SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Lemma C.16. Suppose  $\mathbf{G}$  is generated by first sampling  $\mathbf{G} \sim \mathrm{SBM}_n\left(\frac{\alpha + \beta}{2} \cdot \log n, \frac{\alpha - \beta}{\alpha + \beta}, x\right)$  for some balanced  $x$  and then for each vertex removing at most  $\Delta \leqslant O(\log^2 n)$  adjacent edges arbitrarily. Then on input  $\mathbf{G}$  and a balanced rough estimate  $\tilde{x}$  satisfying  $\mathrm{Ham}(\tilde{x}, x) \leqslant n / 16$ , Algorithm C.13 efficiently outputs  $\hat{x}(\mathbf{G})$  such that for each vertex  $v$ ,
$$
\mathbb {P} \left(\hat {x} (\mathbf {G}) _ {v} \neq x _ {v}\right) \leqslant \exp \left(- \frac {1}{6 4} \cdot \varepsilon (\alpha - \beta) \cdot \log n\right) + 2 \cdot \exp \left(- \frac {1}{1 6 ^ {2}} \cdot \frac {(\alpha - \beta) ^ {2}}{\alpha + \beta} \cdot \log n\right).
$$
Proof. Let us fix an arbitrary vertex  $v$  and analyze the probability  $\mathbb{P}(\hat{x} (\mathbf{G})_v\neq x_v)$ . Let  $r\coloneqq \mathrm{Ham}(\tilde{x},x) / n$ . Then it is not hard to see
$$
\mathbb {P} \left(\hat {x} (\mathbf {G}) _ {v} \neq x _ {v}\right) \leqslant \mathbb {P} \left(\mathbf {B} + \mathbf {A} ^ {\prime} - \mathbf {A} - \mathbf {B} ^ {\prime} + \mathbf {W} > 0\right) \tag {C.8}
$$
-  $\mathbf{A} \sim \mathrm{Binomial}((1/2 - r)n - \Delta, \alpha \frac{\log n}{n})$ , corresponding to the number of neighbors that are from the same community and correctly labeled by  $\tilde{x}$ ,
-  $\mathbf{B}' \sim \mathrm{Binomial}(rn - \Delta, \beta \frac{\log n}{n})$ , corresponding to the number of neighbors that are from the different community but incorrectly labeled by  $\tilde{x}$ ,
-  $\mathbf{B} \sim \mathrm{Binomial}((1/2 - r)n, \beta \frac{\log n}{n})$ , corresponding to the number of neighbors that are from the different community and correctly labeled by  $\tilde{x}$ ,
-  $\mathbf{A}' \sim \mathrm{Binomial}(rn, \alpha \frac{\log n}{n})$ , corresponding to the number of neighbors that are from the same community but incorrectly labeled by  $\tilde{x}$ ,
-  $\mathbf{W} \sim \mathrm{Lap}(0, 2 / \varepsilon)$ , independently.
The  $\Delta$  term appearing in both  $\mathbf{A}$  and  $\mathbf{B}'$  corresponds to the worst case where  $\Delta$  "favorable" edges are removed. If  $r \geqslant \Omega(1)$ , then  $\Delta = O(\log^2 n)$  is negligible to  $rn = \Theta(n)$  and we can safely ignore the effect of removing  $\Delta$  edges. If  $r = o(1)$ , then we can safely assume  $\tilde{x}$  is correct on all vertices and ignore the effect of removing  $\Delta$  edges as well. Thus, we will assume  $\Delta = 0$  in the following analysis.
For any  $t, t'$ , we have
$$
\begin{array}{l} \mathbb {P} \left(\mathbf {A} ^ {\prime} + \mathbf {B} - \mathbf {A} - \mathbf {B} ^ {\prime} + \mathbf {W} > 0\right) \leqslant \mathbb {P} \left(\mathbf {A} ^ {\prime} + \mathbf {B} + \mathbf {W} > t\right) + \mathbb {P} \left(\mathbf {A} + \mathbf {B} ^ {\prime} \leqslant t\right) \\ \leqslant \mathbb {P} \left(\mathbf {A} ^ {\prime} + \mathbf {B} \geqslant t - t ^ {\prime}\right) + \mathbb {P} \left(\mathbf {W} \geqslant t ^ {\prime}\right) + \mathbb {P} \left(\mathbf {A} + \mathbf {B} ^ {\prime} \leqslant t\right). \\ \end{array}
$$
We choose  $t, t'$  by first picking two constants  $a, b > 0$  satisfying  $a + b < 1$  and then solving
-  $\mathbb{E}[\mathbf{A}' + \mathbf{B}] - t = a \cdot (\mathbb{E}[\mathbf{A} + \mathbf{B}'] - \mathbb{E}[\mathbf{A}' + \mathbf{B}])$  and
-  $t' = (1 - a - b) \cdot (\mathbb{E}[\mathbf{A} + \mathbf{B}'] - \mathbb{E}[\mathbf{A}' + \mathbf{B}])$ .
By Fact A.7,
$$
\mathbb {P} \left(\mathbf {W} > t ^ {\prime}\right) \leqslant \exp \left(- \frac {t ^ {\prime} \varepsilon}{2}\right) \leqslant \exp \left(- \frac {(1 / 4 - r) (1 - a - b)}{2} \cdot \varepsilon (\alpha - \beta) \cdot \log n\right).
$$
By Fact E.4 and the assumption  $r \leqslant 1/16$ , we have
$$
\mathbb {P} \left(\mathbf {A} + \mathbf {B} ^ {\prime} \leqslant t\right) \leqslant \exp \left(- \frac {\left(\mathbb {E} [ \mathbf {A} + \mathbf {B} ^ {\prime} ] - t\right) ^ {2}}{2 \mathbb {E} [ \mathbf {A} + \mathbf {B} ^ {\prime} ]}\right) \leqslant \exp \left(- (1 / 4 - r) ^ {2} a ^ {2} \cdot \frac {(\alpha - \beta) ^ {2}}{\alpha + \beta} \cdot \log n\right).
$$
Setting  $b = 1 / 2$ , by Fact E.4 and the assumption  $r \leqslant 1 / 16$ , we have
$$
\mathbb {P} \left(\mathbf {A} ^ {\prime} + \mathbf {B} \geqslant t - t ^ {\prime}\right) \leqslant \exp \left(- \frac {(t - t ^ {\prime} - \mathbb {E} [ \mathbf {A} ^ {\prime} + \mathbf {B} ]) ^ {2}}{t - t ^ {\prime} + \mathbb {E} [ \mathbf {A} ^ {\prime} + \mathbf {B} ]}\right) \leqslant \exp \left(- \frac {2 (1 / 4 - r) ^ {2}}{7} \cdot \frac {(\alpha - \beta) ^ {2}}{\alpha + \beta} \cdot \log n\right).
$$
Further setting  $a = 1 / 3$ , we have
$$
\mathbb {P} \left(\hat {x} (\mathbf {G}) _ {v} \neq x _ {v}\right) \leqslant \exp \left(- \frac {1 / 4 - r}{1 2} \cdot \varepsilon (\alpha - \beta) \cdot \log n\right) + 2 \cdot \exp \left(- \frac {(1 / 4 - r) ^ {2}}{9} \cdot \frac {(\alpha - \beta) ^ {2}}{\alpha + \beta} \cdot \log n\right).
$$
Finally, plugging the assumption  $r \leqslant 1/16$  to conclude.
Then it is not difficult to show the utility guarantee of our priavte exact recovery algorithm.
Lemma C.17 (Utility). Suppose  $\alpha, \beta$  are fixed constants satisfying
$$
\sqrt {\alpha} - \sqrt {\beta} \geqslant 1 6 \quad a n d \quad \alpha - \beta \geqslant \Omega \left(\frac {1}{\varepsilon^ {2}} \cdot \frac {\log (2 / \delta)}{\log n} + \frac {1}{\varepsilon}\right).
$$
Then for any balanced  $x \in \{\pm 1\}^n$ , on input  $\mathbf{G} \sim \mathsf{SBM}_n\left(\frac{\alpha + \beta}{2} \cdot \log n, \frac{\alpha - \beta}{\alpha + \beta}, x\right)$ , Algorithm C.12 efficiently outputs  $\hat{x}(\mathbf{G})$  satisfying  $\hat{x}(\mathbf{G}) \in \{x, -x\}$  with probability  $1 - o(1)$ .
Proof. We will show the probability of a fixed vertex being misclassified is at most  $o(1/n)$ . Then by union bound, exact recovery can be achieved with probability  $1 - o(1)$ .
As the graph-splitting probability is  $1/2$ ,  $\mathbf{G}_1$  follows  $\mathrm{SBM}_n\left(\frac{\alpha}{2} \cdot \frac{\log n}{n}, \frac{\beta}{2} \cdot \frac{\log n}{n}, x\right)$ . By Theorem C.3, the rough estimate  $\tilde{x}\big(\mathbf{G}_1\big)$  satisfies
$$
\operatorname {e r r} \left(\tilde {x} \left(\mathbf {G} _ {1}\right), x\right) \leqslant r := o (1) + \frac {1 4 0 0 0}{(\alpha - \beta) \varepsilon^ {2}} \cdot \frac {\log (2 / \delta)}{\log n}. \tag {C.9}
$$
with probability at least  $1 - \exp(-\Omega(n))$ . Without loss of generality, we can assume  $\mathrm{Ham}(\tilde{x}(\mathbf{G}_1), x) \leqslant rn$ , since we consider  $-x$  otherwise. By Fact E.2, the maximum degree of  $\mathbf{G}_1$  is at most  $\Delta := 2\log^2 n$  with probability at least  $1 - n\exp(-(\log n)^2/3)$ . In the following, we condition our analysis on the above two events regarding  $\tilde{x}(\mathbf{G}_1)$  and  $\mathbf{G}_1$ .
Now, let us fix a vertex and analyze the probability  $p_e$  that it is misclassified after majority voting. With  $G_1$  being fixed,  $\mathbf{G}_2$  can be thought of as being generated by first sampling  $\mathbf{G}$  and then removing  $G_1$  from  $\mathbf{G}$ . To make  $r \leqslant 1/16$ , it suffices to ensure  $\alpha - \beta > \frac{500^2}{\varepsilon^2} \cdot \frac{\log(2/\delta)}{\log n}$  by Eq. (C.9). Then by Lemma C.16, we have
$$
p _ {e} \leqslant \exp \left(- \frac {1}{6 4} \cdot \varepsilon (\alpha - \beta) \cdot \log n\right) + 2 \cdot \exp \left(- \frac {1}{1 6 ^ {2}} \cdot \frac {(\alpha - \beta) ^ {2}}{\alpha + \beta} \cdot \log n\right).
$$
To make  $p_e$  at most  $o(1/n)$ , it suffices to ensure
$$
\frac {1}{6 4} \cdot \varepsilon (\alpha - \beta) > 1 \quad \text {a n d} \quad \frac {1}{1 6 ^ {2}} \cdot \frac {(\alpha - \beta) ^ {2}}{\alpha + \beta} > 1.
$$
28It is easy to make the output of Algorithm C.4 balanced at the cost of increasing the error rate by a factor of at most 2.
Note  $(\alpha -\beta)^2 /(\alpha +\beta) > (\sqrt{\alpha} -\sqrt{\beta})^2$  for  $\alpha >\beta$  . Therefore, as long as
$$
\sqrt {\alpha} - \sqrt {\beta} \geqslant 1 6 \quad \text {a n d} \quad \alpha - \beta \geqslant \frac {5 0 0 ^ {2}}{\varepsilon^ {2}} \cdot \frac {\log (2 / \delta)}{\log n} + \frac {6 4}{\varepsilon},
$$
Algorithm C.12 recovers the hidden communities exactly with probability  $1 - o(1)$ .
Proof of Theorem C.10. By Lemma C.15 and Lemma C.17.
C.3 Inefficient recovery using the exponential mechanism
In this section, we will present an inefficient algorithm satisfying pure privacy which succeeds for all ranges of parameters - ranging from weak to exact recovery. The algorithm is based on the exponential mechanism [43] combined with the majority voting scheme introduced in section Appendix C.2. In particular, we will show
Theorem C.18 (Full version of Theorem 1.4). Let  $\gamma \sqrt{d} \geqslant 12800$  and  $x \in \{\pm 1\}^n$  be balanced. Let  $\zeta \geqslant 2\exp\left(-\frac{\gamma^2d}{512}\right)$ . For any  $\varepsilon \geqslant \frac{64\log(2 / \zeta)}{\gamma d}$ , there exists an algorithm, Algorithm C.19, which on input  $\mathbf{G} \sim \mathsf{SBM}_n(\gamma, d, x^*)$  outputs an estimate  $\hat{x}(\mathbf{G}) \in \{\pm 1\}^n$  satisfying
$$
\operatorname {e r r} \left(\hat {x} (\mathbf {G}), x ^ {*}\right) \leqslant \zeta
$$
with probability at least  $1 - \zeta$ . In addition, the algorithm is  $\varepsilon$ -private. Further, by slightly modifying the algorithm, we can achieve error  $20 / \sqrt{\log(1 / \zeta)}$  with probability  $1 - e^{-n}$ .<sup>29</sup>
A couple of remarks are in order. First, our algorithm works across all degree-regimes in the literature and matches known non-private thresholds and rates up to constants. We remark that for ease of exposition we did not try to optimize these constants. In particular, for  $\gamma^2 d$  a constant we achieve weak recovery. We reiterate, that  $\gamma^2 d > 1$  is the optimal non-private threshold. For the regime, where  $\gamma^2 d = \omega(1)$ , it is known that the optimal error rate is  $\exp\left(-(1 - o(1))\gamma^2 d\right)$  even non-privately [64], where  $o(1)$  goes to zero as  $\gamma^2 d$  tends to infinity. We match this up to constants. Moreover, our algorithm achieves exact recovery as soon as  $\gamma^2 d \geqslant 512\log n$  since then  $\zeta < \frac{1}{n}$ . This also matches known non-private thresholds up to constants [5, 47]. Also, our dependence on the privacy parameter  $\varepsilon$  is also optimal as shown by the information-theoretic lower bounds in Appendix C.4.
We also emphasize, that if we only aim to achieve error on the order of
$$
\frac {1}{\gamma \sqrt {d}} = \Theta \left(\frac {1}{\sqrt {\log (1 / \zeta)}}\right),
$$
we can achieve exponentially small failure probability in  $n$ , while keeping the privacy parameter  $\varepsilon$  the same. This can be achieved, by omitting the boosting step in our algorithm and will be clear from the proof of Theorem C.18. We remark that in this case, we can also handle non-balanced communities.
Again, for an input graph  $G$ , consider the matrix  $Y(G) = \frac{1}{\gamma d} \left( A(G) - \frac{d}{n} J \right)$ . For  $x \in \{\pm 1\}^n$  we define the score function
$$
s _ {G} (x) = \langle x, Y (G) x \rangle .
$$
Since the entries of  $A(G)$  are in [0, 1] and adjacent graphs differ in at most one edge, it follows immediately, that this score function has sensitivity at most
$$
\Delta = \max_{\substack{G\sim G^{\prime},\\ x\in \{\pm 1\}^{n}}}|s_{G}(x) - s_{G^{\prime}}(x)| = \frac{2}{\gamma d}\cdot \max_{\substack{G\sim G^{\prime},\\ x\in \{\pm 1\}^{n}}}| \langle x,(A(G) - A(G^{\prime}))  x\rangle | \leqslant \frac{2}{\gamma d}.
$$
29The first, smaller, error guarantee additionally needs the requirement that  $\zeta \leqslant \exp (-640)$ . The second one does not.
Algorithm C.19 (Inefficient algorithm for SBM).
Input: Graph  $G$ , privacy parameter  $\varepsilon > 0$
1. Graph-splitting: Initialize  $\mathbf{G}_1$  to be an empty graph on vertex set  $V(G)$ . Independently assign each edge of  $G$  to  $\mathbf{G}_1$  with probability  $1/2$ . Let  $\mathbf{G}_2 = G \setminus \mathbf{G}_1$ .
2. Rough estimation on  $\mathbf{G}_1$ : Sample  $\tilde{x}$  from the distribution with density
$$
p (x) \propto \exp \left(\frac {\varepsilon}{2 \Delta} \langle x, Y (\mathbf {G} _ {1}) x \rangle\right),
$$
where  $\Delta = \frac{2}{\gamma a}$
3. Majority voting on  $\mathbf{G}_2$ : Run the  $\varepsilon$ -DP majority voting algorithm (Algorithm C.13) with input  $(\mathbf{G}_2, \tilde{x}(\mathbf{G}_1))$ . Denote its output by  $\hat{\mathbf{x}}$ .
4. Return  $\hat{\mathbf{x}}$ .
We first analyze the privacy guarantees of the above algorithm.
Lemma C.20. Algorithm C.19 is  $\varepsilon$ -DP.
Proof. For simplicity and clarity of notation, we will show that the algorithm satisfies  $2\varepsilon$ -DP. Clearly, the graph splitting step is 0-DP. Step 2 corresponds to the exponential mechanism. Since the sensitivity of the score function is at most  $\Delta = \frac{2}{\gamma d}$  it follows by the standard analysis of the mechanism that this step is  $\varepsilon$ -DP [43]. By Lemma C.14, the majority voting step is also  $\varepsilon$ -DP. Hence, the result follows by composition (cf. Lemma A.4).
Next, we will analyze its utility.
Lemma C.21. Let  $\gamma \sqrt{d} \geqslant 12800$  and  $x \in \{\pm 1\}^n$  be balanced. Let  $\exp(-640) \geqslant \zeta \geqslant 2\exp\left(-\frac{\gamma^2d}{512}\right), \varepsilon \geqslant \frac{64\log(2 / \zeta)}{\gamma d}$ , and  $\mathbf{G} \sim \mathsf{SBM}_n(\gamma, d, x^*)$ , the output  $\hat{x}(\mathbf{G}) \in \{\pm 1\}^n$  of Algorithm C.19 satisfies
with probability at least  $1 - \zeta$ .
Proof. We will first show that the rough estimate  $\tilde{x}$  obtained in step 2 achieves
$$
\operatorname {e r r} \left(\tilde {x}, x ^ {*}\right) \leqslant \frac {2 0}{\sqrt {\log (1 / \zeta)}}
$$
with probability  $e^{-n}$ . This will prove the second part of the theorem - for this we don't need that  $\zeta \leqslant \exp(-640)$ . In fact, arbitrary  $\zeta$  works. The final error guarantee will then follow by Lemma C.16. First, notice that similar to the proof of [28, Lemma 4.1], using Bernstein's inequality and a union bound, we can show that (cf. Fact H.2 for a full proof)
$$
\max  _ {x \in \{\pm 1 \} ^ {n}} \left| \langle x, \left[ Y (\mathbf {G}) - \frac {1}{n} x ^ {*} (x ^ {*}) ^ {\top} \right] x \rangle \right| \leqslant \frac {1 0 0 n}{\gamma \sqrt {d}} \leqslant \frac {5}{\sqrt {\log (1 / \zeta)}}
$$
with probability at least  $1 - \exp^{-10n}$ . Recall that  $s_{\mathbf{G}}(x) = \langle x, Y(\mathbf{G})x \rangle$ . Let  $\alpha = \frac{5}{\sqrt{\log(1 / \zeta)}}$ . We call  $x \in \{\pm 1\}^n$  good if  $s_{\mathbf{G}}(x) \geqslant (1 - 3\alpha)n$ . It follows that for good  $x$  it holds that
$$
\left. \right. \frac {1}{n} \cdot \langle x, x ^ {*} \rangle^ {2} \geqslant \langle x, Y (\mathbf {G}) x \rangle - \left|\left\langle x, \left[ Y (\mathbf {G}) - \frac {1}{n} x ^ {*} (x ^ {*}) ^ {\top} \right] x \right\rangle\right| \geqslant (1 - 4 \alpha) n.
$$
$$
2 \operatorname {e r r} (x, x ^ {*}) \leqslant 1 - \sqrt {1 - 4 \alpha} = 1 - \frac {1 - 4 \alpha}{\sqrt {1 - 4 \alpha}} \leqslant 1 - \frac {1 - 4 \alpha}{1 - 2 \alpha} = \frac {2 \alpha}{1 - 2 \alpha} \leqslant 4 \alpha ,
$$
where we used that  $\alpha \leqslant 1 / 4$  and that  $\sqrt{1 - 4x} \leqslant 1 - 2x$  for  $x \geqslant 0$ . Hence, we have for good  $x$  that
$$
\operatorname {e r r} (x, x ^ {*}) \leqslant \frac {2 0}{\sqrt {\log (1 / \zeta)}}.
$$
Since  $s_{\mathbf{G}}(x^{*}) \geqslant (1 - \alpha)n$ , there is at least one good candidate. Hence, we can bound the probability that we do not output a good  $x$  as
$$
\frac {\exp \left(\frac {\varepsilon}{2 \Delta} (1 - 3 \alpha) n\right) \cdot e ^ {n}}{\exp \left(\frac {\varepsilon}{2 \Delta} (1 - \alpha) n\right) \cdot 1} = \exp \left(\left(1 - \frac {2 \varepsilon \alpha}{\Delta}\right) n\right) \leqslant e ^ {- n},
$$
$$
\frac {2 \varepsilon \alpha}{\Delta} \geqslant \frac {6 4 \log (2 / \zeta)}{\gamma d} \cdot \frac {5 \gamma d}{\sqrt {\log (1 / \zeta)}} \geqslant 3 2 0 \sqrt {\log (1 / \zeta)} \geqslant 2.
$$
We will use Lemma C.16 to proof the final conclusion of the theorem. In what follows, assume without loss of generality that  $\operatorname{Ham}(x,x^{*}) < \operatorname{Ham}(x, - x^{*})$ . The above discussion implies that
$$
\operatorname {H a m} \left(x, x ^ {*}\right) \leqslant 8 \alpha n \leqslant \frac {4 0 n}{\sqrt {\log \left(1 / \zeta\right)}} \leqslant \frac {n}{1 6},
$$
where the last inequality uses  $\zeta \leqslant e^{-640}$ . Further, by Fact E.2 it also follows that the maximum degree of  $\mathbf{G}_2$  is at most  $O\left(\log^2 n\right)$  (by some margin). Recall that  $\mathbf{G}_2 \sim \mathrm{SBM}(d,\gamma,x^*)$ . In the parametrization of Lemma C.16 this means that
$$
\alpha = \frac {(1 + \gamma) d}{\log n}, \quad \beta = \frac {(1 - \gamma) d}{\log n},
$$
$$
\alpha - \beta = \frac {2 \gamma d}{\log n}, \quad \alpha + \beta = \frac {2 d}{\log n}.
$$
Thus, it follows that the output  $\hat{x}$  of the majority voting step satisfies for every vertex  $v$
$$
\begin{array}{l} \mathbb {P} \left(\hat {x} (\mathbf {G}) _ {v} \neq x _ {v}\right) \leqslant \exp \left(- \frac {1}{6 4} \cdot \varepsilon (\alpha - \beta) \cdot \log n\right) + 2 \cdot \exp \left(- \frac {1}{1 6 ^ {2}} \cdot \frac {(\alpha - \beta) ^ {2}}{\alpha + \beta} \cdot \log n\right) \\ \leqslant \exp \left(- \frac {1}{3 2} \cdot \varepsilon \gamma d\right) + \exp \left(- \frac {1}{1 6 ^ {2}} \cdot \gamma^ {2} d\right) \\ \leqslant \zeta^ {2} / 4 + \zeta^ {2} / 4 \leqslant \zeta^ {2}. \\ \end{array}
$$
By Markov's Inequality it now follows that
$$
\mathbb {P} \left(\operatorname {e r r} \left(\hat {x} (\mathbf {G}), x ^ {*}\right) \geqslant \zeta\right) \leqslant \zeta .
$$
C.4 Lower bound on the parameters for private recovery
In this section, we prove a tight lower bound for private recovery for stochastic block models. Recall the definition of error rate,  $\mathrm{err}(u,v)\coloneqq \frac{1}{n}\cdot \min \{\mathrm{Ham}(u,v),\mathrm{Ham}(u, - v)\}$  for  $u,v\in \{\pm 1\} ^n$  . Our main result is the following theorem.
Theorem C.22 (Full version of Theorem 1.5). Suppose there exists an  $\varepsilon$ -differentially private algorithm such that for any balanced  $x \in \{\pm 1\}^n$ , on input  $\mathbf{G} \sim \mathrm{SBM}_n(d, \gamma, x)$ , outputs  $\hat{x}(\mathbf{G}) \in \{\pm 1\}^n$  satisfying
$$
\mathbb {P} \left(\operatorname {e r r} (\hat {x} (\mathbf {G}), x) <   \zeta\right) \geqslant 1 - \eta ,
$$
where  $1 / n \leqslant \zeta \leqslant 0.04$  and the randomness is over both the algorithm and stochastic block models. Then,
$$
e ^ {2 \varepsilon} - 1 \geqslant \Omega \left(\frac {\log (1 / \zeta)}{\gamma d} + \frac {\log (1 / \eta)}{\zeta n \gamma d}\right). \tag {C.10}
$$
30Error rate less than  $1 / n$  already means exact recovery. Thus it does not make sense to set  $\zeta$  to any value strictly smaller than  $1 / n$ . The upper bound  $\zeta \leqslant 0.04$  is just a technical condition our proof needs for Eq. (C.12).
Remark C.23. Both terms in lower bound Eq. (C.10) are tight up to constants by the following argument. Considering typical privacy parameters  $\varepsilon \lesssim 1$ , then  $e^{2\varepsilon} - 1 \approx 2\varepsilon$ . For exponentially small failure probability, i.e.  $\eta = 2^{-\Omega(n)}$ , the lower bound reads  $\varepsilon \geqslant \Omega\left(\frac{1}{\gamma d} \cdot \frac{1}{\zeta}\right)$ , which is achieved by Algorithm C.19 without the boosting step - see the discussion after Theorem C.18. For polynomially small failure probability, i.e.  $\eta = 1/\text{poly}(n)$ , the lower bound Eq. (C.10) reads  $\varepsilon \geqslant \Omega\left(\frac{1}{\gamma d} \cdot \log \frac{1}{\zeta}\right)$ , which is achieved by Theorem C.18.
By setting  $\zeta = 1 / n$  in Theorem C.22, we directly obtain a tight lower bound for private exact recovery as a corollary.
Corollary C.24. Suppose there exists an  $\varepsilon$ -differentially private algorithm such that for any balanced  $x \in \{\pm 1\}^n$ , on input  $\mathbf{G} \sim \mathrm{SBM}_n(d, \gamma, x)$ , outputs  $\hat{x}(\mathbf{G}) \in \{\pm 1\}^n$  satisfying
$$
\mathbb {P} \left(\hat {x} (\mathbf {G}) \in \{x, - x \}\right) \geqslant 1 - \eta ,
$$
where the randomness is over both the algorithm and stochastic block models. Then,
$$
e ^ {2 \varepsilon} - 1 \geqslant \Omega \left(\frac {\log (n) + \log \frac {1}{\eta}}{\gamma d}\right). \tag {C.11}
$$
Remark C.25. The lower bound Eq. (C.11) for priave exact recovery is tight up to constants, since there exists an (inefficient)  $\varepsilon$ -differentially priave exact recovery algorithm with  $\varepsilon \leqslant O\left(\frac{\log n}{\gamma d}\right)$  and  $\eta = 1/\mathrm{poly}(n)$  by Theorem C.18 and [55, Theorem 3.7].
In rest of this section, we will prove Theorem C.22. The proof applies the packing lower bound argument similar to [29, Theorem 7.1]. To this end, we first show  $\mathrm{err}(\cdot ,\cdot)$  is a semimetric over  $\{\pm 1\} ^n$
Lemma C.26.  $\mathrm{err}(\cdot ,\cdot)$  is a semimetric over  $\{\pm 1\} ^n$
Proof. Symmetry and non-negativity are obvious from the definition. We will show  $\mathrm{err}(\cdot, \cdot)$  satisfies triangle inequality via case analysis. Let  $u, v, w \in \{\pm 1\}^n$  be three arbitrary sign vectors. By symmetry, we only need to consider the following four cases.
Case 1:  $\mathrm{Ham}(u,v),\mathrm{Ham}(u,w),\mathrm{Ham}(v,w)\leqslant n / 2$  . This case is reduced to showing Hamming distance satisfies triangle inequality, which is obvious.
Case 2:  $\operatorname{Ham}(u, v), \operatorname{Ham}(u, w) \leqslant n / 2$  and  $\operatorname{Ham}(v, w) \geqslant n / 2$ . We need to check two subcases. First,
$$
\begin{array}{l} \operatorname {e r r} (u, v) \leqslant \operatorname {e r r} (u, w) + \operatorname {e r r} (v, w) \Leftrightarrow \operatorname {H a m} (u, v) + \operatorname {H a m} (v, w) \leqslant \operatorname {H a m} (u, w) + n \\ \Lleftarrow \operatorname {H a m} (u, v) + H (u, v) + H (u, w) \leqslant \operatorname {H a m} (u, w) + n \\ \Leftrightarrow \operatorname {H a m} (u, v) \leqslant n / 2. \\ \end{array}
$$
$$
\begin{array}{l} \operatorname {e r r} (v, w) \leqslant \operatorname {e r r} (u, v) + \operatorname {e r r} (u, w) \Leftrightarrow n \leqslant \operatorname {H a m} (v, w) + \operatorname {H a m} (u, v) + \operatorname {H a m} (u, w) \\ \Leftarrow n \leqslant 2 \operatorname {H a m} (v, w). \\ \end{array}
$$
Case 3:  $\operatorname{Ham}(u, v) \leqslant n / 2$  and  $\operatorname{Ham}(u, w), \operatorname{Ham}(v, w) \geqslant n / 2$ . This case can be reduced to case 1 by considering  $u, v, -w$ .
Case 4:  $\operatorname{Ham}(u, v)$ ,  $\operatorname{Ham}(u, w)$ ,  $\operatorname{Ham}(v, w) \geqslant n / 2$ . This case can be reduced to case 2 by considering  $-u, v, w$ .
Proof of Theorem C.22. Suppose there exists an  $\varepsilon$ -differentially private algorithm satisfying the theorem's assumption.
We first make the following notation. Given a semimetric  $\rho$  over  $\{\pm 1\}^n$ , a center  $v \in \{\pm 1\}^n$ , and a radius  $r \geqslant 0$ , define  $B_{\rho}(v,r) := \{w \in \{\pm 1\}^n : \mathbf{1}^\top w = 0, \rho(w,v) \leqslant r\}$ .
Pick an arbitrary balanced  $x \in \{\pm 1\}^n$ . Let  $M = \{x^1, x^2, \ldots, x^m\}$  be a maximal  $2\zeta$ -packing of  $B_{\mathrm{err}}(x, 4\zeta)$  in semimetric  $\mathrm{err}(\cdot, \cdot)$ . By maximality of  $M$ , we have  $B_{\mathrm{err}}(x, 4\zeta) \subseteq \cup_{i=1}^{m} B_{\mathrm{err}}(x^i, 2\zeta)$ , which implies
$$
\left| B _ {\text {e r r}} (x, 4 \zeta) \right| \leqslant \sum_ {i = 1} ^ {m} \left| B _ {\text {e r r}} \left(x ^ {i}, 2 \zeta\right) \right|
$$