对角线的无限图是否具有无限分量？

14

假设我们使用一组无向边连接的点，使得连接到或连接到，对所有随机独立和均匀地分布。 $V = \mathbb{Z}^2$ $E$ $(i, j)$ $(i + 1, j + 1)$ $(i + 1, j)$ $(i, j + 1)$ $i, j$

（灵感来自本书的标题和封面。）

该图具有无限大的连通分量的概率是多少？类似地，考虑，即图的平面嵌入的补码。补码具有无限个连通分量的概率是多少？ $\mathbb{R}^2 \setminus G$

显然，如果所有对角线都指向相同的方向，则图及其补码都将具有无限分量。上面的均匀随机图怎么样？

graph-theory

— Mathias Rav
source

2

AFAICS，任何平面图的对偶图都是连通的，那么您的第二个问题真的是对偶图是否无穷大？还是您在谈论对偶图的不同概念？

— 埃米尔·杰拉贝克（EmilJeřábek）

2

作为用于有限性：在周期是从图中激发这个问题特别是不存在，预期数是无限的（对于每个

，在正方形的边缘

形成具有概率的周期

i, j

$i,j$

(2 i, 2 j), (2 i, 2 j + 1), (2 i + 1, 2 j), (2 i + 1, 2 j + 1)

$(2i,2j),(2i,2j+1),(2i+1,2j),(2i+1,2j+1)$

，独立地）。

1 / 16

$1/16$

— 克劳斯·德拉格

@EmilJeřábek对不起，我不是古典意义上的对偶-我已进行编辑以阐明我的意思是平面嵌入的补充。

— Mathias Rav 2015年

9

概率为0。

这遵循以下定理（请参见Grimmett的“图论概率”中的定理5.33，http：//www.statslab.cam.ac.uk/~grg/books/USpgs-rev3.pdf）：

Theorem Consider bond percolation on $\mathbb Z^2$ , where each edge between lattice points is open with probability $\frac12$ . The probability that the origin is in an infinite connected component is 0.

我们可以从问题简化为这个问题：基本上，它只是上键渗滤的2个不相交（但从属）的副本。考虑配置，其中边缘形成包含原点的钻石的无限晶格。如果翻转所有边缘，则会得到另一个无限的菱形晶格。考虑实际配置与和的交集。每个这些正是上键渗流模型，只是旋转。因此，任意点在无穷簇中的概率为0（没有边） $\mathbb Z^2$ $D_1$ $D_2$ $D_1$ $D_2$ $\mathbb Z^2$ $45^{\circ}$ $D_1$ 可以连接到的边缘。） $D_2$

总而言之，请注意，概率为0的可数事件的总和的概率为0；否则，概率为0。任何晶格点在无限簇中的概率之和。

（任意大的组件的存在都是一个红色的鲱鱼。应该固定一个点，并询问它是否在无界的组件中。）

— 李荷顿
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If we fix the origin and ask if it is in an unbounded component, then we can disregard all edges in

D_{2}

$D_2$ and we remain with a single instance of bond percolation on

Z^{2}

$\mathbb{Z}^2$ with the edges in

D_{1}

$D_1$ . A useful illustration is Bollobás and Riordan 2008, Figure 2, rotated 45 degrees.

— Mathias Rav

2

Hmm, well, here's one first attempt. Let's observe two important things:

If this graph has an infinitely large connected component, by König's infinity lemma, it has an infinite simple path.
The event that the graph has an infinite simple path is independent of each individual choice of edge orientation (and thus, every finite set of edge choices). Therefore it is a tail event and by Kolmogorov's zero-one law the probability is either zero or one.

So, is it zero or one? It's not immediately clear, though we can make a guess, since by the "infinite monkeys with typewriters" theorem, this graph contain simple paths of arbitrarily large length with probability one. Of course, more is needed to rigorously prove that it actually has an infinite path with probability one.

— Joe Bebel
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3

It’s also a good idea to observe 0. The event that the graph has an infinite connected component is Borel, hence measurable, hence the question makes sense in the first place. (This is not obvious when restated with infinite simple paths.)

— Emil Jeřábek supports Monica

1

Here's some weak empirical evidence that the answer is yes. Let $G_n$ be a random graph on the $2n+1 \times 2n+1$ lattice defined by picking each diagonal randomly. Here's a plot of reachability probability estimates vs. $n$ (ignoring the vertices which are always unreachable due to parity).

If we rescale the square to $[0,1]^2$ , the probability appears to converge to a smooth function independent of scale, which would mean even more: that the probability of the origin reaching infinity is positive.

However, it's also possible that I haven't computed far enough out to see the downwards trend (the $n = 800$ plot does seem a little smaller than the others).

Code here: https://gist.github.com/girving/16a0ffa1f0abb08934c2

— Geoffrey Irving
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1

Update: As pointed out in the comments, the lemma does not imply infinite paths after all, so this answer overall is wrong. Not sure if it can be used in another probabilistic way.

The answer is yes: an infinite path exists. Indeed, an infinite path exists for every such graph; probability is not required.

Lemma: Let $G$ be any diagonal graph on the $n \times n$ lattice, with $n \ge 2$ . Then there is either a path from left to the right through even parity vertices or a path from the top to the bottom through odd parity vertices.

Proof sketch: This is essentially the determinacy theorem in Hex on a different graph. The even and odd parity halves of $G$ are locally dual, so an obstruction in one parity is a connection in the other. However, I will omit the details since they're hard to write down without pictures and/or case analysis.

If the lemma is true, the infinite version follows from König as noted by Joe. (Update: Wrong, see comments)

— Geoffrey Irving
source

2

I think the following graph consisting of nested diamonds contradicts your claim that every such graph has an infinite path: Connecting

(- n, 0)

$(-n, 0)$ to

(0, n)

$(0, n)$ ,

(0, n)

$(0, n)$ to

(n, 0)

$(n, 0)$ ,

(n, 0)

$(n, 0)$ to

(0, - n)

$(0, -n)$ and

(0, - n)

$(0, -n)$ to

(- n, 0)

$(-n, 0)$ by straight lines for all

n > 0

$n > 0$ . This graph has components of arbitrarily large length, but it has no infinite components, since any point is in a finite component of size proportional to its L1 norm. How does your proof sketch handle this graph?

— Mathias Rav

Very true, Koenig does not apply after all.

— Geoffrey Irving

2

Specifically, I believe the lemma still holds, but of course does not imply the desired result.

— Geoffrey Irving