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From: Matt Whitlock <bip@mattwhitlock.name>
To: Peter Todd <pete@petertodd.org>
Date: Fri, 12 Jun 2015 14:52:31 -0400
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Cc: bitcoin-development@lists.sourceforge.net
Subject: Re: [Bitcoin-development] User vote in blocksize through fees
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On Friday, 12 June 2015, at 7:44 pm, Peter Todd wrote:
> On Fri, Jun 12, 2015 at 02:36:31PM -0400, Matt Whitlock wrote:
> > On Friday, 12 June 2015, at 7:34 pm, Peter Todd wrote:
> > > On Fri, Jun 12, 2015 at 02:22:36PM -0400, Matt Whitlock wrote:
> > > > Why should miners only be able to vote for "double the limit" o=
r "halve" the limit? If you're going to use bits, I think you need to u=
se two bits:
> > > >=20
> > > > =090 0 =3D no preference ("wildcard" vote)
> > > > =090 1 =3D vote for the limit to remain the same
> > > > =091 0 =3D vote for the limit to be halved
> > > > =091 1 =3D vote for the limit to be doubled
> > > >=20
> > > > User transactions would follow the same usage. In particular, a=
user vote of "0 0" (no preference) could be included in a block castin=
g any vote, but a block voting "0 0" (no preference) could only contain=
transactions voting "0 0" as well.
> > >=20
> > > Sounds like a good encoding to me. Taking the median of the three=
> > > options, and throwing away "don't care" votes entirely, makes sen=
se.
> >=20
> > I hope you mean the *plurality* of the three options after throwing=
away the "don't cares," not the *median*.
>=20
> Median ensures that voting "no change" is meaningful. If "double" + "=
no
> change" =3D 66%-1, you'd expect the result to be "no change", not "ha=
lve""
> With a plurality vote you'd end up with a halving that was supported =
by
> a minority.
I'm very confused.
Let's say, out of the 12,000 blocks in a voting period:
=E2=80=A2 1000 blocks express no preference,
=E2=80=A2 2000 blocks vote to halve the limit,
=E2=80=A2 3500 blocks vote to double the limit, and
=E2=80=A2 5500 blocks vote to keep the limit the same.
The plurality vote is to keep the limit the same. The median vote is=E2=
=80=A6 well, I'm not sure, since there are four choices, and an average=
of the two "middle" choices doesn't seem to make sense.
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