Xavier Holt edited WSPD_Approximation_For_our_approximation__.md  almost 8 years ago

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We justify this strategy by noting that we're going to use our `diameter` sub-routine. This calculates the optimal diameter with `i',j'` poles. As such, any particular point-pole connection strategy we choose for analysis will never result in a better diameter than our system.  What is the radius of the circle centered at `i'`? By the triangle inequality if If  we set the radius of the `i'`-circle  to be \(|i,i'| + L_r\) where \(L_r\) was the optimal `i`-optimal  radiusthen  we cover allof the same  points. This follows from the triangle inequality. For all points \(p\in P_i\), \(|i',p| \leq |i',i| + |i,p| \leq |i',i| + L_r\).  The same thing applies for `j'`. As such, we have that our diameter is given by the following \(d=\left(|i,i'| + L_r \right) + \left(|j,j'| + R_r \right) + |i',j'|\). We substitute in our WSPD bounds of \(|i,i'| \leq \frac{2}{s}|i,j|\) (resp. \(|j,j'| \leq \dots\) ) and \(|i',j'| \leq (1 + \frac{4}{s}) |i,j|\).