppose first that there is some v ∈ S +
and z ∈ V (H) such that v ∼ z and dH (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 vi+1 and vj+1 in S + , which implies that for any edge xy in H,
vi xyvj vj−1 · · · vi+1 vj+1 vj+2 · · · vi 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 → (Cn , Kt ) or
(2)
` + p + s ≤ t − 1.
Summing (1) and (2) gives 2` + 2p + m − s ≤ 2t − 2. Now, since 2` + p =
n−1
|V (H)| = n + t − 2 − m, we have (n −
s)+ p ≤ t. Note that s ≤ m
3 ≤ 3 , and
2n
2n+1
hence 3 + p ≤ t, contradicting t ≤ 3 and completing the proof.
Theorem 4.2. For t ≥ 3 and n ≥ 4 with t >
2n 3
rpot (Cn , Kt ) = 2t − 2 +
,
lnm
3
.
Proof. The lower bound stems from Theorem 2.2 and/or consideration of π(K(d n3 e−1) ∨
(t − 1)K2 ), which is unigraphic. To show the reverse inequality, let π = (d1 , . . . , dk )
be graphic with k = 2t − 2 + d n3 e. 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 = v1 v2 · · · vn vn+1 v1 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
v1 ∼ v3 and v1 v3 v4 · · · vn+1 v1 isa cycle
of length n, or V (G) − {v2 , v4 , . . . , vn+1 } is
a set of k − (n − 1) = 2t − 1 − 2n
≥
t
indpendent vertices.
3
Case 2: Some realization G of π contains a cycle v1 v2 · · · vn+2 v1 .
Again, the aguments in Case 2 of Theorem 4.1 imply that G = Cn+2 ∪ Kk−n−2
and since n ≥ 4, this graph has an indpendent set of size at least k − n + 1 ≥ t.
Case 3: Some realization of π has circumference m > n + 2.
The proof is identical to the proof of Case 3 of Theorem 4.1.
Case 4: Every realization of π has circumference m ≤ n − 1.
10
Let G be a realization of π containing a longest cycle C = v1 v2 . . . vm 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 = `K2 ∪ pK1
for some integers p and `. Proceeding in the same manner, we define the set
S = {vi | vi x ∈ E(G) for some vertex x with degH (x) = 1}. and note that vi ∈ S
implies that none of vi±1 , vi±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 2K2 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 ` + p + m − 2|S|, so either π → (Cn , Kt )
or ` + p + m − 2s ≤ t − 1. Since |V (H)| = 2` + p = 2t − 2 + 2n
3 − m, we must have
2(` + p + m − 2|S|) ≤ 2t − 2
|V (H)| + p + m − 2|S| ≤ 2t − 2
2n
− m) + p + m − 2|S| ≤ 2t − 2
(2t − 2 +
3
jmk
2n
2n
− 2|S| ≤ p +
−2
p+
≤0
3
3
3
An easy case analysis then showsthat
the left hand side is strictly positive except
when p = 0, m = n − 1, |S| = m
and n ≡ 1 (mod 3). However, this in turn
3
2n+1
requires k = 2t − 2 + n+2
,
and
|V
(H)|
= 2` = 2t − 2n+1
3
3 . As n ≡ 1 (mod 3),
3
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 rpot (Pn , Kt ). In fact, rpot (Pn , Kt ) and rpot (Cn , Kt ) differ by at most one.
Note, however, that r(Pn , Kt ) = (n − 1)(t − 1) + 1 [18], which is the conjectured
value for r(Cn , Kt ) for all n ≥ t ≥ 3.
Theorem 4.3. For n ≥ 6 and t ≥ 3,
rpot (Pn , Kt ) =

 n + t − 2,

2t − 2 +
if t ≤
n
3
,
if t >
2n 3
2n 3
,
.
Proof. When t ≤ 2n
3 , we note that rpot (Pn , Kt ) ≤ rpot (Cn , Kt ) = n + t − 2, and
equality is established via Corollary 2.3 and/or by observing that Kn−1 ∪ Kt−2
contains neither Pn nor Kt .
2n Therefore,
n we may assume t > 3 . Theorem 2.2 establishes that rpot (Pn , Kt ) ≥
2t − 2 + 3 and if n ≡ 0 (mod 3), we observe that rpot (Pn , Kt ) ≤ rpot (Cn , Kt ) =
2t − 2 + n3 , establishing the result.
Finally, for n 6≡ 0 (mod
3), we note that by Theorem 4.1, every degree sequence
π with k ≥ 2t − 2 + n3 terms has a realization G which contains either a subgraph
isomorphic to Cn−2 or an independent set of cardinality t. If the
latter, π →
(Pn , Kt ), and if the former, we note that there are at least 2t − 2 + n3 − (n − 2) =
2t − d 2n
3 e ≥ t vertices in V (G) − V (Cn−2 ). Since we are assuming α(G) < t, there is
RAMSEY-TYPE NUMBERS FOR DEGREE SEQUENCES
11
at least one edge xy ∈ E(G) with x, y ∈
/ V (Cn−2 ). Now, without loss of generality,
either vx ∈ E(G) and G contains a subgraph isomorphic to Pn , or we can select any
edge vi vi+1 ∈ E(Cn−2 ) and exchange the edges vi vi+1 and xy with the non-edges
vi x and vi+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 rpot . Let
(i)
(i)
πi , . . . , πk be graphic sequences, with πi = (d1 , . . . , dn ) for all i (they need not
be monotone). Then, as defined in [1], π1 , . . . , πk pack if there exist edge-disjoint
graphs G1 , . . . , Gk , all with vertex set {v1 , . . . , vn }, such that
(i)
dGi (vj ) = dj
for each i, and
dG1 ∪···∪Gk (vj ) =
n
X
(i)
dj .
i=1
In traditional graph packing, the vertex sets of the graphs in question may be permuted prior to their embedding into Kn . Note that when packing degree sequences,
the ordering of the terms in π1 , . . . , πk remains fixed.
With this definition in mind, we define rpot (G1 , . . . , Gk ) to be the smallest integer n such that for any collection of n-term graphic sequences π1 , . . . , πk that sum,
termwise, to n − 1 and pack, there exist edge disjoint graphs F1 , . . . , Fk all with
(i)
vertex set {v1 , . . . , vn }, such that dFi (vj ) = dj for all i, j and also that Fi contains
Gi 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 rpot (G1 , . . . , Gk ) for k ≥ 3 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, 2011.
[2] Z. Dvorak and B. Mohar. Chromatic number and complete graph substructures for degree
sequences. Arxiv preprint arXiv:0907.1583, 2009.
[3] P. Erdős, R. J. Faudree, C. C. Rousseau, and R. H. Schelp. On cycle-complete graph Ramsey
numbers. J. Graph Theory, 2(1):53–64, 1978.
[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 439–449. Wiley, New York, 1991.
[5] Paul Erdős and Tibor Gallai. Graphs with prescribed degrees of vertices (hungarian). Mat.
Lapok, 11:264–274, 1960.
[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 500–515. Springer, Berlin, 1978.
[7] Michael J. Ferrara and John Schmitt. A general lower bound for potentially H-graphic sequences. SIAM J. Discrete Math., 23(1):517–526, 2008/09.
12
[8] John Frederick Fink and Michael S. Jacobson. On n-domination, n-dependence and forbidden
subgraphs. In Graph theory with applications to algorithms and computer science (Kalamazoo, Mich., 1984), Wiley-Intersci. Publ., pages 301–311. Wiley, New York, 1985.
[9] L. Gerencsér and A. Gyárfás. On Ramsey-type problems. Ann. Univ. Sci. Budapest. Eötvös
Sect. Math., 10:167–170, 1967.
[10] S. Hakimi and Edward Schmeichel. Graphs and their degree sequences: A survey. In Yousef
Alavi and Don Lick, editors, Theory and Applications of Graphs, volume 642 of Lecture Notes
in Mathematics, pages 225–235. Springer Berlin / Heidelberg, 1978. 10.1007/BFb0070380.
[11] S. L. Hakimi. On realizability of a set of integers as degrees of the vertices of a linear graph.
I. J. Soc. Indust. Appl. Math., 10:496–506, 1962.
[12] Frank Harary. Recent results on generalized Ramsey theory for graphs. In Graph theory and
applications (Proc. Conf., Western Michigan Univ., Kalamazoo, Mich., 1972; dedicated to
the memory of J. W. T. Youngs), pages 125–138. Lecture Notes in Math., Vol. 303. Springer,
Berlin, 1972.
[13] Václav Havel. Eine Bemerkung über die Existenz der endlichen Graphen. Časopis Pěst. Mat.,
80:477–480, 1955.
[14] Chunhui Lai. An extremal problem on potentially Kp,1,1 -graphic sequences. Discrete Math.
Theor. Comput. Sci., 7(1):75–79 (electronic), 2005.
[15] Jiong-Sheng Li and Zi-Xia Song. The smallest degree sum that yields potentially Pk -graphical
sequences. J. Graph Theory, 29(2):63–72, 1998.
[16] Rong Luo. On potentially Ck -graphic sequences. Ars Combin., 64:301–318, 2002.
[17] V. Nikiforov. The cycle-complete graph Ramsey numbers. Combin. Probab. Comput.,
14(3):349–370, 2005.
[18] T. D. Parsons. The Ramsey numbers r(Pm , Kn ). Discrete Math., 6:159–162, 1973.
[19] Stanislaw P. Radziszowski. Small Ramsey numbers. Electron. J. Combin., 1:Dynamic Survey
1, 30 pp. (electronic), 1994.
[20] A. R. Rao. The clique number of a graph with a given degree sequence. In Proceedings of the
Symposium on Graph Theory (Indian Statist. Inst., Calcutta, 1976), volume 4 of ISI Lecture
Notes, pages 251–267, New Delhi, 1979. Macmillan of India.
[21] S. B. Rao. A survey of the theory of potentially P -graphic and forcibly P -graphic degree
sequences. In Combinatorics and graph theory (Calcutta, 1980), volume 885 of Lecture Notes
in Math., pages 417–440. Springer, Berlin, 1981.
[22] Neil Robertson and Zi-Xia Song. Hadwiger number and chromatic number for near regular
degree sequences. J. Graph Theory, 64(3):175–183, 2010.
[23] Jian-Hua Yin and Jiong-Sheng Li. Two sufficient conditions for a graphic sequence to have a
realization with prescribed clique size. Discrete Math., 301(2-3):218–227, 2005.
[24] Jian Hua Yin and Jiong Sheng Li. A variation of a conjecture due to Erdős and Sós. Acta
Math. Sin. (Engl. Ser.), 25(5):795–802, 2009.
[25] Jian-Hua Yin, Jiong-Sheng Li, and Guo-Liang Chen. A variation of a classical Turán-type
extremal problem. European J. Combin., 25(7):989–1002, 2004.