\section{Proof}
In this section, we mainly prove the lemmas used in our paper.

\subsection{Proof of Lemma~\ref{lem:camera}}
\camera*

\begin{proof}
    Denote $\mathbf{o} = (o_1,o_2,o_3)$, $\mathbf{\overline{o}} = (\overline{o}_1, \overline{o}_2, \overline{o}_3)$.
    Using the property of uniform distribution and the spherical polar coordinates, we have 
    \begin{align*}
         d(R, \overline{R}) &= E_{\mathbf{r} \in R, \mathbf{\overline{r}}  \in \overline{R}} \left [\int_{ 0}^{T_1} \int_{0}^{T_2} \|\mathbf{r}(t_1)-\mathbf{\overline{r}}(t_2)\|^2_2 dt_2 dt_1 \right] \\
        &= \int_{ 0}^{T_1} \int_{0}^{T_2} E_{\theta, \phi}[(o_1 - \overline{o}_1 + t_1 \cos \theta_1 \cos \phi_1 - t_2 \cos \theta_2 \cos \phi_2)^2  \\
        &+(o_2 - \overline{o}_2 + t_1 \cos \theta_1 \sin \phi_1 - t_2 \cos \theta_2 \sin \phi_2)^2 + (o_3 - \overline{o}_3 + t_1 \sin \theta_1  - t_2 \sin\theta_2)^2]dt_2 dt_1\\
        &= \int_{ 0}^{T_1} \int_{0}^{T_2}[ \|\mathbf{o}-\mathbf{\overline{o}}\|^2_2 + C_1(\mathbf{o}, \mathbf{\overline{o}}, t_1, t_2) + C_2(t_1, t_2) ]dt_2 dt_1 \\
        &= T_1 T_2\|\mathbf{o}-\overline{\mathbf{o}}\|^2_2 + C + \int_{ 0}^{T_1} \int_{0}^{T_2} C_1(\mathbf{o}, \mathbf{\overline{o}}, t_1, t_2) dt_2 dt_1 \,.
    \end{align*}
    where we represent 
    \begin{align*}
        C_1(\mathbf{o}, \mathbf{\overline{o}},t_1,t_2)& =  2(o_1 - \overline{o}_1) E_{\theta, \phi}[t_1 \cos \theta_1 \cos \phi_1 - t_2 \cos \theta_2 \cos \phi_2] \\
        &+ 2(o_2 - \overline{o}_2) E_{\theta, \phi}[t_1 \cos \theta_1 \sin \phi_1 - t_2 \cos \theta_2 \sin \phi_2] + 2(o_3 - \overline{o}_3) E_{\theta, \phi}[t_1 \sin \theta_1 - t_2 \sin \theta_2] \,.
    \end{align*}
    and let $C_2(t_1, t_2)$ include all items that are not related to $\mathbf{o}$ and $\overline{\mathbf{o}}$.
    
    By the symmetry property of $\phi \in U(0,2\pi)$, we know that the $E_{\theta, \phi}[\sin \phi] = E_{\theta, \phi}[\cos \phi] = 0$. Furthermore, by the i.i.d property of $\theta_1$ and $\theta_2$, we have $E_{\theta, \phi}[\sin \theta_1] = E_{\theta, \phi}[\sin \theta_2]$. Observing that the integration over $t_1$ and $t_2$ is also symmetrical, we can deduce that $\int_{ 0}^{T_1} \int_{0}^{T_2}C_1(\mathbf{o}, \mathbf{\overline{o}},t_1,t_2)dt_2 dt_1 = 0$.
    Therefore, we finally get 
    \begin{align*}
        d(R, \overline{R}) =  T_1 T_2\|\mathbf{o}-\overline{\mathbf{o}}\|^2_2 + C \,.
    \end{align*}
    where $C = \int_{ 0}^{T_1} \int_{0}^{T_2} C_2(t_1,t_2) dt_2 dt_1$ represent a constant independent of the camera position $\mathbf{o}$ and $\mathbf{\overline{o}}$.
\end{proof}


\subsection{Proof of Lemma~\ref{lem:greedy}}
\greedy*
\begin{proof}
  Suppose the optimal solution in the primal problem is $R_1, R_2, \ldots, R_N$, the optimal solution obtained by our greedy algorithm is $\Tilde{R}_1, \Tilde{R}_2, \ldots, \Tilde{R}_N$.

We first prove that the optimal value in the $i+1$-th iteration of our method is not larger than the optimal value in the $i$-th iteration. Assume not, $\delta_{i+1} > \delta_i$, then we can find the image $\Tilde{R}_{i+1}$ satisfy $H(\Tilde{R}_{i+1}|R_j) \ge \delta_{i+1}$ for all $j \le i$. Because in the $i$-th iteration we only have the constraints $H(\Tilde{R}|\Tilde{R}_j) \ge \delta_{i}$ for all $j \le i-1$, therefore, we take $\Tilde{R} = \Tilde{R}_{i+1}$ will satisfy this constraints, with the value $\delta_{i+1} > \delta_i$, contradict with the property that $\delta_i$ is the optimal solution in the $i$-th iteration. So the optimal value in the $i+1$-th iteration of our method is not larger than the optimal value in the $i$-th iteration. 

Then we can assume the optimal value we find in each iteration as $\delta_1 \ge \delta_2 \ge \ldots \ge \delta_N$. So we have $\Tilde{\delta} = \min\{\delta_1, \delta_2, \ldots, \delta_N\} = \delta_N$.

\textbf{Then we prove the conclusion by contradiction.} Suppose we have $\Tilde{\delta} < \frac{1}{2}\delta$. Assume we have $n$ common images of the solution of the primal problem and the solution obtained by our greedy algorithm. By the solution $\Tilde{\delta} < \frac{1}{2}\delta$ we know that $n<=N-1$. So there are $N-n$ images in the primal solution that do not appear in our solution. Suppose the different images of primal solution are $R_{i_1}, R_{i_2}, \ldots, R_{i_{N-n}}$ and the different images in our solution are $\Tilde{R}_{i_1}, \Tilde{R}_{i_2}, \ldots, \Tilde{R}_{i_{N-n}}$. Then we consider the optimization problem in the iteration that we choose the last different image. Then we consider the last iteration of our algorithm:
\begin{align*}
        \max_{R \in \mathcal{R}} \delta_N
        \;\;\textrm{s.t.} \; H(R|\Tilde{R}_j)\ge\delta_N,\forall 1\le j\le N-1  \,.
\end{align*}



Then for the different images, $R_{i_1}, R_{i_2}, \ldots, R_{i_{N-n}}$ appear in primal solution but do not appear in our solution, we have that taking these images in the solution will incur a smaller solution. That is, for each $R$ in $R_{i_1}, R_{i_2}, \ldots, R_{i_{N-n}}$, we have a corresponding image $\Tilde{R}$ in $\Tilde{R}_1, \ldots, \Tilde{R}_{N-1}$, incur the relative difference $H(R|\Tilde{R}) \le \delta_N = \Tilde{\delta}$. By the definition of $\delta$, we know that $\Tilde{R}$ can only choose in the difference set $\Tilde{R}_{i_1}, \Tilde{R}_{i_2}, \ldots, \Tilde{R}_{i_{N-n}}$. Then we consider two cases:
\begin{itemize}
    \item Case 1. \textbf{The optimal solution of the last iteration $\Tilde{R}_N$ is not in the set of common image.} In this case, Because we have not selected it in the first $n-1$ iterations, we only have $N-n-1$ images to choose for the corresponding images selected in our algorithm which satisfy $H(R|\Tilde{R}) \le \delta_N = \Tilde{\delta}$. However, we have $R_{i_1}, R_{i_2}, \ldots, R_{i_{N-n}}$ in optimal solution of primal set, there are $N-n$ images satisfy this inequality. Therefore, by the Pigeonhole Principle, there exists two images in $R_{i_1}, R_{i_2}, \ldots, R_{i_{N-n}}$ corresponding to the same image $\Tilde{R_{i_k}}$ in $\Tilde{R}_{i_1}, \Tilde{R}_{i_2}, \ldots, \Tilde{R}_{i_{N-n}}$ that incur $H(R|\Tilde{R}) \le \Tilde{\delta}$. By Assumption~\ref{ass:inverse} we know that $H(R| \overline{R}) \propto d(R, \overline{R})$. By the definition of $d(R, \overline{R}) = E_{\mathbf{r} \in R, \overline{\mathbf{r}}  \in \overline{R}} \left [\int_{ 0}^{T_1} \int_{0}^{T_2} \|\mathbf{r}(t_1)-\mathbf{\overline{r}}(t_2)\|^2_2 dt_2 dt_1 \right]$ we know that it satisfy the triangle inequality: $d(R_{i_1},R_{i_2}) \le d(R_{i_1},\Tilde{R}_{i_k}) + d(R_{i_2},\Tilde{R}_{i_k})$. Therefore we can get the triangle inequality of $H$:
\begin{align*}
    H(R_{i_1}|R_{i_2}) \le H(R_{i_1}|\Tilde{R}_{i_k}) + H(R_{i_2}|\Tilde{R}_{i_k}) < \frac{\delta}{2} + \frac{\delta}{2} = \delta \,.
\end{align*}
This is contradictory to the fact that these two images are in the solution of the primal problem with distance $H(R_{i_1}|R_{i_2}) \ge \delta$.

\item Case 2. \textbf{The optimal solution of the last iteration $\Tilde{R}_N$ is in the set of common image.} In this case, we have $N-n$ images to choose for the corresponding images selected in our algorithm which satisfy $H(R|\Tilde{R}) \le \delta_N = \Tilde{\delta}$. Note that we have $R_{i_1}, R_{i_2}, \ldots, R_{i_{N-n}}$ in optimal solution of primal set, there are $N-n$ images satisfy this inequality. If there are two images in the primal set corresponding to the same image selected by our algorithm, using the analysis of case 1 will get a contradiction. Therefore, we only need to consider the case they are all corresponding to different images in our set, that is, each image $\Tilde{R}_{i_k}$ in our set has a unique corresponding image $R_{i_l}$ in the primal set. However, note that the last iteration solution $\Tilde{R}_N$ is in the set of common images and it also satisfies the constraint, that is, it also corresponds to an image $R$, satisfy the inequality $H(\Tilde{R}_N| R) = \delta_N < \frac{\delta}{2}$. By the definition of $\delta$, we know that $R$ must be in the different sets in our solution, not the common set. But we have proved that each image in $\Tilde{R}_{i_1}, \Tilde{R}_{i_2}, \ldots, \Tilde{R}_{i_{N-n}}$ corresponds to an image in primal set satisfies the inequality. Suppose $H(R|R_{i_k}) < \frac{\delta}{2}$.  By Assumption~\ref{ass:inverse} we know that the relative difference is proportional to the distance metric so it also satisfies the triangle inequality. So we have:
\begin{align*}
    H(\Tilde{R}_N|R_{i_k}) \le H(\Tilde{R}_N|R) + H(R|R_{i_k}) < \frac{\delta}{2} + \frac{\delta}{2} = \delta \,.
\end{align*}
This is contradictory to the fact that these two images $\Tilde{R}_N$ and $R_{i_k}$ are in the solution of the primal problem with distance $H(\Tilde{R}_N|R_{i_k}) \ge \delta$.
\end{itemize}




Therefore, we have proved this lemma by contradiction and show that $\Tilde{\delta} \ge \frac{1}{2} \delta$.


\end{proof}

\subsection{Proof of Lemma~\ref{lem:rgb}}

\rgb*

\begin{proof}
    By the definition of $\hat{C}(\mathbf{r})$, we have
    \begin{align*}
        \|\hat{C}(\mathbf{r})- \hat{C}(\overline{\mathbf{r}})\| &\le \sum^N_{i=1}\|T_i \alpha_i \mathbf{c}_i - \overline{T}_i \overline{\alpha}_i \overline{\mathbf{c}}_i\|\,.
    \end{align*}
    Then we analysis $|T_i - \overline{T}_i|$, $|\alpha_i - \overline{\alpha}_i|$, $\|\mathbf{c}_i - \overline{\mathbf{c}}_i\|$ separately.
    We have 
    \begin{align*}
        |T_i - \overline{T}_i| &= |\exp(-\sum^{i-1}_{j=1} \sigma_j \delta_j) - \exp(-\sum^{i-1}_{j=1} \overline{\sigma}_j \delta_j)| \\
        &= |\exp(-\sum^{i-1}_{j=1} \sigma(\mathbf{o} + t_j\mathbf{d}) \delta_j) - \exp(-\sum^{i-1}_{j=1} \sigma(\mathbf{\overline{o}} + t_j\mathbf{\overline{d}})\delta_j)| \\
        & \le |-\sum^{i-1}_{j=1} \sigma(\mathbf{o} + t_j\mathbf{d}) \delta_j +\sum^{i-1}_{j=1} \sigma(\mathbf{\overline{o}} + t_j\mathbf{\overline{d}})\delta_j| \\
        &\le \sum^{i-1}_{j=1} |\sigma(\mathbf{o} + t_j\mathbf{d}) - \sigma(\mathbf{\overline{o}} + t_j\mathbf{\overline{d}})| \delta_j \\
        &\le \sum^{i-1}_{j=1} L\|\mathbf{o} + t_j\mathbf{d} - \mathbf{\overline{o}} + t_j\mathbf{\overline{d}}\|\delta_j\\
        &\le (\sum^{i-1}_{j=1} \delta_j L) \|\mathbf{o} - \mathbf{\overline{o}}\| + (\sum^{i-1}_{j=1} \delta_j t_j L) \|\mathbf{d} - \mathbf{\overline{d}}\| \\
        &= t_i L\|\mathbf{o} - \mathbf{\overline{o}}\| + (\sum^{i-1}_{j=1} \delta_j t_j L) \|\mathbf{d} - \mathbf{\overline{d}}\| \,.
    \end{align*}
    where the first inequality is because $|e^{-x} - e^{-y}| \le |x-y|$, the second inequality is by the lipschitz property of $\sigma$, the final equality is because $\delta_i = t_{i+1} - t_i$ and $t_1 = 0$.
    We also have 
    \begin{align*}
        |\alpha_i - \overline{\alpha}_i| &= |\exp(- \overline{\sigma}_i \delta_i) - \exp(- \sigma_i \delta_i)| \\
        &\le |\sigma(\mathbf{o} + t_i\mathbf{d}) - \sigma(\mathbf{\overline{o}} + t_i\mathbf{\overline{d}})| \delta_i \\
        &\le L\|\mathbf{o} + t_i\mathbf{d} - \mathbf{\overline{o}} + t_i\mathbf{\overline{d}}\|\delta_i \\
        &\le \delta_i L \|\mathbf{o} - \overline{\mathbf{o}}\| + t_i \delta_i L \|\mathbf{d} - \overline{\mathbf{d}}\| \,.
    \end{align*}
    Finally, we have 
    \begin{align*}
        \|\mathbf{c}_i - \overline{\mathbf{c}}_i\| &= \|\mathbf{c}(\mathbf{o} + t_i\mathbf{d}, \mathbf{d}) - \mathbf{c}(\mathbf{\overline{o}} + t_i\mathbf{\overline{d}}, \overline{\mathbf{d}})\| \\
        &\le \|\mathbf{c}(\mathbf{o} + t_i\mathbf{d}, \mathbf{d}) - \mathbf{c}(\mathbf{\overline{o}} + t_i\mathbf{\overline{d}}, \mathbf{d})\| + \|\mathbf{c}(\mathbf{\overline{o}} + t_i\mathbf{\overline{d}}, \mathbf{d}) - \mathbf{c}(\mathbf{\overline{o}} + t_i\mathbf{\overline{d}}, \overline{\mathbf{d}})\|  \\
        &\le L\|\mathbf{o}-\overline{\mathbf{o}} + t_i(\mathbf{d}-\overline{\mathbf{d}})\| + L\|\mathbf{d} - \overline{\mathbf{d}}\| \\
        &\le L\|\mathbf{o}-\overline{\mathbf{o}}\| + L(t_i + 1)\|\mathbf{d} - \overline{\mathbf{d}}\|  \,.
    \end{align*}
    By the expression of $T_i$ and $\alpha_i$, we know $T_i \le 1$, $\alpha_i \le \delta_i$. As $c_i$ represents the RGB color, the norm of $c_i$ is also bounded by 1. Furthermore, the difference of viewing direction $\|d - \overline{d}\|$ is bounded. Therefore, we finally have the following upper bound:
    \begin{align*}
        \|\hat{C}(\mathbf{r})- \hat{C}(\overline{\mathbf{r}})\| &\le \sum^N_{i=1}\|T_i \alpha_i \mathbf{c}_i - \overline{T}_i \overline{\alpha}_i \overline{\mathbf{c}}_i\|\\
        &\le \sum^N_{i=1}|T_i -\overline{T}_i||\alpha_i|\|\mathbf{c}_i\| + |\overline{T}_i||\alpha_i-\overline{\alpha}_i|\|\mathbf{c}_i\| + |\overline{T}_i \overline{\alpha}_i| \|\mathbf{c}_i - \mathbf{\overline{c}_i}\| \\
        &\le \sum^N_{i=1}(t_i \delta_i L + 2\delta_i L) \|\mathbf{o}-\mathbf{\overline{o}}\| +  C \\
        &\le 3(\sum^N_{i=1}\delta_i)L \|\mathbf{o}- \overline{\mathbf{o}}\| + C \\
        &= 3L\|\mathbf{o} - \mathbf{\overline{o}}\| + C \,.
    \end{align*}
    The first inequality is by definition, the second inequality is by triangle inequality, the third inequality is by the conclusion we have proved and the bounding property of $T_i$, $\alpha_i$ and $\mathbf{c}_i$, the final inequality is by $t_i\le 1$ and the final equality is by $\delta_i = t_{i+1} - t_i$ and $t_1 = 0, t_{N+1} = 1$. $C$ represents a bounding constant of $\|\mathbf{d} - \mathbf{\overline{d}}\|$, independent of $\|\mathbf{o} - \mathbf{\overline{o}}\|$.
\end{proof}



