格伦迪的比赛扩展到两个以上的堆 [英] Grundy's game extended to more than two heaps

问题描述

如何在打破堆分为两个堆在格兰迪的游戏？

How can In break a heap into two heaps in the Grundy's game?

什么破堆到任意数量的堆（没有他们俩是平等的）？

What about breaking a heap into any number of heaps (no two of them being equal)?

推荐答案

这种类型的游戏很详细的书系列分析的致胜之道为你的数学播放。你们中的大多数都在寻找的东西可能是在第1卷。

Games of this type are analyzed in great detail in the book series "Winning Ways for your Mathematical Plays". Most of the things you are looking for are probably in volume 1.

您也可以看看这些链接： Nimbers（维基百科），的的Sprague-格伦迪定理（维基百科）或做搜索组合博弈论。

You can also take a look at these links: Nimbers (Wikipedia), Sprague-Grundy theorem (Wikipedia) or do a search for "combinatorial game theory".

我在这方面的知识是相当生疏，所以我怕我不能帮你自己与此特定问题。我的借口，如果你已经知道的一切，我挂了。

My knowledge on this is quite rusty, so I'm afraid I can't help you myself with this specific problem. My excuses if you were already aware of everything I linked.

编辑：一般情况下，解决这些类型的游戏的方法是建立堆栈大小。所以先从一摞1和决定谁具有最佳的游戏获胜。然后执行相同的一摞2，它可以分割成1和; 1.〜3中的移动，这可以被分成1和; 2.同为4（这里总是存在意外）：3及1或2及2，使用Spague-格伦迪定理＆安培;对于nimbers代数的规则，就可以计算出谁赢。一直走，直到你达到了你需要知道的答案堆栈大小。

In general, the method of solving these types of games is to "build up" stack sizes. So start with a stack of 1 and decide who wins with optimal play. Then do the same for a stack of 2, which can be split into 1 & 1. The move on to 3, which can be split into 1 & 2. Same for 4 (here it gets trickier): 3 & 1 or 2 & 2, using the Spague-Grundy theorem & the algebraic rules for nimbers, you can calculate who will win. Keep going until you reach the stack size for which you need to know the answer.

编辑2：的网站，我在评论中谈到，似乎将下降。这是它的一个备份的链接：的 Wayback机器 - 介绍组合游戏

Edit 2: The website I was talking about in the comments seems to be down. Here is a link of a backup of it: Wayback Machine - Introduction to Combinatorial Games.

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