Nope. It's starting with 34
And the public key starts with 03?  |
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what do you mean payout!
can you give me answer here, thank you
It means that he has a private key or knows someone who does. And that he is large miner who will include that transaction in the next block to be calculated. In other words, the bots are screwed. |
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There is a new NVIDIA GPU that was announced, the GB200 which is exponentially more powerful than the current gpus.
You need GB200 NVL72 rack to solve 130. Find out there how much it costs. I meant when they become available through data centers and gpu servers. I think there will be a long line to wait. |
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There is a new NVIDIA GPU that was announced, the GB200 which is exponentially more powerful than the current gpus.
You need GB200 NVL72 rack to solve 130. Find out there how much it costs. |
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I’m seeing so many variations of software throughout this thread and the other one. If I wanted to randomly let my pc waste electricity on weekends doing a random search for #66, what would be my best bet?
Use the program as you wish, but from puzzle 79, 81, 82, 83 and above. Or, ask someone with knowledge about timings which is the lowest feasible puzzle to be solved that will leave enough time to withdraw without a bot snatchingup. I don't know this answer, could be even puzzle #70?
If someone can answer that, that would be great.
Puzzle 81 is quite safe. I solved Puzzle 80 in about 26 hours (with extremely large memory in BSGS). The bot does not work on that number. |
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But I'm sure (no, I know) that you don't know the private key of #66 yet.  I would hire some block solvers here right away. |
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# ./keyhunt -m bsgs -f 65.txt -t 12 -k 4096 -l compress -b 65 -S -q -S -e - Version 0.2.230519 Satoshi Quest, developed by AlbertoBSD
- Threads : 12
- K factor 4096
- Search compress only
- Quiet thread output
- Endomorphism enabled
- Mode BSGS sequential
- Opening file 65.txt
- Added 1 points from file
- Bit Range 65
- -- from : 0x10000000000000000
- -- to : 0x20000000000000000
- N = 0x100000000000
- Bloom filter for 17179869184 elements : 58890.60 MB
- Bloom filter for 536870912 elements : 1840.33 MB
- Bloom filter for 16777216 elements : 57.51 MB
- Allocating 256.00 MB for 16777216 bP Points
- Reading bloom filter from file keyhunt_bsgs_4_17179869184.blm .... Done!
- Reading bloom filter from file keyhunt_bsgs_6_536870912.blm .... Done!
- Reading bP Table from file keyhunt_bsgs_2_16777216.tbl .... Done!
- Reading bloom filter from file keyhunt_bsgs_7_16777216.blm .... Done!
- Thread Key found privkey 1a838b13505b26867
- Publickey 0230210c23b1a047bc9bdbb13448e67deddc108946de6de639bcc75d47c0216b1b
All points were found It takes about 10 seconds to find after the files are loaded. AMD Ryzen 5 3600 / 64GB RAM *It even depends on how it is compiled for which platform - for example: gcc -Q -march=native --help=target | grep -E '^\s+-.*(sse|march)' -march= znver2 -mfpmath= sse -mno-sse4 [disabled] -msse4.2 [enabled] -msse4a [enabled] -msse5 -mavx -msseregparm [disabled] -mssse3 [enabled] Makefile g++ -m64 -march=native -mtune=native -msse4.2 ...etc... |
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which processor is faster amd ryzen or intel in python intel core i9 14900hx---intel core i9-13980hx---amd ryzen 9 7945hx, does anyone have any experience with how fast your processors with iceland secp256k1 library go
I know it depends on the code settings, but it would be nice to know such approximate information
This is a demonstration and test of how slow Python is. Even if it's a few million keys per second. Random sequence: from multiprocessing.pool import Pool
from subprocess import check_output
from tqdm import tqdm
from tqdm.contrib.concurrent import process_map
import secp256k1 as ice
import math
import random
import sys
div=16384
start=0x20000000000000000
end=0x3ffffffffffffffff
rng=0x3ffffffffffffffff-0x20000000000000000
stepout=int(rng/div)
stepin=0x200000000
right='13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so'
sys.stdout.write("\033[01;33m")
print('[+] target: '+right)
def int_to_bytes3(value, length = None): # in: int out: bytearray(b'\x80...
if not length and value == 0:
result = [0]
else:
result = []
for i in range(0, length or 1+int(math.log(value, 2**8))):
result.append(value >> (i * 8) & 0xff)
result.reverse()
return bytearray(result)
def pvk_to_addr(pvk):
return ice.privatekey_to_address(0, True, pvk)
global c
c = 0
def go(r):
global c
if c % 100 == 0:
print(f'[+] {c:,} Keys\r'.replace(',', ' '), end='')
c = c + 1
by = int_to_bytes3(r, 32)
pvk = int.from_bytes(by, byteorder='big') # Convert bytearray to integer
ad = pvk_to_addr(pvk)
# print('\r'+ad,end='')
if ad == right:
print('found!')
print(r)
print(hex(r))
HEX = "%064x" % int(r)
wifc = ice.btc_pvk_to_wif(HEX)
print(wifc)
print('\a')
with open('found.txt', 'w') as f:
f.write(str(r))
f.write('\n')
f.write(hex(r))
f.write('\n')
f.write(wifc)
f.write('\n')
f.flush()
sys.exit(0)
return
def n(a,b):
return list(range(a,b))
s=int(rng/div)
pool = Pool(10)
u=1048576
while True:
ra=random.randint(start,end-u)
rb=ra+u
print(f'\r[+] from: {hex(ra)} to: {hex(rb)} range: {hex(u)}={u}')
#global c
c=0
pool.map(go, range(ra,rb), chunksize=32768)
pool.close()
pool.join() Sequential sequence: from multiprocessing.pool import Pool
from subprocess import check_output
from tqdm import tqdm
from tqdm.contrib.concurrent import process_map
import secp256k1 as ice
import math
import random
import sys
div=16384
start=0x20000000000000000
end=0x3ffffffffffffffff
rng=0x3ffffffffffffffff-0x20000000000000000
stepout=int(rng/div)
stepin=0x200000000
right='13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so'
sys.stdout.write("\033[01;33m")
print('[+] target: '+right)
def int_to_bytes3(value, length = None): # in: int out: bytearray(b'\x80...
if not length and value == 0:
result = [0]
else:
result = []
for i in range(0, length or 1+int(math.log(value, 2**8))):
result.append(value >> (i * 8) & 0xff)
result.reverse()
return bytearray(result)
def pvk_to_addr(pvk):
return ice.privatekey_to_address(0, True, pvk)
global c
c = 0
def go(r):
global c
if c % 100 == 0:
print(f'[+] {c:,} Keys\r'.replace(',', ' '), end='')
c = c + 1
by = int_to_bytes3(r, 32)
pvk = int.from_bytes(by, byteorder='big') # Convert bytearray to integer
ad = pvk_to_addr(pvk)
# print('\r'+ad,end='')
if ad == right:
print('found!')
print(r)
print(hex(r))
HEX = "%064x" % int(r)
wifc = ice.btc_pvk_to_wif(HEX)
print(wifc)
print('\a')
with open('found.txt', 'w') as f:
f.write(str(r))
f.write('\n')
f.write(hex(r))
f.write('\n')
f.write(wifc)
f.write('\n')
f.flush()
sys.exit(0)
return
def n(a,b):
return list(range(a,b))
s=int(rng/div)
pool = Pool(10)
u = 1048576
for ra in range(start, end - u + 1, u):
rb = ra + u
print(f'\r[+] from: {hex(ra)} to: {hex(rb)} range: {hex(u)}={u}')
c = 0
pool.map(go, range(ra, rb), chunksize=32768)
pool.close()
pool.join() You can test with this... Very slow. About a million keys per core. Just for the sake of comparison, I have about 40 Mkeys/s on BSGS with 12 Cores in address mode..... |
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So how did the person who claims to be a constituent managed to create the HEX address 0000000000000000000000000000000000000000000000000000000000000000000000000000000 000001?
It is simple script If you know what is last key - e.g. '910ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b' Python script as an example key = '910ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b'
last_index = 0
while last_index <= 255:
mask = 1 << last_index
mask_hex = mask + int(key, 16) % mask
last_index += 1
hexa = "%064x" % mask_hex
print (hexa) result 0000000000000000000000000000000000000000000000000000000000000001
0000000000000000000000000000000000000000000000000000000000000003
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000007ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
00000bab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
000013ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
000033ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
000053ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
0000d3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
0001d3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
0002d3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
0006d3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
000ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
001ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
002ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
004ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
008ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
010ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
030ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
050ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
090ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
110ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
310ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
510ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b
910ed3ab5775b97da1a33e38b72eba94f90addff7051f276c9d6c538c369632b But no one knows the exact formula from the last private key to 1. I even think he used logarithms and polynomials as steps between private keys.. So it ends up looking like there is no pattern. But it's not random either according to the polynomial analysis. https://bitcointalk.org/index.php?topic=1306983.msg63730770#msg63730770There is an exact math formula for making this puzzle with some script, errors = ZERO. With high decimal precision (mp.dps = 20 at least) And the formula is in the creator's mind. |
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in short, you have puzzle 66 key, and afraid to create tx ?
Who knows how many of them here are with a key |
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 This is Blackwell |
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So anyone could do millions any time.
It seems so. From Puzzle 66 to 79... It's a waste of electricity to brute force. However, you can wait for someone else to do it and then pick up the money Maybe Satoshi was just waiting for us to come to this conclusion. |
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I did about 10 attempts today, trying to change already broadcasted transaction, sent from bitcoin-core with RBF flag off.
All attempts fail. First transaction always win.
Then we must do kind of kung-fu. Using older Electrum versions than 4.4.0 with RBF flag off. |
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"Do not disturb my kangaroos." -- Archimedes |
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If you look, this has been run on Laptop with 9 year old GPU, still < 1 minute.
What will happen if someone run it on Recent Workstation GPU. It will be few seconds only.
You simply create a script that pings: https://blockchain.info/q/pubkeyaddr/13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so. If the pubkey returned is not 'error', extract the key and run Kangaroo. Upon finding the private key, initiate bitcoin-cli to execute the transaction, all within a 10-second timeframe. You don't have to be at the computer and watch at all. No manual supervision required. |
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This is like throwing one fish into a pond of sharks. Either it will be torn apart and there will be nothing left for anyone, or one will be the fastest.
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Maybe we won't find the puzzle, but the person who finds it will be the victim. People with bitcoin technical background and any kind of knowledge on this subject should share what they know.
Don't get too upset. I think there are at least 1000 bots already set up and waiting. There will be chaos the moment puzzle 66 public key appears. If there is any money left in that bidding. |
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In fact, for block 66, a private key must be generated with the 81129638414569788207641586040831 number, and this process is slow with a CPU, but using a graphics processor speeds it up.
This is a demonstration and test of how slow Python is. Even if it's a few million keys per second. Random sequence: from multiprocessing.pool import Pool
from subprocess import check_output
from tqdm import tqdm
from tqdm.contrib.concurrent import process_map
import secp256k1 as ice
import math
import random
import sys
div=16384
start=0x20000000000000000
end=0x3ffffffffffffffff
rng=0x3ffffffffffffffff-0x20000000000000000
stepout=int(rng/div)
stepin=0x200000000
right='13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so'
sys.stdout.write("\033[01;33m")
print('[+] target: '+right)
def int_to_bytes3(value, length = None): # in: int out: bytearray(b'\x80...
if not length and value == 0:
result = [0]
else:
result = []
for i in range(0, length or 1+int(math.log(value, 2**8))):
result.append(value >> (i * 8) & 0xff)
result.reverse()
return bytearray(result)
def pvk_to_addr(pvk):
return ice.privatekey_to_address(0, True, pvk)
global c
c = 0
def go(r):
global c
if c % 100 == 0:
print(f'[+] {c:,} Keys\r'.replace(',', ' '), end='')
c = c + 1
by = int_to_bytes3(r, 32)
pvk = int.from_bytes(by, byteorder='big') # Convert bytearray to integer
ad = pvk_to_addr(pvk)
# print('\r'+ad,end='')
if ad == right:
print('found!')
print(r)
print(hex(r))
HEX = "%064x" % int(r)
wifc = ice.btc_pvk_to_wif(HEX)
print(wifc)
print('\a')
with open('found.txt', 'w') as f:
f.write(str(r))
f.write('\n')
f.write(hex(r))
f.write('\n')
f.write(wifc)
f.write('\n')
f.flush()
sys.exit(0)
return
def n(a,b):
return list(range(a,b))
s=int(rng/div)
pool = Pool(10)
u=1048576
while True:
ra=random.randint(start,end-u)
rb=ra+u
print(f'\r[+] from: {hex(ra)} to: {hex(rb)} range: {hex(u)}={u}')
#global c
c=0
pool.map(go, range(ra,rb), chunksize=32768)
pool.close()
pool.join() Sequential sequence: from multiprocessing.pool import Pool
from subprocess import check_output
from tqdm import tqdm
from tqdm.contrib.concurrent import process_map
import secp256k1 as ice
import math
import random
import sys
div=16384
start=0x20000000000000000
end=0x3ffffffffffffffff
rng=0x3ffffffffffffffff-0x20000000000000000
stepout=int(rng/div)
stepin=0x200000000
right='13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so'
sys.stdout.write("\033[01;33m")
print('[+] target: '+right)
def int_to_bytes3(value, length = None): # in: int out: bytearray(b'\x80...
if not length and value == 0:
result = [0]
else:
result = []
for i in range(0, length or 1+int(math.log(value, 2**8))):
result.append(value >> (i * 8) & 0xff)
result.reverse()
return bytearray(result)
def pvk_to_addr(pvk):
return ice.privatekey_to_address(0, True, pvk)
global c
c = 0
def go(r):
global c
if c % 100 == 0:
print(f'[+] {c:,} Keys\r'.replace(',', ' '), end='')
c = c + 1
by = int_to_bytes3(r, 32)
pvk = int.from_bytes(by, byteorder='big') # Convert bytearray to integer
ad = pvk_to_addr(pvk)
# print('\r'+ad,end='')
if ad == right:
print('found!')
print(r)
print(hex(r))
HEX = "%064x" % int(r)
wifc = ice.btc_pvk_to_wif(HEX)
print(wifc)
print('\a')
with open('found.txt', 'w') as f:
f.write(str(r))
f.write('\n')
f.write(hex(r))
f.write('\n')
f.write(wifc)
f.write('\n')
f.flush()
sys.exit(0)
return
def n(a,b):
return list(range(a,b))
s=int(rng/div)
pool = Pool(10)
u = 1048576
for ra in range(start, end - u + 1, u):
rb = ra + u
print(f'\r[+] from: {hex(ra)} to: {hex(rb)} range: {hex(u)}={u}')
c = 0
pool.map(go, range(ra, rb), chunksize=32768)
pool.close()
pool.join() |
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Maybe disabling RBF would be a start, and using a high fee.
Yep, that's the solution. It is one-time try only. This has been bugging me and put me off from searching 66. Ok so say i found it can i make the fee high enough like in electrum and broadcast it a few minutes before the next block ? This way i have a few mins to get in next block before the pub key is cracked. Is this possible.? Otherwise what is the point ? That is the enigma. Maybe 66 is already hacked, but no one will make transaction first. 130 is safe |
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