I have 26 out of 24 mnemonic words. Am I able to brute force still?

mcdouglasx

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New ideas will be criticized and then admired.

Quote from: BlackHatCoiner on August 16, 2024, 04:41:04 AM

So, based on your input:

You have a seed phrase where the first 6 words and their order are known. For example, you have: crawl vanish secret like rail blush <and 18 unknown words missing>
You have a list of 26 words which are valid out of the 24, but I'm not sure what this means. Is there a passphrase which extends the 24 word-phrase, or does your 26 word-list contain the 18 other words which you need to complete the 24 word-phrase?

Even without a passphrase, if you know the exact 18 words but not their order, you would still need to search through 18! (6,402,373,705,728,000) permutations to cover the entire space of possible seed phrases. Even with immense computing power capable of searching 1 million seed phrases per second, it would take about 6.4 billion seconds to complete, or roughly 205 years of continuous operation. And that's your best case scenario.

There would be approximately 1,216,451,004,088,320,000 possible combinations, because there are 20 words to combine to form permutations of 18 words.

BlackHatCoiner

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Farewell, Leo

So, based on your input:

You have a seed phrase where the first 6 words and their order are known. For example, you have: crawl vanish secret like rail blush <and 18 unknown words missing>
You have a list of 26 words which are valid out of the 24, but I'm not sure what this means. Is there a passphrase which extends the 24 word-phrase, or does your 26 word-list contain the 18 other words which you need to complete the 24 word-phrase?

Even without a passphrase, if you know the exact 18 words but not their order, you would still need to search through 18! (6,402,373,705,728,000) permutations to cover the entire space of possible seed phrases. Even with immense computing power capable of searching 1 million seed phrases per second, it would take about 6.4 billion seconds to complete, or roughly 205 years of continuous operation. And that's your best case scenario.

NotATether

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bitcoincleanup.com / bitmixlist.org

Quote from: BitcoinBarrel on August 11, 2024, 01:25:42 AM

For $70 you might as well mine some meme coin with the hashpower.

Not to mention, it will probably take several months for all that mining to convert into a profit.

But hey at least it's better than throwing 70$ of compute at BTCRecover and hope to solve a puzzle with a 0.02% success rate for the amount of hardware you are using.

The costs will quickly exceed the amount of money that is to be recovered, even in the event of a bull run.

SilverCryptoBullet

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Probability of seed phrase collision and brute forcing.

There are many math and examples in that topic and it's headache to understand all of them. The information I got from it is, it's very hard to brute fore a wallet mnemonic seed words if you don't have any known word or have very few words of the mnemonic seed. The second case is your case and chance for you to get it done is very small.

Seed splitting is a bad idea because difficulty of brute force like your case.

BitcoinBarrel

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Fill Your Barrel with Bitcoins!

For $70 you might as well mine some meme coin with the hashpower.

nc50lc

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Quote from: cryptorecovery.io on August 10, 2024, 04:33:22 PM

Recovering this 26-word mnemonic phrase, assuming it follows BIP-39 -snip-

OP is trying to recover a 24-word mnemonic phrase from 26 possible words.

And for someone who's offering a recovery service, you're not familiar of BIP39 specs that a 26-word mnemonic isn't following the standards.
Since; 26 * 11 (bits per word) = 286bits can't have an entropy divisible by 32bits when the right length of checksum is removed.

Quote from: graphite on August 10, 2024, 11:14:53 PM

-snip- but I don't think that algorithm would be possible in this instance.

Yes, but that's a weird addition to your post though.

It's improper to correlate the method of an attack with ECDLP on a private key with limited search space (puzzle transactions)
with bruteforce attack to a mnemonic phrase just because the possible number of combinations are similar.

The 'master private key', 'extended private key' or 'private key' pair of a public key that he may have should be in full length.

graphite

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Im assuming its a 24 word seed so should only need 18 words if the first 6 are in order. So you would need test 20!/2! combinations. Which is around 2^60 and should be possible since the bitcoin puzzles have been solved up to 2^65. Also the seeds have checksum bits which could reduce the total combinations to test down to 2^52. 24 words out of a list of 2^11 words is in total 2^264 of entropy and 8 bits of that is a checksum. and If you have the public key you might be able to use pollard's kangaroo algorithm to solve it in 2^30 time but I don't think that algorithm would be possible in this instance.

cryptorecovery.io

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Recovering this 26-word mnemonic phrase, assuming it follows BIP-39 and only knowing the order of 6 words, would take too much time to brute force and would be insufficient. To recover it, you should try to reduce the number of possibilities by, for example, gaining knowledge about the order of the remaining 14 words. This would significantly lower the number of possibilities and increase the chances of successfully recovering it through brute force. Otherwise the numbers are way too big

Grantrocks

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Quote from: greenAlien on April 24, 2024, 02:03:10 AM

Do you know ANY right position of the words? that would reduce the number of possibilities. Otherwise... As the rest of the people have told you, you need many lifetimes to solve it...

I know the first 6

greenAlien

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Do you know ANY right position of the words? that would reduce the number of possibilities. Otherwise... As the rest of the people have told you, you need many lifetimes to solve it...

pooya87

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Quote from: apogio on April 23, 2024, 11:54:52 AM

The first word can be in 20 positions.
The second word can be in 19 positions.
...
The 20th word can be in 1 position.

So it's 20x19x18x...x1 = 20! = 2,432,902,008,176,640,000

You have 20 words to place in 18 positions (the first 6 are already filled and 24-6=18) so it is 20*19*18*...*4*3

Grantrocks

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It's alright Its not a big prize anyway only 70 USD as of rn. Thanks for the advise I'd probably be better off cracking the btc transaction puzzles lol.

philipma1957

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Quote from: hosseinimr93 on April 23, 2024, 12:05:09 PM

26 out of 24 words? Are you saying you have two extra words?

If that's the case and since you say you know the 6 first words in the correct order, there should be 1.2 *10¹⁸ possible combinations. Assuming your seed phrase is BIP39, around 4.7 * 10¹⁵ out of those combinations should be valid on average. That's a very big number and you can't brute-force your seed phrase.

he could attempt brute force but mathematically it would take many many many lifetimes.

any number can be attempted to be brute forced.

but as you said the number is too large to think it is worth while.

at the op if the value of cracking it is high you can stab at it but you won’t hit it.

I would say it is harder to hit than it would be to pick the correct grain of sand out of the Sahara desert on 1 try.

hosseinimr93

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26 out of 24 words? Are you saying you have two extra words?

If that's the case and since you say you know the 6 first words in the correct order, there should be 1.2 *10¹⁸ possible combinations. Assuming your seed phrase is BIP39, around 4.7 * 10¹⁵ out of those combinations should be valid on average. That's a very big number and you can't brute-force your seed phrase.

Quote from: apogio on April 23, 2024, 11:54:52 AM

So it's 20x19x18x...x1 = 20! = 2,432,902,008,176,640,000

If I got OP correctly, there should be (20!)/2 possible combinations.
(Seed phrases with invalid checksum are also included.)

apogio

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You have to place 20 words in the correct order (26 total - 6 in known places).

So how many combinations (the correct word is permutations because order matters) of 20 objects do you get ?

The first word can be in 20 positions.
The second word can be in 19 positions.
...
The 20th word can be in 1 position.

So it's 20x19x18x...x1 = 20! = 2,432,902,008,176,640,000

How fast does this get brute-forced? I will leave this to be answered by the rest of the forum. I have no idea! I was never good in brute-force stuff.

Grantrocks

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So I'm solving a private puzzle and I have a list of 26 valid mnemonic words out of 24. The first 6 are in order and the rest are not. Would it be possible to bruteforce this mnemonic still? If so how can I and how long would it roughly take?

Topic: I have 26 out of 24 mnemonic words. Am I able to brute force still? (Read 318 times)