“…For p 2, the cost of computing d s p (T 1 , T 2 ), for T 1 , T 2 ∈ T n , as the p-th root of d s p (T 1 , T 2 ) p will depend on the accuracy with which this root is computed. For instance, using the Newton method to compute it with an accuracy of an 1/2 h -th of its value has a cost of O(p 2 log(p) log(hp)); see, for instance, [4]. So, in practice, for small p and not too large h, this step will be dominated by the computation of d s p (T 1 , T 2 ) p , and the total cost will be O(n 2 ) (we understand in this case log(p) as part of the constant factor).…”