[chord] problem?

[Baohua Wu]

I read your wonderful paper "Chord: A scalable p2p lookup service for internet applications".
I have a question regarding Figure 6 on Page 6. 

n.update_others() {
for i=1 to m {
    p = find_predecessor(n-2^(i-1));    // should it be find_predecessor(n+1-2^(i-1))?
    p.update_finger_table(n,i);
}
}

Should it be find_predecessor(n+1-2^(i-1))? For example, 0 and 3 are in the ring. 4 joins. To update the 3rd finger in Node 0, we have to use find_predecessor(n+1-2^(i-1)). 

Thanks,

Baohua
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