COLORING NUMBER AND ON-LINE RAMSEY THEORY FOR GRAPHS AND HYPERGRAPHS

COMBINATORICABolyai Society – Springer-Verlag

0209–9683/109/$6.00 c©2009 Janos Bolyai Mathematical Society and Springer-Verlag

Combinatorica 29 (1) (2009) 49–64DOI: 10.1007/s00493-009-2264-1

COLORING NUMBER AND ON-LINE RAMSEY THEORYFOR GRAPHS AND HYPERGRAPHS

H. A. KIERSTEAD, GORAN KONJEVOD

Received December 31, 2005

Let c,s, t be positive integers. The (c,s, t)-Ramsey game is played by Builder and Painter.Play begins with an s-uniform hypergraph G0 =(V,E0), where E0 =∅ and V is determinedby Builder. On the ith round Builder constructs a new edge ei (distinct from previousedges) and sets Gi =(V,Ei), where Ei =Ei−1∪{ei}. Painter responds by coloring ei withone of c colors. Builder wins if Painter eventually creates a monochromatic copy of Kt

s,the complete s-uniform hypergraph on t vertices; otherwise Painter wins when she hascolored all possible edges.

We extend the definition of coloring number to hypergraphs so that χ(G)≤col(G) forany hypergraph G and then show that Builder can win (c,s, t)-Ramsey game while buildinga hypergraph with coloring number at most col(Kt

s). An important step in the proof isthe analysis of an auxiliary survival game played by Presenter and Chooser. The (p,s, t)-survival game begins with an s-uniform hypergraph H0 = (V,∅) with an arbitrary finitenumber of vertices and no edges. Let Hi−1 =(Vi−1,Ei−1) be the hypergraph constructedin the first i − 1 rounds. On the i-th round Presenter plays by presenting a p-subsetPi ⊆Vi−1 and Chooser responds by choosing an s-subset Xi ⊆Pi. The vertices in Pi −Xi

are discarded and the edge Xi added to Ei−1 to form Ei. Presenter wins the survival gameif Hi contains a copy of Kt

s for some i. We show that for positive integers p,s, t with s≤p,Presenter has a winning strategy.

1. Introduction

For a positive integer n and a set S, let [n] denote the set {1, . . . ,n} and(Sn

)

denote {X⊆S : |X|=n}. An s-uniform hypergraph is a structure H =(V,E),where E ⊆

(Vs

). Elements of V are called vertices and elements of E are

Mathematics Subject Classification (2000): 05D10, 05C55, 05C65, 03C13, 03D99

http://dx.doi.org/10.1007/s00493-009-2264-1

50 H. A. KIERSTEAD, GORAN KONJEVOD

called edges or s-edges. At times we will refer to s-uniform hypergraphs ass-graphs. Then ordinary graphs are 2-graphs. Let Kt

s denote the complete s-graph on t vertices defined by Kt

s =([t],

([t]s

)). For s-graphs G and H we write

G→c H, if every c-coloring of the s-edges of G results in a monochromaticcopy of the target H.

For positive integers s,c, t the Ramsey number Ramsc(t) is the least integer

n such that Kns →c Kt

s. Estimating Ramsey numbers, even for graphs, isa notoriously difficult problem. This has led researchers to consider otherversions of the problem. Suppose that f is an increasing graph parameter.For positive integers s,c, t, define the f -Ramsey number, f -Rams

c(t), to bethe least integer n for which there exists an s-graph G such that f(G)= nand G→c Kt

s. Trivially, we have

f(Kts) ≤ f - Rams

c(t) ≤ f(Kns ), where n = Rams

c(t).

Erdos, Faudree, Rousseau and Schelp [2] studied the size Ramsey num-ber, obtained when f(G) = size(G), the number of edges of G. Erdos et al.showed, with a proof attributed to Chvatal, that the size Ramsey number forgraphs is exactly the trivial upper bound

(n2

), n=Ram2

c(t). In other words,allowing more vertices will not reduce the number of edges needed to forcea monochromatic clique. The coloring number of a graph G= (V,E) is theleast integer d such that its vertices can be ordered as v1≺·· ·≺vn so that

∣∣{vi : i < j ∧ {i, j} ∈ E}∣∣ < d for all j ∈ [n].

Let χ(G) be the chromatic number of G. Clearly, χ(G) ≤ col(G), since itsvertices can be colored with col(G) colors using First-Fit on the enumera-tion v1, . . . ,vn. Recently, Kurek and Rucinski [4] considered the chromaticand coloring Ramsey numbers obtained when f is χ and col, respectively.They observed that the trivial upper bound for the chromatic Ramsey num-ber is again tight, and thus the trivial upper bound is also tight for thecoloring Ramsey number.

From a Ramsey theoretic perspective, these results are disappointing.Kurek and Rucinski suggested a more promising line of inquiry might beto study on-line versions of f -Ramsey numbers. This seems encouragingbecause many Ramsey theoretic constructions are inherently on-line.

In the on-line setting we consider a process in which edges (not vertices)are generated one at a time and then immediately and irrevocably colored byan on-line algorithm. This is best understood as a game played between twoplayers, Builder and Painter. For positive integers c,s, t the (c,s, t)-Ramseygame (with target Kt

s) is played as follows. Play begins with an empty s-graph G0 =(V,E0) on an arbitrarily large, but finite, vertex set V determined

COLORING NUMBER AND ON-LINE RAMSEY THEORY FOR GRAPHS 51

by Builder. (So E0 =∅.) The game is played in rounds. At the beginning ofthe ith round Builder will have constructed an s-graph Gi−1 =(V,Ei−1) with|Ei−1|= i−1 and Painter will have constructed a coloring fi−1 :Ei−1 → [c].On the ith round Builder constructs a new edge ei (distinct from previousedges) and sets Gi = (V,Ei), where Ei = Ei−1 ∪{ei}. Painter responds bycoloring ei to obtain a coloring fi : Ei → [c] with fi−1 ⊆ fi. Builder wins ifPainter eventually creates a monochromatic copy of Kt

s; otherwise Painterwins when she has colored all

(|V |s

)edges.

For positive integers s,c, t and an increasing graph parameter f , definethe on-line f -Ramsey number, f -oRams

c(t), to be the least integer n suchthat Builder can force a win in the (c,s, t)-Ramsey game while constructinga graph G with f(G)=n. Trivially,

f(Kts) ≤ f - oRams

c(t) ≤ f -Ramsc(t) ≤ f(Kn

s ), where n = Ramsc(t).

Kurek and Rucinski conjectured:

Conjecture 1. For all positive integers c and t, limt→∞

size-oRam2c(t)

size-Ram2c(t)

(t)=0.

In some sense this appears to be a trick question, since the heart of theproblem still seems to be to improve known estimates of Ram2

c(t). A stan-dard construction for bounding Ram2

c(t), and hence size-Ram2c(t), provides

a much stronger bound for size-oRam2c(t): The game is played on a set of

vertices V0 of cardinality cct−c. During the ith stage of the game Builderpicks a vertex xi ∈ Vi−1 and constructs all possible edges between xi andVi−1\{xi}. Painter must color at least 1

c |Vi−1| of these edges with the samecolor αi. Now Builder will only play on the set Vi of vertices linked to xi byan edge colored αi. At the end of this process the set {xi : i ∈ [ct− c+1]}will contain a monochromatic copy of Kt

2. This shows that Ram2c(t) ≤ cct,

size-Ram2c(t)≤

(cct

2

)and

size-oRam2c(t) ≤

ct−c∑

i=0

ci ≤ cct.

Thus if cct is a good bound on Ram2c(t) then we have a positive answer to

the question of Kurek and Rucinski. Of course, there may exist a differentstrategy for Builder that requires even fewer edges.

Another approach is to change the question. Let us consider the on-linecoloring Ramsey number, col-oRam2

c(t). The construction above shows thatcol-oRam2

c(t)≤ct: Order the vertices so that for all i the vertices in Vi\Vi+1

precede the vertices in Vi+1. Since each vertex in Vi+1 is adjacent to at most


one vertex in each Vj\Vj+1 for j <i, the coloring number is at most ct. Thusby using a well known lower bound on the Ramsey number, we have

limt→∞

col-oRam2c(t)

col-Ram2c(t)

≤ ct√

2t = 0.

The main result of this paper is that much more is true. First we prove thefollowing theorem that shows that the trivial lower bound on col-oRam2

c(t)is tight even though the trivial upper bound is tight for χ-Ram2

c(t).

Theorem 2. For all positive integers c,t, col-oRam2c(t)=χ(Kt

2)=col(Kt2)=

t.

Next we extend the definition of coloring number to hypergraphs in anatural way so that χ(G)≤ col(G) for all hypergraphs G. Finally we proveour main result:

Theorem 3. For all positive integers c,s, t, col-oRamsc(t)=χ(Kt

s)=col(Kts).

Our techniques were first used in [3], where it is shown that χ-oRam22(t)=

t for every positive integer t and col-oRam2c(3)=3 for every positive integer c.

As in [3] our main tool is the analysis of an auxiliary game called survival,which seems to be interesting in its own right. The novelty of the currentpaper is that our previous analysis of survival for graphs is extended tohypergraphs. This is needed, even in the case of graphs (Theorem 2) toextend the results of [3] to arbitrary c and t. This paper is organized asfollows. In Section 2 we introduce the survival game and prove our keytechnical result about it. In Section 3 we use our result on survival to proveTheorem 2. In Section 4 we introduce our generalization of coloring numberto hypergraphs and show how to modify the proof of Theorem 2 to proveTheorem 3.

1.1. Notation and terminology

Let tn denote the sequence t1, . . . , tn. So t0 is the empty sequence λ. It willalso occasionally be convenient to let t0 = λ. Also we may abuse notationand write tn for {t1, . . . , tn}. Let G=(V,E) be a graph and X,Y ⊆V . ThenE (X,Y ) denotes the set of edges with at least one vertex in X and at leastone vertex in Y . Let K(X,Y ) denote the bipartite complete graph withbipartition {X,Y }. Notice that this notation makes sense even when G is ahypergraph. In this case K(X,Y )=

{e∈

(X∪Ys

):e∩X �=∅ �=e∪Y

}. If X ={x}

we may write E(x,Y ) and K(x,Y ) instead of E({x},Y ) and K({x},Y ). If


V has been ordered as v1≺·· ·≺vn and vivj is an edge with 1≤ i<j≤n thenwe say that vi is a back-neighbor of vj and vj is a forward-neighbor of vi. Theback-degree of a vertex is the number of back-neighbors that it has. Thus agraph has coloring number t if there exists an ordering of its vertex set suchthat every vertex has back-degree at most t−1. A forward-edge of a vertex vis an edge to a forward-neighbor of v.

2. The Survival Game

Let p,s, t be positive integers with s ≤ p. The (p,s, t)-survival game isplayed by two players, Presenter and Chooser. Play begins with the s-graphH0 = (S0,E0), where S0 is an arbitrarily large, but finite, set of verticesdetermined at the beginning of the game by Presenter and E0 = ∅. Thegame is played in rounds. At the beginning of the ith round the playerswill have constructed an s-graph Hi−1 =(Si−1,Ei−1). During the ith roundthey construct Hi = (Si,Ei) as follows. Presenter plays by presenting a p-subset Pi ⊆ Si−1. Chooser responds by choosing an s-set Xi ⊆ Pi. The re-maining vertices in Pi−Xi are discarded, leaving Si =Si−1− (Pi−Xi) andEi =(Ei−1∪{Xi})−{Xj ∈Ei−1 :Xj �Si}. The vertices in Si are called surviv-ing vertices. Presenter wins if Hi contains a copy of Kt

s for some i. OtherwiseChooser wins when eventually |Si|<t as then Presenter cannot make a play.

Theorem 4. For all positive integers p,s, t with s ≤ p, Presenter has awinning strategy in the (p,s, t)-survival game.

Our approach to proving the theorem will be to show recursively thatPresenter can meet a certain series of goals. We view the starting position,a graph with a huge number of vertices, but no edges, as having great po-tential, but being very far from the final goal of a complete subgraph. AsPresenter progress through these goals he will be sacrificing potential forgraphs that in a certain usable sense are closer and closer to being com-plete. In order to express these goals we will first develop a new logicalsystem consisting of models (partitioned s-graphs), basic formulas and rulesfor satisfaction.

We assume that the initial set S0 of vertices on which the game is playedis ordered by the relation <. By a partitioned s-graph we mean a structureH =(U,W,E), where (U ∪W,E) is an s-uniform hypergraph and U and Ware disjoint subsets of S0. The set U∪W inherits the order of S0. We call Uthe universal set and W the witness set. The partition of vertices into twosets does not imply any kind of bipartite property. In particular, edges maybe contained in either subset of vertices. The order of a partitioned s-graph,


which measures its potential, is the cardinality of its universal set. Given apartitioned s-graph (U,W,E), we define a relation � on (U ∪W ) by v � v′

iff v < v′ and if v′ ∈W then {u∈U : v < u < v′}= ∅. Also λ�v′ means thatif v′ ∈ W then {u ∈ U : u < v′} = ∅. We view v � v′ as indicating that v′ isbigger than v, but not by too much. Given a partitioned s-graph (U,W,E)and a sequence of vertices vn∈U ∪W , we say that vn is strongly increasingiff λ�v1 and vi �vi+1 for all i∈ [n−1].

The concept of a basic formula is defined recursively as follows. Thelanguage for basic formulas consists of an s-ary predicate P , intended toindicate strongly increasing s-edges, and exactly s variables ξs = ξ1, . . . ,ξs.The variable ξi is only allowed to appear in the ith position of P . There areno logical connectives, but universal and existential quantifiers may appear.

1. P (ξs) is the only basic formula with free variables ξs.2. ϕ=∀ξh+1ψ is a basic formula with free variables ξh iff ψ is a basic formula

with free variables ξh+1.3. ϕ=∃ξh+1ψ is a basic formula with free variables ξh iff ψ is a basic formula

with free variables ξh+1.

Finally we recursively define satisfaction for basic formulas. In doing sowe give a special (and definitely nonstandard) interpretation of the mean-ing of universal and existential quantifiers. Roughly, it says that universalquantification only applies to sufficiently large vertices in the universal set U ,while existential quantification only applies to vertices in a specific range ofthe witness set W . For notational convenience, let v0 =λ=v0. Also u>λ isdefined to be true.

Definition 5. Let H be a partitioned s-graph as above, ϕ be a basic formulawith free variables ξh, 0≤h≤ s, and vh ⊆U ∪W be a sequence of vertices.Then H satisfies ϕ(vh) iff one of the following holds:

1. ϕ=P (ξs), vh∈E, and vh is increasing (so h=s).2. ϕ=∀ξh+1ψ and H satisfies ψ(vh,u) for all u∈U with u>vh.3. ϕ=∃ξh+1ψ and H satisfies ψ(vh,w) for some w∈W with w�vh.

In particular, H satisfies a basic formula ϕ with no free variables, ifH satisfies ϕ(v0), i.e., ϕ(λ). Roughly, the next lemma states that a basicformula satisfied by a partitioned s-graph remains true if we restrict theuniversal set, while expanding the witness set. This is needed for the mainstep in the proof of Lemma 7. The intricate definitions above and hypothesesbelow are all needed to prove a statement that can be applied.


Lemma 6. Suppose H =(U,W,E) and H ′ =(U ′,W ′,E′) are partitioned s-graphs, ϕ is a basic formula with free variables ξh and vh⊆(U∪W )∩(U ′∪W ′)is a sequence of vertices. Suppose the following conditions are all satisfied:

1. If ys∈E then ys∈E′ for all ys⊆(U ∪W )∩(U ′∪W ′).2. U ′−{u∈U ′ :u≤vh}⊆U .3. W −{w∈W :w≤vh}⊆W ′.

If H satisfies ϕ(vh) then H ′ satisfies ϕ(vh).

Proof. We argue by induction on the definition of a basic formula. Forthe base step, take ϕ = P (ξs) and suppose that H satisfies ϕ(vs). Then byDefinition 5.1 and Hypothesis 1, H ′ satisfies ϕ(vs).

Let us now consider the induction step. First suppose that H satis-fies ϕ(vh), where ϕ = ∀ξh+1ψ and consider u ∈ U ′ with u > vh. Then, byHypothesis 2, u∈U . So H satisfies ψ(vh,u) by Definition 5.2. By the induc-tion hypothesis H ′ satisfies ψ(vh,u). Since u was arbitrary, we conclude byDefinition 5.2 that H ′ satisfies ϕ(vh).

Now suppose that H satisfies ϕ(vh), where ϕ=∃ξh+1ψ. By Definition 5.3,there exists a w∈W with w�vh such that H satisfies ψ(vh,w). By Hypoth-esis 3, w ∈ W ′. So, by the induction hypothesis, H ′ satisfies ψ(vh,w). ByHypothesis 2, H ′ satisfies w�vh. Thus by Definition 5.3, H ′ satisfies ϕ(vh).

The rank r(ϕ) of a basic formula ϕ is defined recursively as follows.

r(P (ξs)) = 0;r(∀ξh+1ψ) = 1 + 2r(ψ);r(∃ξh+1ψ) = 2r(ψ).

A sentence is a formula with no free variables. A basic sentence ϕ has theform

ϕ = Q1ξ1 . . . QsξsP (ξs)

where each Qi ∈ {∀,∃}. Note that there are exactly 2s basic sentences andthat each of them has a unique rank between 0 and 2s−1. Let ϕr denote thebasic sentence with rank r. The type of ϕ is the maximum � such that Qi =∀for all i∈ [�]. So ϕ has type 0 if Q1 =∃. Suppose ϕ=∀ξ�∃ξ�+1ψ has type �.Then we set ϕ+ =∃ξ�∀ξ�+1ψ. Notice that ϕ+

r =ϕr+1. Presenters ith goal willbe to force an s-graph with sufficiently large potential that satisfies ϕi.

Let ϕ be a basic sentence and consider the set A of partitioned s-graphsG = (U,W,E) such that G has order at least n and G satisfies ϕ. Recallthat Hi = (Si,Ei) is the s-graph constructed after i rounds of the survivalgame. We say that Presenter can construct a partitioned s-graph in A, if he


can force the play so that eventually Hi contains some subgraph (V,E) suchthat V can be partitioned as {U,W} so that G=(U,W,E)∈A. Let f :N→Nbe a function. The sentence ϕ is f -satisfiable if, starting from f(n) vertices,Presenter can construct a partitioned s-graph of order n that satisfies ϕ.Our plan is to show by induction on rank that for every basic sentence ϕr,there exists a function fr such that ϕr is fr-satisfiable. This will prove thetheorem, since, if H = (U,W,E) satisfies ϕ2s−1 = ∀ξ1 . . .∀ξsP (ξs) and hasorder t then U induces Kt

s. The next lemma provides the induction step.

Lemma 7. Let ϕ = ∀ξ�∃ξ�+1ψ be a basic sentence of type � < s. Supposethat for some function f , the sentence ϕ is f -satisfiable. Then the sentenceϕ+ =∃ξ�∀ξ�+1ψ is F -satisfiable, where F is defined recursively by

F (0) = s

F (j + 1) = f(F (j)), if j ≥ 0.

Proof. Fix a positive integer n. We shall construct a partitioned s-graphH ′=(U ′,W ′,E′) of order n that satisfies ϕ+ from a set U0 of F (n) verticesordered by <.

We claim that, starting from U0, Presenter can construct a sequence(Hi : i∈ [n]) with Hi =(Ui,Wi,Ei) such that for all i∈ [n]:

1. |Ui|=F (n− i),2. Ui∪Wi⊆Ui−1, and3. Hi satisfies ϕ.

By the definition of F we have |U0|= F (n) = f(F (n−1)). Since ϕ is f -satisfiable, Presenter can construct H1 in U0 such that H1 has order F (n−1)and satisfies ϕ. Arguing by induction on i, if |Ui−1|=F (n−i+1)=f(F (n−i)),then Presenter can construct Hi in Ui such that Hi has order F (n− i) andsatisfies ϕ.

Let u�⊆Un be an increasing sequence of length � (if �=0 then u� =λ=u�).Such a sequence exists since |Un| ≥F (0) = s≥ �+1. Since Ui+1 ⊆Ui for alli∈ [n−1], u�⊆Ui for all i∈ [n].

By construction, Hi satisfies ϕ for each i. Hence by Definition 5.2, Hi

satisfies ∃ξ�+1ψ(u�,ξ�+1). Thus by Definition 5.3, for each i there exists awi∈Wi such that Hi satisfies ψ(u�,wi) and u� �wi.

Define H ′=(U ′,W ′,E′) by

U ′ = {w1, . . . , wn},W ′ = u� ∪ (W1 − {w1}) ∪ · · · ∪ (Wn − {wn}), andE′ = E[U ′ ∪ W ′].


By the construction of Hi, Wi⊆Ui−1; so Wi∩Wi−1 =∅ and thus wi−1 �=wi.Moreover, wi−1 <wi, since wi∈Ui−1 and u��wi−1 in Hi−1 implies that thereare no elements of Ui−1 between u� �wi−1. It follows that H ′ has order n.

We claim that H ′ satisfies ϕ+. First note that u� is a strongly increasingsequence in H ′ and that u� ⊆W ′. Thus by Definition 5 it suffices to showthat H ′ satisfies ψ(u�,u) for all u∈U ′.

Consider u ∈ U ′. Then u = wi for some i ∈ [n]. Recall that Hi satis-fies ψ(u�,wi). Furthermore, the partitioned s-graphs Hi and H ′ together withthe sequence u�,wi (= u�,u) satisfy the conditions of Lemma 6: if wi < wj

then as above wj ∈ Ui; if w′ ∈ Wi and wi < w′ then w′ ∈ Wi −{wi} ⊆ W ′.Hence by the lemma, H ′ satisfies ψ(u�,u) as well.

We are finally ready to prove Theorem 4.

Proof of Theorem 4. Define the function F p:

F p(n, r) ={

n + p, r = 0F p(F p(n − 1, r), r − 1), r > 1.

We show by induction on r=r(ϕ) that for all positive integers n and basicsentences ϕ, Presenter can, starting from a set S0 of F p(n,r(ϕ)) vertices,construct partitioned s-graphs of order n that satisfy ϕ.

For the base step r =0 we have ϕ0 =∃ξsP (ξs). Presenter should presentthe first p vertices vp of S0 on his first move. Chooser must return an s-edgews and the vertices in vp−ws are discarded. Setting U =S0−vp and W = ws,we have a partitioned s-graph H =(U,W,{ws}), which satisfies ∃ξsP (ξs) andhas order n.

Now consider the induction step r > 0. By the induction hypothesis,for every k Presenter can construct a partitioned s-graph from F p(k,r−1)vertices that has order k and satisfies ϕr−1 = ∀ξ�∃ξ�+1ψ. Take k = F p(n−1,r). Then by Lemma 7 and the definition of F p, Presenter can construct apartitioned s-graph of order n that satisfies ϕr =∃ξ�∀ξ�+1ψ.

So, starting from F p(t,2s −1) vertices, Presenter can construct a parti-tioned s-graph H =(U,W,E) of order t that satisfies ϕ2s−1 =∀ξ1 . . .∀ξsP (ξs).As remarked above, H[U ]=Kt

s.

For future reference, let h(p,s, t) be the number of vertices and l(p,s, t)be the number of rounds required for Presenter to win the (p,s, t)-survivalgame. From Theorem 4, it follows that h(p,s, t)≤F p(t,2s−1).

The function F p grows very quickly. It is in fact very closely related to theAckermann function [1], which was the first example of a number-theoreticfunction that is not primitive recursive. Many versions of the Ackermann


function have been defined. For comparison, we now give the complete def-inition of one version from Peter [5]:

ψ(0, n) = n + 1ψ(m + 1, 0) = ψ(m, 1)

ψ(m + 1, n + 1) = ψ(m,ψ(m + 1, n)).

The recursive part of the definition of F p is the same as that of ψ, exceptthat the order of arguments is reversed. Thus F p grows faster than anyprimitive recursive function. However, for any fixed r, F p(n,r) is primitiverecursive as a function of n. For example, if p and s are fixed, Theorem 4gives a primitive recursive bound on the number of vertices needed for thePresenter to win the (p,s, t)-survival game.

3. Builder’s Winning Strategy

In this section we prove Theorem 2. We must show that there exists a func-tion R(c,t) such that Builder can win the (c,2, t)-Ramsey game while con-structing a graph with R(c,t) vertices, whose coloring number is t. In thestandard proof of Ramsey’s Theorem for graphs, outlined in the introduc-tion, one begins by finding a vertex v1 which is linked to a huge set V1 ofother vertices by edges of the same color. This is then iterated ct times, al-ways working inside the set Vi constructed at the previous step, to constructa sequence of vertices such that all vertices have monochromatic forward-edges in Vtc. Then the vertices are partitioned according to the common colorof their forward-edges. One part of this partition must contain a monochro-matic Kt

2. However, this produces a graph with coloring number ct.Instead of constructing single vertices with large monochromatic sets of

incident edges, Builder will construct a large set S such that for every (t−1)-subset X⊆S there exists some vertex w such that E(w,X)=K(w,X) andE(w,X) is monochromatic. We call such a vertex w a witness for X; ifall the edges of K(w,X) are colored with β, then we say that w is a β-witness for X. Builder will construct a huge independent set S such thatevery (t−1)-subset X ⊆S has a witness w∈V −S. Moreover, Builder willbe able to order the vertices so that the vertices of S precede V −S and theback-degree of any vertex is at most t−1. Next, by Ramsey’s Theorem for(t−1)-graphs, S has a large subset V1 such that there exists a color β so thatevery (t−1)-subset X ⊆S has a β-witness w∈V −S. This is then iteratedct times, always working inside the set Vi constructed at the previous step.It will then follow that the constructed graph has coloring number t andcontains a monochromatic Kt

2.


Here are the details. Fix c and t.

Definition 8. For a function r : N → N, we say that Builder has an r-strategy, if starting from r(n) vertices he can play so that eventually G :=Gi

has the form G=(V,E), where V can be partitioned as {S,W,T} and orderedby ≺ so that

1. |S|=Ramt−1c (n);

2. for every (t−1)-subset X⊆S there exists a witness w∈W ;3. all of S precedes all of W ∪T in ≺ and every vertex has back-degree at

most t−1;4. S is independent, and so the back-degree of every vertex in S is 0.

Next we use Presenter’s strategy in the survival game to provide Builderwith an r-strategy. Recall that h(p,s,n) and l(p,s,n) are the numbers ofvertices and rounds, respectively, required for Presenter to win the (p,s,n)-survival game.

Lemma 9. Let p=c(t−1)+2 and s= t−1. Builder has an r-strategy, wherer :N→N is defined by

r(n) = h(p, s,Ramt−1

c (n))

+ l(p, s,Ramt−1

c (n)).

Proof. Fix n and set N := Ramt−1c (n), h := h(p,s,N) and l = l(p,s,N).

Builder begins by selecting a set V of r(n) vertices and partitioning it intotwo sets S0 and W = {wi : i∈ [l]} so that |S0|= h and |W |= l. He also setsT0 :=∅. The set W is ordered so that

wl ≺ wl−1 ≺ · · · ≺ w1.

Builder will simulate a play of the (p,s,N)-survival game using Presen-ter’s winning strategy as follows. The vertices of S0 correspond to the verticesthat Presenter uses in a winning play of the survival game. The vertices ofW correspond to the plays that Presenter makes in this game. Suppose thaton the first move of his winning strategy, Presenter would play a p-subsetP1⊆S0. Then on Builder’s first round of play, he simulates Presenter’s moveby constructing the edges in K(w1,P1). After Painter colors these edges,there exists an s-subset X1⊆P1 such that K(w1,X1) is monochromatic, i.e.,w1 is a witness for X1. Builder interprets this as a response of X1 by Chooserin the survival game. He removes the set D1 :=P1−X1 of discarded verticesfrom S0 to form S1 := S0 −D1 and adds D1 to T0 to form T1 := T0 ∪D1.The vertices of D1 are placed in the ordering ≺ so that w1≺D1, where thismeans that w1 comes before every vertex of D1 in ≺.


The game continues in this manner. On his ith round of play, Builderconsiders the move Pi called for by Presenter’s winning strategy and playsthe edges in K(wi,Pi). After Painter colors them, there exists an s-subsetXi such that wi is a witness for Xi. Builder interprets this as a response Xi

by Chooser in the simulated game. He sets Di :=Pi−Xi, Si :=Si−1−Di andTi :=Ti−1∪Di. The vertices of Di are placed between wi and wi−1 in ≺ sothat wi≺Di≺wi−1.

After l rounds the simulated game ends with a win for Presenter. Set S :=Sl and T :=Tl, and place S before wl in the order ≺. By the definition of a winin the (p,s,N)-survival game and our method of simulation, Conditions 1and 2 of Definition 8 hold. Clearly S is independent. Finally, V has beenordered by ≺ so that

S ≺ wl ≺ Dl ≺ wl−1 ≺ Dl−1 ≺ · · · ≺ w1 ≺ D1.

Thus every vertex in S precedes every vertex in W , every vertex in W hasback-degree s and every vertex in T has back-degree 1.

Now we iterate this construction.

Lemma 10. Define R′ :N→N recursively by

R′(0) = 1R′(j + 1) = r(R′(j)).

If Builder starts with a set V of R′(n) vertices then he can construct agraph G′, a sequence (αi : i∈ [n−1]) of colors and a sequence (Vi : i∈ [n]) ofsubsets of V such that for every i∈ [n−1] and every (t−1)-subset X⊆Vi:

1. Vi⊆Vi+1;2. there exists an αi-witness w∈Vi+1 for X; and3. col(G′)= t.

Proof. We argue by induction on n. The base step is trivial so consider theinduction step n = j + 1. By definition R′(j + 1) = r(R′(j)). By Lemma 9Builder has an r-strategy. Thus he can construct a graph G, a set S⊆V andan ordering ≺ on V satisfying Conditions 1–4 of Definition 8. Let Vj+1 :=V .By Condition 1, |S| = Ramt−1

c (R′(j)). Define a function ϕ :(

Vt−1

)→ [c] by

setting ϕ(X) to be the least color α ∈ [c] such that X has an α-witness.The color α exists by Condition 2. By the definition of Ramsey number,there exists Vj ⊆ V such that |Vj | = R′(j) and ϕ is constant on

(V

t−1

); let

αj be this constant value. We may choose ≺ so that Vj precedes S − Vj.By Conditions 3 and 4 all vertices in Vj have back-degree 0 and all other


vertices have back-degree at most t−1. From now on Builder will only playon the vertices of Vj . By the induction hypothesis, he can do this so thatfor the resulting graph there exists a sequence (αi : i∈ [j−1]) of colors anda sequence (Vi : i∈ [j−1]) of subsets of V such that for every i∈ [j−1] andevery (t−1)-subset X⊆Vi Conditions 1–3 of Lemma 10 hold.

We are now ready to prove our main theorem.

Proof of Theorem 2. Let R(c,t) :=R′(c(t−1)+2), where R′ is the functiondefined in the statement of Lemma 10, and set n=c(t−1)+2. We must showthat starting from R(c,t) vertices, Builder has a winning strategy in the(c,2, t)-Ramsey game. By Lemma 10, Builder can construct a graph G′ withcoloring number t for which there exists a sequence (αi : i∈ [n−1]) of colorsand a sequence (Vi : i∈ [n]) of subsets of V such that for every i∈ [n−1], every(t−1)-subset X⊆Vi has an αi-witness wi∈Vi+1. By the pigeonhole principlethere exists a color β and a subsequence (ih :h∈ [t]) such that β =αih . Letv1∈Vi1 . Now suppose recursively that we have constructed (vh :h∈ [j]) suchthat each v ∈ Vih and X = {vh : h∈ [j]} is complete. Choose vj+1 ∈ Vij+1 sothat vj+1 is a β-witness for X. Then {vh :h∈ [t]} is a Kt

2, all of whose edgeshave been colored β. This completes the proof.

According to Theorem 2, Builder has a winning strategy in the (c,2, t)-Ramsey game when starting from R(c,t) vertices.

Recall now the definition of r(n) as r(n) = h(p,s,Ramt−1c (n)) +

l(p,s,Ramt−1c (n)), where h and l stand for the number of vertices and num-

ber of rounds, respectively, used in the winning strategy for the (p,s, t)-survival game. Regardless of the strategy used, in every round Presenteruses p vertices, some of which may survive to be used again. Therefore,p · l(p,s, t) ≥ h(p,s, t). On the other hand, in each round p− s vertices arethrown away, and so h(p,s, t)≥(p−s)l(p,s, t). Therefore, we have

h(p, s,Ramt−1

c (n))≤ r(n) ≤ h

(p, s,Ramt−1

c (n))(

1 +1

p − s

).

Writing the definition of R(c,t) in terms of R′, r and h, we find thatR(c,t) is computed by c(t− 1) + 2 recursive applications of h (recall thatR′(0)=1).

R(c, t) = R′(c(t − 1) + 2)

R′(j + 1) = r(R′(j))

≤(1 +

1(c − 1)(t − 1) + 2

)· h

(c(t − 1) + 2, t − 1,Ramt−1

c (R′(j)).


Thus our bound R(c,t) on the number of vertices needed to win the (c,2, t)-Ramsey game is not primitive recursive even when one of the parameters(c or t) is fixed.

4. Hypergraphs

In this section we give a natural generalization of Theorem 2 to hypergraphs.We first review hypergraph coloring and extend the notion of coloring num-ber to hypergraphs in an appropriate way. A proper c-coloring of a hyper-graph H is a function f : V → [c] such that every edge contains vertices ofat least two different colors. The chromatic number, denoted χ(H), of H isthe least c such that H has a proper c-coloring. Then χ(Kt

s)=⌈

ts−1

⌉.

Let v1≺·· ·≺vn be an ordering of the vertices of an s-graph H =(V,E).We call the largest vertex of an s-edge X the root of X. An edge X is aback-edge of a vertex v if v is the root of X. Two back-edges of v are almostdisjoint if v is their only common vertex. We define the coloring numbercol(H) of H to be the least d such that for some ordering of V every vertexhas fewer than d almost disjoint back-edges. So col(Kt

s)=⌈

ts−1

⌉. Notice that

when s = 2 this is the same definition as the coloring number for graphs.Also, if the vertices of H are colored by First-Fit in an order that achievesthe coloring number d of H then at most d colors will be used. To see thisnote that if First-Fit is forced to use a color k on a vertex v then for eachcolor α∈ [k−1], there is a back-edge X of v such that all s−1 vertices inX−{v} have been colored with α. It follows that v has at least (k−1) almostdisjoint back-edges. Since we started with an optimal ordering, the coloringnumber of H is at least k.

Now we are ready to sketch the proof of Theorem 3. We must show thatthere exists a function R(c,s, t) such that Builder can win the (c,s, t)-Ramseygame while constructing a graph with R(c,s, t) vertices, whose coloring num-ber is t. Fix c, s and t. The argument is essentially the same as in the proofof Theorem 2. We first must prove Lemmas 9 and 10 for s-graphs. Note thatthe definition of a witness still makes sense in the case of hypergraphs. Theonly significant difference from the proof of Lemma 9 is in the details of howBuilder uses a simulation of the survival game. This time he simulates the(p,t−1,Ramt−1

c (n))-survival game, where p=Rams−1c (t−1). When Presenter

would play a p-set Pi on round i, Builder constructs the s-edges of K(wi,Pi).By the choice of p, after Painter colors the edges of K(wi,Pi) there existsa (t−1)-subset Xi ⊆Pi such that K(wi,Xi) is monochromatic. Builder in-terprets this as Chooser’s response. One other change involves checking themaximum number of almost disjoint back-edges of a vertex. For vertices of


W this is at most⌊

t−1s−1

⌋<

⌈t

s−1

⌉. For vertices v∈Di this is at most one since

every back-edge of v contains wi. The proofs of Lemma 10 and Theorem 3proceed as before.

5. Conclusion and Open Questions

We have shown a striking contrast between the on-line and off-line Ramseycoloring number for graphs. We have extended our evaluation of the onlinecoloring Ramsey number for graphs to hypergraphs in a natural way. How-ever the magnitude of the off-line coloring Ramsey number of hypergraphsremains unknown. It would be nice to clear this up.

The Kurek–Rucinski conjecture is still open and of great interest. Oneapproach might be to prove the conjecture for specific classes of graphs. Ourresults may give hope that the on-line size Ramsey number is significantlyless than the size Ramsey number, but our techniques are not directly ap-plicable because they require such an incredibly large vertex set. This leadsto the question of whether Presenter really needs so many vertices to winthe survival game.

If one examines the survival game more closely, it is relatively easy toargue that a winning game (on s-uniform hypergraphs) for Presenter mustpass through stages i = 0,1, . . . ,2s − 1 where the graph built by stage isatisfies the positive sentence of rank i. However, our strategy for the gamedoes ignore many edges that might be useful later in the game and it is byno means clear that the strategy is optimal.

Further, the application of the survival game to the Ramsey game iswhere the numbers really blow up, and it would be interesting to know ifthere is a better strategy for Builder. For example, is it possible to give aprimitive recursive upper bound on the number of vertices needed to winthe (c,2, t)-Ramsey game, even for a fixed small c?

References

[1] W. Ackermann: Zum Hilbertschen Aufbau der reellen Zahlen, Math. Annalen 99(1928), 118–131.

[2] P. Erdos, R. J. Faudree, C. Rousseau and R. H. Schelp: The size Ramseynumber, Period. Math. Hungar. 9 (1978), 145–161.

[3] J. A. Grytczuk, M. Ha�luszczak and H. A. Kierstead: On-line Ramsey theory,Electronic J. of Combinatorics 11 (2004), #R57.

[4] A. Kurek and A. Rucinski: Two variants of the size Ramsey number, Discuss. Math.Graph Theory 25(1–2) (2005), 141–149.

[5] R. Peter: Recursive Functions, Academic Press, New York, 1967.

64 KIERSTEAD, KONJEVOD: COLORING NUMBER AND RAMSEY THEORY. . .

H. A. Kierstead

Department of Mathematics and Statistics

Arizona State University

Tempe, AZ 85287

USA

[email protected]

Goran Konjevod

Department of Computer Science

Arizona State University

Tempe, AZ 85287

USA

[email protected]

mailto:[email protected]

mailto:[email protected]

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COLORING NUMBER AND ON-LINE RAMSEY THEORY FOR GRAPHS AND HYPERGRAPHS