def test_rendezvous_redistribution():
# check that approximately a third of keys move to the new bucket
# when one is added
n_keys = 3000
# count how many keys were moved, which bucket a key started from and
# which bucket a key was moved from (to the new bucket)
n_moved = 0
from_bucket = {"b1": 0, "b2": 0}
start_in = {"b1": 0, "b2": 0}
for i in range(n_keys):
key = f"key-{i}"
two_buckets = utils.rendezvous_rank(["b1", "b2"], key)
start_in[two_buckets[0]] += 1
three_buckets = utils.rendezvous_rank(["b1", "b2", "b3"], key)
if two_buckets[0] != three_buckets[0]:
n_moved += 1
from_bucket[two_buckets[0]] += 1
# should always move to the newly added bucket
assert three_buckets[0] == "b3"
# because of statistical fluctuations we have to leave some room when
# making this comparison
assert 0.31 < n_moved / n_keys < 0.35
# keys should move from the two original buckets with approximately
# equal probability. We pick 30 because it is "about right"
assert abs(from_bucket["b1"] - from_bucket["b2"]) < 30
# the initial distribution of keys should be roughly the same
# We pick 30 because it is "about right"
assert abs(start_in["b1"] - start_in["b2"]) < 30
def test_rendezvous_rank():
# check that a key doesn't move if its assigned bucket remains but the
# other buckets are removed
key = "crazy frog is a crazy key"
first_round = utils.rendezvous_rank(["b1", "b2", "b3"], key)
second_round = utils.rendezvous_rank([first_round[0], first_round[1]], key)
assert first_round[0] == second_round[0], key
def test_rendezvous_independence():
# check that the relative ranking of 80 buckets doesn't depend on the
# presence of 20 extra buckets
key = "k1"
eighty_buckets = utils.rendezvous_rank([f"b{i}" for i in range(80)], key)
hundred_buckets = utils.rendezvous_rank([f"b{i}" for i in range(100)], key)
for i in range(80, 100):
hundred_buckets.remove(f"b{i}")
assert eighty_buckets == hundred_buckets