Vertex set splitting and adjacency matrix

In Algorithm 2, we first randomly partition the vertex set $V$ into two disjoint subsets $Z$ and $Y$ by assigning $+1$ and $-1$ to each vertex independently with equal probability. Let ${\mathbf{B}} \in{\mathbb{R}}^{|Z|\times |Y|}$ denote the submatrix of ${\mathbf{A}}$, while ${\mathbf{A}}$ was defined in (6), where rows and columns of ${\mathbf{B}}$ correspond to vertices in $Z$ and $Y$ respectively. Let $n_i$ denote the number of vertices in $Z\cap V_i$, where $V_i$ denotes the true partition with $|V_i| = \frac{n}{k}$ for all $i\in [k]$, then $n_i$ can be written as a sum of independent Bernoulli random variables, i.e.,

(12) \begin{align} n_i = |Z \cap V_i| = \sum _{v\in V_i} \unicode{x1D7D9}_{ \{v\in Z\} }\,, \end{align}

and $|Y\cap V_i| = |V_i| - |Z\cap V_i| = \frac{n}{k} - n_i$ for each $i \in [k]$.

However, the splitting is imperfect in most cases since the size of $Z$ and $Y$ would not be exactly the same under the independence assumption. The random matrix ${\mathbf{B}}$ is parameterised by $\{\boldsymbol{\mathcal{A}}^{(m)}\}_{m \in{\mathcal M}}$ and $\{n_i\}_{i=1}^{k}$. If we take expectation over $\{\boldsymbol{\mathcal{A}}^{(m)}\}_{m \in{\mathcal M}}$ given the block size information $\{n_i\}_{i=1}^{k}$, then it gives rise to the expectation of the imperfect splitting, denoted by $\widetilde{{\mathbf{B}}}$,

\begin{equation*} \widetilde {{\mathbf{B}}} \,:\!=\, \begin {bmatrix} \alpha {\mathbf{J}}_{n_1 \times ( \frac {n}{k} - n_1)} & \quad \beta {\mathbf{J}}_{n_1 \times (\frac {n}{k} - n_2)} & \quad \dots & \quad \beta {\mathbf{J}}_{n_1 \times (\frac {n}{k}- n_k)} \\[4pt] \beta {\mathbf{J}}_{n_2 \times (\frac {n}{k}- n_1)} & \quad \alpha {\mathbf{J}}_{n_2 \times (\frac {n}{k}- n_2)} & \quad\dots & \quad \beta {\mathbf{J}}_{n_2 \times (\frac {n}{k}- n_k)} \\[4pt] \vdots & \quad \vdots & \quad \ddots & \quad \vdots \\[4pt] \beta {\mathbf{J}}_{n_k \times (\frac {n}{k}- n_1)} & \quad \beta {\mathbf{J}}_{n_k \times (\frac {n}{k}- n_2)} & \quad \dots & \quad \alpha {\mathbf{J}}_{n_k \times (\frac {n}{k}- n_k)} \end {bmatrix}\,, \end{equation*}

where $\alpha$, $\beta$ are defined in (8). In the perfect splitting case, the dimension of each block is $n/(2k)\times n/(2k)$ since ${\mathbb{E}} n_i = n/(2k)$ for all $i \in [k]$, and the expectation matrix $\overline{{\mathbf{B}}}$ can be written as

\begin{equation*} \overline {{\mathbf{B}}} \,:\!=\, \begin {bmatrix} \alpha {\mathbf{J}}_{\frac {n}{2k}} & \quad \beta {\mathbf{J}}_{\frac {n}{2k}} & \quad \dots & \quad \beta {\mathbf{J}}_{\frac {n}{2k}} \\[4pt] \beta {\mathbf{J}}_{\frac {n}{2k}} & \quad \alpha {\mathbf{J}}_{\frac {n}{2k}} & \quad \dots & \quad \beta {\mathbf{J}}_{\frac {n}{2k}} \\[4pt] \vdots & \quad \vdots & \quad \ddots & \quad \vdots \\[4pt] \beta {\mathbf{J}}_{\frac {n}{2k}} & \quad \beta {\mathbf{J}}_{\frac {n}{2k}} & \quad\dots & \quad \alpha {\mathbf{J}}_{\frac {n}{2k}} \end {bmatrix}\,. \end{equation*}

In Algorithm 4, $Y_1$ is a random subset of $Y$ obtained by selecting each element with probability $1/2$ independently, and $Y_2 = Y\setminus Y_1$. Let $n^{\prime }_{i}$ denote the number of vertices in $Y_1\cap V_i$, then $n^{\prime }_{i}$ can be written as a sum of independent Bernoulli random variables,

(13) \begin{align} n^{\prime }_{i} = |Y_1 \cap V_i| = \sum _{v\in V_i} \mathbf{1}_{ \{v\in Y_1\} }\,, \end{align}

and $|Y_2\cap V_i| = |V_i| - |Z\cap V_i| - |Y_1\cap V_i| = n/k - n_i - n^{\prime }_{i}$ for all $i \in [k]$.

Induced sub-hypergraph

Let $H[Y_1 \cup Z]$(resp. $H[Y_2 \cup Z]$) denote the induced sub-hypergraph on vertex set $Y_1\cup Z$ (resp. $Y_2 \cup Z$), and ${\mathbf{B}}_1 \in{\mathbb{R}}^{|Z|\times |Y_1|}$ (resp. ${\mathbf{B}}_2 \in{\mathbb{R}}^{|Z|\times |Y_2|}$) denote the adjacency matrices corresponding to the sub-hypergraphs, where rows and columns of ${\mathbf{B}}_1$ (resp. ${\mathbf{B}}_2$) are corresponding to elements in $Z$ and $Y_1$ (resp., $Z$ and $Y_2$). Therefore, ${\mathbf{B}}_1$ and ${\mathbf{B}}_2$ are parameterised by $\{\boldsymbol{\mathcal{A}}^{(m)}\}_{m \in{\mathcal M}}$, $\{n_i\}_{i=1}^{k}$ and $\{n^{\prime }_{i}\}_{i=1}^{k}$, and the entries in ${\mathbf{B}}_1$ are independent of the entries in ${\mathbf{B}}_2$, due to the independence of hyperedges. If we take expectation over $\{\boldsymbol{\mathcal{A}}^{(m)}\}_{m \in{\mathcal M}}$ conditioning on $\{n_i\}_{i=1}^{k}$ and $\{n^{\prime }_{i}\}_{i=1}^{k}$, then it gives rise to the expectation of the imperfect splitting, denoted by $\widetilde{{\mathbf{B}}}_1$,

(14) \begin{equation} \widetilde{{\mathbf{B}}}_1 \,:\!=\, \begin{bmatrix} \widetilde{\alpha }_{11}{\mathbf{J}}_{n_1 \times n^{\prime }_{1}} & \quad \dots & \quad \widetilde{\beta }_{1k}{\mathbf{J}}_{n_1 \times n^{\prime }_{k}} \\[4pt] \vdots & \quad \ddots & \quad \vdots \\[4pt] \widetilde{\beta }_{k1}{\mathbf{J}}_{n_k \times n^{\prime }_{1}} & \quad \dots & \quad \widetilde{\alpha }_{kk}{\mathbf{J}}_{n_k \times n^{\prime }_{k}} \end{bmatrix}\,, \end{equation}

where

(15a) \begin{align} \widetilde{\alpha }_{ii} \,:\!=\,&\, \sum _{m \in{\mathcal{M}}}\left \{ \binom{n_i + n_i^{\prime } - 2}{m-2} \frac{a_m - b_m}{\binom{n}{m-1}} + \binom{\sum _{l=1}^{k}(n_l + n_l^{\prime }) - 2}{m-2} \frac{b_m}{\binom{n}{m-1}} \right \}\,, \end{align}

(15b) \begin{align} \widetilde{\beta }_{ij} \,:\!=\,&\, \sum _{m \in{\mathcal{M}}} \binom{\sum _{l=1}^{k}(n_l + n_l^{\prime }) - 2}{m-2}\frac{b_m}{\binom{n}{m-1}}\,,\quad i\neq j, \, i, j\in [k]\,. \end{align}

The expectation of the perfect splitting, denoted by $\overline{{\mathbf{B}}}_1$, can be written as

(16) \begin{equation} \overline{{\mathbf{B}}}_1 \,:\!=\, \begin{bmatrix} \overline{\alpha }{\mathbf{J}}_{\frac{n}{2k} \times \frac{n}{4k}} & \quad \overline{\beta }{\mathbf{J}}_{\frac{n}{2k} \times \frac{n}{4k}} & \quad \dots & \quad \overline{\beta }{\mathbf{J}}_{\frac{n}{2k} \times \frac{n}{4k}} \\[4pt] \overline{\beta }{\mathbf{J}}_{\frac{n}{2k} \times \frac{n}{4k}} & \quad \overline{\alpha }{\mathbf{J}}_{\frac{n}{2k} \times \frac{n}{4k}} & \quad \dots & \quad \overline{\beta }{\mathbf{J}}_{\frac{n}{2k} \times \frac{n}{4k}} \\[4pt] \vdots & \quad \vdots & \quad \ddots & \quad \vdots \\[4pt] \overline{\beta }{\mathbf{J}}_{\frac{n}{2k} \times \frac{n}{4k}} & \quad \overline{\beta }{\mathbf{J}}_{\frac{n}{2k} \times \frac{n}{4k}} & \quad\dots & \quad \overline{\alpha }{\mathbf{J}}_{\frac{n}{2k} \times \frac{n}{4k}} \end{bmatrix}\,, \end{equation}

where

(17) \begin{align} \overline{\alpha } \,:\!=\, \sum _{m \in{\mathcal M}}\left \{ \binom{ \frac{3n}{4k} - 2}{m-2} \frac{a_m - b_m}{\binom{n}{m-1}} + \binom{\frac{3n}{4} - 2}{m-2}\frac{b_m}{\binom{n}{m-1}} \right \}\,,\quad \overline{\beta } \,:\!=\, \sum _{m \in{\mathcal M}} \binom{\frac{3n}{4} - 2}{m-2}\frac{b_m}{\binom{n}{m-1}}\,. \end{align}

The matrices $\widetilde{{\mathbf{B}}}_2, \overline{{\mathbf{B}}}_2$ can be defined similarly, since dimensions of $|Y_2\cap V_i|$ are also determined by $n_i$ and $n^{\prime }_{i}$. Obviously, $\overline{{\mathbf{B}}}_2 = \overline{{\mathbf{B}}}_1$ since ${\mathbb{E}} n^{\prime }_{i} ={\mathbb{E}} (n/k- n_{i} - n_i^{\prime }) = n/(4k)$ for all $i\in [k]$.

Fixing Dimensions

The dimensions of $\widetilde{{\mathbf{B}}}_1$ and $\widetilde{{\mathbf{B}}}_2$, as well as blocks they consist of, are not deterministic – since $n_i$ and $n^{\prime }_i$, defined in (12) and (13) respectively, are sums of independent random variables. As such, we cannot directly compare them. In order to overcome this difficulty, we embed ${\mathbf{B}}_1$ and ${\mathbf{B}}_2$ into the following $n\times n$ matrices:

(18) \begin{equation} {\mathbf{A}}_1\,:\!=\, \begin{bmatrix} \mathbf{0}_{|Z|\times |Z|} & \quad{\mathbf{B}}_1 & \quad \mathbf{0}_{|Z|\times |Y_2|} \\[4pt] \mathbf{0}_{|Y|\times |Z|} & \quad \mathbf{0}_{|Y|\times |Y_1|} & \quad \mathbf{0}_{|Y|\times |Y_2|} \end{bmatrix}\,,\quad{\mathbf{A}}_2\,:\!=\, \begin{bmatrix} \mathbf{0}_{|Z|\times |Z|} & \quad \mathbf{0}_{|Z|\times |Y_1|} & \quad{\mathbf{B}}_2 \\[4pt] \mathbf{0}_{|Y|\times |Z|} & \quad \mathbf{0}_{|Y|\times |Y_1|} & \quad \mathbf{0}_{|Y|\times |Y_2|} \end{bmatrix}\,. \end{equation}

Note that ${\mathbf{A}}_1$ and ${\mathbf{A}}_2$ have the same size. Also by definition, the entries in ${\mathbf{A}}_1$ are independent of the entries in ${\mathbf{A}}_2$. If we take expectation over $\{\boldsymbol{\mathcal{A}}^{(m)}\}_{m\in{\mathcal M}}$ conditioning on $\{n_i\}_{i=1}^{k}$ and $\{n^{\prime }_{i}\}_{i=1}^{k}$, then we obtain the expectation matrices of the imperfect splitting, denoted by $\widetilde{{\mathbf{A}}}_1$(resp. $\widetilde{{\mathbf{A}}}_2$), written as

(19) \begin{equation} \widetilde{{\mathbf{A}}}_1\,:\!=\, \begin{bmatrix} \mathbf{0}_{|Z|\times |Z|} & \quad \widetilde{{\mathbf{B}}}_1 & \quad \mathbf{0}_{|Z|\times |Y_2|} \\[4pt] \mathbf{0}_{|Y|\times |Z|} & \quad \mathbf{0}_{|Y|\times |Y_1|} & \quad \mathbf{0}_{|Y|\times |Y_2|} \end{bmatrix}\,,\quad \widetilde{{\mathbf{A}}}_2\,:\!=\, \begin{bmatrix} \mathbf{0}_{|Z|\times |Z|} & \quad \mathbf{0}_{|Z|\times |Y_1|} & \quad \widetilde{{\mathbf{B}}}_2 \\[4pt] \mathbf{0}_{|Y|\times |Z|} & \quad \mathbf{0}_{|Y|\times |Y_1|} & \quad \mathbf{0}_{|Y|\times |Y_2|} \end{bmatrix}\,. \end{equation}

The expectation matrix of the perfect splitting, denoted by $\overline{{\mathbf{A}}}_1$(resp. $\overline{{\mathbf{A}}}_2$), can be written as

(20) \begin{equation} \overline{{\mathbf{A}}}_1\,:\!=\, \begin{bmatrix} \mathbf{0}_{ \frac{n}{2} \times \frac{n}{2}} & \quad \overline{{\mathbf{B}}}_1 & \quad \mathbf{0}_{\frac{n}{2}\times \frac{n}{4}} \\[4pt] \mathbf{0}_{\frac{n}{2}\times \frac{n}{2}} & \quad \mathbf{0}_{\frac{n}{2}\times \frac{n}{4}} & \quad \mathbf{0}_{\frac{n}{2}\times \frac{n}{4}} \end{bmatrix}\,,\quad \overline{{\mathbf{A}}}_2\,:\!=\, \begin{bmatrix} \mathbf{0}_{ \frac{n}{2} \times \frac{n}{2}} & \quad \mathbf{0}_{\frac{n}{2}\times \frac{n}{4}} & \quad \overline{{\mathbf{B}}}_2\\[4pt] \mathbf{0}_{\frac{n}{2}\times \frac{n}{2}} & \quad \mathbf{0}_{\frac{n}{2}\times \frac{n}{4}} & \quad \mathbf{0}_{\frac{n}{2}\times \frac{n}{4}} \end{bmatrix}\,. \end{equation}

Obviously, $\widetilde{{\mathbf{A}}}_i$ and $\widetilde{{\mathbf{B}}}_i$(resp. $\overline{{\mathbf{A}}}_i$ and $\overline{{\mathbf{B}}}_i$) have the same non-zero singular values for $i = 1, 2$. In the remaining of this section, we will deal with $\widetilde{{\mathbf{A}}}_i$ and $\overline{{\mathbf{A}}}_i$ instead of $\widetilde{{\mathbf{B}}}_i$ and $\overline{{\mathbf{B}}}_i$ for $i = 1, 2$.

5.1.1. Proof outline

Recall that ${\mathbf{A}}_1$ is defined as the adjacency matrix of the induced sub-hypergraph $H[Y_1\cup Z]$ in equation (5.2). Consequently, the index set should contain information only from $H[Y_1\cup Z]$. Define the index sets

\begin{align*} \mathcal{I} = \left \{i\in [n]\,:\, \mathrm{row}(i)\leq 20{\mathcal M}_{\max }d \right \}, \quad \mathcal{I}_{1} = \left \{i\in [n]\,:\, \mathrm{row}(i)\Big |_{Y_1 \cup Z} \leq 20{\mathcal M}_{\max }d \right \}\,, \end{align*}

where $d = \sum _{m \in{\mathcal M}}(m-1)a_m$, and $\mathrm{row}(i)\big |_{Y_1 \cup Z}$ is the row sum of $i$ on $H[Y_1\cup Z]$. We say $\mathrm{row}(i)\big |_{Y_1 \cup Z}=0$ if $i\not \in Y_1\cup Z$, and for vertex $i\in Y_1\cup Z$,

\begin{align*} \mathrm{row}(i)\Big |_{Y_1 \cup Z} \,:\!=\, \sum _{j=1}^{n} \sum _{m\in{\mathcal M}}\,\,\sum _{ \substack{e\in E_m[Y_1\cup Z]\\ \{i, j\}\subset e} } \,\, \boldsymbol{\mathcal{A}}_e^{(m)} = \sum _{m\in{\mathcal M}}(m-1)\sum _{ \substack{e\in E_m[Y_1\cup Z]\\ \{i, j\}\subset e} } \,\, \boldsymbol{\mathcal{A}}_e^{(m)}\,. \end{align*}

As a result, the matrix $({\mathbf{A}}_1)_{\mathcal{I}_1}$ is obtained by restricting ${\mathbf{A}}_1$ on index set $\mathcal{I}_1$. The next $4$ steps guarantee that Algorithm 4 outputs a $\nu$-correct partition.

1. (i) Find the singular subspace ${\mathbf{U}}$ spanned by the first $k$ left singular vectors of $({\mathbf{A}}_1)_{\mathcal{I}_1}$.

2. (ii) Randomly pick $s = 2k \log ^2 n$ vertices from $Y_2$ and denote the corresponding columns in ${\mathbf{A}}_2$ by $\boldsymbol{a}_{i_1},\dots,\boldsymbol{a}_{i_s}$. Project each vector $\boldsymbol{a}_i - \overline{\boldsymbol{a}}$ onto the singular subspace ${\mathbf{U}}$, with $\overline{\boldsymbol{a}}\in{\mathbb{R}}^{n}$ defined by $\overline{\boldsymbol{a}}(j) = \unicode{x1D7D9}_{j\in Z} \cdot (\overline{\alpha } + \overline{\beta })/2$, where $\overline{\alpha }$, $\overline{\beta }$ were defined in (17).

3. (iii) For each projected vector $P_{{\mathbf{U}}}(\boldsymbol{a}_i - \overline{\boldsymbol{a}})$, identify the top $n/(2k)$ coordinates in value and place the corresponding vertices into a set $U^{\prime }_i$. Discard half of the obtained $s$ subsets, those with the lowest blue edge densities.

4. (iv) Sort the remaining sets according to blue hyperedge density and identify $k$ distinct subsets $U^{\prime }_1, \cdots, U^{\prime }_k$ such that $|U^{\prime }_i \cap U^{\prime }_j| \lt \lceil (1 - \nu )n/k\rceil$ if $i\neq j$.

Based on the $4$ steps above in Algorithm 4, the proof of Lemma 4.1 is structured in $4$ parts.

1. (i) Let $\widetilde{{\mathbf{U}}}$ denote the subspace spanned by first $k$ left singular vectors of $\widetilde{{\mathbf{A}}}_1$ defined in (19). Subsection 5.1.2 shows that the subspace angle between ${\mathbf{U}}$ and $\widetilde{{\mathbf{U}}}$ is smaller than any $c\in (0, 1)$ as long as $a_m, b_m$ satisfy certain conditions depending on $c$.

2. (ii) The vector $\widetilde{{\boldsymbol \delta }}_i$, defined in (22), reflects the underlying true partition $Z\cap V_{k(i)}$ for each $i\in [s]$, where $k(i)$ denotes the membership of vertex $i$. Subsection 5.1.3 shows that $\overline{{\boldsymbol \delta }}_i$, an approximation of $\widetilde{{\boldsymbol \delta }}_i$ defined in (23), can be recovered by the projected vector $P_{{\mathbf{U}}}(\boldsymbol{a}_{i} - \overline{\boldsymbol{a}})$, since projection error $\|P_{{\mathbf{U}}}(\boldsymbol{a}_{i} - \overline{\boldsymbol{a}}) - \overline{{\boldsymbol \delta }}_{i}\|_2 \lt c\|\overline{{\boldsymbol \delta }}_{i}\|_2$ for any $c\in (0, 1)$ if $a_m, b_m$ satisfy the desired property in part (i).

3. (iii) Subsection 5.1.4 indicates that the coincidence ratio between the remaining sets and the true partition is at least $\nu$, after discarding half of the sets with the lowest blue edge densities.

4. (iv) Lemma 5.13 proves that we can find $k$ distinct subsets $U^{\prime }_i$ within $k\log ^2 n$ trials with high probability.

5.1.2. Bounding the angle between ${\mathbf{U}}$ and $\widetilde{{\mathbf{U}}}$

The angle between subspaces ${\mathbf{U}}$ and $\widetilde{{\mathbf{U}}}$ is defined as

\begin{equation*}\sin \angle ({\mathbf{U}}, \widetilde {{\mathbf{U}}}) \,:\!=\, \|P_{{\mathbf{U}}} - P_{\widetilde {{\mathbf{U}}}}\|.\end{equation*}

A natural idea is to apply Wedin’s $\sin \Theta$ Theorem (Lemma D.7). Lemma 5.3 indicates that the difference of $\sigma _i(\widetilde{{\mathbf{A}}}_1)$ and $\sigma _i(\overline{{\mathbf{A}}}_1)$ is relatively small, compared to $\sigma _i(\overline{{\mathbf{A}}}_1)$.

Lemma 5.3. Let $\sigma _i(\overline{{\mathbf{A}}}_1)$(resp. $\sigma _i(\widetilde{{\mathbf{A}}}_1)$) denote the singular values of $\overline{{\mathbf{A}}}_1$ (resp. $\widetilde{{\mathbf{A}}}_1$) for all $i\in [k]$, where the matrices $\overline{{\mathbf{A}}}_1$ and $\widetilde{{\mathbf{A}}}_1$ are defined in (20) and (19) respectively. Then

\begin{align*} \sigma _1(\overline{{\mathbf{A}}}_1) =&\,\frac{n\left [ \overline{\alpha } + (k-1)\overline{\beta } \right ] }{2\sqrt{2}k}= \frac{n}{2\sqrt{2}k}\sum _{m\in{\mathcal M}} \left [ \binom{\frac{3n}{4k} - 2}{m-2} \frac{a_m - b_m}{ \binom{n}{m-1}} + k\binom{\frac{3n}{4} - 2}{m-2}\frac{b_m}{\binom{n}{m-1}} \right ] \,,&\\[5pt] \sigma _i(\overline{{\mathbf{A}}}_1) =&\, \frac{n( \overline{\alpha } - \overline{\beta })}{2\sqrt{2}k} = \frac{n}{2\sqrt{2}k}\sum _{m\in{\mathcal M}} \binom{\frac{3n}{4k} - 2}{m-2} \frac{a_m - b_m}{ \binom{n}{m-1}}\,, 2 \leq i \leq k\,,&\\[5pt] \sigma _i(\overline{{\mathbf{A}}}_1) =&\, 0\,, k+1 \leq i \leq n\,.& \end{align*}

with $\overline{\alpha }$, $\overline{\beta }$ defined in (17). Moreover, with probability at least $1 - 2k\exp ({-}k\log ^2(n))$,

\begin{align*} \frac{|\sigma _i(\overline{{\mathbf{A}}}_1) - \sigma _i(\widetilde{{\mathbf{A}}}_1)|}{\sigma _i(\overline{{\mathbf{A}}}_1)} = O\left (n^{-\frac{1}{4}}\log ^{\frac{1}{2} }(n)\right ). \end{align*}

Therefore, with Lemma 5.3, we can write $ \sigma _i(\widetilde{{\mathbf{A}}}_1) = \sigma _i(\overline{{\mathbf{A}}}_1)(1 + o(1)).$ Define ${\mathbf{E}}_1 \,:\!=\,{\mathbf{A}}_1 - \widetilde{{\mathbf{A}}}_1$ and its restriction on $\mathcal{I}_1$ as

(21) \begin{equation} ({\mathbf{E}}_1)_{\mathcal{I}_1} \,:\!=\, ({\mathbf{A}}_1 - \widetilde{{\mathbf{A}}}_1)_{\mathcal{I}_1} = ({\mathbf{A}}_1)_{\mathcal{I}_1} - (\widetilde{{\mathbf{A}}}_1)_{\mathcal{I}_1}\,, \end{equation}

as well as ${\boldsymbol \Delta }_{1}\,:\!=\, (\widetilde{{\mathbf{A}}}_1)_{\mathcal{I}_1} - \widetilde{{\mathbf{A}}}_1$. Then $({\mathbf{A}}_1)_{\mathcal{I}_1} - \widetilde{{\mathbf{A}}}_1$ is decomposed as

\begin{align*} ({\mathbf{A}}_1)_{\mathcal{I}_1} - \widetilde{{\mathbf{A}}}_1 = [({\mathbf{A}}_1)_{\mathcal{I}_1} - (\widetilde{{\mathbf{A}}}_1)_{\mathcal{I}_1}] + [(\widetilde{{\mathbf{A}}}_1)_{\mathcal{I}_1} - \widetilde{{\mathbf{A}}}_1 ] = ({\mathbf{E}}_1)_{\mathcal{I}_1} +{\boldsymbol \Delta }_{1}\,. \end{align*}

Lemma 5.4 shows that the number of high-degree vertices is relatively small. Consequently, Corollary 5.5 indicates $\|{\boldsymbol \Delta }_1\| \leq \sqrt{d}$ with high probability.

Proof of Corollary 5.5. Note that $n - |\mathcal{I}| \leq d^{-3}n$ and $\mathcal{I} \subset \mathcal{I}_1$, then $n - |\mathcal{I}_1| \leq d^{-3}n$. From Lemma 5.4, there are at most $d^{-3} n$ vertices with row sum greater than $20{\mathcal M}_{\max }d$ in the adjacency matrix ${\mathbf{A}}_1$, then the matrix ${\boldsymbol \Delta }_1 = (\widetilde{{\mathbf{A}}}_1)_{\mathcal{I}_1} - \widetilde{{\mathbf{A}}}_1$ has at most $2d^{-3} n^{2}$ non-zero entries. Every entry of $\widetilde{{\mathbf{A}}}_1$ in (19) is bounded by $\alpha$, then,

\begin{align*} \|{\boldsymbol \Delta }_1\| \leq &\, \|{\boldsymbol \Delta }_1 \|_{{\mathrm{F}}} = \|(\widetilde{{\mathbf{A}}}_1)_{\mathcal{I}_1} - \widetilde{{\mathbf{A}}}_1\|_{{\mathrm{F}}} \\ \leq &\, \sqrt{2 d^{-3}n^{2}} \, \alpha = \sqrt{2 d^{-3}} n \sum _{m\in{\mathcal M}} \left [ \binom{ \frac{n}{k}-2}{m-2} \frac{a_m - b_m}{\binom{n}{m-1}} +\left({n - 2 \atop m-2}\right) \frac{b_m}{\binom{n}{m-1}} \right ]\\ \leq &\, \sqrt{2d^{-3} } \sum _{m\in{\mathcal M}} (m-1)a_m \leq \sqrt{2d^{-1}} \leq \sqrt{d}\,. \end{align*}

Moreover, taking $\tau =20{\mathcal M}_{\max }, K=3$ in Theorem 3.3, with probability at least $1 - n^{-2}$

(22) \begin{align} \|({\mathbf{E}}_1)_{\mathcal{I}_1}\|\leq C_3\sqrt{d}\,, \end{align}

where constant $C_3$ depends on ${\mathcal M}_{\max }$. Together with upper bounds for $\|({\mathbf{E}}_1)_{\mathcal{I}_1}\|$ and $\|{\boldsymbol \Delta }_1\|$, Lemma 5.6 shows that the angle between ${\mathbf{U}}$ and $\widetilde{{\mathbf{U}}}$ is relatively small with high probability.

Lemma 5.6. For any $c\in (0,1)$, there exists some constant $C_2$ such that, if

\begin{equation*} \sum _{m\in {\mathcal M}}(m-1)(a_m-b_m)\geq C_2k^{{\mathcal M}_{\max } - 1}\sqrt {d}\,, \end{equation*}

then $\sin \angle ({\mathbf{U}}, \widetilde{{\mathbf{U}}}) \leq c$ with probability $1 - n^{-2}$. Here $\angle ({\mathbf{U}}, \widetilde{{\mathbf{U}}})$ is the angle between ${\mathbf{U}}$ and $\widetilde{{\mathbf{U}}}$.

Proof of Lemma 5.6. From (22) and Corollary 5.5, with probability at least $1 - n^{-2}$,

\begin{align*} \|({\mathbf{A}}_1)_{\mathcal{I}_1} - \widetilde{{\mathbf{A}}}_1\| \leq \|({\mathbf{E}}_1)_{\mathcal{I}_1}\| + \|{\boldsymbol \Delta }_1 \|\leq (C_3 + 1)\sqrt{d}. \end{align*}

Since $\sigma _{k+1}(\widetilde{{\mathbf{A}}}_1)=0$, using Lemma 5.3 to approximate $\sigma _k(\widetilde{{\mathbf{A}}}_1)$, we obtain

\begin{align*} &\,\sigma _k(\widetilde{{\mathbf{A}}}_1) - \sigma _{k+1}(\widetilde{{\mathbf{A}}}_1) =\sigma _k(\widetilde{{\mathbf{A}}}_1)=(1+o(1))\sigma _k(\overline{{\mathbf{A}}}_1) \geq \frac{1}{2}\sigma _k(\overline{{\mathbf{A}}}_1)\\[6pt] \geq &\, \frac{n}{4\sqrt{2}k }\sum _{m \in{\mathcal M}}\left({ \frac{3n}{4k} -2 \atop m-2}\right) \frac{a_m - b_m}{\left({n \atop m-1}\right)} \geq \frac{1}{8k}\sum _{m\in{\mathcal M}} \bigg (\frac{3}{4k} \bigg )^{m-2}(m-1)(a_m-b_m)\\[6pt] \geq &\, \frac{1}{8k}\left (\frac{1}{2k}\right )^{{\mathcal M}_{\max } - 2}\sum _{m\in{\mathcal M}} (m-1)(a_m-b_m)\geq \frac{C_2 \sqrt{d}}{2^{{\mathcal M}_{\max } + 1}}\,. \end{align*}

Then for any $c\in (0, 1)$, we can find $C_2 = [2^{{\mathcal M}_{\max } + 2}(C_3 + 1)/c]$ such that $\|({\mathbf{A}}_1)_{\mathcal{I}_1} - \widetilde{{\mathbf{A}}}_1\|\leq (1-1/\sqrt{2}) \sigma _k(\widetilde{{\mathbf{A}}}_1)$. By Wedin’s Theorem (Lemma D.7), the angle $\angle ({\mathbf{U}}, \widetilde{{\mathbf{U}}})$ is bounded by

\begin{align*} \sin \angle ({\mathbf{U}}, \widetilde{{\mathbf{U}}}) \,:\!=\, \|P_{{\mathbf{U}}} - P_{\widetilde{{\mathbf{U}}}}\| \leq &\, \frac{\sqrt{2}\|({\mathbf{A}}_1)_{\mathcal{I}_1} - \widetilde{{\mathbf{A}}}_1\|}{ \sigma _k(\widetilde{{\mathbf{A}}}_1)} \leq \frac{\sqrt{2}(C_3+1)\sqrt{d}}{C_2\sqrt{d}/ 2^{{\mathcal M}_{\max } + 1}} = \frac{\sqrt{2}}{2}c\lt c\,. \end{align*}

5.1.3. Bound the projection error

Randomly pick $s=2k\log ^2 n$ vertices from $Y_2$. Let $\boldsymbol{a}_{i_1},\dots,\boldsymbol{a}_{i_s}$, $\widetilde{\boldsymbol{a}}_{i_1}$,$\dots$, $\widetilde{\boldsymbol{a}}_{i_s}$, $\overline{\boldsymbol{a}}_{i_1},\dots, \overline{\boldsymbol{a}}_{i_s}$ and $\boldsymbol{e}_{i_1}, \dots,\boldsymbol{e}_{i_s}$ be the corresponding columns of ${\mathbf{A}}_2$, $\widetilde{{\mathbf{A}}}_2$, $\overline{{\mathbf{A}}}_2$ and ${\mathbf{E}}_2\,:\!=\,{\mathbf{A}}_2 - \widetilde{{\mathbf{A}}}_2$ respectively, where ${\mathbf{A}}_2$, $\widetilde{{\mathbf{A}}}_2$ and $\overline{{\mathbf{A}}}_2$ were defined in (18), (19) and (20). Let $k(i)$ denote the membership of vertex $i$. Note that entries of vector $\widetilde{\boldsymbol{a}}_{i}$ are $\widetilde{\alpha }_{ii}$, $\widetilde{\beta }_{ij}$ or $0$, according to the membership of vertices in $Z$, where $\widetilde{\alpha }_{ii}$, $\widetilde{\beta }_{ij}$ were defined in (15a), (15b). Then the corresponding vector $\widetilde{{\boldsymbol \delta }}_i \in{\mathbb{R}}^{n}$ with the entries given by

(23) \begin{equation} \widetilde{\boldsymbol{a}}_{i}(j) = \begin{cases} \widetilde{\alpha }_{ii}, & \mathrm{if } j\in Z\cap V_{k(i)}\\[6pt] \widetilde{\beta }_{ij}, & \mathrm{if } j\in Z\setminus V_{k(i)} \\[6pt] 0\,, & \mathrm{if } j\in Y \end{cases}\,,\quad \widetilde{{\boldsymbol \delta }}_i(j) = \begin{cases} (\widetilde{\alpha }_{ii} - \widetilde{\beta }_{ij})/2 \gt 0, & \mathrm{if } j\in Z\cap V_{k(i)} \\[6pt] (\widetilde{\beta }_{ij} - \widetilde{\alpha }_{ii})/2 \lt 0, & \mathrm{if } j\in Z\setminus V_{k(i)} \\[6pt] 0, & \mathrm{if } j\in Y \end{cases}, \end{equation}

can be used to recover the vertex set $Z\cap V_{k(i)}$ based on the sign of elements in $\widetilde{{\boldsymbol \delta }}_i$. However, it is hard to handle with $\widetilde{{\boldsymbol \delta }}_i$ due to the randomness of $\widetilde{\alpha }_{ii}$, $\widetilde{\beta }_{ij}$ originated from $n_i$ and $n^{\prime }_i$. Note that $n_i$ and $n^{\prime }_i$ concentrate around $n/(2k)$ and $n/(4k)$ respectively as shown in Lemma 5.3. Thus a good approximation of $\widetilde{{\boldsymbol \delta }}_i$, which rules out randomness of $n_i$ and $n^{\prime }_i$, was given by $\overline{{\boldsymbol \delta }}_i \,:\!=\, \overline{\boldsymbol{a}}_{i} - \overline{\boldsymbol{a}}$, with entries given by $\overline{\boldsymbol{a}}(j)\,:\!=\, \unicode{x1D7D9}_{j\in Z}\cdot (\overline{\alpha } + \overline{\beta })/2$, where $\overline{\alpha }$ and $\overline{\beta }$ were defined in (17), and

(24) \begin{equation} \overline{\boldsymbol{a}}_i (j) = \begin{cases} \overline{\alpha }, \quad &\mathrm{if } j\in Z\cap V_{k(i)} \\[6pt] \overline{\beta }, \quad &\mathrm{if } j\in Z\setminus V_{k(i)} \\[6pt] 0, \quad &\mathrm{if } j\in Y \end{cases} \,,\quad \overline{{\boldsymbol \delta }}_i(j) = \begin{cases} (\overline{\alpha } - \overline{\beta })/2 \gt 0, \quad & \mathrm{if } j\in Z\cap V_{k(i)} \\[6pt] (\overline{\beta } - \overline{\alpha })/2 \lt 0, \quad &\mathrm{if } j\in Z\setminus V_{k(i)} \\[6pt] 0, \quad &\mathrm{if } j\in Y \end{cases} \,. \end{equation}

By construction, $\overline{{\boldsymbol \delta }}_i$ identifies vertex set $Z \cap V_{k(i)}$ in the case of perfect splitting for any $i\in \{i_1, \cdots, i_s\}\cap Y_2\cap V_{k(i)}$. However, the access to $\overline{{\boldsymbol \delta }}_i$ is limited in practice, thus the projection $P_{{\mathbf{U}}}(\boldsymbol{a}_i - \overline{\boldsymbol{a}})$ is used instead as an approximation of $\overline{{\boldsymbol \delta }}_i$. Lemma 5.7 proves that at least half of the projected vectors have small projection errors.

Lemma 5.7. For any $c\in (0, 1)$, there exist constants $C_1$ and $C_2$ such that if $d\gt C_1$ and

\begin{equation*}\sum _{m \in {\mathcal M}}(m-1)(a_m -b_m) \gt C_2 k^{{\mathcal M}_{\max }}\sqrt {d},\end{equation*}

then among all projected vectors $P_{{\mathbf{U}}}(\boldsymbol{a}_{i} - \overline{\boldsymbol{a}})$ for $i\in \{i_1, \cdots, i_s\}\cap Y_2$, with probability $1 - O(n^{-k})$, at least half of them satisfy

(25) \begin{align} \|P_{{\mathbf{U}}}(\boldsymbol{a}_{i} - \overline{\boldsymbol{a}}) - \overline{{\boldsymbol \delta }}_{i}\|_2 \lt c \, \|\overline{{\boldsymbol \delta }}_{i}\|_2. \end{align}

Proof Lemma 5.7. Note that $\overline{{\boldsymbol \delta }}_i = P_{\overline{{\mathbf{U}}}}\overline{{\boldsymbol \delta }}_i$, where $\overline{{\mathbf{U}}}$ is spanned by the first $k$ left singular vectors of $\overline{{\mathbf{A}}}_1$ with $\mathrm{rank}(\overline{{\mathbf{A}}}_1) = k$, and $\overline{{\mathbf{A}}}_1$, $\overline{{\mathbf{A}}}_2$ preserve the same eigen-subspace. The approximation error between $ P_{{\mathbf{U}}}(\boldsymbol{a}_i - \overline{\boldsymbol{a}})$ and $\overline{{\boldsymbol \delta }}_i$ can be decomposed as

\begin{align*} P_{{\mathbf{U}}}(\boldsymbol{a}_i - \overline{\boldsymbol{a}}) - \overline{{\boldsymbol \delta }}_i =\,& P_{{\mathbf{U}}}\big [ (\boldsymbol{a}_i - \widetilde{\boldsymbol{a}}_i) + (\widetilde{\boldsymbol{a}}_i - \overline{\boldsymbol{a}}_{i}) + (\overline{\boldsymbol{a}}_i - \overline{\boldsymbol{a}}) \big ] - P_{\overline{{\mathbf{U}}}}\overline{{\boldsymbol \delta }}_i\\ =\,& P_{{\mathbf{U}}}\boldsymbol{e}_{i} + P_{{\mathbf{U}}}(\widetilde{\boldsymbol{a}}_i - \overline{\boldsymbol{a}}_{i}) + (P_{{\mathbf{U}}} - P_{\overline{{\mathbf{U}}}})\overline{{\boldsymbol \delta }}_i\,. \end{align*}

Then by triangle inequality,

\begin{align*} \| P_{{\mathbf{U}}}(\boldsymbol{a}_i - \overline{\boldsymbol{a}}) - \overline{{\boldsymbol \delta }}_i\|_2 \leq \|P_{{\mathbf{U}}}\boldsymbol{e}_{i}\|_2 + \|P_{{\mathbf{U}}}(\widetilde{\boldsymbol{a}}_i - \overline{\boldsymbol{a}}_{i})\|_2 + \|P_{{\mathbf{U}}} - P_{\overline{{\mathbf{U}}}}\|\cdot \|\overline{{\boldsymbol \delta }}_i\|_{2}\,. \end{align*}

Note that $\|\overline{{\boldsymbol \delta }}_i\| = O(n^{-\frac{1}{2} })$ and $n^{\prime }_{i}$ concentrates around $n/(4k)$ for each $i\in [k]$ with deviation at most $\sqrt{n}\log (n)$, then by definitions of $\overline{\alpha }$ and $\overline{\beta }$ in (17),

\begin{align*} \|P_{{\mathbf{U}}}(\widetilde{\boldsymbol{a}}_i - \overline{\boldsymbol{a}}_{i})\|_2 \leq \|\widetilde{\boldsymbol{a}}_i - \overline{\boldsymbol{a}}_{i}\|_2 = O\Big (\big [ k\sqrt{n}\log (n)(\overline{\alpha } -\overline{\beta })^2 \big ]^{\frac{1}{2}} \Big ) = O[ n^{-\frac{3}{4}}\log ^{\frac{1}{2}}(n)]=o(\|\overline{{\boldsymbol \delta }}_i\|_2). \end{align*}

Meanwhile, by an argument similar to Lemma 5.6, it can be proved that $\sin \angle ({\mathbf{U}}, \overline{{\mathbf{U}}}) \lt c/2$ for any $c\in (0, 1)$, if constants $C_1, C_2$ are chosen properly, hence $\|P_{{\mathbf{U}}} - P_{\overline{{\mathbf{U}}}}\|\cdot \|\overline{{\boldsymbol \delta }}_i\|_{2}\lt \frac{c}{2} \|\overline{{\boldsymbol \delta }}_i\|_{2}$. Lemma 5.8 shows that at least half of the indices from $\{i_1, \cdots, i_s\}\cap Y_2$ satisfy $\|P_{{\mathbf{U}}}\boldsymbol{e}_i \|_2 \lt \frac{c}{2}\|\overline{{\boldsymbol \delta }}_{i}\|_2$, which completes the proof.

Lemma 5.8. Let $d = \sum _{m\in{\mathcal M}_{\max }} (m-1)a_m$. For any $c\in (0, 1)$, with probability $1 - O(n^{-k\log n})$, at least $\frac{s}{2}$ of the vectors $\boldsymbol{e}_{i_1}, \dots,\boldsymbol{e}_{i_s}$ satisfy

\begin{align*} \|P_{{\mathbf{U}}}\boldsymbol{e}_i \|_2 \leq 2\sqrt{kd({\mathcal M}_{\max } + 2)/n} \lt \frac{c}{2}\|\overline{{\boldsymbol \delta }}_{i}\|_2 \,, \quad i\in \{ i_1, \cdots, i_s \} \subset Y_2\,. \end{align*}

To avoid introducing extra notations, let $\boldsymbol{a}_{i_1}, \dots,\boldsymbol{a}_{i_{s_1}}$ denote good vectors with $i_1, \cdots, i_{s_1}$ denoting good indices. Lemma 5.8 indicates that the number of good vectors is at least $s_1\geq \frac{s}{2} = k\log ^2 n$.

5.1.4. Accuracy

We are going to prove the accuracy of the initial partition obtained from Algorithm 4. Lemmas 5.10, 5.11 and 5.13 are crucial in proving our results. We present the proof logic first and defer the Lemma statements later.

For each projected vector $P_{{\mathbf{U}}}(\boldsymbol{a}_{i} - \overline{\boldsymbol{a}})$, let $U^{\prime }_{i}$ denote the set of its largest $\frac{n}{2k}$ coordinates, where $i\in \{i_1, \cdots, i_s\}$ and $s = 2k \log ^2 n$. Note that vector $\overline{{\boldsymbol \delta }}_{i_j}$ in (23) only identifies blocks $V_{k(i_j)}$ and $V\setminus V_{k(i_j)}$, which can be regarded as clustering two blocks with different sizes. By Lemma 5.7, good vectors satisfy $\|P_{{\mathbf{U}}}(\boldsymbol{a}_{i_j} - \overline{\boldsymbol{a}}) - \overline{{\boldsymbol \delta }}_{i_j}\|_2 \lt c \, \|\overline{{\boldsymbol \delta }}_{i_j}\|_2$ for any $c\in (0, 1)$. Then by Lemma 5.10 (after proper normalisation), for a good index $i_j$, the number of vertices in $U^{\prime }_{i_j}$ clustered correctly is at least $(1 - \frac{4}{3}kc^{2})\frac{n}{k}$. By choosing $c = \sqrt{3(1-\nu )/(8k)}$, the condition $|U^{\prime }_{i_j} \cap V_i| \gt (1+\nu )/2 |U^{\prime }_{i_j}|$ in part (ii) of Lemma 5.11 is satisfied. In Lemma 5.6, we choose

(26) \begin{align} C_2 = 2^{{\mathcal M}_{\max } + 2}(C_3 + 1)/c = 2^{{\mathcal M}_{\max } + 2}(C_3 + 1)\sqrt{8k/(3-3\nu )}\,\,, \end{align}

where $C_3$ defined in (22). Hence, with high probability, all good vectors have at least $\mu _{\mathrm{T}}$ blue hyperedges (we call this “high blue hyperedge density”). From Lemma 5.8, at least half of the selected vectors are good. Then, in Algorithm 4, throwing out half of the obtained sets $U^{\prime }_{i}$ (those with the lowest blue hyperedge density) guarantees that the remaining sets are good.

Recall that, by choosing constant appropriately, we can make the subspace angle $\sin \angle ({\mathbf{U}}, \overline{{\mathbf{U}}}) \lt c$ for any $c\in (0, 1)$ ($\overline{{\mathbf{U}}}$ is spanned by the first $k$ left singular vectors of $\overline{{\mathbf{A}}}_1$). Then for each vector $\overline{{\boldsymbol \delta }}_{i_1}, \cdots, \overline{{\boldsymbol \delta }}_{i_k}$ with each $i_j$ selected from different vertex set $V_j$, there is a vector $P_{{\mathbf{U}}}(\boldsymbol{a}_{i_j} - \overline{\boldsymbol{a}})$ in ${\mathbf{U}}$ arbitrarily close to $\overline{{\boldsymbol \delta }}_{i_j}$, which was proved by Lemma 5.7. From (i) of Lemma 5.11, so obtained $U_{i_j}^{\prime }$ must satisfy $|U_{i_j}^{\prime }\cap V_j|\geq \nu |U_{i_j}^{\prime }|$ for each $j\in [k]$. The remaining thing is to select $k$ different $U_{i_j}^{\prime }$ with each of them concentrating around distinct $V_j$ for each $j\in [k]$. This problem is equivalent to finding $k$ vertices in $Y_2$, each from a different partition class, which can be done with $k\log ^2(n)$ samplings as shown in Lemma 5.13.

To summarise, this section is a more precise and quantitative version of the following argument: with high probability,

\begin{align*} \left \{ i_j \,:\, \exists i\,\mbox{s.t.}\,|U^{\prime}_{i_j} \cap V_i| \geq \frac{1+\nu }{2} \frac{n}{k}\right \} & \subset \left \{ i_j \,:\, U^{\prime}_{i_j} \text{ has $\geq \mu _T$ blue hyperedges}\right \}\\ & \subset \left \{ i_j\,:\, \exists i\,\mbox{s.t.}\,|U^{\prime}_{i_j} \cap V_i| \geq \nu \frac{n}{k}\right \} \,. \end{align*}

Lemma 5.11. Suppose that we are given a set $X \subset Z$ with size $|X| = n/(2k)$. Define

\begin{align*} \mu _1\,:\!=\,&\,\frac{1}{2} \sum _{m\in{\mathcal M}}m(m-1) \left \{ \left [ \binom{\frac{\nu n}{2k}}{m} + \binom{\frac{(1 - \nu )n}{2k}}{m} \right ] \frac{a_m - b_m }{ \binom{n}{m-1} } + \binom{\frac{n}{2k}}{m} \frac{b_{m}}{ \binom{n}{m-1} } \right \}\,,\\ \mu _{2} \,:\!=\,&\, \frac{1}{2} \sum _{m\in{\mathcal M}} m(m-1)\left \{ \left [ \binom{\frac{(1 + \nu ) n}{4k} }{m} + (k-1)\binom{ \frac{(1 - \nu )n}{4k(k-1)}}{m} \right ] \frac{a_m - b_m }{ \binom{n}{m-1} } + \binom{\frac{n}{2k}}{m} \frac{b_{m}}{ \binom{n}{m-1} } \right \}\,, \end{align*}

and $\mu _{\mathrm{T}} \,:\!=\, (\mu _1 + \mu _2)/2 \in [\mu _1, \mu _2]$. There is a constant $c\gt 0$ depending on $k, a_m, \nu$ such that for sufficiently large $n$,

1. (i) If $|X\cap V_i| \leq \nu |X|$ for each $i\in [k]$, then with probability $1 - e^{-cn}$, the number of blue hyperedges in the hypergraph induced by $X$ is at most $\mu _{\mathrm{T}}$.

2. (ii) Conversely, if $|X\cap V_i| \geq \frac{1+ \nu }{2}|X|$ for some $i\in \{1, \dots, k\}$, then with probability $1 - e^{-cn}$, the number of blue hyperedges in the hypergraph induced by $X$ is at least $\mu _{\mathrm{T}}$.

Lemma 5.13. Through random sampling without replacement in Step $6$ of Algorithm 4, we can find at least $k$ indices $i_1,\dots, i_k$ in $Y_2$ among $k\log ^2 n$ samples such that with probability $1 -n^{-\Omega (\log n)}$,

\begin{equation*} |U^{\prime}_{i_j} \cap U^{\prime}_{i_l}| \leq (1-\nu )\frac {n}{k}, \,\text { for any $j,l\in [k]$} \text { with } j\not =l. \end{equation*}