RIKEN Center for Computational Science, Kobe, Japan.
Department of Physics, Pukyong National University, Busan, Korea.
PLoS Comput Biol. 2021 Jan 21;17(1):e1008217. doi: 10.1371/journal.pcbi.1008217. eCollection 2021 Jan.
Repeated interaction promotes cooperation among rational individuals under the shadow of future, but it is hard to maintain cooperation when a large number of error-prone individuals are involved. One way to construct a cooperative Nash equilibrium is to find a 'friendly-rivalry' strategy, which aims at full cooperation but never allows the co-players to be better off. Recently it has been shown that for the iterated Prisoner's Dilemma in the presence of error, a friendly rival can be designed with the following five rules: Cooperate if everyone did, accept punishment for your own mistake, punish defection, recover cooperation if you find a chance, and defect in all the other circumstances. In this work, we construct such a friendly-rivalry strategy for the iterated n-person public-goods game by generalizing those five rules. The resulting strategy makes a decision with referring to the previous m = 2n - 1 rounds. A friendly-rivalry strategy for n = 2 inherently has evolutionary robustness in the sense that no mutant strategy has higher fixation probability in this population than that of a neutral mutant. Our evolutionary simulation indeed shows excellent performance of the proposed strategy in a broad range of environmental conditions when n = 2 and 3.
重复互动可以促进未来背景下理性个体之间的合作，但当涉及大量易错个体时，维持合作就变得困难了。构建合作纳什均衡的一种方法是找到一种“友好竞争”策略，该策略旨在实现完全合作，但从不允许共同参与者获得更好的结果。最近已经表明，在存在错误的迭代囚徒困境中，可以通过以下五个规则设计友好竞争对手：如果每个人都合作，则合作，为自己的错误接受惩罚，惩罚背叛，发现机会时恢复合作，并且在所有其他情况下背叛。在这项工作中，我们通过推广这五个规则，为迭代 n 人公共物品博弈构建了这样一个友好竞争策略。所得到的策略通过参考前 m = 2n - 1 轮做出决策。n = 2 的友好竞争策略在进化上具有稳健性，因为在这种情况下，没有任何突变策略的固定概率比中性突变更高。当 n = 2 和 3 时，我们的进化模拟确实在广泛的环境条件下展示了所提出策略的出色性能。
