Sorry, I totally forgot the checksum. You can take my ops-per-second and multiply it by about 16 (because of the 4 check bits), making a delete + two swaps or 4 swaps, etc. still pretty reasonable.
On Mon, Jul 11, 2022 at 9:11 AM Erik Aronesty <e...@q32.com> wrote: > 1. You can swap two positions, and then your recovery algorithm can > brute-force the result by trying all 132 possible swaps. > 2. You can make a single deletion and only have to brute 2048 > 3. You can keep doing these, being aware that it becomes geometrically > more difficult each time (deletion + swap = 270k ops) > 4. A home PC can make 20k secpk256 operations per second per core, so try > to keep your number under a few million ops and it's still a decent UX > (under a minute) > > > On Sat, Jul 9, 2022 at 8:01 PM Anton Shevchenko via bitcoin-dev < > bitcoin-dev@lists.linuxfoundation.org> wrote: > >> I would say removing ordering from 12-word seed reduces 25 bits of >> entropy, not 29. Additional 4 bits come from checksum (12 words encode 132 >> bits, not 128). >> >> My idea [for developing this project] was to feed its output to some kind >> of AI story generator (GPT-3 based?) so a user can remember a story, not >> ordered words. But as others pointed out, having 12 words without order is >> probably good enough. So at this point there's not much sense of using the >> proposed encoding. Unless a remembered story has wholes/errors. In this >> case recovering few words would be easier with unordered encoding. Any >> thoughts? >> >> -- Anton Shevchenko >> >> >> On Sat, Jul 9, 2022, at 1:31 PM, Zac Greenwood via bitcoin-dev 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 >> >> _______________________________________________ >> bitcoin-dev mailing list >> bitcoin-dev@lists.linuxfoundation.org >> https://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev >> >> >> _______________________________________________ >> bitcoin-dev mailing list >> bitcoin-dev@lists.linuxfoundation.org >> https://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev >> >
_______________________________________________ bitcoin-dev mailing list bitcoin-dev@lists.linuxfoundation.org https://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev
