bastard of the two: the cones are symmetric, but the players get to
choose two interior points.
+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}`:
+
+.. doctest::
+
+ >>> 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}`:
+
+.. doctest::
+
+ >>> 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:
+ [0.8347039]
+ [0.1652961]
+ Player 2 optimal:
+ [0.5000000]
+ [0.5000000]
+
+Note that these solutions are not unique, although the game values are.
+
Requirements
------------
Background
----------
-A linear game is a generalization of a two-person zero-sum matrix
-game~\cite{karlin, parthasarathy-raghavan,
-von_neumann-morgenstern}. Classically, such a game involves a matrix
-:math:`A \in \mathbb{R}^{n \times n}` and a set (the unit simplex) of
-"strategies" :math:`\Delta = \operatorname{conv}\left(
-\left\lbrace e_{1}, e_{2}, \ldots, e_{n} \right\} \right)` from which two
-players are free to choose. If the players choose :math:`x,y \in \Delta`
-respectively, then the game is played by evaluating :math:`y^{T}Ax` as the
-"payoff" to the first player. His payoff comes from the second
-player, so that their total sums to zero.
+A linear game is a generalization of a two-person zero-sum matrix game
+[Karlin]_. Classically, such a game involves a matrix :math:`A \in
+\mathbb{R}^{n \times n}` and a set (the unit simplex) of "strategies"
+:math:`\Delta = \operatorname{conv}\left( \left\lbrace e_{1}, e_{2},
+\ldots, e_{n} \right\} \right)` from which two players are free to
+choose. If the players choose :math:`x,y \in \Delta` respectively,
+then the game is played by evaluating :math:`y^{T}Ax` as the "payoff"
+to the first player. His payoff comes from the second player, so that
+their total sums to zero.
Each player will try to maximize his payoff in this scenario,
or—what is equivalent—try to minimize the payoff of his
symmetric. This simplifies some definitions.
What is a linear game?
-----------------------
+~~~~~~~~~~~~~~~~~~~~~~
In the classical setting, the interpretation of the strategies as
probabilities results in a strategy set :math:`\Delta \subseteq
points :math:`e_{1}` and :math:`e_{2}` belong to
:math:`\operatorname{int}\left(K\right)`.
+**Definition.** The strategy sets for our linear game are
+
+.. math::
+ \begin{aligned}
+ \Delta_{1}\left(L,K,e_{1},e_{2}\right) &=
+ \left\lbrace
+ x \in K\ \middle|\ \left\langle x,e_{2} \right\rangle = 1
+ \right\rbrace\\
+ \Delta_{2}\left(L,K,e_{1},e_{2}\right) &=
+ \left\lbrace
+ y \in K\ \middle|\ \left\langle y,e_{1} \right\rangle = 1
+ \right\rbrace.
+ \end{aligned}
+
+Since :math:`e_{1},e_{2} \in \operatorname{int}\left(K\right)`, these
+are bases for :math:`K`. We will usually omit the arguments and write
+:math:`\Delta_{i}` to mean :math:`\Delta_{i}\left(L,K,e_{1},e_{2}\right)`.
+
+To play the game :math:`\left(L,K,e_{1},e_{2}\right)`, the first
+player chooses an :math:`x \in \Delta_{1}`, and the second player
+independently chooses a :math:`y \in \Delta_{2}`. This completes the
+turn, and the payoffs are determined by applying the payoff operator
+:math:`\left(x,y\right) \mapsto \left\langle L\left(x\right), y
+\right\rangle`. The payoff to the first player is :math:`\left\langle
+L\left(x\right), y \right\rangle`, and since we want the game to be
+zero-sum, the payoff to the second player is :math:`-\left\langle
+L\left(x\right), y \right\rangle`.
+
+The payoff operator is continuous in both arguments because it is
+bilinear and the ambient space is finite-dimensional. We constructed
+the strategy sets :math:`\Delta_{1}` and :math:`\Delta_{2}` to be
+compact and convex; as a result, Karlin's [Karlin]_ general min-max
+Theorem 1.5.1, guarantees the existence of optimal strategies for both
+players.
+
+**Definition.** A pair :math:`\left(\bar{x},\bar{y}\right) \in
+\Delta_{1} \times \Delta_{2}` is an *optimal pair* for the game
+:math:`\left(L,K,e_{1},e_{2}\right)` if it satisfies the *saddle-point
+inequality*,
+
+.. math::
+ \left\langle L\left(x\right),\bar{y} \right\rangle
+ \le
+ \left\langle L\left( \bar{x}\right), \bar{y} \right\rangle
+ \le
+ \left\langle L\left(\bar{x}\right),y \right\rangle
+ \text{ for all }
+ \left(x,y\right) \in \Delta_{1} \times \Delta_{2}.
+
+At an optimal pair, neither player can unilaterally increase his
+payoff by changing his strategy. The value :math:`\left\langle L
+\left( \bar{x} \right) , \bar{y} \right\rangle` is unique (by the same
+min-max theorem); it is shared by all optimal pairs. There exists at
+least one optimal pair :math:`\left(\bar{x},\bar{y}\right)` of the
+game :math:`\left(L,K,e_{1},e_{2}\right)` and its *value* is
+:math:`v\left(L,K,e_{1},e_{2}\right) = \left\langle
+L\left(\bar{x}\right), \bar{y} \right\rangle`.
+
+Thanks to Karlin [Karlin]_, we have an equivalent characterization of
+a game's value that does not require us to have a particular optimal
+pair in mind,
+
+.. math::
+ v\left( L,K,e_{1},e_{2} \right)
+ =
+ \underset{x \in \Delta_{1}}{\max}\
+ \underset{y\in \Delta_{2}}{\min}\
+ \left\langle L\left(x\right),y \right\rangle
+ =
+ \underset{y\in \Delta_{2}}{\min}\
+ \underset{x \in \Delta_{1}}{\max}\
+ \left\langle L\left(x\right),y \right\rangle.
+
+Linear games reduce to two-person zero-sum matrix games in the
+right setting.
+
+**Example.** If :math:`K = \mathbb{R}^{n}_{+}` in :math:`V =
+\mathbb{R}^{n}` and :math:`e_{1} = e_{2} =
+\left(1,1,\ldots,1\right)^{T} \in \operatorname{int}\left(K\right)`,
+then :math:`\Delta_{1} = \Delta_{2} = \Delta`. For any :math:`L \in
+\mathbb{R}^{n \times n}`, the linear game :math:`\left(
+L,K,e_{2},e_{2} \right)` is a two-person zero-sum matrix game. Its
+payoff is :math:`\left(x,y\right) \mapsto y^{T}Lx`, and its value is
+
+.. math::
+ v\left( L,K,e_{1},e_{2} \right)
+ =
+ \underset{x \in \Delta}{\max}\
+ \underset{y\in \Delta}{\min}\
+ \left( y^{T}Lx \right)
+ =
+ \underset{y\in \Delta}{\min}\
+ \underset{x \in \Delta}{\max}\
+ \left( y^{T}Lx \right).
+
+
References
----------
projects / dunshire.git / blobdiff