Topic: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED== - page 49. (Read 47425 times)

I think an algorithm that uses n00by
based on subtract public keys

for example here http://gobittest.appspot.com/VanitySum
use addition rather than subtraction

hey supika, nice software i guess,
smart man you are, but when you transfer it to that another language thing you said,

could you do it a bit faster for us, so we can get used to that piece,

thanks bro

Why put it here if  "...  The first part of the script is useless/generic  ..." 

Doesn't have to delete any video. The first part of the script is useless/generic.

import collections
import time
import gmpy2
import random

EllipticCurve = collections.namedtuple('EllipticCurve', 'name p a b g n h')

curve = EllipticCurve(
    'secp256k1',
    p=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F,
    a=0,
    b=7,
    g=(0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798,0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8),
    n=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141,
    h=1,
)

def pubkey_point(pubkey):

    if len(pubkey) == 130:
        hexpubX = pubkey[2:66]
        hexpubY = pubkey[67:130]
        X = int(hexpubX,16)
        Y = int(hexpubY,16)
        return (X, Y)
    else:
        hexpub = pubkey[2:66]
        prefix = pubkey[1:2]
        X = int(hexpub,16)
        Y = pow(int(X**3+7), int((curve.p+1)//4), int(curve.p))
        if Y%2 !=0:
            if int(prefix)%2 !=0:
                Y = Y
            else:
                Y = curve.p - Y
        else:
            if int(prefix)%2 ==0:
                Y = Y
            else:
                Y = curve.p - Y
        return (X, Y)

def point_neg(point):

    if point is None:
        return None
    x, y = point
    result = (x, -y % curve.p)
    return result

def point_add(point1, point2):

    if point1 is None:
        return point2
    if point2 is None:
        return point1
    x1, y1 = point1
    x2, y2 = point2
   
    if x1 == x2 and y1 != y2:
        return None
    if x1 == x2:
        #m = (3 * x1 *x1 + curve.a) * inverse_mod(2 * y1, curve.p)
        m = (3 * x1 *x1 + curve.a) * gmpy2.invert(2 * y1, curve.p)
    else:
        m = (y1 - y2) * gmpy2.invert(x1-x2, curve.p)
        #m = (y1 - y2) * inverse_mod(x1-x2, curve.p)
    x3 = m * m - x1 - x2
    y3 = y1 + m * (x3 - x1)
    result = (x3 % curve.p, -y3 % curve.p)
    return result

def scalar_mult(k, point):
    if k % curve.n == 0 or point is None:
        return None
    if k < 0:
        return scalar_mult(-k, point_neg(point))
    result = None
    power = ' '
    addend = point
   
    while k:
        if k & 1:
            result = point_add(result, addend)
        addend = point_add(addend, addend)
        k >>= 1
    return result

def make_keypair(intkey, point=curve.g):
    public_key = scalar_mult(intkey, point)   #random.randrange(1, curve.n)
    return public_key

if __name__ == '__main__':
   
    start = time.time()
    target_bit = 45
   
    while True:
        tb = random.randrange(1, 2**target_bit)
        if tb.bit_length()            pass
        else:
            target_pnt = make_keypair(tb)
            break
    print ('Target random:', tb)
    print ('Target random hex:', hex(tb))
   
    target_pnt = '026ecabd2d22fdb737be21975ce9a694e108eb94f3649c586cc7461c8abf5da71a'
    target_pnt = pubkey_point(target_pnt)

I didn't want to say it here by either delete your video or blur out your code. If really don't want to share it yet.

what is B and C points ?

I do not plan to publish the code yet, but as the important parts are transferred to another language, I will publish the entire algorithm in this thread.

@n00by
i have a few questions about your code if you don't mind.
can you post your previous version of the code not the new one you working on ?
I can't send you a private message as i haven't posted anything here but registered a while back.

This is the release date of Python 3.7.5 )

please let us know more...

that video running a python script is dated in october 2019, so have you made any improvements since then?

I am now modifying the code for execution in multiple threads. After that I’ll try to run 110 bit in a cloud server to find a solution.
I don’t think you would show anyone such code

Sharing is caring

Can you send this python file to me in PM for testing?

https://drive.google.com/open?id=1abZRyPMhC0A5W7esdp-odh_-ToB6hHaw

Post the evidence please

which python libraries you using if its no secret ?
And i'm interested in your video as well, please show if you not mind.

My link to the latest post from 57fe. My algorithm is based on a different math. And he is faster

your link not in post at 57fe.
which from your algorimt is faster?

I have a search algorithm that is several times faster than the @57fe https://bitcointalksearch.org/topic/m.52582231 algorithm in single-threaded mode. According to my calculations, in skilled hands, you can keep within a month on a good server in the cloud to calculate 110 tasks. Server costs will reduce the reward by about half. This is more interesting than simple mining.
The algorithm can only be shifted to FPGA. Modular arithmetic does not imply the use of ASIC.

If there are interested people, I can post the evidence in the video.

Python does not assume that code is closed when used by other people, so I can not publish the source code, and the prototype algorithm is implemented in python.

Sorry for my English, but I'm not a native speaker of this language

While not part of the puzzle, but having a bit to do with why it was created in the first place.. an interesting exercise  no reward other than knowledge though I'll probably put a pi with a randomizied python script on it. finding a collision for the first two in the puzzle and see how far apart they are, might give a "clearer view" of the curve if that makes any sense. - Though that would probably end up being harder than finding #160 - Still, worth thinking about.