19

This is not an answer but is too long for a comment. The point is that the distance between any two zombies is non-increasing with time no matter what your strategy.
Change the coordinate system so that you're at the origin at all times and assume that zombies move at speed $1$ (the stupid, non-colluding kind of zombie). If your speed is zero then the ...

15

(1) Look first at the references in Schramm and Smironv and Lowler. They refer to some important physics papers. Also look at the survey of Langlands, Pouliot, Saint-Aubin, Conformal invariance in two-dimensional percolation. Bull. Amer. Math. Soc. (N.S.) 30 (1994), no. 1, 1–61. This is almost a "physics paper", but Langlands is a mathematician, and it has ...

answered Aug 21 '15 at 23:29

Alexandre Eremenko

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Site percolation on the triangular lattice (or, equivalently, face percolation on the hexagonal lattice) is the only case for which conformal invariance of percolation at criticality has been proven. This has to do with very special combinatorial properties of the lattice. For a detailed explanation of this, see this paper by Vincent Beffara.

12

Area grows faster than length, so the zombies eat you, as Joseph Van Name said.
It is sufficient for the zombies to form an uncrossable circular barrier enclosing you and then to shrink the circle till the catch you.
To form an uncrossable circular barrier, there need to be $2\pi r/d$ zombies. For the zombies to reach their position on the barrier before ...

11

The square lattice is self-dual, and it is of interest to study also lattices which are not self-dual. The critical probabilities for bond percolation of dual lattices satisfy $p_c=1-p'_c$, so the square lattice must have $p_c=1/2$, and one needs to go to a non-self-dual lattice for a nontrivial value. The triangular lattice is the most widely studied ...

answered Aug 14 '17 at 10:19

Carlo Beenakker

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7

This paper of Grimmett and Manolescu prove RSW for bond percolation on isoradial graphs with critical weights (see also this one). The critical weights are those for which the model satisfies the star-triangle relation, and they use this relation, essentially, to transform the graph into a regular grid.
Tassion recently proved the RSW bound for a large ...

6

It seems the most reasonable way to formalize the problem is saying that at start the zombies are distributed according to a Poisson process in the plane with density $\mu$. As this is (distributionally) translation invariant we can assume that you start at the origin.
Now we observe that changing the zombie configuration in any finite box will not affect ...

6

In the case of increasing events, the standard proof of the BK inequality works like this. Start with your set $\Omega$, and successively replace each $\Omega_i$ by a disjoint union of two copies $\Omega_i^1$ and $\Omega_i^2$. After step $i$, the event ${(A\circ{B})}(i)$ is defined by saying that $A$ and $B$ need to occur disjointly on $\Omega_{i+1}$ up to $\...

6

This is not true without additional assumptions on your subset. E.g. consider $$V = \{(x,y) \in \mathbb Z^2 : x>0, |y| < \sqrt x \}.$$ It clearly has zero asymptotic density, but on the other hand its critical point is equal to $1/2$ like that of $\mathbb Z^2$ itself. (You can prove that it percolates for every $p>1/2$ using exponential decay of ...

6

This is true for every positive $n$. I assume that for $n$ odd or fractional, the bounding curve is $y=|x|^n$. If $(0,0)$ is not connected to infinity, then there must be a blocking contour in the dual lattice. The exponential decay iof connectivity in subcritical percolation and the Borel Cantelli Lemma preclude that. See e.g. Grimmett’s book on ...

5

Define a random variable $Y\in \{0,1\}^N$ by $P(y_i^{(n)}=1) = \epsilon$ for all $1\leq i\leq N$ and $n\in\mathbb{N}$ and observe that $x_i^{(n)} \geq y_i^{(n)}$ for all $i,n$ (as long as $X,Y$ are being driven by the same random process). With probability 1 there exists a time $n$ such that $y_i^{(n)}=1$ for all $i$ (since all the events are independent ...

answered Sep 26 '12 at 13:39

Vaughn Climenhaga

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5

The papers "Euler integrals for commuting SLEs" by Julien Dubédat and "Logarithmic operator intervals in the boundary theory of critical percolation" by Jacob JH Simmons contain formulas and equations satisfied by probabilities for different crossing events in hexagons for continuum percolation at criticality. It was not immediately clear to me if due to ...

5

I don't know if there are any expressions that take such a simple form as the C-S equation of state. Note that spherocylinders have an additional geometrical parameter $L/D$ relating the length $L$ to the diameter $D$ of the "caps" and there's also additional complexity in that they can undergo multiple phase transitions, e.g. from an isotropic liquid to a ...

3

You still need more information on the structure of the tree. The 1-3-tree (Example 1.2 in Lyons & Peres: Probability on trees and networks) shows that the probability can go to 0, even if the maximum degree is bounded.
Here is a quick proof sketch: It is not hard to see that the subtree hanging off of any other than the rightmost vertex in every layer ...

3

If I understood correctly your question, you are looking to study the probability that a point $x$ is connected to the origin (for example) at time $t$. The probability of being connected depends on the distribution of your walkers.
Since the process of the walkers is ergodic, after a long time, your walkers will be distributed according to the reversible ...

3

Even if all the zombies do is walk toward you, they will win if they are uniformly distributed and know where you are. Common sense says that you do not walk toward a potential infinity of zombies, each of whom can sense you. If they have nonzero speed, they will converge toward you.
Lets assume zombies know less about pursuit curves than I do, but that ...

answered Jul 29 '14 at 20:31

The Masked Avenger

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I think your formula for $P(N_n=k)$ is false because as you said, paths are dependent on one-another. With that said:
Expectation is linear, regardless of dependence. Let $S_n$ be the index of set of SAWs of length $n$ and $\gamma_{ni}$ denote a self avoiding path of length $n$, $i\in S_n$. You write
$$N_n=\sum_{i\in S_n} 1_{\gamma_{ni}}$$
Now you note ...

3

The finite size scaling analysis is described in Appendix B of Thermal metal-insulator transition in a helical topological superconductor. This is for a different type of phase transition (metal-insulator instead of percolation), but the method of analysis is analogous.
In summary, you have a quantity $g$ that depends on system size $L$ through a powerlaw ...

answered Jun 13 '19 at 15:59

Carlo Beenakker

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2

Check this out, I hope it helps.
The van den Berg--Kesten--Reimer operator and inequality for infinite spaces, by Arratia-Garibaldi-Hales
https://arxiv.org/abs/1508.05337

2

I think the detail you missed was that they assumed that $0 \in S$
If $\{x,y\}$ is a pivotal edge, we need that at least one of then connects to $0$,
and one of then needs to connect to $\Lambda^c$. Otherwise, opening this edge wouldn't make $0 \longleftrightarrow \Lambda^c$. Suppose then that $0\longleftrightarrow x$ and $ y \longleftrightarrow \Lambda^c$. ...

2

Simple pursuit
Zombies moving towards you will always catch you, but due to their lack of intelligence, your survival time increases exponentially with your relative speed. In $k=O(1)$ dimensional space ($k=2$ in the problem), the expected survival time is $d⋅(1/Θ(μd^k))^{(1+1/v)(1±o(1))/(k-1)}$ if $v$ is bounded below 1 and $μd^k→0$. I conjecture that ...

2

Smirnov's theorem assert the convergence to SLE in the scaling limit: one discretizes the domain with the triangular lattice of mesh size $\delta$, and lets $\delta$ go to zero. It is only proven for the triangular lattice; it's a major open problem to prove universality of this result (in fact, even to extend it to the square lattice).
Now, in the ...

2

This is just one reference; by no means a comprehensive answer to your broad question.
This paper,
Jacob J.H. Simmons. "Logarithmic operator intervals in the boundary theory of critical percolation." Journal of Physics A: Mathematical and Theoretical 46, no. 49 (2013): 494015. (Journal link.)
(Earlier arXiv abstract.)
uses conformal field
theory (CFT) "...

answered Aug 23 '15 at 0:25

Joseph O'Rourke

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The problem considered in the Island-Mainland paper is site-percolation on a two-dimensional square lattice, with one modification of the conventional problem: two squares of the same colour are considered connected if they are nearest-neighbor (they share an edge) or next-nearest-neighbor (they share a vertex). This modification has no effect on the ...

answered Jan 15 '18 at 19:41

Carlo Beenakker

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Perhaps this is a place to start, from which to search forward in time:
Grimmett, Geoffrey. Percolation. Springer, Berlin, Heidelberg, 1999.
(Springer link.)
(Image from Massimo Franceshetti.)

answered May 1 '18 at 22:40

Joseph O'Rourke

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2

The probability that it is dead, i.e. is surrounded by white stones, is the sum, over all finite connected sets $C$ containing $(0,0)$, of $3^{-(|C|+|C'|)}$. I don't expect this to have a closed form. Simulation seems to indicate it is about $0.0152$. The contribution of $C = \{(0,0)\}$ to this is $3^{-4} \approx 0.0123$.

2

For a reference on the independence of critical exponents on the type of two-dimensional lattice (square, triangular, honeycomb, bond/site percolation, nearest-neighbor+next nearest neighbor, ...) you could cite Staffer and Aharony's Introduction to Percolation Theory (page 53). The reason for this "universality" is that the critical exponent describes the ...

answered Feb 20 '19 at 22:25

Carlo Beenakker

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source
Both striped sites and black sites belong to the percolation cluster, but only the black sites are part of the backbone. The backbone can be defined as the set of current carrying paths from A to B, or equivalently as the set of self-avoiding paths from A to B. As you can see, loops are included in the backbone, according to both definitions. The ...

answered Mar 8 '19 at 9:43

Carlo Beenakker

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Welcome to MathOverflow! The answer to your question is yes. Indeed, let $X$ and $Y$ be independent random elements of the set $\{0,1\}^n$ whose distributions are percolation measures, which latter thus satisfy the Harris--FKG inequality, Proposition 2.3: $Ef(X)g(X)\ge Ef(X)\, Eg(X)$ and $Ef(Y)g(Y)\ge Ef(Y)\, Eg(Y)$ for all bounded nondecreasing functions $f$...

1

I don't see that it has anything to do with Benford's law. That has to do with the frequency of $k$ as the leading digit for lists of a certain sort, like the heights of 100 mountains measured in feet, or inches or meters (in base $10$ one expects $k$ to occur $\log(k+1)-\log(k)$ where the logs are base $10$ so $2$ appears only about $55\%$ as often as $1.$
...

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