SlowGuess/NeurIPS_2023_OCR huggingface dataset:
[1] Ravindra K Ahuja and James B Orlin. A fast and simple algorithm for the maximum flow problem. Operations Research, 37(5):748-759, 1989.
[2] Haris Angelidakis, Yury Makarychev, and Pasin Manurangsi. An improved integrality gap for the calinescu-karloff-rabani relaxation for multiway cut. In International Conference on Integer Programming and Combinatorial Optimization, pages 39-50. Springer, 2017.
[3] Kristof Berczi, Karthekeyan Chandrasekaran, Tamás Király, and Vivek Madan. Improving the integrality gap for multiway cut. Mathematical Programming, 183(1):171-193, 2020.
[4] Avrim Blum and Shuchi Chawla. Learning from labeled and unlabeled data using graph mincuts. In Proceedings of the Eighteenth International Conference on Machine Learning, pages 19-26, 2001.
[5] Avrim Blum, John Lafferty, Mugizi Robert Rwebangira, and Rajashekar Reddy. Semi-supervised learning using randomized minutcuts. In Proceedings of the twenty-first international conference on Machine learning, page 13, 2004.
[6] Gruia Calinescu, Howard Karloff, and Yuval Rabani. An improved approximation algorithm for multiway cut. In Proceedings of the thirtieth annual ACM symposium on Theory of computing, pages 48-52, 1998.
[7] Li Chen, Rasmus Kyng, Yang P Liu, Richard Peng, Maximilian Probst Gutenberg, and Sushant Sachdeva. Maximum flow and minimum-cost flow in almost-linear time. In 2022 IEEE 63rd Annual Symposium on Foundations of Computer Science (FOCS), pages 612-623. IEEE, 2022.
[8] Vincent Cohen-Addad, Chenglin Fan, Silvio Lattanzi, Slobodan Mitrovic, Ashkan Norouzi-Fard, Nikos Parotsidis, and Jakub Tarnawski. Near-optimal correlation clustering with privacy. arXiv preprint arXiv:2203.01440, 2022.
[9] Elias Dahlhaus, David S Johnson, Christos H Papadimitriou, Paul D Seymour, and Mihalis Yannakakis. The complexity of multiway cuts. In Proceedings of the twenty-fourth annual ACM symposium on Theory of computing, pages 241-251, 1992.
[10] Elias Dahlhaus, David S. Johnson, Christos H. Papadimitriou, Paul D. Seymour, and Mihalis Yannakakis. The complexity of multiterminal cuts. SIAM Journal on Computing, 23(4):864-894, 1994.
[11] E.A. Dinits, A.V. Karzanov, and M.V. Lomonosov. On the Structure of a Family of Minimal Weighted Cuts in a Graph. 1991. URL https://books.google.de/books?id=00pOkgAACAAJ.
[12] Cynthia Dwork. Differential privacy. In Automata, Languages and Programming: 33rd International Colloquium, ICALP 2006, Venice, Italy, July 10-14, 2006, Proceedings, Part II 33, pages 1-12. Springer, 2006.
[13] Cynthia Dwork and Jing Lei. Differential privacy and robust statistics. In Proceedings of the forty-first annual ACM symposium on Theory of computing, pages 371-380, 2009.
[14] Cynthia Dwork, Frank McSherry, Kobbi Nissim, and Adam Smith. Calibrating noise to sensitivity in private data analysis. In Theory of Cryptography: Third Theory of Cryptography Conference, TCC 2006, New York, NY, USA, March 4-7, 2006. Proceedings 3, pages 265-284. Springer, 2006.
[15] Cynthia Dwork, Guy N Rothblum, and Salil Vadhan. Boosting and differential privacy. In 2010 IEEE 51st Annual Symposium on Foundations of Computer Science, pages 51-60. IEEE, 2010.
[33] Yapu Zhang, Jianxiong Guo, Wenguo Yang, and Weili Wu. Mixed-case community detection problem in social networks: Algorithms and analysis. Theoretical Computer Science, 854: 94-104, 2021.
[34] Liang Zhao, Hiroshi Nagamochi, and Toshihide Ibaraki. Greedy splitting algorithms for approximating multiway partition problems. Mathematical Programming, 102:167-183, 2005.
A Multiple min  $s - t$  cuts
Let  $G$  and  $G'$  be two neighboring graphs, and let  $\widetilde{G}$  and  $\widetilde{G'}$ , respectively, be their modified versions constructed by Algorithm 1. Algorithm 1 outputs a min  $s-t$  cut in  $G$ . However, what happens if there are multiple min  $s-t$  cuts in  $\widetilde{G}$  and the algorithm invoked on Line 4 breaks ties in a way that depends on whether a specific edge  $e$  appears in  $G$  or not? If it happens that  $e$  is the edge difference between  $G$  and  $G'$ , then such a tie-breaking rule might reveal additional information about  $G$  and  $G'$ . We now outline how this can be bypassed.
Observe that if the random variables  $X_{s,u}$  and  $X_{t,u}$  were sampled by using infinite bit precision, then with probability 1 no two cuts would have the same value. So, consider a more realistic situation where edge-weights are represented by  $O(\log n)$  bits, and assume that the least significant bit corresponds to the value  $2^{-t}$ , for an integer  $t \geq 0$ . We show how to modify edge-weights by additional  $O(\log n)$  bits that have extremely small values but will help obtain a unique min  $s-t$  cut. Our modification consists of two steps.
First step. All the bits corresponding to values from  $2^{-t - 1}$  to  $2^{-t - 2\log n}$  remain 0, while those corresponding to larger values remain unchanged. This is done so that even summing across all – but at most  $\binom{n}{2}$  – edges it holds that no matter what the bits corresponding to values  $2^{-t - 2\log n - 1}$  and less are, their total value is less than  $2^{-t}$ . Hence, if the weight of cut  $C_1$  is smaller than the weight of cut  $C_2$  before the modifications we undertake, then  $C_1$  has a smaller weight than  $C_2$  after the modifications as well.
Second step. We first recall the celebrated Isolation lemma.
Lemma A.1 (Isolation lemma, [26]). Let  $T$  and  $N$  be positive integers, and let  $\mathcal{F}$  be an arbitrary nonempty family of subsets of the universe  $\{1, \ldots, T\}$ . Suppose each element  $x \in \{1, \ldots, T\}$  in the universe receives an integer weight  $g(x)$ , each of which is chosen independently and uniformly at random from  $\{1, \ldots, N\}$ . The weight of a set  $S \in \mathcal{F}$  is defined as  $g(S) = \sum_{x \in S} g(x)$ .
Then, with probability at least  $1 - T / N$  there is a unique set in  $\mathcal{F}$  that has the minimum weight among all sets of  $\mathcal{F}$ .
We now apply Lemma A.1 to conclude our modification of the edge weights in  $\widetilde{G}$ . We let the universe  $\{1, \ldots, T\}$  from that lemma be the following  $2(n - 2)$  elements  $\mathcal{U} = \{(s, v) \mid v \in V(G) \setminus \{s, t\} \} \cup \{(t, v) \mid v \in V(G) \setminus \{s, t\}\}$ . Then, we let  $\mathcal{F}$  represent all min  $s$ - $t$  cuts in  $\widetilde{G}$ , i.e.,  $S \subseteq \mathcal{U}$  belongs to  $\mathcal{F}$  iff there is a min  $s$ - $t$  cut  $C$  in  $\widetilde{G}$  such that for each  $(a, b) \in S$  the cut  $C$  contains  $X_{a, b}$ . So, by letting  $N = 2n^2$ , we derive that with probability  $1 - 1/n$  it holds that no two cuts represented by  $\mathcal{F}$  have the same minimum value with respect to  $g$  defined in Lemma A.1. To implement  $g$  in our modification of weights, we modify the bits of each  $X_{s, v}$  and  $X_{t, v}$  corresponding to values from  $2^{-t - 2\log n - 1}$  to  $2^{-t - 2\log n - \log N}$  to be an integer between 1 and  $N$  chosen uniformly at random.
Only after these modifications, we invoke Line 4 of Algorithm 1. Note that the family of cuts  $\mathcal{F}$  is defined only for the sake of analysis. It is not needed to know it algorithmically.
B Lower Bound for min  $s - t$  cut error
In this section, we prove our lower bound. Our high-level idea is similar to that of [8] for proving a lower bound for private algorithms for correlation clustering.
Proof. For the sake of contradiction, let  $\mathcal{A}$  be a  $(\epsilon, \delta)$ -differential private algorithm for min  $s-t$  cut that on any input  $n$ -node graph outputs an  $s-t$  cut that has expected additive error of less than  $n/20$ . We construct a set of  $2^n$  graphs  $S$  and show that  $\mathcal{A}$  cannot have low expected cost on all of the graphs on this set while preserving privacy.
The node set of all the graphs in  $S$  are the same and consist of  $V = \{s, t, v_1, \ldots, v_n\}$  where  $s$  and  $t$  are the terminals of the graph and  $n > 30$ . For any  $\tau \in \{0, 1\}^n$ , let  $G_{\tau}$  be the graph on node set  $V$ .
with the following edges: For any  $1 \leq i \leq n$ , if  $\tau_i = 1$ , then there is an edge between  $s$  and  $v_i$ . If  $\tau = 0$ , then there is an edge between  $t$  and  $v_i$ . Note that  $v_i$  is attached to exactly one of the terminals  $s$  and  $t$ . Moreover, the min  $s-t$  cut of each graph  $G_{\tau}$  is zero.
Algorithm  $\mathcal{A}$  determines for each  $i$  if  $v_{i}$  is on the  $s$ -side of the output cut or the  $t$ -side. The contribution of each node  $v_{i}$  to the total error is the number of edges attached to  $v_{i}$  that are in the cut. We denote this random variable in graph  $G_{\tau}$  by  $e_{\tau}(v_i)$ . Since there are no edges between any two non-terminal nodes in any of the graphs  $G_{\tau}$ , the total error of the output is the sum of these individual errors, i.e.,  $\sum_{i = 0}^{n}e_{\tau}(v_i)$ . Let  $\bar{e}_{\tau}(v_i)$  be the expected value of  $e_{\tau}(v_i)$  over the outputs of  $\mathcal{A}$  given  $G_{\tau}$ .
Let  $p_{\tau}^{(i)}$  be the marginal probability that  $v_{i}$  is on the  $s$ -side of the output  $s-t$  cut in  $G_{\tau}$ . If  $\tau_{i} = 0$ , then  $v_{i}$  is connected to  $t$  and so  $\bar{e}_{\tau}(v_{i}) = p_{\tau}^{(i)}$ . If  $\tau_{i} = 1$ , then  $v_{i}$  is connected to  $s$  and so  $\bar{e}_{\tau}(v_{i}) = 1 - p_{\tau}^{(i)}$ . By the assumption that  $\mathcal{A}$  has a low expected error on every input, we have that for any  $\tau \in \{0,1\}^n$ ,
$$
(n + 2) / 2 0 > \sum_ {i, \tau_ {i} = 0} p _ {\tau} ^ {(i)} + \sum_ {i, \tau_ {i} = 1} (1 - p _ {\tau} ^ {(i)}) \tag {3}
$$
Let  $S_{i}$  be the set of  $\tau \in \{0,1\}^n$  such that  $\tau_{i} = 0$ , and  $\bar{S}_{i}$  be the complement of  $S_{i}$ , so that  $\tau \in \bar{S}_{i}$  if  $\tau_{i} = 1$ . Note that  $|S_{i}| = |\bar{S}_{i}| = 2^{n - 1}$ . Fix some  $i$ , and for any  $\tau \in \{0,1\}^n$ , let  $\tau'$  be the same as  $\tau$  except for the  $i$ -th entry being different, i.e., for all  $j \neq i$ ,  $\tau_{j} = \tau_{j}'$ , and  $\tau_{i} \neq \tau_{i}'$ . Since  $G_{\tau}$  and  $G_{\tau'}$  only differ in two edges, from  $A$  being  $(\epsilon, \delta)$ -differentially private for any  $j$  we have  $p_{\tau}^{(j)} \leq e^{2\epsilon} \cdot p_{\tau'}^{(j)} + \delta$ . So for any  $i, j$  we have
$$
\sum_ {\tau \in S _ {i}} p _ {\tau} ^ {(j)} \leq \sum_ {\tau \in \bar {S} _ {i}} \left(e ^ {2 \epsilon} p _ {\tau} ^ {(j)} + \delta\right) \tag {4}
$$
From Eq. (3) we have
$$
\begin{array}{l} 2 ^ {n} \cdot 0. 0 5 (n + 2) > \sum_ {\tau \in \{0, 1 \} ^ {n}} \sum_ {i: \tau_ {i} = 1} \left(1 - p _ {\tau} ^ {(i)}\right) \\ = \sum_ {i = 1} ^ {n} \sum_ {\tau \in S _ {i}} (1 - p _ {\tau} ^ {(i)}) \\ \geq \sum_ {i = 1} ^ {n} \sum_ {\tau \in \bar {S} _ {i}} (1 - [ e ^ {2 \epsilon} p _ {\tau} ^ {(i)} + \delta ]) \\ = n 2 ^ {n - 1} (1 - \delta) - e ^ {2 \epsilon} \sum_ {i = 1} ^ {n} \sum_ {\tau \in \bar {S} _ {i}} p _ {\tau} ^ {(i)} \\ \end{array}
$$
Where the last inequality comes from Eq. (4). Using Eq. (3) again, we have that  $\sum_{i=1}^{n} \sum_{\tau \in \bar{S}_i} p_\tau^{(i)} < 2^n \cdot 0.05(n+2)$ , so we have that  $2^n \cdot 0.05(n+2) > n2^{n-1}(1-\delta) - e^{2\epsilon}(2^n \cdot 0.05(n+2))$ . Dividing by  $2^n$  we have
$$
0. 0 5 (n + 2) \left(1 + e ^ {2 \epsilon}\right) > n (1 - \delta) / 2.
$$
Now since  $\epsilon \leq 1$ ,  $\delta \leq 0.1$ , and  $e^2 < 7.4$  we get that
$$
0. 0 5 \cdot 8. 4 (n + 2) > 0. 4 5 n
$$
Hence we have  $n < 28$  which is a contradiction to  $n > 30$ .
C.1 Proof of Claim 1
$$
\frac {f _ {\mathrm {L a p}} (t + \tau)}{f _ {\mathrm {L a p}} (t)} = \frac {\frac {\epsilon}{2} \exp (- \epsilon | t + \tau |)}{\frac {\epsilon}{2} \exp (- \epsilon | t |)} = \exp (- \epsilon | t + \tau | + \epsilon | t |) \leq \exp (\tau \epsilon). \tag {5}
$$
Also by definition, it holds  $F_{\mathrm{Lap}}(t + \tau) = \int_{-\infty}^{t + \tau}f_{\mathrm{Lap}}(x)dx$ . Using Eq. (5) we derive
$$
F _ {\mathrm {L a p}} (t + \tau) \leq \exp (\tau \epsilon) \int_ {- \infty} ^ {t + \tau} f _ {\mathrm {L a p}} (x - \tau) d x = \exp (\tau \epsilon) \int_ {- \infty} ^ {t} f _ {\mathrm {L a p}} (x) d x = \exp (\tau \epsilon) F _ {\mathrm {L a p}} (t).
$$
C.2 Proof of Lemma 3.1
To prove the lower-bound, we observe that if  $x < \alpha$  and  $y < \beta$ , then  $x < \alpha + \tau$  and  $y < \beta + \tau$  as well. Hence, it trivially holds  $P(\alpha + \tau, \beta + \tau, \gamma) \geq P(\alpha, \beta, \gamma)$  and hence
$$
\frac {P (\alpha + \tau , \beta + \tau , \gamma)}{P (\alpha , \beta , \gamma)} \geq 1.
$$
We now analyze the upper-bound. For the sake of brevity, in the rest of this proof, we use  $F$  to denote  $F_{\mathrm{Lap}}$  and  $f$  to denote  $f_{\mathrm{Lap}}$ . We consider three cases depending on parameters  $\alpha, \beta, \gamma$ .
Case  $\gamma \geq \alpha +\beta +2\tau$  . In this case we have  $\operatorname *{Pr}[x + y <   \gamma |x <   \alpha ,y <   \beta ] = 1 =$ $\operatorname *{Pr}[x + y <   \gamma |x <   \alpha +\tau ,y <   \beta +\tau ].$  So we have that
$$
\begin{array}{l} P (\alpha , \beta , \gamma) = \Pr [ x + y <   \gamma | x <   \alpha , y <   \beta ] \cdot \Pr [ x <   \alpha , y <   \beta ] \\ = \Pr [ x <   \alpha , y <   \beta ] \\ = F (\alpha) F (\beta) \\ \end{array}
$$
Similarly,  $P(\alpha + \tau, \beta + \tau, \gamma) = F(\alpha + \tau)F(\beta + \tau)$ . Now using Claim 1, we obtain that  $\frac{P(\alpha + \tau, \beta + \tau, \gamma)}{P(\alpha, \beta, \gamma)} \leq e^{2\tau \epsilon}$ .
Case  $\gamma <  \alpha +\beta$  .We write  $P(\alpha ,\beta ,\gamma)$  as follows.
$$
\begin{array}{l} P (\alpha , \beta , \gamma) = \int_ {- \infty} ^ {\beta} \int_ {- \infty} ^ {\min  (\alpha , \gamma - y)} f _ {x} (x | y) d x f _ {y} (y) d y \\ = \int_ {- \infty} ^ {\beta} F (\min  (\alpha , \gamma - y)) f (y) d y \\ = F (\alpha) \int_ {- \infty} ^ {\gamma - \alpha} f (y) d y + \int_ {\gamma - \alpha} ^ {\beta} F (\gamma - y) f (y) d y \\ = F (\alpha) F (\gamma - \alpha) + \int_ {\gamma - \alpha} ^ {\beta} F (\gamma - y) f (y) d y \tag {6} \\ \end{array}
$$
Similar to Eq. (6) we have
$$
P (\alpha + \tau , \beta + \tau , \gamma) = F (\alpha + \tau) F (\gamma - \alpha - \tau) + \int_ {\gamma - \alpha - \tau} ^ {\beta + \tau} F (\gamma - y) f (y) d y \tag {7}
$$
Now we rewrite Eq. (6) as follows to obtain a lower bound on  $P(\alpha, \beta, \gamma)$ .
$$
\begin{array}{l} P (\alpha , \beta , \gamma) = F (\alpha) F (\gamma - \alpha - 2 \tau) + \int_ {\gamma - \alpha - 2 \tau} ^ {\gamma - \alpha} F (\alpha) f (y) d y + \int_ {\gamma - \alpha} ^ {\beta} F (\gamma - y) f (y) d y \\ \geq F (\alpha) F (\gamma - \alpha - 2 \tau) + e ^ {- 2 \tau \epsilon} \int_ {\gamma - \alpha - 2 \tau} ^ {\beta} F (\gamma - y) f (y) d y \tag {8} \\ \end{array}
$$
In obtaining the inequality, we used the fact that if  $y \in [\gamma - \alpha - 2\tau, \gamma - \alpha]$  then  $0 \leq (\gamma - y) - \alpha \leq 2\tau$  and so by Claim 1 we have  $F(\alpha) \geq e^{-2\tau \epsilon} F(\gamma - y)$ .
Now we compare the two terms of Eq. (8) with Eq. (7). By Claim 1 we have that  $F(\alpha)F(\gamma - \alpha - 2\tau) \geq e^{-2\tau \epsilon} F(\alpha + \tau)F(\gamma - \alpha - \tau)$  and  $\int_{\gamma - \alpha - 2\tau}^{\beta} F(\gamma - y)f(y)dy \geq e^{-\tau \epsilon} \int_{\gamma - \alpha - \tau}^{\beta + \tau} F(\gamma - y)f(y)dy$ . So we have  $P(\alpha, \beta, \gamma) \geq e^{-3\tau \epsilon} P(\alpha + \tau, \beta + \tau, \gamma)$ .
Case  $\alpha +\beta \leq \gamma <  \alpha +\beta +2\tau$  . Then
$$
\begin{array}{l} P (\alpha , \beta , \gamma) = \int_ {- \infty} ^ {\beta} \int_ {- \infty} ^ {\min (\alpha , \gamma - y)} f _ {x} (x | y) d x f _ {y} (y) d y \\ = \int_ {- \infty} ^ {\beta} F (\min  (\alpha , \gamma - y)) f (y) d y \\ = F (\alpha) F (\beta) (9) \\ \geq e ^ {- 2 \tau \epsilon} F (\alpha + \tau) F (\beta + \tau) (10) \\ = e ^ {- 2 \tau \epsilon} (F (\alpha + \tau) F (\beta - \tau) + F (\alpha + \tau) (F (\beta + \tau) - F (\beta - \tau)) \\ \geq e ^ {- 4 \tau \epsilon} (F (\alpha + \tau) F (\beta + \tau) + F (\alpha + \tau) (F (\beta + \tau) - F (\beta - \tau)). (11) \\ \end{array}
$$
Note that Eq. (9) is obtained since for any  $y \leq \beta$  we have  $\alpha \leq \gamma - y$ . Eq. (10) and Eq. (11) are both obtained using Claim 1. One can easily verify that Eq. (7) for  $P(\alpha + 1, \beta + 1, \gamma)$  holds in this case as well. Using the fact that  $\gamma - \alpha - \tau \leq \beta + \gamma$  and  $F$  being a non-decreasing function, we further lower-bound Eq. (7) as
$$
\begin{array}{l} P (\alpha + \tau , \beta + \tau , \gamma) \leq F (\alpha + \tau) F (\beta + \tau) + \int_ {\gamma - \alpha - \tau} ^ {\beta + \tau} F (\gamma - y) f (y) d y \\ \leq F (\alpha + \tau) F (\beta + \tau) + F (\alpha + \tau) \int_ {\gamma - \alpha - \tau} ^ {\beta + \tau} f (y) d y \\ = F (\alpha + \tau) F (\beta + \tau) + F (\alpha + \tau) (F (\beta + \tau) - F (\gamma - \alpha - \tau)) \\ \leq F (\alpha + \tau) F (\beta + \tau) + F (\alpha + \tau) (F (\beta + \tau) - F (\beta - \tau)). \tag {12} \\ \end{array}
$$
Eqs. (11) and (12) conclude the analysis of this case as well.
C.3 Proof of Theorem 4.1
We use induction on  $k$  to prove the theorem. Suppose that Algorithm 2 outputs  $C_{ALG}$ . We show that the  $C_{ALG}$  is a multiway  $k$ -cut and that the value of  $C_{ALG}$  is at most  $w(E(\overline{V})) + \sum_{i=1}^{k} \delta(V_i) + 2 \log(k)e(n)$ . We will first perform the analysis of approximation assuming that  $\mathcal{A}$  provides the stated approximation deterministically, and at the end of this proof, we will take into account that the approximation guarantee holds with probability  $1 - \alpha$ .
Base case:  $k = 1$ . If  $k = 1$ , then  $C_{ALG} = \emptyset$ , and so it is a multiway 1-cut and  $w(C_{ALG}) = 0 \leq \delta(V_1) + w(E(\overline{V}))$ .
Inductive step:  $k \geq 2$ . So suppose that  $k \geq 2$ . Hence,  $k' \geq 1$  and  $k - k' \geq 1$ , where  $k'$  is defined on Line 4. Let  $(A, B)$  be the  $s-t$  cut obtained in Line 6, where  $\widetilde{G}_1$  is the graph induced on  $A$  and  $\widetilde{G}_2$  is the graph induced on  $B$ . Since the only terminals in  $\widetilde{G}_1$  are  $s_1, \ldots, s_{k'}$ , we have that  $V_1 \cap A, \ldots, V_{k'} \cap A$  is a partial multiway  $k'$ -cut on  $\widetilde{G}_1$ . By the induction hypothesis, the cost of the multiway cut that Algorithm 2 finds on  $\widetilde{G}_1$  is at most  $w(E(\overline{V} \cap A)) + \sum_{i=1}^{k'} \delta_{\widetilde{G}_1}(V_i \cap A) + 2 \log(k') e(|A|)$ . Similarly, by considering the partial multiway  $(k - k')$  cut  $V_{k' + 1} \cap B, \ldots, V_k \cap B$  on  $\widetilde{G}_2$ , the cost of the multiway cut that Algorithm 2 finds on  $\widetilde{G}_2$  is at most  $w(E(\overline{V} \cap B)) + \sum_{i=k' + 1}^{k} \delta_{\widetilde{G}_2}(V_i \cap B) + 2 \log(k - k') e(|B|)$ . So the total cost  $w(C_{ALG})$  of the multiway cut that Algorithm 2 outputs is at most
$$
\begin{array}{l} w \left(C _ {A L G}\right) \leq w (E (\bar {V} \cap A)) + w (E (\bar {V} \cap B)) \tag {13} \\ + \sum_ {i = 1} ^ {k ^ {\prime}} \delta_ {\widetilde {G} _ {1}} (V _ {i} \cap A) + \sum_ {i = k ^ {\prime} + 1} ^ {k} \delta_ {\widetilde {G} _ {2}} (V _ {i} \cap B) \\ + w (E (A, B)) \\ + 2 \log (k ^ {\prime}) e (| A |) + 2 \log (k - k ^ {\prime}) e (| B |) \\ \end{array}
$$
First note that  $C_{ALG}$  is a multiway  $k$ -cut: this is because by induction the output of the algorithm on  $\widetilde{G}_1$  is a multiway  $k'$ -cut and the output of the algorithm on  $\widetilde{G}_2$  is a multiway  $(k - k')$ -cut. Moreover,
Figure 2: Node subsets of graph  $G$ . The subsets in asterisks have terminals in them. Red edges indicate the left-hand side edges in Eq. (14), and purple edges indicate the edges on the right-hand side in Eq. (14).
$E(A,B)\in C_{ALG}$ . So the union of these cuts and  $E(A,B)$  is a  $k$ -cut, and since each terminal is in exactly one partition, it is a multiway  $k$ -cut.
Now we prove the value guarantees. Let  $U_{1} = V_{1} \cup \ldots \cup V_{k'}$  and  $U_{2} = V_{k' + 1} \cup \ldots \cup V_{k}$ . So  $U = U_{1} \cup U_{2} = V_{1} \cup \ldots \cup V_{k}$  is the set of nodes that are in at least one partition. Recall that  $\overline{V} = V \setminus U$  is the set of nodes that are not in any partition.
Consider the following cut that separates  $\{s_1, \ldots, s_{k'}\}$  from  $\{s_{k' + 1}, \ldots, s_k\}$ : Let  $A' = [U_1 \cap A] \cup [U_1 \cap B] \cup [\overline{V} \cap A]$ . Let  $B' = [U_2 \cap B] \cup [U_2 \cap A] \cup [\overline{V} \cap B]$ . Since  $(A, B)$  is a min cut that separates  $\{s_1, \ldots, s_{k'}\}$  from  $\{s_{k' + 1}, \ldots, s_k\}$  with additive error  $e(n)$ , we have  $w(E(A, B)) \leq w(E(A', B')) + e(n)$ . Note that  $A = [U_1 \cap A] \cup [U_2 \cap A] \cup [\overline{V} \cap A]$  and  $B = [U_1 \cap B] \cup [U_2 \cap B] \cup [\overline{V} \cap B]$ . So turning  $(A, B)$  into  $(A', B')$  is equivalent to switching  $U_2 \cap A$  and  $U_1 \cap B$  between  $A$  and  $B$ . So we have that
$$
\begin{array}{l} w (E (U _ {2} \cap A, U _ {2} \cap B)) + w (E (U _ {2} \cap A, \bar {V} \cap B)) + w (E (U _ {1} \cap B, U _ {1} \cap A)) + w (E (U _ {1} \cap B, \bar {V} \cap A)) \\ \leq \end{array} \tag {14}
$$
$$
\begin{array}{l} w (E (U _ {2} \cap A, U _ {1} \cap A)) + w (E (U _ {2} \cap A, \overline {{V}} \cap A)) + w (E (U _ {1} \cap B, U _ {2} \cap B)) + w (E (U _ {1} \cap B, \overline {{V}} \cap B)) \\ + e (n) \end{array}
$$
Eq. (14) is illustrated in Figure 2. Using Eq. (14), we obtain that
$$
\begin{array}{l} w (E (A, B)) = w \left(E \left(U _ {2} \cap A, U _ {2} \cap B\right)\right) + w \left(E \left(U _ {2} \cap A, \bar {V} \cap B\right)\right) \\ + w \left(E \left(U _ {1} \cap B, U _ {1} \cap A\right)\right) + w \left(E \left(U _ {1} \cap B, \bar {V} \cap A\right)\right) \\ + w \left(E \left(U _ {2} \cap A, U _ {1} \cap B\right)\right) + w \left(E \left(\left[ U _ {1} \cap A \right] \cup \left[ \bar {V} \cap A \right], \left[ U _ {2} \cap B \right] \cup \left[ \bar {V} \cap B \right]\right)\right) \\ \leq w (E (U _ {2} \cap A, U _ {1} \cap A)) + w (E (U _ {2} \cap A, \bar {V} \cap A)) \\ + w \left(E \left(U _ {1} \cap B, U _ {2} \cap B\right)\right) + w \left(E \left(U _ {1} \cap B, \bar {V} \cap B\right)\right) \\ + w \left(E \left(U _ {2} \cap A, U _ {1} \cap B\right)\right) + w \left(E \left(\left[ U _ {1} \cap A \right] \cup \left[ \bar {V} \cap A \right], \left[ U _ {2} \cap B \right] \cup \left[ \bar {V} \cap B \right]\right)\right) \\ + e (n) \\ \end{array}
$$
So we conclude that
$$
\begin{array}{l} w (E (A, B)) \leq w (E (U _ {1} \cap B, [ U _ {2} \cap B ] \cup [ \bar {V} \cap B ] \cup [ U _ {2} \cap A ])) \tag {15} \\ + w \left(E \left(U _ {1} \cap A, [ U _ {2} \cap B ] \cup [ \bar {V} \cap B ]\right)\right) \\ + w \left(E \left(U _ {2} \cap A, [ U _ {1} \cap A ] \cup [ \bar {V} \cap A ]\right)\right) \\ + w \left(E \left(U _ {2} \cap B, \bar {V} \cap A\right)\right) \\ + w (E (\bar {V} \cap A, \bar {V} \cap B)) \\ + e (n) \\ \end{array}
$$
We substitute  $w(E(A, B))$  in Eq. (13) using Eq. (15). Recall that  $U_1 = \cup_{i=1}^{k'} V_i$ ,  $\delta_{\widetilde{G}_1}(V_i \cap A) = w(E(V_i \cap A, A \setminus V_i))$  and  $\delta_G(V_i) = w(E(V_i, V \setminus V_i))$ . For any  $i \in \{1, \ldots, k'\}$ , we have that
$E(V_{i}\cap B,[U_{2}\cap B]\cup [\overline{V}\cap B]\cup [U_{2}\cap A])$  and  $E(V_{i}\cap A,[U_{2}\cap B]\cup [\overline{V}\cap B])$  are both disjoint from  $E(V_{i}\cap A,A\setminus V_{i})$ . Moreover all these three terms appear in  $E(V_{i},V\setminus V_{i})$ . So we have
$$
\begin{array}{l} w \left(E \left(U _ {1} \cap B, [ U _ {2} \cap B ] \cup [ \bar {V} \cap B ] \cup [ U _ {2} \cap A ]\right)\right) \\ + w \left(E \left(U _ {1} \cap A, [ U _ {2} \cap B ] \cup [ \bar {V} \cap B ]\right)\right) + \sum_ {i = 1} ^ {k ^ {\prime}} \delta_ {\widetilde {G} _ {1}} \left(V _ {i} \cap A\right) \\ \leq \sum_ {i = 1} ^ {k ^ {\prime}} \delta_ {G} (V _ {i}) \\ \end{array}
$$
Note that the first two terms above are the first two terms in Eq. (15). Similarly, we have