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If someone is curious about the bruteforce solution:
def pmix(s, k): d = s[0] for a in range(k): d = s[0] for i in range(0, len(s)): if (i == len(s)-1): if (s[i] == "A"): if (d == "A"): s = s[:-1] + "A" elif (d == "B"): s = s[:-1] + "B" elif (d == "C"): s = s[:-1] + "C" elif (d == "D"): s = s[:-1] + "D" elif (s[i] == "B"): if (d == "A"): s = s[:-1] + "B" elif (d == "B"): s = s[:-1] + "A" elif (d == "C"): s = s[:-1] + "D" elif (d == "D"): s = s[:-1] + "C" elif (s[i] == "C"): if (d == "A"): s = s[:-1] + "C" elif (d == "B"): s = s[:-1] + "D" elif (d == "C"): s = s[:-1] + "A" elif (d == "D"): s = s[:-1] + "B" elif (s[i] == "D"): if (d == "A"): s = s[:-1] + "D" elif (d == "B"): s = s[:-1] + "C" elif (d == "C"): s = s[:-1] + "B" elif (d == "D"): s = s[:-1] + "A" elif(i < (len(s)-1)): if (s[i] == "A"): if (s[i + 1] == "A"): s = s[:i] + "A" + s[i + 1:] elif (s[i + 1] == "B"): s = s[:i] + "B" + s[i + 1:] elif (s[i + 1] == "C"): s = s[:i] + "C" + s[i + 1:] elif (s[i + 1] == "D"): s = s[:i] + "D" + s[i + 1:] elif (s[i] == "B"): if (s[i + 1] == "A"): s = s[:i] + "B" + s[i + 1:] elif (s[i + 1] == "B"): s = s[:i] + "A" + s[i + 1:] elif (s[i + 1] == "C"): s = s[:i] + "D" + s[i + 1:] elif (s[i + 1] == "D"): s = s[:i] + "C" + s[i + 1:] elif (s[i] == "C"): if (s[i + 1] == "A"): s = s[:i] + "C" + s[i + 1:] elif (s[i + 1] == "B"): s = s[:i] + "D" + s[i + 1:] elif (s[i + 1] == "C"): s = s[:i] + "A" + s[i + 1:] elif (s[i + 1] == "D"): s = s[:i] + "B" + s[i + 1:] elif (s[i] == "D"): if (s[i + 1] == "A"): s = s[:i] + "D" + s[i + 1:] elif (s[i + 1] == "B"): s = s[:i] + "C" + s[i + 1:] elif (s[i + 1] == "C"): s = s[:i] + "B" + s[i + 1:] elif (s[i + 1] == "D"): s = s[:i] + "A" + s[i + 1:] return s
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Mixing proteins
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If someone is curious about the bruteforce solution: