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To: Billy Tetrud <billy.tetrud@gmail.com>
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Cc: Bitcoin Protocol Discussion <bitcoin-dev@lists.linuxfoundation.org>
Subject: Re: [bitcoin-dev] Jets (Was: `OP_FOLD`: A Looping Construct For
Bitcoin SCRIPT)
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Good morning Billy,
> > I think we would want to have a cleanstack rule at some point
>
> Ah is this a rule where a script shouldn't validate if more than just a t=
rue is left on the stack? I can see how that would prevent the non-soft-for=
k version of what I'm proposing.=C2=A0
Yes.
There was also an even stronger cleanstack rule where the stack and alt sta=
ck are totally empty.
This is because a SCRIPT really just returns "valid" or "invalid", and `OP_=
VERIFY` can be trivially appended to a SCRIPT that leaves a single stack it=
em to convert to a SCRIPT that leaves no stack items and retains the same b=
ehavior.
>
> > How large is the critical mass needed?
>
> Well it seems we've agreed that were we going to do this, we would want t=
o at least do a soft-fork to make known jet scripts lighter weight (and unk=
nown jet scripts not-heavier) than their=C2=A0non-jet counterparts. So give=
n a situation where this soft fork happens, and someone wants to implement =
a new jet, how much critical mass would be needed for the network to get so=
me benefit from the jet? Well, the absolute minimum for some benefit to hap=
pen is that two nodes that support that jet are connected. In such a case, =
one node can send that jet scripted transaction along without sending the d=
ata of what the jet stands for. The jet itself is pretty small, like 2 or s=
o bytes. So that does impose a small additional cost on nodes that don't su=
pport a jet. For 100,000 nodes, that means 200,000 bytes of transmission wo=
uld need to be saved for a jet to break even. So if the jet stands for a 22=
byte script, it would break even when 10% of the network supported it. If =
the jet stood for a 102 byte script, it would break even when 2% of the net=
work supported it. So how much critical mass is necessary for it to be wort=
h it depends on what the script is.=C2=A0
The math seems reasonable.
> The question I have is: where would the constants table come from? Would =
it reference the original positions of items on the witness stack?=C2=A0
The constants table would be part of the SCRIPT puzzle, and thus not in the=
witness solution.
I imagine the SCRIPT would be divided into two parts: (1) a table of consta=
nts and (2) the actual opcodes to execute.
Regards,
ZmnSCPxj
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