<p>I am trying to prove that this holds
<a href="https://i.sstatic.net/AS2vc.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/AS2vc.png" alt="[cond 1]"></a>  [cond 1]</p>

<p>However this paper shows that the constant should be greater than 2/5
<a href="http://www.ghassankarame.com/ripple.pdf" rel="nofollow noreferrer">http://www.ghassankarame.com/ripple.pdf</a><br>
Section 8, page 10</p>

<p>Could someone please help me determine if the paper is correct, or if the 1/5 holds.</p>

<hr>

<p>So far, I think that the counterexample does not provide us with the UNL list for all nodes, and we do need that to determine if cond. 1  holds for each pair of servers.</p>

CC BY-SA 3.0
2015-10-25T23:26:28.117Z
UNL is Unique Node List, in case anyone else wonders.
Murch
CC BY-SA 3.0
2015-10-28T22:21:23.497Z
An UNL graph would be directied. That is, if some server X appears in the UNL of server Y, server Y does not necessarily appear in the UNL of server X. 
Therefore the example in the paper holds, as it is possible to construct the UNLs for the remaining nodes without breaking the counterexample.
My confusion stemmed from thinking that the relationships were symmetric, ie the graph was not directed.
Albert S
<ripple>
Ripple Consensus
Albert S
# Ripple Consensus

(Asked by: username_0 on 2015-10-25)

I am trying to prove that this holds
[![[cond 1]](https://i.sstatic.net/AS2vc.png)](https://i.sstatic.net/AS2vc.png) [cond 1]

However this paper shows that the constant should be greater than 2/5
<http://www.ghassankarame.com/ripple.pdf>  
Section 8, page 10

Could someone please help me determine if the paper is correct, or if the 1/5 holds.

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So far, I think that the counterexample does not provide us with the UNL list for all nodes, and we do need that to determine if cond. 1 holds for each pair of servers.

**Tags:** ripple

> **Comments:**
> **username_1**: UNL is Unique Node List, in case anyone else wonders.  
> **username_0**: An UNL graph would be directied. That is, if some server X appears in the UNL of server Y, server Y does not necessarily appear in the UNL of server X.  Therefore the example in the paper holds, as it is possible to construct the UNLs for the remaining nodes without breaking the counterexample. My confusion stemmed from thinking that the relationships were symmetric, ie the graph was not directed.  

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## 0 answers

