Exact Community Recovery under Side Information: Optimality of Spectral Algorithms

## Abstract

Abstract In this paper, we study the problem of exact community recovery in general, two-community block models considering both Bernoulli and Gaussian matrix models, capturing the Stochastic Block Model, submatrix localization, and ℤ 2 subscript ℤ 2 \mathbb{Z}_{2} -synchronization as special cases. We also study the settings where side information about community assignment labels is available, modeled as passing the true labels through a noisy channel: either the binary erasure channel (where some community labels are known while others are erased) or the binary symmetric channel (where some labels are flipped). We provide a unified analysis of the effect of side information on the information-theoretic limits of exact recovery, generalizing prior works and extending to new settings. Additionally, we design a simple but optimal spectral algorithm that incorporates side information (when present) along with the eigenvectors of the matrix observation. Using the powerful tool of entrywise eigenvector analysis [Abbe, Fan, Wang, Zhong 2020], we show that our spectral algorithm can mimic the so called genie-aided estimators , where the i th superscript 𝑖 th i^{\mathrm{th}} genie-aided estimator optimally computes the estimate of the i th superscript 𝑖 th i^{\mathrm{th}} label, when all remaining labels are revealed by a genie. This perspective provides a unified understanding of the optimality of spectral algorithms for various exact recovery problems in a recent line of work.

## 1 Introduction

In this paper, we consider inference problems of the following form: there is an unknown partition of the set $[n]:=\{1,2,\dots,n\}$ into two *communities*, denoted $(C_{+},C_{-})$ that can be encoded by the community assignment vector $\sigma^{*}\in\{\pm 1\}^{n}$. We observe a matrix $A\in\mathbb{R}^{n\times n}$, which is specified by three distributions: $\mathcal{P}_{+}$, $\mathcal{P}_{-}$, and $\mathcal{Q}$. The entries of $A$ are independent, such that $A_{ij}\sim\mathcal{P}_{+}$ if $i,j\in C_{+}$, $A_{ij}\sim\mathcal{P}_{-}$ if $i,j\in C_{-}$, and $A_{ij}\sim\mathcal{Q}$ otherwise. One famous example is the Stochastic Block Model (SBM), where $\mathcal{P}_{+}\equiv\text{Bern}(p_{1})$, $\mathcal{P}_{-}\equiv\text{Bern}(p_{2})$, and $\mathcal{Q}\equiv\text{Bern}(q)$. Other prominent examples include $\mathbb{Z}_{2}$-synchronization and submatrix localization, in which $\mathcal{P}_{+},\mathcal{P}_{-},\mathcal{Q}$ are Gaussian distributions. Given the observation $A$, the goal is to recover the unknown $\sigma^{*}$.

While these inference problems have received significant attention in the literature, with many precise results about the fundamental limits of community recovery [, , , , , , , ], less is known about inference in the presence of *side information*. Yet, from a practical standpoint, side information is abundant in inference tasks, such as e-commerce platforms and recommender systems [] and opinion recovery in social networks []. More broadly, network inference under side information is part of a larger research effort on semi-supervised inference [].

In community detection tasks, we may have prior knowledge of the community labels; either (i) the true community assignment of some subset of vertices are known and we want to determine the rest or (ii) we may have a “guess” on community assignment of each vertex and interested in recovering the true assignment. The former can be modeled as a receiving a side information vector $y\in\{-1,0,+1\}^{n}$ by passing $\sigma^{*}$ through a binary erasure channel ($\mathsf{BEC}$) where each label is erased to 0 0, independently with probability $\epsilon\in(0,1]$ to form $y$. Similarly, (ii) can be modeled as receiving a noisy community guess vector $y\in\{\pm 1\}^{n}$, where each ground-truth label is independently flipped with probability $\alpha\in(0,0.5]$, i.e. $\sigma^{*}$ went through a binary symmetric channel $(\mathsf{BSC})$ with crossover probability parameter $\alpha$. The presence of side information turns community detection from an unsupervised learning problem into a semi-supervised learning problem. In this paper, we ask: can we design algorithms that optimally incorporate side information, allowing us to recover $\sigma^{*}$ which would be impossible to recover without side information?

Figure: (a) $\mathsf{BEC}$ side information
Refer to caption: /html/2406.13075/assets/x1.png

Besides the practical motivation, considering exact recovery under side information is also interesting from an information-theoretic viewpoint. To fix ideas, consider the symmetric Stochastic Block Model with two communities $C_{+},C_{-}$ of equal size, and equal intra-community probability parameters $p_{1}=p_{2}=p$. Different notions of recovery (exact, almost exact, and partial) are achievable depending on the regime of the parameters; our focus will be on the exact recovery problem. When the intra- and inter-community probabilities are of the same order, they need to be of order $\Omega(\log n/n)$ for exact recovery to be possible; otherwise there will be an isolated vertex with high probability and hence no hope of exact recovery.
In contrast, if we are willing to relax our goal to recover a $(1-o(1))$-fraction of the community labels (i.e., achieve almost exact recovery), then it is possible to recover the community label of all but vanishing fraction of vertices as long as $\nicefrac{{(a_{n}-b_{n})^{2}}}{{2(a_{n}+b_{n})}}=\omega(1)$ where $p=a_{n}/n$ and $q=b_{n}/n$ [, , , ]. In particular, almost exact recovery does not require connectivity of $C_{+}$ or $C_{-}$ and is possible in much sparser regimes. This suggests that exact recovery is a very demanding condition. To better understand this landscape of inference problems, another way to relax the exact recovery problem is to consider the availability of $\mathsf{BEC}$ or $\mathsf{BSC}$ side information. A natural question is then, how small should the erasure probability $\epsilon\in(0,1]$ or flipping probability $\alpha\in(0,0.5]$ be in order to shift the information-theoretic threshold for exact recovery?

Saad and Nosratinia [, ] provided the answer to this question in several cases, including the symmetric SBM and the Planted Dense Subgraph (PDS) model, under both models of side information. It turns out that side information must be substantial in order to shift the exact recovery threshold. For example, the side information helps only when $\epsilon=O(n^{-\beta})$ and $\alpha=O(n^{-\beta}),$ where $\beta>0$ for the case of partially observed labels ($\mathsf{BEC}$) and noisy labels ($\mathsf{BSC}$) respectively. In particular, $\epsilon$ or $\alpha$ must vanish as $n\rightarrow\infty$ and that too at a sufficient rate. In other words, the side information must tell us all but some $n^{1-\beta}$ vertex labels correctly in order to have an effect on the information-theoretic threshold. Note that the side information vector $y$ already satisfies the almost exact recovery criterion. [, ] further derived sharp information-theoretic thresholds for the symmetric SBM and PDS in terms of the model parameters and constant $\beta$ for both $\mathsf{BEC}$ and $\mathsf{BSC}$ type side information (see also Section [1.2](#S1.SS2) for additional references).

In this paper, our goal is to shed light on the demand of exact recovery in a unified way. We study the effect of side information in two settings:

- 1.
Rank One Spike ($\mathsf{ROS}$): A Gaussian community detection problem, where a vertex belongs to $C_{+}$ with probability $\rho$ and to $C_{-}$ with probability $1-\rho$, and $\mathcal{P}_{+}$, $\mathcal{P}_{-}$, and $\mathcal{Q}$ are Gaussian distributions, which captures $\mathbb{Z}_{2}$-synchronization and submatrix localization as special cases.
- 2.
$\mathsf{SBM}$: The general two community stochastic block model, where $C_{+}$ and $C_{-}$ are sampled as specified in the above, and $\mathcal{P}_{+}$, $\mathcal{P}_{-}$, and $\mathcal{Q}$ are Bernoulli distributions.

For $\beta\geq 0$, we parameterize(^1^11This parameterization ensure that $\epsilon_{n}\in(0,1]$ and $\alpha_{n}\in(0,0.5]$ for $\beta\geq 0$.)

$$ $\epsilon_{n}=\frac{1}{n^{\beta}}\quad\text{and}\quad\alpha_{n}=\frac{1}{n^{\beta}+1}.$ (1) $$

for $\mathsf{BEC}$ and $\mathsf{BSC}$ side information respectively. Note that as $\beta$ increases, the strength of side information increases.

###### Remark 1.1 .

The parameterization is only for convenience and to derive threshold in terms of $\beta$. All of our results hold for general sequences of $\epsilon_{n}$ and $\alpha_{n}$ such that $\lim_{n\rightarrow\infty}\frac{\log(1/\epsilon_{n})}{\log n}=\beta$ and $\lim_{n\rightarrow\infty}\frac{\log(1/\alpha_{n})}{\log n}=\beta$, like [].

In addition to establishing the fundamental statistical limits of exact recovery, we are also interested in computational aspects. Some progress has already been made in this direction: [] developed an efficient two-stage algorithm for the symmetric SBM, which achieves exact recovery. In this paper, we investigate the capabilities of single-stage algorithms, which do not require a clean-up phase. Motivated by the recent success of spectral algorithms [, , , ] in achieving the sharp thresholds for exact community recovery in various settings, we ask: how can we design simple spectral strategies that can optimally make use of the $\mathsf{BSC}$ or $\mathsf{BEC}$ side information?

###### Remark 1.1 .

### 1.1 Our Contribution

Our contributions can be summarized in the following informal result.
Main Result.
(Informal)
Consider the $\mathsf{ROS}$ or $\mathsf{SBM}$ model, with or without side information. Then for each of these models, there is a sharp information-theoretic threshold such that:
•
Below the threshold, with high probability, there exists $i\in[n]$ for which it is not possible to recognize its own label even when the remaining labels are revealed. It follows that any estimator fails to exactly recover the community labels, with high probability.
•
Above the corresponding threshold, there is a simple spectral algorithm that achieves exact recovery with high probability.

In more detail, we establish the information-theoretic limits for exact recovery in both inference tasks, and under both models of side information. We show that impossibility of exact recovery is essentially driven by the failure of the so-called *genie-aided estimators*, whose goal is to determine the $i^{\mathrm{th}}$ label given all of the remaining labels.
Even though these genie-aided estimators are not possible to compute exactly, we show that they can be mimicked using a spectral strategy, which is efficient. When no side information is present, this is done by first computing a carefully chosen linear combination of the leading eigenvectors of the observation matrix $A$, and thresholding the entries of the vector by a prescribed value in order to arrive at the community assignments. Under $\mathsf{BEC}$ side information, we show that the strategy of first computing the estimator ignoring the side information, and then simply replacing the labels with the already provided ones in $y$ is optimal.
Under $\mathsf{BSC}$ side information, we instead propose shifting the linear combination of eigenvectors computed, in the direction of the side information vector $y$ by a carefully chosen scalar in terms of the crossover probability $\alpha$, which we call the trust factor.

Using the powerful technique of entrywise eigenvector approximation developed by the influential work of Abbe, Fan, Wang, and Zhong [], we show that taking an appropriately weighted sum of the leading eigenvectors of $A$ (along with side information use prescribed above when they are available) produces a vector whose $i^{\text{th}}$ entry is well-approximated by the statistic computed by these genie-aided estimators in each of the settings. Thus, the spectral algorithm without any clean-up step is able to mimic the genie, and therefore achieves exact recovery down to the information-theoretic limits. Both the negative and positive results are driven by the genie-aided estimators, so it comes as no surprise that there is a certain *threshold collapse*: the genie-aided estimators, spectral estimator, and MAP estimator all achieve the same recovery threshold.

###### Main Result .

#### Degree-Profiling Algorithm.

The genie-based estimation can also be seen as an extreme form of side information, where for each $i\in[n]$, we receive all but its own label denoted by $\sigma^{*}_{-i}$. A natural question is what happens if we naïvely try to emulate the genie for each label by trusting the side information $y$ instead of the vectors $\{\sigma^{*}_{-i}:i\in[n]\}$? We call this a degree-profiling algorithm and show that in the regimes when side information in substantial to change the thresholds of exact recovery, even this naïve strategy of emulating the genie achieves optimality. While the success of this algorithm crucially relies on the availability of high quality side information with most labels correct, the spectral algorithm continues to succeed even otherwise.

### 1.2 Related Work

#### Spectral algorithms for exact recovery.

Spectral algorithms were popularized by classical works such as McSherry’s algorithm for community detection [] and the planted clique recovery algorithm of Alon, Krivilevich, and Sudakov []. In recent years, attention has shifted to spectral algorithms without the need of a combinatorial cleanup phase. This line of work was initiated by Abbe, Fan, Wang, and Zhong [], who developed the powerful tool of entrywise eigenvector analysis, which led to a new proof technique for exact recovery problems. They show a vanilla spectral algorithm that just sign thresholds entries of the second eigenvector in the symmetric, balanced SBM and the first eigenvector for $\mathbb{Z}_{2}$-synchronization is information-theoretically optimal.

Following [], a series of papers used the entrywise eigenvector technique to give strong guarantees for spectral algorithms. For example, [] showed that, in the symmetric SBM, using the Laplacian instead of the adjacency matrix also yields an optimal algorithm. [] studied another popular special case of PDS and showed that one needs to choose an appropriate linear combination of the top two eigenvectors of the adjacency matrix, and then threshold at a critically chosen value, to achieve optimality. Another line of work [, ] considered the censored variant of the problem, where the status of some edges is unknown. Even for the censored variant, they show that spectral algorithms are optimal, where the encoding of the unknown edges is chosen carefully to achieve optimaliy. Perhaps surprisingly, [] showed that for this censored variant of the problem, to handle cases beyond the symmetric SBM and the PDS, any clustering algorithm based on a single adjacency matrix does not reach the information-theoretic threshold. Instead, the authors devised a spectral algorithm which forms two matrices of the same network and carefully combines their eigenvectors. All these results at their core rely on entrywise behavior of eigenvectors.

This raises a fundamental question: what governs the optimality of these seemingly different problem specific choices? Is there some principle behind designing new algorithms in related settings? For example, for the standard uncensored variant as we study here, it was not clear what strategy would be optimal (beyond the symmetric SBM or PDS)? Do we need two matrices like its censored counterpart []? Our work settles all these closely related questions; these strategies are mimicking the genie-aided estimators in their respective models. Moreover, we do not need two matrices to achieve optimality, unlike the censored variant of general SBM.

#### Community recovery under side information.

Community detection problems with side information have been studied in numerous settings. Saad and Nosratinia [] considered exact recovery in the symmetric, balanced SBM, under the $\mathsf{BEC}$, $\mathsf{BSC}$, and more general side information with $K$ features, where $K$ may grow with $n$. Additionally, an efficient two-stage exact recovery algorithm was proposed. Vector-valued side information was also studied in [], in the recovery of a planted dense subgraph of size $o(n)$. Community detection in the sparse setting under side information has received significant attention- see for example [, , , , ]; we note that [] considers Gaussian side information with either Bernoulli or Gaussian observations. See also [] which includes statistical physics conjectures for recovery thresholds derived from the cavity method. Numerous approaches for clustering have been proposed in the network science literature, such as [, , , , , , , , , , ]; see [] for a survey.

#### Other inference problems with side information

Related problems in the literature include document classification [, ] and text classification []. A recent line of work studies the problem of community detection from correlated graphs [, ], so that the additional graph plays the role of side information. See also [], which considers attributed graph alignment. More broadly, inference with side information falls under the area of semi-supervised learning (see e.g. [, , , , ]).

### 1.3 Organization

Section [2](#S2) contains our models and other preliminary setup. Our main results are stated in Section [3](#S3). Section [4](#S4) contains our proof ideas. Future directions are proposed in Section [5](#S5). The proofs are postponed to the appendices.

## 2 Preliminaries

### 2.1 Models

We first introduce the General Two Community Block Model (GBM), which captures the two special cases that we consider.

###### Definition 2.1 (General Two Community Block Model (GBM)) .

For any $\rho\in(0,1)$ and distributions $\mathcal{P}_{+},\mathcal{P}_{-},$ and $\mathcal{Q}$, we say that $(A,\sigma^{*})\sim\mathsf{GBM}_{n}(\rho,\mathcal{P}_{+},\mathcal{P}_{-},\mathcal{Q})$, where $A\in\mathbb{R}^{n\times n}$ and $\sigma^{*}\in\{\pm 1\}^{n}$ are sampled as follows. Each coordinate of $\sigma^{*}$ is sampled i.i.d. such that $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1)=\rho$ and $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=-1)=1-\rho$. Moreover, we will use the notation $C_{+}:=\{i:\sigma^{*}_{i}=+1\}$ and $C_{-}:=\{i:\sigma^{*}_{i}=+1\}$. Conditioned on $\sigma^{*}$, we sample $A$, a zero diagonal symmetric matrix with independent entries, such that for $1\leq i<j\leq n$, we have

$$ $A_{ij}\sim\begin{cases}\mathcal{P}_{+},&\text{if }i,j\in C_{+};\\ \mathcal{P}_{-},&\text{if }i,j\in C_{-};\\ \mathcal{Q},&\text{otherwise}.\\ \end{cases}$ $$

In the above definition, we restrict to distributions which are either (i) a continuous distribution or (ii) a finite, discrete distribution. We then let $\mathcal{P}_{+}(\cdot),\mathcal{P}_{-}(\cdot),\mathcal{Q}(\cdot)$ identify the corresponding probability density function or probability mass function, respectively. We will consider special cases where the distributions $(\mathcal{P}_{+},\mathcal{P}_{-},\mathcal{Q})$ are either all Gaussian or Bernoulli distributions. The specialized definitions are given below.

###### Definition 2.2 (Rank One Spike (ROS)) .

Fix any $\rho\in(0,1)$ and $a,b\in\mathbb{R}$ such that $\max\{|a|,|b|\}>0$. We say that $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$ if they are sampled as follows. First sample $\sigma^{*}$ as mentioned for $\mathsf{GBM}$. Conditioned on $\sigma^{*}$ consider the vector $v^{*}\in\{a,b\}^{n}$ such that for $i\in[n]$

$$ $v^{*}_{i}=a\cdot\mathbf{1}[\sigma^{*}_{i}=+1]+b\cdot\mathbf{1}[\sigma^{*}_{i}=-1].$ (2) $$

Finally, conditioned on $\sigma^{*}$, we get independent noisy measurements for every $i<j$ of the following form.

$$ $A_{ij}=v^{*}_{i}v^{*}_{j}\sqrt{\frac{\log n}{n}}+W_{ij},\hskip 5.69054pt\text{where}\hskip 5.69054ptW_{ij}\stackrel{{\scriptstyle\mathrm{i.i.d.}}}{{\sim}}\mathcal{N}(0,1).$ $$

Note that the model $\mathsf{ROS}_{n}(\rho,a,b)$ is a special case of $\mathsf{GBM}_{n}(\rho,\mathcal{P_{+}},\mathcal{P}_{-},\mathcal{Q})$, by taking

$$ $\mathcal{P}_{+}\equiv\mathcal{N}\left(a^{2}\sqrt{\frac{\log n}{n}},1\right),\mathcal{P}_{-}\equiv\mathcal{N}\left(b^{2}\sqrt{\frac{\log n}{n}},1\right),\text{ and }\mathcal{Q}\equiv\mathcal{N}\left(ab\sqrt{\frac{\log n}{n}},1\right).$ $$

Taking $b=0$ yields a version of the Gaussian submatrix localization problem [], for which the goal is to recover a submatrix of elevated mean (corresponding to the entries in $C_{+}$). Taking $a=-b$ yields the $\mathbb{Z}_{2}$-synchronization problem [] after rescaling.

###### Remark 2.3 .

We remark that the $\mathbb{Z}_{2}$-synchronization problem is typically formulated as $A_{ij}=x^{*}_{i}{x^{*}_{j}}+\sigma W_{ij}$, where $x^{*}$ is chosen uniformly at random from the set $\{\pm 1\}^{n}$ is an unknown vector,
$W$ is a zero-diagonal symmetric matrix with independent entries sampled from $\mathcal{N}(0,1)$ []. In that case, the relevant parameterization of $\sigma$ is $\sigma=c\sqrt{\frac{n}{\log n}}$, as $\sigma=\sqrt{\frac{n}{2\log n}}$ is the threshold value for exact recovery []. Thus, taking $a=1/\sqrt{c},b=-1/{\sqrt{c}}$ and $\rho=1/2$ in our $\mathsf{ROS}$ model (Definition [2.2](#S2.Thmlemma2)) produces a matrix $A$ such that

$$ $A_{ij}=\frac{1}{c}\sqrt{\frac{\log n}{n}}\,x^{*}_{i}{x^{*}_{j}}+W,\quad x^{*}\sim\textnormal{Uniform}(\{\pm 1\}^{n}).$ $$

After scaling $A$ by $c\sqrt{\nicefrac{{n}}{{\log n}}}$, we achieve the standard model $x^{*}{x^{*}}^{\top}+\sigma W$ (with zero diagonal).

Our scaling choice allows both submatrix localization and $\mathbb{Z}_{2}$-synchronization to be handled under a unified model. We also consider the Stochastic Block Model (SBM) in the logarithmic-degree regime, which is the relevant regime for exact recovery.

###### Definition 2.4 (Stochastic Block Model (SBM)) .

Fix any $\rho\in(0,1)$ and $a_{1},a_{2}>b>0$. Then the model $\mathsf{SBM}_{n}(\rho,a_{1},a_{2},b)$ is a special case of $\mathsf{GBM}_{n}(\rho,\mathcal{P_{+}},\mathcal{P}_{-},\mathcal{Q})$ with

$$ $\mathcal{P}_{+}\equiv\textnormal{Bern}\left(\frac{a_{1}\log n}{n}\right),\mathcal{P}_{-}\equiv\textnormal{Bern}\left(\frac{a_{2}\log n}{n}\right),\text{ and }\mathcal{Q}\equiv\textnormal{Bern}\left(\frac{b\log n}{n}\right).$ $$

Finally, we consider the following noise models for the side information.

###### Definition 2.5 (Binary Erasure Channel (BEC)) .

For any $\sigma^{*}\in\{\pm 1\}^{n}$ and $\epsilon\in(0,1]$, we say $y\sim\mathsf{BEC}(\sigma^{*},\epsilon)$ where each entry of $\sigma^{*}$ is erased to 0, independently with probability $\epsilon$, to form $y\in\{-1,0,+1\}^{n}$.

###### Definition 2.6 (Binary Symmetric Channel (BSC)) .

For any $\sigma^{*}\in\{\pm 1\}^{n}$ and $\alpha\in(0,1/2]$, we say $y\sim\mathsf{BSC}(\sigma^{*},\alpha)$ where each entry of $\sigma^{*}$ is flipped independently with probability $\alpha$, to form $y\in\{\pm 1\}^{n}$.

###### Definition 2.1 (General Two Community Block Model (GBM)) .

###### Definition 2.2 (Rank One Spike (ROS)) .

###### Remark 2.3 .

###### Definition 2.4 (Stochastic Block Model (SBM)) .

###### Definition 2.5 (Binary Erasure Channel (BEC)) .

###### Definition 2.6 (Binary Symmetric Channel (BSC)) .

### 2.2 Exact Recovery

Our goal is to exactly recover the community labels $\sigma^{*}$ given the observation matrix $A$ and the side information $y$ when available, as formalized below.

###### Definition 2.7 (Exact Recovery) .

We say that an estimator $\hat{\sigma}$ *succeeds* if

- (i)
$\hat{\sigma}\in\{\pm\sigma^{*}\}$ when $\mathcal{P}_{+}\equiv\mathcal{P}_{-},\rho=1/2$ and there is no side information (the symmetric case);
- (ii)
$\hat{\sigma}=\sigma^{*},$ when $\mathcal{P}_{+}\not\equiv\mathcal{P}_{-}$ or $\rho\neq 1/2$ or side information (non-symmetric case).

Otherwise, we say $\hat{\sigma}$ fails. We say that $\hat{\sigma}$ *achieves exact recovery* if
$\lim\limits_{n\rightarrow\infty}\operatorname*{\mathbb{P}}(\hat{\sigma}\text{ succeeds})=1.$

Note that in the symmetric setting described in Definition [(i)](#S2.I1.ix1), it is not possible to recover $\sigma^{*}$ with high probability, and we can only hope to recover the partition. In all other cases, we wish to recover the labels and not just the partition. All our positive results will demonstrate recovery in this strong sense, while our negative results will show that even recovering the partition is impossible below the threshold. The optimal predictor for any model and side information is the one which has maximum posterior probability given the observation matrix $A$ and the side information $y$, when it is present.

###### Definition 2.8 (MAP Estimator) .

Consider the observation matrix $A$ and the side information $y$ (either $\mathsf{BSC}$ or $\mathsf{BEC}$). We define the Maximum A Posteriori (MAP) estimator as

$$ $\hat{\sigma}_{\textnormal{\tiny MAP}}=\operatorname*{arg\,max}_{\sigma\in\{\pm 1\}^{n}}\operatorname*{\mathbb{P}}(\sigma^{*}=\sigma\mid A,y).$ $$

When no side information is present, define $\hat{\sigma}_{\textnormal{\tiny MAP}}=\operatorname*{arg\,max}_{\sigma\in\{\pm 1\}^{n}}\operatorname*{\mathbb{P}}(\sigma^{*}=\sigma\mid A)$.

###### Definition 2.7 (Exact Recovery) .

###### Definition 2.8 (MAP Estimator) .

### 2.3 Genie-Aided Estimators

We now define the framework of genie-aided estimation, which plays a crucial role in our analysis. We note that the concept of genie-aided estimation is attributed to the survey of [], while our contribution is a systematic method of connecting spectral algorithms to genie estimators.
In the genie-aided setting, we suppose that all labels but the $i^{\mathrm{th}}$ are known, and the goal is to determine the $i^{\mathrm{th}}$ label. More formally, let $\sigma^{*}_{-i}$ denote the true labels, apart from $\sigma^{*}_{i}$. The optimal estimator for the $i^{\mathrm{th}}$ label is given by

$$ $\hat{\sigma}_{\textnormal{\tiny Gen},i}=\begin{cases}+1,\quad\textnormal{if }\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1\mid A,y,\sigma^{*}_{-i})\geq\,\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=-1\mid A,y,\sigma^{*}_{-i}).\\ -1,\quad\textnormal{otherwise}.\end{cases}$ (3) $$

(where the conditioning on $y$ is omitted when there is no side information is present). Moreover, we say that $\hat{\sigma}_{\textnormal{\tiny Gen},i}$ fails on an instance if the posterior probability of the incorrect label is *strictly* greater than the posterior probability of the correct label. The following lemma rigorously establishes the intuitive claim that the failure of some genie-aided estimator implies the global MAP also fails.

###### Lemma 2.9 .

Let $\rho\in(0,1)$ and $\mathcal{P}_{+},\mathcal{P}_{-},\mathcal{Q}$ be any distributions. Let $(A,\sigma^{*})\sim\mathsf{GBM}_{n}(\rho,\mathcal{P}_{+},\mathcal{P}_{-},\mathcal{Q})$. Optionally, let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$ for $\epsilon_{n}\in(0,1]$ or $\alpha_{n}\in(0,1/2]$ respectively. Define the genie-aided estimators $\{\hat{\sigma}_{\textnormal{\tiny Gen},i}:i\in[n]\}$ and $\hat{\sigma}_{\textnormal{\tiny MAP}}$ for the respective model. Then

$$ $\operatorname*{\mathbb{P}}(\exists i\in[n]:\hat{\sigma}_{\textnormal{\tiny Gen},i}\text{ fails})\leq\operatorname*{\mathbb{P}}(\hat{\sigma}_{\textnormal{\tiny MAP}}\text{ fails}).$ $$

###### Lemma 2.9 .

#### Genie scores.

We form a vector $z^{*}\in(\mathbb{R}\cup\{\pm\infty\})^{n}$, called the *genie score* vector, where $z_{i}^{*}$ records the log of the ratio of posterior probabilities of the label $\sigma_{i}^{*}$ given $\sigma^{*}_{-i}$ by a genie. That is,

$$ $z^{*}_{i}=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1\mid A,y,\sigma^{*}_{-i})}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1}\mid A,y,\sigma^{*}_{-i})}\right)\quad\text{or}\quad z^{*}_{i}=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1\mid A,\sigma^{*}_{-i})}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1}\mid A,\sigma^{*}_{-i})}\right),$ (4) $$

in the cases of side information and no side information respectively. Then the optimal genie-based estimator corresponds to $\hat{\sigma}_{\textnormal{\tiny Gen},i}=\textnormal{sgn}(z^{*}_{i})$. The following lemma gives the form of the genie scores, when there is no side information, under the $\mathsf{BEC}$ channel, and under the $\mathsf{BSC}$ channel, respectively. Here we define $C_{+}^{-i}:=C_{+}\setminus\{i\}$ and $C_{-}^{-i}:=C_{-}\setminus\{i\}.$

###### Lemma 2.10 .

Consider any $\rho\in(0,1)$ and $\mathcal{P}_{+},\mathcal{P}_{-}$ and $\mathcal{Q}$ be either all Gaussian or Bernoulli distributions. Let $(A,\sigma^{*})\sim\mathsf{GBM}_{n}(\rho,\mathcal{P}_{+},\mathcal{P}_{-},\mathcal{Q})$. Optionally, let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$ for $\epsilon_{n}\in(0,1]$ or $\alpha_{n}\in(0,1/2]$. Then for any $i\in[n]$, the genie score for the $i^{\mathrm{th}}$ label is given by

- •
No side information:
$z^{*}_{i}=\sum_{j\in C_{+}^{-i}}\log\left(\frac{\mathcal{P}_{+}(A_{ij})}{\mathcal{Q}(A_{ij})}\right)+\sum_{j\in C_{-}^{-i}}\log\left(\frac{\mathcal{Q}(A_{ij})}{\mathcal{P_{-}}(A_{ij})}\right)+\log\left(\frac{\rho}{1-\rho}\right).$
- •
$\mathsf{BEC}$ side information: If $y_{i}=+1$ then $z_{i}^{*}=+\infty$, and if $y_{i}=-1$ then $z_{i}^{*}=-\infty$. Otherwise
$\hskip 8.53581ptz^{*}_{i}=\sum_{j\in C_{+}^{-i}}\log\left(\frac{\mathcal{P}_{+}(A_{ij})}{\mathcal{Q}(A_{ij})}\right)+\sum_{j\in C_{-}^{-i}}\log\left(\frac{\mathcal{Q}(A_{ij})}{\mathcal{P_{-}}(A_{ij})}\right)+\log\left(\frac{\rho}{1-\rho}\right).$
- •
$\mathsf{BSC}$ side information:
$z^{*}_{i}=\sum_{j\in C_{+}^{-i}}\log\left(\frac{\mathcal{P}_{+}(A_{ij})}{\mathcal{Q}(A_{ij})}\right)+\sum_{j\in C_{-}^{-i}}\log\left(\frac{\mathcal{Q}(A_{ij})}{\mathcal{P_{-}}(A_{ij})}\right)+\log\left(\frac{\rho}{1-\rho}\right)+\log\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)y_{i}.$

We devise a principled method for determining the weights and threshold value in our spectral algorithm such that $i^{\mathrm{th}}$ entry of the vector formed, approximates the the genie score $z^{*}_{i}$ under no side information. The statistic $z^{*}_{i}$ only depends on the $i^{\mathrm{th}}$ row of $A$ but knows the locations $C_{+}^{-i}$ and $C_{-}^{-i}$ due to a genie. In light of Lemma [2.10](#S2.Thmlemma10), it is possible to see the motivation behind the use of $\mathsf{BEC}$ and $\mathsf{BSC}$ side information prescribed in Section [1.1](#S1.SS1). In particular, the genie scores undergo exactly the same transformation when the side information becomes available according to Lemma [2.10](#S2.Thmlemma10); for $\mathsf{BEC}$ side information, overriding the scores with $+\infty$ or $-\infty$, depending on the side information label is $+1$ or $-1$ respectively, and for $\mathsf{BSC}$ side information, shifting the genie score vector in the direction of side information. The trust factor discussed previously in Section [1.1](#S1.SS1) exactly corresponds to the scalar quantity $\log\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)$, which is the log ratio of the probabilities of seeing the correct label and the incorrect label. Note that this is a decreasing function of the flipping probability $\alpha_{n}$.

###### Lemma 2.10 .

## 3 Main Results

### 3.1 Results for ROS

Define $\Psi:=\Psi(\rho,a,b)=(a-b)^{2}(\rho a^{2}+(1-\rho)b^{2})$, the quantity that controls the signal-to-noise ratio in the $\mathsf{ROS}_{n}(\rho,a,b)$. Then the information-theoretic (IT) threshold without or in the presence of side information of strength $\beta$ (see Remark [1.1](#S1.Thmlemma1)) is given by the following table in terms $\Psi$ and $\beta$.

| Model | IT Threshold |
| --- | --- |
| No side information | $\Psi/8=1$ |
| $\mathsf{BEC}$ side information | $\nicefrac{{\Psi}}{{8}}+\beta=1$ |
| $\mathsf{BSC}$ side information | $\frac{(\Psi+2\beta)^{2}}{8\Psi}\mathbbm{1}\{\Psi>2\beta\}+\beta\mathbbm{1}\{\Psi\leq 2\beta\}=1$ |

Table: Table 1: Information-Theoretic Thresholds for $\mathsf{ROS}$.

**Figure 2: Visualization of Information Theoretic Thresholds for $\mathsf{ROS}$. The colored regions in the above $\Psi$ vs $\beta$ plot indicate how $\mathsf{BEC}$ or $\mathsf{BSC}$ side information of strength $\beta$ helps.**
| Possible without needing<br>any side information<br>Trivial just based on<br>side information<br>Possible using either $\mathsf{BEC}$<br>or $\mathsf{BSC}$ but not otherwise<br>Possible using $\mathsf{BEC}$<br>but not with $\mathsf{BSC}$<br>Impossible despite $\mathsf{BEC}$<br>or $\mathsf{BSC}$ side information |  |
| --- | --- |

We show that above the IT threshold (as in, $\Psi/8>1$ in the case of no side information, for example), we can achieve exact recovery with spectral algorithm with high probability. On the other hand, below the IT threshold (as in, $\Psi/8<1$ in the case of no side information), any algorithm fails to achieve exact recovery with high probability. More formally:

###### Theorem 1 .

Fix $\rho\in(0,1)$ and $a,b\in\mathbb{R}$ such that $\max\{|a|,|b|\}>0$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$. Optionally, let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$, where $\epsilon_{n}$ and $\alpha_{n}$ be parameterized as ([1](#S1.E1)) for $\beta\geq 0$. Then for the respective IT threshold given in Table [1](#S3.T1)

- •
Below the IT threshold, no estimator achieves exact recovery. Moreover, for any (sequence of) estimator $\hat{\sigma}_{n}$, we have $\lim\limits_{n\to\infty}\mathbb{P}(\hat{\sigma}_{n}\text{ succeeds})=0.$
- •
Above the IT threshold, there is a spectral algorithm (Algorithm [D.4](#A4.SS4)) that returns the estimator $\hat{\sigma}_{\textnormal{spec}}$ which achieves exact recovery.

Figure [2](#S3.F2) provides a visualization of the IT threshold in the ROS model. Recalling the scaling of the error given by ([1](#S1.E1)) (and Remark [1.1](#S1.Thmlemma1)), we see that $\epsilon_{n}=O(n^{-\beta})$ and $\alpha_{n}=O(n^{-\beta})$, so that the achievable regimes in the two error models are comparable according to the parameter $\beta$. Side information improves the asymptotic IT threshold whenever $\beta>0$ (and makes the problem trivial when $\beta>1$). For any $\beta\in(0,1)$, there is a regime where recovery is possible using the $\mathsf{BEC}$ but not the $\mathsf{BSC}$. The reason that the $\mathsf{BEC}$ is more powerful comes from a coupling argument: consider the $\mathsf{BSC}$ with additional information which reveals whether the side information is true or false. That is, we observe either $(y_{i},\text{true})$ or $(y_{i},\text{false})$. But then the distribution of $(y_{i},\text{true})$ and $(y_{i},\text{false})$ is the same as the distribution of $y_{i}$ (which is always a true observation) and 0 0 in the $\mathsf{BEC}$.

To give some further intuition for the IT thresholds, note that the genie-aided estimator of a given label takes the form of a Gaussian testing problem. Here, the quantity $\Psi/8$ is the Chernoff exponent for testing between two Gaussian distributions, which arise from taking weighted sums of $A_{i\cdot}$ for $i\in C_{+}$ or $i\in C_{-}$. When side information in the form of the $\mathsf{BEC}$ is included, this translates into a reduced need for confidence in the Gaussian testing problem, as only a small (sublinear) number of labels need to be determined. The $\mathsf{BSC}$ threshold arises from two different failure modes, depending on the correctness of the side information for a given vertex.

Finally, when the side information is already substantial to shift the thresholds of recovery ($\beta>0$), then we also propose a degree-profiling algorithm which is optimal in the following sense.

###### Theorem 2 .

Fix $\rho\in(0,1)$ and $a,b\in\mathbb{R}$ such that $\max\{|a|,|b|\}>0$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$. Let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$, where $\epsilon_{n}$ and $\alpha_{n}$ be parameterized as ([1](#S1.E1)) for $\beta>0$. Then there is a degree-profiling algorithm (Algorithm [D.5](#A4.SS5)) that returns the estimator $\hat{\sigma}_{\textnormal{dp}}$ which achieves exact recovery above the IT threshold.

###### Remark 3.1 .

It is important to note that the success of the degree-profiling algorithm crucially relies on the side information because it just uses that as the ground truth to emulate the genie. For example, the algorithm would fail to recover communities under weak side information, even though the recovery was possible using just using the observation matrix $A$. However, the spectral algorithm does not have this limitation as it uses the eigenvectors of $A$ and will continue to succeed even under weak or no side information.

###### Theorem 1 .

###### Theorem 2 .

###### Remark 3.1 .

### 3.2 Results for SBM

The IT thresholds for the SBM are stated in terms of a Chernoff-Hellinger divergence, as introduced by []. For any $\mu,\nu\in\mathbb{R}_{+}^{k}$, define

$$ $D_{t}(\mu,\nu):=\sum_{i\in[k]}t\mu_{i}+(1-t)\nu_{i}-\mu_{i}^{t}\nu_{i}^{1-t}.$ (5) $$

The Chernoff–Hellinger divergence between $\mu$ and $\nu$ is given by $\sup_{t\in[0,1]}D_{t}(\mu,\nu)$.
We also define the profiles of two communities

$$ $\theta_{+}:=(\rho a_{1},(1-\rho)b)\in\mathbb{R}^{2}\quad\text{ and }\quad\theta_{-}:=(\rho b,(1-\rho)a_{2})\in\mathbb{R}^{2}.$ (6) $$

Then the IT thresholds are given in Table [2](#S3.T2).

| Model | IT Threshold |
| --- | --- |
| No side information | $\sup_{t\in[0,1]}D_{t}(\theta_{+},\theta_{-})=1$ |
| $\mathsf{BEC}$ side information | $\beta+\sup_{t\in[0,1]}D_{t}(\theta_{+},\theta_{-})=1$ |
| $\mathsf{BSC}$ side information | $\min\Big{\{}\sup_{t\in[0,1]}\beta t+D_{t}(\theta_{+},\theta_{-}),\sup_{t\in[0,1]}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})\Big{\}}=1$ |

Table: Table 2: Information-Theoretic Thresholds for $\mathsf{SBM}$.

###### Theorem 3 .

Let $\rho\in(0,1)$ and $a_{1},a_{2}>b>0$. Let $(A,\sigma^{*})\sim\mathsf{SBM}_{n}(\rho,a_{1},a_{2},b)$. Optionally, let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$, where $\epsilon_{n}$ and $\alpha_{n}$ are parameterized as ([1](#S1.E1)) for $\beta\geq 0$. Then for the respective IT threshold given in Table [2](#S3.T2),

- •
Below the IT threshold, no estimator achieves exact recovery. Moreover, for any (sequence of) estimator $\hat{\sigma}_{n}$, we have $\lim\limits_{n\to\infty}\mathbb{P}(\hat{\sigma}_{n}\text{ succeeds})=0.$
- •
Above the IT threshold, there is a spectral algorithm (Algorithms [E.3](#A5.SS3) and [E.3](#A5.SS3)) that returns the estimator $\hat{\sigma}_{\textnormal{spec}}$ which achieves exact recovery above the IT threshold.

In the absence of side information, the CH-divergence $\sup_{t\in[0,1]}D_{t}(\theta_{+},\theta_{-})=1$ is the Chernoff exponent when distinguishing a pair of labels, given all the other labels. As in $\mathsf{ROS}$, the IT threshold shifts by $\beta$ in the presence of $\mathsf{BEC}$ side information. The IT threshold in the $\mathsf{BSC}$ requires two conditions to hold (namely, $\sup_{t\in[0,1]}\beta t+D_{t}(\theta_{+},\theta_{-})>1$ and $\sup_{t\in[0,1]}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})>1$), since there are two ways that the genie estimator of a given label $i$ could fail, depending on whether the label of $i$ is correctly transmitted or not.

Finally, we also state the result for the degree profiling algorithm for the SBM, while drawing attention to Remark [3.1](#S3.Thmlemma1).

###### Theorem 4 .

Let $\rho\in(0,1)$ and $a_{1},a_{2},b>0$. Let $(A,\sigma^{*})\sim\mathsf{SBM}_{n}(\rho,a_{1},a_{2},b)$. Let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$, where $\epsilon_{n}$ and $\alpha_{n}$ be parameterized as ([1](#S1.E1)) for $\beta>0$. Then there is a degree-profiling algorithm (Algorithm [E.4](#A5.SS4)) that returns the estimator $\hat{\sigma}_{\textnormal{dp}}$ which achieves exact recovery above the IT threshold.

###### Theorem 3 .

###### Theorem 4 .

## 4 Proof Ideas

All our impossibility results and algorithmic results are proved in a unified way.

### 4.1 Unified proof strategy and the threshold collapse phenomenon

In order to derive the fundamental statistical limits of a recovery problem, it suffices to show that the MAP estimator (Definition [2.8](#S2.Thmlemma8)) fails below the threshold, due to the optimality of MAP. However, here we take a different approach and instead analyze the collection of genie-aided estimators. These individual estimators are clearly optimal for the recovery of each individual label, and thus, the failure $\hat{\sigma}_{\textnormal{\tiny Gen},i}$ for some $i\in[n]$, in recovering the $i^{\mathrm{th}}$ label, implies the failure of the MAP (Lemma [2.9](#S2.Thmlemma9)). The implication $(c)$ in Figure [3](#S4.F3) is a direct contrapositive of this. In turn, the genie scores are sums of independent random variables, conditioned on the community assignments (Lemma [2.10](#S2.Thmlemma10)). Below the threshold, we show that each genie-aided estimator fails with probability $\omega(1/n)$; from there, standard tools [] are used to show that with high probability at least one genie-aided estimator fails.

Conversely, above the IT threshold, the genie-aided estimators all succeed with high probability; moreover, they succeed with a *margin*. Indeed, we show that for both $\mathsf{ROS}$ and $\mathsf{SBM}$, without or with either type of side information, above the respective IT threshold, there exists a constant $\delta>0$ sufficiently small such that

$$ $\min_{i\in[n]}\sigma^{*}_{i}z^{*}_{i}>\delta\log n.$ (7) $$

In order to design optimal algorithms for exact recovery, we exploit this margin, recalling that the $i^{\text{th}}$ genie estimator thresholds $z_{i}^{*}$ at 0 0.
Obviously, exactly computing the genie scores is impossible without the genie. But as we will show, the spectral algorithm computes a score for each vertex $i\in[n]$, which we denote by $z^{\textnormal{spec}}_{i}$ that approximates $z^{*}_{i}$ well. In vector notation, the genie score vector $z^{*}$, is well-approximated *entrywise*, by the score vector $z^{\textnormal{spec}}$ formed by the algorithm:

$$ $\left\lVert z^{*}-z^{\textnormal{spec}}\right\rVert_{\infty}=o(\log n).$ (8) $$

Since the spectral algorithm thresholds $z^{\textnormal{spec}}$ at 0 0 to come up with the estimator $\hat{\sigma}_{\textnormal{spec}}$, the margin property ([7](#S4.E7)) implies the optimality of the spectral algorithm.
In summary, the genie-aided estimators, spectral estimator $\hat{\sigma}_{\textnormal{spec}}$, and MAP estimator $\hat{\sigma}_{\textnormal{\tiny MAP}}$ all achieve the same threshold. The entire discussion can be summarized in Figure [3](#S4.F3).

Figure: Figure 3: Summary of our unified proof framework. Our spectral algorithm is designed such that $(a)$ holds. Note that: $(b)$ follows from the optimality of $\hat{\sigma}_{\textnormal{\tiny MAP}}$ and $(c)$ follows from Lemma [2.9](#S2.Thmlemma9). Finally, we establish the threshold collapse phenomenon by showing: below the IT threshold, the event in the fourth block happens with probability $o(1)$, and above the IT threshold, the event in the first block happens with high probability.
Refer to caption: /html/2406.13075/assets/x4.png

Finally, when the side information is strong enough to shift the IT threshold, the degree profiling algorithm forms $z^{\textnormal{dp}}$ that approximates $z^{*}$ similar to ([8](#S4.E8)). The same phenomenon described by Figure [3](#S4.F3) governs the optimality of degree-profiling algorithm.

### 4.2 Overview of algorithms

We first establish that for both $\mathsf{ROS}$ and $\mathsf{SBM}$ the genie score vector takes a special form

$$ $z^{*}\approx Aw+\gamma\mathbf{1}_{n},$ $$

where the approximation is in the $\ell_{\infty}$ norm, for a certain $\gamma\in\mathbb{R}$ and $w\in\mathbb{R}^{n}$ with entries $(w_{+},w_{-})\in\mathbb{R}^{2}$ in the locations of $C_{+}$ and $C_{-}$ respectively. See Lemmas [D.1](#A4.Thmlemma1) ($\mathsf{ROS}$) and [E.2](#A5.Thmlemma2) ($\mathsf{SBM}$) for precise statements. Note that $(w_{+},w_{-},\gamma)$ are just scalars that can be calculated from the model parameters and do not depend on $\sigma^{*}$. The main power of the genie lies in forming the vector $w$, which requires knowing the locations $C_{+}$ and $C_{-}$. Then the question that remains is how one may come up with a proxy for $w$ such that the genie score is well-approximated.

#### Degree-Profiling Algorithm.

In this, we use the proxy for $w$ where we just trust the side information on the face value and use the locations of $S_{+}:=\{i\in[n]:y_{i}=+1\}$ and $S_{-}:=\{i\in[n]:y_{i}=-1\}$ instead of $C_{+}$ and $C_{-}$. When the side information is such that $|C_{+}\Delta S_{+}|,|C_{-}\Delta S_{-}|=o(n)$ with high probability, these scores are well-approximated as given in ([8](#S4.E8)). We emphasize that this step crucially relies on the availability of side information that already satisfies the almost exact recovery criterion.

#### Spectral Algorithm.

The spectral algorithm affords more versatility than the degree-profiling algorithm, as it emulates the genie score $z^{*}$ without any clean-up step even when no side information is present.
The design of our spectral algorithm is informed by the entrywise eigenvector analysis result of [], which allows us to say that the leading eigenvectors of $A$ satisfy $u\approx\frac{Au^{*}}{\lambda^{*}}$, where $(\lambda^{*},u^{*})$ is the corresponding eigenvector of the expectation matrix $\mathbb{E}[A\mid\sigma^{*}]$, and the approximation is in the $\ell_{\infty}$ norm.
Since the matrix $\mathbb{E}[A\mid\sigma^{*}]$ has a block structure, its eigenvectors do as well. Thus, we can take an appropriate linear combination of the leading eigenvectors of $A$, in order to approximate $Aw$.

## 5 Discussion and Future Work

In this paper, we provide a systematic treatment of two-community matrix inference problems under side information, focused on the Bernoulli and Gaussian cases. When the side information is already substantial, it shifts the thresholds of exact recovery and there are sharp new recovery thresholds. Moreover, we propose simple algorithms that achieve these thresholds. From a technical standpoint, our work makes the precise connection between spectral algorithms and genie-aided estimators, characterizing its effectiveness in achieving sharp thresholds for various exact recovery problems in a recent line of work. We refer the reader to Section [1.2](#S1.SS2) for a detailed discussion. Understanding the capabilities of such vanilla spectral algorithms, without any clean-up stage, is of fundamental interest; we hope this perspective will guide the design and analysis of spectral algorithms for exact community recovery problems moving forward. Some directions for future work include:

- •
Exact recovery in Gaussian Mixture Block Model: In a recent work, Li and Schramm [] proposed an alternative model to better capture real-world networks and sketched out the general landscape for recovery by studying almost exact recovery. What about exact recovery?
Interestingly, [] proposed exactly the same vanilla spectral algorithm and showed it achieves almost exact recovery. Does it also succeed for exact recovery?
- •
A refined analysis of degree-profiling: In the weak side information regimes, i.e. when the noise parameters satisfy $\lim_{n\to\infty}\frac{\log(1/\epsilon_{n})}{\log n}=0$ or $\lim_{n\to\infty}\frac{\log(1/\alpha_{n})}{\log n}=0$, what can we say about the success of degree profiling?
- •
Minimax error rates: In similar spirit to [] but now in the presence of side information, what can we say about the minimax error rates when exact recovery is impossible?
- •
More general settings: What is the IT threshold with more general side information, more general observation distributions, or more than two communities? Can we adapt the algorithms to these more general settings?

#### Acknowledgements.

J.G. was supported in part by NSF CCF-215410, and N.J. was supported in part by the Institute for Data, Econometric, Algorithms, and Learning (NSF TRIPODS HDR). The authors thank Raghav Sinha from the Illinois Mathematics and Science Academy, who simulated the algorithms as they were developed.

## Appendix A Notation and Probability Facts

### A.1 Notation

Let $X$ and $Y$ be any two collections of random variables, then we write $X\perp Y$ if they are independent. For any real numbers $a,b\in R$, we denote $a\vee b=\max\{a,b\}$ and $a\wedge b=\min\{a,b\}$. Let $\textnormal{sgn}:\mathbb{R}\to\{\pm 1\}$ be the function defined by $\textnormal{sgn}(x)=1$ if $x\geq 0$ and $\textnormal{sgn}(x)=-1$ if $x<0$. We also extend the definition to vectors; let $\textnormal{sgn}:\mathbb{R}^{n}\to\{\pm 1\}^{n}$ be the map defined by applying the sign function componentwise. We define $\mathbb{R}_{+}=[0,\infty)$. For $n\in\mathbb{N}$, we write $[n]=\{1,2,\dots,n\}$. We use the standard notation $o(.)$, $O(.)$, $\omega(.)$, $\Omega(.)$, $\Theta(.)$ etc. throughout the paper.
For nonnegative sequences $(a_{n})_{n\geq 1}$
and $(b_{n})_{n\geq 1}$, we write $a_{n}\lesssim b_{n}$ to mean
$a_{n}\leq Cb_{n}$ for some constant $C>0$. The notation $\asymp$ is similar, hiding two constants in upper and lower bounds. Moreover, we denote $a_{n}\approx b_{n}$ as a shorthand for $\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=1$.

For a vector $x\in\mathbb{R}^{n}$, we define $\left\lVert x\right\rVert_{2}=(\sum_{i=1}^{n}x_{i}^{2})^{1/2}$, $\|x\|_{1}=\sum_{i=1}^{n}|x_{i}|$, and $\left\lVert x\right\rVert_{\infty}=\max_{i}|x_{i}|$. Additionally, for any $i\in[n]$, we define $x_{-i}$ as the vector in $\mathbb{R}^{n}$ such that $(x_{-i})_{j}=x_{j}$ for $j\neq i$ and $(x_{-i})_{i}=0$.
For any matrix $M\in\mathbb{R}^{n\times n}$, $M_{i\cdot}$ refers to its $i^{\mathrm{th}}$ row, which is a row
vector, and $M_{\cdot i}$ refers to its $i^{\mathrm{th}}$ column, which is a column vector. The matrix spectral
norm is $\left\lVert M\right\rVert_{2}=\sup_{\left\lVert x\right\rVert_{2}=1}\left\lVert Mx\right\rVert_{2}$, the matrix $2\rightarrow\infty$ norm is $\left\lVert M\right\rVert_{2\rightarrow\infty}=\sup_{\left\lVert x\right\rVert_{2}=1}\left\lVert Mx\right\rVert_{\infty}=\sup_{i}\left\lVert M_{i\cdot}\right\rVert_{2}$. Let $\mathsf{zd}:\mathbb{R}^{n\times n}\rightarrow\mathbb{R}^{n\times n}$ be such that for any $A\in\mathbb{R}^{n\times n}$, $\mathsf{zd}(A)_{ij}=A_{ij}$ if $i\neq j$ and 0 otherwise.

Finally, we will adopt the standard extended real line algebra and let $+\infty$ and $-\infty$ respectively be greater and less than any other real number. Moreover, for any two $a,b\in\mathbb{R}\cup\{+\infty,-\infty\}$, we consider the convention that $a-b=0$ means $a=b$.

### A.2 Standard Probability Lemmas

###### Lemma A.1 .

Let $Z\sim\mathcal{N}(0,1)$. The for any $t>0$,

$$ $\left(\frac{1}{t}-\frac{1}{t^{3}}\right)\frac{1}{\sqrt{2\pi}}e^{-t^{2}/2}\leq\operatorname*{\mathbb{P}}(Z\geq t)\leq\frac{1}{t\sqrt{2\pi}}e^{-t^{2}/2}.$ $$

###### Lemma A.2 .

Let $N\in Z_{+}$ and $p\in[0,1]$ and $X\sim\textnormal{Binom}(N,p)$. Then for any $t\in\mathbb{R}$

$$ $\operatorname*{{\mathbb{E}}}[e^{tX}]\leq\exp(Np(e^{t}-1)).$ $$

###### Proof.

By the moment generating function of the Binomial distribution

$$ $\displaystyle\operatorname*{{\mathbb{E}}}[e^{tX}]=(1-p+pe^{t})^{N}=(1+p(e^{t}-1))^{N}\leq e^{Np(e^{t}-1)},$ $$

where in the last inequality we used that $(1+x)\leq e^{x}$ for any $x\in\mathbb{R}$.

Finally, in our proofs, we will need to analyze a tail of weighted sum of binomials. Formally, we show the following lemma, which is a minor variant of [, Theorem 10].

###### Lemma A.3 .

Let $r$ be positive integers. Let $c_{1},c_{2},\dots,c_{r}$ be real numbers such that $\min_{i\in[r]}c_{i}<0$. Let $X_{i}$ be the random variable distributed as the binomial distribution $\textnormal{Binom}(N_{i},p_{i})$ where

$$ $N_{i}=(1+o(1))\rho_{i}n\quad\text{and}\quad p_{i}=\zeta_{i}\frac{\log n}{n}$ $$

for some positive constant (not depending on n) $\rho_{i}$ and $\zeta_{i}$. Let $X=\sum_{i=1}^{r}c_{i}X_{i}$. Then for any $\delta$ such that $\delta\leq\sum_{i=1}^{r}c_{i}\rho_{i}\zeta_{i}$, we have

$$ $\operatorname*{\mathbb{P}}\left(X\leq(1+o(1))\delta\log n\right)\geq\exp\left(-(1+o(1))I^{*}\log n\right),$ $$

where

$$ $I^{*}=\sup_{t\geq 0}\left(-t\delta+\sum_{i=1}^{r}\zeta_{i}\rho_{i}(1-e^{-tc_{i}})\right).$ $$

If $\delta>\sum_{i=1}^{r}c_{i}\rho_{i}\zeta_{i}$, then

$$ $\operatorname*{\mathbb{P}}\left(X\leq(1+o(1))\delta\log n\right)\geq n^{-o(1)}=n^{-I^{*}-o(1)},\text{ with }I^{*}=0.$ $$

###### Proof.

The proof is directly adapted from []. Let us begin by considering $\delta\leq\sum_{i=1}^{r}c_{i}\rho_{i}\zeta_{i}$. Let $x:=(1+o(1))\delta\log n$. Observe that

$$ $\operatorname*{\mathbb{P}}\left(X\leq x\right)\geq\prod_{i=1}^{r}\operatorname*{\mathbb{P}}(X_{i}=x_{i})=\prod_{i=1}^{r}\binom{N_{i}}{x_{i}}p_{i}^{x_{i}}(1-p_{i})^{N_{i}-x_{i}}.$ (9) $$

for any positive integers $x_{1},\dots,x_{r}$ such that $\sum_{i=1}^{r}c_{i}x_{i}\leq x$. Let $\phi(t):=-\delta t+\sum_{i=1}^{r}\zeta_{i}\rho_{i}(1-e^{-tc_{i}})$.
Then $\phi^{\prime\prime}(t)=-\sum_{i=1}^{r}c_{i}^{2}\rho_{i}\zeta_{i}<0.$ Additionally, we have $\phi^{\prime}(0)=\sum_{i=1}^{r}c_{i}\rho_{i}\zeta_{i}-\delta\geq 0$, and $\lim_{t\rightarrow\infty}\phi^{\prime}(t)=-\infty$, since $\min_{i\in[r]}c_{i}<0$. Therefore, there exists a unique maximizer $t^{*}\geq 0$ such that $\phi^{\prime}(t^{*})=0$.

We now let $\tau_{i}=\rho_{i}\zeta_{i}e^{-c_{i}t^{*}}$ and let $x_{i}=(1-o(1))\tau_{i}\log n$ for $i\in[r]$. It is straightforward to verify that

$$ $\sum_{i=1}^{r}c_{i}x_{i}=(1-o(1))\sum_{i=1}^{r}c_{i}\tau_{i}\log n=(1-o(1))\sum_{i=1}^{r}c_{i}\rho_{i}\zeta_{i}e^{-c_{i}t^{*}}\log n=(1-o(1))\delta\log n\leq x.$ $$

When $\ell\leq\sqrt{N}$, the binomial coefficient bound is

$$ $\binom{N}{\ell}\geq\frac{N^{\ell}}{4\cdot\ell!}\,\,\text{ which gives us }\log\binom{N}{\ell}\geq\ell\log\left(\frac{eN}{\ell}\right)-\log(4e\sqrt{\ell}),$ $$

where in the last step we use The Stirling’s approximation $\ell!\leq e\sqrt{\ell}\left(\frac{\ell}{e}\right)^{\ell}$. Note that $x_{i}=\Theta(\log n)=o(\sqrt{N_{i}})$, thus using the above simplification

$$ $\log\left(\binom{N_{i}}{x_{i}}p_{i}^{x_{i}}(1-p_{i})^{N_{i}-x_{i}}\right)\geq x_{i}\log\left(\frac{eN_{i}p_{i}}{(1-p_{i})x_{i}}\right)+N_{i}\log(1-p_{i})-\log(4e\sqrt{x_{i}}).$ (10) $$

We simplify each term

$$ $\displaystyle x_{i}\log\left(\frac{eN_{i}p_{i}}{(1-p_{i})x_{i}}\right)$ $\displaystyle=(1-o(1))\tau_{i}\log n\log\left(\frac{e\rho_{i}\zeta_{i}}{\tau_{i}}\right).$ $\displaystyle N_{i}\log(1-p_{i})$ $\displaystyle=-(1+o(1))\rho_{i}n\cdot\frac{\zeta_{i}\log n}{n}=-(1+o(1))\rho_{i}\zeta_{i}\log n.$ $\displaystyle\log(4e\sqrt{x_{i}})$ $\displaystyle=o(\log n).$ $$

Combining the above with ([9](#A1.E9)) and ([10](#A1.E10)), we obtain

$$ $\operatorname*{\mathbb{P}}\left(X\leq x\right)\geq\exp\left(-(1+o(1))\log n\sum_{i=1}^{r}\left(\rho_{i}\zeta_{i}-\tau_{i}\log\left(\frac{e\rho_{i}\zeta_{i}}{\tau_{i}}\right)\right)\right).$ $$

Since we had $\tau_{i}=\rho_{i}\zeta_{i}e^{-c_{i}t^{*}}$, we obtain

$$ $\displaystyle\sum_{i=1}^{r}\left(\rho_{i}\zeta_{i}-\tau_{i}\log\left(\frac{e\rho_{i}\zeta_{i}}{\tau_{i}}\right)\right)$ $\displaystyle=\sum_{i=1}^{r}\left(\rho_{i}\zeta_{i}-\rho_{i}\zeta_{i}e^{-c_{i}t^{*}}(1+c_{i}t^{*})\right)$ $\displaystyle=-t^{*}\sum_{i=1}^{r}c_{i}\rho_{i}\zeta_{i}e^{-c_{i}t^{*}}$ $\displaystyle+\sum_{i=1}^{r}\rho_{i}\zeta_{i}(1-e^{-c_{i}t^{*}})=-\delta t^{*}+\sum_{i=1}^{r}\rho_{i}\zeta_{i}(1-e^{-c_{i}t^{*}})=I^{*},$ $$

where we used that $\delta=\sum_{i=1}^{r}c_{i}\rho_{i}\zeta_{i}e^{-c_{i}t^{*}}$. Finally, we obtain

$$ $\operatorname*{\mathbb{P}}\left(X\leq x\right)\geq\exp\left(-(1+o(1))I^{*}\log n\right).$ $$

Finally, we consider the case where $\delta>\sum_{i=1}^{r}c_{i}\rho_{i}\zeta_{i}$. Then $\phi^{\prime}(0)=-\delta+c_{i}\rho_{i}\zeta_{i}<0$. Note that $\phi(t)$ being a strictly concave function, we conclude that $I^{*}=\sup_{t\geq 0}\phi(t)=\phi(0)=0.$
Now consider $\delta^{\prime}=\sum_{i=1}^{r}c_{i}\rho_{i}\zeta_{i}<\delta$. Then

$$ $\displaystyle\operatorname*{\mathbb{P}}(X\leq(1+o(1)\delta\log n)\geq\operatorname*{\mathbb{P}}(X\leq(1+o(1))\delta^{\prime}\log n)\geq\exp(-o(\log n))=n^{-o(1)},$ $$

where the second inequality follows from the already derived tail bound for $\delta^{\prime}$.

###### Lemma A.1 .

###### Lemma A.2 .

###### Proof.

###### Lemma A.3 .

###### Proof.

## Appendix B Omitted Proofs from Prelimanaries

In this section, we prove some preliminary lemmas. We first show that the sampling procedure of the community assignment vector $\sigma^{*}$ in any of the models leads to communities with roughly $\rho$ and $(1-\rho)$ fraction of vertices.

###### Lemma B.1 .

Let $\sigma^{*}\in\{\pm 1\}^{n}$ be a vector whose coordinates are i.i.d. with $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1)=\rho.$ Then, let $C_{+}:=\{i:\sigma^{*}_{i}=+1\}$ and $C_{-}:=\{i:\sigma^{*}_{i}=-1\}$. Define the event $E$ as follow.

$$ $E:=\{||C_{+}|-\rho n|\leq\rho n^{2/3}\quad\textnormal{ and }\quad||C_{-}|-(1-\rho)n|\leq\rho n^{2/3}\}.$ (11) $$

Then $\operatorname*{\mathbb{P}}(E)\geq 1-o(1)$.

###### Proof.

For each $1\leq i\leq n$, since $\operatorname*{\mathbb{P}}(\sigma^{*}_{i})=\rho$ i.i.d., we have that $|C_{+}|=\left|{\{i:\sigma^{*}_{i}=+1\}}\right|$ follows $\textnormal{Bin}(n,\rho)$. The Chernoff bound for binomial random variables [, Theorem 4.4, Theorem 4.5] implies that for any $\delta\in(0,1)$,

$$ $\operatorname*{\mathbb{P}}(||C_{+}|-\operatorname*{{\mathbb{E}}}[|C_{+}|]|\leq\delta n)\geq 1-2\,\textnormal{exp}\left(-\delta^{2}\operatorname*{{\mathbb{E}}}[|C_{+}|]/3\right).$ $$

Note that $\operatorname*{{\mathbb{E}}}[|C_{+}|]=\rho n$. Choosing $\delta=\rho n^{-1/3}=o(1)$ (as $\rho>0$ is a constant)

$$ $\operatorname*{\mathbb{P}}(||C_{+}|-\rho n|\leq\rho n^{2/3})\geq 1-2\,\textnormal{exp}\left(-n^{-2/3}\rho^{3}n/3\right)=1-2\,\textnormal{exp}\left(-\rho^{3}n^{1/3}/3\right).$ $$

Since $n^{1/3}=\omega(\log n)$ and $\rho>0$ is a constant,

$$ $\operatorname*{\mathbb{P}}\left((1-n^{-1/3})\,\rho n\leq|C_{+}|\leq\left(1+n^{-1/3}\right)\rho n\right)\geq 1-O(\,\textnormal{exp}\left(-10\log n/3\right))=1-O(n^{-3}),$ $$

which implies the lemma.

We next show the proof of Lemma [2.9](#S2.Thmlemma9), which establishes that the failure of some genie-aided estimator implies that the global MAP estimator also fails.

###### Proof of Lemma 2.9 .

We first consider the case when side information $y$ (either $\mathsf{BEC}$ or $\mathsf{BSC}$) is provided. Let $\mathcal{S}_{1}=\{(\overline{A},\overline{\sigma},\overline{y}):\exists i\in[n]\text{ such that }\hat{\sigma}_{\textnormal{\tiny Gen},i}\text{ fails}\}$. Similarly, $\mathcal{S}_{2}=\{(\overline{A},\overline{\sigma},\overline{y}):\hat{\sigma}_{\textnormal{\tiny MAP}}\text{ fails}\}$. To show the desired claim, it suffices to show that $\mathcal{S}_{1}\subseteq\mathcal{S}_{2}$.

To this end, fix any instance $(\overline{A},\overline{\sigma},\overline{y})\in\mathcal{S}_{1}$ of $(A,\sigma^{*},y)$.
Then by definition of the failure of the genie-aided estimators give below Equation ([3](#S2.E3)), there exists $i\in[n]$ such that

$$ $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=-\overline{\sigma}_{i}\mid\overline{A},\overline{y},\overline{\sigma}_{-i})>\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=\overline{\sigma}_{i}\mid\overline{A},\overline{y},\overline{\sigma}_{-i}).$ (12) $$

We now consider the community assignment vector $\sigma^{\prime}\in\{\pm 1\}^{n}$ whose labeling agrees with $\overline{\sigma}$ except for the $i^{\mathrm{th}}$ label, for which $\sigma^{\prime}_{i}=-\overline{\sigma}_{i}$. Then

$$ $\displaystyle\operatorname*{\mathbb{P}}(\sigma^{*}=\sigma^{\prime}\mid\overline{A},\overline{y})$ $\displaystyle=\operatorname*{\mathbb{P}}(\sigma^{*}_{-i}=\sigma^{\prime}_{-i},\sigma^{*}_{i}=\sigma^{\prime}_{i}\mid\overline{A},\overline{y})$ $\displaystyle=\operatorname*{\mathbb{P}}(\sigma^{*}_{-i}=\sigma^{\prime}_{-i}\mid\overline{A},\overline{y})\cdot\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=\sigma_{i}^{\prime}\mid\overline{A},\overline{y},\sigma^{*}_{-i}=\sigma^{\prime}_{-i})$ $\displaystyle=\operatorname*{\mathbb{P}}(\sigma^{*}_{-i}=\overline{\sigma}_{-i}\mid\overline{A},\overline{y})\cdot\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=-\overline{\sigma}_{i}\mid\overline{A},\overline{y},\sigma^{*}_{-i}=\overline{\sigma}_{-i})\$ $\displaystyle>\operatorname*{\mathbb{P}}(\sigma^{*}_{-i}=\overline{\sigma}_{-i}\mid A,y)\cdot\operatorname*{\mathbb{P}}(\sigma^{*}=\overline{\sigma}_{i}\mid\overline{A},\overline{y},\sigma^{*}_{-i}=\overline{\sigma}_{-i})$ $\displaystyle=\operatorname*{\mathbb{P}}(\sigma^{*}_{-i}=\overline{\sigma}_{-i},\sigma^{*}_{i}=\overline{\sigma}_{i}\mid\overline{A},\overline{y})$ $\displaystyle=\operatorname*{\mathbb{P}}(\sigma^{*}=\overline{\sigma}\mid\overline{A},\overline{y}).$ (13) $$

By the definition of the MAP estimator $\hat{\sigma}_{\textnormal{\tiny MAP}}=\operatorname*{arg\,max}_{\sigma\in\{\pm 1\}^{n}}\operatorname*{\mathbb{P}}(\sigma^{*}=\sigma|\overline{A},\overline{y})\neq\overline{\sigma}$, which implies $(\overline{A},\overline{\sigma},\overline{y})\in\mathcal{S}_{2}$.

Finally, we consider the case when there is no side information. Define $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ similarly, but after dropping $y$.

- •
Case $\mathcal{P}_{+}\not\equiv\mathcal{P}_{-}$ or $\rho\neq 1/2$. Consider any $(\overline{A},\overline{\sigma})\in\mathcal{S}_{1}$ and follow exactly the same argument used in in deriving ([13](#A2.E13)) (after dropping the conditioning on $y$). This will lead to the conclusion that $(\overline{A},\overline{\sigma})\in\mathcal{S}_{2}$, yielding $\mathcal{S}_{1}\subseteq\mathcal{S}_{2}$.
- •
Case $\mathcal{P}_{+}\equiv\mathcal{P}_{-}$ and $\rho=1/2$. Consider any $(\overline{A},\overline{\sigma})\in\mathcal{S}_{1}$. Follow exactly the same argument as in ([13](#A2.E13)) and conclude that $\operatorname*{\mathbb{P}}(\sigma^{*}=\sigma^{\prime}\mid\overline{A})>\operatorname*{\mathbb{P}}(\sigma^{*}=\overline{\sigma}\mid\overline{A})$. Additionally, due to the symmetry, $\operatorname*{\mathbb{P}}(\sigma^{*}=\overline{\sigma}\mid\overline{A})=\operatorname*{\mathbb{P}}(\sigma^{*}=-\overline{\sigma}|A)$. Combining these two, we obtain that $\hat{\sigma}_{\textnormal{\tiny MAP}}\notin\{\pm\sigma^{*}\}$, which implies $\hat{\sigma}_{\textnormal{\tiny MAP}}$ fails by Definition [2.7](#S2.Thmlemma7). Conclude $(\overline{A},\overline{\sigma})\in\mathcal{S}_{2}$, as desired.

We now derive the expressions for genie scores given by Lemma [2.10](#S2.Thmlemma10), without or with either type of side information.

###### Proof of Lemma 2.10 .

We first recall the definition of genie scores from ([4](#S2.E4)). For any $i\in[n]$, we do the following analysis in each model of side information.
No side information:

$$ $\displaystyle z^{*}_{i}$ $\displaystyle=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1\mid A,\sigma^{*}_{-i})}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1}\mid A,\sigma^{*}_{-i})}\right)=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1)\cdot\mathcal{L}(A,\sigma^{*}_{-i}\mid\sigma^{*}_{i}=+1)}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1})\cdot\mathcal{L}(A,\sigma^{*}_{-i}\mid\sigma^{*}_{i}=-1)}\right)$ $\displaystyle=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1)\cdot\mathcal{L}(A\mid\sigma^{*}_{i}=+1,\sigma^{*}_{-i})}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1})\cdot\mathcal{L}(A\mid\sigma^{*}_{i}=-1,\sigma^{*}_{-i})}\right)$ ( $\sigma^{*}_{-i}$ is independent of $\sigma^{*}_{i}$ ) $\displaystyle=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1)\cdot\mathcal{L}(A_{i\cdot}\mid\sigma^{*}_{i}=+1,\sigma^{*}_{-i})}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=-1)\cdot\mathcal{L}(A_{i\cdot}\mid\sigma^{*}_{i}=-1,\sigma^{*}_{-i})}\right)$ (the likelihood of all but $i^{\mathrm{th}}$ row is same conditioned under $\sigma^{*}_{-i}$ irrespective of $\sigma^{*}_{i}$ ) $\displaystyle=\log\left(\rho\cdot\prod_{j\in C_{+}^{-i}}\mathcal{P}_{+}(A_{ij})\cdot\prod_{j\in C_{-}^{-i}}\mathcal{Q}(A_{ij})\middle/(1-\rho)\cdot\prod_{j\in C_{+}^{-i}}\mathcal{Q}(A_{ij})\cdot\prod_{j\in C_{-}^{-i}}\mathcal{P}_{-}(A_{ij})\right)$ (due to the conditional independence of the entries and the law of $\mathsf{GBM}$ ) $\displaystyle=\sum_{j\in C_{+}^{-i}}\log\left(\frac{\mathcal{P}_{+}(A_{ij})}{\mathcal{Q}(A_{ij})}\right)+\sum_{j\in C_{-}^{-i}}\log\left(\frac{\mathcal{Q}(A_{ij})}{\mathcal{P_{-}}(A_{ij})}\right)+\log\left(\frac{\rho}{1-\rho}\right).$ $$

$\mathsf{BEC}$ side information: From ([4](#S2.E4)), we have

$$ $z^{*}_{i}=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1\mid A,y,\sigma^{*}_{-i})}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1}\mid A,y,\sigma^{*}_{-i})}\right).$ $$

If $y_{i}=+1$ then we have $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={+1}\mid A,y,\sigma^{*}_{-i})=1$ and $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1}\mid A,y,\sigma^{*}_{-i})=0$, so that $z^{*}_{i}=+\infty$. Similarly, when $y_{i}=-1$, we have $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={+1}\mid A,y,\sigma^{*}_{-i})=0$ and
$\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1}\mid A,y,\sigma^{*}_{-i})=1$, which gives us $z^{*}_{i}=-\infty$.
Finally, when $y_{i}=0$:

$$ $\displaystyle z^{*}_{i}$ $\displaystyle=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1\mid A,y,\sigma^{*}_{-i})}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1}\mid A,y,\sigma^{*}_{-i})}\right)=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1\mid A,y_{i}=0,\sigma^{*}_{-i})}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1}\mid A,y_{i}=0,\sigma^{*}_{-i})}\right)$ $\displaystyle=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1\mid A,\sigma^{*}_{-i})}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1}\mid A,\sigma^{*}_{-i})}\right)$ (since $y_{i}=0$ and $\sigma^{*}_{-i}$ already subsumes $y_{-i}$ ) $\displaystyle=\sum_{j\in C_{+}^{-i}}\log\left(\frac{\mathcal{P}_{+}(A_{ij})}{\mathcal{Q}(A_{ij})}\right)+\sum_{j\in C_{-}^{-i}}\log\left(\frac{\mathcal{Q}(A_{ij})}{\mathcal{P_{-}}(A_{ij})}\right)+\log\left(\frac{\rho}{1-\rho}\right),$ $$

where in the last step, we used ([4](#S2.E4)) and the genie score expression without side information.
$\mathsf{BSC}$ side information: Again by ([4](#S2.E4)), we have

$$ $\displaystyle z^{*}_{i}$ $\displaystyle=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1\mid A,y,\sigma^{*}_{-i})}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1}\mid A,y,\sigma^{*}_{-i})}\right)=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1\mid A,y_{i},\sigma^{*}_{-i})}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}={-1}\mid A,y_{i},\sigma^{*}_{-i})}\right)$ $\displaystyle=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1)\cdot\mathcal{L}(A,y_{i},\sigma^{*}_{-i}\mid\sigma^{*}_{i}=+1)}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=-1)\cdot\mathcal{L}(A,y_{i},\sigma^{*}_{-i}\mid\sigma^{*}_{i}=-1)}\right)$ $\displaystyle=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1)\cdot\mathcal{L}(A,\sigma^{*}_{-i}\mid\sigma^{*}_{i}=+1)\cdot\operatorname*{\mathbb{P}}(y_{i}\mid\sigma^{*}_{i}=+1)}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=-1)\cdot\mathcal{L}(A,\sigma^{*}_{-i}\mid\sigma^{*}_{i}=-1)\cdot\operatorname*{\mathbb{P}}(y_{i}\mid\sigma^{*}_{i}=-1)}\right)$ $\displaystyle=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1\mid A,\sigma^{*}_{-i})\cdot\operatorname*{\mathbb{P}}(y_{i}\mid\sigma^{*}_{i}=+1)}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=-1\mid A,\sigma^{*}_{-i})\cdot\operatorname*{\mathbb{P}}(y_{i}\mid\sigma^{*}_{i}=-1)}\right)$ $\displaystyle=\log\left(\frac{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1\mid A,\sigma^{*}_{-i})}{\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=-1\mid A,\sigma^{*}_{-i})}\right)+\log\left(\frac{\operatorname*{\mathbb{P}}(y_{i}\mid\sigma^{*}_{i}=+1)}{\operatorname*{\mathbb{P}}(y_{i}\mid\sigma^{*}_{i}=-1)}\right)$ (14) $$

Note that

$$ $\log\left(\frac{\operatorname*{\mathbb{P}}(y_{i}\mid\sigma^{*}_{i}=+1)}{\operatorname*{\mathbb{P}}(y_{i}\mid\sigma^{*}_{i}=-1)}\right)=\log\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)\mathbf{1}[y_{i}=+1]+\log\left(\frac{\alpha_{n}}{1-\alpha_{n}}\right)\mathbf{1}[y_{i}=-1]=\log\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)y_{i}.$ $$

Substituting this in ([14](#A2.E14)) along with the definition of genie score without side information ([4](#S2.E4)) and using the expression from without side information case

$$ $z^{*}_{i}=\sum_{j\in C_{+}^{-i}}\log\left(\frac{\mathcal{P}_{+}(A_{ij})}{\mathcal{Q}(A_{ij})}\right)+\sum_{j\in C_{-}^{-i}}\log\left(\frac{\mathcal{Q}(A_{ij})}{\mathcal{P_{-}}(A_{ij})}\right)+\log\left(\frac{\rho}{1-\rho}\right)+\log\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)y_{i}.\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

###### Lemma B.1 .

###### Proof.

###### Proof of Lemma 2.9 .

###### Proof of Lemma 2.10 .

## Appendix C Entrywise Behavior of Eigenvectors

Abbe, Fan, Wang, and Zhong [] showed the powerful entrywise behavior of eigenvectors for a general ensemble of random matrices under certain assumptions. Their result [, Theorem 2.1] applies more generally to eigenspaces; below we note a special case of their result when the eigenspace has a single eigenvalue.

Suppose $A\in\mathbb{R}^{n\times n}$ is a symmetric random matrix and $A^{*}=\operatorname*{{\mathbb{E}}}[A]$. Let the eigenvalues of $A$ be $|\lambda_{1}|\geq\dots\geq|\lambda_{n}|$, and their associated eigenvectors be $\{u_{j}\}_{j\in[n]}$ (defined up to rotation if eigenvalues are repeated). Analogously for $A^{*}$, the eigenvalues and eigenvectors are $|\lambda_{1}^{*}|\geq\dots\geq|\lambda_{n}^{*}|$ and $\{u^{*}_{j}\}_{j\in[n]}$, respectively. For any fixed $(\lambda^{*}_{i},u^{*}_{i})$, define the eigengap quantity

$$ $\Delta^{*}:=|\lambda^{*}_{i}|\,\wedge\min_{j\in[n]\setminus\{i\}}|\lambda^{*}_{i}-\lambda^{*}_{j}|.$ (15) $$

Here we define the eigengap for the special case of [, Theorem 2.1] applied to a single eigenvector, rather than for an eigenspace associated with consecutive eigenvalues. For more general definition when the eigenspace contains multiple eigenvalues, see [, Equation (2.1)]. We define $\kappa:=|\lambda^{*}_{i}|/\Delta^{*}$, which is always bounded from below by $1$. For a parameter $\gamma\geq 0$, consider the following four assumptions.

###### A 1 (Incoherence) .

$\left\lVert A^{*}\right\rVert_{2\rightarrow\infty}\leq\gamma\Delta^{*}$.

###### A 2 (Row- and column-wise independence) .

For any $m\in[n]$, the entries in the $m^{\mathrm{th}}$ row and column of $A$ are independent from others, i.e. $\{A_{ij}:i=m\text{ or }j=m\}\perp\{A_{ij}:i\neq m,j\neq m\}$.

###### A 3 (Spectral norm concentration) .

For some $\delta_{0}\in(0,1)$, suppose $\operatorname*{\mathbb{P}}(\left\lVert A-A^{*}\right\rVert_{2}\leq\gamma\Delta^{*})\geq 1-\delta_{0}.$

###### A 4 (Row concentration) .

Suppose $\varphi(x)$ is continuous and non-decreasing in $\mathbb{R}_{+}$ and $\varphi(x)/x$ is non-increasing for $x>0$. Additionally $\varphi(0)=0$ and $32\kappa\max\{\gamma,\varphi(\gamma)\}\leq 1$. Let there be some $\delta_{1}\in(0,1)$ such that for any $m\in[n]$ and $w\in\mathbb{R}^{n}$

$$ $\operatorname*{\mathbb{P}}\left(|(A-A^{*})_{m\cdot}w|\leq\Delta^{*}\left\lVert w\right\rVert_{\infty}\varphi\left(\frac{\left\lVert w\right\rVert_{2}}{\sqrt{n}\left\lVert w\right\rVert_{\infty}}\right)\right)\geq 1-\frac{\delta_{1}}{n}.$ $$

###### Lemma C.1 (Theorem 2.1 [ AFWZ20 ] ) .

Under Assumptions [1](#Thmassumption1) to [4](#Thmassumption4), with probability at least
$1-\delta_{0}-2\delta_{1}$, we have

$$ $\min_{s\in\{\pm 1\}}\left\lVert su_{i}-\frac{Au^{*}_{i}}{\lambda^{*}_{i}}\right\rVert_{\infty}\lesssim\kappa(\kappa+\varphi(1))(\gamma+\varphi(\gamma))\left\lVert u_{i}^{*}\right\rVert_{\infty}+\frac{\gamma\left\lVert A^{*}\right\rVert_{2\rightarrow\infty}}{\Delta^{*}}.$ $$

###### A 1 (Incoherence) .

###### A 2 (Row- and column-wise independence) .

###### A 3 (Spectral norm concentration) .

###### A 4 (Row concentration) .

###### Lemma C.1 (Theorem 2.1 [ AFWZ20 ] ) .

### C.1 Entrywise Analysis for Eigenvectors for ROS

In this subsection, we will show that the top eigenvector of $A$ sampled from $\mathsf{ROS}$ exhibits the entrywise behavior discussed above. More formally, we show the following lemma.

###### Lemma C.2 .

Fix $\rho\in(0,1)$ and $a,b\in\mathbb{R}$ such that $\max\{|a|,|b|\}>0$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$. Condition on $\sigma^{*}$ satisfying $E$ from ([11](#A2.E11)). Let $A^{*}:=\operatorname*{{\mathbb{E}}}[A\mid\sigma^{*}]$. Define $(\lambda_{1},u_{1})$ and $(\lambda_{1}^{*},u_{1}^{*})$ as above. Then with probability $1-o(1)$

$$ $\min_{s\in\{\pm 1\}}\left\lVert su_{1}-\frac{Au_{1}^{*}}{\lambda_{1}^{*}}\right\rVert_{\infty}\leq\frac{C}{\sqrt{n\log n}},$ $$

for some constant $C:=C(\rho,a,b)>0$.

According to the definition of the ROS model, we have $A=\mathsf{zd}\left(\sqrt{\frac{\log n}{n}}\,v^{*}{v^{*}}^{\top}+W\right)$. The entire analysis is done conditioned on $\sigma^{*}$, so the only randomness in this analysis is from the added noise matrix $W$. We verify Assumptions [1](#Thmassumption1)-[4](#Thmassumption4) required to apply Lemma [C.1](#A3.Thmlemma1) using similar ideas as [, Theorem 3.1].

First, observe that $A^{*}=\mathsf{zd}(v^{*}{v^{*}}^{\top}\sqrt{\nicefrac{{\log n}}{{n}}})$. Let $(\lambda_{1}^{*},u_{1}^{*})$ be the top eigenpair. The corresponding eigengap quantity defined in ([15](#A3.E15)) is $\Delta^{*}:=|\lambda_{1}^{*}|\wedge\min_{2\leq i\leq n}|\lambda_{1}^{*}-\lambda_{i}^{*}|$. We begin by characterizing $u_{1}^{*}$, $\lambda_{1}^{*}$, and $\Delta^{*}$.

###### Lemma C.3 .

Let $(\lambda^{*}_{1},u_{1}^{*})$ be the top eigenpair of $A^{*}$. Then

$$ $u_{1}^{*}=\frac{(1+o(1))v^{*}}{\left\lVert v^{*}\right\rVert_{2}}\quad\lambda_{1}^{*}=(1+o(1))\sqrt{\frac{\log n}{n}}\left\lVert v^{*}\right\rVert_{2}^{2},\quad\Delta^{*}\approx|\lambda_{1}^{*}|=\Theta(\sqrt{n\log n}).$ $$

###### Proof.

Note that $v^{*}{v^{*}}^{\top}\sqrt{\nicefrac{{\log n}}{{n}}}$ is a rank-1 matrix. Let $|\tilde{\lambda}_{1}|\geq\dots\geq|\tilde{\lambda}_{n}|$ be its eigenvalues. Then we have that only non-zero eigenvalue is $\tilde{\lambda}_{1}=\sqrt{\frac{\log n}{n}}\left\lVert v^{*}\right\rVert_{2}^{2}=\Theta(\sqrt{n\log n})$ and the corresponding normalized eigenvector is $v^{*}/\left\lVert v^{*}\right\rVert_{2}$ and $\tilde{\lambda}_{2}=\dots=\tilde{\lambda}_{n}=0$. After zeroing out the diagonal, the entries of the corresponding eigenvector $v^{*}/\left\lVert v^{*}\right\rVert_{2}$ will be perturbed by a factor of $(1+o(1))$ since the diagonal correction is of the order of $O(\sqrt{\nicefrac{{\log n}}{{n}}})$. Hence, we obtain $u_{1}^{*}=(1+o(1))v^{*}/\left\lVert v^{*}\right\rVert_{2}$. By Weyl’s inequality, we calculate the effect of zeroing out the diagonal on the eigenvalue:

$$ $|\lambda_{1}^{*}-\tilde{\lambda}_{1}|\leq\left\lVert v^{*}{v^{*}}^{\top}\sqrt{\nicefrac{{\log n}}{{n}}}-\mathsf{zd}\left(v^{*}{v^{*}}^{\top}\sqrt{\nicefrac{{\log n}}{{n}}}\right)\right\rVert_{2}=O\left(\sqrt{\frac{\log n}{n}}\right).$ $$

Therefore,

$$ $\lambda_{1}^{*}=\tilde{\lambda}_{1}+O(\sqrt{\nicefrac{{\log n}}{{n}}})=\sqrt{\frac{\log n}{n}}\left\lVert v^{*}\right\rVert_{2}^{2}+O(\sqrt{\nicefrac{{\log n}}{{n}}})=(1+o(1))\sqrt{\frac{\log n}{n}}\left\lVert v^{*}\right\rVert_{2}^{2}\asymp\sqrt{n\log n}.$ $$

Applying Weyl’s inequality, for $2\leq i\leq n$, we get $|\lambda_{i}^{*}|=O(\sqrt{\nicefrac{{\log n}}{{n}}})$. Hence, $\Delta^{*}\approx|\lambda_{1}^{*}|\asymp\sqrt{n\log n}$.

###### Proof of Lemma C.2 .

We will let
$\gamma:=\frac{3\sqrt{n}}{\Delta^{*}}=1/\Theta(\sqrt{\log n})$, due to Lemma [C.3](#A3.Thmlemma3). Let us now verify Assumption [1](#Thmassumption1). For any $i\in C_{+}$

$$ $\displaystyle\left\lVert A^{*}_{i\cdot}\right\rVert_{2}$ $\displaystyle=\sqrt{|C_{+}^{-i}|\cdot a^{4}\frac{\log n}{n}+|C_{-}|\cdot a^{2}b^{2}\frac{\log n}{n}}$ $\displaystyle=\sqrt{(1+o(1))\rho n\cdot a^{4}\frac{\log n}{n}+(1+o(1))(1-\rho)n\cdot a^{2}b^{2}\frac{\log n}{n}}=\Theta(\sqrt{\log n}),$ $$

where the second step follows from using Lemma [B.1](#A2.Thmlemma1). Similarly, also for any $i\in C_{-}$

$$ $\displaystyle\left\lVert A^{*}_{i\cdot}\right\rVert_{2}$ $\displaystyle=\sqrt{|C_{+}|\cdot a^{2}b^{2}\frac{\log n}{n}+|C_{-}^{-i}|\cdot b^{4}\frac{\log n}{n}}=\Theta(\sqrt{\log n}).$ $$

Overall, combining these two we obtain $\left\lVert A^{*}\right\rVert_{2\rightarrow\infty}=\Theta(\sqrt{\log n})\leq 3\sqrt{n}=\gamma\Delta^{*},$ verifying Assumption [1](#Thmassumption1). Assumption [2](#Thmassumption2) on row and column-wise independence trivially holds due to the i.i.d. noise matrix $W$ (up to symmetry).

To verify Assumption [3](#Thmassumption3) on spectral norm concentration, first observe that $A-A^{*}=W$, where $W$ is the zero diagonal symmetric matrix with i.i.d. $\mathcal{N}(0,1)$ entries. Applying [, Proposition 3.3], we have that with probability at least $1-e^{-n/2}$,

$$ $\left\lVert A-A^{*}\right\rVert_{2}=\left\lVert W\right\rVert_{2}\leq 3\sqrt{n}=\gamma\Delta^{*},$ $$

Therefore, Assumption [3](#Thmassumption3) holds with $\delta_{0}=e^{-n/2}$. We now turn our attention to Assumption [4](#Thmassumption4). Let us choose $\varphi(x)=cx$ for some constant $c>0$ which we will decide later. Clearly, $\varphi$ is continuous, non-decreasing in $\mathbb{R}_{+}$ with $\varphi(0)=0$, and $\varphi(x)/x=c$ is also non-increasing in $(0,\infty)$. Letting $\kappa=1$, it is straightforward to see that $32\kappa\max\{\gamma,\varphi(\gamma)\}=32\gamma\max\{1,c\}=o(1)\leq 1$, as $\gamma=o(1)$.

We now verify the row concentration part of the assumption. Using Lemma [C.3](#A3.Thmlemma3), we have $\Delta^{*}\approx|\lambda_{1}^{*}|\geq\max\{\rho a^{2},(1-\rho)b^{2}\}\sqrt{n\log n}.$ Therefore, it holds that for any $\epsilon>0$, there is a sufficiently large $n$ such that $\Delta^{*}\geq(1-\epsilon)\max\{\rho a^{2},(1-\rho)b^{2}\}\sqrt{n\log n}$. Moreover, for any fixed $w\in\mathbb{R}^{n}$, one can say that

$$ $\displaystyle\Delta^{*}\left\lVert w\right\rVert_{\infty}\varphi\left(\frac{\left\lVert w\right\rVert_{2}}{\sqrt{n}\left\lVert w\right\rVert_{\infty}}\right)$ $\displaystyle\geq\frac{(1-\epsilon)\max\{\rho a^{2},(1-\rho)b^{2}\}\sqrt{n\log n}\cdot\left\lVert w\right\rVert_{\infty}\cdot c\left\lVert w\right\rVert_{2}}{\sqrt{n}\left\lVert w\right\rVert_{\infty}}$ $\displaystyle=(1-\epsilon)c\max\{\rho a^{2},(1-\rho)b^{2}\}\sqrt{\log n}\left\lVert w\right\rVert_{2}.$ $$

Additionally, for any fixed $m\in[n]$, $(A-A^{*})_{m\cdot}w\sim\mathcal{N}(0,\left\lVert w_{-m}\right\rVert_{2}^{2})$. Therefore,

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(|(A-A^{*})_{m\cdot}w|\leq\Delta^{*}\left\lVert w\right\rVert_{\infty}\varphi\left(\frac{\left\lVert w\right\rVert_{2}}{\sqrt{n}\left\lVert w\right\rVert_{\infty}}\right)\right)$ $\displaystyle\geq\operatorname*{\mathbb{P}}\left(|(A-A^{*})_{m\cdot}w|\leq(1-\epsilon)c\max\{\rho a^{2},(1-\rho)b^{2}\}\sqrt{\log n}\left\lVert w\right\rVert_{2}\right)$ $\displaystyle=\operatorname*{\mathbb{P}}\left(|\mathcal{N}(0,\left\lVert w_{-m}\right\rVert_{2}^{2})|\leq(1-\epsilon)c\max\{\rho a^{2},(1-\rho)b^{2}\}\sqrt{\log n}\left\lVert w\right\rVert_{2}\right)$ $\displaystyle\geq\operatorname*{\mathbb{P}}\left(|\mathcal{N}(0,\left\lVert w\right\rVert_{2}^{2})|\leq(1-\epsilon)c\max\{\rho a^{2},(1-\rho)b^{2}\}\sqrt{\log n}\left\lVert w\right\rVert_{2}\right)$ $\displaystyle=\operatorname*{\mathbb{P}}\left(|\mathcal{N}(0,1)|\leq(1-\epsilon)c\max\{\rho a^{2},(1-\rho)b^{2}\}\sqrt{\log n}\right)$ $\displaystyle\geq 1-\frac{2e^{-(1-\epsilon)^{2}c^{2}\max\{\rho a^{2},(1-\rho)b^{2}\}^{2}\log n/2}}{(1-\epsilon)c\max\{\rho a^{2},(1-\rho)b^{2}\}\sqrt{\log n}\sqrt{2\pi}}$ $\displaystyle=1-\frac{2n^{-(1-\epsilon)^{2}c^{2}\max\{\rho a^{2},(1-\rho)b^{2}\}^{2}/2}}{(1-\epsilon)c\max\{\rho a^{2},(1-\rho)b^{2}\}\sqrt{2\pi\log n}}.$ $$

Therefore, letting

$$ $\delta_{1}=\frac{2n^{1-(1-\epsilon)^{2}c^{2}\max\{\rho a^{2},(1-\rho)b^{2}\}^{2}/2}}{(1-\epsilon)c\max\{\rho a^{2},(1-\rho)b^{2}\}\sqrt{2\pi\log n}},\quad\text{and setting}\quad c=\frac{2}{(1-\epsilon)\max\{\rho a^{2},(1-\rho)b^{2}\}},$ $$

we get $\delta_{1}=o(1)$. Finally, applying Lemma [C.1](#A3.Thmlemma1), we obtain that with probability $1-\delta_{0}-2\delta_{1}=1-o(1)$

$$ $\displaystyle\min_{s\in\{\pm 1\}}\left\lVert u_{1}-\frac{Au_{1}^{*}}{\lambda_{1}^{*}}\right\rVert_{\infty}$ $\displaystyle\lesssim\kappa(\kappa+\varphi(1))(\gamma+\varphi(\gamma))\left\lVert u_{1}^{*}\right\rVert_{\infty}+\frac{\gamma\left\lVert A^{*}\right\rVert_{2\rightarrow\infty}}{\Delta^{*}}$ $\displaystyle\leq(1+c)(1+c)\gamma\left\lVert u_{1}^{*}\right\rVert_{\infty}+\frac{\gamma\left\lVert A^{*}\right\rVert_{2\rightarrow\infty}}{\Delta^{*}}$ (since $\kappa=1$ and $\varphi(x)=cx$ ) $\displaystyle=\frac{1}{\Theta(\sqrt{n\log n})}.$ $$

We used $\gamma=\frac{1}{\Theta(\sqrt{\log n})}$, $\left\lVert u_{1}^{*}\right\rVert_{\infty}=O(\frac{1}{\sqrt{n}}),\left\lVert A^{*}\right\rVert_{2\rightarrow\infty}=\sqrt{\log n}$, and $\Delta^{*}=\sqrt{n\log n}$.

###### Lemma C.2 .

###### Lemma C.3 .

###### Proof.

###### Proof of Lemma C.2 .

### C.2 Entrywise Analysis of Eigenvectors for SBM

In this subsection, we show that the similar behavior also holds for eigenvectors of $A$ sampled from the $\mathsf{SBM}$. More specifically, we restrict ourselves to the case when the expectation $A^{*}$ (after the appropriate diagonal correction) has rank 2. This is achieved when

$$ $\frac{a_{1}}{b}\neq\frac{b}{a_{2}}.$ $$

In this case, the eigenvectors that correspond to the top two leading eigenvalues (in magnitude) exhibit the entrywise behavior, which is formalized in the following lemma.

###### Lemma C.4 .

Let $\rho\in(0,1)$ and $a_{1},a_{2},b>0$ such that $a_{1}a_{2}\neq b^{2}$. Let $(A,\sigma^{*})\sim\mathsf{SBM}_{n}(\rho,a_{1},a_{2},b)$. Condition on $\sigma^{*}$ such that the event $E$ from ([11](#A2.E11)) holds and let $A:=\operatorname*{{\mathbb{E}}}[A\mid\sigma^{*}]$. Define $\{(\lambda_{i},u_{i})\}_{i\in[n]}$ and $\{(\lambda_{i}^{*},u_{i}^{*})\}_{i\in[n]}$ as above. Then with probability $1-O(n^{-3})$

$$ $\min_{s_{1}\in\{\pm 1\}}\left\lVert s_{1}u_{1}-\frac{Au_{1}^{*}}{\lambda_{1}^{*}}\right\rVert_{\infty}\leq\frac{C}{\sqrt{n}\log\log n}\quad\text{and}\quad\min_{s_{2}\in\{\pm 1\}}\left\lVert s_{2}u_{2}-\frac{Au_{2}^{*}}{\lambda_{2}^{*}}\right\rVert_{\infty}\leq\frac{C}{\sqrt{n}\log\log n},$ $$

for some constant $C:=C(\rho,a_{1},a_{2},b)$.

This again requires verifying Assumptions [1](#Thmassumption1)-[4](#Thmassumption4) for the top two eigenpairs. We note that [] showed a similar lemma for the special case of Planted Dense Subgraph (PDS), and our proof just generalizes their results. In order to do this, we first note down a couple of important lemmas. The first one directly establishes the spectral norm concentration (Assumption [3](#Thmassumption3)).

###### Lemma C.5 .

Let $\rho\in(0,1)$ and $a_{1},a_{2},b>0$. Sample $(A,\sigma^{*})\sim\mathsf{SBM}_{n}(\rho,a_{1},a_{2},b)$. Condition on $\sigma^{*}$ such that the event $E$ from ([11](#A2.E11)) holds. Let $A^{*}:=\operatorname*{{\mathbb{E}}}[A\mid\sigma^{*}]$, then there exists a constant $c_{1}=c_{1}(\rho,a_{1},a_{2},b)>0$ such that

$$ $\operatorname*{\mathbb{P}}(\left\lVert A-A^{*}\right\rVert_{2}\leq c_{1}\sqrt{\log n})\geq 1-n^{-3}.$ $$

###### Proof.

The lemma is a special case of [, Theorem 5], invoking the theorem with $c={3}$.

The next lemma establishes that the leading two eigenvalues of $A^{*}$, in the rank-2 case, are different in the following sense.

###### Lemma C.6 .

Consider $\rho\in(0,1)$ and $a_{1},a_{2},b>0$ such that $a_{1}a_{2}\neq b^{2}$. Let $(A,\sigma^{*})\sim\mathsf{SBM}_{n}(\rho,a_{1},a_{2},b)$. Condition on a labelling $\sigma^{*}$ such that the event $E$ from ([11](#A2.E11)) holds. Let $A^{*}:=\operatorname*{{\mathbb{E}}}[A\mid\sigma^{*}]$. Then the top two eigenvalues in magnitude are given by $\lambda_{1}^{*}=(1+o(1))\theta_{1}\log n$ and $\lambda_{2}^{*}=(1+o(1))\theta_{2}\log n$, for some non-zero constants $\theta_{1}\neq\theta_{2}$ in terms of $(\rho,a_{1},a_{2},b)$. As a consequence, $|\lambda_{1}^{*}-\lambda_{2}^{*}|=\Theta(\log n)$.

###### Proof.

The proof follows similar arguments as the proof of [, Lemma 3.2]. We note that they prove the special case of the PDS when $a_{2}=b$, but the same argument directly generalizes as long as $a_{1}a_{2}\neq b^{2}$.

###### Proof of Lemma C.4 .

The entire analysis is done conditioned on $\sigma^{*}$ such that $E$ holds. We will verify Assumptions [1](#Thmassumption1)-[4](#Thmassumption4) for the leading two eigenpairs. First note that after adding a diagonal matrix $D$, whose entries are $O(\frac{\log n}{n})$, the matrix $A^{*}+D$ has rank 2, and its remaining eigenvalues satisfy $\tilde{\lambda}_{3}=\dots=\tilde{\lambda}_{n}=0$, where $|\tilde{\lambda}_{1}|\geq\dots|\tilde{\lambda}_{n}|$. Applying Weyl’s inequality for $3\leq i\leq n$,

$$ $|\lambda_{i}^{*}-\tilde{\lambda}_{i}|=|\lambda_{i}^{*}|\leq\left\lVert D\right\rVert_{2}=O(\nicefrac{{\log n}}{{n}}).$ (16) $$

By the definition of the eigengap quantity in ([15](#A3.E15)) for both the eigenvalues respectively

$$ $\Delta_{1}^{*}:=|\lambda_{1}^{*}|\wedge\min_{i\neq 1}|\lambda_{i}^{*}-\lambda_{1}^{*}|=\Theta(\log n)\,\text{ and }\,\Delta_{2}^{*}:=|\lambda_{2}^{*}|\wedge\min_{i\neq 2}|\lambda_{i}^{*}-\lambda_{2}^{*}|=\Theta(\log n),$ $$

where we used ([16](#A3.E16)) and Lemma [C.6](#A3.Thmlemma6). We also define $\kappa_{1}:=\frac{|\lambda_{1}^{*}|}{\Delta_{1}^{*}}$ and $\kappa_{2}:=\frac{|\lambda_{2}^{*}|}{\Delta_{2}^{*}}$. We first make an inportant observation, that to verify Assumptions [1](#Thmassumption1)-[4](#Thmassumption4) for both eigenpairs separately, it suffices to just verify them with $\Delta^{*}$ and $\kappa$ such that

$$ $\Delta^{*}:=\min\{\Delta_{1}^{*},\Delta_{2}^{*}\}=\Theta(\log n)\text{ and }\kappa:=\max\{\kappa_{1},\kappa_{2}\}$ $$

To verify Assumption [1](#Thmassumption1), first let $\tau=2\max\{a_{1},a_{2},b\}$. Fixing any $i\in C_{+}$,

$$ $\displaystyle\left\lVert A_{i\cdot}\right\rVert_{2}$ $\displaystyle=\sqrt{|C_{+}^{-i}|\left(\frac{a_{1}\log n}{n}\right)^{2}+|C_{-}|\left(\frac{b\log n}{n}\right)^{2}}$ $\displaystyle=\sqrt{(1+o(1))\rho\left(\frac{a_{1}^{2}\log^{2}n}{n}\right)+(1+o(1))(1-\rho)\left(\frac{b^{2}\log^{2}n}{n}\right)}\leq\frac{\tau\log n}{\sqrt{n}}.$ $$

Similarly, even for $i\in C_{-}$

$$ $\displaystyle\left\lVert A_{i\cdot}\right\rVert_{2}$ $\displaystyle=\sqrt{|C_{+}|\left(\frac{b\log n}{n}\right)^{2}+|C_{-}^{-i}|\left(\frac{a_{2}\log n}{n}\right)^{2}}\leq\frac{\tau\log n}{\sqrt{n}}$ $$

Combining both bounds, we obtain

$$ $\left\lVert A^{*}\right\rVert_{2\rightarrow\infty}\leq\frac{\tau\log n}{\sqrt{n}}.$ $$

We now define the parameter $\gamma$ in terms of $\Delta^{*}$ and the constant $c_{1}$ from Lemma [C.5](#A3.Thmlemma5):

$$ $\gamma\triangleq\frac{c_{1}\sqrt{\log n}}{\Delta^{*}}=\frac{c_{1}\sqrt{\log n}}{\Theta(\log n)}=o(1).$ $$

Then $\gamma\Delta^{*}=c_{1}\sqrt{\log n}=\Omega(\sqrt{\log n})$, which dominates $\nicefrac{{\tau\log n}}{{\sqrt{n}}}$. This implies $\left\lVert A^{*}\right\rVert_{2\rightarrow\infty}\leq\gamma\Delta^{*}$, verifying Assumption [1](#Thmassumption1). Assumption [2](#Thmassumption2) trivially holds due to the conditional independence of the entries of $A$, conditioned on $\sigma^{*}$. By Lemma [C.5](#A3.Thmlemma5), Assumption [3](#Thmassumption3) holds with $\delta_{0}=n^{-3}$

$$ $\operatorname*{\mathbb{P}}(\left\lVert A-A^{*}\right\rVert_{2}\leq\gamma\Delta^{*})\geq 1-n^{-3}.$ $$

To verify Assumption [4](#Thmassumption4), we let

$$ $\varphi(x)\triangleq\frac{(2\tau+4)\log n}{\Delta^{*}(1\vee\log(1/x))}\text{ for $x>0$ and }\varphi(0)=0.$ $$

It is straightforward to verify that $\varphi$ satisfies the desired property stated in Assumption [4](#Thmassumption4) and $\varphi(\gamma)=O\left(\nicefrac{{1}}{{\log\log n}}\right)$. Also, $\kappa=O(1)$ since both $\Delta_{1}^{*}\asymp\Delta_{2}^{*}\asymp\log n$, and by Lemma [C.6](#A3.Thmlemma6), also $|\lambda_{1}^{*}|\asymp|\lambda_{2}^{*}|\asymp\log n$. This implies $32\kappa\max\{\gamma,\varphi(\gamma)\}=o(1)$ verifying the first part of the assumption.

To verify the row concentration part, we simply apply [, Lemma 7] with $p=\nicefrac{{\tau\log n}}{{n}}$ and $\alpha=4/\tau$. We obtain that for a fixed vector $w\in\mathbb{R}^{n}$ and $m\in[n]$,

$$ $\operatorname*{\mathbb{P}}\left(|(A-A^{*})_{m\cdot}w|\leq\frac{(2\tau+4)\log n}{\max\left\{1,\log\left(\frac{\sqrt{n}\left\lVert w\right\rVert_{\infty}}{\left\lVert w\right\rVert_{2}}\right)\right\}}\left\lVert w\right\rVert_{\infty}\right)\geq 1-2n^{-4}.$ $$

Substituting the definition of $\Delta^{*}$ and $\varphi(\cdot)$,

$$ $\operatorname*{\mathbb{P}}\left(|(A-A^{*})_{m\cdot}w|\leq\Delta^{*}\left\lVert w\right\rVert_{\infty}\varphi\left(\frac{\left\lVert w\right\rVert_{2}}{\sqrt{n}\left\lVert w\right\rVert_{\infty}}\right)\right)\geq 1-2n^{-4},$ $$

which verifies Assumption [4](#Thmassumption4) with $\delta_{1}=2n^{-3}$. Finally, applying Lemma [C.1](#A3.Thmlemma1), with probability $1-\delta_{0}-2\delta_{1}=1-O(n^{-3})$,

$$ $\displaystyle\min_{s_{1}\in\{\pm 1\}}\left\lVert s_{1}u_{1}-\frac{Au_{1}^{*}}{\lambda_{1}^{*}}\right\rVert_{\infty}$ $\displaystyle\lesssim\kappa(\kappa+\varphi(1))(\gamma+\varphi(\gamma))\left\lVert u_{1}^{*}\right\rVert_{\infty}+\frac{\gamma\left\lVert A^{*}\right\rVert_{2\rightarrow\infty}}{\Delta^{*}}=O\left(\frac{1}{\sqrt{n}\log\log n}\right).$ $$

We used $\gamma=\frac{1}{\Theta(\sqrt{\log n})},\varphi(\gamma)=O(\frac{1}{\log\log n}),\left\lVert u_{1}^{*}\right\rVert_{\infty}=O(\frac{1}{\sqrt{n}}),\left\lVert A^{*}\right\rVert_{2\rightarrow\infty}=O\left(\frac{\log n}{\sqrt{n}}\right)$, and $\Delta^{*}=\Theta(\log n)$.

Similarly, with probability $1-O(n^{-3})$, we have

$$ $\displaystyle\min_{s_{2}\in\{\pm 1\}}\left\lVert s_{2}u_{2}-\frac{Au_{2}^{*}}{\lambda_{2}^{*}}\right\rVert_{\infty}$ $\displaystyle=O\left(\frac{1}{\sqrt{n}\log\log n}\right).$ $$

The proof is complete by a union bound.

###### Lemma C.4 .

###### Lemma C.5 .

###### Proof.

###### Lemma C.6 .

###### Proof.

###### Proof of Lemma C.4 .

## Appendix D Proofs and Algorithms for ROS

In Appendix [D.1](#A4.SS1), we first show that the genie scores succeed with margin above the IT threshold. In Appendix [D.2](#A4.SS2), we show some genie-aided estimator fails below the threshold, while in Appendix [D.3](#A4.SS3), we show statistical achievability above the threshold when side information is not present, which will be crucially required to show the optimality of spectral algorithm. Finally, in Appendices [D.4](#A4.SS4) and [D.5](#A4.SS5), we provide algorithms and the proof of main results.

### D.1 Genie scores’ success with margin above the IT threshold

We start by noting the form of genie scores when no side information is present.

###### Lemma D.1 .

Fix $\rho\in(0,1)$ and $a,b\in\mathbb{R}$ such that $\max\{|a|,|b|\}>0$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$. Then for any $i\in[n]$

$$ $\displaystyle z^{*}_{i}=$ $\displaystyle(a-b)\sqrt{\frac{\log n}{n}}\left(a\sum_{j\in C_{+}^{-i}}A_{ij}+b\sum_{j\in C_{-}^{-i}}A_{ij}\right)+\frac{\log n}{2n}(|C_{+}^{-i}|(a^{2}b^{2}-a^{4})+|C_{-}^{-i}|(b^{4}-a^{2}b^{2}))$ $\displaystyle\hskip 5.69054pt+\log\left(\frac{\rho}{1-\rho}\right).$ $$

Moreover, conditioned on the event $E$ from ([11](#A2.E11)), the genie score vector $z^{*}\in\mathbb{R}^{n}$ can be written as

$$ $z^{*}=(a-b)\sqrt{\frac{\log n}{n}}Av^{*}+\left(\gamma+\log\left(\frac{\rho}{1-\rho}\right)\right)\mathbf{1}_{n}+o(1),$ $$

where $\gamma=\left(\rho(a^{2}b^{2}-a^{4})+(1-\rho)(b^{4}-a^{2}b^{2})\right)\log n/2$ and $v^{*}$ is given by ([2](#S2.E2)).

###### Proof.

For ease of notation, we denote $f(n)=\sqrt{\nicefrac{{\log n}}{{n}}}$. First, note that $\mathsf{ROS}$ is a special case of $\mathsf{GBM}$ with $\mathcal{P}_{+}=\mathcal{N}(a^{2}f(n),1),\mathcal{P}_{-}=\mathcal{N}(b^{2}f(n),1)$ and $\mathcal{Q}=\mathcal{N}(abf(n),1)$. Applying Lemma [2.10](#S2.Thmlemma10) for this special case, we obtain the Genie score expressions; for any $i\in[n]$,

$$ $\displaystyle z^{*}_{i}$ $\displaystyle=\sum_{j\in C_{+}^{-i}}\log\left(e^{-\frac{(A_{ij}-a^{2}f(n))^{2}}{2}}\middle/e^{-\frac{(A_{ij}-abf(n))^{2}}{2}}\right)$ $\displaystyle\hskip 5.69054pt+\sum_{j\in C_{-}^{-i}}\log\left(e^{-\frac{(A_{ij}-abf(n))^{2}}{2}}\middle/e^{-\frac{(A_{ij}-b^{2}f(n))^{2}}{2}}\right)+\log\left(\frac{\rho}{1-\rho}\right)$ $\displaystyle=\sum_{j\in C_{+}^{-i}}\left(\frac{(A_{ij}-abf(n))^{2}}{2}-\frac{(A_{ij}-a^{2}f(n))^{2}}{2}\right)$ $\displaystyle\hskip 5.69054pt+\sum_{j\in C_{-}^{-i}}\left(\frac{(A_{ij}-b^{2}f(n))^{2}}{2}-\frac{(A_{ij}-abf(n))^{2}}{2}\right)+\log\left(\frac{\rho}{1-\rho}\right)$ $\displaystyle=(a^{2}-ab)f(n)\sum_{j\in C_{+}^{-i}}A_{ij}+(ab-b^{2})f(n)\sum_{j\in C_{-}^{-i}}A_{ij}$ $\displaystyle\hskip 5.69054pt+\frac{f^{2}(n)}{2}(|C_{+}^{-i}|(a^{2}b^{2}-a^{4})+|C_{-}^{-i}|(b^{4}-a^{2}b^{2}))+\log\left(\frac{\rho}{1-\rho}\right)$ $\displaystyle=(a-b)\sqrt{\frac{\log n}{n}}\left(a\sum_{j\in C_{+}^{-i}}A_{ij}+b\sum_{j\in C_{-}^{-i}}A_{ij}\right)$ $\displaystyle\hskip 5.69054pt+\frac{\log n}{2n}(|C_{+}^{-i}|(a^{2}b^{2}-a^{4})+|C_{-}^{-i}|(b^{4}-a^{2}b^{2}))+\log\left(\frac{\rho}{1-\rho}\right).$ (17) $$

Conditioned on $E$, simplifying a term from ([D.1](#A4.Ex98))

$$ $\displaystyle\frac{\log n}{2n}(|C_{+}^{-i}|$ $\displaystyle(a^{2}b^{2}-a^{4})+|C_{-}^{-i}|(b^{4}-a^{2}b^{2}))$ $\displaystyle=\frac{\log n}{2n}\left(\rho n(a^{2}b^{2}-a^{4})+(1-\rho)n\cdot(b^{4}-a^{2}b^{2})\right)+o(1)$ $\displaystyle=\left(\rho(a^{2}b^{2}-a^{4})+(1-\rho)(b^{4}-a^{2}b^{2})\right)\log n/2+o(1)$ $\displaystyle=\gamma+o(1).$ $$

Therefore, substituting this in ([D.1](#A4.Ex98)), we obtain that for any $i\in[n]$:

$$ $\displaystyle z^{*}_{i}$ $\displaystyle=(a-b)\sqrt{\frac{\log n}{n}}\left(a\sum_{j\in C_{+}^{-i}}A_{ij}+b\sum_{j\in C_{-}^{-i}}A_{ij}\right)+\gamma+\log\left(\frac{\rho}{1-\rho}\right)+o(1).$ $\displaystyle=(a-b)\sqrt{\frac{\log n}{n}}A_{i\cdot}v^{*}+\gamma+\log\left(\frac{\rho}{1-\rho}\right)+o(1).$ $$

Writing the same for all $i\in[n]$ in vector notation, we obtain

$$ $z^{*}=(a-b)\sqrt{\frac{\log n}{n}}Av^{*}+\left(\gamma+\log\left(\frac{\rho}{1-\rho}\right)\right)\mathbf{1}_{n}+o(1).\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

Roughly speaking, these genie scores in absolute value are on a $\log n$ scale for both $\mathsf{ROS}$ and $\mathsf{SBM}$. This is when the exact recovery becomes statistically possible and explains the scaling choices for both models. We next analyze the distribution of the genie scores from two different communities (still without side information).

###### Lemma D.2 .

Fix $\rho\in(0,1)$ and $a,b\in\mathbb{R}$ such that $\max\{|a|,|b|\}>0$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$. Condition on $\sigma^{*}$ such that the event $E$ from ([11](#A2.E11)) holds. Then $z^{*}_{i}\sim\mathcal{N}(\mu_{n}^{+},\sigma^{2}_{n})$ and $z^{*}_{i}\sim\mathcal{N}(\mu_{n}^{-},\sigma^{2}_{n})$ for any $i\in C_{+}$ and $i\in C_{-}$ respectively, where

$$ $\mu_{n}^{+}=\frac{\Psi\log n}{2}+O(1),\quad\mu_{n}^{-}=\frac{-\Psi\log n}{2}+O(1),\hskip 5.69054pt\text{and }\hskip 5.69054pt\sigma^{2}_{n}=\Psi\log n+o(1).$ $$

###### Proof.

Recall that, by Lemma [D.1](#A4.Thmlemma1), we have $z^{*}_{i}=(a-b)\sqrt{\frac{\log n}{n}}A_{i\cdot}v^{*}+\gamma+\log\left(\frac{\rho}{1-\rho}\right)+o(1)$ for any $i\in[n]$. Since the entries of $A_{i\cdot}$ are independent Gaussians conditioned on $\sigma^{*}$, it follows that $z^{*}_{i}$ is a Gaussian random variable. Now consider the following calculation of the mean and variance of $z^{*}_{i}$, in which we substitute the community sizes under the event $E$. For any $i\in C_{+}$,

$$ $\displaystyle\mu_{n}^{+}$ $\displaystyle=\operatorname*{{\mathbb{E}}}[z^{*}_{i}]=(a-b)\sqrt{\frac{\log n}{n}}\operatorname*{{\mathbb{E}}}[A_{i\cdot}v^{*}]+\gamma+\log\left(\frac{\rho}{1-\rho}\right)+o(1)$ $\displaystyle=(a-b)\sqrt{\frac{\log n}{n}}\left(\sum_{i\in C_{+}^{-i}}a^{2}\sqrt{\frac{\log n}{n}}v^{*}_{i}+\sum_{i\in C_{-}}ab\sqrt{\frac{\log n}{n}}v^{*}_{i}\right)+\gamma+O(1)$ $\displaystyle=(a-b)\sqrt{\frac{\log n}{n}}\left(|C_{+}^{-i}|a^{2}\sqrt{\frac{\log n}{n}}\cdot a+|C_{-}|ab\sqrt{\frac{\log n}{n}}\cdot b\right)+\gamma+O(1)$ $\displaystyle=(a-b)\frac{\log n}{n}\left(\rho na^{3}+(1-\rho)n\cdot ab^{2}\right)+\gamma+O(1)$ $\displaystyle=(a-b)(\rho a^{3}+(1-\rho)ab^{2})\log n+\left(\rho(a^{2}b^{2}-a^{4})+(1-\rho)(b^{4}-a^{2}b^{2})\right)\log n/2+O(1)$ $\displaystyle=\left(\rho\left(a^{3}(a-b)+\frac{a^{2}b^{2}-a^{4}}{2}\right)+(1-\rho)\left(ab^{2}(a-b)+\frac{b^{4}-a^{2}b^{2}}{2}\right)\right)\log n+O(1)$ $\displaystyle=\left(\frac{\rho a^{2}}{2}(a-b)^{2}+\frac{(1-\rho)b^{2}}{2}(a-b)^{2}\right)\log n+O(1)=\frac{\Psi}{2}\log n+O(1).$ $$

A very similar calculation gives that for any $i\in C_{-}$,

$$ $\displaystyle\mu_{n}^{-}$ $\displaystyle=(a-b)\sqrt{\frac{\log n}{n}}\operatorname*{{\mathbb{E}}}[A_{i\cdot}v^{*}]+\gamma+\log\left(\frac{\rho}{1-\rho}\right)+o(1)$ $\displaystyle=(a-b)\sqrt{\frac{\log n}{n}}\left(\sum_{i\in C_{+}}abf\sqrt{\frac{\log n}{n}}v^{*}_{i}+\sum_{i\in C_{-}^{-i}}b^{2}\sqrt{\frac{\log n}{n}}v^{*}_{i}\right)+\gamma+O(1)$ $\displaystyle=(a-b)(\rho a^{2}b+(1-\rho)b^{3})\log n+\left(\rho(a^{2}b^{2}-a^{4})+(1-\rho)(b^{4}-a^{2}b^{2})\right)\log n/2+O(1)$ $\displaystyle=\left(\rho\left(a^{2}b(a-b)+\frac{a^{2}b^{2}-a^{4}}{2}\right)+(1-\rho)\left(b^{3}(a-b)+\frac{b^{4}-a^{2}b^{2}}{2}\right)\right)\log n+O(1)$ $\displaystyle=-\left(\frac{\rho a^{2}}{2}(a-b)^{2}+\frac{(1-\rho)b^{2}}{2}(a-b)^{2}\right)\log n+O(1)=-\frac{\Psi}{2}\log n+O(1).$ $$

Finally, for $i\in[n]$ (irrespective of its community assignment)

$$ $\displaystyle\sigma_{n}^{2}$ $\displaystyle=\frac{(a-b)^{2}\log n}{n}\cdot\textnormal{Var}[A_{i\cdot}v^{*}]=\frac{(a-b)^{2}\log n}{n}\left(\sum_{i\in C_{+}^{-i}}(v^{*}_{i})^{2}+\sum_{i\in C_{-}^{-i}}(v^{*}_{i})^{2}\right)$ $\displaystyle=\frac{(a-b)^{2}\log n}{n}\left(a^{2}|C_{+}^{-i}|+b^{2}|C_{-}^{-i}|\right)=\frac{(a-b)^{2}\log n}{n}(\rho na^{2}+(1-\rho)nb^{2})+o(1)$ $\displaystyle=(a-b)^{2}(\rho a^{2}+(1-\rho)b^{2})\log n+o(1)=\Psi\log n+o(1).\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

Finally, we show that the genie scores succeed with $\Omega(\log n)$ margin above the threshold.

###### Lemma D.3 .

Fix $\rho\in(0,1)$ and $a,b\in\mathbb{R}$ such that $\max\{|a|,|b|\}>0$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$. Condition on a labeling $\sigma^{*}$ such that the event $E$ from ([11](#A2.E11)) holds. Optionally, let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$, where $\epsilon_{n}$ and $\alpha_{n}$ be parameterized as ([1](#S1.E1)) for $\beta\geq 0$. Let $z^{*}$ be the genie score vector for the corresponding model of side information. Then above the information-theoretic threshold as specified in Table [1](#S3.T1), there exists some constant $\delta:=\delta(\rho,a,b,\beta)>0$ such that, for any $i\in[n]$:

$$ $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)=o(n^{-1}).$ $$

###### Proof.

Let $Z_{+}\sim\mathcal{N}(\mu_{n}^{+},\sigma_{n}^{2})$ and $Z_{-}\sim\mathcal{N}(\mu_{n}^{-},\sigma^{2}_{n})$, where $\mu_{n}^{+},\mu_{n}^{-},$ and $\sigma^{2}_{n}$ are defined in Lemma [D.2](#A4.Thmlemma2) and $Z\sim\mathcal{N}(0,1)$. Moreover, $Z\sim\mathcal{N}(0,1)$ denotes the standard normal random variable.
No side information: In this case, by Lemma [D.2](#A4.Thmlemma2), for any $i\in C_{+}$:

$$ $\displaystyle\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)$ $\displaystyle=\operatorname*{\mathbb{P}}(Z_{+}<\delta\log n)=\operatorname*{\mathbb{P}}\left(Z<\frac{-\mu_{n}^{+}+\delta\log n}{\sigma_{n}}\right)=\operatorname*{\mathbb{P}}\left(Z>\frac{\mu_{n}^{+}-\delta\log n}{\sigma_{n}}\right)$ $\displaystyle=\operatorname*{\mathbb{P}}\left(Z>\frac{(\Psi/2-\delta+o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)$ $$

Similarly, for any $i\in C_{-}$ by Lemma [D.2](#A4.Thmlemma2),

$$ $\displaystyle\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)$ $\displaystyle=\operatorname*{\mathbb{P}}(-Z_{-}<\delta\log n)=\operatorname*{\mathbb{P}}(Z_{-}>-\delta\log n)=\operatorname*{\mathbb{P}}\left(Z>\frac{-\mu_{n}^{-}-\delta\log n}{\sigma_{n}}\right)$ $\displaystyle=\operatorname*{\mathbb{P}}\left(Z>\frac{(\Psi/2-\delta+o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right).$ $$

Therefore, it suffices to show that the above Gaussian tail is bounded by $o(n^{-1})$ for some $\delta>0$. Towards this, for any constant $\delta<\Psi/2$, observe that

$$ $\frac{(\Psi/2-\delta+o(1))\log n}{\sqrt{\Psi\log n+o(1)}}=\Omega(\sqrt{\log n})=\omega(1).$ $$

Applying Lemma [A.1](#A1.Thmlemma1) then implies that, for a sufficiently large $n$:

$$ $\displaystyle\operatorname*{\mathbb{P}}$ $\displaystyle\left(Z>\frac{(\Psi/2-\delta+o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)\leq\exp\left(-\frac{(\Psi/2-\delta+o(1))^{2}\log n)}{2(\Psi+o(1))}\right)$ $\displaystyle=\exp\left(-\frac{(\Psi-2\delta+o(1))^{2}\log n)}{8\Psi}\right)=n^{-\frac{(\Psi-2\delta)^{2}}{8\Psi}+o(1)}.$ $$

Above the information theoretic threshold, we have $\Psi/8>1$ from Table [1](#S3.T1). Thus, we can choose $\delta(\rho,a,b)>0$ such that $\frac{(\Psi-2\delta)^{2}}{8\Psi}>1$. This finally implies for any $i\in[n]$:

$$ $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)=o(n^{-1}).$ $$

$\mathsf{BEC}$ side information: We now let $z^{*}$ be the genie score vector with $\mathsf{BEC}$ side information and switch to $z^{\prime}$ to denote the Genie score vector without side information. By Lemma [2.10](#S2.Thmlemma10),

$$ $\displaystyle\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)$ $\displaystyle=\operatorname*{\mathbb{P}}(y_{i}=0)\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n\mid y_{i}=0)+\operatorname*{\mathbb{P}}(y_{i}=\sigma^{*}_{i})\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n\mid y_{i}=\sigma^{*}_{i})$ $\displaystyle=\epsilon_{n}\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z_{i}^{\prime}<\delta\log n)\leq n^{-\beta}\cdot n^{-\frac{(\Psi-2\delta)^{2}}{8\Psi}+o(1)}$ (from the previous case) $\displaystyle=n^{-\frac{(\Psi-2\delta)^{2}}{8\Psi}-\beta+o(1)}$ $$

When above the information-theoretic threshold, from Table [1](#S3.T1), we have $\Psi/8>1-\beta.$ Hence we can choose $\delta:=\delta(\rho,a,b,\beta)$ sufficiently small such that $\frac{(\Psi-2\delta)^{2}}{8\Psi}>1-\beta$. Therefore,

$$ $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)=o(n^{-(1-\beta)-\beta})=o(n^{-1}).$ $$

$\mathsf{BSC}$ side information: Let $z^{*}$ and $z^{\prime}$ denote the genie score vector for $\mathsf{BSC}$ and no side information respectively. By Lemma [2.10](#S2.Thmlemma10), recall that $z^{*}=z^{\prime}+\log(\frac{1-\alpha_{n}}{\alpha_{n}})y=z^{\prime}+(\beta\log n)y$, as $\log(\frac{1-\alpha_{n}}{\alpha_{n}})=\log(n^{\beta})=\beta\log n$. Now for any constant $\delta>0$ and any $i\in C_{+}$,

$$ $\displaystyle\operatorname*{\mathbb{P}}$ $\displaystyle(\sigma^{*}_{i}z_{i}^{*}<\delta\log n)=\operatorname*{\mathbb{P}}(y_{i}=+1)\operatorname*{\mathbb{P}}(z_{i}^{*}<\delta\log n\mid y_{i}=+1)+\operatorname*{\mathbb{P}}(y_{i}=-1)\operatorname*{\mathbb{P}}(z_{i}^{*}<\delta\log n\mid y_{i}=-1)$ $\displaystyle=(1-\alpha_{n})\operatorname*{\mathbb{P}}\left(z_{i}^{\prime}+\beta\log n<\delta\log n\right)+\alpha_{n}\operatorname*{\mathbb{P}}\left(z_{i}^{\prime}-\beta\log n<\delta\log n\right)$ $\displaystyle=(1-\alpha_{n})\operatorname*{\mathbb{P}}\left(Z_{+}+\beta\log n<\delta\log n\right)+\alpha_{n}\operatorname*{\mathbb{P}}\left(Z_{+}-\beta\log n<\delta\log n\right)$ $\displaystyle\leq\operatorname*{\mathbb{P}}\left(Z_{+}+\beta\log n<\delta\log n\right)+n^{-\beta}\cdot\operatorname*{\mathbb{P}}\left(Z_{+}-\beta\log n<\delta\log n\right)$ $\displaystyle=\operatorname*{\mathbb{P}}\left(Z>\frac{(\Psi/2+\beta-\delta-o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)+n^{-\beta}\cdot\operatorname*{\mathbb{P}}\left(Z>\frac{(\Psi/2-\beta-\delta-o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)$ $\displaystyle:=P_{1}(\delta)+P_{2}(\delta).$ $$

Similarly, for any $i\in C_{-}$ and constant $\delta>0$

$$ $\displaystyle\operatorname*{\mathbb{P}}$ $\displaystyle(\sigma^{*}_{i}z_{i}^{*}<\delta\log n)$ $\displaystyle=\operatorname*{\mathbb{P}}(y_{i}=-1)\operatorname*{\mathbb{P}}(-z_{i}^{*}<\delta\log n\mid y_{i}=-1)+\operatorname*{\mathbb{P}}(y_{i}=+1)\operatorname*{\mathbb{P}}(-z_{i}^{*}<\delta\log n\mid y_{i}=+1)$ $\displaystyle=\operatorname*{\mathbb{P}}(y_{i}=-1)\operatorname*{\mathbb{P}}(z_{i}^{*}>-\delta\log n\mid y_{i}=-1)+\operatorname*{\mathbb{P}}(y_{i}=+1)\operatorname*{\mathbb{P}}(z_{i}^{*}>-\delta\log n\mid y_{i}=+1)$ $\displaystyle=(1-\alpha_{n})\operatorname*{\mathbb{P}}\left(z_{i}^{\prime}-\beta\log n>-\delta\log n\right)+\alpha_{n}\operatorname*{\mathbb{P}}\left(z_{i}^{\prime}+\beta\log n>-\delta\log n\right)$ $\displaystyle=(1-\alpha_{n})\operatorname*{\mathbb{P}}\left(Z_{-}-\beta\log n>-\delta\log n\right)+\alpha_{n}\operatorname*{\mathbb{P}}\left(Z_{-}+\beta\log n>-\delta\log n\right)$ $\displaystyle\leq\operatorname*{\mathbb{P}}\left(Z_{-}>(\beta-\delta)\log n\right)+n^{-\beta}\cdot\operatorname*{\mathbb{P}}\left(Z_{-}>-(\beta+\delta)\log n\right)$ $\displaystyle=\operatorname*{\mathbb{P}}\left(Z>\frac{(\Psi/2+\beta-\delta-o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)+n^{-\beta}\cdot\operatorname*{\mathbb{P}}\left(Z>\frac{(\Psi/2-\beta-\delta-o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)$ $\displaystyle=P_{1}(\delta)+P_{2}(\delta).$ $$

From the above two equations, it suffices to show that whenever the parameters satisfy the IT threshold conditions, there exists $\delta>0$ such that $P_{1}(\delta)=o(n^{-1})$ and $P_{2}(\delta)=o(n^{-1})$.
We first show that there exists $\delta_{1}>0$ sufficiently small such that $P_{1}(\delta_{1})=o(n^{-1})$. It is straightforward to see that above the information theoretic limit, we always have $\frac{(\Psi+2\beta)^{2}}{8\Psi}>1$. Moreover, for any constant $\delta<\Psi/2+\beta$

$$ $\frac{(\Psi/2+\beta-\delta-o(1))\log n}{\sqrt{\Psi\log n+o(1)}}=\Omega(\sqrt{\log n})=\omega(1).$ $$

Applying the Gaussian tail bound from Lemma [A.1](#A1.Thmlemma1),

$$ $\displaystyle P_{1}(\delta)$ $\displaystyle=\operatorname*{\mathbb{P}}\left(Z>\frac{(\Psi/2+\beta-\delta-o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)\leq\exp\left(\frac{-(\nicefrac{{\Psi}}{{2}}+\beta-\delta-o(1))^{2}\log n}{2(\Psi+o(1))}\right)$ $\displaystyle=\exp\left(\frac{-(\Psi+2\beta-2\delta-o(1))^{2}\log n}{8\Psi}\right)=n^{\frac{-(\Psi+2\beta-2\delta-o(1))^{2}}{8\Psi}}.$ $$

Above the IT threshold, since $\nicefrac{{(\Psi+2\beta)^{2}}}{{8\Psi}}>1$, one can choose $\delta_{1}(\rho,a,b)$ sufficiently small such that $P_{1}(\delta_{1})=o(n^{-1}).$

We now show that there exists $\delta_{2}>0$ small enough such that $P_{2}(\delta_{2})=o(n^{-1})$.

Case 1: ($\Psi\leq 2\beta$). In this case, the threshold is governed by $\beta>1$. For any $\delta_{2}>0$

$$ $P_{2}(\delta_{2})=n^{-\beta}\cdot\operatorname*{\mathbb{P}}\left(Z>\frac{\Psi/2-\beta-\delta_{2}-o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)\leq n^{-\beta}=o(n^{-1}).$ $$

Case 2: ($\Psi>2\beta$). In this case, $\Psi/2>\beta$ and for any constant
$\delta<\nicefrac{{\Psi}}{{2}}-\beta$

$$ $\frac{(\Psi/2-\beta-\delta-o(1))\log n}{\sqrt{\Psi\log n+o(1)}}=\Omega(\sqrt{\log n})=\omega(1).$ $$

Using the Gaussian tail bound from Lemma [A.1](#A1.Thmlemma1), for any constant $\delta<\Psi/2-\beta$

$$ $\displaystyle P_{2}(\delta)$ $\displaystyle=n^{-\beta}\cdot\operatorname*{\mathbb{P}}\left(Z>\frac{(\Psi/2-\beta-\delta-o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)\leq n^{-\beta}\exp\left(\frac{-(\nicefrac{{\Psi}}{{2}}-\beta-\delta-o(1))^{2}\log n}{2\Psi+o(1)}\right)$ $\displaystyle=n^{-\beta}\exp\left(\frac{-(\Psi-2\beta-2\delta-o(1))^{2}\log n}{8\Psi}\right)=n^{-\beta}n^{\frac{-(\Psi-2\beta-2\delta-o(1))^{2}}{8\Psi}}.$ $$

Finally, above the IT threshold, the condition $\nicefrac{{(\Psi+2\beta)^{2}}}{{8\Psi}}>1$ also implies $\nicefrac{{(\Psi-2\beta)^{2}}}{{8\Psi}}>1-\beta$. Therefore, one can choose $\delta_{2}(\rho,a,b,\beta)>0$ sufficiently small such that $\nicefrac{{(\Psi-2\beta-2\delta_{2})^{2}}}{{8\Psi}}>1-\beta$, which gives us

$$ $P_{2}(\delta_{2})\leq n^{-\beta}o(n^{-(1-\beta)})=o(n^{-1}).$ $$

Finally, choosing $\delta:=\min\{\delta_{1},\delta_{2}\}$, we have that both $P_{1}(\delta_{1})$ and $P_{2}(\delta_{2})$ are $o(n^{-1})$.

###### Lemma D.1 .

###### Proof.

###### Lemma D.2 .

###### Proof.

###### Lemma D.3 .

###### Proof.

### D.2 Failure of genie-aided estimation below the threshold

Our proof involves similar techniques from []. We first need a helper lemma along the way.

###### Lemma D.4 .

Consider $\rho\in(0,1)$ and $a,b\in\mathbb{R}$ such that $\max\{|a|,|b|\}>0$ Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$. Condition on $\sigma^{*}$ such that the event $E$ holds. Fix any set $T\subset C_{+}$ or $T\subset C_{-}$ such that $|T|=O(n/\log^{10}n)$. Then for any $i\in[n]$, let us define $Y_{i}=\sqrt{\frac{\log n}{n}}\sum_{j\in T}A_{ij}$.

$$ $\operatorname*{\mathbb{P}}(\forall i\in[n]:|Y_{i}|\leq 1)\geq 1-O(n^{-3}).$ $$

We show the proof of this lemma at the end of this subsection. We now show our desired claim; below the IT threshold, there is some $i\in[n]$, such that its genie score has the incorrect sign. Formally,

###### Lemma D.5 .

Fix $\rho\in(0,1)$ and $a,b$ such that $\max\{|a|,|b|\}>0$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$ and condition on $\sigma^{*}$ such that $E$ from ([11](#A2.E11)) holds. For any $\beta\geq 0$, we optionally sample $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$ for $\epsilon_{n}$ and $\alpha_{n}$ parameterized according to ([1](#S1.E1)). Let $z^{*}$ be the genie score vector for the corresponding model. Then below the information-theoretic threshold given by Table [1](#S3.T1),

$$ $\operatorname*{\mathbb{P}}\left(\exists i\in[n]:\sigma^{*}_{i}z^{*}_{i}<-\frac{\log n}{\log\log n}\right)=1-o(1).$ $$

###### Proof.

We have conditioned on $\sigma^{*}$ such that $E$ happens. First consider the following general analysis with or without side information. Fix any set $T\subseteq C_{+}$ such that $|T|={\left\lceil{\frac{\rho n}{\log^{10}n}}\right\rceil}.$ For each $i\in T$, define the following random variables.

$$ $X_{i}:=a(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in T}A_{ij}\quad\text{and}\quad R_{i}:=z^{*}_{i}-X_{i}.$ $$

Essentially, we defined $R_{i}$ to be the genie score for the $i^{\mathrm{th}}$ label but without the contribution of $\{A_{ij}\}_{j\in T}$. By using Lemma [D.4](#A4.Thmlemma4) for the set $T$ since it is a fixed set with $|T|=\Theta(\nicefrac{{n}}{{\log^{10}n}})$,

$$ $\operatorname*{\mathbb{P}}\left(\forall i\in[n]:|X_{i}|\leq|a(a-b)|\right)\geq 1-O(n^{-3}).$ (18) $$

This is simple to see using the genie score form in Lemma [D.1](#A4.Thmlemma1) when no side information is present. And, also in the cases of either type of side information, Lemma [2.10](#S2.Thmlemma10) implies that $z^{*}_{i}$ only depends on $\{A_{ij}:j\in T\}$ through the sum $X_{i}$. Therefore, subtracting off $X_{i}$ from $z^{*}_{i}$, the resultant random variable $R_{i}$ only depends on $\{A_{ij}:j\in[n]\setminus T\}$, and $y_{i}$ when considering side information. Observe that for any $i,\ell\in T$ with $i\neq\ell$, the sets $\{A_{ij}:j\in[n]\setminus T\}$ and $\{A_{\ell j}:j\in[n]\setminus T\}$ are disjoint. Therefore, condition on $\sigma^{*}$, the random variables $\{R_{i}:i\in T\}$ are independent. This follows from conditional independence of the entries of $A$, and the side information labels $y_{i}$ when they are present.

Therefore, if we manage to show that below the IT threshold for the corresponding model of side information given by Table [1](#S3.T1), there is some constant $\varepsilon>0$ such that

$$ $\operatorname*{\mathbb{P}}\left(\exists i\in[n]:R_{i}\leq\frac{-2\log n}{\log\log n}\right)\geq n^{-1+\varepsilon},$ (19) $$

then the following analysis directly implies the lemma.

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\exists i\in[n]:\sigma^{*}_{i}z^{*}_{i}<\frac{-\log n}{\log\log n}\right)$ $\displaystyle\geq\operatorname*{\mathbb{P}}\left(\exists i\in T:z^{*}_{i}<\frac{-\log n}{\log\log n}\right)$ $\displaystyle=\operatorname*{\mathbb{P}}\left(\exists i\in T:R_{i}+X_{i}<\frac{-\log n}{\log\log n}\right)$ $\displaystyle\overset{\scriptstyle(a)}{\geq}\operatorname*{\mathbb{P}}\left(\exists i\in T:R_{i}<\frac{-\log n}{\log\log n}+O(1)\right)-o(1)$ $\displaystyle\geq\operatorname*{\mathbb{P}}\left(\exists i\in T:R_{i}<\frac{-2\log n}{\log\log n}\right)-o(1)$ $\displaystyle=1-\operatorname*{\mathbb{P}}\left(\forall i\in T:R_{i}\geq-\frac{2\log n}{\log\log n}\right)-o(1)$ $\displaystyle=1-\prod_{i\in T}\left(1-\operatorname*{\mathbb{P}}\left(R_{i}<-\frac{2\log n}{\log\log n}\right)\right)-o(1)$ $\displaystyle\overset{\scriptstyle(b)}{\geq}1-(1-n^{-1+\varepsilon})^{|T|}-o(1)$ $\displaystyle\overset{\scriptstyle(c)}{\geq}1-\exp(-|T|n^{-1+\varepsilon})-o(1)$ $\displaystyle=1-\exp(-n^{1-1+\varepsilon+o(1)})-o(1)=1-\exp(-n^{\varepsilon+o(1)})$ $\displaystyle=1-o(1),$ $$

where $(a)$ follows from ([18](#A4.E18)), $(b)$ from ([18](#A4.E18)), and $(c)$ by using the fact that $(1+x)\leq e^{x}$ for all $x\in\mathbb{R}$. We now show $\eqref{eq:ros-lb-1}$ holds in each case.
No side information: By Lemma [D.1](#A4.Thmlemma1) and the way we defined $R_{i}$, for any $i\in T$

$$ $\displaystyle R_{i}:=$ $\displaystyle(a-b)\sqrt{\frac{\log n}{n}}\left(a\sum_{j\in C_{+}\setminus T}A_{ij}+b\sum_{j\in C_{-}}A_{ij}\right)$ $\displaystyle\hskip 5.69054pt+\frac{\log n}{2n}(|C_{+}^{-i}|(a^{2}b^{2}-a^{4})+|C_{-}|(b^{4}-a^{2}b^{2}))+\log\left(\frac{\rho}{1-\rho}\right).$ $$

Note that each $R_{i}$ is a Gaussian random variable and follows the same distribution. Let it be denoted by denote by $R\sim\mathcal{N}(\mu,\sigma^{2})$. We follow exactly the same steps as in Lemma [D.1](#A4.Thmlemma1) and derive

$$ $\mu=\frac{\Psi\log n}{2}+O(1)\quad\text{and}\quad\sigma^{2}=\Psi\log n+o(1).$ $$

Letting $Z\sim\mathcal{N}(0,1)$, for every $i\in T$

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(R_{i}<-\frac{2\log n}{\log\log n}\right)$ $\displaystyle=\operatorname*{\mathbb{P}}\left(R<-\frac{2\log n}{\log\log n}\right)=\operatorname*{\mathbb{P}}\left(Z<\frac{-\mu-\frac{2\log n}{\log\log n}}{\sigma}\right)$ $\displaystyle=\operatorname*{\mathbb{P}}\left(Z<-\frac{(\Psi+o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)=\operatorname*{\mathbb{P}}\left(Z>\frac{(\Psi+o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)$ $\displaystyle\geq\exp\left(-\frac{(\frac{\Psi}{2}+o(1))^{2}\log n}{2(\Psi+o(1))}\right)$ (applying Lemma A.1 ) $\displaystyle=n^{-\frac{\Psi}{8}+o(1)}.$ $$

Recall from Table [1](#S3.T1), below the IT threshold means $\Psi/8<1$, thus, there exists $\varepsilon>0$ sufficiently small such that

$$ $\operatorname*{\mathbb{P}}\left(R_{i}<-\frac{2\log n}{\log\log n}\right)\geq n^{-1+\varepsilon}.$ $$

$\mathsf{BEC}$ side information: First note that the definition of $R_{i}$ is such that $R_{i}=+\infty$ when $y_{i}=+1$ or otherwise $R_{i}$ has the same distribution as $R$. Therefore,

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(R_{i}<-\frac{2\log n}{\log\log n}\right)$ $\displaystyle=\operatorname*{\mathbb{P}}(y_{i}=0)\operatorname*{\mathbb{P}}\left(R_{i}<-\frac{2\log n}{\log\log n}\mid y_{i}=0\right).$ $\displaystyle=n^{-\beta}\operatorname*{\mathbb{P}}\left(R_{i}<-\frac{2\log n}{\log\log n}\right)$ $\displaystyle\geq n^{-\beta}\cdot n^{-\frac{\Psi}{8}+o(1)}$ $\displaystyle=n^{-\frac{\Psi}{8}-\beta+o(1)}.$ $$

Below the IT threshold, we have $\frac{\Psi}{8}+\beta<1$ from Table [1](#S3.T1). Hence there exists $\varepsilon>0$ small enough such that

$$ $\operatorname*{\mathbb{P}}\left(R_{i}<-\frac{2\log n}{\log\log n}\right)\geq n^{-1+\varepsilon}.$ $$

$\mathsf{BSC}$ side information: Recall Lemma [2.10](#S2.Thmlemma10) as to how genie score changes under $\mathsf{BSC}$ side information.
We consider the two cases $\Psi>2\beta$ and $\Psi\leq 2\beta$ separately.

- •
When $\Psi>2\beta$: From the way we have defined $R_{i}$,
$\displaystyle\operatorname*{\mathbb{P}}$
$\displaystyle\left(R_{i}<-\frac{2\log n}{\log\log n}\right)\geq\operatorname*{\mathbb{P}}(y_{i}=+1)\operatorname*{\mathbb{P}}\left(R_{i}<-\frac{2\log n}{\log\log n}\mid y_{i}=+1\right)$
$\displaystyle\geq(1-\alpha_{n})\operatorname*{\mathbb{P}}\left(R+\beta\log n<-\frac{2\log n}{\log\log n}\right)\geq 0.5\operatorname*{\mathbb{P}}\left(Z<\frac{-\mu-\beta\log n-\frac{2\log n}{\log\log n}}{\sigma}\right)$
$\displaystyle=0.5\operatorname*{\mathbb{P}}\left(Z<-\frac{(\frac{\Psi}{2}+\beta-o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)=0.5\operatorname*{\mathbb{P}}\left(Z<-\frac{(\frac{\Psi}{2}+\beta-o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)$
$\displaystyle=0.5\operatorname*{\mathbb{P}}\left(Z>\frac{(\frac{\Psi}{2}+\beta-o(1))\log n}{\sqrt{\Psi\log n+o(1)}}\right)$
$\displaystyle\geq n^{-o(1)}\exp\left(-\frac{(\frac{\Psi}{2}+\beta-o(1))^{2}\log n}{2\Psi}\right)$
(applying Lemma [A.1](#A1.Thmlemma1))
$\displaystyle\geq n^{-\frac{(\Psi+2\beta)^{2}}{8\Psi}-o(1)}$
Recall from Table [1](#S3.T1), below the IT threshold means $\frac{(\Psi+2\beta)^{2}}{8\Psi}<1$. Therefore, there exists $\delta>0$ small enough such that
$\operatorname*{\mathbb{P}}\left(R_{i}<-\frac{2\log n}{\log\log n}\right)\geq n^{-1+\varepsilon}.$
- •
When $\Psi\leq 2\beta$: From the definition of $R_{i}$,
$\displaystyle\operatorname*{\mathbb{P}}\left(R_{i}<-\frac{2\log n}{\log\log n}\right)$
$\displaystyle\geq\operatorname*{\mathbb{P}}(y_{i}=-1)\operatorname*{\mathbb{P}}\left(R_{i}<-\frac{2\log n}{\log\log n}\mid y_{i}=-1\right)$
$\displaystyle\geq\alpha_{n}\,\operatorname*{\mathbb{P}}\left(R-\beta\log n<-\frac{2\log n}{\log\log n}\right)$
$\displaystyle\geq n^{-\beta+o(1)}\operatorname*{\mathbb{P}}\left(Z<\frac{-\mu+\beta\log n-\frac{2\log n}{\log\log n}}{\sigma}\right)$
$\displaystyle=n^{-\beta+o(1)}\operatorname*{\mathbb{P}}\left(Z<\frac{(\frac{-\Psi}{2}+\beta)\log n-O(1)}{\sqrt{\Psi\log n+o(1)}}\right)$
$\displaystyle\geq n^{-\beta+o(1)}\operatorname*{\mathbb{P}}\left(Z>O(\nicefrac{{1}}{{\sqrt{\log n}}})\right)$
$\displaystyle\geq n^{-\beta+o(1)},$
where in the last step we used that $\operatorname*{\mathbb{P}}(Z>O(\nicefrac{{1}}{{\sqrt{\log n}}}))$ is at least some constant. In this case, by Table [1](#S3.T1), below the IT threshold corresponds to $\beta<1$, and thus, there exists $\varepsilon>0$ sufficiently small such that
$\operatorname*{\mathbb{P}}\left(R_{i}<-\frac{2\log n}{\log\log n}\right)\geq n^{-1+\varepsilon}.\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$

We now return to the deferred proof.

###### Proof of Lemma D.4 .

First of all, observe that $Y_{i}$ is a Gaussian random variable. Let us say $Y_{i}\sim\mathcal{N}(\mu_{i},\sigma^{2}_{i})$, for some $\mu_{i}$ and $\sigma^{2}_{i}>0$. Applying Lemma [A.1](#A1.Thmlemma1) for any $Y_{i}$ (after renormalizing) yields

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\frac{|Y_{i}-\mu_{i}|}{\sigma_{i}}\geq 4\sqrt{\log n}\right)\leq e^{-8\log n}=n^{-8}.$ $$

Rearrangement of the terms using the triangle inequality along with a union bound over all $i\in[n]$ gives us

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\exists i\in[n]:|Y_{i}|\geq|\mu_{i}|+4\sigma_{i}\sqrt{\log n}\right)\leq n^{-7}.$ $$

Therefore it simply suffices to show that for every $i\in[n]$, we have $|\mu_{i}|+4\sigma_{i}\sqrt{\log n}\leq 1$. Indeed, we will show that these terms are $o(1)$. To this end, first consider the term $4\sigma_{i}\sqrt{\log n}$ for any $i\in[n]$. Recall that $Y_{i}$ is the sum of at most $|T|$ i.i.d. Gaussian random variables, all with variance 1, scaled by $\sqrt{\nicefrac{{\log n}}{{n}}}$. Therefore,

$$ $\sigma^{2}_{i}\leq\frac{\log n}{n}|T|=O\left(\frac{1}{\log^{9}n}\right)\implies 4\sigma_{i}\sqrt{\log n}=O\left(\frac{1}{\log^{2}n}\right)=o(1),$ $$

where we used $|T|=O(\nicefrac{{n}}{{\log^{10}n}})$. We now show that $|\mu_{i}|=o(1)$ too for all $i\in[n]$, which requires some casework. First consider $T\subset C_{+}$, then for any $i\in C_{+}$:

$$ $|\mu_{i}|\leq\sqrt{\frac{\log n}{n}}|T|a^{2}\sqrt{\frac{\log n}{n}}=O\left(\frac{1}{\log^{9}n}\right),$ $$

Similarly, for any $i\in C_{-}$:

$$ $|\mu_{i}|=\sqrt{\frac{\log n}{n}}|T||ab|\sqrt{\frac{\log n}{n}}=O\left(\frac{1}{\log^{9}n}\right).$ $$

Exactly following the same arguments, we also get the same bounds on $\mu_{i}$ even when $T\subset C_{-}$. Overall, we established that

$$ $\operatorname*{\mathbb{P}}(\forall i\in[n]:|Y_{i}|\leq 1)\geq 1-O(n^{-7}).\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

###### Lemma D.4 .

###### Lemma D.5 .

###### Proof.

###### Proof of Lemma D.4 .

### D.3 Statistical Achievability without Side Information

In this section, we establish the statistical achivability for $\mathsf{ROS}$ when side information is absent. This claim is crucially used in establishing the optimality of the spectral algorithm. Roughly speaking, the spectral algorithm can compute the leading eigenvector of $A$ only up to a global sign flip, and the spectral algorithm must resolve a sign ambiguity in order to compute the correct estimator. The spectral algorithm will keep two candidates (of which one is correct) and choose the one that maximizes the posterior probability. This will establish that the spectral algorithm succeeds as long as the MAP estimator succeeds. Thus, to complete the argument, we also need to show that the MAP estimator succeeds above the IT threshold. When side information is substantial $(\beta>0)$, the Degree-Profiling Algorithm (Algorithm [D.5](#A4.SS5)) succeeds above the threshold, and hence so does the MAP. However, when side information is not present or weak $(\beta=0)$, the Degree-Profiling Algorithm does not apply. Hence, we will directly establish statistical achievability.

###### Lemma D.6 .

Fix $\rho\in(0,1)$ and $a,b\in\mathbb{R}$ such that $\max\{|a|,|b|\}>0$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$. Let $\hat{\sigma}_{\textnormal{\tiny MAP}}$ be the estimator of $\sigma^{*}$. If $\Psi/8>1$, then $\hat{\sigma}_{\textnormal{\tiny MAP}}$ achieves exact recovery.

The proof uses ideas from [, Theorem 3.2]. Rather than showing that $\hat{\sigma}_{\textnormal{\tiny MAP}}$ achieves exact recovery, we will show that the *restricted MLE* achieves exact recovery. Let $\epsilon>0$ be a constant, to be specified. Let $f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^{2}}$ denote the standard Normal PDF. The restricted MLE $\hat{\sigma}:V\to\{\pm 1\}^{n}$ maximizes

$$ $\displaystyle\ell(A,\sigma)$ $\displaystyle\triangleq\sum_{i:\hat{\sigma}(i)=1}\sum_{j:\hat{\sigma}(j)=1}\log\left(f\left(A_{ij}-a^{2}\right)\right)+\sum_{i:\hat{\sigma}(i)=-1}\sum_{j:\hat{\sigma}(j)=-1}\log\left(f\left(A_{ij}-b^{2}\right)\right)$ $\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ +2\sum_{i:\hat{\sigma}(i)=1}\sum_{j:\hat{\sigma}(j)=-1}\log\left(f\left(A_{ij}-ab\right)\right)$ $$

subject to

$$ $\left|C_{+}(\hat{\sigma})-\rho n\right|\leq\epsilon n,$ (20) $$

where $C_{+}(\hat{\sigma})=|\{i:\hat{\sigma}(i)=1\}|$ is the number of indices classified as $+1$ by $\hat{\sigma}$. Thus, the restricted MLE maximizes (twice) the log-likelihood of $A$, subject to having community sizes that are close to their expectations. Note that $\sigma^{*}$ satisfies ([20](#A4.E20)) with high probability.

The case where $\rho=1/2$ and $a=-b$ corresponds to $\mathbb{Z}_{2}$-synchronization after rescaling. In this case, exact recovery requires determining the partition, but not the labeling. Here we may scale the observation matrix, obtaining $A^{\prime}=\frac{1}{a^{2}}\sqrt{\frac{\log n}{n}}A$. Then $A^{\prime}$ has the same distribution as $xx^{\top}+W^{\prime}$, where $x$ has i.i.d. Rademacher entries, and $W^{\prime}$ is a symmetric Gaussian matrix with independent $\mathcal{N}\left(0,\tau^{2}\right)$ entries, where $\tau=\frac{1}{a^{2}}\sqrt{\frac{\log n}{n}}$. In this case, [, Theorem 3.1] gives that $a^{4}/2>1$ is sufficient for exact recovery, which verifies the threshold of $\frac{\Psi}{8}=1$, as $\frac{\Psi}{8}=\frac{(2a)^{2}\cdot a^{2}}{8}=\frac{a^{4}}{2}$.

In all other cases, exact recovery requires recovering the labels of the vertices, not just the partition. Thus, our goal is to show that with high probability, any labeling $\sigma$ which satisfies ([20](#A4.E20)) but does not exactly coincide with $\sigma^{*}$ satisfies $\ell(\sigma)<\ell(\sigma^{*})$.
Similarly to [], we separately consider labelings $\sigma$ with many or few errors relative to $\sigma^{*}$.

###### Lemma D.7 .

Fix $\rho\in(0,1)$ and $a,b\in\mathbb{R}$ such that $\max\{|a|,|b|\}>0$, excluding the case where $\rho=1/2$ and $a=-b$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$. Then there exists $c_{1}>0$ such that with high probability, $\ell(A,\sigma)<\ell(A,\sigma^{*})$ for all $\sigma$ satisfying $1\leq d_{H}(\sigma,\sigma^{*})\leq c_{1}n$.

###### Lemma D.8 .

Fix $\rho\in(0,1)$ and $a,b\in\mathbb{R}$ such that $\max\{|a|,|b|\}>0$, excluding the case where $\rho=1/2$ and $a=-b$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$. Then for any $0<c_{2}<1$, with high probability, $\ell(A,\sigma)<\ell(A,\sigma^{*})$ for all $\sigma$ satisfying ([20](#A4.E20)) and $d_{H}(\sigma,\sigma^{*})\geq c_{2}n$.

###### Proof of Lemma D.6 .

We omit the case where $\rho=1/2$ and $a=-b$, as discussed above.
Let $c_{1}>0$ be the constant from Lemma [D.7](#A4.Thmlemma7), and set $c_{2}$ in Lemma [D.8](#A4.Thmlemma8) to be equal to $c_{1}$. Let $\epsilon$ in ([20](#A4.E20)) be set as $\epsilon=\frac{1}{6}\min\{|2\rho-1|,c_{2}\}$ if $\rho\neq 1/2$, and otherwise the constraint is not needed. Together, Lemmas [D.7](#A4.Thmlemma7) and [D.8](#A4.Thmlemma8) imply that $\ell(A,\sigma)<\ell(A,\sigma^{*})$ for all $\sigma$ satisfying ([20](#A4.E20)) and $d_{H}(\sigma,\sigma^{*})\geq 1$. Finally, Lemma [B.1](#A2.Thmlemma1) implies that $\sigma^{*}$ itself satisfies ([20](#A4.E20)), completing the proof.

We now return to the proofs of Lemmas [D.7](#A4.Thmlemma7)
and [D.8](#A4.Thmlemma8).

###### Proof of Lemma D.7 .

For a given labeling $\sigma$, we bound the difference $\ell(A,\sigma)-\ell(A,\sigma^{*})$. Since $\sigma$ is assumed to satisfy $d_{H}(\sigma,\sigma^{*})\leq cn$, for small $c>0$ we have

$$ $\displaystyle\ell(A,\sigma)-\ell(A,\sigma^{*})$ $\displaystyle\approx 2\sum_{i\in C^{+}:\sigma(i)=-1}\left[\sum_{j\in C_{+}}\log\left(\frac{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-a^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}\right)+\sum_{j\in C_{-}}\log\left(\frac{f\left(A_{ij}-b^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}\right)\right]$ $\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ +2\sum_{i\in C^{-}:\sigma(i)=1}\left[\sum_{j\in C_{+}}\log\left(\frac{f\left(A_{ij}-a^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}\right)+\sum_{j\in C_{-}}\log\left(\frac{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-b^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}\right)\right].$ (21) $$

Bounding the difference of log-likelihoods essentially reduces to bounding the above right hand side. More precisely, we have

$$ $\displaystyle\ell(A,\sigma)-\ell(A,\sigma^{*})$ $\displaystyle=2\sum_{i\in C^{+}:\sigma(i)=-1}\left[\sum_{j\in C_{+}}\log\left(\frac{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-a^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}\right)+\sum_{j\in C_{-}}\log\left(\frac{f\left(A_{ij}-b^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}\right)\right]$ $\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ +2\sum_{i\in C^{-}:\sigma(i)=1}\left[\sum_{j\in C_{+}}\log\left(\frac{f\left(A_{ij}-a^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}\right)+\sum_{j\in C_{-}}\log\left(\frac{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-b^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}\right)\right]$ $\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ +\sum_{i\in C_{+}:\sigma(i)=-1}\sum_{j\in C_{+}:\sigma(i)=-1}\left(\log\left(\frac{f\left(A_{ij}-b^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-a^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}\right)-2\log\left(\frac{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-a^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}\right)\right)$ $\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ +\sum_{i\in C_{-}:\sigma(i)=1}\sum_{j\in C_{-}:\sigma(i)=1}\left(\log\left(\frac{f\left(A_{ij}-a^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-b^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}\right)-2\log\left(\frac{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-b^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}\right)\right)$ (22) $$

Since $f(\cdot)$ is the standard Normal PDF, we have that for any $x_{1},x_{2}\in\mathbb{R}$,

$$ $\log\left(\frac{f(A_{ij}-x_{1})}{f(A_{ij}-x_{2})}\right)=-\frac{1}{2}\left((A_{ij}-x_{1})^{2}-(A_{ij}-x_{2})^{2}\right)=(x_{1}-x_{2})A_{ij}+\frac{1}{2}(x_{2}^{2}-x_{1}^{2}).$ $$

Therefore, the last two lines in ([22](#A4.E22)), which represent the error in the approximation ([21](#A4.E21)), are bounded by

$$ $\displaystyle\left(C\sqrt{\frac{\log n}{n}}\max_{i,j}|A_{ij}|+C^{\prime}\frac{\log n}{n}\right)\left(\left|\{(i,j):\sigma(i)\neq\sigma^{*}(i),\sigma(j)\neq\sigma^{*}(j)\}\right|\right)$ $\displaystyle=\left(C\sqrt{\frac{\log n}{n}}\max_{i,j}|A_{ij}|+C^{\prime}\frac{\log n}{n}\right)\cdot(d_{H}(\sigma,\sigma^{*}))^{2}$ (23) $$

where $C$ and $C^{\prime}$ are functions of $a$ and $b$.

Next, consider $i\in C^{+}$ such that $\sigma(i)=1$. The contribution to the first line of ([22](#A4.E22)) is given by

$$ $\displaystyle\sum_{j\in C_{+}}\log\left(\frac{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-a^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}\right)+\sum_{j\in C_{-}}\log\left(\frac{f\left(A_{ij}-b^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}\right)$ $\displaystyle=2a(b-a)\sqrt{\frac{\log n}{n}}\sum_{j\in C_{+}}A_{ij}+2b(b-a)\sqrt{\frac{\log n}{n}}\sum_{j\in C_{-}}A_{ij}$ $\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ +(|C_{+}|-1)(a^{4}-a^{2}b^{2})\frac{\log n}{n}+|C_{-}|(a^{2}b^{2}-b^{4})\frac{\log n}{n}$ $\displaystyle=-2\left(z_{i}^{\star}-\log\left(\frac{\rho}{1-\rho}\right)\right),$ $$

where the last step is due to Lemma [D.1](#A4.Thmlemma1).

Similarly, consider $i\in C_{-}$ such that $\sigma(i)=-1$. The contribution to the second line of ([22](#A4.E22)) is given by

$$ $\displaystyle\sum_{j\in C_{+}}\log\left(\frac{f\left(A_{ij}-a^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}\right)+\sum_{j\in C_{-}}\log\left(\frac{f\left(A_{ij}-ab\cdot\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-b^{2}\cdot\sqrt{\frac{\log n}{n}}\right)}\right)$ $\displaystyle=2a(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in C_{+}}A_{ij}+2b(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in C_{-}}A_{ij}$ $\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ +|C_{+}|(a^{2}b^{2}-a^{4})\frac{\log n}{n}+(|C_{-}|-1)(b^{4}-a^{2}b^{2})\frac{\log n}{n}$ $\displaystyle=2\left(z_{i}^{\star}-\log\left(\frac{\rho}{1-\rho}\right)\right)$ $$

By Lemma [D.3](#A4.Thmlemma3), we have $\sigma_{i}^{\star}z_{i}^{\star}>\delta\log n$ with probability $1-o(n^{-1})$ where $\delta=\delta(\rho,a,b)>0$. A union bound gives that $\sigma_{i}^{\star}z_{i}^{\star}>\delta\log n$ for all $i\in[n]$ with probability $1-o(1)$ .

Combining these observations along with a ([23](#A4.E23)), we obtain that with probability $1-o(1)$, for all $\sigma$ such that $1\leq d_{H}(\sigma,\sigma^{*})\leq cn$

$$ $\displaystyle\ell(A,\sigma)-\ell(A,\sigma^{*})$ $\displaystyle\leq-2d_{H}(\sigma,\sigma^{*})\cdot\left(\delta\log n-2\log\left(\frac{\rho}{1-\rho}\right)\right)$ $\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ +\left(C\sqrt{\frac{\log n}{n}}\max_{i,j}|A_{ij}|+C^{\prime}\frac{\log n}{n}\right)\cdot(d_{H}(\sigma,\sigma^{*}))^{2}$ $\displaystyle\leq-d_{H}(\sigma,\sigma^{*})\cdot\delta^{\prime}\log n+C^{\prime\prime}\frac{\log n}{n}\cdot(d_{H}(\sigma,\sigma^{*}))^{2}$ $\displaystyle=-d_{H}(\sigma,\sigma^{*})\log(n)\left(\delta^{\prime}-\frac{C^{\prime\prime}d_{H}(\sigma,\sigma^{*})}{n}\right)$ $$

where the second line is due to $\log\left(\frac{\rho}{1-\rho}\right)=\Theta(1)$ and $\mathbb{E}[\max_{i,j}|A_{ij}|]=\theta(\log n)$, taking $\delta^{\prime}=\frac{\delta}{2}$ and $C^{\prime\prime}=2C^{\prime}$. Finally, setting $c_{1}=\frac{\delta^{\prime}}{2C^{\prime\prime}}$ ensures that the right hand side is bounded by

$$ $-d_{H}(\sigma,\sigma^{*})\log(n)\frac{\delta^{\prime}}{2}<0.\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

###### Proof of Lemma D.8 .

Fix $0<c<1$.
Consider a labeling $\sigma$ with $d_{H}(\sigma,\sigma^{*})>c_{2}n$. For this labeling $\sigma$, let $X_{ij}$ denote the contribution of the $(i,j)$ matrix entry to the difference $\ell(A,\sigma)-\ell(A,\sigma^{*})$. For example, if $i,j\in C_{+}$ with $\sigma(i)=1$ and $\sigma(j)=-1$, then the distribution of $X_{ij}$ is the same as the distribution of

$$ $\log\frac{f\left(A_{ij}-ab\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-a^{2}\sqrt{\frac{\log n}{n}}\right)},$ $$

where $A_{ij}\sim N\left(0,a^{2}\sqrt{\frac{\log n}{n}}\right)$.

In order to bound the difference in likelihoods, we employ a Chernoff bound strategy, letting $\lambda>0$.

$$ $\displaystyle\mathbb{P}\left(\ell(A,\sigma)-\ell(A,\sigma^{*})\geq 0\right)$ $\displaystyle=\mathbb{P}\left(\sum_{i=1}^{n}\sum_{j=1}^{n}X_{ij}\geq 0\right)\leq\mathbb{E}\left[\exp\left(\lambda\sum_{i=1}^{n}\sum_{j=i}^{n}X_{ij}\right)\right]$ $\displaystyle=\prod_{i=1}^{n}\prod_{j=i}^{n}\mathbb{E}\left[e^{\lambda X_{ij}}\right].$ (24) $$

In the above case where $i,j\in C_{+}$, $\sigma(i)=1$ and $\sigma(j)=-1$, observe that

$$ $\displaystyle\mathbb{E}\left[e^{\lambda X_{ij}}\right]$ $\displaystyle=\mathbb{E}\left[\left(\frac{f\left(A_{ij}-ab\sqrt{\frac{\log n}{n}}\right)}{f\left(A_{ij}-a^{2}\sqrt{\frac{\log n}{n}}\right)}\right)^{\lambda}\right]=\int_{-\infty}^{\infty}f\left(x-ab\sqrt{\frac{\log n}{n}}\right)^{\lambda}f\left(x-a^{2}\sqrt{\frac{\log n}{n}}\right)^{1-\lambda}dx,$ $$

where the second equality comes from the observation that the density of $A_{ij}$ evaluated at $x$ is $f\left(x-a^{2}\sqrt{\frac{\log n}{n}}\right)$. We set $\lambda=1/2$ and use the fact that the (squared) Hellinger divergence of $P\equiv N(\mu_{1},1)$, $Q\equiv N(\mu_{2},1)$ is given by

$$ $H^{2}(P,Q)=1-\int_{-\infty}^{\infty}\sqrt{P(x)Q(x)}dx=1-e^{-\frac{(\mu_{1}-\mu_{2})^{2}}{8}}.$ (25) $$

It follows that

$$ $\displaystyle\mathbb{E}\left[e^{\frac{1}{2}X_{ij}}\right]$ $\displaystyle=\exp\left\{-\frac{1}{8}\left(a^{2}-ab\right)^{2}\cdot\frac{\log n}{n}\right\}=\exp\left\{-\frac{a^{2}}{8}\left(a-b\right)^{2}\cdot\frac{\log n}{n}\right\}.$ $$

A similar derivation holds for $i,j\in C_{-}$ with $\sigma(i)=-1$ and $\sigma(j)=1$; in that case,

$$ $\displaystyle\mathbb{E}\left[e^{\frac{1}{2}X_{ij}}\right]$ $\displaystyle=\exp\left\{-\frac{b^{2}}{8}\left(a-b\right)^{2}\cdot\frac{\log n}{n}\right\}.$ $$

Also note that in all cases, $\mathbb{E}\left[e^{\frac{1}{2}X_{ij}}\right]\leq 1$, due to the identity ([25](#A4.E25)).

First suppose $\rho\neq 1/2$. Recalling the constraint ([20](#A4.E20)), set $\epsilon=\frac{1}{6}\min\{|2\rho-1|,c_{2}\}>0$. Observe that one community must have at least $c_{2}n/2$ errors, by the pigeonhole principle. Furthermore, the other community must have at least $c_{2}n/3$ errors; otherwise, ([20](#A4.E20)) would be violated as $\epsilon\leq\frac{c}{6}$. When $\rho\neq 1/2$, the larger community necessarily contains at least $(|2\rho-1|-2\epsilon)n$ correctly labeled indices. Thus, when $\rho\neq 1/2$,
([24](#A4.E24)) is bounded by

$$ $\exp\left\{-Cn^{2}\cdot\frac{\min\{a^{2},b^{2}\}}{8}\left(a-b\right)^{2}\cdot\frac{\log n}{n}\right\}=e^{-\Theta(n\log n)}.$ $$

where $C=\frac{c_{2}}{3}\cdot(|2\rho-1|-2\epsilon)$.

Next, suppose $\rho=1/2$. In that case, we only need to treat the case where $a^{2}\neq b^{2}$. Due to the previous observations, there are at least $\binom{c_{2}n/2}{2}=\Theta(n^{2})$ pairs of vertices which are both misclassified; for such a pair $(i,j)$, we obtain

$$ $\displaystyle\mathbb{E}\left[e^{\frac{1}{2}X_{ij}}\right]$ $\displaystyle=\exp\left\{-\frac{1}{8}\left(a^{2}-b^{2}\right)^{2}\cdot\frac{\log n}{n}\right\}.$ $$

Thus we again obtain that ([24](#A4.E24)) is bounded by $e^{-\Theta(n\log n)}$.

Finally, there are at most $2^{n}$ candidate labelings, and since $2^{n}e^{-\Theta(n\log n)}=o(1)$, the claim follows by a union bound.

###### Lemma D.6 .

###### Lemma D.7 .

###### Lemma D.8 .

###### Proof of Lemma D.6 .

###### Proof of Lemma D.7 .

###### Proof of Lemma D.8 .

### D.4 Spectral Algorithm and Proof of Theorem 1

Below is our spectral algorithm which takes $A$ (and optionally $y$ when available) as input along with the parameters and returns an estimator $\hat{\sigma}_{\textnormal{spec}}$. One of the (two) score vectors formed by the algorithm approximates the genie score $z^{*}$.

Algorithm 1 Spectral recovery algorithm for $\mathsf{ROS}$, without or with $\mathsf{BEC}$ or $\mathsf{BSC}$ side information.

$$ $c_{1}:=\sqrt{n}\log n\cdot(a-b)\cdot(\rho a^{2}+(1-\rho)b^{2})^{3/2}\textnormal{ and }\gamma:=\left(\rho(a^{2}b^{2}-a^{4})+(1-\rho)(b^{4}-a^{2}b^{2})\right)\frac{\log n}{2}.$ $$

- •
No side information:
$z^{(s)}=sc_{1}u_{1}+\gamma\mathbf{1}_{n}.$
- •
$\mathsf{BEC}$ side information: For any $i\in[n]$,
$z^{(s)}_{i}=\begin{cases}sc_{1}(u_{1})_{i}+\gamma&\textnormal{if }y_{i}=0;\\
+\infty,&\textnormal{if }y_{i}=+1;\\
-\infty&\textnormal{if }y_{i}=-1;\end{cases}$
- •
$\mathsf{BSC}$ side information:
$z^{(s)}=sc_{1}u_{1}+\gamma\mathbf{1}_{n}+\ln\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)y$

- •
No side information: Return $\hat{\sigma}_{\textnormal{spec}}=\operatorname*{arg\,max}_{\{\hat{\sigma}^{(s)}:s\in\{\pm 1\}\}}\operatorname*{\mathbb{P}}(\sigma^{*}=\hat{\sigma}^{(s)}\mid A)$.
- •
$\mathsf{BEC}$ or $\mathsf{BSC}$ side information: Return $\hat{\sigma}_{\text{spec}}=\operatorname*{arg\,max}_{\{\hat{\sigma}^{(s)}:s\in\{\pm 1\}\}}\operatorname*{\mathbb{P}}(\sigma^{*}=\hat{\sigma}^{(s)}\mid A,y)$.

The values $c_{1}$ and $\gamma$ are carefully designed to emulate the genie score. Since the eigenvectors are only recovered up to a global direction flip, we need to keep both candidates in the algorithm. One of them is approximating the genie score well. Finally, whichever one has the higher posterior probability is picked in step [6](#algx1.l6). To show the proof of the score approximation guarantee, we need the following lemma, whose proof is included at the end of this subsection.

###### Lemma D.9 .

Fix $\rho\in(0,1)$ and $a,b\in\mathbb{R}$ such that $\max\{|a|,|b|\}>0$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$. Condition on $\sigma^{*}$ satisfying $E$ from ([11](#A2.E11)) holds for it. Then there exists a constant $c:=c(\rho,a,b)$ such that with probability $1-O(n^{-3})$, the following event holds

$$ $E_{1}:=\left\{\sqrt{\frac{\log n}{n}}\left\lVert Av^{*}\right\rVert_{\infty}\leq c\log n\right\}.$ (26) $$

Below is our primary lemma which shows the spectral and genie score vector approximation in $\ell_{\infty}$ norm.

###### Lemma D.10 .

Fix $\rho\in(0,1)$ and $a,b$ such that $\max\{|a|,|b|\}>0$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$ and condition on $\sigma^{*}$ such that $E$ from ([11](#A2.E11)) holds. Optionally, let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$, where $\epsilon_{n}$ and $\alpha_{n}$ is parameterized as ([1](#S1.E1)) for $\beta\geq 0$. Let $z^{*}$ and $z^{(s)}$ be the genie score and the spectral score vectors respectively for the corresponding model. Then (irrespective of the parameter values), with probability $1-o(1)$,

$$ $\min_{s\in\{\pm 1\}}\left\lVert z^{*}-z^{(s)}\right\rVert_{\infty}=o(\log n).$ $$

###### Proof.

First of all, note that conditioned on $E$, we have $\lambda_{1}^{*}=(1+o(1))\sqrt{\frac{\log n}{n}}\left\lVert v^{*}\right\rVert_{2}^{2},$ and $u_{1}^{*}=(1+o(1))\frac{v^{*}}{\left\lVert v^{*}\right\rVert_{2}}$. Additionally, $\left\lVert v^{*}\right\rVert_{2}=\sqrt{|C_{+}|a^{2}+|C_{-}|b^{2}}=(1+o(1))\sqrt{n}\sqrt{\rho a^{2}+(1-\rho)b^{2}}.$ Using these, one can simplify

$$ $\frac{c_{1}Au_{1}^{*}}{\lambda_{1}^{*}}=\frac{(1+o(1))c_{1}Av^{*}}{\lambda_{1}^{*}\left\lVert v^{*}\right\rVert_{2}}\approx\frac{\sqrt{n}\log n(a-b)(\rho a^{2}+(1-\rho)b^{2})^{3/2}Av^{*}}{\sqrt{\nicefrac{{\log n}}{{n}}}\left\lVert v^{*}\right\rVert_{2}^{3}}\approx\sqrt{\frac{\log n}{n}}(a-b)Av^{*},$ (27) $$

where in the last step we substitute $\left\lVert v^{*}\right\rVert_{2}$.

Now the high probability event in this lemma is such that (i) the behavior of eigenvectors as stated in Lemma [C.4](#A3.Thmlemma4) and (ii) the event $E_{1}$ from ([26](#A4.E26)) hold. By Lemma [C.2](#A3.Thmlemma2) and [D.9](#A4.Thmlemma9), they both happen with probability $1-o(1)$. Additionally, let $s$ be the sign for which the conclusion of Lemma [C.2](#A3.Thmlemma2) holds. We now analyze the three models separately.

- •
No side information: For every $i\in[n]$
$\displaystyle|z^{(s)}-z^{*}_{i}|$
$\displaystyle=\left|c_{1}s(u_{1})_{i}+\gamma-z^{*}_{i}\right|,$
(recall Algorithm [D.4](#A4.SS4))
$\displaystyle=\left|\frac{c_{1}(Au_{1}^{*})_{i}}{\lambda_{1}^{*}}+\gamma-z^{*}_{i}\right|+O\left(\frac{c_{1}}{\sqrt{n\log n}}\right)$
(by Lemma [C.4](#A3.Thmlemma4) and the triangle inequality)
$\displaystyle=\left|(1+o(1))\sqrt{\frac{\log n}{n}}(a-b)(Av^{*})_{i}+\gamma-z^{*}_{i}\right|+O(\sqrt{\log n})$
(using ([27](#A4.E27)) and $c_{1}\asymp\sqrt{n}\log n$)
$\displaystyle=\left|(1+o(1))\sqrt{\frac{\log n}{n}}(a-b)(Av^{*})_{i}+\gamma-\sqrt{\frac{\log n}{n}}(a-b)A_{i\cdot}v^{*}-\gamma-O(1)\right|$
$\displaystyle\hskip 5.69054pt+O(\sqrt{\log n})$
(putting $z^{*}_{i}$ from Lemma [D.1](#A4.Thmlemma1))
$\displaystyle=o(1)\sqrt{\frac{\log n}{n}}|(Av^{*})_{i}|+O(\sqrt{\log n})=o(\log n),$
(recall $E_{1}$ from ([26](#A4.E26)))
- •
$\mathsf{BEC}$ side information: By Lemma [2.10](#S2.Thmlemma10) and Algorithm [D.4](#A4.SS4) (Step [5](#algx1.l5)), when $y_{i}=+1$, we have $z^{*}_{i}=z^{(s)}_{i}=+\infty$ and thus by our convention, $|z^{(s)}_{i}-z^{*}_{i}|=0$. Similarly, for $y_{i}=-1$, we have $z^{*}_{i}=z^{(s)}_{i}=-\infty$, which gives us $|z^{(s)}_{i}-z^{*}_{i}|=0$. Finally, when $y_{i}=0$, both $z^{(s)}_{i}$ and $z^{*}_{i}$ are the same as in the case when no side information is provided. Then the analysis in the previous immediately implies that $|z^{(s)}_{i}-z^{*}_{i}|=o(\log n)$.
- •
$\mathsf{BSC}$ side information: By Lemma [2.10](#S2.Thmlemma10) and Algorithm [D.4](#A4.SS4) (step [5](#algx1.l5)), when $\mathsf{BSC}$ side information is provided, both the Genie score vector $z^{*}$ and the spectral score vector $z^{(s)}$ are achieved by adding $\log\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)y$ to their counterpart when no side information is provided. Therefore, the triangle inequality along with the analysis in the no side information case gives us that for every $i\in[n]$, we have $|z^{(s)}_{i}-z^{*}_{i}|=o(\log n).$

In either case, one can equivalently write conclusions for all $i\in[n]$ together in vector notation

$$ $\min_{s\in\{\pm 1\}}\left\lVert z^{(s)}-z^{*}\right\rVert_{\infty}=o(\log n).\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

We are finally set to prove our first main result in Theorem [1](#Thmtheorem1).

###### Proof of Theorem 1 .

Let $z^{*}$ be the genie score vector for the respective model of side information. By Lemma [D.5](#A4.Thmlemma5), below the respective IT threshold given in Table [1](#S3.T1)

$$ $\operatorname*{\mathbb{P}}\left(\exists i:[n]:\sigma^{*}_{i}z^{*}_{i}<\frac{-\log n}{\log\log n}\right)=1-o(1).$ $$

By the definition of the failure genie-aided estimator below ([3](#S2.E3)),

$$ $\operatorname*{\mathbb{P}}(\exists i\in[n]:\hat{\sigma}_{\textnormal{\tiny Gen},i}\textnormal{ fails})=1-o(1).$ $$

Using Lemma [2.9](#S2.Thmlemma9)

$$ $\operatorname*{\mathbb{P}}(\hat{\sigma}_{\textnormal{\tiny MAP}}\textnormal{ fails})\geq\operatorname*{\mathbb{P}}(\exists i\in[n]:\hat{\sigma}_{\textnormal{\tiny Gen},i}\textnormal{ fails})=1-o(1).$ $$

Using the optimality of the MAP estimator; for any estimator $\hat{\sigma}$, we have $\operatorname*{\mathbb{P}}\left(\hat{\sigma}\textnormal{ fails}\right)\geq\operatorname*{\mathbb{P}}(\hat{\sigma}_{\textnormal{\tiny MAP}}\textnormal{ fails})=1-o(1).$ This immediately implies for any estimator $\hat{\sigma}$.

$$ $\lim_{n\rightarrow\infty}\operatorname*{\mathbb{P}}\left(\hat{\sigma}\textnormal{ succeeds}\right)=0.$ $$

To show the positive result, first note that step [6](#algx1.l6) of Algorithm [D.4](#A4.SS4) keeps two candidates $\{\hat{\sigma}^{(s)}:s\in\{\pm 1\}\}$ and chooses the one which has maximum posterior probability. Therefore, to show that $\hat{\sigma}_{\textnormal{spec}}$ achieves exact recovery above the IT threshold, it suffices to show that one of the two candidates achieves exact recovery, and $\hat{\sigma}_{\textnormal{\tiny MAP}}$ also succeeds above the IT threshold, which ensures that the algorithm selects the correct vector by maximizing the posterior probability. We already showed statistical achievability in Lemma [D.6](#A4.Thmlemma6) when no side information. The statistical achievability in the case of $\mathsf{BEC}$ or $\mathsf{BSC}$ side information follows from Theorem [2](#Thmtheorem2), where we showed $\hat{\sigma}_{\textnormal{dp}}$ (and so $\hat{\sigma}_{\textnormal{\tiny MAP}}$) achieves exact recovery. It remains to show that one of $\{\hat{\sigma}^{(s)}:s\in\{\pm 1\}\}$ succeeds. To this end, recall Lemma [D.10](#A4.Thmlemma10) that with probability $1-o(1)$,

$$ $\min_{s\in\{\pm 1\}}\left\lVert z^{*}-z^{(s)}\right\rVert_{\infty}=o(\log n).$ $$

Moreover, above the IT threshold by Lemma [D.3](#A4.Thmlemma3) and union bound over $i\in[n]$, there exists $\delta>0$ such that

$$ $\operatorname*{\mathbb{P}}\left(\min_{i\in[n]}\sigma^{*}_{i}z^{*}_{i}>\delta\log n\right)=1-o(1).$ $$

Taking a union bound over these two events, there exists $\varsigma>0$ and $s^{*}\in\{\pm 1\}$ such that

$$ $\operatorname*{\mathbb{P}}\left(\min_{i\in[n]}\sigma^{*}_{i}z^{(s^{*})}_{i}>\varsigma\log n\right)=1-o(1).$ $$

Since $\hat{\sigma}^{(s^{*})}=\textnormal{sgn}(z^{(s^{*})})$ in step [6](#algx1.l6), we obtain $\hat{\sigma}^{(s^{*})}$ achieves exact recovery. As a consequence, even $\hat{\sigma}_{\textnormal{spec}}$ achieves exact recovery above the IT threshold. In other words,

$$ $\lim_{n\rightarrow\infty}\operatorname*{\mathbb{P}}\left(\hat{\sigma}_{\textnormal{spec}}\textnormal{ succeeds}\right)=1.\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

We finally return to the proof of the lemma already mentioned.

###### Proof of Lemma D.9 .

For each $i\in[n]$, first define $Y_{i}=\sqrt{\frac{\log n}{n}}(Av^{*})_{i}$. We first note that $Y_{i}$ is a Gaussian random variable as it is the sum of at most $n$ independent Gaussian random variables. Therefore, we have $Y_{i}\sim\mathcal{N}(\mu_{i},\sigma^{2}_{i})$ for certain $\mu_{i}$ and $\sigma^{2}_{i}$, which we will calculate later. Applying Lemma [A.1](#A1.Thmlemma1) for any $Y_{i}$ (after normalizing) yields

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\frac{|Y_{i}-\mu_{i}|}{\sigma_{i}}\geq 4\sqrt{\log n}\right)\leq e^{-8\log n}=n^{-8}.$ $$

Rearrangement of the terms using the triangle inequality along with a union bound over all $i\in[n]$ gives us

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\forall i\in[n]:|Y_{i}|\leq|\mu_{i}|+4\sigma_{i}\sqrt{\log n}\right)\geq 1-n^{-7}.$ $$

Therefore it simply suffices to show that for every $i\in[n]$, the quantity $|\mu_{i}|+4\sigma_{i}\sqrt{\log n}=O(\log n)$. To this end, we first observe that $(Av^{*})_{i}$ is the sum of $n-1$ independent Gaussian random variables all with means whose absolute values are $O\left(\sqrt{\frac{\log n}{n}}\right)$. Thus,

$$ $|\mu_{i}|=\sqrt{\frac{\log n}{n}}\operatorname*{{\mathbb{E}}}[(Av^{*})_{i}]=\sqrt{\frac{\log n}{n}}(n-1)\cdot O\left(\sqrt{\frac{\log n}{n}}\right)=O(\log n).$ $$

Similarly, $(Av^{*})_{i}$ is the sum of $n-1$ independent Gaussian random variables with variances $O(1)$. This gives us

$$ $\sigma^{2}_{i}=\textnormal{Var}\left[\sqrt{\frac{\log n}{n}}(Av^{*})_{i}\right]=\frac{\log n\cdot\textnormal{Var}\left[(Av^{*})_{i}\right]}{n}=\frac{\log n\cdot O(n)}{n}=O(\log n),$ $$

which also implies

$$ $4\sigma_{i}\sqrt{\log n}=O(\log n).\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

###### Lemma D.9 .

###### Lemma D.10 .

###### Proof.

###### Proof of Theorem 1 .

###### Proof of Lemma D.9 .

### D.5 Degree-Profiling Algorithm and Proof of Theorem 2

The following is a simple degree-profiling algorithm that tries to mimic the genie naïvely and achieves exact recovery if side information is substantial to shift the thresholds of exact recovery.

Algorithm 2 Degree-Profiling algorithm for $\mathsf{ROS}$ in the presence of $\mathsf{BEC}$ or $\mathsf{BSC}$ side information.

$$ $\gamma:=\left(\rho(a^{2}b^{2}-a^{4})+(1-\rho)(b^{4}-a^{2}b^{2})\right)\log n/2.$ $$

$$ $z_{i}=a(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in S_{+}}A_{ij}+b(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in S_{-}}A_{ij}+\gamma.$ $$

- •
$\mathsf{BEC}$ side information: For any $i\in[n]$,
$z^{\textnormal{dp}}_{i}=\begin{cases}z_{i}&\textnormal{if }y_{i}=0;\\
+\infty,&\textnormal{if }y_{i}=+1;\\
-\infty&\textnormal{if }y_{i}=-1;\end{cases}$
- •
$\mathsf{BSC}$ side information:
$z^{\textnormal{dp}}=z+\ln\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)y$

The following lemma plays a crucial role in the analysis of our degree profiling algorithm which formalizes the notion of receiving most of the labels correct.

###### Lemma D.11 .

Let $\sigma^{*}\in\{\pm 1\}^{n}$ be sampled such that each entry is i.i.d. with $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}=+1)=\rho$. Condition on $\sigma^{*}$ such that the event $E$ from ([11](#A2.E11)) holds. For any $\beta>0$, we let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$ for $\epsilon_{n}$ and $\alpha_{n}$ parameterized as ([1](#S1.E1)). Define $S_{+}=\{i:y_{i}=+1\}$ and $S_{-}=\{i:y_{i}=-1\}$. Then with probability $1-o(1)$

$$ $\max\left\{\left|{C_{+}\setminus S_{+}}\right|,|C_{-}\setminus S_{-}|\right\}=O\left(\frac{n}{\log^{10}n}\right).$ $$

###### Proof.

First, recall that conditioned on the even $E$ about $\sigma^{*}$, we have $|C_{+}|=\Theta(n)$ and $|C_{-}|=\Theta(n)$. We now consider the two types of side information.

- •
$\mathsf{BEC}$ side information: Observe that
$\operatorname*{{\mathbb{E}}}[\left|{C_{+}\setminus S_{+}}\right|]=\sum_{i\in C_{+}}\operatorname*{\mathbb{P}}(y_{i}=0)=|C_{+}|\epsilon_{n}=n^{-\beta}|C_{+}|\leq n^{1-\beta}.$
Then Markov’s inequality immediately implies that, with probability $1-O(n^{-\beta/2})$, we have
$\left|{C_{+}\setminus S_{+}}\right|\leq n^{1-\beta/2}=O\left(\nicefrac{{n}}{{\log^{10}n}}\right).$
Similarly, we also have $\operatorname*{{\mathbb{E}}}[\left|{C_{-}\setminus S_{-}}\right|]=n^{-\beta}\left|{C_{-}}\right|\leq n^{1-\beta}$. Thus, applying Markov’s inequality again implies $\left|{C_{-}\setminus S_{-}}\right|=O(\nicefrac{{n}}{{\log^{10}n}})$ with probability $1-O(n^{-\beta/2})$. A simple union bound over these two events implies, with probability $1-O(n^{-\beta/2})=1-o(1)$
$\max\left\{\left|{C_{+}\setminus S_{+}}\right|,|C_{-}\setminus S_{-}|\right\}=O\left(\frac{n}{\log^{10}n}\right).$
- •
$\mathsf{BSC}$ side information: Under $\mathsf{BSC}$ side information, $\operatorname*{{\mathbb{E}}}[|C_{+}\setminus S_{+}|]=\alpha_{n}|C_{+}|\leq n^{1-\beta}$ and $\operatorname*{{\mathbb{E}}}[|C_{-}\setminus S_{-}|]=\alpha_{n}|C_{-}|\leq n^{1-\beta}$.
Therefore, using the Markov’s inequality for both of these sets along with a union bound immediately implies that with probability $1-O(n^{-\beta/2})=1-o(1)$,
$\max\left\{\left|{C_{+}\setminus S_{+}}\right|,|C_{-}\setminus S_{-}|\right\}=O\left(\frac{n}{\log^{10}n}\right).$

Using this lemma, we now show that the degree profiling vector $z^{\textnormal{dp}}$ is a good approximation to the genie score vector $z^{*}$ in $\ell_{\infty}$ norm.

###### Lemma D.12 .

Fix $\rho\in(0,1)$ and $a,b$ such that $\max\{|a|,|b|\}>0$. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a,b)$ and condition on $\sigma^{*}$ such that $E$ from ([11](#A2.E11)) holds. For $\beta>0$, let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$ where $\epsilon_{n}$ and $\alpha_{n}$ are parameterized as ([1](#S1.E1)). Let $z^{*}$ and $z^{\textnormal{dp}}$ respectively be the genie score vector and the degree-profiling score vector produced by Algorithm [D.5](#A4.SS5) for the corresponding model of side information. Then (irrespective of the parameter values), with probability $1-o(1)$,

$$ $\left\lVert z^{*}-z^{\textnormal{dp}}\right\rVert_{\infty}=O(1).$ $$

###### Proof.

We first start by observing, in the case of $\mathsf{BEC}$ side information $z^{\textnormal{dp}}$ is just formed by overriding the entries of $z$ from step [3](#algx2.l3) of Algorithm [D.5](#A4.SS5) with $+\infty$ or $-\infty$ depending on the side information label being $+1$ or $-1$. Also, for $\mathsf{BSC}$ side information, $z^{\textnormal{dp}}=z+\log\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)y$. By Lemmas [2.10](#S2.Thmlemma10), this is precisely how the genie score vector $z^{*}$ in the respective model of side information relates to the genie score vector without side information which we denote by $z^{\prime}$.

Therefore, to show the lemma, it suffices to show that, with probability $1-o(1)$,

$$ $\left\lVert z^{\prime}-z\right\rVert_{\infty}=O(1).$ $$

- •
$\mathsf{BEC}$ side information:
$\displaystyle\left\lVert z^{\prime}-z\right\rVert_{\infty}=\max_{i\in[n]}|z^{\prime}_{i}-z_{i}|$
$\displaystyle=\max_{i\in[n]}\left|a(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in C_{+}\setminus S_{+}}A_{ij}+b(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in C_{-}\setminus S_{-}}A_{ij}+\log\left(\frac{\rho}{1-\rho}\right)+o(1)\right|$
(substituting $z^{\prime}$ from Lemma [D.1](#A4.Thmlemma1) and $z$ from Algorithm [D.5](#A4.SS5))
$\displaystyle\leq\max_{i\in[n]}\left|a(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in C_{+}\setminus S_{+}}A_{ij}\right|+\left|b(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in C_{-}\setminus S_{-}}A_{ij}\right|+O(1),$
(28)
where the last step follows from the triangle inequality. By Lemma [D.11](#A4.Thmlemma11), both $|C_{+}\setminus S_{+}|$ and $|C_{-}\setminus S_{-}|$ are bounded by $O(n/\log^{10}n)$ with probability $1-o(1)$. Moreover, these sets are chosen only based on the side information $y$ and hence independent of $A$, conditioned on $\sigma^{*}$. Using Lemma [D.4](#A4.Thmlemma4) for these set $C_{+}\setminus S_{+}$ and $C_{-}\setminus S_{-}$ as $T$, and using a union bound, we obtain that with probability $1-o(1)$
$\max_{i\in[n]}\left|a(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in C_{+}\setminus S_{+}}A_{ij}\right|=O(1)\quad\text{and}\quad\max_{i\in[n]}\left|b(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in C_{-}\setminus S_{-}}A_{ij}\right|=O(1).$
Substituting these bounds in ([28](#A4.E28)), with probability $1-o(1)$
$\left\lVert z^{\prime}-z\right\rVert_{\infty}=O(1).$
- •
$\mathsf{BSC}$ side information:
$\displaystyle\left\lVert z^{\prime}-z\right\rVert_{\infty}=\max_{i\in[n]}|z^{\prime}_{i}-z_{i}|$
$\displaystyle=\max_{i\in[n]}\Bigg{|}a(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in C_{+}\setminus S_{+}}A_{ij}-a(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in S_{+}\setminus C_{+}}A_{ij}$
$\displaystyle\hskip 14.22636pt+b(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in C_{-}\setminus S_{-}}A_{ij}-b(a-b)\sqrt{\frac{\log n}{n}}\sum_{j\in S_{-}\setminus C_{-}}A_{ij}+\log\left(\frac{\rho}{1-\rho}\right)+o(1)\Bigg{|}$
(substituting $z^{\prime}$ from Lemma [D.1](#A4.Thmlemma1) and $z$ from Algorithm [D.5](#A4.SS5))
$\displaystyle=\max_{i\in[n]}\Bigg{|}(a-b)^{2}\sqrt{\frac{\log n}{n}}\sum_{j\in C_{+}\setminus S_{+}}A_{ij}-(a-b)^{2}\sqrt{\frac{\log n}{n}}\sum_{j\in C_{-}\setminus S_{-}}A_{ij}+\log\left(\frac{\rho}{1-\rho}\right)+o(1)\Bigg{|}$
$\displaystyle\leq\max_{i\in[n]}\left|(a-b)^{2}\sqrt{\frac{\log n}{n}}\sum_{j\in C_{+}\setminus S_{+}}A_{ij}\right|+\left|(a-b)^{2}\sqrt{\frac{\log n}{n}}\sum_{j\in C_{-}\setminus S_{-}}A_{ij}\right|+O(1),$
(29)
where in the last step, we used the triangle inequality. Again by similar arguments, first using Lemma [D.11](#A4.Thmlemma11), both $|C_{+}\setminus S_{+}|$ and $|C_{-}\setminus S_{-}|$ is $O(n/\log^{10}n)$ with probability $1-o(1)$. Using Lemma [D.4](#A4.Thmlemma4) for these set $C_{+}\setminus S_{+}$ and $C_{-}\setminus S_{-}$ further implies that, with probability $1-o(1)$
$\max_{i\in[n]}\left|(a-b)^{2}\sqrt{\frac{\log n}{n}}\sum_{j\in C_{+}\setminus S_{+}}A_{ij}\right|=O(1)\quad\text{and}\quad\max_{i\in[n]}\left|(a-b)^{2}\sqrt{\frac{\log n}{n}}\sum_{j\in C_{-}\setminus S_{-}}A_{ij}\right|=O(1).$
Substituting these bounds in ([29](#A4.E29)), with probability $1-o(1)$
$\left\lVert z^{\prime}-z\right\rVert_{\infty}=O(1).$

Finally, we prove Theorem [2](#Thmtheorem2).

###### Proof of Theorem 2 .

When $\beta>0$, by Lemma [D.12](#A4.Thmlemma12) that with probability $1-o(1)$,

$$ $\left\lVert z^{*}-z^{\textnormal{dp}}\right\rVert_{\infty}=O(1).$ $$

Above the IT threshold by Lemma [D.3](#A4.Thmlemma3) and union bound over $i\in[n]$, there exists $\delta>0$ such that

$$ $\operatorname*{\mathbb{P}}\left(\min_{i\in[n]}\sigma^{*}_{i}z^{*}_{i}>\delta\log n\right)=1-o(1).$ $$

Taking a union bound, there exists $\varsigma>0$ such that

$$ $\operatorname*{\mathbb{P}}\left(\min_{i\in[n]}\sigma^{*}_{i}z^{\textnormal{dp}}_{i}>\varsigma\log n\right)=1-o(1).$ $$

Observing $\hat{\sigma}_{\textnormal{dp}}=\textnormal{sgn}(z^{\textnormal{dp}})$, we obtain $\hat{\sigma}_{\textnormal{dp}}$ achieves exact recovery, i.e.

$$ $\lim_{n\rightarrow\infty}\operatorname*{\mathbb{P}}\left(\hat{\sigma}_{\textnormal{dp}}\textnormal{ succeeds}\right)=1.\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

###### Lemma D.11 .

###### Proof.

###### Lemma D.12 .

###### Proof.

###### Proof of Theorem 2 .

## Appendix E Proofs and Algorithms for SBM

We follow the same structure: in Appendices [E.1](#A5.SS1) and [E.2](#A5.SS2), respectively, we show the success with margin and failure of genie-based estimation. In Appendices [E.3](#A5.SS3) and [E.4](#A5.SS4), we present the spectral and the degree profiling algorithms respectively, with our main results.

### E.1 Genie scores’ success with margin above the IT threshold

We begin by showing that, with high probability, all the vertices have degrees logarithmic in $n$.

###### Lemma E.1 .

Let $\rho\in(0,1)$ and $a_{1},a_{2},b>0$. Let $(A,\sigma^{*})\sim\mathsf{SBM}_{n}(\rho,a_{1},a_{2},b)$. Condition on $\sigma^{*}$ such that the event $E$ from ([30](#A5.E30)) holds then. For $c=6\max\{1,a_{1},a_{2},b\}$ let

$$ $E_{1}=\left\{\forall i:\sum_{j\in[n]}A_{ij}\leq c\log n\right\};$ (30) $$

then with $\operatorname*{\mathbb{P}}(E_{1})=1-O(n^{-3}).$

###### Proof.

Note that the entries of $A$ (up to symmetry) are independent conditioned on $\sigma^{*}$. Therefore, for any $i\in[n]$, the $i^{\mathrm{th}}$ row has independent Bernoulli entries with means either $p_{1},p_{2}$ or $q$, where $(p_{1},p_{2},q)=(a_{1},a_{2},b)\log n/n$. Therefore, defining $X\sim\text{Binom}(n,\tau\log n/n)$, where $\tau=\max\{a_{1},a_{2},b\}$, we have that $X$ stochastically dominates $\sum_{j\in[n]}A_{ij}$, for any $i\in[n]$. Then applying the Chernoff bound for Binomial random variables [, Theorem 4.4, Equation 4.3] we get, for any $i\in[n]$

$$ $\operatorname*{\mathbb{P}}\left(\sum_{j\in[n]}A_{ij}>6\max\{1,\tau\}\log n\right)\leq\operatorname*{\mathbb{P}}\left(X>6\max\{1,\tau\}\log n\right)\leq 2^{-6\log n}=O(n^{-4}).$ $$

Taking a union bound over all $i\in[n]$ yields the desired claim.

We next analyze the form of genie scores without side information.

###### Lemma E.2 .

Let $\rho\in(0,1)$ and $a_{1},a_{2},b>0$. Let $(A,\sigma^{*})\sim\mathsf{SBM}_{n}(\rho,a_{1},a_{2},b)$. Denote $(p_{1},p_{2},q):=(a_{1},a_{2},b)\,\nicefrac{{\log n}}{{n}}$. Then for any $i\in[n]$, the genie score can be written as

$$ $\displaystyle z^{*}_{i}$ $\displaystyle=\log\left(\frac{p_{1}(1-q)}{q(1-p_{1})}\right)\sum_{j\in C_{+}^{-i}}A_{ij}+\log\left(\frac{q(1-p_{2})}{p_{2}(1-q)}\right)\sum_{j\in C_{-}^{-i}}A_{ij}+\log\left(\frac{\rho}{1-\rho}\right)$ $\displaystyle\hskip 5.69054pt+|C_{+}^{-i}|\log\left(\frac{1-p_{1}}{1-q}\right)+|C_{-}^{-i}|\log\left(\frac{1-q}{1-p_{2}}\right).$ $$

Moreover, conditioned on the event $E$ from ([11](#A2.E11)) and $E_{1}$ from ([30](#A5.E30)),

$$ $\left\lVert z^{*}-Aw-\gamma\mathbf{1}_{n}\right\rVert_{\infty}=O(1),$ $$

where $w\in\mathbb{R}^{n}$ is a vector with entries $(w_{+},w_{-}):=(\log(a_{1}/b),\log(b/a_{2}))$ on locations of $C_{+}$ and $C_{-}$ respectively and $\gamma:=(\rho(b-a_{1})+(1-\rho)(a_{2}-b))\log n$.

###### Proof.

First of all, note that the $\mathsf{SBM}$ is a special case of the $\mathsf{GBM}$ model with $\mathcal{P}_{+}\equiv\textnormal{Bern}(p_{1})$, $\mathcal{P}_{-}\equiv\textnormal{Bern}(p_{2})$ and $\mathcal{Q}\equiv\textnormal{Bern}(q)$.
Using Lemma [2.10](#S2.Thmlemma10) for this special case, for any $i\in[n]$

$$ $\displaystyle z^{*}_{i}$ $\displaystyle=\log\left(\frac{\rho}{1-\rho}\right)+\sum_{i\in C_{+}^{-i}}\log\left(\frac{p_{1}^{A_{ij}}(1-p_{1})^{(1-A_{ij})}}{q^{A_{ij}}(1-q)^{(1-A_{ij})}}\right)+\sum_{j\in C_{-}^{-i}}\log\left(\frac{q^{A_{ij}}(1-q)^{(1-A_{ij})}}{p_{2}^{A_{ij}}(1-p_{2})^{(1-A_{ij})}}\right)$ $\displaystyle=\log\left(\frac{p_{1}(1-q)}{q(1-p_{1})}\right)\sum_{j\in C_{+}^{-i}}A_{ij}+\log\left(\frac{q(1-p_{2})}{p_{2}(1-q)}\right)\sum_{j\in C_{-}^{-i}}A_{ij}+\log\left(\frac{\rho}{1-\rho}\right)$ $\displaystyle\hskip 28.45274pt+|C_{+}^{-i}|\log\left(\frac{1-p_{1}}{1-q}\right)+|C_{-}^{-i}|\left(\frac{1-q}{1-p_{2}}\right).$ (31) $$

To show the second part of the lemma, we further simplify

$$ $\displaystyle\left|\log\left(\frac{1-q}{1-p_{1}}\right)\right|$ $\displaystyle=\left|\log\left(1+\frac{p_{1}-q}{(1-p_{1})}\right)\right|=\left|\log\left(1+\frac{(a_{1}-b)\log n}{(1-p_{1})n}\right)\right|=O\left(\frac{\log n}{n}\right).$ (32) $$

In the last inequality, we used $\frac{x}{x+1}\leq\log(1+x)\leq x$ for $x>-1$. Similarly,

$$ $\displaystyle\left|\log\left(\frac{1-p_{2}}{1-q}\right)\right|$ $\displaystyle=\left|\log\left(1+\frac{q-p_{2}}{(1-q)}\right)\right|=\left|\log\left(1+\frac{(b-a_{2})\log n}{(1-q)n}\right)\right|=O\left(\frac{\log n}{n}\right).$ (33) $$

Recall the definition of event $E$ from ([11](#A2.E11)) and $E_{1}$ from ([30](#A5.E30)). Conditioned on $E\cap E_{1}$, we simplify ([31](#A5.E31)) using ([32](#A5.E32)) and ([33](#A5.E33)).

$$ $\displaystyle\log\left(\frac{p_{1}(1-q)}{q(1-p_{1})}\right)\sum_{j\in C_{+}^{-i}}A_{ij}+\log\left(\frac{q(1-p_{2})}{p_{2}(1-q)}\right)\sum_{j\in C_{-}^{-i}}A_{ij}$ $\displaystyle=\log\left(\frac{a_{1}}{b}\right)\sum_{j\in C_{+}^{-i}}A_{ij}+\log\left(\frac{b}{a_{2}}\right)\sum_{j\in C_{-}^{-i}}A_{ij}+O\left(\frac{\log n}{n}\right)\sum_{j\in[n]}A_{ij}$ $\displaystyle=\log\left(\frac{a_{1}}{b}\right)\sum_{j\in C_{+}^{-i}}A_{ij}+\log\left(\frac{b}{a_{2}}\right)\sum_{j\in C_{-}^{-i}}A_{ij}+o(1),$ (34) $$

where the last equality followed by conditioning on $E_{1}$. We also simplify

$$ $\displaystyle|C_{+}^{-i}|\log\left(\frac{1-p_{1}}{1-q}\right)$ $\displaystyle=|C_{+}^{-i}|\log\left(1+\frac{q-p_{1}}{(1-q)}\right)=|C_{+}^{-i}|\log\left(1+\frac{(b-a_{1})\log n}{(1-q)n}\right)$ $\displaystyle=|C_{+}^{-i}|\left(\frac{(b-a_{1})\log n}{(1-q)n}+O\left(\frac{\log^{2}n}{n^{2}}\right)\right)$ (using a Taylor expansion of $\log(1+x)$ ) $\displaystyle=(1+O(n^{-1/3}))\rho n\left(\frac{(b-a_{1})\log n}{(1-q)n}+O\left(\frac{\log^{2}n}{n^{2}}\right)\right)$ $\displaystyle=\rho(b-a_{1})\log n+o(1)$ (35) $$

Similarly,

$$ $\displaystyle|C_{-}^{-i}|\log\left(\frac{1-q}{1-p_{2}}\right)$ $\displaystyle=|C_{-}^{-i}|\log\left(1+\frac{p_{2}-q}{(1-p_{2})}\right)=(1-\rho)(a_{2}-b)\log n+o(1)$ (36) $$

Substituting ([E.1](#A5.Ex299)), ([35](#A5.E35)) and ([36](#A5.E36)) into ([31](#A5.E31)),

$$ $\displaystyle z^{*}_{i}$ $\displaystyle=\log\left(\frac{a_{1}}{b}\right)\sum_{j\in C_{+}^{-i}}A_{ij}+\log\left(\frac{b}{a_{2}}\right)\sum_{j\in C_{-}^{-i}}A_{ij}+\rho(b-a_{1})+(1-\rho)(a_{2}-b)\log n+O(1)$ $\displaystyle=w_{+}\sum_{j\in C_{+}^{-i}}A_{ij}+w_{-}\sum_{j\in C_{-}^{-i}}A_{ij}+\gamma+O(1)=A_{i\cdot}w+\gamma+O(1).$ $$

Writing the above for all $i\in[n]$ in a vector notation, we obtain

$$ $\left\lVert z^{*}-(Aw+\gamma\mathbf{1}_{n})\right\rVert_{\infty}=O(1).\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

Finally, we turn our attention to the lemma which establishes that above the IT threshold, the genie score succeeds with margin.

###### Lemma E.3 .

Fix $\rho\in(0,1)$ and $a_{1},a_{2},b>0$. Let $(A,\sigma^{*})\sim\mathsf{SBM}_{n}(\rho,a_{1},a_{2},b)$. Condition $\sigma^{*}$ such that the event $E$ from ([11](#A2.E11)) holds. Optionally, let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$, where $\epsilon_{n}$ and $\alpha_{n}$ be parameterized as ([1](#S1.E1)) for $\beta\geq 0$. Let $z^{*}$ be the genie score vector for the corresponding model of side information. Then above the information-theoretic threshold as specified in Table [2](#S3.T2), there exists some constant $\delta:=\delta(\rho,a_{1},a_{2},b,\beta)>0$ such that, for any $i\in[n]$:

$$ $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)=o(n^{-1}).$ $$

In order to prove Lemma [E.3](#A5.Thmlemma3), we need the following concentration result, whose proof is deferred to the end of this subsection. For any $i\in[n]$, we define the random variables $X_{i}$ as follows.

$$ $X_{i}\triangleq\log\left(\frac{a_{1}}{b}\right)\sum_{j\in C_{+}^{-i}}A_{ij}+\log\left(\frac{b}{a_{2}}\right)\sum_{j\in C_{-}^{-i}}A_{ij}+\gamma,$ (37) $$

where $\gamma=(\rho(b-a_{1})+(1-\rho)(a_{2}-b))\log n$. Note that $X_{i}$ is the weighted sum of binomial random variables shifted by $\gamma$. Using the standard Chernoff style analysis, we show the following tail bound. Recall the definition of $D_{t}$ from ([5](#S3.E5)) and the community profile vectors $\theta_{+}$ and $\theta_{-}$ from ([6](#S3.E6)).

###### Lemma E.4 .

Consider any $i\in C_{+}$ and fixed $\varepsilon\in\mathbb{R}$. Then for any $t>0$,

$$ $\operatorname*{\mathbb{P}}(X_{i}<(1+o(1))\varepsilon\log n)\leq n^{-D_{t}(\theta_{-},\theta_{+})+\varepsilon t+o(1)}.$ $$

Similarly, for any $i\in C_{-}$, fixed $\varepsilon\in\mathbb{R}$ and $t>0$,

$$ $\operatorname*{\mathbb{P}}(-X_{i}<(1+o(1))\varepsilon\log n)\leq n^{-D_{t}(\theta_{+},\theta_{-})+\varepsilon t+o(1)}.$ $$

With Lemma [E.4](#A5.Thmlemma4) in hand, we prove Lemma [E.3](#A5.Thmlemma3).

###### Proof of Lemma E.3 .

We discuss different side information cases in order.
No side information: First, note that under the event $E_{1}$, which happens with probability $1-O(n^{-3})$, we have

$$ $z^{*}_{i}=\log\left(\frac{a_{1}}{b}\right)\sum_{j\in C_{+}^{-i}}A_{ij}+\log\left(\frac{b}{a_{2}}\right)\sum_{j\in C_{-}^{-i}}A_{ij}+\gamma+O(1):=X_{i}+O(1).$ $$

For any $i\in C_{+}$ and any fixed $\delta,t>0$, we have

$$ $\displaystyle\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)$ $\displaystyle\leq\operatorname*{\mathbb{P}}\left(X_{i}+O(1)<\delta\log n\right)+O(n^{-3})$ $\displaystyle\leq\operatorname*{\mathbb{P}}\left(X_{i}<(1+o(1))\delta\log n\right)+O(n^{-3})$ $\displaystyle\leq n^{-D_{t}(\theta_{-},\theta_{+})+\delta t+o(1)}+O(n^{-3}),$ (38) $$

where we use Lemma [E.4](#A5.Thmlemma4) in the final step. Recall from Table [2](#S3.T2), above the IT threshold means $\sup_{t\in[0,1]}D_{t}(\theta_{+},\theta_{-})>1$, which also implies $\sup_{t\in[0,1]}D_{t}(\theta_{-},\theta_{+})>1$ by the change of variables from $t$ to $1-t$ and noting that $D_{t}(\theta_{-},\theta_{+})=D_{1-t}(\theta_{+},\theta_{-})$. Moreover, $D_{t}$ is a continuous function $t$ and therefore, there exists $t^{*}\in(0,1)$ and $\delta_{1}>0$ sufficiently small such that $D_{t^{*}}(\theta_{-},\theta_{+})-\delta_{1}t^{*}-o(1)>1.$
Substituting this above we obtain that there exists $\delta_{1}:=\delta_{1}(\rho,a_{1},a_{2},b)>0$ such that

$$ $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}\leq\delta_{1}\log n)=o(n^{-1}).$ $$

Similarly, for any $i\in C_{-}$ and $\delta>0$,

$$ $\displaystyle\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)$ $\displaystyle\leq\operatorname*{\mathbb{P}}\left(-X_{i}-O(1)<\delta\log n\right)+O(n^{-3})$ $\displaystyle\leq\operatorname*{\mathbb{P}}\left(-X_{i}<(1+o(1))\delta\log n\right)+O(n^{-3})$ $\displaystyle\leq n^{-D_{t}(\theta_{+},\theta_{-})+\delta t+o(1)}+O(n^{-3}),$ (39) $$

using Lemma [E.4](#A5.Thmlemma4). Above the IT threshold corresponds to $\sup_{t\in[0,1]}D_{t}(\theta_{+},\theta_{-})>1$, which implies there exists $t^{*}\in(0,1)$ and $\delta_{2}>0$ sufficiently small such that $D_{t^{*}}(\theta_{+},\theta_{-})-\delta_{2}t^{*}-o(1)>1.$
Substituting this above, we obtain that there exists $\delta_{2}:=\delta_{2}(\rho,a_{1},a_{2},b)>0$ such that for any $i\in C_{-}$

$$ $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}\leq\delta_{2}\log n)=o(n^{-1}).$ $$

Choosing $\delta=\min\{\delta_{1},\delta_{2}\}$, we get the desired claim.
$\mathsf{BEC}$ side information: For any $i\in[n]$, let $z^{*}_{i}$ and $z^{\prime}_{i}$, respectively, be the genie score under side information and without any side information. For any $i\in[n]$ and constants $\delta,t>0$,

$$ $\displaystyle\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)$ $\displaystyle=\operatorname*{\mathbb{P}}(y_{i}=0)\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n\mid y_{i}=0)+\operatorname*{\mathbb{P}}(y_{i}=\sigma^{*}_{i})\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}z^{*}_{i}<\delta\log n\mid y_{i}=\sigma^{*}_{i}\right)$ $\displaystyle\leq n^{-\beta}\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{\prime}_{i}<\delta\log n\mid y_{i}=0)$ $\displaystyle\leq\begin{cases}n^{-\beta-D_{t}(\theta_{-},\theta_{+})+\delta t+o(1)}+O(n^{-3}),&\text{ if }i\in C_{+};\\ n^{-\beta-D_{t}(\theta_{+},\theta_{-})+\delta t+o(1)}+O(n^{-3}),&\text{ if }i\in C_{-};\end{cases}$ (40) $$

The last inequality follows from using ([38](#A5.E38)) and ([39](#A5.E39)). Again from Table [2](#S3.T2), above the IT threshold, we have $\beta+\sup_{t\in[0,1]}D_{t}(\theta_{+},\theta_{-})>1$, which also implies $\beta+\sup_{t\in[0,1]}D_{t}(\theta_{-},\theta_{+})>1$. Therefore, in either case $i\in C_{+}$ or $i\in C_{-}$, there exists $t^{*}\in(0,1)$ and constant $\delta>0$ sufficiently small such that the exponent of $n$ in ([40](#A5.E40)) is strictly less than $-1$. This gives us, for any $i\in[n]$, there exists $\delta>0$ small enough for which

$$ $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}\leq\delta\log n)=o(n^{-1}).$ $$

$\mathsf{BSC}$ side information: We now denote the genie score for the $i^{\mathrm{th}}$ label with $\mathsf{BSC}$ side information and without side information by $z_{i}^{*}$ and $z^{\prime}_{i}$ respectively. By Lemma [2.10](#S2.Thmlemma10), for any $i\in[n]$

$$ $z^{*}_{i}=z^{\prime}_{i}+y_{i}\log\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)=z^{\prime}_{i}+y_{i}\beta\log n.$ $$

Consider any $i\in C_{+}$. For any constants $\delta,t>0$,

$$ $\displaystyle\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)$ $\displaystyle=\operatorname*{\mathbb{P}}(y_{i}=+1)\operatorname*{\mathbb{P}}(z^{*}_{i}<\delta\log n\mid y_{i}=+1)+\operatorname*{\mathbb{P}}(y_{i}=-1)\operatorname*{\mathbb{P}}(z^{*}_{i}<\delta\log n\mid y_{i}=-1)$ $\displaystyle=(1-\alpha_{n})\operatorname*{\mathbb{P}}(z^{\prime}_{i}+\beta\log n<\delta\log n)+\alpha_{n}\cdot\operatorname*{\mathbb{P}}(z^{\prime}_{i}-\beta\log n<\delta\log n)$ $\displaystyle\leq\operatorname*{\mathbb{P}}(z^{\prime}_{i}+\beta\log n<\delta\log n)+n^{-\beta}\operatorname*{\mathbb{P}}(z^{\prime}_{i}-\beta\log n<\delta\log n)$ $\displaystyle\leq\operatorname*{\mathbb{P}}(z^{\prime}_{i}<(\delta-\beta)\log n)+n^{-\beta}\operatorname*{\mathbb{P}}(z^{\prime}_{i}<(\delta+\beta)\log n)$ $\displaystyle\leq n^{-D_{t}(\theta_{-},\theta_{+})-\beta t+\delta t+o(1)}+n^{-\beta}n^{-D_{t}(\theta_{-},\theta_{+})+\beta t+\delta t+o(1)}+O(n^{-3})$ $\displaystyle=n^{-D_{t}(\theta_{-},\theta_{+})-\beta t+\delta t+o(1)}+n^{-D_{t}(\theta_{-},\theta_{+})-\beta(1-t)+\delta t+o(1)}+O(n^{-3})$ $\displaystyle:=P_{1}(\delta,t)+P_{2}(\delta,t)+O(n^{-3})$ $$

Above the IT threshold in this case (see Table [2](#S3.T2)), we have

$$ $\sup_{t\in[0,1]}\beta t+D_{t}(\theta_{+},\theta_{-})>1\text{ and }\sup_{t\in[0,1]}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})>1.$ $$

Replacing $t$ with $1-t$ also implies

$$ $\sup_{t\in[0,1]}\beta(1-t)+D_{t}(\theta_{-},\theta_{+})>1\text{ and }\sup_{t\in[0,1]}\beta t+D_{t}(\theta_{-},\theta_{+})>1.$ $$

Therefore, it is possible to choose $t_{1},t_{2}\in(0,1)$ and $\delta_{1},\delta_{2}>0$ small enough such that $P_{1}(\delta_{1},t_{1})=o(n^{-1})$ and $P_{2}(\delta_{2},t_{2})=o(n^{-1})$.

We finally turn our focus to any $i\in C_{-}$. Again by similar calculation, for any $\delta,t>0$

$$ $\displaystyle\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)$ $\displaystyle=(1-\alpha_{n})\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{\prime}_{i}+\beta\log n<\delta\log n)+\alpha_{n}\cdot\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{\prime}_{i}-\beta\log n<\delta\log n)$ $\displaystyle\leq\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{\prime}_{i}+\beta\log n<\delta\log n)+n^{-\beta}\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{\prime}_{i}-\beta\log n<\delta\log n)$ $\displaystyle=\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{\prime}_{i}<(\delta-\beta)\log n)+n^{-\beta}\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{\prime}_{i}<(\delta+\beta)\log n)$ $\displaystyle\leq n^{-D_{t}(\theta_{+},\theta_{-})-\beta t+\delta t+o(1)}+n^{-D_{t}(\theta_{-},\theta_{+})-\beta(1-t)+\delta t+o(1)}+O(n^{-3})$ $\displaystyle:=P_{3}(\delta,t)+P_{4}(\delta,t)+O(n^{-3}).$ $$

The second last line follows from taking a union bound with $E_{1}$ from ([30](#A5.E30)) and using Lemma [E.4](#A5.Thmlemma4). Finally, above the threshold, it is possible to choose $t_{3},t_{4}\in(0,1)$ and $\delta_{3},\delta_{4}>0$ small enough such that both $P_{3}(\delta_{3},t_{3})$ and $P_{4}(\delta_{4},t_{4})$ are $o(n^{-1})$. Choosing $\delta:=\min\{\delta_{1},\delta_{2},\delta_{3},\delta_{4}\}>0$ finally implies that for any $i\in[n]$

$$ $\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{*}_{i}<\delta\log n)=o(n^{-1}).\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

Finally, we prove Lemma [E.4](#A5.Thmlemma4).

###### Proof of Lemma E.4 .

Define $(p_{1},p_{2},q):=(a_{1},a_{2},b)\log n/n$. For any $i\in C_{+}$, note that

$$ $\sum_{j\in C_{+}^{-i}}A_{ij}\equiv\textnormal{Binom}(N_{1},p_{1})\textnormal{ and }\sum_{j\in C_{-}^{-i}}A_{ij}\equiv\textnormal{Binom}(N_{2},q).$ $$

where $N_{1}=(1+o(1))\rho n$ and $N_{2}=(1+o(1))(1-\rho)n$.
Let $Z_{1}\sim\textnormal{Binom}(N_{1},p_{1})$ and $Z_{2}\sim\textnormal{Binom}(N_{2},q)$ be independent.
Then $X_{i}\sim\log(a_{1}/b)Z_{1}+\log(b/a_{2})Z_{2}+\gamma$. For any constant $\varepsilon\in\mathbb{R}$ and $t>0$,

$$ $\displaystyle\operatorname*{\mathbb{P}}(X_{i}<(1+o(1))\varepsilon\log n)$ $\displaystyle\leq\operatorname*{\mathbb{P}}\left(\log(a_{1}/b)Z_{1}+\log(b/a_{2})Z_{2}+\gamma<(1+o(1))\varepsilon\log n\right)$ $\displaystyle\leq e^{(1+o(1))\varepsilon t\log n}\cdot\mathbb{E}\left[\exp\left(-t\left(\log\left(\frac{a_{1}}{b}\right)Z_{1}+\log\left(\frac{b}{a_{2}}\right)Z_{2}+\gamma\right)\right)\right]$ $\displaystyle=e^{(1+o(1))\delta t\log n}\cdot\mathbb{E}\left[e^{-t\log\left(\frac{a_{1}}{b}\right)Z_{1}}\right]\mathbb{E}\left[e^{-t\log\left(\frac{b}{a_{2}}\right)Z_{2}}\right]e^{-t\gamma}$ $\displaystyle=e^{(1+o(1))\varepsilon t\log n}\cdot\exp\left(N_{1}p_{1}(e^{-t\log(a_{1}/b)}-1)\right)$ $\displaystyle\hskip 5.69054pt\cdot\exp\left(N_{2}q(e^{-t\log(b/a_{2})}-1)\right)\cdot\exp(-t\gamma)$ (by Lemma A.2 ) $\displaystyle=e^{(1+o(1))\varepsilon t\log n}\cdot\exp\bigg{(}(1+o(1))\,\rho n\,\frac{a_{1}\log n}{n}\left(\left(\frac{b}{a_{1}}\right)^{t}-1\right)$ $\displaystyle\hskip 5.69054pt+(1+o(1))(1-\rho)n\,\frac{b\log n}{n}\left(\left(\frac{a_{2}}{b}\right)^{t}-1\right)-t\gamma\bigg{)}$ $\displaystyle=e^{(1+o(1))\varepsilon t\log n}\cdot\exp\bigg{(}(1+o(1))\log n\big{[}\rho(a_{1}^{1-t}b^{t}-a_{1})$ $\displaystyle\hskip 5.69054pt+(1-\rho)(b^{1-t}a_{2}^{t}-b)-t(\rho(b-a_{1})+(1-\rho)(a_{2}-b))\big{]}\bigg{)}$ $\displaystyle=e^{(1+o(1))\varepsilon t\log n}\cdot\exp\bigg{(}-(1+o(1))\log n\big{[}(1-t)\rho a_{1}+t\rho b$ $\displaystyle\hskip 5.69054pt+(1-t)(1-\rho)b+t(1-\rho)a_{2}-\rho a_{1}^{1-t}b^{t}-(1-\rho)(b^{1-t}a_{2}^{t})\big{]}\bigg{)}$ $\displaystyle=\exp\left(\log n\left[-D_{t}(\theta_{-},\theta_{+})+\varepsilon t+o(1)\right]\right)$ $\displaystyle=n^{-D_{t}(\theta_{-},\theta_{+})+\varepsilon t+o(1)}.$ (41) $$

Similarly, for any $i\in C_{-}$, we have

$$ $\sum_{j\in C_{+}^{-i}}A_{ij}\equiv\textnormal{Binom}(N_{1},q)\textnormal{ and }\sum_{j\in C_{-}^{-i}}A_{ij}\equiv\textnormal{Binom}(N_{2},p_{2}).$ $$

where $N_{1}=(1+o(1))\rho n$ and $N_{2}=(1+o(1))(1-\rho)n$. Let $Z_{3}\sim\textnormal{Binom}(N_{1},q)$ and $Z_{4}\sim\textnormal{Binom}(N_{2},p_{2})$ be independent.
Then $X_{i}\sim\log(a_{1}/b)Z_{3}+\log(b/a_{2})Z_{4}+\gamma$.

$$ $\displaystyle\operatorname*{\mathbb{P}}(-X_{i}<(1+o(1))\varepsilon\log n)$ $\displaystyle\leq\operatorname*{\mathbb{P}}\left(-\log(a_{1}/b)Z_{3}-\log(b/a_{2})Z_{4}-\gamma<(1+o(1))\varepsilon\log n\right)$ $\displaystyle\leq e^{(1+o(1))\varepsilon t\log n}\cdot\mathbb{E}\left[\exp\left(t\left(\log\left(\frac{a_{1}}{b}\right)Z_{3}+\log\left(\frac{b}{a_{2}}\right)Z_{4}+\gamma\right)\right)\right]$ $\displaystyle=e^{(1+o(1))\delta t\log n}\cdot\mathbb{E}\left[e^{t\log\left(\frac{a_{1}}{b}\right)Z_{3}}\right]\mathbb{E}\left[e^{t\log\left(\frac{b}{a_{2}}\right)Z_{4}}\right]e^{t\gamma}$ $\displaystyle=e^{(1+o(1))\varepsilon t\log n}\cdot\exp\left(N_{1}q(e^{t\log(a_{1}/b)}-1)\right)$ $\displaystyle\hskip 5.69054pt\cdot\exp\left(N_{2}p_{2}(e^{t\log(b/a_{2})}-1)\right)\cdot\exp(t\gamma)$ (by Lemma A.2 ) $\displaystyle=e^{(1+o(1))\varepsilon t\log n}\cdot\exp\bigg{(}(1+o(1))\,\rho n\,\frac{b\log n}{n}\left(\left(\frac{a_{1}}{b}\right)^{t}-1\right)$ $\displaystyle\hskip 5.69054pt+(1+o(1))(1-\rho)n\,\frac{a_{2}\log n}{n}\left(\left(\frac{b}{a_{2}}\right)^{t}-1\right)+t\gamma\bigg{)}$ $\displaystyle=e^{(1+o(1))\varepsilon t\log n}\cdot\exp\bigg{(}(1+o(1))\log n\big{[}\rho(b^{1-t}a_{1}^{t}-b)$ $\displaystyle\hskip 5.69054pt+(1-\rho)(a_{2}^{1-t}b^{t}-a_{2})+t(\rho(b-a_{1})+(1-\rho)(a_{2}-b))\big{]}\bigg{)}$ $\displaystyle=e^{(1+o(1))\varepsilon t\log n}\cdot\exp\bigg{(}-(1+o(1))\log n\big{[}t\rho a_{1}+(1-t)\rho b$ $\displaystyle\hskip 5.69054pt+t(1-\rho)b+(1-t)(1-\rho)a_{2}-\rho a_{1}^{t}b^{1-t}-(1-\rho)(b^{t}a_{2}^{1-t})\big{]}\bigg{)}$ $\displaystyle=\exp\left(\log n\left[-D_{t}(\theta_{+},\theta_{-})+\varepsilon t+o(1)\right]\right)$ $\displaystyle=n^{-D_{t}(\theta_{+},\theta_{-})+\varepsilon t+o(1)}.\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ (42) $$

###### Lemma E.1 .

###### Proof.

###### Lemma E.2 .

###### Proof.

###### Lemma E.3 .

###### Lemma E.4 .

###### Proof of Lemma E.3 .

###### Proof of Lemma E.4 .

### E.2 Failure of genie-aided estimation below the threshold

We begin by showing a helper lemma.

###### Lemma E.5 .

Consider $\rho\in(0,1)$ and $a_{1},a_{2},b>0$. Let $(A,\sigma^{*})\sim\mathsf{SBM}_{n}(\rho,a_{1},a_{2},b)$. Condition on $\sigma^{*}$ such that the event $E$ from ([11](#A2.E11)) holds. Fix any set $T\subset C_{+}$ or $T\subset C_{-}$ such that $|T|=O(n/\log^{10}n)$. Then for any $i\in[n]$, define $Y_{i}:=\sum_{j\in T}A_{ij}$. Then

$$ $\operatorname*{\mathbb{P}}\left(\forall i\in[n]:Y_{i}\leq\frac{\log n}{\log\log n}\right)\geq 1-O(n^{-3}).$ $$

###### Proof.

Fix any set $T$ according to the lemma and define $\{Y_{i}:i\in[n]\}$. Let $\tau=\max\{a_{1},a_{2},b\}$ and $Y\sim\textnormal{Binom}(|T|,\frac{\tau\log n}{n})$. Observe that for any $i\in[n]$, we have $Y_{i}$ is stochastically dominated by $Y$. Applying the following Chernoff bound for Binomial random variable [, Thereom 4.4]: for any $r>1$ and $X\sim\textnormal{Binom}(N,p)$, $\operatorname*{\mathbb{P}}(X\geq rnp)\leq(e/r)^{rnp}$, we obtain for any $i\in[n]$

$$ $\displaystyle\operatorname*{\mathbb{P}}\Big{(}Y_{i}\leq\frac{\log n}{\log\log n}\Big{)}$ $\displaystyle\leq\operatorname*{\mathbb{P}}\Big{(}Y\leq\frac{\log n}{\log\log n}\Big{)}\leq\left(\frac{e|T|\tau\log n/n}{\log n/\log\log n}\right)^{\frac{\log n}{\log\log n}}$ $\displaystyle=O\left(\frac{\log\log n}{\log^{10}n}\right)^{\frac{\log n}{\log\log n}}=n^{-10+o(1)}.$ $$

A simple union bound over all $i\in[n]$ gives us

$$ $\operatorname*{\mathbb{P}}\left(\forall i\in[n]:Y_{i}\leq\frac{\log n}{\log\log n}\right)\geq 1-O(n^{-3}).\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

We will handle the degenerate case of $a_{1}=a_{2}=b$ when we show the failure of MAP. Throughout this section, we focus on the case when they are not equal simultaneously. Below is the lemma that we want to show.

###### Lemma E.6 .

Fix $\rho\in(0,1)$ and $a_{1},a_{2},b>0$ such that not all are equal simultaneously. Let $(A,\sigma^{*})\sim\mathsf{ROS}_{n}(\rho,a_{1},a_{2},b)$ and condition on $\sigma^{*}$ such that $E$ from ([11](#A2.E11)) holds. Let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$ where $\epsilon_{n}$ and $\alpha_{n}$ are parameterized as ([1](#S1.E1)). Let $z^{*}$ be the genie score vector the corresponding model of side information. Then below the IT threshold given by Table [2](#S3.T2)

$$ $\operatorname*{\mathbb{P}}\left(\exists i\in[n]:\sigma^{*}_{i}z^{*}_{i}<\frac{-2\log n}{\log\log n}\right)\geq 1-o(1).$ $$

The proof of this lemma follows similar ideas from the proof of Lemma [D.5](#A4.Thmlemma5).
We have conditioned on $\sigma^{*}$ such that $E$ happens. Choose fixed $T_{+}\subset C_{+}$ and $T_{-}\subset C_{-}$ such that both $|T_{+}|$ and $|T_{-}|$ are $\Theta(n/\log^{10}n)$. It would be possible to choose such sets because of community sizes under $E$. For each $i\in T_{+}\cup T_{-}$, we define two random variables as follows:

$$ $H_{i}:=\log\left(\frac{a_{1}}{b}\right)\sum_{j\in C_{+}\setminus T_{+}}A_{ij}+\log\left(\frac{b}{a_{2}}\right)\sum_{j\in C_{-}\setminus T_{-}}A_{ij}+\gamma,$ $$

where $\gamma$ is defined in Lemma [E.2](#A5.Thmlemma2), and

$$ $Y_{i}:=\log\left(\frac{a_{1}}{b}\right)\sum_{j\in T_{+}}A_{ij}+\log\left(\frac{b}{a_{2}}\right)\sum_{j\in T_{-}}A_{ij}.$ $$

The purpose of defining $\{H_{i}\}_{i\in T_{+}\cup T_{-}}$ in this fashion is that they are all independent now using the same argument used for $\mathsf{ROS}$ in Lemma [D.5](#A4.Thmlemma5); it just follows from the independence of the entries of $A$ and that $H_{i}$ does not depend on $\{A_{ij}\}_{j\in T_{+}\cup T_{-}}$. Under $\mathsf{BEC}$ side information, let $R_{i}$ be the random variables that are formed after applying the same transformation that the genie score would go under from the case of no side information to $\mathsf{BEC}$ side information in Lemma [2.10](#S2.Thmlemma10); overriding with $+\infty$ or $-\infty$ when $y_{i}=+1$ or $y_{i}=-1$. Similarly, in the $\mathsf{BSC}$ channel, $R_{i}=H_{i}+\log\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)y_{i}$. Again, the random variables $\{R_{i}\}_{i\in T}$ are independent since $A$ and $y$ are independent conditioned on $\sigma^{*}$. Then we show the following lemma.

###### Lemma E.7 .

Below the IT threshold, for one of the $T:=T_{+}$ or $T:=T_{-}$, for all $i\in T$, there exists $\varepsilon>0$ such that

$$ $\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)\geq n^{-1+\varepsilon}.$ $$

Using this lemma, we now prove Lemma [E.6](#A5.Thmlemma6).

###### Proof of Lemma E.6 .

Let $z_{i}^{\prime}$ be the genie score without side information. From the definition of $Y_{i}$ and $H_{i}$, Lemma [E.2](#A5.Thmlemma2) implies that with probability $1-O(n^{-3})$, we have $|H_{i}+Y_{i}-z^{\prime}_{i}|=O(1)$ for all $i\in T_{+}\cup T_{-}$.
From the definition of $R_{i}$ and Lemma [2.10](#S2.Thmlemma10), we continue to have $|R_{i}+Y_{i}-z^{*}_{i}|=O(1)$ with probability $1-O(n^{-3})$, where $z^{*}_{i}$ is the genie score for the corresponding model of side information. Let $T$ be the set from Lemma [E.7](#A5.Thmlemma7). Using the lemma, we derive the desired result.

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\exists i\in[n]:\sigma^{*}_{i}z^{*}_{i}<\frac{-\log n}{\log\log n}\right)$ $\displaystyle\geq\operatorname*{\mathbb{P}}\left(\exists i\in T:\sigma^{*}_{i}z^{*}_{i}<\frac{-\log n}{\log\log n}\right)$ $\displaystyle\overset{(a)}{\geq}\operatorname*{\mathbb{P}}\left(\exists i\in T:\sigma^{*}_{i}R_{i}+\sigma^{*}_{i}Y_{i}+O(1)<\frac{-\log n}{\log\log n}\right)-o(1)$ $\displaystyle\overset{(b)}{\geq}\operatorname*{\mathbb{P}}\left(\exists i\in T:\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)-o(1),$ $\displaystyle\overset{(c)}{=}1-\prod_{i\in T}\left(1-\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)\right)-o(1)$ $\displaystyle\geq 1-\left(1-n^{-1+\varepsilon}\right)^{|T|}-o(1)$ (using Lemma E.7 ) $\displaystyle\geq 1-\exp\left(-|T|n^{-1+\varepsilon}\right)-o(1)$ ( $1+x\leq e^{x}$ ) $\displaystyle=1-\exp\left({-n^{\varepsilon-o(1)}}\right)-o(1)=1-o(1),$ $$

where the last step follows from $|T|=\Theta(\nicefrac{{n}}{{\log^{10}n}})=n^{1-o(1)}.$ The inequality (a) follows from the way we defined $R_{i}$ and $Y_{i}$, (b) follows from using Lemma [E.5](#A5.Thmlemma5) and (c) follows from the independence of $\{R_{i}\}_{i\in T}$ as argued already.

Note that each $R_{i}$ is a sum of weighted Binomial random variables up to some constant shift. The goal is to provide a lower bound on the tail of a binomial. To do this end, we will heavily use Lemma [A.3](#A1.Thmlemma3). To show Lemma [E.7](#A5.Thmlemma7), we need equivalent ways of writing the conditions that hold below the IT threshold.

###### Lemma E.8 .

Let $a_{1},a_{2},b>0$ such that not all are equal simultaneously. Refer to the IT thresholds from Table [2](#S3.T2). When no side information is present, below the IT threshold

$$ $\sup_{t\geq 0}D_{t}(\theta_{+},\theta_{-})<1\,\,\,\text{ \text@underline{and} }\,\,\,\sup_{t\geq 0}D_{t}(\theta_{-},\theta_{+})<1.$ $$

Under $\mathsf{BEC}$ side information, below the threshold

$$ $\beta+\sup_{t\geq 0}D_{t}(\theta_{+},\theta_{-})<1\,\,\,\text{ \text@underline{and} }\,\,\,\beta+\sup_{t\geq 0}D_{t}(\theta_{-},\theta_{+})<1.$ $$

Under the $\mathsf{BSC}$ channel, below the IT threshold

- 1.
When $\max\{\log(a_{1}/b),\log(b/a_{2}))\}\leq 0$
$\sup_{t\geq 0}\beta t+D_{t}(\theta_{-},\theta_{+})<1\,\,\text{ \text@underline{or} }\,\,\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{-},\theta_{+})<1.$
- 2.
When $\min\{\log(a_{1}/b),\log(b/a_{2}))\}\geq 0$
$\sup_{t\geq 0}\beta t+D_{t}(\theta_{+},\theta_{-})<1\,\,\text{ \text@underline{or} }\,\,\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})<1.$
- 3.
When $\max\{\log(a_{1}/b),\log(b/a_{2})\}>0$ and $\min\{\log(a_{1}/b),\log(b/a_{2})\}<0$
$\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})<1\text{ or }\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{-},\theta_{+})<1.$

###### Proof.

Let $\phi(t):=D_{t}(\theta_{+},\theta_{-})$. It is easy to verify that

$$ $\phi^{\prime\prime}(t)=-\rho b\left(\frac{a_{1}}{b}\right)^{t}\log^{2}\left(\frac{a_{1}}{b}\right)-(1-\rho)a_{2}\left(\frac{b}{a_{2}}\right)^{t}\log^{2}\left(\frac{b}{a_{2}}\right)<0\,,$ $$

when $(a_{1},a_{2},b)$ are not equal simultaneously. Thus $\phi:\mathbb{R}\rightarrow\mathbb{R}$ is a strictly concave function with a unique maximizer located in $[0,1]$ because $\phi(0)=\phi(1)=0$. Therefore, $\sup_{t\in[0,1]}D_{t}(\theta_{+},\theta_{-})=\sup_{t\in\mathbb{R}}D_{t}(\theta_{+},\theta_{-})=\sup_{t\geq 0}D_{t}(\theta_{+},\theta_{-})=\sup_{t\leq 1}D_{t}(\theta_{+},\theta_{-})=\sup_{t\geq 0}D_{t}(\theta_{-},\theta_{+})$.

Under no side information, below the threshold corresponds to $\sup_{t\in[0,1]}D_{t}(\theta_{+},\theta_{-})<1$. The above calculation implies $\sup_{t\geq 0}D_{t}(\theta_{+},\theta_{-})<1$ and $\sup_{t\geq 0}D_{t}(\theta_{-},\theta_{+})<1$. Under the $\mathsf{BEC}$ channel, below the threshold, we have $\beta+\sup_{t\in[0,1]}D_{t}(\theta_{+},\theta_{-})<1$. The same calculation implies $\beta+\sup_{t\geq 0}D_{t}(\theta_{+},\theta_{-})=\beta+\sup_{t\geq 0}D_{t}(\theta_{-},\theta_{+})<1$.

Finally, we consider the $\mathsf{BSC}$ channel.
It is straightforward to verify that $\phi_{1}(t):=\beta t+D_{t}(\theta_{+},\theta_{-})$ and $\phi_{2}(t):=\beta(1-t)+D_{t}(\theta_{-},\theta_{+})$ are again strictly concave functions, each with a unique minimizer.
$\phi^{\prime}(0)=D^{\prime}_{t}(\theta_{+},\theta_{-})|_{t=0}>0$ and $\phi^{\prime}(1)=D^{\prime}_{t}(\theta_{+},\theta_{-})|_{t=1}<0$. We use these observations below.

- 1.
When both $\log(a_{1}/b)$ and $\log(b/a_{2})$ are non-positive: If $\sup_{t\in[0,1]}\beta t+D_{t}(\theta_{+},\theta_{-})<1,$ we have $\sup_{t\leq 1}\beta t+D_{t}(\theta_{+},\theta_{-})<1$, giving us $\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{-},\theta_{+})<1.$ We now consider $\sup_{t\in[0,1]}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})<1$. Here, we have to consider two further subcases: first consider $0\geq\phi_{2}^{\prime}(0)=-\beta+D^{\prime}_{t}(\theta_{+},\theta_{-})|_{t=0}$. Then using $a_{1}\leq b\leq a_{2}$, it can be deduced that $-D^{\prime}_{t}(\theta_{+},\theta_{-})|_{t=1}\leq D^{\prime}_{t}(\theta_{+},\theta_{-})|_{t=0}\leq\beta$; rather than, this implies $\beta+D^{\prime}_{t}(\theta_{+},\theta_{-})|_{t=1}=\phi_{1}^{\prime}(1)\geq 0$. Thus, $\sup_{t\in[0,1]}\beta t+D_{t}(\theta_{+},\theta_{-})=\sup_{t\leq 1}\beta t+D_{t}(\theta_{+},\theta_{-})=\beta<1$. Again, we conclude $\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{-},\theta_{+})<1$. Lastly, the other subcase is $0<\phi_{2}^{\prime}(0)$. In this case, the unique minimizer of $\phi_{2}$ is located in $[0,1]$ and we have $\sup_{t\leq 1}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})=\sup_{t\in[0,1]}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})<1$, which implies $\sup_{t\geq 0}\beta t+D_{t}(\theta_{-},\theta_{+})<1$.
- 2.
When both $\log(a_{1}/b)$ and $\log(b/a_{2})$ are nonnegative, i.e $a_{1}\geq b\geq a_{2}$: If we have $\sup_{t\in[0,1]}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})<1$, then it immediately implies $\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})<1$. If $\sup_{t\in[0,1]}\beta t+D_{t}(\theta_{+},\theta_{-})<1$, we again need to consider two subcases: the first is $\phi_{1}^{\prime}(1)>0$. We have $\beta\geq-D^{\prime}_{t}(\theta_{+},\theta_{-})|_{t=1}\geq D^{\prime}_{t}(\theta_{+},\theta_{-})|_{t=0}$, where the last inequality follows from $a_{1}\geq b\geq a_{2}$. Therefore,
$\phi_{2}^{\prime}(0)=-\beta+D^{\prime}_{t}(\theta_{+},\theta_{-})|_{t=0}\leq 0.$ This gives us $\sup_{t\in[0,1]}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})=\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})=\beta<1$. Finally, the other subcase is $\phi_{1}^{\prime}(1)\leq 0$. In this case, the function $\phi_{1}(t)$ has its unique minimizer in $[0,1]$ and we have $\sup_{t\geq 0}\beta t+D_{t}(\theta_{+},\theta_{-})<1$.
- 3.
When $(\log(a_{1}/b),\log(b/a_{2}))$ has one each positive and negative: If $\sup_{t\in[0,1]}\beta t+D_{t}(\theta_{+},\theta_{-})<1$, we have $\sup_{t\leq 1}\beta t+D_{t}(\theta_{+},\theta_{-})<1$, which is equivalent to $\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{-},\theta_{+})<1$. In the other case, we have $1>\sup_{t\in[0,1]}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})=\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})$, which concludes the proof.

We return to the pending proof of Lemma [E.7](#A5.Thmlemma7).

###### Proof of Lemma E.7 .

Denote $Z_{1}\sim\textnormal{Binom}(N_{1},p_{1}),Z_{2}\sim\textnormal{Binom}(N_{2},q),Z_{3}\sim\textnormal{Binom}(N_{1},q),$ and $Z_{4}\sim\textnormal{Binom}(N_{2},p_{2})$, where $N_{1}:=(1+o(1))\rho n,N_{2}=(1+o(1))(1-\rho)n$ and $(p_{1},p_{2},q):=(a_{1},a_{2},b)\nicefrac{{\log n}}{{n}}$. We handle each side information separately.
No side information: By our consideration in the lemma, when not all $(a_{1},a_{2},b)$ are equal simultaneously, at least one of $\log(a_{1}/b)$ and $\log(b/a_{2})$ is non-zero. We first consider the case when at least one of them is negative. For any $i\in T_{+}$, first observe that $R_{i}$ has the same distribution as $\log(a_{1}/b)Z_{1}+\log(b/a_{2})Z_{2}+\gamma$.

$$ $\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)\geq\operatorname*{\mathbb{P}}\left(\log(a_{1}/b)Z_{1}+\log(b/a_{2})Z_{2}\leq-(1+o(1))\gamma\right)$ (43) $$

Thus, we will apply Lemma [A.3](#A1.Thmlemma3) with $\delta:=-\gamma/\log n=\rho(a_{1}-b)+(1-\rho)(b-a_{2})$ which is a constant. Using $\log(x)\geq 1-\frac{1}{x}$, it is immediately verifiable that

$$ $\displaystyle\log(a_{1}/b)\rho a_{1}+\log(b/a_{2})(1-\rho)b-\delta\geq 0.$ $$

Applying Lemma [A.3](#A1.Thmlemma3) for the two binomials $Z_{1}$ and $Z_{2}$ for the tail in ([43](#A5.E43)),

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(R_{i}<\frac{-3\log n}{\log\log n}\right)\geq n^{-I^{*}+o(1)},$ $\displaystyle\textnormal{where }I^{*}=\sup_{t\geq 0}\left(-t(a_{1}-b)\rho-t(1-\rho)(b-a_{2})+\rho a_{1}\left(1-\left(\frac{b}{a_{1}}\right)^{t}\right)+(1-\rho)b\left(1-\left(\frac{a_{2}}{b}\right)^{t}\right)\right)$ $\displaystyle=\sup_{t\geq 0}D_{t}(\theta_{-},\theta_{+}).$ (following identical calculation used in deriving ( 41 )) $$

Similarly, when both $\log(a_{1}/b)$ and $\log(b/a_{2})$ are non-negative, at least one of $(-\log(a_{1}/b),-\log(b/a_{2}))$ is negative; recall that they both are not zero simultaneously when $(a_{1},a_{2},b)$ are not all equal. We consider any $i\in T_{-}$, then $R_{i}$ has the same distribution as $\log(a_{1}/b)Z_{3}+\log(b/a_{2})Z_{4}+\gamma$, and so

$$ $\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)\geq\operatorname*{\mathbb{P}}\left(-\log(a_{1}/b)Z_{3}-\log(b/a_{2})Z_{4}\leq(1+o(1))\gamma\right)$ (44) $$

We now apply Lemma [A.3](#A1.Thmlemma3) with $\delta:=\gamma/\log n=\rho(b-a_{1})+(1-\rho)(a_{2}-b)$. The fact $\log(x)\leq x-1$ gives us

$$ $\displaystyle-\log(a_{1}/b)\rho b-\log(b/a_{2})(1-\rho)a_{2}-\delta\geq 0.$ $$

Using Lemma [A.3](#A1.Thmlemma3) for the two binomials $Z_{3}$ and $Z_{4}$ in ([44](#A5.E44)), for any $i\in T_{-}$

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)\geq n^{-I^{*}+o(1)},$ $\displaystyle\textnormal{where }I^{*}=\sup_{t\geq 0}\left(t(a_{1}-b)\rho+t(1-\rho)(b-a_{2})+\rho b\left(1-\left(\frac{a_{1}}{b}\right)^{t}\right)+(1-\rho)a_{2}\left(1-\left(\frac{b}{a_{2}}\right)^{t}\right)\right)$ $\displaystyle=\sup_{t\geq 0}D_{t}(\theta_{+},\theta_{-}).$ (following identical calculation used in deriving ( 41 )) $$

Below the IT threshold, we have $\sup_{t\geq 0}D_{t}(\theta_{+},\theta_{-})<1$ by Lemma [E.8](#A5.Thmlemma8). Finally, we conclude that there exists $\varepsilon>0$ such that

$$ $\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)\geq n^{-1+\varepsilon}.$ $$

Overall, we established that at least for one of $T=T_{+}$ or $T=T_{-}$ the desired property from the lemma holds, under no side information.
$\mathsf{BEC}$ side information: The claim mainly just follows from the fact that $R_{i}$ is $+\infty$ or $-\infty$ (depending on the true label) when side information is given or otherwise just follow the same distribution as its corresponding counterpart when no side information is present. Doing the same case analysis, first suppose that at least one of $\log(a_{1}/b)$ or $\log(b/a_{2})$ is negative. For any $i\in T_{+}$

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)$ $\displaystyle=\epsilon_{n}\operatorname*{\mathbb{P}}\left(R_{i}<\frac{-3\log n}{\log\log n}\mid y_{i}=0\right)+(1-\epsilon_{n})\cdot\operatorname*{\mathbb{P}}\left(R_{i}<0\mid y_{i}=+1\right)$ $\displaystyle=n^{-\beta}\cdot n^{-\sup_{t\geq 0}D_{t}(\theta_{-},\theta_{+})+o(1)}.$ $$

From Lemma [E.8](#A5.Thmlemma8), use $\beta+\sup_{t\geq 0}D_{t}(\theta_{-},\theta_{+})<1$. This implies there exists $\varepsilon>0$ small enough such that

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)$ $\displaystyle\geq n^{-1+\varepsilon}.$ $$

Similarly, below the threshold, we also have $\beta+\sup_{t\geq 0}D_{t}(\theta_{+},\theta_{-})<1$. Thus, even when both $\log(a_{1}/b)$ and $\log(b/a_{2})$ are non-negative, we consider any $i\in T_{-}$, and conclude that

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)$ $\displaystyle=\epsilon_{n}\operatorname*{\mathbb{P}}\left(-R_{i}<\frac{-3\log n}{\log\log n}\mid y_{i}=0\right)+(1-\epsilon_{n})\cdot\operatorname*{\mathbb{P}}\left(-R_{i}<\frac{-3\log n}{\log\log n}\mid y_{i}=-1\right)$ $\displaystyle\geq n^{-\beta}\cdot n^{-\sup_{t\geq 0}D_{t}(\theta_{+},\theta_{-})+o(1)}$ $\displaystyle=n^{-1+\varepsilon}$ $$

for some small constant $\varepsilon>0$, as desired, concluding the $\mathsf{BEC}$ side information case.
$\mathsf{BSC}$ side information:
Refer Lemma [E.8](#A5.Thmlemma8) for the conditions that hold below the IT threshold. We first consider the case when both $\max\{\log(a_{1}/b),\log(b/a_{2})\}\leq 0$ and we have $\sup_{t\geq 0}\beta t+D_{t}(\theta_{-},\theta_{+})<1.$
For any $i\in T_{+}$,

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)$ $\displaystyle\geq\operatorname*{\mathbb{P}}(y_{i}=+1)\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\mid y_{i}=+1\right)$ $\displaystyle\geq(1-\alpha_{n})\operatorname*{\mathbb{P}}\left(\log(a_{1}/b)Z_{1}+\log(b/a_{2})Z_{2}\leq-(1+o(1))\gamma-\beta\log n\right)$ $$

Note that $\min\{\log(a_{1}/b),\log(b/a_{2})\}<0$ because not both are zero simultaneously. We apply Lemma [A.3](#A1.Thmlemma3) with $\delta:=-\gamma/\log n=\rho(a_{1}-b)+(1-\rho)(b-a_{2})-\beta$. Using $\log(x)\geq 1-\frac{1}{x}$, it is immediately verifiable that

$$ $\displaystyle\log(a_{1}/b)\rho a_{1}+\log(b/a_{2})(1-\rho)b-\delta\geq 0.$ $$

Applying Lemma [A.3](#A1.Thmlemma3) for the two binomials $Z_{1}$ and $Z_{2}$ for the tail in ([43](#A5.E43)),

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(R_{i}<\frac{-3\log n}{\log\log n}\right)\geq n^{-I^{*}+o(1)},$ $\displaystyle\textnormal{where }I^{*}=\sup_{t\geq 0}\left(t\beta-t(a_{1}-b)\rho-t(1-\rho)(b-a_{2})+\rho a_{1}\left(1-\left(\frac{b}{a_{1}}\right)^{t}\right)+(1-\rho)b\left(1-\left(\frac{a_{2}}{b}\right)^{t}\right)\right)$ $\displaystyle=\sup_{t\geq 0}\beta t+D_{t}(\theta_{-},\theta_{+}).$ (following identical calculation used in deriving ( 41 )) $$

Thus, using $\sup_{t\geq 0}\beta t+D_{t}(\theta_{-},\theta_{+})<1$, we can choose $\varepsilon>0$ such that

$$ $\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)\geq n^{-1+\varepsilon}.$ $$

The case when $\min\{\log(a_{1}/b),\log(b/a_{2})\}\geq 0$ and $\sup_{t\geq 0}\beta t+D_{t}(\theta_{+},\theta_{-})<1,$
follows exactly symmetrical argument but for $T=T_{-}$.

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)$ $\displaystyle=(1-\alpha_{n})\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\mid y_{i}=-1\right)$ (45) $\displaystyle\geq\operatorname*{\mathbb{P}}\left(-\log(a_{1}/b)Z_{3}-\log(b/a_{2})Z_{4}\leq(1+o(1))\gamma-\beta\log n\right)$ (46) $$

We observe that it must be that $\min\{-\log(a_{1}/b),-\log(b/a_{2})\}<0$. We again apply Lemma [A.3](#A1.Thmlemma3) with $\delta:=\gamma/\log n-\beta=\rho(b-a_{1})+(1-\rho)(a_{2}-b)-\beta$. Using $\log(x)\leq x-1$, we have that

$$ $\displaystyle-\log(a_{1}/b)\rho b-\log(b/a_{2})(1-\rho)a_{2}-\delta\geq 0.$ $$

Using Lemma [A.3](#A1.Thmlemma3) for the two binomials $Z_{3}$ and $Z_{4}$ in ([44](#A5.E44)), for any $i\in T_{-}$

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)\geq n^{-I^{*}+o(1)},$ $\displaystyle\textnormal{where }I^{*}=\sup_{t\geq 0}\left(t\beta+t(a_{1}-b)\rho+t(1-\rho)(b-a_{2})+\rho b\left(1-\left(\frac{a_{1}}{b}\right)^{t}\right)+(1-\rho)a_{2}\left(1-\left(\frac{b}{a_{2}}\right)^{t}\right)\right)$ $\displaystyle=\sup_{t\geq 0}\beta t+D_{t}(\theta_{+},\theta_{-}).$ (following identical calculation used in deriving ( 41 )) $$

Given that $\sup_{t\geq 0}\beta t+D_{t}(\theta_{+},\theta_{-})<1$, we can choose $\varepsilon>0$ such that

$$ $\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)\geq n^{-1+\varepsilon}.$ $$

We now turn to the cases when $\min\{\log(a_{1}/b),\log(b/a_{2})\}<0$ and $\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{-},\theta_{+})<1$. This covers one scenario each from ([1](#A5.Ex386)) and ([3](#A5.Ex388)) in Lemma [E.8](#A5.Thmlemma8).
For any $i\in T_{+}$,

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)$ $\displaystyle\geq\operatorname*{\mathbb{P}}(y_{i}=-1)\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\mid y_{i}=-1\right)$ $\displaystyle\geq\alpha_{n}\operatorname*{\mathbb{P}}\left(\log(a_{1}/b)Z_{1}+\log(b/a_{2})Z_{2}\leq-(1+o(1))\gamma+\beta\log n\right)$ $$

Applying Lemma [A.3](#A1.Thmlemma3) with $\delta:=-\gamma/\log n=\rho(a_{1}-b)+(1-\rho)(b-a_{2})+\beta$ for the two binomials $Z_{1}$ and $Z_{2}$

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(R_{i}<\frac{-3\log n}{\log\log n}\right)\geq n^{-\beta}\cdot n^{-I^{*}+o(1)},$ $\displaystyle\textnormal{where }I^{*}=\sup_{t\geq 0}\left(-t\beta-t(a_{1}-b)\rho-t(1-\rho)(b-a_{2})+\rho a_{1}\left(1-\left(\frac{b}{a_{1}}\right)^{t}\right)+(1-\rho)b\left(1-\left(\frac{a_{2}}{b}\right)^{t}\right)\right)$ $\displaystyle=\sup_{t\geq 0}-\beta t+D_{t}(\theta_{-},\theta_{+}).$ (following identical calculation used in deriving ( 41 )) $$

Thus, when $\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{-},\theta_{+})<1$, there exists $\varepsilon>0$ such that

$$ $\operatorname*{\mathbb{P}}\left(R_{i}<\frac{-3\log n}{\log\log n}\right)\geq n^{-1+\varepsilon}.$ $$

Finally, we turn our attention to $\min\{-\log(a_{1}/b),-\log(b/a_{2})\}<0$ and $\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})<1$. This covers the remaining scenarios from ([1](#A5.Ex386)) and ([3](#A5.Ex388)) in Lemma [E.8](#A5.Thmlemma8). For any $i\in T_{-}$,

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)$ $\displaystyle\geq\operatorname*{\mathbb{P}}(y_{i}=+1)\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\mid y_{i}=+1\right)$ $\displaystyle\geq\alpha_{n}\operatorname*{\mathbb{P}}\left(-\log(a_{1}/b)Z_{3}-\log(b/a_{2})Z_{4}\leq(1+o(1))\gamma+\beta\log n\right)$ $$

We apply Lemma [A.3](#A1.Thmlemma3) with $\delta:=\gamma/\log n+\beta=\rho(b-a_{1})+(1-\rho)(a_{2}-b)+\beta$.

$$ $\displaystyle\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)\geq n^{\beta}\cdot n^{-I^{*}+o(1)},$ $\displaystyle\textnormal{where }I^{*}=\sup_{t\geq 0}\left(-t\beta+t(a_{1}-b)\rho+t(1-\rho)(b-a_{2})+\rho b\left(1-\left(\frac{a_{1}}{b}\right)^{t}\right)+(1-\rho)a_{2}\left(1-\left(\frac{b}{a_{2}}\right)^{t}\right)\right)$ $\displaystyle=\sup_{t\geq 0}-\beta t+D_{t}(\theta_{+},\theta_{-}).$ (following identical calculation used in deriving ( 41 )) $$

Using $\sup_{t\geq 0}\beta(1-t)+D_{t}(\theta_{+},\theta_{-})<1$, we conclude that there exists $\varepsilon>0$ such that

$$ $\operatorname*{\mathbb{P}}\left(\sigma^{*}_{i}R_{i}<\frac{-3\log n}{\log\log n}\right)\geq n^{-1+\varepsilon}.$ $$

Overall, we established that at least for one of $T=T_{+}$ or $T=T_{-}$, the desired property from the lemma holds under $\mathsf{BSC}$ side information, concluding the proof of the lemma.

###### Lemma E.5 .

###### Proof.

###### Lemma E.6 .

###### Lemma E.7 .

###### Proof of Lemma E.6 .

###### Lemma E.8 .

###### Proof.

###### Proof of Lemma E.7 .

### E.3 Spectral Algorithm and Proof of Theorem 3

In this section, we present our spectral algorithm for the $\mathsf{SBM}$ that can emulate the genie. As discussed in Section [4.2](#S4.SS2), this requires taking an appropriate linear combination of eigenvectors such that the top two eigenvectors such that $c_{1}u_{1}^{*}+c_{2}u_{2}^{*}$ approximates $w$ in the $\ell_{\infty}$ norm. The vector $w$ is a block vector with entries $(w_{+},w_{-})$ on the locations of $C_{+}$ and $C_{-}$. Recall by Lemma [E.2](#A5.Thmlemma2) that

$$ $w_{+}=\log\left(\frac{a_{1}}{b}\right)\quad\text{ and }\quad w_{-}=\log\left(\frac{b}{a_{2}}\right).$ $$

We first present a subroutine that finds these coefficients $(c_{1},c_{2})$. We will introduce the formal correctness of the subroutine later, but first we provide informal discussion as to how these coefficients are computed. Roughly speaking, both $u_{1}^{*}$ and $u_{2}^{*}$ also have a block structure, and therefore, finding $(c_{1},c_{2})$ just corresponds to solving a system of $2\times 2$ linear equations. Also, the coefficients do not depend on the locations of $\sigma^{*}$ with $+1$ or $-1$ labels, so exploiting this fact we just do calculation as if $C_{+}$ is on the first ${\left\lfloor{\rho n}\right\rfloor}$ vertices and compute the proxy for actual $A^{*}$. This results into the following subroutine.

Algorithm 3 Find Linear Combination Coefficients

$$ $B_{ij}=\begin{cases}a_{1}\log n/n&\textnormal{if }i,j\in S;\\ b\log n/n&\textnormal{if }i\in S\text{ but }j\notin S;\\ a_{2}\log n/n&\textnormal{if }i\notin S\text{ and }j\notin S.\\ \end{cases}$ $$

$$ $w_{i}=\begin{cases}w_{+}=\log(a_{1}/b),&\textnormal{if }i\in S;\\ w_{-}=\log(b/a_{2})&\textnormal{if }i\notin S.\end{cases}$ $$

$$ $c_{1}\left(\frac{\tilde{v}_{1}}{\tilde{\lambda}_{1}}\right)+c_{2}\left(\frac{\tilde{v}_{2}}{\tilde{\lambda}_{2}}\right)=w.$ (47) $$

We note that in the rank-2 case when $a_{1}a_{2}\neq b^{2}$, the vectors $\tilde{v}_{1}$ and $\tilde{v}_{2}$ have block structure and are linearly independent. Therefore, it is possible to span any vector with block structure, and in particular, even $w$. We now propose our spectral algorithm
in the rank-2 case.

Algorithm 4 Spectral recovery algorithm for $\mathsf{SBM}$ (Rank-2)

$$ $\gamma:=(\rho(b-a_{1})+(1-\rho)(a_{2}-b))\log n.$ $$

- •
No side information:
$z^{(s)}=s_{1}c_{1}u_{1}+s_{2}c_{2}u_{2}+\gamma\mathbf{1}_{n}.$
- •
$\mathsf{BEC}$ side information: For any $i\in[n]$,
$z^{(s)}_{i}=\begin{cases}s_{1}c_{1}(u_{1})_{i}+s_{2}c_{2}(u_{2})_{i}+\gamma&\textnormal{if }y_{i}=0;\\
+\infty,&\textnormal{if }y_{i}=+1;\\
-\infty&\textnormal{if }y_{i}=-1.\end{cases}$
- •
$\mathsf{BSC}$ side information:
$z^{(s)}=s_{1}c_{1}u_{1}+s_{2}c_{2}u_{2}+\gamma\mathbf{1}_{n}+\ln\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)y.$

- •
No side information: Return $\hat{\sigma}_{\text{spec}}=\operatorname*{arg\,max}_{\{\hat{\sigma}^{(s)}:s\in\{\pm 1\}\}}\operatorname*{\mathbb{P}}(\sigma^{*}=\hat{\sigma}^{(s)}\mid A)$.
- •
$\mathsf{BEC}$ or $\mathsf{BSC}$ side information: Return $\hat{\sigma}_{\text{spec}}=\operatorname*{arg\,max}_{\{\hat{\sigma}^{(s)}:s\in\{\pm 1\}\}}\operatorname*{\mathbb{P}}(\sigma^{*}=\hat{\sigma}^{(s)}\mid A,y)$.

Finally, when $a_{1}a_{2}=b^{2}$, from Lemma [E.2](#A5.Thmlemma2) we have $w_{+}=w_{-}=\log\left(\frac{a_{1}}{b}\right).$ In this case, we can just use the deterministic vector along $\mathbf{1}_{n}$ to emulate the genie-score vector. Strictly speaking, in this degenerate case, we do not even need a spectral strategy to achieve optimality. Despite this, we just refer to this as a spectral algorithm in the rest of the analysis for simplicity of exposition.

Algorithm 5 (Spectral) Recovery algorithm for $\mathsf{SBM}$ (Rank-1)

$$ $c:=\log\left(\frac{a_{1}}{b}\right),\quad\gamma:=(\rho(b-a_{1})+(1-\rho)(a_{2}-b))\log n.$ $$

- •
No side information:
$z^{\textnormal{spec}}=cA\mathbf{1}_{n}+\gamma\mathbf{1}_{n}.$
- •
$\mathsf{BEC}$ side information: For any $i\in[n]$,
$z^{\textnormal{spec}}_{i}=\begin{cases}c(A\mathbf{1})_{i}+\gamma&\textnormal{if }y_{i}=0;\\
+\infty,&\textnormal{if }y_{i}=+1;\\
-\infty&\textnormal{if }y_{i}=-1.\end{cases}$
- •
$\mathsf{BSC}$ side information:
$z^{\textnormal{spec}}=cA\mathbf{1}_{n}+\gamma\mathbf{1}_{n}+\ln\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)y.$

We now show that the score vectors formed by these algorithms approximate the genie score vector $z^{*}$ in each case.

###### Lemma E.9 .

Consider $\rho\in(0,1)$ and $a_{1},a_{2},b>0$. Let $(A,\sigma^{*})\sim\mathsf{SBM}_{n}(\rho,a_{1},a_{2},b)$ and condition on $\sigma^{*}$ such that $E$ from ([11](#A2.E11)) holds. Optionally, let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$, where $\epsilon_{n}$ and $\alpha_{n}$ is parameterized as ([1](#S1.E1)) for $\beta\geq 0$. Let $z^{*}$ be the genie score vector for the corresponding model. Then (irrespective of the parameter values), with probability $1-o(1)$

- •
Case $a_{1}a_{2}\neq b^{2}$: for some $s=(s_{1},s_{2})\in\{\pm 1\}^{2}$
$\left\lVert z^{*}-z^{(s)}\right\rVert_{\infty}=o(\log n).$
- •
Case $a_{1}a_{2}=b^{2}$:
$\left\lVert z^{*}-z^{\textnormal{spec}}\right\rVert_{\infty}=o(\log n)$

###### Proof.

Recall from Lemma [2.10](#S2.Thmlemma10), how the genie score changes in the presence of $\mathsf{BEC}$ and $\mathsf{BSC}$ side information. In Algorithm [E.3](#A5.SS3) (step [5](#algx4.l5)) and Algorithm [E.3](#A5.SS3) (step [4](#algx5.l4)), this is precisely how the spectral score vectors are updated from the case when no side information is present. Therefore, it suffices to show that the score approximation holds in the case when no side information is present, which now will be the focus of the proof. The argument is exactly analogous to the one done used in Lemma [D.10](#A4.Thmlemma10). We now discuss the rank 1 and rank 2 cases one by one.
Rank-1 case: $a_{1}a_{2}=b^{2}$. In this case, when no side information is present

$$ $\displaystyle\left\lVert z^{*}-z^{\textnormal{spec}}\right\rVert_{\infty}=\left\lVert z^{*}-\log(\nicefrac{{a_{1}}}{{b}})A\mathbf{1}_{n}-\gamma\mathbf{1}_{n}\right\rVert_{\infty}=O(1),$ $$

where the last equation follows from Lemma [E.2](#A5.Thmlemma2) and using that $\frac{a_{1}}{b}=\frac{b}{a_{2}}$.
Rank-2 case: $a_{1}a_{2}\neq b^{2}$. Fix $(s_{1},s_{2})\in\{\pm 1\}^{n}$ to be the signs for which the high probability event in Lemma [C.4](#A3.Thmlemma4) holds. Define $w$ to be the vector from Lemma [E.2](#A5.Thmlemma2) with entries $(w_{+},w_{-})=(\log(a_{1}/b),\log(b/a_{2}))$ on the locations of $C_{+}$ and $C_{-}$ respectively. In this case, using Lemma [C.4](#A3.Thmlemma4), we will show that with probability $1-o(1)$, we have

$$ $\left\lVert Aw-s_{1}c_{1}u_{1}-s_{2}c_{2}u_{2}\right\rVert_{\infty}=o(\log n).$ (48) $$

Combing this ([48](#A5.E48)) along with Lemma [E.2](#A5.Thmlemma2) implies the desired result: with probability $1-o(1)$,

$$ $\displaystyle\left\lVert z^{*}-z^{(s)}\right\rVert_{\infty}$ $\displaystyle\leq\left\lVert z^{*}-Aw-\gamma\mathbf{1}_{n}\right\rVert_{\infty}+\left\lVert Aw+\gamma\mathbf{1}_{n}-z^{(s)}\right\rVert_{\infty}$ $\displaystyle=\left\lVert z^{*}-Aw-\gamma\mathbf{1}_{n}\right\rVert_{\infty}+\left\lVert Aw-s_{1}c_{1}u_{1}-s_{2}c_{2}u_{2}\right\rVert_{\infty}$ (step 5 of Algorithm E.3 ) $\displaystyle=o(\log n).$ $$

It remains to show ([48](#A5.E48)). Note that, we calculate $(c_{1},c_{2})$ in Algorithm [E.3](#A5.SS3) using the matrix $B$ where community sizes are exactly $\rho n$. But, condition on $E$, we have community sizes $(1+o(1))\rho n$ and $(1+o(1))(1-\rho)n$. Also, in $A^{*}$, we have zero diagonal, where as, the matrix $B$ has diagonal entries of the order of $O(\log n/n)$. These changes only affect the eigenvalues by the multiplicative factor of $(1+o(1))$ by Weyl’s inequality. Therefore,

$$ $\lambda_{1}^{*}=(1+o(1))\tilde{\lambda}_{1}\text{ and }\lambda_{2}^{*}=(1+o(1))\tilde{\lambda}_{2}.$ $$

Moreover, the entries of $u_{1}^{*}$ in location of $C_{+}$ are $(1+o(1))$ factor of the entries of $\tilde{v}_{1}$ in the location $S$. By the same argument, one can say the same for $u_{1}^{*}$ in locations of $C_{-}$ and $\tilde{v}_{1}$ in locations of $[n]\setminus S$, and also for $u_{2}^{*}$ and $\tilde{v}_{2}$. From the way we calculate $(c_{1},c_{2})$ in ([47](#A5.E47)), we have

$$ $c_{1}\left(\frac{\tilde{v}_{1}}{\tilde{\lambda}_{1}}\right)+c_{2}\left(\frac{\tilde{v}_{2}}{\tilde{\lambda}_{2}}\right)=w.$ $$

Then the above discussion implies

$$ $w_{+}=(1+o(1))\left(\frac{c_{1}u_{1,i}^{*}}{\lambda_{1}^{*}}+\frac{c_{2}u_{2,i}^{*}}{\lambda_{2}^{*}}\right),\text{ for $i\in C_{+}$ and }w_{-}=(1+o(1))\left(\frac{c_{1}u_{1,i}^{*}}{\lambda_{1}^{*}}+\frac{c_{2}u_{2,i}^{*}}{\lambda_{2}^{*}}\right),\text{for $i\in C_{-}$}.$ $$

Finally using Lemma [C.4](#A3.Thmlemma4), with probability $1-o(1)$

$$ $\displaystyle\left\lVert Aw-s_{1}c_{1}u_{2}-s_{2}c_{2}u_{2}\right\rVert_{\infty}$ $\displaystyle\leq\left\lVert Aw-c_{1}\frac{Au_{1}^{*}}{\lambda_{1}^{*}}-c_{2}\frac{Au_{2}^{*}}{\lambda_{2}^{*}}\right\rVert_{\infty}+o(1)$ $\displaystyle=\left\lVert(1+o(1))\left(c_{1}\frac{Au_{1}^{*}}{\lambda_{1}^{*}}+c_{2}\frac{Au_{2}^{*}}{\lambda_{2}^{*}}\right)-c_{1}\frac{Au_{1}^{*}}{\lambda_{1}^{*}}-c_{2}\frac{Au_{2}^{*}}{\lambda_{2}^{*}}\right\rVert_{\infty}+o(1)$ $\displaystyle=o(1)\left(\max_{i\in[n]}\lVert A_{i\cdot}\rVert_{1}\right)\cdot\left(\left\lVert\frac{c_{1}u_{1}^{*}}{\lambda_{1}^{*}}\right\rVert_{\infty}\vee\left\lVert\frac{c_{2}u_{2}^{*}}{\lambda_{2}^{*}}\right\rVert_{\infty}\right)$ $\displaystyle=o(\log n),$ $$

where the last step follows by using Lemma [E.1](#A5.Thmlemma1) and $(c_{1},c_{2})$ are chosen in such a way that the entries of vectors $\frac{c_{1}u_{1}^{*}}{\lambda_{1}^{*}}$ and $\frac{c_{2}u_{2}^{*}}{\lambda_{2}^{*}}$ are $O(1)$ by ([47](#A5.E47)).

We finally combine all the pieces and give the proof of Theorem [3](#Thmtheorem3).

###### Proof of Theorem 3 .

We first show the impossibility result below the IT threshold. When $a_{1}=a_{2}=b$, it is clear that the recovery is impossible without side information. Even under side information, when $\beta<1$, for at least $\omega(1)$ vertices, it is impossible to recover the true labels, and exact recovery is impossible. Also, under side information, verify that below the IT threshold simply corresponds to $\beta<1$ when $a_{1}=a_{2}=b$.

We now consider the case when all three $(a_{1},a_{2},b)$ are not same simultaneously. Let $z^{*}$ be the genie score vector (for the model of side information in consideration). By Lemma [E.6](#A5.Thmlemma6), below the IT threshold

$$ $\operatorname*{\mathbb{P}}\left(\exists i:[n]:\sigma^{*}_{i}z^{*}_{i}<\frac{-2\log n}{\log\log n}\right)=1-o(1).$ $$

The definition of genie scores ([3](#S2.E3)) and Lemma [2.9](#S2.Thmlemma9) then implies

$$ $\operatorname*{\mathbb{P}}(\hat{\sigma}_{\textnormal{\tiny MAP}}\textnormal{ fails})\geq\operatorname*{\mathbb{P}}(\exists i\in[n]:\hat{\sigma}_{\textnormal{\tiny Gen},i}\textnormal{ fails})=1-o(1).$ $$

Using the optimality of the MAP estimator, we finally conclude, for any estimator $\hat{\sigma}$,

$$ $\operatorname*{\mathbb{P}}\left(\hat{\sigma}\textnormal{ fails}\right)\geq\operatorname*{\mathbb{P}}(\hat{\sigma}_{\textnormal{\tiny MAP}}\textnormal{ fails})=1-o(1).$ $$

To show the positive result, we break into cases.

- 1.
Rank 1, i.e. $a_{1}a_{2}=b^{2}$: First note that Algorithm [E.3](#A5.SS3) creates a score vector $z^{\textnormal{spec}}\in\mathbb{R}^{n}$ such that $\left\lVert z^{*}-z^{\textnormal{spec}}\right\rVert_{\infty}=o(\log n)$ by Lemma [E.9](#A5.Thmlemma9). Using Lemma [E.3](#A5.Thmlemma3), above the IT threshold, there exists $\vartheta>0$ such that
$\operatorname*{\mathbb{P}}(\sigma^{*}_{i}z^{\textnormal{spec}}_{i}\leq\varsigma\log n)=o(n^{-1}).$
Taking a union bound,
$\operatorname*{\mathbb{P}}(\forall i\in[n],\,\sigma^{*}_{i}z^{\textnormal{spec}}_{i}>\varsigma\log n)=1-o(1).$
Finally, the algorithm outputs $\hat{\sigma}_{\textnormal{spec}}=\textnormal{sgn}(z^{\textnormal{spec}})$ (step [5](#algx5.l5)), this immediately implies
$\operatorname*{\mathbb{P}}(\hat{\sigma}_{\textnormal{spec}}=\sigma^{*})=1-o(1).$
- 2.
Rank 2, i.e. $a_{1}a_{2}\neq b^{2}$: We first note that, above the IT threshold (in the corresponding side information case), the estimator $\hat{\sigma}_{\textnormal{\tiny MAP}}$ succeeds with high probability. That is $\hat{\sigma}_{\textnormal{\tiny MAP}}=\sigma^{*}$, unless $a_{1}=a_{2}$ and no side information, in which case $\hat{\sigma}_{\textnormal{\tiny MAP}}\in\{\pm\sigma^{*}\}$, with high probability. This follows from the achievability result [, Theorem 6] when no/weak side information, and Theorem [2](#Thmtheorem2) when $\beta>1$ under side information cases. Recall that Algorithm [E.3](#A5.SS3) creates four candidates for $\sigma^{*}$ and chooses the one with maximum posterior probability. Due to statistical achievability, it suffices to show that one of the $\{\sigma^{(s)}:s=(s_{1},s_{2})\in\{\pm 1\}^{2}\}$ maintained by the algorithm is $\sigma^{*}$ with high probability.
To this end, recall Lemma [E.9](#A5.Thmlemma9) that with probability $1-o(1)$,
$\min_{s^{*}\in\{\pm 1\}}\left\lVert z^{*}-z^{(s)}\right\rVert_{\infty}=o(\log n).$
Combining this with Lemma [E.3](#A5.Thmlemma3), there exists $s^{*}\in\{\pm 1\}^{2}$ such that, there exists $\varsigma>0$ such that
$\operatorname*{\mathbb{P}}\left(\min_{i\in[n]}\sigma^{*}_{i}z_{i}^{(s^{*})}>\varsigma\log n\right)=1-o(1),$
after taking a union bound. As $\hat{\sigma}^{(s^{*})}=\textnormal{sgn}(z^{(s^{*})})$ in step [6](#algx4.l6), we obtain $\hat{\sigma}^{(s^{*})}=\sigma^{*}$ with high probability. Overall, we established that $\hat{\sigma}_{\textnormal{spec}}$ achieves exact recovery above the IT threshold.
$\lim_{n\rightarrow\infty}\operatorname*{\mathbb{P}}\left(\hat{\sigma}_{\textnormal{spec}}\textnormal{ succeeds}\right)=1.\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$

###### Lemma E.9 .

###### Proof.

###### Proof of Theorem 3 .

### E.4 Degree-Profiling Algorithm and Proof of Theorem 4

Algorithm 6 Degree-Profiling algorithm for $\mathsf{SBM}$ in the presence of $\mathsf{BEC}$ or $\mathsf{BSC}$ side information.

$$ $z_{i}=\log\left(\frac{a_{1}}{b}\right)\sum_{j\in S_{+}}A_{ij}+\log\left(\frac{b}{a_{2}}\right)\sum_{j\in S_{-}}A_{ij}+\gamma.$ $$

- •
$\mathsf{BEC}$ side information: For any $i\in[n]$,
$z^{\textnormal{dp}}_{i}=\begin{cases}z_{i}&\textnormal{if }y_{i}=0;\\
+\infty,&\textnormal{if }y_{i}=+1;\\
-\infty&\textnormal{if }y_{i}=-1;\end{cases}$
- •
$\mathsf{BSC}$ side information:
$z^{\textnormal{dp}}=z+\ln\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)y$

###### Lemma E.10 .

Fix $\rho\in(0,1)$ and $a_{1},a_{2},b>0$. Let $(A,\sigma^{*})\sim\mathsf{SBM}_{n}(\rho,a_{1},a_{2},b)$ and condition on $\sigma^{*}$ such that $E$ from ([11](#A2.E11)) holds. For $\beta>0$, let $y\sim\mathsf{BEC}(\sigma^{*},\epsilon_{n})$ or $y\sim\mathsf{BSC}(\sigma^{*},\alpha_{n})$ where $\epsilon_{n}$ and $\alpha_{n}$ are parameterized as ([1](#S1.E1)). Let $z^{*}$ and $z^{\textnormal{dp}}$ respectively be the genie score vector and the degree-profiling score vector produced by Algorithm [E.4](#A5.SS4) for the corresponding model of side information. Then with probability $1-o(1)$,

$$ $\left\lVert z^{*}-z^{\textnormal{dp}}\right\rVert_{\infty}=O\left(\frac{\log n}{\log\log n}\right).$ $$

###### Proof.

The proof has similar calculations as in Lemma [D.12](#A4.Thmlemma12) for $\mathsf{ROS}$. We again first start by noting that $z^{\textnormal{dp}}$ is just formed by overriding the entries of $z$ from step [3](#algx6.l3) of Algorithm [E.4](#A5.SS4) with $+\infty$ or $-\infty$ depending on the side information label being $+1$ or $-1$ under $\mathsf{BEC}$ channel. Also, in step [4](#algx6.l4) under the $\mathsf{BSC}$ channel, we have $z^{\textnormal{dp}}=z+\log\left(\frac{1-\alpha_{n}}{\alpha_{n}}\right)y$. Recall Lemma [2.10](#S2.Thmlemma10) as to how the genie scores change under both types of side information. It suffices to show that, with probability $1-o(1)$,

$$ $\left\lVert z^{\prime}-z\right\rVert_{\infty}=O\left(\frac{\log n}{\log\log n}\right),$ $$

where $z^{\prime}$ is the genie score vector without side information and $z$ is formed in step [3](#algx2.l3). Recall that $E_{1}$ holds with probability $1-O(n^{-3})$ from ([30](#A5.E30)). Using Lemma [E.2](#A5.Thmlemma2) along with the triangle inequality we obtain the following.

- •
$\mathsf{BEC}$ side information:
$\displaystyle\left\lVert z^{\prime}-z\right\rVert_{\infty}$
$\displaystyle=\max_{i\in[n]}|z^{\prime}_{i}-z_{i}|$
$\displaystyle\leq\max_{i\in[n]}\left|\log\left(\frac{a_{1}}{b}\right)\sum_{j\in C_{+}\setminus S_{+}}A_{ij}+\log\left(\frac{b}{a_{2}}\right)\sum_{j\in C_{-}\setminus S_{-}}A_{ij}\right|+O(1)$
where we substitute $z^{\prime}$ from step [3](#algx6.l3). From Lemma [D.11](#A4.Thmlemma11), both $|C_{+}\setminus S_{+}|$ and $|C_{-}\setminus S_{-}|$ are bounded by $O(n/\log^{10}n)$ with probability $1-o(1)$. These sets are independent of $A$ and are chosen based on $y$. Using Lemma [E.5](#A5.Thmlemma5) for these sets, with probability $1-o(1)$
$\left\lVert z^{\prime}-z\right\rVert_{\infty}\leq\max_{i\in[n]}\left|\log\left(\frac{a_{1}}{b}\right)\sum_{j\in C_{+}\setminus S_{+}}A_{ij}\right|+\left|\log\left(\frac{b}{a_{2}}\right)\sum_{j\in C_{-}\setminus S_{-}}A_{ij}\right|+O(1)=O\left(\frac{\log n}{\log\log n}\right).$
- •
$\mathsf{BSC}$ side information: There are additional error terms caused by the sets $S_{+}\setminus C_{+}$ and $S_{-}\setminus C_{-}$:
$\displaystyle\left\lVert z^{\prime}-z\right\rVert_{\infty}$
$\displaystyle=\max_{i\in[n]}|z^{\prime}_{i}-z_{i}|=\max_{i\in[n]}\Bigg{|}\log\left(\frac{a_{1}}{b}\right)\sum_{j\in C_{+}\setminus S_{+}}A_{ij}-\log\left(\frac{a_{1}}{b}\right)\sum_{j\in S_{+}\setminus C_{+}}A_{ij}$
$\displaystyle\hskip 14.22636pt+\log\left(\frac{b}{a_{2}}\right)\sum_{j\in C_{-}\setminus S_{-}}A_{ij}-\log\left(\frac{b}{a_{2}}\right)\sum_{j\in S_{-}\setminus C_{-}}A_{ij}\Bigg{|}+O(1)$
(substituting $z^{\prime}$ from Lemma [E.2](#A5.Thmlemma2) and $z$ from step [3](#algx6.l3))
$\displaystyle=O(1)\cdot\max_{i\in[n]}\Bigg{|}\sum_{j\in C_{+}\setminus S_{+}}A_{ij}+\sum_{j\in C_{-}\setminus S_{-}}A_{ij}\Bigg{|}+O(1).$
Exactly the same argument of using Lemma [D.11](#A4.Thmlemma11), both $|C_{+}\setminus S_{+}|$ and $|C_{-}\setminus S_{-}|$ is $O(n/\log^{10}n)$ with probability $1-o(1)$. Using Lemma [E.5](#A5.Thmlemma5) for these sets, with probability $1-o(1)$
$\left\lVert z^{\prime}-z\right\rVert_{\infty}=O\left(\frac{\log n}{\log\log n}\right).$

We finally prove Theorem [4](#Thmtheorem4).

###### Proof of Theorem 4 .

When $\beta>0$, by Lemma [E.10](#A5.Thmlemma10), with probability $1-o(1)$,

$$ $\left\lVert z^{*}-z^{\textnormal{dp}}\right\rVert_{\infty}=O(1).$ $$

Above the IT threshold by Lemma [E.3](#A5.Thmlemma3) and union bound over $i\in[n]$, there exists $\delta>0$ such that

$$ $\operatorname*{\mathbb{P}}\left(\min_{i\in[n]}\sigma^{*}_{i}z^{*}_{i}>\delta\log n\right)=1-o(1).$ $$

Taking a union bound of these two events, there exists $\varsigma>0$ such that

$$ $\operatorname*{\mathbb{P}}\left(\min_{i\in[n]}\sigma^{*}_{i}z^{\textnormal{dp}}_{i}>\varsigma\log n\right)=1-o(1).$ $$

Observing $\hat{\sigma}_{\textnormal{dp}}=\textnormal{sgn}(z^{\textnormal{dp}})$ in step [5](#algx6.l5) of Algorithm [E.4](#A5.SS4), we obtain $\hat{\sigma}_{\textnormal{dp}}$ achieves exact recovery, i.e.

$$ $\lim_{n\rightarrow\infty}\operatorname*{\mathbb{P}}\left(\hat{\sigma}_{\textnormal{dp}}\textnormal{ succeeds}\right)=1.\nobreak\hskip 0.0pt\hskip 15.00002pt minus 5.0pt\nobreak\vrule height=7.5pt,width=5.0pt,depth=2.5pt$ $$

###### Lemma E.10 .

###### Proof.

###### Proof of Theorem 4 .

## Peer Reviews

confidence: 3
contribution: 4
initial_rating: 8
presentation: 2
questions: My only concern with this paper has to do with the formatting issue raised in the weaknesses section. If this is a typical approach that is commonplace and accepted in ML conferences, then I am willing to cede that point.
soundness: 4
strengths: This is a well-written paper -- there are quite a few moving parts
involved in this work, but the explanation was clear and easy to
follow. The authors ought to be commended for clear technical
communication. I do have onemajor misgiving about this, but that 
is saved for the weaknesses section.

I am admittedly not too familiar with the literature on exact recovery
of SBMs beyond the paper of Abbe (2018), but the results here seem
quite good, and fit into the surveyed prior results nicely. I
appreciate how the recovery guarantees are accompanied by specific
algorithms, rather than proving detectability in merely an abstract
way.
summary: This paper considers the problem of recovery of two-community block
models. It sits in the context of a broad literature on detectability
limits, exact algorithms, etc. The particular setting considered by
the authors is one of a general sort of block-model random matrix
based on a given partition/distribution model, along with a vector of
"side-information" that is aligned with the block structure in some
way. The authors study the information theoretic limit of this
detection problem via simple spectral methods, and by studying
"genie-aided estimators."
weaknesses: My critique of this paper is not one aimed at the strength of the
results -- indeed, I found the results to be nice, and the exposition
to be quite good. However, I have the sense that the main body of the
paper was merely a rough overview of the "real paper" hidden in the
appendix. I understand that for space reasons, lots of ML papers have
to put substantial content in the supplementary material. I don't have
a deep argument for why this is the case, but I get the feeling that
this work is better-suited for a journal format. For instance, one of
the main constructed algorithms that attains exact recovery
(Algorithm 2) is only given in the supplementary material. It would be
more appropriate to structure the paper so that the algorithm is given
and explained in the body of the paper, with other variants possibly left
to the supplement.

I should note that although I am a reader of theory papers of this
sort, I have not reviewed many of them in the context of ML
conferences, so I am ultimately willing to defer to the other
reviewers on this point.

Other notes:

(I am not holding the following against you in any way -- deadlines
can be tough!)

Please check the paper for spleling mistakes -- I spotted a few such
as "Gaussain Features" and "shifing the eigenvector combination."

---

confidence: 3
contribution: 3
initial_rating: 6
presentation: 3
questions: 1. On Page 6, $L(t)$ and $I^*$ are introduced to characterize the information theoretic limit. Are there any general sufficient conditions that ensure $L(t)$, and consequently $I^*$, are well-defined? 

2. In Section 5.1, degree-profiling algorithm is mentioned as having certain caveat. However, it is unclear to me what this algorithm refers to. Perhaps adding a brief summary of the key points of this algorithm would improve clarity.

3. On page 9, it is stated that an approximate linear combination coefficients $c_i$ such that $w\approx \sum_{i=1}^K c_i/\lambda_i^* u_i^*$. Is there any intuitive explanation for why $w$ is approximately in the span of $u_i^*$? Is this result purely from the computations of the specific model (Gaussian and Bernoulli), or is there a broader rationale suggesting that this holds across more robust models? 

4. Regarding the algorithm, do the parameters $c_i$ and $\gamma$ need be explicitly computed from the model parameters, or is there a possibility to estimate them using only the adjacency matrix $A$ and the side information?

5. This final question may not be directly related to the problem considered in the paper. Dreveton et al (2024) provides  the information theoretic limit of exact recovery in terms of $I^*$.  I am curious whether the regime $I^*<1$ can be further split. Specifically, could we identify a sub-regime where exact recovery is achievable even without the side information, and another sub-regime where side information is essential with some specific conditions?
soundness: 3
strengths: The paper is well-structured, with clearly stated objectives and supported arguments. It provides background definitions and results on related models, which greatly aids in understanding the topic. The novel approach of connecting the eigenvectors to the genie-aided estimator is particularly intriguing, contributing to the design of an efficient spectral algorithm. Additionally, the rigorous mathematical derivations are well-organized in the appendix, offering solid theoretical support.
summary: This paper addresses the problem of exact community recovery for general two-community block models with node-attributed side information. The authors consider both Gaussian and Bernoulli matrix models for the edge observations, as well as a general side information channel, including Gaussian Features and Binary Erasure Channel. Based on the recent work of Dreveton et al. (2024) on the information-theoretic limit on this problem, the authors demonstrate the algorithmic achievability using a spectral algorithm that incorporates side information with the eigenvectors of the observed edge matrix. The main technical novelty lies in establishing a rigorous connection between the spectral estimator and the genie-aided estimator by the results on the first-order approximation of the eigenvectors in Abbe et al. (2020).
weaknesses: Due to the page limit, I feel that some important details have been deferred to the appendix. For instance, I would like to see a more rigorous discussion in the main context on the special form of the genie score vector, $z^* \approx Aw + \gamma I_n$; how $w$ can be derived from the model, any regularity conditions needed for this approximation, and whether this form extends to models beyond Gaussian and Bernoulli. I suggest moving part of the theoretical discussion into the main text and possibly relocating some of the preliminaries to the appendix.

 Additionally, I am uncertain about the robustness of this algorithm, as the coefficients $c_i$ relating $w$ to eigenvectors must be computed from model parameters. However, in practice, we typically observe only the adjacency matrix and side information, without access to underlying parameters like $a_1, a_2, b$ in ${\rm SBM}_n(\rho, a_1,a_2, b)$. A discussion or clarification on this would be beneficial.
 
  Another limitation of this paper is the lack of a numerical study. I suggest including simulation examples that compare the proposed algorithm’s performance (in terms of accuracy and computation cost) with the existing two-stage algorithms. This would strengthen the paper by demonstrating the practical advantages of this approach.

---

confidence: 4
contribution: 2
initial_rating: 3
presentation: 3
questions: - I understand this is a theoretical work, but have the authors verified their algorithm on any dataset? 
- What is the computational complexity in terms of n?
- Is it possible to generalize this work to the case of multiple communities? If not, what are the technical difficulties?
soundness: 2
strengths: - The paper is well organized and easy to follow. The notations and definitions are mostly self contained.
- The authors provide the statistical bounds for exact recovery in the case of the Rank One Spike (ROS) model and the Stochastic Blockmodel (SBM).
summary: The paper studies block models with two communities, where each community has a general form of edge distribution. Side information on nodes is also considered. The authors specifically focuses on the case of Bernoulli and Gaussian distributions. The paper then proposes a variant of spectral algorithm to recover the community structure in the model, and provides statistical bounds for exact recovery of the signal.
weaknesses: My biggest concern is that some of the language in the paper seems overclaiming. For example, the authors state that the paper proposes a "unified model" and "unified proof framework". The amount of novelty and contribution is unclear to me. Can the authors specify, preferably in bullet points, their new results, algorithm, bounds, proof techniques, or experiments?

Beyond that, I have some more detailed questions:
- The paper aims to provide a unified framework of analysis for at-most-two-community block models, but the results in Theorem 1 and 2 only target two specific and arguably "simpler" classic models. Why is that? Is it possible to give a bound based on the definition of P+, P-, and Q, for instance? 
- Section 3.1: Does the information-theoretic bounds in Prop 3.1 contain anything novel, or just cited as is from [Dreveton et al., 2024]? If it is the latter, I strongly believe it should be moved to the preliminary section and not put in the "Main Results".
- Theorem 1 and 2: The exact recovery condition depends on I* > 1. Is there any way to calculate this efficiently? From a reader's perspective, I'd love to see many examples of two community models and know how their parameters are related to this I*. For example, can the authors relate I* to p and q in the SBM?
- The proofs like the concentration inequalities and the spectral analysis in the paper seem standard. In fact the union bound strategy feels similar to the ones in E. Abbe's and Dreveton's work. What are the technical contribution of this paper? Is there any proof technique that can be used in the future?

Minor:
- Bern() used but not defined on the first page.

---

confidence: 4
contribution: 3
initial_rating: 6
presentation: 3
questions: See weaknesses above. Also, the paper could benefit from a careful proofreading. For example:

1. Gaussain -> Gaussian on page 3.
2. In Algorithm 7,  $\alpha_n$ is not defined. Should it be $\alpha$?
3. In Algorithm 6, the input section ends with two periods.
soundness: 3
strengths: 1. The paper is well-organized, and the arguments flow logically.

2. The claims are solid, and the proposed framework covers a broad range of interesting models.

3. The results provide theoretical insights into community detection with side information and demonstrate the optimality of spectral methods.
summary: This paper addresses the problem of exact community recovery in two-community block models with node-attributed side information. The primary contribution is the design of an optimal spectral algorithm that achieves the information-theoretic limits in various settings. The general block model in this paper encompass  rank-one spike Gaussian model and SBM, and the side information incorporates GMM and two types of "noisy'' labels.
weaknesses: 1. Some related literature is missing, e.g., [1]. Specifically, the SBM model with $\rho = 1/2$ and Gaussian Features aligns with Section 4 in [1] and should be cited and compared. Additionally, for the SBM with GF where $\rho = 1/2$ and $a_1 = a_2 = a$, the expression for $I^*$ takes a simple form in Equation (4.4) of [1]. Providing explicit forms of $I^*$ for other models would enhance interpretability of the results.

2. For SBM with BEC/BSC, the algorithms assume prior knowledge of parameters $(\rho,\varepsilon, \alpha)$. The paper should mention this clearly and discuss why it's reasonable to assume this prior knowledge, or suggest how these parameters could be estimated. From a theory standpoint, achieving optimality might require paying a cost for adapting to these parameters, especially in regimes $\varepsilon = n^{-\beta}$ and $\alpha = n^{-\beta}$ for $\beta>0$. 

3. The two-community assumption limits the algorithm’s practical use. While I understand that the optimality of the spectral algorithm depends on this assumption, the statement in Appendix A, "The entire framework of genie-aided estimation naturally generalizes to the multi-community case," is confusing. Can the authors elaborate on how the spectral algorithm could be modified, for example, to handle $K=3$ communities?

4. There are no experiments. It would be helpful to include numerical experiments demonstrating the algorithm’s performance on the specific models considered.

[1] Abbe, E., Fan, J., & Wang, K. (2022). An $\ell_p$ theory of PCA and spectral clustering. The Annals of Statistics, 50(4), 2359-2385.

Meta-review: Three of the reviews are positive, commenting favorably on the contributions, writing, and other aspects, with some minor reservations such as missing references and no experiments.  The authors said they will add some experiments, and it's not ideal that those would not be reviewed, but overall I'm OK with taking this as a theory paper.

One reviewer gave a much lower score, but from what I can tell, the comments themselves would be more aligned with at most 'borderline reject' rather than such a strong rejection.  This reviewer did not acknowledge the rebuttal and did not respond even after repeated requests.  From my own viewing of the rebuttal, I think the concerns are addressed adequately.  But I ask the authors to very carefully revise the final version to address the feedback from all 4 reviewers.

Decision: Accept (Poster)