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Author Topic: Shuffled Mnemonic  (Read 641 times)
otaviobps
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June 08, 2017, 05:17:23 PM
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Hey everyone.

I have all 12 words, but they are shuffled. Does somebody know how can I brute force them to find the correct order? I know one of the firsts addresses if it helps.

I'm no coder at all, so I don't know how to code or create programs.

Thanks!
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June 08, 2017, 06:42:27 PM
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Your looking bitcoin dave,  https://walletrecoveryservices.com/

goodluck

heres a coindesk article about him. http://www.coindesk.com/meet-man-will-hack-long-lost-bitcoin-wallet-money/
otaviobps
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June 08, 2017, 08:03:13 PM
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Thanks!
adaseb
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June 08, 2017, 09:05:08 PM
 #4

Hey everyone.

I have all 12 words, but they are shuffled. Does somebody know how can I brute force them to find the correct order? I know one of the firsts addresses if it helps.

I'm no coder at all, so I don't know how to code or create programs.

Thanks!

Just do it one by one, only 144 possible scenarios.

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DannyHamilton
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June 09, 2017, 12:22:43 AM
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Just do it one by one, only 144 possible scenarios.

Are you sure about that?

  • There are 12 possibilities for the first word
  • You have to try each of the 11 other words with each of the choices that you make for the first word
That's 12 X 11 = 132

For each of those 132 possibilities, there are 10 remaining words to try.

That's 132 X 10 = 1320

For each of those 1320 different sets of 3 words, there are 9 remaining words to try.

That's 132 X 9 = 11880

For each of those 11880 different sets of 4 words, there are 8 remaining words to try.

That's 11880 X 8 = 95040

For each of those 95040 different sets of 5 words, there are 7 remaining words to try.

That's 95040 X 7 = 665280

For each of those 665280 different sets of 6 words, there are 6 remaining words to try.

That's 665280 X 6 = 3991680

For each of those 3991680 different sets of 7 words, there are 5 remaining words to try.

That's 3991680 X 5 = 19958400

For each of those 19958400 different sets of 8 words, there are 4 remaining words to try.

That's 19958400 X 4 = 79833600

For each of those 79833600 different sets of 9 words, there are 3 remaining words to try.

That's 79833600 X 3 = 239500800

For each of those 239500800 different sets of 10 words, there are 2 remaining words to try.

That's 239500800 X 2 = 479001600

For each of those 479001600 different sets of 11 words, there is only 1 remaining word. Since there aren't multiple options, that remaining word doesn't increase the number of possibilities.

That's nearly 480 million possibilities.  (A bit more than 144).

Just doing it one-by-one by hand, might take a bit of time.

jeroenn13
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June 09, 2017, 12:25:17 AM
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8916100400000 is the number of possibilities, no? Is it not 12 ^ 12?
cr1776
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June 09, 2017, 12:32:57 AM
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8916100400000 is the number of possibilities, no? Is it not 12 ^ 12?

No, 12!  (12 factorial)
Or as Danny pointed out:
479,001,600
DannyHamilton
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June 09, 2017, 01:27:14 AM
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Or as Danny pointed out:
479,001,600

I can't think of any way to make it more clear to understand than I already have.

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June 09, 2017, 02:05:51 AM
 #9

8916100400000 is the number of possibilities, no? Is it not 12 ^ 12?

Its permutation because it has an order. The formula is n!/(n-r)! which (n-r)! is not shown because it is understandable that it is one because n and r are same in value hence if subtracted would result to 0! or 1.

8916100400000 is the number of possibilities, no? Is it not 12 ^ 12?

No, 12!  (12 factorial)
Or as Danny pointed out:
479,001,600

Your equation 12^12, I don't know where you get it but the right one has been pointed out. Danny's explanation is much easier to understand if you want to.

Combination on the other hand doesn't care for arrangement or order and the formula is (n!)/(r!(n-r)!).

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