Pathways to Equilibria, Pretty Pictures and Diagrams (PPAD)

Transcription

1 1 Pathways to Equilibria, Pretty Pictures and Diagrams (PPAD) Bernhard von Stengel partly joint work with: Marta Casetti, Julian Merschen, Lászlo Végh Department of Mathematics London School of Economics

2 2 2-player game: find one Nash equilibrium 2-NASH PPAD (Polynomial Parity Argument with Direction) Implicit digraph with indegrees and outdegrees 1 is a set of [nodes], paths and cycles: 0 Parity argument: number of sources of paths = number of sinks Comput. problem: given one source 0, find another source or sink [Chen/Deng 2006] 2-NASH is PPAD-complete.

3 Symmetric Nash equilibria of symmetric games square game matrix A = payoffs to row player A =

4 Symmetric Nash equilibria of symmetric games equilibrium: only optimal strategies are played A = 1/ 2/

5 Symmetric Nash equilibria of symmetric games z plot polytope with strategy weights z 1, z 2, z z 0, A = z 1 z 2 z z 2 z 1

6 Symmetric Nash equilibria of symmetric games with payoffs (scaled to 1) and labels for binding inequalities z 2 1 (back) 2 z 1 1 { z z 0, A z 1 } z 1 z 2 z 0 0 A = (bottom) z 2

7 Symmetric Nash equilibria of symmetric games z equilibrium = completely labeled point 2 1 (back) 2 z 1 1 { z z 0, A z 1 } (bottom) z 2 A = 1/6 1/ /2

8 Symmetric Nash equilibria of symmetric games start path with artificial equilibrium z=0 z 1 (back) { z z 0, A z 1 } 2 2 z 1 1 (bottom) z 2 A =

9 Symmetric Nash equilibria of symmetric games start path with artificial equilibrium z=0, choose e.g. z missing label 1 1 (back) { z z 0, A z 1 } 2 2 z 1 1 (bottom) z 2 A =

10 Symmetric Nash equilibria of symmetric games leave facet with label 1, find duplicate label z missing label 1 1 (back) { z z 0, A z 1 } 2 2 z 1 1 (bottom) z 2 A =

11 Symmetric Nash equilibria of symmetric games leave facet with old label, find duplicate label 2 z missing label 1 1 (back) { z z 0, A z 1 } 2 2 z 1 1 (bottom) z 2 A =

12 Symmetric Nash equilibria of symmetric games leave facet with old label 2, find duplicate label z missing label 1 1 (back) { z z 0, A z 1 } 2 2 z 1 1 (bottom) z 2 A =

13 Symmetric Nash equilibria of symmetric games leave facet with old label, find missing label 1 z 1 found label 1 (back) { z z 0, A z 1 } 2 2 z 1 1 (bottom) z 2 A =

14 Symmetric Nash equilibria of symmetric games equilibria (including artificial equilibrium) = endpoints of paths z missing label (back) 2 z 1 1 { z z 0, A z 1 } (bottom) z 2 A = 1/6 1/ /2

15 The castle where each room has at most two doors 4

16 The castle where each room has at most two doors 4

17 The castle where each room has at most two doors 4

18 The castle where each room has at most two doors 4

19 5 Path of almost completely labeled edges two completely labeled vertices

20 5 Path of almost completely labeled edges path because at most two neighbours ( doors in castle)

21 5 Path of almost completely labeled edges orientation of edges: 2 on left, on right

22 5 Path of almost completely labeled edges opposite orientation ( sign ) of endpoints

23 5 Path of almost completely labeled edges equilibrium sign or + does not depend on path

24 5 Path of almost completely labeled edges equilibrium sign or + does not depend on path

25 5 Path of almost completely labeled edges equilibrium sign or + does not depend on path

26 5 Path of almost completely labeled edges equilibrium sign or + does not depend on path

27 5 Path of almost completely labeled edges equilibrium sign or + does not depend on path

28 6 Let a j R m, β j R, Labeled polytope P P = {x R m a j x β j, 1 j n }, let facet label F j = { x P a j x = β j } have l(j) {1,..., m}. Assume P is a simple polytope (no x P on > m facets) each vertex x on m facets = m linearly independent equations. x completely labeled {l(j) x F j } = {1,..., m}.

29 7 Completely labeled points come in pairs Theorem [ Parity Argument ] Let P be a labeled polytope. Then P has an even number of completely labeled vertices.

30 7 Completely labeled points come in pairs of opposite sign Theorem [ Parity Argument with Direction ] Let P be a labeled polytope. Then P has an even number of completely labeled vertices. Half of these have sign, half have sign +.

31 7 Completely labeled points come in pairs of opposite sign Theorem [ Parity Argument with Direction ] Let P be a labeled polytope. Then P has an even number of completely labeled vertices. Half of these have sign, half have sign +. sign of completely labeled x is sign of determinant of facet normal vectors in order of their labels: if (e.g.) facet a i x = β i has label i = 1, 2,..., m, then sign(x) = sign a 1 a 2 a m

32 8 Pivoting changes signs Lemma Let x, y R m be adjacent vertices of a simple polytope P x y

33 8 Pivoting changes signs Lemma Let x, y R m be adjacent vertices of a simple polytope P with facet normals c, a 2,..., a m for x and d, a 2,..., a m for y. a 2 c x y d a

34 8 Pivoting changes signs Lemma Let x, y R m be adjacent vertices of a simple polytope P with facet normals c, a 2,..., a m for x and d, a 2,..., a m for y. Then c a 2 a m and d a 2 a m have opposite sign. a 2 c x y d a

35 9 Proof : Pivoting changes signs cx = β 0 dy = β 1 a 2 x = β 2 a 2 y = β 2.. a m x = β m a m y = β m

36 9 Proof : Pivoting changes signs cx = β 0 dy = β 1 a 2 x = β 2 a 2 y = β 2.. a m x = β m a m y = β m Let (γ, δ, α 2,..., α m ) (0, 0, 0,..., 0) with γc + δd + α 2 a α m a m = 0

37 9 Proof : Pivoting changes signs cx = β 0 dy = β 1 a 2 x = β 2 a 2 y = β 2.. a m x = β m a m y = β m Let (γ, δ, α 2,..., α m ) (0, 0, 0,..., 0) with γc + δd + α 2 a α m a m = 0 γ 0, δ 0, (γc + δd)x = (γc + δd)y

38 9 Proof : Pivoting changes signs cx = β 0 cy < β 0 dx < β 1 dy = β 1 a 2 x = β 2 a 2 y = β 2. a m x = β m. a m y = β m Let (γ, δ, α 2,..., α m ) (0, 0, 0,..., 0) with γ 0, δ 0, γc + δd + α 2 a α m a m = 0 (γc + δd)x = (γc + δd)y, γ(cx cy) = δ(dy dx)

39 9 Proof : Pivoting changes signs cx = β 0 cy < β 0 dx < β 1 dy = β 1 a 2 x = β 2 a 2 y = β 2. a m x = β m. a m y = β m Let (γ, δ, α 2,..., α m ) (0, 0, 0,..., 0) with γ 0, δ 0, γc + δd + α 2 a α m a m = 0 (γc + δd)x = (γc + δd)y, γ and δ have same sign γ(cx cy) = δ(dy dx)

40 9 Proof : Pivoting changes signs cx = β 0 cy < β 0 dx < β 1 dy = β 1 a 2 x = β 2 a 2 y = β 2. a m x = β m. a m y = β m Let (γ, δ, α 2,..., α m ) (0, 0, 0,..., 0) with γ 0, δ 0, γc + δd + α 2 a α m a m = 0 (γc + δd)x = (γc + δd)y, γ and δ have same sign, γ(cx cy) = δ(dy dx) (γc + δd) a 2 a m = γ c a 2 a m + δ d a 2 a m = 0

41 Proof : Pivoting changes signs cx = β 0 cy < β 0 dx < β 1 dy = β 1 a 2 x = β 2 a 2 y = β 2. a m x = β m. a m y = β m Let (γ, δ, α 2,..., α m ) (0, 0, 0,..., 0) with γ 0, δ 0, γc + δd + α 2 a α m a m = 0 (γc + δd)x = (γc + δd)y, γ and δ have same sign, γ(cx cy) = δ(dy dx) (γc + δd) a 2 a m = γ c a 2 a m + δ d a 2 a m = 0 c a 2 a m and d a 2 a m have opposite sign, QED. 9

42 10 General Parity Argument with Direction Facet normal vectors a 1 a 2 a c 1 c 2 c, labels a 2 c c 2 a 1 c 1 a

43 10 General Parity Argument with Direction Start with a 1 a 2 a, sign a 1 a 2 a a 2 c c 2 a 1 c 1 a

44 10 General Parity Argument with Direction Start with a 1 a 2 a, sign, label 1 missing, a 1 c gives sign + a 1 a 2 a + c a 2 a a 2 c c 2 a 1 c 1 a

45 10 General Parity Argument with Direction Switch columns c and a in determinant: back to sign a 1 a 2 a + c a 2 a a 2 c c 2 a 1 c 1 a a 2 c a

46 10 General Parity Argument with Direction next pivot a c 2 gives sign + a 1 a 2 a + c a 2 a c 2 a a 2 c c2 a 2 c a 2 a 1 c c 1 a

47 10 General Parity Argument with Direction Switch columns c 2 and a 2 in determinant: back to sign a 2 c 2 a 1 a 2 a a a 2 c a 1 c c a c c c a 2 a c 2 a 2 c a

48 10 General Parity Argument with Direction next pivot a 2 a gives sign + a 1 a 2 a + c a 2 a a 2 a 1 c c a a c 1 2 c 2 c c 2 a c 2 a a 2 c c 2 a 2 c

49 10 General Parity Argument with Direction Switch columns a and c in determinant: back to sign a 2 c a 1 c 1 a c 2 a 1 a 2 a a a 2 c a2 c 2 c c c 2 a + c a 2 a c 2 a 2 c a c 2 c

50 10 General Parity Argument with Direction Last pivot c c 1 gives sign +, opposite to starting sign. a 2 c c 2 a 1 c 1 a 1 a 2 a a a 2 c a2 c 2 c + c a 2 a c 2 a 2 c a c 2 c a c c 2 a c 1 c 2 a

51 10 General Parity Argument with Direction Only need: sign-switching of pivots and column exchanges a 2 c c 2 a 1 c 1 a 1 a 2 a a a 2 c a2 c 2 c + c a 2 a c 2 a 2 c a c 2 c a c c 2 a c 1 c 2 a

52 11 a 1 a 2 a a 4 a 5 A more abstract example

53 11 A more abstract example + a 1 a 2 a a 4 a 5 c a 2 a a 4 a 5

54 11 A more abstract example + a 1 a 2 a a 4 a 5 c a 2 a a 4 a 5 a a 2 c a 4 a 5

55 11 A more abstract example + a 1 a 2 a a 4 a 5 a a 2 c a 4 a 5 c a 2 a a 4 a 5 c4 a 2 c 4 a a 5

56 11 A more abstract example + a 1 a 2 a a 4 a 5 a a 2 c a 4 a 5 c a 2 a a 4 a 5 c4 a 2 c 4 a a 5 a 4 a 2 c c 4 a 5

57 11 A more abstract example + a 1 a 2 a a 4 a 5 a a 2 c a 4 a 5 a 4 a 2 c c 4 a 5 c a 2 a a 4 a 5 c4 a 2 c 4 a a 5 c 5 a 2 c c 4 a 5

58 11 A more abstract example + a 1 a 2 a a 4 a 5 a a 2 c a 4 a 5 a 4 a 2 c c 4 a 5 c a 2 a a 4 a 5 c4 a 2 c 4 a a 5 c 5 a 2 c c 4 a 5 a 5 a 2 c c 4 c 5

59 11 A more abstract example + a 1 a 2 a a 4 a 5 a a 2 c a 4 a 5 a 4 a 2 c c 4 a 5 a 5 a 2 c c 4 c 5 c a 2 a a 4 a 5 c4 a 2 c 4 a a 5 c 5 a 2 c c 4 a 5 c 1 a 2 c c 4 c 5

60 12 Nash equilibria of bimatrix games Recall: m m matrix C, P = {z R m z 0, Cz 1 } with 2m inequalities labeled 1,..., m, 1,..., m.

61 12 Nash equilibria of bimatrix games Recall: m m matrix C, P = {z R m z 0, Cz 1 } with 2m inequalities labeled 1,..., m, 1,..., m. Completely labeled z 0 Nash equilibrium (z, z) of game (C, C )

62 12 Nash equilibria of bimatrix games Recall: m m matrix C, P = {z R m z 0, Cz 1 } with 2m inequalities labeled 1,..., m, 1,..., m. Completely labeled z 0 Nash equilibrium (z, z) of game (C, C ) Normalize sign of artificial equilibrium 0 to, in general index(z) = sign(z) ( 1) m+1

63 12 Nash equilibria of bimatrix games Recall: m m matrix C, P = {z R m z 0, Cz 1 } with 2m inequalities labeled 1,..., m, 1,..., m. bimatrix game (A, B): C = ( ) 0 A B, z = (x, y) : 0 Completely labeled (x, y) (0, 0) Nash equilibrium (x, y) of game (A, B)

64 1 Index of an equilibrium Theorem [Shapley 1974] A nondegenerate bimatrix game (A, B) has an odd number of equilibria, one more of index + than of index.

65 1 Index of an equilibrium Theorem [Shapley 1974] A nondegenerate bimatrix game (A, B) has an odd number of equilibria, one more of index + than of index. [Proof: Endpoints of pivoting paths have opposite index and +.]

66 1 Index of an equilibrium Theorem [Shapley 1974] A nondegenerate bimatrix game (A, B) has an odd number of equilibria, one more of index + than of index. [Proof: Endpoints of pivoting paths have opposite index and +.] Equilibria of index + include every pure-strategy equilibrium unique equilibrium dynamically stable equilibrium [Hofbauer 200]

67 Dynamically stable equilibrium: only if index

68 Dynamically stable equilibrium: only if index

69 14 Dynamically stable equilibrium: only if index

70 14 Dynamically stable equilibrium: only if index

71 14 Dynamically stable equilibrium: only if index

72 14 Dynamically stable equilibrium: only if index

73 Strategic characterization of the index Theorem [von Schemde / von Stengel 2004] An equilibrium of a nondegenerate bimatrix game has index + it is the unique equilibrium in a larger game that has suitable additional strategies for one player. 15

74 Strategic characterization of the index Theorem [von Schemde / von Stengel 2004] An equilibrium of a nondegenerate bimatrix game has index + it is the unique equilibrium in a larger game that has suitable additional strategies for one player. Theorem [Balthasar / von Stengel 2009] A symmetric equilibrium of a nondegenerate symmetric bimatrix game has symmetric index + it is the unique equilibrium in a larger symmetric game that has suitable additional strategies for both players. 15

75 16 Signed perfect matchings Graph G = (V, E), V = {1,..., n} orient each edge ab E as (a, b) or (b, a) perfect matching M E of G for the edges ab of M (in any sequence), write down endpoints a, b in the order of the orientation of the edge. Define sign(m) = parity of the resulting permutation of 1,..., n.

76 16 Signed perfect matchings Graph G = (V, E), V = {1,..., n} orient each edge ab E as (a, b) or (b, a) perfect matching M E of G for the edges ab of M (in any sequence), write down endpoints a, b in the order of the orientation of the edge. Define sign(m) = parity of the resulting permutation of 1,..., n

77 16 Signed perfect matchings Graph G = (V, E), V = {1,..., n} orient each edge ab E as (a, b) or (b, a) perfect matching M E of G for the edges ab of M (in any sequence), write down endpoints a, b in the order of the orientation of the edge. Define sign(m) = parity of the resulting permutation of 1,..., n

78 16 Signed perfect matchings Graph G = (V, E), V = {1,..., n} orient each edge ab E as (a, b) or (b, a) perfect matching M E of G for the edges ab of M (in any sequence), write down endpoints a, b in the order of the orientation of the edge. Define sign(m) = parity of the resulting permutation of 1,..., n

79 16 Signed perfect matchings Graph G = (V, E), V = {1,..., n} orient each edge ab E as (a, b) or (b, a) perfect matching M E of G for the edges ab of M (in any sequence), write down endpoints a, b in the order of the orientation of the edge. Define sign(m) = parity of the resulting permutation of 1,..., n

80 16 Signed perfect matchings Graph G = (V, E), V = {1,..., n} orient each edge ab E as (a, b) or (b, a) perfect matching M E of G for the edges ab of M (in any sequence), write down endpoints a, b in the order of the orientation of the edge. Define sign(m) = parity of the resulting permutation of 1,..., n

81 17 Euler graph Euler graphs every node has even degree (= number of neighbours)

82 17 Euler graph Euler graphs every node has even degree (= number of neighbours) has Eulerian orientation (indegree = outdegree)

83 17 Euler graph Euler graphs... have tours every node has even degree (= number of neighbours) has Eulerian orientation (indegree = outdegree)... and tour

84 18 Signs of matchings in Euler graphs Theorem A graph with an Eulerian orientation has as many perfect matchings of sign + as of sign.

85 18 Signs of matchings in Euler graphs Theorem A graph with an Eulerian orientation has as many perfect matchings of sign + as of sign. Proof : Any two perfect matchings are connected by a pivoting path which connects matchings of opposite sign.

86 19 Finding a second perfect matching in an Euler graph

87 19 Finding a second perfect matching in an Euler graph

88 19 Finding a second perfect matching in an Euler graph

89 19 Finding a second perfect matching in an Euler graph

90 19 Finding a second perfect matching in an Euler graph

91 19 Finding a second perfect matching in an Euler graph

92 19 Finding a second perfect matching in an Euler graph

93 19 Finding a second perfect matching in an Euler graph

94 19 Finding a second perfect matching in an Euler graph

95 19 Finding a second perfect matching in an Euler graph

96 19 Finding a second perfect matching in an Euler graph

97 19 Finding a second perfect matching in an Euler graph

98 19 Finding a second perfect matching in an Euler graph

99 19 Finding a second perfect matching in an Euler graph

100 19 Finding a second perfect matching in an Euler graph

101 19 Finding a second perfect matching in an Euler graph

102 19 Finding a second perfect matching in an Euler graph

103 19 Finding a second perfect matching in an Euler graph

104 19 Finding a second perfect matching in an Euler graph

105 20 Input: A computational problem Graph G = (V, E) with Eulerian orientation and perfect matching of sign +. Output: A perfect matching with sign.

106 20 Input: A computational problem Graph G = (V, E) with Eulerian orientation and perfect matching of sign +. Output: A perfect matching with sign. The pivoting algorithm finds this in linear time for bipartite graphs but may take exponential time in general [Morris 1994]

107 20 Input: A computational problem Graph G = (V, E) with Eulerian orientation and perfect matching of sign +. Output: A perfect matching with sign. The pivoting algorithm finds this in linear time for bipartite graphs but may take exponential time in general [Morris 1994] Note: A second matching can be found in polynomial time [Edmonds 1965], but not with sign. Related difficult problem: Pfaffian orientations of graphs.

108 21 Finding a second matching of opposite sign Theorem [Végh / von Stengel 2014] Given a graph G = (V, E) with an Eulerian orientation and a perfect matching of sign +, a matching of sign can be found in time near-linear in E.

109 21 Finding a second matching of opposite sign Theorem [Végh / von Stengel 2014] Given a graph G = (V, E) with an Eulerian orientation and a perfect matching of sign +, a matching of sign can be found in time near-linear in E. Suffices to find sign-switching cycle.

110 21 Finding a second matching of opposite sign Theorem [Végh / von Stengel 2014] Given a graph G = (V, E) with an Eulerian orientation and a perfect matching of sign +, a matching of sign can be found in time near-linear in E. Suffices to find sign-switching cycle. up to factor given by inverse Ackermann function α.

111 22 Sign-switching cycle (SSC) Given an oriented graph and a perfect matching M, a sign-switching cycle is a cycle C with every other edge in M and an even number of forward-pointing edges. M C is a matching of opposite sign to M

112 2 Finding a SSC in near-linear time Two reductions which preserve Euler and matching property: 1. contract node of indegree = outdegree = 1 with its two edges

113 2 Finding a SSC in near-linear time Two reductions which preserve Euler and matching property: 1. contract node of indegree = outdegree = 1 with its two edges

114 2 Finding a SSC in near-linear time Two reductions which preserve Euler and matching property: 1. contract node of indegree = outdegree = 1 with its two edges

115 2 Finding a SSC in near-linear time Two reductions which preserve Euler and matching property: 1. contract node of indegree = outdegree = 1 with its two edges 2. delete directed cycle of unmatched edges

116 2 Finding a SSC in near-linear time Two reductions which preserve Euler and matching property: 1. contract node of indegree = outdegree = 1 with its two edges 2. delete directed cycle of unmatched edges

117 2 Finding a SSC in near-linear time Two reductions which preserve Euler and matching property: 1. contract node of indegree = outdegree = 1 with its two edges 2. delete directed cycle of unmatched edges

118 2 Finding a SSC in near-linear time Two reductions which preserve Euler and matching property: 1. contract node of indegree = outdegree = 1 with its two edges 2. delete directed cycle of unmatched edges until trivial SSC found

119 2 Finding a SSC in near-linear time Two reductions which preserve Euler and matching property: 1. contract node of indegree = outdegree = 1 with its two edges 2. delete directed cycle of unmatched edges until trivial SSC found

120 2 Finding a SSC in near-linear time Two reductions which preserve Euler and matching property: 1. contract node of indegree = outdegree = 1 with its two edges 2. delete directed cycle of unmatched edges until trivial SSC found, re-insert contracted edge pairs

121 2 Finding a SSC in near-linear time Two reductions which preserve Euler and matching property: 1. contract node of indegree = outdegree = 1 with its two edges 2. delete directed cycle of unmatched edges until trivial SSC found, re-insert contracted edge pairs

122 2 Finding a SSC in near-linear time Two reductions which preserve Euler and matching property: 1. contract node of indegree = outdegree = 1 with its two edges 2. delete directed cycle of unmatched edges until trivial SSC found, re-insert contracted edge pairs, switch.

123 24 Bimatrix games and signed matchings Lemke Howson for bimatrix games can take exponential time [Savani / von Stengel 2004], like complementary pivoting on labeled polytopes [Morris 1994].

124 24 Bimatrix games and signed matchings Lemke Howson for bimatrix games can take exponential time [Savani / von Stengel 2004], like complementary pivoting on labeled polytopes [Morris 1994]. Proved using dual cyclic polytopes, defined by n inequalities in d-space with vertices characterized by Gale evenness: Gale string = bitstring of length n with d bits 1 with forbidden odd runs of 1 s such as 010, 01110, ,... Examples: , ,

125 24 Bimatrix games and signed matchings Lemke Howson for bimatrix games can take exponential time [Savani / von Stengel 2004], like complementary pivoting on labeled polytopes [Morris 1994]. Proved using dual cyclic polytopes, defined by n inequalities in d-space with vertices characterized by Gale evenness: Gale string = bitstring of length n with d bits 1 with forbidden odd runs of 1 s such as 010, 01110, ,... Examples: , , Gale string = vertex, bit 1 = facet (tight inequality).

126 25 Labeled Gale strings Dual cyclic d-polytope with n facets has vertices encoded as Gale strings.

127 25 Labeled Gale strings Dual cyclic d-polytope with n facets has vertices encoded as Gale strings. Take n-string of facet labels in {1,..., d}, e.g

128 25 Labeled Gale strings Dual cyclic d-polytope with n facets has vertices encoded as Gale strings. Take n-string of facet labels in {1,..., d}, e.g labels of Gale string (completely labeled) (not completely labeled) (completely labeled)

129 25 Labeled Gale strings Dual cyclic d-polytope with n facets has vertices encoded as Gale strings. Take n-string of facet labels in {1,..., d}, e.g labels of Gale string (completely labeled) (not completely labeled) (completely labeled) Theorem [Casetti / Merschen / von Stengel 2010] The completely labeled Gale strings are the perfect matchings of the graph with nodes 1,..., d and an Euler tour given by the label string. This preserves pivoting and signs.

130 25 Labeled Gale strings Dual cyclic d-polytope with n facets has vertices encoded as Gale strings. Take n-string of facet labels in {1,..., d}, e.g labels of Gale string (completely labeled) (not completely labeled) (completely labeled) Theorem [Casetti / Merschen / von Stengel 2010] The completely labeled Gale strings are the perfect matchings of the graph with nodes 1,..., d and an Euler tour given by the label string. This preserves pivoting and signs

131 25 Labeled Gale strings Dual cyclic d-polytope with n facets has vertices encoded as Gale strings. Take n-string of facet labels in {1,..., d}, e.g labels of Gale string (completely labeled) (not completely labeled) (completely labeled) Theorem [Casetti / Merschen / von Stengel 2010] The completely labeled Gale strings are the perfect matchings of the graph with nodes 1,..., d and an Euler tour given by the label string. This preserves pivoting and signs

132 26 Definition [Edmonds 2009] Oiks and pivoting (V, R) d-oik (Euler complex) V finite set of nodes, R multiset of rooms R with R = d, any wall W = R {v} for v R R is contained in an even number of rooms

133 26 Definition [Edmonds 2009] Oiks and pivoting (V, R) d-oik (Euler complex) V finite set of nodes, R multiset of rooms R with R = d, any wall W = R {v} for v R R is contained in an even number of rooms (in two rooms: R called manifold).

134 26 Definition [Edmonds 2009] Oiks and pivoting (V, R) d-oik (Euler complex) V finite set of nodes, R multiset of rooms R with R = d, any wall W = R {v} for v R R is contained in an even number of rooms (in two rooms: R called manifold). Example d = 2: Euler graph with nodes in V and edges in R.

135 26 Definition [Edmonds 2009] Oiks and pivoting (V, R) d-oik (Euler complex) V finite set of nodes, R multiset of rooms R with R = d, any wall W = R {v} for v R R is contained in an even number of rooms (in two rooms: R called manifold). Example d = 2: Euler graph with nodes in V and edges in R. manifold, d = d = 2

136 26 Definition [Edmonds 2009] Oiks and pivoting (V, R) d-oik (Euler complex) V finite set of nodes, R multiset of rooms R with R = d, any wall W = R {v} for v R R is contained in an even number of rooms (in two rooms: R called manifold). Example d = 2: Euler graph with nodes in V and edges in R. manifold, d = pivoting d = 2

137 27 Room partitions come in pairs Given an oik R with node set V, a room partition is a partition of V into rooms. Theorem [Edmonds 2009] The number of room partitions is even.

138 28 Room partition for -manifold w

139 28 Room partition for -manifold w

140 28 w-almost room partition w

141 28 w-almost room partition w

142 28 w-almost room partition w

143 28 w-almost room partition w

144 28 w-almost room partition w

145 28 w-almost room partition w

146 28 w-almost room partition w

147 28 w-almost room partition w

148 28 Found second room partition w

149 28 [Edmonds / Sanità 2010]: exponentially long path w

150 [Edmonds / Sanità 2010]: exponentially long path 28

151 [Edmonds / Sanità 2010]: exponentially long path 28

152 [Edmonds / Sanità 2010]: exponentially long path 28

153 [Edmonds / Sanità 2010]: exponentially long path 28

154 [Edmonds / Sanità 2010]: exponentially long path 28

155 6 extra nodes, 12 extra rooms 28

156 Path length more than doubles 28

157 Path length more than doubles 28

158 Forward recursion 28

159 Forward recursion 28

160 Forward recursion 28

161 Forward recursion 28

162 Forward recursion 28

163 Forward recursion 28

164 Forward recursion 28

165 Forward recursion 28

166 Forward recursion 28

167 Forward recursion 28

168 Backward recursion 28

169 Backward recursion 28

170 Backward recursion 28

171 Backward recursion 28

172 Backward recursion 28

173 Backward recursion 28

174 Backward recursion 28

175 Backward recursion 28

176 Backward recursion 28

177 Backward recursion 28

178 Backward recursion 28

179 Final steps 28

180 Final steps 28

181 Final steps 28

182 General construction: exponentially long path 28

183 29 Orienting oiks W = R {v} for v R is called a wall of a room R

184 29 Orienting oiks W = R {v} for v R is called a wall of a room R A d-manifold is orientable if each room has a sign + or so that any two rooms with a common wall W induce opposite orientation on W ( pivoting changes sign). 2 W =

185 29 Orienting oiks W = R {v} for v R is called a wall of a room R A d-manifold is orientable if each room has a sign + or so that any two rooms with a common wall W induce opposite orientation on W ( pivoting changes sign). 2 W 4 W =

186 Orienting oiks W = R {v} for v R is called a wall of a room R A d-manifold is orientable if each room has a sign + or so that any two rooms with a common wall W induce opposite orientation on W ( pivoting changes sign). 2 W 4 W = A d-oik is orientable if half of the rooms with a common wall W induce sign + on W, the other half sign on W. 29

187 0 How to orient room partitions? Example: orientable manifold 1 B D A 5 d 4 C c b 6 a 2

188 0 How to orient room partitions? Example: orientable manifold 1 B D A 5 d 4 C c b 6 a 2

189 0 How to orient room partitions? Room partition A, a = {1,, 5}, {2, 4, 6} 1 B D A, a A 5 d 4 C c b 6 a 2

190 0 How to orient room partitions? Room partition A, a : drop node 1 1 B D A, a 1 c, a A 5 d 4 C c b 6 a 2

191 0 How to orient room partitions? Room partition A, a, sign + : drop node 1 leads to c, C, sign 1 B D + A, a 1 c, a c, C A 5 d 4 C c b 6 a 2

192 0 How to orient room partitions? Room partition A, a, sign + : drop node 1 B D + A, a 1 c, a c, C A 5 d 4 C B, a c b 6 a 2

193 0 How to orient room partitions? Room partition A, a, sign + : drop node leads to B, b, sign 1 B D + A, a 1 c, a c, C A c 5 b d 6 4 C a 2 B, a B, b

194 0 How to orient room partitions? Room partition c, C, sign : drop node 5 1 B D + A, a 1 c, a c, C A c 5 b d 6 4 C a 2 B, a B, b

195 0 How to orient room partitions? Room partition c, C, sign : drop node 5 1 B D + A, a 1 c, a c, C 5 A c 5 b d 6 4 C a 2 B, a b, C B, b

196 How to orient room partitions? Room partition c, C, sign : drop node 5 leads to b, B, sign + 1 D + A, a 1 c, a c, C B 5 A c 5 b d 6 4 C a 2 + B, a b, B? b, C B, b 0

197 How to orient room partitions? Which sign for {b, B}? 1 D + A, a 1 c, a c, C B 5 A c 5 b d 6 4 C a 2 + B, a b, B? b, C B, b 0

198 How to orient room partitions? Which sign for {b, B}? + for b, B, for B, b! 1 D + A, a 1 c, a c, C B 5 A c 5 b d 6 4 C a 2 + B, a b, B? b, C B, b 0

199 0 How to orient room partitions? for odd dimension (here d = ), order of rooms matters: permutations (for b, B) and (for B, b) have opposite parity. 1 B D A c 5 b d 6 4 C a 2 b, B + B, b

200 1 Ordered room partitions Theorem [Végh / von Stengel 2014] Let R be an oriented d-oik with node set V. Then the number of ordered room partitions (R 1,..., R V /d ) is even. Any two ordered room partitions connected by a pivoting path have opposite sign, and the respective unordered partitions are distinct.

201 1 Ordered room partitions Theorem [Végh / von Stengel 2014] Let R be an oriented d-oik with node set V. Then the number of ordered room partitions (R 1,..., R V /d ) is even. Any two ordered room partitions connected by a pivoting path have opposite sign, and the respective unordered partitions are distinct. If d is even, the order of rooms in a room partition is irrelevant.

202 1 Ordered room partitions Theorem [Végh / von Stengel 2014] Let R be an oriented d-oik with node set V. Then the number of ordered room partitions (R 1,..., R V /d ) is even. Any two ordered room partitions connected by a pivoting path have opposite sign, and the respective unordered partitions are distinct. If d is even, the order of rooms in a room partition is irrelevant. Proof uses pivoting systems with labels = nodes. Pivoting systems generalize labeled polytopes, Lemke s algorithm, Sperner s lemma, room partitions in oiks, and more.

203 2 Summary of results complementary pivoting paths on polytopes find equilibria in games

204 2 Summary of results complementary pivoting paths on polytopes find equilibria in games if pivoting is sign-switching (orientability) endpoints of paths have opposite signs +

205 2 Summary of results complementary pivoting paths on polytopes find equilibria in games if pivoting is sign-switching (orientability) endpoints of paths have opposite signs + opposite-signed matching in Euler graph found in linear time

206 2 Summary of results complementary pivoting paths on polytopes find equilibria in games if pivoting is sign-switching (orientability) endpoints of paths have opposite signs + opposite-signed matching in Euler graph found in linear time exponentially long paths for matchings in Euler graph emulate exponentially long Lemke Howson paths in games we can orient oiks and room partitions (in odd dimension: need ordered partition).

207 Thank you!