Constant Price of Anarchy in Network Creation Games via Public Service Advertising Presented by Sepehr Assadi Based on a paper by Erik D. Demaine and Morteza.

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Constant Price of Anarchy in Network Creation Games via Public Service Advertising

Presented bySepehr Assadi

Based on a paper byErik D. Demaine and Morteza Zadimoghaddam

[Demaine et. al’]

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Intro. To Game Theoretic Concepts

• Game: – A Set of selfish agents– A strategy profile for each agent• Showing possible actions

– A cost function for each agent• Showing cost of each agent in the game

• Dynamic:– Each agents try to minimize her cost

selfishly!

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Game Theoretic Concepts (cont.)

• Stable State:– A state which no one wants to change!– Has different mathematical notions

• Nash equilibrium ( pure / approximate / strong …)

• Pair wise stability• …

• Social cost:– A cost defined for the game in a state

• Usually the sum of all the agents cost.

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Nash equilibrium

• Definition:– State of the game where no agent can

change its strategy and gain lower cost• Somehow everyone are satisfied of their

current state!

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Price of Stability & Anarchy

• Optimum social cost:– The minimum social cost in a state

designed by an authority• Maybe not a stable state!

• Price of Stability:– The ratio of minimum social cost in a

stable state over optimum social cost

• Price of Anarchy:– The ratio of maximum social cost in a

stable state over optimum social cost

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Network Creation Games

• Network Creation Games:– Agents:

• nodes of a graph

– Strategy profile: • creating edges to other nodes and accepting/rejecting the edge connected to them.

– Cost function: • cost of establishing network + cost of using

network

– Social Cost:• Usually sum/max of the cost of the agents

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(n,k)-uniform bounded budget connection game

• Definition:– n agent creating the nodes of graph– Each agent strategy is subset of at most k other

nodes• Each node create edges to other and accepts/rejects the edge connected to it

– Agent cost:• Sum of its distances to all other nodes in the underlying

graph

– Social cost:• Sum of the cost of all agents in a state

– Stable State• Nash equilibrium

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Known Results

• Price of Stability:

• Price of Anarchy:

• What is the meaning?– Good Nash equilibrium vs Bad Nash

equilibrium!

• Natural question:– How can we shift the decision of the agents

towards the better Nash equilibrium?

Adver

tise!

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Public Service Advertising (PSA)

• Introduced in SODA 2009 [Balcan et.al’]• Previously studied on:– Fair cost sharing, selfish routing, scheduling

games

• Idea:– Use an advertising campaign – Introduce better strategies to agents

• Hopefully they will accept!

– Agents may share only a fraction of their budgets in the campaign • Even a small fraction can goes a long way!

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Public Service Adverting (cont.)

• Formal Definition:– Each agent is advertised a strategy

• Strategies might differs with each other!

– Each agent accept the advertised strategy with probability of α independently.• These nodes are called Receptive Nodes.

– Each receptive node is willing to share β fraction of its budget. • Therefore in this game there are βk edge for each

receptive node.

– The final state again should be Stable state.• In this game Nash equilibrium.

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Motivation

• Social Networks!• Political Campaigns!• Peer-to-peer Networks!

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Result of this work

• Result:–With proper method of advertising the

price of anarchy would be O(1/α) – Even without knowing α and β

beforehand• Not mentioned in this lecture

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Overview of method

1. Partition nodes to different set – Each set is advertised uniquely

2. Create a core graph using receptive nodes– Examine the cost in this graph

3. Wait for the others to reach a stable state– Examine the cost of this newly created

graph

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Method of Advertising

• Let:– K” = αβ /c log(n) • for sufficiently large c, i.e. c ≥ 5.

• Partitioning:– Partition nodes to l ≤ logk (n) sets S1, S2,

…,Sl

• Such that: – |S1| = βk / 2

– |Si+1| / |Si| = k”

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Method of Advertising (Cont.)

• Advertised strategy:– S1:

• Connect βk / 2 -1 edges to other nodes in S1.

– Si for i > 1:• Choose randomly c log(n)/2α from Si-1

– c log(n)/2α ≤ βk / 2

– Si i ≥ 1:• Accepts βk / 2 edges from Si+1

• Notes:– If a node is non-receptive node we assume that it deletes

the edge!– Even if a node is receptive it might be overwhelmed!

• Gets more than βk edge

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Lemma: Hierarchical Tree

• Lemma: – Above strategy creates a hierarchical

tree shaped subgraph– Diameter of the subgraph is 2logk” (n)

– Covers all the receptive nodes with high probability• High probability: 1 – 1/nc for some c ≥ 1

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Lemma: Hierarchical Tree (Example)

• Sketch: S1

S3

S2

Sl

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Lemma: Hierarchical Tree (Example)

• Sketch:

• Diameter!– Shown in green!

S1

S3

S2

Sl

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Lemma: Hierarchical Tree (Proof)

• Proof:– Every receptive node is connected to a

receptive node in higher level– For each node in Si : • expected number of receptive nodes it

connects is c log(n)/2α * α = c log(n) / 2• Chernouff bound: At least log (n) of them are

receptive!

–What if they delete this edge? • Happens if it is overwhelmed!

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Lemma: Hierarchical Tree (Proof)

• Proof:– Probability of a receptive node being

overwhelmed:– Expected number of edges from higher

level:• α|Si| (c log(n)/2α ) / |Si-1|

• |Si| / |Si-1| = k” = αβk / (c log(n)/2α )

• = αβk / 2.

–Markov inequality: • P(Overwhelmed) = P(edge > βk ) < α/2 < 1/2.

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Lemma: Hierarchical Tree (Proof)

• Proof:– Probability of a receptive node being

overwhelmed:– Expected number of edges from higher

level:• α|Si| (c log(n)/2α ) / |Si-1|

• |Si| / |Si-1| = k” = αβk / (c log(n)/2α )

• = αβk / 2.

–Markov inequality: • P(Overwhelmed) = P(edge > βk ) < α/2 < 1/2.

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Lemma: Hierarchical Tree (Proof)

• Proof:– Probability of a receptive node being

overwhelmed:– Expected number of edges from higher

level:• α|Si| (c log(n)/2α ) / |Si-1|

• |Si| / |Si-1| = k” = αβk / (c log(n)/2α )

• = αβk / 2.

–Markov inequality: • P(Overwhelmed) = P(edge > βk ) < α/2 < 1/2.

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Lemma: Hierarchical Tree (Proof)

• Proof:– Probability of a receptive node being

overwhelmed:– Expected number of edges from higher

level:• α|Si| (c log(n)/2α ) / |Si-1|

• |Si| / |Si-1| = k” = αβk / (c log(n)/2α )

• = αβk / 2.

–Markov inequality: • P(Overwhelmed) = P(edge > βk ) < α/2 < 1/2.

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Lemma: Hierarchical Tree (Proof)

• Proof:– Probability of a receptive node being

overwhelmed:– Expected number of edges from higher

level:• α|Si| (c log(n)/2α ) / |Si-1|

• |Si| / |Si-1| = k” = αβk / (c log(n)/2α )

• = αβk / 2.

–Markov inequality: • P(Overwhelmed) = P(edge > βk ) < α/2 < 1/2.

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Lemma: Hierarchical Tree (Proof)

• Proof:– Probability of a node not connected to any

receptive and not overwhelmed node• X: number of receptive and not overwhelmed

nodes• xi: i-th receptive node it connects is not

overwhelmed

• X = x1 + x2 + x3 + … + x log(n) & E[X] >= log(n)/2

• P(X <= 1) ? • Chernouff bound ?

– They are not independent!

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Lemma: Hierarchical Tree (Proof)

• Proof:– What should we do?

• xi are negatively correlated

• Negative correlation: – E[XY] – E[X]E[Y] < 0– If one happens, the probability of the other became lower

• Generalized Chernouff bound: holds for negative correlated xi (and some more families too)

– Thus, every receptive node is connected to at least one another receptive node in lower level• And it is not overwhelmed!

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Lemma: Hierarchical Tree (Proof)

• Figure:

A node in Si

Set Si-1

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Lemma: Hierarchical Tree (Proof)

• Figure:• Requesting initial edges!

A node in Si

Set Si-1

(c log(n) / 2α ) random node!

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Lemma: Hierarchical Tree (Proof)

• Figure:• Expected of receptive nodes

A node in Si

Set Si-1

(c log(n) / 2α ) * α = c log(n) / 2

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Lemma: Hierarchical Tree (Proof)

• Figure:• High probable number of receptive nodes

A node in Si

Set Si-1

Chernouff Bound:log(n)

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Lemma: Hierarchical Tree (Proof)

• Figure:• Some are overwhelmed

A node in Si

Set Si-1

Probability of overwhelming 1/2

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Lemma: Hierarchical Tree (Proof)

• Figure:• One remains at least!

A node in Si

Set Si-1

Negative CorrelationChernouff BoundOne node!

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Lemma: Hierarchical Tree (Proof)

• Proof:–We have:• Each receptive node is connected to at least

one on the top level• All the nodes in first level are connected to

each other• Therefore, all receptive nodes are in the

subgraph with diameter 2*l (l is number of Si)

• End of proof ☐

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Handling Non-receptive nodes

• Now, what about others?• Lemma:– In any stable graph maximum distance of

other nodes from any receptive node v, is O(l + logk(n)/α)

– Diameter of stable graph now is at most O(logk”(n)/α)

• Proof: – Proof is completely combinatorial. (not of our

interest here)

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Final Result

• Theorem:– Price of anarchy is at most

O(logk”(n)/α logk(n)) = O(logk”(n)/α)

• Proof:

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Final Result

• Theorem:– Price of anarchy is at most

O(logk”(n)/α logk(n)) = O(logk”(n)/α)

• Proof:– Previous lemma,

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Final Result

• Theorem:– Price of anarchy is at most

O(logk”(n)/α logk(n)) = O(logk”(n)/α)

• Proof:– Previous lemma,– Average distant in the optimal graph is

Ω(logk(n)) [Laotaris et.al’]

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Final Result

• Theorem:– Price of anarchy is at most

O(logk”(n)/α logk(n)) = O(logk”(n)/α)

• Proof:– Previous lemma,– Average distant in the optimal graph is

Ω(logk(n)) [Laotaris et.al’]

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Final Result (Cont.)

• Corollary:– If k is Ω(log1+ε(n)), the price of anarchy is

O(1/αε)

• Finally:– Public service advertising leads to

O(1/α) Price of Anarchy!

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Any Question?

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References

• [Demaine et. al’]:– “Constant Price of Anarchy in Network Creation Games via Public Service

Advertising”, Erik D. Demaine and Morteza Zadimoghaddam . Algorithms

and Models for the Web-Graph.

• [Balcan et.al’]:– “Improved equilibria via public service advertising”, Balcan, Blum,

Mansour. In Proceedings of the 20th Annual ACM-SIAM Symposium on Discrete Algorithms.

• [Laotaris et.al’]:– “Bounded budget connection (BBC) games or how to make friends and

influence people, on a budget”, Laoutaris, Poplawski, Rajamaran, Sundaram, Teng. In Proceedings of the 27th ACM Symposium on Principles of Distributed Computing.

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Thank you!