Comment by sweezyjeezy
1 year ago
for 10^5, to get the same collision probability (~2 * exp(-10)), you would just need to compute the 10 maximum/minimum candidates and check against those.
1 year ago
for 10^5, to get the same collision probability (~2 * exp(-10)), you would just need to compute the 10 maximum/minimum candidates and check against those.
No comments yet
Contribute on Hacker News ↗