This module contains the main :class:`SymmetricLinearGame` class that
knows how to solve a linear game.
"""
-
from cvxopt import matrix, printing, solvers
from .cones import CartesianProduct
from .errors import GameUnsolvableException, PoorScalingException
-from .matrices import append_col, append_row, condition_number, identity
-from . import options
+from .matrices import (append_col, append_row, condition_number, identity,
+ inner_product, norm, specnorm)
+from .options import ABS_TOL, FLOAT_FORMAT, DEBUG_FLOAT_FORMAT
+
+printing.options['dformat'] = FLOAT_FORMAT
-printing.options['dformat'] = options.FLOAT_FORMAT
class Solution:
"""
--------
>>> print(Solution(10, matrix([1,2]), matrix([3,4])))
- Game value: 10.0000000
+ Game value: 10.000...
Player 1 optimal:
[ 1]
[ 2]
----------
L : list of list of float
- A matrix represented as a list of ROWS. This representation
- agrees with (for example) SageMath and NumPy, but not with CVXOPT
- (whose matrix constructor accepts a list of columns).
-
- K : :class:`SymmetricCone`
+ A matrix represented as a list of **rows**. This representation
+ agrees with (for example) `SageMath <http://www.sagemath.org/>`_
+ and `NumPy <http://www.numpy.org/>`_, but not with CVXOPT (whose
+ matrix constructor accepts a list of columns). In reality, ``L``
+ can be any iterable type of the correct length; however, you
+ should be extremely wary of the way we interpret anything other
+ than a list of rows.
+
+ K : dunshire.cones.SymmetricCone
The symmetric cone instance over which the game is played.
e1 : iterable float
[ 1],
e2 = [ 1]
[ 2]
- [ 3],
- Condition((L, K, e1, e2)) = 31.834...
+ [ 3]
Lists can (and probably should) be used for every argument::
e1 = [ 1]
[ 1],
e2 = [ 1]
- [ 1],
- Condition((L, K, e1, e2)) = 1.707...
+ [ 1]
The points ``e1`` and ``e2`` can also be passed as some other
enumerable type (of the correct length) without much harm, since
e1 = [ 1]
[ 1],
e2 = [ 1]
- [ 1],
- Condition((L, K, e1, e2)) = 1.707...
+ [ 1]
However, ``L`` will always be intepreted as a list of rows, even
if it is passed as a :class:`cvxopt.base.matrix` which is
e1 = [ 1]
[ 1],
e2 = [ 1]
- [ 1],
- Condition((L, K, e1, e2)) = 6.073...
+ [ 1]
>>> L = cvxopt.matrix(L)
>>> print(L)
[ 1 3]
e1 = [ 1]
[ 1],
e2 = [ 1]
- [ 1],
- Condition((L, K, e1, e2)) = 6.073...
+ [ 1]
"""
def __init__(self, L, K, e1, e2):
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):
' L = {:s},\n' \
' K = {!s},\n' \
' e1 = {:s},\n' \
- ' e2 = {:s},\n' \
- ' Condition((L, K, e1, e2)) = {:f}.'
- indented_L = '\n '.join(str(self._L).splitlines())
- indented_e1 = '\n '.join(str(self._e1).splitlines())
- indented_e2 = '\n '.join(str(self._e2).splitlines())
+ ' e2 = {:s}'
+ indented_L = '\n '.join(str(self.L()).splitlines())
+ indented_e1 = '\n '.join(str(self.e1()).splitlines())
+ indented_e2 = '\n '.join(str(self.e2()).splitlines())
return tpl.format(indented_L,
- str(self._K),
+ str(self.K()),
indented_e1,
- indented_e2,
- self.condition())
+ indented_e2)
+
+
+ def L(self):
+ """
+ Return the matrix ``L`` passed to the constructor.
+
+ Returns
+ -------
+
+ matrix
+ The matrix that defines this game's :meth:`payoff` operator.
+
+ Examples
+ --------
+
+ >>> from dunshire import *
+ >>> K = NonnegativeOrthant(3)
+ >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
+ >>> e1 = [1,1,1]
+ >>> e2 = [1,2,3]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> print(SLG.L())
+ [ 1 -5 -15]
+ [ -1 2 -3]
+ [-12 -15 1]
+ <BLANKLINE>
+
+ """
+ return self._L
+
+
+ def K(self):
+ """
+ Return the cone over which this game is played.
+
+ Returns
+ -------
+
+ SymmetricCone
+ The :class:`SymmetricCone` over which this game is played.
+
+ Examples
+ --------
+
+ >>> from dunshire import *
+ >>> K = NonnegativeOrthant(3)
+ >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
+ >>> e1 = [1,1,1]
+ >>> e2 = [1,2,3]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> print(SLG.K())
+ Nonnegative orthant in the real 3-space
+
+ """
+ return self._K
+
+
+ def e1(self):
+ """
+ Return player one's interior point.
+
+ Returns
+ -------
+
+ matrix
+ The point interior to :meth:`K` affiliated with player one.
+
+ Examples
+ --------
+
+ >>> from dunshire import *
+ >>> K = NonnegativeOrthant(3)
+ >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
+ >>> e1 = [1,1,1]
+ >>> e2 = [1,2,3]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> print(SLG.e1())
+ [ 1]
+ [ 1]
+ [ 1]
+ <BLANKLINE>
+
+ """
+ return self._e1
+
+
+ def e2(self):
+ """
+ Return player two's interior point.
+
+ Returns
+ -------
+
+ matrix
+ The point interior to :meth:`K` affiliated with player one.
+
+ Examples
+ --------
+
+ >>> from dunshire import *
+ >>> K = NonnegativeOrthant(3)
+ >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
+ >>> e1 = [1,1,1]
+ >>> e2 = [1,2,3]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> print(SLG.e2())
+ [ 1]
+ [ 2]
+ [ 3]
+ <BLANKLINE>
+
+ """
+ return self._e2
+
+
+ def payoff(self, strategy1, strategy2):
+ r"""
+ Return the payoff associated with ``strategy1`` and ``strategy2``.
+
+ The payoff operator takes pairs of strategies to a real
+ number. For example, if player one's strategy is :math:`x` and
+ player two's strategy is :math:`y`, then the associated payoff
+ is :math:`\left\langle L\left(x\right),y \right\rangle \in
+ \mathbb{R}`. Here, :math:`L` denotes the same linear operator as
+ :meth:`L`. This method computes the payoff given the two
+ players' strategies.
+
+ Parameters
+ ----------
+
+ strategy1 : matrix
+ Player one's strategy.
+
+ strategy2 : matrix
+ Player two's strategy.
+
+ Returns
+ -------
+
+ float
+ The payoff for the game when player one plays ``strategy1``
+ and player two plays ``strategy2``.
+
+ Examples
+ --------
+
+ The value of the game should be the payoff at the optimal
+ strategies::
+
+ >>> from dunshire import *
+ >>> K = NonnegativeOrthant(3)
+ >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
+ >>> e1 = [1,1,1]
+ >>> e2 = [1,1,1]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> soln = SLG.solution()
+ >>> x_bar = soln.player1_optimal()
+ >>> y_bar = soln.player2_optimal()
+ >>> SLG.payoff(x_bar, y_bar) == soln.game_value()
+ True
+
+ """
+ return inner_product(self.L()*strategy1, strategy2)
+
+
+ def dimension(self):
+ """
+ Return the dimension of this game.
+
+ The dimension of a game is not needed for the theory, but it is
+ useful for the implementation. We define the dimension of a game
+ to be the dimension of its underlying cone. Or what is the same,
+ the dimension of the space from which the strategies are chosen.
+
+ Returns
+ -------
+
+ int
+ The dimension of the cone :meth:`K`, or of the space where
+ this game is played.
+
+ Examples
+ --------
+
+ The dimension of a game over the nonnegative quadrant in the
+ plane should be two (the dimension of the plane)::
+
+ >>> from dunshire import *
+ >>> K = NonnegativeOrthant(2)
+ >>> L = [[1,-5],[-1,2]]
+ >>> e1 = [1,1]
+ >>> e2 = [1,4]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> SLG.dimension()
+ 2
+
+ """
+ return self.K().dimension()
def _zero(self):
-------
matrix
- A ``K.dimension()``-by-``1`` column vector of zeros.
+ A ``self.dimension()``-by-``1`` column vector of zeros.
Examples
--------
<BLANKLINE>
"""
- return matrix(0, (self._K.dimension(), 1), tc='d')
+ return matrix(0, (self.dimension(), 1), tc='d')
- def _A(self):
- """
+ def A(self):
+ r"""
Return the matrix ``A`` used in our CVXOPT construction.
- This matrix ``A`` appears on the right-hand side of ``Ax = b``
- in the statement of the CVXOPT conelp program.
+ This matrix :math:`A` appears on the right-hand side of
+ :math:`Ax = b` in the `statement of the CVXOPT conelp program
+ <http://cvxopt.org/userguide/coneprog.html#linear-cone-programs>`_.
.. warning::
-------
matrix
- A ``1``-by-``(1 + K.dimension())`` row vector. Its first
- entry is zero, and the rest are the entries of ``e2``.
+ A ``1``-by-``(1 + self.dimension())`` row vector. Its first
+ entry is zero, and the rest are the entries of :meth:`e2`.
Examples
--------
>>> e1 = [1,1,1]
>>> e2 = [1,2,3]
>>> SLG = SymmetricLinearGame(L, K, e1, e2)
- >>> print(SLG._A())
+ >>> print(SLG.A())
[0.0000000 1.0000000 2.0000000 3.0000000]
<BLANKLINE>
"""
- return matrix([0, self._e2], (1, self._K.dimension() + 1), 'd')
+ return matrix([0, self.e2()], (1, self.dimension() + 1), 'd')
- def _G(self):
+ def G(self):
r"""
Return the matrix ``G`` used in our CVXOPT construction.
- Thus matrix ``G`` appears on the left-hand side of ``Gx + s = h``
- in the statement of the CVXOPT conelp program.
+ Thus matrix :math:`G` appears on the left-hand side of :math:`Gx
+ + s = h` in the `statement of the CVXOPT conelp program
+ <http://cvxopt.org/userguide/coneprog.html#linear-cone-programs>`_.
.. warning::
-------
matrix
- A ``2*K.dimension()``-by-``1 + K.dimension()`` matrix.
+ A ``2*self.dimension()``-by-``(1 + self.dimension())`` matrix.
Examples
--------
>>> e1 = [1,2,3]
>>> e2 = [1,1,1]
>>> SLG = SymmetricLinearGame(L, K, e1, e2)
- >>> print(SLG._G())
+ >>> print(SLG.G())
[ 0.0000000 -1.0000000 0.0000000 0.0000000]
[ 0.0000000 0.0000000 -1.0000000 0.0000000]
[ 0.0000000 0.0000000 0.0000000 -1.0000000]
<BLANKLINE>
"""
- I = identity(self._K.dimension())
- return append_row(append_col(self._zero(), -I),
- append_col(self._e1, -self._L))
+ identity_matrix = identity(self.dimension())
+ return append_row(append_col(self._zero(), -identity_matrix),
+ append_col(self.e1(), -self.L()))
- def _c(self):
- """
+ def c(self):
+ r"""
Return the vector ``c`` used in our CVXOPT construction.
- The column vector ``c`` appears in the objective function
- value ``<c,x>`` in the statement of the CVXOPT conelp program.
+ The column vector :math:`c` appears in the objective function
+ value :math:`\left\langle c,x \right\rangle` in the `statement
+ of the CVXOPT conelp program
+ <http://cvxopt.org/userguide/coneprog.html#linear-cone-programs>`_.
.. warning::
-------
matrix
- A ``K.dimension()``-by-``1`` column vector.
+ A :meth:`dimension`-by-``1`` column vector.
Examples
--------
>>> e1 = [1,2,3]
>>> e2 = [1,1,1]
>>> SLG = SymmetricLinearGame(L, K, e1, e2)
- >>> print(SLG._c())
+ >>> print(SLG.c())
[-1.0000000]
[ 0.0000000]
[ 0.0000000]
return matrix([-1, self._zero()])
- def _C(self):
+ def C(self):
"""
Return the cone ``C`` used in our CVXOPT construction.
- The cone ``C`` is the cone over which the conelp program takes
- place.
+ This is the cone over which the `CVXOPT conelp program
+ <http://cvxopt.org/userguide/coneprog.html#linear-cone-programs>`_
+ takes place.
Returns
-------
>>> e1 = [1,2,3]
>>> e2 = [1,1,1]
>>> SLG = SymmetricLinearGame(L, K, e1, e2)
- >>> print(SLG._C())
+ >>> print(SLG.C())
Cartesian product of dimension 6 with 2 factors:
* Nonnegative orthant in the real 3-space
* Nonnegative orthant in the real 3-space
"""
return CartesianProduct(self._K, self._K)
- def _h(self):
- """
+ def h(self):
+ r"""
Return the ``h`` vector used in our CVXOPT construction.
- The ``h`` vector appears on the right-hand side of :math:`Gx + s
- = h` in the statement of the CVXOPT conelp program.
+ The :math:`h` vector appears on the right-hand side of :math:`Gx
+ + s = h` in the `statement of the CVXOPT conelp program
+ <http://cvxopt.org/userguide/coneprog.html#linear-cone-programs>`_.
.. warning::
-------
matrix
- A ``2*K.dimension()``-by-``1`` column vector of zeros.
+ A ``2*self.dimension()``-by-``1`` column vector of zeros.
Examples
--------
>>> e1 = [1,2,3]
>>> e2 = [1,1,1]
>>> SLG = SymmetricLinearGame(L, K, e1, e2)
- >>> print(SLG._h())
+ >>> print(SLG.h())
[0.0000000]
[0.0000000]
[0.0000000]
return matrix([self._zero(), self._zero()])
- def _b(self):
- """
+
+ @staticmethod
+ def b():
+ r"""
Return the ``b`` vector used in our CVXOPT construction.
- The vector ``b`` appears on the right-hand side of :math:`Ax =
- b` in the statement of the CVXOPT conelp program.
+ The vector :math:`b` appears on the right-hand side of :math:`Ax
+ = b` in the `statement of the CVXOPT conelp program
+ <http://cvxopt.org/userguide/coneprog.html#linear-cone-programs>`_.
+
+ This method is static because the dimensions and entries of
+ ``b`` are known beforehand, and don't depend on any other
+ properties of the game.
.. warning::
>>> e1 = [1,2,3]
>>> e2 = [1,1,1]
>>> SLG = SymmetricLinearGame(L, K, e1, e2)
- >>> print(SLG._b())
+ >>> print(SLG.b())
[1.0000000]
<BLANKLINE>
return matrix([1], tc='d')
- def _try_solution(self, tolerance):
+ def player1_start(self):
"""
- Solve this linear game within ``tolerance``, if possible.
+ Return a feasible starting point for player one.
- This private function is the one that does all of the actual
- work for :meth:`solution`. This method accepts a ``tolerance``,
- and what :meth:`solution` does is call this method twice with
- two different tolerances. First it tries a strict tolerance, and
- then it tries a looser one.
+ This starting point is for the CVXOPT formulation and not for
+ the original game. The basic premise is that if you scale
+ :meth:`e2` by the reciprocal of its squared norm, then you get a
+ point in :meth:`K` that makes a unit inner product with
+ :meth:`e2`. We then get to choose the primal objective function
+ value such that the constraint involving :meth:`L` is satisfied.
- .. warning::
+ Returns
+ -------
- If you try to be smart and precompute the matrices used by
- this function (the ones passed to ``conelp``), then you're
- going to shoot yourself in the foot. CVXOPT can and will
- clobber some (but not all) of its input matrices. This isn't
- performance sensitive, so play it safe.
+ dict
+ A dictionary with two keys, ``'x'`` and ``'s'``, which
+ contain the vectors of the same name in the CVXOPT primal
+ problem formulation.
+
+ The vector ``x`` consists of the primal objective function
+ value concatenated with the strategy (for player one) that
+ achieves it. The vector ``s`` is essentially a dummy
+ variable, and is computed from the equality constraing in
+ the CVXOPT primal problem.
+
+ """
+ p = self.e2() / (norm(self.e2()) ** 2)
+ 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}
- Parameters
- ----------
- tolerance : float
- The absolute tolerance to pass to the CVXOPT solver.
+ def player2_start(self):
+ """
+ Return a feasible starting point for player two.
+
+ This starting point is for the CVXOPT formulation and not for
+ the original game. The basic premise is that if you scale
+ :meth:`e1` by the reciprocal of its squared norm, then you get a
+ point in :meth:`K` that makes a unit inner product with
+ :meth:`e1`. We then get to choose the dual objective function
+ value such that the constraint involving :meth:`L` is satisfied.
Returns
-------
- :class:`Solution`
- A :class:`Solution` object describing the game's value and
- the optimal strategies of both players.
+ dict
+ A dictionary with two keys, ``'y'`` and ``'z'``, which
+ contain the vectors of the same name in the CVXOPT dual
+ problem formulation.
- Raises
- ------
- GameUnsolvableException
- If the game could not be solved (if an optimal solution to its
- associated cone program was not found).
+ The ``1``-by-``1`` vector ``y`` consists of the dual
+ objective function value. The last :meth:`dimension` entries
+ of the vector ``z`` contain the strategy (for player two)
+ that achieves it. The remaining entries of ``z`` are
+ essentially dummy variables, computed from the equality
+ constraint in the CVXOPT dual problem.
+
+ """
+ q = self.e1() / (norm(self.e1()) ** 2)
+ 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))
+
+ return {'y': y, 'z': z}
- PoorScalingException
- If the game could not be solved because CVXOPT crashed while
- trying to take the square root of a negative number.
+
+ def _L_specnorm(self):
+ """
+ Compute the spectral norm of :meth:`L` and cache it.
+
+ The spectral norm of the matrix :meth:`L` is used in a few
+ places. Since it can be expensive to compute, we want to cache
+ its value. That is not possible in :func:`specnorm`, which lies
+ outside of a class, so this is the place to do it.
+
+ Returns
+ -------
+
+ float
+ A nonnegative real number; the largest singular value of
+ the matrix :meth:`L`.
Examples
--------
- This game can be solved easily, so the first attempt in
- :meth:`solution` should succeed::
-
>>> from dunshire import *
- >>> from dunshire.matrices import norm
- >>> from dunshire.options import ABS_TOL
- >>> K = NonnegativeOrthant(3)
- >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
- >>> e1 = [1,1,1]
- >>> e2 = [1,1,1]
- >>> SLG = SymmetricLinearGame(L, K, e1, e2)
- >>> s1 = SLG.solution()
- >>> s2 = SLG._try_solution(options.ABS_TOL)
- >>> abs(s1.game_value() - s2.game_value()) < ABS_TOL
- True
- >>> norm(s1.player1_optimal() - s2.player1_optimal()) < ABS_TOL
- True
- >>> norm(s1.player2_optimal() - s2.player2_optimal()) < ABS_TOL
+ >>> from dunshire.matrices import specnorm
+ >>> L = [[1,2],[3,4]]
+ >>> K = NonnegativeOrthant(2)
+ >>> e1 = [1,1]
+ >>> e2 = e1
+ >>> SLG = SymmetricLinearGame(L,K,e1,e2)
+ >>> specnorm(SLG.L()) == SLG._L_specnorm()
True
"""
- try:
- solvers.options['show_progress'] = options.VERBOSE
- solvers.options['abs_tol'] = tolerance
- soln_dict = solvers.conelp(self._c(),
- self._G(),
- self._h(),
- self._C().cvxopt_dims(),
- self._A(),
- self._b())
- except ValueError as e:
- if str(e) == '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.
- raise PoorScalingException(self)
- else:
- raise e
+ if self._L_specnorm_value is None:
+ self._L_specnorm_value = specnorm(self.L())
+ return self._L_specnorm_value
- # The optimal strategies are named ``p`` and ``q`` in the
- # background documentation, and we need to extract them from
- # the CVXOPT ``x`` and ``z`` variables. The objective values
- # :math:`nu` and :math:`omega` can also be found in the CVXOPT
- # ``x`` and ``y`` variables; however, they're stored
- # conveniently as separate entries in the solution dictionary.
- p1_value = -soln_dict['primal objective']
- p2_value = -soln_dict['dual objective']
- p1_optimal = soln_dict['x'][1:]
- p2_optimal = soln_dict['z'][self._K.dimension():]
- # The "status" field contains "optimal" if everything went
- # according to plan. Other possible values are "primal
- # infeasible", "dual infeasible", "unknown", all of which mean
- # we didn't get a solution. The "infeasible" ones are the
- # worst, since they indicate that CVXOPT is convinced the
- # problem is infeasible (and that cannot happen).
- if soln_dict['status'] in ['primal infeasible', 'dual infeasible']:
- raise GameUnsolvableException(self, soln_dict)
- elif soln_dict['status'] == 'unknown':
- # When we get a status of "unknown", we may still be able
- # to salvage a solution out of the returned
- # dictionary. Often this is the result of numerical
- # difficulty and we can simply check that the primal/dual
- # objectives match (within a tolerance) and that the
- # primal/dual optimal solutions are within the cone (to a
- # tolerance as well).
- #
- # The fudge factor of two is basically unjustified, but
- # makes intuitive sense when you imagine that the primal
- # value could be under the true optimal by ``ABS_TOL``
- # and the dual value could be over by the same amount.
- #
- if abs(p1_value - p2_value) > tolerance:
- raise GameUnsolvableException(self, soln_dict)
- if (p1_optimal not in self._K) or (p2_optimal not in self._K):
- raise GameUnsolvableException(self, soln_dict)
-
- return Solution(p1_value, p1_optimal, p2_optimal)
+ def tolerance_scale(self, solution):
+ r"""
+
+ Return a scaling factor that should be applied to
+ :const:`dunshire.options.ABS_TOL` for this game.
+
+ When performing certain comparisons, the default tolerance
+ :const:`dunshire.options.ABS_TOL` may not be appropriate. For
+ example, if we expect ``x`` and ``y`` to be within
+ :const:`dunshire.options.ABS_TOL` of each other, than the inner
+ product of ``L*x`` and ``y`` can be as far apart as the spectral
+ norm of ``L`` times the sum of the norms of ``x`` and
+ ``y``. Such a comparison is made in :meth:`solution`, and in
+ many of our unit tests.
+
+ The returned scaling factor found from the inner product
+ mentioned above is
+
+ .. math::
+
+ \left\lVert L \right\rVert_{2}
+ \left( \left\lVert \bar{x} \right\rVert
+ + \left\lVert \bar{y} \right\rVert
+ \right),
+
+ where :math:`\bar{x}` and :math:`\bar{y}` are optimal solutions
+ for players one and two respectively. This scaling factor is not
+ formally justified, but attempting anything smaller leads to
+ test failures.
+
+ .. warning::
+
+ Optimal solutions are not unique, so the scaling factor
+ obtained from ``solution`` may not work when comparing other
+ solutions.
+
+ Parameters
+ ----------
+
+ solution : Solution
+ A solution of this game, used to obtain the norms of the
+ optimal strategies.
+
+ Returns
+ -------
+
+ float
+ A scaling factor to be multiplied by
+ :const:`dunshire.options.ABS_TOL` when
+ making comparisons involving solutions of this game.
+
+ Examples
+ --------
+
+ The spectral norm of ``L`` in this case is around ``5.464``, and
+ the optimal strategies both have norm one, so we expect the
+ tolerance scale to be somewhere around ``2 * 5.464``, or
+ ``10.929``::
+
+ >>> from dunshire import *
+ >>> L = [[1,2],[3,4]]
+ >>> K = NonnegativeOrthant(2)
+ >>> e1 = [1,1]
+ >>> e2 = e1
+ >>> SLG = SymmetricLinearGame(L,K,e1,e2)
+ >>> SLG.tolerance_scale(SLG.solution())
+ 10.929...
+
+ """
+ norm_p1_opt = norm(solution.player1_optimal())
+ norm_p2_opt = norm(solution.player2_optimal())
+ scale = self._L_specnorm()*(norm_p1_opt + norm_p2_opt)
+
+ # Don't return anything smaller than 1... we can't go below
+ # out "minimum tolerance."
+ return max(1, scale)
def solution(self):
Returns
-------
- :class:`Solution`
+ Solution
A :class:`Solution` object describing the game's value and
the optimal strategies of both players.
>>> e2 = [1,1,1]
>>> SLG = SymmetricLinearGame(L, K, e1, e2)
>>> print(SLG.solution())
- Game value: -6.1724138
+ Game value: -6.172...
Player 1 optimal:
- [ 0.551...]
- [-0.000...]
- [ 0.448...]
+ [0.551...]
+ [0.000...]
+ [0.448...]
Player 2 optimal:
[0.448...]
[0.000...]
>>> e2 = [4,5,6]
>>> SLG = SymmetricLinearGame(L, K, e1, e2)
>>> print(SLG.solution())
- Game value: 0.0312500
+ Game value: 0.031...
Player 1 optimal:
[0.031...]
[0.062...]
[0.156...]
[0.187...]
+ This is another Gowda/Ravindran example that is supposed to have
+ a negative game value::
+
+ >>> from dunshire import *
+ >>> from dunshire.options import ABS_TOL
+ >>> L = [[1, -2], [-2, 1]]
+ >>> K = NonnegativeOrthant(2)
+ >>> e1 = [1, 1]
+ >>> e2 = e1
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> SLG.solution().game_value() < -ABS_TOL
+ True
+
+ The following two games are problematic numerically, but we
+ should be able to solve them::
+
+ >>> from dunshire import *
+ >>> L = [[-0.95237953890954685221, 1.83474556206462535712],
+ ... [ 1.30481749924621448500, 1.65278664543326403447]]
+ >>> K = NonnegativeOrthant(2)
+ >>> e1 = [0.95477167524644313001, 0.63270781756540095397]
+ >>> e2 = [0.39633793037154141370, 0.10239281495640320530]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> print(SLG.solution())
+ Game value: 18.767...
+ Player 1 optimal:
+ [0.000...]
+ [9.766...]
+ Player 2 optimal:
+ [1.047...]
+ [0.000...]
+
+ ::
+
+ >>> from dunshire import *
+ >>> L = [[1.54159395026049472754, 2.21344728574316684799],
+ ... [1.33147433507846657541, 1.17913616272988108769]]
+ >>> K = NonnegativeOrthant(2)
+ >>> e1 = [0.39903040089404784307, 0.12377403622479113410]
+ >>> e2 = [0.15695181142215544612, 0.85527381344651265405]
+ >>> SLG = SymmetricLinearGame(L, K, e1, e2)
+ >>> print(SLG.solution())
+ Game value: 24.614...
+ Player 1 optimal:
+ [6.371...]
+ [0.000...]
+ Player 2 optimal:
+ [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:
- # First try with a stricter tolerance. Who knows, it might
- # work. If it does, we prefer that solution.
- return self._try_solution(options.ABS_TOL / 10)
+ opts = {'show_progress': False}
+ soln_dict = solvers.conelp(self.c(),
+ self.G(),
+ self.h(),
+ self.C().cvxopt_dims(),
+ 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'] = DEBUG_FLOAT_FORMAT
+ raise PoorScalingException(self)
+ else:
+ raise error
- except (PoorScalingException, GameUnsolvableException):
- # Ok, that didn't work. Let's try it with the default
- # tolerance, and whatever happens, happens.
- return self._try_solution(tolerance = options.ABS_TOL)
+ # The optimal strategies are named ``p`` and ``q`` in the
+ # background documentation, and we need to extract them from
+ # the CVXOPT ``x`` and ``z`` variables. The objective values
+ # :math:`nu` and :math:`omega` can also be found in the CVXOPT
+ # ``x`` and ``y`` variables; however, they're stored
+ # conveniently as separate entries in the solution dictionary.
+ p1_value = -soln_dict['primal objective']
+ p2_value = -soln_dict['dual objective']
+ p1_optimal = soln_dict['x'][1:]
+ p2_optimal = soln_dict['z'][self.dimension():]
+
+ # The "status" field contains "optimal" if everything went
+ # according to plan. Other possible values are "primal
+ # infeasible", "dual infeasible", "unknown", all of which mean
+ # we didn't get a solution.
+ #
+ # The "infeasible" ones are the worst, since they indicate
+ # that CVXOPT is convinced the problem is infeasible (and that
+ # cannot happen).
+ if soln_dict['status'] in ['primal infeasible', 'dual infeasible']:
+ 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
+ # candidates are good enough, we take them. We do the same
+ # check for "optimal" results.
+ #
+ # First we check that the primal/dual objective values are
+ # close enough because otherwise CVXOPT might return "unknown"
+ # and give us two points in the cone that are nowhere near
+ # optimal. And in fact, we need to ensure that they're close
+ # for "optimal" results, too, because we need to know how
+ # lenient to be in our testing.
+ #
+ 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'] = DEBUG_FLOAT_FORMAT
+ raise GameUnsolvableException(self, soln_dict)
+
+ return soln
def condition(self):
can show up. We define the condition number of this game to be
the average of the condition numbers of ``G`` and ``A`` in the
CVXOPT construction. If the condition number of this game is
- high, then you can expect numerical difficulty (such as
- :class:`PoorScalingException`).
+ high, you can problems like :class:`PoorScalingException`.
+
+ Random testing shows that a condition number of around ``125``
+ is about the best that we can solve reliably. However, the
+ failures are intermittent, and you may get lucky with an
+ ill-conditioned game.
Returns
-------
>>> 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
+ return (condition_number(self.G()) + condition_number(self.A()))/2
def dual(self):
[ 3],
e2 = [ 1]
[ 1]
- [ 1],
- Condition((L, K, e1, e2)) = 44.476...
+ [ 1]
"""
- # We pass ``self._L`` right back into the constructor, because
+ # We pass ``self.L()`` right back into the constructor, because
# it will be transposed there. And keep in mind that ``self._K``
# is its own dual.
- return SymmetricLinearGame(self._L,
- self._K,
- self._e2,
- self._e1)
+ return SymmetricLinearGame(self.L(),
+ self.K(),
+ self.e2(),
+ self.e1())
projects / dunshire.git / blobdiff