This module contains the main :class:`SymmetricLinearGame` class that
knows how to solve a linear game.
"""
-from math import sqrt
-
from cvxopt import matrix, printing, solvers
-from .cones import CartesianProduct, IceCream, NonnegativeOrthant
+from .cones import CartesianProduct
from .errors import GameUnsolvableException, PoorScalingException
from .matrices import (append_col, append_row, condition_number, identity,
inner_product, norm, specnorm)
-from . import options
+from .options import ABS_TOL, FLOAT_FORMAT, DEBUG_FLOAT_FORMAT
+
+printing.options['dformat'] = FLOAT_FORMAT
-printing.options['dformat'] = options.FLOAT_FORMAT
class Solution:
"""
if not self._e2 in K:
raise ValueError('the point e2 must lie in the interior of K')
+ # Initial value of cached method.
+ self._L_specnorm_value = None
def __str__(self):
:meth:`L` is satisfied.
"""
p = self.e2() / (norm(self.e2()) ** 2)
-
- # Compute the distance from p to the outside of K.
- if isinstance(self.K(), NonnegativeOrthant):
- # How far is it to a wall?
- dist = min(list(self.e1()))
- elif isinstance(self.K(), IceCream):
- # How far is it to the boundary of the ball that defines
- # the ice-cream cone at a given height? Now draw a
- # 45-45-90 triangle and the shortest distance to the
- # outside of the cone should be 1/sqrt(2) of that.
- # It works in R^2, so it works everywhere, right?
- height = self.e1()[0]
- radius = norm(self.e1()[1:])
- dist = (height - radius) / sqrt(2)
- else:
- raise NotImplementedError
-
- nu = - specnorm(self.L())/(dist*norm(self.e2()))
- x = matrix([nu,p], (self.dimension() + 1, 1))
+ dist = self.K().ball_radius(self.e1())
+ nu = - self._L_specnorm()/(dist*norm(self.e2()))
+ x = matrix([nu, p], (self.dimension() + 1, 1))
s = - self._G()*x
return {'x': x, 's': s}
Return a feasible starting point for player two.
"""
q = self.e1() / (norm(self.e1()) ** 2)
-
- # Compute the distance from p to the outside of K.
- if isinstance(self.K(), NonnegativeOrthant):
- # How far is it to a wall?
- dist = min(list(self.e2()))
- elif isinstance(self.K(), IceCream):
- # How far is it to the boundary of the ball that defines
- # the ice-cream cone at a given height? Now draw a
- # 45-45-90 triangle and the shortest distance to the
- # outside of the cone should be 1/sqrt(2) of that.
- # It works in R^2, so it works everywhere, right?
- height = self.e2()[0]
- radius = norm(self.e2()[1:])
- dist = (height - radius) / sqrt(2)
- else:
- raise NotImplementedError
-
- omega = specnorm(self.L())/(dist*norm(self.e1()))
+ dist = self.K().ball_radius(self.e2())
+ omega = self._L_specnorm()/(dist*norm(self.e1()))
y = matrix([omega])
z2 = q
z1 = y*self.e2() - self.L().trans()*z2
- z = matrix([z1,z2], (self.dimension()*2, 1))
+ z = matrix([z1, z2], (self.dimension()*2, 1))
return {'y': y, 'z': z}
+ def _L_specnorm(self):
+ """
+ Compute the spectral norm of ``L`` and cache it.
+ """
+ if self._L_specnorm_value is None:
+ self._L_specnorm_value = specnorm(self.L())
+ return self._L_specnorm_value
+
+ def tolerance_scale(self, solution):
+ # Don't return anything smaller than 1... we can't go below
+ # out "minimum tolerance."
+ norm_p1_opt = norm(solution.player1_optimal())
+ norm_p2_opt = norm(solution.player2_optimal())
+ scale = self._L_specnorm()*(norm_p1_opt + norm_p2_opt)
+ return max(1, scale/2.0)
+
+
def solution(self):
"""
Solve this linear game and return a :class:`Solution`.
[2.506...]
[0.000...]
+ This is another one that was difficult numerically, and caused
+ trouble even after we fixed the first two::
+
+ >>> from dunshire import *
+ >>> L = [[57.22233908627052301199, 41.70631373437460354126],
+ ... [83.04512571985074487202, 57.82581810406928468637]]
+ >>> K = NonnegativeOrthant(2)
+ >>> e1 = [7.31887017043399268346, 0.89744171905822367474]
+ >>> e2 = [0.11099824781179848388, 6.12564670639315345113]
+ >>> SLG = SymmetricLinearGame(L,K,e1,e2)
+ >>> print(SLG.solution())
+ Game value: 70.437...
+ Player 1 optimal:
+ [9.009...]
+ [0.000...]
+ Player 2 optimal:
+ [0.136...]
+ [0.000...]
+
+ And finally, here's one that returns an "optimal" solution, but
+ whose primal/dual objective function values are far apart::
+
+ >>> from dunshire import *
+ >>> L = [[ 6.49260076597376212248, -0.60528030227678542019],
+ ... [ 2.59896077096751731972, -0.97685530240286766457]]
+ >>> K = IceCream(2)
+ >>> e1 = [1, 0.43749513972645248661]
+ >>> e2 = [1, 0.46008379832200291260]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> print(SLG.solution())
+ Game value: 11.596...
+ Player 1 optimal:
+ [ 1.852...]
+ [-1.852...]
+ Player 2 optimal:
+ [ 1.777...]
+ [-1.777...]
+
"""
try:
opts = {'show_progress': False}
self.A(),
self.b(),
primalstart=self.player1_start(),
+ dualstart=self.player2_start(),
options=opts)
except ValueError as error:
if str(error) == 'math domain error':
# Oops, CVXOPT tried to take the square root of a
# negative number. Report some details about the game
# rather than just the underlying CVXOPT crash.
- printing.options['dformat'] = options.DEBUG_FLOAT_FORMAT
+ printing.options['dformat'] = DEBUG_FLOAT_FORMAT
raise PoorScalingException(self)
else:
raise error
# that CVXOPT is convinced the problem is infeasible (and that
# cannot happen).
if soln_dict['status'] in ['primal infeasible', 'dual infeasible']:
- printing.options['dformat'] = options.DEBUG_FLOAT_FORMAT
+ printing.options['dformat'] = DEBUG_FLOAT_FORMAT
raise GameUnsolvableException(self, soln_dict)
+ # For the game value, we could use any of:
+ #
+ # * p1_value
+ # * p2_value
+ # * (p1_value + p2_value)/2
+ # * the game payoff
+ #
+ # We want the game value to be the payoff, however, so it
+ # makes the most sense to just use that, even if it means we
+ # can't test the fact that p1_value/p2_value are close to the
+ # payoff.
+ payoff = self.payoff(p1_optimal, p2_optimal)
+ soln = Solution(payoff, p1_optimal, p2_optimal)
+
# The "optimal" and "unknown" results, we actually treat the
# same. Even if CVXOPT bails out due to numerical difficulty,
# it will have some candidate points in mind. If those
# close enough (one could be low by ABS_TOL, the other high by
# it) because otherwise CVXOPT might return "unknown" and give
# us two points in the cone that are nowhere near optimal.
- if abs(p1_value - p2_value) > 2*options.ABS_TOL:
- printing.options['dformat'] = options.DEBUG_FLOAT_FORMAT
+ #
+ if abs(p1_value - p2_value) > self.tolerance_scale(soln)*ABS_TOL:
+ printing.options['dformat'] = DEBUG_FLOAT_FORMAT
raise GameUnsolvableException(self, soln_dict)
# And we also check that the points it gave us belong to the
# cone, just in case...
if (p1_optimal not in self._K) or (p2_optimal not in self._K):
- printing.options['dformat'] = options.DEBUG_FLOAT_FORMAT
+ printing.options['dformat'] = DEBUG_FLOAT_FORMAT
raise GameUnsolvableException(self, soln_dict)
- # For the game value, we could use any of:
- #
- # * p1_value
- # * p2_value
- # * (p1_value + p2_value)/2
- # * the game payoff
- #
- # We want the game value to be the payoff, however, so it
- # makes the most sense to just use that, even if it means we
- # can't test the fact that p1_value/p2_value are close to the
- # payoff.
- payoff = self.payoff(p1_optimal, p2_optimal)
- return Solution(payoff, p1_optimal, p2_optimal)
+ return soln
def condition(self):
>>> e1 = [1]
>>> e2 = e1
>>> SLG = SymmetricLinearGame(L, K, e1, e2)
- >>> actual = SLG.condition()
- >>> expected = 1.8090169943749477
- >>> abs(actual - expected) < options.ABS_TOL
- True
+ >>> SLG.condition()
+ 1.809...
"""
return (condition_number(self._G()) + condition_number(self.A()))/2
projects / dunshire.git / blobdiff