Random graphs with a fixed maximum degree

TL;DR

This paper analyzes the component structure of random graphs with a fixed maximum degree, identifying a threshold for the emergence of a giant component based on the number of edges.

Contribution

It establishes a phase transition threshold for the size of the largest component in random graphs with bounded maximum degree.

Findings

Below the threshold, all components are logarithmic in size.

Above the threshold, a unique giant component emerges.

The maximum degree constraint influences the phase transition behavior.

Abstract

We study the component structure of the random graph $G=G_{n,m,d}$. Here $d=O(1)$ and $G$ is sampled uniformly from ${\mathcal G}_{n,m,d}$, the set of graphs with vertex set $[n]$, $m$ edges and maximum degree at most $d$. If $m=\mu n/2$ then we establish a threshold value $\mu_\star$ such that if $\mu<\mu_\star$ then w.h.p. the maximum component size is $O(\log n)$. If $\mu>\mu_\star$ then w.h.p. there is a unique giant component of order $n$ and the remaining components have size $O( \log n)$.

Tables

Table 1. Table 1: Numerical values of μ ⋆ ​ ( d ) subscript 𝜇 ⋆ 𝑑 \mu_{\star}(d) for small d 𝑑 d .

$d$	${\mu }_{\star }\left(d\right)$
$2$	$\infty$
$3$	$3\left(\sqrt{2}-1\right)=1.23264\dots$
$4$	$1.05783$
$5$	$1.01309$
$6$	$1.00259$
$7$	$1.00044$
$8$	$1.00006$

Equations

$$s_d(\lambda )=\sum _{j=0}^d\frac{{\lambda }^j}{j!}\text{ and }{f}_d(\lambda )=\lambda \frac{s_{d-1}(\lambda )}{s_d(\lambda )}.$$

$${\mu }_{\star }(d)=f_d(f_{d-1}^{-1}(1)),\text{functional inverse being used here},$$

$$f_d(\lambda )=\mu .$$

$${\nu }_i\approx {\lambda }_{i}n\text{ where }{\lambda }_i=\frac{1}{s_d(\lambda )}\frac{{\lambda }^i}{i!}.$$

$$g(x)=D-2x-\sum _{i=1}^{L}i{\lambda }_i{\left(1-\frac{2x}{D}\right)}^{i/2}.$$

$$\Theta =1-\sum _{i=1}^L{\lambda }_i{\left(1-\frac{2\psi }{D}\right)}^{i/2}.$$

$${\mu }_{\star }(d)=1+\frac{1}{e(d-1)!}-\frac{1}{ed!}+O\left(\frac{1}{(d-1){!}^2}\right).$$

$$f_d(1)=\frac{s_{d-1}(1)}{s_d(1)}=1-\frac{1}{d!s_d(1)}=1-\frac{1}{ed!}+O\left(\frac{1}{d{!}^2}\right)$$

$$f_d^{\prime }(1)=\frac{(s_{d-1}(1)+s_{d-2}(1))s_d(1)-s_{d-1}(1{)}^2}{s_d(1{)}^2}=1-\frac{1}{ed!}+O\left(\frac{1}{d{!}^2}\right),$$

$$1=f_{d-1}(1+\epsilon )=f_{d-1}(1)+f_{d-1}^{\prime }(1)\epsilon +O({\epsilon }^2),$$

$$\epsilon +O({\epsilon }^2)=\frac{1-f_{d-1}(1)}{f_{d-1}^{\prime }(1)}=\frac{1}{e(d-1)!}+O\left(\frac{1}{(d-1)!d!}\right).$$

$${\mu }_{\star }(d)=f_d(1+\epsilon )=f_d(1)+f_d^{\prime }(1)\frac{1-f_{d-1}(1)}{f_{d-1}^{\prime }(1)}+O({\epsilon }^2).$$

$$\mathbb{P}\left(Z_i=k\right)=c_{\lambda }\frac{{\lambda }^k}{k!},k=0,1,\dots ,d,$$

$$c_{\lambda }=\frac{1}{s_d(\lambda )}.$$

$$\displaystyle\mathbb{P}\left((Z_{1},\ldots,Z_{n})=z\ \Big{|}\ \sum_{j=1}^{n}Z_{j}=m\right)$$

$$=\frac{{\prod }_{j=1}^n{c}_{\lambda }\frac{{\lambda }^{z_j}}{z_j!}}{{\sum }_{z\in A}{\prod }_{j=1}^n{c}_{\lambda }\frac{{\lambda }^{z_j}}{z_j!}}=\frac{\frac{1}{z_1!\cdot \dots \cdot z_n!}}{{\sum }_{z\in A}\frac{1}{z_1!\cdot \dots \cdot z_n!}}.$$

$$\mathbb{P}\left((x_{1},\ldots,x_{n})=z\right)=\frac{\frac{m!}{z_{1}!\cdot\ldots\cdot z_{n}!}}{\sum_{z\in A}\frac{m!}{z_{1}!\cdot\ldots\cdot z_{n}!}}=\mathbb{P}\left((Z_{1},\ldots,Z_{n})=z\ \Big{|}\ \sum_{j=1}^{n}Z_{j}=m\right).$$

$$\underset{m=0,1,2,\dots }{\sup }\sqrt{n}\left|\mathbb{P}\left(Z_1+\dots +Z_n=m\right)-\frac{1}{\sqrt{2\pi n{\sigma }^2}}\exp \left\{-\frac{(m-\mu n{)}^2}{2n{\sigma }^2}\right\}\right|\underset{n\to \infty }{\to }0,$$

$$e^{-\lambda }{s}_d(\lambda )={\int }_{\lambda }^{\infty }\frac{t^d}{d!}{e}^{-t}\mathrm{d}t.$$

$$f_k(x)=k\frac{x\Gamma (k,x)}{\Gamma (k+1,x)}.$$

$$\frac{\Gamma (k+1,x{)}^2}{k}\frac{\mathrm{d}}{\mathrm{d}x}{f}_{k+1}(x)=(\Gamma (k,x)-x^k{e}^{-x})\Gamma (k+1,x)+x^{k+1}{e}^{-x}\Gamma (k,x).$$

$$k\Gamma (k,x{)}^2+x^k{e}^{-x}(x-k+1)\Gamma (k,x)-x^{2k}{e}^{-2x}>0,$$

$${\left(\Gamma (k,x)+\frac{x^k{e}^{-x}(x-k+1)}{2k}\right)}^2>\frac{x^{2k}{e}^{-2x}}{k}+{\left(\frac{x^k{e}^{-x}(x-k+1)}{2k}\right)}^2$$

$$\Gamma (k,x)>\frac{x^k{e}^{-x}}{2k}(\sqrt{(x-k+1{)}^2+4k}-(x-k+1)).$$

$$\Gamma (k,x)\approx x^{k-1}{e}^{-x}\left(1+\frac{k-1}{x}+\frac{(k-1)(k-2)}{x^2}+\dots \right),\text{as }x\to \infty ,$$

$$h^{\prime }(x)$$

$$\displaystyle=-\frac{x^{k-1}e^{-x}}{2k\sqrt{(x-k+1)^{2}+4k}}\Big{(}2k\sqrt{(x-k+1)^{2}+4k}+(k-x)\big{(}(x-k+1)$$

$$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad-\sqrt{(x-k+1)^{2}+4k}\big{)}\Big{)}$$

$$=-\frac{x^{k-1}{e}^{-x}}{2k\sqrt{(x-k+1{)}^2+4k}}\left((k+x)\sqrt{(x-k+1{)}^2+4k}+(k-x)(x-k+1)\right),$$

$$(k+x)\sqrt{(x-k+1{)}^2+4k}>(x-k)(x-k+1).$$

$$(k+x{)}^2((x-k+1{)}^2+4k)>(x-k{)}^2(x-k+1{)}^2$$

$$|{\mathcal{G}}_{n,x}|\approx e^{-{\alpha }_x({\alpha }_x+1)}\frac{(2m)!}{{\prod }_i{x}_i!}$$

Full text

2010 Mathematics Subject Classification. 05C80.

Key words. Random Graphs, Maximum Degree.

1 Introduction

We study the evolution of the component structure of the random graph $G_{n,m,d}$. Here $d=O(1)$ and $G$ is sampled uniformly from ${\mathcal{G}}_{n,m,d}$, the set of graphs with vertex set $[n]$, $m$ edges and maximum degree at most $d$. In the past the first author has studied properties of sparse random graphs with a lower bound on minimum degree, see for example [6]. In this paper we study sparse random graphs with a bound on the maximum degree. The model we study is close to, but distinct from that studied by Alon, Benjamini and Stacey [1] and Nachmias and Peres [12]. They studied the following model: begin with a random $d$-regular graph and then delete edges with probability $1-p$. They show in [1] that for $d\geq 3$ there is a critical probability $p_{c}=\frac{1}{d-1}$ such that w.h.p. there is a “double jump” from components of maximum size $O(\log n)$ for $p<p_{c}$, a unique giant for $p>p_{c}$ and a mximum component size of order $n^{2/3}$ for $p=p_{c}$. The paper [12] does a detailed analysis of the scaling window around $p=p_{c}$.

Naively, one might think that this analysis covers $G_{n,m,d}$. We shall see however that $G_{n,m,d}$ and random subgraphs of random regular graphs have distinct degree sequence distributions. In the latter the number of vertices of degree $i=0,1,2,\ldots,d$ will be $n$ times a binomial random variable, whereas in $G_{n,m,d}$ this number will be asymptotic to $n$ times a Poisson random variable, truncated from above.

We will write that $A_{n}\approx B_{n}$ if $A_{n}=(1+o(1))B_{n}$ and $A_{n}\lesssim B_{n}$ if $A_{n}\leq(1+o(1))B_{n}$ as $n\to\infty$.

For $d\geq 1$ and $\lambda>0$ define

[TABLE]

Theorem 1**.**

Let $d\geq 2$ and $\mu\in(0,d)$. Let $m=\lceil\frac{\mu n}{2}\rceil$. Let $G=G_{n,m,d}$ be a random graph chosen uniformly at random from the graphs with $n$ vertices, $m$ edges and maximum degree at most $d$. Let

[TABLE]

where the functions $f_{k}$ are defined in (1) and let $\lambda$ satisfy

[TABLE]

The following hold w.h.p.

1. (a)

The number $\nu_{i},i=0,1,\ldots,d$ of vertices of degree $i$ in $G$ satisfies

[TABLE] 2. (b)

If $\mu<\mu_{\star}(d)$, then $G$ has all components of size $O(\log n)$. 3. (c)

If $\mu>\mu_{\star}(d)$, then $G$ has a unique giant component of linear size $\Theta n$, where $\Theta$ is defined as follows: let $D=\sum_{i=1}^{L}i\lambda_{i}$ and

[TABLE]

Let $\psi$ be the smallest positive solution to $g(x)=0$. Then

[TABLE]

All the other components are of size $O(\log n)$.

Remark 2*.*

Numerical values of the threshold point $\mu_{\star}(d)$ for the average degree for small values of $d$ are gathered in Table 1. Note that we have an exact expression for the case $d=3$. We use $f_{2}(\lambda)=\frac{\lambda(1+\lambda)}{1+\lambda+\lambda^{2}/2}$ to see that $f_{2}^{-1}(1)=\sqrt{2}$. And then $\mu_{\star}(3)=\frac{\lambda(1+\lambda+\lambda^{2}/2)}{1+\lambda+\lambda^{2}/2+\lambda^{3}/6}=3(\sqrt{2}-1)$.

Moreover, if we consider large $d$, then we have, as a function of $d$,

[TABLE]

Comparing to the percolation model considered in [1] and [12], where $\mu_{\star}(d)=1+\frac{1}{d-1}$, we see that in our model a giant occurs * significantly earlier* for large $d$. Approximation (5) can be justified as follows. We have

[TABLE]

and

[TABLE]

(Express here $s_{d-1}$ and $s_{d-2}$ in terms of $s_{d}$ and use $s_{d}(1)=e-O(1/d!)$).

If $f_{d-1}^{-1}(1)=1+\varepsilon$, then

[TABLE]

which gives

[TABLE]

Consequently,

[TABLE]

and (5) follows.

2 Proof of Theorem 1

The main idea is to estimate the degree distribution of $G_{n,m,d}$ and then apply the results of Molloy and Reed [10], [11].

2.1 Technical Lemmas

The following lemmas will be needed for the proof of part (a).

Lemma 3**.**

Let $\lambda>0$, $d\geq 1$. Let $Z_{1},Z_{2},\ldots$ be i.i.d. random variables with

[TABLE]

where

[TABLE]

(a truncated Poisson distribution). Let $(x_{1},\ldots,x_{n})$ be a random vector of occupancies of boxes when $m$ distinguishable balls are placed uniformly at random into $n$ labelled boxes, each with capacity $d$. Then the vector $(Z_{1},\ldots,Z_{n})$ conditioned on $\sum_{j=1}^{n}Z_{j}=m$ has the same distribution as $(x_{1},\ldots,x_{n})$.

Proof.

Let $A$ be the set of vectors $z=(z_{1},\ldots,z_{n})$ of non-negative integers $z_{j}$ such that $\sum_{j=1}^{n}z_{j}=m$ and $z_{j}\leq d$ for every $j$. Fix $z\in A$. We have

[TABLE]

On the other hand, there are $\frac{m!}{z_{1}!\cdot\ldots\cdot z_{n}!}$ ways to place $m$ balls into $n$ labelled boxes in such a way that the $j$th box gets $z_{j}$ balls. Therefore,

[TABLE]

∎

Remark 4*.*

The same argument can be adapted to different constraints for the occupancies of the boxes. In general, we can replace $k\in\{0,1,\ldots,d\}$ by $k\in I$ for some set of non-negative integers $I$. For example, instead of restricting the maximal occupancy, we can require a minimal occupancy (which has appeared in Lemma 4 in [2]), or that the occupancy is even, etc.

A straightforward consequence of a standard i.i.d. case of the local central limit theorem (see, e.g. Theorem 3.5.2 in [5]) is the following lemma which will help us get rid of the conditioning from Lemma 3.

Lemma 5**.**

Let $\lambda>0$, $d\geq 1$. Let $Z_{1},Z_{2},\ldots$ be i.i.d. truncated Poisson random variables defined by (6) and (7). Then

[TABLE]

where $\mu=\mathbb{E}Z_{1}$ and $\sigma^{2}=\operatorname{Var}(Z_{1})$.

We shall also need two lemmas concerning the function $s_{d}$ from (1). A function $f$ is log-concave if $\log f$ is concave.

Lemma 6**.**

For every $\lambda>0$, the sequence $(s_{d}(\lambda))_{d=0}^{\infty}$ defined by (1) is log-concave, that is $s_{d-1}(\lambda)s_{d+1}(\lambda)\leq s_{d}(\lambda)^{2}$, $d\geq 1$.

Proof.

First note that the product of log-concave functions is log-concave. Integration by parts yields

[TABLE]

Given this integral representation, the log-concavity of $(s_{d}(\lambda))_{d=0}^{\infty}$ follows from a more general result saying that if $f:(0,\infty)\to[0,\infty)$ is log-concave, then the function $(0,+\infty)\ni p\mapsto\int_{0}^{\infty}\frac{t^{p}}{\Gamma(p+1)}f(t)\mathrm{d}t$ is also log-concave (apply to $f(t)=e^{-t}\textbf{1}_{(\lambda,\infty)}(t)$). This result goes back to Borell’s work [4] (for this exact formulation see, e.g. Corollary 5.13 in [8] or Theorem 5 in [13] containing a direct proof). ∎

Remark 7*.*

The above theorem and proof uses two related notions of log-concavity. They are reconciled by the fact that if $f:(0,\infty)\to[0,\infty)$ is log-concave then the sequence $f(i),i=0,1,\ldots$ is also log-concave.

Lemma 8**.**

For every $k\geq 1$, the function $f_{k}$ is strictly increasing on $(0,\infty)$ and onto $(0,k)$. In particular, the functional inverse, $f_{k}^{-1}:(0,k)\to(0,\infty)$ is well-defined, also strictly increasing.

Proof.

Fix $k\geq 1$ and consider $f_{k}$: rewriting (9) in terms of the upper incomplete gamma function $\Gamma(s,x)=\int_{x}^{\infty}t^{s-1}e^{-t}\mathrm{d}t$, we have

[TABLE]

Differentiating,

[TABLE]

Using $\Gamma(k+1,x)=k\Gamma(k,x)+x^{k}e^{-x}$ we can express the condition $\frac{\mathrm{d}}{\mathrm{d}x}f_{k+1}(x)>0$ as a quadratic inequality for $\Gamma(k,x)$:

[TABLE]

or

[TABLE]

or

[TABLE]

Let $h(x)$ be the left hand side minus the right hand side of (10). Clearly, $h(0)=(k-1)!>0$. Moreover, using a standard asymptotic expansion

[TABLE]

we can check that $h(x)\approx x^{k-1}e^{-x}(\frac{1}{x^{2}}+\ldots)$, so $h(x)\to 0$ as $x\to\infty$. Thus to see that $h(x)>0$ for $x>0$, it suffices to check that $h^{\prime}(x)<0$ for $x>0$. We have,

[TABLE]

so $h^{\prime}(x)<0$ is equivalent to

[TABLE]

When $k-1<x<k$, the right hand side is negative, so the inequality is clearly true. Otherwise, squaring it, we equivalently get

[TABLE]

which is clearly true because $(k+x)^{2}>(x-k)^{2}$ for $x>0$.

It is clear from (7) and (1) that $f_{k}$ is a ratio of two polynomials, each of degree $k$ and $f_{k}(x)=\frac{\frac{x^{k}}{(k-1)!}+\ldots}{\frac{x^{k}}{k!}+\ldots}$, so $f_{k}(x)\to k$ as $x\to\infty$. This combined with the monotonicity and $f_{k}(0)=0$ justifies that $f_{k}$ is a bijection onto $(0,k)$. ∎

2.2 Main elements of the proof

Let $\mathcal{D}$ be the set of all sequences of nonnegative integers $x_{1},\ldots,x_{n}\leq d$ such that $\sum x_{i}=2m$ (possible degrees). For $x\in\mathcal{D}$, let $\mathcal{G}_{n,x}$ be the set of all simple graphs on vertex set $[n]$ such that vertex $i$ has degree $x_{i},\,i=1,2,\ldots,n$. We study graphs in ${\mathcal{G}}_{n,x}$ via the Configuration Model of Bollobás [3]. We do this as follows: let $Z_{x}$ be the multi-set consisting of $x_{i}$ copies of $i$, for $i=1,2,\ldots,n$ and let $z=z_{1},z_{2},\ldots,z_{2m}$ be a random permutation of $Z_{x}$. We then define $\Gamma_{z}$ to be the (configuration) multigraph with vertex set $[n]$ and edges $\left\{z_{2i-1},z_{2i}\right\}$ for $i=1,2,\ldots,m$. It is a classical fact that conditional on being simple, $\Gamma_{z}$ is distributed as a uniform random member of ${\mathcal{G}}_{n,x}$, see for example Section 11.1 of [7].

Let $\alpha_{x}=\frac{\sum_{i}x_{i}(x_{i}-1)}{2m}$. Note that $0\leq\alpha_{x}\leq d$. It is known that

[TABLE]

as $n\to\infty$ with the $o(1)$ term being uniform in $x$ (in fact, depending only on $\Delta=\max_{i}x_{i}$). Here the term $e^{-\alpha_{x}(\alpha_{x}+1)}$ is the asymptotic probability that $\Gamma_{z}$ is simple. Therefore, for any $x\in\mathcal{D}$, we have

[TABLE]

which by Lemma 3 gives

[TABLE]

where $Z_{1},\ldots,Z_{n}$ are i.i.d. truncated Poisson random variables defined in (6).

For any graph property $\mathcal{P}$, we thus have

[TABLE]

where $G_{n,x}$ denotes a random graph selected uniformly at random from $\mathcal{G}_{n,x}$.

To handle the conditioning, we have chosen $\lambda$ so that $\mu=\mathbb{E}Z_{1}$, that is the value of $\lambda$ given by (2).

From Lemma 5 we get that for arbitrary $\delta>0$, for sufficiently large $n$,

[TABLE]

Since $2m-\mu n=2\lceil\frac{\mu n}{2}\rceil-\mu n\leq 2$ and $\sigma^{2}=\operatorname{Var}(Z_{1})$ depends only on $\lambda$ and $d$, hence only on $\mu$ and $d$, for sufficiently large $n$, the exponential factor is greater than, say $1/2$. Adjusting $\delta$ appropriately and using that $\sigma^{2}\leq\mu$, in fact,

[TABLE]

which by Lemma 6 is bounded by $\mathbb{E}Z_{1}=\mu$, we get for sufficiently large $n$,

[TABLE]

Thus, for every $x\in\mathcal{D}$,

[TABLE]

The next step is to break the sum in (11) into likely and unlikely degree sequences. Note that $\mathbb{E}\sum_{j=1}^{d}\textbf{1}_{\{Z_{j}=i\}}=n\mathbb{P}\left(Z_{1}=i\right)=n\lambda_{i}$. By Hoeffding’s inequality,

[TABLE]

Put $\varepsilon=n^{-1/3}\frac{1}{\max_{i}\lambda_{i}}$. The union bound yields

[TABLE]

This proves (a). It also shows that w.h.p. $n\lambda_{i},i=0,1,\ldots,d$ asymptotically defines the degree distribution of $G_{n,m,d}$. Also, given that $x$ is chosen uniformly at random from ${\mathcal{D}}$, we see that the distribution of $G_{n,x}$ in this case is the same as the distribution of the configuration model for the given degree sequence.

To prove (b) and (c), we will use the Molloy-Reed criterion (see [10],[11] and Theorem 11.11 in [7] for the exact formulation we shall use). First define

[TABLE]

Then, using (13) and (14),

[TABLE]

It remains to handle the typical terms $x\in\mathcal{D}\setminus\mathcal{A}$ in (11). For such $x$, we now estimate $p_{x}=\mathbb{P}\left(G_{n,x}\in\mathcal{P}\right)$ in two cases: for $\mathcal{P}$ being the complement of (i) “there are only small components”, and (ii) “there is a giant” depending on the behaviour of the degree sequences.

Let $Q=\sum_{i=0}^{d}i(i-2)\lambda_{i}$. Note that by the definition of $\mathcal{A}$, for every $x\in\mathcal{D}\setminus\mathcal{A}$, the number of vertices in $G_{n,x}$ is $n\lambda_{i}+O(n^{2/3})$, so it is justified to use the Molloy-Reed criterion and we obtain that: if $Q<0$, then $\max_{x}p_{x}\to 0$ in the case (i), and the same if $Q>0$ in the case (ii). Finally note that

[TABLE]

and Lemma 8 together with the definition of $\lambda$, that is (2), finishes the proof. The expression for $\Theta$ is in [11]. (One can also find a simplified proof of the Molloy-Reed results in [7], Theorem 11.11.)

3 Conclusions

We have found tight expressions for the degree sequence of $G_{n,m,d}$ and we have used the Molloy-Reed results to exploit them. In future work, we plan to study the scaling window around $Q$ close to zero. Hatami and Molloy [9] consider this case and their results show that we can expect a maximum component size close to $n^{2/3}$ in this case. They deal with a general degree sequence and perhaps we can prove tighter results for our specific case.