\documentclass[12pt,ifthen]{article} \usepackage{url} \usepackage{amsmath} \usepackage{amssymb} \usepackage{amsthm} \usepackage{html} \usepackage{hyperref} \newcommand{\spec}{{\rm spec}} \newcommand{\D}{{\sf D}} \newcommand{\N}{{\sf N}} \newcommand{\Z}{{\sf Z}} \newcommand{\Q}{{\sf Q}} \newcommand{\into}{{\rightarrow}} \newcommand{\ceil}[1]{\lceil #1 \rceil} \newcommand{\COL}{{\rm COL}} \newtheorem{theorem}{Theorem} \begin{document} \centerline{\bf TAKE HOME MIDTERM FOR CMSC 752} \centerline{\bf Morally Due April 1} \centerline{\bf Dead Cat Day April 3} \bigskip {\bf Rules} Same as the HW: You can get help but you must understand and hand in your own work. You can use ChatGPT if you want to hand in answers that are really bad. \bigskip There are four problems and they add up to 100 points. \newpage \begin{enumerate} \item (25 points) Let $L$ be an ordered set. Let $\COL\colon L\into [d]$. A set $H$ is {\it $c$-homog (relative to $\COL$)} if $\COL$ restricted to $H$ takes on $\le c$ values. \begin{enumerate} \item (15 points) Find $c$ such that the following is true: \begin{itemize} \item For all $d$, for all $\COL\colon \Q \into [d]$ there exists a $c$-homog set $H\equiv \Q$. \item There exists $\COL\colon \Q \into [c]$ such that there are no $c-1$-homog sets $H\equiv \Q$. \end{itemize} \item (10 points) Let $L_1= \Z + \cdots + \Z$ (there are $n$ copies of $\Z$). Find $c$ such that the following is true: \begin{itemize} \item For all $d$, for all $\COL\colon L\into [d]$ there exists a $c$-homog set $H\equiv L_1$. \item There exists $\COL\colon L\into [c]$ such that there are no $c-1$-homog sets $H\equiv L_1$. \end{itemize} \end{enumerate} \newpage \item (25 points) For this problem you may assume the following $R_a(k)$ is the least $n$ such that every $\COL \colon \binom{[n]}{a}\into[2]$ has a homog set of size $k$. For this problem you may assume the following: $R_1(k)\le 2k$ $R_2(k)\le 2^{2k}$ $R_3(k)\le 2^{2^{4k}}$. (These are not the best known upper bounds but use them as the math will be neater.) Fill in the function $XXX(k)$ in the statement below so that the statement is true. Prove your statement. Try to use an $XXX(k)$ that is small. {\it For all $k$ there exists $n=XXX(k)$ such that, for all triples of colorings: $\COL_1\colon [n] \times\into [2]$ $\COL_2\colon \binom{[n]}{2} \times\into [2]$ $\COL_3\colon \binom{[n]}{3} \times\into [2]$ there exists $H$ of size $k$ such that $H$ is homog for $\COL_1$ (all elements of $H$ map to same color). $H$ is homog for $\COL_2$ (all elements of $\binom{H}{2}$ map to the same color). $H$ is homog for $\COL_3$ (all elements of $\binom{H}{3}$ map to the same color). } \newpage \item (30 points) This problem has three parts: $a$, $b$, $c$. In this problem we will guide you through a proof of a theorem; however, you will need to fill in the XXX. Here is the theorem: {\bf Theorem} {\it Let $c\ge 2$. Let $n= XXX(c)$. For all $\COL\colon \binom{[n]}{2}\into[c]$ there exists a monochromatic $C_4$. That is, four vertices $a,b,c,d$ such that $\COL(a,b)=\COL(b,c)=\COL(c,d)=\COL(d,1)$. ($XXX(c)$ will be a polynomial.) } \bigskip {\bf Proof} {\it Plan} We will find a polynomial $XXX(c)$ such that any graph with $\ge XXX(n)$ edges (where $n$ is the number of vertices) has a $C_4$. We will then show that, for all $\COL\colon \binom{[n]}{2}\into[c]$, some color occurs in $XXX(n)$ edges. That subgraph must have a $C_4$, which will be the desired monochromatic $C_4$. Let $G=(V,E)$ where $V=[n]$. Assume $G$ does not have $C_4$. (We are not going towards a contradiction. We will instead get an upper bound on $|E|$.) {\it Notation} \begin{equation} e_{ij}= \begin{cases} 1 & \hbox{ if $(i,j)\in E$}\\ 0 & \hbox{ if $(i,j)\notin E$ or $i=j$}\\ \end{cases} \end{equation} \begin{enumerate} \item (10 points) Let $a,b\in V$. Show that $$\sum_{i=1}^n e_{ai}e_{ib} \le 1.$$ \newpage \item (10 points) Assuming the result of Part 1 we have $$\sum_{i=1}^n e_{ai}e_{ib} \le 1.$$ We now take a summation over $a$ and over $b\ne a$ on both sides. $$ \sum_{a=1}^n \quad \sum_{b=1,b\ne a}^n \quad \sum_{i=1}^n e_{ai}e_{ib} \le \quad \sum_{a=1}^n\sum_{b=1,b\ne a}^n 1.$$ $$\sum_{i=1}^n \sum_{a=1}^n \quad \sum_{b=1,b\ne a}^n e_{ai}e_{ib} \le n(n-1).$$ $$\sum_{i=1}^n \sum_{a=1}^n e_{ai}\quad \sum_{b=1,b\ne a}^n e_{ib}\le n(n-1).$$ Note that $\sum_{b=1,b\ne a}^n e_{ib}\in \{\deg(i),\deg(i)-1\}$, so $$\sum_{b=1,b\ne a}^n e_{ib}\ge \deg(i)-1.$$ Hence we have $$\sum_{i=1}^n \sum_{a=1}^n e_{ai} (\deg(i)-1)\le n(n-1).$$ $$\sum_{i=1}^n (\deg(i)-1)\sum_{a=1}^n e_{ai} \le n(n-1).$$ $$(\sum_{i=1}^n (\deg(i)-1))\deg(i) \le n(n-1).$$ $$(\sum_{i=1}^n (\deg(i)-1))\deg(i) \le n(n-1).$$ $$\sum_{i=1}^n \deg(i)^2 - \sum_{i=1}^n \deg(i) \le n(n-1).$$ Finally, here is the problem: Show that $$\sum_{i=1}^n \deg(i) \le 0.5n + n\sqrt{n-0.75}$$ {\bf Hint} Use the Cauchy-Schwartz Inequality AND the Quadratic Formula. \newpage \item (10 points) We have $$\sum_{i=1}^n \deg(i) \le 0.5n + n\sqrt{n-0.75}$$ Since $\sum_{i=1}^n \deg(i) = 2|E|$ we have $$2|E| \le 0.5n + n\sqrt{n-0.75}$$ $$|E| \le 0.25n + 0.5n\sqrt{n-0.75}$$ So we have the following: {\it Let $G=(V,E)$. If $G$ has no $C_4$ subgraph then $|E| \le 0.25n + 0.5n\sqrt{n-0.75}$} Take the contrapositive: {\it Let $G=(V,E)$. If $|E| > 0.25n + 0.5n\sqrt{n-0.75}$ then $G=(V,E)$ has no $C_4$ subgraph then $|E| \le 0.25n + 0.5n\sqrt{n-0.75}$} Finally, here is the problem: Find a polynomial $XXX(c)$ such that the following is true, and prove it: {\it Let $c\ge 2$. Let $n=XXX(c)$. Then for all $\COL\colon \binom{[n]}{2}\into[c]$ there exists a monochromatic $C_4$.} \end{enumerate} \newpage \item (20 points) Recall that $5\omega = \omega + \omega + \omega + \omega + \omega$. Let $\COL\colon \binom{5\omega}{3} \into [10^{10}]$. A set $H$ is {\it $c$-homog (relative to COL)} if $\COL$ restricted to $\binom{H}{3}$ takes only $c$ colors. Find a number $c$ such that the following is true, and prove both statements. \begin{itemize} \item For all $\COL\colon \binom{5\omega}{3} \into [10^{10}]$ there exists a $c$-homog set $H\equiv 5\omega$. \item There exists $\COL\colon \binom{5\omega}{3} \into [c]$ there is no $c-1$-homog set $H\equiv 5\omega$. \end{itemize} \end{enumerate} \end{document}