SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Robustness experiment. The base graph  $G_{0} = (V, E)$  is generated from SBM with  $n = 3000$ ,  $k = 3$ ,  $p = 0.05$ ,  $q = 0.002$ . Note that randomly deleting some edges in each cluster is equivalent to reducing  $p$  and randomly deleting some edges between different clusters is equivalent to reducing  $q$ . So we consider the worst case. We randomly choose one vertex from each cluster; for each selected vertex  $x_{i}$ , we randomly delete delNum edges connected to  $x_{i}$  in cluster  $C_{i}$ . If  $x_{i}$  has fewer than delNum neighbors within  $C_{i}$ , then we delete all the edges incident to  $x_{i}$  in  $C_{i}$ . We run WHICHCLUSTER queries for all vertices in  $V$  on the resulting graph. We repeated this process for five times for each parameter delNum and recorded the average misclassification error (Table 2). The results show that our oracle is robust against a few number of random edge deletions.
Table 2: The average misclassification error with respect to delNum random edge deletions
delNum	25	32	40	45	50	55	60	65
error	0.00007	0.00007	0.00013	0.00047	0.00080	0.00080	0.00080	0.00087
We also conducted experiments to report the query complexity and running time of our oracle. See Appendix F for more details.
We have devised a new spectral clustering oracle with sublinear preprocessing and query time. In comparison to the approach presented in [14], our oracle exhibits improved preprocessing efficiency, albeit with a slightly higher misclassification error rate. Furthermore, our oracle can be readily implemented in practical settings, while the clustering oracle proposed in [14] poses challenges in terms of implementation feasibility. To obtain our oracle, we have established a property regarding the spectral embeddings of the vertices in  $V$  for a  $d$ -bounded  $n$ -vertex graph  $G = (V, E)$  that exhibits a  $(k, \varphi, \varepsilon)$ -clustering  $C_1, \ldots, C_k$ . Specifically, if  $x$  and  $y$  belong to the same cluster, the dot product of their spectral embeddings (denoted as  $\langle f_x, f_y \rangle$ ) is approximately  $O\left(\frac{k}{n}\right)$ . Conversely, if  $x$  and  $y$  are from different clusters,  $\langle f_x, f_y \rangle$  is close to 0. We also show that our clustering oracle is robust against a few random edge deletions and conducted experiments on synthetic networks to validate our theoretical results.
The work is supported in part by the Huawei-USTC Joint Innovation Project on Fundamental System Software, NSFC grant 62272431 and "the Fundamental Research Funds for the Central Universities".
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A Complete preliminaries
Let  $G = (V, E)$  denote a  $d$ -regular undirected and unweighted graph, where  $V := \{1, \ldots, n\}$ . Throughout the paper, we use  $i \in [n]$  to denote  $1 \leq i \leq n$  and all the vectors will be column vectors unless otherwise specified or transposed to row vectors. For a vertex  $x \in V$ , let  $\mathbb{1}_x \in \mathbb{R}^n$  denote the indicator of  $x$ , which means  $\mathbb{1}_x(i) = 1$  if  $i = x$  and 0 otherwise. For a vector  $\mathbf{x}$ , we let  $\| \mathbf{x} \|_2 = \sqrt{\sum_i \mathbf{x}(i)^2}$  denote its  $\ell_2$  norm. For a matrix  $A \in \mathbb{R}^{n \times n}$ , we use  $\| A \|$  to denote the spectral norm of  $A$ , and we use  $\| A \|_F$  to denote the Frobenius norm of  $A$ . For any two vectors  $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$ , we let  $\langle \mathbf{x}, \mathbf{y} \rangle = \mathbf{x}^T \mathbf{y}$  denote the dot product of  $\mathbf{x}$  and  $\mathbf{y}$ . For a matrix  $A \in \mathbb{R}^{n \times n}$ , we use  $A_{[i]} \in \mathbb{R}^{n \times i}$  to denote the first  $i$  columns of  $A$ ,  $1 \leq i \leq n$ .
Let  $A \in \mathbb{R}^{n \times n}$  denote the adjacency matrix of  $G$  and let  $D \in \mathbb{R}^{n \times n}$  denote a diagonal matrix. For the adjacency matrix  $A$ ,  $A(i,j) = 1$  if  $(i,j) \in E$  and 0 otherwise,  $u, v \in [n]$ . For the diagonal matrix  $D$ ,  $D(i,i) = \deg(i)$ , where  $\deg(i)$  is the degree of vertex  $i$ ,  $i \in [n]$ . We denote with  $L$  the normalized Laplacian of  $G$  where  $L = D^{-1/2}(D - A)D^{-1/2} = I - \frac{A}{d}$ . For  $L$ , we use  $0 \leq \lambda_1 \leq \dots \leq \lambda_n \leq 2$  [7] to denote its eigenvalues and we use  $u_1, \ldots, u_n \in \mathbb{R}^n$  to denote the corresponding eigenvectors. Note that the corresponding eigenvectors are not unique, in this paper, we let  $u_1, \ldots, u_n$  be an orthonormal basis of eigenvectors of  $L$ . Let  $U \in \mathbb{R}^{n \times n}$  be a matrix whose  $i$ -th column is  $u_i$ ,  $i \in [n]$ , then for every vertex  $x \in V$ ,  $f_x = U_{[k]}^T \mathbb{1}_x$ . For any two sets  $S_1$  and  $S_2$ , we let  $S_1 \triangle S_2$  denote the symmetric difference between  $S_1$  and  $S_2$ .
B  $d$ -bounded graphs to  $d$ -regular graphs
For a  $d$ -bounded graph  $G' = (V, E)$ , we can get a  $d$ -regular graph  $G$  from  $G'$  by adding  $d - \deg(x)$  self-loops with weight  $1/2$  to each vertex  $x \in V$ . Note that the lazy random walk on  $G$  is equivalent to the random walk on  $G'$ , with the random walk satisfying that if we are at vertex  $x$ , then we jump to a random neighbor with probability  $\frac{1}{2d}$  and stay at  $x$  with probability  $1 - \frac{\deg(x)}{2d}$ . We use  $w_{self}(x) = (d - \deg(x)) \cdot \frac{1}{2}$  to denote the weight of all self-loops of  $x \in V$ .
C Formal statement of Lemma 2.1 and description of dot-product algorithms in [14]
Lemma C.1 (Lemma 2.1, Formal; Theorem 2, [14]). Let  $\varepsilon, \varphi \in (0,1)$  with  $\varepsilon \leq \frac{\varphi^2}{10^5}$ . Let  $G = (V,E)$  be a  $d$ -regular graph that admits a  $(k,\varphi,\varepsilon)$ -clustering  $C_1,\ldots,C_k$ . Let  $\frac{1}{n^5} < \xi < 1$ . Then INITIALIZEORACLE $(G,1/2,\xi)$  computes in time  $\left(\frac{k}{\xi}\right)^{O(1)} \cdot n^{1/2 + O(\varepsilon/\varphi^2)} \cdot (\log n)^3 \cdot \frac{1}{\varphi^2}$  a sublinear space data structure  $\mathcal{D}$  of size  $\left(\frac{k}{\xi}\right)^{O(1)} \cdot n^{1/2 + O(\varepsilon/\varphi^2)} \cdot (\log n)^3$  such that with probability at least  $1 - n^{-100}$  the following property is satisfied:
For every pair of vertices  $x, y \in V$ , SPECTRALDOTPRODUCT  $(G, x, y, 1/2, \xi, \mathcal{D})$  computes an output value  $\langle f_x, f_y \rangle_{\mathrm{apx}}$  such that with probability at least  $1 - n^{-100}$
$$
\left| \langle f _ {x}, f _ {y} \rangle_ {\mathrm {a p x}} - \langle f _ {x}, f _ {y} \rangle \right| \leq \frac {\xi}{n}.
$$
The running time of SPECTRALDOTPRODUCT $(G, x, y, 1/2, \xi, \mathcal{D})$  is  $\left(\frac{k}{\xi}\right)^{O(1)} \cdot n^{1/2 + O(\varepsilon/\varphi^2)} \cdot (\log n)^2 \cdot \frac{1}{\varphi^2}$ .
Algorithm 4: RUNRANDOMWALKs(G,R,t,x)
1 Run  $R$  random walks of length  $t$  starting from  $x$
2 Let  $\widehat{m}_x(y)$  be the fraction of random walks that ends at  $y$
3 return  $\widehat{m}_x$
Algorithm 5: ESTIMATE TRANSITIONMATRIX  $(G, I_S, R, t)$
1 for each sample  $x\in I_S$  do   
2  $\begin{array}{rlr}{\widehat{m}_x:=}\end{array}$  RUNRANDOMWALKs  $(G,R,t,x)$    
3 end   
4 Let  $\widehat{Q}$  be the matrix whose columns are  $\widehat{m}_x$  for  $x\in I_S$    
5 return  $\widehat{Q}$    
Algorithm 6: ESTIMATECOLLISIONPROBABILITIES(G,IS,R,t)   
1 for  $i = 1$  to  $O(\log n)$  do   
2  $\widehat{Q}_i\coloneqq$  ESTIMATETRANSITIONMATRIX  $(G,I_S,R,t)$    
3  $\widehat{P}_i\coloneqq$  ESTIMATETRANSITIONMATRIX  $(G,I_S,R,t)$    
4  $\mathcal{G}_i\coloneqq \frac{1}{2} (\widehat{P}_i^T\widehat{Q}_i + \widehat{Q}_i^T\widehat{P}_i)$    
5 end   
6 Let  $\mathcal{G}$  be a matrix obtained by taking the entrywise median of  $\mathcal{G}_i^\prime s$    
7 return  $\mathcal{G}$    
Algorithm 7: INITIALIZERACLE(G,δ,ξ) Need:ε/φ²≤105   
1 t := 20-log n /φ²   
2 Rinit := O(n1-δ+980·ε/φ²·k17/ξ²)   
3 s := O(n480·ε/φ²·log n·k8/ξ²)   
4 Let IS be the multiset of s indices chosen independently and uniformly at random from {1,...,n}   
5 for i = 1 to O(log n) do   
6  $\widehat{Q}_i\coloneqq$  ESTIMATETRANSITIONMATRIX(G,IS,Rinit,t)   
7 end   
8 G:=ESTIMATECOLLISIONPROBABILITIES(G,IS,Rinit,t)   
9 Let  $\frac{n}{s}\cdot \mathcal{G}\coloneqq \widehat{W\Sigma W^{T}}$  be the eigendecomposition of  $\frac{n}{s}\cdot \mathcal{G}$    
10 if  $\widehat{\Sigma}^{-1}$  exists then   
11  $\Psi \coloneqq \frac{n}{s}\cdot \widehat{W}_{[k]}\widehat{\Sigma}_{[k]}^{-2}\widehat{W}_{[k]}^{T}$    
12 return D := {Ψ, Q1,..., QO(log n)}   
13 end   
Algorithm 8: SPECTRALDOTPRODUCTORACLE(G,x,y,δ,ξ,D) Need:ε/φ² ≤ 1/105   
1  $R_{\mathrm{query}}\coloneqq O(n^{\delta +500\cdot \varepsilon /\varphi^2}\cdot k^9 /\xi^2)$    
2 for i = 1 to O(log n) do   
3  $\widehat{m}_x^i\coloneqq$  RUNRANDOMWALKs(G,Rquery,t,x)   
4  $\widehat{m}_y^i\coloneqq$  RUNRANDOMWALKs(G,Rquery,t,y)   
5 end   
6 Let  $\alpha_{x}$  be a vector obtained by taking the entrywise median of  $(\widehat{Q}_i)^T (\widehat{m}_x^i)$  over all runs   
7 Let  $\alpha_{y}$  be a vector obtained by taking the entrywise median of  $(\widehat{Q}_i)^T (\widehat{m}_y^i)$  over all runs   
8 return  $\langle f_x,f_y\rangle_{\mathrm{apx}}\coloneqq \alpha_x^T\Psi \alpha_y$
D Deferred proofs
Lemma D.1 (Restatement of Lemma 3.1). Let  $\alpha \in (0,1)$ . Let  $k \geq 2$  be an integer,  $\varphi \in (0,1)$ , and  $\varepsilon \in (0,1)$ . Let  $G = (V,E)$  be a d-regular  $(k,\varphi,\varepsilon)$ -clusterable graph with  $|V| = n$ . There exists a subset  $\widehat{V} \subseteq V$  with  $|\widehat{V}| \geq (1 - \alpha)|V|$  such that for all  $x \in \widehat{V}$ , it holds that  $\|f_x\|_2 \leq \sqrt{\frac{1}{\alpha} \cdot \frac{k}{n}}$ .
Proof. Recall that  $u_{1}, \ldots, u_{k}$  are an orthonormal basis of eigenvectors of  $L$ , so  $\| u_{i}\|_{2}^{2} = 1$  for all  $i\in [k]$ . So  $\sum_{i = 1}^{k}\| u_{i}\|_{2}^{2} = \sum_{i = 1}^{n}\| f_{x_{i}}\|_{2}^{2} = k$ . Let  $X$  be a random variable such that  $X = \| f_{x_i}\| _2^2$  with probability  $\frac{1}{n}$ , for each  $i\in [n]$ . Then we have  $\operatorname {E}[X] = \frac{1}{n}\sum_{i = 1}^{n}\| f_{x_i}\| _2^2 = \frac{k}{n}$ . Using Markov's inequality, we have  $\operatorname *{Pr}[X\geq \frac{1}{\alpha}\cdot \operatorname {E}[X]] = \operatorname *{Pr}[X\geq \frac{1}{\alpha}\cdot \frac{k}{n} ]\leq \alpha$ . This gives us that  $\operatorname *{Pr}[X\leq \frac{1}{\alpha}\cdot \frac{k}{n} ]\geq 1 - \alpha$ , which means that at least  $(1 - \alpha)$  fraction of vertices in  $V$  satisfies  $\| f_x\|_2^2\leq \frac{1}{\alpha}\cdot \frac{k}{n}$ . We define  $\widehat{V} := \{x\in V:\| f_x\|_2^2\leq \frac{1}{\alpha}\cdot \frac{k}{n}\}$ , then we have  $|\widehat{V} |\geq (1 - \alpha)|V|$ . This ends the proof.
Lemma D.2 (Restatement of Lemma 3.2). Let  $\beta \in (0,1)$ . Let  $k \geq 2$  be an integer,  $\varphi \in (0,1)$ , and  $\varepsilon \in (0,1)$ . Let  $G = (V,E)$  be a  $d$ -regular graph that admits a  $(k,\varphi,\varepsilon)$ -clustering  $C_1, \ldots, C_k$  with  $|V| = n$ . There exists a subset  $\widetilde{V} \subseteq V$  with  $|\widetilde{V}| \geq (1 - \beta)|V|$  such that for all  $x \in \widetilde{V}$ , it holds that  $\|f_x - \mu_x\|_2 \leq \sqrt{\frac{4k\varepsilon}{\beta\varphi^2} \cdot \frac{1}{n}}$ .
The following result will be used in our proof:
Lemma D.3 (Lemma 6, [14]). Let  $k \geq 2$  be an integer;  $\varphi \in (0,1)$ , and  $\varepsilon \in (0,1)$ . Let  $G = (V,E)$  be a  $d$ -regular graph that admits a  $(k,\varphi,\varepsilon)$ -clustering  $C_1, \ldots, C_k$ . Then for all  $\alpha \in \mathbb{R}^k$ , with  $\| \alpha \|_2 = 1$  we have
$$
\sum_ {i = 1} ^ {k} \sum_ {x \in C _ {i}} \left\langle f _ {x} - \mu_ {i}, \alpha \right\rangle^ {2} \leq \frac {4 \varepsilon}{\varphi^ {2}}.
$$
Proof of Lemma 3.2. By summing over  $\alpha$  in an orthonormal basis of  $\mathbb{R}^k$ , we can get
$$
\sum_ {x \in V} \| f _ {x} - \mu_ {x} \| _ {2} ^ {2} \leq k \cdot \frac {4 \varepsilon}{\varphi^ {2}} = \frac {4 k \varepsilon}{\varphi^ {2}},
$$
where  $\mu_x$  is the cluster center of the cluster that  $x$  belongs to. Define  $V^* = \{x \in V : \| f_x - \mu_x\|_2^2 \geq \frac{4k\varepsilon}{\beta\varphi^2} \cdot \frac{1}{n}\}$ . Then,
$$
\frac {4 k \varepsilon}{\varphi^ {2}} \geq \sum_ {x \in V} \| f _ {x} - \mu_ {x} \| _ {2} ^ {2} \geq \sum_ {x \in V ^ {*}} \| f _ {x} - \mu_ {x} \| _ {2} ^ {2} \geq \sum_ {x \in V ^ {*}} \frac {4 k \varepsilon}{\beta \varphi^ {2}} \cdot \frac {1}{n} = | V ^ {*} | \cdot \frac {4 k \varepsilon}{\beta \varphi^ {2}} \cdot \frac {1}{n}.
$$
So, we can get  $|V^{*}| \leq \beta n$ . We define  $\widetilde{V} = V \backslash V^{*} = \{x \in V : \| f_{x} - \mu_{x}\|_{2}^{2} \leq \frac{4k\varepsilon}{\beta\varphi^{2}} \cdot \frac{1}{n}\}$ . Therefore, we have  $|\widetilde{V}| \geq (1 - \beta)n = (1 - \beta)|V|$ . This ends the proof.
Lemma D.4 (Restatement of Lemma 3.3). Let  $k \geq 2$  be an integer,  $\varphi \in (0,1)$ , and  $\frac{\varepsilon}{\varphi^2}$  be smaller than a sufficiently small constant. Let  $G = (V,E)$  be a d-regular graph that admits a  $(k,\varphi ,\varepsilon)$ -clustering  $C_1,\ldots ,C_k$ . Let  $C_i$  denote the cluster corresponding to the center  $\mu_{i}$ ,  $i \in [k]$ . Then for every  $C_i$ ,  $i \in [k]$ , there exists a subset  $\widetilde{C_i} \subseteq C_i$  with  $|\widetilde{C_i}| \geq (1 - \frac{10^4\varepsilon}{\varphi^2})|C_i|$  such that for all  $x \in \widetilde{C_i}$ , it holds that  $\langle f_x,\mu_i\rangle \geq 0.96\| \mu_i\| _2^2$ .
Lemma D.5 (Lemma 31, [14]). Let  $k \geq 2$ ,  $\varphi \in (0,1)$ , and  $\frac{\varepsilon}{\varphi^2}$  be smaller than a sufficiently small constant. Let  $G = (V,E)$  be a d-regular graph that admits a  $(k,\varphi ,\varepsilon)$ -clustering  $C_1,\ldots ,C_k$ . If  $\mu_i^\prime s$  are cluster means then the following conditions hold. Let  $S \subset \{\mu_1,\dots ,\mu_k\}$ . Let  $\Pi$  denote the orthogonal projection matrix on to the span  $(S)^{\perp}$ . Let  $\mu \in \{\mu_1,\dots ,\mu_k\} \backslash S$ . Let  $C$  denote the cluster corresponding to the center  $\mu$ . Let
$$
\widehat {C} := \left\{x \in V: \left\langle \Pi f _ {x}, \Pi \mu \right\rangle \geq 0. 9 6 \| \Pi \mu \| _ {2} ^ {2} \right\}
$$
then we have:
$$
| C \backslash \widehat {C} | \leq \frac {1 0 ^ {4} \varepsilon}{\varphi^ {2}} | C |.
$$
Proof of Lemma 3.3. We apply  $S = \emptyset$  in Lemma D.5 so that  $\Pi$  is an identity matrix and we will have  $|C_i\backslash \widehat{C_i} |\leq \frac{10^4\varepsilon}{\varphi^2} |C_i|$ , where  $\widehat{C_i} := \{x\in V:\langle f_x,\mu_i\rangle \geq 0.96\| \mu_i\| _2^2\}$ ,  $i\in [k]$ . So
$$
\left| C _ {i} \cap \widehat {C _ {i}} \right| \geq \left(1 - \frac {1 0 ^ {4} \varepsilon}{\varphi^ {2}}\right) \left| C _ {i} \right|.
$$
We define  $\widetilde{C_i} = C_i\cap \widehat{C_i}$ $i\in [k]$ . Therefore, for every  $C_i$ ,  $i\in [k]$ , there exists a subset  $\widetilde{C_i}\subseteq C_i$  with  $|\widetilde{C_i} |\geq (1 - \frac{10^4\varepsilon}{\varphi^2})|C_i|$  such that for all  $x\in \widetilde{C_i}$ , it holds that  $\langle f_x,\mu_i\rangle \geq 0.96\| \mu_i\| _2^2$ .
Lemma D.6 (Restatement of Lemma 3.4). Let  $k \geq 2$ ,  $\varphi \in (0,1)$  and  $\frac{\varepsilon}{\varphi^2}$  be smaller than a sufficiently small constant. Let  $G = (V,E)$  be a d-regular  $n$ -vertex graph that admits a  $(k,\varphi,\varepsilon)$ -clustering  $C_1, \ldots, C_k$ . Suppose that  $x,y \in V$  are in the same cluster  $C_i, i \in [k]$  and both of them are good vertex with respect to  $\alpha = 2\sqrt{k} \cdot (\frac{\varepsilon}{\varphi^2})^{1/3}$  and  $\beta = 2\sqrt{k} \cdot (\frac{\varepsilon}{\varphi^2})^{1/3}$ , the following holds:
$$
\langle f _ {x}, f _ {y} \rangle \geq 0. 9 6 \left(1 - \frac {4 \sqrt {\varepsilon}}{\varphi}\right) \frac {1}{| C _ {i} |} - \frac {\sqrt {k}}{n} \cdot \left(\frac {\varepsilon}{\varphi^ {2}}\right) ^ {1 / 6}.
$$
Lemma D.7 (Lemma 7, [14]). Let  $k \geq 2$  be an integer,  $\varphi \in (0,1)$ , and  $\varepsilon \in (0,1)$ . Let  $G = (V,E)$  be a  $d$ -regular graph that admits a  $(k,\varphi,\varepsilon)$ -clustering  $C_1, \ldots, C_k$ . Then we have
1. for all  $i\in [k]$ $\left|\| \mu_i\| _2^2 -\frac{1}{|C_i|}\right|\leq \frac{4\sqrt{\varepsilon}}{\varphi}\frac{1}{|C_i|},$
2. for all  $i \neq j \in [k]$ ,  $|\langle \mu_i, \mu_j \rangle| \leq \frac{8\sqrt{\varepsilon}}{\varphi} \frac{1}{\sqrt{|C_i||C_j|}}$ .
Proof of Lemma 3.4. According to the distributive law of dot product, we have
$$
\langle f _ {x}, f _ {y} \rangle = \langle f _ {x}, f _ {y} - \mu_ {i} + \mu_ {i} \rangle = \langle f _ {x}, f _ {y} - \mu_ {i} \rangle + \langle f _ {x}, \mu_ {i} \rangle .
$$
By using Cauchy-Schwarz Inequality, we have  $|\langle f_x,f_y - \mu_i\rangle |\leq \| f_x\| _2\cdot \| f_y - \mu_i\| _2$  . Since  $x$  and  $y$  are both good vertices with respect to  $\alpha = 2\sqrt{k}\cdot (\frac{\varepsilon}{\varphi^2})^{1 / 3}$  and  $\beta = 2\sqrt{k}\cdot (\frac{\varepsilon}{\varphi^2})^{1 / 3}$  , we have
$$
\left| \left\langle f _ {x}, f _ {y} - \mu_ {i} \right\rangle \right| \leq \| f _ {x} \| _ {2} \cdot \| f _ {y} - \mu_ {i} \| _ {2} \leq \sqrt {\frac {1}{\alpha} \cdot \frac {k}{n}} \cdot \sqrt {\frac {4 k \varepsilon}{\beta \varphi^ {2}} \cdot \frac {1}{n}} = \frac {\sqrt {k}}{n} \cdot \left(\frac {\varepsilon}{\varphi^ {2}}\right) ^ {1 / 6},
$$
which gives us that  $\langle f_x, f_y - \mu_i \rangle \geq -\frac{\sqrt{k}}{n} \cdot \left( \frac{\varepsilon}{\varphi^2} \right)^{1/6}$ . Recall that  $x$  is a good vertex, we have  $\langle f_x, \mu_i \rangle \geq 0.96 \| \mu_i \|_2^2$ . Hence, it holds that
$$
\begin{array}{l} \langle f _ {x}, f _ {y} \rangle = \langle f _ {x}, f _ {y} - \mu_ {i} \rangle + \langle f _ {x}, \mu_ {i} \rangle \\ \geq 0. 9 6 \| \mu_ {i} \| _ {2} ^ {2} - \frac {\sqrt {k}}{n} \cdot \left(\frac {\varepsilon}{\varphi^ {2}}\right) ^ {1 / 6} \\ \geq 0. 9 6 \left(1 - \frac {4 \sqrt {\varepsilon}}{\varphi}\right) \frac {1}{| C _ {i} |} - \frac {\sqrt {k}}{n} \cdot \left(\frac {\varepsilon}{\varphi^ {2}}\right) ^ {1 / 6}. \\ \end{array}
$$
The last inequality is according to item 1 in Lemma D.7.
Lemma D.8 (Restatement of Lemma 3.5). Let  $k \geq 2$ ,  $\varphi \in (0,1)$  and  $\frac{\varepsilon}{\varphi^2}$  be smaller than a sufficiently small constant. Let  $G = (V,E)$  be a d-regular  $n$ -vertex graph that admits a  $(k,\varphi,\varepsilon)$ -clustering  $C_1, \ldots, C_k$ . Suppose that  $x \in C_i$ ,  $y \in C_j$ ,  $(i,j \in [k], i \neq j)$  and both of them are good vertex with respect to  $\alpha = 2\sqrt{k} \cdot (\frac{\varepsilon}{\varphi^2})^{1/3}$  and  $\beta = 2\sqrt{k} \cdot (\frac{\varepsilon}{\varphi^2})^{1/3}$ , the following holds:
Proof. According to the distributive law of dot product, we have
$$
\begin{array}{l} \langle f _ {x}, f _ {y} \rangle = \langle f _ {x}, f _ {y} - \mu_ {j} + \mu_ {j} \rangle \\ = \left\langle f _ {x}, f _ {y} - \mu_ {j} \right\rangle + \left\langle f _ {x}, \mu_ {j} \right\rangle \\ = \left\langle f _ {x}, f _ {y} - \mu_ {j} \right\rangle + \left\langle f _ {x} - \mu_ {i} + \mu_ {i}, \mu_ {j} \right\rangle \\ = \langle f _ {x}, f _ {y} - \mu_ {j} \rangle + \langle f _ {x} - \mu_ {i}, \mu_ {j} \rangle + \langle \mu_ {i}, \mu_ {j} \rangle . \\ \end{array}
$$
By Cauchy-Schwarz Inequality, we have
$$
\left| \left\langle f _ {x}, f _ {y} - \mu_ {j} \right\rangle \right| \leq \left\| f _ {x} \right\| _ {2} \cdot \left\| f _ {y} - \mu_ {j} \right\| _ {2}
$$
$$
| \langle f _ {x} - \mu_ {i}, \mu_ {j} \rangle | \leq \| \mu_ {j} \| _ {2} \cdot \| f _ {x} - \mu_ {i} \| _ {2}.
$$
Since  $x$  and  $y$  are both good vertices with respect to  $\alpha = 2\sqrt{k} \cdot \left(\frac{\varepsilon}{\varphi^2}\right)^{1/3}$  and  $\beta = 2\sqrt{k} \cdot \left(\frac{\varepsilon}{\varphi^2}\right)^{1/3}$ , we have
$$
\left| \left\langle f _ {x}, f _ {y} - \mu_ {j} \right\rangle \right| \leq \left\| f _ {x} \right\| _ {2} \cdot \left\| f _ {y} - \mu_ {j} \right\| _ {2} \leq \sqrt {\frac {1}{\alpha} \cdot \frac {k}{n}} \cdot \sqrt {\frac {4 k \varepsilon}{\beta \varphi^ {2}} \cdot \frac {1}{n}} = \frac {\sqrt {k}}{n} \cdot \left(\frac {\varepsilon}{\varphi^ {2}}\right) ^ {1 / 6}
$$
$$
\left| \left\langle f _ {x} - \mu_ {i}, \mu_ {j} \right\rangle \right| \leq \| \mu_ {j} \| _ {2} \cdot \| f _ {x} - \mu_ {i} \| _ {2} \leq \| \mu_ {j} \| _ {2} \cdot \sqrt {\frac {4 k \varepsilon}{\beta \varphi^ {2}} \cdot \frac {1}{n}} = \frac {\sqrt {2} k ^ {1 / 4}}{\sqrt {n}} \cdot \left(\frac {\varepsilon}{\varphi^ {2}}\right) ^ {1 / 3} \cdot \| \mu_ {j} \| _ {2}.
$$
So we have