网络创造博弈中的树纳什均衡

Q3 Mathematics Internet Mathematics Pub Date : 2013-10-30 DOI:10.1080/15427951.2015.1016248
A. Mamageishvili, Matús Mihalák, D. Müller
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引用次数: 34

摘要

在有n个顶点的网络创建游戏中,每个顶点(玩家)创建一个(相邻的)边,并决定创建的边应该去哪些其他顶点。每条创建的边花费固定的α > 0。每个玩家的目标都是与其他顶点保持良好的联系,同时尽可能少花钱。正式地说,在生成的(创建的)图中,每个玩家的成本被定义为α乘以玩家创建的边数加上到所有其他顶点的距离之和。我们已经推测,当α≥n时,这个博弈的每个纳什均衡都是一棵树,并且对于每个α≥273·n都得到了证实。我们改进了这个边界,并证明对于每个α≥65·n都是如此。我们还表明,我们的方法不能用来表示期望的边界,但我们推测可以实现一个稍差的边界α≥1.3·n。针对这一猜想,我们证明了如果纳什均衡的循环长度最多为10,那么α确实< 1.3·n。我们研究了纳什均衡的联盟变体的方法,其中两个参与者的联盟不能共同改进,并证明了如果α≥41·n,那么每个这样的纳什均衡都是树。
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Tree Nash Equilibria in the Network Creation Game
In the network creation game with n vertices, every vertex (player) creates an (adjacent) edge and decides to which other vertices the created edge should go. Each created edge costs a fixed amount α > 0. Each player aims to have a good connection with the rest of the vertices and, at the same time, to pay as little as possible. Formally, the cost of each player in the resulting (created) graph is defined as α times the number of edges created by the player plus the sum of the distances to all other vertices. It has been conjectured that for α ≥ n, every Nash equilibrium of this game is a tree and has been confirmed for every α ≥ 273 · n. We improve on this bound and show that this is true for every α ≥ 65 · n. We also show that our approach cannot be used to show the desired bound, but we conjecture that a slightly worse bound α ≥ 1.3 · n can be achieved. Toward this conjecture, we show that if a Nash equilibrium has a cycle of length at most 10, then indeed α < 1.3 · n. We investigate our approach for a coalitional variant of a Nash equilibrium, which coalitions of two players cannot collectively improve, and show that if α ≥ 41 · n, then every such Nash equilibrium is a tree.
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Internet Mathematics
Internet Mathematics Mathematics-Applied Mathematics
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