The payoff to the first player in this case is
:math:`v\left(A\right)`. Corresponding to :math:`v\left(A\right)` is an
-optimal strategy pair :math:`\left(\xi, \gamma\right) \in \Delta
+optimal strategy pair :math:`\left(\bar{x}, \bar{y}\right) \in \Delta
\times \Delta` such that
.. math::
- \gamma^{T}Ax
+ \bar{y}^{T}Ax
\le
v\left(A\right)
=
- \gamma^{T}A\xi
+ \bar{y}^{T}A\bar{x}
\le
- y^{T}A\xi
+ y^{T}A\bar{x}
\text{ for all } \left(x,y\right) \in \Delta \times \Delta.
-The relationship between :math:`A`, :math:`\xi`, :math:`\gamma`,
+The relationship between :math:`A`, :math:`\bar{x}`, :math:`\bar{y}`,
and :math:`v\left(A\right)` has been studied extensively [Kaplansky]_
[Raghavan]_. Gowda and Ravindran [GowdaRav]_ were motivated by these
results to ask if the matrix :math:`A` can be replaced by a linear
Theorem 1.5.1, guarantees the existence of optimal strategies for both
players.
-**Definition.** A pair :math:`\left(\xi,\gamma\right) \in
+**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),\gamma \right\rangle
+ \left\langle L\left(x\right),\bar{y} \right\rangle
\le
- \left\langle L\left( \xi\right), \gamma \right\rangle
+ \left\langle L\left( \bar{x}\right), \bar{y} \right\rangle
\le
- \left\langle L\left(\xi\right),y \right\rangle
+ \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( \xi \right) , \gamma \right\rangle` is unique (by the same
+\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(\xi,\gamma\right)` of the
+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(\xi\right), \gamma \right\rangle`.
+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
ordering; that is, :math:`x \succcurlyeq y \iff x - y \in K`.
**Theorem.** In the game :math:`\left(L,K,e_{1},e_{2}\right)`, we have
-:math:`L^{*}\left( \gamma \right) \preccurlyeq \nu e_{2}` and
-:math:`L \left( \xi \right) \succcurlyeq \nu e_{1}` for
+:math:`L^{*}\left( q \right) \preccurlyeq \nu e_{2}` and
+:math:`L \left( p \right) \succcurlyeq \nu e_{1}` for
:math:`\nu \in \mathbb{R}` if and only if :math:`\nu` is the
value of the game :math:`\left(L,K,e_{1},e_{2}\right)` and
-:math:`\left(\xi,\gamma\right)` is optimal for it.
+:math:`\left(p,q\right)` is optimal for it.
The proof of this theorem is not difficult, and the version for
:math:`e_{1} \ne e_{2}` can easily be deduced from the one given by
.. math::
\begin{aligned}
&\text{ maximize } &\nu &\\
- &\text{ subject to }& \xi &\in K&\\
- & & \left\langle \xi,e_{1} \right\rangle &= 1&\\
+ &\text{ subject to }& p &\in K&\\
& & \nu &\in \mathbb{R}&\\
- & & L\left(\xi\right) &\succcurlyeq \nu e_{1}.&
+ & & \left\langle p,e_{1} \right\rangle &= 1&\\
+ & & L\left(p\right) &\succcurlyeq \nu e_{1}.&
\end{aligned}
Player two, on the other hand, would like to,
\begin{aligned}
&\text{ minimize } &\omega &\\
- &\text{ subject to }& \gamma &\in K&\\
- & & \left\langle \gamma,e_{2} \right\rangle &= 1&\\
+ &\text{ subject to }& q &\in K&\\
& & \omega &\in \mathbb{R}&\\
- & & L^{*}\left(\gamma\right) &\preccurlyeq \omega e_{2}.&
+ & & \left\langle q,e_{2} \right\rangle &= 1&\\
+ & & L^{*}\left(q\right) &\preccurlyeq \omega e_{2}.&
\end{aligned}
-The `CVXOPT <http://cvxopt.org/>` library can solve symmetric cone
+The `CVXOPT <http://cvxopt.org/>`_ library can solve symmetric cone
programs in the following primal/dual format:
.. math::
We will now pull a rabbit out of the hat, and choose the
matrices/vectors in these primal/dual programs so as to reconstruct
-exactly the goals of the two players. Let,
+the goals of the two players. Let,
.. math::
\begin{aligned}
C &= K \times K\\
- x &= \begin{bmatrix} \nu & \xi \end{bmatrix} \\
+ x &= \begin{bmatrix} \nu \\ p \end{bmatrix} \\
+ y &= \begin{bmatrix} \omega \end{bmatrix} \\
b &= \begin{bmatrix} 1 \end{bmatrix} \\
h &= 0 \\
- c &= \begin{bmatrix} -1 & 0 \end{bmatrix} \\
+ c &= \begin{bmatrix} -1 \\ 0 \end{bmatrix} \\
+ z &= \begin{bmatrix} z_{1} \\ q \end{bmatrix} \\
A &= \begin{bmatrix} 0 & e_{2}^{T} \end{bmatrix} \\
G &= \begin{bmatrix}
0 & -I\\