scholarly journals On the Chvátal-Erdős Triangle Game

10.37236/559 ◽  
2011 ◽  
Vol 18 (1) ◽  
Author(s):  
József Balogh ◽  
Wojciech Samotij
Keyword(s):  
Complete Graph ◽  
Winning Strategy ◽  
Seminal Paper ◽  
Positional Games ◽  
Positive Integers

Given a graph $G$ and positive integers $n$ and $q$, let ${\bf G}(G;n,q)$ be the game played on the edges of the complete graph $K_n$ in which the two players, Maker and Breaker, alternately claim $1$ and $q$ edges, respectively. Maker's goal is to occupy all edges in some copy of $G$; Breaker tries to prevent it. In their seminal paper on positional games, Chvátal and Erdős proved that in the game ${\bf G}(K_3;n,q)$, Maker has a winning strategy if $q < \sqrt{2n+2}-5/2$, and if $q \geq 2\sqrt{n}$, then Breaker has a winning strategy. In this note, we improve the latter of these bounds by describing a randomized strategy that allows Breaker to win the game ${\bf G}(K_3;n,q)$ whenever $q \geq (2-1/24)\sqrt{n}$. Moreover, we provide additional evidence supporting the belief that this bound can be further improved to $(\sqrt{2}+o(1))\sqrt{n}$.

On Biased Positional Games

1998 ◽  
Vol 7 (4) ◽  
pp. 339-351 ◽  
Author(s):  
MAŁGORZATA BEDNARSKA
Keyword(s):  
Complete Graph ◽  
Binary Tree ◽  
Winning Strategy ◽  
Positional Games

Let TBin(N, n, q) be the game on the complete graph KN in which two players, the Breaker and the Maker, alternately claim one and q edges, respectively. The Maker's aim is to build a binary tree on n<N vertices in n−1 turns while the Breaker tries to prevent him from doing so. It is shown that, for every constant ε>0, there exists n0 such that, for every n[ges ]n0, the Breaker has a winning strategy in TBin(N, n, q) if q>(1+ε)N/logn, while, for q<(1−ε)N/logn, the game TBin(N, n, q) can be won by the Maker provided that n=o(N).

scholarly journals Biased Positional Games and Small Hypergraphs with Large Covers

10.37236/794 ◽  
2008 ◽  
Vol 15 (1) ◽  
Author(s):  
Michael Krivelevich ◽  
Tibor Szabó
Keyword(s):  
Complete Graph ◽  
Winning Strategy ◽  
Hamilton Cycle ◽  
Covering Number ◽  
New Approach ◽  
Positional Games

We prove that in the biased $(1:b)$ Hamiltonicity and $k$-connectivity Maker-Breaker games ($k>0$ is a constant), played on the edges of the complete graph $K_n$, Maker has a winning strategy for $b\le(\log 2-o(1))n/\log n$. Also, in the biased $(1:b)$ Avoider-Enforcer game played on $E(K_n)$, Enforcer can force Avoider to create a Hamilton cycle when $b\le (1-o(1))n/\log n$. These results are proved using a new approach, relying on the existence of hypergraphs with few edges and large covering number.

scholarly journals Upper density of monochromatic infinite paths

2019 ◽  
Author(s):  
Jan Corsten ◽  
Louis DeBiasio ◽  
Ander Lamaison ◽  
Richard Lang
Keyword(s):  
Complete Graph ◽  
Edge Coloring ◽  
Classical Case ◽  
Infinite Graphs ◽  
Edge Colorings ◽  
Vertex Set ◽  
Upper Density ◽  
Countably Infinite ◽  
Positive Integers ◽  
Red Path

Ramsey Theory investigates the existence of large monochromatic substructures. Unlike the most classical case of monochromatic complete subgraphs, the maximum guaranteed length of a monochromatic path in a two-edge-colored complete graph is well-understood. Gerencsér and Gyárfás in 1967 showed that any two-edge-coloring of a complete graph Kn contains a monochromatic path with ⌊2n/3⌋+1 vertices. The following two-edge-coloring shows that this is the best possible: partition the vertices of Kn into two sets A and B such that |A|=⌊n/3⌋ and |B|=⌈2n/3⌉, and color the edges between A and B red and edges inside each of the sets blue. The longest red path has 2|A|+1 vertices and the longest blue path has |B| vertices. The main result of this paper concerns the corresponding problem for countably infinite graphs. To measure the size of a monochromatic subgraph, we associate the vertices with positive integers and consider the lower and the upper density of the vertex set of a monochromatic subgraph. The upper density of a subset A of positive integers is the limit superior of |A∩{1,...,}|/n, and the lower density is the limit inferior. The following example shows that there need not exist a monochromatic path with positive upper density such that its vertices form an increasing sequence: an edge joining vertices i and j is colored red if ⌊log2i⌋≠⌊log2j⌋, and blue otherwise. In particular, the coloring yields blue cliques with 1, 2, 4, 8, etc., vertices mutually joined by red edges. Likewise, there are constructions of two-edge-colorings such that the lower density of every monochromatic path is zero. A result of Rado from the 1970's asserts that the vertices of any k-edge-colored countably infinite complete graph can be covered by k monochromatic paths. For a two-edge-colored complete graph on the positive integers, this implies the existence of a monochromatic path with upper density at least 1/2. In 1993, Erdős and Galvin raised the problem of determining the largest c such that every two-edge-coloring of the complete graph on the positive integers contains a monochromatic path with upper density at least c. The authors solve this 25-year-old problem by showing that c=(12+8–√)/17≈0.87226.

scholarly journals Manipulative Waiters with Probabilistic Intuition

2015 ◽  
Vol 25 (6) ◽  
pp. 823-849 ◽  
Author(s):  
MAŁGORZATA BEDNARSKA-BZDȨGA ◽  
DAN HEFETZ ◽  
MICHAEL KRIVELEVICH ◽  
TOMASZ ŁUCZAK
Keyword(s):  
Phase Transition ◽  
Random Graphs ◽  
Complete Graph ◽  
Absolute Constant ◽  
Perfect Information ◽  
Graph Theoretic ◽  
Graph Property ◽  
Positive Integers ◽  
Monotone Graph ◽  
Transition Type

For positive integersnandqand a monotone graph property$\mathcal{A}$, we consider the two-player, perfect information game WC(n,q,$\mathcal{A}$), which is defined as follows. The game proceeds in rounds. In each round, the first player, called Waiter, offers the second player, called Client,q+ 1 edges of the complete graphKnwhich have not been offered previously. Client then chooses one of these edges which he keeps and the remainingqedges go back to Waiter. If, at the end of the game, the graph which consists of the edges chosen by Client satisfies the property$\mathcal{A}$, then Waiter is declared the winner; otherwise Client wins the game. In this paper we study such games (also known as Picker–Chooser games) for a variety of natural graph-theoretic parameters, such as the size of a largest component or the length of a longest cycle. In particular, we describe a phase transition type phenomenon which occurs when the parameterqis close tonand is reminiscent of phase transition phenomena in random graphs. Namely, we prove that ifq⩾ (1 + ϵ)n, then Client can avoid components of ordercϵ−2lnnfor some absolute constantc> 0, whereas forq⩽ (1 − ϵ)n, Waiter can force a giant, linearly sized component in Client's graph. In the second part of the paper, we prove that Waiter can force Client's graph to be pancyclic for everyq⩽cn, wherec> 0 is an appropriate constant. Note that this behaviour is in stark contrast to the threshold for pancyclicity and Hamiltonicity of random graphs.

scholarly journals Vertex Covering with Monochromatic Pieces of few Colours

10.37236/7469 ◽  
2018 ◽  
Vol 25 (3) ◽  
Author(s):  
Marlo Eugster ◽  
Frank Mousset
Keyword(s):  
Chromatic Number ◽  
Complete Graph ◽  
Independence Number ◽  
Vertex Cover ◽  
Constant Factor ◽  
Regular Graphs ◽  
Kneser Graph ◽  
Vertex Covering ◽  
Vertex Graph ◽  
Positive Integers

In 1995, Erdös and Gyárfás proved that in every $2$-colouring of the edges of $K_n$, there is a vertex cover by $2\sqrt{n}$ monochromatic paths of the same colour, which is optimal up to a constant factor. The main goal of this paper is to study the natural multi-colour generalization of this problem: given two positive integers $r,s$, what is the smallest number $pc_{r,s}(K_n)$ such that in every colouring of the edges of $K_n$ with $r$ colours, there exists a vertex cover of $K_n$ by $pc_{r,s}(K_n)$ monochromatic paths using altogether at most $s$ different colours?For fixed integers $r>s$ and as $n\to\infty$, we prove that $pc_{r,s}(K_n) = \Theta(n^{1/\chi})$, where $\chi=\max{\{1,2+2s-r\}}$ is the chromatic number of the Kneser graph $KG(r,r-s)$. More generally, if one replaces $K_n$ by an arbitrary $n$-vertex graph with fixed independence number $\alpha$, then we have $pc_{r,s}(G) = O(n^{1/\chi})$, where this time around $\chi$ is the chromatic number of the Kneser hypergraph $KG^{(\alpha+1)}(r,r-s)$. This result is tight in the sense that there exist graphs with independence number $\alpha$ for which $pc_{r,s}(G) = \Omega(n^{1/\chi})$. This is in sharp contrast to the case $r=s$, where it follows from a result of Sárközy (2012) that $pc_{r,r}(G)$ depends only on $r$ and $\alpha$, but not on the number of vertices.We obtain similar results for the situation where instead of using paths, one wants to cover a graph with bounded independence number by monochromatic cycles, or a complete graph by monochromatic $d$-regular graphs.

scholarly journals An Explicit Edge-Coloring of $K_n$ with Six Colors on Every $K_5$

10.37236/7852 ◽  
2019 ◽  
Vol 26 (4) ◽  
Author(s):  
Alex Cameron
Keyword(s):  
Lower Bound ◽  
Complete Graph ◽  
Upper Bound ◽  
Edge Coloring ◽  
Minimum Number ◽  
Positive Integers

Let $p$ and $q$ be positive integers such that $1 \leq q \leq {p \choose 2}$. A $(p,q)$-coloring of the complete graph on $n$ vertices $K_n$ is an edge coloring for which every $p$-clique contains edges of at least $q$ distinct colors. We denote the minimum number of colors needed for such a $(p,q)$-coloring of $K_n$ by $f(n,p,q)$. This is known as the Erdös-Gyárfás function. In this paper we give an explicit $(5,6)$-coloring with $n^{1/2+o(1)}$ colors. This improves the best known upper bound of $f(n,5,6)=O\left(n^{3/5}\right)$ given by Erdös and Gyárfás, and comes close to matching the order of the best known lower bound, $f(n,5,6) = \Omega\left(n^{1/2}\right)$.

scholarly journals On Winning Fast in Avoider-Enforcer Games

10.37236/328 ◽  
2010 ◽  
Vol 17 (1) ◽  
Author(s):  
János Barát ◽  
Miloš Stojaković
Keyword(s):  
Graph Theory ◽  
Lower Bound ◽  
Complete Graph ◽  
Upper Bound ◽  
Extremal Graph ◽  
Extremal Graph Theory ◽  
Additive Constant ◽  
Main Interest ◽  
Free Graph ◽  
Positional Games

We analyze the duration of the unbiased Avoider-Enforcer game for three basic positional games. All the games are played on the edges of the complete graph on $n$ vertices, and Avoider's goal is to keep his graph outerplanar, diamond-free and $k$-degenerate, respectively. It is clear that all three games are Enforcer's wins, and our main interest lies in determining the largest number of moves Avoider can play before losing. Extremal graph theory offers a general upper bound for the number of Avoider's moves. As it turns out, for all three games we manage to obtain a lower bound that is just an additive constant away from that upper bound. In particular, we exhibit a strategy for Avoider to keep his graph outerplanar for at least $2n-8$ moves, being just $6$ short of the maximum possible. A diamond-free graph can have at most $d(n)=\lceil\frac{3n-4}{2}\rceil$ edges, and we prove that Avoider can play for at least $d(n)-3$ moves. Finally, if $k$ is small compared to $n$, we show that Avoider can keep his graph $k$-degenerate for as many as $e(n)$ moves, where $e(n)$ is the maximum number of edges a $k$-degenerate graph can have.

scholarly journals The Random Graph Intuition for the Tournament Game

2015 ◽  
Vol 25 (1) ◽  
pp. 76-88 ◽  
Author(s):  
DENNIS CLEMENS ◽  
HEIDI GEBAUER ◽  
ANITA LIEBENAU
Keyword(s):  
Lower Bound ◽  
Random Graph ◽  
Complete Graph ◽  
Upper Bound ◽  
Winning Strategy ◽  
Constant Factor ◽  
Additive Constant ◽  
Underlying Graph ◽  
Straightforward Application

In the tournament game two players, called Maker and Breaker, alternately take turns in claiming an unclaimed edge of the complete graph Kn and selecting one of the two possible orientations. Before the game starts, Breaker fixes an arbitrary tournament Tk on k vertices. Maker wins if, at the end of the game, her digraph contains a copy of Tk; otherwise Breaker wins. In our main result, we show that Maker has a winning strategy for k = (2 − o(1))log2n, improving the constant factor in previous results of Beck and the second author. This is asymptotically tight since it is known that for k = (2 − o(1))log2n Breaker can prevent the underlying graph of Maker's digraph from containing a k-clique. Moreover, the precise value of our lower bound differs from the upper bound only by an additive constant of 12.We also discuss the question of whether the random graph intuition, which suggests that the threshold for k is asymptotically the same for the game played by two ‘clever’ players and the game played by two ‘random’ players, is supported by the tournament game. It will turn out that, while a straightforward application of this intuition fails, a more subtle version of it is still valid.Finally, we consider the orientation game version of the tournament game, where Maker wins the game if the final digraph – also containing the edges directed by Breaker – possesses a copy of Tk. We prove that in that game Breaker has a winning strategy for k = (4 + o(1))log2n.

scholarly journals On Two Problems Regarding the Hamiltonian Cycle Game

10.37236/117 ◽  
2009 ◽  
Vol 16 (1) ◽  
Author(s):  
Dan Hefetz ◽  
Sebastian Stich
Keyword(s):  
Complete Graph ◽  
Hamiltonian Cycle ◽  
Winning Strategy ◽  
Minimum Number

We consider the fair Hamiltonian cycle Maker-Breaker game, played on the edge set of the complete graph $K_n$ on $n$ vertices. It is known that Maker wins this game if $n$ is sufficiently large. We are interested in the minimum number of moves needed for Maker in order to win the Hamiltonian cycle game, and in the smallest $n$ for which Maker has a winning strategy for this game. We prove the following results: (1) If $n$ is sufficiently large, then Maker can win the Hamiltonian cycle game within $n+1$ moves. This bound is best possible and it settles a question of Hefetz, Krivelevich, Stojaković and Szabó; (2) If $n \geq 29$, then Maker can win the Hamiltonian cycle game. This improves the previously best bound of $600$ due to Papaioannou.

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