Topic: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED== - page 54. (Read 54326 times)

so why not 1FFFFFFFFFFFFFFFF or just 1A838B13505B26867 as end of keyspace or i am seeing this wrong.?
anyone care to enlighten.? i say that cause most threads around the 32 or 100 btc puzzle are so silent recently.
still thinking the FFF thing is wrong, only one left with pubkey 110 bits,

vanitygen not fast enough, and ndv is paid only for upgrades? bitcrack not fast enough but great i guess,
so these programs to use for all this, and all the others i know are for public key search. no one giving updates,
Telariust hi, any news soon? Zielar, maybe sample of app with gpu pollard ready somewhere.?

So nooby where's you'r app. Finished already with the memory issues.

I think that next several thousands years we are going to need it...

@zielar

I noticed that the PubKeys for 155 and 160 are not listed on the first page.

here they are:


155:   035cd1854cae45391ca4ec428cc7e6c7d9984424b954209a8eea197b9e364c05f6


160:   02e0a8b039282faf6fe0fd769cfbc4b6b4cf8758ba68220eac420e32b91ddfa673




//radd66

Hello everyone
I am glad that the creation of a new topic has been received positively and remind you to submit any suggestions on what should appear in the main thread. From the current information and suggestions to remove a part consisting of already found keys - I allowed myself to move this part to the next topic segment and updated with the missing part (which was automatically cut due to too many characters in the previous version of the topic). Complete removal in my opinion will not have any positive impact, while maintaining it in its current form will provide a permanent source of information about the challenge - the more so because the keys solved so far are also part of the challenge, so it should probably remain so.
I also allowed myself to delete recent posts due to the fact that they referred to an unreliable script offer that was not part of the challenge ... and finally - I updated some of the "important links" about the pool that was created for this challenge and currently working on address 64.
While my activity has been low for some time for various reasons - I can assure you that in my free time I follow what is happening here :-) Regards to all!

I still can not lay out a single-threaded code, because it is not optimized by memory. Soon
[/quote]

We wait!

Thanks for pointing to the github. I learned a lot from Telariust code.
Now I’ve modified the algorithm (I’ll write the whole algorithm here soon) in terms of performing operations specific to python and productivity has increased by 14%. And this is only single-threaded execution.
I also have a multithreaded part ready.
I still can not lay out a single-threaded code, because it is not optimized by memory. Soon

Is not my script. I only wrote the first part of the script from the video, which is the generator but the second part which is most important was not revealed.
Will be released when the creator of the script will decide to do so.
Regarding speed there is on github pollard-kangaroo-c99 from Telariust with multicore ..up to 128 cores.

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?