Nash Equilibria In Graphical
Games On Trees Revisited
(To appear in ACM EC’06)
Edith Elkind
Leslie Ann Goldberg
Paul Goldberg
(University of Warwick)
Normal Form Games
(with 2 actions per player)
• finite set of players {1, …, n}
• each player has 2 actions
(pure strategies): 0 and 1
• payoffs of the ith player: Pi: {0, 1}n → R
0
Row
player:
1
0
2
0
1
0
1
0
Column
player:
1
0
1
0
1
0
3
Mixed Strategies
• pure strategy = action
• mixed strategy = probability distribution
over actions
• pi = Prob [i plays 1]
• expected payoff of the ith player
for a strategy profile p = (p1, …, pn):
EP(i) = E[Pi(a) | Prob[ai=1] = pi]
Nash Equilibrium
• Nash equilibrium: a strategy profile such that
noone wants to deviate given other players’
strategies, i.e., each player’s strategy is a best
response to others’ strategies:
– (0, 0) and (1, 1) are both NE.
– any other NE?
0
Row
player:
1
0
2
0
1
0
1
0
Column
player:
1
0
1
0
1
0
3
Finding NE in 2-player 2-action
Games
0
Row
player:
1
0
2
0
1
0
1
0
Column
player:
BR(C)
Suppose
1 c
R plays 1 w.p. r
EP(C) from playing 0: 2/3
(1-r)*1
EP(C) from playing 1:
r*3
1-r > 3r iff r < ¼
BR(R)
r
1/4
1
NE: r=1/4, c=2/3
1
0
1
0
1
0
3
Suppose
C plays 1 w.p. c
EP(R) from playing 0:
(1-c)*2
EP(R) from playing 1:
c*1
(1-c)*2 > c iff c < 2/3
NE for n-player 2-action games
• (poly-time) algorithm for NE in n-player games?
• representation: payoffs to each player for every
action profile (vector in {0, 1}n): n2n numbers
• graphical games:
– players are associated with the vertices of a graph;
– each player’s payoff depends on his own action and
actions of his neighbors
– n players, max degree d => n2d+1 numbers
W
T
V
U
W’s payoffs
(16 cases):
t=0, u=0, v=0, w=0: 12
t=1, u=0, v=0, w=0: 31
….
t=1, u=1, v=1, w=1: -6
Related Work
• Bounded-degree trees:
– Exp-time algorithm/poly-time approximation algorithm
to find all NE (Kearns, Littmann, Singh, UAI 2001)
– ??? poly-time algorithm to find a single NE (Kearns,
Littmann, Singh, NIPS’2001)
• General graphs:
– can it be NP-hard? no: NE always exists
– hardness notion for total functions: PPAD-hardness
– NE in graphical games with d ≥ 3 is PPAD-complete
(GP, DGP, STOC’06)
Our Results
• Algorithm in NIPS’01 paper is incorrect (does not
always output a NE)
• We fix the NIPS’01 algorithm, but…
– our algorithm runs in poly-time on paths
• with a trick, also on cycles
• can be used to find all NE (rather than a single one)
– there is a graph of pathwidth 2 on which our algorithm
(and all algorithms that use the basic approach of the
UAI’01 paper) runs in exp time
• The problem remains PPAD-complete for
bounded pathwidth graphs
• Open question: what if pathwidth = 1?
Algorithm for Trees
• Recall from 2-player case: best response
c
function
c = BR(r)
r
• Potential best response: v is a PBR to w iff
when W plays w, there is a NE
W
for T’ in which V plays v.
v=PBRV(w)
V
• Bottom-up approach:
information propagates
T’
U
U
from the leaves to the root
U
3
1
2
Computing PBR: Example
U
V
W
• Payoffs to U: 0 if U=V, 1 if U≠V
– EP(U) from playing 0: v; EP(U) from playing 1: 1-v
• Payoffs to V:
– P000=1, P001=-9, P100=9, P101=-1, PU1W=0 for all U, W
– EP(V) from playing 0:
(1-u)(1-w)*1+(1-u)w*(-9)+u(1-w)*9+uw*(-1)
– V is indifferent btw 0 and 1 iff w = (8u+1)/10 = f(u)
v
u
1
1
(v, u) → (f(u), v)
.5
.5
1
v
.1
.9 1
w
Computing PBR: General Case
U
V
W
• EP(V) from playing 0: a1uw+a2u+a3w+a4
• EP(V) from playing 1: b1uw+b2u+b3w+b4
• V is indifferent between 0 and 1 iff
w = f(u) = Au+B/Cu+D
• PBRV(W)=L0 U f(PBRU(V)) U L1
• For paths, we can show that for any V,
PBRV(W) consists of polynomially many
segments (rectangles if degenerate)
Computing PBR on Trees
• Trees: similar algorithm
• Indifference function: w = L1(u1, …, un)/L2(u1, …, un)
• Potential best response can be exponential in size!