Brad Fitzpatrick wrote:
Now, because of this one unlucky distribution where A only has 1 unit of
space between it and B, you're not saved by the multiple points.
Now item A's second point is at 2654435761+10, and item B is at
2654435761+11, etc...
So A overall ends up owning n/2**32 of the space, and B ends up owning
(2**32-n)/32 of the space. Even though they have the same weight!
The whole point of multiple points is to cancel out the unfairness you'd
get if you just did one point.
That said, I'm no expert in this, but your proposal doesn't seem like
it'd work.
Yeah, this is what we found too - the way Ketama does it is by hashing
40 different string variations per server (using MD5, currently, which
is 128bits), then splitting each hash value into four 32-bit values, and
putting each of these on the continuum. This results in the 160 points
per server which I mentioned earlier, and ends up with a relatively even
distribution.
--
Russ Garrett
Last.fm Ltd.
[EMAIL PROTECTED]
