SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \operatorname {div}(\operatorname {SOL})\geq \sum_{x,y\in \operatorname {OPT}:  \operatorname {dist}(x,y)\geq 5\max \{r_{x},r_{y}\}}\operatorname {dist}(\mu (x),\mu (y)) \\ \geq \sum_ {x, y \in \mathrm {O P T}: \operatorname {d i s t} (x, y) \geq 5 \max  \{r _ {x}, r _ {y} \}} \operatorname {d i s t} (x, y) - \operatorname {d i s t} (x, \mu (x)) - \operatorname {d i s t} (y, \mu (y)) \\ \geq \sum_ {\substack {x, y \in \mathrm {OPT}: \operatorname {dist} (x, y) \geq 5 \max  \{r _ {x}, r _ {y} \}}} \operatorname {dist} (x, y) - 2 r _ {x} - 2 r _ {y} \\ \geq \sum_ {x, y \in \mathrm {O P T}: \operatorname {d i s t} (x, y) \geq 5 \max  \{r _ {x}, r _ {y} \}} \operatorname {d i s t} (x, y) / 5 \\ \geq B / 5 \geq \operatorname {d i v} (\mathrm {O P T}) / 1 0 \\ \end{array}
$$
A.2 Proof of Observation 3.1
Proof. One can brute force on each possible subset  $B'$ , and count the number of points from each color that falls in  $B'$ , and thus compute the number of remaining points for each color, i.e.,  $|P_{\ell} \setminus B'|$  in  $O(|B'| \cdot m)$  time and check whether the number exceeds  $k_{\ell}$ . So the total runtime is at most  $\sum_{z=1}^{k_i} \left( \binom{t(j)}{z} \cdot z \cdot m \right) \leq m \cdot k_k^{k_i}$ .
Proof. First one can check if there exists a subset  $B_0'$  of size  $2 = O(1)$  in polynomial time using Approach 1.
Next we construct a solution  $B_{1}$ . For each  $\ell \in [m]$ , let  $d_{\ell} = k_{\ell}$  be the demand of color  $\ell$ , that is the total number of points of that color which we want to exist outside of our solution  $B_{1}$ . First WLOG we assume that all points of  $P$  lie inside one of the balls in  $B$  (if this is not the case, one can subtract the number of points of each color that are outside of all the balls from the demands  $d_{\ell}$ ). Then we start with  $B_{1} = B$  and repeat the following process as long as there are non-zero demands left.
- We let  $A$  be an arbitrary subset of half of the balls in  $B_1$  and let  $\bar{A} = B_1 \setminus A$ .
- Note that for each color  $\ell \in [m]$  with non-zero demand  $d_{\ell} > 0$ , either  $A$  or  $\bar{A}$  contains at least  $d_{\ell}$  points of color  $\ell$  in them. This is initially true as one of  $A$  or  $\bar{A}$  should contain half of the points that is  $n_{\ell} / 2 \geq k_{\ell} \geq d_{\ell}$ . We show this remains true at the end of the process.
- Let  $t_A$  be the number of colors  $\ell \in [m]$  with non-zero demands (i.e.,  $d_\ell > 0$ ) such that  $A$  contains more points of color  $\ell$  than  $\bar{A}$ , and define  $t_{\bar{A}}$  similarly as the number of colors  $\ell \in [m]$  with non-zero demands such that  $\bar{A}$  contains more points of color  $\ell$  than  $A$ .
- Now we let  $B_{1} = A$  if  $t_{\bar{A}} \geq t_{A}$  and let  $B_{1} = \bar{A}$  otherwise. This ensures that the demand of at least half of the colors with non-zero demand has been satisfied as the set we are excluding from  $B_{1}$  satisfies the demand of at least half of the colors with non-zero demands. This means that the total number of iterations is at most  $\log m$ .
- We will update the demands of all the colors based on the set that was excluded from  $B_{1}$ . WLOG assume that  $t_{A} \geq t_{\bar{A}}$ . Then for each  $\ell$  that  $d_{\ell} > 0$  we set  $d_{\ell} = \max \{0, d_{\ell} - |A \cap P_{\ell}|\}$ .
Let  $\ell$  be a color that has still a non-zero demand. Note that this means that  $|A \cap P_{\ell}| \leq d_{\ell}$ . Therefore we will still have the property that  $|\bar{A} \cap P_{\ell}| = |(B_1 \setminus A) \cap P_{\ell}| \geq 2d_{\ell} -$
$|A \cap P_{\ell}| \geq 2(d_{\ell} - |A \cap P_{\ell}|)$ , where here we used  $B_{1}$  and  $d_{\ell}$  to denote their values before we apply the update corresponding to this iteration.
Finally, we return the best of  $B_0$  and  $B_1$  as  $B'$ , i.e., the one with maximum size. Note that if  $\mathrm{OPT} \leq 2m$ , then  $B_0$  itself is an  $O(m)$  approximation. Otherwise, if  $\mathrm{OPT} > 2m$ , then as the total number of iterations to obtain  $B_1$  is at most  $\log m$ , we have  $|B'| = |B_1| \geq |B| / 2^{\log m} = |B| / m \geq \mathrm{OPT} / m$ .
A.4 Proof of Claim 3.3
Proof. Divide the interval  $[1, k_i]$  into logarithmically many numbers of intervals such that for each  $1 \leq \ell \leq \lceil \log (k_i + 1) \rceil$ ,  $I_\ell = [2^{\ell - 1}, \min \{2^\ell, k_i + 1\})$ . Then for each  $\ell$ , we have that
$$
\sum_ {j \in I _ {\ell}} d _ {j} ^ {i} \leq 2 ^ {\ell - 1} \cdot d _ {2 ^ {\ell - 1}} ^ {i} \leq j ^ {*} (i) \cdot d _ {j ^ {*} (i)} ^ {i}
$$
and therefore,  $\sum_{j\leq k_i}d_j^i\leq \lceil \log (k_i + 1)\rceil \cdot j^* (i)\cdot d_{j^* (i)}^i$
Proof. First note that, it can be easily checked that the algorithm runs in polynomial time by Lemma 3.2. Next, consider the iteration of Algorithm 3 corresponding to  $i$  and  $j$  where  $i = i^{*}$  and  $j$  is chosen as follows,
1.  $j$  is the largest value in the range  $j^{*} \leq j \leq k$  such that  $r_{j} \geq d_{j^{*}}^{i^{*}}$ .
2. If no such  $j$  exists, we let  $j = j^{*}$ .
3. Finally, if  $j^* < j < k$  and  $r_{j+1} \geq d_{j*}^{i^*} / 2$ , increase  $j$  by one, i.e.,  $j = j + 1$ .
In all of the above cases, either by Property 3 of the GMM algorithm, or by definition, we have that  $r_j \geq d_{j*}^i / 2$ . Therefore,  $r_{t(j)} \geq d_{j*}^i / 4$ . Therefore the balls in  $B$  defined in Line 7 of the algorithm, have radius at least  $d_{j*}^i / 8$ .
Claim A.2. Let  $B'' \subseteq B$  be the maximum size subset of the balls s.t.  $P \setminus B''$  contains  $k_{i} - |B''|$  points of color  $i$ , and at least  $k_{\ell}$  number of points from color  $\ell$ , for each  $\ell \neq i$ . Then  $|B''| \geq j^{*}$ .
Proof. Note that  $\mathrm{OPT}_{i^*}$  contains a subset  $R$  of  $j^*$  points from  $P_{i^*}$  such that there are no other points from OPT in the ball of radius  $d_{j^*}^{i^*}$  around the points in  $R$ . Now, we consider three cases separately.
Case 1:  $j = k$ . Note that in this case we also have that  $t(j) = k$ . Therefore,  $B$  contains  $k$  disjoint balls. Obtain  $B''$  from  $B$  as follows. First set  $B'' = B$ . Then for each color  $\ell \neq i$ , remove at most  $k_{\ell}$  balls from  $B''$  that contain the points of color  $\ell$  in the optimal solution. Since  $B$  has size  $k = \sum_{\ell} k_{\ell}$ , in the end  $B''$  will have size at least  $k_{i} \geq j^{*}$ .
Case 2:  $j < k$  and  $t(j) = k$ . Again in this case  $B$  contains  $k$  disjoint balls and one can obtain  $B''$  with size at least  $j^*$  similar to the above case.
Case 3:  $j < k$  and  $t(j) < k$ . For this case, let us first prove the following simple claim.
Claim A.3.  $r_{t(j) + 1} <   d_{j^*}^{t^*} / 2$
Proof. First note that by the definition of  $t(j)$ , we know that  $r_{t(j) + 1} < r_j / 2$ . Now if we have  $r_j \leq d_{j^*}^{i^*}$  (this happens when we increased the value of  $j$  in the last step), then we have that  $r_{t(j) + 1} < r_j / 2 \leq d_{j^*}^{i^*} / 2$  and the claim is proved. Otherwise, it means that  $r_{j + 1} < d_{j^*}^{i^*} / 2$  (since we did not increase  $j$  in Item 3). But then since  $t(j) \geq j$ , we have that  $t(j) + 1 \geq j + 1$ . Therefore,  $r_{t(j) + 1} < d_{j^*}^{i^*} / 2$  and the claim is proved again.
Therefore, all the points in  $P$  lie within a distance of at most  $r_{t(j) + 1} < d_{j^*}^{i^*} / 2$  of one of the centers of the balls in  $B$ . For each point  $q \in R$  let  $N(q)$  be the closest center among the centers of  $B$  to  $q$ . Note that for two points  $q_1, q_2 \in R$ ,  $N(q_1) \neq N(q_2)$  as otherwise their distance is less than  $2r_{t(j) + 1} < d_{j^*}^{i^*}$  contradicting the definition of the set  $R$ . Let  $B''$  be the subset of the  $j^*$  balls centered at  $N(R)$ . Note that for any other point in the optimal solution, i.e.,  $q \in \mathrm{OPT} \setminus R$ , it cannot lie inside any of the balls in the subset  $B''$ , as again it contradicts the definition of  $R$ . This means that the set  $P \setminus B''$  contains enough points: i.e.,  $k_i - |R|$  points from color  $i$ , and  $k_\ell$  points from color  $\ell$  for each  $\ell \neq i$ .
Therefore, having Claim A.2 and using Lemma 3.2, we get that  $|B'| \geq j^* / m$ . Now, let  $S_1, \ldots, S_m$  be the solution created at the iteration corresponding to  $i$  and  $j$  as specified above. That is,  $S_i$  contains the centers of the balls in  $B'$  returned by Approach 2, together with an arbitrary set of  $k_i - |B'|$  points from  $P_i \setminus B'$ . Further, for each  $\ell \neq i$ , let  $S_\ell$  contain an arbitrary set of  $k_\ell$  points from  $P_\ell \setminus B'$ . This solution contains a set of  $|B'| \geq j^* / m$  points such that no other point in  $S$  lies within a distance of  $r_{t(j)} / 2 \geq d_{j*}^{i^*} / 8$ . Now, to see the approximation, note that
$$
\begin{array}{l} \operatorname {d i v} (\mathrm {S O L}) \geq \operatorname {d i v} (S) \\ \geq \left(j ^ {*} / m\right) \cdot d _ {j ^ {*}} ^ {i ^ {*}} / c _ {1} \quad \text {b y w h a t w e j u s t p r o v e d} \\ \geq \frac {1}{m \cdot c _ {1} \cdot \lceil \log \left(k _ {i ^ {*}} + 1\right) \rceil} \sum_ {j ^ {\prime} \leq k _ {i ^ {*}}} d _ {j ^ {\prime}} ^ {*} \quad \text {b y} \\ \geq \frac {1}{m ^ {2} \cdot c _ {1} \cdot \lceil \log \left(k _ {i ^ {*}} + 1\right) \rceil} \sum_ {i ^ {\prime} \leq m} \sum_ {j ^ {\prime} \leq k _ {i ^ {\prime}}} d _ {j ^ {\prime}} ^ {i ^ {\prime}} \quad \text {b y} \\ = \frac {1}{m ^ {2} \cdot c _ {1} \cdot \lceil \log k \rceil} \cdot \operatorname {d i v} (\mathrm {O P T}) \quad \text {b y} d _ {j ^ {\prime}} ^ {i ^ {\prime}}, \\ \end{array}
$$
A.6 Proof of Theorem 4.1
Proof. We show the lemma by comparing the result of running Algorithm 3 on the union of the core-sets  $S = \bigcup_{i} S_{i}$ , and running it on the whole dataset  $P = \bigcup_{i} P_{i}$ . In particular, in Line 3 of Algorithm 3, running the GMM algorithm on  $S_{i}$  would return the same set of points as running it on  $P_{i}$ , as  $S_{i}$  itself contains  $G_{i}$ . Therefore, both runs of the algorithm end with the same set of balls  $B$ . We will now show the following claim.
Claim A.4. Let  $B$  be a set of at most  $k$  disjoint balls of radius  $r/2$  such that there exists a subset  $B' \subseteq B$  of size at most  $|B'| \leq k$  of them such that  $P_{\ell} \setminus B'$  contains  $k_{\ell}$  points. Further let  $C$  be a set of  $|B|$  balls with the same centers as  $B$  but with radius  $r/6$  instead, and let  $C' \subseteq C$  be the subset of them corresponding to  $B'$ . Then  $S_{\ell} \setminus C'$  contains  $k_{\ell}$  points.
Proof. Note that  $S_{\ell}$  is the result of running the GMM algorithm  $k$  times on  $P_{\ell}$  each for  $k + 1$  iterations. Now, for  $i \leq k$ , let us use  $T^{i}$  to denote the set of  $k + 1$  points returned by the  $i$ th run of GMM. Thus,  $T^{i}$  is the result of running GMM on the set  $P_{\ell} \setminus \bigcup_{j < i} T^{j}$ . Let  $r^{i}$  be the minimum pairwise distance between the points in  $T^{i}$ . Now if for all  $i \leq k_{\ell}$ ,  $T^{i}$  contains at least one point outside of the balls in  $C'$  then clearly  $T_{\ell}$  contains  $k_{\ell}$  points outside  $C'$  and the claim is proved.
Otherwise consider the first iteration  $i$  such that  $T^i \subset C'$ . This means that since  $|T^i| = k + 1$  and  $|C'| = |B'| \leq k$ , there exists a pair of points in  $T^i$  whose distance is at most  $2 \cdot r/6 = r/3$ . But then by Property 2 of GMM, this means that all the other points in  $P_\ell \setminus \bigcup_{j \leq i} T^i$  are within a distance of  $r/3$  from one of the balls in  $C'$  and therefore, they are all within  $B'$ . This means that  $P_\ell \setminus B'$  can only contain points from  $\bigcup_{j < i} T^j \subseteq S_\ell$ , and thus the lemma is proved.
This means that if we replace the set of balls  $B$  in Algorithm 3 with  $C$ , both Approach 1 and Approach 2, will be able to find the set  $C'$  (either exactly or approximately depending on which Approach we use). This means that corresponding to each solution SOL found by Algorithm 3.2, there will be a corresponding solution in the core-set of diversity at least  $\mathrm{div}(\mathrm{SOL}) / 3$  (as the radius of the balls in
$C$  is  $1/3$  of the radius of the balls in  $B$ . Therefore, as Theorem 3.5 shows that using approach 1, one gets a solution with approximation factor of  $O(m \cdot \log k)$ , then this means that Algorithm 4 provides a composable core-set of size  $O(k^2)$  for each color, with an approximation factor of  $O(m \cdot \log k)$ .
B Generalization of Section 2 to the case where  $k_{i}\geq 1$
In Section 2 of the paper, for simplification of the proof, we assumed that for all  $i$ , we have  $k_{i} > 1$ . Here, we remove this assumption, and show that a simple modification of the algorithm works even if for some  $i$ 's we have  $k_{i} = 1$ . The modified algorithm is shown in Algorithm 5. The only modification made is that in Line 1, the GMM will be run for at least two iterations even if  $k_{i} = 1$ .
Algorithm 5 Core-set Construction Algorithm for SUM-PAIRWISE
Input a point set  $P_{i}$ , together with parameters  $k_{i}$  and  $k$  (where  $k = k_{1} + \dots + k_{m}$ )  
Output a subset  $S_{i} \subseteq P_{i}$
1:  $S_{i} = \{p_{1},\dots ,p_{k_{i}}\} \gets \mathrm{GMM}(P_{i},\max \{k_{i},2\})$
2:  $T\gets \emptyset$
3: for  $p \in S_i$  do
4: for  $j = 1$  to  $k_{i}$  do
5:  $T\gets T\cup$  the closest point  $p_j\in P_i\setminus T$  to p s.t.  $\mathrm{argmin}_{q\in S_i}\mathrm{dist}(p_j,q) = p.$
6: end for
7: end for
8:  $S_{i}\gets S_{i}\cup T$
9: return  $S_{i}$
Theorem B.1. Algorithm 5 produces a core-set with size  $O(k_i^2)$  and a constant factor approximation:  $\mathrm{div}_{k_1,\dots,k_m}(S) \geq \frac{1}{C} \cdot \mathrm{div}_{k_1,\dots,k_m}(P)$  for a constant  $c$ .
Proof. Again it is easy to see that the size of each  $S_{i}$  is at most  $k_{i}^{2}$ , and we only need to show that the approximation factor is  $O(1)$ . We use the same notation as in the proof of Theorem 2.1. Note the only modification is that now for colors  $i$  with  $k_{i} = 1$ , we have that  $r_{i}$  is the distance between the two points in  $S_{i}$  returned by  $\mathrm{GMM}(P_{i},\max \{k_{1},2\})$  in Line 1 of the algorithm. Again, we divide the optimal value of the diversity into two parts  $A$  and  $B$  defined as before. Again we consider the two cases separately. The case of  $A < B$  (i.e., Case 2 in the proof of Theorem 2.1), holds similar to before. So we only need to prove the case of  $A \geq B$ , which we show below.
In this case as before, we have that  $\mathrm{div}(\mathrm{OPT})\leq 20\cdot k\sum_{i\in [m]}k_i r_i$
Now let us divide the colors into two groups based on the value of their  $k_{i}$ , i.e., let  $G = \{i\colon k_i > 1\}$  and thus  $[m]\setminus G$  will contain colors  $i$  with  $k_{i} = 1$ . WLOG, let us rename the colors so that the first  $|G|$  colors of  $[m]$  are the colors in  $G$ , and moreover, colors  $|G| + 1$  to  $m$  are sorted such that  $r_{j_1}\leq r_{j_2}$  whenever  $|G| + 1\leq j_{1} < j_{2}\leq m$ . Finally, we define  $t = \sum_{j\leq |G|}k_j$  to be the total number of points in any solution that are from groups with at least one point in them.
Constructing the solution. Now let us define a solution as follows.
- For  $i \leq |G|$ , define  $\mathrm{SOL}_i \subseteq S_i$  as before, to be the first set of  $k_i$  points added to  $S_i$  in Line 1 of the algorithm using GMM. Let  $\mathrm{SOL}^G = \bigcup_{i \in G} \mathrm{SOL}_i$  be the union of all these solutions.
- To define  $\mathrm{SOL}_i$  for  $|G| + 1 \leq i \leq m$ , note that  $k_i = 1$ , and thus we need to decide which of the two points added in Line 1 of the algorithm to  $S_i$  should be picked in  $\mathrm{SOL}_i$ . Note that these two points are at distance exactly  $r_i$ .
We go over the colors  $|G| + 1$  to  $m$  one by one and decide which of the two points to choose for  $\mathrm{SOL}_i$ . When at color  $i$ , let  $p_i, p_i' \in S_i$  be the two points returned in Line 1 of the algorithm using GMM which are at distance  $r_i$  from each other. We let  $\mathrm{SOL}_i = \{p_i\}$  if  $\sum_{x \in \bigcup_{j < i} \mathrm{SOL}_j} \mathrm{dist}(x, p_i) \geq \sum_{x \in \bigcup_{j < i} \mathrm{SOL}_j} \mathrm{dist}(x, p_i')$ , and let  $\mathrm{SOL}_i = \{p_i'\}$  otherwise.
This way of constructing the solution gives us the property that
$$
\sum_ {x \in \bigcup_ {j <   i} \mathrm {S O L} _ {j}} \operatorname {d i s t} (x, \mathrm {S O L} _ {i}) \geq \frac {r _ {i}}{2} \cdot (t + i - 1 - | G |)
$$
where we abused the notation  $\mathrm{SOL}_i$  to refer to the only point in  $\mathrm{SOL}_i$ .
Analysis. To show that the constructed solution has large diversity, let us divide the participating distances into three parts as below.
$$
\operatorname {div} (\mathrm {S O L}) = \sum_ {x, y \in \mathrm {S O L} ^ {G}} \operatorname {d i s t} (x, y) + \sum_ {x \in \mathrm {S O L} ^ {G}, y \in \mathrm {S O L} \backslash \mathrm {S O L} ^ {G}} \operatorname {d i s t} (x, y) + \sum_ {x, y \in \mathrm {S O L} \backslash \mathrm {S O L} ^ {G}} \operatorname {d i s t} (x, y)
$$
Let us refer to the above three terms as  $\mathrm{div}_1$ ,  $\mathrm{div}_2$ , and  $\mathrm{div}_3$ . Now by the construction of  $\mathrm{SOL}_i$  for  $i \leq |G|$ , similar to Case 1 in the proof of Theorem 2.1, one can show that
$$
\operatorname {div} _ {1} + \operatorname {div} _ {2} \geq \frac {k}{4} \sum_ {i \leq | G |} k _ {i} r _ {i}
$$
Claim B.2.  $\mathrm{div}_2 + \mathrm{div}_3 \geq \frac{k}{8} \sum_{i = |G| + 1}^{m} r_i$ .
Proof. Note that we have
$$
\operatorname {div} _ {2} + \operatorname {div} _ {3} = \sum_ {x \in \operatorname {S O L} \backslash \operatorname {S O L} ^ {G}, y \in \operatorname {S O L}} \operatorname {d i s t} (x, y) \geq \sum_ {i = | G | + 1} ^ {m} \frac {r _ {i}}{2} \cdot (t + i - 1 - | G |)
$$
Now we consider two cases separately, if  $t \geq k / 2$  then clearly  $(t + i - 1 - |G|) \geq k / 2$  for  $i \geq |G| + 1$ , and we have
$$
\operatorname {d i v} _ {2} + \operatorname {d i v} _ {3} \geq \sum_ {i = | G | + 1} ^ {m} \frac {r _ {i}}{2} \cdot (t + i - 1 - | G |) \geq \sum_ {i = | G | + 1} ^ {m} \frac {r _ {i}}{2} \cdot \frac {k}{2} = \frac {k}{4} \cdot \sum_ {i = | G | + 1} ^ {m} r _ {i}
$$
Otherwise, if  $t < k / 2$  then since we assumed that the  $r_i$  for  $|G| + 1 \leq i \leq m$  are sorted, and noting that the last term in the summation is  $\frac{r_m}{2} \cdot (k - 1)$ , we have
$$
\operatorname {d i v} _ {2} + \operatorname {d i v} _ {3} = \sum_ {i = | G | + 1} ^ {m} \frac {r _ {i}}{2} \cdot (t + i - 1 - | G |) \geq \sum_ {i = m - k / 2} ^ {m} \frac {r _ {i}}{2} \cdot \frac {k}{2} = \frac {k}{4} \cdot \sum_ {i = m - k / 2} ^ {m} r _ {i} \geq \frac {k}{8} \sum_ {i = | G | + 1} ^ {m} r _ {i}
$$
Thus, noting that  $k_{i} = 1$  for  $i \geq |G| + 1$ , and using the upper bound we stated on  $\mathrm{div}(\mathrm{OPT})$ , we get that
$$
\operatorname {div} (\operatorname {S O L}) = \operatorname {d i v} _ {1} + \operatorname {d i v} _ {2} + \operatorname {d i v} _ {3} \geq \frac {1}{2} \left(\frac {k}{4} \sum_ {i \leq | G |} k _ {i} r _ {i} + \frac {k}{8} \sum_ {i = | G | + 1} ^ {m} k _ {i} r _ {i}\right) \geq \frac {1}{1 6} \sum_ {i} k _ {i} r _ {i} \geq \frac {1}{3 2 0} \operatorname {d i v} (\mathrm {O P T})
$$
C Arbitrary Partitioning and Fully Composable Core-sets
In this section, we show what our results imply when the partitioning of the point sets into multiple parts is not necessarily based on the colors. This will provide a composable core-set as defined in [20] as opposed to the color-abiding composable core-set as defined in 1.4.
Formally, the goal is to present a summarization algorithm  $\mathcal{A}$  that now receives a colored point set  $P = \bigcup_{i\leq m}P_{i}$  and the goal is to summarize it into  $T\subseteq P$  with the following composability property. For any collection of colored point sets  $P^{(1)},\dots ,P^{(N)}$ , we have that
$$
\operatorname {d i v} _ {k _ {1}, \dots , k _ {m}} (\bigcup_ {j \leq N} \mathcal {A} (P ^ {(j)})) \geq \frac {1}{\alpha} \operatorname {d i v} _ {k _ {1}, \dots , k _ {m}} (\bigcup_ {j \leq N} P ^ {(j)})
$$
Corollary C.1. We have an algorithm that given a colored point set  $P = \bigcup_{i\leq m}P_{i}$  computes a (not necessarily color-abiding) composable core-set of size  $k^2$  for the Fair Diversity Maximization problem under the SUM-PAIRWISE DIST notion of diversity.
Proof. The algorithm simply runs Algorithm 2 for each of  $P_{i}$  and returns their union. The size of the core-set is clearly  $\sum_{i\leq m}k_i^2\leq k^2$ . Moreover, by Theorem 2.1 it is easy to verify that this is in fact a composable core-set, because for any collection of colored point sets  $P^{(1)},\dots ,P^{(N)}$ , one can treat them as having  $N\cdot m$  different groups with  $k_{i}^{(j)}\leq k_{i}$  (more precisely, one can set  $k_{i}^{(j)}$  to be the number of points the optimal solution on  $\bigcup_{j\leq N}P^{(j)}$  picks from  $P^{(j)}$ ). Now note that although one does not know the value of  $k_{i}^{(j)}$  at the time of running Algorithm 2 on  $P_{i}^{(j)}$ , however since  $k_{i}^{(j)}\leq k_{i}$ , it is easy to verify that running Algorithm 2 on  $P_{i}^{(j)}$  with parameter  $k_{i}$  as opposed to  $k_{i}^{(j)}$  still works.
Similarly, one can get a composable core-set for the SUM-NN DIST notion of diversity.
Corollary C.2. We have an algorithm that given a colored point set  $P = \bigcup_{i\leq m}P_{i}$  computes a (not necessarily color-abiding) composable core-set of size  $O(m\cdot k^2)$  for the Fair Diversity Maximization problem under the SUM-NN DIST notion of diversity.
Proof. The algorithm simply runs Algorithm 4 for each of  $P_{i}$  and returns their union. The size of the core-set is clearly  $\sum_{i\leq m}k^{2}\leq mk^{2} = O(k^{3})$ . Moreover, by Theorem 4.1 it is again easy to verify that this is in fact a composable core-set, because for any collection of colored point sets  $P^{(1)},\dots ,P^{(N)}$ , again one can treat them as having  $N\cdot m$  different groups with  $k_{i}^{(j)}\leq k_{i}$ . Again, although one does not know the values of  $k_{i}^{(j)}$  at the time of running Algorithm 4, however since  $k_{i}^{(j)}\leq k_{i}$ , it is easy to verify that the point set picked by running Algorithm 4 on  $P_{i}^{(j)}$  with parameter  $k_{i}$  is a super set of the core-set that would have been picked if the algorithm was run with parameter  $k_{i}^{(j)}$  as opposed to  $k_{i}$ .
Remark C.3 (Applications). Now that we showed how to get a (not necessarily color-abiding) composable core-set, the result of [20] readily implies that we have a streaming algorithm and an MPC algorithm for the Fair Diversity Maximization under the SUM-PAIRWISE DIST and SUM-NN DIST notions of diversity.
D Data Preparation and Additional Experiments
In order to demonstrate the effectiveness of the proposed approximation algorithms, we run simulations on public and timed datasets. In this section, we give details on the used datasets, data preparation, pre-processing, and additional experiment results.
D.1 Datasets, data preparation, and experiments setup
Datasets and data preparation. We use Reddit public dataset [42] of text messages that are semantically embedded into a metric space (e.g., 748-dimensional metric space with BERT embeddings [12]). We use this collection<sup>3</sup> of Reddit messages as it provides the creation time stamp in the schema, which we need to mark the relative recency of the messages, presented by its color
<sup>3</sup>https://github.com/henghuiz/MaskedHierarchicalTransformer
within a given month<sup>4</sup>. We want to remark other commonly used Reddit datasets [37, 22]<sup>5</sup> are not considered in this work as they do not have the creation time in their schema. Since we have the messages metadata included in the creation timestamp  $t_i$  of a message  $i$ , we can assign a color color  $i$  of message  $i$  as
$$
\operatorname {c o l o r} _ {i} = \left\lfloor m \cdot \frac {t _ {i} - t _ {\operatorname* {m i n}}}{t _ {\operatorname* {m a x}} - t _ {\operatorname* {m i n}}} \right\rfloor
$$