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 by Sepehr Assadi Based on a paper by Erik D. Demaine and Morteza Zadimoghaddam [Demaine et. al’] Sharif University Of Technology 1

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! Sharif University Of Technology 2

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. Sharif University Of Technology 3

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! Sharif University Of Technology 4

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 Sharif University Of Technology 5

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 Sharif University Of Technology 6

(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 Sharif University Of Technology 7

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? Sharif University Of Technology 8

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! Sharif University Of Technology 9

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. Sharif University Of Technology 10

Motivation Social Networks! Political Campaigns! Peer-to-peer Networks! Sharif University Of Technology 11

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 Sharif University Of Technology 12

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 Sharif University Of Technology 13

Method of Advertising Let: – K” = αβ /c log(n) for sufficiently large c, i.e. c ≥ 5. Partitioning: – Partition nodes to l ≤ log k (n) sets S 1, S 2,…,S l Such that: – |S 1 | = βk / 2 – |S i+1 | / |S i | = k ” Sharif University Of Technology 14

Method of Advertising (Cont.) Advertised strategy: – S1: Connect βk / 2 -1 edges to other nodes in S1. – S i for i > 1: Choose randomly c log(n)/2α from S i-1 – c log(n)/2α ≤ βk / 2 – S i i ≥ 1: Accepts βk / 2 edges from S i+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 Sharif University Of Technology 15

Lemma: Hierarchical Tree Lemma: – Above strategy creates a hierarchical tree shaped subgraph – Diameter of the subgraph is 2log k” (n) – Covers all the receptive nodes with high probability High probability: 1 – 1/n c for some c ≥ 1 Sharif University Of Technology 16

Lemma: Hierarchical Tree (Example) Sketch: S1S1 S3S3 S2S2 SlSl Sharif University Of Technology 17

Lemma: Hierarchical Tree (Example) Sketch: Diameter! – Shown in green! S1S1 S3S3 S2S2 SlSl Sharif University Of Technology 18

Lemma: Hierarchical Tree (Proof) Proof: – Every receptive node is connected to a receptive node in higher level – For each node in S i : 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! Sharif University Of Technology 19

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 ) / |S i-1 | |S i | / |S i-1 | = k” = αβk / (c log(n)/2α ) = αβk / 2. – Markov inequality: P(Overwhelmed) = P(edge > βk ) < α/2 < 1/2. Sharif University Of Technology 20

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 ) / |S i-1 | |S i | / |S i-1 | = k” = αβk / (c log(n)/2α ) = αβk / 2. – Markov inequality: P(Overwhelmed) = P(edge > βk ) < α/2 < 1/2. Sharif University Of Technology 21

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 ) / |S i-1 | |S i | / |S i-1 | = k” = αβk / (c log(n)/2α ) = αβk / 2. – Markov inequality: P(Overwhelmed) = P(edge > βk ) < α/2 < 1/2. Sharif University Of Technology 22

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 ) / |S i-1 | |S i | / |S i-1 | = k” = αβk / (c log(n)/2α ) = αβk / 2. – Markov inequality: P(Overwhelmed) = P(edge > βk ) < α/2 < 1/2. Sharif University Of Technology 23

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 ) / |S i-1 | |S i | / |S i-1 | = k” = αβk / (c log(n)/2α ) = αβk / 2. – Markov inequality: P(Overwhelmed) = P(edge > βk ) < α/2 < 1/2. Sharif University Of Technology 24

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 x i: i-th receptive node it connects is not overwhelmed X = x 1 + x 2 + x 3 + … + x log(n) & E[X] >= log(n)/2 P(X <= 1) ? Chernouff bound ? – They are not independent! Sharif University Of Technology 25

Lemma: Hierarchical Tree (Proof) Proof: – What should we do? x i 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 x i (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! Sharif University Of Technology 26

Lemma: Hierarchical Tree (Proof) Figure: Sharif University Of Technology 27 A node in S i Set S i-1

Lemma: Hierarchical Tree (Proof) Figure: Requesting initial edges! Sharif University Of Technology 28 A node in S i Set S i-1 (c log(n) / 2α ) random node!

Lemma: Hierarchical Tree (Proof) Figure: Expected of receptive nodes Sharif University Of Technology 29 A node in S i Set S i-1 (c log(n) / 2α ) * α = c log(n) / 2

Lemma: Hierarchical Tree (Proof) Figure: High probable number of receptive nodes Sharif University Of Technology 30 A node in S i Set S i-1 Chernouff Bound: log(n)

Lemma: Hierarchical Tree (Proof) Figure: Some are overwhelmed Sharif University Of Technology 31 A node in S i Set S i-1 Probability of overwhelming 1/2

Lemma: Hierarchical Tree (Proof) Figure: One remains at least! Sharif University Of Technology 32 A node in S i Set S i-1 Negative Correlation Chernouff Bound One node!

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 S i ) End of proof ☐ Sharif University Of Technology 33

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 + log k (n)/α) – Diameter of stable graph now is at most O(log k” (n)/α) Proof: – Proof is completely combinatorial. (not of our interest here) Sharif University Of Technology 34

Final Result Theorem: – Price of anarchy is at most O(log k” (n)/α log k (n)) = O(log k” (n)/α) Proof: Sharif University Of Technology 35

Final Result Theorem: – Price of anarchy is at most O(log k” (n)/α log k (n)) = O(log k” (n)/α) Proof: – Previous lemma, Sharif University Of Technology 36

Final Result Theorem: – Price of anarchy is at most O(log k” (n)/α log k (n)) = O(log k” (n)/α) Proof: – Previous lemma, – Average distant in the optimal graph is Ω(log k (n)) [Laotaris et.al’] Sharif University Of Technology 37

Final Result Theorem: – Price of anarchy is at most O(log k” (n)/α log k (n)) = O(log k” (n)/α) Proof: – Previous lemma, – Average distant in the optimal graph is Ω(log k (n)) [Laotaris et.al’] Sharif University Of Technology 38

Final Result (Cont.) Corollary: – If k is Ω(log 1+ε (n)), the price of anarchy is O(1/αε) Finally: – Public service advertising leads to O(1/α) Price of Anarchy! Sharif University Of Technology 39

Any Question? Sharif University Of Technology 40

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 20 th 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 27 th ACM Symposium on Principles of Distributed Computing. Sharif University Of Technology 41

Thank you! Sharif University Of Technology 42