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?
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1 Answer
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.
answered Mar 3, 2018 at 0:26
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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. Mar 3, 2018 at 16:19
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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 ♦Mar 3, 2018 at 16:35 -
I'm not sure I follow. Is the 12 word mnemonic just for one language? Mar 3, 2018 at 16:40
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1Ahh, whoops. I typed that wrong. It's
2048^11*2048/(11-4). Same pattern for other mnemonic lengths.– Andrew Chow ♦Mar 3, 2018 at 17:12 -
1@JoelAZ Duplicate words are allowed. Mnemonics are not generated by randomly choosing words, they are generated by choosing a large random number, splitting up the data into chunks, converting each chunk to a number, and looking up the word that corresponds. It is possible for multiple chunks to be the same value. Otherwise the seed would not be uniformly random and would reduce the search space necessary for brute forcing seeds.– Andrew Chow ♦Jan 12, 2022 at 1:46