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From: James MacWhyte <macwhyte@gmail.com>
Date: Fri, 8 Jul 2022 16:08:54 +0200
Message-ID: <CAH+Axy4X+uQG5Vw0Efiz6AtNyK=++h-jDeZL1ZxpVJus8BVKeA@mail.gmail.com>
To: Paul Sztorc <truthcoin@gmail.com>,
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Subject: Re: [bitcoin-dev] No Order Mnemonic
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Content-Type: text/plain; charset="UTF-8"
> What do you do if the "first" word (of 12), happens to be the last word in
> the list alphabetically?
>
That couldn't happen. If one word is the very last from the wordlist, it
would end up at the end of your mnemonic once you rearrange your 12 words
alphabetically.
However!
(@vjudeu) Choosing 11 random words and then sorting them alphabetically
before assigning a checksum would reduce entropy considerably. If you think
about it, to bruteforce the entire keyspace one would only need to come up
with every possible combination of 11 words + 1 checksum. I'm not the best
at napkin math, but I think that leaves you with around 10 trillion
combinations, which would only take a couple months to exhaust with
hardware that can do 1 million guesses per second.
James
--000000000000a051bb05e34bbfd0
Content-Type: text/html; charset="UTF-8"
Content-Transfer-Encoding: quoted-printable
<div dir=3D"ltr"><div dir=3D"ltr"><br></div><div class=3D"gmail_quote"><blo=
ckquote class=3D"gmail_quote" style=3D"margin:0px 0px 0px 0.8ex;border-left=
:1px solid rgb(204,204,204);padding-left:1ex"><div dir=3D"auto">What do you=
do if the "first" word (of 12), happens to be the last word in t=
he list alphabetically?</div></blockquote><div><br></div><div>That couldn&#=
39;t happen. If one word is the very last from the wordlist, it would end u=
p at the end of your mnemonic=C2=A0once you rearrange your 12 words alphabe=
tically.<br><br>However!=C2=A0</div><div><br>(@vjudeu) Choosing 11 random w=
ords and then sorting them alphabetically before assigning=C2=A0a checksum =
would reduce entropy considerably. If you think about it, to bruteforce the=
entire keyspace one would only need to come up with every possible combina=
tion of 11 words=C2=A0+ 1 checksum. I'm not the best at napkin math, bu=
t I think that leaves you with around=C2=A010 trillion combinations, which =
would only take a couple months to exhaust with hardware that can do 1 mill=
ion guesses per second.<br></div><div><br></div><div>James</div></div></div=
>
--000000000000a051bb05e34bbfd0--
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