Exponentially many steps for finding a NE in a bimatrix game

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Exponentially many steps for finding a NE in a
bimatrix game

• Rahul Savani, Bernhard von Stengel (2004)

Presentation Angelina Vidali
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Previous Work

• Morris, W. D., Jr. (1994), Lemke paths on simple
  polytopes. Math. of Oper. Res. 19, 780789.
• uses duals of cyclic polytopes
• but with different labeling
• The LH method for finding a symmetric equilibrium
  of a bimatrix game can take exponentially long.
• But these Games have non-symmetric equilibria
  that can be found very quickly.

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Our Result

• A family of dxd games with a unique equilibrium
  for which
• the LH algorithm takes an exponential number of
  steps, for any dropped label.

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Why Duals of cyclic polytopes?

• Idea Look for examples for long Lemke Paths in
  polytopes with many vertices
• Duals of cyclic polytopes have the maximum number
  of vertices for d-polytopes with a
  fixed number of facets.
• a convinient combinatorial description!

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Gale Evenness

• A bitstring represents a vertex iff any substring
  of the form 0110 has an even length.
• Not allowed 010, 001, 0101, 01110,
• Allowed 0110, 0011, 1110001, 011110,
• Note we use cyclic symmetry.

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Dual cyclic polytopes

• space dimension d d even
• The vertices of the dual cyclic polytopes are
  bitstrings (u1,,ud, ud1,,u2d) that
• Fullfill the Gale evenness condition
• Have exactly d ones and d zeros.
• i.e. 00001111, 11000011, 11001100,
• Each vertex is on exactly d facets.
• (simple polytope)

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Facet labels

• Pemutations l (for P) and l (for Q), of 1,,2d
• l the identity permutation (l(k)k)
• Fixed points l (1)1, l (d)d
• Otherwise exchanges adjacent numbers

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Artificial equilibrium e0(1d0d ,0d1d)

• The artificial equilibrium e0 is a vertex pair
  (u,v) such that
• u has labels 1,,d
• v has labels d1,,2d

(u,v) Completely labeled
Labels 1 d d1 2d 1 d d2 2d-1
(u1,,ud, ud1,,u2d)
(v1,,vd, vd1,,v2d)
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Lemke-Howson on dual cyclic polytopesdropping
label 1
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Unique NE (0d1d,1d0d)

• Lemma Let e1(0d1d,1d0d) This is the only NE of
  the Game.
• Proof outline
• let (u,v) be a completely labeled vertex pair
• (u0,,ud,,v0,) (u,v)e1(0d1d,1d0d)
• (u0,,ud,,v0,) (u,v) e0(1d0d ,0d1d)
• using Definition of l

01?0
Gale Evenness
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Names for LH paths

• p(d,l)LH path dropping label l in dim d
• p(d,1)LH path dropping label 1 in dim dA(d)
• p(d,2d)LH path dropping label 2d in dim dB(d)

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Symmetries of p(d,1)
Just mirror the image down!
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Symmetries of p(d,2d)
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A(4)p(4,1)
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B(6)p(d6,12)
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A(4) is preffix of B(6)
A(d) is preffix of B(d2)? -Yes!
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A(6)
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B(6) is preffix of A(6)
B(d) is preffix of A(d2)? -Yes!
B(6)
A(6)B(6)C(6)
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A(6)
C(6)A(4)
B(6)
A(4)
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Solution to the Recurrences Fibonacci numbers

• Length(A(d))Length(B(d))Length(C(d))
• Length(C(d))Length(A(d-2))Length(C(d-2))
• Length(B(d))Length(A(d-2))Length(C(d-2))
• d even
• lengths of
• B(2) C(2) A(2) B(4) C(4) A(4) B(6) C(6) A(6) . .
  .
• are the Fibonacci numbers
• 2 3 5 8 13 21 34 55 89 . . .
• We know their order of growth is exponential.

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• Longest paths drop label 1 or 2d, paths A(d),
  B(d)
• path length O( q 3d/2 )
• Shortest path drop label 3d/2
• We can write it as
• B(d/2)B(d/21)
• path length O( q3d/4 ) (1.434...d )

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p(4,4) is transformed to p(4,1) using a shift
and a reversal
p(4,1)
The transformation shows that the paths have
equal length.
dropping label 1 we dont use cyclic symmetry
Here we do
p(4,4)
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Endnote

• a construction of dxd games with a unique
  equilibrium which is found by the LH algorithm
  using an exponential number of steps, for any
  dropped label.
• But it easily guessed since it has full support.

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It is the end!