Bitcoin puzzle transaction ~32 BTC prize to who solves it

[WanderingPhilospher]

Thank you, but that is not what I asked for, I asked about script for public key  

Public key ??  

Here you go

puzzle 65

Code:

import sys, secrets, secp256k1 as ice
while True:
    A0 = secrets.SystemRandom().randrange(18446744073709551615, 36893488147419103231)
    A1 = ice.scalar_multiplication(A0);B0 = ice.to_cpub(A1.hex())
    message = "\r{}".format(B0);messages = []
    messages.append(message);output = "\033[01;33m" + ''.join(messages) + "\r"
    sys.stdout.write(output);sys.stdout.flush()
    if B0 == "0230210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1b":
        HEX = "%064x" % dec;wifc = ice.btc_pvk_to_wif(HEX)
        with open("KEYFOUNDKEYFOUND.txt", "a") as f:
          f.write(f'Private key (wif) Compressed : {wifc}\n')
        break

It's the same idea, it generate public from private then compare. It's not public key addition

Maybe provide step by step of what you are wanting.

Sounds like you want to add public key a to public key b and look at the resulting hash160.

So you'd have to provide 2 public keys, if that's the case.

[mabdlmonem]

Thank you, but that is not what I asked for, I asked about script for public key  

Public key ??  

Here you go

puzzle 65

Code:

import sys, secrets, secp256k1 as ice
while True:
    A0 = secrets.SystemRandom().randrange(18446744073709551615, 36893488147419103231)
    A1 = ice.scalar_multiplication(A0);B0 = ice.to_cpub(A1.hex())
    message = "\r{}".format(B0);messages = []
    messages.append(message);output = "\033[01;33m" + ''.join(messages) + "\r"
    sys.stdout.write(output);sys.stdout.flush()
    if B0 == "0230210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1b":
        HEX = "%064x" % dec;wifc = ice.btc_pvk_to_wif(HEX)
        with open("KEYFOUNDKEYFOUND.txt", "a") as f:
          f.write(f'Private key (wif) Compressed : {wifc}\n')
        break

It's the same idea, it generate public from private then compare. It's not public key addition

Maybe provide step by step of what you are wanting.

Sounds like you want to add public key a to public key b and look at the resulting hash160.

So you'd have to provide 2 public keys, if that's the case.

I want to make sure from something about binary and public key. So I need a script to see what happens when we add 2 to the public key A and see the results of B . If what I think works I will share with everyone my idea, that's all

[WanderingPhilospher]

I want to make sure from something about binary and public key. So I need a script to see what happens when we add 2 to the public key A and see the results of B . If what I think works I will share with everyone my idea, that's all

Ok, so again, still confusing, at least to me.

You want to take a known public key "A" and add decimal number 2, to it, and see what the results are, "B"?

[mabdlmonem]

I want to make sure from something about binary and public key. So I need a script to see what happens when we add 2 to the public key A and see the results of B . If what I think works I will share with everyone my idea, that's all

Ok, so again, still confusing, at least to me.

You want to take a known public key "A" and add decimal number 2, to it, and see what the results are, "B"?

Yes, something like this

I want to make sure from something about binary and public key. So I need a script to see what happens when we add 2 to the public key A and see the results of B . If what I think works I will share with everyone my idea, that's all

Ok, so again, still confusing, at least to me.

You want to take a known public key "A" and add decimal number 2, to it, and see what the results are, "B"?

Basepoint G:   (55066263022277343669578718895168534326250603453777594175500187360389116729240L, 32670510020758816978083085130507043184471273380659243275938904335757337482424L)

Alice's secret key:   17436825491055586112755527818298542034755947930418580382030036978914692463183
Alice's public key:   (105679268965026450260338478364512614840272341529552480851333141374575362812020L, 16626610969520910407950657067133267950619549449019363437330088722436240164467L)

Bob's secret key:   32291818723468099298317452759803795679231875044644200373977689553806184529332
Bob's public key:   (91972152888645730114115627070016955368567223430226022010551427574648772204461L, 38446745167120461740412855329188667802416705205856357237503216472671667494148L)

=========Now we compare if same =================
(Alice Private+Bob Private)*G (101985652621362431772155823262934243650512147323486167826619770748482854554391L, 55338466074231142921426831718703469030196983642662166495910251269532062902125L)
Alice Public+Bob Public (101985652621362431772155823262934243650512147323486167826619770748482854554391L, 55338466074231142921426831718703469030196983642662166495910251269532062902125L)

[nomachine]

Python

Code:

def point_addition(p, q, a, p_curve):
    if p == "infinity":
        return q
    if q == "infinity":
        return p

    x_p, y_p = p
    x_q, y_q = q

    if p != q:
        m = ((y_q - y_p) * pow(x_q - x_p, -1, p_curve)) % p_curve
    else:
        m = ((3 * x_p**2 + a) * pow(2 * y_p, -1, p_curve)) % p_curve

    x_r = (m**2 - x_p - x_q) % p_curve
    y_r = (m * (x_p - x_r) - y_p) % p_curve

    return (x_r, y_r)

# Elliptic curve parameters
p_curve = 2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 - 1
a_curve = 0
g_basepoint = (
    55066263022277343669578718895168534326250603453777594175500187360389116729240,
    32670510020758816978083085130507043184471273380659243275938904335757337482424
)

# Alice's private and public keys
alice_private_key = 17436825491055586112755527818298542034755947930418580382030036978914692463183
alice_public_key = (
    105679268965026450260338478364512614840272341529552480851333141374575362812020,
    16626610969520910407950657067133267950619549449019363437330088722436240164467
)

# Bob's private and public keys
bob_private_key = 32291818723468099298317452759803795679231875044644200373977689553806184529332
bob_public_key = (
    91972152888645730114115627070016955368567223430226022010551427574648772204461,
    38446745167120461740412855329188667802416705205856357237503216472671667494148
)

# Calculate (Alice Private + Bob Private) mod p
combined_private_key = (alice_private_key + bob_private_key) % p_curve

# Calculate combined public key
combined_public_key = "infinity"
for i in range(combined_private_key.bit_length()):
    if (combined_private_key >> i) & 1:
        combined_public_key = point_addition(combined_public_key, g_basepoint, a_curve, p_curve)

print("(Alice Private + Bob Private) * G:", combined_public_key)
print("Alice Public + Bob Public:", point_addition(alice_public_key, bob_public_key, a_curve, p_curve))

Result:

(Alice Private + Bob Private) * G: (42781960159024299958163639356878453190501432691827926213851649420102986506453, 26209845635735715128050436215405957538148584136159453271268754283021438363749)
Alice Public + Bob Public: (101985652621362431772155823262934243650512147323486167826619770748482854554391, 55338466074231142921426831718703469030196983642662166495910251269532062902125)


This is off-topic and not directly related to brute-forcing Bitcoin (BTC).
The focus here is on brute-forcing (trial-and-error guessing) BTC puzzles.
And this code is demonstration of elliptic curve cryptography operations (a combined public key from the sum of two private keys)
rather than an attempt to crack or exploit BTC.
There doesn't appear to be any straightforward way to solve the presented challenge without resorting to years of brute force.

[citb0in]

I have digaran on ignore but it seems I see him again posting under the new user account mabdlmonem, humm?

[nomachine]

I have digaran on ignore but it seems I see him again posting under the new user account mabdlmonem, humm?

It really reminds me of his additions/subtractions  

[mabdlmonem]

@nomachine, can you have this work with points without using ice? I'm just a newbie needing help, a totally stranger newbie.😂

Code:

from sympy import mod_inverse

N = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141

def ters(scalar, target):
    k = mod_inverse(2, N)
    scalar_bin = bin(scalar)[2:]
    for i in range(len(scalar_bin)):
        if scalar_bin[i] == '0':
            result = (target * k) % N
        else:
            result = ((target * k) % N + N - 57896044618658097711785492504343953926418782139537452191302581570759080747169 ) % N
        target = result
    return result

target1 = 1361129467683753853853498429727072845824
target2 = 961437616415839130310402076835011931977

print("Target results:")
for x in range(1, 256):
    result1 = ters(x, target1)
    print(f"T1: {result1:x}")
for x in range(1, 256):
    result2 = ters(x, target2)
    print(f"T2: {result2:x}")
for x in range(1, 256):
    result1 = ters(x, target1)
    result2 = ters(x, target2)
    subtraction = (result1  - result2) % N
    print(f"S: {subtraction:x}")

could you please explain what is target1 and target2 ,, range (1,256) its length ?

[citb0in]

I have digaran on ignore but it seems I see him again posting under the new user account mabdlmonem, humm?

It really reminds me of his additions/subtractions  

yeah not only that but also ...

I am just good at math but very bad at programming so I couldn't understand the code honestly

 

[Woz2000]

@mods: is there a rule against multiple accounts? same ip = ban???

I guess my ignore list gets a little longer.

I have digaran on ignore but it seems I see him again posting under the new user account mabdlmonem, humm?

It really reminds me of his additions/subtractions  

yeah not only that but also ...

I am just good at math but very bad at programming so I couldn't understand the code honestly

 

Unbelievable!!! The A hole sends me a PM quoting some of the rules. I've reported him to the mods, not sure if something will be done, but maybe some others that are annoyed can do the same???

@mods: is there a rule against multiple accounts? same ip = ban???

I guess my ignore list gets a little longer.

I have digaran on ignore but it seems I see him again posting under the new user account mabdlmonem, humm?

It really reminds me of his additions/subtractions  

yeah not only that but also ...

I am just good at math but very bad at programming so I couldn't understand the code honestly

 

[brainless]

Hello all, is there any python code for public key addition and checking it's hash ?

Yes, look at the iceland2k14/secp256k1 script and create your script to fit within his script.

Also, what do you mean by "checking it's hash"? H160?

Yes , hash160 . I saw his code , I am just good at math but very bad at programming so I couldn't understand the code honestly

Simplest

Code:

import sys, secrets, secp256k1 as ice
while True:
    dec = secrets.SystemRandom().randrange(36893488147419103231, 73786976294838206463)
    h160 = ice.privatekey_to_h160(1, True, dec).hex()
    message = "\r{}".format(h160);messages = []
    messages.append(message);output = "\033[01;33m" + ''.join(messages) + "\r"
    sys.stdout.write(output);sys.stdout.flush()
    if h160 == "20d45a6a762535700ce9e0b216e31994335db8a5":
        HEX = "%064x" % dec;wifc = ice.btc_pvk_to_wif(HEX)
        with open("KEYFOUNDKEYFOUND.txt", "a") as f:
          f.write(f'Private key (wif) Compressed : {wifc}\n')
        break

Thank you, but that is not what I asked for, I asked about script for public key addition, this script is for generating address from private

I am traveling, tomorrow I will post friendly script for your easy work

[nomachine]

About the public keys in Python.....
You can use fastecdsa SEC1Encode  to do this all in one call (from Point X and Y).

Here is link

https://github.com/AntonKueltz/fastecdsa/blob/main/fastecdsa/encoding/sec1.py#L37

or

https://github.com/AntonKueltz/fastecdsa/blob/main/fastecdsa/tests/encoding/test_sec1.py#L47

AntonKueltz does very interesting things here

Let's get back to the topic.
With the same encoder, the same speed can be achieved as with the secp256x1 as ice.....

Code:

import sys, secrets, hashlib, binascii
from fastecdsa.encoding.sec1 import SEC1Encoder
from fastecdsa.curve import secp256k1
while True:
    dec = secrets.SystemRandom().randrange(36893488147419103231, 73786976294838206463)
    h160 = hashlib.new('ripemd160', hashlib.sha256(bytes.fromhex(binascii.hexlify(SEC1Encoder.encode_public_key(secp256k1.G * dec, compressed=True)).decode('utf-8'))).digest()).digest()
    message = "\r{}".format(h160.hex());messages = []
    messages.append(message);output = "\033[01;33m" + ''.join(messages) + "\r"
    sys.stdout.write(output);sys.stdout.flush()
    if h160.hex() == "20d45a6a762535700ce9e0b216e31994335db8a5":
        with open("KEYFOUNDKEYFOUND.txt", "a") as f:
          f.write(f'Private key (dec) : {dec}\n')
        break

Maybe they use the same encoder?  

You have managed to search a total random keys of
9213815776839680  total 66 bit range is
73786976294838206463  now if you wanted to search sequentially it would take you around 8008 months to completely search the whole range, and you don't know how many keys you have generated multiple times with random mode.
You have better chances if you buy lottery tickets. Don't waste your money on it. Lets see what santa has for us.

There is also a third option. Something between sequential and random.To incorporate randomness into the script, you can shuffle the order in which the sequential chunks are processed. Script divides the key search range into cpu_count chunks and assigns each chunk to a separate process using concurrent.futures.ProcessPoolExecutor. The more cpu cores you have, the better. (don't do this on a smartphone)

Code:

import time
import os
import sys
import hashlib
import binascii
import concurrent.futures
from concurrent.futures import ProcessPoolExecutor
from multiprocessing import cpu_count
from fastecdsa.encoding.sec1 import SEC1Encoder
from fastecdsa.curve import secp256k1
import random

def find_key_range(lower, upper, target_binary):
    for dec in range(lower, upper):
        h160 = hashlib.new('ripemd160', hashlib.sha256(bytes.fromhex(binascii.hexlify(SEC1Encoder.encode_public_key(secp256k1.G * dec, compressed=True)).decode('utf-8'))).digest()).digest()
        message = "\r[+] {}".format(h160.hex())
        sys.stdout.write(message)
        sys.stdout.flush()
        if h160 == target_binary:
            with open("KEYFOUNDKEYFOUND.txt", "a") as f:
                f.write(f'Private key (dec) : {dec}\n')
            return dec

def main():
    os.system("clear")
    t = time.ctime()
    sys.stdout.write(f"\033[?25l")
    sys.stdout.write(f"\033[01;33m[+] {t}\n")

    puzzle = 66
    target_binary = bytes.fromhex('20d45a6a762535700ce9e0b216e31994335db8a5')
    lower_range_limit = 2 ** (puzzle - 1)
    upper_range_limit = (2 ** puzzle) - 1

    sys.stdout.write(f"[+] Puzzle: {puzzle}\n")
    sys.stdout.write(f"[+] Lower range limit: {lower_range_limit}\n")
    sys.stdout.write(f"[+] Upper range limit: {upper_range_limit}\n")

    process_count = cpu_count()
    num_processes = process_count
    sys.stdout.write(f"[+] Using {process_count} CPU cores for parallel search\n")

    chunks = [(lower_range_limit + i * (upper_range_limit - lower_range_limit) // num_processes,
               lower_range_limit + (i + 1) * (upper_range_limit - lower_range_limit) // num_processes)
              for i in range(num_processes)]

    random.shuffle(chunks)

    for chunk in chunks:
        lower, upper = chunk
        find_key_range(lower, upper, target_binary)

    sys.stdout.write("\n[+] Key not found in the specified range.\n")

if __name__ == "__main__":
    main()

[alek76]

I am traveling, tomorrow I will post friendly script for your easy work

I checked this script. He's very slow. These scripts are all a waste of time. You need to look towards the GPU, you can increase the speed by 1.5 times if you use spin 32/64, exactly the same as in the GPU kangaroo code. I've already added a loop to the GPU for cheking. Python scripts are kindergarten I found a 32-bit key in 3 seconds.

[nomachine]

I checked this script. He's very slow. These scripts are all a waste of time. You need to look towards the GPU, you can increase the speed by 1.5 - 2 times if you use spin 32/64, exactly the same as in the GPU kangaroo code. I've already added a loop to the GPU for cheking. Python scripts are kindergarten

I agree. But what to do with those who do not have a PC. Digaran runs scripts on the phone. It needs poor man scripts.

[alek76]

I checked this script. He's very slow. These scripts are all a waste of time. You need to look towards the GPU, you can increase the speed by 1.5 - 2 times if you use spin 32/64, exactly the same as in the GPU kangaroo code. I've already added a loop to the GPU for cheking. Python scripts are kindergarten

I agree. But what to do with those who do not have a PC. Digaran runs scripts on the phone. It needs poor man scripts.

Well, of course, I exaggerated this by 2 times But the speed increases significantly in the cycle. Additionally, I removed everything from the GPU except P2PKH.

[yellowstripes]

Has anyone come across the private keys of these two public keys?
02767761d4b43b130134b0fd8348a020c43f4d247a00aa488cb0f6fdb1584a202f
03bd303397e428bf036db49510b70b86d0025f73744b6324a579de16dc591e2056

If yes, Pm me... I shall offer a token of appreciation for this.

[nomachine]

Has anyone come across the private keys of these two public keys?
02767761d4b43b130134b0fd8348a020c43f4d247a00aa488cb0f6fdb1584a202f
03bd303397e428bf036db49510b70b86d0025f73744b6324a579de16dc591e2056

If yes, Pm me... I shall offer a token of appreciation for this.

Why would anyone do this for addresses with a balance $0.00 ? 

[yellowstripes]

Has anyone come across the private keys of these two public keys?
02767761d4b43b130134b0fd8348a020c43f4d247a00aa488cb0f6fdb1584a202f
03bd303397e428bf036db49510b70b86d0025f73744b6324a579de16dc591e2056

If yes, Pm me... I shall offer a token of appreciation for this.

Oh hey there little one! Do you think we are public key archives? Lol, you have to give us some clues like, the bit range, how did you obtain them? did you subtracted or divided, otherwise there are nearly 2**256 public keys, those 2 could be any of them. And we only accept Bitcoin, no tokens around these woods. 😂

Both of them look familiar, don't know where I have seen them before.🤔

I asked in very polite manner, not too sure why the sarcasm is applied. Nonetheless...The range of the two public keys is 116 bit. If you have come across them, PM me and lets see if you wont get bitcoin as a thank you.

[mabdlmonem]

Is there any c# code to make combinations of large list [0 to 100] ?

[yellowstripes]

Has anyone come across the private keys of these two public keys?
02767761d4b43b130134b0fd8348a020c43f4d247a00aa488cb0f6fdb1584a202f
03bd303397e428bf036db49510b70b86d0025f73744b6324a579de16dc591e2056

If yes, Pm me... I shall offer a token of appreciation for this.

Oh hey there little one! Do you think we are public key archives? Lol, you have to give us some clues like, the bit range, how did you obtain them? did you subtracted or divided, otherwise there are nearly 2**256 public keys, those 2 could be any of them. And we only accept Bitcoin, no tokens around these woods. 😂

Both of them look familiar, don't know where I have seen them before.🤔

I asked in very polite manner, not too sure why the sarcasm is applied. Nonetheless...The range of the two public keys is 116 bit. If you have come across them, PM me and lets see if you wont get bitcoin as a thank you.

I bolded the (sarcasm, humor), the rest is serious, if you take that as insult, apology.
Yet you haven't given us any clues, how do you know they are in 116 bit range?, please convince us with proof and show us how you know and are sure they exist in 116 bit range. Nevertheless, you can use kangaroo to search for them. Here is the topic on kangaroo.
Welcome to the land where, when you walk in, there will be no going back.

I appreciate the apology and thank you for the link to Kangaroo.