Enable the dual starting point and fix the test tolerance.

[dunshire.git] / dunshire / games.py

diff --git a/dunshire/games.py b/dunshire/games.py
index a1ac0f54c2982cd59b25cc479293c4e3dc558e44..cfb62a33d8a2666828323019729af0f1b5ab99d7 100644 (file)
--- a/dunshire/games.py
+++ b/dunshire/games.py
@@ -13,6 +13,7 @@ from . import options
 
 printing.options['dformat'] = options.FLOAT_FORMAT
 
+
 class Solution:
     """
     A representation of the solution of a linear game. It should contain
@@ -853,6 +854,15 @@ class SymmetricLinearGame:
             self._L_specnorm_value = specnorm(self.L())
         return self._L_specnorm_value
 
+    def epsilon_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)
+
+
     def solution(self):
         """
         Solve this linear game and return a :class:`Solution`.
@@ -977,6 +987,7 @@ class SymmetricLinearGame:
                                        self.A(),
                                        self.b(),
                                        primalstart=self.player1_start(),
+                                       dualstart=self.player2_start(),
                                        options=opts)
         except ValueError as error:
             if str(error) == 'math domain error':
@@ -1011,6 +1022,20 @@ class SymmetricLinearGame:
             printing.options['dformat'] = options.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
@@ -1021,7 +1046,8 @@ class SymmetricLinearGame:
         # 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:
+        #
+        if abs(p1_value - p2_value) > self.epsilon_scale(soln)*options.ABS_TOL:
             printing.options['dformat'] = options.DEBUG_FLOAT_FORMAT
             raise GameUnsolvableException(self, soln_dict)
 
@@ -1031,19 +1057,7 @@ class SymmetricLinearGame:
             printing.options['dformat'] = options.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):