Return-Path: Received: from smtp4.osuosl.org (smtp4.osuosl.org [140.211.166.137]) by lists.linuxfoundation.org (Postfix) with ESMTP id DABB8C002D for ; Fri, 8 Jul 2022 09:12:23 +0000 (UTC) Received: from localhost (localhost [127.0.0.1]) by smtp4.osuosl.org (Postfix) with ESMTP id B3F4A424B3 for ; Fri, 8 Jul 2022 09:12:23 +0000 (UTC) DKIM-Filter: OpenDKIM Filter v2.11.0 smtp4.osuosl.org B3F4A424B3 Authentication-Results: smtp4.osuosl.org; dkim=pass (2048-bit key) header.d=gmail.com header.i=@gmail.com header.a=rsa-sha256 header.s=20210112 header.b=qsCCeb/Z X-Virus-Scanned: amavisd-new at osuosl.org X-Spam-Flag: NO X-Spam-Score: -2.098 X-Spam-Level: X-Spam-Status: No, score=-2.098 tagged_above=-999 required=5 tests=[BAYES_00=-1.9, DKIM_SIGNED=0.1, DKIM_VALID=-0.1, DKIM_VALID_AU=-0.1, DKIM_VALID_EF=-0.1, FREEMAIL_FROM=0.001, HTML_MESSAGE=0.001, RCVD_IN_DNSWL_NONE=-0.0001, SPF_HELO_NONE=0.001, SPF_PASS=-0.001] autolearn=ham autolearn_force=no Received: from smtp4.osuosl.org ([127.0.0.1]) by localhost (smtp4.osuosl.org [127.0.0.1]) (amavisd-new, port 10024) with ESMTP id kI2ATpJAeQvu for ; Fri, 8 Jul 2022 09:12:20 +0000 (UTC) X-Greylist: whitelisted by SQLgrey-1.8.0 DKIM-Filter: OpenDKIM Filter v2.11.0 smtp4.osuosl.org 363F9424AB Received: from mail-lf1-x12d.google.com (mail-lf1-x12d.google.com [IPv6:2a00:1450:4864:20::12d]) by smtp4.osuosl.org (Postfix) with ESMTPS id 363F9424AB for ; Fri, 8 Jul 2022 09:12:20 +0000 (UTC) Received: by mail-lf1-x12d.google.com with SMTP id m18so16857500lfg.10 for ; Fri, 08 Jul 2022 02:12:19 -0700 (PDT) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=20210112; h=mime-version:references:in-reply-to:from:date:message-id:subject:to :cc; bh=FB1MNKjSOw2IfazrPnPYwfkp9zvEW+8QEhgQCIAniYI=; b=qsCCeb/ZHnhwTFNrZhWw0zmKTbiPzhE2Cdiex4ZqJJWpBIsbCo0O8ZwEHz0H6R276K amlnBFx6TRGWRfN6JHjkLImRwEokAqV6KE2gao0sLhDgk8oxGbaKe+VF4EZSYyZdSnHP FsQJx9IxICh5TaXZnLCn1tkaICi1nEuqzabAFvm1p79zRCBnufYHLu1boJwJJLcBIw7P 7uHUCCVr9lkRia6GKPaLsFLZ0gA+LI9O7O82zCK+d6tHLgCIi0lSghX9foZrv1NMjhgF rwlShATX3c5hpPM40bZYUsGLDoHjO6LKq4uOPX23Y8yUglmuTft3OFUl6hK/8PJRJMPI K/TQ== X-Google-DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=1e100.net; s=20210112; h=x-gm-message-state:mime-version:references:in-reply-to:from:date :message-id:subject:to:cc; bh=FB1MNKjSOw2IfazrPnPYwfkp9zvEW+8QEhgQCIAniYI=; b=xXF8wRTasdk6S5J2q9i8f1V/XE9VLO7sJgtjvlzSUfmVIN2zkCsCsTvbg49HQAwcV5 HKv1asf0MrK4oZEE7eZb09+qX8SSYvl/y5xxSDBdxmc58hVUHRzo2AksRyvEeR88gCob RnUMMLQ8sBYEur+6t08iJPQ+cRTzQAEnegqzk7uzgfWUremYOX7YMccbqfchCe7cBIDo sr0OoTHLHABh+lbIsecUrBdL48hfI5Bjoqmeuvhx3JBgMD5BxOd9BULz/SKjOkienM7R YzqADy7rQYtgkMvirNek0hVns7rnmgn5bqT/fVnlyw7U/hd9HrHLsceVqreNT+Kumq4z x3tA== X-Gm-Message-State: AJIora9VF+b1XR3hLgPvh6cCyrpGkoAuPVUQLujDf5g+OtL2ne59TpJX jHd1VpWOI0ZseDQlOgh+DN42WIR0/NpvVFs9Wb2ibnR2 X-Google-Smtp-Source: AGRyM1uLjtWmZAdeLuJjOQbMpOgOkogtdiD01LdU63n/+CmIe0pqArteZzByon66gi8RHbqKVpTAo2c6uBPZQg50xmw= X-Received: by 2002:ac2:54a2:0:b0:489:57b0:7adc with SMTP id w2-20020ac254a2000000b0048957b07adcmr1760420lfk.271.1657271537926; Fri, 08 Jul 2022 02:12:17 -0700 (PDT) MIME-Version: 1.0 References: <3D3BFE9C-CFF3-49FF-840F-063B52C69A42@voskuil.org> <164256450-0ee6752f92c0be297952fc72b59076df@pmq5v.m5r2.onet> In-Reply-To: <164256450-0ee6752f92c0be297952fc72b59076df@pmq5v.m5r2.onet> From: Paul Sztorc Date: Fri, 8 Jul 2022 05:12:06 -0400 Message-ID: To: vjudeu@gazeta.pl, Bitcoin Protocol Discussion Content-Type: multipart/alternative; boundary="0000000000005084fb05e347990c" X-Mailman-Approved-At: Fri, 08 Jul 2022 09:26:42 +0000 Subject: Re: [bitcoin-dev] No Order Mnemonic X-BeenThere: bitcoin-dev@lists.linuxfoundation.org X-Mailman-Version: 2.1.15 Precedence: list List-Id: Bitcoin Protocol Discussion List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , X-List-Received-Date: Fri, 08 Jul 2022 09:12:23 -0000 --0000000000005084fb05e347990c Content-Type: text/plain; charset="UTF-8" Content-Transfer-Encoding: quoted-printable What do you do if the "first" word (of 12), happens to be the last word in the list alphabetically? So that seems like a dead end. Since users are never expected to memorize the "whole list" (of 2048 words) in any case, it seems that the smarter thing to do (if this "order" criterion is desirable) may have been to just make the whole list 12x longer and cut it into 12 sections. Each of the 12 slots would have 2048 distinct words. Then the computer would handle the order; the user could neglect it. I can guess why people weren't particularly interested in this: words always have to be written down in some order or another. Even if you write them down in a 3x4 grid, there are very few combinations needed to guess the one true ordering. I wonder how obscure the words would have to be, by the 12th list of 2048? But still it might be fun - the 4th word might always be a nautical word, the 5th word a farm word, etc. And no one would confuse it with a bip39 phrase -- in fact since they are just lists of integers 1 to 2048, it would be pretty easy to make them interoperable. Very easy but perhaps still not worth doing. Paul On Fri, Jul 8, 2022, 4:48 AM vjudeu via bitcoin-dev < bitcoin-dev@lists.linuxfoundation.org> wrote: > Isn't it enough to just generate a seed in the same way as today, then > sort the words alphabetically, and then use that as a seed? I know, the > last word is a checksum, but there are only 2048 words, so it is not a bi= g > deal to get any checksum we want. If that is insecure, because of lower > possible combinations, then it is always possible to increase the number = of > words to compensate that. > > > On 2022-07-08 04:27:21 user Eric Voskuil via bitcoin-dev < > bitcoin-dev@lists.linuxfoundation.org> wrote: > > > Without a performance requirement there is no reason you can=E2=80=99t st= ore the > BIP39 words in any order you want. So it=E2=80=99s certainly possible, ju= st brute > force the recovery. If you have less than a second vs. a few days then it= =E2=80=99s > a different question. > > > e > > > On Jul 7, 2022, at 18:48, Bram Cohen via bitcoin-dev < > bitcoin-dev@lists.linuxfoundation.org> wrote: > Part of the rules of my challenge is that the 'new' words need to be in > the same pool as the 'old' words, so any ordering is okay. Without that > requirement it's mathematically very straightforward. > > > On Thu, Jul 7, 2022 at 10:52 AM Pavol Rusnak > wrote: > There is. Just encode the index of permutation used to scramble the > otherwise sorted list. For 12 words you need to store 12! =3D ~32 bits so= 3 > words should be enough. > > > Repetitions make this more difficult, though. > > > On Thu 7. 7. 2022 at 19:41, Bram Cohen via bitcoin-dev < > bitcoin-dev@lists.linuxfoundation.org> wrote: > On Thu, Jul 7, 2022 at 7:43 AM Anton Shevchenko via bitcoin-dev < > bitcoin-dev@lists.linuxfoundation.org> wrote: > I made a python implementation for a different mnemonic encoding. The > encoding requires user to remember words but not the order of those words= . > The code is open (MIT license) at https://github.com/sancoder/noomnem > > > > Thanks Anton. There's an interesting mathematical question of whether it'= s > possible to make a code like this which always uses the BIP-39 words for > the same key as part of its encoding, basically adding a few words as err= or > correction in case the order is lost or confused. If the BIP-39 contains = a > duplicate you can add an extra word. > _______________________________________________ > bitcoin-dev mailing list > bitcoin-dev@lists.linuxfoundation.org > https://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev > _______________________________________________ > bitcoin-dev mailing list > bitcoin-dev@lists.linuxfoundation.org > https://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev > --0000000000005084fb05e347990c Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable
What do you do if the "first" word (of 12), hap= pens to be the last word in the list alphabetically? So that seems like a d= ead end.

Since users are never= expected to memorize the "whole list" (of 2048 words) in any cas= e, it seems that the smarter thing to do (if this "order" criteri= on is desirable) may have been to just make the whole list 12x longer and c= ut it into 12 sections. Each of the 12 slots would have 2048 distinct words= . Then the computer would handle the order; the user could neglect it.

I can guess why people weren= 't particularly interested in this: words always have to be written dow= n in some order or another. Even if you write them down in a 3x4 grid, ther= e are very few combinations needed to guess the one true ordering. I wonder= how obscure the words would have to be, by the 12th list of 2048? But stil= l it might be fun - the 4th word might always be a nautical word, the 5th w= ord a farm word, etc. And no one would confuse it with a bip39 phrase -- in= fact since they are just lists of integers 1 to 2048, it would be pretty e= asy to make them interoperable. Very easy but perhaps still not worth doing= .

Paul

On Fri, Jul = 8, 2022, 4:48 AM vjudeu via bitcoin-dev <bitcoin-dev@lists.linuxfoundation.org> wro= te:
Isn't it enough to just gen= erate a seed in the same way as today, then sort the words alphabetically, = and then use that as a seed? I know, the last word is a checksum, but there= are only 2048 words, so it is not a big deal to get any checksum we want. = If that is insecure, because of lower possible combinations, then it is alw= ays possible to increase the number of words to compensate that.


On 2022-07-08 04:27:21 user Eric Voskuil via bitcoin-dev <bitcoin-dev@lists.linuxfoundation.org> wrote:


Without a performance requirement there is no reason you can=E2=80=99t stor= e the BIP39 words in any order you want. So it=E2=80=99s certainly possible= , just brute force the recovery. If you have less than a second vs. a few d= ays then it=E2=80=99s a different question.


e


On Jul 7, 2022, at 18:48, Bram Cohen via bitcoin-dev <bitcoin-dev@lists.linuxfoundation.org> wrote:
Part of the rules of my challenge is that the 'new' words need to b= e in the same pool as the 'old' words, so any ordering is okay. Wit= hout that requirement it's mathematically very straightforward.


On Thu, Jul 7, 2022 at 10:52 AM Pavol Rusnak <stick@satoshilabs.com> wrote:
There is. Just encode the index of permutation used to scramble the otherwi= se sorted list. For 12 words you need to store 12! =3D ~32 bits so 3 words = should be enough.=C2=A0


Repetitions make this more difficult, though.=C2=A0


On Thu 7. 7. 2022 at 19:41, Bram Cohen via bitcoin-dev <bitcoin-dev@lists.linuxfoundation.org> wrote:
On Thu, Jul 7, 2022 at 7:43 AM Anton Shevchenko via bitcoin-dev <bitcoin-dev@lists.linuxfoundation.org> wrote:
I made a python implementation for a different mnemonic encoding. The encod= ing requires user to remember words but not the order of those words.
The code is open (MIT license) at https://github.com/s= ancoder/noomnem



Thanks Anton. There's an interesting mathematical question of whether i= t's possible to make a code like this which always uses the BIP-39 word= s for the same key as part of its encoding, basically adding a few words as= error correction in case the order is lost or confused. If the BIP-39 cont= ains a duplicate you can add an extra word.
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