lecture19.tex

\chapter{Zero-sum Games, Incompleteness}
``My loss is your gain.''
\section{Rock-paper-scissors}
One zero-sum game is rock-paper-scissors:
	\begin{align}
	\begin{matrix}
	\text{(payoff to \textsc{row player})}& \text{\textsc{Column player}} \\
	\text{\textsc{Row player}} & 
	\begin{tabular}{l|rrrr}
	 & r & p & s \\\hline
	 r & 0 & -1 & 1 \\
	 p & 1 & 0 & -1 \\
	 s & -1 & 1 & 0
	\end{tabular}
	\end{matrix}
	\end{align}
If \textsc{row player} plays with probabilities \(x_1, x_2, x_3\) and \textsc{column player} with probabilities \(y_1, y_2, y_3\), then the objective is 
%\begin{align}
%\sum_{i,j} G_{ij}x_iy_j,
%\end{align}
\(\sum_{i,j} G_{ij}x_iy_j\), which \textsc{row player} wants to maximize and \textsc{column player} wants to minimize.
The objective is equivalently \(\sum_i x_i \sum_j G_{ij}y_j\) 
or \(\sum_j y_j \sum_x G_{ij}x_i\).
If \textsc{row player} plays uniformly, then expected payoff is a multiple of \(\frac{1}{3}\sum_i G_{ij} = 0\).

What if \textsc{row player} announces his strategy? Even then, \textsc{column player} can do no better.
Consider choosing between strategies:
\begin{align}
\begin{matrix}
& c_1 & c_2 \\
r_1 & 3 & -1 \\
r_2 & -2 & 1 
\end{matrix}
\end{align}
Suppose \textsc{row player} plays \(x_1, x_2\).
\textsc{Column payer}'s payoffs can be selected from \(3x_1 - 2x_2\) and \(-x_1 + x_2\), so his best payoff is
\begin{align}
\max_{x_1, x_2} \min \left\{3x_1 - 2x_2, -x_1, + x_2\right\} &\\
\intertext{Likewise, the column player's optimum payoff is}
\min_{y_1, y_2} \max \left\{3y_1 - y_2, -2y_1 + y_2\right\} &
\end{align}
We will frame the two strategy specifications as dual linear programs. A linear program for the first strategy (\(z =  \min \{\ldots\}\)):
\begin{align}
\begin{matrix}
\max & z \\
\text{subj.~to} & \begin{aligned}
-3x_1 + 2x_2 + z &\leq 0 \\
x_1 - x_2 + z &\leq 0 \\
x_1 + x_2 &= 1 \\
x_1, x_2 &\geq 0
\end{aligned}
\end{matrix} &
\intertext{and for the second:}
\begin{matrix}
\min & w \\
\text{subj.~to} & \begin{aligned}
-3y_1 + y_2 + w &\geq 0 \\
2y_1 - y_2 + w &\geq 0 \\
y_1 + y_2 &= 1 \\
y_1, y_2 &\geq 0
\end{aligned}
\end{matrix}
\end{align}
Applying multipliers \(y_1, y_2, w\) (resp.~\(x_1, x_2, z\)) one gets the duality. The expected payoff is \(\frac{1}{7}\).
\section{Hard Problems (e.g.~SAT)}
There exist problems for which we do not have efficient algorithms. It is not, however, a pure tragedy, for hardness helps with cryptography. One such problem is SAT: you are given a boolean formula such as
\(\left(x\lor y\lor z\right)\left(x\lor\overline{y}\right)\left(y\lor \overline{z}\right) \left(z\lor\overline{x}\right)\left(\overline{x}\lor \overline{y} \lor \overline{z}\right)\), and are asked whether any \(\left(x, y, z\right)\) make this true. (This is in conjunctive normal form: an \textsc{and} of \textsc{or}s.) Sadly, we know no better algorithm than brute-force (but there is no lower bound yet).

SAT is called a \emph{search problem}. Generally, a search problem is of the form: given an instance \(I\) find a solution \(S\), or say no. A search problem is specified by an algorithm \(C\) that takes an instance and a solution and decides in time \(\operatorname{polynomial}\left(\left|I\right|\right)\) whether the solution is correct. \(C_\text{SAT}\) would be simple evaluation.
\subsection{TSP as a search problem}
How do you validate a TSP solution? In the ``best solution'' form \(\left(n, d[i,j]\right)\) it is not easy to validate.
But you can formulate a search problem in the same spirit as follows: ``find a permutation \(\tau\) such that the induced path has cost less than \(b\).'' Then computing the cost of a candidate solution can be done in polynomial time. (To get a solution to standard TSP, binary search \(b\) bounds.)