SlowGuess/NeurIPS_2023_OCR huggingface dataset:
On the other hand, in the ALG histogram the distance of  $p$  from the right edge of the histogram is at most  $|R_{t}|$ , therefore for the point  $p$  to be inside the OPT $_o$  histogram we require
$$
\alpha_ {o} \leq 1 - \gamma . \tag {6}
$$
Observe that throughout the proof we did not use the fact that we change the opening cost to 0, therefore the bound on our cost works even if we re-pay the boxes that are reopened.
The fact that the opening cost becomes 0 is not directly used in the analysis (i.e. inequalities (4) and (5)). Our analysis gives an upper bound on the cost of the algorithm, even if the algorithm never changes the cost of an opened box to 0. That is the reason in (4) and (5) the cost appears unchanged but the analysis still works for the algorithm since we just want an upper bound (and if we changed the cost to 0 this would only lower the cost of the algorithm).
Bounding the values cost. By the construction of  $v^{*}$ , the point in the OPT $_{v}$  histogram that has value  $v^{*}$  is at distance at least  $|R_{t}| (1 - \beta)\gamma$  from the right hand side. This means that in the rescaled histogram, the point that has value at least  $v^{*}$  is at distance at least  $(1 - \beta)\gamma |R_{t}| / \alpha_{v}$  from the right hand side.
On the other hand, in the ALG histogram the distance of  $p$  from the right edge of the histogram is at most  $|R_t|$ , therefore for the point  $p$  to be inside the OPT $_o$  histogram we require
$$
\alpha_ {v} \leq (1 - \beta) \gamma . \tag {7}
$$
We optimize the constants  $\alpha_{o},\alpha_{v},\beta ,\gamma$  by ensuring that inequalities (6) and (7) hold. We set  $\alpha_{o} = 1 - \gamma$  and  $\alpha_{v} = (1 - \beta)\gamma$ , and obtain that  $\mathrm{ALG}\leq \mathrm{OPT}_o / (\beta \cdot \gamma \cdot (1 - \gamma)) + \mathrm{OPT}_v / ((1 - \beta)\gamma)$ . Requiring these to be equal we get  $\beta = 1 / (2 - \gamma)$ , which is minimized for  $\beta = 1 / \sqrt{2}$  and  $\gamma = 2 - \sqrt{2}$  for a value of  $3 + 2\sqrt{2}$ .
3.2 Conditioning on  $V_{b} = v$
In this section we switch gears to our second variant of Algorithm 1, where in each step we update the prior  $\mathcal{D}$  conditioning on the event  $V_{b} = v$ . We state our result in Theorem 3.3. In this case, the conditioning on  $\mathcal{D}$  implies that the algorithm at every step removes the scenarios that are inconsistent with the value realized. In order to understand better the differences of the two variants and their conditioning we included an example and a discussion in section A.1 of the Appendix.
Theorem 3.3. Algorithm 1 is a  $3 + 2\sqrt{2} \approx 5.828$ -approximation for PANDORA's BOX against the partially-adaptive optimal, when conditioning on  $V_{b} = v$ .
The main challenge was that the algorithm's solution is now a tree with respect to scenarios instead of a line as in the case of  $\mathcal{D}|_{V_b > \sigma_b}$ . Specifically, in the  $D|_{V_b > \sigma_b}$  variant at every step all scenarios that had  $V_b \leq \sigma_b$  were covered and removed from consideration. However in the  $D|_{V_b = v}$  variant the remaining scenarios are split into different cases, based on the realization of  $V$ , as shown in the example of Figure 4, which is deferred to Section A.3 of the Appendix due to space constraints.
This results into the ALG histogram not being well defined, since there is no unique order of covering the scenarios. We overcome this by generalizing the histogram approach to trees.
Proof of Theorem 3.3. The proof follows similar steps to that of Theorem 3.2, thus we only highlight the differences. The algorithm is presented below, the only change is line 5 where we remove the inconsistent with the value revealed scenarios, which also leads to our solution branching out for different scenarios and forming a tree.
Algorithm 3: Weitzman's rule for Full Updates
Input: Boxes with costs  $c_{i}\in \mathbb{R}$  , set of scenarios S.   
1 Define a root node  $u$  corresponding to the set S   
2  $R_{u}\gets S$  the set of scenarios still uncovered   
3 while  $R_{u}\neq \emptyset$  do   
4 Let  $\sigma_u\gets \min_{b\in \mathcal{B},A\subseteq R_u}\frac{c_b|R_u| + \sum_{s\in A}v_b^s}{|A|}$    
5 Let  $b_{u}$  and  $A_{u}$  be the box and the set of scenarios that achieve the minimum   
6 Open box  $b_{u}$  paying  $c_{b_u}$  and observe value  $v$    
7 Stop and choose the value at box  $b_{u}$  if it is less than  $\sigma_{u}$  : this holds iff  $s\in A_u$    
8 Set  $c_{b_u}\gets 0$    
9 Let  $u^{\prime}$  be a vertex corresponding to the set of consistent scenarios with  $R_{u^{\prime}}\triangleq R_{u}\backslash (A_{u}\cup \{s\in R_{u}:v_{b_{u}}^{s}\neq v\}) / /$  Remove inconsistent scenarios   
10 Set  $u\gets u^{\prime}$    
11 end
Bounding the opening cost Consider the tree  $\mathcal{T}$  of ALG where at every node  $u$  a set  $A_u$  of scenarios is covered. We associate this tree with node weights, where at every node  $u$ , we assign  $|A_u|$  weights  $(\sigma_u,\dots,\sigma_u)$ . Denote, the weighted tree by  $\mathcal{T}_{\mathrm{ALG}}$ . As before, the total cost of ALG is equal to the sum of the weights of the tree.
We now consider two alternative ways of assigning weights to the nodes, forming trees  $\mathcal{T}_{\mathrm{OPT}_o}$ ,  $\mathcal{T}_{\mathrm{OPT}_v}$  using the following process.
-  $\mathcal{T}_{\mathrm{OPT}_o}$ . At every node  $u$  we create a vector of weights  $w_{u}^{\mathrm{OPT}_{o}} = (c_{s}^{\mathrm{OPT}})_{s \in A_{u}}$  where each  $c_{s}^{\mathrm{OPT}}$  is the opening cost that scenario  $s \in A_{u}$  has in the optimal solution.
-  $\mathcal{T}_{\mathrm{OPT}_v}$ . At every node  $u$  we create a vector of weights  $\boldsymbol{w}_u^{\mathrm{OPT}_v} = (v_s^{\mathrm{OPT}})_{s \in A_u}$  where each  $v_s^{\mathrm{OPT}}$  is the value the optimal uses to cover scenario  $s \in A_u$ .
We denote by  $\mathrm{cost}(\mathcal{T}_{\mathrm{ALG}})$  the sum of all weights in every node of the tree  $\mathcal{T}$ . We have that  $\mathrm{cost}(\mathcal{T})$  is equal to the total cost of ALG, while  $\mathrm{cost}(\mathcal{T}_{\mathrm{OPT}_o})$  and  $\mathrm{cost}(\mathcal{T}_{\mathrm{OPT}_v})$  is equal to the optimal opening cost  $\mathrm{OPT}_o$  and optimal value  $\mathrm{OPT}_v$  respectively. Intuitively, the weighted trees correspond to the histograms in the previous analysis of Theorem 3.2.
We want to relate the cost of ALG, to that of  $\mathcal{T}_{\mathrm{OPT}_o}$  and  $\mathcal{T}_{\mathrm{OPT}_v}$ . To do this, we define an operation similar to histogram scaling, which replaces the weights of every node  $u$  in a tree with the top  $\rho$ -percentile of the weights in the subtree rooted at  $u$ . As the following lemma shows, this changes the cost of a tree by a bounded multiplicative factor.
Lemma 3.3.1. Let  $\mathcal{T}$  be a tree with a vector of weights  $w_{u}$  at each node  $u\in \mathcal{T}$ , and let  $\mathcal{T}^{(\rho)}$  be the tree we get when we substitute the weights of every node with the top  $\rho$ -percentile of all the weights in the subtree of  $\mathcal{T}$  rooted at  $u$ . Then
$$
\rho \cdot c o s t (\mathcal {T} ^ {(\rho)}) \leq c o s t (\mathcal {T}).
$$
We defer the proof of Lemma 3.3.1 to Section A.3 of the Appendix. To complete the proof of Theorem 3.3, and bound cost  $(\mathcal{T}_{\mathrm{ALG}})$ , we show as before that the weights at every node  $u$ , are bounded by the weights of  $\mathcal{T}_{\mathrm{OPT}_o}^{(1 - \gamma)}$  scaled by  $\frac{1}{\beta\gamma}$  plus the weights of  $\mathcal{T}_{\mathrm{OPT}_v}^{((1 - \beta)\gamma)}$ , for the constants  $\beta, \gamma \in (0,1)$  chosen in the proof of Theorem 3.2. This implies that
$$
\operatorname {c o s t} \left(\mathcal {T} _ {\mathrm {O P T} _ {o}}\right) \leq \frac {1}{\beta \gamma} \operatorname {c o s t} \left(\mathcal {T} _ {\mathrm {O P T} _ {o}} ^ {(1 - \gamma)}\right) + \operatorname {c o s t} \left(\mathcal {T} _ {\mathrm {O P T} _ {v}} ^ {((1 - \beta) \gamma)}\right)
$$
$$
\leq \frac {1}{\beta \gamma (1 - \gamma)} \operatorname {c o s t} \left(\mathcal {T} _ {\mathrm {O P T} _ {o}}\right) + \frac {1}{(1 - \beta) \gamma} \operatorname {c o s t} \left(\mathcal {T} _ {\mathrm {O P T} _ {v}}\right)
$$
which gives  $\mathrm{ALG} \leq 5.828$  OPT for the choice of  $\beta$  and  $\gamma$ . The details of the proof are similar to the one of Theorem 3.1, and are deferred to section A.3 of the Appendix.
Note on the approximation factors. Observe that Variant 2, where we condition on  $V_{b} = v$  has a worse approximation factor than Variant 1 where we only condition on  $V_{b} > \sigma_{b}$ . Intuitively someone might expect that with more information the approximation factor will improve. However, it is challenging to argue about this formally. It is also plausible that such monotonicity may not hold as more information might lead the greedy algorithm to make wrong decisions. Instead of making any such claims, we analyze this case directly by showing that our proof approach extends to the full update variant with a generalization of the histogram method to work on trees. Our technique for improving the approximation for the partial updates variant could not be generalized however and thus we only obtain the worse approximation guarantee.
4 Learning from Samples
In this section we show that our algorithm also works when we are only given sample access to the correlated distribution  $\mathcal{D}$ .
We will mainly focus on the first variant with partial updates  $\mathcal{D}|_{V > v}$ . The second variant with full Bayesian updates  $\mathcal{D}|_{V = v}$  requires full knowledge of the underlying distribution and can only work with sample access if one can learn the full distribution. To see this consider for example an instance where the values are drawn uniformly from  $[0,1]^d$ . No matter how many samples one draws, it is impossible to know the conditional distribution  $\mathcal{D}|_{V = v}$  after opening the first box for fresh samples  $v$ , and the Bayesian update is not well defined<sup>4</sup>.
Variant 1 does not face this problem and can be learned from samples if the costs of the boxes are polynomially bounded by  $n$ , i.e. if there is a constant  $c > 0$  such that for all  $b \in \mathcal{B}$ ,  $c_b \in [1, n^c]$ . If the weights are unbounded, it is impossible to get a good approximation with few samples. To see this consider the following instance. Box 1 has cost  $1 / H \to 0$ , while every other box has cost  $H$  for a very large  $H > 0$ . Now consider a distribution where with probability  $1 - \frac{1}{H} \to 1$ , the value in the first box is 0, and with probability  $1 / H$  is  $+\infty$ . In this case, with a small number of samples we never observe any scenario where  $v_1 \neq 0$  and believe the overall cost is near 0. However, the true cost is at least  $H \cdot 1 / H \geq$  and is determined by how the order of boxes is chosen when the scenario has  $v_1 \neq 0$ . Without any such samples it is impossible to pick a good order.
Therefore, we proceed to analyze Variant 1 with  $\mathcal{D}|_{V > \sigma}$  in the case when the box costs are similar. We show that polynomial, in the number of boxes, samples suffice to obtain an approximately-optimal algorithm, as we formally state in the following theorem. We present the case where all boxes have cost 1 but the case where the costs are polynomially bounded easily follows.
Theorem 4.1. Consider an instance of Pandora's Box with opening costs equal to 1. For any given parameters  $\varepsilon, \delta > 0$ , using  $m = \text{poly}(n, 1 / \varepsilon, \log(1 / \delta))$  samples from  $\mathcal{D}$ , Algorithm 1 (Variant 1) obtains a  $4.428 + \varepsilon$  approximation policy against the partially-adaptive optimal, with probability at least  $1 - \delta$ .
To prove the theorem, we first note that variant 1 of Algorithm 1 takes a surprisingly simple form, which we call a threshold policy. It can be described by a permutation  $\pi$  of visiting the boxes and a vector of thresholds  $\pmb{\tau}$  that indicate when to stop. The threshold for every box corresponds to the reservation value the first time the box is opened. To analyze the sample complexity of Algorithm 1, we study a broader class of algorithms parameterized by a permutation and vector of thresholds given in Algorithm 4.
Our goal now is to show that polynomially many samples from the distribution  $\mathcal{D}$  suffice to learn good parameters for Algorithm 4. We first show a Lemma that bounds the cost of the algorithm calculated in the empirical  $\hat{\mathcal{D}}$  instead of the original  $\mathcal{D}$  (Lemma 4.1.1), and a Lemma 4.1.2 that shows how capping the reservation values by  $n / \varepsilon$  can also be done with negligible cost.
4For a discrete distribution example see Section A.5 of the appendix.
Algorithm 4: General format of PANDORA's BOX algorithm.
Input: Set of boxes, permutation  $\pi$  , vector of thresholds  $\pmb {\tau}\in \mathbb{R}^n$    
1 best  $\leftarrow \infty$    
2 foreach  $i\in [n]$  do   
3 if best  $>\tau_{i}$  then   
4 Open box  $\pi_i$  , see value  $v_{i}$    
5 best  $\leftarrow \min (\text{best},v_{i})$    
6 else   
7 Accept best   
8 end
Lemma 4.1.1. Let  $\varepsilon, \delta > 0$  and let  $\mathcal{D}'$  be the empirical distribution obtained from poly $(n, 1/\varepsilon, \log(1/\delta))$  samples from  $\mathcal{D}$ . Then, with probability  $1 - \delta$ , it holds that
$$
\left| \mathbb {E} _ {\hat {D}} \left[ A L G (\pi , \tau) - \underset {b \in \mathcal {B}} {\min } v _ {b} \right] - \mathbb {E} _ {D} \left[ A L G (\pi , \tau) - \underset {b \in \mathcal {B}} {\min } v _ {b} \right] \right| \leq \varepsilon
$$
for any permutation  $\pi$  and any vector of thresholds  $\mathbf{v} \in \left[0, \frac{n}{\varepsilon}\right]^n$ .
We defer the proof of Lemmas 4.1.1, 4.1.2 and that of Theorem 4.1 to Section A.5 of the Appendix.
Lemma 4.1.2. Let  $\mathcal{D}$  be any distribution of values. Let  $\varepsilon >0$  and consider a permutation  $\pi$  and thresholds  $\pmb{\tau}$ . Moreover, let  $\tau^{\prime}$  be the thresholds capped to  $n / \varepsilon$ , i.e. setting  $\tau_b^\prime = \min \{\tau_b,n / \varepsilon \}$  for all boxes  $b$ . Then,
$$
\mathbb {E} _ {v \sim D} \left[ A L G \left(\pi , \tau^ {\prime}\right) \right] \leq (1 + \varepsilon) \mathbb {E} _ {v \sim D} \left[ A L G \left(\pi , \tau\right) \right].
$$
Note on Continuous vs Discrete Distributions. The results of Section 4 apply for general distributions (discrete or continuous) and show that the partial updates variant leads to good approximation when run on the empirical distribution obtained just with polynomially many samples. In contrast, the full updates variant requires a complete description of the distribution. However, as the approximation factor does not depend on the support size, It can also apply even for continuous distributions with arbitrary large support by taking a limit over a very fine discretization
We present a summary of our results with a comparison to previous work on Table 1. Our main contribution was to improve the approximation factor for Pandora's Box with correlations given by Chawla et al. [2020], while also greatly simplifying their approach. Our algorithm also directly extends the independent case algorithm, giving us a unified way to solve this problem. An interesting open question is to try and improve their results for more complex combinatorial constraints, like selecting  $k$  boxes (instead of one) or for selecting a basis of size  $k$ , when the boxes are part of a matroid.
Approx. Factor	Learnable from Samples
Algorithm of Chawla et al. [2020]	9.22	Yes
Variant 1 (DVb>σb)	4.428 (Thm 3.2)	Yes (Thm 4.1)
Variant 2 (DVb=v)	5.828 (Thm 3.3)	No (Sec. 4)
Table 1: Summary of our results (in bold) and comparison to previous work.
Observe also that the more natural Variant 2 seems worse than Variant 1 even though the algorithm has more accurate information through the update of the prior. Intuitively we would expect a better factor, however since the algorithm is greedy approximation, and not the optimal, the factor may not necessarily be monotone on the amount of information given. We leave as an open problem whether our analysis in Variant 2 is tight or this greedy algorithm cannot perform better under full information.
A.1 Supplemental Preliminaries
Updating the prior. We include an example showing the process of updating the prior for our two different updating rules. The (correlated) distribution is a set of vectors of size  $n$ , where each is drawn with some probability. When we open a box and see a value, some scenarios are not "possible" anymore, i.e. we know they cannot be the ones realized. We illustrate in the following example. Assume there are 3 of these vectors (scenarios).
b1	b2	b3
S1	3	4	7
S2	6	4	2
S3	7	7	2
Table 2: Example with 3 scenarios and 3 boxes.
The rows in the matrix above are the scenarios, and the columns are the boxes. For example, if scenario  $S_{2}$  is the one realized (i.e. drawn from the distribution) then the values inside boxes  $b_{1}, b_{2}$  and  $b_{3}$  are 6, 4 and 2 respectively. The distribution  $\mathcal{D}$  is essentially drawing one of the scenarios with some probability.
To see what the conditioning means: assume we open box  $b_{1}$  and we see the value 6 (and assume for the sake of the example that the reservation value of box 1 is  $\sigma_{1} = 5$ ).
- Variant 1: we condition on  $6 = V_{b} > \sigma_{1} = 5$  meaning that scenario  $S_{1}$  is not possible anymore (because if  $S_{1}$  was the one drawn from  $\mathcal{D}$ , then we would have seen a value less than  $\sigma_{1} = 5$  when opening the box), and is removed from the set S the algorithm considers (line 9, Alg 2)
- Variant 2: we condition on  $V_{b} = 6$ , which means that scenarios  $S_{1}$  and  $S_{3}$  are both removed (similarly, because if any of these were drawn, we would not have seen 6 upon opening the box)
Differences in the variants. As a result of the different conditioning, the solution for the  $V_{b} > \sigma$  variant is partially adaptive meaning that the next box the algorithm opens, only depends on the scenarios that remain. However, for the  $V_{b} = v$  variant the solution is fully adaptive (meaning that the next box opened, depends on the exact value seen). This is illustrated in Figures 2 and 4 in the Appendix, where Variant 1's solution can be represented by a line graph (Figure 2), while Variant 2's solution is a tree (Figure 4).
A.2 Proofs from Section 3
Figure 2: Algorithm's solution when  $\mathcal{D} \gets \mathcal{D}_{V > \sigma}$ , for an instance with scenarios  $\mathcal{S} = \{s_1, s_2, s_3\}$  and boxes  $\mathcal{B} = \{b_1, b_2, b_3, b_4\}$ . The circles contain the scenarios that have not stopped at each step. Scenario  $s_1$  stopped at box  $b_2$ , scenario  $s_2$  stopped at box  $b_1$  and  $s_3$  at box  $b_4$ .
The tighter guarantee proof follows the steps of the proof in section 3.1 for the opening cost, but provides a tighter analysis for the values of cost.
Tight proof of Theorem 3.2. Denote by  $\sigma_s$  the reservation value for scenario  $s$  when it was covered by ALG and by  $\mathcal{T}$  the set of boxes opened i.e. the steps taken by the algorithm. Then we can write the cost paid by the algorithm as follows
$$
\mathrm {A L G} = \frac {1}{| \mathcal {S} |} \sum_ {s \in \mathcal {S}} \sigma_ {s} = \frac {1}{| \mathcal {S} |} \sum_ {p \in \mathcal {T}} \left| A _ {t} \right| \sigma_ {p}. \tag {8}
$$
We use the same notation as section 3.1 which we repeat here for convenience. Consider any point  $p$  in the ALG histogram, and let  $s$  be its corresponding scenario and  $t$  be the time this scenario is covered.
-  $R_{t}$  : set of uncovered scenarios at step  $t$
-  $A_{t}$  : set of scenarios that ALG chooses to cover at step  $t$
-  $c^*$ : the opening cost such that  $\gamma |R_t|$  of the scenarios in  $R_t$  have opening cost less than  $c^*$ .
-  $R_{\mathrm{low}} = \{s \in R_t : c_s^{\mathrm{OPT}} \leq c^*\}$  the set of these scenarios
-  $v^{*}$ : the value of scenarios in  $R_{\mathrm{low}}$  such that  $b|R_{\mathrm{low}}|$  of the scenarios have value less than  $v^{*}$ .
-  $L = \{ s \in R_{\mathrm{low}} : v_s^{\mathrm{OPT}} \leq v^* \}$  the set of scenarios with value at most  $v^*$ .
-  $B_L$ : set of boxes the optimal uses to cover the scenarios in  $L$  of step  $t$
The split described in the definitions above is again shown in Figure 3, and the constants  $1 > \beta, \gamma > 0$  will be determined in the end of the proof.
Figure 3: Split of scenarios in  $R_{t}$ .
Continuing from equation (8) we obtain the following.
$$
\begin{array}{l} \mathrm {A L G} \leq \frac {1}{| \mathcal {S} |} \sum_ {t \in \mathcal {T}} \left| A _ {t} \right| \frac {\left| R _ {t} \right| \sum_ {b \in B _ {L}} c _ {b} + \sum_ {s \in L} v _ {s} ^ {\mathrm {O P T}}}{| L |} \quad \text {I n e q u a l i t y 5} \\ \leq \frac {1}{| \mathcal {S} |} \sum_ {t \in \mathcal {T}} \left(| A _ {t} | \frac {c ^ {*}}{\beta \gamma} + \frac {\sum_ {s \in L} v _ {s} ^ {\mathrm {O P T}}}{| L |}\right) \quad \text {I n e q .} 5 \text {a n d} | L | = \gamma \beta | R _ {t} | \\ \leq \frac {\mathrm {O P T} _ {o}}{\beta \gamma (1 - \gamma)} \sum_ {t \in \mathcal {T}} \frac {| A _ {t} |}{| \mathcal {S} |} + \sum_ {t \in \mathcal {T}} \frac {| A _ {t} |}{| \mathcal {S} |} \frac {\sum_ {s \in L} v _ {s} ^ {\mathrm {O P T}}}{| L |} \quad \text {S i n c e} c ^ {*} \leq \mathrm {O P T} _ {o} / (1 - \gamma) \\ = \frac {\mathrm {O P T} _ {o}}{\beta \gamma (1 - \gamma)} + \sum_ {p \in \mathcal {T}} \frac {| A _ {t} |}{| S |} \frac {\sum_ {s \in L} v _ {s} ^ {\mathrm {O P T}}}{| L |} \quad \text {S i n c e} \sum_ {t} | A _ {t} | = | S | \\ \end{array}
$$
Where in the second to last inequality we used the same histogram argument from section 3.1, to bound  $c^*$  by  $\mathrm{OPT}_o / (1 - \gamma)$ .
To bound the values term, observe that if we sorted the optimal values  $v_{s}^{\mathrm{OPT}}$  that cover each scenario by decreasing order, and denote  $j_{s}$  the index of  $v_{s}^{\mathrm{OPT}}$  in this ordering, we add  $v_{s}^{\mathrm{OPT}}$  multiplied by
the length of the interval every time  $j_{s} \in \left[(1 - \beta)\gamma |R_{t}|, \gamma |R_{t}|\right]$ . This implies that the length of the intervals we sum up for  $v_{s}^{\mathrm{OPT}}$  ranges from  $j_{s} / \gamma$  to  $j_{s} / ((1 - \beta)\gamma)$ , therefore the factor for each  $v_{s}^{\mathrm{OPT}}$  is
$$
\frac {1}{\gamma} \sum_ {i = j _ {s} / \gamma} ^ {j _ {s} / (1 - \beta) \gamma} \frac {1}{i} \leq \frac {1}{\gamma} \log \left(\frac {1}{1 - \beta}\right)
$$
We want to balance the terms  $1 / (\beta \gamma (1 - \gamma))$  and  $1 / \gamma \log (1 / (1 - \beta))$  which gives that
$$
\gamma = 1 - \frac {1}{\beta \log \left(\frac {1}{1 - \beta}\right)}.
$$
Since we balanced the opening cost and value terms, by substituting the expression for  $\gamma$  we get that the approximation factor is
$$
\frac {1}{\beta \gamma (1 - \gamma)} = \frac {\beta \log^ {2} \left(\frac {1}{1 - \beta}\right)}{\beta \log \left(\frac {1}{1 - \beta}\right) - 1}.
$$
Numerically minimizing that ratio for  $\beta$  and ensuring that  $0 < \beta, \gamma < 1$  we get that the minimum is 4.428 obtained at  $\beta \approx 0.91$  and  $\gamma \approx 0.55$ .
A.3 Proofs from Section 3.2
Figure 4: Algorithm's solution when conditioning on  $V = v$ , for an instance with scenarios  $S = \{s_1, s_2, s_3\}$ , and boxes  $\mathcal{B} = \{b_1, b_2\}$ . The nodes contain the consistent scenarios at each step, and the values  $V$  are revealed once we open the corresponding box.
Continued proof of Theorem 3.3. We now proceed to give the bound on the weights of the nodes of  $\mathcal{T}_{\mathrm{ALG}}$ . Consider any node  $u$ . We have that the weights at this node are equal to
$$
\sigma_ {u} = \frac {c _ {b _ {u}} | R _ {u} | + \sum_ {s \in A _ {t}} v _ {b _ {u}} ^ {s}}{| A _ {t} |} \leq \frac {c _ {b} | R _ {u} | + \sum_ {s \in A} v _ {b} ^ {s}}{| A |}
$$
where the last inequality holds for all  $A \subseteq R_u$  and any  $b \in \mathcal{B}$ .