Subject: Re: [bitcoin-dev] No Order Mnemonic

From: Eric Voskuil <[email-masked]>
Date: Thu, 7 Jul 2022 19:19:43 -0700

Without a performance requirement there is no reason you can’t store the BIP39 words in any order you want. So it’s certainly possible, just brute force the recovery. If you have less than a second vs. a few days then it’s a different question.

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> On Jul 7, 2022, at 18:48, Bram Cohen via bitcoin-dev <[email-masked]> 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 <[email-masked]> wrote:
>> There is. Just encode the index of permutation used to scramble the otherwise sorted list. For 12 words you need to store 12! = ~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 <[email-masked]> wrote:
>>>> On Thu, Jul 7, 2022 at 7:43 AM Anton Shevchenko via bitcoin-dev <[email-masked]> 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 error correction in case the order is lost or confused. If the BIP-39 contains a duplicate you can add an extra word.
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>> -- 
>> Best Regards / S pozdravom,
>> 
>> Pavol "stick" Rusnak
>> Co-Founder, SatoshiLabs
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