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Ingy döt Net 2023-07-01 11:58:00 -04:00
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from itertools import permutations
given = '''ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA
CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB'''.split()
allPerms = [''.join(x) for x in permutations(given[0])]
missing = list(set(allPerms) - set(given)) # ['DBAC']

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def missing_permutation(arr):
"Find the missing permutation in an array of N! - 1 permutations."
# We won't validate every precondition, but we do have some basic
# guards.
if len(arr) == 0: raise Exception("Need more data")
if len(arr) == 1:
return [arr[0][1] + arr[0][0]]
# Now we know that for each position in the string, elements should appear
# an even number of times (N-1 >= 2). We can use a set to detect the element appearing
# an odd number of times. Detect odd occurrences by toggling admission/expulsion
# to and from the set for each value encountered. At the end of each pass one element
# will remain in the set.
missing_permutation = ''
for pos in range(len(arr[0])):
s = set()
for permutation in arr:
c = permutation[pos]
if c in s:
s.remove(c)
else:
s.add(c)
missing_permutation += list(s)[0]
return missing_permutation
given = '''ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA
CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB'''.split()
print missing_permutation(given)

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>>> from collections import Counter
>>> given = '''ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA
CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB'''.split()
>>> ''.join(Counter(x).most_common()[-1][0] for x in zip(*given))
'DBAC'
>>>

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>>> from pprint import pprint as pp
>>> pp(list(zip(*given)), width=120)
[('A', 'C', 'A', 'D', 'B', 'A', 'A', 'C', 'D', 'B', 'C', 'C', 'C', 'A', 'A', 'B', 'D', 'B', 'B', 'B', 'C', 'D', 'D'),
('B', 'A', 'C', 'A', 'C', 'C', 'D', 'D', 'A', 'C', 'A', 'D', 'B', 'B', 'D', 'D', 'C', 'A', 'A', 'D', 'B', 'B', 'C'),
('C', 'B', 'D', 'C', 'D', 'B', 'C', 'A', 'B', 'A', 'D', 'B', 'A', 'D', 'B', 'C', 'B', 'C', 'D', 'A', 'D', 'C', 'A'),
('D', 'D', 'B', 'B', 'A', 'D', 'B', 'B', 'C', 'D', 'B', 'A', 'D', 'C', 'C', 'A', 'A', 'D', 'C', 'C', 'A', 'A', 'B')]
>>> pp([Counter(x).most_common() for x in zip(*given)])
[[('C', 6), ('B', 6), ('A', 6), ('D', 5)],
[('D', 6), ('C', 6), ('A', 6), ('B', 5)],
[('D', 6), ('C', 6), ('B', 6), ('A', 5)],
[('D', 6), ('B', 6), ('A', 6), ('C', 5)]]
>>> pp([Counter(x).most_common()[-1] for x in zip(*given)])
[('D', 5), ('B', 5), ('A', 5), ('C', 5)]
>>> pp([Counter(x).most_common()[-1][0] for x in zip(*given)])
['D', 'B', 'A', 'C']
>>> ''.join([Counter(x).most_common()[-1][0] for x in zip(*given)])
'DBAC'
>>>

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'''Find the missing permutation'''
from functools import reduce
from operator import xor
print(''.join([
chr(i) for i in reduce(
lambda a, s: map(
xor,
a,
[ord(c) for c in list(s)]
), [
'ABCD', 'CABD', 'ACDB', 'DACB',
'BCDA', 'ACBD', 'ADCB', 'CDAB',
'DABC', 'BCAD', 'CADB', 'CDBA',
'CBAD', 'ABDC', 'ADBC', 'BDCA',
'DCBA', 'BACD', 'BADC', 'BDAC',
'CBDA', 'DBCA', 'DCAB'
],
[0, 0, 0, 0]
)
]))