<<  >> (p.249)
    Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 324876 times)

    • nomachine
    Full Member
    ***
    Offline Offline

    Activity: 714
    Merit: 110


    View Profile
    May 04, 2024, 08:28:11 PM
    Last edit: May 04, 2024, 08:47:03 PM by nomachine
     #4961
    Quote from: Tepan on May 04, 2024, 03:00:12 PM
    wish me luck, current speed is 100K keys per sec.

    No luck here.
    BTC: bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8

  • ccinet
    Jr. Member
    *
    Offline Offline

    Activity: 44
    Merit: 1


    View Profile
    May 04, 2024, 08:37:54 PM
     #4962
    Quote from: Tepan on May 04, 2024, 03:00:12 PM

    wish me luck, current speed is 100K keys per sec.

    It must be a joke! You're joking, aren't you? Grin

  • abdenn0ur
    Newbie
    *
    Offline Offline

    Activity: 23
    Merit: 1


    View Profile
    May 04, 2024, 08:40:59 PM
     #4963
    Quote from: Tepan on May 04, 2024, 03:00:12 PM
    wish me luck, current speed is 100K keys per sec.

    100kk/s is very low. 
    You're better off using keyhunt by alberto
    You may get +1Mk/s even on a potato CPU

  • Tepan
    Jr. Member
    *
    Offline Offline

    Activity: 82
    Merit: 1


    View Profile
    May 05, 2024, 03:49:51 AM
     #4964
    Quote from: ccinet on May 04, 2024, 08:37:54 PM
    Quote from: Tepan on May 04, 2024, 03:00:12 PM

    wish me luck, current speed is 100K keys per sec.

    It must be a joke! You're joking, aren't you? Grin

    😂🥶🤫

  • Tepan
    Jr. Member
    *
    Offline Offline

    Activity: 82
    Merit: 1


    View Profile
    May 05, 2024, 03:52:29 AM
     #4965
    Quote from: abdenn0ur on May 04, 2024, 08:40:59 PM
    Quote from: Tepan on May 04, 2024, 03:00:12 PM
    wish me luck, current speed is 100K keys per sec.

    100kk/s is very low. 
    You're better off using keyhunt by alberto
    You may get +1Mk/s even on a potato CPU



    Nice info, but i already know from year past about that Alberto's BSGS.

  • viljy
    Legendary
    *
    Offline Offline

    Activity: 2198
    Merit: 1599



    View Profile
    May 05, 2024, 08:11:31 AM
     #4966
    Quote from: kTimesG on May 03, 2024, 05:30:44 PM
    I'd say the proportion of "unlikely patterns" in a private key of size n is more like very close to 0.00% (zero percent) the higher n gets. And it goes towards 0 really fast as n grows exponentially.

    You and I are probably talking about different things. I mean, it is extremely unlikely that among the remaining undisclosed keys there are patterns such as ffff or 8888 or more repetitions of the same digits.
    At the same time, I doubt that such combinations as, for example, c5ec5e or dd4dd4, etc., can be considered unlikely. Therefore, I think it is possible to discard numbers containing patterns of more than three identical digits.
    Without any calculations, I roughly assumed that there could not be more than 20% of such numbers in the range.

  • maylabel
    Newbie
    *
    Offline Offline

    Activity: 24
    Merit: 0


    View Profile
    May 05, 2024, 09:55:27 AM
     #4967
    Quote from: viljy on May 05, 2024, 08:11:31 AM
    Quote from: kTimesG on May 03, 2024, 05:30:44 PM
    I'd say the proportion of "unlikely patterns" in a private key of size n is more like very close to 0.00% (zero percent) the higher n gets. And it goes towards 0 really fast as n grows exponentially.

    You and I are probably talking about different things. I mean, it is extremely unlikely that among the remaining undisclosed keys there are patterns such as ffff or 8888 or more repetitions of the same digits.
    At the same time, I doubt that such combinations as, for example, c5ec5e or dd4dd4, etc., can be considered unlikely. Therefore, I think it is possible to discard numbers containing patterns of more than three identical digits.
    Without any calculations, I roughly assumed that there could not be more than 20% of such numbers in the range.

    True, this is another pattern

    but has something i found yesterday. I sincerely just want to have some code to run in Mac.

    Any one knows a code for mac? Python?

    thx in advance

  • kTimesG
    Full Member
    ***
    Offline Offline

    Activity: 546
    Merit: 166


    View Profile
    May 05, 2024, 03:33:23 PM
     #4968
    Quote from: viljy on May 05, 2024, 08:11:31 AM
    You and I are probably talking about different things. I mean, it is extremely unlikely that among the remaining undisclosed keys there are patterns such as ffff or 8888 or more repetitions of the same digits.
    At the same time, I doubt that such combinations as, for example, c5ec5e or dd4dd4, etc., can be considered unlikely. Therefore, I think it is possible to discard numbers containing patterns of more than three identical digits.
    Without any calculations, I roughly assumed that there could not be more than 20% of such numbers in the range.
    Sorry to ruin your intuition, but we were talking about the exact same thing

    The effort to discard unlikely patterns is massively submined by the fact that such unlikely patterns are close to 0% of the total possible different patterns (e.g. the ones your mind sees as without a pattern).

    42 is a totally random number, isn't it? Until you find a gazillion corelations, starting with the fact that it's now no longer random, because I mentioned it in this thread. What's the chance a key will contain c5ec5e? Lower now, because you mentioned it? This a classic fallacy.

    A random number generator does NOT care about any patterns, repetitions, weirdnesses of the human perception.

    Just because a found key doesn't contain a sequence of 30 consecutive same digits is not because it was impossible, it is because there's billions over billions of other possible combinations that had exactly the same equal chance of being selected by this thing called nature reality. Repeat the experiment many billions times and it WILL appear. It can appear the first time or the Nth time, the chances are the same at any time...

    There are algorithms that are actually getting the uniform real randomness (not some deterministic PRNG bullshit like someone mentioned earlier, I mean really lol? Haven't they heard about os.urandom and how it works?) as a benefit to speed up results, not as something that cripples the search. So if you're working against randomness, it's a lost battle on all possible fronts. You need to embrace it instead.
    Off the grid, training pigeons to broadcast signed messages.

  • k3ntINA
    Newbie
    *
    Offline Offline

    Activity: 27
    Merit: 0


    View Profile
    May 05, 2024, 06:33:12 PM
     #4969
    Discovered keys up to number 65 in hex format without spaces and back to back.
    In the setting of 180 degrees and the distance between each character of 550, we see a special order.
    -Professors and elders of science, can we know the reason for this?
    -If these keys are random, why should this order occur?
    -Write a script that will give us the same output from the keys?
    Friends, the shape of a circle is not an omen, nor imagination, nor nonsense!
    The circle is mathematics. The formula in the formula...
    https://www.talkimg.com/images/2024/05/05/roBa2.gif

  • nomachine
    Full Member
    ***
    Offline Offline

    Activity: 714
    Merit: 110


    View Profile
    May 05, 2024, 06:33:21 PM
    Last edit: May 05, 2024, 09:57:46 PM by nomachine
     #4970
    Quote from: kTimesG on May 05, 2024, 03:33:23 PM
    There are algorithms that are actually getting the uniform real randomness (not some deterministic PRNG bullshit like someone mentioned earlier, I mean really lol? Haven't they heard about os.urandom and how it works?)

    Code:
    import os, sys, random
    import time

    min_range = 18446744073709551615
    max_range = 36893488147419103231
    counter = 0  # Initialize counter
    start_time = time.time()

    while True:
        random_bytes = os.urandom(9)
        initial_bytes = b'\x00' * 23
        full_bytes = initial_bytes + random_bytes
        dec = int.from_bytes(full_bytes, byteorder='big')
        counter += 1  # Increment counter
        message = "\rPrivate Keys per second: {:.2f}".format(counter (time.time() - start_time))
        messages = []
        messages.append(message)
        output = "\033[01;33m" + ''.join(messages) + "\r"
        sys.stdout.write(output)
        sys.stdout.flush()
        if min_range <= dec <= max_range:
           if dec == 30568377312064202855:
              print("\nSeed :", random_bytes)
              print("Generated number:", dec)
              break

    This is Python script that will test os.urandom speed.
    The example works for Puzzle 65. There is no hash or secp256k1 operations here - just numbers.
    Result is (on my PC) :
    Private Keys per second: 170893.39

    Do you know how many numbers need to be generated per second to find Puzzle 65 in 10 minutes?

    30744573456182586  Private Keys per second !

    It's an mission impossible . Even in C++

    We need Grey aliens hardware to solve this. From Zeta Reticuli  Grin
    BTC: bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8

  • viljy
    Legendary
    *
    Offline Offline

    Activity: 2198
    Merit: 1599



    View Profile
    May 06, 2024, 12:04:04 PM
     #4971
    Quote from: kTimesG on May 05, 2024, 03:33:23 PM
    are close to 0% of the total possible different patterns

    Interestingly, the content of such numbers increases with the growth and height of the range.

    Quote
    ============================= RESTART: D:\NumHex.py ============================
    In the range from 0x1000 to 0x2000, 1 hexadecimal numbers were found with repeating digits occurring four or more times in a row.
    This accounts for 0.02% of the total hexadecimal numbers in the range.

    ============================= RESTART: D:\NumHex.py ============================
    In the range from 0x10000 to 0x20000, 32 hexadecimal numbers were found with repeating digits occurring four or more times in a row.
    This accounts for 0.05% of the total hexadecimal numbers in the range.

    ============================= RESTART: D:\NumHex.py ============================
    In the range from 0x100000 to 0x200000, 737 hexadecimal numbers were found with repeating digits occurring four or more times in a row.
    This accounts for 0.07% of the total hexadecimal numbers in the range.

    ============================= RESTART: D:\NumHex.py ============================
    In the range from 0x1000000 to 0x2000000, 15617 hexadecimal numbers were found with repeating digits occurring four or more times in a row.
    This accounts for 0.09% of the total hexadecimal numbers in the range.

    ============================= RESTART: D:\NumHex.py ============================
    In the range from 0x10000000 to 0x20000000, 311282 hexadecimal numbers were found with repeating digits occurring four or more times in a row.
    This accounts for 0.12% of the total hexadecimal numbers in the range.

    ============================= RESTART: D:\NumHex.py ============================
    In the range from 0x100000000 to 0x200000000, 5963072 hexadecimal numbers were found with repeating digits occurring four or more times in a row.
    This accounts for 0.14% of the total hexadecimal numbers in the range.

    Anyone who wishes can calculate independently in the 130 range, but it will be a very long time:

    Code:
    def count_repeated_digits_hex(start, end):
        count = 0
        total_nums = int(end, 16) - int(start, 16) + 1

        def has_repeated_digits(num_str):
            prev_digit = None
            consecutive_count = 0

            for digit in num_str:
                if digit == prev_digit:
                    consecutive_count += 1
                else:
                    consecutive_count = 1

                if consecutive_count >= 4:
                    return True

                prev_digit = digit

            return False

        for num in range(int(start, 16), int(end, 16) + 1):
            num_str = hex(num)[2:]  # Remove '0x' prefix
            if has_repeated_digits(num_str):
                count += 1

        percentage = (count total_nums) * 100

        return count, percentage

    start = "0x100000000"
    end = "0x200000000"
    result, percentage = count_repeated_digits_hex(start, end)
    print(f"In the range from {start} to {end}, {result} hexadecimal numbers were found with repeating digits occurring four or more times in a row.")
    print(f"This accounts for {percentage:.2f}% of the total hexadecimal numbers in the range.")

  • albert0bsd
    Hero Member
    *****
    Offline Offline

    Activity: 1120
    Merit: 718



    View Profile
    May 06, 2024, 12:25:59 PM
     #4972
    Quote from: nomachine on May 05, 2024, 06:33:21 PM
    Private Keys per second: 170893.39

    Can you guys write a decent code atleast....

    Quote from: nomachine on May 05, 2024, 06:33:21 PM
    It's an mission impossible . Even in C++

    I took that personal

    I write a C program to do the same tha your python script I get at least 100 times more numbers per second in a single thread in my laptop core i5

    Code:
    Total 639630811 numbers in 39 seconds: 16400790.000000 numbers/s
    Total 650116562 numbers in 40 seconds: 16252914.000000 numbers/s
    Total 671088064 numbers in 41 seconds: 16368001.000000 numbers/s
    Total 681573815 numbers in 42 seconds: 16227947.000000 numbers/s

    I know it is still far away for the required number, but is 100x Times faster....
     Roll Eyes Roll Eyes

  • citb0in
    Hero Member
    *****
    Offline Offline

    Activity: 1036
    Merit: 783


    Bitcoin g33k


    View Profile
    May 06, 2024, 01:18:00 PM
     #4973
    Quote from: albert0bsd on May 06, 2024, 12:25:59 PM
    Quote from: nomachine on May 05, 2024, 06:33:21 PM
    Private Keys per second: 170893.39

    Can you guys write a decent code atleast....

    Quote from: nomachine on May 05, 2024, 06:33:21 PM
    It's an mission impossible . Even in C++

    I took that personal

    I write a C program to do the same tha your python script I get at least 100 times more numbers per second in a single thread in my laptop core i5

    Code:
    Total 639630811 numbers in 39 seconds: 16400790.000000 numbers/s
    Total 650116562 numbers in 40 seconds: 16252914.000000 numbers/s
    Total 671088064 numbers in 41 seconds: 16368001.000000 numbers/s
    Total 681573815 numbers in 42 seconds: 16227947.000000 numbers/s

    I know it is still far away for the required number, but is 100x Times faster....
     Roll Eyes Roll Eyes
    I would have been very surprised if it had been the other way around Wink C is of course more performant and much more efficient in these computing areas. Python can hardly keep up.
    Some signs are invisible, some paths are hidden - but those who see, know what to do. Follow the trail - Follow your intuition - [bc1qqnrjshpjpypepxvuagatsqqemnyetsmvzqnafh]

  • albert0bsd
    Hero Member
    *****
    Offline Offline

    Activity: 1120
    Merit: 718



    View Profile
    May 06, 2024, 01:27:48 PM
     #4974
    Quote from: citb0in on May 06, 2024, 01:18:00 PM
    I would have been very surprised if it had been the other way around Wink C is of course more performant and much more efficient in these computing areas. Python can hardly keep up.

    agree, on the other hand the only advantage of python is that is easy to use and easy to write it, there are external packages for almost everything.

    Any way guys good luck finding the puzzles, I am still here but focusing in another life projects and family.

  • citb0in
    Hero Member
    *****
    Offline Offline

    Activity: 1036
    Merit: 783


    Bitcoin g33k


    View Profile
    May 06, 2024, 01:36:49 PM
     #4975
    Quote from: albert0bsd on May 06, 2024, 01:27:48 PM
    Any way guys good luck finding the puzzles, I am still here but focusing in another life projects and family.

    Dito! Smiley there are more important things in my life, too. The BTC puzzle ist just 4 fun and education purposes. Nothing more
    Some signs are invisible, some paths are hidden - but those who see, know what to do. Follow the trail - Follow your intuition - [bc1qqnrjshpjpypepxvuagatsqqemnyetsmvzqnafh]

  • nomachine
    Full Member
    ***
    Offline Offline

    Activity: 714
    Merit: 110


    View Profile
    May 06, 2024, 01:55:40 PM
    Last edit: May 06, 2024, 02:23:47 PM by nomachine
     #4976
    Quote from: albert0bsd on May 06, 2024, 12:25:59 PM

    I know it is still far away for the required number, but is 100x Times faster....
     Roll Eyes Roll Eyes


    I have not tried the same script in C++  Grin

    I also deal with other things. More with hardware than software.  Wink

    p.s.

    i have almost the same result

    Total 653409355 numbers in 39 seconds: 16754086 numbers/s
    Total 670648654 numbers in 40 seconds: 16766216 numbers/s
    Total 687812315 numbers in 41 seconds: 16775910 numbers/s
    Total 705075122 numbers in 42 seconds: 16787502 numbers/s


    Code:
    #include <iostream>
    #include <gmp.h>
    #include <gmpxx.h>
    #include <cstdlib>
    #include <ctime>
    #include <iomanip>

    int main() {
        mpz_class min_range("18446744073709551615");
        mpz_class max_range("36893488147419103231");
        mpz_class counter = 0;
        mpz_class dec;
        gmp_randstate_t state;

        gmp_randinit_default(state);
        std::time_t start_time = std::time(nullptr);
        double total_time = 0;

        mpz_t range;
        mpz_sub(range, max_range.get_mpz_t(), min_range.get_mpz_t());

        while (true) {
            mpz_urandomm(dec.get_mpz_t(), state, range);
            mpz_add(dec.get_mpz_t(), dec.get_mpz_t(), min_range.get_mpz_t());
            counter++;

            std::time_t current_time = std::time(nullptr);
            double elapsed_time = difftime(current_time, start_time);

            if (elapsed_time > total_time) {
                std::cout << "Total " << counter << " numbers in " << elapsed_time << " seconds: " << std::setprecision(0) << std::fixed << counter elapsed_time << " numbers/s" << std::endl;
                total_time = elapsed_time;
            }

        }

        gmp_randclear(state);
        mpz_clear(range);
        return 0;
    }
    BTC: bc1qdwnxr7s08xwelpjy3cc52rrxg63xsmagv50fa8

  • hongpeng945
    Newbie
    *
    Offline Offline

    Activity: 1
    Merit: 0


    View Profile
    May 06, 2024, 02:17:22 PM
     #4977
    Quote from: arulbero on November 07, 2018, 08:41:15 PM
    Quote from: BurtW on November 07, 2018, 08:19:15 PM
    My answer was a general answer to a general question about the two forms of Bitcoin address and was not meant to be a technical paper or exact description of how to create Bitcoin addresses (that is what the wiki is for).  

    So, I started out with "Leaving out some small details:".

    Thanks for filling in a few of the technical details.

    Ok, but anyway I don't understand "257 bit", this value is not correct at all. Or 512 bit and 256 bit, or 8+512 bit and 8+256 bit.


    Quote from: BurtW on November 07, 2018, 08:19:15 PM
    BTW, even with the additional details, the description is still incomplete because you left out the checksum in the hashing description.

    No, the address in the blockchain's blocks (and in the UTXO data set) are stored exactly this way: ripemd160(sha256('02' or '03' + 'x'), 160 bit, no checksum, no base58 encode.

    Example:

    private key = 01
    address = 91B24BF9F5288532960AC687ABB035127B1D28A5  (step 3, address for the blockchain)
    address = 1EHNa6Q4Jz2uvNExL497mE43ikXhwF6kZm (checksum + base58) (step 9, address for human people)

    Check with http://gobittest.appspot.com/Address  or https://www.blockchain.com/it/btc/address/91B24BF9F5288532960AC687ABB035127B1D28A5

    Now if we look at a tx that funds the address 91B24BF9F5288532960AC687ABB035127B1D28A5  1EHNa6Q4Jz2uvNExL497mE43ikXhwF6kZm :

    https://www.blockchain.com/btc/tx/6797afc4d9b91fb9b283fedddec4e35b00d54063d73bb0d3e97f3537ed8fff3c?show_adv=true

    output script: DUP HASH160 PUSHDATA(20)[91b24bf9f5288532960ac687abb035127b1d28a5] EQUALVERIFY CHECKSIG

    The address in the "1EHNa6Q4Jz2uvNExL497mE43ikXhwF6kZm" format is only for human people.
    Hi, I have a question for you pls. Does it means that a miner could hack the bitcoins from who really worked out the puzzles? Because once he signed the transaction, the miners knows the public key and if they know this is for the puzzles, they can utilize your method to find corresponding private key and hacked the funds.

  • pbies
    Full Member
    ***
    Offline Offline

    Activity: 380
    Merit: 203



    View Profile
    May 06, 2024, 03:59:35 PM
     #4978
    Quote from: hongpeng945 on May 06, 2024, 02:17:22 PM
    Hi, I have a question for you pls. Does it means that a miner could hack the bitcoins from who really worked out the puzzles? Because once he signed the transaction, the miners knows the public key and if they know this is for the puzzles, they can utilize your method to find corresponding private key and hacked the funds.

    For what all guys are saying here - seems like each and every random transaction should be possible to be stolen just knowing the public key. At any time when it is still in mempool. Not only puzzle addresses.
    BTC: bc1qmrexlspd24kevspp42uvjg7sjwm8xcf9w86h5k

  • citb0in
    Hero Member
    *****
    Offline Offline

    Activity: 1036
    Merit: 783


    Bitcoin g33k


    View Profile
    May 06, 2024, 05:42:21 PM
     #4979
    Quote from: pbies on May 06, 2024, 03:59:35 PM
    Quote from: hongpeng945 on May 06, 2024, 02:17:22 PM
    Hi, I have a question for you pls. Does it means that a miner could hack the bitcoins from who really worked out the puzzles? Because once he signed the transaction, the miners knows the public key and if they know this is for the puzzles, they can utilize your method to find corresponding private key and hacked the funds.

    For what all guys are saying here - seems like each and every random transaction should be possible to be stolen just knowing the public key. At any time when it is still in mempool. Not only puzzle addresses.

    No, it isn't! Read carefully the details to understand. If you have further questions don't hesitate to put them. This is true only for low-bit ranges. Let's say I make an outgoing transaction from my own bitcoin address which I generated using a 256bit key. You will know the pubkey but you won't be able to bruteforce because you'll never survive it Wink But if I used a 66bit key then you can crack the privkey in less than 10 seconds.
    Some signs are invisible, some paths are hidden - but those who see, know what to do. Follow the trail - Follow your intuition - [bc1qqnrjshpjpypepxvuagatsqqemnyetsmvzqnafh]

  • kTimesG
    Full Member
    ***
    Offline Offline

    Activity: 546
    Merit: 166


    View Profile
    May 06, 2024, 11:19:52 PM
     #4980
    Quote from: viljy on May 06, 2024, 12:04:04 PM
    Quote from: kTimesG on May 05, 2024, 03:33:23 PM
    are close to 0% of the total possible different patterns

    Interestingly, the content of such numbers increases with the growth and height of the range.

    Anyone who wishes can calculate independently in the 130 range, but it will be a very long time:
    1. Sounds normal, as you increase bit length each existing constraint gets factorized, and new ones also appear. Do you want to exclude 99.97% of the search space because they have 2 consecutive hex digits out of, let's say, 32 hex digits? Because that is where this series will end up.

    2. It's computed wrong anyways. Something like 0x7f80 has 2 consecutive hex characters. And I didn't make a typo, maybe think about it in bits. Or should we only exclude byte-aligned bit patterns? See, any seemingly random string can be made into a pattern depending on how you look at it.

    Quote from: nomachine on May 05, 2024, 06:33:21 PM
    Do you know how many numbers need to be generated per second to find Puzzle 65 in 10 minutes?

    It's an mission impossible . Even in C++
    It is very funny that you fell into the same trap as everyone else. Here's the reality:dev/urandom is not a secure random number generator. It falls back to a PRNG when too much data gets read from it.
    Off the grid, training pigeons to broadcast signed messages.
    Pages: « 1 ... 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 [249] 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 ... 568 »
      Print  
Page 248
Viewing Page: 249

IRLBTC™ © 2025 IRLBTC.com