Hw10 solutions

Solution in code form

The important differences between Gillespie's first reaction and direct methods:

In the direct method, we consider the time and identity of the next reaction independently. First, we picked one random number (uniformly on (0,1)) to generate the time of the next reaction (with exponential distribution parametrized by  a = sum(a_i) ):

# how long until the next reaction r1 = random.random tau = (1.0/(a+eps))*math.log(1.0/r1)

Then we picked a second random number uniformly on (0,a) and identified which reaction's "bin" it fell into (where the width of each reactions's bin is equal to its current propensity).

r2a = random.random*a a_sum = 0 for i in a_i: if r2a < (a_i[i]+a_sum): mu = i           break a_sum += a_i[i] Where a_sum is the "right edge" of the current bin you're considering.

In the first reaction method, one picks a random number (uniform on 0,1) and uses it to generate a a time (exponentially distributed with parameter a_i) for each reaction, and selects the reaction with the smallest time to fire at that time. Tau's for all reactions are recalculated at every step, never saved. Note that the behavior of the first and direct methods is equivalent.

mintau = t_max eps = math.exp(-200) # which reaction will happen first? # caluculate each reaction's time based on a different random number for rxn in a_i.keys: ai = a_i[rxn] ri = random.random taui = (1.0/(ai+eps)) * math.log(1.0/ri) if taui < mintau: # "sort as you go" mu = rxn    # the putative first rxn tau = taui  # the putative first time mintau = tau # reset the min

This method of sorting seemed fastest to us, but here's another way of tackling the problem:

tau_i = {} # which reaction will happen first? # caluculate each reaction's time based on a different random number for rxn in a_i.keys: ai = a_i[rxn] ri = random.random taui = (1.0/(ai+eps)) * math.log(1.0/ri) tau_i[taui] = rxn tau = min(tau_i.keys) mu = tau_i[tau]