time one phase consists of one half of the remaining merge steps. Furthermore, we are able to replace the volume argument from Lemma \[volume\_lem\] by a simpler bound. More precisely, as long as there are more than $k$ clusters left, we are able to find a pair of clusters whose centers lie in the same cluster of an optimal $k$-clustering. That is, the distance between the centers is at most two times the discrete radius of the common cluster in the optimal clustering. The following lemma bounds the increase of the cost during a single phase.

\[drad\_rem\_phaselem\] Let $m\in{\mathbb{N}}$ with $k<m\leq|X|$. Then, $$\operatorname{cost_{drad}}({\mathcal{C}}_{k+\left\lfloor\frac{m-k}{2}\right\rfloor})<\operatorname{cost_{drad}}({\mathcal{C}}_m)+2\operatorname{opt}_k.$$

Let $R:=\operatorname{cost_{drad}}({\mathcal{C}}_m)$. From every cluster $C\in{\mathcal{C}}_m$, we fix a center $p_C\in C$ with $C\subset\operatorname{B}_R^d(p_C)$.

Let $t:=k+\left\lfloor\frac{m-k}{2}\right\rfloor$. Then, ${\mathcal{C}}_m\cap {\mathcal{C}}_{t+1}$ is the set of clusters from ${\mathcal{C}}_m$ that still exist $\left\lceil\frac{m-k}{2}\right\rceil-1$ merge steps after the computation of ${\mathcal{C}}_m$. In each iteration of its loop, the algorithm can merge at most two clusters from ${\mathcal{C}}_m$. Thus, $|{\mathcal{C}}_m\cap {\mathcal{C}}_{t+1}|>k$.

Let $P:=\{p_C{\,\vline~}C\in C_m\cap {\mathcal{C}}_{t+1}\}$. Since $X$ is $(k,\operatorname{opt}_k)$-coverable, so is $P\subset X$. Therefore, using $|P|>k$ it follows that there exist distinct $C_1,C_2\in{\mathcal{C}}_m\cap{\mathcal{C}}_{t+1}$ such that $p_{C_1}$ and $p_{C_2}$ are contained in the same ball of radius $\operatorname{opt}_k$, i.e. $\|p_{C_1}-p_{C_2}\|\leq2\operatorname{opt}_k$. Then, the distance 