5 ms·
You can rank people by the likely benefit of testing them. Interestingly, it goes up as the square of the number of people they interact with daily. (Because th
by tlb 7y ago
You can rank people by the likely benefit of testing them. Interestingly, it goes up as the square of the number of people they interact with daily. (Because their risk of having it increases, and also the number they are likely to spread it to.)
So you can allocate tests by sorting by (# of daily contacts in a closed space) ^ 2.
But as PB says, it should be practical to test everyone every day.
- saeranv 7y agoCurious to know how you are getting power of 2 exactly. Are you just saying it's some sort of power law growth and approximately 2, or is there an actual way of deriving it?
- texasbigdata 7y agoWouldn't it also vary by percent of population without immunity, perhaps demographics in the underlying population, and other environmental things (weather?). Seems like a (useful) aproximation.
- tonyarkles 7y agoIf you have N nodes and completely connect the nodes, there are O(N^2) edges. This comes from the formula for the sum(1...N). Edit: I’m not sure I interpreted the original statement correctly.
- dr_zoidberg 7y agoThat's what I was about to answer. That's where squaring comes from, in this case. However, as others have noted, you could just sort by # of people contacted, since squaring just gives you larger numbers for no benefit in analysis here.
- scythe 7y agoIf we assume (key) that the probability of becoming infected is directly proportional to the number of people you interact with, and consider that the probability of spreading infection given that you are infected is obviously directly proportional to the number of people you interact with, the product gives a quadratic function: [probability of becoming infected and spreading] = [probability of becoming infected] * [probability of spreading] ~ [people you meet]^2
- saeranv 7y agoOkay so something like: probability of spreading = a * n probability of infected = b * n where: a,b = some blackbox function for spreading/infection ratio? n = number of people you meet Assuming both are independent events we get: p = a * b * n^2 I can see how we get n^2 with that. The way I'm using a, and b seems clumsy though, is there a better way?
- scythe 7y agoIn math and (especially) physics it is common to express "proportionality" laws usually with a symbol that looks like LaTeX \propto. So for instance the activity A of a radioactive sample is written A \propto e^(-t/T) where t is time and T is the mean lifetime of a single particle. For convenience \propto is often transliterated as ~ when typing. So that becomes p_catch ~ n p_spread ~ n p_relay = p_spread * p_catch ~ n^2.
- saeranv 7y agoThanks, that's a much clearer way to think of it.
- deleted 7y ago[deleted]
- argonaut 7y agoThis isn't quite right. For one, the probability of being infected is not linear (it's capped at 1). If you do the math, the expected number of people you infect (given you are not infected) is roughly linear-ish.
- scythe 7y ago> the probability of being infected is not linear Correct, it's not. But it is roughly linear in the limit of small numbers of people with a small constant probability of becoming infected per interaction. (This assumption becomes problematic when you see "clustering" of highly social people with other highly social people.) To be specific, if P is the probability of becoming infected when interacting with a single person, then the probability of becoming infected after interacting with N people is 1-(1-P)^N = NP - O(N^2 P^2). It's easy to see that the limiting infection probability is 1 in this simplified model, and that if N*P < 1 you're looking at close-to-linear growth.
- tmoertel 7y agoSince squaring is a monotone transform over nonnegative numbers, why not just sort by (# of daily contacts in a closed space)?