SlowGuess/NeurIPS_2023_OCR huggingface dataset:
where the last inequality is due to  $\tau \leq \lambda$ . Recall that  $1 \leq \lambda \leq 2$  and  $\alpha \in (2, \frac{\sqrt{17} + 1}{2})$ . So the R.H.S. of Eqn. (24) is minimized when  $\lambda = 2$ . Therefore, we get
$$
f (S) \geq f (H) - f \left(u ^ {*}\right) \geq (\alpha + 1 - 4 / \alpha) f \left(u ^ {*}\right) \tag {25}
$$
Combining Eqn. (18), Eqn. (21) and Eqn. (25), we get
$$
\mathrm {O P T} \leq 2 f (S) + (2 + \alpha) f \left(u ^ {*}\right) \leq \left(2 + \frac {2 + \alpha}{\alpha + 1 - 4 / \alpha}\right) f (S) \tag {26}
$$
3.  $|K \cap H| = 1$ : In this case, we must have  $A = K \cup S$ ,  $K \cap S = \emptyset$  and  $H \setminus K = S$  according to the definition of  $H$ . Moreover, due to Eqn. (22) we have  $f(u^{*}) \geq \delta(K \cap H) \geq \rho \alpha f(u^{*})$  and hence  $\rho \leq 1 / \alpha$ . Note that all the elements in  $S$  is selected using the pricing rule of Algorithm 2. Therefore, using similar reasoning as Lemma 1, we get
$$
f (S) \geq (1 - \rho) \left(f (K) + \alpha \cdot f \left(u ^ {*}\right)\right) \geq (1 - 1 / \alpha) (\lambda + \alpha) f \left(u ^ {*}\right) \tag {27}
$$
Recall that  $\lambda \leq 2$  and  $A = K\cup S$ . Combining this with Eqn. (18) and Eqn. (27), we get
$$
\begin{array}{l} \mathrm {O P T} \leq 2 f (A) + \alpha f (u ^ {*}) \leq 2 f (K) + 2 f (S) + \alpha f (u ^ {*}) \\ = 2 f (S) + (2 \lambda + \alpha) f \left(u ^ {*}\right) \\ \leq \left(2 + \frac {2 \lambda + \alpha}{(1 - 1 / \alpha) (\lambda + \alpha)}\right) f (S) \\ = \left(2 + \frac {2}{1 - 1 / \alpha} - \frac {\alpha}{(1 - 1 / \alpha) (\lambda + \alpha)}\right) f (S) \\ \leq \left(2 + \frac {4 + \alpha}{\alpha + 1 - 2 / \alpha}\right) f (S) \tag {28} \\ \end{array}
$$
The lemma then follows according to the above discussion.
It can be easily seen that Algorithm 4 offers one price to each seller and incurs at most  $3n$  value oracle queries ( $n$  queries for finding  $u^{*}$  and  $2n$  queries for building  $A_{1}$  and  $A_{2}$ ). The approximation ratio of Algorithm 4 can be proved by directly combining Lemma 7 and Lemma 8:
Lemma 7 If Line 10 of Algorithm 4 is executed, then we have  $\mathrm{OPT} \leq 12f(S)$  when  $\alpha = 3$ .
Proof: In this case, all the elements in  $O$  must have been processed by Lines 3-6. Let  $O_{1}$  (resp.  $O_{2}$ ) denote the sellers in  $O$  who reject the offered prices that are calculated using  $A_{1}$  (resp.  $A_{2}$ ) in Line 4. For convenience, we define  $\delta_{i}(u) = f(u \mid A_{i,u})$  for any  $u \in \mathcal{N}$ , where  $A_{i,u}$  denotes the set of elements already in  $A_{i}$  at the moment that  $u$  is processed by Algorithm 4. It can be seen that, for each  $u \in A_{i} (i \in \{1,2\})$ , we have  $\delta_{i}(u) \geq 0$ , because otherwise we have  $p(u) = \frac{B\delta_{i}(u)}{f(A_{i,u}) + \alpha f(u^{*})} < 0$  and hence  $p(u)$  should be rejected by  $u$  due to  $c(u) \geq 0$ , which implies that  $u$  is not in  $A_{i}$ ; a contradiction. So we have  $f(X) \leq f(A_{i})$  for any prefix  $X$  of  $A_{i}$ .
In the following, we consider the sets  $A_{1}, A_{2}, A^{*}$  after Line 10 of Algorithm 4 is executed. For each  $i \in \{1, 2\}$  and each  $u \in O_{i}$ , we can use similar reasoning as Lemma 1 to get
$$
\frac {f (u \mid A _ {i})}{c (u)} \leq \frac {\delta_ {i} (u)}{c (u)} <   \frac {f \left(A _ {i , u}\right) + \alpha f \left(u ^ {*}\right)}{B} \leq \frac {f \left(A _ {i}\right) + \alpha f \left(u ^ {*}\right)}{B} \leq \frac {f \left(A ^ {*}\right) + \alpha f \left(u ^ {*}\right)}{B} \tag {29}
$$
Moreover, according to submodularity and the greedy rule of Line 3, for all  $i,j\in \{1,2\}$  and  $i\neq j$  we have
$$
\sum_ {u \in A _ {i} \cap O} f (u \mid A _ {j}) \leq \sum_ {u \in A _ {i} \cap O} \delta_ {j} (u) \leq \sum_ {u \in A _ {i} \cap O} \delta_ {i} (u) \leq \sum_ {u \in A _ {i}} \delta_ {i} (u) \leq f (A _ {i}) \leq f \left(A ^ {*}\right), \tag {30}
$$
where the third inequality is due to  $\delta_i(u) \geq 0$  for all  $u \in A_i$ . Using the above inequalities and submodularity, we get
$$
\begin{array}{l} f (O \cup A _ {1}) - f (A _ {1}) \leq \sum_ {u \in O \cap A _ {2}} f (u \mid A _ {1}) + \sum_ {u \in O _ {1} \cup O _ {2}} f (u \mid A _ {1}) \\ \leq f \left(A ^ {*}\right) + \sum_ {u \in O _ {1}} \delta_ {1} (u) + \sum_ {u \in O _ {2}} \delta_ {1} (u) \leq f \left(A ^ {*}\right) + \sum_ {u \in O _ {1}} \delta_ {1} (u) + \sum_ {u \in O _ {2}} \delta_ {2} (u) \\ \leq f \left(A ^ {*}\right) + \left[ c \left(O _ {1}\right) + c \left(O _ {2}\right) \right] \cdot \frac {f \left(A ^ {*}\right) + \alpha f \left(u ^ {*}\right)}{B} \leq 2 f \left(A ^ {*}\right) + \alpha f \left(u ^ {*}\right) \tag {31} \\ \end{array}
$$
where the third inequality is due to the greedy rule in Line 3. Similarly, we get
$$
f \left(O \cup A _ {2}\right) - f \left(A _ {2}\right) \leq 2 f \left(A ^ {*}\right) + \alpha f \left(u ^ {*}\right) \tag {32}
$$
Combining Eqns. (31)-(32) gives us
$$
\mathrm {O P T} \leq \sum_ {i \in \{1, 2 \}} f (O \cup A _ {i}) \leq 6 f \left(A ^ {*}\right) + 2 \alpha f \left(u ^ {*}\right), \tag {33}
$$
where the first inequality is due to submodularity and  $A_1 \cap A_2 = \emptyset$ . Therefore, if  $p(A^*) \leq B$ , then we have  $S = A^*$ ,  $f(S) \geq f(u^*)$  and hence
$$
\mathrm {O P T} \leq 6 f (S) + 2 \alpha f \left(u ^ {*}\right) \leq (6 + 2 \alpha) f (S) \tag {34}
$$
If  $p(A^{*}) > B$ , then let  $H$  be the smallest suffix of  $A^{*}$  satisfying  $p(H) > B$ , and we can use similar reasoning as Lemma 1 to get
$$
f (S) + f \left(u ^ {*}\right) \geq f (H) \geq f \left(A ^ {*} \backslash H\right) + \alpha f \left(u ^ {*}\right) \tag {35}
$$
Combining Eqn. (33) and Eqn. (35), we get
$$
\begin{array}{l} \mathrm {O P T} \leq 6 f (H) + 6 f \left(A ^ {*} \backslash H\right) + 2 \alpha f \left(u ^ {*}\right) \\ \leq 6 \left[ f (S) + f \left(u ^ {*}\right) \right] + 6 \left[ f (S) + f \left(u ^ {*}\right) - \alpha f \left(u ^ {*}\right) \right] + 2 \alpha f \left(u ^ {*}\right) \\ \leq 1 2 f (S) + (1 2 - 4 \alpha) f \left(u ^ {*}\right) \tag {36} \\ \end{array}
$$
The lemma then follows by combining Eqn. (34) and Eqn. (36).
Lemma 8 If Lines 12-19 of Algorithm 4 are executed, then we have  $\mathrm{OPT} \leq 12\mathbb{E}[f(S)]$  when  $\alpha = 3$ .
Proof: For clarity, we use  $J_{1}$  (resp.  $J_{2}$ ) to denote the set  $A_{1}$  (resp.  $A_{2}$ ) at the moment that Algorithm 4 executes Line 12, and define  $J^{*} = \arg \max_{X\in \{J_{1},J_{2}\}}f(X)$ . So we have  $f(J^{*}) < f(u^{*})$  due to Line 7. It can be seen from Line 18 that  $u^{*}$  is added into  $A_{q}$  only if (1)  $\max_{i\in \{1,2\}}f(u^{*}|J_{i})\geq 0$  and (2)  $u^{*}$  accepts the price of  $B - p(J_q)$ . Note that the set  $T$  generated by Line 19 must satisfy  $f(T)\geq f(J^{*})$ . In the sequel, we prove that  $\mathrm{OPT}\leq 6f(T) + 2\alpha f(u^{*})$  by a discussion:
1.  $u^{*} \notin O$ : In this case, we can use similar reasoning as Lemma 7 to get
$$
\mathrm {O P T} = f (O \setminus \{u ^ {*} \}) \leq 6 f (J ^ {*}) + 2 \alpha f (u ^ {*}) \leq 6 f (T) + 2 \alpha f (u ^ {*}) \tag {37}
$$
2.  $u^{*} \in O$  and  $u^{*}$  is added into  $A_{q}$  in Line 18; or  $u^{*} \in O$  and  $\max_{i \in \{1,2\}} f(u^{*} \mid J_{i}) < 0$ : In these situations, after Line 19 is executed, we can use similar reasoning as Lemma 7 to get
$$
\forall i \in \{1, 2 \}: f (O \cup A _ {i}) - f (A _ {i}) \leq 2 f (T) + \alpha f \left(u ^ {*}\right) \tag {38}
$$
Note that  $f(A_{i}) \leq f(T)$  due to Line 19. So we get
$$
\mathrm {O P T} \leq \sum_ {i \in \{1, 2 \}} f (O \cup A _ {i}) \leq 6 f (T) + 2 \alpha f \left(u ^ {*}\right) \tag {39}
$$
3.  $u^{*} \in O$  and  $\max_{i \in \{1,2\}} f(u^{*} \mid J_{i}) \geq 0$  and  $\{u^{*}\}$  rejects the price in Line 18: In this case, we must have  $c(u^{*}) > B - p(J_{q})$  and hence
$$
c \left(O \backslash \left\{u ^ {*} \right\}\right) \leq p \left(J _ {q}\right) \leq \frac {B \cdot f \left(J _ {q}\right)}{f (\emptyset) + \alpha f \left(u ^ {*}\right)} \leq \frac {B f \left(J _ {q}\right)}{\alpha f \left(u ^ {*}\right)} \leq \frac {B f \left(J ^ {*}\right)}{\alpha f \left(u ^ {*}\right)} \leq \frac {B}{\alpha} \tag {40}
$$
where the second inequality is got by Lemma 1. Note that all the elements in  $O \setminus \{u^{*}\}$  has been processed by Lines 3-6. Let  $O_{1}$  (resp.  $O_{2}$ ) denote the set of sellers in  $O \setminus \{u^{*}\}$  who reject the offered prices that are calculated using  $A_{1}$  (resp.  $A_{2}$ ) in Line 4. By similar reasoning with Lemma 7, we get
$$
\forall i \in \{1, 2 \}, \forall u \in O _ {i}: \frac {f (u \mid J _ {i})}{c (u)} \leq \frac {\delta_ {i} (u)}{c (u)} \leq \frac {f (J ^ {*}) + \alpha f (u ^ {*})}{B}; \tag {41}
$$
$$
\forall i, j \in \{1, 2 \} \wedge i \neq j: \sum_ {u \in O \backslash \left\{u ^ {*} \right\} \cap J _ {i}} f (u \mid J _ {j}) \leq f \left(J ^ {*}\right); \tag {42}
$$
where  $\delta_i(u)$  is defined in the same way as that in the proof of Lemma 7. Therefore, we can use Eqns. (40)-(42) to get
$$
\begin{array}{l} f \left(O \backslash \{u ^ {*} \} \cup J _ {1}\right) - f \left(J _ {1}\right) \leq \sum_ {u \in O \backslash \{u ^ {*} \} \cap J _ {2}} f (u \mid J _ {1}) + \sum_ {u \in O _ {1} \cup O _ {2}} f (u \mid J _ {1}) \\ \leq f \left(J ^ {*}\right) + \sum_ {u \in O _ {1}} \delta_ {1} (u) + \sum_ {u \in O _ {2}} \delta_ {2} (u) (43) \\ \leq f (J ^ {*}) + c \left(O \backslash \{u ^ {*} \}\right) \cdot \frac {f (J ^ {*}) + \alpha f \left(u ^ {*}\right)}{B} \\ \leq \left(1 + \frac {1}{\alpha}\right) f \left(J ^ {*}\right) + f \left(u ^ {*}\right), (44) \\ \end{array}
$$
where Eqn. (43) is due to submodularity and the greedy rule in Line 4. Similarly, we get
$$
f \left(O \backslash \left\{u ^ {*} \right\} \cup J _ {2}\right) - f \left(J _ {2}\right) \leq \left(1 + \frac {1}{\alpha}\right) f \left(J ^ {*}\right) + f \left(u ^ {*}\right) \tag {45}
$$
Combining the above equations, we get
$$
f (O \setminus \{u ^ {*} \}) \leq \sum_ {i = 1} ^ {2} f (O \setminus \{u ^ {*} \} \cup J _ {i}) \leq 2 \left(2 + \frac {1}{\alpha}\right) f (J ^ {*}) + 2 f \left(u ^ {*}\right) \tag {46}
$$
$$
\mathrm {O P T} \leq f (O \setminus \{u ^ {*} \}) + f (u ^ {*}) \leq 2 \left(2 + \frac {1}{\alpha}\right) f (J ^ {*}) + 3 f (u ^ {*}), \tag {47}
$$
Therefore, when  $\alpha \geq 1.5$ , we have  $\mathrm{OPT} \leq 6f(J^{*}) + 2\alpha f(u^{*}) \leq 6f(T) + 2\alpha f(u^{*})$ .
According to Lines 12-19 of Algorithm 4, the algorithm returns  $T$  or  $\{u^{*}\}$  with probability of  $\frac{3}{3 + \alpha}$  or  $\frac{\alpha}{3 + \alpha}$ , respectively. So we get
$$
\begin{array}{l} \mathbb {E} [ f (S) ] = \Pr [ S = T ] \cdot f (T) + \Pr [ S = \{u ^ {*} \} ] \cdot f (u ^ {*}) \\ = \frac {3 f (T)}{3 + \alpha} + \frac {\alpha f \left(u ^ {*}\right)}{3 + \alpha} \geq \frac {\mathrm {O P T}}{1 2} \tag {48} \\ \end{array}
$$
when  $\alpha = 3$ . This completes the proof.
In this section, we conduct experiments using a crowdsourcing application similar to that in [8, 16, 20, 22], where each worker (seller) in  $\mathcal{N}$  owns an image, and a buyer with a budget  $B$  needs to crowdsource a set  $S$  of representative images from the workers. Following [8, 16, 20, 22], we use the CIFAR-10 dataset [8, 16, 20, 22] containing ten thousands  $32 \times 32$  color images, where each image is associated with a label denoting its category such as "Airplane", and the valuation (utility) of  $S$  is calculated as
$$
f (S) = \sum_ {u \in \mathcal {N}} \max  _ {v \in S} s _ {u, v} - \frac {1}{| \mathcal {N} |} \sum_ {u \in S} \sum_ {v \in S} s _ {u, v}, \tag {49}
$$
where  $s_{u,v}$  denotes the similarity between any two images  $u$  and  $v$  and is measured by the cosine similarity of the 3,072-dimensional pixel vectors of image  $u$  and image  $v$ . Intuitively, the first and second additive factors in Eqn. (49) represent the "coverage" and "diversity" of  $S$ , respectively, and it is known that such a valuation function  $f(\cdot)$  is non-monotone and submodular [16]. We assume that each worker  $u$  has a private cost  $c(u)$  for providing its image, and set  $c(u)$  to be in proportional to the standard deviation of its pixel intensities. Following the same setting as that in [22], the costs of all images are normalized with the average value of 0.1, and the images with labels in {Airplane, Automobile, Bird} are considered. The other experimental settings are the same with those in Fig. 1.
Figure 2: Comparing the BFMs for the crowdsourcing application
In Fig. 2, we compare our TripleEagleNm (TENM) algorithm with two state-of-the-art BFMs for non-monotone submodular valuations: (1) the Simultaneous-Iterative-Pruning (SIP) algorithm in
[9] with an approximation ratio of 64; and (2) the RandomClockAuction (RCA) algorithm in [22] with an approximation ratio of  $(3 + \sqrt{5})^2$ . The experimental results in Fig. 2 show that, our TENM algorithm consistently outperforms the other two baselines on utility, while incurring about three orders of magnitude fewer value oracle queries than the SIP algorithm. It is also noted that TENM and RCA perform similarly on the number of value oracle queries; this can be explained by the fact that RCA has a nearly-linear query complexity of  $\mathcal{O}(n\log n)$  and hence is also efficient for this problem instance (but with poorer performance on utility than TENM). These experimental results demonstrate the superiorities of our approach once again for non-monotone submodular valuations.
The code of our paper can be found at:
https://anonymous.4open.science/r/TripleEagle-4D1B/README.md