def detect_cycles(iterable):
result = []
i = iter(iterable)
try:
# Point to the tortoise
tortoise = 0
result.append(i.next())
# Point to the hare
hare = 1
result.append(i.next())
# Start looking for cycles
power = 1
while True:
# Use Richard P. Brent's algorithm to find an element that
# repeats.
while result[tortoise] != result[hare]:
if power == (hare - tortoise):
tortoise = hare
power *= 2
hare += 1
result.append(i.next())
# Brent assumes the sequence is stateless, but since we're
# dealing with a DFS tree, we need to make sure that all
# the items between `tortoise` and `hare` are identical.
cycle = True
for j in xrange(0, hare - tortoise + 1):
tortoise += 1
hare += 1
result.append(i.next())
if result[tortoise] != result[hare]:
# False alarm: no cycle here.
cycle = False
power = 1
tortoise = hare
hare += 1
result.append(i.next())
break
# Both loops are done, so we must have a cycle
if cycle:
raise exceptions.CircularDependency(result[tortoise:hare + 1])
except StopIteration:
# Return when `iterable` is exhausted. Obviously, there are no cycles.
return result
def _dfs(start, get_children, path):
if (start, get_children) in dependency_cache:
return dependency_cache[(start, get_children)]
results = []
if start in path:
raise exceptions.CircularDependency(path[path.index(start):] + [start])
path.append(start)
results.append(start)
children = sorted(get_children(start), key=lambda x: str(x))
# We need to apply all the migrations this one depends on
for n in children:
results = _dfs(n, get_children, path) + results
path.pop()
results = list(SortedSet(results))
dependency_cache[(start, get_children)] = results
return results