Download Algorithmic Game Theory by Noam Nisan, Tim Roughgarden, Eva Tardos, Vijay V. Vazirani PDF

By Noam Nisan, Tim Roughgarden, Eva Tardos, Vijay V. Vazirani

Within the previous couple of years online game conception has had a considerable impression on laptop technology, in particular on net- and e-commerce-related concerns. greater than forty of the head researchers during this box have written chapters that cross from the principles to the state-of-the-art. uncomplicated chapters on algorithmic equipment for equilibria, mechanism layout and combinatorial auctions are by way of chapters on incentives and pricing, rate sharing, info markets and cryptography and defense. scholars, researchers and practitioners alike have to research extra approximately those interesting theoretical advancements and their frequent functional program.

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12 (Ultimatum game) Assume that a seller S is trying to sell a good to buyer B. Assume that the interaction has two steps: first seller S offers a price p, and then buyer B reacts to the price. We assume the seller has no value for the good, his payoff is p if the sale occurs, and 0 otherwise. The buyer has a value v for the good, so his payoff is v − p if he buys, and 0 if he does not. Here we are considering a full information game in which seller S is aware of the buyer’s value v, and hence we expect that the seller offers price p just under v, and the buyer buys.

It is not hard to construct a symmetric game whose strategies are all literals (variables and their negations) and whose Nash equilibria are all truth assignments. In other words, if we choose, for each of the n variables, either the variable itself or its negation, and play it with probability n1 , then we get a symmetric Nash equilibrium, and all Nash equilibria of the game are of this sort. It is also easy to add to this game a new pure Nash equilibrium (d, d), with lower utility, where d (for “default”) is a new strategy.

Furthermore, in the subgraph of G induced on A − S ∗ and B − (S ∗ ), all vertices have nonzero degree. proof In the max-flow computed in N for x = x ∗ , the flow going through nodes in S ∗ completely uses up the capacity of edges from (S ∗ ) to t. Therefore, all the flow going through nodes in A − S ∗ must exit via nodes in B − (S ∗ ). 15. Furthermore, a good j ∈ A − S ∗ must have nonzero degree to B − (S ∗ ). Finally, since each buyer i ∈ (B − (S ∗ )) has nonzero degree in G and has no edges to S ∗ , it must have nonzero degree to A − S∗.

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