Thanks, Zac! I indeed did get the napkin math very wrong. I now get around 10^30 total possible phrases, which would take an impossibly long time to brute force. So, it is less entropy but probably still sufficient for low-stakes usage.
James On Sat, Jul 9, 2022 at 10:31 PM Zac Greenwood <zach...@gmail.com> wrote: > 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@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@lists.linuxfoundation.org >> https://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev >> >
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