📅 Original date posted:2022-07-08
📝 Original message: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 at 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.
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On Jul 7, 2022, at 18:48, Bram Cohen via bitcoin-dev <bitcoin-dev at 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 at 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 at lists.linuxfoundation.org> wrote:
On Thu, Jul 7, 2022 at 7:43 AM Anton Shevchenko via bitcoin-dev <bitcoin-dev at 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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vjudeu at gazeta.pl [ARCHIVE] /
npub135…nssga
2023-06-07 23:11:16
in reply to nevent1q…7n4e
Author Public Key
npub1357006afyypkgz03lmq8fnuvlkyjt0rukx8rt56ck8xv396jaceqmnssgaPublished at
2023-06-07 23:11:16Event JSON
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"created_at": 1686179476,
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"content": "📅 Original date posted:2022-07-08\n📝 Original message: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.\n\n\nOn 2022-07-08 04:27:21 user Eric Voskuil via bitcoin-dev \u003cbitcoin-dev at lists.linuxfoundation.org\u003e wrote:\n\n\nWithout 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.\n\n\ne\n\n\nOn Jul 7, 2022, at 18:48, Bram Cohen via bitcoin-dev \u003cbitcoin-dev at lists.linuxfoundation.org\u003e wrote:\nPart 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.\n\n\nOn Thu, Jul 7, 2022 at 10:52 AM Pavol Rusnak \u003cstick at satoshilabs.com\u003e wrote:\nThere 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. \n\n\nRepetitions make this more difficult, though. \n\n\nOn Thu 7. 7. 2022 at 19:41, Bram Cohen via bitcoin-dev \u003cbitcoin-dev at lists.linuxfoundation.org\u003e wrote:\nOn Thu, Jul 7, 2022 at 7:43 AM Anton Shevchenko via bitcoin-dev \u003cbitcoin-dev at lists.linuxfoundation.org\u003e wrote:\nI made a python implementation for a different mnemonic encoding. The encoding requires user to remember words but not the order of those words.\nThe code is open (MIT license) at https://github.com/sancoder/noomnem\n\n\n\nThanks 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.\n_______________________________________________\nbitcoin-dev mailing list\nbitcoin-dev at lists.linuxfoundation.org\nhttps://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev",
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}