Permutations without cycles
Question:
I want to generate all possible permutations of a list, where cyclic permutations (going from left to right) should only occur once.
Here is an example:
Let the list be [A, B, C]. Then I want to have permutations such as [A, C, B] but not [B, C, A] as this would be a circular permutation of the original list [A, B, C]. For the list above, the result should look like
[A, B, C]
[A, C, B]
[B, A, C]
[C, B, A]
Here is a minimal working example that uses permutations() from itertools.
from itertools import permutations
def permutations_without_cycles(seq: list):
# Get a list of all permutations
permutations_all = list(permutations(seq))
print("nAll permutations:")
for i, p in enumerate(permutations_all):
print(i, "t", p)
# Get a list of all cyclic permutations
cyclic_permutations = [tuple(seq[i:] + seq[:i]) for i in range(len(seq))]
print("nAll cyclic permutations:")
for i, p in enumerate(cyclic_permutations):
print(i, "t", p)
# Remove all cyclic permutations except for one
cyclic_permutations = cyclic_permutations[1:] # keep one cycle
permutations_cleaned = [p for p in permutations_all if p not in cyclic_permutations]
print("nCleaned permutations:")
for i, item in enumerate(permutations_cleaned):
print(i, "t", item)
def main():
seq = ["A", "B", "C"]
permutations_without_cycles(seq=seq)
if __name__ == "__main__":
main()
I would like to know if there is a method in itertools to solve this problem for efficiently?
Answers:
See this post: Python: Generate All Permutations Subject to Condition
The idea is to pull the first element from the list, generate all permutations of the remaining elements and append those permutations to that first element.
That’s unusual, so no, that’s not already in itertools. But we can optimize your way significantly (mainly by filtering out the unwanted cyclics by using a set instead of a list, or even by just the single next unwanted one). Even more efficiently, we can compute the indexes of the unwanted permutations[*] and islice between them. See the full code at the bottom.
[*] Using a simplified version of permutation_index from more-itertools.
Benchmark results, using list(range(n)) as the sequence. Ints compare fairly quickly, so if the sequence elements were some objects with more expensive comparisons, my efficient solution would have an even bigger advantage, since it’s the only one that doesn’t rely on comparing permutations/elements.
8 elements:
1.76 ± 0.07 ms efficient
3.60 ± 0.76 ms optimized_iter
4.65 ± 0.81 ms optimized_takewhile
4.97 ± 0.43 ms optimized_set
8.19 ± 0.31 ms optimized_generator
21.42 ± 1.19 ms original
9 elements:
13.11 ± 2.39 ms efficient
34.37 ± 2.83 ms optimized_iter
40.87 ± 4.49 ms optimized_takewhile
46.74 ± 2.27 ms optimized_set
78.79 ± 3.43 ms optimized_generator
237.72 ± 5.76 ms original
10 elements:
160.61 ± 4.58 ms efficient
370.79 ± 14.71 ms optimized_iter
492.95 ± 2.45 ms optimized_takewhile
565.04 ± 9.68 ms optimized_set
too slow optimized_generator
too slow original
Code (Attempt This Online!):
from itertools import permutations, chain, islice, filterfalse, takewhile
from timeit import timeit
from statistics import mean, stdev
from collections import deque
# Your original, just without the prints/comments, and returning the result
def original(seq: list):
permutations_all = list(permutations(seq))
cyclic_permutations = [tuple(seq[i:] + seq[:i]) for i in range(len(seq))]
cyclic_permutations = cyclic_permutations[1:]
permutations_cleaned = [p for p in permutations_all if p not in cyclic_permutations]
return permutations_cleaned
# Your original with several optimizations
def optimized_set(seq: list):
cyclic_permutations = {tuple(seq[i:] + seq[:i]) for i in range(1, len(seq))}
return filterfalse(cyclic_permutations.__contains__, permutations(seq))
# Further optimized to filter by just the single next unwanted permutation
def optimized_iter(seq: list):
def parts():
it = permutations(seq)
yield next(it),
for i in range(1, len(seq)):
skip = tuple(seq[i:] + seq[:i])
yield iter(it.__next__, skip)
yield it
return chain.from_iterable(parts())
# Another way to filter by just the single next unwanted permutation
def optimized_takewhile(seq: list):
def parts():
it = permutations(seq)
yield next(it),
for i in range(1, len(seq)):
skip = tuple(seq[i:] + seq[:i])
yield takewhile(skip.__ne__, it)
yield it
return chain.from_iterable(parts())
# Yet another way to filter by just the single next unwanted permutation
def optimized_generator(seq: list):
perms = permutations(seq)
yield next(perms)
for i in range(1, len(seq)):
skip = tuple(seq[i:] + seq[:i])
for perm in perms:
if perm == skip:
break
yield perm
yield from perms
# Compute the indexes of the unwanted permutations and islice between them
def efficient(seq):
def parts():
perms = permutations(seq)
yield next(perms),
perms_index = 1
n = len(seq)
for rotation in range(1, n):
index = 0
for i in range(n, 1, -1):
index = index * i + rotation * (i > rotation)
yield islice(perms, index - perms_index)
next(perms)
perms_index = index + 1
yield perms
return chain.from_iterable(parts())
funcs = original, optimized_generator, optimized_set, optimized_iter, optimized_takewhile, efficient
#--- Correctness checks
seq = ["A", "B", "C"]
for f in funcs:
print(*f(seq), f.__name__)
seq = 3,1,4,5,9,2,6
for f in funcs:
assert list(f(seq)) == original(seq)
for n in range(9):
seq = list(range(n))
for f in funcs:
assert list(f(seq)) == original(seq)
#--- Speed tests
def test(seq, funcs):
print()
print(len(seq), 'elements:')
times = {f: [] for f in funcs}
def stats(f):
ts = [t * 1e3 for t in sorted(times[f])[:5]]
return f'{mean(ts):6.2f} ± {stdev(ts):5.2f} ms '
for _ in range(25):
for f in funcs:
t = timeit(lambda: deque(f(seq), 0), number=1)
times[f].append(t)
for f in sorted(funcs, key=stats):
print(stats(f), f.__name__)
test(list(range(8)), funcs)
test(list(range(9)), funcs)
test(list(range(10)), funcs[2:])
Claim: The set permutations of N elements that are not related to each other via cyclic permutation is in one-to-one correspondence with all permutations of N-1 elements.
Not rigorous proof: Any permutation {i1,i2,…,iN}, with the k’th element ik=1, can be transformed by cyclic permutations to the permutation {1,j2,…,jN} (with j2=i(k+1),j3=i(k+2),…). The permutation {1,j2,…,jN} uniquely represents an orbit of cyclic permutations. Hence, any permutation of N-1 elements {j2,…,jN} represents the orbit of permutation of N elements not related by cycles.
The easiest way to convince yourself is to count dimensions (number of elements):
Dim[All permutation of N] = N!
Dim[Cyclic permutations] = N
Dim[Permutation of N not related by cycles] = N!/N = (N-1)! = Dim[All permutation of N-1]
Then what you need to do:
0. Given list ["1","2",…,"N"]
- Pop the first element ["2",…,"N"]
- Compute all permutations ["2","3"…,"N"], ["3","2"…,"N"]…
- Append "1" to each list ["1","2","3"…,"N"], ["1","3","2"…,"N"]…
I want to generate all possible permutations of a list, where cyclic permutations (going from left to right) should only occur once.
Here is an example:
Let the list be [A, B, C]. Then I want to have permutations such as [A, C, B] but not [B, C, A] as this would be a circular permutation of the original list [A, B, C]. For the list above, the result should look like
[A, B, C]
[A, C, B]
[B, A, C]
[C, B, A]
Here is a minimal working example that uses permutations() from itertools.
from itertools import permutations
def permutations_without_cycles(seq: list):
# Get a list of all permutations
permutations_all = list(permutations(seq))
print("nAll permutations:")
for i, p in enumerate(permutations_all):
print(i, "t", p)
# Get a list of all cyclic permutations
cyclic_permutations = [tuple(seq[i:] + seq[:i]) for i in range(len(seq))]
print("nAll cyclic permutations:")
for i, p in enumerate(cyclic_permutations):
print(i, "t", p)
# Remove all cyclic permutations except for one
cyclic_permutations = cyclic_permutations[1:] # keep one cycle
permutations_cleaned = [p for p in permutations_all if p not in cyclic_permutations]
print("nCleaned permutations:")
for i, item in enumerate(permutations_cleaned):
print(i, "t", item)
def main():
seq = ["A", "B", "C"]
permutations_without_cycles(seq=seq)
if __name__ == "__main__":
main()
I would like to know if there is a method in itertools to solve this problem for efficiently?
See this post: Python: Generate All Permutations Subject to Condition
The idea is to pull the first element from the list, generate all permutations of the remaining elements and append those permutations to that first element.
That’s unusual, so no, that’s not already in itertools. But we can optimize your way significantly (mainly by filtering out the unwanted cyclics by using a set instead of a list, or even by just the single next unwanted one). Even more efficiently, we can compute the indexes of the unwanted permutations[*] and islice between them. See the full code at the bottom.
[*] Using a simplified version of permutation_index from more-itertools.
Benchmark results, using list(range(n)) as the sequence. Ints compare fairly quickly, so if the sequence elements were some objects with more expensive comparisons, my efficient solution would have an even bigger advantage, since it’s the only one that doesn’t rely on comparing permutations/elements.
8 elements:
1.76 ± 0.07 ms efficient
3.60 ± 0.76 ms optimized_iter
4.65 ± 0.81 ms optimized_takewhile
4.97 ± 0.43 ms optimized_set
8.19 ± 0.31 ms optimized_generator
21.42 ± 1.19 ms original
9 elements:
13.11 ± 2.39 ms efficient
34.37 ± 2.83 ms optimized_iter
40.87 ± 4.49 ms optimized_takewhile
46.74 ± 2.27 ms optimized_set
78.79 ± 3.43 ms optimized_generator
237.72 ± 5.76 ms original
10 elements:
160.61 ± 4.58 ms efficient
370.79 ± 14.71 ms optimized_iter
492.95 ± 2.45 ms optimized_takewhile
565.04 ± 9.68 ms optimized_set
too slow optimized_generator
too slow original
Code (Attempt This Online!):
from itertools import permutations, chain, islice, filterfalse, takewhile
from timeit import timeit
from statistics import mean, stdev
from collections import deque
# Your original, just without the prints/comments, and returning the result
def original(seq: list):
permutations_all = list(permutations(seq))
cyclic_permutations = [tuple(seq[i:] + seq[:i]) for i in range(len(seq))]
cyclic_permutations = cyclic_permutations[1:]
permutations_cleaned = [p for p in permutations_all if p not in cyclic_permutations]
return permutations_cleaned
# Your original with several optimizations
def optimized_set(seq: list):
cyclic_permutations = {tuple(seq[i:] + seq[:i]) for i in range(1, len(seq))}
return filterfalse(cyclic_permutations.__contains__, permutations(seq))
# Further optimized to filter by just the single next unwanted permutation
def optimized_iter(seq: list):
def parts():
it = permutations(seq)
yield next(it),
for i in range(1, len(seq)):
skip = tuple(seq[i:] + seq[:i])
yield iter(it.__next__, skip)
yield it
return chain.from_iterable(parts())
# Another way to filter by just the single next unwanted permutation
def optimized_takewhile(seq: list):
def parts():
it = permutations(seq)
yield next(it),
for i in range(1, len(seq)):
skip = tuple(seq[i:] + seq[:i])
yield takewhile(skip.__ne__, it)
yield it
return chain.from_iterable(parts())
# Yet another way to filter by just the single next unwanted permutation
def optimized_generator(seq: list):
perms = permutations(seq)
yield next(perms)
for i in range(1, len(seq)):
skip = tuple(seq[i:] + seq[:i])
for perm in perms:
if perm == skip:
break
yield perm
yield from perms
# Compute the indexes of the unwanted permutations and islice between them
def efficient(seq):
def parts():
perms = permutations(seq)
yield next(perms),
perms_index = 1
n = len(seq)
for rotation in range(1, n):
index = 0
for i in range(n, 1, -1):
index = index * i + rotation * (i > rotation)
yield islice(perms, index - perms_index)
next(perms)
perms_index = index + 1
yield perms
return chain.from_iterable(parts())
funcs = original, optimized_generator, optimized_set, optimized_iter, optimized_takewhile, efficient
#--- Correctness checks
seq = ["A", "B", "C"]
for f in funcs:
print(*f(seq), f.__name__)
seq = 3,1,4,5,9,2,6
for f in funcs:
assert list(f(seq)) == original(seq)
for n in range(9):
seq = list(range(n))
for f in funcs:
assert list(f(seq)) == original(seq)
#--- Speed tests
def test(seq, funcs):
print()
print(len(seq), 'elements:')
times = {f: [] for f in funcs}
def stats(f):
ts = [t * 1e3 for t in sorted(times[f])[:5]]
return f'{mean(ts):6.2f} ± {stdev(ts):5.2f} ms '
for _ in range(25):
for f in funcs:
t = timeit(lambda: deque(f(seq), 0), number=1)
times[f].append(t)
for f in sorted(funcs, key=stats):
print(stats(f), f.__name__)
test(list(range(8)), funcs)
test(list(range(9)), funcs)
test(list(range(10)), funcs[2:])
Claim: The set permutations of N elements that are not related to each other via cyclic permutation is in one-to-one correspondence with all permutations of N-1 elements.
Not rigorous proof: Any permutation {i1,i2,…,iN}, with the k’th element ik=1, can be transformed by cyclic permutations to the permutation {1,j2,…,jN} (with j2=i(k+1),j3=i(k+2),…). The permutation {1,j2,…,jN} uniquely represents an orbit of cyclic permutations. Hence, any permutation of N-1 elements {j2,…,jN} represents the orbit of permutation of N elements not related by cycles.
The easiest way to convince yourself is to count dimensions (number of elements):
Dim[All permutation of N] = N!
Dim[Cyclic permutations] = N
Dim[Permutation of N not related by cycles] = N!/N = (N-1)! = Dim[All permutation of N-1]
Then what you need to do:
0. Given list ["1","2",…,"N"]
- Pop the first element ["2",…,"N"]
- Compute all permutations ["2","3"…,"N"], ["3","2"…,"N"]…
- Append "1" to each list ["1","2","3"…,"N"], ["1","3","2"…,"N"]…