4 ms·
If anyone wants to dig into the math, this is a non linear recurrence relation, with an external dependency, that is n_d+1 = f(a,n_d) n^(d+1) bad news: How th
by FranceBacce 7y ago
If anyone wants to dig into the math, this is a non linear recurrence relation, with an external dependency, that is
n_d+1 = f(a,n_d) n^(d+1)
bad news: How this behaves will depend a lot on a (containment measures etc...)
good news: If we can make a grow faster than an exponential, we can win