how to simulate the situation in DNA evolution for finding the minimum 
population needed and minimum samples selected to mating in order to no 
extinction in any one of original species and new species

assume mating are randomly selected to become a couple,
how to keep species good which means no extinction in original species

i use i+j to get new species, 
but i do not know whether there is limit in evolution, i assume
limit is 7, then extra evolution will go back to past species and 
cycle again

import matplotlib.pyplot as plt
import random
dict = {}
dist = {}
maxnum = 5
allowedmax = 7
for i in range(1,allowedmax+1):
    dist[str(i)] = 0

rr = range (1,maxnum)
for i in range (1,maxnum):
    for j in range (1,maxnum):
        if i < j:
            print("(" +str(i) + "," + str(j) + ")");
            dict[str(i) + str(j)] = 1;
            dist[str(i+j)] = dist[str(i+j)] + 1
            if i+j > max(rr or [0]):
                 rr = rr + [i+j];

original = rr;
for numberofevolutions in range(1,10):
    rr2 = []
    samples = random.sample(original, len(original)-2)
    print("total rr")
    print(str(rr))
    print("samples")
    print(str(samples))
    for i in samples:
        for j in samples:
            if i < j:
                print("(" +str(i) + "," + str(j) + ")");
                if i+j > allowedmax:
                    dict[str(i) + str(j)] = (i+j) % allowedmax;
                    dist[str((i+j) % allowedmax)] = dist[str((i+j) % 
allowedmax)] + 1
                    if ((i+j) % allowedmax) > max(rr2 or [0]):
                        rr2 = rr2 + [((i+j) % allowedmax)];
                else:
                    dict[str(i) + str(j)] = i+j;
                    dist[str(i+j)] = dist[str(i+j)] + 1
                    if i+j > max(rr2 or [0]):
                        rr2 = rr2 + [i+j];
    temp = rr
    rr = rr2
    rr2 = temp

plt.bar(range(len(dist)), dist.values(), align='center')
plt.xticks(range(len(dist)), dist.keys())
plt.show()
-- 
https://mail.python.org/mailman/listinfo/python-list

Reply via email to