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import itertools
A = 'A'
B = 'B'
C = 'C'
D = 'D'
def Q1():
options = {
A : A,
B : B,
C : C,
D : D,
}
option = options[answer[1]]
return option == answer[1]
def Q2():
options = {
A : C,
B : D,
C : A,
D : B,
}
option = options[answer[2]]
return option == answer[5]
def Q3():
options = {
A : answer[3],
B : answer[6],
C : answer[2],
D : answer[4],
}
option = options[answer[3]]
values = options.values()
values.remove(option)
return option not in values
def Q4():
options = {
A : (answer[1], answer[5]),
B : (answer[2], answer[7]),
C : (answer[1], answer[9]),
D : (answer[6], answer[10]),
}
option = options[answer[4]]
return option[0] == option[1]
def Q5():
options = {
A : answer[8],
B : answer[4],
C : answer[9],
D : answer[7],
}
option = options[answer[5]]
return option == answer[5]
def Q6():
options = {
A : (answer[2], answer[4]),
B : (answer[1], answer[6]),
C : (answer[3], answer[10]),
D : (answer[5], answer[9]),
}
option = options[answer[6]]
return len(set(list(option) + [answer[8]])) == 1
def Q7():
options = {
A : answer.count(C),
B : answer.count(B),
C : answer.count(A),
D : answer.count(D),
}
option = options[answer[7]]
return option == min(options.values())
def Q8():
options = {
A : answer[7],
B : answer[5],
C : answer[2],
D : answer[10],
}
option = options[answer[8]]
return abs(ord(option) - ord(answer[1])) != 1
def Q9():
options = {
A : answer[6],
B : answer[10],
C : answer[2],
D : answer[9],
}
option = options[answer[9]]
assume1 = (answer[1] == answer[6])
assume2 = (option == answer[5])
return assume1 != assume2
def Q10():
options = {
A : 3,
B : 2,
C : 4,
D : 1,
}
option = options[answer[10]]
keys = options.keys()
counts = [answer.count(key) for key in keys]
return option == abs(max(counts) - min(counts))
answers = itertools.product([A, B, C, D], repeat=10)
for answer in answers:
answer = [''] + list(answer)
if Q1() and Q2() and Q3() and Q4() and Q5() and \
Q6() and Q7() and Q8() and Q9() and Q10():
print ''.join(answer)
break
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