├── .gitignore ├── Combinatorics.bib ├── CombinatoricsNotes.pdf ├── CombinatoricsNotes.png ├── CombinatoricsNotes.tex ├── License.txt ├── Readme.md ├── chap_inputs.tex ├── chapters ├── .DS_Store ├── ch0_intro.tex ├── ch10_algebraic.tex ├── ch1_sperner.tex ├── ch2_LO.tex ├── ch3_intersecting_hypergraphs.tex ├── ch4_shadows.tex ├── ch5_turan.tex ├── ch6_hypergraph_turan.tex ├── ch7_ramsey.tex ├── ch7p5_density_VdW.tex ├── ch8_convexity.tex └── ch9_incidence.tex ├── graphics ├── C5p=0.pdf ├── C5p=p0.pdf ├── C5p=p25.pdf ├── C5p=p3.pdf ├── C5p=p45.pdf ├── C5p=p5.pdf ├── C5p=p6.pdf ├── C5p=p75.pdf ├── C5p=pp6.pdf ├── L1f1.eps ├── L1f1.pdf ├── L1f1.svg ├── LOproof.pdf ├── be-contents.pdf ├── be-title.pdf ├── ei-contents.pdf ├── ei-title.pdf ├── figLOproof.tex ├── figLOproofPDF.pdf ├── figLOproofv2.tex ├── helix.asy ├── helix.pdf ├── hilbertcurves.pdf ├── hilbertrecursive.tex ├── nasa_vision_sm.png ├── satir_graph.png ├── sine.asy ├── sine.pdf ├── standalone_fig.tex ├── vdqi-contents.pdf ├── vdqi-title.pdf ├── ve-contents.pdf └── ve-title.pdf ├── tufte-book.cls ├── tufte-common.def ├── tufte-handout.cls └── tufte.bst /.gitignore: -------------------------------------------------------------------------------- 1 | *.ilg 2 | *.log 3 | *.aux 4 | *.bbl 5 | *.bcf 6 | *.blg 7 | *.fdb_latexmk 8 | *.fls 9 | *.idx 10 | *.ind 11 | *.out 12 | *.run.xml 13 | *.synctex.gz 14 | *.toc 15 | -------------------------------------------------------------------------------- /Combinatorics.bib: -------------------------------------------------------------------------------- 1 | @book{matouvsek2002lectures, 2 | title={Lectures on discrete geometry}, 3 | author={Matou{\v{s}}ek, J.}, 4 | volume={212}, 5 | publisher={Springer}, 6 | year = {2002}, 7 | series = {Graduate Texts in Mathematics}, 8 | doi = {10.1007/978-1-4613-0039-7} 9 | } 10 | 11 | @Article{Radon1921, 12 | author="Radon, J.", 13 | title="Mengen konvexer K{\"o}rper, die einen gemeinsamen Punkt enthalten", 14 | journal="Mathematische Annalen", 15 | year="1921", 16 | volume="83", 17 | number="1", 18 | pages="113--115", 19 | issn="1432-1807", 20 | doi="10.1007/BF01464231", 21 | } 22 | 23 | @article{Helly1923, 24 | author = {Helly, E.}, 25 | journal = {Jahresbericht der Deutschen Mathematiker-Vereinigung}, 26 | pages = {175-176}, 27 | title = {\"Uber Mengen konvexer K\"orper mit gemeinschaftlichen Punkte.}, 28 | url = {http://eudml.org/doc/145659}, 29 | volume = {32}, 30 | year = {1923}, 31 | } 32 | @article{sarkaria1992tverberg, 33 | title={Tverberg's theorem via number fields}, 34 | author={Sarkaria, {K. 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Soc.", 691 | month = "12", 692 | number = "12", 693 | pages = "898--902", 694 | publisher = "American Mathematical Society", 695 | title = "On a lemma of Littlewood and Offord", 696 | url = "http://projecteuclid.org/euclid.bams/1183507531", 697 | volume = "51", 698 | year = "1945" 699 | } 700 | 701 | 702 | @article{KLEITMAN1970, 703 | title = "On a lemma of Littlewood and Offord on the distributions of linear combinations of vectors", 704 | journal = "Advances in Mathematics", 705 | volume = "5", 706 | number = "1", 707 | pages = "155 - 157", 708 | year = "1970", 709 | note = "", 710 | issn = "0001-8708", 711 | doi = "http://dx.doi.org/10.1016/0001-8708(70)90038-1", 712 | author = "Kleitman, {D. J.}", 713 | } 714 | 715 | @book{tao_vu, 716 | author = "Tao, T. and Vu, {V. H.} ", 717 | title = {Additive Combinatorics}, 718 | publisher = {Cambridge University Press}, 719 | year = {2006}, 720 | isbn = {9780511755149}, 721 | note = {Cambridge Books Online}, 722 | url = {http://dx.doi.org/10.1017/CBO9780511755149} 723 | } 724 | @Article{erd64, 725 | author="\erdos, P.", 726 | title="On extremal problems of graphs and generalized graphs", 727 | journal="Israel Journal of Mathematics", 728 | volume="2", 729 | number="3", 730 | pages="183--190", 731 | issn="1565-8511", 732 | doi="10.1007/BF02759942", 733 | year="1964" 734 | } 735 | -------------------------------------------------------------------------------- /CombinatoricsNotes.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/CombinatoricsNotes.pdf -------------------------------------------------------------------------------- /CombinatoricsNotes.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/CombinatoricsNotes.png -------------------------------------------------------------------------------- /CombinatoricsNotes.tex: -------------------------------------------------------------------------------- 1 | \documentclass[nobib,nohyper,final]{tufte-handout} 2 | % \usepackage[T1]{fontenc} 3 | % \usepackage[utf8]{inputenc} 4 | % \usepackage{fbb} 5 | 6 | \title{A survey of extremal combinatorics\thanks{McGill \smallcaps{MATH} 550 notes, Winter 2016.}\smallskip}%\thanks{Based on lectures by Prof. Sergey Norin, Winter 2016.}} 7 | 8 | \author{Written by Eric Hanson,\\based on lectures by Prof. Sergey Norin\smallskip} 9 | 10 | \date{April 2016} % without \date command, current date is supplied 11 | 12 | 13 | % Generates the index 14 | \usepackage{imakeidx} 15 | 16 | 17 | \usepackage{adjustbox} 18 | %\geometry{showframe} % display margins for debugging page layout 19 | \usepackage{amsmath} % extended mathematics 20 | 21 | \usepackage{amsthm, amssymb} 22 | \usepackage{braket} 23 | \usepackage{graphicx} % allow embedded images 24 | \setkeys{Gin}{width=\linewidth,totalheight=\textheight,keepaspectratio} 25 | \graphicspath{{graphics/}} % set of paths to search for images 26 | 27 | \usepackage{booktabs} % book-quality tables 28 | \usepackage{units} % non-stacked fractions and better unit spacing 29 | \usepackage{multicol} % multiple column layout facilities 30 | \usepackage{lipsum} % filler text 31 | \usepackage{fancyvrb} % extended verbatim environments 32 | 33 | \usepackage{mathdots} 34 | % \usepackage{schemata} 35 | \usepackage{accents} 36 | \usepackage{tikz-cd} 37 | \usepackage{mathtools} 38 | 39 | 40 | %%%FIX for hyperref changing captions 41 | \makeatletter 42 | \let\tufte@caption\@caption 43 | \usepackage[unicode,hyperfootnotes=false,hypertexnames=false,colorlinks]{hyperref} 44 | \let\@caption\tufte@caption 45 | 46 | \hypersetup{% 47 | pdfborder = {0 0 0}, 48 | bookmarksdepth = section, 49 | citecolor = DarkGreen, 50 | linkcolor = DarkBlue, 51 | urlcolor = DarkGreen, 52 | } 53 | % \usepackage[colorlinks]{hyperref} 54 | % \usepackage{caption} 55 | \usepackage[dvipsnames]{xcolor} 56 | 57 | 58 | \usepackage{soul} 59 | \usepackage{calc} 60 | \newcommand{\ditto}[1]{% 61 | \rule[\heightof{e} * \real{0.8}]{(\widthof{#1} - \widthof{\,\raisebox{-.2ex}{\textquotedbl}\,}) * \real{0.5}}{.2pt}% 62 | \,\raisebox{-.2ex}{\textquotedbl}\,% 63 | \rule[\heightof{e} * \real{0.8}]{(\widthof{#1} - \widthof{\,\raisebox{-.2ex}{\textquotedbl}\,}) * \real{0.5}}{.2pt}% 64 | } 65 | 66 | %% 67 | %Bibliography 68 | % \usepackage[style=alphabetic,maxalphanames = 4,maxcitenames = 3,minalphanames = 3,minnames = 1,backref=true, backend = biber]{biblatex} 69 | \usepackage[style=authoryear,backref=true, backend = biber,autocite=footnote,sortcites=true,defernumbers]{biblatex} 70 | \addbibresource{Combinatorics.bib} 71 | 72 | %% 73 | %Names for subibs 74 | % \usepackage{nameref} 75 | \defbibheading{subbibliography}{% 76 | \section*{References for Section \therefsegment.}} 77 | 78 | \defbibheading{fullbib}{% 79 | \section*{Collected references}} 80 | 81 | \usepackage{cleveref} 82 | 83 | 84 | 85 | 86 | \usepackage{layouts} 87 | 88 | \usepackage{pgfplots} 89 | \pgfplotsset{compat=newest} 90 | \usetikzlibrary{calc,decorations,decorations.pathreplacing,shapes,fit} 91 | \usepackage{tikz-dimline} 92 | \usepackage{tkz-euclide} 93 | 94 | 95 | \definecolor{halfgray}{gray}{0.55} % chapter numbers will be semi transparent .5 .55 .6 .0 96 | \definecolor{webgreen}{rgb}{0,.5,0} 97 | \definecolor{webbrown}{rgb}{.6,0,0} 98 | %\definecolor{Maroon}{cmyk}{0, 0.87, 0.68, 0.32} 99 | %\definecolor{RoyalBlue}{cmyk}{1, 0.50, 0, 0} 100 | %\definecolor{Black}{cmyk}{0, 0, 0, 0} 101 | 102 | 103 | 104 | % TODO NOTES STUFF 105 | \usepackage{xargs} 106 | \usepackage[colorinlistoftodos,prependcaption,textsize=small,linecolor=OliveGreen,backgroundcolor=OliveGreen!25,bordercolor=OliveGreen,obeyFinal]{todonotes} 107 | \newcommandx{\unsure}[2][1=]{\todo[inline,prepend,caption={Unsure},linecolor=red,backgroundcolor=red!25,bordercolor=red,#1]{#2}} 108 | \newcommandx{\change}[2][1=]{\todo[inline,prepend,caption={Change},linecolor=blue,backgroundcolor=blue!25,bordercolor=blue,#1]{#2}} 109 | \newcommandx{\outline}[2][1=]{\todo[inline,prepend,caption={Outline},linecolor=OliveGreen,backgroundcolor=OliveGreen!25,bordercolor=OliveGreen,#1]{#2}} 110 | \newcommandx{\improvement}[2][1=]{\todo[inline,prepend,caption={Improvement},linecolor=Plum,backgroundcolor=Plum!25,bordercolor=Plum,#1]{#2}} 111 | \newcommandx{\thiswillnotshow}[2][1=]{\todo[inline,disable,#1]{#2}} 112 | \newcommandx{\add}[2][1=]{\todo[inline,prepend,caption={Add},linecolor=Plum,backgroundcolor=Plum!25,bordercolor=Plum,#1]{#2}} 113 | \newcommandx{\missing}[2][1=]{\todo[inline,prepend,caption={Missing},linecolor=blue,backgroundcolor=blue!25,bordercolor=blue,#1]{#2}} 114 | 115 | 116 | \newcommandx{\flavor}[2][1=]{\todo[inline,prepend,caption={Flavor},linecolor=pink,backgroundcolor=pink!25,bordercolor=pink,#1]{#2}} 117 | 118 | \newcommandx{\checkthis}[1][1=]{\todo[inline,prepend,linecolor=red,backgroundcolor=red!25,bordercolor=red,#1]{Check this.}} 119 | 120 | \newcommandx{\understand}[1][1=]{\todo[inline,prepend,caption={To do},linecolor=blue,backgroundcolor=blue!25,bordercolor=blue,#1]{Understand this.}} 121 | 122 | 123 | \fvset{fontsize=\normalsize}% default font size for fancy-verbatim environments 124 | 125 | 126 | \def\mathnote#1{% 127 | \tag*{\rlap{\hspace\marginparsep\smash{\parbox[t]{\marginparwidth}{% 128 | \footnotesize#1}}}} 129 | } 130 | 131 | 132 | %colored dots 133 | \newcommand{\rdot}{\textcolor{red}{\textbullet}} 134 | \newcommand{\gdot}{\textcolor{green}{\textbullet}} 135 | \newcommand{\ydot}{\textcolor{yellow}{\textbullet}} 136 | \newcommand{\bdot}{\textcolor{blue}{\textbullet}} 137 | 138 | 139 | 140 | \newcommand{\rdotm}{{\color{red} \bullet}} 141 | \newcommand{\gdotm}{{\color{green} \bullet}} 142 | \newcommand{\ydotm}{{\color{yellow} \bullet}} 143 | \newcommand{\bdotm}{{\color{blue} \bullet}} 144 | 145 | 146 | \newcommand\vartextvisiblespace{% 147 | \makebox[.5em]{\color{black}{% 148 | \kern.07em 149 | \vrule height.3ex 150 | \hrulefill 151 | \vrule height.3ex 152 | \kern.07em 153 | }}% <-- don't forget this one! 154 | } 155 | 156 | 157 | 158 | \renewcommand{\restriction}{\mathord{\upharpoonright}} 159 | 160 | 161 | \usepackage{xparse} 162 | % \DeclareDocumentCommand \defn { m o }{% 163 | % \IfNoValueTF {#2} {% 164 | % \emph{#1}\index{#1}% 165 | % }{% 166 | % \emph{#1} \index{#2}% 167 | % }% 168 | % } 169 | 170 | \NewDocumentCommand{\defn}{m o}{% 171 | {\emph{#1}}% 172 | \IfNoValueTF{#2} 173 | {\index{#1}\label{#1}} 174 | % {\index{#2@#1}\label{#2}}% 175 | {\index{#2}\label{#2}}% 176 | } 177 | % \newcommand{\defn}[1]{\emph{#1}\index{#1}} 178 | % \newcommand{\Pr}{\emph} 179 | \DeclareMathOperator{\prob}{Pr} 180 | \DeclareMathOperator{\reg}{Reg} 181 | \DeclareMathOperator{\crossings}{cr} 182 | 183 | % \renewcommand{\cros}{\operatorname{cr}} 184 | 185 | \DeclareMathOperator{\per}{per} 186 | \newcommand{\field}{\mathbb{F}} 187 | \DeclareMathOperator{\HJ}{HJ} 188 | 189 | \DeclareMathOperator{\ex}{ex} 190 | 191 | \DeclareMathOperator{\area}{Area} 192 | \DeclareMathOperator{\diam}{Diam} 193 | \DeclareMathOperator{\dist}{dist} 194 | 195 | \DeclareMathOperator{\perim}{Perimeter} 196 | \DeclareMathOperator{\aff}{aff} 197 | \DeclareMathOperator{\conv}{conv} 198 | 199 | % \newcommand{\H}{\mathcal{H}} 200 | 201 | 202 | \DeclarePairedDelimiter\ceil{\lceil}{\rceil} 203 | \DeclarePairedDelimiter\floor{\lfloor}{\rfloor} 204 | % \theoremstyle{plain} is the default. it sets the text in italic and adds extra space above and below the \newtheorems listed below it in the input. it is recommended for theorems, corollaries, lemma*s, propositions, conjectures, criteria, and (possibly; depends on the subject area) algorithms. 205 | 206 | % \theoremstyle{definition} adds extra space above and below, but sets the text in roman. it is recommended for definitions, conditions, problems, and examples; i've alse seen it used for exercises. 207 | 208 | % \theoremstyle{remark} is set in roman, with no additional space above or below. it is recommended for remarks, notes, notation, claims, summaries, acknowledgments, cases, and conclusions. 209 | \theoremstyle{definition} 210 | \newtheorem*{xdefinition}{Def} 211 | 212 | \NewDocumentEnvironment{definition}{oo} 213 | {\IfNoValueTF{#1} 214 | % no optional argument to definition 215 | {\begin{xdefinition}} 216 | % at least one optional argument 217 | {\IfNoValueTF{#2} 218 | % only one optional argument 219 | {\begin{xdefinition}\index{Definition!#1}\label{def-#1}} 220 | % two optional arguments 221 | {\begin{xdefinition}\index{Definition!#1@#2}\label{def-#1}}% 222 | }% 223 | } 224 | {\end{xdefinition}} 225 | \makeindex 226 | 227 | 228 | \newtheorem*{examples}{Examples} 229 | \newtheorem*{example}{Example} 230 | \newtheorem*{exercise}{Exercise} 231 | \newtheorem*{solution}{Solution} 232 | 233 | \newtheorem*{problem}{Problem} 234 | \newtheorem*{conjecture*}{Conjecture} 235 | 236 | \theoremstyle{remark} 237 | \newtheorem*{remark}{Remark} 238 | \newtheorem*{claim}{Claim} 239 | 240 | 241 | \theoremstyle{plain} 242 | 243 | \newtheorem*{theorem*}{Theorem} 244 | \newtheorem{theorem}{Theorem}[section] 245 | \newtheorem*{proposition}{Proposition} 246 | \newtheorem*{corollary*}{Corollary} 247 | \newtheorem{corollary}[theorem]{Corollary} 248 | \newtheorem*{lemma*}{Lemma} 249 | \newtheorem{lemma}[theorem]{Lemma} 250 | 251 | \DeclareMathOperator{\cl}{cl} 252 | \DeclareMathOperator{\symd}{\triangle} 253 | 254 | \newcommand{\R}{\mathbb{R}} 255 | \newcommand{\N}{\mathbb{N}} 256 | \newcommand{\Q}{\mathbb{Q}} 257 | \newcommand{\M}{\mathcal{M}} 258 | 259 | \let\Haccent\H 260 | 261 | 262 | \newcommand{\erdos}{Erd\Haccent os} 263 | \renewcommand{\H}{\mathcal{H}} 264 | 265 | \newcommand{\Cset}{\mathcal{C}} 266 | 267 | \newcommand{\Z}{\mathbb{Z}} 268 | \newcommand{\E}{\mathbb{E}} 269 | 270 | \renewcommand{\C}{\mathbb{C}} 271 | \renewcommand{\P}{\mathcal{P}} 272 | \newcommand{\A}{\mathcal{A}} 273 | \newcommand{\B}{\mathcal{B}} 274 | 275 | \newcommand{\F}{\mathcal{F}} 276 | 277 | \newcommand{\imp}{\rightarrow} 278 | 279 | \newcommand{\limp}{\leftarrow} 280 | 281 | \renewcommand{\setminus}{-} 282 | 283 | \newcommand{\inv}{^{-1}} 284 | \newcommand{\ubar}[1]{\underaccent{\bar}{#1}} 285 | 286 | %%% semantic combinatorics commands 287 | \newcommand{\edges}{\mathcal{E}} 288 | 289 | 290 | \newenvironment{subproof}[1][\proofname]{\begin{proof}[#1]\renewcommand{\qedsymbol}{\scalebox{.6}{$\blacksquare$}}}{\end{proof}} 291 | 292 | %%% For Tikz-cd Drawing: 293 | 294 | %% How to use: put \snakesetup earlier in the line which will end with a snaking line. Then put \snakearrow{lld} say at the end of the line. If you want a label, you could do \snakearrow{lld,"label"}. 295 | %I don't entirely understand where you put \snakesetup. I think there has to be an entry below it? 296 | \newcounter{snakecount} 297 | 298 | \newcommand{\snakesetup}{ 299 | \noexpand{\stepcounter{snakecount}\arrow[d, phantom, ""{coordinate, name={Z\the\value{snakecount}}}]}} 300 | \newcommand{\snakearrow}[1]{ 301 | \arrow[#1,rounded corners, 302 | to path={ -- ([xshift=2ex]\tikztostart.east) 303 | |- (Z\the\value{snakecount}) [near end]\tikztonodes 304 | -| ([xshift=-2ex]\tikztotarget.west) 305 | -- (\tikztotarget)}] 306 | } 307 | %%%% 308 | 309 | 310 | %%% 311 | \colorlet{green}{Green} 312 | \colorlet{red}{Red} 313 | \colorlet{blue}{Blue} 314 | \colorlet{pink}{Pink} 315 | 316 | % \colorlet{red}{Plum} 317 | 318 | % \usetikzlibrary{external} 319 | % \tikzexternalize 320 | 321 | \usepackage[american,showdow]{datetime2} 322 | \newcounter{lecture} 323 | \setcounter{lecture}{2} 324 | % \newcommand{\lect}[2]{\marginnote{Lecture \thelecture: \DTMdate{2016-#1-#2}.}\stepcounter{lecture}} 325 | \newcommand{\lect}[2]{} 326 | 327 | 328 | %change sidenotes 329 | \usepackage{perpage} 330 | \MakePerPage{footnote} 331 | % Works up to nine sidenotes 332 | \renewcommand*{\thefootnote}{\fnsymbol{footnote}} 333 | % Define up to 18 footnote symbols 334 | \makeatletter 335 | \def\@fnsymbol#1{% 336 | \ensuremath{% 337 | \ifcase#1\or 338 | \dagger\or 339 | \ddagger\or 340 | *\or 341 | \mathsection\or 342 | \mathparagraph\or 343 | \|\or 344 | **\or 345 | \dagger\dagger\or 346 | \ddagger\ddagger\or 347 | \mathsection\mathsection\or 348 | \mathparagraph\mathparagraph\or 349 | \|\|\or 350 | ***\or 351 | \dagger\dagger\dagger\or 352 | \ddagger\ddagger\ddagger\or 353 | \mathsection\mathsection\mathsection\or 354 | \mathparagraph\mathparagraph\mathparagraph\or 355 | \|\|\| 356 | \else 357 | % We've run out of footnote symbols for this page. 358 | % We would normally emit an error here, 359 | % but because the perpage package won't work until the 360 | % second pass, we need to do something less abrupt. 361 | \fi 362 | }% 363 | } 364 | \makeatother 365 | 366 | 367 | 368 | % \renewcommand*{\thefootnote}{\alph{footnote}} 369 | 370 | %%for pictures in Ramsey theory section 371 | \newcounter{numint} 372 | 373 | 374 | 375 | % remove bullet 376 | % \def\labelitemi{} 377 | 378 | \setcounter{secnumdepth}{2} 379 | \setcounter{tocdepth}{2} 380 | 381 | % The default minimum amount of space between \marginnotes is 10 pt. 382 | \setlength\marginparpush{20pt} 383 | 384 | 385 | % \includeonly{chapters/ch1_sperner} 386 | % \includeonly{chapters/ch2_LO} 387 | % \includeonly{chapters/ch3_intersecting_hypergraphs} 388 | % \includeonly{chapters/ch4_shadows} 389 | % \includeonly{chapters/ch5_turan} 390 | % \includeonly{chapters/ch6_hypergraph_turan} 391 | % \includeonly{chapters/ch7_ramsey} 392 | % \includeonly{chapters/ch8_convexity} 393 | % \includeonly{chapters/ch9_incidence} 394 | 395 | 396 | \begin{document} 397 | \setuplayouts 398 | 399 | % \frontmatter 400 | 401 | \maketitle% this prints the handout title, author, and date 402 | 403 | % \mainmatter 404 | 405 | % \section{Introduction} \marginnote{Lecture 1: Thursday, Jan. 7, 2016.} 406 | 407 | Combinatorics is the study of finite discrete structures\sidenote{Say, collections of sets, or integers, or geometric sets.}. There are several subfields of combinatorics: 408 | \begin{itemize} 409 | \item Enumeratative combinatorics: try to understand how many objects exist with certain properties. 410 | \item Extremal combinatorics: what are the largest or smallest objects with certain properties. This is what we will focus on in this class. 411 | \end{itemize} 412 | \begin{figure} 413 | \begin{center} 414 | \begin{tikzcd} 415 | \text{probability} \arrow[rd] & & \text{number theory}\\ 416 | & \text{combinatorics} \arrow[ru] \arrow[rd] & \\ 417 | \text{algebra} \arrow[ru] & & \text{discrete geometry} 418 | \end{tikzcd} 419 | \end{center} 420 | \caption{The flow of combinatorial arguments. Probabalistic and algebraic tools are used in combinatorial arguments, which yields results in number theory and discrete geometry.} 421 | \end{figure} 422 | 423 | \tableofcontents 424 | 425 | % \listoftodos 426 | 427 | % \renewcommand\thesection{} 428 | \renewcommand\thesubsection{\Roman{section}.\Roman{subsection}} 429 | 430 | 431 | \input{chap_inputs} 432 | 433 | 434 | % \backmatter 435 | \clearpage 436 | \begin{fullwidth} 437 | \printbibliography[heading=fullbib] 438 | 439 | \printindex 440 | \end{fullwidth} 441 | 442 | \end{document} 443 | -------------------------------------------------------------------------------- /License.txt: -------------------------------------------------------------------------------- 1 | MIT License 2 | 3 | Copyright (c) 2016 Eric P Hanson 4 | 5 | Permission is hereby granted, free of charge, to any person obtaining a copy 6 | of this software and associated documentation files (the "Software"), to deal 7 | in the Software without restriction, including without limitation the rights 8 | to use, copy, modify, merge, publish, distribute, sublicense, and/or sell 9 | copies of the Software, and to permit persons to whom the Software is 10 | furnished to do so, subject to the following conditions: 11 | 12 | The above copyright notice and this permission notice shall be included in all 13 | copies or substantial portions of the Software. 14 | 15 | THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR 16 | IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, 17 | FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE 18 | AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER 19 | LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, 20 | OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE 21 | SOFTWARE. 22 | -------------------------------------------------------------------------------- /Readme.md: -------------------------------------------------------------------------------- 1 | These are notes on extremal combinatorics from McGill University's Winter 2016 class Math 550. I hope they are helpful! 2 | 3 | If you find any typos, please let me know, or better yet, submit a pull request. Thanks to Sage Kira for pointing out a typo. 4 | 5 | 6 | [![first page](CombinatoricsNotes.png)](CombinatoricsNotes.pdf) -------------------------------------------------------------------------------- /chap_inputs.tex: -------------------------------------------------------------------------------- 1 | \clearpage 2 | \input{chapters/ch0_intro} 3 | 4 | 5 | 6 | \renewcommand\thesection{\arabic{section}} 7 | \renewcommand\thesubsection{\thesection.\arabic{subsection}} 8 | \setcounter{section}{0} 9 | 10 | 11 | 12 | \clearpage 13 | \newrefsegment 14 | \input{chapters/ch1_sperner} 15 | \begin{fullwidth} 16 | \printbibliography[segment=\therefsegment,heading=subbibliography] 17 | \end{fullwidth} 18 | 19 | 20 | \clearpage 21 | \newrefsegment 22 | \input{chapters/ch2_LO} 23 | \begin{fullwidth} 24 | \printbibliography[segment=\therefsegment,heading=subbibliography] 25 | \end{fullwidth} 26 | 27 | 28 | \clearpage 29 | \newrefsegment 30 | \input{chapters/ch3_intersecting_hypergraphs} 31 | \begin{fullwidth} 32 | \printbibliography[segment=\therefsegment,heading=subbibliography] 33 | \end{fullwidth} 34 | 35 | 36 | \clearpage 37 | \newrefsegment 38 | \input{chapters/ch4_shadows} 39 | \begin{fullwidth} 40 | \printbibliography[segment=\therefsegment,heading=subbibliography] 41 | \end{fullwidth} 42 | 43 | 44 | 45 | \clearpage 46 | \newrefsegment 47 | \input{chapters/ch5_turan} 48 | \begin{fullwidth} 49 | \printbibliography[segment=\therefsegment,heading=subbibliography] 50 | \end{fullwidth} 51 | 52 | 53 | \clearpage 54 | \newrefsegment 55 | \input{chapters/ch6_hypergraph_turan} 56 | \begin{fullwidth} 57 | \printbibliography[segment=\therefsegment,heading=subbibliography] 58 | \end{fullwidth} 59 | \clearpage 60 | 61 | %\input{chapters/midterm} 62 | 63 | \clearpage 64 | \newrefsegment 65 | \input{chapters/ch7_ramsey} 66 | \begin{fullwidth} 67 | \printbibliography[segment=\therefsegment,heading=subbibliography] 68 | \end{fullwidth} 69 | 70 | 71 | 72 | \clearpage 73 | \newrefsegment 74 | \input{chapters/ch8_convexity} 75 | \begin{fullwidth} 76 | \printbibliography[segment=\therefsegment,heading=subbibliography] 77 | \end{fullwidth} 78 | 79 | 80 | \clearpage 81 | \newrefsegment 82 | \input{chapters/ch9_incidence} 83 | \begin{fullwidth} 84 | \printbibliography[segment=\therefsegment,heading=subbibliography] 85 | \end{fullwidth} 86 | 87 | 88 | \clearpage 89 | \newrefsegment 90 | \input{chapters/ch10_algebraic} 91 | \begin{fullwidth} 92 | \printbibliography[segment=\therefsegment,heading=subbibliography] 93 | \end{fullwidth} -------------------------------------------------------------------------------- /chapters/.DS_Store: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/chapters/.DS_Store -------------------------------------------------------------------------------- /chapters/ch0_intro.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = ../CombinatoricsNotes.tex 2 | 3 | \section{Set systems as an introduction to extremal combinatorial results}\marginnote{The following is a brief introduction to some of the results we will see in the course. 4 | 5 | Note that the superscripted symbols $^\dagger$, $^\ddagger,$ and $^*$ are reserved to point towards notes in this margin. 6 | 7 | In this introduction, full references will be included in the margin; afterwards, they will appear at the end of each section.} 8 | A \emph{set system} is any $\F\subset \P(X)$. 9 | Let $X$ be a finite set, and $\P(X)$ the collection of all subsets of $X$. Then $|\P(X)| = 2^{|X|}$. We'll denote $X^{(r)}$ as the collection of all $r$-element subsets of $X$. 10 | \begin{example} 11 | If $X = \{1,2,3\}$, then $\P(X)=$ 12 | \begin{center} 13 | \begin{tikzcd} 14 | X^{(3)} & &\{1,2,3\} \\ 15 | X^{(2)} &\{1,2\} & \{1,3\} & \{2,3\} \\ 16 | X^{(1)} & \{1\} & \{2\}&\{3\} \\ 17 | X^{(0)} & & \emptyset 18 | \end{tikzcd} 19 | \end{center} 20 | \end{example} 21 | 22 | 23 | \begin{definition} 24 | $\F$ is an \defn{intersecting set system} if $A\cap B\neq\emptyset$ for all $A,B\in \F$. 25 | \end{definition} 26 | 27 | \subsection*{Largest intersecting set system} 28 | Given $X$ with $|X| = n$, what is the maximum $|\F|$ with $\F\subset \P(X)$ intersecting? Let's start with $n=3$ (returning to our example). We can select at least 4: every set that contains 1\marginnote{In this case, we could also choose $X^{(2)}\cup X^{(3)}$, but that only works for $n$ odd.}. This method gives us $2^{n-1}= |\P(X \setminus\{1\})|$\sidenote{since the number of subsets without 1 is in bijection with the number of subsets with 1. For a subset without 1, we add in 1, and get a new valid subset with 1. On the other hand, for a subset with 1, we take out 1 and get a valid subset without 1. These are clearly injective.}. Why can't we do better? We can subdivide $\P(X)$ into pairs $\{Y, X\setminus Y\}$. But $Y\cap (X\setminus Y) = \emptyset$, so any intersecting $\F$ contains at most one element of each pair. Thus, we can't do better than $2^{n-1}$. 29 | 30 | \subsection*{Largest intersecting set system with fixed number of elements} 31 | How large can $\F$ intersecting be if $\F\subset X^{(r)}$ for some $r$? Our answer will depend on $r$ and $|X| = n$. Since $|X^{(r)}| = {n \choose r}$, that is an upper bound. When can we achieve it, i.e. when is $X^{(r)}$ intersecting? When $r > n/2$. In this case, every subset has more than half of the elements of $X$, so no two subsets can be disjoint. 32 | 33 | If $r = n/2$, then because of complement pairing, the maximum is ${n \choose r}/2$. By choosing $\F$ as all of the sets that contain a particular element, we get ${n -1\choose r-1}$. So for $r=n/2$, we have 34 | \[ 35 | {n-1\choose r-1} \leq \max |\F| \leq \frac{1}{2}{n\choose r}. 36 | \] 37 | But for $r=n/2$, these two bounds are equal\sidenote{Why? the LHS is the number of sets containing 1. But half of the sets contain 1 and half don't, by the complement argument.}. 38 | For $r< n/2$, the answer is ${n -1 \choose r-1}$; this is a lower bound by selecting all sets containing a given element. That this is an upper bound is the content of the following non-trivial theorem. 39 | \begin{theorem*}[\erdos-Ko-Rado] \marginnote{\fullcite{erdos-ko-rado-1961}} 40 | Let $r\leq n/2$. Let $\F\subset X^{(r)}$ with $|X| = n$ be intersecting. Then $|F| \leq { n-1\choose r-1}$. 41 | \end{theorem*} 42 | 43 | Next, let's consider other types of set systems. 44 | \subsection*{Littlewood-Offord problem} 45 | We'll say $\F$ is a \defn{Sperner system} if $A\subset B$ for $A,B\in \F$, then $A=B$. 46 | Given $X$, $|X|=n$, what is $\max |\F|$ such that $\F\subset \P(X)$ is Sperner? Well, any $X^{(r)}$ is Sperner\sidenote{If you take any element out, you reduce the number of elements, so aren't in $X^{(r)}$ anymore.}. So we can achieve ${n \choose \floor{n/2} }$. It turns out that one cannot do better. For $n=3$, we may make a graph by connecting subsets by edges. Then we need the largest independent set of this graph; it's easy to show in this case that this is ${n \choose \floor{n/2} }$. This is in fact the general result. 47 | \begin{problem}[Littlewood-Offord 1938] \marginnote{\fullcite{LO1943}} 48 | Suppose we have non-zero numbers $a_1,a_2,\ldots,a_n \in \C$. For $I\subset \{1,2,\dotsc,n\}$, consider $\sum_{i\in I} a_i$. There are then $2^n$ possible such sums. What is the maximal number of them that can be equal to be zero? 49 | \end{problem} 50 | Clearly we can shift to any number $z$ instead of zero without changing the answer. \erdos\ solved this in 1945 for real numbers. Let's consider the case of positive numbers. 51 | First, if $a_1=\dotsm = a_n=1$, then the maximum number of equal sums is ${n \choose \floor{n/2} }$. 52 | Now for general positive numbers, consider $\F_z = \{ I: \sum_{i\in I} a_i = z\}$. Then $\F_z$ is a Sperner system, so by the previously quoted result its maximum size is ${n \choose \floor{n/2} }$. 53 | The Littlewood-Offord problem was treated in general in 1966 by Kleitman\footfullcite{KLEITMAN1970}. 54 | 55 | \subsection*{\erdos-Szemer\'edi conjcture} 56 | % \begin{conjecture}[] 57 | Let $A\subset \Z_+$, $|A| = n$. Let $A+A = \{ a+b: a,b\in A\}$. We wish to compare $|A+A|$ to $|A|$. 58 | Now, $\max |A+A| = {n \choose 2} + n = {n+1 \choose 2}$ by choosing our elements so no two sums are the same unless they have to be\sidenote{Intuitively then, we've achieved the maximum by a set with very little structure.}. 59 | 60 | What is $\min|A+A|$? 61 | We'll choose a set with a lot of structure: $A= \{1,\dotsc,n\}$. Then $A+A = \{2,3,\dotsc,2n\}$, and $|A+A| = 2n-1$. We'll prove that this is actually the minimum. 62 | % \end{conjecture} 63 | \begin{proof} 64 | Let $a_1 < a_2 <\dotsm < a_n$\marginnote{The elements are distinct because $A$ is a set.}. Then 65 | \[ 66 | \underbrace{a_1 + a_1 < a_1 + a_2 < \dotsm < a_1 + a_n}_{n} < \underbrace{a_2 + a_n < a_3 + a_n < \dotsm < a_n + a_n}_{n-1} 67 | \] 68 | so we found $2n-1$ different terms already. 69 | \end{proof} 70 | 71 | What about $|A\cdot A|$? Then $\max |A\cdot A| = {n\choose 2} +n$, and $\min |A\cdot A| = 2n-1$. The second is by $A = \{1,2,2^2,\dotsm, 2^{n-1}\}$. Generally we can transform additive problems into multiplicative by exponentiation, and back by the logarithm. 72 | Now, what's $\min ( |A+A| + |A\cdot A|)$? 73 | \begin{conjecture*}[\erdos-Szemer\'edi] \marginnote{\fullcite{Erdos-Szemeredi-1983}} 74 | \marginnote{This qualitatively means the additive and multiplicative structures do not get along; you reduce $|A+A|$ only at the cost of increasing $|A\cdot A|$.} 75 | \[ 76 | |A+A| + |A\cdot A| = \Omega_\epsilon (|A|^{2-\epsilon}) 77 | \] 78 | That is, for all $\epsilon>0$, there exists $c>0$ such that for all $A\subset \Z_+ \setminus \{0\}$, 79 | \[ 80 | |A + A| + |A\cdot A| \geq c_\epsilon |A|^{2- \epsilon}. 81 | \] 82 | \end{conjecture*} 83 | Elekes showed that $|A +A| + |A\cdot A| \geq c |A|^{5/4}$ (\cite{elekes1997number}) and Solymosi improved this to bound to $|A|^{4/3 - \epsilon}$~(\cite{Solymosi2009}). 84 | In fact, this is related to the following geometric problem (see \cref{fig:IPLexample} for an example). 85 | \begin{theorem*}[Szemer\'edi-Trotter theorem] \marginnote{\fullcite{Szemeredi-Trotter1983}} 86 | \begin{marginfigure}[2cm] 87 | \includegraphics{L1f1.pdf} 88 | \caption{An example of a collection $L$ of lines and $P$ of points. We are interested in the number of \defn{incidences} $I(P,L)$ between points in $P$ and lines in $L$. Here, $I(P,L) = 5$.} \label{fig:IPLexample} 89 | \end{marginfigure} 90 | Suppose we have a collection $L$ of lines in the plane, and $P$ a collection of points in the plane. The set of incidences is $\{(p,\ell): p\in P, \ell\in L, p\in \ell\}$. Let $I(P,L)$ denote the number of incidences for $P$ and $L$\sidenote{Clearly, $I(P,L) \leq |P| \cdot |L|$, as a subset of $P\times L$.}. Then 91 | \[ 92 | I(P,L) \leq O ( |P|^{2/3} |L|^{2/3} + |P| + |L| ). 93 | \] 94 | \end{theorem*} 95 | Elekes used this theorem to make progress on the \erdos-Szemer\'edi conjecture by mapping the set $A$ to a collection of points and lines. He showed that if both $|A\cdot A|$ and $|A+A|$ are small, then you find too many incidences. 96 | The Szemer\'edi-Trotter theorem uses the following result: 97 | \begin{lemma*}[Crossing lemma] 98 | Let $G$ be a graph drawn in the plane with crossings. Suppose that $G$ has $m$ edges and $n$ verticies\marginnote{If $m > 3n-6$, then there is at least one crossing, by the result stated at the beginning of the lecture, which comes from Euler's formula. In fact, this is how one proves the crossing lemma.}. If $m\geq 4n$, then there are at least $\frac{m^3}{64 n^2}$ crossings. 99 | \end{lemma*} 100 | From $P$ and $L$, one makes a graph and applies the Crossing Lemma to prove Szemer\'edi-Trotter. 101 | -------------------------------------------------------------------------------- /chapters/ch10_algebraic.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = ../CombinatoricsNotes.tex 2 | 3 | \section{Algebraic methods} 4 | % \marginnote{Lecture X: Wednesday, April 6, 2016.} 5 | \lect{2}{6} 6 | 7 | \subsection{Combinatorial Nullstellensatz (\cite{AlonTarsi89})} 8 | \newthought{For background} we will first consider the following result about zeros of multivariate polynomials. 9 | \begin{theorem*}[Hilbert Nullstellensatz (\cite{HilbertNull})] 10 | Let $\F$ be an algebraically closed field. Let $g_1,g_2,\dotsc,g_m \in \F[x_1,x_2,\dotsc,x_n]$ be polynomials of $n$ variables over $\F$. Suppose $f\in \F[x_1,\dotsc,x_n]$ is such that $f=0$ whenever $g_1=g_2=\dotsm = g_m = 0$. 11 | \marginnote{In one variable: $g = (x-a_1)\cdot (x-a_2)\dotsm (x-a_z)$. Assume all roots are distinct. Then $f=0$ at zeros of $g$ iff $f = gh$ for some $h$. If not all roots are distinct, then $f=0$ at zeros of $g$ iff $f^k =gh$ for some $k\in \N$.} 12 | Then there exists $n\in \N$ and polynomials $h_1,\dotsc,h_m \in \F[x_1,\dotsc,x_n]$ such that $f^n = h_1 g_1 + h_2 g_2 + \dotsm + h_m g_m$. 13 | \end{theorem*} 14 | 15 | 16 | 17 | \newthought{Let us proceed} to \defn{Combinatorial Nullenstellensatz}. 18 | Recall the degree of a multivariable monomial is the sum of degrees in each variable (e.g. $\deg (x_1^2 x_2^3) = 5$), and the degree of a polynomial is as usual the maximum degree of its monomials. 19 | \begin{theorem} \label{thm:comb_nullstell} 20 | Let $\F$ be a field, $S_1,S_2,\dotsc,S_n\subset \F$, 21 | \[ 22 | g_i = \prod_{s\in S_i}(x_i - s) 23 | \] 24 | for $i=1,2,\dotsc, n$. 25 | If $f\in \F[x_1,\dotsc,x_n]$ such that $f$ vanishes on all common zeros of $g_1,g_2,\dotsc,g_n$ (i.e., $f(x_1,\dotsc,x_n) = 0$ if $x_i\in S_i$ for all $i$), then there exist polynomials $h_1,h_2,\dotsc,h_n\in \F[x_1,\dotsc,x_n]$ with $\deg h_i \leq \deg f - \deg g_i$ and $f = h_1g_1 + h_2g_2 + \dotsm + h_n g_n$. 26 | \end{theorem} 27 | \begin{remark} 28 | The $g_i$ are monic, single variable, with no double roots. This is very restrictive compared to Hilbert's Nullstellensatz, but we do obtain a stronger conclusion. 29 | \end{remark} 30 | \marginnote{The degree bound in \cref{thm:comb_nullstell} means that each $h_ig_i$ has degree at most that of $f$.} 31 | Given a multivariable polynomial $p(x_1,\dotsc,x_n)$, let us denote $\deg_{x_i} p$ as the ``degree of $p$ over $x_i$'', i.e. the degree of $p$ when considered as a polynomial in the variable $x_i$ only, treating $x_1,\dotsc,x_{i-1},x_{i+1},\dotsc,x_n$ as constants. To prove \cref{thm:comb_nullstell}, we'll need the following lemma. 32 | % \marginnote{} 33 | \begin{lemma} \label{lem:comb_null_p_vanish} 34 | Let $P\in \F[x_1,\dotsc,x_n]$ with $\deg_{x_i} P \leq t_i$ for $i=1,\dotsc,n$. Let $S_1,\dotsc, S_n \subset \F$ , $|S_i|\geq t_i+1$ for $i\in[n]$. If $P (x_1,\dotsc,x_n) = 0$ when $x_i\in S_i$ for each $i\in [n]$, then $P \equiv 0$. 35 | \end{lemma} 36 | \begin{proof}[Proof by induction on $n$.] 37 | Base case $(n=1)$: a degree $t$ polynomial has at most $t$ roots. 38 | Induction step: we write 39 | \begin{fullwidth} 40 | \begin{align*} 41 | P(x_1,\dotsc,x_n) &= P_{t_n}(x_1,\dotsc,x_{n-1}) x_n^{t_n} + \dotsm + P_1 (x_1,\dotsc,x_{n-1}) x_n + P_0 (x_1,\dotsc,x_{n-1}) \\ 42 | &= \sum_{i=1}^{t_n} P_{t_i}(x_1,\dotsc,x_{n-1}) x_n^i. 43 | \end{align*} 44 | \end{fullwidth} 45 | for some polynomials $P_{t_n},\dotsc,P_0$ in variables $x_1,\dotsc,x_{n-1}$. \marginnote{I.e., the $P_{t_i}$ are the coefficients of $x_n^i$, and do not depend on $x_n$.} 46 | Now, fix $x_1\in S_1$, \ldots, $x_{n-1} \in S_{n-1}$. Then $P(x_1,\dotsc,x_n)$ becomes a one-variable polynomial in $x_n$ of degree at most $t_n$ with at least $t_n+1$ roots (one for each $x_n \in S_n$). So by the base case, each $P_i(x_1,\dotsc,x_{n-1}) = 0$ for $i\in[t_n]$. Since we chose $x_1\in S_1$, \ldots, $x_{n-1}\in S_{n-1}$ arbitrarily, by the induction hypothesis, $P_i \equiv 0$. Hence $P\equiv 0$. 47 | \end{proof} 48 | We may proceed to prove the theorem. 49 | \begin{proof}[Proof of \cref{thm:comb_nullstell}] 50 | 51 | Let $t_i = |S_i| - 1$. For $i\in [n]$, we write 52 | \[ 53 | g_i = x_i^{t_i+1} - \sum_{j=0}^{t_i} a_{ij} x_i^j 54 | \] 55 | for some coefficients $a_{ij}$. 56 | Given a polynomial $f$ in variables $x_1,\dotsc, x_n$, we ``divide with remainder'' by each $g_i$, yielding 57 | \[ 58 | f = \left( \sum_{i=1}^n h_i g_i \right) + P. 59 | \] 60 | with $\deg(h_i) \leq \deg(f) - \deg (g_i)$ and $\deg_{x_i} (P)\leq t_i$. Now, if $f$ vanishes when all $x_i\in S_i$, then $P$ does too, so \cref{lem:comb_null_p_vanish} we have $P\equiv 0$, giving the result. 61 | \end{proof} 62 | 63 | We may reformulate this result into a perhaps more helpful form. 64 | 65 | \begin{theorem}[Combinatorial Nullstellensatz] \label{thm:comb_nullstell_nonzero} 66 | Let $\F$ be a field, $f\in \F[x_1,\dotsc,x_n]$, $\deg(f) = t_1 + t_2 + \dotsm + t_n$ and the coefficient of $x_1^{t_1}x_2^{t_2}\dotsm x_n^{t_n}$ in $f$ non-zero. Let $S_1,S_2,\dotsc,S_n\subset \F$ with $|S_i| \geq t_i+1$. Then there exists $s_1\in S_1$, $s_2\in S_2$,\ldots, $s_n \in S_n$ such that $f(s_1,s_2,\dotsc,s_n) \neq 0$. 67 | \end{theorem} 68 | \begin{proof} 69 | Suppose not. Then $f$ vanishes whenever $x_i \in S_i$, so by \cref{thm:comb_nullstell}, there exist polynomials $h_1,\dotsc,h_n$ with $\deg(h_i) \leq \deg (f) - |S_i| \leq \deg (f) - t_i -1$ such that 70 | \[ 71 | f = \sum_{i=1}^n h_i g_i 72 | \] 73 | where $g_i = \prod_{s\in S_i}(x_i-s)$. Then the monomial $x_1^{t_1}x_2^{t_2}\dotsm x_n^{t_n}$ must appear with non-zero coefficient in some $h_ig_i$. Since $\deg f \geq \deg h_i + \deg g_i$, this term must be obtained by multiplying a monomial in $h_i$ by the highest degree monomial in $g_i$, namely $x_i^{|S_i|}$. Since $|S_i| > t_i$, this is a contradiction. 74 | \end{proof} 75 | 76 | 77 | \begin{theorem*}[Cauchy-Davenport theorem] 78 | Let $A,B \subset \Z/ p\Z =: \Z_p$ for $p$ prime. Then $|A+B| \geq \min\{ p, |A|+|B|-1\}$. 79 | \end{theorem*} 80 | % \begin{remark} 81 | % \missing{put in ref to old C-D thoerem} 82 | % \end{remark} 83 | \begin{proof} 84 | Suppose $|A|+|B| \leq p+1$ without loss of generality\sidenote{by throwing away elements of $A$ and $B$. If the minimum $m$ is acheived by $p$, we throw away elements until $|A|+|B|-1 = p = m$. Then we show that the now-smaller $|A+B|$ has $|A+B| \geq m$. Then the original $|A+B|$ must be at least $m$ as well.}. Suppose, for contradiction, that $|A+B| \leq |A|+|B|- 2$. Then select $C\supseteq A+B$ such that $|C| = |A| + |B| - 2$. 85 | 86 | Let $a= |A|$ and $b=|B|$. Set $f(x,y) = \prod_{c\in C} (x+y-c)$. Then 87 | \[ 88 | \deg f = |C| = a+b-2 = \overbrace{(a-1)}^{t_1} + \overbrace{(b-1)}^{t_2} 89 | \] 90 | The coefficient of $x_1^{t_1}x_2^{t_2}$ in $f$ is ${a+b-2\choose a-1}$. Note since $a+b-2 \leq p+1-2 = p-1 < p$ so the binomial coefficient is not zero. Set $S_1= A$, $S_2=B$. By \cref{thm:comb_nullstell_nonzero}, there exists $s_1\in A$, $s_2\in B$ such that $f(s_1,s_2) \neq 0$. Then $s_1+s_2\not\in C$ which is a contradiction. 91 | \end{proof} 92 | 93 | Let us consider a conjecture by \cite{ErdosHeilbronn}, proven in 1994. 94 | \begin{theorem}[\cite{SilvaHamidoune1994}] 95 | Let $A,B\subset \Z_p$, with $A\oplus B = \{a+b: a\in A, b\in B, a\neq b\}$. Then 96 | \[ 97 | |A\oplus B| \geq \min \{p, |A| +|B|-2 \} 98 | \] 99 | if $|A|\neq |B|$. 100 | \end{theorem} 101 | \begin{remark} 102 | The assumption $|A|\neq|B|$ is necessary. Say $A= \{0,1,\dotsc,k\}$ with $k< p/2$, then $|A| = k+1$ and 103 | \[ 104 | A\oplus A = \{1,2,\dotsc,2k-1 \} 105 | \] 106 | so $|A\oplus A| = 2k-1 = 2|A|-3$. 107 | \end{remark} 108 | \begin{proof} 109 | Let $a=|A|$, $b=|B|$. WLOG, assume $a+b\leq p+2$. Suppose for a contradiction that there exists $C \supset A\oplus B$ such that $|C| = a+b-3$. Let 110 | \[ 111 | f(x,y) = (x-y)\prod_{c\in C} (x+y-c). 112 | \] 113 | Then $\deg f = |C|+1 = a+b-2 = (a-1) + (b-1)$. Assume for now that the coefficient of $x^{a-1}y^{b-1}$ is non-zero. Then by \cref{thm:comb_nullstell_nonzero} there exists $s_1\in A$, $s_2\in B$ such that $f(s_1,s_2) \neq 0$. But then $s_1\neq s_2$ and $s_1+s_2\not \in C$. This is a contradiction. 114 | 115 | It remains to find the coefficient of $x^{a-1}y^{b-1}$ in $f(x,y)$. The coefficient must be the same as in the polynomial $(x-y)(x+y)^{a+b-3}$. We use binomial expansion and only keep the terms of the right degree: $(x-y) (\alpha_1 x^{a-2}y^{b-1} + \alpha_2 x^{a-1} y^{b-2})$. Then the coefficient is 116 | \[ 117 | \alpha_1 - \alpha_2 = {a+b-3 \choose a-2} - {a+b-3 \choose a-1} \neq 0 \pmod{p} 118 | \] 119 | if and only if 120 | \[ 121 | \frac{(a+b-3)!}{(a-2)!(b-1)!} \neq \frac{(a+b-3)!}{(a-1)!(b-2)!} \pmod{p} 122 | \] 123 | which we may simplify to obtain 124 | \[ 125 | a-1 \neq b-1 \pmod{p}.\qedhere 126 | \] 127 | \end{proof} 128 | 129 | 130 | \paragraph{$n$-dimensional hypercube} 131 | $Q_n:=\{0,1\}^n = \{(x_1,\dotsc,x_n) \in \R^n: x_i \in \{0,1\}\}$. $2^n$ points. See \cref{fig:cube} for $Q_3$. 132 | We'd like to cover $Q_n$ by affine hyperplanes; this can be done with $2$: one on top and one on bottom. 133 | Suppose we'd like to cover exactly all verticies except the origin. 134 | 135 | 136 | 137 | \begin{marginfigure} 138 | \begin{center} 139 | \begin{tikzpicture} 140 | \newcommand{\Depth}{2} 141 | \newcommand{\Height}{2} 142 | \newcommand{\Width}{2} 143 | \def\radius{2pt} 144 | \coordinate (O) at (0,0,0); 145 | \coordinate (A) at (0,\Width,0); 146 | \coordinate (B) at (0,\Width,\Height); 147 | \coordinate (C) at (0,0,\Height); 148 | \coordinate (D) at (\Depth,0,0); 149 | \coordinate (E) at (\Depth,\Width,0); 150 | \coordinate (F) at (\Depth,\Width,\Height); 151 | \coordinate (G) at (\Depth,0,\Height); 152 | 153 | 154 | 155 | \draw[black!50] (O) -- (C) -- (G) -- (D) -- cycle;% Bottom Face 156 | \draw[black!50] (O) -- (A) -- (E) -- (D) -- cycle;% Back Face 157 | \draw[black!50] (O) -- (A) -- (B) -- (C) -- cycle;% Left Face 158 | \draw[black!50] (D) -- (E) -- (F) -- (G) -- cycle;% Right Face 159 | \draw[black!50] (C) -- (B) -- (F) -- (G) -- cycle;% Front Face 160 | \draw[black!50] (A) -- (B) -- (F) -- (E) -- cycle;% Top Face 161 | 162 | \filldraw (O) circle(\radius); 163 | \filldraw (A) circle(\radius); 164 | \filldraw (B) circle(\radius); 165 | \filldraw (C) circle(\radius); 166 | \filldraw (D) circle(\radius); 167 | \filldraw (E) circle(\radius); 168 | \filldraw (F) circle(\radius); 169 | \filldraw (G) circle(\radius); 170 | 171 | %% Following is for debugging purposes so you can see where the points are 172 | %% These are last so that they show up on top 173 | %\foreach \xy in {O, A, B, C, D, E, F, G}{ 174 | % \node at (\xy) {\xy}; 175 | %} 176 | \end{tikzpicture} 177 | \end{center} 178 | \caption{The cube $Q_3$.}\label{fig:cube} 179 | \end{marginfigure} 180 | \begin{theorem}[\cite{AlonFuredi1993}] 181 | Let $H_1,H_2,\dotsc,H_m$ be affine hyperplanes in $\R^n$ with cover exactly $2^n-1$ vertices of the $n$-dimensinal hypercube $Q_n$. Then $m\geq n$. 182 | \end{theorem} 183 | \lect{4}{11} 184 | \begin{proof} 185 | Let $H_i = \{x: \braket{a_i,x} = b_i\}$ for some $a_i = (a_{i_1},\dotsc,a_{i_n})\in \R^n$ and $b_i\in \R$. That is, if $x=(x_1,\dotsc,x_n)\in H_i$, 186 | \[ 187 | a_{i_1}x_1 + a_{i_2}x_2 + \dotsm + a_{i_n}x_n = b_i. 188 | \] 189 | Assume for contradiction that $m] (0,-.5) -- (0,.5); 296 | \end{tikzpicture} 297 | \hfill 298 | \begin{tikzpicture}[scale=2] 299 | \def\x{2/1.732} 300 | \coordinate (a) at (0,0); 301 | \coordinate (b) at (\x/2,1); 302 | \coordinate (c) at (\x,0); 303 | \draw[opacity=.5] (a)--(b) -- (c) -- cycle; 304 | \draw[<->] (\x/2,0) node[label=left:$b$]{} -- (\x/2,.95)node[label=left:$a$]{}; 305 | \end{tikzpicture} 306 | \hfill 307 | \caption{Left: we may simply rotate our needle of length about its middle, using a circle area of $\pi/4$. Right: we could rotate our needle by pulling end $b$ to the bottom left corner of the triangle, then end $a$ down the right side of the triangle. Then we could pull $b$ up the left side of the triangle, and finally $a$ to the middle, therby completing the rotation. This uses an equilateral triangle of altitude $1$, which then has area $\frac{1}{\sqrt{3}}$.} \label{fig:needle_circ_tri} 308 | \end{marginfigure} 309 | 310 | Let us try to restate this in terms of finite fields. First, we have the equivalent\sidenote{It turns out these are equivalent, by adding regions of area zero connecting the segments in each direction.} question: What is the minimum area of a region in the plane in which contains a segment of length 1 in every direction? 311 | 312 | Now, we have the finite field Kakeya problem: What is the minimum $|E|$ such that $E\subset \field^d$ and $E$ contains a \emph{line} in every direction. That is, for every $v\in \field^d\setminus\{0\}$ there exists $x_v$ such that $x_v + tv\in E$ for every $t\in \field$. 313 | In particular, does there exist $c_d>0$ (depending on $d$ but not $\field$) such that 314 | \[ 315 | |E| \geq c_d |\field|^d 316 | \] 317 | for all finite fields $\field$? 318 | % This is in a dual sense to Combinatorial Nullenstallsatz, which says roughly that if $f$ is a low-degree polynomial is cannot vanish everywhere on a set of a certain structure. Here, we will show that with a set of a different st 319 | \begin{lemma}\label{lem10p8:polynomial_vanishes_on_E} 320 | Let $E\subset \F^d$ such that $|E|< {k+d \choose k}$. Then there exists non-zero polynomial $f$ of $d$ variables such $f(x)=0$ for every $x\in E$, and $\deg f \leq k$. 321 | \end{lemma} 322 | \begin{proof} 323 | Let $V$ be the vector space of polynomials in $d$ variables over $\F$ of degree at most $k$. Then $\dim V$ is the number of monomials $x_1^{k_1}x_2^{k_2}\dotsm x_d^{k_d}$ such that $k_1+k_2+\dotsc+k_d \leq k$. This is the same as the number of decompositions $k_1 + k_2 + \dotsm + k_d + k_{d+1} = k$ such that $k_i\geq 0$, $k_i \in \Z$. This is ${k+d \choose k}$ as follows: 324 | if we write $k$ stars 325 | \[ 326 | \star \star \star \star \star \dotsm \star\] 327 | then our task is to put $d$ bars between the stars and thus partition the elements into $k_1,k_2,\dotsc,k_{d+1}$ 328 | \[ 329 | \operatorname*{\star}_{k_1}|\operatorname*{\star \star \star}_{k_2} | \star \dotsm \star 330 | \] 331 | sowe must choose $k$ stars from the $k+d$ total number of symbols. 332 | 333 | Let $E = \{x_1,\dotsc,x_m\}\subset\field^d$. For $p\in V$, the map 334 | \[ 335 | p\mapsto ( p(x^1),p(x^2),\dotsc,p(x^m)) 336 | \] 337 | is a linear map from $V$ to $|E|$-dimensional space. 338 | If $\dim(V) > |E|$, then there exists $p\in V$, $p\neq 0$ such that $p$ is mapped to zero, i.e. $p$ is zero everywhere on $E$. 339 | \end{proof} 340 | \lect{4}{13} 341 | % \marginnote{Lecture X: Wednesday, April 13, 2016} 342 | Now, given $\field$ a finite field, we will say $E\subset \field^d$ is a \defn{Kakeya set} if for every $v\in \field^d \setminus \{0\}$ there exists $x\in \field^d$ such that $x+tv \in E$ for every $t$. 343 | \begin{lemma} 344 | If $E\subset \field^d$ is Kakeya, then there exists no $p\in \field[x_1,\dotsc,x_d]$, $p\neq 0$, $\deg p\leq |F| - 1$ such that $p(x) = 0$ for every $x\in E$. 345 | \end{lemma} 346 | \begin{proof} 347 | Suppose such $p$ exists. Then 348 | \[ 349 | p(x) = p_k(x) + p_{k-1}(x) + \dotsm + p_0(x) 350 | \] 351 | where $p_i(x)$ is homogenous polynomial of degree $i$ and $p_k(x)$ not identically $0$ ($\deg p = k$). 352 | By definition, for every $v\in \field^d$ there exists $x$ such that $p(x+tv) =0$ for every $t\in \field$. Fix $x$ and $v$, $p(x+tv)$ is a polynomial in $t$ of degree $\leq |F| - 1$. If $x= ( x_1,\dotsc,x_d)$ and $v = ( v_1,\dotsc,v_d)$, then $x+tv = (x_1+tv_1,\dotsc,x_d+ tv_d)$. The coefficient of $t^k$ in $p$ is $p_k(v)$. 353 | 354 | On the other hand, $p$ has $|F|$ roots so $p$ is identically zero as a polynomial in $t$. In particular, $p_k(v) = 0$. But this is true for every $v$, contradicting the assumption that $p_k(v) $ is not identically zero. 355 | \end{proof} 356 | 357 | \begin{corollary}[\cite{dvir2009size}] 358 | If $E\subset \field^d$ is Kakeya, then 359 | \[ 360 | |E| \geq {|F| - 1 + d \choose d} = \frac{|\field|^d}{d!} + o(|\field|^d). 361 | \] 362 | \end{corollary} 363 | 364 | 365 | 366 | \subsection{Shannon capacity (\cite{shannonGraph})} 367 | We are transmitting messages in alphabet $V$ over a noisy channel. Some symbols can be confused with each other during transmission. 368 | 369 | Let $G$ be a graph with vertex set $V$ with a pair of symbols joined by an edge if they can be confounded. 370 | 371 | \begin{example} 372 | $V = \{a,b,d,p,q\}$. 373 | 374 | \begin{figure} 375 | \begin{center} 376 | \begin{tikzpicture}%[color=DarkBlue] 377 | \def\n{5} 378 | \def\radius{1} 379 | \def\rotation{-30} 380 | \def\circrad{1pt} 381 | \def\deg{360/\n} 382 | \pgfmathsetmacro\nminusone{\n-1} 383 | 384 | % \foreach \x in {1,...,\n} 385 | % { 386 | \def\x{1} 387 | \pgfmathsetmacro\myangle{\x*\deg+\rotation}; 388 | \filldraw (\myangle:\radius cm) circle (\circrad) node[label=$a$](a\x){}; 389 | \def\x{2} 390 | \pgfmathsetmacro\myangle{\x*\deg+\rotation}; 391 | \filldraw (\myangle:\radius cm) circle (\circrad) node[label=$d$](a\x){}; 392 | \def\x{3} 393 | \pgfmathsetmacro\myangle{\x*\deg+\rotation}; 394 | \filldraw (\myangle:\radius cm) circle (\circrad) node[label=left:$q$](a\x){}; 395 | \def\x{4} 396 | \pgfmathsetmacro\myangle{\x*\deg+\rotation}; 397 | \filldraw (\myangle:\radius cm) circle (\circrad) node[label=below:$p$](a\x){}; 398 | \def\x{5} 399 | \pgfmathsetmacro\myangle{\x*\deg+\rotation}; 400 | \filldraw (\myangle:\radius cm) circle (\circrad) node[label=right:$b$](a\x){}; 401 | 402 | 403 | \foreach \y in {1,...,\n} 404 | { 405 | 406 | \pgfmathsetmacro\z{\y+1} 407 | % { 408 | % \ifthenelse{\y < \n+1} 409 | % { 410 | \pgfmathsetmacro\myangley{\y*\deg+\rotation} 411 | \pgfmathsetmacro\myanglez{\z*\deg+\rotation} 412 | \draw (\myangley:\radius cm) -- (\myanglez:\radius cm); 413 | % }{} 414 | 415 | } 416 | 417 | \end{tikzpicture} 418 | \end{center} 419 | \caption{A graph representing transmitting the alphabet $V = \{a,b,d,p,q\}$. Adjacency indicates symbols which could be confused for each other. This graph is $C_5$, the cycle of length five.} 420 | \end{figure} 421 | % Write $a,d,q,p,b$ clockwise, connect each subsequent one. 422 | We'd like to send a set $S_n$ of $n$ letter messages such that no two messages in $S_n$ can be confused with each other. One solution: we can send $2^n$ messages which can't be confused just by using sympols $a$ and $p$. 423 | \end{example} 424 | 425 | 426 | A \emph{strong product} of graphs $G$ and $H$ denoted $G\boxtimes H$ is a graph with vertex set $V(G)\times V(H)$ and $(u_1,v_1)\sim (u_2,v_2)$ if 427 | \begin{itemize} 428 | \item[\phantom{and}] $u_1=u_2$ or $u_1$ is adjacent to $u_2$ 429 | \item[and] $v_1 = v_2$ or $v_1$ is adjacent to $v_2$. 430 | \end{itemize} 431 | Let $\boxtimes^n G$ denote $\underbrace{G\boxtimes G\boxtimes \dotsm \boxtimes G}_{n \text{ times}}$. 432 | 433 | We are interested in the maximum size of an independent set in $\boxtimes^n G$, which we will call $\alpha(\boxtimes^n G)$. Let $\Theta(G) = \lim_{n\to\infty} \sqrt[n]{\alpha(\boxtimes^n G)}$ be the \emph{Shannon capacity}. This exists because $\alpha(G\boxtimes H) \geq \alpha(G) \alpha(H)$, as the product of two independent sets produces an independent set in the product graph. 434 | 435 | \begin{lemma} \label{lem:chrom_num_shannon_capacity} 436 | For any graph, $\alpha(G) \leq \Theta(G) \leq \chi(\bar G)$.\marginnote{$\bar G$ denotes the complement of $G$ in $V(G)^{(2)}$, i.e. two vertices are adjacent in $\bar G$ iff they are not adjacent in $G$.} 437 | \end{lemma} 438 | For brevity, let us write $G^n:=\boxtimes^n G$. 439 | \begin{proof} 440 | By the product inequality for $\alpha$, we have $\alpha(G^n) \geq (\alpha(G))^n$, hence $\alpha(G) \leq \Theta(G)$. 441 | 442 | Now, note $\alpha(G)\leq \chi(\bar G)$, since $\alpha(G)$ is the size of the largest complete subgraph of $\bar G$. Further, $\chi(\overline{G\boxtimes H}) \leq \chi(\bar G)\chi(\bar H)$. 443 | 444 | We may think of coloring $\bar G$ as coloring $G$ such that any two vertices which are the same color are connected. To see $\chi(\overline{G\boxtimes H}) \leq \chi(\bar G)\chi(\bar H)$, color each vertex $(u,v)$ of $\overline{G\boxtimes H}$ by the pair of color $(c_1(u),c_2(v))$ where $c_1$ is a coloring of $\bar G$ and $c_2$ as a coloring of $\bar H$. 445 | 446 | Now every color class is used on $K_l\boxtimes K_s$ for some $l$ and $s$, but the strong product of complete graphs is complete. Therefore 447 | \[ 448 | \alpha(G^n) \leq \chi(\bar G^n) \leq (\chi(\bar G))^n 449 | \] 450 | and we have the result as derised. 451 | \end{proof} 452 | 453 | In fact, $\alpha(G) = \chi(\bar G)$ for many graphs, including perfect graphs (hence all graphs on $4$ vertices). The smallest graph in which $\alpha(G) \neq \chi(\bar G)$ is the cycle of length five, $C_5$, considered earlier. There, we saw we could achieve $2$, and $\chi(\bar C_5) = 3$, so $2\leq \Theta(C_5)\leq 3$. 454 | \begin{marginfigure} 455 | \begin{center} 456 | \begin{tikzpicture}%[color=DarkBlue] 457 | \def\n{5} 458 | \def\radius{1} 459 | \def\rotation{-30} 460 | \def\circrad{1pt} 461 | \def\deg{360/\n} 462 | \pgfmathsetmacro\nminusone{\n-1} 463 | 464 | % \foreach \x in {1,...,\n} 465 | % { 466 | \def\x{1} 467 | \pgfmathsetmacro\myangle{\x*\deg+\rotation}; 468 | \filldraw (\myangle:\radius cm) circle (\circrad) node[label=$1$](a\x){}; 469 | \def\x{2} 470 | \pgfmathsetmacro\myangle{\x*\deg+\rotation}; 471 | \filldraw (\myangle:\radius cm) circle (\circrad) node[label=$2$](a\x){}; 472 | \def\x{3} 473 | \pgfmathsetmacro\myangle{\x*\deg+\rotation}; 474 | \filldraw (\myangle:\radius cm) circle (\circrad) node[label=left:$3$](a\x){}; 475 | \def\x{4} 476 | \pgfmathsetmacro\myangle{\x*\deg+\rotation}; 477 | \filldraw (\myangle:\radius cm) circle (\circrad) node[label=below:$4$](a\x){}; 478 | \def\x{5} 479 | \pgfmathsetmacro\myangle{\x*\deg+\rotation}; 480 | \filldraw (\myangle:\radius cm) circle (\circrad) node[label=right:$5$](a\x){}; 481 | 482 | 483 | \foreach \y in {1,...,\n} 484 | { 485 | 486 | \pgfmathsetmacro\z{\y+1} 487 | % { 488 | % \ifthenelse{\y < \n+1} 489 | % { 490 | \pgfmathsetmacro\myangley{\y*\deg+\rotation} 491 | \pgfmathsetmacro\myanglez{\z*\deg+\rotation} 492 | \draw (\myangley:\radius cm) -- (\myanglez:\radius cm); 493 | % }{} 494 | 495 | } 496 | 497 | \end{tikzpicture} 498 | \end{center} 499 | \caption{$C_5$, the cycle of length 5.} \label{fig:C5} 500 | \end{marginfigure} 501 | 502 | We also have $\alpha(C_5^2)\geq 5$, that is, you can send 5 different 2 symbol messages. These messages are: $\{(1,1),(2,3)(3,5),(4,2),(5,4)\}$, using the labelling of \cref{fig:C5}. 503 | In particular, $2<\sqrt{5} \leq \Theta(C_5)$, so our naive bound of $2$ is not optimal. 504 | 505 | To prove an upper bound on $\Theta$, we'd like a function $ 506 | \theta(G)$ such that 507 | \begin{enumerate} 508 | \item $\alpha(G) \leq \theta(G)$ \label{enum:theta1} 509 | \item $\theta(G\boxtimes H)\leq \theta(G) \theta(H)$. \label{enum:theat2} 510 | \end{enumerate} 511 | Then, as in \cref{lem:chrom_num_shannon_capacity}, any such function will upper bound $\Theta$, i.e., $\Theta(G) \leq \theta(G)$. Ideally, $\theta(C_5)= \sqrt{5}$, to prove the lower bound of $\sqrt{5}$ is optimal. This would confirm that we've found a sharper upper bound than $\chi (\bar G)$. 512 | 513 | One function satisfying all of these is the \defn{Lov\'asz theta}\sidenote{\cite{Lovasztheta}}, defined as follows. 514 | Let $G$ be a graph with $V(G) = [n]$. A collection $v_1,v_2,\dotsc,v_n\in \R^d$ is an \defn{orthonormal representation} of $G$ if 515 | \begin{itemize} 516 | \item $\|v_i\| = 1$, 517 | \item $v_i \perp v_j$ if $i$ and $j$ are not adjacent. 518 | \end{itemize} 519 | \marginnote{Every graph has an orthonormal representation in dimension $n$.} 520 | Let the \defn{value}[orthonormal representation!value] of an orthonormal representation be 521 | \[ 522 | \min_{\|c\|=1} \max_{v_i} \frac{1}{\braket{v_i,c}^2} 523 | \] 524 | where $\braket{\cdot,\cdot}$ is the Euclidean inner product (dot product). 525 | The vector $c$ acheiving the minimum is called a \emph{handle}. 526 | Finally, define $\theta(G)$ as the minimal value over orthonormal representations. 527 | 528 | That $\theta(G)$ satisfies points \labelcref{enum:theta1,enum:theat2} is the content of \cref{lem:alpha_less_than_theta,lem:shannon_cap_theta_submult}. That $\theta(C_5) \leq \sqrt{5}$ is the content of \cref{lem:theta_C5}. This will conclude our discussion of Shannon capacity. 529 | 530 | \begin{lemma} \label{lem:alpha_less_than_theta} 531 | $\alpha(G)\leq \theta(G)$ 532 | \end{lemma} 533 | \begin{proof} 534 | Let $v_1,\dotsc,v_n$ be the optimal orthonormal representation of $G$ with handle $c$. Let $\alpha(G) = k$, and suppose $\{1,2,\dotsc,k\}$ form an independent set. Then $v_1,\dotsc,v_k$ can be completed to an orthonormal basis, so 535 | \[ 536 | 1=\|c\|^2 = \braket{c,c} \geq \sum_{i=1}^k (\braket{c,v_i})^2 \geq \frac{k}{\theta(G)} 537 | \] 538 | since 539 | % \[ 540 | $\theta(G) \geq \frac{1}{\braket{c,v_i}^2}$. 541 | % \] 542 | \end{proof} 543 | 544 | 545 | \begin{lemma}\label{lem:shannon_cap_theta_submult} 546 | \[ 547 | \theta(G\boxtimes H)\leq \theta(G)\theta(H). 548 | \] 549 | \end{lemma} 550 | \begin{proof} 551 | Observation: if $u,v\in U$ and $x,y \in V$, then 552 | \[ 553 | \braket{u\otimes x, v \otimes y} = \braket{u,v}\braket{x,y} 554 | \] 555 | \marginnote{We can see this by working in components: if $u = (u_i)$, $v = (v_i)$, $x = (x_i)$, $y = (y_i)$, then $\braket{u\otimes x, v\otimes y} = \sum_{i,j} u_i x_j v_i y_j$ which we may regroup into the product of the two inner products.} 556 | 557 | % \begin{remark} 558 | If $u_1,\dotsc,u_n$ is an o.r. of $G$ with handle $c$, and $v_1,\dotsc,v_n$ is an o.r. of $H$ with handle $d$, then $(u_i\otimes v_j)$ is an o.r. of $G\boxtimes H$, by the above identity. 559 | 560 | Let $c\otimes d$ be a handle. Then 561 | \begin{align*} 562 | \theta(G\otimes H) &\leq \max_{u_i,v_j}\frac{1}{ (\braket{c\otimes d, u_i\otimes v_j})^2}\\ &=\max_{u_i,v_j} \frac{1}{(\braket{c,u_i})^2 (\braket{d,v_j})^2}\leq \theta(G)\theta(H). 563 | \end{align*} 564 | % \end{remark} 565 | \end{proof} 566 | 567 | \begin{lemma} \label{lem:theta_C5} 568 | $\theta(C_5)\leq\sqrt{5}$. 569 | \end{lemma} 570 | \begin{proof} 571 | Start with 572 | \begin{align*} 573 | v_{i+1} = \frac{(\cos( \tfrac{2\pi}{5}i),(\sin( \tfrac{2\pi}{5}i), p)}{\sqrt{1+p^2}} 574 | \end{align*} 575 | and increase $p$ from zero until each $v_i \perp v_{i+2},v_{i+3}$. 576 | \begin{figure*}[ht] 577 | % \includegraphics[width=.25\linewidth]{graphics/C5p=0.pdf}\includegraphics[width=.25\linewidth]{graphics/C5p=p3.pdf}\includegraphics[width=.25\linewidth]{graphics/C5p=p6.pdf}\includegraphics[width=.25\linewidth]{graphics/C5p=p0.pdf} 578 | \includegraphics[width=.33\linewidth]{graphics/C5p=0.pdf}\includegraphics[width=.33\linewidth]{graphics/C5p=p45.pdf}\includegraphics[width=.33\linewidth]{graphics/C5p=p0.pdf} 579 | \caption{From left to right, the vectors $v_i$ with $p=0,0.45,\frac{1+\sqrt{5}}{4}\approx 0.89$. The intuition is that we are starting with our five vectors equally spaced on the unit circle. Then we ``close the umbrella'', increasing $p$, and pointing the vectors towards the $z$-axis, until each vector is orthogonal to the two vectors across from it.} 580 | \end{figure*} 581 | 582 | We have $(1,0,p) \perp (\cos \tfrac{4\pi}{5}, \sin \tfrac{4\pi}{5},p)$ if $-p^2 = \cos \tfrac{4\pi}{5}$. Then $p^2 = \frac{1+\sqrt{5}}{4}$. Now, we may check that at this value of $p$, each $v_i \perp v_{i+2},v_{i+3}$. That is, the $v_i$ form an orthonormal representation of $C_5$. Take $c = (0,0,1)$. Then $\braket{c,v_i} = \frac{p}{\sqrt{1+p^2}}$ for all $i$, and $\frac{1}{\big( \frac{p}{\sqrt{1+p^2}} \big)^2} = \dotsm = \sqrt{5}$. 583 | \end{proof} -------------------------------------------------------------------------------- /chapters/ch1_sperner.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = ../CombinatoricsNotes.tex 2 | 3 | \section{Representing Sets and Sperner Systems.} 4 | \lect{1}{11} 5 | % \marginnote{Lecture 2: Monday, January 11, 2016. } 6 | \marginnote{We begin our systematic investigation into extremal combinatorics.} 7 | Let's start with some definitions. 8 | \begin{itemize}[] 9 | \item[Power set:] 10 | Let $X$ be a finite set with cardinality $|X| = n$. The \defn{power set} is $\P(X)$, the collection of subsets of $X$. $|\P(X)|= 2^n$. 11 | \item[$r$-element subsets:] 12 | The set $X^{(r)}$ is the collection of all $r$-element subsets of $X$. The cardinality $|X^{(r)}| = {n\choose r}$. 13 | \item[Set system:] 14 | A \defn{set system} $\F \subset \P(X)$ on $X$ is a collection of subsets of $X$. For example, $\F = \{\emptyset, \{1\}, \{1,2\}, \{2,3\},\{3\}\}$ is a set system on $\{1,2,3\}$. 15 | \item[$r$-graph:] 16 | If $\F\subset X^{(r)}$, then we call $\F$ an $r$-graph or \defn{hypergraph}. 2-graphs are just ordinary graphs: $\F\subset X^{(2)}$ can be thought of as a graph with vertex set $X$, edge set $\F$. 17 | \end{itemize} 18 | We'll also frequently use the notation $[n] = \{1,2,\dotsc,n\}$. 19 | 20 | \newthought{Now, given a set system} $\{A_1,A_2,\dotsc,A_m\}$, we want to ``reduce'' the sets such that distinct sets remain distinct. 21 | That is, we want to find $S\subset X$ as small as possible such that $\{A_1\cap S, A_2\cap S, \dotsc, A_m\cap S\}$ are all distinct. \marginnote{Note that $S$ acts by deleting elements from our base set $X$.} 22 | Let's start with two sets, $\{A_1,A_2\}$. We want $S$ as small as possible such that $A_1\cap S\neq A_2\cap S$. In this case, we can simply choose $S$ to be a singleton of an element which is in one set but not the other\sidenote{which always exists because $A_1\neq A_2$.}. 23 | If we use our example earlier, $\{\emptyset, \{1\}, \{1,2\}, \{2,3\},\{3\}\}$ on $[3]$, we cannot remove any; $|\P([2])| = 4$ and we have five elements. 24 | If our set system is $\{\emptyset, \{1,2\}, \{2,3\},\{3\}\}$ on $[3]$, then we may remove $1$; that is, $S = \{2,3\}$. 25 | This motivates a question: How small can $S$ be as a function of $m$? 26 | \begin{theorem} 27 | Let $\F = \{A_1,A_2,\dotsc,A_m\}$ be a set system. Then there exists $S$, $|S|\leq m-1$ such that $A_1\cap S$, $A_2\cap S$, \ldots, $A_m\cap S$ are all distinct. 28 | \end{theorem} 29 | \begin{remark} 30 | The bound is tight: the set system $\{\emptyset, \{1\}, \{2\}, \dotsc, \{m-1\}\}$ has the property that if we remove any element, we collapse two sets to the empty set. 31 | \end{remark} 32 | \begin{proof} 33 | Choose $S$ as small as possible such that $A_1\cap S$, $A_2\cap S$, \ldots, $A_m\cap S$ are all distinct, and assume that $|S| \geq m$. 34 | Let $A_i' = A_i\cap S$. By the minimality of $S$ for every $x\in S$ there exists $i,j\in[m]$ such that $A_i'\setminus \{x\} = A_j'\setminus \{x\}$, with $i\neq j$. 35 | 36 | Now, construct a graph on the vertex set $[m]$ as follows. For each $x\in S$, choose one pair $i$ and $j$ ($i\neq j$) such that $A_i'\setminus \{x\} = A_j'\setminus \{x\}$, and join $i$ and $j$ by an edge\sidenote[][-2cm]{$A_i'\setminus \{x\} = A_j'\setminus \{x\}$ is equivalent to $A_i \symd A_j =\{x\}$, where the symmetric difference $X\symd Y = (X\cup Y)\setminus (X\cap Y)$. 37 | Because of this, we will make a new edge each time: if $x,y\in S$ yielded the same edge, then $\{x\}=A_i \symd A_j = \{y\} $.}. 38 | This graph has $m$ verticies and $|S|\geq m$ edges, so it contains a cycle\sidenote{Easy to see by picture; draw $m-1$ edges on $m$ vertices, and then if you don't have a cycle yet, you have a line, and no matter how you place the last edge, you get a cycle.}. Without loss of generality, assume there is a cycle on verticies $1,2,\dotsc,k$ in order. Then there exists distinct $x_1,x_2,\dotsc,x_k\in S$ such that $A_1\symd A_2 = \{x_1\}$, $A_2\symd A_3 = \{x_2\}$, \ldots, $A_{k-1}\symd A_k = \{x_{k-1}\}$, and $A_k \symd A_1 = \{x_k\}$. 39 | 40 | We can take the symmetric difference of all of them: 41 | \[ 42 | \emptyset = (A_1\symd A_2) \symd (A_2 \symd A_3) \symd \dotsm \symd (A_k \symd A_1) = \{x_1,x_2,\dotsc, x_k\} 43 | \] 44 | \marginnote{Note the symmetric difference is commutative and associative.} 45 | On the left, we have two of each set, so we can regroup and commute to obtain the empty set, using $A\symd A = \emptyset$. On the right, we have the symmetric difference of distinct singletons, which is just the union. This is a contradiction, so our minimal $S$ must have $|S|\leq m-1$. 46 | \end{proof} 47 | Before we continue finding ways to represent sets, we'll need some graph theoretic tools. First, some definitions. 48 | % \begin{definition} 49 | \begin{itemize}[] 50 | \item[Bipartite:] 51 | A graph $G$ is \defn{bipartite}[graph!bipartite] with bipartition $(V_1,V_2)$ if every edge of $G$ contains one vertex of $V_1$ and one vertex of $V_2$. 52 | 53 | \item[Matching: ]A collection of edges $M$ of $G$ is a \defn{matching}[graph!matching] of $V_1$ into $V_2$ if for every $v\in V_1$, $M$ contains exactly one edge containing $v$, and for $v\in V_2$, at most one edge. This is illustrated in \cref{fig:bipartite_matching}. 54 | 55 | \begin{marginfigure} 56 | \begin{center} 57 | \begin{tikzpicture}[color=black] 58 | 59 | \node (rect) at (0,0)[draw,thick,minimum width=3cm, minimum height = 1cm, rounded corners=3pt,label=left:$V_2$]{}; 60 | 61 | \node (rect2) at (0,1.5)[draw,thick,minimum width=3cm, minimum height = 1cm, rounded corners=3pt,label=left:$V_1$]{}; 62 | 63 | \pgfmathsetseed{2} 64 | % \def\z{rand} 65 | \foreach \x in {1,...,6} 66 | { 67 | \filldraw (rand*1.4,rand*.4) circle (0.4pt) node(a\x){}; 68 | \filldraw (rand*1.4,1.5+rand*.4) circle (0.4pt) node(b\x){}; 69 | } 70 | 71 | %some extra nodes in the bottom one 72 | \foreach \x in {1,...,6} 73 | { 74 | \filldraw (rand*1.4,rand*.4) circle (0.4pt); 75 | } 76 | 77 | \foreach \x in {1,...,6} 78 | { 79 | % \ifthenelse{\x > 1}{\draw[dashed] (a\x) -- (u)}{}; 80 | % \foreach \y in {1,...,\x} 81 | % { 82 | \draw (a\x) -- (b\x); 83 | % } 84 | } 85 | 86 | \end{tikzpicture} 87 | \end{center} 88 | \caption{Think of elements of $V_1$ as job applicants, and $V_2$ as positions. Then a matching of $V_1$ into $V_2$ is an arrangement so that every job applicant has a position, but some positions could be unfilled. \label{fig:bipartite_matching}} 89 | \end{marginfigure} 90 | 91 | 92 | \item[Neighborhood:] For $S\subset V_1$, define the \defn{neighborhood}[graph!neighborhood] $N(S)$ as the set of verticies adjacent to at least one vertex in $S$. 93 | \end{itemize} 94 | The following result\sidenote{\cite{Hallmarriage}} connects these ideas. 95 | \begin{theorem}[Hall's marriage theorem] 96 | Let $G$ be a bipartite graph with bipartition $(V_1,V_2)$. Then $G$ contains a matching of $V_1$ into $V_2$ if and only if 97 | \begin{equation} \label{eq:Hall_condition} 98 | |N(S)| \geq |S| \text{ for every } S\subset V_1. \tag{Hall's condition} 99 | \end{equation} 100 | \end{theorem} 101 | \begin{proof} 102 | The condition is necessary because you need to have enough verticies available in $N(S)$ for elements of $S$ to match into. We will prove sufficiency by induction on $|V_1|$. The base case is immediate. For the induction step, we will split into two cases. 103 | \begin{enumerate}[{Case }1:] 104 | \item For every $S\subset V_1$ with $S\neq \emptyset$ and $S\neq V_1$, we have that $|N(S)| > |S|$. 105 | % \end{enumerate} 106 | In this case, choose $v\in V_1$; then $v$ has a $w\in V_2$ adjacent to it, because $|N(\{v\})|> |\{v\}|=1$. Apply the induction hypothesis to $G \setminus \{v,w\}$. 107 | 108 | We then just need to check Hall's condition on $G' =G \setminus \{v,w\}$. For every $S\subset V_1 \setminus\{v\}$, we have 109 | \[ 110 | |N'(S)| \geq |N(S)| -1 \geq |S|, 111 | \] 112 | where $N'$ is the neighborhood with respect to $G'$. The first inequality holds because we removed at most one neighbor by removing $w$. The second inequality holds from our assumption in this case. Then the induction hypothesis yields a matching on $V_1\setminus \{v\}$ into $V_2\setminus \{w\}$, which we can extend to a matching on $V_1$ into $V_2$ by matching $v$ to $w$. 113 | % \begin{enumerate}[{Case }1:]\setcounter{enumi}{1} 114 | \item There exists $S\subset V_1$ with $S\neq \emptyset$ and $S\neq V_1$, such that $|N(S)| = |S|$. 115 | % \end{enumerate} 116 | By induction hypothesis, there exists a matching $M_1$ of $S$ into $N(S)$. 117 | 118 | It remains to find a matching from $V_1\setminus S$ into $V_2 \setminus N(S)$. By induction hypothesis, it is enough to show that for every $T\subset V_1\setminus S$, we have 119 | \[ 120 | |N(T)\cap (V_2 \setminus N(S))| \geq |T|. 121 | \] 122 | \marginnote{Here's the trick.}Since $S\cup T\subset V_1$, by assumption, we have Hall's condition 123 | \[ 124 | |N(S\cup T)|\geq |S\cup T| = |S| + |T|. 125 | \] 126 | 127 | We know $N(S\cup T) = N(S) \cup N(T) = N(S) \cup (N(T)\setminus N(S))$. So $|N(S\cup T)| = |N(S)| + |N(T)\setminus N(S)| $. So Hall's condition becomes 128 | \[ 129 | |N(T)\setminus N(S)| \geq |T| 130 | \] 131 | as desired.\qedhere 132 | \end{enumerate} 133 | \end{proof} 134 | 135 | 136 | Let's employ Hall's theorem to represent sets. Let 137 | \[ 138 | \F = \{A_1,A_2,\dotsc,A_m\} 139 | \] be a set system. 140 | % \begin{definition} 141 | A \defn{system of distinct representatives} for $\F$ is a collection $\{x_1,\dotsc,x_m\}$ of elements such that $x_1,\dotsc, x_m$ are pairwise distinct, and $x_i \in A_i$ for $i\in[m]$. 142 | % \end{definition} 143 | Given $\F$, one can consider the bipartite graph $G$ with bipartition $(V_1,V_2)$ such that $V_1 = \F$ and $V_2= \bigcup_{i\in[m]} A_i$. We join $A_i$ to $x$ iff $x\in A_i$. 144 | Then a system of distinct representatives for $\F$ is exactly a matching on $G$ from $V_1$ into $V_2$. Hall's theorem then immediately implies the following result. 145 | \begin{corollary} 146 | A set system $\F = \{A_1,\dotsc,A_m\}$ has a system of distinct representatives if and only if for every $\F'\subset \F$, 147 | \[ 148 | |\F'| \leq \left|\bigcup_{A\in \F'}A\right|. 149 | \] 150 | \end{corollary} 151 | \newthought{Given a set $X$}, there is a natural bipartite graph and matching which will prove useful. 152 | \begin{corollary} \label{cor:Xr_matching_exists} 153 | Let $X$ be a set with $|X|=n$. Let $G$ be a bipartite graph with bipartition $(X^{(r)}, X^{(r-1)})$ such that $A\in X^{(r)}$ is adjacent to $B \in X^{(r-1)}$ if $B\subset A$. Then if $r> n/2$, the graph $G$ has a matching of $X^{(r)}$ into $X^{(r-1)}$. 154 | \begin{marginfigure} 155 | \begin{center} 156 | \begin{tikzcd}[column sep=tiny] 157 | X^{(3)} & &\{1,2,3\} \\ 158 | X^{(2)} &\{1,2\} \arrow[dash,green]{d} \arrow[dash]{rd}& \{1,3\} \arrow[dash]{ld}\arrow[dash]{d} \arrow[dash,green]{rd} & \arrow[dash,green]{ld} \arrow[dash]{d}\{2,3\} \\ 159 | X^{(1)} & \{1\} & \{2\} &\{3\} \\ 160 | X^{(0)} & & \emptyset 161 | \end{tikzcd} 162 | \end{center} 163 | \caption{Example of the graph relation on $G= (X^{(2)}, X^{(3)})$, with a matching highlighted in green.} 164 | \end{marginfigure} 165 | \end{corollary} 166 | \begin{remark} 167 | This corollory implicitly shows ${n \choose r} \leq {n \choose r-1}$ if $r> n/2$. 168 | \end{remark} 169 | \begin{proof} 170 | It suffices to check \ref{eq:Hall_condition}. For every $\A\subset X^{(r)}$, we want $|N(\A)| \geq |\A|$. Label 171 | \[ 172 | \B:= N(\A) = \{B\in X^{(r-1)}: B\subset A, \text{ for some }A\in \A\}. 173 | \] 174 | Every element of $X^{(r)}$ is incident to $r$ edges of $G$\sidenote{Each edge corresponds to taking an element away from the set.}. So we have $|\A|r$ edges leaving $\A$, ending in $\B$. 175 | Every element of $X^{(r-1)}$ is incident to $n-r+1$ edges of $G$, which can be seen by the fact that there are $n-(r-1)$ possible elements to add to a set $B\in X^{(r-1)}$ to obtain a superset in $X^{(r)}$. So we have at most $|\B|(n-r+1)$ edges leaving $\B$, ending in $\A$\sidenote{Since not every edge leaving $\B$ needs to reach something in $\A$ (it could reach something in $X^{(r)} \setminus \A$), it is only ``at most.''}. 176 | So $|\A|r \leq |\B|(n-r+1)$. But by assumption $r \geq (n-r+1)$, so $|\B| \geq |\A|$ as desired. 177 | \end{proof} 178 | 179 | \lect{1}{13} 180 | % \marginnote{Lecture 3: Wednesday, January 13, 2016.} 181 | 182 | Recall that $\F\subset \P(X)$ is a \emph{Sperner system} if for all $A,B\in \F$, if $A\leq B$, then $A=B$. 183 | We wish to find $\max |\F|$ such that $\F$ is a Sperner system, as a function $|X| = n$. Note that $X^{(r)}$ is always Sperner, and $|X^{(r)}| = {|X| \choose r}$, which is maximized when $r = \floor{n/2}$. 184 | \begin{theorem}[\cite{sperner1928}] \label{thm:sperner} 185 | If $\F\subset \P(X)$ is Sperner, then $|\F| \leq {n \choose \floor{n/2}}$, where $n=|X|$. 186 | \end{theorem} 187 | \begin{proof} 188 | \marginnote{A Sperner system is a system of sets such that no two are comparable. A dual notion is a system of sets such that all are comparable. } 189 | An ordered collection $(A_1,A_2,\dotsc, A_k)$ of sets in $\P(X)$ is a \defn{chain} if $A_1\subsetneqq A_2 \subsetneqq A_3 \subsetneqq\dotsm \subsetneqq A_k$. 190 | \begin{marginfigure} 191 | \begin{center} 192 | \begin{tikzcd}[column sep=tiny] 193 | X^{(3)} & &\{1,2,3\} \arrow[dash]{d}\arrow[dash]{rd}\arrow[dash,green]{ld} \\ 194 | X^{(2)} &\{1,2\} \arrow[dash,green]{d} \arrow[dash]{rd}& \{1,3\} \arrow[dash]{ld}\arrow[dash]{d} \arrow[dash,blue]{rd} & \arrow[dash,red]{ld} \arrow[dash]{d}\{2,3\} \\ 195 | \mathbf{X^{(1)}} & \mathbf{\{1\}} \arrow[dash,green]{rd} & \mathbf{\{2\}} \arrow[dash]{d}&\mathbf{\{3\}}\arrow[dash]{ld} \\ 196 | X^{(0)} & & \emptyset 197 | \end{tikzcd} 198 | \end{center} 199 | \caption{Consider $X^{(1)}$, a natural maximal Sperner system. Note that each element of $X^{(1)}$ can form a distinct chain, such as $\emptyset \subsetneqq \{1\} \subsetneqq \{1,2\} \subsetneqq \{1,2,3\}$.}\label{fig:Sperner_chains} 200 | \end{marginfigure} 201 | It is enough to show that $\P(X)$ can be partitioned into ${n \choose \floor{n/2}}$ chains. Indeed, every Sperner system can contain $\leq 1$ element from each chain in the partition; see \cref{fig:Sperner_chains} for an example. 202 | Better yet, we partition $\P(X)$ into chains such that every chain contains an element of $X^{(\floor{n/2})}$. 203 | 204 | Let's begin by partitioning all subsets of $X$ of size $\geq \floor{n/2}$. 205 | We will first do this inductively starting from $X^{(\floor{n/2})}$, and extending the partition to $X^{(\floor{n/2})}\cup X^{(\floor{n/2}+1)}\cup\dotsm \cup X^{(k)}$ to $X^{(k+1)}$ using the matching obtained in \cref{cor:Xr_matching_exists} from $X^{(k)}$ to $X^{(k+1)}$ by adding each element of $X^{(k+1)}$ to the chain of the set it's matched to. Note that are chains are not maximal; some (all but one) truncate before they reach the top, $X^{(n)}$. 206 | Then we can extend the partition to sets of size $< \floor{n/2}$ by symmetry. 207 | \end{proof} 208 | \begin{remark} 209 | This proof is instructive and provides the useful technique of partitioning into chains. But we can prove stronger results with slicker proofs. 210 | \end{remark} 211 | Suppose $k< n/2$ and we want to find the maximum size Sperner system such that every set in the system has size $\leq k$. 212 | As one may guess, the maximum size will be $|X^{(k)}| = {n\choose k}$. To show this, we'll use the following result. 213 | \begin{theorem}[Lubell, Meshalkin, Yamamoto, Boll\'obas, and possibly others, A.K.A. the LYM inequality] \marginnote{\cite{LUBELL_LYM,Meshalkin_LYM,yamamoto1954_LYM,Bollab_LYM}} 214 | \label{thm:LYM_inequality} 215 | Let $\F\subset \P(X)$, $|X|=n$ be a Sperner system. 216 | Let $\F_k = \F\cap X^{(k)}$ be the set of $k$ element sets in $\F$, and let $f_k = |\F_k|$. Then 217 | \begin{equation} \tag{LYM} 218 | \label{eq:LYM} 219 | \sum_{k=0}^n \frac{f_k}{{n\choose k}} \leq 1. 220 | \end{equation} 221 | \end{theorem} 222 | \begin{remark} 223 | We have 224 | \[ 225 | 1 \geq \sum_{k=0}^n \frac{f_k}{{n\choose k}} \geq \sum_{k=0}^n \frac{f_k}{{n\choose \floor{n/2}}}, 226 | \] 227 | thus 228 | \[ 229 | {n\choose \floor{n/2}} \geq \sum_{k=0}^n f_k = |\F| 230 | \] 231 | which is Sperner's theorem. 232 | \end{remark} 233 | \begin{proof} Let's assume $X=[n]$ for convenience. 234 | Consider all maximum chains in $\P(X)$ and count how many chains an element of $\F$ belongs to. Each of the maximal chains is of the form $\emptyset = A_0$, $A_1$, \ldots, $A_n = X = [n]$, and $|A_i| = i$. So each maximal chain corresponds to an ordering $a_1,a_2,\dotsc,a_n$ of $[n]$, where $\{a_i\} = A_i \setminus A_{i-1}$, is the element you add to $A_i$ to get the next set in the chain. 235 | 236 | Thus, there are $n!$ maximal chains (the number of re-orderings of $[n]$). Consider 237 | $F \in X^{(k)}$. 238 | How many maximal chains is $F$ in? If $k=0,n$, $F$ is in every chain, so $n!$. If $k=1$, then $F$ is in $(n-1)!$ chains. If $k=2$, then $2!(n-2)!$. In general, $F$ is in $k!(n-k)!$ maximal chains\sidenote{We choose $k$ to get to the set, then $(n-k)$ to finish the chain.} Each maximal chain contains $\leq 1$ element of $\F$. The total number of elements of $\F$ in all maximum chains is 239 | \[ 240 | \sum_{k=0}^n f_k k!(n-k)!\leq n! 241 | \] 242 | Dividing by $n!$, we obtain the LYM inequality. 243 | \end{proof} 244 | \begin{remark} 245 | Let's consider an alternate proof. Let $C$ be a uniformly randomly chosen maximal chain, and consider the expectation value of the number of elements of $C\cap \F$. Of course, there is at most 1 element, since $\F$ is a Sperner system. On the other hand, 246 | \begin{align*} 247 | \E ( | C\cap \F|) &=\sum_{k=0}^n \E( | C \cap \F_k|) = \sum_{k=0}^n f_k \cdot (\text{probability that a set of size $k$ is in $C$})\\ 248 | &= \sum_{k=0}^n f_k \cdot \frac{1}{(\text{number of sets of size $k$)}} = \sum_{k=0}^n \frac{f_k}{{n \choose k}} \leq 1. 249 | \end{align*} 250 | \end{remark} 251 | 252 | 253 | When does equality hold in LYM? Certainly when $\F= X^{(r)}$ for any $r$. We'd like to show this condition is necessary as well, but to do so, we'll first prove a more refined inequality in which equality is easier to check. Then we'll use this to show sufficiency for equality in the LYM inequality. 254 | 255 | \begin{theorem}[Local LYM inequality] \label{thm:local_LYM} 256 | Let $\A\subset X^{(r)}$, and $|X|=n$. Define 257 | $\partial \A \subset X^{(r-1)}$ 258 | the \defn{shadow} of $A$ by 259 | \[ 260 | \partial \A := \{B\in X^{(r-1)}: B\supseteq A \text{ for some }A\in \A \}. 261 | \] 262 | Then 263 | \begin{equation} \label{eq:local_LYM} \tag{Local LYM} 264 | \frac{|\partial \A|}{{n\choose r-1}} \geq \frac{|\A|}{{n \choose r}} 265 | \end{equation} 266 | or equivalently, 267 | \[ 268 | r| \A| \leq |\partial \A| (n-r+1). 269 | \] 270 | Moreover, equality holds if and only if $\A= \emptyset$ or $\A= X^{(r)}$. 271 | \end{theorem} 272 | \begin{proof} 273 | \[ 274 | r|\A| = | \{ (B,A): B\in \partial A, \, A\in \A, \, B\subset A\}| \leq |\partial A| (n-r+1) 275 | \] as seen in \cref{cor:Xr_matching_exists}. 276 | If equality holds, then $\A$ contains all supersets in $X^{(r)}$ of all sets in $\partial A$. 277 | 278 | 279 | \begin{marginfigure} 280 | \begin{center} 281 | \begin{tikzcd}[column sep=tiny] 282 | \A & \{x_1,\dotsc,x_r\} \arrow[dash]{d} \arrow[dash]{rd} & \arrow[dash]{ld} \\ 283 | \partial \A & \{x_2,\dotsc,x_n\} & \{x_1,x_2,\dotsc,x_n\} 284 | \end{tikzcd} 285 | \end{center} 286 | \caption{If $\A$ contains all supersets in the $X^{(r)}$ layer of sets in the shadow $\partial \A$, then as long as $\A\neq \emptyset$, it must contain every set; here, for example, $\A$ has to include the endpoint of the edge leaving the bottom left vertex. } 287 | \end{marginfigure} 288 | 289 | Consider the graph as in \cref{cor:Xr_matching_exists}: there are no edges from $A\cup \partial A$ to the remaining verticies, so since $G$ is connected\sidenote{as is easy to check}, we have equality in \eqref{eq:local_LYM}. 290 | \end{proof} 291 | 292 | \begin{theorem} 293 | The equality in the LYM inequality \eqref{eq:LYM} holds iff $\F = X^{(r)}$ for some $r$. 294 | \end{theorem} 295 | \begin{proof} 296 | Inductively define $G_n = \F_n$, and for $k < n$, $G_k = \partial G_{k+1} \cup \F_k$. \marginnote{$G_k$ is the set of all $k$-element sets which are subsets of sets in $\F$.} 297 | 298 | Let $\phi_k = \frac{f_k}{{n \choose k}}$ be the proportion of sets of $\F$ in $X^{(k)}$. Similarly, set $\gamma_k = \frac{|G_k|}{{n\choose k}}$. By the local LYM, 299 | \[ 300 | \frac{|\partial G_{k+1}|}{{n\choose k}} \geq \frac{|G_{k+1}|}{{n\choose k+1}} = \gamma_{k+1} 301 | \] 302 | So, 303 | \[ 304 | \gamma_k = \frac{|\partial G_{k+1} \cup \F_k|}{{n\choose k}} = \frac{|\partial G_{k+1}|}{{n\choose k}} + \frac{|\F_k|}{{n\choose k}} \geq \gamma_{k+1} + \phi_k 305 | \] 306 | where we are using that $\F$ is Sperner, so that $\partial G_{k+1}$ is disjoint from $\F_k$. Thus, we have $\gamma_k \geq \gamma_{k+1} + \phi_k$, with equality iff $\gamma_{k+1} = 1$ or $\gamma_{k+1} = 0$. \marginnote{Since that is when we have equality in \cref{eq:local_LYM}.} 307 | 308 | Note $\gamma_n = \phi_n$. Then 309 | \begin{gather*} 310 | \gamma_{n-1} \geq \gamma_n + \phi_{n-1} = \phi_n + \phi_{n-1}\\ 311 | \gamma_{n-2} \geq \gamma_{n-1} + \phi_{n-2} = \phi_n + \phi_{n-1} + \phi_{n-2}\\ 312 | \text{etc.}\\ 313 | \gamma_k \geq \phi_n + \phi_{n-1} + \dotsm + \phi_k. 314 | \end{gather*} 315 | Hence, $1 \geq \gamma_0 \geq \phi_n + \phi_{n-1} + \dotsb + \phi_0$. This is the LYM inequality. But equality holds if $\gamma_k = \gamma_{k+1} +\phi_k$ for each $k$, i.e. $\gamma_{k+1} = 1$ or $\gamma_{k+1} = 0$ for each $k$. Assuming we have equality, we use that $\gamma_k$ is non-increasing with $k$, so there must exist $k_0$ such that $\gamma_{k_0} = 1$, and $\gamma_{k_0+1} = 0$ (writing $\gamma_{n+1} = 0$). Thus, there are no sets in $\F$ of size at least $k+1$. In other words, $G_{k+1} = \emptyset$, and $G_k = \F_k = X^{(k)}$. But then we must have $\F = X^{(k)}$ as desired, since $\F$ may not have any super sets or subsets of $X^{(k)}$, i.e., any other set in $\P(X)$. 316 | \end{proof} 317 | \begin{remark} 318 | The matchings with $X^{(r)}$ and $X^{(r-1)}$ are the essential objects here in proving the local LYM, and hence LYM and its equality. 319 | \end{remark} 320 | \begin{exercise} 321 | Prove Sperner's theorem using the original way: partitioning into ${n\choose \floor{n/2}}$ chains by induction on $n$ instead of Hall's theorem. 322 | \marginnote{This is \cref{thm:sym_part}.} 323 | \end{exercise} 324 | 325 | -------------------------------------------------------------------------------- /chapters/ch2_LO.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = ../CombinatoricsNotes.tex 2 | 3 | \section{The Littlewood-Offord problem} 4 | \lect{1}{18} 5 | % \marginnote{Lecture 4: Monday, January 18, 2014.} 6 | 7 | \begin{problem}[\cite{LO1943}] 8 | Given $(z_1,\dotsc,z_n)$ complex numbers, with $|z_i| \geq 1$. Consider the $2^n$ possible sums formed by the $z_i$'s. How many sums can have pairwise distances less than 1 from each other? If $z_1 = z_2 = \dotsc =z_n=1$, then we get ${n \choose \floor{n/2}}$ equal sums. 9 | 10 | Given $(z_1,\dotsc,z_n)$, and subset of indicies $A\subset [n]$, let $Z_A = \sum_{i\in A} z_i$, where we define $Z_\emptyset = 0$ and $Z_{\{i\}} = z_i$. We say $\F \subset \P([n])$ is \defn{Littlewood-Offord}[Littlewood-Offord!sets of numbers] (LO) with respect to $(z_1,\dotsc,z_n)$ if $|Z_A- Z_B| < 1$ for all $A,B \in \F$. In this notation, we are looking for the largest collection $\F$ such that $\F$ is LO with respect to $(z_1,\dotsc,z_n)$. 11 | \end{problem} 12 | 13 | \begin{conjecture*} 14 | If $(z_1,\dotsc,z_n) \in \C^n$ with $|z_i|\geq 1$, and $\F$ is LO with respect to $(z_1,\dotsc,z_n)$, then $|\F| \leq {n \choose \floor{n/2}}$. 15 | \end{conjecture*} 16 | \begin{remark} 17 | Littlewood and Offord were not combinatorialists; instead they were interested in studying the roots of random polynomials. 18 | \end{remark} 19 | 20 | \begin{theorem}[\cite{erdos1945}] 21 | If $x_1,\dotsc,x_n$ are real, $|x_i| \geq 1$ for every $i$, and $\F$ is LO with respect to $(x_1,\dotsc,x_n)$, then $|\F| \leq { n\choose \floor{n/2}}$. 22 | \end{theorem} 23 | \begin{proof} 24 | We'll change notation to $X_A$ instead of $Z_A$ for this real case. If $x_1,\dotsc,x_n \geq 1$, then $\F$ is Sperner. Suppose $A\subsetneq B$; then 25 | \[ 26 | |X_A - X_B| = X_B - X_A = X_{B\setminus A} \geq |B-A| \geq 1. \qquad \checkmark 27 | \] 28 | 29 | The general problem for reals can be reduced to this case of $x_i>0$ as follows. 30 | Suppose that $\F$ is LO with respect to $(x_1,\dotsc,x_n)$. Then we will construct a family $\F'$ which is LO with respect to \\$(x_1,\dotsc, x_{i-1},-x_i, x_{i+1}, \dotsc, x_n)$ with $|\F'| = |\F|$. This will suffice to prove the theorem. 31 | 32 | Let $\F' = \{ A \symd \{i\}: A\in \F\}$\sidenote[][]{We remove $i$ if it's there, and add it otherwise.}. Clearly $|\F'| = |\F|$. We've reduced all sums by $x_i$, as shown by the following. If $i \in A$, then $X'_{A \symd \{i\}} = \left(\sum_{j\in A} x_j \right)-x_i$. if $i \not \in B$, then $X'_{B \symd \{i\}} = \left( \sum_{j\in B} x_j \right) - x_i$. 33 | Hence, all sums are still within $1$ of each other, completing the proof. 34 | \end{proof} 35 | 36 | 37 | 38 | \begin{definition} 39 | A chain $A_1 \subset A_2 \subset \dotsc \subset A_k$ in $\P([n])$ is \defn{symmetric}[symmetric!chain] if $|A_{i+1}| = |A_i| +1$ for $i=1,\dotsc,k-1$. Additionally, we require $|A_1| + |A_k| = n$. 40 | \end{definition} 41 | Note that in particular, a symmetric chain intersects $[n]^{(\floor{n/2})}$. An example of symmetric chains is shown in \cref{fig:sym_chains}. 42 | \begin{figure}[ht] 43 | \begin{center} 44 | \begin{tikzcd}[column sep=tiny] 45 | {[3]}^{(3)} & &\{1,2,3\}\arrow[green,dash]{ld} \\ 46 | {[3]}^{(2)} &\{1,2\} \arrow[dash,green]{d}& \{1,3\} \arrow[dash,blue]{rd} & \arrow[dash,red]{ld} \{2,3\} \\ 47 | \mathbf{[3]^{(1)}} & \mathbf{\{1\}} \arrow[dash,green]{rd} & \mathbf{\{2\}}&\mathbf{\{3\}}\\ 48 | {[3]}^{(0)} & & \emptyset 49 | \end{tikzcd} 50 | \begin{tikzcd}[column sep=tiny] 51 | &\{1,2,3\}\arrow[red,dash]{rd} \\ 52 | \{1,2\} \arrow[dash,green]{d}& \{1,3\} \arrow[dash,blue]{rd} & \arrow[dash,red]{ld} \{2,3\} \\ 53 | \mathbf{\{1\}} \arrow[dash,green]{rd} & \mathbf{\{2\}}&\mathbf{\{3\}}\\ 54 | & \emptyset 55 | \end{tikzcd} 56 | \end{center} 57 | \caption{\textit{Left:} We draw our favorite $\P([n])$, partioned into symmetric chains, which thus each intersect $[n]^{(\floor{n/2})}$. \textit{Right:} A partition of $\P([n])$ into chains which intersect $[n]^{(\floor{n/2})}$, which we could obtain from the proof of \cref{thm:sperner}. But these chains are not symmetric.} \label{fig:sym_chains} 58 | \end{figure} 59 | 60 | Our previous method, when proving Sperner's theorem, didn't actually guarantee symmetric chains; see \cref{fig:sym_chains}. We will do this now, which provides a new proof of Sperner's theorem. 61 | 62 | \pagebreak 63 | \begin{theorem} \label{thm:sym_part} 64 | $\P([n])$ can be partitioned into symmetric chains. 65 | \end{theorem} 66 | \begin{proof}[Proof by induction on $n$.] 67 | The base case $n=1$ is immediate. 68 | Induction step: Let $C_1,C_2,\dotsc,C_L$ be symmetric chains forming a partition of $\P{[n-1]}$. Consider $C_i = (A_1,A_2,\dotsc,A_k)$. Form 69 | \begin{align*} 70 | C_i' &= (A_1,A_2,\dotsc,A_k,A_k\cup \{n\})\\ 71 | C_i'' &= (A_1 \cup \{n\}, A_2 \cup \{n\}, \dotsc, A_{k-1} \cup \{n\}). 72 | \end{align*} 73 | % \improvement{Add picture of $n=3$ of $C_i'$ and $C_i''$.} 74 | \begin{figure} 75 | \begin{center} 76 | \begin{tikzcd}[column sep=tiny] 77 | % {[n]}^{(3)} & &\{1,2,3\}\arrow[green,dash]{ld} \\ 78 | \phantom{{[n]}^{(3)}}& & \\ 79 | {[2]}^{(2)} &\{1,2\} \arrow[dash,OliveGreen]{d}& \\ 80 | \mathbf{[2]^{(1)}} & \mathbf{\{1\}} \arrow[dash,OliveGreen]{rd} & \mathbf{\{2\}}&\\ 81 | {[2]}^{(0)} & & \emptyset 82 | \end{tikzcd}\hfill 83 | \begin{tikzcd}[column sep=tiny] 84 | {[3]}^{(3)} & &\{1,2,3\}\arrow[DarkGreen,dash]{ld} \\ 85 | {[3]}^{(2)} &\{1,2\} \arrow[dash,DarkGreen]{d}& \{1,3\} \arrow[dash,LightGreen]{rd} & \arrow[dash,red]{ld} \{2,3\} \\ 86 | \mathbf{[3]^{(1)}} & \mathbf{\{1\}} \arrow[dash,DarkGreen]{rd} & \mathbf{\{2\}}&\mathbf{\{3\}}\\ 87 | {[3]}^{(0)} & & \emptyset 88 | \end{tikzcd} 89 | \end{center} 90 | \caption{An illustration of the induction step, from $n=2$ on the left to $n=3$ on the right. On the left, two chains: $C_1 = (\emptyset, \{1\}, \{1,2\})$ in green, and $C_2 = (\{2\})$. 91 | We create $C_1' = (\emptyset, \{1\},\{1,2\},\{1,2,3\})$ in dark green, and $C_1'' = (\{3\}, \{1,3\})$ in light green. We also create $C_2' = (\{2\},\{2,3\})$ in red, and $C_2'' = ()$, an empty chain.% 92 | }% 93 | \end{figure} 94 | We can easily see that given that $C_i$ was a symmetric chain on $\P([n-1])$, we have $C_i'$ and $C_i''$ are both symmetric chains on $\P([n])$. Moreover, 95 | \[ 96 | \{C_1', \dotsc, C_L'\}\cup \{C_1'', \dotsc, C_L''\} 97 | \] 98 | form a partition of $\P([n])$, as follows: For any set $A\in \P([n])$, if $n\not \in A$, then $A\in C_i$ for some $i$, and hence $A\in C_i'$. If $n\in A$, then $A\setminus \{n\} \in C_i = (A_1,A_2,\dotsc,A_k)$ for some $i$, so $A = A_j$ for some $j$. If $j=k$, then $A\in C_i'$, otherwise $A\in C_i''$. In any case, if $A\in \P([n])$, $A$ is in one of these chains. Finally, $A$ cannot be in two chains, otherwise we have that $\{C_i\}$ was not a partition of $\P([n-1])$. 99 | \end{proof} 100 | 101 | Any partition of $\P([n])$ symmetric chains has ${n\choose \floor{n/2}}$ sets, because each chain must intersect with one point in $[n]^{(\floor{n/2})}$. 102 | In the proof, we took a partition into ${n-1 \choose \floor{(n-1)/2}}$ chains, and seem to have created $2 {n-1 \choose \floor{(n-1)/2}} \neq {n\choose \floor{n/2}}$ chains. The catch is that $C_i''$ is an empty chain if $k=1$. 103 | 104 | 105 | \newthought{Let us count} the sizes of symmetric chains. Suppose $C_1,\dotsc,C_L$ is a partition of $\P([n])$ into symmetric chains, where $L = {n\choose \floor{n/2}}$. How many chains are there of length $r=n+1$ in the partition? Exactly one, the chain which contains the empty set and must therefore contain the set of $n$ elements. 106 | 107 | What about $r=n$? Zero. 108 | 109 | What about $r= n-1$? $n-1$ chains, because it must start the collection of 1 element sets and go to the collection of $n-1$ element sets. There are $n$ 1-element sets, but one is already in the maximal chain, so we are left with $n-1$. 110 | 111 | What about $r= (n+1)-2i$? These are the symmetric chains which start at the $i$th level $[n]^{(i)}$ and go to the $n-i$th level $[n]^{(n-i)}$. The result is ${n\choose i} - {n\choose i-1}$. That is because there are ${n\choose i}$ elements in $[n]^{(i)}$, but ${n\choose i-1}$ of them are part of longer chains, the number of which is the number of elements in the level $[n]^{(i-1)}$. 112 | 113 | Let us formulate analogues of these ideas to solve the Littlewood-Offord problem in $\R^d$. 114 | \begin{theorem}[\cite{KLEITMAN1970}] \label{thm:kleit70} 115 | Let $(x_1,\dotsc,x_n)$ be vectors in $\R^d$, with $\|x_i\| \geq 1$. As before, define for $A\subset [n]$, $X_A = \sum_{i\in A} x_i$, and say that $\F\subset \P([n])$ is \defn{LO}[Littlewood-Offord!sets of vectors] with respect to $(x_1,\dotsc,x_n)$ if for each $A,B\in \F$, we have $\|X_A - X_B\| < 1$. 116 | Let $\F$ be LO with respect to $(x_1,\dotsc,x_n)$. Then $|\F| \leq { n\choose \floor{n/2}}$. 117 | \end{theorem} 118 | \begin{proof} 119 | We will say $\Cset\subset \P([n])$ is \defn{sparse}[sparse!set of indicies of vectors] (with respect to $(x_1,\dotsc,x_n)$) if $\|X_A-X_B\| > 1$ for all $A,B\in \Cset$ with $A\neq B$.x 120 | It is enough to show that $\P([n])$ can be partitioned into ${n\choose \floor{n/2}}$ sparse sets\sidenote{A LO family may only contain one element of each sparse set.}. 121 | 122 | We say that a partition $C_1,\dotsc,C_L$ of $\P([n])$ is \defn{symmetric}[symmetric!partition] if it contains exactly ${n\choose i} - {n\choose i-1}$ ``chains'' of order $(n+1)-2i$ for $i=0,1,\dotsc,\floor{n/2}$, and no chains with sizes not congruent to $n+1 \mod 2$. \marginnote{Here, ``chain'' simply means a set $C_i$ for some $i\in [L]$.} In particular, $L= {n\choose \floor{n/2}}$. 123 | 124 | We will show that $\P[n]$ has a symmetric partition into sparse sets. Then we will have found $L = {n\choose \floor{n/2}}$ sparse sets, completing the proof. 125 | Let's proceed by induction on $n$. For $n=1$, we have $X_\emptyset = 0$, and $X_{\{1\}} = x_1$. Then $\{\emptyset, \{1\}\}$ is sparse if $\|X_\emptyset - X_{\{1\}}\| > 1$, which holds because $\|x_1\| > 1$. 126 | Induction step: Let $C_1,C_2,\dotsc, C_L$ be a symmetric partition of $\P([n-1])$ into sets sparse with respect to $(x_1,\dotsc,x_{n-1})$. 127 | 128 | Let $C_i = \{A_1,\dotsc,A_k\}$. We'd like to form sparse sets 129 | \begin{align*} 130 | C_i' &= \{A_1,A_2,\dotsc,A_k,A_k \cup \{n\}\}\\ 131 | C_i'' &= \{A_1\cup \{n\}, A_2\cup\{n\},\dotsc,A_{k-1}\cup \{n\}\}. 132 | \end{align*} 133 | Our sets $A_1,\dotsc,A_k$ do not have an ordering this time, and we may choose any set to be $A_k$. We will have to use this freedom; consider translating the sums in some chain $C_i = (A_1,\dotsc,A_k)$ by adding the vector $x_n$. We want $C_i'$ to be sparse, but then we need $X_{A_k\cup\{n\}}$ far from each $X_{A_j}$, which does not always need to happen, as illustrated in \cref{fig:LO_proof_prob}. 134 | \begin{marginfigure}[-3cm] 135 | \input{graphics/figLOproofv2} 136 | 137 | % \includegraphics[scale=.5]{graphics/LOproof.pdf} 138 | \caption{With the choice of $A_k$ shown here, we have that $X_{A_k \cup \{n\}} = X_A + x_n$ (in red) is near to $X_{A_1}$ (in blue), so $C_i'$ is not a sparse chain, even though $C_i$ is a sparse chain (all the start points of the arrows are far from each other).} \label{fig:LO_proof_prob} 139 | 140 | \end{marginfigure} 141 | % \improvement{Draw better figure, show that the distance is less than 1 in that one spot.} 142 | Assume WLOG that $x_n = (\alpha,0,0,\dotsc,0)$ only has non-zero first coordinate with $\alpha>0$. Let $p(v)$ denote the first coordinate of $v$. Then $p$ is a linear function, and $p(v) \leq \|v\|$ for each $v$. 143 | 144 | Let $A_k$ in the construction above be chosen so that $p(X_{A_k})\geq p(X_{A_j})$ for each $j$\sidenote{So we are choosing $A_k$ to be the furthest in the direction $x_n$. Then if we add $x_n$, we are only moving it further away from the others, so the conflict in \cref{fig:LO_proof_prob} doesn't occur.}. Then to show that $C_i'$ is sparse, it suffices to see that 145 | \begin{align*} 146 | \|X_{A_k \cup\{n\}} - X_{A_j}\| &\geq p( X_{A_k \cup \{n\}} - X_{A_j})\\ 147 | &= p(x_{A_k}) + p(x_n) - p(X_{A_j})\geq p(x_n)\geq 1. 148 | \end{align*} 149 | Hence, we follow the same proceedure as in \cref{thm:sym_part} to conclude that indeed we did produce a symmetric partition. 150 | % \improvement{Add more words to that proof, 2.2.} 151 | \end{proof} 152 | 153 | 154 | \lect{1}{20} 155 | % \marginnote{Lecture 5: Wednesday, January 20, 2016.} 156 | \newthought{Let us now consider} $\Z_p:=\Z/p\Z$, the integers modulo a prime $p$. Let $x_1,\dotsc,x_n\in \Z_p\setminus\{0\}$. We will say $\Cset \subset \P([n])$ is \defn{sparse}[sparse!set of indicies of integers modulo $p$] with respect to $x_1,\dotsc,x_n$ if $X_A\neq X_B$ for all $A,B\in \C$ with $A\neq B$. 157 | Let us define 158 | \[ 159 | \Sigma(z):= | \{A\subset \P([n]): X_A=z\}|. 160 | \] 161 | We are interested in $\max \Sigma(z)$. First, if each $x_1=\dotsb=x_n=1$, then we see that may achieve ${n \choose \floor{n/2}}$. 162 | On the other hand, since there are $2^n$ possible sums and only $p$ values, we have by the pigeonhole principle that $\max \Sigma(z) \geq 2^n/p$. To determine $\max \Sigma(z)$ we will use a partition into sparse sets, just as in the proof of \cref{thm:kleit70}. 163 | \begin{theorem} 164 | Given $x_1,\dotsc,x_n\in \Z_p\setminus\{0\}$ for prime $p$, with $n\leq p-1$, then there exists a symmetric partition of $\P([n])$ into sparse sets with respect to $x_1,\dotsc,x_n$. 165 | \end{theorem} 166 | \begin{proof} 167 | By induction on $n$. If $\Cset = \{A_1,\dotsc, A_k\}$ is sparse with respect to $x_1,\dotsc,x_{n-1}$, we want to show that 168 | \begin{align*} 169 | \Cset' &= \{A_1,\dotsc, A_k, A_{\ell} \cup \{n\}\},\\ 170 | \Cset'' &= \{A_1\cup \{n\}, A_2\cup\{n\}, \dotsc, A_k \cup \{n\}\}\setminus \{A_\ell \cup \{n\}\} 171 | \end{align*} 172 | are sparse for some $\ell$. Set $Y_i = X_{A_i}$ and consider the sums $Y_1,\dotsc,Y_k$. We wish to find $1\leq \ell\leq k$ such that $Y_\ell+x_n \not\in \{Y_1,\dotsc,Y_k\}$. Note that $k\leq (n-1)+1 \leq p-1$. Suppose there is no such $\ell$. Then wlog, $Y_1 + x_n = Y_2$, $Y_2 + x_n = Y_3$, \ldots, $Y_j + x_n = Y_1$ for some $j$. 173 | 174 | Then $Y_1 + j x_n = Y_1$, so $jx_n=0$. Then $j=0$ or $x_n = 0$, which is a contradiction, since $j\leq k\leq p-1$. 175 | 176 | In other words, if we have a set $A\subset \Gamma$ for some group $\Gamma$, and for some element $x\in \Gamma$ we have $x+A \subset A$. Then $A$ is a union of cosets of a cyclic subgroup of $\Gamma$ generated by $\{x\}$. 177 | \end{proof} 178 | 179 | \begin{corollary} Let $x_1,\dotsc,x_n\in \Z_p\setminus\{0\}$ with $n\leq p-1$. 180 | ~\begin{enumerate} 181 | \item Then $\Sigma(z) \leq {n \choose \floor{n/2}}$. 182 | \item (Cauchy-Davenport). Let $S(x_1,\dotsc,x_n) = \{ X_A: A\in \P([n])\}$. Then $|S(x_1,\dotsc,x_n)| \geq \min\{ p,n+1\}$. 183 | \end{enumerate} 184 | 185 | \end{corollary} 186 | \begin{proof} 187 | ~ \begin{enumerate} 188 | \item A symmetric partition has exactly ${n\choose \floor{n/2}}$ parts, and within each part there is at most one set with corresponding sum $z$. 189 | \item If $n\leq p-1$, then we have a symmetric partition, which contains a sparse set of size $n+1$. If $n\geq p$, then $S(x_1,\dotsc,x_n) = \Z_p$. This is because we can just use any $p-1$ elements to get all the elements by sums.\qedhere 190 | \end{enumerate} 191 | \end{proof} -------------------------------------------------------------------------------- /chapters/ch3_intersecting_hypergraphs.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = ../CombinatoricsNotes.tex 2 | 3 | \section{Intersecting hypergraphs} 4 | 5 | A family $\F\subset \P([n])$ is \defn{intersecting}[intersecting set system] if every two sets $A,B\in \F$ have $A\cap B\neq \emptyset$. If $\F$ is intersecting, then $|\F| \leq 2^{n-1}$ because $\F$ can contain at most one set in each pair $\{A,A^c\}$ for every $A\subset [n]$. On the other hand, $2^{n-1}$ may be acheived by taking $\F = \{A\subset [n]: x\in A\}$ for some $x\in [n]$. 6 | What if $\F\subset [n]^{(r)}$ for some $r$? If $r> n/2$, then $\F = [n]^{(r)}$ is intersecting. 7 | 8 | \begin{theorem}[\cite{erdos-ko-rado-1961}] \label{thm:erdos-ko-rado} 9 | Let $r \leq n/2$ and $\F\subset[n]^{(r)}$ be intersecting. Then 10 | \[ 11 | |\F| \leq {n-1\choose r-1} 12 | \] 13 | which can be acheived by $\F = \{A\in [n]^{(r)}: x\in A\}$ for some $x\in [n]$. 14 | \end{theorem} 15 | \begin{proof} 16 | Let us consider a particular circular ordering of $[n]$ and upper bound the number of sets in $\F$ which are intervals of size $r$ in this order. 17 | \begin{figure}[ht] 18 | \usetikzlibrary{fit} 19 | \begin{center} 20 | \begin{tikzpicture} 21 | \begin{scope} 22 | \node[left] at (0,0) {$\circlearrowleft$}; 23 | 24 | \foreach \x in {1,...,8} 25 | { 26 | \pgfmathsetmacro\myangle{\x*45} 27 | \filldraw[black] (\myangle:1cm) circle (0.4pt) node[left]{\x}; 28 | } 29 | \end{scope} 30 | \node[xshift=45pt] at (0,0) {$=$}; 31 | % \end{tikzpicture} 32 | % $=$ 33 | % \begin{tikzpicture} 34 | \begin{scope}[xshift=100pt] 35 | \node[left] at (0,0) {$\circlearrowleft$}; 36 | \foreach \x in {1,...,8} 37 | { 38 | \pgfmathsetmacro\myangle{\x*45+45} 39 | \filldraw[black] (\myangle:1cm) circle (0.4pt) node[left]{\x}; 40 | } 41 | \end{scope} 42 | \node[xshift=145pt] at (0,0) {$\neq$}; 43 | 44 | % \end{tikzpicture} 45 | % $\neq$ 46 | % \begin{tikzpicture} 47 | \begin{scope}[xshift=200pt] 48 | \node[left] at (0,0) {$\circlearrowleft$}; 49 | \foreach \x in {1,...,8} 50 | { 51 | \pgfmathsetmacro\myangle{-1*\x*45+90} 52 | \filldraw[black] (\myangle:1cm) circle (0.4pt) node[left]{\x}; 53 | } 54 | \end{scope} 55 | \end{tikzpicture} 56 | \end{center} 57 | \caption{An illustration of circular orders on $[8]$. We define our order counter-clockwise, and so the order is invariant under rotations, as the equality between the left and center orders demonstrates. However, the right order was obtained by reversing the order of the left, and thus is a new order.}\label{fig:circ_order} 58 | \end{figure} 59 | Let us prove there are at most $r$ intervals. 60 | Let us fix an interval $I= (a_1,a_2,\dotsc,a_r)$, and count the number of intervals which intersect it. For each pair of consecutive points $(a_{i-1},a_i)$\sidenote{Corresponding to gaps between points} in this interval $I$, there is an interval $I_1$ with first element $a_{i+1}$ and an interval $I_2$ with last element $a_i$, yielding $2(r-1)$ intervals intersecting it. But $\F$ can contain at most one interval in this pair $(I_1,I_2)$, because $I_1\cap I_2=\emptyset$. So we are left with $r-1$ intervals intersecting $I$, along with $I$ itself. Thus, we've found $r$ intersecting intervals in this circular order. 61 | 62 | How many circular orders are there on $[n]$? Every circular order corresponds to $n$ permutations (all rotations of each other), and there are $n!$ total permutations, yielding $(n-1)!$ circular orders. 63 | 64 | \begin{marginfigure} 65 | \begin{center} 66 | \begin{tikzpicture} 67 | \begin{scope} 68 | \node[left] at (0,0) {$\circlearrowleft$}; 69 | 70 | \foreach \x in {1,...,3} 71 | { 72 | \pgfmathsetmacro\myangle{\x*45} 73 | \filldraw[black] (\myangle:1cm) circle (0.4pt) node[left]{$x_{\x}$}; 74 | } 75 | \foreach \x in {1,...,5} 76 | { 77 | \pgfmathsetmacro\myangle{(\x+3)*45} 78 | % \pgfmathsetmacro\y{\x-3} 79 | \filldraw[black] (\myangle:1cm) circle (0.4pt) node[left]{$y_{\x}$}; 80 | } 81 | \end{scope} 82 | \end{tikzpicture} 83 | \end{center} 84 | \caption{An order in which $X = \{x_1,x_2,x_3\} \in [8]^{(3)}$ is an interval, where we've enumerated $[8]\setminus X = \{y_1,\dotsc,y_5\}$.} \label{fig:X_interval_circ_order} 85 | \end{marginfigure} 86 | In how many circular orders is a given $X\in [n]^{(r)}$ an interval? To obtain an interval, we order the elements of $X$ in $r!$ ways, and then order the rest of the set in $(n-r)!$ ways, yielding $r!(n-r)!$ orders in which $X$ is an interval. See \cref{fig:X_interval_circ_order} for an illustration. 87 | 88 | We have at most $r$ sets of $\F$ as intervals per circular order, and $(n-1)!$ circular orders, so at most $r(n-1)!$ sets in $\F$ over all circular orders. 89 | 90 | On the other hand, each set of $\F$ has only $r!(n-r)!$ orders in which it is an interval. Thus, we have $|\F| r!(n-r)!$ intervals corresponding to sets in $\F$, counted over all orders. Hence, 91 | \[ 92 | |\F| r! (n-r)! \leq r(n-1)! 93 | \] 94 | which completes the proof. 95 | \end{proof} 96 | 97 | \begin{remark} 98 | For $r < n/2$, if $\F\subset [n]^{(r)}$ is intersecting and has maximal size, i.e. $|\F| = {n-1\choose r-1}$, then for some $x\in n$, we have 99 | \[ 100 | \F = \{A\in [n]^{(r)}: x\in A \}. 101 | \] 102 | The proof is left as an exercise. In particular, there are only $n$ possible extremal families. 103 | 104 | Consider $r=n/2$. For each set $A \in [n]^{(r)}$, there is only one forbidden set $A^c$. So $\F$ may be formed by choosing one set $A$ from each pair $\{A,A^c\}$ arbitrarily. This yields a doubly-exponential number of extremal families. 105 | \end{remark} 106 | 107 | \newthought{We will now consider} a generalization of intersecting. 108 | % \begin{definition} 109 | We say $\F\subset [n]^{(r)}$ is \defn{$t$-intersecting}[t-intersecting] if $|A\cap B|\geq t$ for all $A,B\in \F$. 110 | % \end{definition} 111 | We wish to find $m(n,r,t)$, the maximum size of a $t$-intersecting family $\F$ which is a subset of $[n]^{(r)}$. 112 | A natural guess of an optimal family would be 113 | \[ 114 | \F_0 := \F_0(n,r,t) = \{ A \in [n]^{(r)}: L\subset A\} 115 | \] for some $L\subset [n]$ with $|L| = t$. 116 | Note $|\F_0| = {n-t \choose r-t}$. 117 | On the other hand, for which $n,r,t$ is $m(n,r,t) = {n\choose r}$? I.e., when is the family $[n]^{(r)}$ itself $t$-intersecting? 118 | We may use the inequality 119 | \[ 120 | t\leq |A\cap B| \leq |A| + |B| - |A\cup B| = 2r - n 121 | \] 122 | to obtain the bound $2r-t \geq n$. So, as with the $1$-intersecting case, for large enough $r$ (compared to $t$ and $n$), $[n]^{(r)}$ itself is intersecting. 123 | 124 | What about $n=2r-t+1$? Then $|\F_0| = {2r-2t+1 \choose r-t}$. But we could still take all $r$ element subsets of $[n-1]$ to obtain 125 | \[ 126 | m(n,r,t) \geq {2r -t \choose r}. 127 | \] 128 | But 129 | \[ 130 | {2r -t \choose r} > {2r-t+1 \choose r} \geq {2r -2t+1 \choose r-t}. 131 | \] 132 | So $\F_0$ does not achieve the maximum in this case either. 133 | 134 | \begin{exercise} 135 | Prove that $\F_0$ is optimal when $n$ is large compared to $r$ and~$t$.\marginnote{This is the content of \cref{thm:F0max}.} 136 | \end{exercise} 137 | 138 | \lect{1}{26} 139 | % \marginnote{Lecture 6: Monday, January 26, 2016.} 140 | 141 | 142 | Now, consider the family 143 | \[ 144 | \F_{r-t} := \{ A\in [n]^{(r)}: |L\cap A|\geq r\} 145 | \] 146 | for some $L$ with $|L| = 2r-t$. 147 | % maybe: 148 | % When $n \leq 2r-t$, we may choose all of $[n]^{(r)}$, and this family is optimal. 149 | Define an interpolation between $\F_0$ and $\F_{r-t}$ by 150 | \[ 151 | \F_k := \{A\in [n]^{(r)}: |A\cap L|\geq k+t\} 152 | \] 153 | for $|L| = 2k+t$. 154 | Then for $A,B\in \F_k$ we have 155 | \[ 156 | |A\cap B\cap L| \geq |A\cap L| + |B\cap L| - |L| \geq 2(k+t) - (2k+t) = t 157 | \] 158 | so for every $0\leq k \leq r-t$, the family $\F_k$ is $t$-intersecting. 159 | ~\begin{margintable}[4cm] 160 | \begin{center} 161 | \begin{tabular}{lccc}\toprule 162 | & \multicolumn{3}{c}{$k$} \\ \cmidrule(r){2-4} 163 | $n$& $0$ &$1$&$2$\\\midrule 164 | % & & 0 & 1 & 2\\ \midrule 165 | $8$ &10 & 16 & \boxed{21}\\ 166 | $11$ & 28 & \boxed{31} & 21\\ 167 | $13$ & \boxed{45}& 41& 21 \\\bottomrule 168 | \end{tabular} 169 | \end{center} 170 | \caption{The size of $\F_k(n)$, given $r=5$ and $t=3$, tabulated over several choices of $k$ and $n$. We see that the maximal $\F_k(n)$ changes based on both $k$ and $n$.}\label{tab:Fks} 171 | \end{margintable} 172 | \begin{example} 173 | Consider $r=5$, $t=3$. Then 174 | \begin{gather*} 175 | |\F_0(n) | = {n-3 \choose 2}, \\ 176 | |\F_1(n)| = {5\choose 4}{n-5 \choose 1} + 1 = 5(n-5)+1,\\ 177 | |\F_2(n)| = {7\choose 5} = 21. 178 | \end{gather*} 179 | By choosing different values of $n$ (see \cref{tab:Fks}), we see that the maximum size family occurs at different $k$'s: it is a more complicated situation than the $1$-intersecting case. 180 | \end{example} 181 | In general, for $n\geq 2k+t$, 182 | \[ 183 | | \F_k(n)| = \sum_{s=k+t}^{2k+t} {2k+t \choose s} {n-2k-t \choose r-s} 184 | \] 185 | where we are summing over possible sizes $s$ of $|A\cap L|$. The first combination comes from choosing within $L$, and the second from choosing outside of $L$. 186 | 187 | We may think of $|\F_k(n)|$ as a function of $n$. Then it is a polynomial of degree $r-k-t$ in which the coefficient of the monomial of largest degree is positive. Among the $\F_k(n)$'s, the family $\F_0(n)$ is eventually of the largest size, because it is the polynomial of highest degree. In fact, $\F_0(n)$ is eventually the largest family overall, as the following result shows. 188 | \begin{theorem} \label{thm:F0max} 189 | For all $r,t$ there exists $n_0$ such that for $n\geq n_0$, 190 | \[ 191 | m(n,r,t) = |\F_0| = {n-t \choose r-t}. 192 | \] 193 | \end{theorem} 194 | \begin{proof} 195 | Let $\F\subset [n]^{(r)}$ be $t$-intersecting such that $|\F| = m(n,r,t)$. Then there exists $A,B\in \F$ such that 196 | \[ 197 | |A\cap B| = t 198 | \] 199 | for $n\geq 2r-t$. Assume not. If $\min_{A,B\in \F} |A\cap B| = t+\ell$ for some $\ell\geq 1$, then choose some $L\subset A\cap B$ with $|L|=\ell$, and consider the set $A'=A\setminus L$. If $A'\in \F$, then we would have $|A'\cap B|=t < t+\ell$, so $A'\not \in \F$. But for any $C\in \F$, we have 200 | \[ 201 | t+\ell \leq |A\cap C| = |A'\cap C| + |L\cap C| \leq |A'\cap C| +\ell, 202 | \] 203 | so $|A'\cap C|\geq t$. Thus, $\F\cup \{A'\}$ is $t$-intersecting and $|\F\cup \{A'\}| > |\F|$, which contradicts the maximality of $\F$. 204 | 205 | Now, let $Z=A\cap B$, where $|A\cap B|=t$. If $Z\subset C$ for all $C\in \F$, then $|\F| \leq {n-t \choose r-t}$. So we may assume there exists $C\in \F$ such that $Z\not \subset C$. We will show 206 | \begin{equation} \label{eq:XcapAcupBcupC_big} \tag{$\star$} 207 | |X \cap (A\cup B\cup C)|\geq t+1 208 | \end{equation} for all $X\in \F$. Since each set is of size $r$, the size $L:=|A\cup B\cup C| \leq 3r$. This is enough as it implies 209 | \[ 210 | |\F| \leq \sum_{s=t+1}^r {L \choose s} {n-L \choose r-s} 211 | \] 212 | which is a polynomial of degree at most $r-t-1$, and so is eventually less than ${n-t\choose r-t}$, which is a polynomial of degree $n-t$. 213 | 214 | Let us show \eqref{eq:XcapAcupBcupC_big}. We know 215 | \[ 216 | X\cap (A\cup B\cup C) = (X\cap A)\cup (X\cap B)\cup (X\cap C). 217 | \] 218 | Each set is of size at least $t$; for $|X\cap (A\cup B\cup C)|\leq t$, then $|X\cap (A\cup B\cup C)| = t$, and in particular, $Y:=X\cap A= X\cap B = X\cap C$. Then $Y\subset Z$, but since $|Y| = |Z| = t$, we have $Y=Z$. Then we have $Z=Y = X\cap C \subset C$, which is a contradiction. 219 | \end{proof} 220 | 221 | The following theorem, presented here without proof, resolves our question. 222 | \begin{theorem}[\cite{Ahlswede-Khachatrian-1997}] 223 | For all $n,r,t$, 224 | \[ 225 | m(n,r,t) = \max_k |\F_k|. 226 | \] 227 | \end{theorem} 228 | 229 | \newthought{Let us consider} a consequence of \erdos-Ko-Rado\sidenote{\Cref{thm:erdos-ko-rado}}. Let $Z_1,\dotsc,Z_n$ be independent Bernoulli random variables each with expectation value $p > \tfrac{1}{2}$. Then $\Pr[Z_i=1] = p$, $\Pr[Z_i = 0] = 1-p$. 230 | Suppose the $Z_i$'s are stocks, and the price of each is $\tfrac{1}{2}$. If we invest \$0.50, then our expected return is \$$p$, no matter how we invest. 231 | 232 | Suppose we are very conservative and our goal is to have at least $\tfrac{1}{2}$ in the end. A good strategy is to diversify and invest uniformly in each stock; then the law of large numbers says that as the number of stocks goes to infinity, we almost surely recover our $1/2$. 233 | What's the worst possible strategy? It should be to invest in only 1 stock; in that case, our probability of success is $p$. 234 | \begin{theorem}[\cite{LIGGETT197715}] 235 | Let $Z_1,\dotsc,Z_n$ be independent Bernoulli random variables with expectation value $p$. Let $c_1,\dotsc,c_n\geq 0$ with $\sum c_i =1$. Then 236 | \[ 237 | \prob \Big[ \sum_{i=1}^nc_i Z_i \geq \tfrac{1}{2} \Big] \geq p. 238 | \] 239 | \end{theorem} 240 | \begin{proof} 241 | Assume $c_i >0$ for all $i$, and $n$ odd for simplicity. Let $\F = \{A\subset [n]: \sum_{i\in A} c_i \geq \tfrac{1}{2} \}$. That is, $\F$ is the collection of all sets of r.v. such that if exactly those random variables obtain return 1, we did not lose. 242 | Let $\F_k = \F \cap [n]^{(k)}$ and $f_k = |\F_k|$. Then 243 | \[ 244 | \prob \Big[ \sum_{i=1}^n c_i Z_i \geq \tfrac{1}{2}\Big] = \sum_{A\in \F} \prob [ \text{exactly }Z_i \text{ with indicies }i\in A \text{ have value 1}]. 245 | \] 246 | If we fix some $A$ of size $k$, what is the probability that these $k$ trials succeed? $p^k(1-p)^{n-k}$. Thus, 247 | \[ 248 | \prob \Big[ \sum_{i=1}^n c_i Z_i \geq \tfrac{1}{2}\Big] = \sum_{k=0}^n f_k p^k (1-p)^{n-k}. 249 | \] 250 | Our goal is to show that the LHS is larger than $p$. 251 | \begin{enumerate}[{Fact }1.] 252 | \item $\F_k$ is intersecting for each $k < n/2$. 253 | 254 | \begin{subproof} 255 | Suppose not: then there exists $A,B\in \F_k$ such that $A\cap B = \emptyset$. Then $\sum_{a\in A} c_i \geq \frac{1}{2}$, and $\sum_{a\in B} c_i \geq \frac{1}{2}$. But since each $c_i>0$, we must have $A\cup B = [n]$, but we know $|A\cup B|< 2\cdot n/2=n$. 256 | \end{subproof} 257 | 258 | \item[Corollary.] For $kn/2$, swapping $n-k$ with $k$ to get} 281 | &= \sum_{k>n/2} p^{n-k} \left[ {n-1\choose k-1} \right](1-p)^k \\ 282 | &\qquad+\sum_{kn/2} \Bigg({n\choose k}- {n-1 \choose k} \Bigg) p^k(1-p)^{n-k}\mathnote{Using ${n-1\choose k-1}={n\choose k}- {n-1 \choose k}$.}\\ 285 | &=\sum_{k=1}^n {n-1 \choose k-1} p^k (1-p)^{n-k}\\ 286 | &= p \sum_{\ell=0}^{n-1} {n-1 \choose \ell} p^\ell (1-p)^{n-1-\ell} \\ 287 | &= p( p + (1-p))^{n-1} =p. \qedhere 288 | \end{align*} 289 | \end{proof} -------------------------------------------------------------------------------- /chapters/ch4_shadows.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = ../CombinatoricsNotes.tex 2 | 3 | \section{Compression and Shadows} %\marginnote{Lecture 7: Wednesday, January 27, 2016.} 4 | \lect{1}{27} 5 | As an aside, let's consider Steiner symmetrization\sidenote{\cite{steiner1838einfache}}, a technique in geometry. 6 | Given a shape $K\subset \R^2$, with a line of ``symmetry'' $L$, we obtain $S_L(K)$ from $K$ by replacing $K\cap L'$ for every line $L'$ orthogonal to $L$ by an interval of length equal to $|K\cap L'|$ centered at $L$. 7 | \paragraph{Properties:} 8 | \begin{enumerate} 9 | \item $\area(S_L(K)) = \area(K)$ 10 | \item $\diam(S_L(K)) \leq \diam (K)$ \marginnote{The diameter is the largest distance between two points on $K$. We may prove this by drawing trapezoids between two lines $L'$ and $L''$, one of which passes through each point (close to) achieving the diameter.} 11 | \item $\perim(S_L(K)) \leq \perim(K)$. 12 | \end{enumerate} 13 | Given a shape in $\R^2$ with area 1, what is the smallest diameter? That's the diameter of the area 1 disc. Sketch of proof: take the shape acheiving minimum diameter. By a compactness argument, we could show that if we ``repeatedly'' symmetrize, eventually it is symmetric across every line. 14 | 15 | If some bounded shape is symmetric with respect to reflection about three lines $L_1, L_2,$ and $L_3$, then $L_1,L_2,L_3$ all go through the same point. Why? The center of mass must lie on each line. 16 | This will show us that we have a disk. Let us consider a discrete to this symmetrization process: \defn{compression}. 17 | For $A\in [n]^{(r)}$, set 18 | \[ 19 | R_{ij}(A) = \begin{cases} 20 | (\A\setminus \{j\})\cup \{i\}, & \text{if } j\in A, i\not \in A\\ 21 | A, & \text{otherwise}. 22 | \end{cases} 23 | \] 24 | For example, 25 | \begin{gather*} 26 | R_{15}(\{2,3,5\}) = \{1,2,3\}, \qquad 27 | R_{15}(\{2,3,4\}) = \{2,3,4\},\\ 28 | R_{15}(\{1,3,5\}) = \{1,3,5\}. 29 | \end{gather*} 30 | Let $\tilde{R}_{ij}(\A) = \{R_{ij}(A): A\in \A\} \cup \{A: R_{ij}(A)\in \A \}$. \marginnote{Intuition: $\tilde{R}_{ij}$ applies $R_{ij}$ unless the resulting set is already in $\A$, to prevent collapse.} 31 | 32 | \paragraph{Properties:} 33 | \begin{enumerate} 34 | \item $|\tilde{R}_{ij}(\A)| = |\A|$. 35 | 36 | Let $P_{ij} \subset [n]^{(r)}$ denote the collection of all sets containing $j$ but not $i$. Then $R_{ij}: P_{ij} \to P_{ji}$ is bijection. 37 | \end{enumerate} 38 | 39 | 40 | \noindent We will say $\A$ is \defn{compressed} if $\tilde{R}_{ij}(\A)= \A$ for all $ia_{r-1}>\dotsb> a_1$ then 239 | \[ 240 | i(A) = {a_{r}-1\choose r} + {a_{r-1}-1 \choose r-1} + \dotsb + {a_1-1 \choose 1} + 1. 241 | \] 242 | \end{lemma} 243 | \begin{proof} 244 | ${a_{k} -1\choose k}$ counts sets $B\subset A$ which coincide with $A$ over 245 | \[ 246 | a_r,a_{r-1},\dotsc,a_{k+1} 247 | \] 248 | but whose $k$th largest element is smaller. 249 | \end{proof} 250 | 251 | \begin{lemma} We have that the shadow of the inital segment $I(A)$ is the initial segment of $A\setminus\{\min A\}$. That is, 252 | \[ 253 | \partial I(A) = I( A\setminus \{\min A\}). 254 | \] 255 | \end{lemma} 256 | \begin{proof} 257 | If $A = \{a_r,\dotsc,a_2,a_1\}$, then we wish to show that $B\subset \N^{(r-1)}$ is in $\partial I(A)$ if and only if $B \leq \{a_r,\dotsc,a_2\} = A\setminus \{\min A\}$ in colex order. This is left as an exercise. One way to proceed is by induction on $r$, splitting into cases where $B$ contains $a_r$ or not. 258 | % \add{proof.} 259 | \end{proof} 260 | 261 | \begin{corollary} 262 | If 263 | \[ 264 | i(A) = {a_r-1\choose r} + {a_{r-1} \choose r-1} + \dotsb + {a_1-1\choose 1} + 1, 265 | \] 266 | then 267 | \[ 268 | |\partial (I(A))| = {a_r -1\choose r-1} + {a_{r-1} -1 \choose r-2} + \dotsb + {a_2-1 \choose 1} + 1. 269 | \] 270 | \end{corollary} 271 | 272 | Let us restate \cref{thm:kruskal_katona}: 273 | \begin{theorem*}[Kruskal--Katona] 274 | Let $\A\subset \N^{(r)}$ with 275 | \[ 276 | |\A| = {a_r-1\choose r} + {a_{r-1} -1\choose r-1} + \dotsb + {a_1-1\choose 1} + 1 277 | \] 278 | for some $a_r>\dotsc>a_1$. Then 279 | \[ 280 | |\partial \A| \geq{a_r -1\choose r-1} + {a_{r-1} -1 \choose r-2} + \dotsb + {a_2-1 \choose 1} + 1. 281 | \] 282 | \end{theorem*} 283 | \begin{remark} 284 | For every positive integer $m$, we may write $m$ as 285 | \[ 286 | m={a_r-1\choose r} + {a_{r-1} \choose r-1} + \dotsb + {a_1-1\choose 1} + 1 287 | \] 288 | for some $a_r> a_{r-1} > \dotsm > a_1$. We have shown this already by considering the $m$th element in colex order of $\N^{(r)}$. 289 | \end{remark} 290 | 291 | \begin{proof}[Proof by induction on $r$; for fixed $r$, by induction on $|\A|$.] The base case $r=1$ is trivial. For the induction step, we will assume that $\A$ is compressed, by \cref{lem:compression_decreases_size_of_shadow}. Let 292 | \begin{align*} 293 | A_1 :\!&= \{A\in \A: 1\in A\}, & A_0 :\!&= \A\setminus \A_1. 294 | \end{align*} 295 | 296 | \begin{enumerate}[{Claim} 1:] 297 | \item $|\A_1| \geq |\partial A_0|$.\marginnote{Compression pushes us towards smaller elements, so $\A_1$ should be large.} 298 | 299 | \begin{subproof} 300 | The map $A\mapsto A\cup\{1\}$ is an injection from $\partial \A_0$ to $\A_1$, by compression. 301 | \end{subproof} 302 | \item 303 | \[ 304 | |\A_1|\geq {a_r-2\choose r-1} + {a_{r-1}-2\choose r-2} + \dotsb + {a_2-2\choose 1} + 1. 305 | \] 306 | \begin{subproof}Assume not. Since $|\A_0| =|\A| - |\A_1|$, we have then have 307 | \begin{align*} 308 | |\A_0|&> \left( {a_r-1\choose r} - {a_r-2\choose r-1} \right) + \left( {a_{r-1}-1\choose r-1} - {a_{r-1}-2\choose r-2} \right) + \dotsb + {a_1-1 \choose 1}\\ 309 | &= {a_{r}-2 \choose r} + {a_{r-1}-2\choose r-1} + \dotsb + {a_1-2 \choose 1}+1. 310 | \end{align*} 311 | By the IH\sidenote{Since $\A$ is compressed it must contain 1}, 312 | \[ 313 | |\partial \A_0| \geq {a_r-2\choose r-1} + \dotsb + {a_2-2\choose 1}+1 314 | \] 315 | Then claim 1 yields the result. 316 | \end{subproof} 317 | \end{enumerate} 318 | Let $B = \{A-\{1\}: A\in \A_1\}\subset \N^{(r-1)}$. Note $|B| = |\A_1|$. 319 | \begin{enumerate}[{Claim} 1:] 320 | \setcounter{enumi}{2} 321 | \item 322 | $|\partial \A|\geq |B| + |\partial B|$. 323 | 324 | \begin{subproof} 325 | Note $B\subset \partial \A$. Let $B' = \{B\cup \{1\}: B\in \partial B\} \subset \partial A$. Then $B\cup B' \subset \partial \A$. But since sets in $B'$ contain $1$ and sets in $B$ do not contain $1$, $|B\cup B'| = |B|+|B'|$. 326 | \end{subproof} 327 | \end{enumerate} 328 | By claims 2, 3, and the IH, we have 329 | \begin{align*} 330 | |\partial \A| &\geq |B| + |\partial B| \\ 331 | &\geq {a_r-2 \choose r-1} + {a_{r-1}-2\choose r-2} + \dotsb + {a_2-2\choose 1}+1 \\ 332 | &\qquad + {a_r-2 \choose r-2} + {a_{r-1}-2 \choose r-3} + \dotsb + {a_3-2\choose 1} + 1\\ 333 | &= {a_r-1\choose r-1} + \dotsb {a_3-1\choose 2} + {a_2-1\choose 1}+1. \qedhere 334 | \end{align*} 335 | \end{proof} 336 | \begin{theorem}[\cite{lovaszbook_hyper}] 337 | Let $\A\subset \N^{(r)}$, where $|\A| = {x\choose r}$ for $x\in \R$.\marginnote{We define ${x\choose r} := \frac{x(x-1)\dotsm(x-r+1)}{r!}$. Since the polynomials 338 | ${x\choose r}$ and ${x-1\choose r} + {x-1\choose r-1}$ agree on the integers, they agree everywhere.} 339 | Then $|\partial \A| \geq {x\choose r-1}$.\sidenote{If $x$ is an integer, then equality can be achieved by taking $\A=[x]^{(r)}$.}. 340 | \end{theorem} 341 | \begin{proof} 342 | The proof follows that of the reformulated Kruskal--Katona. Claims 1 \& 3 have the same proof. For claim 2, there is a much simpler proof: 343 | \begin{enumerate}[{Claim }1.]\setcounter{enumi}{1} 344 | % \item $|\partial \A_0|\leq|\A_1|$ 345 | \item $|\A_1|\geq {x-1\choose r-1}$. 346 | \begin{subproof} 347 | If not, $|\A_0| = |\A| - |\A_1| \geq {x\choose r} - {x-1 \choose r-1} = {x-1\choose r}$. Then $|\partial \A_0| \geq {x-1\choose r-1} > |\A_1|$, a contradiction. 348 | \end{subproof} 349 | % \item 350 | \end{enumerate} 351 | The rest of the proof is the same. 352 | \end{proof} 353 | 354 | 355 | \begin{corollary} \label{cor:ell_shadows} 356 | Let $\A\subset \N^{(r)}$ with $|\A| = {x\choose r}$ for $x\in \R$. Let 357 | \[ 358 | \partial^{(\ell)}\A = \{B\in \N^{(r-\ell)}: B\subset A \text{ for some }A\in\A\}. 359 | \] 360 | Then 361 | \[ 362 | |\partial^{(\ell)}A|\geq {x\choose r-\ell}. 363 | \] 364 | \end{corollary} 365 | \begin{proof}[Proof by induction on $\ell$.] The base case is Lor\'asz's theorem. Then 366 | \[ 367 | \partial^{(\ell)}\A = \partial ( \partial^{(\ell-1)}\A) 368 | \] 369 | so the IH and Lor\'asz's theorem yield the result. 370 | \end{proof} 371 | 372 | \begin{corollary} 373 | Let $G$ be a 2-graph. Then \marginnote{$\edges (G)$ denotes the edge set of $G$. We identify $G$ with $\edges (G)$.} 374 | \[ 375 | |G| = |\edges (G)| = {x\choose 2} 376 | \] 377 | for some $x\in \R$. Then $G$ contains at most ${x\choose k}$ complete subgraphs\sidenote{$k$-tuples of verticies pairwise joined by edges.} of size $k$. 378 | \end{corollary} 379 | \begin{remark} 380 | We may reformulate this as follows, in a special case. Let $G$ be a graph with ${n\choose 2}$ edges (but possibly more than $n$ verticies). Then $G$ contains a maximum number of triangles when $G$ is a complete graph of $n$ verticies. We think of this as we are given a budget of edges, and are trying to maximize the number of triangles we make. Here it is intuitive that to do this, we make a complete graph. 381 | \end{remark} 382 | \begin{proof} 383 | We may assume that $x\geq k$; otherwise we may not form any complete $k$-subgraphs. If the number of complete subgraphs of $G$ is strictly larger than ${x\choose k}$, then it is equal to ${x' \choose k}$ for some $x' > x$. So we choose $\A$ to be the family of verticies of complete subgraphs of $G$: 384 | \[ 385 | \A = \{V(H): H \text{ is a complete subgraph of }G\} \subset V(G)^{(k)}. 386 | \] Then each $\{x,y\} \in \partial^{(k-2)} \A$ is a two element subset of a complete subgraph of $G$, so $\partial^{(k-2)} \A\subset \edges(G)$. Then by \cref{cor:ell_shadows}, we must have $|\edges (G)| \geq {x'\choose 2} > {x\choose 2}$, a contradiction. 387 | % \understand 388 | % Everything that lies in the shadow of order $k-2$ must 389 | \end{proof} -------------------------------------------------------------------------------- /chapters/ch7p5_density_VdW.tex: -------------------------------------------------------------------------------- 1 | 2 | \lect{3}{30} 3 | \marginnote{Guest lecture} 4 | % \marginnote{Lecture X: Wednesday March 30, 2016. Guest lecture.} 5 | 6 | \section*{Density version of VdW} 7 | Consider $[0,N]\subset \Z$. We will color these numbers with $m$ colors. At least one color-class will contain more than $\frac{N}{m}$ elements by pigeonhole. Now, is the following density version of VdW theorem true? 8 | 9 | \begin{theorem}[Density VdW] 10 | \label{thm:density_VdW} 11 | For every $k\in \N$ and for every $\delta>0$, there exists $N_0$ such that if $A\subset\{1,\dotsc,N\}$ and $N>N_0$, and $|A|\geq \delta N$, then $A$ contains a $k$-AP (arithmetic progression of length $k$). 12 | \end{theorem} 13 | 14 | 15 | In 1963, Roth proved this theorem for $k=3$ using the ``circle method'' of number theory. Later, Szemeredi proved the statement for arbitrary $k$. His proof, however, gives terrible bounds for $N_0$ and was extremely technical. Next, F\"urstenberg used ergodic theory (developing a whole theory) to prove the result in a clear way, but also has very soft bounds. In 1998, Tim Gowers developed a different proof using higher-order Fourier analysis (giving more reasonable bounds); this won him a fields medal. 16 | 17 | Let $A\subset \{1,\dotsc,N\}$ be chosen randomly: every element is chosen independently with probability $\delta$\marginnote{For each $x\in \{1,\dotsc,N\}$ we include $x$ in $A$ with probability $\delta$ and don't include it with probability $1- \delta$ (say by flipping a biased coin).}. 18 | Then $|A|\approx \delta N$. How many $3$-AP's are in $A$? $\delta^3 \times$ \# of $3$-AP's in $\{1,\dotsc,N\}$. This is then $c \delta^3 N^2$ for some $c>0$. 19 | 20 | So random sets contain many $3$-APs. If $A$ does not have any 3-AP's, then ``it is far from looking random.'' Then we will try to find some structure in $A$ and exploit it. We will come back to this later. 21 | 22 | \paragraph{Triangles in graphs.} How many triangles are in $G(n, \frac{1}{2})$? This is the \erdos-Ramsey (?) random graph, where each edge is decided independently by a fair coin flip. The probability of a triangle existing is $\left( \frac{1}{2} \right)^3 = \frac{1}{8}$, and there are ${n\choose 3}$ possible triangles, so we expect $\frac{1}{8}{n\choose 3}$ triangles. 23 | 24 | The edge density is $\frac{1}{2}$ meaning the number of edges is $\frac{1}{2}{n\choose 2}$. 25 | 26 | If we consider the adjacency matrix of $G(n,\frac{1}{2})$, besides symmetry, it just looks ``gray,'' meaning half filled, half not. 27 | 28 | If we consider the complete bipartite graph on $n$ vertices, we have the same edge density, and the adjacency matrix looks like 29 | \[ 30 | \begin{pmatrix} 31 | 0 & 1 \\ 32 | 1 & 0 33 | \end{pmatrix} 34 | \] 35 | as a block matrix. We can think of each block as uniform, either with density $0$ or $1$, but the whole matrix is certainly not uniform. 36 | 37 | For $G(n,\frac{1}{2})$, the whole matrix is uniform, with density $\frac{1}{2}$. 38 | 39 | We could also consider a graph with adjacency matrix 40 | \[ 41 | \begin{pmatrix} 42 | \text{choose edge with prob. }\frac{2}{3} &\text{choose edge with prob. } \frac{1}{3}\\ 43 | \phantom{\text{choose edge with prob. }}\frac{1}{3} & \phantom{\text{choose edge with prob. }}\frac{2}{3} 44 | \end{pmatrix} 45 | \] 46 | in some division into blocks. This graph would yield a different number of triangles. 47 | \begin{definition} 48 | Bipartite graph on $(A,B)$ is called $\epsilon$-regular if for all $A'\subset A$, for all $B'\subset B$, $|A'|\geq \epsilon |A|$, $|B'|\geq \epsilon B$ implies 49 | \[ 50 | |d(A',B') - d(A,B)| \leq \epsilon, 51 | \] 52 | where $d(X,Y) = \frac{\edges (X,Y)}{|X|\cdot |Y|} $ is the edge density between $X$ and $Y$.\marginnote{$\edges(X,Y)$ denotes the number of edges between vertex sets $X$ and $Y$.} 53 | \end{definition} 54 | \begin{remark} 55 | This means if we restrict to any large enough subsets, the edge density of those subsets is close to the edge density of the whole bipartite graph. 56 | \end{remark} 57 | 58 | \begin{lemma} \label{lem:eregular_yields_triangles} 59 | If $A,B,C$ are vertex sets so that $d(A,B) = a$, $d(A,C)=b$, and $d(B,C)=c$, and $a,b,c\geq 2 \epsilon$, and $(A,B)$, $(A,C)$, $(B,C)$ are $\epsilon$-regular, then $(A,B,C)$ contains at least \[ 60 | (1-2 \epsilon)(a- \epsilon)(b- \epsilon)(c- \epsilon)\cdot |A|\cdot |B|\cdot|C| 61 | \] 62 | triangles. 63 | \end{lemma} 64 | \begin{proof} 65 | First, $A$ has at most $\epsilon |A|$ vertices whose number of neighbors in $B$ is less than $(a- \epsilon)|B|$. Otherwise, take $B'=B$ and $A' = \{v\in A: |N_B(v)| < (a - \epsilon)|B|\}$. If $|A'|\geq \epsilon |A|$, then $d(A',B') \geq a - \epsilon$, contradicting the definition of $A'$. 66 | 67 | In the same way, $A$ has at most $\epsilon|A|$ vertices whose neighbors in $C$ is less than $(b - \epsilon)|C|$. 68 | 69 | Likewise, $B$ has at most $\epsilon |B|$ vertices whose neighbors in $C$ is less than $(c- \epsilon)|C|$. 70 | 71 | \missingfigure{Oval for each of $A,B,C$ far away from each other. Cross out two patches of $A$ corresponding to vertices with few neighbors in $B$ and few in $C$. Connect $A$ to paches in $B$ and $C$. Patch in $B$ has size $(a - \epsilon) |B|$ and pach in $C$ has size at least $(b- \epsilon)|C|$. } 72 | \end{proof} 73 | 74 | 75 | 76 | 77 | 78 | \begin{theorem}[Regularity lemma] \label{thm:regularity_lemma} 79 | For every $\epsilon>0$ there exists $T = T(\epsilon)$ such that the vertices of every graph $G$ can be partitioned into almost equal size parts $V_1,\dotsc, V_T$ such that at least $(1-\epsilon)$ fraction of pairs $(V_i,V_j)$ are $\epsilon$-regular. 80 | \end{theorem} 81 | 82 | \begin{remark} 83 | $T$ does not depend on the number of vertices, only on $\epsilon$. So instead of considering an arbitrarily large graph, we may simply think of a $T\times T$ matrix of pairs $(V_i,V_j)$ such that most pairs are regular. The bad news is $T(\epsilon)\geq 2^{2^{2^{\iddots^2}}}$ where there are approximately $\frac{1}{\epsilon^2}$ $2$'s. This bound can't be substationally improved, as shown by examples. 84 | \end{remark} 85 | 86 | \begin{definition} 87 | A graph is \emph{$\epsilon$-far} from being triangle free if one needs to remove $\geq \epsilon n^2$ edges to make $G$ triangle free. 88 | \end{definition} 89 | \begin{theorem}[Triangle removal lemma] \label{thm:triangle_removal_lemma} 90 | If $G$ is $\epsilon$-far from being triangle-free, then it contains at least $\delta( \epsilon)n^3$ triangles. 91 | \end{theorem} 92 | \begin{proof} 93 | Let $G$ be $\epsilon$-far from being triangle free. Consider an $\frac{\epsilon}{4}$-regularization of $G$ via \cref{thm:regularity_lemma}. We get a partition of $V(G)$ into $T(\epsilon/4)$ parts such that most pairs are $\epsilon/4$-regular. 94 | % \begin{tabular}{c|c|c|c} 95 | % &&&\\ 96 | % \hline\\ 97 | % \hline\\ 98 | % \hline\\ 99 | % \hline 100 | % \end{tabular} 101 | 102 | Let us clean up $G$. Remove all the edges from irregular $(V_i,V_j)$'s. Remove all the edges from sparse\sidenote{where here, sparse means the density is less than $\epsilon/2$.} pairs $(V_i,V_j)$, and remove the edges from the diagonal pairs $(V_i,V_i)$. This is less than $\frac{3}{4}\epsilon n^2$ edges in total. 103 | 104 | Since $G$ was $\epsilon$-far from being triangle free, the cleaned-up version still has triangles. Since we have at least one triangle, there is $V_i,V_j,V_k$ with a triangle\sidenote{Recall, we removed all edges from $(V_i,V_i)$, so each vertex of the triangle is in a different $V_i$.}. But then each pair has density at least $\epsilon/2$, is $\epsilon/4$-regular and \cref{lem:eregular_yields_triangles} yields $ \epsilon^3 \left( \frac{n}{T} \right)^3$ triangles. 105 | \end{proof} 106 | 107 | 108 | 109 | Let $A\subset \{1,\dotsc,N\}$ with density at least $\epsilon$. Let $U,V,W$ be... 110 | 111 | For every $s\in A$ we add a triangle $x\in U$, $x+s \in V$, $x+2s\in W$ for each $x$. Label the edges by $s$. 112 | \missingfigure{3 ovals, $U=\{1,\dotsc,N\},V,W$. Label $x\in U$, connect to $x+s\in V$, $x+2s \in W$ to make a triangle. Label the edges with $s$.} 113 | 114 | These are $N\cdot |A|$ edge-disjoint triangles, since each $(x,s)$ yields a new triangle: any one edge uniquely determines the triangle because it determines $x$ and $s$. We need to remove at least $|A|N \geq \epsilon N^2$ edges to make this graph triangle free. So by \cref{thm:triangle_removal_lemma}, there are $\sim N^3$ triangles, so we can find a new triangle different from any determined by $(x,s)$. 115 | 116 | Then the labels $(s_1,s_2,s_3)$ of this new triangle form a 3-AP in $A$. We can check by the way we constructed these triangles that $s_1+s_3=2s_2$. 117 | 118 | Check: we have an edge $(x,x+s_1)$, $(x+s_1,y+2s')$, and $(y+2s',x)$. Then $s_2 = y+2s'- x - s_1$, and $s_3 = \frac{1}{2}(y+2s'-x)$. So $s_1 + s_2 = y+ 2s' - x = 2s_3$, as we wanted. -------------------------------------------------------------------------------- /chapters/ch9_incidence.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = ../CombinatoricsNotes.tex 2 | 3 | \section{Incidence problems} 4 | \lect{4}{4} 5 | % \marginnote{Lecture X: Monday, April 4, 2016.} 6 | \begin{theorem}[Euler's formula] 7 | Let $G$ be a connected graph drawn in the plane without crossings. Then 8 | \[ 9 | |V(G)| - |\edges (G)| + \reg (G) =2 10 | \] 11 | where $\reg (G)$ is the number of regions in which the drawing divides the plane. 12 | \end{theorem} 13 | 14 | \begin{corollary} \label{cor:bound_edges_of_planar_3VG} 15 | Let $G$ be a graph drawn in the plane without crossings. Then 16 | \[ 17 | |\edges (G) |\leq 3 |V(G)|. 18 | \] 19 | \end{corollary} 20 | \begin{proof} 21 | We assume $|V(G)|\geq 3$. By adding additional edges we can ensure that each region of the drawing has a cycle of length 3 as its boundary. Then 22 | \[ 23 | 3 \reg(G) = 2 |\edges (G)| 24 | \] 25 | by double counting pairs (edge, region) such that each edge belongs to two regions boundary. Substituting into Euler's formula, we have 26 | \[ 27 | |V(G)| - |\edges (G)| + \frac{2}{3}|\edges (G)| = 2 28 | \] 29 | so 30 | \[ 31 | |\edges (G)| = 3 | V(G)| - 6 \leq 3 |V(G)|. 32 | \] 33 | \end{proof} 34 | 35 | Let $\crossings(G)$ denote the minimal number of pairs of crossing edges taken over all drawing of $G$ in the plane where vertices are represented by points, edges by curves joining corresponding points, and the drawing of edges are allowed to intersect, and each intersection is a \emph{crossing}\sidenote{locally looks like an $X$, not two curves bouncing off each other.}. 36 | 37 | Then $\crossings(G)=0$ iff $G$ can be drawn in the plane without crossings, and $\crossings(K_5)=1$, by the drawing 38 | \begin{center} 39 | \begin{tikzpicture} 40 | 41 | \end{tikzpicture} 42 | \end{center} 43 | 44 | 45 | \begin{corollary} \label{cor:bound_crossings_planar_graph93} 46 | $\crossings(G) \geq |\edges (G)| - 3 |V(G)|$. 47 | \end{corollary} 48 | \begin{proof} 49 | Let $G$ be a graph with $\crossings(G) = c$. In a drawing of $G$ with $c$ pairs of crossing edges, remove $c$ edges to obtain a graph drawin without crossings. Then by \cref{cor:bound_edges_of_planar_3VG}, 50 | \[ 51 | |\edges (G)| - c \leq 3 |V(G)| 52 | \] 53 | as we wanted. 54 | \end{proof} 55 | 56 | \begin{theorem}[Crossing number lemma] 57 | \label{thm:crossing_number_lemma} 58 | 59 | \[ 60 | \crossings(G) \geq \frac{1}{64}\frac{m^3}{n^2} 61 | \] 62 | for every graph $G$ with $m$ edges and $n$ vertices such that $m\geq 4n$. 63 | \end{theorem} 64 | \begin{proof} 65 | Let $p \in [0,1]$ which we will choose later. Let $X\subset V(G)$ be obtained by choosing to include each vertex independently at random with probability $p$. Let $G'$ be the random subgraph of $G$ induced by $X$, namely $V(G') = X$, $\edges(G') = \edges(G| X)$. Let $m = |\edges(G)|$, $n= |V(G)|$, $x= \crossings(G)$, $m' = |\edges(G')|$, $n' = |V(G')|$, $x' = \crossings(G')$. Then by construction $\E[n'] = np$. Since each edge of $G$ survives with probability $p^2$, by linearity $\E[m']= mp^2$. Next, $\E[x'] \leq xp^4$ because the probability that a particular pair of crossing edges survives is $p^4$. 66 | 67 | By \cref{cor:bound_crossings_planar_graph93}, $x'\geq m' - 3n'$. Taking expectation values, we have the bound $x p^4 \geq mp^2 - 3np$. That is, 68 | \[ 69 | x\geq \frac{m}{p^2} - 3 \frac{n}{p^3}. 70 | \] 71 | Now, we choose optimal $p$: 72 | \begin{gather*} 73 | \frac{\partial}{\partial p} \left( \frac{m}{p^2}- \frac{3n}{ p^3} \right) = - \frac{2m}{p^3} + \frac{9n}{p^4} = 0 \\ 74 | \implies p=\frac{1}{4.5}\frac{n}{m}. 75 | \end{gather*} 76 | Let us simply take $p = \frac{1}{4}\frac{n}{m}.$ Then 77 | \begin{gather*} 78 | \crossings (G):=x \geq \frac{m}{\frac{16 n^2}{m^2}} - \frac{3n}{\frac{64 n^3}{m^3}}p = \frac{m^3}{n^2} \left( \frac{1}{16} - \frac{3}{64} \right) = \frac{1}{64}\frac{m^3}{n^2}. 79 | \end{gather*} 80 | \end{proof} 81 | 82 | Let $P$ be a set of points in the plane. Let $L$ be a set of lines. Then define 83 | \[ 84 | I(P,L):= |\{ ( p,L ): p\in P, l\in L, p\in L \}|. 85 | \] 86 | Then $I(P,L)$ is the number of \emph{incidences} of $P$ and $L$. Let $I(m,n)$ denote the maximum $I(P,L)$ over sets $P,L$ with $|P| = m$, $|L| = n$. Clearly, $I(m,n)\leq mn$. 87 | \begin{example} We see $I(3,3) \leq 6$ by considering cases: if the three lines are parallel, the number of incidences is at most three. If two lines are parallel, there are at most five incidences. If none of the lines are parallel, there are at most six incidences. On the other hand, \cref{fig:I33} shows $I(3,3)\geq 6$. 88 | \begin{marginfigure} 89 | \begin{center} 90 | \begin{tikzpicture}[scale=.7,rotate=180] 91 | \node (x) at (2,0) {$\bullet$}; 92 | \node (y) at (1,2) {$\bullet$}; 93 | 94 | \node (z) at (0,0) {$\bullet$}; 95 | 96 | % \node (w) at (1,0) {$w$}; 97 | 98 | \draw [shorten >=-1cm,shorten <=-1cm](x) -- (y); 99 | \draw[shorten >=-1cm,shorten <=-1cm] (y) -- (z); 100 | \draw[shorten >=-1cm,shorten <=-1cm] (z) -- (x); 101 | \end{tikzpicture} 102 | \vspace{1em} 103 | \end{center} 104 | \caption{We see $I(3,3)\geq 6$.} \label{fig:I33} 105 | \end{marginfigure} 106 | \end{example} 107 | 108 | 109 | 110 | 111 | \begin{example} 112 | Let $n=4k^3$. Set 113 | \[ 114 | P = \{ (x,y): 0\leq x \leq k-1, 0 \leq y\leq 4k^2 - 1, x,y\in \Z\} 115 | \] 116 | and 117 | \[ 118 | L = \{ y = ax+b: 0\leq a \leq 2k-1, 0\leq b\leq 2k^2-1, a,b\in \Z \} 119 | \] 120 | Then $|P| = |L| =n$. We have $I(P,L)= k \cdot |L| = kn \geq \frac{1}{2}n^{4/3}$ 121 | 122 | every line in $L$ incident with $k$ points in $P$. 123 | 124 | 125 | 126 | \begin{gather*} 127 | (x,ax+b)\in P \\ 128 | \text{for } 0\leq x \leq k-1, 0\leq a \leq 2k-1, 0\leq b \leq 2k^2- 1 129 | \end{gather*} 130 | $ax+b \leq (k-1)(2k-1) + 2k^2 - 1 < 4k^2$. So $I(n,n) \geq c n^{4/3}$ for all $n$ and some $c>0$. 131 | \end{example} 132 | 133 | \begin{theorem}[\cite{Szemeredi-Trotter1983}] \label{thm:bound_I_m_n95} \label{thm:Szemerdi_Trotter} 134 | \[ 135 | I(m,n) \leq 4 m^{2/3} n^{2/3} + 4m + n. 136 | \] 137 | \end{theorem} 138 | 139 | \begin{proof} 140 | Let $P,L$ be sets of points and lines such that $|P| = m$, $|L| = n$ and $I(P,L) = I(n,m)$. Let $G$ be a graph drawn in the plane with crossings such that $V(G) = P$, and $\edges(G)$ are drawn as line segments joining consecutive points on lines in $L$. \marginnote{I.e., follow one line at a time, connecting consecutive points.} 141 | Then 142 | \[ 143 | \crossings(G)\leq n^2 144 | \] 145 | because every crossing in $G$ is an intersection of two lines in $L$ (loose estimate). Then 146 | \[ 147 | I:=I(P,L) = |\edges(G)| + n 148 | \] 149 | if there is a point of $P$ on every line in $L$ (if not, delete that line). We see this by following each line in $L$, and noting between every two incidences, we have an edge: if there are $k$ points in $P$ on a single line, there are $k-1$ edges connecting them. 150 | 151 | Then $|\edges(G)| = I -n$. We have $|\edges(G)|\leq 4 |V(G)|$ if $I - n \leq 4m$, that is $I\leq 4m + n$. 152 | 153 | Otherwise, by \cref{thm:crossing_number_lemma}, 154 | \[ 155 | n^2 \geq \crossings (G) \geq \frac{1}{64} \frac{|\edges(G)|^3}{|V(G)|^2} = \frac{1}{64} \frac{(I-n)^3}{m^2}. 156 | \] 157 | That is, 158 | \begin{gather*} 159 | 64 m^2 n^2 \geq (I-n)^3\\ 160 | 4 m^{2/3} n^{2/3}\geq I- n\\ 161 | I \leq 4m^{2/3}n^{2/3}+n.\qedhere 162 | \end{gather*} 163 | \end{proof} 164 | 165 | \begin{conjecture*}[\cite{Erdos-Szemeredi-1983}] 166 | For every $\epsilon>0$ there exists $c_{\epsilon} > 0$ s.t. for every $A\subset \Z$, 167 | \[ 168 | |A+A| + |A\cdot A| \geq c_{\epsilon} |A|^{2 - \epsilon}, 169 | \] 170 | where $A+A = \{a+b: a,b\in A\}$ and $A\cdot A = \{a\cdot b: a,b\in A\}$. 171 | \end{conjecture*} 172 | 173 | 174 | \begin{theorem}[\cite{elekes1997number}] \label{thm:96_elekes} 175 | There exists $c>0$ such that for every $A\subset \Z$, 176 | \[ 177 | |A+A|\cdot |A\cdot A| \geq c |A|^{5/2}. 178 | \] 179 | In particular, by the arithmetic-geometric inequality, 180 | \[ 181 | |A+A| + |A\cdot A| \geq c |A|^{5/4}. 182 | \] 183 | \end{theorem} 184 | \begin{proof} 185 | Let 186 | \[ 187 | P = \{(a,b):a \in A+A, b\in A\cdot A\}. 188 | \] Then $|P| =|A+A|\cdot |A\cdot A| $.Choose 189 | \[ 190 | L = \{y = a(x-b): a,b\in A\}. 191 | \] 192 | Then $|L| = |A|^2$. Next, 193 | \[ 194 | I(P,L) \geq |A|^3 195 | \] since each line in $L$ contains $|A|$ points in $P$ as follows: $y = a(x-b)$, so choose $x= b+a' \in A+A$ for any $a' \in A$. Then $y = aa' \in A\cdot A$, so $(x,y) \in P$. 196 | 197 | Therefore, by \cref{thm:bound_I_m_n95}, 198 | \begin{align*} 199 | |A|^3 &\leq I(P,L) \leq 4 |A|^{4/3} |A+A|^{2/3}|A\cdot A|^{2/3} + 4|A+A| \cdot |A\cdot A| + |A|^2. 200 | \end{align*} 201 | At least one of these three terms is $\frac{1}{3}$ of the LHS. Easy if it is not the first one. Otherwise: $\frac{1}{3}|A|^3 \leq 4 |A|^{4/3} |A+A|^{2/3} |A\cdot A|^{2/3}$, so 202 | \[ 203 | \left(\frac{1}{12}\right)^{3/2}|A|^{5/2} \leq |A+A| \cdot |A\cdot A|. \qedhere 204 | \] 205 | \end{proof} -------------------------------------------------------------------------------- /graphics/C5p=0.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/graphics/C5p=0.pdf -------------------------------------------------------------------------------- /graphics/C5p=p0.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/graphics/C5p=p0.pdf -------------------------------------------------------------------------------- /graphics/C5p=p25.pdf: 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0.754 l S 76 | 1 g 77 | 320.199 54.039 m 0.199 38.039 l f* 78 | 0 g 79 | 320.199 54.039 m 0.199 38.039 l S 80 | 152.059 139.695 m 152.059 142.098 148.457 142.098 148.457 139.695 c 148.457 81 | 137.297 152.059 137.297 152.059 139.695 c h 82 | 152.059 139.695 m f* 83 | 54.328 292.5 m 54.328 294.898 50.73 294.898 50.73 292.5 c 50.73 290.102 84 | 54.328 290.102 54.328 292.5 c h 85 | 54.328 292.5 m f* 86 | 54.328 292.5 m 54.328 294.898 50.73 294.898 50.73 292.5 c 50.73 290.102 87 | 54.328 290.102 54.328 292.5 c h 88 | 54.328 292.5 m f* 89 | 181.113 76.836 m 181.113 79.238 177.516 79.238 177.516 76.836 c 177.516 90 | 74.438 181.113 74.438 181.113 76.836 c h 91 | 181.113 76.836 m f* 92 | 189.008 47.391 m 189.008 49.793 185.406 49.793 185.406 47.391 c 185.406 93 | 44.992 189.008 44.992 189.008 47.391 c h 94 | 189.008 47.391 m f* 95 | 154.805 100.516 m 154.805 102.914 151.207 102.914 151.207 100.516 c 151.207 96 | 98.117 154.805 98.117 154.805 100.516 c h 97 | 154.805 100.516 m f* 98 | 129.035 70.355 m 129.035 72.754 125.434 72.754 125.434 70.355 c 125.434 99 | 67.953 129.035 67.953 129.035 70.355 c h 100 | 129.035 70.355 m f* 101 | 236.242 49.754 m 236.242 52.152 232.645 52.152 232.645 49.754 c 232.645 102 | 47.352 236.242 47.352 236.242 49.754 c h 103 | 236.242 49.754 m f* 104 | Q Q 105 | showpage 106 | %%Trailer 107 | end restore 108 | %%EOF 109 | -------------------------------------------------------------------------------- /graphics/L1f1.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/graphics/L1f1.pdf -------------------------------------------------------------------------------- /graphics/LOproof.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/graphics/LOproof.pdf 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-------------------------------------------------------------------------------- /graphics/ei-title.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/graphics/ei-title.pdf -------------------------------------------------------------------------------- /graphics/figLOproof.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = ../Combinatorics.tex 2 | \begin{tikzpicture}[darkstyle/.style={circle,draw,fill=gray!40,minimum size=20}] 3 | \foreach \x in {0,...,4} 4 | \foreach \y in {0,...,4} 5 | {\pgfmathtruncatemacro{\label}{\x - 5 * \y +21} 6 | \node [darkstyle] (\x\y) at (1.5*\x,1.5*\y) {\label};} 7 | 8 | \foreach \x in {0,...,4} 9 | \foreach \y [count=\yi] in {0,...,3} 10 | \draw (\x\y)--(\x\yi) (\y\x)--(\yi\x) ; 11 | 12 | \end{tikzpicture} 13 | \begin{tikzpicture}[darkstyle/.style={circle,draw,fill=gray!40,minimum size=20}] 14 | %[xscale=.8,yscale=.75] 15 | 16 | %\draw (-3,3) ellipse (2 and 3); 17 | %\draw (3,3) ellipse (2 and 3); 18 | 19 | % %V side 20 | % \def\vx{-1.5} 21 | % \draw (2.8341+\vx,3.8393) ellipse (1.1649 and 1.3675); 22 | % \node (v2) at (2.6853+\vx+1,5.5201) {$V$}; 23 | 24 | % \draw (2.4996+\vx,4.0144) node[](v4){\textbullet}; 25 | % \draw (2.4996+\vx,4.0144) node [right] {\footnotesize $v_0$}; 26 | 27 | % % \draw (2.4996+\vx,4.0144) ellipse (0.7 and 0.7) ; 28 | % % \node(vr1) at (2.4996+\vx,4.0144) {}; 29 | % % \node(vr2) at (2.4996+\vx,4.0144+.7) {}; 30 | % % \draw (vr1) edge[<->] node[auto]{\footnotesize $r$}(vr2); 31 | 32 | % % \draw (2.4996+\vx,4.0144-.4) node[](v){\textbullet}; 33 | % % \draw (2.4996+\vx,4.0144-.4) node [right] {\footnotesize $v$}; 34 | 35 | % %U side 36 | % \draw (-3.1,3.6) node (v5) {}ellipse (1.8533 and 2.5917); 37 | % \node (v1) at (-3.3313,6.4532+.3) {$U$}; 38 | 39 | % \draw[opacity=.5] (v5) ellipse (1.6 and 1.5); 40 | % \node[opacity=.5] at (-3.0112,1.728) {$\Omega$}; 41 | 42 | % \draw (-3.616,3.5459) node[](v3){\textbullet}; 43 | % \draw (-3.616,3.5459) node [right] {\footnotesize $u_0$}; 44 | 45 | % % \draw (-3.616,3.5459) ellipse (1 and 1); 46 | % % \node(vCr1) at (-3.616,3.5459) {}; 47 | % % \node(vCr2) at (-3.616,3.5459+1) {}; 48 | % % \draw (vCr1) edge[->,{<->}] node[auto]{\footnotesize $Cr$}(vCr2); 49 | 50 | % % \draw (-3.616-.5,3.5459-.4) node[] (pv){\textbullet}; 51 | % % \draw (-3.616-.5,3.5459-.4) node [right] {\footnotesize $wr+u_0$}; 52 | 53 | % %Bridge 54 | % \draw (v3) edge[bend left,<-,dashed, shorten >=-3] node[pos=.65, auto]{$T$}(v4); 55 | 56 | \def\slope{.5} 57 | 58 | % \foreach \x in {0,...,4} 59 | % \foreach \y in {0,...,4} 60 | % \node at (1.5*\x,1.5*\y) {}; 61 | 62 | \node (s1) at (0,0) {}; 63 | \node (e1) at (2,1) {}; 64 | \draw (s1) edge[->] node[auto] {$x_n$} (e1); 65 | 66 | 67 | % \draw (v) edge[bend left,->,dashed, shorten >= 2, shorten <= -3] node[near start, auto]{$T$}(pv); 68 | 69 | \end{tikzpicture} -------------------------------------------------------------------------------- /graphics/figLOproofPDF.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/graphics/figLOproofPDF.pdf -------------------------------------------------------------------------------- /graphics/figLOproofv2.tex: -------------------------------------------------------------------------------- 1 | %!TEX root = ../Combinatorics.tex 2 | 3 | \newcommand{\makearrow}[2]{ 4 | \begin{scope} 5 | \coordinate (A) at (#1); 6 | \coordinate (B) at ([xshift=+120pt,yshift=+60pt]A); 7 | \draw (A) edge[->] (B); 8 | \end{scope} 9 | } 10 | 11 | \newdimen\XCoord 12 | \newdimen\YCoord 13 | 14 | \newcommand*{\ExtractCoordinate}[1]{\path (#1); \pgfgetlastxy{\XCoord}{\YCoord};}% 15 | \newcommand*{\LabelCurrentCoordinate}[2]{\fill [#1] ($(\XCoord,\YCoord)$) circle (2pt) node [right] {#2}}% 16 | 17 | \pgfmathsetseed{3} 18 | 19 | \begin{center} 20 | \begin{tikzpicture}[scale=.25] 21 | \def\scale{.1}; 22 | \def\x{2}; 23 | \def\y{5}; 24 | 25 | \coordinate (A) at (\x,\y); 26 | \coordinate (B) at ([xshift=+120pt,yshift=+60pt]A); 27 | \draw (A) edge[->,Red] node[auto=right, pos=-0.1]{{\tiny $X_{A_k}$}} node[ auto=right, near end]{{\tiny $X_{A_k\cup\{n\}}$}} (B); 28 | 29 | \coordinate (C) at ($ (A) + (4,4) $); 30 | \coordinate (D) at ([xshift=+120pt,yshift=+60pt]C); 31 | 32 | \draw (C) edge[->,Blue] node[auto, near start]{{\tiny $X_{A_1}$}} node[ auto=right, near end]{{\tiny $X_{A_1\cup\{n\}}$}} (D); 33 | 34 | 35 | 36 | \dimline[label style={above=0.5ex},extension start length=.24,extension end length=.24]{($ (A) + .24*(-4,4) $)}{($ (C) + .24*(-4,4) $)}{{\tiny $>1$}}; 37 | 38 | \coordinate (new) at ($ (C) - (B) $); 39 | 40 | \ExtractCoordinate{$(new)$}; 41 | 42 | \coordinate (newer) at ($(\YCoord,-\XCoord)$); 43 | %(\YCord, -\XCord) 44 | 45 | 46 | \dimline[label style={ above=.2ex,right, rotate=90},extension start length=-.6,extension end length=-.6]{($ (B) + 3*(newer) $)}{($ (C) + 3*(newer) $)}{{\tiny $<1$}}; 47 | 48 | 49 | % \end{scope} 50 | 51 | % \begin{scope}[scale=\scale] 52 | % \coordinate (C) at $(\x + 10, \y)$; 53 | % \coordinate (D) at ([xshift=+120pt,yshift=+60pt]C); 54 | % \draw (C) edge[->] (D); 55 | % \end{scope} 56 | 57 | \foreach \x in {1,2,...,5}{ 58 | \makearrow{rand*5+4,rand*5-2.5}{\scale};} 59 | 60 | \makearrow{3,12}{\scale}; 61 | \end{tikzpicture} 62 | \end{center} -------------------------------------------------------------------------------- /graphics/helix.asy: -------------------------------------------------------------------------------- 1 | import graph3; 2 | 3 | size(0,200); 4 | 5 | usepackage("palatino"); 6 | usepackage("mathpazo"); 7 | 8 | currentprojection=orthographic(4,6,3); 9 | 10 | real x(real t) {return cos(2pi*t);} 11 | real y(real t) {return sin(2pi*t);} 12 | real z(real t) {return t;} 13 | 14 | defaultpen(overwrite(SuppressQuiet)); 15 | 16 | path3 p=graph(x,y,z,0,2.7,operator ..); 17 | bbox3 b=autolimits(min(p),max(p)); 18 | aspect(b,1,1,1); 19 | 20 | xaxis(rotate(X)*"$x$",all=true,b,red,RightTicks(rotate(X)*Label,2,2)); 21 | yaxis(rotate(Y)*"$y$",all=true,b,red,RightTicks(rotate(Y)*Label,2,2)); 22 | zaxis("$z$",all=true,b,red,RightTicks); 23 | 24 | draw(p,Arrow); 25 | 26 | -------------------------------------------------------------------------------- /graphics/helix.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/graphics/helix.pdf -------------------------------------------------------------------------------- /graphics/hilbertcurves.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/graphics/hilbertcurves.pdf -------------------------------------------------------------------------------- /graphics/hilbertrecursive.tex: -------------------------------------------------------------------------------- 1 | % Draw Hilbert curves. 2 | % Credits: Based on code by Marc van Dongen 3 | % See: http://www.fauskes.net/pgftikzexamples/hilbert-curve/ 4 | \title{Hilbert Curves} 5 | \author{} 6 | \date{} 7 | 8 | \documentclass{tufte-handout} 9 | 10 | \usepackage{tikz} 11 | 12 | \usepackage[active,tightpage]{preview} 13 | \PreviewEnvironment{tikzpicture} 14 | \usetikzlibrary{positioning} 15 | 16 | \begin{document} 17 | 18 | \newdimen\HilbertLastX 19 | \newdimen\HilbertLastY 20 | \newcounter{HilbertOrder} 21 | 22 | \def\DrawToNext#1#2{% 23 | \advance \HilbertLastX by #1 24 | \advance \HilbertLastY by #2 25 | \pgfpathlineto{\pgfqpoint{\HilbertLastX}{\HilbertLastY}} 26 | % Alternative implementation using plot streams: 27 | % \pgfplotstreampoint{\pgfqpoint{\HilbertLastX}{\HilbertLastY}} 28 | } 29 | 30 | % \Hilbert[right_x,right_y,left_x,left_x,up_x,up_y,down_x,down_y] 31 | \def\Hilbert[#1,#2,#3,#4,#5,#6,#7,#8] { 32 | \ifnum\value{HilbertOrder} > 0% 33 | \addtocounter{HilbertOrder}{-1} 34 | \Hilbert[#5,#6,#7,#8,#1,#2,#3,#4] 35 | \DrawToNext {#1} {#2} 36 | \Hilbert[#1,#2,#3,#4,#5,#6,#7,#8] 37 | \DrawToNext {#5} {#6} 38 | \Hilbert[#1,#2,#3,#4,#5,#6,#7,#8] 39 | \DrawToNext {#3} {#4} 40 | \Hilbert[#7,#8,#5,#6,#3,#4,#1,#2] 41 | \addtocounter{HilbertOrder}{1} 42 | \fi 43 | } 44 | 45 | 46 | % \hilbert((x,y),order) 47 | \def\hilbert((#1,#2),#3){% 48 | \advance \HilbertLastX by #1 49 | \advance \HilbertLastY by #2 50 | \pgfpathmoveto{\pgfqpoint{\HilbertLastX}{\HilbertLastY}} 51 | % Alternative implementation using plot streams: 52 | % \pgfplothandlerlineto 53 | % \pgfplotstreamstart 54 | % \pgfplotstreampoint{\pgfqpoint{\HilbertLastX}{\HilbertLastY}} 55 | \setcounter{HilbertOrder}{#3} 56 | \Hilbert[1mm,0mm,-1mm,0mm,0mm,1mm,0mm,-1mm] 57 | \pgfusepath{stroke}% 58 | } 59 | 60 | \def\scalefac{1} 61 | \maketitle 62 | 63 | \begin{figure}[h!] 64 | \begin{tikzpicture}[node distance=0cm,every path/.style={thin}] 65 | \node[label={above:$n=1$}] (n1) {\tikz[scale=\scalefac*18] \hilbert((0mm,0mm),1);}; 66 | \node[label={above:$n=2$},right=of n1] (n2) {\tikz[scale=\scalefac*6] \hilbert((0mm,0mm),2);}; 67 | \node[label={above:$n=3$},right=of n2] (n3) {\tikz[scale=\scalefac*2.6] \hilbert((0mm,0mm),3);}; 68 | \node[label={above:$n=4$},right=of n3] (n4) {\tikz[scale=\scalefac*1.2] \hilbert((0mm,0mm),4);}; 69 | \node[label={above:$n=5$},right=of n4] (n5) {\tikz[scale=\scalefac*0.58] \hilbert((0mm,0mm),5);}; 70 | \end{tikzpicture} 71 | \end{figure} 72 | \end{document} 73 | -------------------------------------------------------------------------------- /graphics/nasa_vision_sm.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/graphics/nasa_vision_sm.png -------------------------------------------------------------------------------- /graphics/satir_graph.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/graphics/satir_graph.png -------------------------------------------------------------------------------- /graphics/sine.asy: -------------------------------------------------------------------------------- 1 | import graph; 2 | 3 | usepackage("palatino"); 4 | usepackage("mathpazo"); 5 | 6 | size(7inches,3inches); 7 | scale(true); 8 | 9 | real f(real x) { return sin(x); } 10 | pair F(real x) { return (x, f(x)); } 11 | 12 | xaxis("$x$"); 13 | yaxis("$y$"); 14 | 15 | draw(graph(f,-10.,10,operator ..), red); 16 | 17 | -------------------------------------------------------------------------------- /graphics/sine.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/graphics/sine.pdf -------------------------------------------------------------------------------- /graphics/standalone_fig.tex: -------------------------------------------------------------------------------- 1 | \documentclass{standalone} 2 | 3 | \usepackage{tikz} 4 | \usepackage{graphicx} % allow embedded images 5 | 6 | \usepackage{pgfplots} 7 | \pgfplotsset{compat=newest} 8 | \usetikzlibrary{calc,decorations,decorations.pathreplacing} 9 | \usepackage{tikz-dimline} 10 | 11 | \usepackage{optparams} 12 | \usepackage{pgfplots} 13 | \pgfplotsset{compat=newest} 14 | \usetikzlibrary{calc,decorations,decorations.pathreplacing} 15 | \usepackage{tikz-dimline} 16 | 17 | \renewcommand\normalsize{% 18 | \@setfontsize\normalsize\@xpt{14}% 19 | \abovedisplayskip 10\p@ \@plus2\p@ \@minus5\p@ 20 | \abovedisplayshortskip \z@ \@plus3\p@ 21 | \belowdisplayshortskip 6\p@ \@plus3\p@ \@minus3\p@ 22 | \belowdisplayskip \abovedisplayskip 23 | \let\@listi\@listI} 24 | \normalbaselineskip=14pt 25 | \normalsize 26 | \renewcommand\small{% 27 | \@setfontsize\small\@ixpt{12}% 28 | \abovedisplayskip 8.5\p@ \@plus3\p@ \@minus4\p@ 29 | \abovedisplayshortskip \z@ \@plus2\p@ 30 | \belowdisplayshortskip 4\p@ \@plus2\p@ \@minus2\p@ 31 | \def\@listi{\leftmargin\leftmargini 32 | \topsep 4\p@ \@plus2\p@ \@minus2\p@ 33 | \parsep 2\p@ \@plus\p@ \@minus\p@ 34 | \itemsep \parsep}% 35 | \belowdisplayskip \abovedisplayskip 36 | } 37 | \renewcommand\footnotesize{% 38 | \@setfontsize\footnotesize\@viiipt{10}% 39 | \abovedisplayskip 6\p@ \@plus2\p@ \@minus4\p@ 40 | \abovedisplayshortskip \z@ \@plus\p@ 41 | \belowdisplayshortskip 3\p@ \@plus\p@ \@minus2\p@ 42 | \def\@listi{\leftmargin\leftmargini 43 | \topsep 3\p@ \@plus\p@ \@minus\p@ 44 | \parsep 2\p@ \@plus\p@ \@minus\p@ 45 | \itemsep \parsep}% 46 | \belowdisplayskip \abovedisplayskip 47 | } 48 | \renewcommand\scriptsize{\@setfontsize\scriptsize\@viipt\@viiipt} 49 | \renewcommand\tiny{\@setfontsize\tiny\@vpt\@vipt} 50 | \renewcommand\large{\@setfontsize\large\@xipt{15}} 51 | \renewcommand\Large{\@setfontsize\Large\@xiipt{16}} 52 | \renewcommand\LARGE{\@setfontsize\LARGE\@xivpt{18}} 53 | \renewcommand\huge{\@setfontsize\huge\@xxpt{30}} 54 | \renewcommand\Huge{\@setfontsize\Huge{24}{36}} 55 | 56 | \begin{document} 57 | \input{figLOproofv2} 58 | \end{document} -------------------------------------------------------------------------------- /graphics/vdqi-contents.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/ericphanson/CombinatoricsNotes/6b369a77b77cf6f0281b59f227aaa31e6903079c/graphics/vdqi-contents.pdf 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-------------------------------------------------------------------------------- /tufte-book.cls: -------------------------------------------------------------------------------- 1 | \NeedsTeXFormat{LaTeX2e}[1994/06/01] 2 | 3 | \ProvidesClass{tufte-book}[2015/06/30 v3.5.3 Tufte-book class] 4 | 5 | %% 6 | % Declare we're tufte-book 7 | \newcommand{\@tufte@class}{book}% the base LaTeX class (defaults to the article/handout style) 8 | \newcommand{\@tufte@pkgname}{tufte-book}% the name of the package (defaults to tufte-handout) 9 | 10 | %% 11 | % Load the common style elements 12 | \input{tufte-common.def} 13 | 14 | 15 | %% 16 | % Set up any book-specific stuff now 17 | 18 | %% 19 | % The front matter in Tufte's /Beautiful Evidence/ contains everything up 20 | % to the opening page of Chapter 1. The running heads, when they appear, 21 | % contain only the (arabic) page number in the outside corner. 22 | %\newif\if@mainmatter \@mainmattertrue 23 | \renewcommand\frontmatter{% 24 | \if@openright% 25 | \cleardoublepage% 26 | \else% 27 | \clearpage% 28 | \fi% 29 | \@mainmatterfalse% 30 | \pagenumbering{arabic}% 31 | %\pagestyle{plain}% 32 | \fancyhf{}% 33 | \ifthenelse{\boolean{@tufte@twoside}}% 34 | {\fancyhead[LE,RO]{\thepage}}% 35 | {\fancyhead[RE,RO]{\thepage}}% 36 | } 37 | 38 | 39 | %% 40 | % The main matter in Tufte's /Beautiful Evidence/ doesn't restart the page 41 | % numbering---it continues where it left off in the front matter. 42 | \renewcommand\mainmatter{% 43 | \if@openright% 44 | \cleardoublepage% 45 | \else% 46 | \clearpage% 47 | \fi% 48 | \@mainmattertrue% 49 | \fancyhf{}% 50 | \ifthenelse{\boolean{@tufte@twoside}}% 51 | {% two-side 52 | \renewcommand{\chaptermark}[1]{\markboth{##1}{}}% 53 | \fancyhead[LE]{\thepage\quad\smallcaps{\newlinetospace{\plaintitle}}}% book title 54 | \fancyhead[RO]{\smallcaps{\newlinetospace{\leftmark}}\quad\thepage}% chapter title 55 | }% 56 | {% one-side 57 | \fancyhead[RE,RO]{\smallcaps{\newlinetospace{\plaintitle}}\quad\thepage}% book title 58 | }% 59 | } 60 | 61 | 62 | %% 63 | % The back matter contains appendices, indices, glossaries, endnotes, 64 | % biliographies, list of contributors, illustration credits, etc. 65 | \renewcommand\backmatter{% 66 | \if@openright% 67 | \cleardoublepage% 68 | \else% 69 | \clearpage% 70 | \fi% 71 | \@mainmatterfalse% 72 | } 73 | 74 | %% 75 | % Only show the chapter titles in the table of contents 76 | \setcounter{tocdepth}{0} 77 | 78 | %% 79 | % If there is a `tufte-book-local.sty' file, load it. 80 | 81 | \IfFileExists{tufte-book-local.tex}{% 82 | \@tufte@info@noline{Loading tufte-book-local.tex}% 83 | \input{tufte-book-local}% 84 | }{} 85 | 86 | %% 87 | % End of file 88 | \endinput 89 | -------------------------------------------------------------------------------- /tufte-handout.cls: -------------------------------------------------------------------------------- 1 | \NeedsTeXFormat{LaTeX2e}[1994/06/01] 2 | 3 | \ProvidesClass{tufte-handout}[2015/06/21 v3.5.2 Tufte-handout class] 4 | 5 | %% 6 | % Declare we're tufte-handout 7 | \newcommand{\@tufte@class}{article}% the base LaTeX class (defaults to the article/handout style) 8 | \newcommand{\@tufte@pkgname}{tufte-handout}% the name of the package (defaults to tufte-handout) 9 | 10 | %% 11 | % Load the common style elements 12 | \input{tufte-common.def} 13 | 14 | 15 | %% 16 | % Set up any handout-specific stuff now 17 | 18 | %% 19 | % Abstract 20 | 21 | % TODO The abstract should be printed on its own page of the `titlepage' 22 | % option was specified. 23 | \renewenvironment{abstract} 24 | {\begin{quotation}\if@tufte@sfsidenotes\begin{sffamily}\fi} 25 | {\if@tufte@sfsidenotes\end{sffamily}\fi\end{quotation}} 26 | 27 | 28 | %% 29 | % If there is a `tufte-handout-local.tex' file, load it. 30 | 31 | \IfFileExists{tufte-handout-local.tex}{% 32 | \input{tufte-handout-local}% 33 | \@tufte@info@noline{Loading tufte-handout-local.tex}% 34 | }{} 35 | 36 | 37 | %% 38 | % End of file 39 | \endinput 40 | --------------------------------------------------------------------------------