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301

Bitcoin / Bitcoin Discussion / Re: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED==

on: July 28, 2023, 06:14:28 PM

I use poor man's ideas. One of them.... Grin

First, i generate list of 500.000 (more is better) WIFs in equal parts of range....
All WIFs are with first 42 caracters.....

Code:

Import secp256k1 as ice

num_parts = int(input("Enter the number of equal parts you want: "))

def calculate_puzzle66_parts():
    start = 39199331156632797184
    end = 73786976294838206464
    step = (end - start) / num_parts

    wifs = []

    for i in range(num_parts):
        decimal_num = start + (i + 1) * step
        int_num = int(decimal_num)
        hex_value = "%064x" % int_num
        wif_compressed = ice.btc_pvk_to_wif(hex_value)
        wifs.append(wif_compressed[:42])
        print(wif_compressed[:42])

    with open("Puzzle66Wifs.txt", "w") as file:
        for wif in wifs:
            file.write(wif + "\n")

if __name__ == "__main__":
    calculate_puzzle66_parts()

Then I try to solve them like this

Code:

#!/bin/bash

# Initialize the attempt counter
attempts=0

tail -n +1 Puzzle66Wifs.txt | while read -r wif; do
    # Increment the attempt counter
    ((attempts++))

    # Create or overwrite 66.conf with END and the current WIF
    echo "END" > 66.conf
    echo "$wif" >> 66.conf
    echo "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so" >> 66.conf
    java -jar wifSolver.jar 66.conf

    # Display the number of attempts after each try
    echo "Attempts so far: $attempts"

    # Check if any file with the desired pattern was found and break the loop if it was
    if [ -n "$(find . -maxdepth 1 -type f -name 'END_result_*.txt' -print -quit)" ]; then
        echo "WIF found! Stopping the loop."
        break
    fi
done

echo "Total attempts: $attempts"

It's been going on for months...

302

Bitcoin / Bitcoin Discussion / Re: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED==

on: July 26, 2023, 07:03:27 PM

Speaking of patterns...I use this method to see if I can spot anything. Code: import decimal import secp256k1 as ice def calculate_puzzle66_parts(): start = decimal.Decimal("36893488147419103231") end = decimal.Decimal("73786976294838206463") step = (end - start) / 16 for i in range(16): decimal_num = start + (i + 1) * step - 1 hex_value = "%064x" % int(decimal_num) wif_compressed = ice.btc_pvk_to_wif(hex_value) print(f"Part {i + 1} Start point dec:", decimal_num) print(f"Part {i + 1} Start point WIF:", wif_compressed) quarter_dec = start + (i + 1) * step // 4 quarter_hex = "%064x" % int(quarter_dec) wif_quarter = ice.btc_pvk_to_wif(quarter_hex) print(f"Part {i + 1} 1/4 dec from 66:", quarter_dec) print(f"Part {i + 1} 1/4 WIF from 66:", wif_quarter) half_dec = start + (i + 1) * step // 2 half_hex = "%064x" % int(half_dec) wif_half = ice.btc_pvk_to_wif(half_hex) print(f"Part {i + 1} 1/2 dec from 66:", half_dec) print(f"Part {i + 1} 1/2 WIF from 66:", wif_half) three_quarter_dec = start + (i + 1) * step * 3 // 4 three_quarter_hex = "%064x" % int(three_quarter_dec) wif_three_quarter = ice.btc_pvk_to_wif(three_quarter_hex) print(f"Part {i + 1} 3/4 dec from 66:", three_quarter_dec) print(f"Part {i + 1} 3/4 WIF from 66:", wif_three_quarter) print(f"Part {i + 1} End point dec:", start + (i + 1) * step) print(f"Part {i + 1} End point WIF:", ice.btc_pvk_to_wif("%064x" % int(start + (i + 1) * step)), "\n") if __name__ == "__main__": calculate_puzzle66_parts() Part 1 Start point dec: 39199331156632797182 Part 1 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZWzeyikXcEhFs6rMa1V Part 1 1/4 dec from 66: 37469948899722526719 Part 1 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZUziJujmDznaXnGPQSV Part 1 1/2 dec from 66: 38046409652025950207 Part 1 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZVfMsBQggk69993Lj3p Part 1 3/4 dec from 66: 38622870404329373695 Part 1 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZWL1RT5c9VPhkZ8H9oW Part 1 End point dec: 39199331156632797183 Part 1 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZWzeyikXcEhGMwnHdn7 Part 2 Start point dec: 41505174165846491134 Part 2 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZZfFCoSDTDuWJc1i5sr Part 2 1/4 dec from 66: 38046409652025950207 Part 2 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZVfMsBQggk69993Lj3p Part 2 1/2 dec from 66: 39199331156632797183 Part 2 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZWzeyikXcEhGMwnHdn7 Part 2 3/4 dec from 66: 40352252661239644159 Part 2 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZYKx6G6NXjJPak7TotN Part 2 End point dec: 41505174165846491135 Part 2 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZZfFCoSDTDuWoVrghLp Part 3 Start point dec: 43811017175060185086 Part 3 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZcKqRt7uJD7kk9AvAt8 Part 3 1/4 dec from 66: 38622870404329373695 Part 3 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZWL1RT5c9VPhkZ8H9oW Part 3 1/2 dec from 66: 40352252661239644159 Part 3 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZYKx6G6NXjJPak7TotN Part 3 3/4 dec from 66: 42081634918149914623 Part 3 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZaKtm578uyD5QqCf8om Part 3 End point dec: 43811017175060185087 Part 3 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZcKqRt7uJD7mF4zwEkF Part 4 Start point dec: 46116860184273879038 Part 4 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZezRexob9CL1BfTmF23 Part 4 1/4 dec from 66: 39199331156632797183 Part 4 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZWzeyikXcEhGMwnHdn7 Part 4 1/2 dec from 66: 41505174165846491135 Part 4 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZZfFCoSDTDuWoVrghLp Part 4 3/4 dec from 66: 43811017175060185087 Part 4 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZcKqRt7uJD7mF4zwEkF Part 4 End point dec: 46116860184273879039 Part 4 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZezRexob9CL1gZKDxzH Part 5 Start point dec: 48422703193487572990 Part 5 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZhf1t3VGzBYFdHuq1cQ Part 5 1/4 dec from 66: 39775791908936220671 Part 5 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZXfJXzRT4yzpyMhSS32 Part 5 1/2 dec from 66: 42658095670453338111 Part 5 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZazYKLn4NiWe2KmF7j8 Part 5 3/4 dec from 66: 45540399431970455551 Part 5 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZeKn6h8fgT2T5BG1zyQ Part 5 End point dec: 48422703193487572991 Part 5 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZhf1t3VGzBYG87a7NRw Part 6 Start point dec: 50728546202701266942 Part 6 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZkKc78AxqAkW4otaiui Part 6 1/4 dec from 66: 40352252661239644159 Part 6 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZYKx6G6NXjJPak7TotN Part 6 1/2 dec from 66: 43811017175060185087 Part 6 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZcKqRt7uJD7mF4zwEkF Part 6 3/4 dec from 66: 47269781688880726015 Part 6 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZgKimW9S4gw8uNS9xwk Part 6 End point dec: 50728546202701266943 Part 6 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZkKc78AxqAkWZdisLwg Part 7 Start point dec: 53034389211914960894 Part 7 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZnzCLCreg9xkWGorWC6 Part 7 1/4 dec from 66: 40928713413543067647 Part 7 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZYzbeXmHzUbxC59ow7z Part 7 1/2 dec from 66: 44963938679667032063 Part 7 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZdf8YRTkDhitTp71pAW Part 7 3/4 dec from 66: 48999163945790996479 Part 7 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZiKfSKACSvqpjW95JbD Part 7 End point dec: 53034389211914960895 Part 7 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZnzCLCreg9xm1FxSeJp Part 8 Start point dec: 55340232221128654846 Part 8 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZqenZHYLX9AzwtzyNbB Part 8 1/4 dec from 66: 41505174165846491135 Part 8 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZZfFCoSDTDuWoVrghLp Part 8 1/2 dec from 66: 46116860184273879039 Part 8 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZezRexob9CL1gZKDxzH Part 8 3/4 dec from 66: 50728546202701266943 Part 8 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZkKc78AxqAkWZdisLwg Part 8 End point dec: 55340232221128654847 Part 8 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZqenZHYLX9B1Sh6RD2t Part 9 Start point dec: 57646075230342348798 Part 9 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZtKNnNE2N8PFPMTB21Z Part 9 1/4 dec from 66: 42081634918149914623 Part 9 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZaKtm578uyD5QqCf8om Part 9 1/2 dec from 66: 47269781688880726015 Part 9 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZgKimW9S4gw8uNS9xwk Part 9 3/4 dec from 66: 52457928459611537407 Part 9 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZnKYmwBjDQfCPra6THT Part 9 End point dec: 57646075230342348799 Part 9 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZtKNnNE2N8PFtGjKJj8 Part 10 Start point dec: 59951918239556042750 Part 10 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZvyy1SuiD7bVpxgmdfN Part 10 1/4 dec from 66: 42658095670453338111 Part 10 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZazYKLn4NiWe2KmF7j8 Part 10 1/2 dec from 66: 48422703193487572991 Part 10 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZhf1t3VGzBYG87a7NRw Part 10 3/4 dec from 66: 54187310716521807871 Part 10 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZpKVSkCVbeZtE1H9iQ8 Part 10 End point dec: 59951918239556042751 Part 10 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZvyy1SuiD7bWKqCb6cT Part 11 Start point dec: 62257761248769736702 Part 11 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZyeZEXbQ46okGY7gWo9 Part 11 1/4 dec from 66: 43234556422756761599 Part 11 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZbfBscSyqTpCdf9Admy Part 11 1/2 dec from 66: 49575624698094419967 Part 11 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZizJzaq7ug9PLvewANu Part 11 3/4 dec from 66: 55916692973432078335 Part 11 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZrKS7ZDFytUa4BmMTS9 Part 11 End point dec: 62257761248769736703 Part 11 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZyeZEXbQ46okmPYKfxL Part 12 Start point dec: 64563604257983430654 Part 12 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa2K9TcH5u61zi2oYhET Part 12 1/4 dec from 66: 43811017175060185087 Part 12 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZcKqRt7uJD7mF4zwEkF Part 12 1/2 dec from 66: 50728546202701266943 Part 12 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZkKc78AxqAkWZdisLwg Part 12 3/4 dec from 66: 57646075230342348799 Part 12 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZtKNnNE2N8PFtGjKJj8 Part 12 End point dec: 64563604257983430655 Part 12 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa2K9TcH5u621CsjkSLq Part 13 Start point dec: 66869447267197124606 Part 13 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa4yjggxmk5EF9cf1wrW Part 13 1/4 dec from 66: 44387477927363608575 Part 13 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZczUz9npkxRKrQ27nPS Part 13 1/2 dec from 66: 51881467707308113919 Part 13 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZmeuDfWokfMdnPcTtgB Part 13 3/4 dec from 66: 59375457487252619263 Part 13 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZvKKTBEnkNHwiSqAUkY Part 13 End point dec: 66869447267197124607 Part 13 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa4yjggxmk5EFeVWWKPF Part 14 Start point dec: 69175290276410818558 Part 14 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa7eKumeTb4SVb6skSxw Part 14 1/4 dec from 66: 44963938679667032063 Part 14 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZdf8YRTkDhitTp71pAW Part 14 1/2 dec from 66: 53034389211914960895 Part 14 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZnzCLCreg9xm1FxSeJp Part 14 3/4 dec from 66: 61104839744162889727 Part 14 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZxKG7zFZ8cCdYZASpQa Part 14 End point dec: 69175290276410818559 Part 14 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa7eKumeTb4SW635PmjP Part 15 Start point dec: 71481133285624512510 Part 15 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qaAJv8rL9S3ek2d8DWnX Part 15 1/4 dec from 66: 45540399431970455551 Part 15 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZeKn6h8fgT2T5BG1zyQ Part 15 1/2 dec from 66: 54187310716521807871 Part 15 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZpKVSkCVbeZtE1H9iQ8 Part 15 3/4 dec from 66: 62834222001073160191 Part 15 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZzKCnoGKWr7KNn5yNok Part 15 End point dec: 71481133285624512511 Part 15 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qaAJv8rL9S3ekXXbZqcM Part 16 Start point dec: 73786976294838206462 Part 16 Start point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qaCyWMw1qH2rzUAU5Hx5 Part 16 1/4 dec from 66: 46116860184273879039 Part 16 1/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZezRexob9CL1gZKDxzH Part 16 1/2 dec from 66: 55340232221128654847 Part 16 1/2 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZqenZHYLX9B1Sh6RD2t Part 16 3/4 dec from 66: 64563604257983430655 Part 16 3/4 WIF from 66: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qa2K9TcH5u621CsjkSLq Part 16 End point dec: 73786976294838206463 Part 16 End point WIF: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qaCyWMw1qH2rzy6joVM6 ------------------------------------------------------------------------------------------------------------- But nothing. Just a glass ball, pure luck or brute force...

303

Bitcoin / Bitcoin Discussion / Re: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED==

on: July 23, 2023, 03:39:47 AM

Determined are the adversaries, toiling with 512 GPUs akin to the heralded 3090. To unravel a 130-bit key, these diligent souls expend near a solar cycle, traversing through 306 days of unyielding pursuit. But even with such a multitude of computational might, the 130-bit realm does not relent easily. A mighty task to conquer the 2^130-bit range beckons for 512 valiant GPUs donning the attire of the 3090. Their efforts encompass days and nights, amid soaring costs, for the serenity of cryptographic victory.The pursuit of knowledge and foresight, the wily Pollard's kangaroo method seeks to challenge Bitcoin's cryptographic prowess.

304

Bitcoin / Development & Technical Discussion / Re: BitCrack - A tool for brute-forcing private keys

on: July 22, 2023, 12:48:24 PM

Quote from: negate7o4 on April 21, 2023, 09:45:15 AM Quote from: garlonicon on March 02, 2023, 05:08:45 PM I wonder if keys were generated by some deterministic wallet first, and then truncated to N bits. Because in that case, it could be possible to recover the master key, and then sweep coins from all of them, while it would not mean that ECDSA is broken (because HD keys could be non-hardened). Interesting puzzle, I checked for patterns between binary values, decimal values, floor values. Just cant find anything I updated an excel sheet puzzle bitcoin.xlsx with allot more data of each puzzle. Like decimal, binary, floor, and allot more data per puzzle. Even multipliers of the data to the next puzzle. Since this range is just so huge, I think a random generator would maybe work best to try and find the needle. I attached below my python script trying the next puzzle randomly if someone might benefit to https://github.com/negate7o4/Bitcoin-Puzzle On my rig, I7 if I run 5 X instances, it checks about 11 500 possibilities per second per thread, but eats 80% cpu. So the code chews threw 16.5 Mil random possibilities per day If you want to try a different puzzle, just change these values in the script : address = '1FeexV6bAHb8ybZjqQMjJrcCrHGW9sb6uF' start_key = '0000000000000000000000000000000000000000000000040000000000000000' stop_key = '000000000000000000000000000000000000000000000007ffffffffffffffff' Also take note of the cpu threads the code uses, update according to your threads. Their is also a text file in my repository of the correct pip's you need to install for the code to work We can utilize multiprocessing instead of threading. Multiprocessing allows us to take advantage of multiple CPU cores and achieve better parallelism. Also, we can optimize the search_for_key function by reducing unnecessary calculations within the loop. There is a potential bottleneck that can be accelerated. The bottleneck lies in the line where a random integer is generated using os.urandom(32). We can generate a batch of random numbers in advance and use these numbers in the search_for_key function instead of generating a new random number for each iteration. Code: import os import multiprocessing as mp from gmpy2 import mpz, powmod from bitcoinlib.encoding import addr_to_pubkeyhash, to_hexstring from bitcoinlib.keys import HDKey from ecdsa import SigningKey, SECP256k1 def save_key_to_file(private_hex: str): with open(f"{private_hex}.txt", "w") as f: f.write(private_hex) def generate_random_numbers(num_numbers): return [os.urandom(32) for _ in range(num_numbers)] def search_for_key(start_key_int, stop_key_int, target_address, attempts_per_process, random_numbers): for i in range(attempts_per_process): current_key_int = mpz(powmod(int.from_bytes(random_numbers[i], 'big'), 1, stop_key_int - start_key_int)) + start_key_int ecdsa_private_key = SigningKey.from_secret_exponent(int(current_key_int), curve=SECP256k1) key = HDKey(key=ecdsa_private_key.to_string()) if key.address() == target_address: save_key_to_file(key.private_hex) print(f"Private key found: {key.private_hex}") return key return None def search_in_range(start_key_int, stop_key_int, target_address, attempts_per_process, random_numbers): process_start_key_int = mpz(powmod(int.from_bytes(random_numbers[0], 'big'), 1, stop_key_int - start_key_int)) + start_key_int process_stop_key_int = (process_start_key_int + attempts_per_process) % stop_key_int search_for_key(process_start_key_int, process_stop_key_int, target_address, attempts_per_process, random_numbers) if __name__ == "__main__": address = '1FeexV6bAHb8ybZjqQMjJrcCrHGW9sb6uF' start_key = '0000000000000000000000000000000000000000000000040000000000000000' stop_key = '000000000000000000000000000000000000000000000007ffffffffffffffff' start_key_int = mpz(start_key, 16) stop_key_int = mpz(stop_key, 16) attempts_per_process = 2000000 num_processes = 8 random_numbers = generate_random_numbers(attempts_per_process * num_processes) processes = [] for _ in range(num_processes): process = mp.Process(target=search_in_range, args=(start_key_int, stop_key_int, address, attempts_per_process, random_numbers)) processes.append(process) process.start() for process in processes: process.join() print(f"Private key not found in the given range after {num_processes * attempts_per_process} attempts.") But no matter how you turn it and perfect it, it takes a couple of million years to reach the result.

305

Bitcoin / Bitcoin Discussion / Re: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED==

on: July 21, 2023, 03:32:51 PM

Quote from: davidpro on July 21, 2023, 03:16:38 PM @nomachine if you can share with ppls, it would be great! I don't know what I didn't do. I tried everything possible. There is no pattern here. I even targeted the last eight characters at the end of WIF for seven whole months trying to reconstruct the private key. Whichever option you choose and you don't know the public key, this is unsolvable. Or it can be solved in a couple of thousand years. The numbers in question are the size of the entire universe. Quantum computing is needed to solve this. And a very good one at that.

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Bitcoin / Bitcoin Discussion / Re: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED==

on: July 21, 2023, 03:14:56 PM

Quote from: davidpro on July 21, 2023, 01:09:03 PM Whoopse! I've found some similiar wallets and hope it will helps others! I have about half a million addresses that start with the first 8 characters. It's best if someone tries to look into the glass ball and tell me exactly what range to aim for.

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Bitcoin / Bitcoin Discussion / Re: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED==

on: July 19, 2023, 02:30:41 PM

Quote from: nomachine on July 19, 2023, 02:19:49 PM Quote from: ripemdhash on July 16, 2023, 08:18:30 PM 64 bit privatekey when I know public key. what is the time from start to find? if you know the public key 10 minutes. if not a couple of million years.When all the public keys from this puzzle are gone, we're at a dead end. more precisely https://bitcointalk.org/index.php?topic=1306983.msg51485099#msg51485099 He write "It took almost 12 hours to find the #70 on my system (Intel® Xeon® CPU E3-1505M v6 @ 3.00GHz + 32 GB RAM). No GPU, only CPU (1 core, my program is not multithread). Only 6 minutes to retrieve the #65."

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Bitcoin / Bitcoin Discussion / Re: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED==

on: July 19, 2023, 02:19:49 PM

Quote from: ripemdhash on July 16, 2023, 08:18:30 PM 64 bit privatekey when I know public key. what is the time from start to find? if you know the public key 10 minutes. if not a couple of million years.When all the public keys from this puzzle are gone, we're at a dead end.

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Bitcoin / Project Development / Re: Keyhunt - development requests - bug reports

on: July 16, 2023, 01:42:38 PM

Quote from: BS0D on May 04, 2023, 04:28:11 PM Will there be any speed improvement using AVX2 and hugepages for CPUs? As well as L3 cache usage of ryzen cpus which in some cases may replace ram reads increasing the speed by times. It is happen to be using a xmrig on the same machine as Keyhunt. Is it possible to boost command from here increase the number of keys per second on AMD EPYC https://raw.githubusercontent.com/xmrig/xmrig/dev/scripts/randomx_boost.sh I have strange results.

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Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

on: July 14, 2023, 05:25:45 PM

Quote from: ripemdhash on July 09, 2023, 10:58:11 AM This is puzzle created by someone from first BTC development. Maybe Satoshi, Maybe Finn, it is created for people to check that is possible to crack secp256k1 so it is not theft. But the problem is someone has knowledge - not power as milion GPU's. it has been cracked by math not by pollard , not by BSGS, not by GPUS. and it is the problem . And BSGS is not a math ? it's all math here bro.

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Bitcoin / Bitcoin Discussion / Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

on: July 14, 2023, 10:18:12 AM

Quote from: GR Sasa on July 14, 2023, 09:28:38 AM Nevertheless recently puzzle 120 and 125 got solved by the same persons or group of people which make us doubt that there are indeed a hidden formula that is being used to solve the puzzles that have their pub keys revealed by those unknown group of people. We believe the same solvers are currently trying to solve 130 now too. Which has over 13 BTC now. There is no hidden formula here. BSGS

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