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:
"""
----------
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}.'
+ ' 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()),
indented_e1,
- indented_e2,
- self.condition())
+ indented_e2)
def L(self):
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
+ 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.
strategies::
>>> from dunshire import *
- >>> from dunshire.options import ABS_TOL
>>> K = NonnegativeOrthant(3)
>>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
>>> e1 = [1,1,1]
>>> soln = SLG.solution()
>>> x_bar = soln.player1_optimal()
>>> y_bar = soln.player2_optimal()
- >>> abs(SLG.payoff(x_bar, y_bar) - soln.game_value()) < ABS_TOL
+ >>> SLG.payoff(x_bar, y_bar) == soln.game_value()
True
"""
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 + self.dimension())`` row vector. Its first
- entry is zero, and the rest are the entries of ``e2``.
+ entry is zero, and the rest are the entries of :meth:`e2`.
Examples
--------
- 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::
>>> 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]
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 ``self.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 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
-------
"""
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::
>>> 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]
@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
Return a feasible starting point for player one.
This starting point is for the CVXOPT formulation and not for
- the original game. The basic premise is that if you normalize
- :meth:`e2`, 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.
+ 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.
+
+ Returns
+ -------
+
+ 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)
-
- # 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))
- s = - self._G()*x
+ 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}
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
+ -------
+
+ dict
+ A dictionary with two keys, ``'y'`` and ``'z'``, which
+ contain the vectors of the same name in the CVXOPT dual
+ problem formulation.
+
+ 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)
-
- # 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 :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
+ --------
+
+ >>> from dunshire import *
+ >>> 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
+
+ """
+ if self._L_specnorm_value is None:
+ self._L_specnorm_value = specnorm(self.L())
+ return self._L_specnorm_value
+
+
+ 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):
"""
Solve this linear game and return a :class:`Solution`.
Returns
-------
- :class:`Solution`
+ Solution
A :class:`Solution` object describing the game's value and
the optimal strategies of both players.
[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}
- soln_dict = solvers.conelp(self._c(),
- self._G(),
- self._h(),
+ 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'] = 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
# candidates are good enough, we take them. We do the same
- # check (perhaps pointlessly so) for "optimal" results.
+ # check for "optimal" results.
#
# First we check that the primal/dual objective values are
- # 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
+ # 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'] = 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):
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
projects / dunshire.git / blobdiff