for VOT on balanced and unbalanced measures. They follow the same trend and converge at almost the same rate. The resulting clusters are exactly the same.[]{data-label="fig:uot"}](iter.pdf){width="0.85\columnwidth"}

**Case 1:** $\NUL{\IdxCentroid} = \frac{1}{\K}$. It is trivial to verify that minimizing the second term in  over $\bm{\Region}$ under its equality constraint yields all $w_{\RegionL{\IdxCentroid}}$’s equal to each other, i.e. $w_{\RegionL{\IdxCentroid}} = \frac{1}{\K}w$. On the other hand, the gradient of the VOT energy  has the form $\int_{\RegionL{\IdxCentroid}} d\MU(\x) - \NUL{\IdxCentroid} \equiv w_{\RegionL{\IdxCentroid}} - \frac{1}{\K}w$. The question now is whether $w_{\RegionL{\IdxCentroid}} = \frac{1}{\K}w$ minimizes . If so, then we can instead solve  to minimize the second term in .

When $w_{\RegionL{\IdxCentroid}} = \frac{1}{\K}w$, the gradient $\nabla\EnergyL{2}[\bm{\h}] = \{w_{\RegionL{\IdxCentroid}} - \NUL{\IdxCentroid}\}_{\IdxCentroid}$ becomes constant and thus $\bm{\h}$ is being translated at a constant rate. Certainly, translation does not modify a power Voronoi diagram as specified in . Therefore, $\EnergyL{2}[\bm{\h}]$ saturates

. For any other partition such that $\exists\ \RegionL{\IdxCentroid}', w_{\RegionL{\IdxCentroid}'} \neq \frac{1}{\K}w$, we have $$\begin{split}
\sum_{\IdxCentroid=1}^{\K} \int_{\RegionL{\IdxCentroid}} & \left( \|\x - \yL{\IdxCentroid}\|_{2}^{2} + \hL{\IdxCentroid} \right) d\MU(\x) \\
&\leq \sum_{\IdxCentroid=1}^{\K} \int_{\RegionL{\IdxCentroid}'} \left( \|\x - \yL{\IdxCentroid}\|_{2}^{2}  + \hL{\IdxCentroid} \right) d\MU(\x).
\nonumber
\end{split}$$ Therefore, $w_{\RegionL{\IdxCentroid}} = \frac{1}{\K}w$ indeed minimizes . Meanwhile, we know that an unweighted Voronoi diagram ($\hL{\IdxCentroid} = 