arXivDaily arXiv每日学术速递 周一至周五更新
arXiv 1208.0722cs.DMcs.GTmath.CO
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图上的顶点Nim

Vertex Nim played on graphs

Eric Duchêne, Gabriel Renault

摘要:

给定一个顶点上具有正整数权重的图G,以及放置在某个当前顶点u上的令牌,两个玩家交替地从u移除正整数权重,然后将令牌移动到与u相邻的新的当前顶点。当一个顶点的权值设置为0时,它将被移除,其邻域将成为一个团。走到最后一步的玩家获胜。Nim在图上的这种改编被称为Vertexnim,与Stockman在2004年推出的游戏Vertex NimG略有不同。顶点可以在有向图或无向图上播放。在本文中,我们研究的复杂性,决定是否一个给定的游戏位置的顶点是第一个或第二个玩家获胜。特别是,我们表明,对于无向图,这个问题可以在二次时间内解决。我们的算法也适用于游戏顶点NimG,从而改进了Stockman的exptime算法。在有向情况下,我们能够在多项式时间内计算出几个实例的获胜策略,包括具有自循环的回路或有向图。

英文摘要:

Given a graph G with positive integer weights on the vertices, and a token placed on some current vertex u, two players alternately remove a positive integer weight from u and then move the token to a new current vertex adjacent to u. When the weight of a vertex is set to 0, it is removed and its neighborhood becomes a clique. The player making the last move wins. This adaptation of Nim on graphs is called Vertexnim, and slightly differs from the game Vertex NimG introduced by Stockman in 2004. Vertexnim can be played on both directed or undirected graphs. In this paper, we study the complexity of deciding whether a given game position of Vertexnim is winning for the first or second player. In particular, we show that for undirected graphs, this problem can be solved in quadratic time. Our algorithm is also available for the game Vertex NimG, thus improving Stockman's exptime algorithm. In the directed case, we are able to compute the winning strategy in polynomial time for several instances, including circuits or digraphs with self loops.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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