* The optimal strategy of player one.
* The optimal strategy of player two.
- """
+ EXAMPLES:
+
+ >>> print(Solution(10, matrix([1,2]), matrix([3,4])))
+ Game value: 10.0000000
+ Player 1 optimal:
+ [ 1]
+ [ 2]
+ Player 2 optimal:
+ [ 3]
+ [ 4]
+ """
tpl = 'Game value: {:.7f}\n' \
'Player 1 optimal:{:s}\n' \
- 'Player 2 optimal:{:s}\n'
+ 'Player 2 optimal:{:s}'
p1_str = '\n{!s}'.format(self.player1_optimal())
p1_str = '\n '.join(p1_str.splitlines())
def __str__(self):
"""
Return a string representatoin of this game.
+
+ EXAMPLES:
+
+ >>> from cones import NonnegativeOrthant
+ >>> K = NonnegativeOrthant(3)
+ >>> L = [[1,-1,-12],[-5,2,-15],[-15,-3,1]]
+ >>> e1 = [1,1,1]
+ >>> e2 = [1,2,3]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> print(SLG)
+ The linear game (L, K, e1, e2) where
+ L = [ 1 -5 -15]
+ [ -1 2 -3]
+ [-12 -15 1],
+ K = Nonnegative orthant in the real 3-space,
+ e1 = [ 1]
+ [ 1]
+ [ 1],
+ e2 = [ 1]
+ [ 2]
+ [ 3].
+
"""
- return "a game"
+ tpl = 'The linear game (L, K, e1, e2) where\n' \
+ ' L = {:s},\n' \
+ ' K = {!s},\n' \
+ ' e1 = {:s},\n' \
+ ' e2 = {:s}.'
+ L_str = '\n '.join(str(self._L).splitlines())
+ e1_str = '\n '.join(str(self._e1).splitlines())
+ e2_str = '\n '.join(str(self._e2).splitlines())
+ return tpl.format(L_str, str(self._K), e1_str, e2_str)
+
def solution(self):
"""
could *not* be solved -- which should never happen -- then a
GameUnsolvableException is raised. It can be printed to get the
raw output from CVXOPT.
+
+ EXAMPLES:
+
+ This example is computed in Gowda and Ravindran in the section
+ "The value of a Z-transformation":
+
+ >>> from cones import NonnegativeOrthant
+ >>> K = NonnegativeOrthant(3)
+ >>> L = [[1,-1,-12],[-5,2,-15],[-15,-3,1]]
+ >>> e1 = [1,1,1]
+ >>> e2 = [1,1,1]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> print(SLG.solution())
+ Game value: -6.1724138
+ Player 1 optimal:
+ [ 0.5517241]
+ [-0.0000000]
+ [ 0.4482759]
+ Player 2 optimal:
+ [0.4482759]
+ [0.0000000]
+ [0.5517241]
+
+ The value of the following game can be computed using the fact
+ that the identity is invertible:
+
+ >>> from cones import NonnegativeOrthant
+ >>> K = NonnegativeOrthant(3)
+ >>> L = [[1,0,0],[0,1,0],[0,0,1]]
+ >>> e1 = [1,2,3]
+ >>> e2 = [4,5,6]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> print(SLG.solution())
+ Game value: 0.0312500
+ Player 1 optimal:
+ [0.0312500]
+ [0.0625000]
+ [0.0937500]
+ Player 2 optimal:
+ [0.1250000]
+ [0.1562500]
+ [0.1875000]
+
"""
# The cone "C" that appears in the statement of the CVXOPT
# conelp program.
# The matrix "A" that appears on the right-hand side of Ax = b
# in the statement of the CVXOPT conelp program.
- A = matrix([0, self._e1], (1, self._K.dimension() + 1), 'd')
+ A = matrix([0, self._e2], (1, self._K.dimension() + 1), 'd')
# Actually solve the thing and obtain a dictionary describing
# what happened.
def dual(self):
"""
Return the dual game to this game.
+
+ EXAMPLES:
+
+ >>> from cones import NonnegativeOrthant
+ >>> K = NonnegativeOrthant(3)
+ >>> L = [[1,-1,-12],[-5,2,-15],[-15,-3,1]]
+ >>> e1 = [1,1,1]
+ >>> e2 = [1,2,3]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> print(SLG.dual())
+ The linear game (L, K, e1, e2) where
+ L = [ 1 -1 -12]
+ [ -5 2 -15]
+ [-15 -3 1],
+ K = Nonnegative orthant in the real 3-space,
+ e1 = [ 1]
+ [ 2]
+ [ 3],
+ e2 = [ 1]
+ [ 1]
+ [ 1].
+
"""
return SymmetricLinearGame(self._L.trans(),
self._K, # Since "K" is symmetric.