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From: Dave Hudson <dave@hashingit.com>
In-Reply-To: <CABsx9T0yad_2NOYEdrJbqX48wCvuQiYb=zP1eC3YwOCc1ODN9w@mail.gmail.com>
Date: Tue, 4 Aug 2015 16:37:57 -0700
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To: Gavin Andresen <gavinandresen@gmail.com>
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Cc: Bitcoin Dev <bitcoin-dev@lists.linuxfoundation.org>
Subject: Re: [bitcoin-dev] "A Transaction Fee Market Exists Without a Block
	Size Limit"--new research paper suggests
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> On 4 Aug 2015, at 14:30, Gavin Andresen <gavinandresen@gmail.com> =
wrote:
>=20
> On Tue, Aug 4, 2015 at 2:41 PM, Dave Hudson via bitcoin-dev =
<bitcoin-dev@lists.linuxfoundation.org =
<mailto:bitcoin-dev@lists.linuxfoundation.org>> wrote:
> Fundamentally a block maker (pool or aggregation of pools) does not =
orphan its own blocks.
>=20
> Unless the block maker has an infinitely fast connection to it's =
hashpower OR it's hashpower is not parallelized at all, that's not =
strictly true -- it WILL orphan its own blocks because two hashing units =
will find solutions in the time it takes to communicate that solution to =
the block maker and to the rest of the hashing units.
>=20
> That's getting into "how many miners can dance on the head of a pin" =
territory, though. I don't think we know whether the communication =
advantages of putting lots of hashing power physically close together =
will outweigh the extra cooling costs of doing that (or maybe some other =
tradeoff I haven't thought of). That would be a fine topic for another =
paper....

Yes, but the block maker won't publish the second block it finds for the =
same set of transactions. It won't orphan its own block. In fact even if =
it does it still doesn't matter because the block maker still gets the =
block reward irrespective of which of the two solutions are published.

It's not about which hash wins, the issue is who gets paid as a result.


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<html><head><meta http-equiv=3D"Content-Type" content=3D"text/html =
charset=3Dus-ascii"></head><body style=3D"word-wrap: break-word; =
-webkit-nbsp-mode: space; -webkit-line-break: after-white-space;" =
class=3D""><br class=3D""><div><blockquote type=3D"cite" class=3D""><div =
class=3D"">On 4 Aug 2015, at 14:30, Gavin Andresen &lt;<a =
href=3D"mailto:gavinandresen@gmail.com" =
class=3D"">gavinandresen@gmail.com</a>&gt; wrote:</div><br =
class=3D"Apple-interchange-newline"><div class=3D""><div dir=3D"ltr" =
class=3D""><div class=3D"gmail_extra"><div class=3D"gmail_quote">On Tue, =
Aug 4, 2015 at 2:41 PM, Dave Hudson via bitcoin-dev <span dir=3D"ltr" =
class=3D"">&lt;<a href=3D"mailto:bitcoin-dev@lists.linuxfoundation.org" =
target=3D"_blank" =
class=3D"">bitcoin-dev@lists.linuxfoundation.org</a>&gt;</span> =
wrote:<br class=3D""><blockquote class=3D"gmail_quote" style=3D"margin:0 =
0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">Fundamentally a =
block maker (pool or aggregation of pools) does not orphan its own =
blocks.</blockquote></div><br class=3D"">Unless the block maker has an =
infinitely fast connection to it's hashpower OR it's hashpower is not =
parallelized at all, that's not strictly true -- it WILL orphan its own =
blocks because two hashing units will find solutions in the time it =
takes to communicate that solution to the block maker and to the rest of =
the hashing units.</div><div class=3D"gmail_extra"><br =
class=3D""></div><div class=3D"gmail_extra">That's getting into "how =
many miners can dance on the head of a pin" territory, though. I don't =
think we know whether the communication advantages of putting lots of =
hashing power physically close together will outweigh the extra cooling =
costs of doing that (or maybe some other tradeoff I haven't thought of). =
That would be a fine topic for another =
paper....</div></div></div></blockquote><br class=3D""></div><div>Yes, =
but the block maker won't publish the second block it finds for the same =
set of transactions. It won't orphan its own block. In fact even if it =
does it still doesn't matter because the block maker still gets the =
block reward irrespective of which of the two solutions are =
published.</div><div><br class=3D""></div><div>It's not about which hash =
wins, the issue is who gets paid as a result.</div><br =
class=3D""></body></html>=

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