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1 """
2 Symmetric linear games and their solutions.
3
4 This module contains the main :class:`SymmetricLinearGame` class that
5 knows how to solve a linear game.
6 """
7 from cvxopt import matrix, printing, solvers
8 from .cones import CartesianProduct
9 from .errors import GameUnsolvableException, PoorScalingException
10 from .matrices import (append_col, append_row, condition_number, identity,
11 inner_product, norm, specnorm)
12 from .options import ABS_TOL, FLOAT_FORMAT, DEBUG_FLOAT_FORMAT
13
14 printing.options['dformat'] = FLOAT_FORMAT
15
16
17 class Solution:
18 """
19 A representation of the solution of a linear game. It should contain
20 the value of the game, and both players' strategies.
21
22 Examples
23 --------
24
25 >>> print(Solution(10, matrix([1,2]), matrix([3,4])))
26 Game value: 10.000...
27 Player 1 optimal:
28 [ 1]
29 [ 2]
30 Player 2 optimal:
31 [ 3]
32 [ 4]
33
34 """
35 def __init__(self, game_value, p1_optimal, p2_optimal):
36 """
37 Create a new Solution object from a game value and two optimal
38 strategies for the players.
39 """
40 self._game_value = game_value
41 self._player1_optimal = p1_optimal
42 self._player2_optimal = p2_optimal
43
44 def __str__(self):
45 """
46 Return a string describing the solution of a linear game.
47
48 The three data that are described are,
49
50 * The value of the game.
51 * The optimal strategy of player one.
52 * The optimal strategy of player two.
53
54 The two optimal strategy vectors are indented by two spaces.
55 """
56 tpl = 'Game value: {:.7f}\n' \
57 'Player 1 optimal:{:s}\n' \
58 'Player 2 optimal:{:s}'
59
60 p1_str = '\n{!s}'.format(self.player1_optimal())
61 p1_str = '\n '.join(p1_str.splitlines())
62 p2_str = '\n{!s}'.format(self.player2_optimal())
63 p2_str = '\n '.join(p2_str.splitlines())
64
65 return tpl.format(self.game_value(), p1_str, p2_str)
66
67
68 def game_value(self):
69 """
70 Return the game value for this solution.
71
72 Examples
73 --------
74
75 >>> s = Solution(10, matrix([1,2]), matrix([3,4]))
76 >>> s.game_value()
77 10
78
79 """
80 return self._game_value
81
82
83 def player1_optimal(self):
84 """
85 Return player one's optimal strategy in this solution.
86
87 Examples
88 --------
89
90 >>> s = Solution(10, matrix([1,2]), matrix([3,4]))
91 >>> print(s.player1_optimal())
92 [ 1]
93 [ 2]
94 <BLANKLINE>
95
96 """
97 return self._player1_optimal
98
99
100 def player2_optimal(self):
101 """
102 Return player two's optimal strategy in this solution.
103
104 Examples
105 --------
106
107 >>> s = Solution(10, matrix([1,2]), matrix([3,4]))
108 >>> print(s.player2_optimal())
109 [ 3]
110 [ 4]
111 <BLANKLINE>
112
113 """
114 return self._player2_optimal
115
116
117 class SymmetricLinearGame:
118 r"""
119 A representation of a symmetric linear game.
120
121 The data for a symmetric linear game are,
122
123 * A "payoff" operator ``L``.
124 * A symmetric cone ``K``.
125 * Two points ``e1`` and ``e2`` in the interior of ``K``.
126
127 The ambient space is assumed to be the span of ``K``.
128
129 With those data understood, the game is played as follows. Players
130 one and two choose points :math:`x` and :math:`y` respectively, from
131 their respective strategy sets,
132
133 .. math::
134 \begin{aligned}
135 \Delta_{1}
136 &=
137 \left\{
138 x \in K \ \middle|\ \left\langle x, e_{2} \right\rangle = 1
139 \right\}\\
140 \Delta_{2}
141 &=
142 \left\{
143 y \in K \ \middle|\ \left\langle y, e_{1} \right\rangle = 1
144 \right\}.
145 \end{aligned}
146
147 Afterwards, a "payout" is computed as :math:`\left\langle
148 L\left(x\right), y \right\rangle` and is paid to player one out of
149 player two's pocket. The game is therefore zero sum, and we suppose
150 that player one would like to guarantee himself the largest minimum
151 payout possible. That is, player one wishes to,
152
153 .. math::
154 \begin{aligned}
155 \text{maximize }
156 &\underset{y \in \Delta_{2}}{\min}\left(
157 \left\langle L\left(x\right), y \right\rangle
158 \right)\\
159 \text{subject to } & x \in \Delta_{1}.
160 \end{aligned}
161
162 Player two has the simultaneous goal to,
163
164 .. math::
165 \begin{aligned}
166 \text{minimize }
167 &\underset{x \in \Delta_{1}}{\max}\left(
168 \left\langle L\left(x\right), y \right\rangle
169 \right)\\
170 \text{subject to } & y \in \Delta_{2}.
171 \end{aligned}
172
173 These goals obviously conflict (the game is zero sum), but an
174 existence theorem guarantees at least one optimal min-max solution
175 from which neither player would like to deviate. This class is
176 able to find such a solution.
177
178 Parameters
179 ----------
180
181 L : list of list of float
182 A matrix represented as a list of ROWS. This representation
183 agrees with (for example) SageMath and NumPy, but not with CVXOPT
184 (whose matrix constructor accepts a list of columns).
185
186 K : :class:`SymmetricCone`
187 The symmetric cone instance over which the game is played.
188
189 e1 : iterable float
190 The interior point of ``K`` belonging to player one; it
191 can be of any iterable type having the correct length.
192
193 e2 : iterable float
194 The interior point of ``K`` belonging to player two; it
195 can be of any enumerable type having the correct length.
196
197 Raises
198 ------
199
200 ValueError
201 If either ``e1`` or ``e2`` lie outside of the cone ``K``.
202
203 Examples
204 --------
205
206 >>> from dunshire import *
207 >>> K = NonnegativeOrthant(3)
208 >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
209 >>> e1 = [1,1,1]
210 >>> e2 = [1,2,3]
211 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
212 >>> print(SLG)
213 The linear game (L, K, e1, e2) where
214 L = [ 1 -5 -15]
215 [ -1 2 -3]
216 [-12 -15 1],
217 K = Nonnegative orthant in the real 3-space,
218 e1 = [ 1]
219 [ 1]
220 [ 1],
221 e2 = [ 1]
222 [ 2]
223 [ 3]
224
225 Lists can (and probably should) be used for every argument::
226
227 >>> from dunshire import *
228 >>> K = NonnegativeOrthant(2)
229 >>> L = [[1,0],[0,1]]
230 >>> e1 = [1,1]
231 >>> e2 = [1,1]
232 >>> G = SymmetricLinearGame(L, K, e1, e2)
233 >>> print(G)
234 The linear game (L, K, e1, e2) where
235 L = [ 1 0]
236 [ 0 1],
237 K = Nonnegative orthant in the real 2-space,
238 e1 = [ 1]
239 [ 1],
240 e2 = [ 1]
241 [ 1]
242
243 The points ``e1`` and ``e2`` can also be passed as some other
244 enumerable type (of the correct length) without much harm, since
245 there is no row/column ambiguity::
246
247 >>> import cvxopt
248 >>> import numpy
249 >>> from dunshire import *
250 >>> K = NonnegativeOrthant(2)
251 >>> L = [[1,0],[0,1]]
252 >>> e1 = cvxopt.matrix([1,1])
253 >>> e2 = numpy.matrix([1,1])
254 >>> G = SymmetricLinearGame(L, K, e1, e2)
255 >>> print(G)
256 The linear game (L, K, e1, e2) where
257 L = [ 1 0]
258 [ 0 1],
259 K = Nonnegative orthant in the real 2-space,
260 e1 = [ 1]
261 [ 1],
262 e2 = [ 1]
263 [ 1]
264
265 However, ``L`` will always be intepreted as a list of rows, even
266 if it is passed as a :class:`cvxopt.base.matrix` which is
267 otherwise indexed by columns::
268
269 >>> import cvxopt
270 >>> from dunshire import *
271 >>> K = NonnegativeOrthant(2)
272 >>> L = [[1,2],[3,4]]
273 >>> e1 = [1,1]
274 >>> e2 = e1
275 >>> G = SymmetricLinearGame(L, K, e1, e2)
276 >>> print(G)
277 The linear game (L, K, e1, e2) where
278 L = [ 1 2]
279 [ 3 4],
280 K = Nonnegative orthant in the real 2-space,
281 e1 = [ 1]
282 [ 1],
283 e2 = [ 1]
284 [ 1]
285 >>> L = cvxopt.matrix(L)
286 >>> print(L)
287 [ 1 3]
288 [ 2 4]
289 <BLANKLINE>
290 >>> G = SymmetricLinearGame(L, K, e1, e2)
291 >>> print(G)
292 The linear game (L, K, e1, e2) where
293 L = [ 1 2]
294 [ 3 4],
295 K = Nonnegative orthant in the real 2-space,
296 e1 = [ 1]
297 [ 1],
298 e2 = [ 1]
299 [ 1]
300
301 """
302 def __init__(self, L, K, e1, e2):
303 """
304 Create a new SymmetricLinearGame object.
305 """
306 self._K = K
307 self._e1 = matrix(e1, (K.dimension(), 1))
308 self._e2 = matrix(e2, (K.dimension(), 1))
309
310 # Our input ``L`` is indexed by rows but CVXOPT matrices are
311 # indexed by columns, so we need to transpose the input before
312 # feeding it to CVXOPT.
313 self._L = matrix(L, (K.dimension(), K.dimension())).trans()
314
315 if not self._e1 in K:
316 raise ValueError('the point e1 must lie in the interior of K')
317
318 if not self._e2 in K:
319 raise ValueError('the point e2 must lie in the interior of K')
320
321 # Initial value of cached method.
322 self._L_specnorm_value = None
323
324
325 def __str__(self):
326 """
327 Return a string representation of this game.
328 """
329 tpl = 'The linear game (L, K, e1, e2) where\n' \
330 ' L = {:s},\n' \
331 ' K = {!s},\n' \
332 ' e1 = {:s},\n' \
333 ' e2 = {:s}'
334 indented_L = '\n '.join(str(self.L()).splitlines())
335 indented_e1 = '\n '.join(str(self.e1()).splitlines())
336 indented_e2 = '\n '.join(str(self.e2()).splitlines())
337
338 return tpl.format(indented_L,
339 str(self.K()),
340 indented_e1,
341 indented_e2)
342
343
344 def L(self):
345 """
346 Return the matrix ``L`` passed to the constructor.
347
348 Returns
349 -------
350
351 matrix
352 The matrix that defines this game's :meth:`payoff` operator.
353
354 Examples
355 --------
356
357 >>> from dunshire import *
358 >>> K = NonnegativeOrthant(3)
359 >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
360 >>> e1 = [1,1,1]
361 >>> e2 = [1,2,3]
362 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
363 >>> print(SLG.L())
364 [ 1 -5 -15]
365 [ -1 2 -3]
366 [-12 -15 1]
367 <BLANKLINE>
368
369 """
370 return self._L
371
372
373 def K(self):
374 """
375 Return the cone over which this game is played.
376
377 Returns
378 -------
379
380 SymmetricCone
381 The :class:`SymmetricCone` over which this game is played.
382
383 Examples
384 --------
385
386 >>> from dunshire import *
387 >>> K = NonnegativeOrthant(3)
388 >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
389 >>> e1 = [1,1,1]
390 >>> e2 = [1,2,3]
391 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
392 >>> print(SLG.K())
393 Nonnegative orthant in the real 3-space
394
395 """
396 return self._K
397
398
399 def e1(self):
400 """
401 Return player one's interior point.
402
403 Returns
404 -------
405
406 matrix
407 The point interior to :meth:`K` affiliated with player one.
408
409 Examples
410 --------
411
412 >>> from dunshire import *
413 >>> K = NonnegativeOrthant(3)
414 >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
415 >>> e1 = [1,1,1]
416 >>> e2 = [1,2,3]
417 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
418 >>> print(SLG.e1())
419 [ 1]
420 [ 1]
421 [ 1]
422 <BLANKLINE>
423
424 """
425 return self._e1
426
427
428 def e2(self):
429 """
430 Return player two's interior point.
431
432 Returns
433 -------
434
435 matrix
436 The point interior to :meth:`K` affiliated with player one.
437
438 Examples
439 --------
440
441 >>> from dunshire import *
442 >>> K = NonnegativeOrthant(3)
443 >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
444 >>> e1 = [1,1,1]
445 >>> e2 = [1,2,3]
446 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
447 >>> print(SLG.e2())
448 [ 1]
449 [ 2]
450 [ 3]
451 <BLANKLINE>
452
453 """
454 return self._e2
455
456
457 def payoff(self, strategy1, strategy2):
458 r"""
459 Return the payoff associated with ``strategy1`` and ``strategy2``.
460
461 The payoff operator takes pairs of strategies to a real
462 number. For example, if player one's strategy is :math:`x` and
463 player two's strategy is :math:`y`, then the associated payoff
464 is :math:`\left\langle L\left(x\right),y \right\rangle` \in
465 \mathbb{R}. Here, :math:`L` denotes the same linear operator as
466 :meth:`L`. This method computes the payoff given the two
467 players' strategies.
468
469 Parameters
470 ----------
471
472 strategy1 : matrix
473 Player one's strategy.
474
475 strategy2 : matrix
476 Player two's strategy.
477
478 Returns
479 -------
480
481 float
482 The payoff for the game when player one plays ``strategy1``
483 and player two plays ``strategy2``.
484
485 Examples
486 --------
487
488 The value of the game should be the payoff at the optimal
489 strategies::
490
491 >>> from dunshire import *
492 >>> from dunshire.options import ABS_TOL
493 >>> K = NonnegativeOrthant(3)
494 >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
495 >>> e1 = [1,1,1]
496 >>> e2 = [1,1,1]
497 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
498 >>> soln = SLG.solution()
499 >>> x_bar = soln.player1_optimal()
500 >>> y_bar = soln.player2_optimal()
501 >>> abs(SLG.payoff(x_bar, y_bar) - soln.game_value()) < ABS_TOL
502 True
503
504 """
505 return inner_product(self.L()*strategy1, strategy2)
506
507
508 def dimension(self):
509 """
510 Return the dimension of this game.
511
512 The dimension of a game is not needed for the theory, but it is
513 useful for the implementation. We define the dimension of a game
514 to be the dimension of its underlying cone. Or what is the same,
515 the dimension of the space from which the strategies are chosen.
516
517 Returns
518 -------
519
520 int
521 The dimension of the cone :meth:`K`, or of the space where
522 this game is played.
523
524 Examples
525 --------
526
527 The dimension of a game over the nonnegative quadrant in the
528 plane should be two (the dimension of the plane)::
529
530 >>> from dunshire import *
531 >>> K = NonnegativeOrthant(2)
532 >>> L = [[1,-5],[-1,2]]
533 >>> e1 = [1,1]
534 >>> e2 = [1,4]
535 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
536 >>> SLG.dimension()
537 2
538
539 """
540 return self.K().dimension()
541
542
543 def _zero(self):
544 """
545 Return a column of zeros that fits ``K``.
546
547 This is used in our CVXOPT construction.
548
549 .. warning::
550
551 It is not safe to cache any of the matrices passed to
552 CVXOPT, because it can clobber them.
553
554 Returns
555 -------
556
557 matrix
558 A ``self.dimension()``-by-``1`` column vector of zeros.
559
560 Examples
561 --------
562
563 >>> from dunshire import *
564 >>> K = NonnegativeOrthant(3)
565 >>> L = identity(3)
566 >>> e1 = [1,1,1]
567 >>> e2 = e1
568 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
569 >>> print(SLG._zero())
570 [0.0000000]
571 [0.0000000]
572 [0.0000000]
573 <BLANKLINE>
574
575 """
576 return matrix(0, (self.dimension(), 1), tc='d')
577
578
579 def A(self):
580 """
581 Return the matrix ``A`` used in our CVXOPT construction.
582
583 This matrix ``A`` appears on the right-hand side of ``Ax = b``
584 in the statement of the CVXOPT conelp program.
585
586 .. warning::
587
588 It is not safe to cache any of the matrices passed to
589 CVXOPT, because it can clobber them.
590
591 Returns
592 -------
593
594 matrix
595 A ``1``-by-``(1 + self.dimension())`` row vector. Its first
596 entry is zero, and the rest are the entries of ``e2``.
597
598 Examples
599 --------
600
601 >>> from dunshire import *
602 >>> K = NonnegativeOrthant(3)
603 >>> L = [[1,1,1],[1,1,1],[1,1,1]]
604 >>> e1 = [1,1,1]
605 >>> e2 = [1,2,3]
606 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
607 >>> print(SLG.A())
608 [0.0000000 1.0000000 2.0000000 3.0000000]
609 <BLANKLINE>
610
611 """
612 return matrix([0, self.e2()], (1, self.dimension() + 1), 'd')
613
614
615
616 def G(self):
617 r"""
618 Return the matrix ``G`` used in our CVXOPT construction.
619
620 Thus matrix ``G`` appears on the left-hand side of ``Gx + s = h``
621 in the statement of the CVXOPT conelp program.
622
623 .. warning::
624
625 It is not safe to cache any of the matrices passed to
626 CVXOPT, because it can clobber them.
627
628 Returns
629 -------
630
631 matrix
632 A ``2*self.dimension()``-by-``(1 + self.dimension())`` matrix.
633
634 Examples
635 --------
636
637 >>> from dunshire import *
638 >>> K = NonnegativeOrthant(3)
639 >>> L = [[4,5,6],[7,8,9],[10,11,12]]
640 >>> e1 = [1,2,3]
641 >>> e2 = [1,1,1]
642 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
643 >>> print(SLG.G())
644 [ 0.0000000 -1.0000000 0.0000000 0.0000000]
645 [ 0.0000000 0.0000000 -1.0000000 0.0000000]
646 [ 0.0000000 0.0000000 0.0000000 -1.0000000]
647 [ 1.0000000 -4.0000000 -5.0000000 -6.0000000]
648 [ 2.0000000 -7.0000000 -8.0000000 -9.0000000]
649 [ 3.0000000 -10.0000000 -11.0000000 -12.0000000]
650 <BLANKLINE>
651
652 """
653 identity_matrix = identity(self.dimension())
654 return append_row(append_col(self._zero(), -identity_matrix),
655 append_col(self.e1(), -self.L()))
656
657
658 def c(self):
659 """
660 Return the vector ``c`` used in our CVXOPT construction.
661
662 The column vector ``c`` appears in the objective function
663 value ``<c,x>`` in the statement of the CVXOPT conelp program.
664
665 .. warning::
666
667 It is not safe to cache any of the matrices passed to
668 CVXOPT, because it can clobber them.
669
670 Returns
671 -------
672
673 matrix
674 A ``self.dimension()``-by-``1`` column vector.
675
676 Examples
677 --------
678
679 >>> from dunshire import *
680 >>> K = NonnegativeOrthant(3)
681 >>> L = [[4,5,6],[7,8,9],[10,11,12]]
682 >>> e1 = [1,2,3]
683 >>> e2 = [1,1,1]
684 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
685 >>> print(SLG.c())
686 [-1.0000000]
687 [ 0.0000000]
688 [ 0.0000000]
689 [ 0.0000000]
690 <BLANKLINE>
691
692 """
693 return matrix([-1, self._zero()])
694
695
696 def C(self):
697 """
698 Return the cone ``C`` used in our CVXOPT construction.
699
700 The cone ``C`` is the cone over which the conelp program takes
701 place.
702
703 Returns
704 -------
705
706 CartesianProduct
707 The cartesian product of ``K`` with itself.
708
709 Examples
710 --------
711
712 >>> from dunshire import *
713 >>> K = NonnegativeOrthant(3)
714 >>> L = [[4,5,6],[7,8,9],[10,11,12]]
715 >>> e1 = [1,2,3]
716 >>> e2 = [1,1,1]
717 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
718 >>> print(SLG.C())
719 Cartesian product of dimension 6 with 2 factors:
720 * Nonnegative orthant in the real 3-space
721 * Nonnegative orthant in the real 3-space
722
723 """
724 return CartesianProduct(self._K, self._K)
725
726 def h(self):
727 r"""
728 Return the ``h`` vector used in our CVXOPT construction.
729
730 The ``h`` vector appears on the right-hand side of :math:`Gx + s
731 = h` in the statement of the CVXOPT conelp program.
732
733 .. warning::
734
735 It is not safe to cache any of the matrices passed to
736 CVXOPT, because it can clobber them.
737
738 Returns
739 -------
740
741 matrix
742 A ``2*self.dimension()``-by-``1`` column vector of zeros.
743
744 Examples
745 --------
746
747 >>> from dunshire import *
748 >>> K = NonnegativeOrthant(3)
749 >>> L = [[4,5,6],[7,8,9],[10,11,12]]
750 >>> e1 = [1,2,3]
751 >>> e2 = [1,1,1]
752 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
753 >>> print(SLG.h())
754 [0.0000000]
755 [0.0000000]
756 [0.0000000]
757 [0.0000000]
758 [0.0000000]
759 [0.0000000]
760 <BLANKLINE>
761
762 """
763
764 return matrix([self._zero(), self._zero()])
765
766
767 @staticmethod
768 def b():
769 r"""
770 Return the ``b`` vector used in our CVXOPT construction.
771
772 The vector ``b`` appears on the right-hand side of :math:`Ax =
773 b` in the statement of the CVXOPT conelp program.
774
775 This method is static because the dimensions and entries of
776 ``b`` are known beforehand, and don't depend on any other
777 properties of the game.
778
779 .. warning::
780
781 It is not safe to cache any of the matrices passed to
782 CVXOPT, because it can clobber them.
783
784 Returns
785 -------
786
787 matrix
788 A ``1``-by-``1`` matrix containing a single entry ``1``.
789
790 Examples
791 --------
792
793 >>> from dunshire import *
794 >>> K = NonnegativeOrthant(3)
795 >>> L = [[4,5,6],[7,8,9],[10,11,12]]
796 >>> e1 = [1,2,3]
797 >>> e2 = [1,1,1]
798 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
799 >>> print(SLG.b())
800 [1.0000000]
801 <BLANKLINE>
802
803 """
804 return matrix([1], tc='d')
805
806
807 def player1_start(self):
808 """
809 Return a feasible starting point for player one.
810
811 This starting point is for the CVXOPT formulation and not for
812 the original game. The basic premise is that if you normalize
813 :meth:`e2`, then you get a point in :meth:`K` that makes a unit
814 inner product with :meth:`e2`. We then get to choose the primal
815 objective function value such that the constraint involving
816 :meth:`L` is satisfied.
817 """
818 p = self.e2() / (norm(self.e2()) ** 2)
819 dist = self.K().ball_radius(self.e1())
820 nu = - self._L_specnorm()/(dist*norm(self.e2()))
821 x = matrix([nu, p], (self.dimension() + 1, 1))
822 s = - self.G()*x
823
824 return {'x': x, 's': s}
825
826
827 def player2_start(self):
828 """
829 Return a feasible starting point for player two.
830 """
831 q = self.e1() / (norm(self.e1()) ** 2)
832 dist = self.K().ball_radius(self.e2())
833 omega = self._L_specnorm()/(dist*norm(self.e1()))
834 y = matrix([omega])
835 z2 = q
836 z1 = y*self.e2() - self.L().trans()*z2
837 z = matrix([z1, z2], (self.dimension()*2, 1))
838
839 return {'y': y, 'z': z}
840
841
842 def _L_specnorm(self):
843 """
844 Compute the spectral norm of :meth:`L` and cache it.
845
846 The spectral norm of the matrix :meth:`L` is used in a few
847 places. Since it can be expensive to compute, we want to cache
848 its value. That is not possible in :func:`specnorm`, which lies
849 outside of a class, so this is the place to do it.
850
851 Returns
852 -------
853
854 float
855 A nonnegative real number; the largest singular value of
856 the matrix :meth:`L`.
857
858 """
859 if self._L_specnorm_value is None:
860 self._L_specnorm_value = specnorm(self.L())
861 return self._L_specnorm_value
862
863
864 def tolerance_scale(self, solution):
865 r"""
866 Return a scaling factor that should be applied to ``ABS_TOL``
867 for this game.
868
869 When performing certain comparisons, the default tolernace
870 ``ABS_TOL`` may not be appropriate. For example, if we expect
871 ``x`` and ``y`` to be within ``ABS_TOL`` of each other, than the
872 inner product of ``L*x`` and ``y`` can be as far apart as the
873 spectral norm of ``L`` times the sum of the norms of ``x`` and
874 ``y``. Such a comparison is made in :meth:`solution`, and in
875 many of our unit tests.
876
877 The returned scaling factor found from the inner product mentioned
878 above is
879
880 .. math::
881
882 \left\lVert L \right\rVert_{2}
883 \left( \left\lVert \bar{x} \right\rVert
884 + \left\lVert \bar{y} \right\rVert
885 \right),
886
887 where :math:`\bar{x}` and :math:`\bar{y}` are optimal solutions
888 for players one and two respectively. This scaling factor is not
889 formally justified, but attempting anything smaller leads to
890 test failures.
891
892 .. warning::
893
894 Optimal solutions are not unique, so the scaling factor
895 obtained from ``solution`` may not work when comparing other
896 solutions.
897
898 Parameters
899 ----------
900
901 solution : Solution
902 A solution of this game, used to obtain the norms of the
903 optimal strategies.
904
905 Returns
906 -------
907
908 float
909 A scaling factor to be multiplied by ``ABS_TOL`` when
910 making comparisons involving solutions of this game.
911
912 """
913 norm_p1_opt = norm(solution.player1_optimal())
914 norm_p2_opt = norm(solution.player2_optimal())
915 scale = self._L_specnorm()*(norm_p1_opt + norm_p2_opt)
916
917 # Don't return anything smaller than 1... we can't go below
918 # out "minimum tolerance."
919 return max(1, scale)
920
921
922 def solution(self):
923 """
924 Solve this linear game and return a :class:`Solution`.
925
926 Returns
927 -------
928
929 :class:`Solution`
930 A :class:`Solution` object describing the game's value and
931 the optimal strategies of both players.
932
933 Raises
934 ------
935 GameUnsolvableException
936 If the game could not be solved (if an optimal solution to its
937 associated cone program was not found).
938
939 PoorScalingException
940 If the game could not be solved because CVXOPT crashed while
941 trying to take the square root of a negative number.
942
943 Examples
944 --------
945
946 This example is computed in Gowda and Ravindran in the section
947 "The value of a Z-transformation"::
948
949 >>> from dunshire import *
950 >>> K = NonnegativeOrthant(3)
951 >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
952 >>> e1 = [1,1,1]
953 >>> e2 = [1,1,1]
954 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
955 >>> print(SLG.solution())
956 Game value: -6.172...
957 Player 1 optimal:
958 [0.551...]
959 [0.000...]
960 [0.448...]
961 Player 2 optimal:
962 [0.448...]
963 [0.000...]
964 [0.551...]
965
966 The value of the following game can be computed using the fact
967 that the identity is invertible::
968
969 >>> from dunshire import *
970 >>> K = NonnegativeOrthant(3)
971 >>> L = [[1,0,0],[0,1,0],[0,0,1]]
972 >>> e1 = [1,2,3]
973 >>> e2 = [4,5,6]
974 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
975 >>> print(SLG.solution())
976 Game value: 0.031...
977 Player 1 optimal:
978 [0.031...]
979 [0.062...]
980 [0.093...]
981 Player 2 optimal:
982 [0.125...]
983 [0.156...]
984 [0.187...]
985
986 This is another Gowda/Ravindran example that is supposed to have
987 a negative game value::
988
989 >>> from dunshire import *
990 >>> from dunshire.options import ABS_TOL
991 >>> L = [[1, -2], [-2, 1]]
992 >>> K = NonnegativeOrthant(2)
993 >>> e1 = [1, 1]
994 >>> e2 = e1
995 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
996 >>> SLG.solution().game_value() < -ABS_TOL
997 True
998
999 The following two games are problematic numerically, but we
1000 should be able to solve them::
1001
1002 >>> from dunshire import *
1003 >>> L = [[-0.95237953890954685221, 1.83474556206462535712],
1004 ... [ 1.30481749924621448500, 1.65278664543326403447]]
1005 >>> K = NonnegativeOrthant(2)
1006 >>> e1 = [0.95477167524644313001, 0.63270781756540095397]
1007 >>> e2 = [0.39633793037154141370, 0.10239281495640320530]
1008 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
1009 >>> print(SLG.solution())
1010 Game value: 18.767...
1011 Player 1 optimal:
1012 [0.000...]
1013 [9.766...]
1014 Player 2 optimal:
1015 [1.047...]
1016 [0.000...]
1017
1018 ::
1019
1020 >>> from dunshire import *
1021 >>> L = [[1.54159395026049472754, 2.21344728574316684799],
1022 ... [1.33147433507846657541, 1.17913616272988108769]]
1023 >>> K = NonnegativeOrthant(2)
1024 >>> e1 = [0.39903040089404784307, 0.12377403622479113410]
1025 >>> e2 = [0.15695181142215544612, 0.85527381344651265405]
1026 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
1027 >>> print(SLG.solution())
1028 Game value: 24.614...
1029 Player 1 optimal:
1030 [6.371...]
1031 [0.000...]
1032 Player 2 optimal:
1033 [2.506...]
1034 [0.000...]
1035
1036 This is another one that was difficult numerically, and caused
1037 trouble even after we fixed the first two::
1038
1039 >>> from dunshire import *
1040 >>> L = [[57.22233908627052301199, 41.70631373437460354126],
1041 ... [83.04512571985074487202, 57.82581810406928468637]]
1042 >>> K = NonnegativeOrthant(2)
1043 >>> e1 = [7.31887017043399268346, 0.89744171905822367474]
1044 >>> e2 = [0.11099824781179848388, 6.12564670639315345113]
1045 >>> SLG = SymmetricLinearGame(L,K,e1,e2)
1046 >>> print(SLG.solution())
1047 Game value: 70.437...
1048 Player 1 optimal:
1049 [9.009...]
1050 [0.000...]
1051 Player 2 optimal:
1052 [0.136...]
1053 [0.000...]
1054
1055 And finally, here's one that returns an "optimal" solution, but
1056 whose primal/dual objective function values are far apart::
1057
1058 >>> from dunshire import *
1059 >>> L = [[ 6.49260076597376212248, -0.60528030227678542019],
1060 ... [ 2.59896077096751731972, -0.97685530240286766457]]
1061 >>> K = IceCream(2)
1062 >>> e1 = [1, 0.43749513972645248661]
1063 >>> e2 = [1, 0.46008379832200291260]
1064 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
1065 >>> print(SLG.solution())
1066 Game value: 11.596...
1067 Player 1 optimal:
1068 [ 1.852...]
1069 [-1.852...]
1070 Player 2 optimal:
1071 [ 1.777...]
1072 [-1.777...]
1073
1074 """
1075 try:
1076 opts = {'show_progress': False}
1077 soln_dict = solvers.conelp(self.c(),
1078 self.G(),
1079 self.h(),
1080 self.C().cvxopt_dims(),
1081 self.A(),
1082 self.b(),
1083 primalstart=self.player1_start(),
1084 dualstart=self.player2_start(),
1085 options=opts)
1086 except ValueError as error:
1087 if str(error) == 'math domain error':
1088 # Oops, CVXOPT tried to take the square root of a
1089 # negative number. Report some details about the game
1090 # rather than just the underlying CVXOPT crash.
1091 printing.options['dformat'] = DEBUG_FLOAT_FORMAT
1092 raise PoorScalingException(self)
1093 else:
1094 raise error
1095
1096 # The optimal strategies are named ``p`` and ``q`` in the
1097 # background documentation, and we need to extract them from
1098 # the CVXOPT ``x`` and ``z`` variables. The objective values
1099 # :math:`nu` and :math:`omega` can also be found in the CVXOPT
1100 # ``x`` and ``y`` variables; however, they're stored
1101 # conveniently as separate entries in the solution dictionary.
1102 p1_value = -soln_dict['primal objective']
1103 p2_value = -soln_dict['dual objective']
1104 p1_optimal = soln_dict['x'][1:]
1105 p2_optimal = soln_dict['z'][self.dimension():]
1106
1107 # The "status" field contains "optimal" if everything went
1108 # according to plan. Other possible values are "primal
1109 # infeasible", "dual infeasible", "unknown", all of which mean
1110 # we didn't get a solution.
1111 #
1112 # The "infeasible" ones are the worst, since they indicate
1113 # that CVXOPT is convinced the problem is infeasible (and that
1114 # cannot happen).
1115 if soln_dict['status'] in ['primal infeasible', 'dual infeasible']:
1116 printing.options['dformat'] = DEBUG_FLOAT_FORMAT
1117 raise GameUnsolvableException(self, soln_dict)
1118
1119 # For the game value, we could use any of:
1120 #
1121 # * p1_value
1122 # * p2_value
1123 # * (p1_value + p2_value)/2
1124 # * the game payoff
1125 #
1126 # We want the game value to be the payoff, however, so it
1127 # makes the most sense to just use that, even if it means we
1128 # can't test the fact that p1_value/p2_value are close to the
1129 # payoff.
1130 payoff = self.payoff(p1_optimal, p2_optimal)
1131 soln = Solution(payoff, p1_optimal, p2_optimal)
1132
1133 # The "optimal" and "unknown" results, we actually treat the
1134 # same. Even if CVXOPT bails out due to numerical difficulty,
1135 # it will have some candidate points in mind. If those
1136 # candidates are good enough, we take them. We do the same
1137 # check (perhaps pointlessly so) for "optimal" results.
1138 #
1139 # First we check that the primal/dual objective values are
1140 # close enough (one could be low by ABS_TOL, the other high by
1141 # it) because otherwise CVXOPT might return "unknown" and give
1142 # us two points in the cone that are nowhere near optimal.
1143 #
1144 if abs(p1_value - p2_value) > self.tolerance_scale(soln)*ABS_TOL:
1145 printing.options['dformat'] = DEBUG_FLOAT_FORMAT
1146 raise GameUnsolvableException(self, soln_dict)
1147
1148 # And we also check that the points it gave us belong to the
1149 # cone, just in case...
1150 if (p1_optimal not in self._K) or (p2_optimal not in self._K):
1151 printing.options['dformat'] = DEBUG_FLOAT_FORMAT
1152 raise GameUnsolvableException(self, soln_dict)
1153
1154 return soln
1155
1156
1157 def condition(self):
1158 r"""
1159 Return the condition number of this game.
1160
1161 In the CVXOPT construction of this game, two matrices ``G`` and
1162 ``A`` appear. When those matrices are nasty, numerical problems
1163 can show up. We define the condition number of this game to be
1164 the average of the condition numbers of ``G`` and ``A`` in the
1165 CVXOPT construction. If the condition number of this game is
1166 high, then you can expect numerical difficulty (such as
1167 :class:`PoorScalingException`).
1168
1169 Returns
1170 -------
1171
1172 float
1173 A real number greater than or equal to one that measures how
1174 bad this game is numerically.
1175
1176 Examples
1177 --------
1178
1179 >>> from dunshire import *
1180 >>> K = NonnegativeOrthant(1)
1181 >>> L = [[1]]
1182 >>> e1 = [1]
1183 >>> e2 = e1
1184 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
1185 >>> SLG.condition()
1186 1.809...
1187
1188 """
1189 return (condition_number(self.G()) + condition_number(self.A()))/2
1190
1191
1192 def dual(self):
1193 r"""
1194 Return the dual game to this game.
1195
1196 If :math:`G = \left(L,K,e_{1},e_{2}\right)` is a linear game,
1197 then its dual is :math:`G^{*} =
1198 \left(L^{*},K^{*},e_{2},e_{1}\right)`. However, since this cone
1199 is symmetric, :math:`K^{*} = K`.
1200
1201 Examples
1202 --------
1203
1204 >>> from dunshire import *
1205 >>> K = NonnegativeOrthant(3)
1206 >>> L = [[1,-5,-15],[-1,2,-3],[-12,-15,1]]
1207 >>> e1 = [1,1,1]
1208 >>> e2 = [1,2,3]
1209 >>> SLG = SymmetricLinearGame(L, K, e1, e2)
1210 >>> print(SLG.dual())
1211 The linear game (L, K, e1, e2) where
1212 L = [ 1 -1 -12]
1213 [ -5 2 -15]
1214 [-15 -3 1],
1215 K = Nonnegative orthant in the real 3-space,
1216 e1 = [ 1]
1217 [ 2]
1218 [ 3],
1219 e2 = [ 1]
1220 [ 1]
1221 [ 1]
1222
1223 """
1224 # We pass ``self.L()`` right back into the constructor, because
1225 # it will be transposed there. And keep in mind that ``self._K``
1226 # is its own dual.
1227 return SymmetricLinearGame(self.L(),
1228 self.K(),
1229 self.e2(),
1230 self.e1())