Percolation: Slipping through the Cracks（渗透问题）

目录

一，二叉树场景

1，求存在长度为n的路径的概率的递推式

2，根据递推式，用不动点法求极限

二，阿米巴的生存

三，网格图场景

1，问题背景

2，感性认知

3，Menshikov's theorem

4，The Harris-Kesten Theorem

四，六边形网格场景

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原文链接：AMS :: Feature Column from the AMS

在一个无限大的图中，每条边都有p的概率是可通行的，有1-p的概率是不可通行的，问从起点出发存在无限长路径的概率。

一，二叉树场景

Let's begin with a simple example that illustrates some general principles. We will study percolation on the infinite binary tree, a portion of which is shown below.

  

在一个无限大的二叉树中，每条边都有p的概率是可通行的，有1-p的概率是不可通行的，问从根节点出发存在无限长路径的概率。

1，求存在长度为n的路径的概率的递推式

If p is the probability that each edge is open, we want to find θ(p), the probability that there is an infinite open path containing the root of the tree v0. We begin by letting Pn be the probability that there is an open path from v0 to a vertex n levels below. Notice that we have P0 = 1. Shown in red below is a path from v0 to a vertex three levels below.

This path moves through another vertex v1. The probability that an open path from v0 to a vertex n levels below passes through v1 is pPn-1, the probability p that the edge from v0 to v1 is open times the probability Pn-1 that a path from v1 to a vertex n - 1 levels below it is open. Therefore, the probability that there is no open path from v0 to a vertex n levels below passing through v1 is 1 - pPn-1.

Since any path from v0 to a vertex n levels below must pass through one of the children of v0, we find that the probability that there is not an open path from v0 to a vertex n levels below is  

至此，我们就得到递推式 

2，根据递推式，用不动点法求极限

If we define the function fp(x) = 1 - (1 - px)2, we have Pn = fp(Pn-1). The graph of fp is shown below in two different cases, depending on the derivative fp'(0) = 2p.

In the first case, 2p > 1 , we see that there is a fixed point

 

 

Here we see that the critical probability, pc = 1/2; above pc, θ(p) > 0 and we have percolation. Below the critical probability, θ(p) = 0 a