Transcript Selected Topics in Algorithmic Game Theory

The Computational Complexity
of Finding a Nash Equilibrium
Edith Elkind, U. of Warwick
Based On…
• Reducibility Among Equilibrium Problems
(Goldberg, Papadimitriou): Aug 2005
• The Complexity of Computing a Nash Equilibrium
(Goldberg, Daskalakis, Papadimitriou): Sep 2005
• 3-NASH is PPAD-Complete
(Chen, Deng): Nov 2005
• Three-Player Games Are Hard
(Daskalakis, Papadimitriou): Nov 2005
• Settling the Complexity of 2-Player NashEquilibrium (Chen, Deng): Dec 2005
Normal Form Games
• finite set of players {1, …, n}
• each player has k actions
(pure strategies): 1, …, k
• payoffs of the ith player: Pi: {1, …, k}n → R
0
Row
player:
1
0
2
0
1
0
1
0
Column
player:
1
0
1
0
1
0
3
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
0
Row
player:
1
0
2
0
1
0
1
0
Column
player:
1
0
1
0
1
0
3
Pure vs. Mixed Strategies
• NE in pure strategies may not exist!
– “matching pennies”
• Mixed strategy: a probability distribution
over actions
– 50% tail, 50% head
H
Row
player:
T
H
1
-1
T
-1
1
Column
player:
H
T
H
-1
1
T
1
-1
Existence of NE
• Theorem (Nash 1951):
any game in normal form has an
equilibrium in mixed strategies
$1 000 000 question:
how to find one?
Finding mixed NE in 2 x 2 Games
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
Suppose
R plays 1 w.p. r
EP(C) from playing 0: (1-r)*1
EP(C) from playing 1: r*3
1-r = 3r iff r = ¼
NE: r=1/4, c=2/3
0
Row
player:
1
0
2
0
1
0
1
0
Column
player:
1
0
1
0
1
0
3
2 players, k actions
• Representation: two k x k matrices
• Checking for pure NE: easy
– at most k2 of them
• Checking for mixed NE:
– all straightforward methods are exptime
– Lemke-Howson algorithm is exptime, too
(previous talk)
• For 2 players all NE are rational
– but not for 3 and more players…
n players, 2 actions
• 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
Algorithms for NE in
Graphical Games
• 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):
• shown to be incorrect in E., Goldberg, Goldberg, ACM EC’06
• Graphs of max degree 2:
– poly-time algorithm (EGG’06)
Is Finding NE NP-hard?
• Reminder: a problem P is NP-hard if you can
reduce 3-SAT to it:
– “yes”-instance 3-SAT → “yes”-instance of P
– “no”-instance 3-SAT → “no”-instance of P
• Problem:
each instance of NASH is a “yes”-instance!
– every game has a NE
– need complexity theory for search problems
• Side note: pure Nash for n players, NE of total
value > K are NP-hard
Reducibility Among Search
Problems
S: X
Y
f
T: X’
g
Y’
• S associates x in X with a solution set S(x)
• Total search problem: for any x, S(x) is not empty
• S is reducible to T if:
– f, g easy to compute
– g(T(f(x))) is in S(x)
If T is easy, so is S
Equivalences: GP’05
r-player game G
NE of G
f
g
deg 3 graphical game G’
NE of G’
deg d graphical game G
NE of G
f
d2-player game G’
g
NE of G’
d-Graphical Game GG →
d2-Player Game G
• Color the graph of GG:
d(u,v) ≤ 2  color(u) ≠ color(v)
• Each color is a player of G
• RED chooses a red vertex in GG
and an action for that vertex in GG
• payoff=payoff1+payoff2
– payoff1: BLUE tries to guess which vertex RED chose;
RED pays a penalty if BLUE guesses correctly
– payoff2: if all neighbors of a chosen vertex are also
chosen, it gets same payoff as in GG, else 0
r-Player Game G →
3-Graphical Game GG
•
•
•
•
•
Si: space of pure strategies of player i
S- i = S1 * … Si-1*Si+1 *.. * Sr
xij: the probability that ith player uses jth strategy
xs: x1s1 * x2s2 … * xrsr (for s in S-i)
uijs: utility of the ith player when he plays j and others play
according to s
NE: 0 ≤ xij ≤ 1
Sj xij =1
Ss in S uijsxs > Ss in S uij’sxs implies xij’ = 0
-p
-p
r-Player Game G →
3-Graphical Game GG
• Vertex Vij for any pair (player=i, action=j)
• Want: Pr[Vij plays 1] = Pr [i plays j in G]=xij
• Idea: graphical games can do math!
– Enforce constraints from the previous slide…
v1
v2
v3
Need gadgets for
+, *, c, =, min, max, …
u
Set payoffs to u, v3 so that p[v3]=p[v1] * p[v2]
Equivalences: GP’05
r-player game G
NE of G
f
g
deg 3 graphical game G’
NE of G’
deg d graphical game G
NE of G
f
d2-player game G’
g
NE of G’
Combining Reductions: GP’05
r-player game G
4
X9-player game G’
NE of G
f
g
NE of G’
Finding NE in a 4-player game is
as hard as
finding NE in a r-player game
for any constant r
Completeness Results?
• Can we prove that any total search problem is
reducible to r-NASH?
• Not really: the class T of all total search
problems is a semantic class
– not known how to find complete problems for these
• Want to pick a large subclass S of T s.t.
– S includes some natural problems
– there are problems that are complete for S
– in particular, r-NASH is complete for S
END OF THE LINE
• Input: Boolean circuits
S (Successor), P (Predecessor):
– n inputs, n outputs
– S(0n) ≠ 0n, P(0n) = 0n
• Output: x ≠ 0n s.t.
– S(P(x)) ≠ x or P(S(x)) ≠ x
Intuition: G=(V, E):
– V = Sn;
– E = {(x,y) | y=S(x), x=P(y)}
00000
11001
01011
01011
PPAD
• PPAD: Polynomial Parity Argument,
Directed version
• PPAD is the class of all search problems
that are reducible to END OF THE LINE
search problem
solution
f
circuits S, T
g
“end of the line”
r-NASH is in PPAD
• Proof on Nash’s theorem:
– existence of NE reduces to Brouwer’s fixpoint
theorem
– Brouwer’s fixpoint theorem reduces to
Sperner’s lemma
– Sperner’s lemma is proven by a parity
argument (similar to END OF THE LINE)
• Reduction of r-NASH to END OF THE
LINE can be extracted from these proofs
(Papadimitriou 94)
Brouwer’s Fixpoint Theorem
• Brouwer’s Theorem: Any continuous
mapping from the simplex to itself has a
fixpoint.
• Nash  Brouwer proof sketch:
– set of all strategy profiles → simplex
– mapping: (s1, …, sn) → (s1+d1, …, sn+dn),
where di is a shift in the direction of best
response to (s1, …, si-1, si+1, …, sn)
– NE is a point where noone wants to deviate,
i.e., a fixpoint
Sperner’s Lemma
• Proper coloring:
B
– vertices on BC are not blue
– vertices on AC are not green
– vertices on AB are not yellow
A
• Sperner’s Lemma:
there exists a trichromatic triangle
• Brouwer’s theorem  Sperner’s Lemma:
– x is blue if the grad(F) at x points away from A, etc.
– trichromatic triangle “has no direction”
– repeat at increased resolution…
C
Opposite Direction: 3D-BROUWER
• Input:
– 3D unit cube divided into 23n cubelets
– cijk is the center of Kijk
– f(cijk)=cijk+dijk, dijk is in {d0, d1, d2, d3}, where
• d1=(a, 0, 0), d2=(0, a, 0)
• d3=(0, 0, a), d0=(-a, -a, -a)
– circuit C: {0, 1} 3n → {0, 1, 2, 3} selects dijk
• Output:
– a panchromatic cubelet, i.e., one that has all
of d0, d1, d2, d3 among its 8 neighbors
3D-BROUWER is PPAD-complete
• Papadimitriou (1994) shows that a more
complicated version of 3D-BROUWER is
PPAD-complete
• This version was proven hard in DGP’05
• Reduction from END OF THE LINE
– embed the line L into 3d cube
– “protect” L from color 0 using three other colors
– color the rest of inner cubelets with 0
r-NASH vs 3D BROUWER
• Existence of NE follows from Brouwer’s fixpoint
theorem
• NE are special cases of Brouwer’s fixpoints
– just how special?
• Can any fixpoint be represented as a NE of a
game?
• DGP’05: YES!  4-NASH is PPAD complete
• Proof:
– 4-NASH  deg 3 Graphical Nash
– graphical games can compute fixpoints
4-NASH to 3-NASH
• Daskalakis, Papadimitriou: modify
arithmetic gadgets so that the graph
is 3-colorable
• Chen, Deng: same gadgets, but allow
for small error
2-NASH
• Chen, Deng:
– avoid graphical games
– reduce directly from 3D-BROUWER to
2-NASH using arithmetic gadgets similar to
graphical game gadgets
• Game over?
Graphical Games: Open Problems
• Degree:
– deg 3 PPAD-complete (DGP’05b)
– deg 2 polynomial time solvable (EGG’06)
• Pathwidth:
–
–
–
–
paths: poly-time
pathwidth 1: maybe algorithm from EGG’06 still works
pathwidth 2: any KLS-style algo is exptime (EGG’06)
pathwidth > r, r constant: PPAD-complete (EGG’06)
• Finding NE on trees?