computing nash-equilibria in 2-player games and beyond ...akmenon/nashequilibria...finding...

OutlineIntroductionComplexity

Finding equilibria in 2-player games

Computing Nash-equilibria in 2-player games andbeyond

Algorithms Reading Group

Aditya Menon

June 3, 2007

Aditya Menon Computing Nash-equilibria in 2-player games and beyond



IntroductionNash equilibriumA formal definition

ComplexityCharacterizing complexityPPAD

Finding equilibria in 2-player gamesLemke-Howson algorithmOther approaches




Nash equilibriumA formal definition

Nash-equilibrium - Recap

I Refers to a special kind of state in an n-player gameI No player has an incentive to unilaterally deviate from his

current strategyI A kind of “stable” solution

I Existence depends on the type of gameI If strategies are “pure” i.e. deterministic, does not have to

exist in the gameI If strategies are “mixed” i.e. probabilistic, then it always exists

I Yet how do we find it!?!





Notation

I Suppose that player p follows the mixed strategyxp = (xp1, . . . , xpnp)

I The ith entry gives the probability that player p plays move i

I Let x := (x1, . . . , xn) be the collection of strategies for allplayers

I Let the function Up(x) denote the expected utility or payoffthat player p gets when each player uses the strategy dictatedin x

I Based on up(s1, . . . , sn), which is the utility when player qplays sq





Formal definition

I We say that x∗ = (x1∗, . . . , xn

∗) is a Nash equilibrium if...I “No player has an incentive to unilaterally deviate from his

current strategy”

I If player p decides to switch to a strategy yp, then write theresulting strategy set as x−p; yp

I So, x∗ is a N.E. if, for every player p, and for any mixedstrategy yp for that player, we have

Up(x∗) ≥ Up(x

∗−p; yp)

I A more symmetric version:

Up(x∗−p; x

∗p) ≥ Up(x

∗−p; yp)





Why the equilibrium exists

I Based on Kakutani’s fixed point thereom

I Generalizes the following: take two pieces of paper (which are3D objects) and imagine placing them on top of each other,so that the “points” align. Crumple the top sheet withouttearing it, and place it on top of the other sheet. Then, thereis always a point that has not changed position on the twosheets.




Characterizing complexityPPAD

Questions about finding Nash equilibria

I Proof of existence is non-constructiveI So how do we find it?

I Can we find it?I Can we find it efficiently?





Do we really care?

I Obviously an “interesting” problem, but is there a significantbenefit in being able to compute it?

I Arguably, equilibrium is not useful unless we can find itI Consider the case of a marketI “If your laptop can’t find it [the equilibrium], then neither can

the market” [Papadimitriou]





Importance of the problem

I “Together with factoring, the complexity of finding a Nashequilibrium is in my opinion the most important concrete openquestion on the boundary of P today” [Papadimitriou]





Complexity of the problem

I Need to define a special class, PPAD, for this problem

I Turns out that finding the Nash-equilibrium is PPAD-completeI What about NP?

I Probably not NP-completeI The decision version is in P

I Why?

I Because the equilibrium always exists!





Complexity of the problem

I Need to define a special class, PPAD, for this problem

I Turns out that finding the Nash-equilibrium is PPAD-completeI What about NP?

I Probably not NP-completeI The decision version is in P

I Why?I Because the equilibrium always exists!





Polynomial parity argument

I Simple fact: Every graph has an even number of odd-degreenodes

I Proof: Let W = {v ∈ V : v has odd degree }

2|E | =∑v∈W

deg(v) +∑v /∈W

deg(v)

=∑v∈W

odd + even

I Corollary: If a graph has maximum degree 2, then it musthave an even number of leaves





The class TFNP

I PPAD is based on the class TFNP

I Suppose we have a set of binary predicates P(x , y) where

(∀x)(∃y) : P(x , y) = TRUE

I Problems in TFNP: Given an x , find a y so that P(x , y) isTRUE

I Subclasses defined based on how we decide (∃y) : P(x , y) isTRUE





PPAD in terms of TFNP

I PPAD is defined by the following problemI Suppose we have an exponential-size graph where the

in-degree and out-degree of each node is at most 1I Given any node v , we have a polynomial-time algorithm that

finds the neighbours of vI Problem: Given a leaf node, output another leaf node

I Existence is guaranteed by the parity argument





Completeness of finding Nash equilibrium

I Finding a Nash equilibrium is PPAD-completeI For 4-player games... [Papadimitriou]

I ...and 3-player games [Papadimitriou]I ...and even for 2-player games! [Chen]

I So finding the Nash equilibrium even for 2-player games is noeasier than doing it for n-players!

I At the moment, however, not much known about how “hard”a class PPAD is






I Finding a Nash equilibrium is PPAD-completeI For 4-player games... [Papadimitriou]I ...and 3-player games [Papadimitriou]

I ...and even for 2-player games! [Chen]








I Finding a Nash equilibrium is PPAD-completeI For 4-player games... [Papadimitriou]I ...and 3-player games [Papadimitriou]I ...and even for 2-player games! [Chen]






Lemke-Howson algorithmOther approaches

Approaches to finding equilibria

I No P algorithms known!

I Most approaches are based on solving non-linear programs(for general n)

I Completeness result means even 2-player games are not (yet)“easy” to solve

I One of the earliest algorithms for finding equilibria in 2-playergames: Lemke-Howson algorithm





Lemke-Howson algorithm

I An algorithm for finding the Nash equilibrium for a game with2 players

I Developed in 1964I Independent proof of why equilibrium must exist

I Worst-case exponential time, but in practise quite goodperformance

I Still viewed favourably, although it has been beaten of late





How we proceed

I We need to redefine a Nash equilibrium for 2-playersI Try and make a graph that lets us find equilibria easily

I Exploiting the convenience of the alternate definition





Utility for 2-players

I Suppose that for a 2-player game, we have the mixedstrategies x = (s, t)

I Label the strategies by I = {1, . . . ,m} for player 1, andJ = {m + 1, . . . ,m + n} for player 2

I Expected utility for player p must be

Up(x) =∑

i

∑j

Pr[player 1 chooses i ]× Pr[player 2 chooses j ]

× Payoff for player p when 1 plays i and 2 plays j

=∑

i

∑j

s(i)t(j)up(i , j)





Nash equilibria for two players

I We call x∗ = (s∗, t∗) a Nash equilibrium iff

(∀s)∑

i

∑j

s∗(i)t∗(j)u1(i , j) ≥∑

i

∑j

s(i)t∗(j)u1(i , j)

(∀t)∑

i

∑j

s∗(i)t∗(j)u2(i , j) ≥∑

i

∑j

s∗(i)t(j)u2(i , j)





Claim

I Claim: If in a Nash equilibrium player p can play strategy i(non-zero probability), then if player p chooses to playstrategy i , it is the best possible strategy

I Mathematically,

s∗(i) > 0 =⇒ (∀i0)∑

j

t∗(j)u1(i , j) ≥∑

j

t∗(j)u1(i0, j) (1)

t∗(j) > 0 =⇒ (∀j0)∑

i

s∗(i)u2(i , j) ≥∑

i

s∗(i)u2(i , j0) (2)





Lemma

I Lemma: Let πp,i denote the mixed strategy (0, . . . , 1, . . . , 0) i.e. wedeterministically choose strategy i for player p. Then, x is a NashEquilibrium iff

(∀p, πp,i)Up(x) ≥ Up(x−p;πp,i)

I Proof: (note that =⇒ direction is by definition)

Up(x−p; yp) =∑ip

yp(ip)

{∑i1...in

x1(i1) . . . xn(in)up(x−p;πp,ip)

}=

∑ip

yp(ip)Up(x−p;πp,ip)

≤ Up(x) since∑

yp(i) = 1





Proof of claim

I Use the lemma: Up(x∗) ≥ Up(x−p;πp,i)

Up(x∗) =

∑x∗p(i)Up(x

∗) since∑

x∗p(i) = 1

≥∑

x∗p(i)Up(x∗−p;πp,i)

= Up(x∗)

I So, we have to conclude that∑x∗p(i)Up(x

∗) =∑

x∗p(i)Up(x∗−p;πp,i)

I Taking terms to one side,

x∗p(i) > 0 =⇒ Up(x∗) = Up(x

∗−p;πp,i)





Reformulation of Nash equilibrium - I

I So, x∗ is a Nash equilibrium iffI For player 1, equation 1 holds or Pr[strategy i ] = 0I For player 2, equation 2 holds or Pr[strategy j ] = 0





Reformulation of Nash equilibrium - II

I Define

S i = {s : s(i) = 0},S j = {s :∑

i

s(i)u2(i , j) ≥∑

i

s(i)u2(i , j0)}

T j = {t : t(j) = 0},T i = {t :∑

j

t(j)u1(i , j) ≥∑

j

t(j)u1(i0, j)}

I Then, x∗ is a Nash equilibrium iff

(∀i)s ∈ S i ∨ t ∈ T i

(∀j)s ∈ S j ∨ t ∈ T j





Labelling

I We are claiming that x = (s, t) is an equilibrium iff...I For any k ∈ I ∪ J, either s or t (or maybe both) is in the

appropriate region Sk or T k

I Can think of these k’s as labels of pointsI Labels(s) = {k ∈ I ∪ J : s ∈ Sk}I Labels(t) = {k ∈ I ∪ J : t ∈ T k}I Note: This definition also makes sense even when s, t aren’t

valid strategies





Reformulation of Nash equilibrium - III

I Natural label for x = Labels(s)∪ Labels(t)

I So, x is a Nash equilibrium iff it is completely labelled





Strategy simplex

I m strategies =⇒ valid strategy space is an (m − 1)dimensional simplex

(1, 0, 0)

(0, 0, 1)

(0, 1, 0)

(0, 0, 0)

I With labelling, we can split up the simplex into regions





Labelling - example

I For the payoff matrix A = [0 6; 2 5; 3 3]

I Label the strategy space for player 2:

1, 4

1, 2

2, 3

3, 5

4

5

(13, 2

3

)

(23, 1

3

)

(1, 0)

(0, 1)





Reformulation of problem

I Using the labelling definition, x is a Nash equilibrium iff it iscompletely labelled

I ∼ Point that lie in the intersection of many regionsI But here there are two variables, and two sets of spacesI Can “sacrifice” one for the sake of the other

I New problem: How do we find points that are completelylabelled?





High-level solution

I With labelling, we have divide the strategy spaces intosub-spaces

I We want to end up with a pair of strategies that together “liein the intersection of the spaces”

I Search problem: starting off with some initial pair ofstrategies, we want to find the equilibrium





High-level solution

I Think of the space as a graph

I Vertices should correspond to strategy pairsI Edges should define some well-defined change in the strategies

I Which makes it easy to find equilibria

I Following a well-defined path on this graph should take us toequilibria

I Problem: How do we make such a graph?





Graph construction

I Form the graphs GS = (VS ,ES), GT = (VT ,ET ) where:I VS ← {s ∈ Rm

+ :∑

si ≤ 1 ∧ s has exactly m labels }I ES between s1, s2 if they differ in exactly one labelI Similarly for VT ,ET

I Note: This is now “filling” the strategy simplex





Example

I Payoff B = [1 0; 0 2; 4 3]

I Vertices lie on simplex, except for 0

(0, 0, 0) (1, 0, 0)

(0, 1, 0)

(0, 0, 1)

(23, 1

3, 0

)

(0, 1

3, 2

3

)

L = 1, 3, 5

L = 1, 2, 3

L = 1, 2, 4

L = 3, 4, 5

L = 1, 2, 4

L = 2, 3, 5





Graph construction

I Form the product graph G = GS × GT

I V = {(s, t) : s ∈ VS , t ∈ VT}I E = {(s1, t1)→ (s1, t2) : t1 → t2} ∪ {(s1, t1)→ (s2, t1) :

s1 → s2}





Graph and equilibria

I So now, have a graph where vertices are strategy pairs

I Know that the equilibria is one of the vertices

I Question: How do we find this vertex?





Two definitions

I Define:I L−(k) := vertices that have all labels except, possibly, kI L := vertices that have all labels

I By definition, L ⊆ L−(k)

I L = (0, 0) ∪ {Equilibria}I We can prove some properties about these sets...





Fact 1

I Fact: For any k, every member of L is adjacent to exactly onemember of L−(k)− L

I For any label, every (pseudo) equilibrium is adjacent to exactlyone strategy pair that is missing that label

I Proof:I Let (s, t) ∈ L. Then the label k must apply to either s or t, by

definitionI Suppose that s is labelled with k. Then, there must be an

edge between (s, t) and the point (s′, t) where s′ is missing thelabel k

I There is only one such s′ that is missing the label k - hencethe neighbour is unique

I Similar argument if t is labelled with k





Fact 2

I Fact: For any k, every member of L−(k)− L is adjacent toexactly two members of L−(k)

I Every strategy pair missing exactly one label is adjacent toexactly two other strategy pairs that are potentially missingthe same label

I Proof:I Since|Labels(s, t)| = m + n − 1 6= |Labels(s)|+ |Labels(t)| = m + n,there must be a duplicate label, `

I In the graph GS , we must have an edge from s to some otherpoint s′, where s′ does not have the label `

I Then, the edge (s, t)→ (s′, t) must belong to EI Similarly for GT - this means that the graph G has exactly two

edges that change the labelling





Putting the facts together

I L−(k) describes a subgraph of G containing (disjoint) pathsand loops of G

I The endpoints of a path in G are (pseudo) equilibriaI Problem: How do we find this set quickly?

I Look at this later...





Finding equilibria

I Start off at the “artificial equilibrium” (0, 0)

I Choose an arbitrary label ` ∈ I ∪ J

I Follow the path generated by the set L−(`)

I When we reach the end of the path, we necessarily havestopped at an equilibria





Example

I Payoff matrices

A =

0 62 53 3

B =

1 00 24 3

I Choose the label 2 to be dropped i.e. move along L−(2)





Example

I Start off at the artificial equilibrium, ((0, 0, 0), (0, 0))→ labels{1, 2, 3}, {4, 5}

(13, 2

3

)

(23, 1

3

)

(1, 0)

(0, 1)

(0, 0)

(0, 0, 0) (1, 0, 0)

(0, 1, 0)

(0, 0, 1)

(23, 1

3, 0

)

(0, 1

3, 2

3

)

L = 1, 3, 5

L = 1, 2, 3

L = 1, 2, 4

L = 3, 4, 5

L = 1, 2, 4

L = 2, 3, 5

L = 4, 5

L = 1, 4

L = 1, 2

L = 2, 3

L = 3, 5





Example

I Step 1: ((0, 1, 0), (0, 0))→ labels {1, 3, 5}, {4, 5}; duplicate is5

(13, 2

3

)

(23, 1

3

)

(1, 0)

(0, 1)

(0, 0)

(0, 0, 0) (1, 0, 0)

(0, 1, 0)

(0, 0, 1)

(23, 1

3, 0

)

(0, 1

3, 2

3

)

L = 1, 3, 5

L = 1, 2, 3

L = 1, 2, 4

L = 3, 4, 5

L = 1, 2, 4

L = 2, 3, 5

L = 4, 5

L = 1, 4

L = 1, 2

L = 2, 3

L = 3, 5





Example

I Step 2: ((0, 1, 0), (0, 1))→ labels {1, 3, 5}, {1, 4}; duplicate is1

(13, 2

3

)

(23, 1

3

)

(1, 0)

(0, 1)

(0, 0)

(0, 0, 0) (1, 0, 0)

(0, 1, 0)

(0, 0, 1)

(23, 1

3, 0

)

(0, 1

3, 2

3

)

L = 1, 3, 5

L = 1, 2, 3

L = 1, 2, 4

L = 3, 4, 5

L = 1, 2, 4

L = 2, 3, 5

L = 4, 5

L = 1, 4

L = 1, 2

L = 2, 3

L = 3, 5





Example

I Step 3: ((23 , 1

3 , 0), (0, 1))→ labels {3, 4, 5}, {1, 4}; duplicate is4

(13, 2

3

)

(23, 1

3

)

(1, 0)

(0, 1)

(0, 0)

(0, 0, 0) (1, 0, 0)

(0, 1, 0)

(0, 0, 1)

(23, 1

3, 0

)

(0, 1

3, 2

3

)

L = 1, 3, 5

L = 1, 2, 3

L = 1, 2, 4

L = 3, 4, 5

L = 1, 2, 4

L = 2, 3, 5

L = 4, 5

L = 1, 4

L = 1, 2

L = 2, 3

L = 3, 5





Example

I Step 4: ((23 , 1

3 , 0), (13 , 2

3))→ labels {3, 4, 5}, {1, 2}

I Completely labelled, and so an equilibria

(13, 2

3

)

(23, 1

3

)

(1, 0)

(0, 1)

(0, 0)

(0, 0, 0) (1, 0, 0)

(0, 1, 0)

(0, 0, 1)

(23, 1

3, 0

)

(0, 1

3, 2

3

)

L = 1, 3, 5

L = 1, 2, 3

L = 1, 2, 4

L = 3, 4, 5

L = 1, 2, 4

L = 2, 3, 5

L = 4, 5

L = 1, 4

L = 1, 2

L = 2, 3

L = 3, 5





Example

I Step 4: ((23 , 1

3 , 0), (13 , 2

3))→ labels {3, 4, 5}, {1, 2}I Completely labelled, and so an equilibria

(13, 2

3

)

(23, 1

3

)

(1, 0)

(0, 1)

(0, 0)

(0, 0, 0) (1, 0, 0)

(0, 1, 0)

(0, 0, 1)

(23, 1

3, 0

)

(0, 1

3, 2

3

)

L = 1, 3, 5

L = 1, 2, 3

L = 1, 2, 4

L = 3, 4, 5

L = 1, 2, 4

L = 2, 3, 5

L = 4, 5

L = 1, 4

L = 1, 2

L = 2, 3

L = 3, 5





Algorithm summary

I Consider strategies sk that have all labels except, possibly, somelabel k

I Clearly, every equilibrium belongs to this setI So too does the “artificial equilibrium”, 0

I Construct a graph from all strategies

I Then, one can show:

I Each strategy missing a label is adjacent to exactly two suchstrategies

I Each equilibrium is adjacent to only one strategy sk

I It follows that:

I Equilibria are endpoints of paths on the graph





Generating L−(`)

I Second problem...I How do we find adjacent strategies?





Generating L−(`)

I “Pivoting” approach

I Able to implicitly generate the graph, on-the-fly





Optimization problem

I Suppose we know player 2’s strategy to be t

I Suppose we have payoff matrices A,B

I We need to choose our strategy to maximize the payoff, i.e.

U1(x) =∑

i

∑j

s(i)t(j)A(i , j)

=∑

i

s(i)A(i , :)t

= sT (At)





Optimization problem

I Primal:maximize sT (At) :

∑s(i) = 1, s ≥ 0

I Dual:minimize u : u1 ≥ At





Linear system

I We need to solves.1 = 1

t.1 = 1

u1− At ≥ 0

v1− BT s ≥ 0

s, t ≥ 0

I Equilibrium iff the above, and

sT (u1− At) = 0

tT (v1− BT s) = 0





Linear system with slackness

I Introduce slack variables z,w:

s.1 = 1

t.1 = 1

−u1 + At + w.1 = 0

−v1 + BT s + z.1 = 0

s, z ≥ 0

t,w ≥ 0





System in matrix form

I Can write system as a matrix equation

I Consider player 1’s equations - same as:

s(1) + . . . + s(m) = 1

bTi .s− v + z(i) = 0, i = 1, . . . , n

s, z ≥ 0





System in matrix form

I Can write as Dr = c, where

D =

1 1 . . . 1 0 0 . . . 0

b11 b21 . . . bm1 −1 1 . . . 0...

b1n b2n . . . bmn −1 0 . . . 1

r =

s(1)...

s(m)v

z(1)...

z(n)

, and c =

10...0





Connection with labelling

I Let the basis of the column space of D be {Dβ}I Call rj ...

I A basic variable if Dj is in the basisI A non-basic variable otherwise; let rj = 0 in this case

I Labels correspond to non-basic variablesI r1, . . . , rm = 0 =⇒ s(i) = 0 i.e. strategy is not playedI rm+2, . . . , rm+n+1 = 0 =⇒ z(i) = 0 i.e. strategy is

best-response

I So, removing a label corresponds to making a variable strictlypositive





Rewriting system

I System is the same as

D−1β (Dr) = D−1

β c

I Fact: D−1β D = I +

∑j /∈β D−1

β Dj

I System is the same as

rβ = D−1β c−

∑j /∈β

D−1β Dj rj





Why rewrite?

I What is wrong with the standard representation of thesystem?

I Nothing “wrong” with it...I Just that the new system makes it easier to find strategies

with different labelsI i.e. Generating the set L−(`)I This is the “pivot” operation





Example

I Use earlier payoff matrix

I Matrix D for player 1 is

D =

1 1 1 0 0 01 0 4 −1 1 00 2 3 −1 0 1

I Choose column-space {2, 4, 5}

I Setting x1, x3 = 0I Setting z2 = 0

Dβ =

1 0 00 −1 12 −1 0





Example

I Solution:

rβ =

122

−1

23

r1 −

1−13

r3 −

0−1−1

r6

I Or equivalently:

x2 = 1− x1 − x3

v = 2− 2x1 + x3 + z2

z1 = 2− 3x1 − 3x3 + z2





Example

I Can read off solution from hereI s = (0, 1, 0)I v = 2I z1 = 2, z2 = 0

I Can read the labels: 1, 3, 5

I Do the same thing for player 2

I Result: We get a pair of strategies (s, t), and we know whattheir labels are

I Question: How do we find the “adjacent” strategy, as per thegraph?





Finding adjacent strategy

I Traversal step in the algorithm was...I Find the duplicate label in the two strategiesI Drop the label from player 1 and player 2 alternately

I Not hard to find duplicate (just read off from solution)I How to drop label?

I Just change the appropriate variable to be positiveI As large as possible while satisfying constraints








I Just change the appropriate variable to be positive

I As large as possible while satisfying constraints








I Just change the appropriate variable to be positiveI As large as possible while satisfying constraints





Example of pivoting

I Starting equation for player 1:

x2 = 1− x1 − x3

v = 2− 2x1 + x3 + z2

z1 = 2− 3x1 − 3x3 + z2

I Starting equation for player 2:

y2 = 1− y1

u = 6− 6y1 + w1

w2 = 1− 3y1 + w1

w3 = 3− 6y1 + w1





Example of pivoting

I Solve to get ((0, 1, 0), (0, 1))I s = (0, 1, 0), z = (2, 0)I w = (0, 1, 3), t = (0, 1)

I x1 = 0 and w1 = 0, so label 1 is duplicateI So we look to get rid of itI =⇒ Modify x1 while satisfying non-negativity constraints

I Relevant equations:x2 = 1− x1

z1 = 2− 3x1





Example of pivoting

I For first equation: x1 ∈ (0, 1]

I For second equation: x1 ∈(0, 2

3

]I Clearly, x1 → 2

3I x2 → 1

3I But, z1 → 0!I As a result, we see that label 4 is duplicate

I Another duplicate =⇒ we repeat the process!





Updating system

I Now x1 should appear on the LHS, since it is no longer zero

I New column basis is {D1,D2,D4}

Dβ =

1 1 01 0 −10 2 −1





Updating system

I New system is

v =2

3− 3x3 −

2

3z1 +

1

3z2

x1 =1

3− x3 −

1

3z1 +

1

3z2

x2 =1

3+

1

3z1 −

1

3z2





Complementary pivoting

I Newly added label will be duplicate

I e.g. Label 4 is added, and is a duplicateI Make y1 positive

I y2 = 1− y1 =⇒ y1 ∈ (0, 1]I w2 = 1− 3y1 + w1 =⇒ y1 ∈

(0, 1

3

]I So, new strategy is

(13 , 2

3

)I Now completely labelledI Hence, we have found the equilibria





Performance of Lemke-Howson

I Worst-case exponential running timeI In practise, reasonably fast

I Of late, been beaten by a few competing algorithms e.g.[SHNP]





Problems?

I Does not generalize to n > 2 players

I Sometimes, equilibria may be out of reach





Other approaches

I Many more techniques...I Local-search techniques [SHNP]I Mixed integer programmingI Global Newton optimizationI Computer algebraI Markov Random FieldsI etc...

I Quite a few generalize to more than 2 players

I Nothing (as yet) tells us about the boundary of P!





Xi Chen, Xiaotie Deng, and Shang hua Teng.Settling the complexity of computing two-player nashequilibria.Technical report, 2007.

Ryan Porter, Eugene Nudelman, and Yoav Shoham.Simple search methods for finding a nash equilibrium.In AAAI, pages 664–669, 2004.

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computing nash-equilibria in 2-player games and beyond ...akmenon/nashequilibria...finding...

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