Topic: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED== - page 55. (Read 56520 times)

wich file is this step? or wrong question?

Yes, I think everyone wants to try Pollard at GPU
Everyone else is embarrassed to admit

hi
anybody try to use p106 with bitcrack?

wow alek76,
just come in here, and add what all are waiting for in someone else's code. you must be a great programmer,
where are you from,? what kind of work you do.
thanks again for the showing of the gpu code.

If with this code you still need to waste an electricity for #105, then this code is still useless. In order to have it profitable it must solve #105 in a few seconds... ok, few minutes... ok, few hours... )))

Bro, show us the result of finding the address # 105, othervise all this has no sense at all...

What is the meaning of this post?
As I understand it, something to brag about, or maybe you're waiting for potential investors?

@bigvito19

Guess the gpu pollard multi-gpu 2021 maybe, and any other bitcrack multi-gpu edition 2021 maybe, and vgen ultraspeed only in dev's hands. but jean_luc is still the bomb all else said oclvgen.
thanks for your response big

Of course they're all not fast enough and slow, because you're trying to brute force it, what do you expect. By the time someone else comes up with a new or modified program, the next key will be cracked by then.

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)