+In this game, we have two players who are competing for a "payoff."
+There is a symmetric cone :math:`K`, a linear transformation :math:`L`
+on the space in which :math:`K` lives, and two points :math:`e_{1}`
+and :math:`e_{2}` in the interior of :math:`K`. The players make their
+"moves" by choosing points from two strategy sets. Player one chooses
+an :math:`\bar{x}` from
+
+.. math::
+ \Delta_{1} =
+ \left\lbrace
+ x \in K\ \middle|\ \left\langle x,e_{2} \right\rangle = 1
+ \right\rbrace
+
+and player two chooses a :math:`\bar{y}` from
+
+.. math::
+ \Delta_{2} &=
+ \left\lbrace
+ y \in K\ \middle|\ \left\langle y,e_{1} \right\rangle = 1
+ \right\rbrace.
+
+That ends the turn, and player one is paid :math:`\left\langle
+L\left(\bar{x}\right),\bar{y}\right\rangle` out of player two's
+pocket. As is usual to assume in game theory, we suppose that player
+one wants to maximize his worst-case payoff, and that player two wants
+to minimize his worst-case *payout*. In other words, player one wants
+to solve the optimization problem,
+
+.. math::
+ \text{find }
+ \underset{x \in \Delta_{1}}{\max}\
+ \underset{y\in \Delta_{2}}{\min}\
+ \left\langle L\left(x\right),y \right\rangle
+
+while player two tries to (simultaneously) solve a similar problem,
+
+.. math::
+ \text{find }
+ \underset{y\in \Delta_{2}}{\min}\
+ \underset{x \in \Delta_{1}}{\max}\
+ \left\langle L\left(x\right),y \right\rangle.
+
+There is at least one pair :math:`\left(\bar{x},\bar{y}\right)` that
+solves these problems optimally, and Dunshire can find it. The optimal
+payoff, called *the value of the game*, is unique. At the moment, the
+symmetric cone :math:`K` can be either the nonnegative orthant or the
+Lorentz "ice cream" cone in :math:`\mathbb{R}^{n}`. Here are two of
+the simplest possible examples, showing off the ability to solve a
+game over both of those cones.
+
+First, we use the nonnegative orthant in :math:`\mathbb{R}^{2}`:
+
+>>> from dunshire import *
+>>> K = NonnegativeOrthant(2)
+>>> L = [[1,0],[0,1]]
+>>> e1 = [1,1]
+>>> e2 = e1
+>>> G = SymmetricLinearGame(L,K,e1,e2)
+>>> print(G.solution())
+Game value: 0.5000000
+Player 1 optimal:
+ [0.5000000]
+ [0.5000000]
+Player 2 optimal:
+ [0.5000000]
+ [0.5000000]
+
+Next we try the Lorentz ice-cream cone in :math:`\mathbb{R}^{2}`:
+
+>>> from dunshire import *
+>>> K = IceCream(2)
+>>> L = [[1,0],[0,1]]
+>>> e1 = [1,1]
+>>> e2 = e1
+>>> G = SymmetricLinearGame(L,K,e1,e2)
+>>> print(G.solution())
+Game value: 0.5000000
+Player 1 optimal:
+ [0.5000000]
+ [0.5000000]
+Player 2 optimal:
+ [0.5000000]
+ [0.5000000]
+
+(The answer when :math:`L`, :math:`e_{1}`, and :math:`e_{2}` are so
+simple is independent of the cone.)
+