--- Log opened Fri Nov 12 00:00:30 2021
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05:09 < andytoshi> nickler: can you join #bitcoin-rust
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05:15 < db> Rusty Bitcoins. I'll show myself out.
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14:59 < andytoshi> what is the origin of the bip39 checksum :(
15:04 < andytoshi> musing about how to deal with it if you're hand-generating keys ... i guess just keep regenerating the last word until you get one that a HWW accepts
15:09 < andytoshi> i guess i could say "when loading the key onto hardware, keep incrementing the last word til it works" which would bias the resulting key, but only by 8 bits, and wouldn't require a HWW at generation time
15:11 < _aj_> it'd bias it by 3 bits (24 word seed) to 7 bits (12 word see) wouldn't it?
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15:12 < _aj_> for the 24 word seed case, the HWW could just prompt for which of the valid 8 final words you want?
15:13 < andytoshi> for a 12-word seed the checksum is 4 bits, and for a 24-word seed the checksum is 8 bits
15:13 < andytoshi> _aj_: well, if the HWW could do custom thinsg like that then it could just not require the stupid checksum in the first place
15:14 < andytoshi> _aj_: on further reflection though i think "just increment til it works" won't actually bias the key (at least, not in a computationally accessible way)
15:14 < andytoshi> since every choice of the initial 23 words results in a new uniform distribution of allowable 24th words
15:14 < andytoshi> err not uniform
15:14 < andytoshi> but independent
15:15 < _aj_> i don't think so? the 24th word splits the word list into groups of 8 {w0..w255, w256..w511, ..} and exactly one word in each of those is valid
15:16 < _aj_> so incrementing will always pick a word from the first 256 in the word list
15:16 < _aj_> unless i'm missing some trickery?
15:18 < _aj_> oh, maybe it groups them by MSB rather than LSB
15:19 < andytoshi> _aj_: i'm not incrementing from 0
15:19 < andytoshi> i'm generating 24 random words then incrementing the 24th from wherever it initially landed
15:21 < andytoshi> also, even if i were incrementing from 0, i don't think the checksum actually splits the wordlist into groups of 8, but rather randomly sized groups
15:21 < andytoshi> because the checksum has no algebraic properties whatsoever
15:22 < andytoshi> so if i were to start incrementing from 0, i'd expect my last word to be drawn from a poisson-like distribution whose mean was 256
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17:12 < roconnor> andytoshi: If you did want to be clearly unbaised, you can split the word list into groups of 256 words.
17:13 < roconnor> and try replacing the last word with every word in the group that your random last word fell into.
17:13 < roconnor> Only one word from that randomly selected group will work.
17:13 < roconnor> Or equivalently instead of randomly generating the last word, randomly select one of these groups of 256 words, try each word.
17:14 < roconnor> only one word in the group will work.
17:14 < roconnor> That said, I ended up just computing the checksum on a freshly formated rasberrypi. :/
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18:21 < _aj_> andytoshi: oh, of course, then yeah, i agree
18:24 < roconnor> _aj_: I'm pretty sure you are right, the word list is split into 8 groups of 256 words.
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18:49 < roconnor> and for every prefix of 23 words, there are only 8 words that can be a valid 24th word.
18:50 < roconnor> and each of those 8 words lies in a distict group of 256 words.
18:52 < _aj_> roconnor: yep, but if you randomly choose a word, you'll randomly pick one of those distinct groups, and iterate until you get a matching checksum as apoelstra says
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19:47 < roconnor> as long as you loop back and stay within the randomly choosen group and not iterate into the next group.
19:48 < roconnor> I mean, if SHA256 is a random function, then even if you spill over into the next group, then it is still fair.
19:48 < roconnor> but looping withing th choosen group is fair even if SHA256 isn't a random function.
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--- Log closed Sat Nov 13 00:00:31 2021