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 big
> 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’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.
>
>
> 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 <stick@satoshilabs.com>
> 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 <
> 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 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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