Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 282.

Jolly Jocker

jr. member

Activity: 49

Merit: 1

A fast scanner that completely covers the 64-bit range with more than 12000 keys/s (depending on the CPU clock) with Intel Xeon E3-1240 V2 @ 3.40GHz (13500 keys/s)

Code:

import sys
import random
from bit import *
from time import sleep
from tqdm import tqdm
print('\n              ================16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN============== \n\n')
x=1048576
list=[]
print(' WAIT!....', end='\r')
while x:
    b01 = random.choice("0123456789abcdef")
    b02 = random.choice("0123456789abcdef")
    b03 = random.choice("0123456789abcdef")
    b04 = random.choice("0123456789abcdef")
    b05 = random.choice("0123456789abcdef")
    b06 = random.choice("0123456789abcdef")
    b07 = random.choice("0123456789abcdef")
    b08 = random.choice("0123456789abcdef")
    b09 = random.choice("0123456789abcdef")
    b10 = random.choice("0123456789abcdef")
    b11 = random.choice("0123456789abcdef")
    b12 = random.choice("0123456789abcdef")
    b13 = random.choice("0123456789abcdef")
    b14 = random.choice("0123456789abcdef")
    b15 = random.choice("0123456789abcdef")
    
    hex = (b01+b02+b03+b04+b05+b06+b07+b08+b09+b10+b11+b12+b13+b14+b15)
    hex8 = '8'+ hex
    list.append(hex8)
    hex9 = '9'+ hex
    list.append(hex9)
    hexa = 'a'+ hex
    list.append(hexa)
    hexb = 'b'+ hex
    list.append(hexb)
    hexc = 'c'+ hex
    list.append(hexc)
    hexd = 'd'+ hex
    list.append(hexd)
    hexe = 'e'+ hex
    list.append(hexe)
    hexf = 'f'+ hex
    list.append(hexf)
    if len(list)==x:    
        line_count=0
        for line0 in tqdm(list):
            line0!='\n'             
            line_count+=1
            line = int(line0, 16)
            a = len(bin(line))-2
            key = Key.from_int(line)
            addr = key.address                                       
         #  sys.stdout.write('\r scanning.... [' + (line0) + '] ' + str(a) + ' bit ')
                        
            if addr.startswith('16jY7'): # 16jY7q
                sys.stdout.write('\n\n ' + line0 + ' | ' + addr)
                sleep(1)
                sys.stdout.write('\n\n continue...\n\n')
                sleep(1)

            if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
                print('\n\n Target found!!' + line0 + ' \n ' + addr,'\n')
                file=open(u"16j.Target.Info.txt","a")
                file.write('\n ' + line0 + '| ' + addr)
                file.close()
                sleep(2)
                wait = input("Press Enter to Exit.")
                sleep(1)
                exit()
                
            if line_count==x:
                list.clear()
                
    if len(list)==0:
        print('\n WAIT!....\n')

wipall

newbie

Activity: 10

Merit: 0

Quote from: itod on February 22, 2022, 05:25:17 AM

Quote from: wipall on February 22, 2022, 05:09:51 AM

Quote from: cookiebro on February 22, 2022, 03:52:30 AM

Quote from: wipall on February 22, 2022, 02:12:52 AM

Is there a python code that proceeds sequentially with the binary number system and can specify the range, guys?

Why would you want the binary value over hex?

I think the scanning process will be faster.

No it won't, python interpreter represents these numbers in the exact same way.

arrow. So do you have a python code that I can scan in order? a sequential code where I can specify the scan interval.

itod

legendary

Activity: 1988

Merit: 1077

Honey badger just does not care

Quote from: wipall on February 22, 2022, 05:09:51 AM

Quote from: cookiebro on February 22, 2022, 03:52:30 AM

Quote from: wipall on February 22, 2022, 02:12:52 AM

Is there a python code that proceeds sequentially with the binary number system and can specify the range, guys?

Why would you want the binary value over hex?

I think the scanning process will be faster.

No it won't, python interpreter represents these numbers in the exact same way.

wipall

newbie

Activity: 10

Merit: 0

Quote from: cookiebro on February 22, 2022, 03:52:30 AM

Quote from: wipall on February 22, 2022, 02:12:52 AM

Is there a python code that proceeds sequentially with the binary number system and can specify the range, guys?

Why would you want the binary value over hex?

I think the scanning process will be faster.

cookiebro

newbie

Activity: 7

Merit: 1

Quote from: wipall on February 22, 2022, 02:12:52 AM

Is there a python code that proceeds sequentially with the binary number system and can specify the range, guys?

Why would you want the binary value over hex?

wipall

newbie

Activity: 10

Merit: 0

Is there a python code that proceeds sequentially with the binary number system and can specify the range, guys?

cookiebro

newbie

Activity: 7

Merit: 1

A revised effort to consolidate our compute power by means of a pool:

https://discord.gg/YeUYPDUmVV

salah98

newbie

Activity: 3

Merit: 0

Quote from: wipall on February 21, 2022, 04:45:28 PM

Hi,
Where can I find python software to create private key with publickey?
It may be in other software.

thank you.

I wonder if there is bruteforce software available for python by giving short distances (hex or binnary).

yes this fast one https://github.com/albertobsd/keyhunt#readme

wipall

newbie

Activity: 10

Merit: 0

Hi,
Where can I find python software to create private key with publickey?
It may be in other software.

thank you.

I wonder if there is bruteforce software available for python by giving short distances (hex or binnary).

fxsniper

member

Activity: 406

Merit: 47

I think all method with "random" any technic or idea all it never works.
but method calculate or algorithm may be can work possible

Feron

jr. member

Activity: 67

Merit: 1

Someone has to break 64 puzzles because the creator sent and published from 65 up publick keys which seems a little strange to me that up to 63 it could creater data after 1 bit and from 64 can be fully full protect 64 hexadecimal address
Simply put, the puzzle could have been created to determine the maximum possible protection against the Kangaroo guess method
therefore I will probably look for a 120 bit address, because I am sure that there should be 100% in the range of 120 bits
therefore, someone should finally 64 puzzle complete the of the puzzle to show in doubt that the whole pyramid is correct and each address up increases by one bit.

jhonjhon

full member

Activity: 1316

Merit: 126

Quote from: Erkallys on December 28, 2015, 01:12:52 PM

I don't know why but I'm smelling a big scam. Because a newbie that offer more than 12 000€ to solve a following of numbers this is strange...

Thats what i am thinking too. I smells something's fishy..Well, lets just ignore this if we dont know how it works.. Grin

Its better to be safe than sorry.

devjorge

newbie

Activity: 1

Merit: 0

this little update will loop infinite to solution and less output without tqdm

Code:

import time
import random
import bit
from bit import *
from time import sleep

print('',time.ctime())
print('\n ============== ================= ==RANDOM HUNTER== ================= ==============\n\n')

i = 1000000

keys=0
while True:
   t = int(time.time())
   while keys<=i:
   #a1 = random.randrange(0x20000000000000000,0x3ffffffffffffffff)
   ## "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so"

   a1 = random.randrange(0x8000000000000000,0xffffffffffffffff)
   ## "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN"

   keys+=1
   key = Key.from_int(a1)
   y1 = key.address
   #print('y1=',y1,' hex=',hex(a1))
   #print('a1=',a1)
   #filetest=open("16j.Target.test.txt","a")
   #filetest.write('\n '+str(hex(a1))+' | a1='+ str(a1) + ' | addr=' + str(y1) + '\n')
   #filetest.close()
   if y1.startswith('16jY7q'):
   print('' + hex(a1) + ' | ' + y1 + ' \n')
   file=open("16j.Target.Info.partial.txt","a")
   file.write('\n '+str(hex(a1))+' | a1='+ str(a1) + ' | addr=' + str(y1) + '\n')
   file.close()
   if y1 == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
   print('Target found!!' + hex(line) + ' | ' + y1 + '\n')
   file=open("16j.Target.Info.txt","a")
   file.write('\n '+str(hex(a1))+' | a1='+ str(a1) + ' | addr=' + str(y1) + '\n')
   file.close()
   wait = input("Press Enter to Exit.")
   sleep(8640000)
   exit()

   #if keys == i:
   # took = int(time.time() - t)
   # pps = int(keys / took)
   # #print('',i,' keys took',took,'seconds @ ',pps,' pps')
   # keys = 0
   # t = int(time.time())

hooked my 3080ti up to ttdsales.com/64bit and time to solution went down from 85 to 39 years now in last 2 days.

come and crack #64!

download bitcrack64, search cudart64_101.dll in your windows folder and copy it to bitcrack.
move ttdclient to bitcrack folder

settings.ini

Code:

devjorge
cu
-b 80 -t 640 -p 1024

8

-b 80 -t 640 -p 1024

settings for geforce rtx 3080ti with good cooling and powerlimit @ 113% mem +1000 core +100 = 2050 MKey/s @ 400-450W (Eff: 4,55 Mkey/W)

without overclocking and powerlimit 80% = 1600-1700 Mkey/s @ 300-325W (Eff: 5,2 Mkey/W)

ccqt

newbie

Activity: 1

Merit: 0

zahid888 is NOT the creator and I can prove it. This is a puzzle, and I've found a major piece of it. I need someone with advanced programming skills. I'm willing to share what I've learned with the right person, for half of the coins. I've figured out how to reduce the possibilities infinitely, like literally down to thousands, from such an incomprehensibly large number that no one can really understand.

I'm thinking about doing a youtube video on it. If I get enough support here I will.

zahid888

member

Activity: 282

Merit: 20

the right steps towerds the goal

Quote from: saatoshi_rising on April 27, 2017, 01:41:08 AM

I am the creator.

You are quite right, 161-256 are silly. I honestly just did not think of this. What is especially embarrassing, is this did not occur to me once, in two years. By way of excuse, I was not really thinking much about the puzzle at all.

I will make up for two years of stupidity. I will spend from 161-256 to the unsolved parts, as you suggest. In addition, I intend to add further funds. My aim is to boost the density by a factor of 10, from 0.001*length(key) to 0.01*length(key). Probably in the next few weeks. At any rate, when I next have an extended period of quiet and calm, to construct the new transaction carefully.

A few words about the puzzle. There is no pattern. It is just consecutive keys from a deterministic wallet (masked with leading 000...0001 to set difficulty). It is simply a crude measuring instrument, of the cracking strength of the community.

Finally, I wish to express appreciation of the efforts of all developers of new cracking tools and technology. The "large bitcoin collider" is especially innovative and interesting!

Need more hints for puzzle 64 Sad

Jolly Jocker

jr. member

Activity: 49

Merit: 1

Hi Dude, I am aware of this problem and it has the following cause..

The range is limited by random (9223372036854775807).. so 66 bit can only be partially scanned...
In this case from 30000000000000000 - 37ffffffffffffffff, then 37ffffffffffffffff + 9223372036854775807 (7fffffffffffffff), etc...

this is the maximum of random Range..

Code:

print('\n > The range is limited by random (9223372036854775807).. so 66 bit can only be partially scanned! <')
print('\n > In this case from 30000000000000000 - 37ffffffffffffffff ! < \n\n')
import gc
import tqdm
import time
import random
import bit
from bit import *
from time import sleep
t = time.ctime()
print('',t)
list=[]
i = 1000000
while True:
   start = time.time()
   print('\n\n generate',str(i),'Keys and sort..... ')
   del list
   gc.collect(generation=2)
   sleep(1)
   list = random.sample(range(55340232221128654848,64563604257983430655), i)
   list.sort()
   n = len(list)
   ende = time.time()
   if n == i:
   print ('\n',n,'Keys DONE !!','{:5.3f}s'.format(ende-start))
   sleep(1)
   print ('\n Start Scanning..!! \n')
   sleep(2)
   for line in tqdm.tqdm(list):
   line != "\n"
   key1 = Key.from_int(line)
   y1 = key1.address

   if y1.startswith('13zb1h'):
   print('\n\n ' + hex(line) + ' \n ' + y1,'\n')
   file=open(u"Addr.Info.txt","a")
   file.write('\n '+hex(line)+' | '+ y1)
   file.close()
   sleep(3)
   print('\n continue...\n')
   sleep(1)

   if y1 == "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so":
   print('\n\n Target found!!' + hex(line) + ' \n ' + y1,'\n')
   file=open(u"16j.Target.Info.txt","a")
   file.write('\n '+hex(line)+' | '+ y1)
   file.close()
   sleep(3)
   print('\n\n STOP!')
   sleep(1)
   break

Similar structure, without "random.sample" limit, with "random.randrange" there is no such limit...

Code:

import tqdm
import time
import random
import bit
from bit import *
from time import sleep
from tqdm import tqdm
t = time.ctime()
print('',t)
print('\n ============== ================= ==RANDOM HUNTER== ================= ==============\n\n')
list=[]
list.clear()
i = 4000000
print(' generate',str(i),'Random Keys..... ')
while len(list)<=i:
a1 = random.randrange(0x20000000000000000,0x3ffffffffffffffff)
list.append(a1)
if len(list)==i:
line_count=0
print ('',i,'Keys DONE !!')
sleep(1)
print ('\n Start Scanning..!! \n')
sleep(2)
for line in tqdm(list):
line != "\n"
seed=str(line)
key = Key.from_int(line)
y1 = key.address
line_count+=1

if y1.startswith('13zb1h'):
print('\n\n ' + hex(line) + ' | ' + y1,'\n')
print(' continue...\n')
sleep(1)

if y1 == "13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so":
print('\n\n Target found!!' + hex(line) + ' \n ' + y1,'\n')
file=open(u"16j.Target.Info.txt","a")
file.write('\n '+hex(line)+' | '+ y1)
file.close()
sleep(3)
wait = input("Press Enter to Exit.")
sleep(1)
exit()

if line_count==i:
list.clear()

if len(list)==0:
print('\n generate',str(i),'Random Keys..... ')

solvepuzzle

newbie

Activity: 2

Merit: 0

Quote from: Jolly Jocker on January 29, 2022, 07:46:44 AM

Code:

import bit
import tqdm
import time
import random
from bit import *
from tqdm import tqdm
from time import sleep
t = time.ctime()
print('',t)
sleep(1)
print('\n ================= ================= ================= ================= =================')
print('\n\n RANDOM BITCOIN PREDATOR \n #64: 8000000000000000...ffffffffffffffff\n\n')
f=[]
i=0
a=4000000
print('\n Generate',str(a),'random Keys and sort....')
while i < a:
f.clear()
print('\n\n WAIT! ', end='\r')
sleep(1)
start = time.time()
f = random.sample(range(9223372036854775808,18446744073709551615), a)
f.sort()
end = time.time()
print('',str(a),'Keys DONE !!','{:5.3f}s'.format(end-start),'\n')
sleep(1)
print (' start scanning... ', end='\r')
sleep(2)
for line in tqdm(f):
line!='\n'
pass
key = Key.from_int(line)
addr = key.address

if addr.startswith('16jY7') or addr.endswith('jyXQN'):
print('\n\n',hex(line),'|', addr)
sleep(1)
print('\n continue...\n')
sleep(1)

if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
print('\n\n Target found!!' + hex(line) + ' \n ' + addr,'\n')
file=open(u"16j.Target.Info.txt","a")
file.write('\n ' + hex(line) + ' | ' + addr)
file.close()
sleep(2)
wait = input("Press Enter to Exit.")
sleep(1)
exit()

Hello thanks for the code. I tried to modify it for puzzle 66. This error pops up

RANDOM BITCOIN PREDATOR
#66: 20000000000000000...3ffffffffffffffff

Generate 4000000 random Keys and sort....

Traceback (most recent call last):
File "main.py", line 22, in
f = random.sample(range(36893488147419103232,73786976294838206463), a)
File "/usr/lib/python3.8/random.py", line 361, in sample
n = len(population)
OverflowError: Python int too large to convert to C ssize_t

Any ideas how to modify it to work with other puzzles

Jolly Jocker

jr. member

Activity: 49

Merit: 1

# Python 3.9.7
view, way of working.. happy hunting! BTC Grin

Code:

Sat Jan 29 13:32:58 2022

================= ================= ================= ================= =================

RANDOM BITCOIN PREDATOR
#64: 8000000000000000...ffffffffffffffff

Generate 4000000 random Keys and sort....

4000000 Keys DONE !! 6.426s

37%|█████████████████? | 1462908/4000000 [02:09<03:26, 12289.05it/s]

0xaed60f700efb2d13 | 16jY79d6zdGihLwnJ6to9XsUpBGhx7DJzh

continue...

100%|█████████████████████████████████████████████████| 4000000/4000000 [05:39<00:00, 11784.16it/s]

4000000 Keys DONE !! 6.582s

100%|█████████████████████████████████████████████████| 4000000/4000000 [05:27<00:00, 12211.18it/s]

4000000 Keys DONE !! 6.422s

24%|███████████? | 957290/4000000 [01:20<04:05, 12390.64it/s]

0x9ea1c04c218e3225 | 16jY7Bw3QAKqEUaipMpoJuo8Bk6yYQhgPA

continue...

100%|█████████████████████████████████████████████████| 4000000/4000000 [05:29<00:00, 12132.94it/s]

4000000 Keys DONE !! 6.578s

36%|█████████████████? | 1457119/4000000 [01:57<03:20, 12651.31it/s]

0xaeace2bf4b35cdf3 | 16jY7TpoxQJurJUD4JjNNVYDqR1mbf9JUn

continue...

100%|█████████████████████████████████████████████████| 4000000/4000000 [05:50<00:00, 11414.73it/s]

WAIT!

Jolly Jocker

jr. member

Activity: 49

Merit: 1

Code:

import bit
import tqdm
import time
import random
from bit import *
from tqdm import tqdm
from time import sleep
t = time.ctime()
print('',t)
sleep(1)
print('\n ================= ================= ================= ================= =================')
print('\n\n RANDOM BITCOIN PREDATOR \n #64: 8000000000000000...ffffffffffffffff\n\n')
f=[]
i=0
a=4000000
print('\n Generate',str(a),'random Keys and sort....')
while i < a:
f.clear()
print('\n\n WAIT! ', end='\r')
sleep(1)
start = time.time()
f = random.sample(range(9223372036854775808,18446744073709551615), a)
f.sort()
end = time.time()
print('',str(a),'Keys DONE !!','{:5.3f}s'.format(end-start),'\n')
sleep(1)
print (' start scanning... ', end='\r')
sleep(2)
for line in tqdm(f):
line!='\n'
pass
key = Key.from_int(line)
addr = key.address

if addr.startswith('16jY7') or addr.endswith('jyXQN'):
print('\n\n',hex(line),'|', addr)
sleep(1)
print('\n continue...\n')
sleep(1)

if addr == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
print('\n\n Target found!!' + hex(line) + ' \n ' + addr,'\n')
file=open(u"16j.Target.Info.txt","a")
file.write('\n ' + hex(line) + ' | ' + addr)
file.close()
sleep(2)
wait = input("Press Enter to Exit.")
sleep(1)
exit()

dextronomous

full member

Activity: 431

Merit: 105

so wow puzzle 20 works,
but you have anything else for the 64th puzzle.
thanks

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 282. (Read 253799 times)