> 0. Assume a 10 million message archive.
> 1. What percentage of permalinks need another click?
> 2. What percentage of permalinks will result in a list of more than 10
> matches?
Ignoring cross posts, for a 4 character hash:
1. Approximately 90%
2. Approximately 50%
Ignoring cross posts, for a 13 character hash:
1. Effectively 0%
2. Effectively 0%
Pick message count and collision tolerance, and hash size will follow.
-Jeff
========== simulation code
#!/usr/bin/python
import random
hashlength = 4
message_count = 10000000
database = {}
collisions = 0
for i in range(message_count):
n = random.randint(0, pow(2, 5 * hashlength))
if n in database:
collisions += 1
database[n] += 1
else:
database[n] = 1
over_ten_collisions = 0
for i in database:
if database[i] > 10:
over_ten_collisions += database[i]
p1 = (100.0 * collisions) / float(message_count)
p2 = (100.0 * over_ten_collisions) / float(message_count)
print("Percent coliisions %f" % p1)
print("Percent over ten collisions %f" % p2)
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