Topic: BitCrack - A tool for brute-forcing private keys - page 3. (Read 76581 times)

Sorry for confusion. I just try to understand the whole thing, that is why i am asking. For instance a wallet 1BgGZ9tcN4rm9KBzDn7KprQz87SZ26SAMH had a very simple private key of 0x0000000000000000000000000000000000000000000000000000000000000001? This means it is easy to crack it as the only thing is needed is checksum?
So if i would ever find an address with a privkey of 0x0000000000000000000000000000000000000000000000000000000000000001 but no public key - is it something i would have to run 2^65 keys to open it?

Can you show us what you mean by zero bit private key? If it has "zero" bits then how can you derive an address from it? But there is a solution for you, it's not related to zero or 256, you can find 2^96 different private keys for every single address out there, under one condition, you'd have to brute force 2^80 keys to find one collision.(in theory)
Without public key, only having the address and knowing the exact range, you can go ahead and try puzzle #66, even if you halve the range, you'd still need to grind 2^65 keys.😉

it will take you 1.42 million years to find the solution to your zero-bit question

Hi guys,

I was reading this thread and found it interesting. Just a few questions get in my mind:
What if I find a zero-bit privkey address? How long would it take me to find a private key for this address, knowing only the address? Is it a million years, assuming that I don't have a public key?

What were you developing in 30 years? If it was anything crypto/ security related, you should contact your clients and tell them they are compromised. 😅
Kidding.

---

Welcome on board!
I can't really understand your goal here, why are you trying to reverse a sha256 of a hex private key, which is BTW 17 characters and it's hash has no use in at least in Bitcoin.
Private keys are irrelevant to hashes and addresses, once you generate your priv key, then derive your public key from it, after that you perform sha256 on the public key and then rmd160 on that sha256 hash, it's not over yet,  When you have rmd160 you'd add version/network byte then use a double sha256 to derive the checksum in order to have an address.

So if you want to reverse anything, try by reversing the rmd160 hash of puzzle #66, when you find which sha256 hash outputs the rmd160, you have either found a collision or the original hash, but knowing that is impossible, anyways after having the sha256 hash, you'd need to reverse it back to the public key, and guess what?

There is no need to reverse 2 hash functions to find the public key, you already have access to public keys of puzzle 130, 135 etc which are housing more coins than 66.
I suggest you focus on public keys if you are good with math.😉

You're mistaken. P64 was depleted only 1 year ago.

P65 was depleted by different algo.

So, we are not really STUCK with 66

I suggest you visit this useful site,  https://learnmeabitcoin.com it has useful introduction about bitcoin system, elliptic curve etc, now you are trying to learn running before walking, that can't happen, but if you are interested in learning, first do the basics and when you understand how everything works, you will come here for more advanced technics, then those who know things will try to explain the real deal stuff to you.

For bitcoin, you can generate all addresses/keys starting from 1 to 115792089237316195423570985008687907852837564279074904382605163141518161494336

So this is 1 which is a 1 bit key, all the standard wallet generated keys have 78 digits, they are all 256 bit, in hex they have 64 characters.  Just visit the site, no one could explain these things better than the site author.😉

I realize that the combination or possible hex numbers are very large, but of all the btc addresses created, 100% are 256 bits? until the first btc addresses created? A few weeks ago I saw someone talking about trying to crack a 2015 satoshi wallet or something similar with a lot of btc having only the pubkey, assuming it was less than 256 bits, right? I'm not confused about this


Regarding the bip39 seeds, if I restored 1 million bip 39 phrases of 12 words, and another million bip 39 phrases of 24 words (I am not referring to creating a new wallet, but to restore and use 12 random words from the bip39 and expect to find 1 used address) would there be the same chances of finding 1 address with at least 1 transaction as through hex and btc? That is, if there are 2048 words, and I only base it on bip39 phrases of 12 words, there are a total of 12,980,005,607,406,348,644,160,768,000 possible combinations in total, let's say that only 50% are valid and the rest are not, it is still a huge number, but a lot smaller than that of btc, also with the words bip39, there are possibilities of obtaining more than 1 transaction in eth bnb or matic, from hex of btc only 1 network is checked, that of btc

Likewise, BTC is a huge number.

0% = 000000000000000000000000000000000000000000000000000000000000001

0.00005% = 000008b18610932ab6b2906ea3dfeaa8da073a862d7e0d7ffffffffffffffffc4

0.001% = 0000a7c5ac471b4784230fcf80dc33721d52f8aac2ffe738776930d94f9175a8


It really is a more than astronomical number.


So, hahaha the chances of getting puzzle 66 through random mode and GPU, and the chances of getting 130 through kangaroo or bsgs, between the two which has more chances?



PD: I have tested 231 possible combinations of 12 bip39 words (12-word phrases) and the result is 142 invalid phrases and 89 valid phrases.
Even so, it is still a huge number, close to 4 trillion possible combinations, and I also doubt that any bayc hold uses 12 bip39 words instead of 24

They are huge numbers but no comparison with btc

You can't search for different address types at once, there is is no need to do that, if addresses have their public keys revealed, you can just generate public keys and compare with your list, tools such as kangaroo and BSGS work with one input at a time, though I know this for sure to be true for kangaroo, but even if a tool operates with several keys in parallel, performance will get lower the more input you have.

But to show you an estimation, you'd need 1000 GPUs to run at 1B key/s for around 850 days non-stop to exhaust the entire 66 bit range, maybe half of that to search 130 public key if you know how to reduce the bit range though.

You should try out with small ranges and input your own selected keys and then start searching, generate 1 billion keys sequentially in 60 bit range and run all the tools you have.

There are already so many discussions about the probability of finding a match, something about using the energy of the sun for billions of years "only to count from 1 to 2^256" and lose to the death of the universe before finishing the simple counting.

Private keys start with 63 zeros and a single 1 like this :

First key:
0x0000000000000000000000000000000000000000000000000000000000000001

To this point, last key  minus 1 :
0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141

Go to a site like private key dot pw, and go to different pages to see it yourself, but never paste or enter your own private key because you will lose your funds.

I understand... I'll think twice about that idea again... Hahaha



Changing the subject radically, to try to try your luck with puzzle 66 and puzzle 130 since 125 was opened, are you trying to use these codes, do not modify main or anything, any improvement that you can make or do you do wrong?


VBCr.exe -begr 20000000000000000 -endr 3ffffffffffffffff -t 6 -dis 4 -r 30000 -drk 1 -i addresses.txt -m 1 -o key.txt -gpu

cuBitCrack.exe -b 96 -t 512 -p 1024 --keyspace 26000000000000000:2ffffffffffffffffff -i addresses.txt -o FOUNDKEY.txt

BSGS
40000000
20000000000000000000000000000000000
3ffffffffffffffffffffffffffffffffff
03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852


 And another question, from the 1st bitcoin address created to the last, both compressed and decompressed, are 100% of those addresses 100% 256 bits?

There are bitcoin addresses with more than 1 transaction 22,285,749,153

and there are 10,527,367,033 that start with 1 (Compressed and decompressed).
The rest start with bc1 and 3

Let's say I get 10,000 random hex keys per second, if I convert each hex key to its respective address (Compressed, decompressed, bc1 and 3)
What is the chance of getting 1 match against puzzle 66 or 130 using bsgs or kangaroo?



PD From the range 004189374bc6a7ef9db22d0e56041893746922865feb562d1da1bf120f9175e4 it would be 0.09999%
and the 66 puzzle is SO difficult, now I'm realizing what a big number it is.

It doesn't matter what your intentions are, even if you were trying to rob all bitcoiners of their coins, nobody would touch you, actually they would gladly help you through the process, because as you mentioned, puzzle keys are extremely small compared with actual randomly generated keys holding people's funds, to give you something for understanding, just imagine the entire universe and all it's atoms, now all the 160 puzzle keys combined can only occupy something around 1/89th of earth.

Ever since 2015, we are stuck at puzzle 66, you should double the key space size each bit you go up.
If you have been reading these topics for the past few month, you should already know about the size of 2^256.

I just want to save your time, don't bother searching because you will not find a single match, you'd have a chance to win in lottery 1000 times in a row in your life time, but finding a single match? Not a chance.  To test your luck just go buy lottery ticket.

However, there is a +$25m prize, almost as 25 nobel prizes ready for grab, if anyone in the world claims they can hack bitcoin keys, they should try these legitimate puzzles because they are the easiest ones.

A few month from now, I will see you with an electricity bill in your hands.😉

Hello ! Thank you for answering me first of all, and Hahaha, with a lot of luck, I will be able to obtain 1 or 2 private keys from the BTC.TXT addresses, and in btc.txt all the addresses there are compressed and decompressed, I don't know which ones are decompressed and which ones are compressed

Another thing, in the 19GB, there are all the addresses that have ever received a balance, not those that contain a balance, that would be around 50% or 60% less, I want to focus the search on all the addresses that have ever received btc and that begin with 1 regardless of whether they are compressed or decompressed, to check what the probability would be that someone would find an address already used

But with that 45% data I will then try to specify in btcC.txt, all the compressed bitcoin addresses and thus the file will weigh less, and the speed will increase by not having decompressed addresses


I don't know how to use C++, only python node and some already created programs such as bitcrack VBCr and VanBitCrackenS1, work perfectly and find matches, but as long as the text file weighs less than 2 GB.

I forgot the last question but thank you so much for answering.


Another question, it would be easier to obtain all the public keys of all the btc.txt addresses, which have spent btc, and create pubbtc.txt with all the public keys of their respective wallets, there would be more chances of obtaining a match. Or is it the same if you don't know the hex range where to search? To say that I have been watching these forums, especially the BTC puzzle ones, and I have learned a lot over time about Kangaroo Baby Giants and the other programs, although I believe the account practically 1 day or 2 days ago, I have been seeing these things for months and I am somewhat informed.




PD: Note that I am not trying to hack or do illegal things or steal funds from people's wallets. I want to know when it is said: BTC is safe, how safe it is. My intention is to search for 100% random keys until one matches an address. from btc.txt, whether or not you have a balance, having more than 1 transaction is enough

First of all, don't empty all our addresses when you brute force them all.🤣
Second, why would you want to search for both address types? Searching for uncompressed addresses reduces your speed by 45%, and I don't think any txt file could handle 19GB of text, I mean for windows is 2GB max, larger than that, it won't open, maybe that's the problem. You better to use keyhunt with all ever used public keys, just any key.
Note that to achieve same speed, you can't use python, best is C++. Anyways I like your enthusiasm thinking it can hit a key, maybe you should try....  wait, how can all the addresses with balance occupy 19GB?  How many are there? Can't you make it 2GB to see if it works or not?


What's that? Here the post ended what happened?😅

Good afternoon, I hope everything is well.

My idea is the following:

I have a BTC.txt file, which has all addresses with a balance greater than 0 starting with 1 (both compressed and uncompressed) (BTC.TXT IS 19 GB AND has 1 btc address in each line)

I want to run bitcrack, VanBitCracken or VBCr which work perfectly, but when I put -i btc.txt, in bitcrack it stays loading for about 6 hours and is removed, and in VanBitCracken and VBCr it doesn't seem to read it directly

I would like to know if I can modify the exe file from either bitcrack, VanBitCracken or VBCr to perform my search or is there another program?


PS: Create a code in python that more or less does the following:

Creates a random 32 byte hex key, then converts it to its respective WIF key both compressed and decompressed, then checks if the compressed and decompressed address of that address matches any address in btc.txt, if so it saves it to found. txt, if it does not find anything it will continue searching until it finds a match OR until it stops the program. What I mean is that in python I have achieved my goal and I would like to do the same but with my gpu

This is the result of my python file WITHOUT GPU and using 4 processors out of 16:

Generated: 20400000 (SHA-256 - HEX)
Last generated key in hexadecimal: 13d0f03fab9c75a97fa6492b8b832b0ce7b11b8cc30f25e8eb1164b9fbc57ded
Compressed Bitcoin address: 1LZfGfy88QjBVkB3DTqpjFwonStb93Bed4
Decompressed Bitcoin address: 1DUXqLAJch8pncVKf63HZBefaKT3b81p5E
Generation rate: 4497.490059165694 keys per second

Generated: 20500000 (SHA-256 - HEX)
Last key generated in hexadecimal: f55599ce4720d4061894e644b99e190ffebb758c7780572e43651e9d622c2f69
Compressed Bitcoin address: 1Nr4BuyWgxS6Bmd4WzUZrkRFZHM6FQcqXp
Decompressed Bitcoin address: 1FgYyC5RAQwr8AUqHH7CP7Sx52avAJYtb
Generation rate: 4572.502992815531 keys per second



PD: I am aware of the difficulty and luck it takes to get 1 match that is very very very unlikely, but I know what I want and that is it Smiley

PD2: If necessary, I could search O randomly OR sequentially, for example I would put from 20% to 40% of the total range, I would scan that range randomly and sequentially for a few days or weeks, and then I would go to the range 40% 60% and so and one last question



Excuse my English, I am from BCN (SPAIN) and I only speak Spanish and Catalan, so my English is translated

you understand me right

I asking about time need to....

thank you for you answer

I don't really understand what you trying to say, but from what I did understand,  here is 2^55 * 2000 = 72057594037927936000   and this is 2^66 =  73786976294838206464  difference between them is = 1729382256910270464

Before I loose my mind, let me give you what you initially asked, searching  2000 ranges with size of 2^55, with speed rate of 1 billion per second will take around 2284 years to brute force.

Edit: with 1000Bk/s, you can search that range in 833 days, and that's not even the entire 66 bit range.

no, I talk about 2000 pcs of 2**55. "2**55 * 2000 " = ?

??
...


What I'm talk about.

66 puzzle, I know what one of 2000 ranges, differ from 66 ouzzle range for 10 bit, and difference in range 2**55 . My idea was take 2000 end ranges (one by one) and  subtract 2**55 from all of them and search in this 2000 ranged and get 66 puzzzle.....


this example with difference from 66 bit to one of end range is 2**45:

4473851742395 5093917
6674995201175590393
6674990727323847998
25033953356734 5329065
6675015761277204732
6674990727323847998
17273917703686 6208005
6675008001241551684
6674990727323847998
18833788691815 6540667
6675009561112539813
6674990727323847998


4473851742395 -difference betwen end range and 2**66 bit puz ,she is <=2**45;

 5093917 range number
6674995201175590393 -  END  of range

6674990727323847998 - target 2**66 priv

You mean dividing 2^55 by 2000 ranges? Well in that case it would take exactly the same as searching 2^55,  total range, if your ranges have specific size, then you should tell us their size or just multiply your range by 2000 and then divide the result by your speed per second, the answer would be seconds taking to search your range. Then divide 60 to get minutes, again divide by 60 to get hours, then divide by 24 to get days, /365 to get years etc, but you already knew that right?

Hello

how many time will take search 2000 ranges in 55 bit ?

Thank You

Not cool to post a suspicious link to some unknown blog, you should highlight the important parts and post them instead. But something tells me, if who ever wrote that paper knew how to crack private keys, he wouldn't publish it.😉