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