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Lecture 19: Poa for Strong Nash Equilibrium CS 6840, Spring 2020 Lecture 19: PoA for Strong Nash Equilibrium CS 6840 Algorithmic Game Theory March 9, 2020 Lecture 19: PoA for Strong Nash Equilibrium Instructor: Eva´ Tardos Scribe: Abhishek Shetty 1 Price of Anarchy for Strong Nash Equilibrium Consider a cost sharing game with the set of \shareable" elements E and n players with strategies Si ⊆ E. The cost of player i for a strategies given by S = (S1;:::;Sn) is given by X ce ci (S) = χe e2Si where ce is the cost of element e and χe is the number of players j such that e 2 Sj. The social cost SC is defined as n n X X X ce SC (S) = ci = : χe i=1 i=1 e2Si Recall that this game is a potential game with potential Φ given by χe X X ce Φ(S) = : k e2E k=1 Definition 1.1 (Strong Nash Equilibrium, See Notes). Let G be a game with n players and costs ci.A strategy s = (s1; : : : ; sn) is said to be a strong Nash equilibrium if for every subset A ⊆ [n] and every 0 alternate strategy vector sA, we have 0 ci (s) ≤ ci (sA; s−A) for all i 2 A. We will show that in such games social cost of a strong Nash equilibrium, if one exists, is bounded by a Hn factor of the optimal social cost, where Hn is the nth Harmonic number. Recall that Hn ∼ ln n. Also, recall the example given by the following graph. There is a unique Nash equilibrium which is also a strong Nash equilibrium and it has a price of anarchy of Hn, which shows that the result of the following theorem is tight. Theorem 1.1. Consider a cost sharing game as defined above. Let S = (S1;:::;Sn) be a strong Nash ∗ ∗ ∗ equilibrium and let S = (S1 ;:::;Sn) be a strategy of minimum social cost. Then, ∗ SC (S) ≤ Hn · SC (S ) Pn 1 where Hn = k=1 =k is the nth Harmonic number. 1 CS 6840, Spring 2020 Lecture 19: PoA for Strong Nash Equilibrium Proof. From the definition of strong Nash equilibrium, we have that there is a player (which we label n without loss of generality) such that ∗ cn (S) ≤ cn (S ) : Now let An−1 = f1; 2; : : : n − 1g denote the remaining set. We will inductively renumber of players and define the sets A = f1; 2; : : : ig (backwards induction from n). By considering S∗ as the alternative i Ai strategy for the set of players Ai. From the definition of strong Nash equilibrium, there is a player, which we will relabel at player i, such that has at least as large a cost by moving: c (S) ≤ c S∗ ;S : i i Ai −Ai Summing this over i gives us n n X X c (S) ≤ c S∗ ;S : i i Ai −Ai i=1 i=1 We consider a possible extension of the game where certain subsets of players participate in none of the congestible elements (equivalently, they `leave' the game). This can be realized by including the empty set in the set of allowed strategies of every player. For any strategy T and subset B ⊆ [n], denote by TB the strategy that agrees with T on the players indexed by B while the rest of the players play the empty set. Lemma 1.2. For every player i, we have c S∗ ;S ≤ c S∗ : i Ai −Ai i Ai Remark. This lemma can be seen to capture the fact that the game has positive `externalities'. Proof. The proof follows by observing that for every element e, S∗ ;S has more player accessing Ai −Ai e than in S∗ and the cost of each element is monotone decreasing in the number of player accessing Ai each element. Lemma 1.3. For every player i, we have c S∗ = Φ S∗ − Φ S∗ : i Ai Ai Ai−1 Proof. Since the game is a potential game, we have c S∗ − c S∗ = Φ S∗ − Φ S∗ : i Ai i Ai−1 Ai Ai−1 The claim follows by noting that c S∗ = 0, since the strategy of player i here is the empty set. i Ai−1 From the two lemmas above, we get n n X X h i c (S) = Φ S∗ − Φ S∗ i Ai Ai−1 i=1 i=1 = Φ (S∗) ∗ ≤ Hn · SC (S ) where the equation uses the fact that Φ(;) = 0. The last inequality follows from the proof of the bound on the price of stability for cost sharing as shown in class on February 3, 2020. It is an interesting question to construct a potential game with a strong Nash equilibrium, such that the minimizer of the potential is not a strong Nash equilibrium. 2 CS 6840, Spring 2020 Lecture 19: PoA for Strong Nash Equilibrium 2 Core of a Cost Sharing Game In this section, we shall look at a different solution concept for games. In order to motivate this, recall the following game. Note that this game has no strong Nash equilibria. We shall look at a solution concept which does indeed exist for this game. Definition 2.1. Consider a game with n players. Let c : 2[n] ! R be a function such that for every A ⊆ [n], c (A) denotes the cost suffered if players in A cooperate. A vector α is said to be in the core if Pn • j=1 αi = c ([n]) P • For every subset S ⊆ [n], j2S αi ≤ c (S) Remark. The condition for a vector being a core is a set of 2n linear inequalities. Note that in the above game (10; 10) is in the core of the game, thus serving as an example of a game where the core is non-empty but there is no strong Nash equilibrium. Unfortunately, not all games (or even cost sharing games) have a non-empty core. To see this, consider the following example. A B a b c C Figure 1: The cost of each red edge is 1, the black edges are 0. The player source-sink pairs are: (a; A); (b; B); (c; C). To show that the core is empty notice that the min-cost for any two players is only 1. Adding these constraints the three of them total can only pay 1.5, but the total cost of the min-cost solution is the paths 2. 3.
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