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def next_permutation(a):
"""Generate the lexicographically next permutation inplace.
https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order
Return false if there is no next permutation.
"""
# Find the largest index i such that a[i] < a[i + 1]. If no such
# index exists, the permutation is the last permutation
for i in reversed(range(len(a) - 1)):
if a[i] < a[i + 1]:
break # found
else: # no break: not found
return False # no next permutation
# Find the largest index j greater than i such that a[i] < a[j]
j = next(j for j in reversed(range(i + 1, len(a))) if a[i] < a[j])
# Swap the value of a[i] with that of a[j]
a[i], a[j] = a[j], a[i]
# Reverse sequence from a[i + 1] up to and including the final element a[n]
a[i + 1:] = reversed(a[i + 1:])
return True
def perm_to_str(a):
return ''.join(map(str, a))
def str_to_perm(s):
return [int(c) for c in s]
t = [0,1,2,3]
perm_to_idx = dict()
def bbin(x):
return bin(x)[2:]
cnt = 23
while True:
perm_to_idx[perm_to_str(t)] = cnt
cnt -= 1
if not next_permutation(t):
break
for p in perm_to_idx.keys():
for i in range(4):
t = str_to_perm(p)
for j in range(4):
if t[j] > t[i]:
t[j] -= 1
t[i] = 3
print(p, 'x', i, '->', perm_to_str(t), ':\t', bbin(perm_to_idx[p]), '\t', bbin(i), '\t', bbin(perm_to_idx[perm_to_str(t)]), end='\t')
print(perm_to_idx[p], '\t', i, '\t', perm_to_idx[perm_to_str(t)])
print('\n-----\n')
for p in perm_to_idx.keys():
for i in range(4):
t = str_to_perm(p)
for j in range(4):
if t[j] > t[i]:
t[j] -= 1
t[i] = 3
print(p, 'x', i, '->', perm_to_str(t), ':\t', bbin(perm_to_idx[p]), '\t', bbin(i), '\t', bbin(perm_to_idx[perm_to_str(t)]), end='\t')
print(perm_to_idx[p], '\t', i, '\t', perm_to_idx[perm_to_str(t)])
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