
%Communication Complexity Scribe Notes template

\ifx\CompleteCourse\relax
\ClassScribeSetupA
\else
\documentclass[11pt]{article}
\usepackage{url}
\usepackage{amsmath}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{homework}{Homework}
\newtheorem{case}{Case}
\newenvironment{definition}{\begin{trivlist}\item[]{\bf Definition}\ }%
{\end{trivlist}}
\newenvironment{fact}{\begin{trivlist}\item[]{\bf Fact}\ }%
{\end{trivlist}}
\newenvironment{example}{\begin{trivlist}\item[]{\bf Example}\ }%
{\end{trivlist}}
\newenvironment{proof}{\begin{trivlist}\item[]{\bf Proof}\ }%
{\end{trivlist}}
% Make the page large
\addtolength{\textwidth}{1.50in}
\addtolength{\textheight}{1.00in}
\addtolength{\evensidemargin}{-0.75in}
\addtolength{\oddsidemargin}{-0.75in}
\addtolength{\topmargin}{-.50in}

% \vdashsub{X}  makes a turnstyle with subscript "X"
% \vdashsup{X}  makes a turnstyle with superscript "X"

\newdimen\srbdimenA
\newcommand{\vdashsupsub}[2]{ \mathop{
  \setbox251 = \hbox{$\scriptstyle #1$}
  \setbox252 = \hbox{$\scriptstyle #2$}
  \ifdim \wd251<\wd252 \srbdimenA = \wd252 \else \srbdimenA = \wd251 \fi
  \setbox255 = \hbox {${\srbAvdash \vphantom( \kern -\srbdimenA \kern +.05em}
                  ^{\hbox to\srbdimenA{\hfill \box251\hfill}}
                  _{\hbox to\srbdimenA{\hfill \box252\hfill}}$}
  \box255 \kern .05em}}
\newcommand{\srbAvdash}{\hbox{ \vrule height1.4ex width0.02em
                                \dimen255 = \srbdimenA
                                \advance\dimen255 by 0.1em
                              \vbox{\hrule width\srbdimenA height0.02em
                                    \kern .65ex  }}}
\newcommand{\vdashsup}[1]{\vdashsupsub{{#1}}{\mbox{~}}}
\newcommand{\vdashsub}[1]{\vdashsupsub{\mbox{~}}{#1}}
\fi

%   FOR THE SCRIBE: CUSTOMIZE THE ENTRIES BELOW:
%   Fill in the following information particular to these scribe notes:
\usepackage{amsmath}
\usepackage{amssymb}
\newcommand{\disc}{\mathrm{disc}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\cv}[1]{\mathbf{1}_{#1}}
\newcommand{\abs}[1]{\left\lvert#1\right\rvert}
\newcommand{\norm}[1]{\left\lVert#1\right\rVert}
\newcommand{\p}[1]{\left(#1\right)}
\newcommand{\bra}[1]{\left[#1\right]}
\newcommand{\T}[1]{{#1}^{\mathrm{T}}}
\newcommand{\IP}{\mathrm{IP}}
\newcommand{\sdeg}{\mathrm{sign\text{-}deg}}
\newcommand{\edeg}[1][\epsilon]{#1\mathrm{\text{-}deg}}
\newcommand{\DTC}{\mathrm{DTC}}
\newcommand{\qedd}{\hspace*{\fill} \square}
\DeclareMathOperator*{\Exp}{\mathbb{E}}

\def\scribeone{Jaiganesh Balasundaram and Tyrone Strangway}    % Who is the scribe?
\def\classdate{Fall, 2014} % Date of the class
\def\classnumber{9}     % Is this the first, second, ...?

% Here are some commands that stay the same for the whole class.

\def\classinstructor{Jaiganesh Balasundaram and Tyrone Strangway}
\def\classtitle{Applications of Communication Complexity for Game Theory}
\def\doctitle{\textup{CSC 2429 - Comm Complexity: Applications and New Directions}}
\def\classid{\textup{Lecture \#\classnumber: \classdate}}

%   Put your macros for these scribe notes HERE
%  It is best to use as few as possible.
%  environments for "theorem", "corollary", "lemma", "fact" "definition"
%   "homework", "proof", "example", "proposition"
%   are already defined above.

% Start the document

\ifx\CompleteCourse\relax
\ClassScribeSetupB
\else
\def\makeatletter{\catcode`\@=11\relax}
\def\makeatother{\catcode`\@=12\relax}
\makeatletter
\def\ps@scribeheadings{\let\@mkboth\@gobbletwo
\def\@oddhead{\sl\doctitle \hfill \classid
}\def\@oddfoot{\hfil \rm \thepage \hfil}\def\@evenhead{\@oddhead}%
\def\@evenfoot{\@oddfoot}\def\sectionmark##1{}\def\subsectionmark##1{}}
\makeatother
\pagestyle{scribeheadings}
\begin{document}
\bibliographystyle{siam}
\fi

\begin{center}
\Large\bf\classtitle\\[1em]
%\Large\bf\classid\\[1em]
{\large\bf Lecturers: \classinstructor}\\[.5em]
%{\large\bf Scribe Notes by: \scribeone}
\end{center}

\vspace*{.4in}

This document is divided into two sections. The first section deals with the communication complexity of reaching Nash equilibrium. The second section focuses on the applications of communication complexity to algorithmic mechanism design.

\section{Nash Equilibrium}

\subsection{Introduction}
\subsubsection{An example of a 2 player game}
Imagine the following scenario. Two prisoners, prisoner $A$ and prisoner $B$, who are in the prison for the same crime, are being interrogated simultaneously in separate rooms. They are each given 2 options: $(a)$ Either cooperate with the other prisoner by staying silent or $(b)$ Betray the other prisoner by snitching him out. They are each given deals (reduced prison sentences) for betraying the other prisoner. However, if it turns out that both of them had betrayed each other, then they will both serve longer sentences than if they had both stayed silent. So, there are four possible scenarios: $(a)$ Both stay silent $(b)$ Both betray each other $(c)$ Prisoner $A$ remains silent and prisoner $B$ betrays and $(d)$ Prisoner $B$ remains silent and prisoner $A$ betrays. Below is a matrix representing the ``payoffs" for each prisoner for the four different scenarios (By ``payoffs", we mean the quantification of preferences for different scenarios, depending on the prison sentences they each get for each scenario).

\begin{table}[h]
\centering
\begin{tabular}{|l|l|l|}
\hline
\textbf{}          & \textbf{Cooperate} & \textbf{Betray} \\ \hline
\textbf{Cooperate} & \textbf{3,3}       & \textbf{0,5}    \\ \hline
\textbf{Betray}    & \textbf{5,0}       &  \textbf{2,2}    \\ \hline
\end{tabular}
\end{table}

\noindent The first entries of the matrix represent the payoffs for prisoner $A$, and the second entries of the matrix represent the payoffs for prisoner $B$. Lower numbers should be interpreted as higher prison sentences and higher numbers as reduced prison sentences. Now, both prisoners are better off by both staying silent, than by both betraying each other (because in the first case both get a payoff of 3, whereas in the latter case both get a payoff of 2).
%However, consider this. For prisoner $A$, assuming that prisoner $B$ is going to cooperate, the best course of action is \textit{not} to cooperate, but rather to betray, since betraying prisoner $B$ (assuming $B$ is going to cooperate) gives him a better payoff than remaining silent. A similar argument can be made for prisoner $B$. However, for prisoner $A$, assuming that prisoner $B$ is going to betray, the best course of action is to betray prisoner $B$, since betraying prisoner $B$ (assuming $B$ is going to betray) gives him a better payoff than remaining silent. 
However, the scenario where both prisoners betray each other is called the \textit{equilibrium} point for this 2 person game because of the following reason. In this scenario, both of them have no incentive to deviate from their actions because, for each prisoner, deviating from his action while the other person doesn't (that is, choosing to cooperate while the other person still betrays) puts him in a worse position. It is easy to see that the same cannot be said for the remaining three scenarios. This is actually a quite famous example called the ``Prisoners' Dilemma". The two prisoners are the \textit{players} of this game, where each player has 2 possible \textit{actions} or \textit{strategies} forming their action set: \{cooperate, betray\}. Each of the 4 possible scenario is termed as a \textit{joint action}. The entries of the matrix form the \textit{payoff or utility function} for each player, which as we mentioned earlier, is nothing but the quantification of their preferences for the different possible scenarios. Finally, the joint action (betray,betray), which we described above as the \textit{equilibrium} point of this game, is formally termed the ``Nash equilibrium" of the game. We will be formalizing these notions in the next section. \\

\subsubsection{The Problem}
In this paper, we will be discussing the problem of complexity of \textit{reaching} equilibrium. More specifically, we will be looking at the communication model (which we will be formally defining in the next section), in which players initially only know their own payoff functions \footnote{This is usually called the ``uncoupled" setting.}, and analyse how much information must be transferred between them to jointly compute the equilibrium point. We will also assume that the players follow a predetermined protocol, that is, we abstract away the incentives of each player and are only concerned about the amount of information that needs to be transferred between them to attain equilibrium. The motivation for this model has been well studied \cite{CS04}. \\

\noindent We will be discussing results from two papers. First, we will be discussing results from the paper ``How long to equilibrium? The communication complexity of uncoupled equilibrium procedures" by Hart and Mansour \cite{HM06}, where will be looking at the communication complexity lower bounds for reaching \textit{pure} Nash equilibrium. Then, we will be discussing results from the paper ``On the communication complexity of approximate Nash equilibria" by Goldberg and Pastink \cite{GP12}, where we will be discussing the communication complexity results for \textit{mixed} Nash equilibrium.\\      

\noindent Now, we present some of the results regarding reaching pure Nash equilibrium. We will be using the notations and definitions of \cite{HM06}.
\subsection{The Setting for Pure Action Games}
There are $n \geq 2$ players. Each player $i$, $1 \leq i \leq n$, has a finite set of actions (or strategies) $A_i$, with $|A_i| \geq 2$. For the analysis of pure action games, we will be only considering binary action games, that is, for each $i$, $A_i=\{0,1\}$. Let $A=\prod_{i=1}^{n}A_i$ denote the joint action space (for binary action games, $A=\{0,1\}^n$, the set of all $n-$ bit strings). Each player has a private payoff (or utility) function $u_i: A \rightarrow \{0,1\}^n$ (For simplicity we are assuming that the utilities are finitely represented, that is, are rational numbers with $n-$ bit numerator and denominator). Let's denote the game by $G=(n,\{A_i\},\{u_i\})$.\\

\noindent For a joint action $a=(a_1,...,a_n) \in A$ (for binary action games, a joint action is an $n-$ bit binary string), let $a^{-i}$ denote the joint action of all players except player $i$. For each player $i$ (with utility function $u_i$), the \textit{best response} to a joint action
 $a^{-i}$, denoted by $BR(a^{-i},u_i)$ is $arg\,max_{a_i \in A_i}u_i(a_i,a^{-i})$. A joint action $a \in A$ is a \textbf{Pure Nash Equilibrium} if $u_i(a) \geq u_i(b_i,a^{-i})$ for \textit{every} player $i$ and any action $b_i \in A_i$ (equivalently, $a_i \in BR(a^{-i},u_i)$ for every $i$).


%\begin{example}
%We give another example of a 2 player game, called \textit{Matching Pennies}. This game is the binary action equivalent of the popular ``Rock, Paper and Scissors" game. There are two players $A$ and $B$. Each has a penny and they secretly turn it to either heads or tails. Then both players reveal their pennies simultaneously. If both pennies have the same side up (either both are heads or both are tails), then player $A$ wins and gets the penny of player $B$. If the pennies have opposite sides up (either head-tail or tail-head), then player $B$ wins and gets the penny of player $A$. This is a binary action game since each player has 2 actions: \{Head,Tail\}. Below is a payoff matrix for this game. 
%
%
%\begin{table}[h]
%\begin{tabular}{|l|l|l|}
%\hline
%\textbf{}          & \textbf{Head} & \textbf{Tail} \\ \hline
%\textbf{Head} & \textbf{1,-1}       & \textbf{-1,1}    \\ \hline
%\textbf{Tail}    & \textbf{-1,1}       & \textbf{1,-1}    \\ \hline
%\end{tabular}
%\end{table}
%
%\noindent The first entries of the matrix are the payoff of player $A$ for the four possible joint actions, and the second entries of the matrix are the payoff of player $B$ for the four possible joint actions. It is interesting to note that this game has no pure Nash Equilibrium (easy to see), but has a unique ``mixed" Nash equilibrium, where the players, instead of playing a specific action, play each of their actions with a probability $1/2$ (so they each actually choose a probability distribution on their action set). We will see more of mixed Nash equilibria in the later part of the notes. 
%
%\end{example} 
 
\subsection{The Communication Model: Pure Nash Equilibrium Procedures}
A Pure Nash Equilibrium Procedure is a dynamic process by which the players reach a Pure Nash Equilibrium of the game (if one exists). Recall that a game is denoted by $G=(n,\{A_i\},\{u_i\})$. Fix $n$ and the action spaces, so a game is identified by its payoff function set $(u_1,...,u_n)$. Let $\mathcal{G}$ be a family of games to which the procedure should apply. The input to the procedure is a game $G=(u_1,...,u_n)$ in the family $\mathcal{G}$. Initially, each player $i$ has access only to his own payoff function $u_i$. In each round $t=1,2,...,$ every player outputs a bit (say writes the bit on a blackboard). This bit is a function of the the player's payoff function and what he has seen so far in the procedure. At the end of round $t$, all the players observe the joint output.\\

\noindent In a Pure Nash Equilibrium Procedure $\Pi$ for a game $G$, the ``output" of player $i$ is either
\begin{itemize}
\item a pure action $a_i \in A_i$ (which is a bit) or
\item a declaration of ``no pure nash equilibrium"
\end{itemize}
In the first case, the joint output $(a_1,...,a_n) \in A$ is a pure Nash equilibrium of $G$. In the second case, $G$ has no pure Nash equilibrium. \\


\noindent The communication complexity of a Pure Nash equilibrium procedure $\Pi$ applied to a game $G$, denoted by $CC(\Pi,G)$, is the number of bits communicated until $\Pi$ terminates when the input is $G$. Given a family of games $\mathcal{G}$, the communication complexity of a Pure Nash equilibrium procedure $\Pi$ for a family $\mathcal{G}$, denoted by $CC(\Pi, \mathcal{G})$, is the worst case communication complexity of $\Pi$ over all games $G \in \mathcal{G}$, that is,  $CC(\Pi,\mathcal{G})=max_{G \in \mathcal{G}}CC(\Pi,G)$. The communication complexity of pure Nash equilibrium procedures for a family of games $\mathcal{G}$, denoted by $CC(\mathcal{G})$, is the minimal communication complexity of any Pure Nash equilibrium procedure $\Pi$ for the family $\mathcal{G}$, that is, $CC(\mathcal{G})=min_{\Pi}CC(\Pi,\mathcal{G})$. Finally, when the games in $\mathcal{G}$ are chosen according to a probability distribution \textbf{P}, then the expected communication complexity of Pure Nash Equilibrium Procedures for the family $\mathcal{G}$ is $min_{\Pi} \textbf{E} [CC(\Pi, \mathcal{G})]$, where the expectation is taken with respect to \textbf{P}.

\subsection{Communication Complexity of Pure Nash Equilibrium Procedures}
We are interested in the asymptotic behaviour of the communication complexity of Nash equilibrium procedures as the number of player $n$ increases, while the size of the action sets is fixed. Specifically, let $\Gamma_2^n$ be the family of all $n-$ person  binary action games. We prove the following theorem.

\begin{theorem}
\label{worst}
Any Nash equilibrium procedure for the family $\Gamma_2^n$ has communication complexity $\Omega(2^n)$. That is,
\[ CC(\Gamma_2^n)=\Omega(2^n)\]
\end{theorem}
Note that since the communication complexity is defined as the worst case over all games, $\Omega(2^n)$ is also a lower bound for the communication complexity of any Nash equilibrium procedure for the family $\Gamma_s^n$ for every $s \geq 2$. 

\begin{proof}
We give a simple reduction from set disjointness. Say $n$ is even and divide the player set $\{1,2,...,n\} $ into two sets $T_1$ and $T_2$, each of size $n/2$, where $T_1=\{1,...,n/2\}$ and $T_2=\{n/2+1,...,n\}$. Also for notational convenience, each player in set $T_l$ is referred to as player $(l,i)$ where $l \in \{1,2\}$ and $i \in \{1,2,...,n/2\}$ \footnote{So for example, the first player in $T_1$ is referred to as player $(1,1)$, and last player in $T_2$ is referred to as player $(2,n/2)$}. Using similar notations, let us denote a $n-$ bit joint action $a$ as $(a_{11},...,a_{1n/2},a_{21},...,a_{2n/2})$, where each $a_{li}$, $l \in \{1,2\}$ and $i \in \{1,2,...,n/2\}$, is a bit.\\

\noindent Let $S=\{0,1\}^n$, the set of all $n-$ bit strings \footnote{Equivalently a $2^n$ bit string with all 1s. So any subset of $S$ is also a set of strings (or equivalently a $2^n$ bit string with 1s representing the strings in the subset).}. For any two sets $S_1,S_2 \subseteq S$, the input to the set disjointness problem, we define a binary action game $G$ in $\Gamma_2^n$, such that $S_1 \cap S_2 \neq \emptyset$ iff $G$ has a pure Nash equilibrium, and for every $a \in A$, the value of the payoff function of player $(l,i)$ on $a$, that is, $u_{l,i}(a)$, is computable, by a finite algorithm, from $a, S_l, i$. So, we will have that if the players reach a pure Nash equilibrium in $G$, then the sets $S_1$ and $S_2$ are not disjoint, and if they do not reach a pure Nash equilibrium then the sets are disjoint. Given a pure Nash equilibrium procedure $\Pi_{NE}$, we will be able to generate a protocol $\Pi_{D}$ for the set disjointness problem, with the same communication complexity \footnote{More specifically, we will have Alice simulate all players in $T_1$ and Bob simulate all players in $T_2$.}. \\

\noindent Now we define the payoff function for each player $(l,i)$ where $l \in \{1,2\}$ and $i \in \{1,2,...,n/2\}$. The idea is that, for each player $(l,i)$ and each joint action $a \in A$, the the payoff $u_{l,i}(a)$ will be ``high'' if $a$ lies in the set $S_l$ and ``low" if it doesn't. Also, the payoff functions for players $i \geq 3$ in both sets $T_1,T_2$ would be identical, and for players $i=1,2$ in both sets $T_1,T_2$, the payoff functions would be slightly different than the rest of the players. The reason we are singling out these 4 players $(1,1), (1,2), (2,1)$ and $(2,2)$ is because, we would like to show that if the sets are disjoint, then there exists a player (among these four) who would benefit by deviating on any joint action set (and hence we would show that there is no pure Nash equilibrium for the game).\\

\noindent Formally, for $l=1,2$ and $i \geq 3$, the payoff functions are
\[ u_{l,i}(a)=
\begin{cases}
2, & \text{if}\,\, a \in S_l \\
0, & \text{if} \,\, a \notin S_l
\end{cases}   
\]
\noindent That is, for a player $(l,i)$ where  $l=1,2$ and $i \geq 3$, the value of $u_{l,i}(a)$ is just a function of whether $a$ lies in $S_l$ or not.\\

\noindent We give the payoff functions for players $(1,1)$ and $(1,2)$. The payoff functions for players $(2,1)$ and $(2,2)$ are similar \footnote{Just replace $S_1$ by $S_2$, $a_{11}$ by $a_{21}$, and $a_{12}$ by $a_{22}$.} (Recall that the $n-$ bit joint action is represented as $a=(a_{11},...,a_{1n/2},a_{21},...,a_{2n/2})$).

\[ u_{1,1}(a)=
\begin{cases}
2, & \text{if}\,\, a \in S_1 \\
1, & \text{if} \,\, a \notin S_1\,\, \text{and}\,\,a_{11}=a_{12} \\
0, & \text{if} \,\, a \notin S_1\,\, \text{and}\,\,a_{11} \neq a_{12}
\end{cases}   
\]


\[ u_{1,2}(a)=
\begin{cases}
2, & \text{if}\,\, a \in S_1 \\
1, & \text{if} \,\, a \notin S_1\,\, \text{and}\,\,a_{11} \neq a_{12} \\
0, & \text{if} \,\, a \notin S_1\,\, \text{and}\,\,a_{11} = a_{12}
\end{cases}   
\]

\noindent For player $(1,1)$, the value $u_{1,1}(a)$ is not just a function of whether $a$ lies in $S_1$, but is also a function of what the first two bits of $a$ are (Similar thing holds for the other three players).  Now, we prove that $S_1$ and $S_2$ are disjoint if and only if the game $G=\{u_{l,i}\}_{l,i}$ has no pure Nash equilibrium.\\

\noindent Say $S_1$ and $S_2$ are not disjoint. Let $a \in S_1 \cap S_2$. We claim that $a$ is indeed a pure Nash equilibrium for the game $G$. This is because, since $a \in S_1 \cap S_2$, by definition, the payoff for each player on $a$ is 2, which is the maximal payoff. So no player has any reason to deviate from the joint action $a$.\\

\noindent Say $S_1$ and $S_2$ are disjoint. For every $a \in A$, if $a \notin S_l$ for some $l \in \{1,2\}$, then either $(l,1)$ or $(l,2)$ benefits by deviating. To see this, consider the case that $a \in S_2$, but $a \notin S_1$. Now since $a \in S_2$, by definition, the payoff of all the players in the set $T_2$ on the joint action $a$ is 2. Also, since $a \notin S_1$, by definition, the payoff of every player in the set $T_1$, except players $(1,1)$ and $(1,2)$, is zero on the joint action $a$. Now the payoff of players  $(1,1)$ and $(1,2)$ depends on the the first two bits of $a$, namely $a_{11},a_{12}$ \footnote{For this case, we can think of players $(1,1)$ and $(1,2)$ as playing the ``matching pennies game", which is the binary action equivalent of the popular ``Rocks, Papers and Scissors" game, where the two pennies are represented by the bits $a_{11}$ and $a_{12}$. Indeed, \cite{HM06} term this reduction as the ``matching pennies reduction''.}. There are two cases. If $a_{11}=a_{12}$, the the payoff of player $(1,1)$ is 1 and the payoff of player $(1,2)$ is zero. In this case, $a$ is not a pure Nash equilibrium because player $(1,2)$ benefits by negating his action, since for this modified joint action say $a'$ (which is $a$, but with the second bit negated), the player $(1,2)$ gets a payoff of at least \footnote{We say ``at least" because $a'$ might now be in $S_1$ and not $S_2$, in which case, the payoff will be 2.} 1 (since $a'_{11} \neq a'_{12}$). A similar argument can be made for the case $a_{11} \neq a_{12}$. So, we have proved that if $S_1$ and $S_2$ are disjoint, then no pure Nash equilibrium exists for the game $G$. This completes the proof of the theorem.\\ $\qedd$
\end{proof}

\subsection{Average Case Communication Complexity of Pure Nash Equilibrium Procedures}
While the communication complexity of Pure Nash Equilibrium procedures is exponential (in the number of players) in the worst case, it is possible that the \textit{expected} communication complexity (when the games are chosen according to a probability distribution) will be smaller. However, we will show a simple probability distribution over $\Gamma_2^n$, such that the expected communication complexity of pure Nash equilibrium procedures (where the expectation is over the choice of games in $\Gamma_2^n$ according to the probability distribution) is also exponential in the number of players. More specifically, we will prove the following theorem.
\begin{theorem}
\label{average}
There exists a probability distribution \textbf{P} over $\Gamma_2^n$ such that
\[ min_{\Pi}\textbf{E}(CC(\Pi,\Gamma_2^n))=\Omega(2^n) \]
\end{theorem}
Note that Theorem \ref{average} implies Theorem \ref{worst}. We will provide a direct proof here, rather than applying a reduction. Before proceeding with the proof, we will define what we mean by a \textit{combinatorial rectangle} in this setting.
\begin{definition}[Combinatorial Rectangle]
A \textit{combinatorial rectangle} is a set $U=U_1 \times ... \times U_n $ where each $U_i$ is a set of payoff functions of player $i$.
\end{definition}
Let $V_i$ represent the set of all possible payoff functions of player $i$. Then, we have that a combinatorial rectangle is a set $U \subseteq V_1 \times ... \times V_n$ \footnote{This is actually $\Gamma_2^n$.}, such that $U=U_1 \times ... \times U_n $, where for each $i$, $U_i \subseteq V_i$.
\begin{definition}[Monochromatic Combinatorial Rectangle]
 A \textit{monochromatic} rectangle is labelled by either 
\begin{itemize}
\item a pure joint action $a \in A$, where $a$ is the pure Nash Equilibrium for every game $(u_1,...,u_n) \in U$ or
\item ``No pure Nash equilibrium" (when no game $(u_1,...,u_n) \in U$ has a pure Nash equilibrium)
\end{itemize}
\end{definition}
Now, we proceed to prove Theorem \ref{average}.
\begin{proof}
The outline of the proof is as follows. The idea is that it will be ``hard" for the players to agree that there is no pure Nash equilibrium. Specifically, we will construct a probability distribution over $\Gamma_2^n$ such that 
\begin{itemize}
\item The probability that there is no pure Nash equilibrium is bounded \textit{away} from 0 as $n$ increases.
\item Any combinatorial rectangle that is labelled ``No pure Nash equilibrium'' has a low probability.
\end{itemize}
Now, we give the simple probability distribution \textbf{P}, over $\Gamma_2^n$. The payoff function $u_i$ of player $i$ is selected randomly as follows. For each $a^{-i} \in \{0,1\}^{n-1}$ (recall that $a^{-i}$ denotes the joint action of all players except player $i$), we toss a fair coin. Let $c$ be the outcome of the coin toss. We let $u_i(0,a^{-i})=c$ and  
$u_i(1,a^{-i})=1-c$. The coin tosses are independent over all $a^{-i}$ and over all $i$. More formally, 
for every $a^{-i} \in \{0,1\}^{n-1}$, with probability $1/2$ we let\\
$u_i(0,a^{-i})=0$ and\\
$u_i(1,a^{-i})=1$,\\
and with probability $1/2$ we let\\
$u_i(0,a^{-i})=1$ and\\
$u_i(1,a^{-i})=0$,\\
These choices are made independently over all $a^{-i}$ and over all $i$. It is easy to that, for all $a \in \{0,1\}^n$, for all $i$, we have that 


$$Pr[u_i\,\,:\,\,a_i \in BR(a^{-i},u_i)]=Pr[u_i\,\,:\,\,a_i \notin BR(a^{-i},u_i)]=1/2$$


\noindent where $a_i$ is the action of player $i$ (which is a bit). Also, recall that for each player $i$ with utility $u_i$, $BR(a^{-i},u_i)$ is the best response to a joint action $a^{-i}$, and is  $arg\,max_{a_i \in A_i}u_i(a_i,a^{-i})$.\\

\noindent Suppose we sample from the set $\Gamma_2^n$ according to the distribution \textbf{P}. Let $(u_1,...,u_n)$ represent the corresponding random variable.\\

\noindent Now we proceed to prove that the probability that there is no pure Nash equilibrium for $(u_1,...,u_n)$ is bounded \textit{away} from 0 as $n$ increases. Formally,
\begin{lemma}
\label{firstlemma}
There exists a constant $\alpha>0$ such that

$$Pr[(u_1,...,u_n)\,\, \text{has no pure Nash equilibira} ]\geq \alpha$$
for all $n\geq 2$.

\end{lemma}

\paragraph{Proof of Lemma \ref{firstlemma}} Let $N$ be the random variable representing the number of pure Nash equilibria for $(u_1,...,u_n)$. We have to prove that the probability $Pr[N=0]$ is bounded away from zero. To this end, we calculate the expectation of the random variable $N$. First, it is easy to see that, for every joint action $a \in \{0,1\}^n$, the probability that $a$ is a pure Nash equilibrium for $G$ is $1/2^n$. This is because, for a joint action $a$ to be a pure Nash equilibrium, we should have that for each $i$, the payoff function $u_i$ is such that $a_i \in BR(a^{-i},u_i)$. As we saw above, this happens with probability $1/2$, independently, for each player $i$. So, the probability that $a$ is a pure Nash equilibrium is $1/2^n$. So, we have that, each joint action $a \in \{0,1\}^n$ is a pure Nash equilibria with probability $1/2^n$, and since there are $2^n$ possible joint actions, we have that $\textbf{E}(N)=1$.\\


%A little more formally, let $I_j$ (where $1 \leq j \leq 2^n$) be the indicator random variable which says whether the $j^{th}$ joint action in the set $\{0,1\}^n$ is a pure Nash equilibrium or not. That is $I_j$ is 1 if the $j^{th}$ joint action in the set $\{0,1\}^n$ is a pure Nash equilibrium and 0 if not. It is easy to see that for each $j$, where $1 \leq j \leq 2^n$, $\textbf{E}(I_j)=1/2^n$ (because for each $j$, $I_j=1$  with probability $1/2^n$ and $I_j=0$ with probability $1-1/2^n$). Now, the random variable $N$ can be represented as a sum of these indicator random variables, that is, $N=\sum_{j=1}^{2^n}I_j$. Taking expectation on both sides, and by linearity of expectation, we get the $\textbf{E}(N)=1$.\\

\noindent Now since $\textbf{E}(N)=\sum_{k \geq 1}kPr[N=k]$, we get that

\begin{equation*}
\begin{split}
E[N]  &\geq Pr[N=1] + 2 Pr[N \geq 2]\\
&= (1-Pr(N=0)-Pr(N \geq 2))+2Pr(N \geq 2)\\
&=1-Pr(N=0)+Pr(N \geq 2)
\end{split}
\end{equation*}

%
%Pr[N=0] \geq Pr[N \geq 2] \geq Pr[N=2].



\noindent Since we have that $\textbf{E}(N)=1$, we get

\begin{equation*}
\begin{split}
Pr[N=0] &\geq Pr[N \geq 2]\\
&\geq Pr[N=2]
\end{split}
\end{equation*}
So, we just need a lower bound on the probability that the $(u_1,...,u_n)$ has exactly 2 pure Nash equilibria. It turns out that this probability is tricky to calculate. The reason is as follows. Consider the joint actions $a=1^n$ and $b=1^{n-1}0$. The probability that $a$ is a pure Nash equilibrium is $1/2^n$. However, given that $a$ is a pure Nash equilibrium, that probability that $b$ is a pure Nash equilibrium is 0. This is because of the definition of the probability distribution \textbf{P}. So, we have that for every pair of distinct strings $a,b$ in $\{0,1\}^n$, it is not necessarily the case that the event that $a$ is a pure Nash equilibrium is independent of the event that $b$ is a pure Nash equilibrium. We would like to get around this problem.\\

\noindent To this end, consider the set $Z=\{a \in \{0,1\}^n\,\,:\,\,a\,\,\text{has an even number of 1s} \}$. Now, we claim that  for every pair of distinct strings $a,b \in Z$, it \textit{is} the case that the event that $a$ is a pure Nash equilibrium is independent of the event that $b$ is a pure Nash equilibrium. To see this, consider two arbitrary but distinct strings $a,b \in Z$. It must be the case that $a,b$ must differ in at least two different positions. This is because, if they differ in only one position, then one of the two strings would end up containing an odd number of 1s, which is a contradiction, since $a,b \in Z$. Hence, for each player $i$, we have that $a^{-i} \neq b^{-i}$ (Note that this wouldn't have been necessarily true, if $a,b$ were arbitrary but distinct strings from $\{0,1\}^n$). Now it is easy to calculate the probability $\gamma$, that exactly 2 elements of $Z$ are pure Nash equilibria for $(u_1,...,u_n)$. Since the size of $Z$ is $2^{n-1}$, we have that
\[\gamma={2^{n-1} \choose 2} (1/2^n)^2 (1-1/2^n)^{2^{n-1}-2}\] 

\noindent which is positive for all $n \geq 2$. Moreover, we have that $\gamma \rightarrow (1/8)e^{-1/2}>0$ as $n \rightarrow \infty$. Since, $Z \subset \{0,1\}^n$, we have that $Pr[N=0] \geq Pr[N=2] \geq \gamma$.  Hence we conclude that the probability that there is no pure Nash equilibrium for $(u_1,...,u_n)$ is bounded away from 0. This completes the proof of the lemma.\\ $\qedd$ \\

\noindent Now we proceed to prove that the probability that $(u_1,..,u_n)$ belongs to a rectangle labelled ``no pure Nash equilibrium" is small. Formally, 


\begin{lemma}
\label{secondlemma}
Let $U=U_1 \times...\times U_n$ be a monochromatic combinatorial rectangle labelled ``no pure Nash equilibria". Then,
$$Pr[(u_1,...,u_n) \in U]\leq 2^{-2^{n-1}}$$
\end{lemma}


\paragraph{Proof of Lemma \ref{secondlemma}} First, we claim that for every $a \in \{0,1\}^n$, there exists a player $i$ such that, $a_i \notin BR(a^{-i},u_i)$ for every $u_i \in U_i$. To see this, let us assume this is not the case. That is, there exists an $a \in \{0,1\}^n$ such that, for every player $i$, $a_i \in BR(a^{-i},u_i')$, for some $u_i' \in U_i$. Then this would mean that $a$ is a pure Nash equilibrium of the game $(u_1',...,u_n')$, where $(u_1',...,u_n') \in U$. But this is a contradiction, since $U$ is a monochromatic combinatorial rectangle labelled ``no pure Nash equilibrium".\\

\noindent Now let us consider the set $Z=\{a \in \{0,1\}^n\,\,:\,\,a\,\,\text{has an even number of 1s} \}$ (for a similar reason which will be clear shortly). Since for every $a \in \{0,1\}^n$, there exists a player $i$ such that, $a_i \notin BR(a^{-i},u_i)$ for every $u_i \in U_i$, the same is true for every $a \in Z$. Let us divide the set $Z$ based on $i$. That is, $Z_i$ will contain all the $a \in Z$ such that $a_i \notin BR(a^{-i},u_i)$ for all $u_i \in U_i$. More formally, let
\[Z_i=\{a \in Z\,\,:\,\, a_i \notin BR(a^{-i},u_i) \,\, \forall u_i \in U_i \}\]
\noindent We have that $Z=\cup_{i=1}^{n}Z_i$ (Note that a single $a \in \{0,1\}^n$ can be in multiple $Z_i$s). Now consider the following set
\[\{u_i \,\, : \,\, a_i \notin BR(a^{-i},u_i) \,\, \forall a \in Z_i  \}\]

\noindent That is, this set is the set of all payoff functions $u_i$ of player $i$ such that, for every $a \in Z_i$, we have that $a_i \notin BR(a^{-i},u_i)$. But since the definition of the set $Z_i$ is the set of all $a \in Z$ such that $a_i \notin BR(a^{-i},u_i)$ for all $u_i \in U_i$, it is not difficult to see that

\[U_i \subseteq \{u_i \,\, : \,\, a_i \notin BR(a^{-i},u_i) \,\, \forall a \in Z_i  \}\]

\noindent So, we have that if a payoff function of player $i$, $u_i \in U_i$, then $u_i \in \{u_i \,\, : \,\, a_i \notin BR(a^{-i},u_i) \,\, \forall a \in Z_i  \}$. Now, for every $i$, for every distinct pair $(a,b) \in Z_i$, we have that the event $\{ u_i \,\, : \,\, a_i \notin BR(a^{-i},u_i)\}$ is independent of the event that $\{ u_i \,\, : \,\, b_i \notin BR(b^{-i},u_i)\}$ (for the same reason as discussed in the previous lemma) and each event has a probability of $1/2$. So, it is easy to see that
\[Pr[u_i \in U_i] \leq \prod_{a \in Z_i} Pr[u_i \,\, : \,\, a_i \notin BR(a^{-i},u_i)]=1/2^{|Z_i|}\]
\noindent So, since the coin tosses for choosing $u_i$ are independent over $i$, we have that

\[Pr[(u_1,...,u_n) \in U]\leq \prod_{i=1}^n Pr[u_i \in U_i] \leq 1/2^{\sum_{i=1}^n |Z_i|}\]

\noindent Since we have that $Z=\cup_{i=1}^{n}Z_i$, it follows that $\sum_{i=1}^{n} |Z_i| \geq |Z| = 2^{n-1}$. Hence, we have that, 

\[ Pr[(u_1,...,u_n) \in U]\leq 1/2^{2^{n-1}} \]

\noindent This completes the proof of the lemma.\\ $\qedd$ \\

\noindent Now we complete the proof of Theorem \ref{average} as follows. By Lemma \ref{firstlemma} we have that the probability that $(u_1,...,u_n)$ has no pure Nash equilibrium is bounded from below by $\alpha>0$. However, by Lemma \ref{secondlemma}, we have that the probability that $(u_1,...,u_n)$ belongs to a monochromatic combinatorial rectangle labelled ``no pure Nash equilibrium" is small, namely at most $1/2^{2^{n-1}}$. So the number of such monochromatic rectangles should be $\geq \alpha 2^{2^{n-1}}$. This gives us a lower bound of log$(\alpha 2^{2^{n-1}})=\Omega(2^n)$ on the expected communication complexity. This completes the proof of Theorem \ref{average}. \\ $\qedd$
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Approximate Nash start %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%




\subsection{Approximate Nash Equilibrium}

\subsubsection{Setup}

\noindent Here we will present some results from \cite{GP12}. The setup to this problem is slightly different from the previous section.

\noindent We will only focus on the case of two agents, let us call them Row and Column. Furthermore each agent will have a strategy set of n actions, because of this we can represent the payoffs to Row and Column as n by n matrices R and C respectively. Assume that all the entries of R and C are normalised to lie in [0,1]. As before Row and Column know their own payoff functions but not the other players payoff functions.

\noindent The communication protocol proceeds as follows, Row will send a message, based on R, to Column. Column sends a message based on C and Rows message back to Row. They alternate in this fashion, each player sending a message based on his matrix and the previous messages, until each player knows a strategy he wishes to play. At this point they will each play their strategy on the payoff matrix [R,C], where $[R,C]_{i,j}=(R_{i,j},C_{i,j})$ . We stress that the player only need to know the strategy they wish to play and not the strategy his counterpart wishes to play. 

\noindent We start by showing deciding if a pure Nash equilibrium exists in our setup is still hard, this proof can also be found on Noam Nissan's blog \cite{NN}.

\begin{theorem}
The communication complexity of finding a pure Nash equilibrium for two players with large strategy sets is near maximal (in terms of the size of their strategy sets). This is true even if the payoffs are binary.
\end{theorem}

\begin{proof}
\noindent We will show this via a set disjointness reduction. Say Row has input string x and Column y, each of length n. Assume that n is a perfect square, if it is not we can just add 0's to the end of x and y until they are of appropriate length. We will form payoff matrices X, from x and Y, from y, such that dim(X)=dim(Y)=($\sqrt{n}+2$,$\sqrt{n}+2$), We will be playing our game on the matrix [X,Y]. Here we show [X,Y] for n=4.

\[ \left( \begin{array}{cccc}
(x_{1}, y_{1}) & (x_{2}, y_{2}) & (0, 1) &  (0, 1) \\
(x_{3}, y_{3}) & (x_{4}, y_{4}) & (0, 1) & (0, 1) \\
(1, 0) & (1, 0) & (0, 1) & (1, 0) \\
(1, 0) & (1, 0) & (1, 0) & (0, 1)\end{array} \right) \]

\noindent The bits of x and y constitute the elements of the first $\sqrt{n}$ columns and rows of [X,Y] in order. If x and y do intersect about some bit i we see the corresponding element in the matrix must be (1,1) a Nash equilibrium since each player is receiving maximum payoff. The reason the last 2 rows and columns include items of the form (0,1) and (1,0) is to ensure if there is no intersection there will be no Nash equilibrium. Observe if we are currently in a position where Row has 0 payoff he can deviate to one of the last two Rows and get 1 payoff. Column has a similar option. Thus if x and y do not intersect, there is no (1,1) element in the matrix, so the players will never converge to equilibrium. So we require $\Omega(n)$ communication where the player inputs are of size $(\sqrt{n}+2)^{2}$. \\ $\qedd$

\end{proof}

\subsubsection{Mixed Strategies}

\noindent We can think of a mixed strategy for a player as a probability distribution over his action set. We can represent a mixed strategy as a length n vector. 

\begin{definition}
A mixed strategy is a vector $[p_{1},...,p_{n}]$, s.t. $\sum_{i}p_{i}=1$. The intuition is we play strategy i with probability $p_{i}$.
\end{definition}


\begin{definition}
The utility of a mixed strategy pair x played by Row and y played by Column under payoff matrix [R,C], are $x^{T}Ry$ to Row and $x^{T}Cy$ to Column. That is they aim to maximise their expected payoff.
\end{definition}

\noindent Nash proved that any game has a mixed strategy Nash equilibrium. Still finding one, in the setting of several agents with binary strategy sets, requires near maximal communication \cite{HM06}.

\subsubsection{$\epsilon$-Nash Equilibrium}

\noindent We will now weaken the requirements of finding an equilibrium by only requiring we find an approximate equilibrium. We say an agent will only deviate if by doing so he can gain a factor of at least $\epsilon$ ($0 \leq \epsilon \leq 1$) to his expected utility. We need the following definitions. 

\begin{definition}
If Column is playing strategy y, the best response of Row is a strategy $e_{best}$ such that for any strategy x, $e_{best}^{T}Ry \geq x^{T}Ry$. We can define the best response of Column similarly. It is easy to see that a best response is always a pure strategy.
\end{definition}

\begin{definition}
The regret of Row playing x and Column playing y is the difference between between his expected utility and the expected utility of his best response. That is Row's regret is $e_{best}^{T}Ry-x^{T}Ry$. We can define Column's regret in a similar way.
\end{definition}

\begin{definition} If both Column and Row have regret of at most $\epsilon$ we say they are at a $\epsilon$-Nash Equilibrium. 

\end{definition}

\subsubsection{Mini-Max}

\noindent We will often use the Mini-Max result in our proofs. This result was first proven by Von-Neumann. 

\begin{theorem}
Consider the zero sum game [A,-A] for some square matrix A. Say [A,-A] has a Nash equilibrium strategy pair (x,y) for Row and Column, and expected payoff of v for Row and -v for column.

By playing y Column limits Rows payoffs to at most v, equivalently Column guarantees himself a payment of at least -v. Similarly by playing x Row limits Columns payoffs to at most -v, and guarantees himself a payoff of at least v.
\end{theorem}

\noindent The Mini-Max name comes from fact players try to minimise the maximum payment given to the other player, equivalently they minimise their maximum losses.

\subsubsection{No communication}

\begin{theorem}
It is possible to achieve a 0.75-Nash Equilibrium using no communication.
\end{theorem}

\begin{proof}
\noindent For Row let his i'th strategy be the best response to Column taking column 1 of C. For Column let his j'th strategy be the best response to Row taking row 1 of R.

\noindent Row will now play row 1 with probability 0.5 and row i with probability 0.5. Column is defined similarly for column 1 and j.

\noindent Let us now look at the regret of Row. Let his best response to Column be row b.

\noindent Regret of Row is:

\noindent $(\frac{1}{2}R_{b,1} + \frac{1}{2}R_{b,j}) - (\frac{1}{4}R_{1,1} + \frac{1}{4}R_{1,j} + \frac{1}{4}R_{i,1} + \frac{1}{4}R_{i,j})$

\noindent Since i is Row's best response to Column taking 1 it must be that  $R_{b,1} \leq R_{i,1}$, we get the above value must be at most:

\noindent $(\frac{1}{4}R_{b,1} + \frac{1}{2}R_{b,j}) - (\frac{1}{4}R_{1,1} + \frac{1}{4}R_{1,j} + \frac{1}{4}R_{i,j})$

\noindent Furthermore since each of the entries of R is in [0,1] the term on the right of the difference can't be negative. Also the term on the left is maximised when each of the matrix entries is 1. Thus we get it must be at most:

$(\frac{1}{4}1 + \frac{1}{2}1)=\frac{3}{4}$, which completes the proof. \\ $\qedd$

\end{proof}

\begin{theorem}
$\forall \delta>0$ it is impossible to achieve a (0.5-$\delta$)-Nash Equilibrium using no communication.
\end{theorem}

\begin{proof}
The reader is referred to \cite{GP12} for the proof.
\end{proof}

\subsubsection{0.438-Nash Equilibrium using little communication}

\begin{theorem}
There is a protocol using $\mathop{O}(log^{2}(n))$ 2-way communication that finds a 0.438-Nash Equilibrium.
\end{theorem}

\begin{proof}

\noindent We will first show how to find an $\alpha$-equilibrium and later see the minimum value of $\alpha$ is 0.438.
 
\noindent First Row will find a solution $(x^{*}_{r},y^{*}_{r})$ to the zero sum game [R,-R]. Column will find a solution $(x^{*}_{c},y^{*}_{c})$ to [-C,C], also a zero sum game. Let us say the values specified by mini-max to these games are $v_{r}$ and  $v_{c}$ for Row and Column respectively. We have 2 cases to consider.

\begin{case}
If $v_{r}$, $v_{c}$ $\leq$ $\alpha$:
\end{case}

\noindent Row will communicate to Column $y^{*}_{r}$, Column will communicate to Row $x^{*}_{c}$. We will now play the game with strategy pair $(x^{*}_{c},y^{*}_{r})$. We will show how to communicate these strategies in the next section.

\noindent The intuition is each player is playing a strategy that mini-max guarantees will limit his counterpart to a payoff of most  $v_{r}$ or  $v_{c}$ which are both less than $\alpha$. Since they can't get better than an $\alpha$ payoff their regret must also be at most $\alpha$. Thus we are at an $\alpha$-NE

\begin{case}
If at least one of $v_{r},v_{c} > \alpha$ (Say w.l.o.g. $v_{r} > \alpha$):
\end{case}

\noindent Row sends to Column $x^{*}_{r}$. Column will send back to Row his best response, a pure strategy, $e_{c}$ to $x^{*}_{r}$. If we were to play the game using ($x^{*}_{r}$,$e_{c}$), we see that Column would have no regret since he is playing his best response. Row may have regret larger than $\alpha$ so we need to modify his strategy a bit.

\noindent Say that Row's best response to $e_{c}$ is $e_{b}$ with payoff $\beta$. Take $\frac{1}{2}\alpha$ probability from the pure strategy that has the worst payoff and give it to $e_{b}$. If the worst payoff strategy did not have enough probability repeat the process until we shift enough probability. The intuition is we are still getting at least $(1-\frac{1}{2}\alpha)\alpha$ payoff from Row's non best response strategies and $(\frac{1}{2}\alpha)\beta$ more payoff from his best response.

\noindent At worst the regret of column is now $\alpha$. This is because Row shifting $\frac{1}{2}\alpha$ probability at most made his current strategy appear $\frac{1}{2}\alpha$  worse and some other strategy appear at most $\frac{1}{2}\alpha$ better. It turns out the minimum $\alpha$ where we can guarantee Row will not deviate is $\alpha$=$\frac{1}{2}(5-\sqrt{17})$, which is roughly 0.438. \\ $\qedd$

\end{proof}

\subsubsection{How to communicate information for the algorithm}

\noindent By sending one bit each we can identify which of the two cases we are in. To communicate a pure strategy we can just send $log(n)$ bits as the index for the pure strategy. Instead of communicating a mixed strategy we will instead communicate an approximation of it with some nice properties. We will get a 0.438+$\delta$ approximation for any $\delta > 0$.

\noindent Let k=$\frac{ln(n)}{\delta^{2}}$ Let's say that we wish to send mixed strategy x. Let A be a multiset of pure strategies formed by sampling from x k times according to the probability distribution x. Now we will approximate x with $x'$. For each pure strategy in our action set let us play it with probability $\frac{c}{k}$ where c is the number of times that pure strategy appears in A.

\noindent We want the expected utility of $x'$ to be not too far from x. Let us consider any mixed strategy y and payoff matrix R. Let the event $\phi$ = \{$|x'^{T}Ry - x^{T}Ry|$ $\leq$ $\delta $\}. As in \cite{LMEMA} using some probability theory, and applying a tail inequality, we get Pr[$\phi$] $\geq$ $1 - \frac{1}{2^{n}}$. Thus if our estimation is too far off we can just re-sample until we succeed.

\noindent To send $x'$ we can just send every element of A to our counterpart and let him reconstruct $x'$. Each element takes log(n) bits to specify and $|A| \in \mathop{O}(log(n))$. Thus we need to use $\mathop{O}(log^{2}(n))$ bits.

\noindent In the first case where each zero sum game had a solution value of at most 0.438, when we communicate our limiting strategy to our counterpart at worst this strategy limits us to a payoff of 0.438+$\delta$. Thus our regret is at most 0.438+$\delta$. In the second case where Row's expected payoff was greater than 0.438 we still achieve a 0.438 equilibrium. This is because we just relied on Row playing $x^{*}_{r}$ having a guaranteed expected payoff of at least 0.438, so we can always find an $x'$ such that it gets at least 0.438 payoff by taking a small enough $\delta$. 

\subsubsection{Nash Equilibrium Conclusion}

In the approximate Nash equilibrium setting the biggest open question is how good are the factors of $\epsilon$. In the communication free setting it would be interesting to see if one could increase the 0.5 lower bound or find a protocol that beats 0.75. As for the 2-way communication model no known lower bounds exist. 

The 0.438 upper bound is inspired by an algorithm for the uncoupled model that starts with a solution to the zero sum game [R-C,C-R]. Here it is possible to achieve a 0.36395 approximation \cite{B2} in polynomial time. It would be interesting to see if a algorithm that is not based on a polynomial time algorithm could achieve a better $\epsilon$ than 0.438.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Mechanism Design Start %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Algorithmic Mechanism Design}

Before we talk about algorithmic mechanism design it will be helpful to introduce a problem we wish to study, so our first section will focus on the combinatorial auctions problem.

\subsection{Combinatorial Auctions}

\noindent In the Combinatorial Auctions problem there is a set of $m$ items $M$, and $n$ agents $N$. There is also a valuation profile $v=(v_{1},...,v_{n})$ where agent $i$ is given valuation function $v_{i}$ which is only known to him. Here for each $i$, $1 \leq i \leq n$, $v_{i}:2^{[m]} \rightarrow \mathbb{R}^{\geq 0}$. We require that $v_{i}$ is submodular and monotone (see definition below). Also, we assume that for every $i \leq n$, $v_i(\emptyset)=0$. We restrict our attention to submodular functions since they are widely studied in economics and they capture the notion of diminishing marginal returns.

\noindent We wish to pick a partition $A={A_{1},...,A_{n}}$ of $M$ (which we call an \textit{allocation}) s.t. $\sum_{i}v_{i}(A_{i})$ is maximised, that is, we think of $A_i$ as a set of items allocated to agent $i$, and our goal is to maximize the social welfare  $\sum_{i}v_{i}(A_{i})$.

\begin{definition}
A valuation function $v_{i}:2^{[m]} \rightarrow \mathbb{R}^{\geq 0}$ is submodular if $\forall S,T \subseteq M$ such that $S \subseteq T$, and all $x \in M \backslash T$, $v_{i}(x \cup T)-v_{i}(T)$ $\leq$ $v_{i}(x \cup S)-v_{i}(S)$. It is monotone if $v_{i}(S) \leq v_{i}(T)$, for all $S,T \subseteq M$ such that $S \subseteq T$.
\end{definition}

%\noindent We note that the problem as we have described it, with no game theoretic considerations, is NP-Hard.

\subsection{The Mechanism}

\noindent For a Combinatorial Auction problem with item set $M$, agent set $N$, and valuation profile $v=(v_1,...,v_n)$, we would like to design a  centralized algorithm, with certain properties (which we will state shortly). This algorithm, called the mechanism $\mathcal{M}$, will communicate with the agents over a series of rounds in an effort to learn about their underlying valuation functions. At the conclusion of this process, $\mathcal{M}$ will pick a partition $A={A_{1},...,A_{n}}$ of $M$, where agent $i$ is assigned  set $A_{i}$, in an effort to maximise $\sum_{i}v_{i}(A_{i})$.

\noindent The game theoretic aspect to our problem is each agent $i$ wants to be allocated a subset $A_{i}$ s.t. $v_{i}(A_{i})$ is maximised. Thus the agents will provide incorrect information to the mechanism, regarding their valuation function, if they will receive a better subset at the conclusion of the protocol.

\noindent Thus we would like our mechanism to satisfy three criteria, quality of solution, efficient to compute and incentive compatible (immune to agent manipulation). We formally define these criteria in the next three sections.

\subsection{Quality of Solution}

\noindent We define the approximation ratio of the mechanism as the ratio between what it returns and the optimum. 

\begin{definition}
$\mathcal{M}$ is a $\alpha$ approximation if $\forall$ valuation profiles $v=(v_{1},...,v_{n})$ if the mechanism returns partition $A=(A_{1},...,A_{n})$ and the optimal partition is $O=(O_{1},...,O_{n})$ then $\frac{\sum_{i}v_{i}(O_{i})}{\sum_{i}v_{i}(A_{i})} \leq \alpha$.
\end{definition}

\subsection{Efficiency}

\noindent Our notion of efficiency will be in the communication model, where we place no restrictions on the computational powers of the mechanism and the agents, and are only interested in the amount of communication that occurs between them. We will now define the model more formally. Recall that $M$ is the set of items, $N$ is the set of agents, and each agent, as input, is given a private valuation function $v_{i}$ as defined previously. 

\noindent The communication will proceed in a series of rounds. In each round the mechanism is allowed to query one of the agents for information about $v_{i}$, we allow the query to be based on the previous queries and responses to them. We allow the agents' reply, which will be seen by all other agents and the mechanism, to be based on the query, his underlying valuation function $v_{i}$ and the previous queries and responses of any agent. 

\noindent Formally if agent $i$ is queried in round $l$ his response is some function $r_{i}(v_{i}, Q_{l}, R_{l-1})$, where $v_{i}$ is his valuation function, $Q_{l}$ is the set of queries asked so far (including the current one) and $R_{l-1}$ is the responses to the queries (up to the last one). %We place no restrictions on the agents' computational power. %function $r_{i}$ may even be uncomputable.  

\noindent At the conclusion of this protocol the mechanism will announce a partition $A={A_{1},...,A_{n}}$ of $M$, based on the responses to his queries, where agent $i$ gets set $A_{i}$.

\begin{itemize}
\item One of the most common queries that appears in the literature is the valuation query, where the mechanism asks agent $i$, the value of $v_{i}(S)$, for some $S \subseteq M$. 
\item In our lower bound proofs we allow the queries to be anything.
\end{itemize}

\begin{definition}
The communication complexity of a mechanism $\mathcal{M}$, with agent set $N$ and item set $M$, on valuation profile $v=(v_{1},...,v_{n})$ is the number of bits communicated over the execution of the protocol. The number of bits is a function of $|N|$ and $|M|$.

\noindent The communication complexity of a mechanism is the maximum number of bits communicated over all valuation profiles $v=(v_{1},...,v_{n})$.
\end{definition}

\noindent The goal in algorithmic game theory is to design mechanisms whose communication complexity is polynomial in $|N|$ and $|M|$.

\noindent One may question why we even study this from a communication standpoint. In fact a divergent avenue of research assumes the agents, using some succinct representation, hand their entire valuation functions to the mechanism. In this non communication model the mechanism must run in time $poly(|M|,|N|)$. % In our setting we do not place any time or computation restrictions on the mechanism or agents.

\noindent There are two main reasons for studying the communication model. First the valuation functions can be exponentially large in $|M|$, even if we only consider submodular functions. It would be unreasonable to assume that the agents must send these large objects to the mechanism. Also the agents may be fine revealing a bit of their private valuations but not all of it, this is especially true in settings where knowledge of their valuations could prove advantageous to the mechanism or other agents in future games.

\subsection{Truthfulness}

\noindent In order to motivate the agents to be honest with their valuations we allow the mechanism to charge payments to the agents at the end of communication protocol. That is, in addition to deciding upon an allocation $A=(A_1,...,A_n)$, the mechanism $\mathcal{M}$ will also pick a payment scheme $p=(p_{1},...,p_{n})$ where agent $i$ is charged $p_{i} \in \mathbb{R}^{\geq 0}$. Each agent now has the goal of maximising his valuation utility minus his payment, that is agent $i$ wishes to maximise $u_{i}(A_{i}) = v_{i}(A_{i}) - p_{i}$.

\noindent Now we will define incentive compatibility, our notion of truthfulness. 

\begin{definition}
A mechanism $\mathcal{M}$ is incentive compatible if for each valuation profile $v=(v_1,...,v_n)$ the following holds. For each agent $i$, his utility ($u_i$) is maximized if he answers all his queries using his actual valuation function $v_i$. %$u_{i}(\mathcal{M}(v_{-i},v_{i})) \geq u_{i}(\mathcal{M}(v_{-i},v'_{i}))$.
\end{definition}

%\noindent We are being a bit abusive with the notation when we say $\mathcal{M}(v_{-i},v_{i})$, by this we mean the partition, $A=(A_{1},...,A_{n})$, and the payment scheme, $p=(p_{1},...,p_{n})$, the mechanism picks after the communication protocol concludes where the agents behave as if they had valuation profile $v =(v_{-i},v_{i})$. Thus $u_{i}(\mathcal{M}(v_{-i},v_{i}))$ is just $v_{i}(A_{i})-p_{i}$.

\noindent Intuitively, if $\mathcal{M}$ is incentive compatible, then for any agent, it is in his best interest to respond to the queries using his true valuation function, regardless of the other agents' valuation functions.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Start of Jai's VCG stuff %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{VCG, VCG based mechanisms, and MIR algorithms}
From the previous sections, it is clear that a mechanism can be thought of as an (\textit{allocation algorithm, payment scheme}) pair, where the allocation algorithm picks an allocation (which is a partition of $M$ into $n$ sets) from the set of all possible allocations (by communicating with the agents), and the payment scheme charges payments to each agent, based on the allocation picked by the allocation algorithm.\\


\noindent A major achievement in the field of mechanism design was the introduction of a general method of construction of incentive compatible mechanisms called the Vickrey-Clarke-Groves (VCG) method \cite{VIC61} \cite{CLA71} \cite{GRO73}. VCG method has a specific payment rule, that when applied to the output of an allocation algorithm that computes the optimal allocation (allocation which maximizes social welfare), guarantees incentive compatibility. A little more formally, a VCG mechanism is an (\textit{allocation algorithm, payment scheme}) pair, $(K,P)$, where for every valuation profile $v=(v_1,...,v_n)$, the allocation algorithm $K$ returns the allocation that maximizes the social welfare (that is, for every $v=(v_1,...,v_n)$, $K$ on $v$ returns  $arg\,max_{A \in O} \sum_{i}v_i(A_i)$, where $O$ is the set of all possible allocations), and the payment scheme $P=(P_1,...,P_n)$ is as follows: For each $i$, $P_i(v)=\sum_{j \neq i}v_j(A_j)+h_i(v_{-i})$ (where $A=(A_1,...,A_n)$ is the allocation returned by the allocation algorithm $K$ on the valuation profile $v$, and $h_i$ is an arbitrary function of $v_{-i}$, where $v_{-i}=(v_{1},...,v_{i-1},v_{i+1},...v_{n})$). We won't be going into detail about the payment scheme $P$, but the intuition is that each agent is being charged the ``social cost" of him receiving $A_i$, that is incurred by the rest of the agents, and it is not difficult to prove that VCG mechanisms are indeed incentive compatible \cite{GRO73}.\\

\noindent However, the problem is apparent when we apply this scheme to combinatorial auctions: The allocation algorithm of a VCG mechanism is required to compute \textit{optimal allocation}, and hence is not efficient, since the range of possible allocations for a combinatorial auction problem is huge, and we should maximize the social welfare over this huge range (the allocation algorithm is not efficient in the sense that it is not feasible to compute the optimal allocation with, say polynomial (in $|M|,|N|$) communication). \\

\noindent So, a natural idea is to  retain the VCG payment scheme and replace the optimal algorithm by an algorithm that is approximately optimal but efficient (say that requires only polynomial (in $|M|,|N|$) communication). This idea was introduced in \cite{NR00} and they termed such mechanisms as \textit{VCG-based} mechanisms. \cite{NR00} showed that if we replace the optimal algorithm by an arbitrary sub-optimal (but efficient) algorithm (and apply the VCG payment scheme to the output of this algorithm), then the resulting mechanism is not necessarily incentive compatible. However, they proved that the following family of sub optimal allocation algorithms, called the Maximal in Range (MIR) algorithms, do yield incentive compatible VCG-based mechanisms.
\begin{definition}
An allocation algorithm is called Maximal In Range (MIR), if there exists a subset $R \subseteq O$ (where $O$ is the set of all possible allocations), such that for every $v=(v_1,...,v_n)$, the algorithm outputs an allocation that has the maximum social welfare, among all the allocations in $R$. That is, for all $v=(v_1,...,v_n)$, the algorithm outputs $arg\,max_{A \in R} \sum_{i}v_i(A_i)$. $R$ is called the range of the MIR algorithm.
\end{definition}
\noindent So, an MIR algorithm maximizes the social welfare over some subset $R \subseteq O$. It is not difficult to see that the VCG-based mechanisms where the allocation algorithm is an $MIR$ algorithm, are indeed incentive compatible \cite{NR00}, however, the main result of \cite{NR00} is that this is essentially it, no other VCG-based mechanisms are incentive compatible. Note that the algorithm that outputs the optimal allocation is an MIR algorithm with range $R=O$ (so the approximation ratio is 1, however the algorithm is not efficient). The algorithm which always chooses only one allocation, say the allocation where we allocate all items to agent 1, is also an MIR algorithm, with the range $R$ containing only that allocation. Though such an algorithm is very efficient, it provides a very poor approximation ratio.

%\paragraph{A small note on the notations} For a combinatorial auction with $m$ items and $n$ bidders, recall that an allocation $A$ is a partition of the $m$ items into $n$ sets. We use the notation $A_i$ to represent the items allocated to bidder $i$. Also if $v_i$ is the valuation of bidder $i$, we use $v_i(A)$ to represent the welfare that the bidder $i$ got from allocation $A$, that is by $v_i(A)$, we actually mean $v_i(A_i)$.


\subsection{Limitations of VCG based Mechanisms}
In this section, we will present the results from the paper ``Limitations of VCG-based mechanisms" by Dobzinski and Nisan \cite{DN06}. The main result of this paper provides a lower bound for the approximation factor that can be achieved by incentive compatible VCG-based mechanisms (that make a sub-exponential number of queries to the bidders) for combinatorial auctions with submodular valuations. The best known mechanism for this case is the $O(\sqrt{m})$-approximation mechanism of \cite{DNS05}. \cite{DN06} show that this is almost the best approximation one can get by showing a $m^{1/6}$ lower bound. We first state their theorem,
\begin{theorem}
\label{md1}
Every incentive compatible VCG-based mechanism for approximating the social welfare in combinatorial auctions with submodular bidders that uses a sub-exponential number of queries to the bidders achieves an approximation factor of $\Omega(min(n,m^{1/6}))$.
\end{theorem}
\noindent The proof proceeds by combinatorially analysing MIR algorithms (this is sufficient because of the main Theorem of \cite{NR00}). The main idea is, if the range is ``small", then we get a bad approximation. So the range should be ``large". However, if the range is ``large", then optimizing over it needs exponential communication. A little more formally, we define two complexity measures for the range $R$ of an MIR algorithm $A$, called $cover(R)$ and $intersect(R)$, and adopt the following outline.
\begin{enumerate}
\item Show, using probabilistic method, that if $cover(R)$ is ``small", then there exists an instance such that $A$ fails to provide a good approximation. Therefore the range must be ``large".
\item Show that if the $cover(R)$ is large, then $intersect(R)$ is exponential.
\item Show that $intersect(R)$ serves as a lower bound to the communication complexity of $A$.
\end{enumerate}
\noindent $cover(R)$ roughly corresponds to the size of the range $R$ \footnote{It is not exactly the size of $R$. We exclude certain ``degenerate" allocations which don't affect the guaranteed approximation ration of the MIR algorithm. However, we won't be formally defining $cover(R)$, so for the purpose of this document, $cover(R)$ can be thought of as representing the size of the range $R$}. We will shortly define what $intersect(R)$ means. We first state the lemma corresponding to (1) of the proof outline, without the proof.
\begin{lemma}
\label{md2}
Let $A$ be an MIR algorithm with range $R$. If $cover(R)<e^{m/300n}$, then there is an instance in which A provides no more than $1.01/n$-fraction of the welfare.
\end{lemma} 
\noindent The high level idea of this proof is to construct a random instance of combinatorial auctions and prove that the value of optimal  solution in the random instance is exactly $m$, however with non-negligible probability, the welfare provided by the MIR algorithm $A$ with range $R$ (with small cover number), is only $1.01m/n$. We direct the readers to \cite{DN06} for the full proof.\\

\noindent Next, we state the lemma corresponding to (2) of the proof outline, without the proof.
\begin{lemma}
\label{md3}
Let $R$ be a set of allocations with $intersect(R)\leq d$. Then, we have that
\[ cover(R)<(8d)^{m^{3n/5}}m^{4m^{2/5}n^{2}} \]
\end{lemma}
This is a pure combinatorial lemma, which relates the two measures defined on the range $R$ of an MIR algorithm. This lemma is rather difficult to prove, and we direct the readers to \cite{DN06} for the proof.\\

\noindent We now proceed to state and prove the lemma corresponding to (3) of the proof outline, which is the relevant result to us from a communication complexity perspective. Before that, we state some definitions, including the definition of the intersection number.

\begin{definition}
We say that a set of allocations $R$ is \textit{regular} if there exists constants $s_1,...,s_n$ such that for all $S \in R$ and for all $1 \leq i \leq n$, it holds that $|S_i|=s_i$.

\end{definition}

\noindent So, if the allocations are chosen from a set of allocations $R$ which is regular, then no matter what the allocation is, for each $i$, agent $i$ ends up getting the same number of items, namely $s_i$.\\

\noindent Now, we proceed to define what be mean by an intersection set and the intersection number of a set of allocations.


\begin{definition}
A set of allocations $\mathcal{D}$ is called an $(i,j)$ intersection set if for each $D,D' \in \mathcal{D}$ such that $D \neq D'$, it holds that $D_i \cap D_j' \neq \emptyset$.
\end{definition}

\begin{definition}
Define the intersection number of a set of allocations $R$, denoted by $intersect(R)$ to be the maximum cardinality regular $(i,j)$ intersecting set which is a subset of $R$, where the maximum is taken over all pairs of bidders $i$ and $j$.
\end{definition}

\noindent So, for a set of allocations $R$, we consider the power set of $R$. From this power set, we only consider the sets (which are nothing but the subsets of $R$) which are regular. Out of these, for each distinct $i,j$, $1 \leq i,j \leq n$, we consider the sets (which now are subsets of $R$ that are regular), that are $(i,j)$ intersection sets, and find the set the has maximum size among these (where the maximum is taken over all distinct $i,j$), and the size of this set is the intersection number of $R$ \footnote{We acknowledge that this measure looks somewhat artificial, and unlike the cover number, there is no clear intuition as to what characteristic/property of the range this measure exactly captures.}. \\

\noindent Now, we proceed to prove that the intersection number of the range of a MIR algorithm serves as a lower bound to the communication complexity of the algorithm.

\begin{lemma}
\label{md4}
Let $A$ be a MIR algorithm for combinatorial auctions with submodular bidders, with range $R$. Let $intersect(R)=d$. Then the communication complexity of $A$ is $\Omega(d)$.
\end{lemma}
\begin{proof}
The outline of the proof is as follows. We reduce from set disjointness. Alice and Bob each hold a $d-$ bit string, $a_1,...,a_d$ and $b_1,..,b_d$ respectively (where $d$ is the intersection number of the range $R$ of the MIR algorithm $A$), the input to the set disjointness problem. Given, the input to the set disjointness problem, we construct a combinatorial auction with $m$ items and $n$ submodular bidders such that, the MIR algorithm $A$ with range $R$ (with intersection number $d$) outputs a specific fixed value (which we will specify shortly) for this auction if there exists $k$ such that $a_k=b_k=1$ (that is, if the sets represented by the inputs to the set disjointness problem are not disjoint) and outputs a value strictly less than the specific fixed value if the sets represented by the inputs to the set disjointness problem are disjoint. More formally,\\

\noindent Let $\mathcal{D}=\{D^1,...,D^d\}$ be an $(i,j)$ intersection set of the range $R$ (we know such a set exists for some distinct $i,j$, $1 \leq i,j \leq n$, since the intersection number of $R$ is $d$). By definition, $\mathcal{D}$ is regular. So, for each bidder $t$,  there exists a constant $s_t$ such that $|D_t|=s_t$ for every $D \in \mathcal{D}$.\\

\noindent We will construct a combinatorial auction with $m$ items as follows. Alice will play the role of bidder $i$. Bob will play the role of rest of the bidders, especially bidder $j$. We now define the valuation functions for each of the bidders. We specify the valuation function of bidder $i$ played by Alice. 
\[ v_{i}(S)=
\begin{cases}
|S|, & \text{if}\,\, |S| \leq s_i-1 \\
s_i, & \text{if} \,\, |S| \geq s_i\,\, \text{and}\,\, \exists k \,\, \text{s.t.}\,\,D^k \subseteq S \,\,\text{and}\,\, a_k=1 \\
s_i-1/2^{|S|-s_i+1} , & \text{otherwise}
\end{cases}   
\]
\noindent The valuation $v_j$ is defined in an analogous way. The valuations of the rest of the bidders is zero on any input.\\

\noindent First, we claim that the valuations are submodular. This is not difficult to see, and we just give the outline of the proof. To say that $v_i$ is submodular, we prove that for every $S \subseteq M$, and $x_1,x_2 \in M \backslash S$,
 \[ v_i(S \cup \{x_1,x_2\}) + v_i(S) \leq v_i(S \cup \{x_1\})+ v_i(S \cup \{x_2\}) \] 
 
\noindent We select an arbitrary $S$, and we do case analysis based on the size of $S$ to show that the above equation is true for each of the cases \footnote{Based on the definition of the $v_i$ and the definition of submodularity we chose to use, we can think of four different cases, $|S| < s_i-2$, $|S| = s_i-2$, $|S| = s_i-1$ , and $|S| > s_i-1$.}.\\

\noindent Now we claim that if there exists $k$ such that $a_k=b_k=1$, then the optimal welfare is exactly $s_i+s_j$, otherwise the optimal welfare is strictly less than $s_i+s_j$. To see this, note that if there exists $k$ such that $a_k=b_k=1$, then the valuation $v_i$ on any superset of $D_i^k$ is $s_i$ and the valuation $v_j$ on any superset of $D_j^k$ is $s_j$. This is by definition of the valuation functions. So, in particular, the allocation $D^k$ yields the social welfare of $s_i+s_j$, which is indeed optimal \footnote{By definition, the maximum value that $v_i$ can evaluate to is $s_i$, and for $v_j$ it is $s_j$ and zero for rest of the bidders}.\\

\noindent Now we prove the other direction. Observe that on an allocation, if bidder $i$ gains $s_i$ (say his allocation is $S_1$) and bidder $j$ gets $s_j$ \footnote{That is, the optimal welfare is $s_i+s_j$.} (say his allocation is $S_2$), then by definition there exists $k,k'$ such that $D_i^k \subseteq S_1$ and $a_k=1$, and $D_j^{k'} \subseteq S_2$ and $b_{k'}=1$. However since             $\mathcal{D}$ is an intersection set, we have that $D_i^k \cap D_j^{k'} \neq \emptyset$. So it must be the case that $S_1 \cap S_2 \neq \emptyset$, so the allocation is not a valid allocation, and hence $k$ must equal $k'$.\\

\noindent Finally, by construction, if the optimal welfare is $s_i+s_j$, then it can be achieved by an allocation in $R$. Since $A$ is an MIR algorithm with range $R$, $A$ must return $s_i+s_j$ in this case (and obviously $A$ returns value strictly less than $s_i+s_j$ if the optimal welfare is strictly less than $s_i+s_j$). So, we will be able to decide if there if there exists $k$ such that $a_k=b_k=1$ based on the value returned by the MIR algorithm $A$. Hence, we conclude that the communication complexity of $A$ is at least as that of the set disjointness problem, which is $\Omega(d)$. This concludes the proof of Lemma \ref{md4}. \\ $\qedd$
\end{proof}

\noindent It is not difficult to see that we can combine Lemmas \ref{md2}, \ref{md3}, \ref{md4} in a straightforward way to prove Theorem \ref{md1}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Start of Tyrone's  CPPP stuff %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 

\subsection{The Combinatorial Public Projects Problem (CPPP)}
We now present the main result from \cite{PSS08} showing that no mechanism for the CPPP that satisfies all three of our criteria exists. The setup to the CPPP is similar to the Combinatorial Auction problem.

\noindent We have a set $N$ of $n$ players, and a set $M$ of $m$ resources with a constant $k$. We also have a valuation profile $v=(v_{1},...,v_{n})$, player i is given valuation $v_{i}$ which is only known to him. Here for each $i$, $v_{i}: 2^{[m]} \rightarrow \mathbb{R}^{\geq 0}$. 

\noindent We require that each $v_{i}$ is submodular. That is $\forall S,T \subseteq M$ where $S \subseteq T$, and for all $x \in M \backslash T$, $v_{i}(x \cup T)-v_{i}(T)$ $\leq$ $v_{i}(x \cup S)-v_{i}(S)$. We also require that each $v_{i}$ is monotone, that is $v_{i}(S) \leq v_{i}(T)$, for all $S,T \subseteq M$ such that $S \subseteq T$. Finally we require that each $v_{i}$ is normalized, that is $v_i(\emptyset)=0$.

\noindent The goal of the mechanism $\mathcal{M}$ is to pick a set of items $A \subseteq M$ s.t. $|A| = k$ and maximise $\sum_{i=1}^{n}v_{i}(A)$. The way to perceive this problem is we have a set of potential resources M, say potential hospital locations, and the budget to build $k$ of them that will be shared amongst the agents of $N$. We wish to maximise the social welfare of the agents, that is the sum of their valuations on the shared subset of $M$.

\noindent We define the quality of the solution in the same way as before, the ratio between the optimal and the algorithm for any valuation profile.

\noindent The communication model, and our notion of communication complexity, is exactly the same as in the combinatorial auction setting. That is the mechanism will make a series of queries to the agents regarding their valuation functions. After some number number of rounds the mechanism will pick $A \subseteq M$ s.t. $|A| = k$ as the solution. 

\noindent As in the combinatorial auction case to ensure his queries are answered honestly the mechanism will also create a payment scheme $p=(p_{1},...,p_{n})$ where each agent is charged their corresponding payment. As before each agent $i$ aims to maximise $v_{i}(A) - p_{i}$.

%\noindent In a non game theoretic setting the CPPP is NP-Hard \cite{F98} %we cannot apply the VCG payment rule, but there is a simple greedy algorithm that achieves a $1-\frac{1}{e}$ approximation \cite{N78}.

%\noindent \cite{PSS08} also showed similar lower bounds in the non communication complexity, functions with succinct representations, model.

\begin{theorem}\label{thm:CPPP}
Any incentive compatible mechanism that achieves a $m^{\frac{1}{2}-\epsilon}$ approximation for any $\epsilon>0$ requires communication $\Omega(2^{m^{\epsilon}})$\footnote{There is a simple greedy incentive compatible mechanism that achieves an $\mathop{O}(\sqrt{m})$ approximation for the CPPP, thus this lower bound is tight \cite{SMSY}.}
\end{theorem}

\begin{proof}
The proof proceeds in three steps. We will only outline the first two steps since they are not very interesting from a communication standpoint. See \cite{PSS08} for the full proof. We first need to define the following two concepts.

\begin{definition}
The range of a mechanism $\mathcal{M}$ is $R_{\mathcal{M}}$=\{$A \subseteq M$ : $|A|=k$ and $\exists$ a valuation profile $v=(v_1,...,v_n)$ such that the mechanism $\mathcal{M}$ on $v$ outputs $A$ \}. 
\end{definition}

\begin{definition}
A mechanism $\mathcal{M}$ is an affine maximiser if for every valuation profile $v=(v_{1},...,v_{n})$, it pick $argmax_{A \in R_{\mathcal{M}}} (\sum_{i}w_{i}v_{i}(A) + C_{A})$. 
\end{definition}

\noindent For $1 \leq i \leq n$, $w_{i}$ is a non  negative constant associated with player $i$. We require at least one of them is non zero.

\noindent For every $ A \in R_{\mathcal{M}}$, $C_{A}$ is a non negative constant associated with outcome A. We do not require at least one of them to be non zero.

\noindent The definition of affine maximiser mechanism (as described above) can be seen as a weighted version of maximal in range algorithms (defined in the Section 2.6).


\begin{lemma}(Affine Maximiser)
If mechanism $\mathcal{M}$ is incentive compatible for CPPP, $\mathcal{M}$ is an affine maximiser.
\end{lemma}

\begin{proof}
The proof of the lemma is a modification of the proof of Roberts' theorem \footnote{See \cite{LN} for two recent proofs of Roberts' theorem.}, which proved the same results for general problems and valuation functions.
\end{proof}


\begin{lemma}(Large Range)
Any affine maximiser that achieves an approximation ratio of $m^{\frac{1}{2}-\epsilon}$ requires a range of size $\Omega(2^{m^{\epsilon}})$. For this lemma we will fix $k$ to be $\sqrt{m}$.
\end{lemma}

\begin{proof}
This will be done via a probabilistic argument. Informally we construct a set $V \subset M$, by adding each element $a \in M$ to $V$ independently with probability $m^{-\frac{1}{2} + \epsilon}$. We then consider the game consisting of 1 agent with a valuation function $v(S)=|V \cap S|$. Using Chernoff bounds we get that with high probability no solution contained in $R_{\mathcal{M}}$ will serve as a good approximation to the optimal solution, unless $|R_{\mathcal{M}}|$ $\in$ $\Omega(2^{m^{\epsilon}})$.
\end{proof}

\noindent Now we can prove that for 2 players we require communication of size $\Omega(|R_{\mathcal{M}}|)$ via a reduction from 2 player set disjointness.

\noindent Let us say each player has a private set of outcomes he prefers $R_{i} \subset R_{\mathcal{M}}$ for $i \in \{1,2\}$. Define his valuation as $v_{i}(T)= \beta_{i} (\prod_{S \in R_{i}}|S| - \prod_{S \in R_{i}}(|S| - |S \cap T|)$), where $\beta_{i}$ is some constant. To show this function is submodular we can perform induction on $|R_{i}|$. 

\noindent First we note for the affine maximiser can assume that the agent constants $w_{1}=w_{2}=1$. If this were not the case we could just scale the $\beta_i$ value of the agents. Furthermore we can assume that each of the outcome constants $C_{j}$ is 0 since we can just scale up each $\beta_i$ to make $C_{j}$ negligible. 

\noindent For each $T \in R_{i}$, $v_{i}$ is maximised. This is because the first product is independent of the solution set $T$ so it is just a constant for the game. On the other hand since $|T \cap T|$ = $|S|$ the second product, a non positive value, will become 0. If $T \notin R_{i}$ we note the value to agent $i$ must be smaller. Again because in the second product all of the item in it will have value at least 1, thus it must be negative.

\noindent Thus if $|R_{1} \cap  R_{2}| > 0$ the set the affine maximiser picks must lie in the intersection of the two, and the sum of the 2 utilities must be $k^{|R_{1}|} + k^{|R_{2}|}$ where $k$ is the number of elements we activate (each S is of size $k$ since it is an outcome).

\noindent Now the reduction from set disjointness is obvious. If Alice gets $x$ and Bob gets $y$ each of size $|R_{\mathcal{M}}|$. They form sets $R_{1}$ where we include the $i^{th}$ outcome iff $x$ has a 1 in the $i^{th}$ bit, and do the same for $R_{2}$ and $y$. They both simulate the mechanism asking the queries it would ask of each other until they find a solution. If both Alice and Bob have achieved an optimal value for their set functions they conclude that their sets must intersect, otherwise they conclude they must not intersect.

\noindent Thus we see we need $\Omega(|R_{\mathcal{M}}|)$ communication. Assuming $\mathcal{M}$ achieves an approximation ratio of $m^{\frac{1}{2}-\epsilon}$ we require $\Omega(2^{m^{\epsilon}})$ communication. \\ $\qedd$

\end{proof}

\subsubsection{Mechanism Design Conclusion}
The biggest open question now is are the techniques of \cite{PSS08} applicable to different problems. It would be interesting to see if we can use similar arguments to get strong lower bounds on incentive compatible mechanisms for the combinatorial auctions problem, not just VCG based ones.

\bibliography{main}

\end{document}

