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# CST  ## Monopolar  Consider every point as a potential pole. For each of these points, find the two points furthest away from ### Algorithm  * \(d\leftarrow \infty; u\leftarrow\) None 
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* u \(\leftarrow p\)  * Return u.  Runtime: \(O(n^2)\).  ### Correctedness  We consider all possible poles. The MST and diameter are unique given a pole. Clearly when we return the minimum over all of these trees we find the overall minimum diameter MST.  ## Dipolar  We want to do a process similar to above with all pair of points. In order to do so, we need an efficient way of calculating the diameter of a MST given poles \(i,j\), which we refer to as \(T_{ij}\).  Let \(P_i, P_j\) be the points connected to \(i,j\) respectively in \(T_{ij}\). The diameter of the MST \(d(T_{ij}) = \max_{p \in P_i} |ip| + \max_{p \in P_j} |jp| + |ij|\). We can view our task as essentially centering a circle at each pole such that all points are covered by one circle, and the sum of the radii is minimum. We define \(L_r:=\max_{p \in P_i} |ip|; R_r:=\max_{p \in P_j} |jp|\)  Our radii are always going to be defined by points \(\in P-\{i,j\}\). As such, we could simply iterate through all possible pairs and see which gets us the best diameter. This takes us to \(O(n^4)\) however, so we have to try a bit harder. It turns out we can get away with checking far fewer candidate radii.