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word list #2
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My slow brute forcer goes with ~1500 mnemonics per second, i guess i have
not updated the github code yet
…On Sun, Oct 8, 2023 at 8:56 AM Steve502r ***@***.***> wrote:
i actually scanned the article for all bip39 words but the list was way to
big for me to manually try to guess. I cant write code but i’m gonna look
for that list. i’ll get back to ya
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The text contains 448 total words out of the 2048 bip39 mnemonic words. I also tried brute-forcing for 12, 18 and 24 word mnemonics from the 448 words but all in vain. |
If either one of you have all the words i have code that will do all combination possible with the private key, public key, and address along with it. Even checks the checksums are correct. Im sure yall did this already but just in case. Sent from my iPhoneOn Jun 1, 2024, at 8:59 AM, SmartArt09 ***@***.***> wrote:
The text contains 448 total words out of the 2048 bip39 mnemonic words.
You can make your code a little better by just starting off with that many words.
I also tried brute-forcing for 12, 18 and 24 word mnemonics from the 448 words but all in vain.
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If either one of you have all the words i have code that will do all combination possible with the private key, public key, and address along with it. Even checks the checksums are correct. Im sure yall did this already but just in case. Sent from my iPhoneOn Jun 1, 2024, at 8:59 AM, SmartArt09 ***@***.***> wrote:
The text contains 448 total words out of the 2048 bip39 mnemonic words.
You can make your code a little better by just starting off with that many words.
I also tried brute-forcing for 12, 18 and 24 word mnemonics from the 448 words but all in vain.
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i also made a code to seek combinations of 12 words taken sequentially but
with different shifts, like 1,2,3,4,5,6,7,8,9,10,11,12;
1,3,4,5,6,7,8,9,10,11,12,13;
1,4,5,6,7,8,9,10,11,12,13,14;....,1,2,4,5,6,7,8,9,10,11,12,13;
but anyway i will not be able to check all
…On Mon, Jun 3, 2024 at 11:26 PM Steve502r ***@***.***> wrote:
If either one of you have all the words i have code that will do all
combination possible with the private key, public key, and address along
with it. Even checks the checksums are correct. Im sure yall did this
already but just in case. Sent from my iPhoneOn Jun 1, 2024, at 8:59 AM,
SmartArt09 ***@***.***> wrote:
The text contains 448 total words out of the 2048 bip39 mnemonic words.
You can make your code a little better by just starting off with that many
words.
I also tried brute-forcing for 12, 18 and 24 word mnemonics from the 448
words but all in vain.
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Sorry, there's actually 264 words from the whole story that are valid BIP39 words. |
Gotta do everything all over again because my laptop is now gone, so...no, I don't have all the words now. But you can write a simple script yourself. Just remove all the punctuations in the article except the spaces and split the entire string, then see if any of the words in this list are there in the original 2048 BIP39 English wordlist. I mentioned the correct word count here in this thread in another reply. |
BIP 44Sent from my iPhoneOn Sep 9, 2024, at 9:31 PM, Aarbi99 ***@***.***> wrote:
what is the derivation path for the seeds ?
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The normal Bitcoin derivation path, at least that's what I used when I checked it. |
Anyone found anything? |
nothing so far as i know
…On Sun, Nov 3, 2024 at 4:49 AM SmartArt09 ***@***.***> wrote:
Anyone found anything?
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Damn, so all of the puzzles are really luck based... |
Well... some of them i guess. Those that are seem easy - AR puzzles - are
also could be considered as lucky baseed. Cause some of us easily miss 1
correct words and some other would be lucky to catch that words
specifically.
…On Wed, Dec 25, 2024 at 3:55 PM SmartArt09 ***@***.***> wrote:
Damn, so all of the puzzles are really luck based...
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That's true. Any new story for this puzzle? Like somehow eliminating any of the total words? There's 266 words in my code that are in the BIP39 English wordlist and following is the permutation/combination calculation for total population "n" = 266, if a sample "r" = 12, is chosen (most likely at random): Permutations, nPr = 266! / (266 - 12)! = 97,551,751,568,636,575,002,304,512,000 Doesn't seem so solvable now... :( |
i actually scanned the article for all bip39 words but the list was way to big for me to manually try to guess. I cant write code but i’m gonna look for that list. i’ll get back to ya
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