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As I understand for BIP32. There are 2048 words.
12 words would give you 2048^12 combinations. However, not all combinations are valid because of the checksum, so how do i calculate the actual space?
How many combinations are there from the BIP32 mnemonic list?
Answers
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BIP 32 is not a mnemonic specification, so it has 0 possible combinations.
I assume you are talking about BIP 39. In that case there are are 2048^12 combinations for 12 word mnemonics, 2048^18 for 18 word mnemonics, and 2048^24 for 24 word mnemonics. However BIP 39 currently supports 8 languages, so there are (2048^12 + 2048^18 + 2048^24)*8 possible combinations.
The checksum doesn't matter because BIP 39 does not specify that the checksum must be enforced. The checksum can be invalid, and all that BIP 39 compliant software should do is show a warning.
Thank you for the response. I understand that there are 2048^12 combinations. Is there a way to know what combinations are invalid out of that space according to the checksum? Thanks for the correction on the BIP. – Kyle Graham – 2018-03-03T16:19:54.113
For a 12 word mnemonic, there are
2048^11*2048^(11-4)valid combinations, for 18 it's2048^11*2048^(11-6)and for 24 it's2048^11*2048^(11-8)– Andrew Chow – 2018-03-03T16:35:32.917I'm not sure I follow. Is the 12 word mnemonic just for one language? – Kyle Graham – 2018-03-03T16:40:57.507
Yes. All are for one language. Multiply by 8 for all combinations with all current BIP 39 wordlists. – Andrew Chow – 2018-03-03T16:44:13.283
So all possible combinations for 12 words is 2048^12 and for just valid words it is 2048^11*2048^(11-4) which is equal to 2048^18 , I think I am missing something as there seems to be more valid combinations than there are combinations – Kyle Graham – 2018-03-03T17:03:05.657
Ahh, whoops. I typed that wrong. It's
2048^11*2048/(11-4). Same pattern for other mnemonic lengths. – Andrew Chow – 2018-03-03T17:12:44.230Ahh that makes sense to me now, thank you for writing back and I saw someone with the same name as yours make a interesting proposal for partial signatures. Thanks again for the formula 2048^11 / 2048^(11-4) – Kyle Graham – 2018-03-03T17:57:08.083