Author

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 154. (Read 231001 times)

copper member
Activity: 1330
Merit: 899
🖤😏
how can we map secp256k1 points to a small size curve???

You already got an answer to the same question here

https://bitcointalk.org/index.php?topic=5469636;wap

You still hope that someone will tell you precise method before they grab the prize for themselves ?  Grin
I think you got it backwards, I'm trying to level the playing field for everyone to have equal opportunity/ time.
I'm the one trying to find the precise algorithm.

Lets correct the sentences above:
We are trying to level the playing field, and
We are trying to find the precise algorithm to solve this puzzle.

As long as everybody thinks like the first version of wording above, no results will be achieved, only by team work this can have a result in much less time than going solo.

I suggest you to watch Jigsaw movies once again, the key to the *victims success was always working as a team.

In that topic, there is no mention of changing G to a smaller size G.
Let me spit explain what I'm hallucinating talking about :
Let's change p to this one :
Code:
0xcdf15ce5b341762d
Now what I'm having a problem to find, is to shrink down secp256k1's G to be the same size while acting as before, meaning if the size is changed, multiplying the new G by 10 should give me a point similar/distinguishable from secp256k1's 0xa public key. If that is even possible and how much more speed we could gain by having a smaller G?


*OMG, are we really like the victims of jigsaw puzzle? Lol.
member
Activity: 499
Merit: 38
how can we map secp256k1 points to a small size curve???

You already got an answer to the same question here

https://bitcointalk.org/index.php?topic=5469636;wap

You still hope that someone will tell you precise method before they grab the prize for themselves ?  Grin
copper member
Activity: 1330
Merit: 899
🖤😏
Warning, random spits ahead, watch out!😉

So, as I was spitting out some random thoughts coming to my mind and having fun with myself for no apparent reason, what if we reduce the size of our curve down to for example 2^130, then we would change G to something like e.g,  0x1, 0x2 of course with a much much smaller prime (P), would that somehow help us to compute much faster if the size of our points is considerably small?
There is one small problem, how can we map secp256k1 points to a small size curve???

Stay tuned for more spits, I'm dry right now.🤣
member
Activity: 499
Merit: 38
If you mean by “stride” jumps of 2-2 like 2,4,6.........

Nope....
About CUDA Programming
https://www.math.wsu.edu/math/kcooper/CUDA/c09Shuffle.php
https://nanxiao.gitbooks.io/cuda-little-book/content/posts/grid-stride-loops.html


How do you imagine that he can achieve 1,200 MKey/s per card if he doesn't hack CUDA itself ?
member
Activity: 239
Merit: 53
New ideas will be criticized and then admired.
My problem was, I could not create a stride function inside of CUDA. Or else I would have found 130's key already lol.

Is there a built-in "stride" function specifically in the CUDA kernel? Or maybe there is a workaround in a grid-stride loop? Or do we have to write a new kernel? Smiley

If you mean by “stride” jumps of 2-2 like 2,4,6, or jumps of any other number, you simply have to modify G by the pub that corresponds to the number of jumps you want, this does not affect the speed of the scalar multiplication, BSGS, etc.
jr. member
Activity: 149
Merit: 7

You could be working in a 50 bit range or a 256 bit range; and you can subtract, divide, multiply, etc. whatever you want to do, or you can use whatever stride you can some up with, 14, 87, 1234344564, 47398573854734834, etc., it won't work because again, you could go around the curve and you won't know where the key lies or where to start your stride function from.

There is truly only one way to take advantage of a stride function; and I stated that months ago. My problem was, I could not create a stride function inside of CUDA. Or else I would have found 130's key already lol.



If you write the exact step code you want, there will be friends who will try to help. If you draw the algorithm flow diagram they can help you

I wrote my own brute-forcer that implements stride on CUDA. actually I'm running it 24/7 for #66.

glad to see that people are focusing on new algos and scripts, I feel that someone is closer to solution.
jr. member
Activity: 56
Merit: 1

You could be working in a 50 bit range or a 256 bit range; and you can subtract, divide, multiply, etc. whatever you want to do, or you can use whatever stride you can some up with, 14, 87, 1234344564, 47398573854734834, etc., it won't work because again, you could go around the curve and you won't know where the key lies or where to start your stride function from.

There is truly only one way to take advantage of a stride function; and I stated that months ago. My problem was, I could not create a stride function inside of CUDA. Or else I would have found 130's key already lol.



If you write the exact step code you want, there will be friends who will try to help. If you draw the algorithm flow diagram they can help you
member
Activity: 499
Merit: 38
My problem was, I could not create a stride function inside of CUDA. Or else I would have found 130's key already lol.

Is there a built-in "stride" function specifically in the CUDA kernel? Or maybe there is a workaround in a grid-stride loop? Or do we have to write a new kernel? Smiley
jr. member
Activity: 47
Merit: 12
gmaxwell creator of 1000 BTC puzzl + Pinapple fund
If we are talking speed, I have a program (a variation of BitCrack) that will get around 400 MKey/s (per GPU) but honestly, that is to slow. I would like to get around 1,200 MKey/s (per card, low end 30xx card) and multiple cards per instance.

Ok I can help you with coding if I think it works. I also modified BitCrack and solved some of the early puzzles with pubkey.
full member
Activity: 1162
Merit: 237
Shooters Shoot...
I know you are used to spitting out random things and some are like ooooohhhhh ahhhhhhhhh, but you seldom listen lol.

He is just having a good time. Writing everything that comes to his mind here and creating alt accounts etc for additional drama Cheesy. Why not.

How many keys/s do you need for your approach to work?
If we are talking speed, I have a program (a variation of BitCrack) that will get around 400 MKey/s (per GPU) but honestly, that is to slow. I would like to get around 1,200 MKey/s (per card, low end 30xx card) and multiple cards per instance.
jr. member
Activity: 47
Merit: 12
gmaxwell creator of 1000 BTC puzzl + Pinapple fund
I know you are used to spitting out random things and some are like ooooohhhhh ahhhhhhhhh, but you seldom listen lol.

He is just having a good time. Writing everything that comes to his mind here and creating alt accounts etc for additional drama Cheesy. Why not.

How many keys/s do you need for your approach to work?
full member
Activity: 1162
Merit: 237
Shooters Shoot...
Like you said, we do not know the target, only the range.
So let's say the target's key is 199,999
Alrighty then, to be exact, when it comes to slapping me suddenly the secure random generator gives us 199,999? And when we are working with 6 digits then as an example use 14, 15 etc, suddenly we go high as 2^130 and compare the numbers?
So why not scaling up all the values and use actual 2^130 range/keys? Though you forgot that I was asking if the stride idea could be tweaked to find a perfect stride or not.

Even if the key is e.g, 189776, we could still divide the end range and subtract from our target by different subranges, like 200,000 - 189776 = 10224, we could then try subtracting the result from endrange/4, er/6, er/8 etc.

Nvm that, I don't know anything about math, I'm not even working actively on 130, my target is much bigger.

Btw, Legends_Never_Die is my alter ego account, time for a paint job on trust wall.😉
I know you are used to spitting out random things and some are like ooooohhhhh ahhhhhhhhh, but you seldom listen lol.

You could be working in a 50 bit range or a 256 bit range; and you can subtract, divide, multiply, etc. whatever you want to do, or you can use whatever stride you can some up with, 14, 87, 1234344564, 47398573854734834, etc., it won't work because again, you could go around the curve and you won't know where the key lies or where to start your stride function from.

There is truly only one way to take advantage of a stride function; and I stated that months ago. My problem was, I could not create a stride function inside of CUDA. Or else I would have found 130's key already lol.
member
Activity: 499
Merit: 38
the secure random generator gives us 199,999

Try non-secure random generator

Code:
import random
puzzle = 18
lower_range_limit = 2 ** (puzzle - 1)
upper_range_limit = (2 ** puzzle) - 1
seed = b'\x97h6\xd9\x7f\xcbh\xc4'
random.seed(seed)
dec = random.randint(lower_range_limit, upper_range_limit)
print(f"{dec}, {seed}")

 Grin
copper member
Activity: 1330
Merit: 899
🖤😏
Like you said, we do not know the target, only the range.
So let's say the target's key is 199,999
Alrighty then, to be exact, when it comes to slapping me suddenly the secure random generator gives us 199,999? And when we are working with 6 digits then as an example use 14, 15 etc, suddenly we go high as 2^130 and compare the numbers?
So why not scaling up all the values and use actual 2^130 range/keys? Though you forgot that I was asking if the stride idea could be tweaked to find a perfect stride or not.

Even if the key is e.g, 189776, we could still divide the end range and subtract from our target by different subranges, like 200,000 - 189776 = 10224, we could then try subtracting the result from endrange/4, er/6, er/8 etc.

Nvm that, I don't know anything about math, I'm not even working actively on 130, my target is much bigger.

Btw, Legends_Never_Die is my alter ego account, time for a paint job on trust wall.😉
member
Activity: 499
Merit: 38
Here is new script....
Bytea HASH160 Search.
Bitcoin addresses and hashes are typically represented as byte sequences. When you work with bytes, you can directly perform binary operations and comparisons, which is faster and more efficient than dealing with hexadecimal or string representations.
In this context, the script generates private keys as integers and converts them into bytes to derive Bitcoin addresses. It then calculates the Hash160 of these addresses and compares it to the target Hash160.
Script uses puzzle creator method - the b'\x00' * 32 line creates a 32-byte value composed entirely of zero bytes. This is just an initial placeholder value, and as the script progresses, it replaces parts of these zeros with the random bytes generated to create a valid private key for Bitcoin.
You can also play with patterns and random seed values at the same time.

Code:
import sys
import os
import time
import random
import binascii
import base58
import hashlib
import ecdsa
from multiprocessing import cpu_count
import threading

# Set the Hash 160 address you want to check for
target_hex = "20d45a6a762535700ce9e0b216e31994335db8a5"
add_set = bytes.fromhex(target_hex)

puzzle = 66
min_number = 2 ** (puzzle - 1)
max_number = (2 ** puzzle) - 1

# Clear the terminal screen
if os.name == 'nt':
    os.system('cls')
else:
    os.system('clear')

# Print information about the program
t = time.ctime()
sys.stdout.write("\033[01;33m")
sys.stdout.write("\033[?25l")
sys.stdout.write(f"[+] [Bytea HASH160 Search] {t}" + "\n")
sys.stdout.write(f"[+] [Puzzle]: {puzzle}" + "\n")
sys.stdout.write(f"[+] [Lower range limit]: {min_number}" + "\n")
sys.stdout.write(f"[+] [Upper range limit]: {max_number}" + "\n")
sys.stdout.flush()

# Define the check_private_key function
def check_private_key(add_set):
    while True:
        constant_prefix = b''  #Can be b'\xbc\x9b' or any random_bytes
        prefix_length = len(constant_prefix)
        length = 8
        ending_length = length - prefix_length
        ending_bytes = os.urandom(ending_length)
        random_bytes = constant_prefix + ending_bytes
        random.seed(random_bytes)
        dec = random.randint(min_number, max_number)
        dec_bytes = dec.to_bytes((dec.bit_length() + 7) // 8, 'big')
        private_key_bytes = b'\x00' * 32
        private_key_bytes = private_key_bytes[:-len(dec_bytes)] + dec_bytes
        signing_key = ecdsa.SigningKey.from_string(private_key_bytes, curve=ecdsa.SECP256k1)
        compressed_public_key = signing_key.get_verifying_key().to_string("compressed")
        sha256_hash = hashlib.sha256(compressed_public_key).digest()
        HASH160 = hashlib.new('ripemd160', sha256_hash).digest()
        extended_private_key = b'\x80' + private_key_bytes + b'\x01'
        extended_ripe160_hash = b'\x00' + HASH160
        checksum = hashlib.sha256(hashlib.sha256(extended_private_key).digest()).digest()[:4]
        checksum_ripe160 = hashlib.sha256(hashlib.sha256(extended_ripe160_hash).digest()).digest()[:4]
        private_key_with_checksum = extended_private_key + checksum
        HASH160_wif = base58.b58encode(private_key_with_checksum).decode()
        address_bytes = extended_ripe160_hash + checksum_ripe160
        bitcoin_address = base58.b58encode(address_bytes).decode()
        message = "\r[+] Public Key Hash (Hash 160): {}             ".format(HASH160.hex())
        messages = []
        messages.append(message)
        output = "\033[01;33m" + ''.join(messages) + "\r"
        sys.stdout.write(output)
        sys.stdout.flush()
        if bitcoin_address.startswith("13zb1h"):
            sys.stdout.write(f"\n[+]\033[32m Pattern Found: {bitcoin_address}, {dec}, {random_bytes}              \033[0m")
            sys.stdout.write(f"\n[+] continue...\n")
        if HASH160 == add_set:
            dec_to_hex = hex(dec).split('x')[-1]
            t = time.ctime()
            sys.stdout.write(f"\033[0m\n\n")
            sys.stdout.write(f"[+] SOLVED:    |\033[32m{t}                                                                 \033[0m\n")
            sys.stdout.write(f"[+] Key Found: |\033[32m {dec_to_hex}                                               \033[0m\n"
                                        f"[+] WIF:       |\033[32m {HASH160_wif}                                                 \033[0m\n"
                                        f"[+] Address:   |\033[32m {bitcoin_address}                                          \033[0m\n"
                                        f"[+] Seed:      |\033[32m {random_bytes}                                               \033[0m\n\n")
            sys.stdout.flush()
            with open("KEYFOUNDKEYFOUND.txt", "a") as f:
                f.write("SOLVED: " + str(t) + '\n' + "HEX: " + str(dec_to_hex) + '\n' + "Address: " + str(bitcoin_address) + '\n' + "Private Key:  " + str(HASH160_wif)+ '\n' + "Random Seed:  " + str(random_bytes) + '\n\n')
                f.flush()
                f.close()
            sys.stdout.write("\033[?25h")
            sys.stdout.flush()
            return

# Create multiple threads for generating addresses
num_threads = cpu_count()
threads = []

for _ in range(num_threads):
    thread = threading.Thread(target=check_private_key, args=(add_set,))
    thread.start()
    threads.append(thread)

# Wait for all threads to finish
for thread in threads:
    thread.join()
full member
Activity: 1162
Merit: 237
Shooters Shoot...
Guys/gals, don't waste your time with random search, random mode etc, there is no such a thing as "random" all events follow a pattern,  we as humans have to interfere and change these patterns, by following logic and the power of our minds.

Instead of "hoping" for a "lucky" hit in a random search, start working on ways to find a stride for public key brute force.
Example:
Target = 127654 (unknown) we only know it's between 100,000 and 200,000.
So we subtract it from 200,000 - 127654 = 72,346.  Now we are certain our result is 100% smaller than 100,000 (start range), but since we don't know how much smaller, we'd just subtract it from half of start range, 50,000 - 72,346 = 22,346. And now we are 100% certain the result is smaller than 50,000 but we don't know how much, again we subtract it from 1/4 of start range 25,000 - 22,346 = 2,654.

Now tell me what do you see?
127654
2654
127654 - 2654 = 125000
What you need to work on is figuring out a stride to add to 125,000 till we reach our target, whether we add 13 at each step, 14 or 15, what happens when we reach 127698? Can we save the keys between 127654 and 127700 so when we are adding stride and land on one of the saved keys, we know right away and easily solve the key or not?

These things should be your priority, enough of running this tool/script and that tool/script, come up with an algorithm which doesn't require "random" and "luck" but requires math equations and numbers.

Sorry man, this either will not work because you went around the curve, or you will have trillions^trillions of strides to go through/take.

Your example:
Target = 127654
range = 100,000 - 200,000

Like you said, we do not know the target, only the range.
So let's say the target's key is 199,999
Watch the math:
200,000 - 199,999 = 1
1 - 50,000 = -49,999
-49,999 - 25,000 = -74,999

You have now went backwards around the curve. So now, you can't even start your program to start striding from 0 or 1.

But let's deal with your math above and talk about the strides.

It's ok when a range is only 200,000 but now imagine a much larger range, say 2^130. If you have a stride of 13, 14, 15, etc., it will take you more than a lifetime to still get through that range.
jr. member
Activity: 64
Merit: 1
34Sf4DnMt3z6XKKoWmZRw2nGyfGkDgNJZZ
member
Activity: 499
Merit: 38
Except these safes are publicly reachable unlike private safes inside houses.

ATMs are publicly reachable. Try to hack them.  I'm kidding...  
🤣
I understand the whole story.    Grin
newbie
Activity: 4
Merit: 1
this isn't conceptually different from picking up someone's safe and finding its lock combination, then claiming the safe's contents for the sake of science.   Grin
Except these safes are publicly reachable unlike private safes inside houses. Consider yourself a security team member working on security issues, your payment is in puzzle addresses, find a hole and grab your payment, if the designer was a professional programmer, he wouldn't have made this puzzle, he is a good scientist but not a good programmer, as early bitcoin clients were full of bugs/code related issues, but the concept was perfect and it is to this day.  Scientists act based on science, puzzle hunters act based on monetary incentive.

"They asked someone, imagine you are in the middle of the ocean and sharks are approaching, what will you do?"
"He said I will climb a tree"
-They said but there are no trees in the middle of the ocean-
"He said I know, but I don't have any other choice"
🤣
member
Activity: 499
Merit: 38
I'm delusional by relying on mathematics, whilst you and a few others are delusional relying on "random" search

It doesn't matter what method did you use if you succeed in your intention. And no matter how crazy the way may seem,
this isn't conceptually different from picking up someone's safe and finding its lock combination, then claiming the safe's contents for the sake of science.   Grin
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