diff --git a/anki_header.tex b/anki_header.tex index 96ac4ee..19f1fb2 100644 --- a/anki_header.tex +++ b/anki_header.tex @@ -73,7 +73,6 @@ \newcommand{\imp}{\implies} \newcommand{\for}{\forall} -\newcommand{\mor}{\rightarrow} \newcommand{\nin}{\notin} \newcommand{\comp}{\circ} \newcommand{\union}{\cup} @@ -85,11 +84,11 @@ \newcommand{\aeeq}{\overset{\text{ae}}=} \newcommand{\lexlt}{\overset{\text{lex}}<} \newcommand{\colexlt}{\overset{\text{colex}}<} -\newcommand{\wtendsto}{\overset w\mor} -\newcommand{\wstartendsto}{\overset{w*}\mor} +\newcommand{\wto}{\overset w\to} +\newcommand{\wstarto}{\overset{w*}\to} \renewcommand{\vec}[1]{\boldsymbol{\mathbf{#1}}} \renewcommand{\bar}[1]{\overline{#1}} - +\newcommand{\compc}[1]{\mathrm{\mathbf{#1}}} \let\Im\relax \let\Re\relax diff --git a/header.tex b/header.tex index 0fa63bf..e206b93 100644 --- a/header.tex +++ b/header.tex @@ -140,7 +140,6 @@ \newcommand{\imp}{\implies} \newcommand{\for}{\forall} -\newcommand{\mor}{\rightarrow} \newcommand{\nin}{\notin} \newcommand{\comp}{\circ} \newcommand{\union}{\cup} @@ -152,8 +151,8 @@ \newcommand{\aeeq}{\overset{\text{ae}}=} \newcommand{\lexlt}{\overset{\text{lex}}<} \newcommand{\colexlt}{\overset{\text{colex}}<} -\newcommand{\wtendsto}{\overset w\mor} -\newcommand{\wstartendsto}{\overset{w*}\mor} +\newcommand{\wto}{\overset w\to} +\newcommand{\wstarto}{\overset{w*}\to} \renewcommand{\vec}[1]{\boldsymbol{\mathbf{#1}}} \renewcommand{\bar}[1]{\overline{#1}} diff --git a/ramsey_theory.pdf b/ramsey_theory.pdf index f483a8e..9643775 100644 Binary files a/ramsey_theory.pdf and b/ramsey_theory.pdf differ diff --git a/ramsey_theory.tex b/ramsey_theory.tex index 4e6b7f6..7590133 100644 --- a/ramsey_theory.tex +++ b/ramsey_theory.tex @@ -8,21 +8,7 @@ \def\ncourse{Ramsey Theory on Graphs} \def\draft{Incomplete} -\usepackage{mathrsfs} -\usepackage{imakeidx} -\usepackage{marginnote} -\usepackage{mathdots} -\usepackage{tabularx} -\usepackage{ifthen} - \input{header} -\swapnumbers -\reversemarginpar - -\usetikzlibrary{positioning, decorations.pathmorphing, decorations.text, calc, backgrounds, fadings} -\tikzset{node/.style = {circle,draw,inner sep=0.8mm}} - -\makeindex[intoc] \setcounter{section}{-1} @@ -45,8 +31,8 @@ \section{Introduction} \item For $X$ a set, $r \in \N$, $X^{(r)} = \{S \subseteq X | |S| = r\}$ \item $\chi$ for a $k$-coloring of the edges of $K_n$ \begin{align*} - \chi : E(K_n) \mor & [k] \\ - \chi : E(K_n) \mor & \{\text{red}, \text{blue}\} & (\text{if } k = 2) + \chi : E(K_n) & \to [k] \\ + \chi : E(K_n) & \to \{\text{red}, \text{blue}\} & (\text{if } k = 2) \end{align*} \end{itemize} \end{notation} @@ -88,7 +74,7 @@ \section{Old bounds on \texorpdfstring{$R(k)$}{R(k)}} \begin{lemma} For all $k, \ell \ge 3$, - $$R(\ell, k) \le \underbrace{R(\ell - 1, k)}a + \underbrace{R(\ell, k - 1)}b$$ + $$R(\ell, k) \le \underbrace{R(\ell - 1, k)}_a + \underbrace{R(\ell, k - 1)}_b$$ \end{lemma} \begin{proof} Let $n = a + b$. Pick a vertex $v$. By pigeonhole, either @@ -167,19 +153,20 @@ \subsection{Lower bounds} $$R(3, k) \ge c\left(\frac k{\log k}\right)^{\frac 32}$$ for some constant $c > 0$. \end{thm} +\begin{idea} + We will look at a binomial random graph and choose the parameters so that there are very few red $K_k$ and the number of blue $K_3$ is at most some fixed proportion of $n$. Then we will remove one vertex from each red $K_k$ and one vertex from each blue $K_3$. The resulting graph will have neither and most likely will still contain a fixed proportion of the vertices we started with. +\end{idea} \begin{proof} Change the language. Discuss the blue graph. We are now looking for the maximum number of edges of a graph with no triangles and no independent set of size $k$. \\ - Take $n = \left(\frac k{\log k}\right)^{\frac 32}, p = n^{-\frac 23} = \frac{\log k}k$. Now sample $G \sim G(n, p)$ and define $\tilde G$ to be $G$ with one vertex removed from each triangle and independent set of size $k$. By construction, $K_3 \not\subseteq \tilde G$ and $\alpha(\tilde G) < k$. We will show $\E[\abs{\tilde G}] \ge \frac n2$ using - $$\abs{\tilde G} \ge n - \#\text{triangles in } G - \#\text{independent sets of size $k$ in } G$$ + Take $n = \left(\frac k{\log k}\right)^{\frac 32}, p = n^{-\frac 23} = \frac{\log k}k$. Now sample $G \sim G(n, p)$ and define $\tilde G$ to be $G$ with one vertex removed from each triangle and independent set of size $k$. By construction, $K_3 \not\subseteq \tilde G$ and $\alpha(\tilde G) < k$. We will show $\bbE |\tilde G| \ge \frac n2$ using + $$|\tilde G| \ge n - \#\text{triangles in } G - \#\text{independent sets of size $k$ in } G$$ First, - \begin{align*} - \E[\#\text{triangles in } G] - & = \sum_{T \in [n]^(3)} \P(T \text { triangle in } G) \\ - & = \binom n3 p^3 \le \frac{(np)^3}6 = \frac n6 - \end{align*} + $$\bbE \#\text{triangles in } G + = \sum_{T \in [n]^(3)} \P(T \text { triangle in } G) + = \binom n3 p^3 \le \frac{(np)^3}6 = \frac n6$$ Second, \begin{align*} - \E[\#\text{independent sets of size $k$ in } G] + \bbE \#\text{independent sets of size $k$ in } G & = \binom nk (1 - p)^{\binom k2} \\ & \le \left(\frac{en}k\right)^k e^{-p\binom k2} \\ & \sim \left(\frac{en}k e^{-\frac{pk}2}\right)^k \\ @@ -187,9 +174,13 @@ \subsection{Lower bounds} & = \left(\frac e{\log^{\frac 32} k}\right)^k \longrightarrow 0 \end{align*} Hence, for large enough $k$, - $$\E[\abs{\tilde G}] \ge n - \frac n6 - 1 \ge \frac n2 = \frac 12 \left(\frac k{\log k}\right)^{\frac 32}$$ + $$\bbE |\tilde G| \ge n - \frac n6 - 1 \ge \frac n2 = \frac 12 \left(\frac k{\log k}\right)^{\frac 32}$$ By adjusting $c > 0$, we have proved the theorem. \end{proof} +\begin{rmk} + The values of $n$ and $p$ come from the constraints + $$n^3p^3 \ll n, \quad p^{-1}\log p^{-1} \ll k$$ +\end{rmk} We are being wasteful here. Why throw an entire vertex away when we could get away with removing a single edge? Because we might accidentally create an independent set of size $k$. But we can be smarter... @@ -203,7 +194,7 @@ \subsection{Lower bounds} \end{thm} \begin{lemma} - Let $\mathcal F = \{A_1, \dots, A_m\}$ be a family of events in a probability space. Let $\Eps_t$ be the even that $t$ {\it independent} events from $\mathcal F$ occur. then + Let $\mathcal F = \{A_1, \dots, A_m\}$ be a family of events in a probability space. Let $\Eps_t$ be the event that $t$ {\it independent} events from $\mathcal F$ occur. then $$\P(\Eps_t) \le \frac 1{t!}\left(\sum_{i = 1}^m \P(A_i)\right)^t$$ \end{lemma} \begin{proof} @@ -220,6 +211,10 @@ \subsection{Lower bounds} \newlec +\begin{idea} + Instead of killing vertices, we will kill edges. We only need to kill edges from a maximal set of edge-disjoint triangles. For this killing to create an independent set $I$ of size $k$, we must have killed all edges in $I$. But with high probability all sets of size $k$ contain a fixed fraction of the expectation $p\binom k2$ of the number of edges, and having so many edge-disjoint triangles each with two vertices among $k$ fixed vertices is unlikely. +\end{idea} + \begin{lemma} Let $n, k \in \N, p \in [0, 1]$ be such that $pk \ge 16\log n$. Then with high probability every subset of size $k$ of $G \sim G(n, p)$ contains at least $\frac{pk^2}8$ edges. \end{lemma} @@ -228,15 +223,16 @@ \subsection{Lower bounds} We fix $n = \left(\frac{c_1 k}{\log k}\right)^2, p = c_2n^{-\frac 12} = \frac{c_2}{c_1}\frac{\log k}k$. Let $G \sim G(n, p), \mathcal T$ a maximal collection of edge-disjoint triangles in $G$, $\tilde G$ be $G$ with all edges of $\mathcal T$ removed. Note, $\tilde G$ contains no triangle. We show $$\P(\alpha(\tilde G) \ge k) < 1$$ Let $Q$ be the event that every set of $k$ vertices of $G$ contains $\ge \frac{pk^2}8$ edges. Setting $\frac{c_2}{c_1} = 48$, we get - $$pk = \frac{c_2}{c_1}\frac{\log k}k k = 48\log k > 16\log n$$ By the lemma, we know so that $\P(Q) = 1 + o(1)$ by the lemma. Now note that + $$pk = \frac{c_2}{c_1}\frac{\log k}k k = 48\log k > 16\log n$$ + so that $\P(Q) = 1 - o(1)$ by the lemma. Now note that $$\P(\alpha(\tilde G) \ge k) \le \P(\alpha(\tilde G) \ge k, Q) + \cancelto 0{\P(Q^c)}$$ - So we focus on $\P(\alpha(\tilde G) \ge k, Q)$. Set $t = \frac{pk^2}{24}$. + So we focus on $\P(\alpha(\tilde G) \ge k, Q)$. Observe that if $\tilde G$ contains an independent set $I$ of size $k$ and $Q$ holds, then $I$ contains $\ge \frac{pk^2}8$ edges of $G$ by assumption. But $I$ is an independent set in $\tilde G$, so all $\frac{pk^2}8$ edges must have belonged to some triangle $T \in \mathcal T$ and been removed. Therefore \begin{align*} \P(\alpha(\tilde G) \ge k, Q) - & \le \P\left(\exists S \in [n]^{(k)}, \mathcal T \text{ meets $S$ in } \ge \frac{pk^2}8\right) \\ + & \le \P\left(\exists S \in [n]^{(k)}, \mathcal T \text{ meets $S$ in $\ge \frac{pk^2}8$ edges}\right) \\ & \le \binom n k \P\left(\underbrace{\substack{\text{at least $t$ triangles of $\mathcal T$} \\ \text{meet $[k]$ in at least two vertices}}}_B\right) \end{align*} - Let $\{T_i\}$ be the collection of triangles in $K_n$ that meet $[k]$ in at least two vertices. Let $A_i = \{T_i \subseteq G\}$. Note that if $T_{i_1}, \dots, T_{i_k}$ are edge-disjoint, then $A_{i_1}, \dots, A_{i_k}$ are independent. So + where $t = \frac{pk^2}{24}$. Let $\{T_i\}$ be the collection of triangles in $K_n$ that meet $[k]$ in at least two vertices. Let $A_i = \{T_i \subseteq G\}$. Note that if $T_{i_1}, \dots, T_{i_k}$ are edge-disjoint, then $A_{i_1}, \dots, A_{i_k}$ are independent. So \begin{align*} \P(B) & \le \P(\Eps_t) \\ @@ -248,15 +244,15 @@ \subsection{Lower bounds} by choosing $c_2 = \frac 1{\sqrt{24} e}$. To finish, observe that $$t = \frac{pk^2}{24} = 2k\log k \ge k\log n$$ Hence - $$\binom n k \P(B) \le \binom n k e^{-t} \le \left(\frac{en}k e^{-\log n}\right)^k = \left(\frac ek\right)^k \mor 0$$ + $$\binom n k \P(B) \le \binom n k e^{-t} \le \left(\frac{en}k e^{-\log n}\right)^k = \left(\frac ek\right)^k \to 0$$ \end{proof} \subsection{Large deviation inequalities} Let $Z$ be a gaussian random variable. -$$\P(Z - \E[Z] \ge t) \le e^{\frac{-t}{2\Var Z}}$$ -Let $X_1, \dots, X_n$ be iid Bernoulli random variables. We denote this $X_i \sim \Ber(p)$. Write $S_n = X_1 + \dots + X_n$. Note $\E[S_n] = np$, $\Var(S_n)=np(1 - p)$. +$$\P(Z - \bbE Z \ge t) \le e^{\frac{-t}{2\Var Z}}$$ +Let $X_1, \dots, X_n$ be iid Bernoulli random variables. We denote this $X_i \sim \Ber(p)$. Write $S_n = X_1 + \dots + X_n$. Note $\bbE S_n = np, \Var(S_n) = np(1 - p)$. \begin{idea} Often, the tail of $S_n$ looks like a gaussian tail. @@ -264,19 +260,23 @@ \subsection{Large deviation inequalities} \begin{thm}[Chernoff inequality] Let $X_1, \dots, X_n \sim \Ber(p)$. Then - $$\P\left(\abs{S_n - pn} \ge t\right) \le 2\exp\left(\underbrace{-\frac t{2pn}}_{\text{meat}} + \underbrace{\frac{t^3}{(pn)^2}}_{\text{error term}}\right)$$ + $$\P\left(\abs{S_n - pn} \ge t\right) \le 2\exp\left(\underbrace{-\frac{t^2}{2pn}}_{\text{meat}} + \underbrace{\frac{t^3}{(pn)^2}}_{\text{error term}}\right)$$ \end{thm} \newlec -\begin{proof}[Proof of the $\frac{pk^8}8$ lemma] +\begin{proof}[Proof of the $\frac{pk^2}8$ lemma] + Using Chernoff on $e(G[[k]])$, namely with $p := p, n := \binom k2, t := \frac{pk^2}4$, we get \begin{align*} \P(G \text{ fails the statement}) & = \P\left(\exists S \in [n]^{(k)}, e(G[s]) < \frac{pk^2}8\right) \\ & \le \binom nk \P\left(e(G[[k]]) < \frac{pk^2}8\right) \\ - & \le \binom nk \P\left(\frac{pk^2}4 < \abs{e(G[[k]]) - p\binom k2}\right) + & \le \binom nk \P\left(\frac{pk^2}4 < \abs{e(G[[k]]) - p\binom k2}\right) \\ + & \le 2\left(\frac{en}k\right)^k \exp\left(-\frac{pk^2}{16} + \frac 18\right) \\ + & \ll \left(\frac{en}k\right)^k \exp\left(-k\log n\right) \\ + & = \left(\frac ek\right)^k \end{align*} - We are now done by Chernoff. TODO: Are we really? The numbers don't seem to work out. + which tends to 0 as $k$ tends to infinity. \end{proof} \subsection{The Local Lemma} @@ -337,7 +337,7 @@ \subsection{The Local Lemma} \begin{enumerate} \item $p^3 = \P(A_T) \le x_T \prod_{\abs{T \inter T'} = 2} (1 - x_T) \prod_{\abs{I \inter T} \ge 2} (1 - x_I) = 3p^3 (1 - 3p^3)^{3n}(1 - n^{-k})^{3n^{k - 2}}$ Indeed, - $$(1 - 3p^3)^{3n}(1 - n^{-k})^{3n^{k - 2}} \ge \exp(-18p^3n - 6n^{-2}) = \exp(-18\eps^2 p - 6n^{-2}) \mor 1$$ + $$(1 - 3p^3)^{3n}(1 - n^{-k})^{3n^{k - 2}} \ge \exp(-18p^3n - 6n^{-2}) = \exp(-18\eps^2 p - 6n^{-2}) \to 1$$ \newlec @@ -484,7 +484,7 @@ \subsection{Upper bounds on \texorpdfstring{$R(3, k)$}{R(3, k)}} \begin{proof}[Second proof of AKS] Let $I$ be an independent set in $G$ sampled uniformly among all independent sets of $G$. We will show - $$\E\abs I \ge c\frac nd\log n$$ + $$\bbE\abs I \ge c\frac nd\log n$$ Let $v$ be a vertex. We define the random variable $$X_v = d1_{v \in I} + \abs{N(v) \inter I}$$ For any independent set $I$, @@ -595,9 +595,9 @@ \subsection{Off-diagonal \texorpdfstring{$3$}{3}-uniform hypergraph Ramsey} What about lower bounds? Let's try the probabilistic method. Color triples blue with probability $p$. -$$\E\left[\#\text{red }K_k^{(3)}\right] = \binom nk (1 - p)^{\binom k3} \le \left(\frac{en}k\right)^k e^{-p\binom k3} = \left(\frac{en}k e^{-\frac{pk^2}6}\right)^k$$ +$$\bbE\left[\#\text{red }K_k^{(3)}\right] = \binom nk (1 - p)^{\binom k3} \le \left(\frac{en}k\right)^k e^{-p\binom k3} = \left(\frac{en}k e^{-\frac{pk^2}6}\right)^k$$ This is nontrivial if $p \gg \frac 1{k^2}$. Then -$$\E\left[\#\text{blue }K_4^{(3)}\right] = \binom n4 p^4 \ge \left(\frac n4\right)^4 \gg \left(\frac n{k^2}\right)^4$$ +$$\bbE\left[\#\text{blue }K_4^{(3)}\right] = \binom n4 p^4 \ge \left(\frac n4\right)^4 \gg \left(\frac n{k^2}\right)^4$$ So the (naïve) probabilistic approach looks useless for anything better than polynomial in $k$. \begin{thm} @@ -620,10 +620,10 @@ \subsection{Off-diagonal \texorpdfstring{$3$}{3}-uniform hypergraph Ramsey} \end{obs} But \begin{align*} - \E[\#\text{transitive tournament of size } k] + \bbE \#\text{transitive tournament of size } k & = \binom nk k! 2^{-\binom k2} \\ & \le n^k 2^{-\binom k2} \\ - & = \left(n2^{-\frac{k - 1}2}\right)^n \mor 0 + & = \left(n2^{-\frac{k - 1}2}\right)^n \to 0 \end{align*} Hence $$\P\left(\chi \text{ has no red } K_k^{(3)}\right) = \P(T \text{ has no transitive tournament of size } k) > 0$$ @@ -655,7 +655,7 @@ \subsection{Off-diagonal \texorpdfstring{$3$}{3}-uniform hypergraph Ramsey} Call such a set $K \in [n]^{[k]}$ {\bf sad}. We consider \begin{align*} - \E[\#\text{ sad sets}] + \bbE \#\text{ sad sets} & \le n^k \prod_{i = 1}^k \binom r{\frac k4} \left(\frac k{4r}\right)^{k - i} \\ & = n^k \binom r{\frac k4}^k \left(\frac k{4r}\right)^{\sum_{i = 1}^k k - i} \\ & \le n^k \left(\frac{4er}k\right)^{\frac{k^2}4} \left(\frac k{4r}\right)^{\frac{k^2}4} \\