SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Proof. Define the following potential function:
$$
\Phi_ {t} (v) = \mathbf {1} (v \notin U _ {t}) \cdot X _ {t} (v) - K _ {t} (v) \cdot \left(| N (v) | - X _ {t} (v)\right),
$$
where  $\mathbf{1}(v\notin U_t)$  is the indicator of event  $\{v\notin U_t\}$ . Let  $\mathcal{F}_t$  be the state of the algorithm at the beginning of step  $t$ . We claim that  $\Phi_t(u)$  is a submartingale i.e.,  $\mathbf{E}[\Phi_{t + 1}(u)\mid \mathcal{F}_t]\geq \Phi_t(u)$ .
Claim 2.4.  $\Phi_t(v)$  is a submartingale.
Proof. Consider step  $t$  of the algorithm. If  $v$  is already clustered (i.e.,  $v \notin U_t$ ), then all edges incident on  $v$  are settled and terms  $\mathbf{1}(v \notin U_t)$ ,  $K_t(v)$ , and  $X_t(v)$  no longer change. Hence, the potential function  $\Phi_t(v)$  does not change as well.
Suppose that  $v \in U_t$ . If at step  $t$ , the algorithm picks  $w_t$  not from the neighbourhood  $N(v)$ , then  $v$  does not get clustered at this step and also  $K_t(v)$  does not change. The value of  $X_t(v)$  may increase as some neighbours of  $v$  may get clustered. However,  $X_t(v)$  cannot decrease. Therefore,  $\Phi_{t+1}(v) \geq \Phi_t(v)$ , and  $\mathbf{E}\left[\Phi_{t+1}(v) - \Phi_t(v) \mid w_t \notin N(v); \mathcal{F}_t\right] \geq 0$ .
Let us now assume that at step  $t$ , the algorithm chooses  $w_{t}$  from the neighbourhood of  $v$ . Then,  $v$  is clustered with conditional probability at least
$$
\operatorname * {P r} \left(w _ {t} \in N (v) \cap U _ {t} \mid w _ {t} \in N (v); \mathcal {F} _ {t}\right) = \frac {| N (v) \cap U _ {t} |}{| N (v) \cap V _ {t} |},
$$
because if the pivot is chosen in  $N(v) \cap U_t$ , then  $v$  is put in the cluster of  $w_t$ . We have  $|N(v) \cap U_t| = |N(v)| - X_t(v)$ , since every clustered neighbour of  $v$  contributes 1 to the number of cut edges  $X_t(v)$ , and the total number of  $v$ 's positive neighbours equals  $|N(v)|$ . Also, note that  $|N(v) \cap V_t| \leq |N(v)|$ . Hence,
$$
\Pr \left(w _ {t} \in N (v) \cap U _ {t} \mid w _ {t} \in N (v); \mathcal {F} _ {t}\right) \geq \frac {| N (v) | - X _ {t} (v)}{| N (v) |}.
$$
With this conditional probability,  $\mathbf{1}(v \notin U_t)$  gets equal to 1 at step  $t$ , and, as a result,  $\Phi_t(v)$  increases by  $X_{t+1}(v) \geq X_t(v)$ . With the remaining probability,  $K_t(v)$  increases by 1, and, as a result,  $\Phi_t(v)$  decreases by at most  $|N(v)| - X_{t+1}(v) \leq |N(v)| - X_t(v)$ . Thus,
$$
\mathbf {E} \big [ \Phi_ {t + 1} (v) - \Phi_ {t} (v) \mid w _ {t} \in N (v); \mathcal {F} _ {t} \big ] \geq \frac {| N (v) | - X _ {t} (v)}{| N (v) |} \cdot X _ {t} (v) - \frac {X _ {t} (v)}{| N (v) |} \cdot (| N (v) | - X _ {t} (v)) = 0.
$$
Since  $\Phi_1(v) = 0$  and  $\Phi_t(v)$  is a submartingale, we have  $\mathbf{E}[\Phi_{n + 1}(v)]\geq 0$ . Therefore,
$$
\mathbf {E} [ X _ {n + 1} (v) ] = \mathbf {E} [ \mathbf {1} (v \notin U _ {n + 1}) \cdot X _ {n + 1} (v) ] \geq \mathbf {E} [ K _ {n + 1} (v) \cdot (| N (v) | - X _ {n + 1} (v)) ].
$$
If  $v$  is in a singleton cluster, then  $K_{n+1}(v) = k$  and  $X_{n+1}(v) + S(v) = |N(v)|$  (because all positive edges incident to  $v$  are cut). Hence,
$$
k S (v) = K _ {n + 1} (v) \cdot \left(| N (v) | - X _ {n + 1} (v)\right).
$$
If  $v$  is not in a singleton cluster, then  $S(v) = 0$ . Thus, we always have  $kS(v) \leq K_{n+1}(v) \cdot (|N(v)| - X_{n+1}(v))$ . Consequently,  $\mathbf{E}[X_{n+1}(v)] \geq k\mathbf{E}[S(v)]$ .
Konstantin Makarychev is supported by NSF Awards CCF-1955351, CCF-1934931, EECS-29 2216970. The authors would like to thank the anonymous reviewers for their valuable comments.
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A Figures
Figure 2: The cost of the clustering obtained by our algorithm comparing with the cost of the clustering obtained by the PIVOT algorithm for different values of  $k$ . The input graph is the  $K$ -nearest neighbour graph for the Gaussian mixture model.