Bitcoin puzzle transaction ~32 BTC prize to who solves it

[Cricktor]

it’s not very obvious to me what you’re talking about, but I noticed transfers of 1000 satoshi to some wallets. (?)

Puzzle addresses which are multiples of 5 have a spending transaction which exposes the public key. That's the main difference compared to puzzles which are not multiples of 5.

An exposed public key allows other more efficient attacks like Kangaroo to find the private key. That's why #75, #80, #85, ... #130 are all already solved while determined solvers still grind on #71 where the public key is not exposed. No better (actually working!) method has been proposed than brute-force search in the key space range of such a puzzle without known public key.

If the public key of #71 were to be exposed by a silly public withdrawal of a retarded solver, bots would solve it within seconds and try to steal the funds via full-RBF replacement of the withdrawal transaction waiting in mempools for confirmation.


Can you stop violating rule #32 of Unofficial list of (official) Bitcointalk.org rules, guidelines, FAQ!

[0x6667d]

it’s not very obvious to me what you’re talking about, but I noticed transfers of 1000 satoshi to some wallets. (?)

Puzzle addresses which are multiples of 5 have a spending transaction which exposes the public key. That's the main difference compared to puzzles which are not multiples of 5.

An exposed public key allows other more efficient attacks like Kangaroo to find the private key. That's why #75, #80, #85, ... #130 are all already solved while determined solvers still grind on #71 where the public key is not exposed. No better (actually working!) method has been proposed than brute-force search in the key space range of such a puzzle without known public key.

If the public key of #71 were to be exposed by a silly public withdrawal of a retarded solver, bots would solve it within seconds and try to steal the funds via full-RBF replacement of the withdrawal transaction waiting in mempools for confirmation.


Can you stop violating rule #32 of Unofficial list of (official) Bitcointalk.org rules, guidelines, FAQ!

Thx a lot!
Oh, I'm sorry. I will definitely read these rules

[Anas35]

Hey guys, instead of wasting your eye sight on long and useless base58 WIFs which literally represent 0s in hexadecimal, let me share a little secret regarding public keys.

Here is how you can find half of your public key, it's not straight forward method but I bet many of you didn't know about it.

First we need to extract 1 and half of our public key then we can subtract our p which is 1 from it's 1.5 to get it's 0.5 half. Though we could just divide it by 2 without all this trouble, this is a hint to make you dive deeper in to this vast ocean of numbers and equations.

Target  pub:

Code:

03219b4f9cef6c60007659c79c45b0533b3cc9d916ce29dbff133b40caa2e96db8

Target priv:

Code:

0x800000000000

Our multiplier inverse or not doesn't matter, +n will give you +n result and vice versa.

Scalar : aka n/2+1 half n +1 or 1 and half of n.

Code:

7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a2

If you multiply our target pub with scalar above, you will get this :
Pub2 :

Code:

02161c6cbee1483deaf6f9b395c817eb019228cda5afac5857295ba10959dffc96

Priv :

Code:

0xc00000000000

Now if we subtract target from pub2, we will get :

Pub3, half of target :

Code:

0313d1ffc481509beee68f17d8ff41c2590f4c85f15268605087eda8bab4e218da

Priv :

Code:

0x400000000000

We didn't even use division, *chop chop and good luck diving.😅


* = hurry! Get to work.

 I rather be use the target public key is then (using ecdsa arithmetic) reduced subseqentially until hitting one of the rendezvous point. Given G is the generator point, and  R=x*G the rendezvous point (which private key is known).

and BSGS.

u just like a teacher on class math in 4th grade lmao.


----------

There's i got brute the WIF and it's manually not any application or program i use.

I just use called MATH.

i just make key range more better for fast scan at the lower and upper range.

Puzzle 130 addresses 1Fo65aKq8s8iquMt6weF1rku1moWVEd5Ua


   2baf7cf4d51574f5deee49961d9609ac6 # 1FHMVBF6Z5ebm5LJKRpkZs8RjNJXUMyUVV # KwDiBf89QgGbjEhKnhXJuH85q VZKQivFjnRMPNnqDyFoSqrSaKfd
   2baf78fec51be4f59eaf6966bd9709ac6 # 1FXUvZnrcqRnJV1Cvre2ribDz6yKqQugjR # KwDiBf89QgGbjEhKnhXJuH85q VMiSmxd4F8urBwimS72wrJD2XYS
   2baf78fac5b3e4f6feaf6962bd9709ac6 # 1FqZVS2u9iqcycye9BmtVEcCGM6wCYXHM2 # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvjexRvRbBdvgeEZd
   2baf78fac5b3e4f6feaf6942bd9709ac6 # 1Fe7X3dKymfuae7xLDZ9Hk5dn9W5pf4SE8 # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvjAC4AFCc3v8Pooi
   2baf78fac5b3e4f6feaf69c2bf1709ac6 # 1FnUH4z8cgLuWBLFFWvXBDdQ1VfFc4SM1k # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvm9FrvdFhuoEb9k1
   2baf78fac5b3e4f6feaf69f2bf9a8a46c # 1FfUFwHRS4cEESP3JDBMQMaZxLCZTN8tY  # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPyuPyvmtQXdpe3Zfy6UtD
   2baf78fac5b3e4d018100000000000000 # 1F2nhqcHQRKwjjokgkKkcSpLH4uucuQZSV # KwDiBf89QgGbjEhKnhXJuH85q VMfnuJPuxaPiBGv4PL2aFAnuFR8
   3baf70fac037e51011500040000000000 # 1FpLqZyycLiaLWQ6Lbe5zcXuypbX6F7q4C # KwDiBf89QgGbjEhKnhXJuH8Ma WVpmffaLXxzgafdFBXKhAAvtpYn
   3baf7efac999e3c95c4ea7660b9946c94 # 1Fcmac1xrLNMMr286BsFnWN35pFNgZeWWb # KwDiBf89QgGbjEhKnhXJuH8Ma XCqFfLS8nKhgd7RpHR2XZB5AKhx
   3baf7efac999e3cddc6ba764fb9b66cb4 # 1FqfgJmp8AKvDkAKbUcncgMA7kmDn9dJ2v # KwDiBf89QgGbjEhKnhXJuH8Ma XCqFfLS9Eo6LP4p7NDXN5sx82Js
   3bbff6cba4bd242d6ceb2777fb9866ca4 # 1FfPWje1WcZ47fAKuGBijKsqhvF8bHzf3f # KwDiBf89QgGbjEhKnhXJuH8Me C7XhwLTtDcX5XKNNEHm7o8DUj98
   3bbff6cb95bca56cdee22756fb970bcaa # 1F2UYUqkShEp8aa9CCE6mKN3yLHbKpEPud # KwDiBf89QgGbjEhKnhXJuH8Me C7XfgWcWBrqsVyALfRFgYf7PCe9

i just hunt the pub and address by privK variation with 16 bit per each line of HEX.

just reminder, if you BSGS the full range this 130 Puzzle, it takes 13.8 billion years lmao.

Hey,

I would like to meet you on WhatsApp.  If you can please inbox me  I have a pattern but really confuses me in the recovery but with your help we definitely can find the right solution.

Thanks
Anaz

[pbies]

All these puzzle pvks could be generated this way:

(Python 3 script)

Code:

#!/usr/bin/env python3

from multiprocessing import Pool, cpu_count
from tqdm import tqdm
import sys, base58, random

from hdwallet import HDWallet
from hdwallet.cryptocurrencies import Bitcoin as BTC
from hdwallet.hds import BIP32HD

_hdwallet = HDWallet(cryptocurrency=BTC, hd=BIP32HD)

def pvk_to_wif2(key_hex: str) -> str:
	return base58.b58encode_check(b'\x80' + bytes.fromhex(key_hex)).decode()

def go(key_hex: str):
	_hdwallet.from_private_key(private_key=key_hex)
	wif = pvk_to_wif2(key_hex)
	out = (
		f"{key_hex}\n"
		f"{wif}\n"
		f"{_hdwallet.wif()}\n"
		f"{_hdwallet.address('P2PKH')}\n"
		f"{_hdwallet.address('P2SH')}\n"
		f"{_hdwallet.address('P2TR')}\n"
		f"{_hdwallet.address('P2WPKH')}\n"
		f"{_hdwallet.address('P2WPKH-In-P2SH')}\n"
		f"{_hdwallet.address('P2WSH')}\n"
		f"{_hdwallet.address('P2WSH-In-P2SH')}\n"
		f"\n"
	)
	return out

def gen():
	for i in range(0,256):
		j=2**i
		r=random.randint(j,j*2)
		yield hex(r)[2:].zfill(64)

def main():
	o = open('output.txt', 'w')

	for s in tqdm(range(0, 2049)):
		random.seed(s)
		with Pool(cpu_count()) as pool:
			for res in pool.imap_unordered(go, gen()):
				o.write("".join(res))

	print('\a', end='', file=sys.stderr)

if __name__ == '__main__':
	main()

[skedarve]

The truth is that I have a pattern and im willing to share it with someone who has a Voucher and can implement it in CUDA.

[snaz3d]

The truth is that I have a pattern and im willing to share it with someone who has a Voucher and can implement it in CUDA.

If there was one you wouldn't need a GPU implementation