SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Let  $b$  be any point on the boundary of  $Q$  and  $L$  be the set of points between  $b$  and  $c$  excluding endpoints, we show that  $\pi_c$  also sorts  $w^{(b)}$ . Let  $i,j \in E$  where  $w_i^{(c)} < w_j^{(c)}$ , then  $w_i^{(b)} > w_j^{(b)}$  implies  $w_i^{(d)} = w_j^{(d)}$  for some  $d \in L$ , meaning  $L$  meets a hyperplane in  $A$ , a contradiction as  $L \subseteq \operatorname{Int}(Q)$ . For all pairs  $i,j \in E$  where  $w_i^{(c)} = w_j^{(c)}$ ,  $\delta_{i,j} = 0$  so  $w_i^{(b)} = w_j^{(b)}$ . With this, every pair is accounted for, so  $\pi_c$  sorts  $w^{(b)}$ . Therefore, since Greedy guarantees optimality, any base minimizing  $w^{(c)}$  also minimizes  $w^{(b)}$ , yielding the claim.
Proof of Corollary 1. We see that  $|A'| = |H_A|$ , which is upper bounded by the number of half-space intersections from hyperplanes in  $A$ . Since these are  $(k - 2)$ -dimensional hyperplanes, applying the formula in [32] gives  $|H_A| \leq \sum_{i=1}^{k} \binom{|A|}{i-1}$  which is increasing in  $|A|$ , so the claim follows from  $|A| \leq m(m - 1)/2$ . We have  $A'$  is a sufficient trade-off set following from Lemma 1 and  $\bigcup_{Q \in H_A} Q = U$ .
Proof of Lemma 2. Let  $0 < \lambda_c, < \lambda_d < 1$  such that  $b := (1 - b)a + ba' \in A^*$  and for all  $\lambda \in [\lambda_c, \lambda_b)$ ,  $(1 - \lambda)a + \lambda a' \notin A^*$ , and let  $c := (1 - \lambda_c)a + \lambda_c a'$ , then elements sharing weight in  $w^{(b)}$  must be mapped to consecutive positions in  $\pi_c$ . Indeed, let  $p, q \in E$  ( $\pi_c(p) < \pi_c(q)$ ) where  $w_p^{(b)} = w_q^{(b)}$ , if there is  $o \in E$  where  $\pi_c(o) \in (\pi_c(p), \pi_c(q))$  and  $w_o^{(b)} \neq w_p^{(b)}$ , then since the former implies  $w_o^{(c)} \in (w_p^{(c)}, w_q^{(c)})$ , we have  $w_o^{(d)} = w_p^{(d)}$  or  $w_o^{(d)} = w_q^{(d)}$  for some  $d$  in the open line segment connecting  $b$  and  $c$  which implies  $d \in A^*$ , a contradiction. Each such consecutive sequence of  $l$  positions contains  $l(l - 1)/2$  pairs. From here, we consider two cases:
- If such a sequence contains no pair  $(i,j)$  where  $\delta_{i,j} = 0$ , then the aforementioned pairs correspond to  $l(l - 1) / 2$  duplicates of  $b$  in  $A^*$ . Furthermore, since the weights are transformed linearly w.r.t. trade-off, for all sufficiently small  $\epsilon > 0$ , these sequences are reversed between  $\pi_c$  and  $\pi_{b + \epsilon (b - c)}$ , whereas positions not in these sequences are stationary. Reversing  $l$  consecutive positions requires  $l(l - 1) / 2$  adjacent swaps, so the Kendall distance between  $\pi_c$  and  $\pi_{b + \epsilon (b - c)}$  equals the multiplicity of  $b$  in  $A^*$ .
$^6$ If  $|Z|\geq k^{\prime}$  for all solutions  $z$ , the instance is reducible to  $(k - k^{\prime})$ -objective instances, and the guarantee factor is likewise tight.
- If such a sequence contains  $h > 1$  elements with the same weight at all trade-off, then these must occupy consecutive positions in  $\pi_c$ . As we assumed, the relative ordering among these elements is fixed, so exactly  $h(h - 1) / 2$  swaps are saved. Furthermore, any pair  $(i,j)$  among these elements is such that  $\Delta_{i,j} \notin A$ , meaning these  $h(h - 1) / 2$  pairs are already subtracted from  $A^*$ .
In any case, we can assign to each duplicate of  $b$  in  $A^*$  a permutation sorting  $w^{(b)}$  so that these form a sequence of adjacent swap from  $\pi_c$  to  $\pi_{b + \epsilon (b - c)}$  including  $\pi_{b + \epsilon (b - c)}$  and not  $\pi_c$ . This directly yields the claim if  $a$  and  $a^*$  are not in  $A^*$ .
Assume otherwise, then for all hyperplanes  $\Delta_{i,j}$  containing  $a$ ,  $w_i^{(a)} = w_j^{(a)}$ , so for every such pair  $(i,j)$ , we arrange  $\pi_a$  so that their pairwise ordering in  $\pi_a$  is the reverse of that in  $\pi_{a'}$ . We likewise give  $a'$  the same treatment<sup>7</sup>. With this, the Kendall distance between  $\pi_a$  and  $\pi_{a'}$  is maximized and equal to  $|A^*|$ .
Proof of Lemma 3. Let  $E_{o} \coloneqq \{a \in E : \tau(a) < \tau(o)\}$  be the set of elements Greedy considers adding to  $x$  before  $o \in E$  when run on  $\tau$ , we have  $x \cap E_{u} = x' \cap E_{u}$ . If  $v \in x$  or  $v \notin x'$  or  $u \notin x$  or  $u \in x'$  then  $x = x'$ , as can be seen from how Greedy selects elements:
- If  $v \in x$ , then  $v \in x'$  since Greedy observes  $v$  before  $u$  when run on  $\tau'$ . Whether Greedy adds  $u$  to  $x$  only depends on whether there is a circuit in  $(x \cap E_u) \cup \{u\} = (x' \cap E_u) \cup \{u\}$ , so it makes the same decision when run on  $\tau'$ . Afterwards, it proceeds identically on both permutations, leading to the same outcome, so  $x = x'$ . By symmetry, the same follows from  $u \in x'$ .
- If  $u \notin x$ , then there is a circuit in  $(x \cap E_u) \cup \{u\} = (x' \cap E_u) \cup \{u\}$ , so  $u \notin x'$ . By the same argument, Greedy makes the same decision regarding  $v$  on both permutations, leading to  $x = x'$ . By symmetry, the same follows from  $v \notin x'$ .
Assume otherwise, it is a known property of bases [20] that  $x \cup \{v\}$  contains a unique circuit  $C$  and that  $v \in C$ . Greedy not adding  $v$  to  $x$  implies that  $C \subseteq (x \cap E_v) \cup \{v\} = (x' \cap E_v) \cup \{u, v\}$ . Let  $v'$  be the first element after  $v$  that  $x$  and  $x'$  differ at and assume w.l.o.g.  $v' \in x \setminus x'$ , we have  $(x' \cap E_{v'}) \cup \{u\} = (x \cap E_{v'}) \cup \{v\}$  and since  $v'$  is not added into  $x'$  before Greedy terminates, there must be another circuit in  $(x' \cap E_{v'}) \cup \{v'\} \subset x \cup \{v\}$  containing  $v'$ , which is distinct from the unique circuit  $C$ . The contradiction implies that  $x$  and  $x'$  do not differ after  $v$ , so  $x \otimes x' = \{u, v\}$ .
Proof of Theorem 2. We define  $l_{c} \coloneqq (1 - c)a + cb$  for  $c \in [0,1]$ , let  $0 \leq \theta \leq \theta' \leq 1$  where  $\pi_{l_{\theta}}$  and  $\pi_{l_{\theta'}}$  are an adjacent swap apart<sup>8</sup> and the Greedy solutions on them,  $x$  and  $x'$ , are such that  $|x \otimes x'| = 2$ . Let  $u, v \in E$  where  $x \cap \{u, v\} = \{u\}$  and  $x' \cap \{u, v\} = \{v\}$ , Lemma 3 implies  $\pi_{l_{\theta}}(u) < \pi_{l_{\theta}}(v)$  and  $\pi_{l_{\theta'}}(u) > \pi_{l_{\theta'}}(v)$ , so  $\pi_{a}(u) < \pi_{a}(v)$ . This means as the trade-off moves from  $a$  to  $b$ , the Greedy solution minimizing the scalarized weight changes incrementally by having exactly one element shifted to the right on  $\pi_{a}$  (to a position not occupied by the current solution). Since at most  $hm - h(h + 1)/2$  such changes can be done sequentially, Greedy produces at most  $hm - h(h + 1)/2 + 1$  distinct solutions in total across all trade-offs between  $a$  and  $b$ .
To show this upper bound, we keep track of the following variables as the trade-off moves from  $a$  to  $b$ . Since each solution contains  $n$  elements, let  $p_i$  be the  $i$ th leftmost position on  $\pi_a$  among those occupied by the current Greedy solution for  $i = 1, \dots, n$ , we see that upon each change, there is at least a  $j \in \{1, \dots, n\}$  where  $p_j$  increases. Furthermore, for all  $i$ ,  $p_i$  can increase by at most  $m - n$  since it cannot be outside of  $[i, m - n + i]$ , so the quantity  $p := \sum_{i=1}^{n} p_i$  can increase by at most  $n(m - n)$ . We see that  $p$  increases by  $l$  when the change is incurred by a swap in the Greedy solution such that the added element is positioned  $l$  to the right of the removed element on  $\pi_a$ , we call this a  $l$ -move. Furthermore, each element pair participates in at most one swap, so  $p$  can be increased by at most  $m - l$ $l$ -moves for every  $l = 1, \dots, m - 1$ . Therefore, to upper bound the number of moves, we can assume smallest possible distance in each move, and the increase in  $p$  from using all possible  $l$ -moves for all  $l = 1, \dots, h$  is  $\sum_{j=1}^{h} j(m - j) \geq n(m - n)$ . This means no more than  $\sum_{j=1}^{h} (m - j) = hm - h(h + 1)/2$  moves can be used to increase  $p$  by at most  $n(m - n)$ .
This is also done for any  $\Delta_{i,j}$  containing both  $a$  and  $a^\prime$
If  $\theta = \theta^{\prime}$ , we assume  $\pi_{l_{\theta}}$  is closer to  $\pi_{a}$  in Kendall distance.
Proof of Lemma 4. First, we observe that a set supersets a base iff its rank is  $n$ . We see that for all  $\lambda \in [0,1]$  and  $x,y \in \{0,1\}^m$ ,  $r(x) > r(y)$  implies  $f_{\lambda}(x) < f_{\lambda}(y)$ . Thus, for each  $\lambda \in \Lambda$ , MOEA/D performs (1+1)-EA search toward a base's superset with fitness  $f_{\lambda}$ , which concludes in  $O(m\log n)$  expected steps [26]. The claim follows from the fact that MOEA/D produces  $|\Lambda|$  search points in each step.
Proof of Theorem 3. We assume each solution in  $P_{\lambda}$  supersets a base for all  $\lambda \in \Lambda$ ; this occurs within expected time  $O(|\Lambda|m\log n)$ , according to Lemma 4. Since for each  $\lambda \in \Lambda$ , the best improvement in  $f_{\lambda}$  is retained in each step, the expected number of steps MOEA/D needs to minimize  $f_{\lambda}$  is at most the expected time  $(1 + 1)$ -EA needs to minimize  $f_{\lambda}$ . We thus fix a trade-off  $\lambda$  and assume the behaviors of  $(1 + 1)$ -EA. Note that we use  $d_{\lambda} \cdot w^{(\lambda)}$  in the analysis instead for integral weights; we scale  $f_{\lambda}$  and  $OPT_{\lambda}$  accordingly.
We call the bit flips described in Lemma 5 good flips. Let  $s$  be the current search point, if good 1-bit flips incur larger total weight reduction than good 2-bit flips on  $s$ , we call  $s$  1-step, and 2-step otherwise. If at least half the steps from  $s$  to the MWB  $z$  are 1-steps, Lemma 5 implies the optimality gap of  $s$  is multiplied by at most  $1 - 1/2(m - n)$  on average after each good 1-bit flip. Therefore, from  $f_{\lambda}(s) \leq d_{\lambda}(m - n)w_{max} + OPT_{\lambda}$ , the expected difference  $D_L$  after  $L$  good 1-bit flips is  $E[D_L] \leq d_{\lambda}(m - n)w_{max}(1 - 1/2(m - n))^L$ . At  $L = \lceil (2\ln 2)(m - n)\log (2d_{\lambda}(m - n)w_{max} + 1)\rceil$ ,  $E[D_L] \leq 1/2$  and by Markov's inequality and the fact that  $D_L \geq 0$ ,  $\operatorname*{Pr}[D_L < 1] \geq 1/2$ . Since weights are integral,  $D_L < 1$  implies that  $z$  is reached. The probability of making a good 1-bit flip is  $\Theta((m - n)/m)$ , so the expected number of steps before  $L$  good 1-bit flips occurs is  $O(Lm/(m - n)) = O(m(\log(m - n) + \log w_{max} + \log d_{\lambda}))$ . Since 1-steps take up most steps between  $s$  and  $z$ , the bound holds.
If at least half the steps from  $s$  to  $z$  are 2-steps, Lemma 5 implies the optimality gap of  $s$  is multiplied by at most  $1 - 1/2n$  on average after each good 2-bit flip. Repeating the argument with  $L = \lceil (2\ln 2)n\log(2d_{\lambda}(m-n)w_{max}+1)\rceil$  and the probability of making a good 2-bit flip being  $\Theta(n/m^2)$ , we get the bound  $O(m^2(\log(m-n)+\log w_{max}+\log d_{\lambda}))$ . Summing this over all  $\lambda \in \Lambda$  gives the total bound.
Proof of Theorem 4. From Corollary 2, to collect a new point in  $C$ , it is sufficient to perform a 2-bit flip on some supported solution. In worst-case, there can be only one trade-off  $\lambda \in \Lambda$  such that all non-extreme supported solutions minimize  $w^{(\lambda)}$ , so the correct solution is mutated with probability at least  $1 / l$  in each iteration, where  $l$  is the number of already collected points. Since  $|\Lambda|$  search points are generated in each iteration, the expected number of search points required to enumerate  $C$  is  $O(|\Lambda|m^2\sum_{l=1}^{|C|}l) = O(|\Lambda||C|^2m^2)$ .