The way that I understand it (and that seems to be consistent with what was 
said in this discussion) is that each DC has its own data space. Using your 
simplified 1-10 system:
   DC1   DC2
0  D1R1  D2R2
1  D1R1  D2R1
2  D1R1  D2R1
3  D1R1  D2R1
4  D1R1  D2R1
5  D1R2  D2R1
6  D1R2  D2R2
7  D1R2  D2R2
8  D1R2  D2R2
9  D1R2  D2R2

Each node is responsible for half of the ring in its own DC.

----- Original Message -----
From: "Eric tamme" <eta...@gmail.com>
To: user@cassandra.apache.org
Sent: Wednesday, May 4, 2011 1:58:19 PM
Subject: Re: Replica data distributing between racks

>        Jonathan is suggesting the approach Jeremiah was using.
>
>        Calculate the tokens the nodes in each DC independantly, and then add 
> 1 to the tokens if there are two nodes with the same tokens.
>
>        In your case with 2 DC's with 2 nodes each.
>
> In DC 1
> node 1 = 0
> node 2 = 85070591730234615865843651857942052864
>
> In DC 2
> node 1 = 1
> node 2 =  85070591730234615865843651857942052865
>
> This will evenly distribute the keys in each DC, which is what the 
> NetworkTopologyStrategy is trying to do.



Okay - I appreciate the direct solution, but I am still really
confused. I think I am missing some thing conceptual here... it just
isn't "clicking".

If I have nodes 4 nodes, in two data centers, each in it's own rack:
DC1R1, DC1R2, DC2R1, DC2R2

Tokens:

DC1R1: N
DC1R2: M
DC2R1: N+1
DC2R1: M+1

Who is responsible for what in primary distribution and in
replication?  Is DC1R2 responsible for M-M+1 (aka 1 token, M)??? that
doesn't make any sense... or am I supposed to be making primary
distribution uneven so that the uneven replication then balances it?

I am trying to conceptualize this... I drew up a graph of the range
responsibility based on this token assignment based on a simplified
token range of 0-9 http://dl.dropbox.com/u/19254184/tokens.jpg

I must be missing some thing, I just don't know what.  Please if some
one can please explain or point me to resources that clearly explain
this.

Thanks for everyones time

-Eric

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