📅 Original date posted:2022-07-09
📝 Original message:Sorting a seed alphabetically reduces entropy by ~29 bits.
A 12-word seed has (12, 12) permutations or 479 million, which is ln(469m)
/ ln(2) ~= 29 bits of entropy. Sorting removes this entropy entirely,
reducing the seed entropy from 128 to 99 bits.
Zac
On Fri, 8 Jul 2022 at 16:09, James MacWhyte via bitcoin-dev <
bitcoin-dev at lists.linuxfoundation.org> wrote:
>
> 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
> _______________________________________________
> bitcoin-dev mailing list
> bitcoin-dev at lists.linuxfoundation.org
> https://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev
>
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Zac Greenwood [ARCHIVE] /
npub1gy…dvff7
2023-06-07 23:11:17
in reply to nevent1q…w3wy
Author Public Key
npub1gymmksd9tgwzc5w33umlx08sc2ggys3v2cucmpvl7yy9720wh49s8dvff7Published at
2023-06-07 23:11:17Event JSON
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"content": "📅 Original date posted:2022-07-09\n📝 Original message:Sorting a seed alphabetically reduces entropy by ~29 bits.\n\nA 12-word seed has (12, 12) permutations or 479 million, which is ln(469m)\n/ ln(2) ~= 29 bits of entropy. Sorting removes this entropy entirely,\nreducing the seed entropy from 128 to 99 bits.\n\nZac\n\n\nOn Fri, 8 Jul 2022 at 16:09, James MacWhyte via bitcoin-dev \u003c\nbitcoin-dev at lists.linuxfoundation.org\u003e wrote:\n\n\u003e\n\u003e What do you do if the \"first\" word (of 12), happens to be the last word in\n\u003e\u003e the list alphabetically?\n\u003e\u003e\n\u003e\n\u003e That couldn't happen. If one word is the very last from the wordlist, it\n\u003e would end up at the end of your mnemonic once you rearrange your 12 words\n\u003e alphabetically.\n\u003e\n\u003e However!\n\u003e\n\u003e (@vjudeu) Choosing 11 random words and then sorting them alphabetically\n\u003e before assigning a checksum would reduce entropy considerably. If you think\n\u003e about it, to bruteforce the entire keyspace one would only need to come up\n\u003e with every possible combination of 11 words + 1 checksum. I'm not the best\n\u003e at napkin math, but I think that leaves you with around 10 trillion\n\u003e combinations, which would only take a couple months to exhaust with\n\u003e hardware that can do 1 million guesses per second.\n\u003e\n\u003e\n\u003e James\n\u003e _______________________________________________\n\u003e bitcoin-dev mailing list\n\u003e bitcoin-dev at lists.linuxfoundation.org\n\u003e https://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev\n\u003e\n-------------- next part --------------\nAn HTML attachment was scrubbed...\nURL: \u003chttp://lists.linuxfoundation.org/pipermail/bitcoin-dev/attachments/20220709/fade4b82/attachment.html\u003e",
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