Dynamics of Nearest-Neighbour Competitions on Graphs

Abstract

Considering a collection of agents representing the vertices of a graph endowed with integer points, we study the asymptotic dynamics of the rate of the increase of their points according to a very simple rule: we randomly pick an an edge from the graph which unambiguously defines two agents we give a point the the agent with larger point with probability p and to the lagger with probability q such that $p+q=1$. The model we present is the most general version of the nearest-neighbour competition model introduced by Ben-Naim, Vazquez and Redner. We show that the model combines aspects of hyperbolic partial differential equations—as that of a conservation law—graph colouring and hyperplane arrangements. We discuss the properties of the model for general graphs but we confine in depth study to d-dimensional tori. We present a detailed study for the ring graph, which includes a chemical potential approximation to calculate all its statistics that gives rather accurate results. The two-dimensional torus, not studied in depth as the ring, is shown to possess critical behaviour in that the asymptotic speeds arrange themselves in two-coloured islands separated by borders of three other colours and the size of the islands obey power law distribution. We also show that in the large d limit the d-dimensional torus shows inverse sine law for the distribution of asymptotic speeds.

Appendix: The Chemical Potential Approach for the Ring Graph

In this appendix we shall develop an approximation for the statistics of our model on the ring graph. We shall assume that all configurations with the same number of M-type agents are equally probable. But we shall weigh them with respect to the full LS checkerboard configuration via a chemical potential $\mu$. For instance if a configuration contains k many M-type agents we shall weigh it with $\mu ^{k}$.

As well known, to obtain the statistics of the ring diagram one will need to solve the path diagram first. But since we shall ultimately be considering the limit of infinite number of vertices we do not really need to solve for the ring diagram’s partition function: Using a well known trick, whenever we need to calculate the expectation value of a given finite size subconfiguration we can place it in the middle of an infinitely long path diagram.

Now let us remember that the configurations we are interested in are all 3-colourings where the colours are labelled as L, S and M respectively and where tiles of type LML and SMS are forbidden.

We therefore have the following partition function for the path graph

$$\begin{aligned} Z_{N}\equiv \sum _{\{ \sigma _{i}\}}\prod _{i=1}^{N} {\mathcal {Q}}_{\sigma _{i-1},\sigma _{i},\sigma _{i+1}}, \end{aligned}$$

(A.1)

where the measure $\mathcal {Q}$ is given as

$$\begin{aligned} {\mathcal {Q}}_{\sigma _{i-1},\sigma _{i},\sigma _{i+1}}= & {} (1-\delta _{\sigma _{i},\sigma _{i+1}})(1-(\mu -1)\delta _{\sigma _{i},M})(1-\delta _{\sigma _{i-1},L}\delta _{\sigma _{i},M}\delta _{\sigma _{i+1},L})\nonumber \\&\times (1-\delta _{\sigma _{i-1},S}\delta _{\sigma _{i},M}\delta _{\sigma _{i+1},S}). \end{aligned}$$

(A.2)

Summing over the first agent we get the following recursion relation

$$\begin{aligned} Z_{N}=(\mu +1)Z_{N-1}-\mu M_{N}Z_{N}, \end{aligned}$$

(A.3)

where

$$\begin{aligned} M_{N}Z_{N}\equiv \sum _{\{ \sigma _{i}\}}\prod _{i=1}^{N} \delta _{\sigma _{1},M}{\mathcal {Q}}_{\sigma _{i-1},\sigma _{i},\sigma _{i+1}}, \end{aligned}$$

(A.4)

and thus $M_{N}$ is the probability that the first agent on the path has colour M. Summing over the first agent we get the following recursion relation

$$\begin{aligned} M_{N}Z_{N}=\mu \left( 1-M_{N-1}\right) Z_{N-1}, \end{aligned}$$

(A.5)

Let us also define the following, since we shall eventually need them

$$\begin{aligned} L_{N}Z_{N}\equiv & {} \sum _{\{ \sigma _{i}\}}\prod _{i=1}^{N} \delta _{\sigma _{1},L}{\mathcal {Q}}_{\sigma _{i-1},\sigma _{i},\sigma _{i+1}}, \end{aligned}$$

(A.6a)

$$\begin{aligned} S_{N}Z_{N}\equiv & {} \sum _{\{ \sigma _{i}\}}\prod _{i=1}^{N} \delta _{\sigma _{1},S}{\mathcal {Q}}_{\sigma _{i-1},\sigma _{i},\sigma _{i+1}}, \end{aligned}$$

(A.6b)

and thus $L_{N}$ and $S_{N}$ are the probabilities that the first agent has colour L and S respectively. Since these are exclusive we of course have $M_{N}+L_{N}+S_{N}=1$.

One can, summing the first agent out, show that $L_{N}$ and $S_{N}$ obey the following

$$\begin{aligned} L_{N}Z_{N}= & {} Z_{N-1}-L_{N-1}Z_{N-1}-\mu L_{N-1}Z_{N-2}, \end{aligned}$$

(A.7a)

$$\begin{aligned} S_{N}Z_{N}= & {} Z_{N-1}-S_{N-1}Z_{N-1}-\mu S_{N-1}Z_{N-2}. \end{aligned}$$

(A.7b)

One can by direct enumeration show that the initial conditions for $L_{N}$ and $S_{N}$ are identical and thus we must have $L_{N}=S_{N}$ which means that they are equal to $(1-M_{N})/2$. This is expected since the constraints in the measure are symmetric under $L\leftrightarrow S$.

Now one can, by rearranging the equations involving $Z_{N}$ and $M_{N}$, show that they are equivalent to

$$\begin{aligned}&\boxed {Z_{N}=Z_{N-1}-\mu Z_{N-2}} \end{aligned}$$

(A.8)

$$\begin{aligned}&\boxed {M_{N}=1-\frac{Z_{N-1}}{Z_{N}}\;\;\;\;\mathrm{and}\;\;\;\;L_{N}=S_{N}=\frac{1}{2}\frac{Z_{N-1}}{Z_{N}}} \end{aligned}$$

(A.9)

To resolve Eq. A.8 we need to provide two initial conditions. But these initial conditions must be such that they cover all the relevant constraints and thus $Z_{3}$ must be in the list. With these we get the following solution

$$\begin{aligned} Z_{N}=A_{+}\lambda _{+}^{N}+A_{-}\lambda _{-}^{N}, \end{aligned}$$

(A.10)

with

$$\begin{aligned} \lambda _{\pm }= & {} \frac{1}{2}\left( 1\pm \sqrt{1+4\mu }\right) , \end{aligned}$$

(A.11a)

$$\begin{aligned} A_{\pm }= & {} 1\pm \frac{1+2\mu }{\sqrt{1+4\mu }}. \end{aligned}$$

(A.11b)

Note that $\vert \lambda _{+} \vert >\vert \lambda _{-}\vert$ and hence we have

$$\begin{aligned} Z_{N\rightarrow \infty }= & {} A\lambda _{+}^N \end{aligned}$$

(A.12a)

$$\begin{aligned} M_{N\rightarrow \infty }= & {} \frac{\lambda _{+}-1}{\lambda _{+}} \end{aligned}$$

(A.12b)

$$\begin{aligned} L_{N\rightarrow \infty }= & {} \frac{1}{2\lambda _{+}} \end{aligned}$$

(A.12c)

1.1 Mean Number of M-Type Agents

To get the average number of M-type agents we consider the probability that the agent in the middle of a path graph is of type M. This means we need to calculate,

$$\begin{aligned} \bar{M}_{2N+1}Z_{2N+1}=\sum _{\{ \sigma _{i}\}}\prod _{i=-N}^{N} \delta _{\sigma _{0},M}{\mathcal {Q}}_{\sigma _{i-1},\sigma _{i},\sigma _{i+1}} \end{aligned}$$

(A.13)

By summing over the agent $\sigma _{0}$ we shall end up with

$$\begin{aligned} \bar{M}_{2N+1}Z_{2N+1}=2\mu L_{N}^{2}Z_{N}^{2} \end{aligned}$$

(A.14)

In the limit $N\rightarrow \infty$ the quantity $\bar{M}_{2N+1}$ must converge to $n_{M}$ of the paper and we recover

$$\begin{aligned} \boxed {n_{M}=\frac{1}{2}\left( 1-\frac{1}{\sqrt{1+4\mu }}\right) } \end{aligned}$$

(A.15)

as advertized before via other means.

1.2 The Distance Distribution Between Two M-Type Agents

To study the probability distribution P(k) of distances between two M-type agents we again place the region of interest in the middle of a path graph of length $2N+k+2$ with $k\ge 1$. A straightforward calculation yields

$$\begin{aligned} P(k)=\frac{2}{Z_{2N+k+2}}\mu ^{2}L_{N}^{2}Z_{N}^{2}, \end{aligned}$$

(A.16)

which in the large N limit will give us

$$\begin{aligned} \boxed {P(k)=\frac{\mu ^{2} A}{2\lambda _{+}^{4}}\;\lambda _{+}^{-k}} \end{aligned}$$

(A.17)

As expected and advertized in the paper via other means this probability distribution adds up to $n_{M}$

$$\begin{aligned} \sum _{k=1}^{\infty }P(k)=\frac{1}{2}\left( 1-\frac{1}{\sqrt{1+4\mu }}\right) =n_{M}. \end{aligned}$$

(A.18)

Furthermore again as advertized in the paper via other means we see that $\lambda _{+}=e^{\alpha }$ with $\alpha =\ln (2n_{L}/(4n_{L}-1))$.

1.3 The Pair Correlation Function of Asymptotic Speeds

The quantity of interest is

$$\begin{aligned} C_{k+1}\equiv & {} \langle z_{i}z_{i+k+1}\rangle =4\langle L_{i}L_{i+k+1}\rangle +2\langle M_{i}L_{i+k+1}\rangle +2\langle L_{i}M_{i+k+1}\rangle \nonumber \\&+\langle M_{i}H_{i+k+1}\rangle . \end{aligned}$$

(A.19)

But due to the flip symmetry the middle two terms are equal and thus we have

$$\begin{aligned} \langle z_{i}z_{i+k+1}\rangle =4\langle L_{i}L_{i+k+1}\rangle +4\langle L_{i}M_{i+k+1}\rangle -\langle M_{i}M_{i+k+1}\rangle . \end{aligned}$$

(A.20)

Furthermore since $L_{i+k+1}+M_{i+k+1}+S_{i+k+1}=1$ we can recast this as

$$\begin{aligned} \langle z_{i}z_{i+k+1}\rangle =4 n_{L}+\langle M_{i}M_{i+k+1}\rangle -4\langle L_{i}S_{i+k+1}\rangle . \end{aligned}$$

(A.21)

Some tedious algebra will yield the following equations

$$\begin{aligned} \langle M_{i}M_{i+k+1}\rangle= & {} 2\frac{\mu ^{2}L_{N}^{2}Z_{N}^{2}Z_{k}}{Z_{2N+2+k}}\left[ B_{LL}(k)+B_{LS}(k)\right] ,\end{aligned}$$

(A.22a)

$$\begin{aligned} \langle L_{i}S_{i+k+1}\rangle= & {} \frac{Z_{K}B_{LS}(k+2)}{Z_{2N+2+k}}\left( \mu ^{2}L_{N-1}^{2}Z_{N-1}^{2}+L_{N}^{2}Z_{N}^{2}+2\mu L_{N}L_{N-1}Z_{N}Z_{N-1}\right) .\nonumber \\ \end{aligned}$$

(A.22b)

The quantity $B_{LL}(k)$ is the probability that a path graph starts with an L-type agent and ends with an L-type agent. Similarly the quantity $B_{LS}$ is the probability that a path graph starts with an L-type agent and ends with an S-type agent. They both satisfy the following equation

$$\begin{aligned} f_{k}=\frac{1}{2}Z_{k-2}-f_{k-1}-\mu f_{k-2}, \end{aligned}$$

(A.23)

where $f_{k}$ stands either for $B_{LS}(k)Z(k)$ or for $B_{LL}(k)Z(k)$.

However as one can easily show that their initial conditions, which one can explicitly find by direct enumeration of graphs starting with $k=2$, are not the same. Therefor they are not equal, and in fact there is no a priori expectation for them to be equal anyways.

One can readily find the particular solution to be $f^{P}_{k}=Z_{k}/2$. Therefor the general solution can be cast as $f_{k}=f^{P}_{k}+g_{k}$ where $g_{k}$ obeys

$$\begin{aligned} g_{k}=-g_{k-1}-\mu g_{k-2}, \end{aligned}$$

(A.24)

which can be readily solved to give

$$\begin{aligned} g_{k}={\tilde{A}}_{+}{\tilde{\lambda }_{+}}^{k}+{\tilde{A}}_{-}{\tilde{\lambda }_{-}}^{k}, \end{aligned}$$

(A.25)

where one has

$$\begin{aligned} {\tilde{\lambda }}_{\pm }=-\frac{1}{2}\left( 1\pm \sqrt{1-4\mu }\right) . \end{aligned}$$

(A.26)

Note that $\vert \lambda _{+}\vert > \vert \tilde{\lambda }_{\pm }\vert$. All one now needs are the initial conditions and we have $B_{LL}(2)=0$ and $B_{LL}(3)=1/Z_{3}$ along with $B_{LS}(2)=1/Z_{2}$ and $B_{LS}(3)=\mu /Z_{3}$.

With these and taking the $N\rightarrow \infty$ limit we finally have

$$\begin{aligned} C_{k+1}=4n_{L}+\alpha \frac{Z_{k}}{{\lambda _{+}}}^{k}\left( B_{LL}(k)+B_{LS}(k)\right) -\beta \frac{Z_{k+2}}{{\lambda _{+}}^{k+2}}B_{LS}(k+2), \end{aligned}$$

(A.27)

with

$$\begin{aligned} \alpha= & {} \frac{\mu ^{2}}{2{\lambda _{+}}^{4}}A_{+},\end{aligned}$$

(A.28a)

$$\begin{aligned} \beta= & {} A_{+}. \end{aligned}$$

(A.28b)

We have thus achieved an exact expression for the pair correlation functions in the limit $N\rightarrow \infty$. One can furthermore show that as $k\rightarrow \infty$ one has $C_{k}\rightarrow 1$. This expected since $\langle z_{i}z_{i+k}\rangle$ is expected to factorize as $\langle z_{i}\rangle \langle z_{i+k}\rangle$ and $\langle z_{i}\rangle =1$ via the constraints on the system.