Does this greedy algorithm minimize the sum of the diameters of two clusters?

Question

Suppose given $2n$ points in the plane and we want to partition points into two clusters $C_1$ , $C_2$ such that each cluster contains exactly $n$ points and we want to minimize the sum of the diameters of two clusters. The diameter of a cluster is the maximum distance between any two points in that cluster.

I have a greedy algorithm for this problem:

Find diametral pair $(a,b)$ of $2n$ points, and then consider $a$ as the center of $C_1$ and $b$ as the center of $C_2$. Now, we make two lists $L_a$ and $L_b$ such that $L_a$ contains all points sorted by increasing distance from $a$ and the other by increasing distance from $b$. I keep the first $n$ points in $L_a$ that are closer to $a$ and remove others and then set $C_1=L_a$. I do the same for list $L_b$ and then set $C_2=L_b$.

My question is, the above idea is optimal or not?

I think above idea return optimal solution because I can use Exchange argument as follows:

Suppose in the optimal partitions $C'_{1}$ and $C'_2$ isn't the same as $C_1$ and $C_2$ so there exists two points $c,d$ such that $c\in C_1, d\in C_2$ and $c\in C'_2, d\in C'_1$ but according to the algorithm if we exchange $c ,d$ we get $d\in C'_2, c\in C'_1$ so we achieve a new optimal solution that it's no worse than the optimal solution. So we get contradiction.

Answer 1

No, the algorithm does not return the optimal sets.

Consider the case of the $4$ points $P=\{(0,0), (1,0), (0.5, 0.5), (0.5, 0.5)\}$. The diametrical points are $(0,0)$ and $(1,0)$ with distance one. So your algorithm puts them in their own clusters. The closest other-point to $(0,0)$ is $(0.5, 0.5)$ so your algorithm returns the set $\{(0,0), (0.5, 0.5)\}$ and $\{(1,0), (0.5, 0.5)\}$ with cost $\sqrt{0.5^2+0.5^2} + \sqrt{0.5^2+0.5^2}=\sqrt{2}$.

However, the optimal sets are $\{(0,0),(1,0)\}$ and $\{(0.5,0.5),(0.5,0.5)\}$ with cost $1+0=1 < \sqrt{2}$.