In summary, my search is 70%
Is somehow now near of 100%?
Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 133. (Read 245170 times)
I think he's at 101%
Is somehow now near of 100%?
Is somehow now near of 100%?
In summary, my search is 70%
Is somehow now near of 100%?
I wonder, if we somehow know the starting number, which we are doubling, can we determine or calculate number of doubles it requires to reach value 1?
This is surely related to EC calculations and I believe it certainly will help minimize size of private key bay many folds, depending on doubles!!!
This is surely related to EC calculations and I believe it certainly will help minimize size of private key bay many folds, depending on doubles!!!
This task has an analytical solution.
Try to ask Gemeni Pro Chat https://chat.googlegemini.co/ about it.
If (a⋅n) mod b = 1
n = 2*ceil(LCM(a, b)/b)
LCM stands for Least Common Multiple. It is the smallest positive integer that is divisible by both of two or more integers
I've noticed a strange phenomenon in mod calculations. I wonder whether this particular question can be answered by someone fluent with maths:
4 8 16 3 7 11 15 2 6 10 14 1 5
5 10 15 3 8 13 1 6
6 12 1 7
7 14 4 11 1 8
8 16 7 15 6 14 5 13 4 12 3 11 2 10 1 9
Above numbers are running on mod 17. If we double a number, say 4, we double it 11 times until it results in 1.
Similarly if we pick number 5 and start doubling it, we need to double it 6 times to reach 1. and two times double for 6, and 4 time doubling with 7 and so on.
I wonder, if we somehow know the starting number, which we are doubling, can we determine or calculate number of doubles it requires to reach value 1?
This is surely related to EC calculations and I believe it certainly will help minimize size of private key bay many folds, depending on doubles!!!
4 8 16 3 7 11 15 2 6 10 14 1 5
5 10 15 3 8 13 1 6
6 12 1 7
7 14 4 11 1 8
8 16 7 15 6 14 5 13 4 12 3 11 2 10 1 9
Above numbers are running on mod 17. If we double a number, say 4, we double it 11 times until it results in 1.
Similarly if we pick number 5 and start doubling it, we need to double it 6 times to reach 1. and two times double for 6, and 4 time doubling with 7 and so on.
I wonder, if we somehow know the starting number, which we are doubling, can we determine or calculate number of doubles it requires to reach value 1?
This is surely related to EC calculations and I believe it certainly will help minimize size of private key bay many folds, depending on doubles!!!
Is there any code for scanning address in sequence using GPU?
if you trying to bruteforce the range,plz dont
Is there any code for scanning address in sequence using GPU?
You can count on one hand the number of people who have the knowledge to deal with this matter seriously.
Is there any code for scanning address in sequence using GPU?
Quote
There is no instant solution and it all boils down to the fact that you have to know C++ or some other compiled language like Rust........
Khm... Is there Python ECDSA (for secp256k1) library works with CUDA cores? I searched and didn't find.
You have to write your own.
I don't like the existing solutions in C++ either.
Whether it is SECP256K1.cpp or OpenSSL
There are too many calls through the various libraries.
My goal is to do it all from one single cpp file using only GMP.
hi there,
could we get a copy? hmm if not ok, but want to try your nicely made software piece,
you have a github.. or so. anyway thanks a lot.
could we get a copy? hmm if not ok, but want to try your nicely made software piece,
you have a github.. or so. anyway thanks a lot.
This involves many third-party commands, here the output of a local server, .exe, and several self-made modules are being taken on a one display, so it is difficult to share it at the moment. I will make it easier and upload on GitHub in the future
Quote
There is no instant solution and it all boils down to the fact that you have to know C++ or some other compiled language like Rust........
Khm... Is there Python ECDSA (for secp256k1) library works with CUDA cores? I searched and didn't find.
Zahid888, is the expert at prefix searches...probably has the world's largest prefix database, he can attest to what I am saying to you.
Yes, perhaps I have the world's highest number of prefixes for puzzle66. And maybe I have conducted the most experiments on prefixes worldwide, whether it be in the form of base58 or hash160.
Through these experiments, I have consistently encountered a 50-50 probability of outcomes.
Sometimes, a few bytes of a private key match, and at other times, they do not match at all.
So, there is no difference between a 'private keys with matched prefixes' and a 'totally random private keys'.
I save addresses with prefixes only as a 'proof of work'; beyond that, these addresses serve no further purpose.
If your objective is solely to locate addresses with matching prefixes, it would be advisable to opt for the vanity search implemented by WanderingPhilosopher, which is based on multi-GPU counting, rather than attempting to enhance speed in Python, which never ever compete with GPU-counting.
Edit :
Anyway, I have scanned about 6.5% of puzzle 66 keys individually, I need your best wishes, and a lot of good wishes from my side to all of you.
could we get a copy? hmm if not ok, but want to try your nicely made software piece,
you have a github.. or so. anyway thanks a lot.
Thank you, colleagues, for suggesting Rust and C 
, but for educational purposes, what is specifically required is Python code for CUDA that can brute-force keys.
, but for educational purposes, what is specifically required is Python code for CUDA that can brute-force keys. @numba.jit....To use Numba to compile the performance-critical parts of code into machine code. But this will not work
with large numbers over puzzle 64
example
@njit
def add_numba(P, Q, p=modulo):
@njit
^
This error may have been caused by the following argument(s):
- argument 0: Int value is too large: 110560903758971929709743161563183868968201998016819862389797221564458485814982
- argument 2: Int value is too large: 115792089237316195423570985008687907853269984665640564039457584007908834671663
Numba does not support big int. It is essentially limited to integer types that are supported by numpy. The max integer width is currently limited to 64-bit.
GMP is the best option for now.
Numba have limitations regarding the types it supports for GPU computations. If you need to work with big integers or other types not supported by Numba directly, you might need to use PyCUDA or other libraries specifically designed for CUDA programming in Python.
And you’ll still need to write the kernel in CUDA C++.
There is no instant solution and it all boils down to the fact that you have to know C++ or some other compiled language like Rust........
And even if you know C++ very well, it's not easy to play with such big numbers. I ran into a lot of bugs in my early attempts.
Thank you, colleagues, for suggesting Rust and C , but for educational purposes, what is specifically required is Python code for CUDA that can brute-force keys.
, but for educational purposes, what is specifically required is Python code for CUDA that can brute-force keys. 
if you like the maximum possible speed than you need to go for C. This is primarily because C is a compiled language, while Python is an interpreted language. That means C code is compiled directly into machine code, which can be executed directly by the computer's hardware. On the other hand, Python code is interpreted, meaning that it's translated into bytecode by the Python interpreter and executed line by line at runtime. Additionally Python utilizes automatic memory management via garbage collection which can introduce overhead.
Frankly speaking, go for C if you like to achieve the best performnce. While C tends to be faster than Python in terms of raw performance, Python offers advantages such as ease of use. And as already mentioned, Rust is also interesting because it combines the performance of languages like C with the safety and concurrency features of modern languages.
Frankly speaking, go for C if you like to achieve the best performnce. While C tends to be faster than Python in terms of raw performance, Python offers advantages such as ease of use. And as already mentioned, Rust is also interesting because it combines the performance of languages like C with the safety and concurrency features of modern languages.
For educational purposes. Can anyone provide Python code for iterating through private keys in a specified range that works with CUDA? Or a link to such a resource.
Thank you.
Thank you.
It is 20 times faster in Rust then Python/Numba
https://github.com/Rust-GPU/Rust-CUDA
https://github.com/Rust-GPU/Rust-CUDA/blob/master/guide/src/guide/getting_started.md
But there is a 0 puzzle script written in Rust.
There is this one here as an example
https://bitcointalksearch.org/topic/m.63539973
Wow I just did the math to see how long it would really take if you went from start to finish with no stride for puzzle 66
With the start and end params of 20000000000000000...3ffffffffffffffff and a 3080 at 1100 MKeys/s or 1.1 billion keys a second
I got
33,539,534,679 seconds
558,992,244 minutes
9,316,537 hours
388,189 days
1,063 years
Then 1000 billion keys a second would take one year (31,536,000 seconds) ? With the start and end params of 20000000000000000...3ffffffffffffffff and a 3080 at 1100 MKeys/s or 1.1 billion keys a second
I got
33,539,534,679 seconds
558,992,244 minutes
9,316,537 hours
388,189 days
1,063 years
For educational purposes. Can anyone provide Python code for iterating through private keys in a specified range that works with CUDA? Or a link to such a resource.
Thank you.
Thank you.
Has anyone ever figured out what the #64 and #120 keys were, up to now?
Obviously not, this was just my wishful thinking when I had less experience ⬇
16jY7qLJXAVFd7AJXJ5N8xT9DEs24NDaXV F7051F24C01D5BB2
16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN F7051F27B09112D4
feeling too much unlucky
I should have taken these green letters seriously. I hope Tepan has some new idea
Has anyone ever figured out what the #64 and #120 keys were, up to now?
sure, i'll dm you, btw your codes looks cool and more advanced!
wow!, i'm beginner can you teach ?!
Sure, I'll see what else i can do to improve your code.
i already make some display looks clean and more likely like your display codes on terminal
i'll send the python files, i hope you can help me
, thanks! Jump to: