Subject: [Din] draft-mazieres-dinrg-scp-04 (page 16) setting preparedPrime

From: Piers Powlesland <[email-masked]>
Date: Thu, 27 Sep 2018 14:54:43 +0100

In the section "preparedPrime", I'm wondering if the "highest accepted prepared
ballot", is referring to a concrete ballot that has been seen or a ballot
within the conceptual space of ballots?

For example, if we consider a system with 3 possible values x, y and z and
x<y<z. If you imagine we have p == (4,z) and we see blocking threshold for
(3,x) and so set p' == (3,x) if you consider the ballots that have been
accepted aborted through p and p', you can see that all the ballots required to
be aborted in order for (3,y) to be considered prepared have been aborted. Do
we now set p' to (3,y) since (3,y) is greater than (3,x) ?

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From: David Mazieres <[email-masked]>
Date: Fri, 28 Sep 2018 18:46:06 -0700

Piers Powlesland <[email-masked]> writes:

> In the section "preparedPrime", I'm wondering if the "highest accepted
> prepared ballot", is referring to a concrete ballot that has been seen
> or a ballot within the conceptual space of ballots?
>

Thanks for raising the point.  The short answer is that it doesn't
matter--the protocol works either way.

The slightly longer answer, which someone recently pointed out to me, is
that because of this fact, it is actually wasteful to send around the
value of preparedPrime, rather than just the counter.  So the next
revision of the draft will probably address this question...

David