Constructing Provably Robust Scale-Free Networks

Abstract

Scale-free networks have been described as robust to random failures but vulnerable to targeted attacks. We show that their degree sequences admit realizations that are, in fact, provably robust against any vertex removal strategy. We propose an algorithm that constructs such realizations almost surely, requiring only linear time and space. Our experiments confirm the robustness of the networks generated by this algorithm against adaptive and non-adaptive vertex removal strategies.

A Appendix: Omitted Proofs

Lemma 1. Let $(d_1,d_2,\dots , d_n)$ be an extremely scaling integer sequence with scaling factor $\gamma$. If $\gamma <7$ and $d_n\ge 3$, then $\sum _{i=1}^n (d_i-3)\in \varTheta (n)$ and there exists a constant $\kappa \in (0,1)$ such that $\sum _{i=1}^{d_1} d_i\in O(n^\kappa )$.

Proof

Note that Definition 1 and the assumptions of the lemma imply that $\gamma \in (1,7)$ and $d_n\ge 3$ are constants. Moreover, based on Definition 1 we have

$$\begin{aligned} -0.5 < d_i-n^{\frac{1}{\gamma }}d_n i^{-\frac{1}{\gamma }}\le 1.5, \end{aligned}$$

(1)

for all $i\in \{1,2,\dots , n\}$. Let $\zeta (z)$ be the Euler’s generalized constant defined for $z\in (0,1)$[18]. It is shown in [3] that

$$\begin{aligned} \sum _{i=1}^t i^{-\frac{1}{\gamma }}=\frac{t^{1-\frac{1}{\gamma }}}{1-\frac{1}{\gamma }}+\zeta (\frac{1}{\gamma }) + O(t^{-\frac{1}{\gamma }}). \end{aligned}$$

(2)

Let $\alpha :=\frac{1}{n}\sum _{i=1}^{n}(d_i-n^{\frac{1}{\gamma }}d_n i^{-\frac{1}{\gamma }})$. From (1), we know that $\alpha \in (-0.5,1.5]$. Thus, we can conclude from (2) that

$$\begin{aligned} \sum _{i=1}^n (d_i-3) = n^{\frac{1}{\gamma }}d_n\sum _{i=1}^n i^{-\frac{1}{\gamma }}-3n+\alpha n = (\frac{\gamma }{\gamma -1}d_n-3+\alpha )n+o(n)\in \varTheta (n). \end{aligned}$$

(3)

In the last step of the derivation above, we used the fact that $\frac{\gamma }{\gamma -1}d_n$ is a constant strictly larger than 3.5. Using (1) and (2), we have

$$\begin{aligned} \sum _{i=1}^{d_1} d_i\le n^{\frac{1}{\gamma }}d_n\sum _{i=1}^{d_1}i^{-\frac{1}{\gamma }}+1.5d_1\le (d_1+0.5)\sum _{i=1}^{d_1}i^{-\frac{1}{\gamma }}+1.5d_1\in O(d_1^{2-\frac{1}{\gamma }}). \end{aligned}$$

(4)

From (1), we know that $d_1\in O(n^\frac{1}{\gamma })$. Therefore, $\sum _{i=1}^{d_1} d_i\in O(n^\kappa )$, where $\kappa :=\frac{2}{\gamma }-\frac{1}{\gamma ^2}=\frac{\gamma ^2-(\gamma -1)^2}{\gamma ^2}\in (0,1)$ is a constant. This concludes the proof.    $\square$

Lemma 2. Let $(d_1,d_2, \dots , d_n)$ be a non-increasing integer sequence with even sum. If there exists $k\in \{0,1,\dots , d_n\}$ such that $\sum _{i=1}^{d_1} (d_i+k)\le \sum _{i=d_1+1}^{n} (d_i-k)$ and nk is even, then $(d_1-k, d_2-k, \dots , d_n-k)$ is graphical.

Proof

Based on the lemma’s assumptions, $D_{k}:=(d_1-k,d_2-k, \dots , d_n-k)$ is a non-increasing sequence of non-negative integers. Based on a sufficient and necessary condition for graphicality proposed in [7], $D_{k}$ is graphical if and only if the following conditions are satisfied: (I) $\sum _{i=1}^n(d_i-k)$ is even, and (II) $2\sum _{i=1}^t (d_i-k)\le \sum _{i=1}^n (d_i-k) + \sum _{i=1}^t \min \{d_i-k, t-1\}$ for all $t\in \{1,\dots , n-1\}$.

Note that condition (I) is satisfied since $\sum _{i=1}^n d_i$ and nk are both even under lemma’s assumptions. If $t\ge d_{1}-k+1$, then condition (II) is equivalent to $\sum _{i=1}^t (d_i-k)\le \sum _{i=1}^n (d_i-k)$, which is trivial since $D_k$ has non-negative elements.

Therefore, it suffices to show that the inequality in condition (II) is satisfied for all $t\in \{1,\dots , d_1-k\}$. For all such t, we know due to the assumptions of the lemma that $\sum _{i=1}^{t}(d_i-k)\le \sum _{i=1}^{d_1}d_i$ and $\min \{d_i-k, t-1\}\ge 0$ for all $i\in \{1,\dots , t\}$. Therefore, it suffices to show that $2\sum _{i=1}^{d_1}d_i\le \sum _{i=1}^n (d_i-k)$, but this is just the rearranged form of $\sum _{i=1}^{d_1} (d_i+k)\le \sum _{i=d_1+1}^n (d_i-k)$, which was assumed in the statement of the lemma. This concludes the proof.    $\square$

Lemma 3. Let $G_{\textsf {reg}}$ be a random k-regular graph with n vertices, where $k\in \mathbb {N}$ is constant. Furthermore, let $G_{\textsf {res}}$ be a graph with the same vertices as $G_{\textsf {reg}}$. If $E(G_{\textsf {res}})\in O(n)$, then there exists a constant $C>0$ such that

$$ \mathsf {Pr}\left[ |E(G_{\textsf {res}})\cap E(G_{\textsf {reg}})|\ge \frac{n^{1-\kappa }}{\log n}\right] \le \frac{C\log n}{n^{1-\kappa }}, $$

for any $\kappa \in (0,1)$.

Proof

Let m denote the number of edges in $G_{\textsf {res}}$, where $e_1, e_2, \dots , e_m$ is an arrangement of them. Moreover, let $X=\sum _{i=1}^{m} X_i$, where $X_i=1$ if $e_i\in E(G_{\textsf {reg}})$ and $X_i=0$ otherwise. It is easy to verify that $X=|E(G_{\textsf {res}})\cap E(G_{\textsf {reg}})|$.

The argument in the proof of Lemma 2.4 in [13] implies that $\mathbb {E}[X_i]=\frac{k}{n-1}$ for all $i\in \{1,2,\dots , n\}$, and thus by linearity of expectation $\mathbb {E}[X]=\frac{mk}{n-1}$. Since $m\in O(n)$ and $k\in \mathbb {N}$ is constant, we can derive that $\mathbb {E}[X]\le C$ for some constant $C > 0$. Since X is by definition non-negative, Markov’s inequality implies that

$$ \mathsf {Pr}\left[ X\ge \frac{n^{1-\kappa }}{\log n}\right] \le \frac{C\log n}{n^{1-\kappa }}. $$

Since $X=|E(G_{\textsf {res}})\cap E(G_{\textsf {reg}})|$, the above derivation completes the proof.    $\square$