Bitcoin puzzle transaction ~32 BTC prize to who solves it

[nomachine]

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

No luck here.

[ccinet]

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

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

[abdenn0ur]

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]

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

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

😂🥶🤫

[Tepan]

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]

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]

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]

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.

[k3ntINA]

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]

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 

[viljy]

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.

============================= 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]

Private Keys per second: 170893.39

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

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....
 

[citb0in]

Private Keys per second: 170893.39

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

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....
 

I would have been very surprised if it had been the other way around C is of course more performant and much more efficient in these computing areas. Python can hardly keep up.

[albert0bsd]

I would have been very surprised if it had been the other way around 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]

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

Dito! there are more important things in my life, too. The BTC puzzle ist just 4 fun and education purposes. Nothing more

[nomachine]

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

I have not tried the same script in C++  

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

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;
}

[hongpeng945]

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.

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]

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.

[citb0in]

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 But if I used a 66bit key then you can crack the privkey in less than 10 seconds.

[kTimesG]

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.

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.