On 17 June 2010 20:08, Lev Stesin wrote:
> Hi,
>
> What is the correct procedure to create a well balanced cluster (in
> terms of key distribution). From what I understand whenever I add a
> new node its takes half from its neighbor. How can I make each node to
> contain 1/3 of the keys in a 3 no
so bc and ruby agree ;)
On Thu, Jun 17, 2010 at 10:20 PM, Benjamin Black wrote:
> in ruby:
>
> def token(nodes) 1.upto(nodes) {|i| p (2**127/nodes) * i}; end
>
> >> token(3)
> 56713727820156410577229101238628035242
> 113427455640312821154458202477256070484
> 1701411834604692317316873037158841057
in ruby:
def token(nodes) 1.upto(nodes) {|i| p (2**127/nodes) * i}; end
>> token(3)
56713727820156410577229101238628035242
113427455640312821154458202477256070484
170141183460469231731687303715884105726
On Thu, Jun 17, 2010 at 12:08 PM, Lev Stesin wrote:
> Hi,
>
> What is the correct procedure
+ user, - dev (bcc actually)
If you use a random partitioner use the following InitialToken for your
nodes:
$ bc
bc 1.06
Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'.
(2^127)/3
*5671372782015641
Hi,
What is the correct procedure to create a well balanced cluster (in
terms of key distribution). From what I understand whenever I add a
new node its takes half from its neighbor. How can I make each node to
contain 1/3 of the keys in a 3 node cluster? Thanks.
--
Lev
